Distributions: Theory and Applications (Cornerstones) 0817646728, 9780817646721

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Cornerstones Series Editors Charles L. Epstein, University of Pennsylvania, Philadelphia Steven G. Krantz, Washington University, St. Louis

Advisory Board Anthony W. Knapp, State University of New York at Stony Brook, Emeritus

J.J. Duistermaat J.A.C. Kolk

Distributions Theory and Applications

Translated from Dutch by J.P. van Braam Houckgeest

J.J. Duistermaat† Mathematical Institute Utrecht University

J.A.C. Kolk Mathematical Institute Utrecht University P.O. Box 80.010 3508 TA Utrecht The Netherlands [email protected]

Translated from Dutch by J.P. van Braam Houckgeest

ISBN 978-0-8176-4672-1 e-ISBN 978-0-8176-4675-2 DOI 10.1007/978-0-8176-4675-2 Springer New York Dordrecht Heidelberg London Library of Congress Control Number: 2010932757 Mathematics Subject Classification (2010): 46-01, 42-01, 35-01, 28-01, 34-01, 26-01

© Springer Science+Business Media, LLC 2010 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Birkhäuser is part of Springer Science+Business Media www.birkhauser-science.com

To V. S. Varadarajan A True Friend and Source of Inspiration

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix Standard Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv 1

Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2

Test Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3

Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4

Differentiation of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5

Convergence of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6

Taylor Expansion in Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7

Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

8

Distributions with Compact Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

9

Multiplication by Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

10 Transposition: Pullback and Pushforward . . . . . . . . . . . . . . . . . . . . . . . . 91 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 vii

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11 Convolution of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 12 Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 13 Fractional Integration and Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 153 13.1 The Case of Dimension One . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 13.2 Wave Family . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 13.3 Appendix: Euler’s Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 14 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 15 Distribution Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 16 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 17 Fundamental Solutions and Fourier Transform . . . . . . . . . . . . . . . . . . . . 271 17.1 Appendix: Fundamental Solution of .I
/k . . . . . . . . . . . . . . . . . . 279 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 18 Supports and Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 19 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 20 Appendix: Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 21 Solutions to Selected Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 Index of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

Preface

I am sure that something must be found. There must exist a notion of generalized functions which are to functions what the real numbers are to the rationals (G. Peano, 1912) Not that much effort is needed, for it is such a smooth and simple theory (F. Tr`eves, 1975)

In undergraduate physics a lecturer will be tempted to say on certain occasions: “Let ı.x/ be a function on the line that equals 0 away from 0 and is infinite at 0 in such a way that its total integral is 1. The most important property of ı.x/ is exemplified Z 1 by the identity .x/ı.x/ dx D .0/; 1

where  is any continuous function of x.” Such a function ı.x/ is an object that one frequently would like to use, but of course there is no such function, because a function that is 0 everywhere except at one point has integral 0. All the same, it is important to realize what our lecturer is trying to accomplish: to describe an object in terms of the way it behaves when integrated against a function. It is for such purposes that the theory of distributions, or “generalized functions,” was created. It can be formulated in all dimensions, its mathematical scope is vast, and it has revolutionized modern analysis. One way to elaborate on the distributional point of view1 is to note that a pointwise definition of functions is not very relevant to many situations arising in engineering or physics. This is due to the fact that physical observations often do not represent sharp computations at a single point in space-time but rather averages of fluctuations in small but finite regions in space-time. This is an essential point in signal theory, where there are limitations to the determination of pulse lengths, and 1

Here we follow the masterly exposition of Varadarajan [22, p. 185]. ix

x

Preface

in quantum theory, where the electromagnetic fields of elementary particles cannot be measured unless one uses a macroscopic test body. From the mathematical point of view one can say that a measurement of a physical quantity by means of a test body yields an average of the values of that quantity in a very small region, the latter being represented by a smooth function that is zero outside a small domain. One replaces the test bodies by these functions, which are naturally called test functions. The value thus measured is a function on the space of test functions, and the interpretation of the measurement as an average makes it clear that this function must be linear. Thus, if T is the space of test functions (unspecified at this point), physical quantities assign real or complex values to functions in T . In keeping with our idea that measurements are averages, we recognize that sometimes things are not so bad and that actual point measurements are possible. Thus ordinary functions are also allowed to be viewed as functionals on T . If f is such an ordinary function, it represents the following functional on T : Z  7! f .x/.x/ dx 2 C: However, since we admit measurements that are too singular to be represented by ordinary functions, we refer to the general functionals on T as generalized functions or distributions. We have been vague about what the space is in which we are operating and also what functions are chosen as test functions. This actually is a great strength of these ideas, because the methods evidently apply without any restriction on the nature or the dimensions of the space. In this book, however, we restrict ourselves to the most important case, that of open subsets of the Euclidean spaces Rn . Distributions are to functions what the real numbers R are to the rational numbers Q. In R, the cube root of any number also belongsp to R, as does the logarithm of the absolute value of a nonzero number; by contrast, 3 2 and log 2 do not belong to Q. Moreover, R is the smallest extension of Q having such properties, while every real number can be approximated by rationals with arbitrary precision. Similarly, distributions are always infinitely differentiable, which is not true of all functions. Here, too, distributions are the smallest possible extension of the test functions satisfying this property, while every distribution can be approximated in the appropriate sense by test functions with arbitrary precision. Continuing the analogy, we mention that differential equations may have distributional solutions in situations where there are no classical solutions, that is, given by differentiable functions. In numerous problems it is of great advantage that solutions exist, even at the penalty of introducing new objects such as distributions, because the solutions can be subject to further study. The theory of distributions provides many tools for the investigation of these so-called weak solutions; for example, these tools enable one to determine when and where distributions are actually functions. One of the early triumphs of distribution theory was the result that every partial differential equation with constant coefficients has a fundamental solution in the sense of distributions: classically, nothing comparable is available.

Preface

xi

Fourier theory is another branch of analysis in which a suitable subclass of all distributions helps to clarify many issues. This theory is a far-reaching generalization of writing a vector x D .x1 ; : : : ; xn / in Rn as xD

n X

xk ek ;

kD1

that is, as a superposition of a finite sum of multiples of the basis vectors ek . Analogously, in Fourier analysis one attempts to write functions or even distributions as superpositions of basic functions. In this case, finitely many functions do not suffice, but the collection of all bounded exponential functions turns out to be a good choice: bounded, because unbounded exponentials grow too fast at infinity, and exponential, because such functions are simultaneous eigenvectors of all partial derivatives. The sense in which the infinite superposition represents the original object then becomes an important issue: is the convergence pointwise or uniform, or in a smeared sense? Fourier analysis in the distributional setting enables one to handle problems that classically were out of reach, as well as many new ones. So one obtains, working modulo 2, 1 1 X i kx ı.x/ D e : 2 kD 1

This formula goes back to Euler, except that he found the sum to be equal to 0 when x is away from 0. H¨ormander’s monumental treatise [11] on linear partial differential equations and Harish-Chandra’s pioneering work [10] on harmonic analysis on semisimple Lie groups over the fields of real, complex, or p-adic numbers are but two of the rich fruits borne by Schwartz’s text [20], which gave birth to the theory of distributions. This book aims to be a thorough, yet concise and application-oriented, introduction to the theory of distributions that can be covered in one semester. These constraints forced us to make choices: we try to be rigorous but do not construct a complete theory that prepares the reader for all aspects and applications of distributions. It supplies a certain degree of rigor for a kind of calculation that people long ago did completely heuristically, and it establishes what is legitimate and what is not. The amount of functional analysis that is needed in our treatment is reduced to a bare minimum: only the principle of uniform boundedness is used, while the Hahn–Banach theorems are applied to give alternative proofs, with one exception, of results obtained by different methods. On the other hand, in our exposition of the theory and, in particular, in the problems, we stress applications and interactions with other parts of mathematics. As a result of this approach our text is complementary to the books [13] and [14] by A.W. Knapp, also published in the Cornerstones series. Building on firm foundations in functional analysis and measure theory, Knapp develops the theory rigorously and in greater depth and wider context than we do, by treating pseudodifferential operators on manifolds, for instance. In many ways our text is introductory;

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on the other hand, it presents students of (theoretical) physics or electrical engineering with an idea of what distributions are all about from the mathematical point of view, while giving applied or pure mathematicians a taste of the power of distributions as a natural method in analysis. Our aim is to make the reader familiar with the essentials of the theory in an efficient and fairly rigorous way, while emphasizing the applications. Solutions of important ordinary and partial differential equations, such as the equation for an electrical LRC network, those of Cauchy–Riemann, Laplace, and Helmholtz and the heat and wave equations, are studied in great detail. Tools for the investigation of the regularity of the solution, that is, its smoothness, are developed. Topics in signal reconstruction have also been treated, such as the mathematical theory underlying CT (= computed tomography) scanners as well as results on bandlimited functions. The fundamentals of the theory of complex-analytic functions in one variable are efficiently derived in the context of distributions. In order to make the book self-contained, various results on special functions that are used in our treatment are deduced as consequences of the theory itself, wherever possible. A large number of problems is included; they are found at the end of each chapter. Some of these illustrate the theory itself, while others explore its relevance to other parts of mathematics. They vary from straightforward applications of the theory to theorems or projects examining a topic in some depth. In particular, important aspects of multidimensional real analysis are studied from the point of view of distributions. Complete solutions to 146 of the 281 problems are provided; problems for which solutions are available are marked by the symbol  . A great number of the remaining exercises are supplied with copious hints, and many of the more difficult problems have been tested in take-home examinations. In more technical terms, the first eleven chapters cover the basics of general distributions. Specifically, Chap. 10 presents a systematic calculus of pullback and pushforward for the transformation of distributions under a change of variables, whereas Chap. 13 considers complex-analytic one-parameter families of distributions with the aim of obtaining fundamental solutions of certain partial differential operators. Chap. 14 then goes on to treat the Fourier transform of the subclass of tempered distributions in the general, aperiodic case, which is of fundamental importance for the subsequent Chaps. 15–19. Chap. 15 discusses the notion of a distribution kernel of a continuous linear mapping. This notion enables an elegant verification of many properties of such mappings. More generally, it enables aspects of the theory of distributions to be surveyed from a fresh and unifying point of view, as is exemplified by many of the problems in the chapter. The Fourier inversion formula is used in a novel proof of the Kernel Theorem. The Fourier transform is applied in Chap. 16 to study the periodic case and in Chap. 17 to construct additional fundamental solutions. Chap. 18 deals with the Fourier transforms of compactly supported distributions, and Chap. 19 considers rudiments of the theory of Sobolev spaces. Mathematically sophisticated readers, having perused the first ten chapters, might prefer to proceed immediately to Chaps. 14 and 15.

Preface

xiii

Important characteristics of the present treatment of the theory of distributions are the following. The theory as presented provides a highly coherent context with a strong potential for unification of seemingly distant parts of analysis. A systematic use of the operations of pullback and pushforward enables the development of a very clean and concise notation. A survey of distribution theory in the framework of distribution kernels allows a description that is algebraic rather than analytic in nature, and makes it possible to study distributions with a minimal use of test functions. In particular, within this framework some more advanced aspects of distribution theory can be developed in a highly efficient manner and transparent proofs can be given. The treatment emphasizes the role of symmetry in obtaining short arguments. In addition, distributions invariant under the actions of various groups of transformations are investigated. Our preferred theory of integration is that of Riemann, because it will be more familiar to most readers than that of Lebesgue. In some instances, however, our arguments might be slightly shortened by the use of Lebesgue’s theory. In the very limited number of cases in which Lebesgue integration is essential, we mention this explicitly. The reader who is not familiar with measure theory may safely skip these passages. On the other hand, in the theory of distributions Radon measures arise naturally as linear forms defined on compactly supported continuous functions, and therefore the Daniell approach to the theory of integration, which emphasizes linear forms acting on functions instead of functions acting on sets, is very natural in this setting. In the Appendix, Chap. 20, we survey the theory of Lebesgue integration with respect to a measure from this point of view. Although the approach seems very appropriate in our context, we are aware of the fact that it is of limited value to the mathematical probabilist, who primarily requires a theory of integration on function spaces, which are not usually locally compact. We strongly feel that a mathematical style of writing is appropriate for our purposes, so the book contains a certain amount of theorem–proof text. The reader of a text at this level of mathematical sophistication rightly expects to find all the information needed to follow the argument as well as clear expositions of difficult points, and the theorem–proof format is a time-honored vehicle for conveying these. Furthermore, in theorems one summarizes useful information for future application. Important results (for instance, the Fourier inversion formula) often get several proofs; in this manner different aspects or unexpected relations are brought to the fore. The present text has evolved from a set of notes for courses taught at Utrecht University over the last twenty years, mainly to bachelor-degree students in their third year of theoretical physics and/or mathematics. In those courses, familiarity with measure theory, functional analysis, or even some of the more theoretical aspects of real analysis, such as compactness, could not be assumed. Since this book addresses the same type of audience, the present text was therefore designed to be essentially self-contained: the reader is assumed to have merely a working knowledge of linear algebra and of multidimensional real analysis (see [7], for instance), while only a

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few of the problems also require some acquaintance with the residue calculus from complex analysis in one variable. In some cases, the notion of a group will be encountered, mainly in the form of a (one-parameter) group of transformations acting on Rn . Each time the course was taught, the notes were corrected and refined, with the help of the students; we are grateful to them for their remarks. In particular, J.J. Kuit made a considerable number of original contributions and we benefitted from fruitful discussions with him. M.A. de Reus suggested many improvements. Also, we express our gratitude to our colleagues E.P. van den Ban, for making available the notes for his course in 1987 on distributions and Fourier transform and for very constructive criticism of a preliminary draft, and R.W. Bruggeman, for the improvements and additional problems that he contributed over the past few years. In addition, T.H. Koornwinder read substantial parts of the manuscript with great care when preparing a course on distributions, and contributed significantly, by many valuable queries and comments, to the accuracy of the final version. Furthermore, we wish to acknowledge our indebtedness to A.W. Knapp, who played an essential role in the publication of this book, for his generous advice and encouragement. The enthusiasm and wisdom of Ann Kostant, our editor at Birkh¨auser, made it all possible, and we are very grateful to her for this. Jessica Belanger saw the manuscript through its final stages of production. The original Dutch text has been translated with meticulous care by J.P. van Braam Houckgeest. In addition, his comments have led to considerable improvement in formulation. The second author is very thankful to M.J. Suttorp and H.W.M. Plokker, cardiologists, and their teams: their intervention was essential for the completion of this book. The responsibility for any imprecisions remains entirely ours; we would be grateful to be told of them, at [email protected]. Utrecht, February 2010

Hans Duistermaat Johan Kolk

Standard Notation

The symbol  set against the right margin signifies the end of a proof. Furthermore, the symbol ˛ marks the end of a definition, example or remark. Item ; o, O i x 2 X or X 3 x x…X fx 2 X j P g @X , X X  Y or Y  X X [Y,X \Y,X nY X Y f W X ! Y , x 7! f .x/ f .
; y/ f .X /, f 1 .X / f jX gıf Z, Q, R, C Za N R>a jxj Œx sgn x  a; b Œ Œ a; b  Œ a; b Œ,  a; b  .a; b/ .x1 ; : : : ; xn /

Meaning empty set little p and big O symbol of Landau 1 x an element of X x not an element of X the set of x in X such that P holds boundary and closure of the set X X a subset of Y union, intersection, difference of sets Cartesian product of sets mapping, effect of mapping mapping x 7! f .x; y/ direct and inverse image under f of the set X restriction to X composition of f and g, or of g following f integers, rationals, reals, complex numbers integers greater than or equal to a D Z1 , natural numbers reals larger than a absolute value of x 2 R greatest integer  x sign of x open interval from a to b closed interval from a to b half-open intervals column vector in R2 column vector xv

xvi

kxk h x; y i R n , Cn Re z, Im z z jzj f .a
/ f 0 , f 00 f .k/ Df @j f P# Q 0 , I det A t A '

Standard Notation

norm of vector x inner product of vectors x and y spaces of column vectors real and imaginary parts of complex z complex conjugate of z absolute value of z 2 C function f W Rn ! C given by x 7! f .ax/ derivative and second derivative of f W R ! R kth-order derivative of f (total) derivative of mapping f W Rn ! Rp partial derivative of f with respect to jth variable  approaches 0 through positive values sum and product, possibly with a limit operation identity matrix or operator determinant of matrix or operator A transpose of matrix or operator A is isomorphic to, is equivalent to

Chapter 1

Motivation

Distributions form a class of objects that contains the continuous functions as a subset. Conversely, every distribution can be approximated by infinitely differentiable functions, and for that reason one also uses the term “generalized functions” instead of distributions. Even so, not every distribution is a function. In several respects, the calculus of distributions can be developed more readily than the theory of continuous functions. For example, every distribution has a derivative, which itself is also a distribution (see Chap. 4). Hence, every continuous function considered as a distribution has derivatives of all orders. Conversely, we shall prove that every distribution can locally be written as a linear combination of derivatives of some continuous function (see Theorem 13.1 or Example18.2). If every continuous function is to be infinitely differentiable as a distribution, no proper subset of the space of distributions can therefore be adequate. In this sense, the extension of the concept of functions to that of distributions is as economical as it possibly can be. This may be compared with the extension of the system Z of integers to the system Q of rational numbers, where to any x and y 2 Z with y ¤ 0 corresponds the quotient yx 2 Q. In this case, too, Q is the smallest extension of Z having the desired properties. We now discuss some more concrete types of problem and show how they are solved by the calculus of distributions. We also indicate some typical contexts in which these questions arise. It should be pointed out that the reader will not be assumed to be familiar with the nonmathematical concepts used in those contexts. Likewise, in Examples 1.1 through 1.5 we will occasionally use some mathematical tools that are not yet assumed to be known by the reader. The point of these examples is to provide insight into where the subject is going, rather than to give all details about each example. Example 1.1. Here we consider the second-order derivative of a function that is nondifferentiable at one point. The function f defined by f .x/ D jxj for x 2 R is continuous on R. It is differentiable on R0 , with derivative equaling 1 and C1 on these J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_1, © Springer Science+Business Media, LLC 2010

1

2

1 Motivation

intervals, respectively. f is not differentiable at 0. So it seems natural to say that the derivative f 0 .x/ equals the sign sgn.x/ of x, the value of f 0 .0/ not being defined and, intuitively speaking, being of little importance. But beware, we obviously require that the second-order derivative f 00 .x/ equal 0 for x < 0 and for x > 0, while f 00 .x/ must have an essential contribution at x D 0. Indeed, if f 00 .x/  0, the conclusion is that f 0 .x/  c for a constant c 2 R; in other words, f .x/ D c x for all x 2 R, which is different from the function x 7! jxj, whatever choice is made for c. The correct description turns out to be that f 00 D 2ı, with ı as in the preface; see Problem 4.1 for more details. ˛ Example 1.2. Now we are concerned with the electrical field of a point charge. In Maxwell’s theory of electromagnetism there are physical difficulties with the concept of a point charge, and in its mathematical description a problem occurs as well. Let v W R3 ! R3 be a continuously differentiable mapping, interpreted as a vector field on R3 . Further, let 3 X @vj .x/ x 7! div v.x/ D @xj j D1

be the divergence of v; this is a continuous real-valued function on R3 . Suppose that X is a bounded and open set in R3 having a smooth boundary @X and lying at one side of @X ; we write the outer normal to @X at the point y 2 @X as .y/. The Divergence Theorem then asserts that Z Z div v.x/ dx D hv.y/; .y/i dyI (1.1) X

@X

see for instance Duistermaat–Kolk [7, Theorem 7.8.5]. Here the right-hand side is interpreted as an amount of volume that flows outward across the boundary, while div v is rather like a local expansion (= source strength) in a motion whose velocity field equals v. Traditionally, one also wishes to allow R R point (mass) sources at a point p, for which X div v.x/ dx D c if p 2 X and X div v.x/ dx D 0 if p … X , where X is the closure of X in R3 . (We make no statement for the case that p 2 @X .) Here c is a positive constant, the strength of the point source in p. These conditions cannot be realized by a function div v continuous everywhere on R3 . More specifically, the divergence of the special vector field v.x/ D

1 x kxk3

(1.2)

vanishes at every point x ¤ 0; verify this. This implies that the left-hand side of (1.1) equals 0 if 0 … X and 4 if 0 2 X ; the latter result is obtained when we replace the set X by X n B, where B is a closed ball around 0, having a radius sufficiently small that B  X , and then compute the right-hand side of (1.1) (see [7, Example 7.9.4]). Thus we would like to conclude that div v in this case equals the point source at the point 0 with strength 4; in mathematical terms, div v D 4 ı,

1 Motivation

3

with ı the generalization to R3 of the ı from the preface (see Problem 4.6 and its solution for the details). ˛ Example 1.3. The underlying mathematical background of this example is the theory of the Hilbert transform H , which gives a way of describing a negative phase shift of signals by 90ı (see Problem 14.52), that is,     ; .H sin/.x/ D cos x D sin x : .H cos/.x/ D sin x D cos x 2 2

In addition, in the example we come across interesting new distributions that play a role in quantum field theory. The function x 7! x1 is not absolutely integrable on any bounded interval around 0, so it is not immediately clear what Z .x/ dx R x is to mean if  is a continuous function that vanishes outside a bounded interval. Even if .0/ D 0, the integrand is not necessarily absolutely integrable. For example, let 0 <  < c < 1 and define (see Fig. 1.1) .x/ D

˚0

1 j log xj

if

x  0;

if

0 < x  c:

This function  is continuous on  1; c  and can be extended to a continuous 1.4

50

0.5

Fig. 1.1 Graphs of  and x 7!

0.5

.x/ x

function on R that vanishes outside a bounded interval. Then Z c .x/ dx D log j log j log j log cj; x  and the right-hand side converges to 1 as  # 0. But if  is continuously differentiable and vanishes outside a bounded interval, integration by parts and the estimate ./ . / D O./ as  # 0, which is a

4

1 Motivation

consequence of the Mean Value Theorem, give Z Z .x/ dx D lim  0 .x/ log jxj dx: #0 RnŒ ;   x R

(1.3)

Note the importance of the excluded intervals being symmetric about the origin. The left-hand side is called the principal value of the integral of x 7! .x/ x and is also Z written as   1 .x/ PV dx DW PV ./: (1.4) x R x More generally, if c 2 R and  is a continuous function on R, define Z  1  .x/ PV ./ D lim dx; x c #0 RnŒ c ; cC  x c

provided that this limit exists. Other, equally natural, propositions can also be made. Indeed, always assuming  to be continuously differentiable and to vanish outside a bounded interval, Z .x/ dx x Ci R converges as  # 0, or  " 0, respectively; see Problem 1.3. The limit is denoted by Z .x/ dx; (1.5) x Ci0 R or

Z

R

.x/ dx; x i0

(1.6)

1 1 respectively. Clearly, the “functions” PV x1 , xCi 0 , and x i 0 differ only at x D 0, that is, integration against a function  will produce identical results if .0/ D 0. In Problem 1.3 and its solution as well as in Examples 3.3 and 5.7 one may find more information. ˛

Example 1.4. We will now make plausible that the theory of distributions provides limits in cases where these do not exist classically and that it also enables more freedom in the interchange of analytic operations. x 2 R, summation of the geometric series leads to P For 0 i xr n < 1 and 1 n2Z0 .re / D 1 re i x . By taking the real parts in this identity we obtain X

n2Z0

r n cos nx D

1 r cos x DW Ar .x/: 1 C r 2 2r cos x

Our interest is in the behavior of the preceding identity when r D 1 or under taking limits for r " 1. First consider the series for r D 1. Then it is clearly divergent, to 1, if x 2 2Z. In fact, the series is divergent everywhere on R. This follows from

1 Motivation

5

the fact that for no x 2 R do we have cos nx ! 0 as n ! 1. Indeed, otherwise we would have sin2 nx D 1 cos2 nx ! 1, but also sin2 nx D 21 .1 cos 2nx/ ! 12 . On the other hand, lim Ar .x/ D r"1

1 2

.x 2 R n 2Z/

Ar .x/ D

and

1 1

r P

.x 2 2Z/:

Abel’s Theorem (see [13, Theorem 1.48]), which would imply n2Z0 cos nx D 12 for x 2 Rn2Z, does not apply, because of the divergence everywhere of the series.

-0.1

20

20

10

10

-0.05

0.05

Fig. 1.2 Graph of Ar , for r D 9

-2 Π

0.1

Pk

j D1

10

j

2Π

with 2  k  5, and of A0:99999

Nevertheless the numerical evidence in Fig. 1.2 above strongly suggests that the sum of the series is given by the function having value 12 on R n 2Z and 1 on 2Z. However, in the theory of integration as discussed in Chap. 20 such a function would be identified with the constant function 12 on R, which ignores the serious divergence of the series on 2Z. Therefore, P it might be more reasonable to describe the limit of the series as A1 WD 12 C c k2Z ı2k on R. Here c 2 C is a suitable constant and ı2k denotes the Dirac function located at 2k. We determine c by Z  Z  demanding lim Ar .x/ dx D A1 .x/ dx: r"1





For 0  r < 1, an antiderivative Ir of Ar is given by (see Fig. 1.3 below) Z  1 C r X x x tan I so Ir .x/ D C arctan Ar .x/ dx D ˙Ir .˙/ D 2; 2 1 r 2  ˙

whichRis also directly obvious by termwise integration of the series. On the other  hand,  A1 .x/ dx D  C c, and so c D ; phrased differently, X

n2Z0

cos n
D

X 1 C ı2k 2

on R:

k2Z

The validity of this equality in the sense of distributions is rigorously verified in Problem 16.8.

6

1 Motivation Π

Π 2

-Π

Π

-

Π 2

-Π

Fig. 1.3 Graph of Ir , for r D

k 10

with 0  k  10, and of an antiderivative of

1 2

R x Note P that I1 .x/ WD limr"1 Ir .x/ D 2 , for x ≶ 0. Accordingly  A1 .x/ dx D ˙ ˙I1 .˙/. In addition, A1 is the derivative in the sense of distributions of I1 (classically the latter is nondifferentiable at 0), as will be shown in Example 4.2. According to the theory of integration we would have the limit of functions limr"1 Ar D 21 (the union of the lower solid line segment and the dashed line segment in Fig. 1.3 is the graph of an antiderivative of 12 ) and then lim r"1

Z

 

Ar .x/ dx D 2 ¤  D

Z



lim Ar .x/ dx:

 r"1

(1.7)

In view of the Dominated Convergence Theorem, Theorem 20.26.(iv), of Lebesgue or that of Arzel`a (see [7, Theorem 6.12.3]), the fact that interchange of limit and integration in (1.7) is invalid means that the family .Ar /0r 0, the function v .x/ D

arctan.x=/ ; arctan.1=/

for x 2 Œ 1; 1 . The denominator has been included in order to guarantee that v .˙1/ D ˙1. For x < 0, or x > 0, we have that arctan.x=/ converges to =2, or C=2, respectively, as  # 0. Therefore, v converges to the sign function sgn as  # 0. To study the behavior of F .v / we write v0 .x/ D

1 1 :  arctan.1=/ 1 C .x=/2

The change of variables x D  y leads to Z 1 F .v / D  arctan2 .1=/

1= 1=

y2 dy: 2.1 C y 2 /2

1 Motivation

9

Note that in this expression the factor has 2y 2 D  0 .y/ if .1 C y 2 /2

1 arctan2 .1=/

converges to 4= 2 as  # 0. One

.y/ D

y C arctan y: 1 C y2

That makes the integral equal to ..1=/ . 1=//=4, from which we can see that the integral converges to =4 when  # 0. The conclusion is that F .v / D  ./, where ./ converges to 1= as  # 0. In particular, F .v / converges to zero as  # 0. Thus we see that the infimum of F on C 11; 1 equals zero; indeed, even the infimum on the subspace C 11; 1 equals zero. However, if u is a C 1 function with F .u/ D 0, we have du .x/  0, which means that u is constant. But then u cannot dx satisfy the boundary conditions u. 1/ D 1 and u.1/ D 1. In other words, the restriction of F to the space C 11; 1 does not attain its minimum in this example. In the beginning of the twentieth century the following discovery was made. Let H.1/ be the space of the square-integrable functions v on  a; b Œ whose derivatives v 0 are also square-integrable on  a; b Œ . Actually, this is not so easy to define. A correct definition is given in Chap. 19: v 2 H.1/ if and only if v is square-integrable and the distribution v 0 is also square-integrable. In order to understand this definition, we have to know how a square-integrable function can be interpreted as a distribution. Next, we use that the derivative of any distribution is another distribution, which may or may not equal a square-integrable function. If v is a continuously differentiable function on Œ a; b , application of the Cauchy–Schwarz inequality (see [7, Exercise 6.72]) gives jv.x/

ˇZ ˇ v.y/j D ˇ

y 0

x

ˇ Z ˇ v 0 .z/ dz ˇ 

 kv kL2 jx

y

yj

1=2

;

x

1=2 Z v 0 .z/2 dz

y

x

1=2 dz

(1.9)

where kv 0 kL2 is the L2 norm of v 0 . This can be used to prove that every v 2 H.1/ can be interpreted as a continuous function on Œ a; b  (also compare Example 19.3), with the same estimate jv.x/

v.y/j  kv 0 kL2 jx

yj1=2 :

The continuity of the functions v 2 H.1/ implies that one can meaningfully speak ˛;ˇ of the subspace H.1/ of the v 2 H.1/ for which v.a/ D ˛ and v.b/ D ˇ. Also, for every v 2 H.1/ the number F .v/ is well-defined. Now assume that p.x/ > 0 for every x 2 Œ a; b ; this excludes the example of Weierstrass. The assumption implies the existence of a constant c with the property kv 0 k2 2  c F .v/; L

for all v 2 H.1/ . In combination with the estimate for jv.x/ v.y/j this tells us ˛;ˇ that every sequence .vj /j 2N in H.1/ with bounded values F .vj / is an equicon-

10

1 Motivation

tinuous and uniformly bounded sequence of continuous functions. By the Arzel`a– Ascoli Theorem (see Knapp [13, Theorem 10.48]), a subsequence .vj.k/ /k2N then converges uniformly to a continuous function u as k ! 1. A second fact here of˛;ˇ fered without proof is that u 2 H.1/ and that the values F .vj.k/ / converge to F .u/ as k ! 1. This is now applied to a sequence of vj for which F .vj / converges to ˛;ˇ ˛;ˇ the infimum i of F on H.1/ . Thus one can show the existence of a u 2 H.1/ with

˛;ˇ . F .u/ D i . In other words, F attains its minimum on H.1/ This looked promising, but one then ran into the problem that initially all one could say about this minimizing u was that u0 is square-integrable. This does not even imply that u is differentiable under the classical definition that the limit of the difference quotients exists. Because so far we do not even know that u 2 C 2 , the integration by parts is problematic and, as a consequence, so is the conclusion that u is a solution of the Euler–Lagrange equation. What we can do is to integrate by parts with the roles of u and  interchanged and thereby conclude that Z b u.x/ .L/.x/ dx D 0; (1.10) a

1

for every  2 C that vanishes identically in a neighborhood of the boundary points a and b. For this statement to be meaningful, u need only be a locally integrable function on the interval I D  a; b Œ. In that case the function u is said to satisfy the differential equation Lu D 0 in a distributional sense. Historically, a somewhat older term is in a weak sense, but this is not very specific. Assume that p and q are sufficiently differentiable and that p has no zeros in the interval I . In this text we will show by means of distribution theory that if u is a locally integrable function and satisfies the equation Lu D 0 in the distributional sense, u is in fact infinitely differentiable in I and satisfies the equation Lu D 0 on I in the usual sense. See Theorem 9.4. In this way, distribution theory makes a contribution to the calculus of variations: by application of the Arzel`a–Ascoli Theorem it demonstrates the existence of a ˛;ˇ ˛;ˇ minimizing function u 2 H.1/ . Every minimizing function u 2 H.1/ satisfies the differential equation Lu D 0 in the distributional sense; distribution theory yields the result that u is in fact infinitely differentiable and satisfies the differential equation Lu D 0 in the classical sense. This application may be extended to a very broad class of variational problems, also including functions of several variables, for which the Euler–Lagrange variation equation then becomes a partial differential equation. ˛ Some of the interesting phenomena in the preceding examples form our starting point for the development of the theory of distributions. The estimation result from the next lemma, Lemma 1.6, will play an important role in what follows. The functions in Examples 1.1, 1.2, and 1.3 are not continuous, or differentiable, respectively, at a special point. Singularities in functions can be mitigated by translating the function f back and forth and averaging the functions thus obtained with

1 Motivation

11

a weight function .y/ that depends on the translation y applied to the original function. Let us assume that  is sufficiently differentiable on R, that .x/  0 for all x 2 R, that a constant m > 0 exists such that .x/ D 0 if jxj  m, and finally, that Z .x/ dx D 1: (1.11) R

For the existence of such , see Problem 1.4. The averaging procedure is described by the formula Z Z .f  /.x/ D f .x y/ .y/ dy D f .z/ .x z/ dz: (1.12) R

R

The minus sign is used to obtain symmetric formulas; in particular, f   D   f . The function f  is called the convolution of f and , because one of the functions is reflected and translated, then multiplied by the other one, following which the result is integrated. Another interpretation is that of a measuring device recording a signal f around the position x, where .y/ represents the sensitivity of the device at displacement y. In practice, this  is never completely concentrated at y D 0; because of built-in inertia, .y/ will have one or more bounded derivatives. Yet another interpretation is obtained by defining Ty f , the function translated by y, via .Ty f /.x/ WD f .x y/: (1.13)

Here we use the rule that .Ty f /.x C y/, the value of the translated function at the translated point, equals f .x/, the value of the function at the original point. In other words, under Ty the graph translates to the right if y > 0. If we now read the first equality in (1.12) as an identity between functions of x, we have Z f  D .y/ Ty f dy: (1.14) R

Here the right-hand side is defined as the limit of Riemann sums in the space of continuous functions (of x), where the limit is taken with respect to the supremum norm. Thus, the functions f translated by y are superposed, with application of a weight function .y/, similar to a photograph that becomes softer (blurred) if the camera is moved during the exposure. Indeed, differentiation with respect to x under the integral sign in the right-hand side in (1.12) yields, even in the case that f is merely continuous, that f   is differentiable, with derivative .f  /0 D f   0 : In obtaining this result, we have not used the normalization (1.11). We can therefore repeat this and conclude that f   is equally often continuously differentiable as . For more details, see the proof of Lemma 2.18 below. How closely does the smoothed signal f   approximate the true signal f ? If a  f .z/  b for all z 2 Œ x m; x C m , we conclude from (1.12) and (1.11)

12

1 Motivation

that a  .f  /.x/  b as well. This can be improved upon if we can bring the positive number m closer to 0. To achieve this, we replace the function  by  (see Fig. 1.5), for an arbitrary constant  > 0, with 1 x   .x/ D  : (1.15)   Furthermore,  is equally often continuously differentiable as .

Fig. 1.5 Graph of  as in (1.15) with  equal to 1 and 1/2, respectively

Lemma 1.6. If f is continuous on R, the function f   converges uniformly to f on every bounded interval Œ a; b  as  # 0. And for every  > 0, the function f   on R is equally often continuously differentiable as . Proof. We have

from which

Z

R

 .y/ dy D

.f   /.x/

Z

 R

f .x/ D D

Z

Z

 y  dy

D

f .x

y/





R m

f .x

Z

R

.z/ dz D 1;


f .x/  .y/ dy y/

m


f .x/  .y/ dy;

where in the second identity we have used  .y/ D 0 if jyj >  m. This leads to the estimate Z m j.f   /.x/ f .x/j  jf .x y/ f .x/j  .y/ dy m

 sup jf .x

y/

f .x/j;

jyj m

where in the first inequality we have applied  .y/  0 and in the second inequality we have once again used the fact that the integral of  equals 1. The continuity of f gives that for every ı0 > 0 the function f is uniformly continuous on the bounded interval Œ a ı0 ; b C ı0  (if necessary, see [7, Theorem

1 Problems

13

1.8.15] taken in conjunction with Theorem 2.2 below). This implies that for every  > 0 there exists a 0 < ı  ı0 with the property that jf .x y/ f .x/j <  if x 2 Œ a; b  and jyj < ı. From this we may conclude j.f   /.x/ f .x/j  , if x 2 Œ a; b  and 0 <  < ı=m.  Bochner [3] has called the mapping f 7! f   an approximate identity, and Weyl [24], a mollifier.

Problems 1.1. For a < c < b, the integral PV

1.2. Calculate PV

R

R

.x/ x

Z

Rb

1 a x c

b a

x

dx is divergent. Prove

1 c

dx D log

b c

c : a

dx, for the following choices of :

1 .x/ D

x 1 C x2

and

2 .x/ D

1 : 1 C x2

Which of these two integrals converges absolutely as an improper integral?  1.3. Determine the difference between (1.4) and (1.5), and between (1.5) and (1.6). Each is a complex multiple of .0/. See Example 14.30 and Problem 12.14 for different approaches.  1.4. Determine a polynomial function p on R of degree six for which p.a/ D 35 32

-1

1

Fig. 1.6 Illustration for Problem 1.4

R1 p 0 .a/ D p 00 .a/ D 0 for a D ˙1, while in addition, 1 p.x/ dx D 1. Define .x/ D p.x/ for jxj  1 and .x/ D 0 for jxj > 1. Prove that  is twice continuously differentiable. Sketch the graph of  (see Fig. 1.6).  1.5. Let be twice continuously differentiable on R and let equal 0 outside a bounded interval. Set f .x/ D jxj. Calculate the second-order derivative of g D

14

1 Motivation 200

1

-0.0015

-0.01

0.0015 -1

0.01

0.003

-0.003

0.003

00

Fig. 1.7 Illustration for Problem 1.5. Graphs of g D 2p , for  D 1=100, and of g 0 and g, for  D 1=1000

f  by first differentiating under the integral sign, then splitting the integration at the singular point of f , and finally eliminating the differentiations in every subintegral. Now take equal to  with  as in Problem 1.4 and  as in (1.15). Draw a sketch of g00 for small , and, by finding antiderivatives, of g0 and g. Show the sketches of g, g 0 , g 00 next to those of f , f 0 , and f 00 (?), respectively (see Fig. 1.7).  1.6. We consider integrable functions f and g on R that vanish outside the interval Œ 1; 1 .

(i) Determine the interval outside which the convolution f  g certainly vanishes. (ii) Using simple examples of your own choice for f and g, calculate f  g, and sketch the graphs of f , g, and f  g. (iii) Try to choose f and g such that f and g are not continuous while f  g is. (iv) Try to choose f and g such that f  g is not continuous. Hint: let ˛ > 1, f .x/ D g.x/ D x ˛ if 0 < x < 1, f .x/ D g.x/ D 0 if x  0 or x  1. Verify that f and g are integrable. Prove the existence of a constant c > 0 such that .f  g/.x/ D c x 2˛C1 if 0 < x  1. For what values of ˛ is f  g discontinuous at the point 0?

Π

-4 Π

-3 Π

-2 Π

-Π

Π

Fig. 1.8 Illustration for Problem 1.7. Graph of arccos B cos

2Π

3Π

4Π

1 Problems

15

1.7. Set I.x/ D x, for all jxj  1, and let triangle W R ! R denote the unit triangle function given by triangle.x/ D 1 jxj, for jxj  1 and triangle.x/ D 0, for jxj > 1. Prove (see Fig. 1.8) in the notation of Definition 2.17 that cos B arccos D I

on Œ 1; 1  ; 
 X X arccos B cos D  D T.2kC1/ triangle T2k .1Œ 0;   1Œ 0;  / on R:  k2Z

k2Z

Hint: show that arccos B cos is continuous on R, while on R n  Z one has .arccos B cos/0 D

X sin D . 1/k Tk 1 0; Œ : j sin j k2Z

Chapter 2

Test Functions

We will now introduce test functions and do so by specializing the testing of f as in (1.12). If we set x D 0 and replace .y/ by . y/, the result of testing f by means of the weight function  becomes equal to the “integral inner product” Z hf; i D f .x/ .x/ dx: (2.1) R

(For real-valued functions this is in fact an inner product; for complex-valued functions one uses the Hermitian inner product hf; i.) In Chap. 1 we went on to vary , by translating and rescaling. The idea behind the definition of distributions is that we consider, (2.1) as a function of all possible test functions , in other words, we will be considering the Z mapping test f W  7! f .x/ .x/ dx: R

Before we can do so, we first have to specify what functions will be allowed as test functions. The first requirement is that all these functions be complex-valued. Definition 2.5 below, of test functions, refers to compact sets. In this text we will be frequently encountering such sets; therefore we begin by collecting some information on them.

Definition 2.1. An open cover of a set K in Rn is a collection U of open sets in Rn such that their union contains K. That is, for every x 2 K there exists a U 2 U with x 2 U . A subcover is a subcollection E of U still covering K. In other words, E  U and K is contained in the union of the sets U with U 2 E. The set K is said to be compact if every open cover of K has a finite subcover. This concept is applicable in very general topological spaces. Next, recall the concept of a subsequence of an infinite sequence .x.j //j 2N . This is a sequence having terms of the form y.j / D x.i.j // where i.1/ < i.2/ <


; in particular, limj !1 i.j / D 1. Note that if the sequence .x.j //j 2N converges to x, every subsequence of this sequence also converges to x. ˛ J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_2, © Springer Science+Business Media, LLC 2010

17

18

2 Test Functions

For the sake of completeness we prove the following theorem, which is known from analysis (see [7, Sect. 1.8]). Theorem 2.2. For a subset K of Rn the following properties (a) – (c) are equivalent. (a) K is bounded and closed. (b) Every infinite sequence in K has a subsequence that converges to a point of K. (c) K is compact. Q Proof. (a) ) (c). We begin by proving that a cube B D jnD1 Ij is compact. Here Ij denotes a closed interval in R of length l, for every 1  j  n. Let U be an open cover of B; we assume that it does not contain a finite cover of B and will show that this assumption leads to a contradiction. When we bisect a closed interval I of length l, we obtain I D I .l/ [ I .r/ , where .l/ I and I .r/ are closed intervals of length l=2. Consider the cubes of the form Q B 0 D jnD1 Ij0 , where for every 1  j  n we have made a choice Ij0 D Ij.l/ or

Ij0 D Ij.r/ . Then B equals the union of the 2n subcubes B 0 . If it were possible to cover each of these by a finite subcollection E of U, the union of these E would be a finite subcollection of U covering B, in contradiction to the assumption. We conclude that there is a B 0 that is not covered by a finite subcollection of U. Applying mathematical induction, we thus obtain a sequence .B .t / / t 2N of cubes with the following properties: (i) B .1/ D B and B .t /  B .t 1/ for every t 2 Z2 . Q (ii) B .t / D jnD1 Ij.t / , where Ij.t / denotes a closed interval of length 2 (iii) B .t / is not covered by a finite subcollection of U. From (i) we now have, for every j , Ij.t /  Ij.t

1/

t

l.

, that is, the left endpoints lj.t / of the

Ij.t / , considered as a function of t, form a monotonically nondecreasing sequence

in R. This sequence is bounded; indeed, lj.t / 2 Ij.s/ when t  s. As t ! 1, the

sequence therefore converges to an lj 2 R; we have lj 2 Ij.s/ because Ij.s/ is closed. Conclusion: the limit point l WD .l1 ; : : : ; ln / belongs to B .s/ , for every s 2 N. Because U is a cover of B and l 2 B, there exists a U 2 U for which l 2 U . Since U is open, there exists an  > 0 such that x 2 Rn and jxj lj j <  for all j implies that x 2 U . Choose s 2 N with 2 s < . Because l 2 B .s/ , the fact that x 2 B .s/ implies that jxj lj j  2 s <  for all j ; therefore x 2 U . As a consequence, B .s/  U , in contradiction to the assumption that B .s/ was not covered by a finite subcollection of U. Now let K be an arbitrary bounded and closed subset of Rn and U an open cover of K. Because K is bounded, there exists a closed cube B that contains K. Because K is closed, the complement C WD Rn n K of K is open. The collection e WD U [ fC g covers K and C , and therefore Rn , and certainly B. In view of the U e Removing C from E, z we foregoing, B is covered by a finite subcollection Ez of U. obtain a finite subcollection E of U; this covers K. Indeed, if x 2 K, there exists

2 Test Functions

19

U 2 Ez with x 2 U . Since U cannot equal C , we have U 2 E. (c) ) (b). Suppose that .x.j // is an infinite sequence in K that has no subsequence converging in K. This means that for every x 2 K there exist an .x/ > 0 and an N.x/ for which kx x.j /k  .x/ whenever j > N.x/. Let U.x/ D f y 2 K j ky

xk < .x/ g:

The U.x/ with x 2 K form an open cover of K; condition (c) implies the existence of a finite subset F of K such that for every x 2 K there is an f 2 F with x 2 U.f /. Let N be the maximum of the N.f / with f 2 F ; then N is well-defined because F is finite. For every j we find that an f 2 F exists with x.j / 2 U.f /, and therefore j  N.f /  N . This is in contradiction to the unboundedness of the indices j . (b) ) (a). Suppose that K satisfies (b). If K is not bounded, we can find a sequence .x.j //j 2N with kx.j /k  j for all j . There is a subsequence .x.j.k///k2N that converges and that is therefore bounded, in contradiction to kx.j.k//k  j.k/  k for all k. In order to prove that K is closed, suppose limj !1 x.j / D x for a sequence .x.j // in K. This contains a subsequence that converges to a point y 2 K. But the subsequence also converges to x, and in view of the uniqueness of limits we conclude that x D y 2 K.  The preceding theorem contains the Bolzano–Weierstrass Theorem, which states that every bounded sequence in Rn has a convergent subsequence; see [7, Theorem 1.6.3]. The implication (a) ) (c) is also referred to as the Heine–Borel Theorem; see [7, Theorem 1.8.18]. However, linear spaces consisting of functions are usually of infinite dimension. In normed linear spaces of infinite dimension, “compact” is a much stronger condition than “bounded and closed,” while in such spaces (b) and (c) are still equivalent. As a first application of compactness we obtain conditions that guarantee that disjoint closed sets in Rn possess disjoint open neighborhoods; see Lemma 2.3 below and its corollary. To do so, we need some definitions, which are of independent interest. Introduce the set of sums A C B of two subsets A and B of Rn by means of A C B WD f a C b j a 2 A; b 2 B g:

(2.2)

It is clear that A C B is bounded if A and B are bounded. Also, A C B is closed whenever A is closed and B compact. Indeed, suppose that the sequence .cj /j 2N in A C B converges in Rn to c. One then has cj D aj C bj for some aj 2 A, bj 2 B. By the compactness of B, a subsequence .bj.k/ /k2N converges to a b 2 B. Consequently, the sequence with terms aj.k/ D cj.k/ bj.k/ converges to a WD c b as k ! 1. Because A is closed, a lies in A. The conclusion is that c 2 A C B. In particular, A C B is compact whenever A and B are both compact. An example of two closed subsets A and B of R for which A C B is not closed is the pair A D Z m, while m C 1=n converges to m as n ! 1. Furthermore, the distance d.x; U / from a point x 2 Rn to a set U  Rn is defined by d.x; U / D inff kx uk j u 2 U g: (2.3) Note that d.x; U / D 0 if and only if x 2 U , the closure of U in Rn . The ıneighborhood Uı of U is given by (see Fig. 2.1) Uı D f x 2 Rn j d.x; U / < ı g:

(2.4)

Fig. 2.1 Example of a ı-neighborhood

Observe that x 2 Uı if and only if a u 2 U exists with kx uk < ı. Using the notation B.uI ı/ for the open ball of center u and radius ı, this gives [ Uı D B.uI ı/; u2U

which implies that Uı is an open set. Also, B.uI ı/ D fug C B.0I ı/, and therefore Uı D U C B.0I ı/: Finally, we define U ı as the set of all x 2 U for which the ı-neighborhood of x is contained in U . Note that U ı equals the complement of .Rn n U /ı and that consequently, U ı is a closed set. Now we are prepared enough to obtain the following two results on separation of sets. Lemma 2.3. Let K  Rn be compact and A  Rn closed, while K \ A D ;. Then there exists ı > 0 such that Kı \ Aı D ;. Proof. Assume the negation of the conclusion. Then there exists an element x.j / 2 K1=j \ A1=j , for every j 2 N. Therefore, one can select y.j / 2 K and a.j / 2 A satisfying ky.j /

x.j /k
0 and ˛.x/ D 0 for x  0. Then ˛ 2 C 1 .R/ with ˛.x/ > 0 for x > 0, and supp ˛ D R0 . Proof. The only problem is the differentiability at 0; see Fig. 2.2. From the power series for the exponential function one obtains, for every n 2 N, the estimate e y  yn nŠ for all y  0. Hence 1 nŠ ˛.x/ D 1=x  D nŠ x n .x > 0/: 1=x n e This tells us that ˛ is differentiable at 0, with ˛ 0 .0/ D 0. As regards the higher-order derivatives, we note that for x > 0 the function ˛ satisfies the differential equation ˛.x/ ˛ 0 .x/ D 2 : x By applying this in the induction step we obtain, with mathematical induction on k, ˛ .k/ .x/ D pk

1 x

˛.x/;

where the pk are polynomial functions inductively determined by p0 .y/ D 1

and

pkC1 .y/ D pk .y/


pk 0 .y/ y 2 :

In particular, pk is of degree 2k and therefore satisfies an estimate of the form jpk .y/j  c.k/ y 2k From this we derive the estimate

.y  1/:

2 Test Functions

23

j˛ .k/ .x/j  c.k/ nŠ x n

2k

.0 < x  1/:

If we then choose n  2k C 2, we obtain, with mathematical induction on k, that ˛ 2 C k .R/ and ˛ .k/ .0/ D 0.  Lemma 2.8. Let ˛ 2 C 1 .R/ be as in the preceding lemma. Let a and b 2 R with a < b. Define the function ˇ D ˇa;b by ˇ.x/ D ˇa;b .x/ D ˛.x

a/ ˛.b

x/:

One then has ˇ 2 C 1 .R/ with ˇ > 0 on  a; b Œ and supp ˇ D Œ a; b . FurtherZ more, I.ˇ/ WD ˇ.x/ dx > 0: R

The function D a;b WD R R .x/ dx D 1.

1 I.ˇ /

ˇ has the same properties as ˇ (see Fig. 2.2), while

1 ã2

0.598

1

0

1

-1

2

2

2

Fig. 2.2 Graphs of ˛ as in Lemma 2.7 on Œ 0; 1=2  and of scales adjusted

1;2

as in Lemma 2.8, with the

Lemma 2.9. Let aj and bj 2 R with aj < bj and define aj ;bj 2 C01 .R/ as in the preceding lemma, for 1  j  n. Write x D .x1 ; : : : ; xn / 2 Rn . For a and b 2 Rn , define the function a;b W Rn ! R by (see Fig. 2.3) a;b .x/

D

Then we have a;b

supp

2 C 1 .Rn /;

a;b

D

n Y

Œ aj ; bj ;

j D1

n Y

aj ;bj .xj /:

j D1

a;b

> 0 on Z

n Y 

aj ; bj

j D1 Rn

a;b .x/ dx



;

D 1:

For a complex number c, the notation c  0 means that c is a nonnegative real number. For a complex-valued function f , f  0 means that f .x/  0 for every x

24

2 Test Functions

Fig. 2.3 Graph of

. 1;2/;.2;3/

as in Lemma 2.9

in the domain space of f . If g is another function, one writes f  g or g  f if f g  0.

Corollary 2.10. For every point p 2 Rn and every neighborhood U of p in Rn there exists a  2 C01 .Rn / with the following properties:

(a)   0 and .p/ > 0. (b) Rsupp   U . (c) Rn .x/ dx D 1.

By superposition and taking limits of the test functions thus constructed we obtain a wealth of new test functions. For example, consider  as in Corollary 2.10 and set 1 1   .x/ WD n  x : (2.5)   Further, let f be an arbitrary function in C0 .Rn /, the space of all continuous functions on Rn with compact support; these are easily constructed in abundance. By straightforward generalization of Lemma 1.6 to Rn , the functions f WD f   converge uniformly on Rn to f , as  # 0. The f are test functions, in other words, f 2 C01 .Rn /;

(2.6)

as one can see from Lemma 2.18 below. Consequently, for every f 2 C0 .Rn / there exists a family of functions in C01 .Rn / that converges to f uniformly on compact subsets. We say that C01 .Rn / is dense in C0 .Rn /; see Definition 8.3 below for the general definition of dense sets. Lemma 2.11. For every a 2 Rn and r > 0 there exists  2 C01 .Rn / satisfying supp   B.aI 2r/;

0    1;

D1

on B.aI r/:

Proof. By translation and rescaling we see that it is sufficient to prove the assertion for a D 0 and r D 1. By Lemma 2.8 we can find ˇ 2 C 1 .R/ such that ˇ > 0 on R3  1; 3 Œ and supp ˇ D Œ 1; 3 , while I D 1 ˇ.x/ dx > 0. Hence we may write

2 Test Functions

.x/ WD

1 I

Z

25 3

ˇ.t/ dt:

x

Then  2 C 1 .R/, 0    1, while  D 1 on  1; 1  and  D 0 on Œ 3; 1 Œ . Now set .x/ D .kxk2 / D .x12 C


C xn2 /.  We now review notation that will be needed for Definition 2.13 and Lemma 2.18 below, among other things. In this text we use the following notation for higherorder derivatives. A multi-index is a sequence ˛ D .˛1 ; : : : ; ˛n / 2 .Z0 /n of n nonnegative integers. The sum j˛j WD

n X

˛j

j D1

is called the order of the multi-index ˛. For every multi-index ˛ we write @˛x WD

@˛ WD @˛1 1 B


B @˛nn ; @x ˛

where

@j WD

@ : @xj

(2.7)

Furthermore, we use the shorthand notation @˛ D

@˛ @x ˛

when we want to differentiate only with respect to the variables xj . The crux is that the Theorem on the interchangeability of the order of differentiation (see for instance [7, Theorem 2.7.2]), which holds for functions sufficiently often differentiable, allows us to write every higher-order derivative in the form (2.7); also refer to the introduction to Chap. 6. Finally, in the case of n D 1, we define @ as @.1/ . Remark 2.12. In (2.7) we defined the partial derivatives @˛ f of arbitrary order of a function f depending on an arbitrary number of variables. For the kth-order derivatives of the product f g of two functions f and g that are k times continuously differentiable, we have Leibniz’s formula: ! X ˛ @˛ .f g/ D @˛ ˇ f @ˇ g; (2.8) ˇ ˇ ˛

for j˛j D k. Here ˛ D .˛1 ; : : : ; ˛n / and ˇ D .ˇ1 ; : : : ; ˇn / are multi-indices, while ˇ  ˛ means that for every 1  j  n one has ˇj  ˛j . The n-dimensional binomial coefficients in (2.8) are given by ! ! ! n Y pŠ ˛ ˛j p WD ; ; where D ˇ ˇj .p q/Š qŠ q j D1

26

2 Test Functions

for p and q 2 Z with 0  q  p. Formula (2.8) is obtained with mathematical induction on the order k D j˛j of differentiation, using Leibniz’s rule @j .f g/ D g @j f C f @j g

(2.9) ˛

in the induction step.

Definition 2.5 is supplemented by the following, which introduces a notion of convergence in the infinite-dimensional linear space C01 .X /: Definition 2.13. Let j and  2 C01 .X /, for j 2 N and X an open subset of Rn . The sequence .j /j 2N is said to converge to  in the space C01 .X / of test functions as j ! 1, notation lim j D  in C01 .X /; j !1

if the following two conditions are both met: (a) there exists a compact subset K of X such that supp j  K for all j ; (b) for every multi-index ˛ the sequence .@˛ j /j 2N converges uniformly on X to @˛ . ˛ Observe that the data above imply that supp   K. The notion of convergence introduced in the definition above is very strong. The stronger the convergence, the fewer convergent sequences there are, and the more readily a function defined on C01 .X / will be continuous. Now we combine compactness and test functions in order to introduce the useful technical tool of a partition of unity over a compact set. 1 2 3

-2

2

-2

2

Fig. 2.4 Example of a partition of unity

Definition 2.14. Let K be a compact subset of an open subset X of Rn and U an open cover of K. A C01 .X / partition of unity over K subordinate to U is a finite sequence 1 ; : : : ; l 2 C01 .X / with the following properties (see Fig. 2.4): P (i) j  0, for every 1  j  l, and jl D1 j  1 on X ; P (ii) there exists a neighborhood V of K in X with jl D1 j .x/ D 1, for all x 2 V ;

(iii) for every j there is a U D U.j / 2 U for which supp

j

 U.

˛

2 Test Functions

27

Given a function f on X , write fj D j f in the notation above. Then we obtain P functions fj with compact support contained in U.j /, while f D jl D1 fj on V . Furthermore, all fj 2 C k if f 2 C k . In the applications, the U 2 U are small neighborhoods of points of K with the property that we can reach certain desired conclusions for functions with support in U . For example, partitions of unity were used in this way in [7, Theorem 7.6.1] to prove the integral theorems for open sets X  Rn with C 1 boundary. Theorem 2.15. For every compact set K contained in an open subset X of Rn and every open cover U of K there exists a C01 .X / partition of unity over K subordinate to U. Proof. For every a 2 K there exists an open set Ua 2 U such that a 2 Ua . Select ra > 0 such that B.aI 2ra /  Ua \ X . By criterion (c) in Theorem 2.2 for compactness, there exist finitely many a.1/; : : : ; a.l/ such that K is contained in the union V of the B.a.j /; ra.j / /, for 1  j  l. Now select the corresponding j 2 C01 .X / as in Lemma 2.11 and set j Y D  I D  .1 i / .1  j < l/: (2.10) 1 1 j C1 j C1 i D1

Then the conditions (i) and (iii) for a C01 .X / partition of unity subordinate to U are satisfied by the 1 ; : : : ; l . The relation j X i D1

i

D1

j Y

.1

i /

(2.11)

i D1

is trivial for j D 1. If (2.11) is true for j < l, then summing (2.10) and (2.11) yields (2.11) for j C 1. Consequently (2.11) is valid for j D l, and this implies that the 1 ; : : : ; l satisfy condition (ii) for a partition of unity with V as defined above.  Corollary 2.16. Let K be a compact subset in Rn . For every open neighborhood X of K in Rn there exists a  2 C01 .Rn / with 0    1, supp   X and  D 1 on an open neighborhood of K. In particular, for ı > 0 sufficiently small, we can find such a function  with  D 1 on Kı . Proof. Consider the open cover fX g of K and let 1 ; : : : ; lPbe a subordinate partition of unity over K as in the preceding theorem. Then  D j j satisfies all requirements. For the second assertion, apply Corollary 2.4 and the preceding result with K replaced by C as in the corollary.  The function  is said to be a cut-off function for the compact subset K of Rn . Through multiplication by  we can replace a function f defined on X by a function g with compact support contained in X . Here g D f on a neighborhood of K and g 2 C k if f 2 C k .

28

2 Test Functions

We still have to verify the claim in (2.6); it follows from Lemma 2.18 below. In the case of k equal to 1, another proof will be given in Theorem 11.2. Later on, in demonstrating Theorem 11.22, we will need an analog of Corollary 2.16 in the case of not necessarily compact sets. To that end, we derive Lemma 2.19 below. In preparation, we introduce some concepts that are useful in their own right. Definition 2.17. Let X  Rn be an open subset. A function f W X ! C is said to be locally integrable if for every a 2 X , there exists an open rectangle B  X with the properties that a 2 B and that f is integrable on B. The characteristic function or indicator function 1U of a subset U of Rn is defined by 1U .x/ D 1

if

x 2 U;

1U .x/ D 0

U is said to be measurable if 1U is locally integrable.

if x 2 Rn n U: ˛

For the purposes of this book it will almost invariably be sufficient to interpret the concept of integrability, as we use it here, in the sense of Riemann. However, for distributions it is common to work with Lebesgue integration, which leads to a more comprehensive theory. Loosely speaking, Lebesgue’s theory is more powerful than Riemann’s, in the sense that it leads to a process of integration for more functions and to a simpler treatment of singular behavior of functions. On the other hand, a thorough treatment of Lebesgue integration is technically more demanding than that of Riemann integration. The distinction between the two concepts rarely arises in the case of the functions that will be encountered in this text. It is primarily in the description of spaces of all functions satisfying certain properties that the difference becomes important. Readers who are not familiar with Lebesgue integration can find a way around this by restricting themselves to locally integrable functions with an absolute value whose improper Riemann integral exists, and otherwise taking our assertions about Lebesgue integration for granted. Some of these assertions do not apply to Riemann integration, but this need not be a reason for serious concern; we will discuss this issue when the need arises. Nonetheless, for the benefit of readers who are interested in the relation between the theory of distributions and that of (Lebesgue) integration we concisely but fairly completely discuss integration in Chap. 20. In particular, local integrability is introduced in Definition 20.37. Lemma 2.18. Let f be locally integrable on Rn and g 2 C0k .Rn /. Then f  g 2 C k .Rn / and supp .f  g/  supp f C supp g:

Here supp f C supp g is a closed subset of Rn , compact if f , too, has compact support; in that case f  g 2 C0k .Rn /. Proof. We study .f  g/.x/ for x 2 U , where U  Rn is bounded and open. Define h.x; y/ WD f .y/ g.x y/. Then the function x 7! h.x; y/ belongs to C k .U / for every y 2 Rn , because for every multi-index ˛ 2 .Z0 /n with j˛j  k,

2 Test Functions

29

@˛ h .x; y/ D f .y/ @˛ g.x @x ˛

y/:

Let B.r/ be a ball about 0 of radius r > 0 such that supp g  B.r/. Then there exists an r 0 > 0 with B.r/ C U  B.r 0 /; furthermore, the characteristic function @˛ h  of B.r 0 / is integrable on Rn . For every x 2 U the function @x ˛ .x;
/ vanishes outside B.r 0 /; consequently, the latter function does not change upon multiplication by . In addition, we have ˇ @˛ h ˇ ˇ ˇ ˇ ˛ .x; y/ˇ  sup j@˛ g.x/j jf .y/j .y/ @x x2Rn

..x; y/ 2 U  Rn /;

where jf j  is an absolutely integrable function on Rn . In view of a well-known theorem on changing the order of differentiation and integration (inRthe context of Riemann integration, see [7, Theorem 6.12.4]) we then know that Rn h.x; y/ dy is a C k function of x whose derivatives equal the integral with respect to y of the corresponding derivatives according to x of the integrand h.x; y/. Furthermore, h.x; y/ D 0 if x 2 U and y … KU , where KU WD .supp f / \ .U C . supp g//: Now suppose u … supp f C supp g. Then there exists a neighborhood U of u in Rn such that x … supp f C supp g for all x 2 U , because the complement of supp f Csupp g is open. But this means KU D ;, which implies that .f g/.x/ D 0 for all x 2 U .  R Lemma 2.19. Let  2 C01 .Rn /,   0, .x/ dx D 1, and kxk  1 if x 2 supp . Suppose that the subset U of Rn is measurable; see Definition 2.17. Select ı > 0 arbitrarily and define, for 0 <  < ı,  .x/ D Then

1 1  x  n 

U;  2 C 1 .Rn /;

and

0  U;  1;

Finally, U;  D 1 on a neighborhood of U

U;  WD 1U   : supp U;   Uı :

ı.

Proof. We have U;  2 C 1 .Rn / by Lemma 2.18. Because   0, we obtain 0 D 0    1U    1   D 1. / D 1: Furthermore, if B denotes the -neighborhood of 0, the support of U;  is contained in supp 1U C supp   U C B , and therefore also in the ı-neighborhood of U as  < ı. The latter conclusion is reached when we replace U by V D Rn n U ; note that 1 U;  D 1   1U   D .1 1U /   D V;  . 

30

2 Test Functions

Usually in applications of Lemma 2.19, the set U is either open or closed, but even then its characteristic function 1U will not always be locally integrable in the sense of Riemann (see [7, Exercise 6.1]), whereas it is in the sense of Lebesgue; see Proposition 20.36. The only occasion in the text where this issue might play a role is in the proof of Theorem 11.22. Finally, there are many situations in which one prefers to use, instead of  2 C01 .Rn /, functions like (see Fig. 2.5)

.x/ D n .x/ D 

n 2

e

kxk2

:

(2.12)

The numerical factor is chosen such that the integral of over Rn equals 1; this

Fig. 2.5 Graph of 2 , with different horizontal and vertical scales

is the Gaussian density or the probability density of the normal distribution, with expectation 0 and variance Z n (2.13) kxk2 n .x/ dx D : n 2 R For larger values of kxk the values .x/ are so extremely small that in many situations we may just as well consider as having compact support. Naturally, this is 2 only relative: if we were to use to test a function that grows at least like e kxk as kxk ! 1, this would utterly fail. R For the sake of completeness we recall the calculation of I WD

n .x/ dx. n Q Because n .x/ D jnD1 1 .xj /, we have In D .I1 /n . The change of variables x D r.cos ˛; sin ˛/ in a dense open subset of R2 now yields

2 Problems

I2 D

31

1 

or

Z

e R2

.x12 Cx22 /

d.x1 ; x2 / D Z

e R

x2

1 

dx D

Z

R>0

Z



e

r2



p :

r d˛ dr D 1; (2.14)

We refer to [7, Exercises 2.73 and 6.15, or 6.41] for other proofs of this identity. For the computation of (2.13) introduce spherical coordinates in Rn by x D r! with r > 0 and ! belonging to the unit sphere f x 2 Rn j kxk D 1 g in Rn ; see [7, Example 7.4.12] and (13.37). Next use the substitution r 2 D s and formulas (13.30) and (13.31) below.

Problems  2.1. Let U  Rn be a closed set. Prove that the corresponding distance function satisfies jd.x; U / d.y; U /j  kx yk, for all x and y 2 Rn .  2.2. Let  2 C01 .R/,  ¤ 0, and 0 … supp . Decide whether the sequence .j /j 2N converges to 0 in C01 .R/ if:

(i) j .x/ D j (ii) j .x/ D j (iii) j .x/ D e

1

p j

.x j /. .j x/. Here p is a given positive integer. .j x/.

In each of these cases verify that for every x 2 R and every k 2 Z0 , the sequence .j.k/ .x//j 2N converges to 0, and in addition, that in case (i) the convergence is even uniform on R.  2.3. Let  and  be as in Lemma 2.19. Prove that for every 2 C01 .X /, the 1 function   converges to in C0 .X / as  # 0.

2.4. ConsiderR ˇ 2 C01 .R/ with ˇ  0 and ˇ.x/ D 0 if and only if jxj  1. Further assume that ˇ.x/ dx D 1. Let 0 <  < 1, ˇ .x/ D 1 ˇ. x /, I D Œ 1; 1   R, and let D 1I  ˇ . Determine where one has D 0, where 0 < < 1, and where D 1, and in addition, where 0 D 0, 0 > 0, and 0 < 0, respectively. Now let .x/ D ˇ.x1 / ˇ.x2 / for x 2 R2 and let U D I  I , a square in the plane. Consider  D U;  as in Lemma 2.19. Prove that .x/ D .x1 / .x2 /. Determine where one has  D 0, or 0 <  < 1, or  D 1, and in addition, for j D 1 and 2, where @j  D 0, @j  > 0, @j  < 0. Verify that if 0 <  < 1, there is a j such that @j  ¤ 0. Prove by the Submersion Theorem (see [7, Theorem 4.5.2]) that for every 0 < c < 1 the level set N.c/ WD f x 2 R2 j .x/ D c g is a C 1 curve in the plane. Is this also true for the boundary of the support of  and of 1 ? Give a description, as detailed as possible, of the level curves of , including a sketch.  2.5. For  > 0, define  2 C 1 .R/ by 1 2 2

 .x/ D p e x = :  

32

2 Test Functions

Calculate j
j   . Prove that this function is analytic on R and examine how closely it approximates the function j
j. Also calculate its derivatives of first and second order. See Fig. 2.6.

25 0.5

-0.05

0.05

-0.5

0.5

Fig. 2.6 Illustration for Problem 2.5. Graphs of  and j
j   with  D 1=50

2.6. Let U be a proper open subset of Rn . Let .x/ be as in (2.12) and  .x/ D 1 1  n .  x/, for  > 0. Denote the “probability of distance to 0 larger than r” by Z .r/ D

.x/ dx: kxk>r

Give an estimate of the r for which .r/ < 10 6 . Prove that  WD 1U   is analytic and that 0 <  < 1. Further prove that  .x/  .ı=/ if d.x; U / D ı > 0; finally, show that  .x/  1 .ı=/ if d.x; Rn n U / D ı > 0. 2.7. Show, for a > 0, that Z  2  n2 2 e akxk =2 dx D a Rn

and

Z

Rn

kxk2 e

akxk2 =2

dx D

n  2  n2 : a a

Chapter 3

Distributions

For an arbitrary linear space E over C, a C-linear mapping W E ! C is also called a linear form or linear functional on E. Definition 3.1. Let X be an open subset of Rn . A distribution on X is a linear form u on C01 .X / that is also continuous in the sense that lim u.j / D u./

j !1

as

lim j D 

j !1

in

C01 .X /:

Phrased differently, in this case continuity means preservation of convergence of sequences. The space of all distributions on X is denoted by D 0 .X /. ˛ The notation derives from the notation D.X / used by Schwartz for the space C01 .X / of test functions equipped with the notion of convergence from Definition 2.13. (The letter D denotes differentiable.) By the linearity of u, the assertion u.j / ! u./ is equivalent to u.j / D u.j / u./ ! 0, while j !  is equivalent to j  ! 0. This implies that the continuity of a linear form u is equivalent to the assertion lim u.

j !1

j/

D0

as

lim

j !1

j

D 0 in

C01 .X /:

R Example 3.2. We have u 2 D 0 .R/ if u./ D R .x/ dx, for all  2 C01 .R/. Indeed, u is well-defined because  is continuous with compact support; the linearity of u is well-known and its continuity can be proved as follows. If limj !1 j D 0 in C01 .R/, then there exists m > 0 such that supp j  Œ m; m  for all j , while the convergence of the j to the zero function is uniform on Œ m; m . Therefore we may interchange taking the limit and integration to obtain Z m Z m Z m lim u.j / D lim j .x/ dx D lim j .x/ dx D 0 dx D 0: ˛ j !1

j !1

m

m j !1

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_3, © Springer Science+Business Media, LLC 2010

m

33

34

3 Distributions

Example 3.3. We have PV x1 2 D 0 .R/ if we define this linear form, in the notation of Example 1.3, by Z  1 1 .x/ 1 PV W C0 .R/ ! C ./ D PV dx: with PV x x R x

1 Indeed, consider arbitrary R m  2 C0 .R/. Then there exists m > 0 with supp   Œ m; m . We may write m j log jxjj dx DW c.m/ > 0 on account of the convergence of this integral. Hence, (1.3) leads to

ˇ 1  ˇˇ ˇ ./ˇ  c.m/ sup j 0 .x/j; ˇ PV x jxjm

and this implies that PV x1 is a continuous linear form on C01 .R/. See Example 5.7 for another proof. ˛ For every complex linear space E with a notion of convergence, it is customary to denote the space of continuous linear forms on E by E 0 ; this space is also referred to as the topological dual of E. If E is of finite dimension, every linear form on E is automatically continuous and E 0 is a complex linear space of the same dimension as E. For function spaces E of infinite dimension, this does not apply, and it is therefore sensible also to require continuity of the linear forms. This will open up a multitude of conclusions that one could not obtain otherwise, while the condition remains sufficiently weak to allow a large space of linear forms. Remark 3.4. The study of general linear spaces E with a notion of convergence is referred to as functional analysis; this is outside the scope of this text. For the benefit of readers who (justifiably) find the preceding paragraph too vague, we add some additional clarification. We require that addition and scalar multiplication in E be continuous with respect to the notion of convergence in E. That is, xj C yj ! x C y and cj xj ! c x if xj ; x; yj ; y 2 E, cj 2 C, and xj ! x, yj ! y and cj ! c as j ! 1. Furthermore, we impose the condition that limits of convergent sequences be uniquely determined. (The latter is a consequence of the usual requirement that E be a topological space having the Hausdorff property, which means that each two distinct points have nonintersecting neighborhoods.) In that case, E with its notion of convergence is also known as a topological linear space. A linear mapping u W E ! C
is said to be continuous if u xj ! u.x/ as xj ! x in E. If E is of finite dimension, there exists a basis .ei /1i n of E, where n is the dimension of E. ThePmapping that assigns to x 2 E the coordinates .x1 ; : : : ; xn / 2 Cn for which x D niD1 xi ei is a linear isomorphism from E to Cn . The assertion is that via this linear isomorphism, convergence in E is equivalent to the usual coordinatewise convergence in Cn . Proof. The property that E is a topological linear space immediately leads to the conclusion that coordinatewise convergence implies convergence in E. We now prove the converse by mathematical induction on n.

3 Distributions

35

Let xj ! x in E as j ! 1. Select 1  i  n and let cj , or c, be the ith coordinate of xj , or x, respectively. Suppose that the complex numbers cj do not converge to c as j ! 1; we will show that this assumption leads to a contradiction. By passing ˇ to aˇ subsequence if necessary, we can arrange for the existence of a ı > 0 with
ˇcj c ˇ  ı for all j . This implies that the sequence with terms dj D 1= cj c is bounded. Passing to a subsequence once again if necessary, we can arrange that there is a d 2 C for which dj ! d in C; here we apply Theorem 2.2.(b). With respect to the E-convergence, this leads to yj WD dj .xj x/ ! d 0 D 0 as j ! 1. On the other hand, for every j , the ith coordinate of yj equals 1. This means that for every j and k, the vector yj yk lies in the .n 1/-dimensional linear subspace E0 of E consisting of the elements of E whose ith coordinates vanish. With respect to the E-convergence in E0 , the yj yk converge to zero if both j and k go to infinity; therefore, by the induction hypothesis, the yj yk converge coordinatewise to zero as j; k ! 1. In other words, the yj form a Cauchy sequence in E ' Cn with respect to coordinatewise convergence. This implies that y 2 E exists for which yj ! y by coordinates as j ! 1, which in turn implies that yj ! y in E. Because we had already obtained that yj ! 0 in E, it follows from the uniqueness of limits that y D 0. This leads to a contradiction to the fact that the yj converge coordinatewise to y, because the ith coordinate of yj equals 1 and the ith coordinate of y D 0 equals 0. This negates the assumption that the complex numbers cj do not converge to c. The conclusion therefore is that for every 1  i  n, the ith coordinate of xj converges to the ith coordinate of x as j ! 1. In other words, xj ! x in E implies the coordinatewise convergence of the xj to x as j ! 1.  For an arbitrary linear function u on a linear space E with basis .ei /1i n , one X  X n n has u.x/ D u xi ei D xi u .ei / : i D1

i D1

From this we can see that u is continuous with respect to coordinatewise convergence and is determined by its “matrix coefficients” ui D u .ei /, for 1  i  n. The conclusion is that for a topological linear space E of finite dimension, E 0 equals the space of all linear forms on E and this is a complex-linear space of the same dimension as E. ˛ D 0 .X / is a complex-linear space with respect to addition and scalar multiplication defined by .u C v/./ D u./ C v./

and

. u/./ D  u./;

for u and v 2 D 0 .X /,  2 C01 .X / and  2 C. It is a mere matter of writing out the definitions to show that u C v and  u 2 D 0 .X / if u; v 2 D 0 .X / and  2 C, and likewise, that with these definitions D 0 .X / does indeed become a linear space over the complex numbers C.

36

3 Distributions

Many functions can be interpreted as distributions. To explain this, we recall Definition 2.17 of local integrability of a function. Theorem 3.5. For every locally integrable function f on X , Z .test f /./ D f .x/ .x/ dx . 2 C01 .X //

(3.1)

X

defines a distribution u D test f on X . Indeed, let j 2 C01 .X /, K  X compact, and supp j  K for all j . Then limj !1 u.j / D 0 if limj !1 j D 0 uniformly on K. Proof. First we show that u is well-defined. By the compactness of K WD supp , there are finitely many a.i / 2 X such that K is contained in the union of the rectangles Ba.i / . This means that the sum of the characteristic functions R of the Ba.i / is a majorant of the characteristic function of K. Using the identity K g.x/ dx D R 1K .x/ g.x/ dx, we deduce the absolute convergence of the integral in (3.1) from Z XZ jf .x/ .x/j dx  jf .x/j j.x/j dx  c sup j.x/j; K

Ba.i /

i

where

cD

XZ i

Ba.i /

x2K

jf .x/j dx > 0:

The continuity of u now follows from the fact that for all j , ju.j /j can be estimated by a constant times the supremum norm of j . Indeed, the supports of the j are contained in a fixed compact K  X as j ! 0 in C01 .X /.  Lemma 3.6. The mapping f 7! test f is linear and injective from C.X /, the linear space of all continuous functions on X , to D 0 .X /, the space of distributions on X . Proof. By writing out the definitions one immediately verifies that the mapping is linear. A linear mapping is known to be injective if and only if its null space equals 0. In other words, assuming .test f /./ D 0 for all test functions , we have to prove that f .x/ D 0 for all x 2 X . With  as in (2.5), we obtain, for x 2 X and  > 0 sufficiently small, .f   /.x/ D .test f /.Tx S /: In the right-hand side the reflection S of a function .S /.z/ D

(3.2)

is defined by

. z/:

We have already come across the translation Tx , in (1.13). This now yields .Tx S /.y/ D .S /.y

x/ D  .x

y/;

3 Distributions

37

which enables us to see that (3.2) is another way of writing (1.12), with  replaced by  . The assumption .test f /./ D 0, for all test functions , now implies that we obtain .f   /.x/ D 0, for every  > 0 and for all x. But in the text preceding (2.6) we have already seen that f   converges uniformly on compact sets to f as  # 0; and certainly, therefore, f .x/ D lim.f   /.x/ D 0: #0



Remark 3.7. This lemma justifies the usual identification of the continuous function f with the distribution test f . In other words, C.X / is identified with the linear subspace f test f j f 2 C.X / g of D 0 .X /, and the premodifier “test” preceding the continuous function f is omitted when f is regarded as a distribution. A small amusement: 1 D test 1 is integration of test functions. In this case the left-hand side is identified with the function that constantly equals 1; when we are piling identifications on top of each other, we should not be surprised that the notation becomes ambiguous. For arbitrary locally integrable functions f the situation is a little subtler. A function f is said to be equal to g almost everywhere if the set f x 2 X j f .x/ ¤ g.x/ g has Lebesgue measure equal to 0; see Definition 20.14. In this case test f D test g, while f D g is not necessarily true in the strict sense that f .x/ D g.x/ for all x 2 X . Conversely, if test f D test g, the functions f and g are equal almost everywhere. Below we outline the proof. ˛ Proof. Write h D f g. If  2 C0 .X / there exists a sequence j 2 C01 .X / that converges uniformly to . This gives Z h.x/ .x/ dx D lim .test h/.j / D 0: j !1

From the theory of Lebesgue integration it is known that the validity of this identity, for all  2 C0 .X /, implies that h vanishes almost everywhere; see Theorem 20.38 and Corollary 20.39. While not all readers may be familiar with the latter result, it will be realized that h vanishes almost everywhere if the integral of h over any rectangle U equals 0. 1 If R h is tested with the functions U; 2 C0 .X / from Lemma 2.19, we obtain that  U h.x/ dx D 0 by taking the limit as  # 0.

It is customary in the theory of Lebesgue integration, see (20.18), to identify functions with each other when they are equal almost everywhere. We can therefore say that this custom amounts to interpreting locally integrable functions f as distributions, via the mapping f 7! test f . Again, the word “test” is omitted. An example of a distribution not of the form test f for a locally integrable function f is the Dirac function, or point measure, ıa , situated at the point a 2 X . This is defined by ıa ./ WD .a/ . 2 C01 .X //;

38

3 Distributions

in other words, by evaluating the test function  at the point a. When a D 0, one simply uses the term Dirac function, without further specification, denoting it by ı. If ıa D test f , for a locally integrable function f , the restriction of f to X n fag equals 0 almost everywhere. Because fag has measure 0, we conclude that f equals 0 almost everywhere; thus, test f D 0. This leads to a contradiction, because there exists a test function  with .a/ ¤ 0; consequently ıa ¤ 0. If one equates locally integrable functions with distributions, one should not be fussy about using function notation for distributions. In particular, for arbitrary u 2 D 0 .X / one encounters the notation Z u.x/ .x/ dx D hu; i D h; ui . 2 C01 .X //: (3.3) u./ D X

This will not lead to problems, provided one does not conclude that there is any meaning in the phrase “the value u.x/ of the distribution u at the point x.” In the case of the Dirac function, for example, this certainly entails some problems at x D 0. So much for comments on the Dirac notation Z Z .a/ D ı.x a/ .x/ dx D ı.x/ .x C a/ dx: Rn

Rn

(This formula describes the sifting property of the Dirac function.) Normally, the continuity of a linear form u on C01 .X / is most easily proved by giving an estimate for u./ in terms of a so-called C k norm of  2 C01 .X /; see Example 3.3, for instance. In order to formulate this, we introduce, for every compact subset K of X , the linear space C01 .K/ consisting of all  2 C01 .X / with supp   K. On this space we have, for every k 2 Z0 , the C k norm, defined by kkC k D sup j@˛ .x/j: j˛jk; x2X

In Chap. 8 we establish, in a general context, a relation between the continuity of a linear mapping and estimates in terms of (semi)norms. For linear forms on C01 .X / this looks as follows. Theorem 3.8. Let X be an open subset of Rn . A linear form u on C01 .X / belongs to D 0 .X / if and only if for every compact subset K of X there exist a constant c > 0 and an order of differentiation k 2 Z0 with the property ju./j  c kkC k

. 2 C01 .K//:

(3.4)

Proof. (. j !  in C01 .X / means that a compact subset K of X exists for which j and  2 C01 .K/ for all j , while limj !1 kj kC k D 0 for all k. If u satisfies (3.4), it then follows that u.j / u./ D u.j / converges to 0 as j ! 1.

3 Distributions

39

). Now suppose that u does not satisfy condition (3.4). This means that a compact subset K of X exists such that for every c > 0 and k 2 Z0 , there is a c;k 2 C01 .K/ for which ju.c;k /j > c kc;k kC k . This implies k

c;k kC k


0 such that K  Œ R m; m . If x 2 R2 satisfies m kxk 2 K, then jxi j  kxk  m, for 1  i  2. Since m jxj dx D m2 , we get ju./j  m4

sup

t 2K; jj j2

j .j / .t/j D m4 kkC 2

. 2 C01 .K//:

The assertion now follows from Theorem 3.8. Suppose that u were defined without absolute signs. Then we would have u D 0. This is directly verified by introducing polar coordinates in R2 (see Example 3.14 below). ˛ Example 3.11. If V is a closed k-dimensional C 1 submanifold contained in an open subset X of Rn and f a continuous function on V , then Z f ıV ./ D f .y/ .y/ dy . 2 C01 .X // V

defines a distribution f ıV 2 D 0 .X /. Here the integration over V is Euclidean k-dimensional integration, introduced in [7, Sect. 7.3], for example. Indeed, if K is an arbitrary compact subset of X , there exists a closed ball B in Rn containing K. Then R m 2 R if m D supy2V \B jf .y/j on account of the continuity of f , and also V \B dy < 1. Hence

40

3 Distributions

jf ıV ./j  m kkC 0

Z

dy V \B

. 2 C01 .K//:

(3.5)

If f D 1 we write f ıV D ıV . For V D fpg this notation is consistent with the use of ı that we have seen before. If k D n, the set V is open in X and ıV D 1V , the characteristic function of V . For k D n 1, the distribution f ıV is said to be a layer in X , with density function f . For n D 3 this may be visualized as a charge distribution over a surface V . ˛ Definition 3.12. If u 2 D 0 .X /, the minimal k 2 Z0 for which (3.4) holds, for certain c > 0, is known as the order of u on K. The supremum over all compact subsets K of X of the orders of u on K is called the order of the distribution u on X . In other words, u is of order k if and only if for every compact subset K of X there exists a constant c D c.K/ > 0 such that ju./j  c kkC k

. 2 C01 .K//:

(3.6) ˛

Example 3.13. The estimate (3.5) implies that the order of f ıV 2 D 0 .X / as in Example 3.11 equals 0. Both PV x1 2 D 0 .R/ from Example 3.9 (use Problem 3.3) and u 2 D 0 .R2 / from Problem 3.5.(ii) are distributions of order 1. Furthermore, see Problem 3.2 for an example of a distribution on R of order k, for arbitrary k 2 Z0 . The order of a distribution u can be 1, as for the distribution u on R defined by X u./ D . 1/j  .j / .j / . 2 C01 .R//: (3.7) j 2Z0

˛

To see why this is so, use Problem 3.2 again. Example 3.14. For a > ua ./ D

Z

1, consider R2

jx1 x2 ja  00 .kxk/ dx

. 2 C01 .R//:

R R Then ua is a well-defined distribution, because R2 jx1 x2 ja dx D . R jxja dx/2 is convergent. In addition, ua is of order  2, see also Example 3.10. By introducing polar coordinates in R2 , we obtain that Z  Z 1 j sin 2˛ja d˛ ua ./ D r 2aC1  00 .r/ dr: 2  R>0 Let ca denote a negative constant depending on a. We may then write u c

1 2

 0 .0/. If a >

1 , 2

integration by parts is permitted. This leads to

ua ./ D ca

Z

r 2a  0 .r/ dr R>0

 a>

1 : 2

1 2

./ D

3 Distributions

41

In particular, then, ua is of order  1, for a  12 . Furthermore we obtain u0 D  ı, which is of order 0. Finally, suppose that a > 0. Upon integrating by parts once more, we see that also in this case ua is of order 0. ˛ In the next theorem we demonstrate that distributions of finite order k can be identified with the continuous linear forms on the space C0k .X / of C k functions with compact support in X . In this context, the convergence in C0k .X / is that of Definition 2.13, where condition (b) is required only for all multi-indices ˛ with j˛j  k. Note that C01 .X /  C0k .X /  C0l .X / if k > l, in which there are no equalities. Furthermore, if 1  k > l and the sequence .j / in C0k .X / converges in C0k .X / to , it also converges in C0l .X /. This implies that the restriction to C0k .X / of a continuous linear form on C0l .X / defines a continuous linear form on C0k .X /. In particular, the restriction to C01 .X / of a continuous linear form on C0l .X / is a distribution on X of order  l. A converse assertion can also be proved: Theorem 3.15. Let X be an open subset of Rn , k 2 Z0 , and u a distribution on X of order  k. Then u has a unique extension to a continuous linear form v on C0k .X /. Proof. Let f 2 C0k .X /. With the functions  as in (2.5) write f D f   . We obtain, for every multi-index ˛ with j˛j  k, that @˛ f D .@˛ f /   converges uniformly to @˛ f as  # 0. From Lemma 2.18 it follows that for sufficiently small  > 0, the supports of the f 2 C01 .Rn / are contained in a fixed compact subset of X . For these , therefore, f 2 C01 .X / and these functions converge to f in C0k .X / as  # 0. Thus, if v exists at all, it is given by v.f / D lim u.f /: #0

This implies the uniqueness of a continuous linear extension of u to C0k .X /. From this uniqueness it follows in turn that the right-hand side (provided it exists) does not depend on the choice of the family of functions  as in (2.5). Now the right-hand side in ju.f /

u.f /j D ju.f

f /j  c kf

f kC k

converges to 0 as  and  # 0. This implies that .u.f //>0 is a Cauchy sequence in C (with  running through n1 for n 2 N), and therefore converges to a complex number that we will denote by v.f /. Because the mapping f 7! f   is linear, we find that this defines a linear form v on C0k .X /. Furthermore, taking the limit as  # 0 in jv.f /j  jv.f /  jv.f /

u.f /j C ju.f /j  jv.f / u.f /j C c kf

u.f /j C c kf kC k

f kC k C c kf kC k

42

3 Distributions

yields the result that jv.f /j  c kf kC k for all f 2 C0k .X / with support in a fixed compact subset of X , which implies that v is continuous on C0k .X /. This also gives v.f / D u.f / if f 2 C01 .X /.  It is common to write v D u, in other words, to identify u with its continuous extension to C0k .X /. A continuous linear form on C0 .X / D C00 .X /, the space of continuous functions with compact support in X , is also known as a measure on X , or preferably, a Radon measure on X , because one wants to distinguish it from the general set-theoretic concept of measure. Thus, distributions of order 0 are identified with Radon measures; for a more detailed discussion of this matter as well as a succinct but fairly complete discussion of integration theory, we refer to Chap. 20. Example 3.16. If f is a locally integrable function on X , the second assertion in Theorem 3.5 implies that test f is a distribution on X of order 0. We conclude that test f has a unique extension to a Radon measure  on X . This  is called the measure with density function f ; but in view of the identifications made, we can also write f D test f D . An example of a Radon measure without locally integrable density function is the Dirac function, which is therefore also called the Dirac measure. The estimate (3.5) for f ıV 2 D 0 .X / shows that f ıV is a Radon measure on X . Because a finite linear combination of Radon measures is again a Radon measure, one may obtain examples of Radon measures for which some summands possess a density function and others not. ˛ Definition 3.17. A distribution u 2 D 0 .X / is said to be positive if  2 C01 .X / and   0 imply that u./  0. In this case we write u  0. Here we use the convention that for a complex number c the notation c  0 means that c 2 R and c  0. A Radon measure u on X is said to be positive if u.f /  0 for every nonnegative f 2 C0 .X /. ˛ Theorem 3.18. Every positive distribution is a positive Radon measure. Proof. Suppose u 2 D 0 .X / and u  0. We begin by showing that u has order 0; on account of Theorem 3.15 it then has an extension to a Radon measure on X . Let K be a compact subset of X . Corollary 2.16 yields a  2 C01 .X / with 0    1 and  D 1 on K. This implies that u./  0; in particular, u./ is real. Let  2 C01 .K/ be real-valued. With the notation c D kk D supx j.x/j we then get c    0, and therefore c u./ u./ D u.c  /  0: This implies that c u./ u./ is a nonnegative real number, so u./ is a real number and u./  u./ kk. Now let  2 C01 .K/ be complex-valued. Then u./ D u.Re  C i Im / D u.Re / C i u.Im /

3 Distributions

43

with u.Re / and u.Im / 2 R. In particular, u./ D u.Re / if u./ 2 R. Using the notation ˛ D arg u./ and D Re .e i ˛ / we now obtain ju./j D e

i˛

u./ D u.e

i˛

/ D u. /  u./ k k  u./ kk:

(3.8)

This shows that u is of order 0. If f 2 C0 .X / and f  0, then also f D f    0 because   0. Therefore u.f /  0 and, consequently, also u.f / D lim#0 u.f /  0. Hence, the Radon measure u is positive.  Remark 3.19. In constructing the proof we have derived that u  0 implies that u./ 2 R, for every continuous real-valued function  with compact support. If both  and are real-valued and .x/  .x/ for all x, then   0, whence u. / u./ D u. /  0, or u./  u. /. The conclusion is that the linear form u is monotone in the sense that u./  u. /, for  and real-valued and   , if and only if u  0. ˛

Example 3.20. A function f may be written as f D fC f with f˙ D 21 .jf j ˙ f /  0, and these functions are continuous if f is. In particular, the distributions f˙ ıV as in Example 3.11 are positive. Theorem 3.18 then implies that they are positive Radon measures. This confirms that fact known from Example 3.16 that f ıV is a Radon measure. Note that the argument does not require an estimate, like the one in (3.5). ˛ Remark 3.21. One writes j " 1 as j ! 1 for a sequence of functions j if for every x we have j .x/ " 1 as j ! 1. A positive Radon measure  is said to be a probability measure if a sequence of test functions j exists with the property that j " 1 and .j / " 1 as j ! 1. This implies that for every sequence of test functions j with j " 1 as j ! 1, one has . j / " 1 as j ! 1; one simply writes .1/ D 1. For a (test) function , this ./ is called the expectation of  with Rrespect to the probability measure . If f is an integrable function with f  0 and f .x/ dx D 1, then  D test f is a probability measure. In this case f is called a probability density. A finite linear combination l X D pj ıa.j / j D1

of point measures, where the a.j / are different points in Rn , is a probability measure P if and only if pj  0 for all j and jl D1 pj D 1. The number pj is then called the probability of the alternative a.j /. Thus we arrive at a distributional interpretation of the calculus of probabilities. One should remain aware, however, that many results for probability measures do not hold for more general distributions. ˛

44

3 Distributions

Problems 3.1. Prove that u./ D  .k/ .0/

. 2 C01 .R//

defines a distribution u 2 D 0 .R/ of order  k 2 Z0 .

 3.2. Show that the distribution u from the preceding problem is not of order strictly lower than k on any neighborhood of 0. Hint: highly oscillatory test functions have large derivatives.

3.3. Show that PV x1 , lem 1.3.

1 xCi 0

and

1 x i0

all are distributions of order 1. Hint: use Prob-

3.4. Verify that u, v, and w below are distributions on R2 : (i) u./ D @˛ .x/, where ˛ is the multi-index .1; 1/ and x the point .1; 1/. R (ii) v./ D R .t; 0/ dt. R 2 (iii) w./ D R2 e kxk .x/ dx.  3.5. We consider functions and distributions on R2 .

(i) r.x/ D kxk defines a function on R2 . Verify that log r and 1=r define distributions on R2 . What is the order of these distributions? Z  (ii) (a) Define u./ D .cos t @1 C sin t @2 /.cos t; sin t/ dt: 0

Show that this defines a distribution u on R2 . What can you say about the order of u? (b) The same questions for Z  v./ D . sin t @1 C cos t @2 /.cos t; sin t/ dt: 0

Observe that cos t @1 C sin t @2 and sin t @1 C cos t @2 is a directional derivative in a direction perpendicular and tangential, respectively, to the curve t 7! .cos; sin t/. In terms of vector analysis, u and v represent the flux of a gradient vector field across the curve and the work along it, respectively. 3.6. Demonstrate that for a continuous function f one has f  0 as a distribution if and only if f  0 as a function.

Chapter 4

Differentiation of Distributions

If f is continuously differentiable on an open set X in Rn , one obtains by means of integration by parts that for every test function  the following holds: .test @j f /./ D .test f /.@j /: Note that the boundary term on the right-hand side is absent because .x/ D 0 for x sufficiently large; see [7, Corollary 7.6.2]. For an arbitrary distribution u on X we now define .@j u/./ D u.@j /

.1  j  n;  2 C01 .X //:

Writing out the definition gives @j u 2 D 0 .X /, which is called the distributional derivative of u with respect to the jth variable. The form of the definition is such that @j .test f / D test.@j f / for every continuously differentiable function f , so that we do not run into difficulties when we simply speak of “functions” when referring to distributions that are of the form test f . By mathematical induction on the number of derivatives we find that partial derivatives of u of arbitrary order are also distributions. As in the case of C 1 functions, the order of differentiation may be changed arbitrarily: Lemma 4.1. For every distribution u on the open subset X of Rn and for every pair of indices 1  j; k  n, one has @j .@k u/ D @k .@j u/. Proof. For every  2 C01 .X / one obtains .@j B @k u/./ D @k u.@j / D u.@k B @j / D u.@j B @k / D @j u.@k / D .@k B @j u/./:



With respect to the examples of continuous functions u for which @j @k u ¤ @k @j u we may therefore note that the two sides are, in fact, distributionally equal. It is interesting to see an example where this happens. For instance, consider J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_4, © Springer Science+Business Media, LLC 2010

45

46

4 Differentiation of Distributions

u W R2 ! R

u.x/ D x1 x2

given by

x12 x22 : x12 C x22

From [7, Theorem 2.7.2 and Example 2.7.1] it follows that @1 @2 u.x/ D @2 @1 u.x/, except if x D 0. However, @1 @2 u and @2 @1 u coincide as locally integrable functions on R2 , and accordingly as distributions. Lemma 4.1 makes it possible to write all higher-order derivatives in the form @˛ u WD

@˛ u : @x ˛

Note that this distribution is of order  k C m when u is of order  k and j˛j D m. If U is open in R and u 2 D 0 .U /, we use the shorthand u0 for the distributional derivative @u, and u.k/ for @k u WD @.k/ u, where k 2 Z>1 . In particular, every continuous function considered as a distribution has partial derivatives of all orders. Conversely, we shall prove in Examples 13.1 or 18.2 below that every distribution can locally be written as a linear combination of derivatives of some continuous function. If every continuous function is to be infinitely differentiable as a distribution, no proper subset of the space of distributions can therefore be adequate. In this sense, the distribution extension of the function concept is as economical as it possibly can be. Example 4.2. H WD 1Œ 0; 1 Œ , the characteristic function of R0 , when interpreted as a distribution on R, is known as the Heaviside function. (At the end of the nineteenth century, Heaviside introduced a kind of distribution calculus for computations on electrical networks. H may be interpreted in terms of the sudden switching on of a current.) One has H 0 D ı. Indeed, for every  2 C01 .R/, Z H 0 ./ D H. 0 / D  0 .x/ dx D .0/ D ı./: R>0

More generally, for a < b, the characteristic function 1Πa;b Πof the interval Πa; b Πsatisfies, in the notation (1.13), 1Πa;b ΠD Ta H

Tb H;

10Œ a;b Œ D ıa

and so

For x 2 R, we define Œx as the integer such that x of x. Then the sawtooth function sD

   1 1 C arctan tan 
2 

ıb :

1 < Œx  x, the integer part

1  W x 7! x 2

Œx

is locally integrable on R and defines s 2 D 0 .R/ (note the slight abuse of notation: the equality is not one of functions, because the right-hand side P is not defined everywhere). With the notation I.x/ D x, we have s D I k2Z k 1Πk;kC1 Π, and therefore we have the following identity in D 0 .R/:

4 Differentiation of Distributions

s0 D 1

47

X

X

ıkC1 / D 1

k.ık

k2Z

ık :

k2Z

In Example 16.24 below this formula will lead to the Fourier series of every polynomial function made periodic with period 1. ˛ The next theorem asserts that every distribution on R has a distributional antiderivative and that this is uniquely determined up to an additive constant. Theorem 4.3. Let I be an open interval in R and f 2 D 0 .I /. Then there exists u 2 D 0 .I / for which u0 D f . If, in addition, v 2 D 0 .I / and v 0 D f , then v D u C c for some constant c 2 C. In particular, v 2 D 0 .I / and v 0 D 0 imply that v equals a constant function. R Proof. Choose  2 C01 .I / such that 1./ D I .x/ dx D 1. For  2 C01 .I /, define the function p./ on R by Z x Z x p./.x/ WD .t/ dt 1./ .t/ dt: 1

1

Then p./ 2 C01 .I /. Indeed, to see that p./ has compact support we note that there exist a and b 2 I for which .t/ D .t/ D 0 if t < a or if t > b. This immediately implies that p./.x/ D 0 if x < a, while in the case x > b one has p./.x/ D 1./ 1./ 1 D 0. Denoting the derivative of  by  0 , we have 1. 0 / D 0 and therefore p. 0 / D . In other words, the mapping p W C01 .I / ! C01 .I / is a left inverse of differentiation. Furthermore, p is linear and sequentially continuous, that is, lim p.j / D 0 in

j !1

C01 .I /

if

lim j D 0 in

j !1

C01 .I /:

This implies that u./ WD f .p.//, for  2 C01 .I /, defines a distribution u on I . That distribution satisfies u0 D f , because u0 ./ D

u. 0 / D f .p. 0 // D f ./

. 2 C01 .I //:

To prove the second assertion in the theorem, observe that w WD v w 0 D 0, or w. 0 / D 0 for all  2 C01 .I /. In particular, 0 D w.p./0 / D w.

1./ / D w./

u satisfies

1./ w./;

for all  2 C01 .I /, which implies that w D w./ test 1 D test w./. Here we denote a constant function with value c by the same symbol c.  Example 4.4. Let I be an open interval in R and m 2 Z0 . Then u 2 D 0 .I / satisfies u.mC1/ D 0 if and only if there exists a polynomial function p on I of degree  m such that u D p.

48

4 Differentiation of Distributions

The proof is by means of Theorem 4.3 and mathematical induction on m. Thus, we obtain that 0 D u.mC1/ D .u.m/ /0 iff there exists c 2 C such that u.m/ D c iff m .u c xmŠ /.m/ D 0 iff there exists a polynomial function q of degree  m 1 such m m that u c xmŠ D q iff u D p with p D c xmŠ C q. ˛

Problems  4.1. Prove that j
j0 D sgn and that j
j00 D 2 ı in D 0 .R/. 00  4.2. Show that .log j
j/0 D PV x1 in D 0 .R/. (For .log j
j/ , see Problem 13.6.) P 4.3. Verify .tan B arctan/0 D 1 and .arctan B tan/0 D 1  k2Z ı.kC 1 / in D 0 .R/. 2 Compute .arccos B cos/0 ; see Problem 1.7.

4.4. Let  2 C and define f .x/ D e  x for x > 0 and f .x/ D 0 for x  0. Prove that the derivatives of f satisfy k X1 f .k/ D k f C k 1 j ı .j / .k 2 Z0 /: j D0

Now let p be a polynomial of degree m > 0 and p./ D 0. Does p.@/f vanish? Calculate the order of p.@/f . 4.5. Determine a continuous function f on Rn and a multi-index ˛ for which @˛ f D ı. Establish how much smaller ˛ may be chosen if f is merely required to be locally integrable.  4.6. Let p 2 Rn and vj .x/ D kx pk n .xj pj /, for 1  j  n. For n D 3 this is the vector field v in (1.2). Verify that the vj are locally integrable on Rn and thus define distributions on Rn . Prove n X div v WD @j vj D cn ıp ; j D1

where cn denotes the .n 1/-dimensional volume of the sphere S n kxk D 1 g in Rn . (See (13.37) for an explicit formula for cn .)  4.7. (Sequel to Problem 4.6.) For x 2 Rn n f0g, define E.x/ D

‚

1 n/ cn kxkn

.2 1 log kxk 2

2

if

n ¤ 2;

if

n D 2:

1

WD f x 2 Rn j

Prove that E is locally integrable on Rn and therefore defines a distribution on Rn . Demonstrate the existence of a constant c for which @j E D c vj as distributions. Then show that
E D ı, where

4 Problems

49


D

n X

j D1

@j2 D div B grad

denotes the Laplace operator in Rn . 0 4.8. Suppose Pq c1 ; : : : ; cq 2 C and a1 <


< aq 2 R. Find the solutions u 2 D .I / 0 of u D j D1 cj ıaj . In which case does one find u D 1I as a solution, for an interval I in R?

Chapter 5

Convergence of Distributions

Definition 5.1. We now introduce a notion of convergence in the linear space D 0 .X / of distributions on an open set X in Rn . Let .uj /j 2N be an infinite sequence in D 0 .X / and let u 2 D 0 .X /. One then writes lim uj D u in

j !1

D 0 .X /

if

. 2 C01 .X //:

lim uj ./ D u./

j !1

(5.1) This is also referred to as weak convergence or convergence of distributions. Instead of a sequence we can also take a family u of distributions that depend on one or more real-valued parameters . We then say that u converges in D 0 .X / to u as  tends to a special value 0 if for every test function  the complex numbers u ./ converge to u./ as  ! 0 . ˛ Example 5.2. Let .fj /j 2N be an infinite sequence of locally integrable functions on X and suppose that f is a locally integrable function on X with the property that for every compact subset K of X , one has Z lim jfj .x/ f .x/j dx D 0: j !1 K

Then fj ! f in D 0 .X /, in the sense that test fj ! test f in D 0 .X / as j ! 1. Indeed, if  2 C01 .X /, application of standard inequalities for integrals yields Z
j test fj test f ./j  jfj .x/ f .x/j dx sup j.x/j: x

supp 

In fact, this leads to convergence in the space of distributions of order 0, also referred to as weak convergence of measures. ˛ Example 5.3. For t > 0 and x 2 R, set ut .x/ D t H.x/ e i tx with H as in Example 4.2. Then Z Z i tx ut ./ D t e .x/ dx D i .0/ C i e i tx  0 .x/ dx R>0

R>0

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_5, © Springer Science+Business Media, LLC 2010

51

52

5 Convergence of Distributions

D i .0/

 0 .0/ t

1 t

Z

R>0

e i tx  00 .x/ dx ! i .0/ as t ! 1;

for  2 C01 .R/. Hence lim t !1 u t D i ı in D 0 .R/.

˛

The following principle of uniform boundedness will not be proved here. It is based on the Banach–Steinhaus Theorem, applied to the Fr´echet space C01 .K/. See for example Rudin [19, Theorem 2.6] or Bourbaki [4, Livre V, Chap. III, ÷3, No 6 and ÷1, No 1, Corollaire]. Maybe the reader will see this principle explained as part of a course in functional analysis. Lemma 5.4. Let .uj /j 2N be a sequence in D 0 .X / with the property that for every  2 C01 .X /, the sequence .uj .//j 2N in C is bounded. Then, for every compact subset K of X , there exist constants c > 0 and k 2 Z0 such that juj ./j  c kkC k

.j 2 N;  2 C01 .K//:

(5.2)

Note the analogy in formulation with Theorem 3.8. The lemma above leads to the following property of completeness of the space of distributions. For the benefit of the reader who is not familiar with the principle of uniform boundedness, the proof of Theorem 5.5 is followed by a second proof of its part (i) that is based on the method of the gliding hump. Theorem 5.5. Let X be open in Rn and .uj /j 2N a sequence in D 0 .X / with the property that for every  2 C01 .X / the sequence .uj .//j 2N converges in C as j ! 1; denote the limit by u./. (i) Then u W  7! u./ defines a distribution on X . Furthermore, limj !1 uj D u in D 0 .X /. (ii) If j !  in C01 .X /, then limj !1 uj .j / D u./ in C.

Proof. Writing out the definitions, we find that u defines a linear form on C01 .X /. From the starting assumption it follows that the sequence .uj .// is bounded for every  2 C01 .X /, and thus we obtain, for every compact K  X , an estimate of the form (5.2). Taking the limit in ju./j  ju./

uj ./j C juj ./j  ju./

uj ./j C c kkC k

as j ! 1, we get ju./j  c kkC k for all  2 C01 .K/. According to Theorem 3.8 this proves that u 2 D 0 .X /, and uj ! u in D 0 .X / now holds by definition. Regarding the last assertion we observe that if j !  in C01 .X /, there exists a compact set K  X such that j and  2 C01 .K/ for all j . Applying Lemma 5.4 once again, we obtain from this juj .j / u./j  juj .j

/j C juj ./ u./j  c kj

kC k C juj ./ u./j;

5 Convergence of Distributions

53

which converges to 0 as j ! 1.



Proof. Suppose that u does not belong to D 0 .X /. Then there exists a sequence .j / in C01 .X / that converges to 0 in C01 .X /, while .u.j // does not converge to 0 as j ! 1. Hence, by passing to a subsequence if necessary, we can arrange that there exists c > 0 such that ju.j /j  c. Recall from Definition 2.13 that j ! 0 in C01 .X / is the case if there exists a compact set K  X such that supp j  K, while kj kC j  41j if we replace .j / by a suitable subsequence if necessary. Accordingly, upon writing j for 2j j , we obtain that j ! 0 in C01 .X /, while ju.j /j ! 1 as j ! 1. Next, we define a subsequence of .j /, say . j / in C01 .X /, and a subsequence of .uj /, say .vj / in D 0 .X /, as follows. Select 1 such that ju. 1 /j > 2. As uj . 1 / ! u. 1 /, we may choose v1 such that jv1 . 1 /j > 2. Now proceed by mathematical induction on j . Thus, assume that k and vk have been chosen, for 1  k < j . Then select j from the sequence .j / such that k

.i/

j kC j


.ii/

(5.3)

1k jvj . k /j C j C 1: (5.4) 1k0

converges and calculate E.x/. Show that E is locally integrable on Rn and can therefore be interpreted as a distribution on Rn . Prove, for every  2 C01 .Rn /,
E./ D

lim

Z

s#0; T "1 s

T

Z

Rn

u t .x/ .
/.x/ dx dt D .0/;

and therefore
E D ı. Verify that E equals the E from Problem 4.7.

5.7. Define f t .x/ D t a e i tx . For what a 2 R does f t converge to 0 as t ! 1 (i) with respect to the C k norm? (ii) in D 0 .R/?

5.8. Define f t .x/ D t e i tx log jxj, for x 2 R. Calculate the limit of f t in D 0 .R/ as t ! 1. 5.9. (Riemann’s “nondifferentiable” function.) Let f W R ! R be defined by X sin.n2 x/ f .x/ D : n2 n2N Weierstrass reported that Riemann suggested f as an example of a continuous function that is nowhere differentiable. (Actually, f is differentiable at points of the form p=q where p and q both are odd integers, with derivative equal to 1=2, but at no other points; see [5], for instance.) In Fig. 5.3, the first ten thousand terms have been summed for ten thousand different values of 0  x  . Prove that lim

N !1

N X

nD1

cos.n2 x/ D f 0 .x/

in D 0 .R/:

5 Problems

57

1

Π

Π

2

Fig. 5.3 The second graph is an enlargement of the first one, in a neighborhood of

 2

In the following problems the reader is asked to derive estimates of the form (5.2), so that we can use a fully proved variant of Theorem 5.5. 5.10. Let cn 2 C for all n 2 Z and suppose that positive constants c and m exist such that jcn j  c jnjm .n 2 Z n f0g/: Let ! > 0 and write uj .x/ D

j X

nD j

cn e i n!x

.j 2 N; x 2 R/:

58

5 Convergence of Distributions

Derive estimates of the form (5.2). Prove that u 2 D 0 .R/ exists for which uj ! u in D 0 .R/. Finally, prove that the order of u is less than or equal to m C 2. 5.11. Let .a.j //j 2N be a sequence in Rn with limj !1 ka.j /k D 1 and let .cj /j 2N be an arbitrary sequence in C. Give estimates of the form (5.2) for uk WD

k X

cj ıa.j / :

j D1

Prove the existence of u 2 D 0 .Rn / such that limk!1 uk D u in D 0 .Rn /. 5.12. Let f be a continuous function on R. Does 2

1 uk WD k

k X

f

j D k2

j  k

ıj k

converge in D 0 .R/ as k ! 1? If so, what is the limit? 5.13. Show that for every  2 C01 .R/ the series in (3.7) converges; also, that (3.7) defines a distribution u on R of infinite order; and finally, that lim

k!1

k X1

j D0

@j ıj D u in

D 0 .R/:

5.14. Let .uj /j 2N be a sequence of positive Radon measures on X with the property that limj !1 uj ./ D u./, for every  2 C01 .X /. Use (3.8) to demonstrate that for every compact subset K of X there is a constant c > 0 with juj ./j  c sup j.x/j x

.j 2 N;  2 C01 .K//:

Prove that u is a positive Radon measure. Furthermore, show that the uj converge weakly as measures to u. That is, limj !1 uj .f / D u.f /, for every f 2 C0 .X /.

Chapter 6

Taylor Expansion in Several Variables

Many classical asymptotic expansions imply interesting distributional limits. An example is the Taylor expansion of functions of several variables. Because we will be using this expansion elsewhere in this text, we begin by repeating its basic properties. A mapping f from an open subset U of Rn to Rp is given by p real-valued functions of n variables, that is, f .x/ D f .x1 ; : : : ; xn / D .f1 .x1 ; : : : ; xn /; : : : ; fp .x1 ; : : : ; xn //: Many properties of vector-valued functions can be derived from those of real-valued functions by proving them for each of the coordinate functions fi .x1 ; : : : ; xn /, with 1  i  p. For instance, the mapping f is continuously differentiable if for every i and every j , the function fi is partially differentiable with respect to the jth variable and if the function @j fi is continuous on U . In that case one writes f 2 C 1 .U; Rp /, or f 2 C 1 .U; V / if f .U /  V . The @j fi .x/, for 1  j  n and 1  i  p, form the matrix of the linear n mapping X Df .x/ W v 7! @j f .x/ vj W Rn ! Rp : j D1

This is the linear approximation at h D 0 of the increase h 7! f .x C h/ f .x/ of f and is also known as the total derivative of f at the point x. If p D 1, one normally writes df .x/ instead of Df .x/. The functions @j fi taken together can again be interpreted as a vector-valued function. This allows us to define, with mathematical induction on k, that f 2 C k if D k 1 f 2 C 1 , where D k f is inductively defined by D k f .x/ D D.D k 1 f /.x/. A theorem already mentioned is that @i .@j f / D @j .@i f / if f 2 C 2 . For f 2 C k this enables every higher-order derivative of f , where differentiation with respect to xj takes place ˛j times in total, to be rearranged in the form @˛ f , as in (2.7).

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_6, © Springer Science+Business Media, LLC 2010

59

60

6 Taylor Expansion in Several Variables

An important result that we will often be using is the chain rule. For describing this, consider f 2 C 1 .U; Rp / and g 2 C 1 .V; Rq / where U  Rn and V  Rp are open subsets. The composition g Bf of f and g is defined by .g Bf /.x/ D g.f .x// for all x 2 U with f .x/ 2 V ; the continuity of f guarantees that the set of these x forms an open subset U \ f 1 .V / of Rn . The chain rule asserts that we have g B f 2 C 1 .U \ f 1 .V /; Rq /, while D.g B f /.x/ D Dg.f .x// B Df .x/I

D.g B f / D .Dg/ B f Df:

or

With mathematical induction on k one finds that gBf 2 C k if f 2 C k and g 2 C k ; for the higher-order derivatives inductive formulas are obtained that rapidly become very cumbersome to write out in explicit form. Considering the matrix coefficients occurring in the chain rule, one obtains the following identity of functions on U \ f 1 .V /, for 1  j  n and 1  i  q: @j .gi B f / D

p X

.@k gi / B f @j fk D

kD1

p X

kD1

@j fk .@k gi / B f:

(6.1)

A practical way of studying f .x/ for x near a point a 2 U is to consider the function on Œ 0; 1   R given by t 7! g.t/ WD f .a C t .x

a//:

We have g.0/ D f .a/ and g.1/ D f .x/; what we actually do is to consider f on the straight line from a to x. The difference vector h D x a is usually small. In the following lemma we use the abbreviations x˛ D

n Y

˛

xj j

and

j D1

˛Š D

n Y

˛j Š:

j D1

Lemma 6.1. If f 2 C k .U; Rp /, we have, for 0  j  k and a C th 2 U , X h˛ 1 dj @˛ f .a C th/: f .a C th/ D j Š dt j ˛Š

(6.2)

j˛jDj

Proof. Using the chain rule we obtain n

X d f .a C th/ D hj @j f .a C th/: dt j D1

By mathematical induction on j this yields X dj f .a C th/ D m˛ h˛ @˛ f .a C th/; dt j j˛jDj

(6.3)

6 Taylor Expansion in Several Variables

61

where the m˛ 2 Z are determined by recurrence relations; in particular, they are independent of the choices made for f , a, and t. To determine the m˛ , we take f .x/ D x ˇ with jˇj D j , a D 0 and differentiate at the point t D 0. Then f .a C th/ D .t h/ˇ D t jˇ j hˇ D t j hˇ ; the jth-order derivative with respect to t of this function equals j Šhˇ . But we also have, for every ˛ with j˛j D j , ( ˇŠ if ˛ D ˇ; ˛ ˇ @ x D (6.4) 0 if ˛ ¤ ˇ: This proves that mˇ D j Š=ˇŠ. Actually, the combinatorics associated with the recurrence relation are simple. Hence, one may alternatively finish the proof by noting that according to the induction hypothesis the right-hand side of (6.3) after division by .j 1/Š can be written as n X X X h˛ Pn ˛i hi hˇ i D1 @i @ˇ f .a C th/ D @˛ f .a C th/:  ˇŠ ˛Š jˇ jDj 1 i D1

j˛jDj

k For f 2 C k .U; Rp /, the Taylor polynomial T D Tf;a of order k at the point a 2 U is defined by ˛ X .x a/ T .x/ D @˛ f .a/: (6.5) ˛Š j˛jk

k of order k is defined as R D f T . Then, by definiThe remainder R D Rf;a tion, f D T C R; we want to find out what information can be obtained about the remainder. The following formulation may be somewhat heavier than necessary for most applications, but it follows directly from the integral formula used for the remainder.

Theorem 6.2. Let f be a C l mapping defined on an open ball U in Rn . For every k 0  k  l, the mapping .a; x/ 7! Rf;a .x/ is a C l k mapping on U  U . Furthermore, for every compact K  U and every  > 0, there exists a ı > 0 such that k kRf;a .x/k   kx

akk

if

a; x 2 K

and kx

ak < ı:

k Finally, R D Rf;a is of class C l and @˛ R.a/ D 0, for all multi-indices ˛ with j˛j  k.

Proof. The Taylor expansion of a C k function g on an interval I containing 0 reads g.t/ D

k X1 j

j D0

t .j / g .0/ C jŠ

Z

t 0

.t s/k 1 .k/ g .s/ ds .k 1/Š

.t 2 I /I

the proof is obtained by mathematical induction on k. When we apply this formula, taking g.t/ D f .a C t .x a// and substituting (6.2) for the derivatives, we obtain,

62

6 Taylor Expansion in Several Variables

k for t D 1, the identity f D T C R, with R D Rf;a and k .x/ Rf;a

D

X .x

j˛jDk

a/˛ ˛Š

Z

1

k .1

s/k

0

1

@˛ f .a C s .x

a//


@˛ f .a/ ds:

On account of the locally uniform continuity of the kth-order derivatives of f , the integral converges locally uniformly to 0 as kx ak ! 0. The fact that the mapping k .a; x/ 7! Rf;a .x/ is of class C l k follows by means of differentiation under the integral sign. k The function Rf;a is of class C l , being the difference of f and the Taylor polynok . From (6.4) we see that @˛ f .a/ D @˛ T .a/ for all ˛ with j˛j  k.  mial T WD Tf;a By way of application we give a “higher-order version” of Problem 5.2. Proposition 6.3. Let f be an integrable function on Rn with compact support. Let a 2 Rn and set  1 1 f .x/ D n f .x a/ ;   for  > 0. Furthermore, we define, for each multi-index ˛, the coefficient c˛ 2 C by Z x ˛ f .x/ dx; c˛ D Rn

and we introduce, for every j 2 Z0 , the distribution uj 2 D 0 .Rn / by uj D We then have, for all k 2 Z0 , lim  #0

k

.f

k X

X c˛ @˛ ıa : ˛Š

j˛jDj

. /j uj / D 0 in

D 0 .Rn /:

j D0

Proof. For every  2 C01 .Rn / we have Z Z  1 n .x a/ .x/ dx D f ./ D  f f .h/ .a C h/ dh:  Rn Rn The assertion now follows by substituting the Taylor expansion of  at the point a and of order k.  The coefficient c˛ is known as the ˛th moment of the function f . A consequence of the proposition is that  k f converges distributionally to . 1/k uk as  # 0 if c˛ D 0 for all multi-indices ˛ with j˛j < k.

6 Problems

63

Problems 6.1. Formulate a variant of Proposition 6.3 in which f is not required to have compact support.  6.2. Let u t be as in Problem 5.5. Prove, for every k 2 Z0 , 1 lim k t #0 t

 ut

 1 t j
ı D
k ı jŠ kŠ

k X1 j

j D0

D 0 .Rn /:

in

Discuss the notation u t D e t
ı, for t > 0.  6.3. (Sequel to Problem 6.2.) For  2 C01 .Rn / and r  0, define the mean value S .r/ of  over the sphere of radius r about 0 by Z 1 S .r/ D .r y/ dy: cn S Here S is the unit sphere in Rn and cn its Euclidean .n which is evaluated in (13.37). (i) Show that S 2 C 1 .R/ and that Z n u t ./ D cn .4 t/ 2

e

r2 4t

rn

1

1/-dimensional volume,

S .r/ dr

.t > 0/:

R>0

(ii) Substitute the Taylor series of S at 0 in this identity and express the coefficients of the series in terms of the
j .0/. Under the assumption of convergence of the series, deduce Pizzetti’s formula, for r  0, S .r/ D D

n X 2

k2Z0


k f .0/  r 2k kŠ . n2 C k/ 2

 n  r p
1 2 2

n 2

J n2

1 .r

p


/f .0/:

Here denotes the Gamma function as in Sect. 13.3 below,
denotes the Laplace operator, and J n2 1 the Bessel function of order n2 1 as in [7, Exercise 6.66] for instance. Furthermore, the last equality is obtained by a formal substitution in the power series for the Bessel function. See [7, Exercises 7.22 and 7.54] for proofs along classical lines. ] (iii) Suppose that ˛j 2 2Z0 , for 1  j  n and consider the monomial function Q ˛ y 7! y ˛ D 1j n yj j . Define .2a 1/ŠŠ D 1
3


.2a 1/, for a 2 N and furthermore . 1/ŠŠ D 1. Prove (compare with [7, Exercise 7.21.(x)]) Q Z 1/ŠŠ 1 1j n .˛j ˛ 2 Q: y dy D Q cn S .n C 2l/ 0l< j˛j 2

Chapter 7

Localization

Generally speaking, distributions cannot be restricted to a point, that is, evaluated at that point. Restriction to an open set, however, is possible. Following Definition 2.5 we have discussed how, for open subsets U and V of Rn with U  V , the space C01 .U / is interpreted as a linear subspace of C01 .V /. More explicitly, we have the linear injection V U W C01 .U / ! C01 .V /:

(7.1)

Suppose that j and  belong to C01 .U /. Since a compact subset K of U is a compact subset of V too, it follows that lim j D 

j !1

in

C01 .U /

H)

lim V U j D V U 

j !1

in

C01 .V /:

Phrased differently, V U preserves convergence of sequences. Now consider v in D 0 .V / and define U V v W C01 .U / ! C

by

.U V v/./ D v.V U /

. 2 C01 .U //:

Then it follows that U V v is a continuous linear form on C01 .U /. In other words, U V v 2 D 0 .U /; this distribution is said to be the restriction of v to U . This restriction of distributions defines a linear mapping U V W D 0 .V / ! D 0 .U /;

(7.2)

the mapping induced by the inclusion C01 .U /  C01 .V /. Note that upon changing to the dual spaces the sense of the arrows is reversed, as is visible in the following diagram: /V U C01 .U / D 0 .U / o

V U U V

/ C01 .V / D 0 .V /

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_7, © Springer Science+Business Media, LLC 2010

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66

7 Localization

The distribution v is said to be of class C k , or of order k, respectively, on U if U V v has this property. If w 2 D 0 .W / and U  W , we say that v D w on U if U V v D U W w. Theorem 7.1. Let X be an open subset of Rn and u 2 D 0 .X /. Suppose that for every x 2 X , there exists an open neighborhood Ux of x in X such that u D 0 on Ux . Then we have u D 0 (on X). Proof. Let  2 C01 .X /. The Ux with x 2 X form an open cover of the compact 1 set supp P . Theorem 2.15 yields 1 ; : : : ; l 2 C0 .X / with supp 1j  Ux.j / and j j D 1 onP supp . Writing j D P j , we have j 2 C0 .Ux.j / /, so u.j / D 0 and  D j j ; therefore u./ D j u.j / D 0. 

Applying Theorem 7.1 to u v, we further obtain that u D v on X if for every x 2 X , there exists an open neighborhood U of x in X such that u D v on U . Definition 7.2. The support of u 2 D 0 .X /, denoted by supp u, is the set of all x 2 X for which there exists no open neighborhood U of x in X such that u D 0 on U . ˛ Let V WD X n supp u be the complement of supp u in X . This is an open subset of X . Indeed, by definition, for every x 2 V there exists an open neighborhood Ux of x in X with u D 0 on Ux , which implies Ux  V . Theorem 7.1 now implies that u D 0 on V . Furthermore, if U is open in X and u D 0 on U , then U \supp u D ;, which is equivalent to U  V . In other words, V is the largest open subset of X on which u D 0. One has u./ D 0

u 2 D 0 .X /;  2 C01 .X /

supp u \ supp  D ;: (7.3) This characterization of supp u immediately implies that supp @˛ u  supp u, for every multi-index ˛ 2 .Z0 /n . if

and

Example 7.3. Let V be a closed k-dimensional C 1 submanifold contained in an open subset X of Rn and f a continuous function on V . Consider the Radon measure f ıV in D 0 .X / as in Example 3.11; then supp f ıV D V \supp f D supp f , because f is a function on V . Furthermore, supp @˛ .f ıV /  V , even though some of the differentiations in @˛ might be in directions that are not tangential to V . ˛ Example 7.4. Let U  X be open subsets of Rn and u 2 D 0 .X /. Suppose u./ D 0, for all  2 C01 .X / satisfying supp  \ U D ;. Then supp u  U while in general it is not true that supp u  U . For a counterexample, consider U D R>0  R and u D 1U 2 D 0 .R/. Then supp u D R0 and not supp u  R>0 . ˛ If U  X and U ¤ X , the restriction mapping UX is not injective. Indeed, if a 2 X n U and u D ıa , we have u D 0 on U . Surjectivity of UX would mean that every distribution u on U is the restriction to U of a distribution v on X . If UX v D u, then v is said to be a distributional

7 Localization

67

extension of u to X . An example: for every k 2 N, the function x 7! x which is locally integrable on R>0 but not on all of R, has the extension

k

on R>0 ,

. 1/k 1 k @ .log j
j/ 2 D 0 .R/: .k 1/Š However, there are many cases in which extension is impossible. Example 7.5. The function f W x 7! e 1=x on R>0 has no distributional extension to any neighborhood of 0 in R. Suppose  2 C01 .R>0 / satisfies   0 and 1./ > 0 and set  .x/ D 1 . x /. We see that for every N 2 N there exists a constant c > 0 such that f . /  c  N , for all 0 <   1. (Hint: use the fact that e 1=x  x N =N Š if x > 0 and apply the change of variables x D  y in the integral.) On the other hand, if u 2 D 0 .  ı; 1 Œ / with ı > 0 were a distributional extension of f , Theorem 3.8 asserts that there would exist c 0 and c 00 > 0 and k 2 Z0 with the property that ju. /j  c 0 k kC k  c 00  k 1 for all 0 <   1. Thus, the assumption that u D f on R>0 leads to a contradiction. ˛ There exists a counterpart of Theorem 7.1 in the form of an existence theorem: Theorem 7.6. Let U be a collection of open subsets of Rn with union X . Assume that for every U 2 U a distribution uU 2 D 0 .U / is given and that uU D uV on U \ V , for all U , V 2 U. Then there exists a unique u 2 D 0 .X / with the property that u D uU on U , for every U 2 U. Proof. Let K be a compact subset of X and j a partition of unity over K subordinate to the cover U. In particular, there exists U.j / 2 U such that supp j  U.j / 2 U. If  2 C01 .X / and supp   K, then  equals the sum of the finitely many j  2 C01 .X /. We therefore have to take u./ D

l X

uU.j / .

j

/:

(7.4)

j D1

The uniqueness of u follows. Before continuing we first prove that the right-hand side of (7.4) is independent of the choice of the partition j of unity over K subordinate to U. Suppose k is another partition of unity over K with supp k  V .k/ 2 U. Then supp .

j

k /  U.j / \ V .k/:

Consequently, uU.j / and uV .k/ assume the same value on l X

j D1

uU.j / .

j

/ D D

l X

j;kD1 l X

kD1

uU.j / .

j

k / D

uV .k/ .k /:

l X

j;kD1

j

k . Therefore

uV .k/ .

j

k /

68

7 Localization

We now interpret the right-hand side in (7.4) as the definition of u. If ˛, ˇ 2 C01 .X / and a, b 2 C, we take K D supp ˛ [ supp ˇ above and thus obtain that u.˛/, u.ˇ/, or u.a ˛ C b ˇ/ equals the right-hand side in (7.4), replacing  by ˛, ˇ, or a ˛ C b ˇ, respectively. From this it is immediate that u.a ˛ C b ˇ/ D a u.˛/ C b u.ˇ/. In other words, u is a linear form on C01 .X /. If k !  in C01 .X /, there exists a compact subset K of X with the property that for every k one has that supp k  K; this also implies that supp   K. Then, for every k, the u.k / equal the right-hand side of (7.4) with  replaced by k . From the continuity of the uU .j / it now follows that u.k / ! u./ as k ! 1. We have now proved that .7:4/ defines a continuous linear form u on C01 .X /. Finally, if supp   U 2 U, we have supp . j /  U.j / \ U , and therefore uU.j / . j / D uU . j /. Summation over j gives u./ D uU ./.  Remark 7.7. Thanks to their properties described in Theorem 7.6, the linear spaces D 0 .U /, as functions of the open subsets U of X , form a system known in the literature as a (pre)sheaf. In many cases, sheaves consist of spaces of functions on U , with the usual restriction of functions to open subsets. One example is the sheaf U 7! C 1 .U / of the infinitely differentiable functions. For these, the sheaf properties are a direct consequence of the local nature of the definition of the functions in the sheaf. Distributions form an example of a sheaf in which the linear spaces contain elements that are not functions. ˛

Definition 7.8. Let u 2 D 0 .X /. The singular support of u, denoted by sing supp u, is the set of all x 2 X that do not possess any open neighborhood U in X such that UX u 2 C 1 .U /. Here UX is as in (7.2). ˛

The complement C of sing supp u in X has an open cover U such that UX u D uU 2 C 1 .U /, for every U 2 U. This follows from the definition of sing supp . If U and V 2 U, then uU D uV on U \ V , which implies that the uU possess a common extension to a C 1 function f on C . In view of Theorem 7.1 we have u D f on C . In other words, u is of class C 1 on C ; every open subset of X on which u is of class C 1 is contained in C . Obviously, sing supp u  supp u. In Problem 7.3 we use the following definition. In subsequent parts of this text we will repeatedly encounter linear partial differential operators with constant coefficients. Definition 7.9. Let P ./ D

X

c˛  ˛

j˛jm

be a polynomial in n variables of degree m with coefficients c˛ 2 C. Replacing  by @ D @x we obtain a differential operator P .@/ D P .@x / D

X

j˛jm

c˛ @˛ D

X

j˛jm

c˛

@˛ ; @x ˛

a linear partial differential operator of order m and with constant coefficients.

(7.5) ˛

7 Problems

69

Note that P .@/ maps D 0 .X / into itself, for any open subset X of Rn ; see also Problem 7.3.

Problems 7.1. Show that for a continuous function the support equals the support in the distributional sense. 7.2. Determine the supports of the Dirac function, the Heaviside function, and PV x1 . 7.3. Prove that supp P .@/ u  supp u, for every u 2 D 0 .X / and every linear partial differential operator with constant coefficients P .@/.  7.4. (Integration of partial derivative.) Let U be a measurable subset of X . Prove that the support of @j 1U is contained in the boundary @U D X \ U n U int of U in X . Here U int is the set of the interior points of U . Now, let U be an open subset of X with C 1 boundary @U and lying at one side of @U . Prove that @j 1U D j ı@U . Here j denotes the jth component of the outer normal to @U . Also discuss the case n D 1.  7.5. Let a and f 2 C 1 .X / and let f be real-valued. A point p 2 X is said to be a stationary point of f if @j f .p/ D 0, for all 1  j  n; let Sf be the set of the stationary points of f in X . Prove that X n Sf is an open subset of X and that for every m 2 R, lim t m a e i tf D 0 in D 0 .X n Sf /: t !1

Hint: use the formula e i tf D @j .e i tf /=.i t@j f /, wherever @j f ¤ 0.  a a 7.6. Let a 2 C and define xC D x a D e a log x if x > 0 and xC D 0 if x < 0. Prove the following assertions: a (i) If Re a > 1, then xC is locally integrable in R and can therefore be interpreted as a distribution. (ii) If a is not a negative integer, a xC D

1 @k x aCk .a C k/


.a C 1/ x C

with

k>

Re a

1

a defines a distribution on R that is an extension of xC on Rnf0g. This distribution is independent of the choice of k. Determine the support and the order of this distribution. What happens if the real part of a is a negative integer (and the imaginary part does not vanish)? (iii) Let lC .x/ D log x if x > 0 and lC .x/ D 0 if x < 0. Then lC is locally integrable on R and thus defines a distribution on R. For every k 2 N,

. 1/k 1 k @ lC .k 1/Š

70

7 Localization

is a distribution on R that forms an extension of xCk on R n f0g. Again, determine its support and its order. (iv) Formulate similar results starting from . x/aC or lC . x/. 7.7. Show that for every C 1 function f on R>0 a real-valued C 1 function g on R>0 can be found with the property that f e ig has an extension to a distribution of order  1 on R. Hint: show that there exists a real-valued C 1 function g on R>0 such that g0 .x/ ¤ 0, f .x/=g0 .x/ converges to zero as x # 0, and f 0 =g 0 is absolutely integrable on a neighborhood of 0. Prove that Z d  f .x/ .x/  u./ D i dx . 2 C01 .R// e i g.x/ dx g 0 .x/ R>0 defines a distribution u on R of order  1 and that u D f e ig on R>0 . 7.8. A subset D of X is said to be discrete if for every a 2 D, there is an open neighborhood Ua of a in X such that D \ Ua D fag. Let X be an open subset of Rn and D a discrete and closed subset of X . For every a 2 D let there be a partial differential operator Pa .@/ of order ma . Write ua D Pa .@/ ıa . Demonstrate the following: (i) There exists exactly one distribution u in X with supp u  D such that for every a 2 D there P is an open neighborhood Ua of a in X on which u D ua . One writes u D a2D ua . (ii) supp u D D, and the order of u equals the supremum of the ma , for a 2 D. (iii) D is countable and for every enumeration j 7! a.j / of D we have that u D P limk!1 jkD1 ua.j / .

7.9. Let .uj /j 2N be a sequence in D 0 .X /. Suppose that for every x 2 X there is an open neighborhood U of x in X with the property that the sequence .uj .//j 2N in C converges for every  2 C01 .U / as j ! 1. Show that there exists u 2 D 0 .X / such that limj !1 uj D u in D 0 .X /.

7.10. Determine the singular support of j
j, the Dirac function, the Heaviside function, PV x1 , 1=.x C i 0/, and 1=.x i 0/. And also of the components vj of the vector field v in (1.2). 7.11. Prove that sing supp .@j u/  sing supp u, for 1  j  n and u 2 D 0 .X /.

Chapter 8

Distributions with Compact Support

If u is locally integrable on an open set X in Rn and has compact support, the Z integral u./ D u.x/ .x/ dx X

is absolutely convergent for every  2 C 1 .X /, as follows from a slight adaptation of the proof of Theorem 3.5. More generally, every distribution with compact support can be extended to a continuous linear form on C 1 .X /. Before we can give a precise formulation of this result, we have to define convergence in C 1 .X /; clearly, Definition 2.13 has to be modified. Definition 8.1. Let .j /j 2N be a sequence in C 1 .X / and let  2 C 1 .X /. We say that lim j D  in C 1 .X / j !1

if for every multi-index ˛ and for every compact subset K of X , the sequence .@˛ j /j 2N converges uniformly on K to @˛  as j ! 1. A linear form u on C 1 .X / is said to be continuous if u.j / ! u./ whenever j !  in C 1 .X / as j ! 1. The space of continuous linear forms on C 1 .X / is denoted by E 0 .X /. One says that for .uj /j 2N and u in E 0 .X /, lim uj D u in

j !1

E 0 .X /

if limj !1 uj ./ D u./ in C, for every  2 C 1 .X / (compare with Definition 5.1). ˛ The notation E 0 .X / echoes the notation E.X / used by Schwartz to indicate the space C 1 .X / endowed with the notion of convergence introduced in Definition 8.1. Note that for a linear form u on C 1 .X / to be continuous it is sufficient that u.j / ! 0 whenever j ! 0 in C 1 .X /. Furthermore, if j ! 0 in C01 .X /, then certainly j ! 0 in C 1 .X /. It follows that for every u 2 E 0 .X /, the restriction  u of u to the linear subspace C01 .X / of C 1 .X / belongs to D 0 .X /. J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_8, © Springer Science+Business Media, LLC 2010

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8 Distributions with Compact Support

A sequence .Kj /j 2N of compact subsets of X is said to absorb the set X if for every compact subset K of X there exists an index j with K  Kj . The sequence is said to be increasing if Kj  Kj C1 for all j . Lemma 8.2. Let X be an open subset of Rn . Then the following hold: (a) There exists an increasing sequence .Kj /j 2N of compact subsets of X that absorbs X . (b) There is a sequence .j /j 2N in C01 .X / with the property that for every  2 C 1 .X /, the sequence with terms j WD j  2 C01 .X / converges in C 1 .X / to  as j ! 1. (c) The restriction mapping  W E 0 .X / ! D 0 .X / is injective. Proof. (a). Using the notation C WD Rn n X , the complement of X in Rn , we define n 1o .j 2 N/: Kj D x 2 X j kxk  j and d.x; C /  j

These are bounded and closed subsets of Rn , and therefore compact on account of Theorem 2.2. Furthermore, every Kj  Kj C1  X . If K is a compact subset of X , there exists an R with kxk  R for all x 2 K. In addition, the ı-neighborhood of K is contained in X for sufficiently small ı > 0 in view of Corollary 2.4; this implies d.x; C /  ı, for all x 2 K. If we choose j 2 N sufficiently large that R  j and ı  j1 , then K  Kj . (b). On the strength of Corollary 2.16, we can find j 2 C01 .X / with j D 1 on an open neighborhood of Kj . Let  2 C 1 .X /. For every compact subset K of X there is a j.K/ such that K  Kj.K/ ; this implies K  Kj if j  j.K/. But then one has j D j  D  on an open neighborhood of K, and therefore also @˛ j D @˛  on K, for every multi-index ˛. It follows that certainly j !  in C 1 .X / as j ! 1. (c). This is a consequence of (b). Indeed, consider u and v 2 E 0 .X / with  u D  v. For every  2 C 1 .X / we have u./ D lim u.j / D lim v.j / D v./; j !1

j !1

where j 2 C01 .X / and j !  in C 1 .X /.



Definition 8.3. A subset D of a topological space E is said to be dense in E if the closure of D is equal to E. ˛ Lemma 8.2.(b) implies that C01 .X / is dense in C 1 .X /. Lemma 8.2.(c) is a special case of the general principle that a continuous extension to the closure of a set is uniquely determined. Because of the injectivity of , every u 2 E 0 .X / can be identified with the distribution  u 2 D 0 .X /; accordingly, we will write  u D u and consider E 0 .X / as a linear subspace of D 0 .X /. Further note that uj ! u in E 0 .X / implies that uj ! u in D 0 .X /.

8 Distributions with Compact Support

73

In order to determine which elements in D 0 .X / belong to E 0 .X /, we first take a closer look at convergence in C 1 .X /. Definition 8.4. For every k 2 Z0 and compact subset K of X , we define kkC k ; K D

sup j˛jk; x2K

j@˛ .x/j

. 2 C k .X //:

(8.1)

For a sequence .j /j 2N and  in C k .X /, we say that limj !1 j D  in C k .X / if we have limj !1 kj kC k ; K D 0, for every compact subset K of X . With this

definition, j !  in C 1 .X / is equivalent to the assertion that j !  in C k .X / for every k 2 Z0 . ˛

Because (8.1) is not subject to the condition supp   K, (8.1) does not define a norm on either C k .X / or C 1 .X /; for every k and K there exist  2 C 1 .X / with kkC k ; K D 0 and  ¤ 0. However, the nk;K W  7! kkC k ; K do form a separating collection of seminorms. We now discuss the definition and properties of seminorms for arbitrary linear spaces. In Theorem 8.8 we will resume the characterization of the distributions in E 0 .X /. Definition 8.5. For a linear space E over C, a seminorm on E is defined as a function n W E ! R with the following properties. For every x and y 2 E and c 2 C, one has (a) n.x/  0, (b) n.x C y/  n.x/ C n.y/ (subadditivity), (c) n.c x/ D jcj n.x/ (absolute homogeneity). In other words, n has all properties of a norm, except that there is no requirement that n.x/ D 0 implies x D 0. (We note that condition (a) follows from (b) and (c).) Instead of a norm on E, one may consider a collection N of seminorms on E that is separating in the following sense: (d) if n.x/ D 0 for all n 2 N , then x D 0, (e) if n and m 2 N there exists p 2 N such that n.x/  p.x/

and

m.x/  p.x/

.x 2 E/:

˛

An example is E D C k .X / with the collection of seminorms N k WD f nk;K j K  X; K compact g: S On C 1 .X / we use N D k2Z0 N k . A pair .E; N / where E is a linear space and N a separating collection of seminorms on E is said to be a locally convex topological linear space. A subset U of E is said to be a neighborhood of a in .E; N / if there exist n 2 N and  > 0 such that B.n; a; / WD f x 2 E j n.x a/ <  g  U:

74

8 Distributions with Compact Support

The word “topological” relates to “neighborhoods.” Furthermore, a subset C of E is said to be convex if for every x and y 2 C the line segment from x to y, the set of x C t .y x/ with 0  t  1, is contained in C . For every n 2 N and  > 0, one has that B.n; a; / is convex; this is the origin of the term “locally convex.” The assertion about convexity follows from n.x C t.y

x/ a/ D n..1 t/.x

a/ C t.y

a//  .1 t/ n.x

a/ C t n.y

a/:

Readers familiar with topology will have observed that the B.n; a; / form a basis. A subset U of E is said to be open if U is a neighborhood of all of its elements. The open sets form a topology in E. For a deeper study of distribution theory we recommend that the reader peruse the general theory of locally convex topological linear spaces, as can be found in Bourbaki [4], for example. Let xj , for j 2 N, and x 2 E. One then says that limj !1 xj D x in .E; N / if lim n.xj

j !1

x/ D 0

.n 2 N /:

If .F; M/ is another locally convex topological linear space, a linear mapping A from E to F is said to be sequentially continuous from .E; N / to .F; M/ if limj !1 A.xj / D A.x/ in .F; M/ whenever limj !1 xj D x in .E; N /. A mapping A W E ! F is said to be continuous from .E; N / to .F; M/ if for every a 2 E and every neighborhood V of A.a/ in .F; M/ there exists a neighborhood U of a in .E; N / such that A.U /  V . For a linear mapping this is equivalent to the assertion that for every m 2 M there exist a constant c > 0 and n 2 N with m.A.x//  c n.x/

.x 2 E/:

(8.2)

For the sake of completeness we give a proof of this general result from functional analysis. Proof. Let a 2 E and let V be a neighborhood of A.a/ in .F; M/, that is, there exist m 2 M and  > 0 such that B.m; A.a/; /  V . Suppose that there are a constant c > 0 and n 2 N such that (8.2) holds. In that case, x 2 B.n; a; ı/ implies m.A.x/ A.a// D m.A.x a//  c n.x a/ < c ı:

Choosing ı D =c, we have ı > 0, and we see that A maps the neighborhood B.n; a; ı/ of a into B.m; A.a/; /, and therefore into V . From this we conclude that assertion (8.2) implies the continuity of A. Conversely, suppose that A is continuous and that m 2 M. From the continuity of A we use only that there is some a 2 E, that there are constants  > 0 and ı > 0, and that there exists n 2 N with the property that n.x a/ < ı implies m.A.x/ A.a// < . (This is a weak form of the continuity of the mapping A at the point a.) Because A.x/ A.a/ D A.x a/, we see that this is equivalent to the property that n.y/ < ı implies m.A.y// < . (This is a weak form of the continuity of the mapping A at the point 0.) Now define c D 2=ı; we are going to prove that for every x 2 E one has m.A.x//  c n.x/. If n.x/ > 0, one has, for y D  x

8 Distributions with Compact Support

75

where  D ı=.2n.x//, n.y/ D n. x/ D  n.x/ D ı=2 < ıI consequently  > m.A.y// D m.A. x// D m. A.x// D  m.A.x//; which implies m.A.x// < c n.x/. If, on the other hand, n.x/ D 0, then for every  > 0 one has n.y/ D n. x/ D  n.x/ D 0 < ı; it follows that  > m.A.y// D  m.A.x//, which implies m.A.x//  0  c n.x/.  Example 8.6. Consider open sets X  Rn and Y  Rm , and a linear mapping A W C01 .X / ! C 1 .Y /. According to (8.2), Definition 2.13, and (8.1), the mapping A is continuous if for any two compact sets K  X and K 0  Y and any order of differentiation k 0 2 Z0 , there exist a constant c > 0 and an order of differentiation k 2 Z0 such that kAk

0 C k ; K0

 c kkC k

. 2 C01 .K//:

˛

For E as in the theory above and F D C, with the absolute value as the only (semi)norm, we thus obtain the continuous linear forms on E, together forming the topological dual E 0 of E. The definition as given here is more exact than that in Chap. 3, because the definition there speaks rather vaguely about a “notion of convergence” in E. In E 0 we use finite sums of the seminorms u 7! ju.x/j for all x 2 E. For these, limj !1 uj D u in E 0 is equivalent to limj !1 uj .x/ D u.x/, for every x 2 E; this is the weak convergence in E 0 that we introduced before, in Definitions 5.1 and 8.1. If the separating collection N of seminorms is finite, we may, as far as the notion of convergence is concerned, as well replace the collection N by the unique norm given by the maximum of the n 2 N . We then find ourselves in the familiar context of linear spaces endowed with a norm, where linear mappings that satisfy estimates as in (8.2) are also said to be bounded linear mappings. For C k .X / and C 1 .X /, the collection N cannot be replaced by a single norm. It is possible, however, without changing the notion of convergence, to replace the collection of seminorms by the increasing countable sequence with terms ni WD ni;Ki , where .Ki / is an increasing sequence of compact subsets of X that absorbs X . Indeed, for every compact K  X there exists a j with K  Kj ; therefore nk;K  ni if we choose i to be the maximum of k and j . The following lemma generalizes a familiar result about linear mappings between linear spaces endowed with a norm. Lemma 8.7. Let .E; N / and .F; M/ be locally convex topological linear spaces. Then every continuous linear mapping A W E ! F is sequentially continuous. If N

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8 Distributions with Compact Support

is countable, the converse is also true, that is, every sequentially continuous linear mapping A W E ! F is continuous. Proof. Suppose that A is continuous and xj ! x in E. For every m 2 M we then have, on account of (8.2), m.A.xj /

A.x// D m.A.xj

x//  c n.xj

x/ ! 0

as

j ! 1:

Now suppose that .nj /j 2N is an enumeration of N . By changing to k WD max nj 1j k

.k 2 N/;

we obtain an increasing sequence of seminorms k that defines the same notion of neighborhood in E. Suppose A is not continuous. Then there exist an m 2 M and, for every c > 0 and k 2 N, an x D xc;k 2 E such that m.A.x// > c k .x/. Next, take c D k and introduce yk D  x with  D 1=m.A.x//. We then find that 1 D m.A.yk // > k k .yk /, from which we conclude that yk ! 0 in .E; N /, without A.yk / ! 0 in .F; M/. It follows that A is not sequentially continuous.  The proof of Theorem 3.8 is the special case of the proof of the second assertion in Lemma 8.7 with E D C01 .K/ and F D C. We have formulated Lemma 8.7 in this general form to avoid having to rewrite the same argument each time we work with another collection of seminorms. Theorem 8.8. Consider u 2 D 0 .X /. One has u 2 E 0 .X / if and only if supp u is a compact subset of X . If that is the case, u is of finite order and there exist constants c > 0 and k 2 Z0 satisfying ju./j  c k kC k

(8.3)

for every  2 C 1 .X / and every  2 C01 .X / such that  D 1 on an open neighborhood of supp u. Proof. Let u 2 E 0 .X /. By Lemma 8.7, with E D C 1 .X / and F D C, it follows from the sequential continuity of u that u is continuous in the sense of (8.2). That is, there exist constants c > 0 and k 2 Z0 and a compact subset K of X such that ju./j  c kkC k ; K

. 2 C 1 .X //:

(8.4)

This implies, in particular, that u./ D 0 if  2 C01 .X n K/, in other words, supp u  K. The conclusion is that supp u is a compact subset of X and that u is of finite order; see Definition 3.12. The estimate (8.3) now follows from .8:4/. Indeed, supp .  / \ supp u D ; implies u.  / D 0; in turn, this gives ju./j D ju. /j  c k kC k ; K  c k kC k :

8 Distributions with Compact Support

77

Next, suppose that supp u is compact. Corollary 2.16 yields a  2 C01 .X / with  D 1 on a neighborhood of supp u. Define v./ D u. / for every  2 C 1 .X /; then v is a linear form on C 1 .X /. Now, j ! 0 in C 1 .X / implies that  j ! 0 in C01 .X / and therefore also v.j / ! 0. As a result, v 2 E 0 .X /. Finally, if  2 C01 .X /, then supp .  / \ supp u D ;, and so 0 D u.  / D u./ v./. This means that u D  v.  In view of Theorem 8.8, E 0 .X / is said to be the space of distributions with compact support in X . Lemma 8.9. Let U and V be open subsets of Rn with U  V . For every u 2 E 0 .U / there exists exactly one v D i.u/ 2 E 0 .V / such that v D u on U and supp v D supp u. This defines an injective continuous linear mapping i from E 0 .U / to E 0 .V /. For v 2 E 0 .V / one has that v 2 i.E 0 .U // if and only if supp v  U . The mapping i is used to identify E 0 .U / with the space of the v 2 E 0 .V / such that supp v  U , with the notation i.u/ D u. Proof. The mapping U V from (7.2) defines a continuous linear mapping from C 1 .V / to C 1 .U /. Thus, v WD u B U V 2 E 0 .V /. It will be clear that v D u on U , and that v D 0 on C WD V n supp u. Because V D U [ C , we conclude that w D v whenever w 2 D 0 .V /, w D u on U and w D 0 on C , which proves the uniqueness. It is also evident that u D 0 if v D 0, which yields the injectivity of the linear mapping i .  The estimate (8.4) is of the same form as (3.4) in Theorem 3.8; however, in the case of (3.4) the test functions were restricted to C01 .K/. On C01 .K/, sequential continuity of linear forms is equivalent to continuity, and Theorem 3.8 asserts that the distributions on X are precisely those linear forms u on C01 .X / for which the restriction of u to C01 .K/, for every compact subset K of X , is continuous from C01 .K/ to C. This is the reason that we took the liberty, in discussing the definition of distributions, to speak of continuous linear forms, although in fact we had introduced them as sequentially continuous linear forms. We may add that it is also possible to endow C01 .X / with an (uncountable) collection of seminorms N , in such a way that the distributions are precisely the continuous linear forms on C01 .X / with respect to N ; see for example H¨ormander [12, Thm. 2.1.5]. Contrary to what might perhaps be expected, (8.4) does not generally hold for K D supp u; a counterexample is given in H¨ormander [12, Example 2.3.2]; see Problem 8.3. But on the other hand, we have the following result. Theorem 8.10. Let a 2 Rn and let U be an open neighborhood of a in Rn . Consider u 2 D 0 .U / with supp u D fag. Then u is of finite order, say k, and one has X u.x 7! .a x/˛ / : uD c˛ @˛ ıa with c˛ D ˛Š j˛jk

In particular, a Radon measure on U with support in a is of the form c ıa with c D u.1/. Finally, there exists c > 0 such that ju./j  c kkC k ; f a g , for all  2 C01 .U /.

78

8 Distributions with Compact Support

Proof. Considering that the conclusion relates only to the behavior of u near a, we may assume that U is a ball with center a. Furthermore, we can reduce the problem to the case that a D 0, by means of a translation. Because u 2 E 0 .U /, it follows from Theorem 8.8 that u has a uniquely determined continuous extension to C 1 .U /, again denoted by u. Based on the estimate (8.3), we will first prove u./ D 0

if

 2 C 1 .U /

and

@˛ .0/ D 0

.j˛j  k/:

Applying Taylor expansion, with an estimate for the remainder according to Theorem 6.2, we obtain for such a function , lim #0

1 sup j.x/j D 0:  k kxk

At 0, all derivatives to order k j j of the function @  vanish, and so the preceding formula implies 1 lim k j j sup j@ .x/j D 0: (8.5) #0  kxk Furthermore, we specify the functions  to be used in (8.3). In fact, choose  2 C01 .Rn / with  D 1 on a neighborhood of 0 and .x/ D 0 if kxk > 1. Write  .x/ D . 1 x/, for  > 0. We then have  D 1 on a neighborhood of 0 and  .x/ D 0 if kxk > ; also, for every multi-index ˇ there exists a constant cˇ > 0 with the property cˇ j@ˇ  .x/j  jˇ j .x 2 Rn /:  Since 0 … supp ..1  //, we have u./ D u. /Cu..1  // D u. /. Next we substitute D   into the estimate (8.3). In doing so, we write @˛ .x/ using Leibniz’s formula as a sum of a constant times terms of the form @ˇ  .x/ @ .x/, with ˇ C D ˛. Each of these terms can be estimated uniformly in x, as cˇ sup j@ .x/j;  jˇ j kxk which converges to 0 as  # 0 on account of (8.5), because jˇj D j˛j j j  k j j. The conclusion is that ju./j converges to 0 as  # 0. This is possible only if u./ D 0. For general  2 C 1 .U /, Taylor expansion to order k at 0 now gives .x/ D

X @˛ .0/ x ˛ C R.x/; ˛Š

j˛jk

where R 2 C 1 .U / and @˛ R.0/ D 0 whenever j˛j  k. The latter implies u.R/ D 0, and therefore X @˛ .0/ u./ D u.x 7! x ˛ /: ˛Š j˛jk

Applying the definition of derivative of ı leads to the formula in the theorem.



8 Distributions with Compact Support

79

The preceding theorem is definitely a local result. Still, its proof had to wait until the present chapter dealing with C 1 functions, on account of the use of the Taylor expansion. Generally speaking, a remainder in the Taylor expansion of a function in C01 .X /, with X open in Rn , does not belong to C01 .X /, because Taylor polynomials do not belong to that space. A characteristic application of the theorem occurs in the initial part of the proof of Theorem 13.3 below. An important role in the theory of locally convex topological linear spaces is played by the Hahn–Banach Theorems. The plural is used here because the terminology is customarily applied to several closely related results. One of these will now be used to derive an interesting property of distributions; the result is the following Dominated Extension Theorem; see [19, Theorem 3.3]. Its verification requires transfinite induction, and the theorem is therefore nonconstructive. Theorem 8.11. Suppose E is a linear space over C and n W E ! R0 a seminorm on E, while L is a linear subspace of E and v a linear form on L such that jv.y/j  n.y/

.y 2 L/:

Then v extends to a linear form u on E (i.e., the restriction of u to L equals v) that satisfies ju.x/j  n.x/ .x 2 E/: For later use, in Remark 11.12, Example 14.31, and the proof of Theorem 15.4, we also mention the following, closely related, result; see [19, Theorem 3.5]. Theorem 8.12. Suppose E is a locally convex topological linear space and L is a linear subspace of E. Then L is a dense subspace of E if and only if every continuous linear form on E that vanishes on L also vanishes on E. Relative to the filtration of distributions by order, those of order 0, the Radon measures, are the least singular ones. In addition, under the operation of differentiation these generate all distributions of higher order. Theorem 8.13. Suppose X  Rn is an open set, u 2 D 0 .X /, and k 2 Z0 . Then u is of order  k if and only if for every open neighborhood U of supp u in X , there exist Radon measures u˛ 2 D 0 .X /, with j˛j  k, such that X @˛ u˛ and supp u˛  U: uD j˛jk

Proof. Only ) requires a proof. Let m 2 N be the number of multi-indices ˛ 2 .Z0 /n such that j˛j  k. Then there exists a natural injection  of C0k .X / into the product space .C0 .X //m ; more specifically,  assigns to each  2 C0k .X / the m-tuple .@˛ /j˛jk of all derivatives of  of order  k. Obviously  is linear and

80

8 Distributions with Compact Support

injective. For k > 0 it is, however, not surjective, on account of Problem 12.12 or the equality of mixed partial derivatives of a C k function (see [7, Theorem 2.7.2]). The image of C0k .X / under  is a linear subspace, say L, of .C0 .X //m . Then the mapping C0k .X / ! L is an isomorphism of locally convex topological linear spaces. Indeed, the j converge to 0 in C0k .X / as j ! 1 if and only if all @˛ j converge to 0 in C0 .X /, for j˛j  k. We may now transfer any continuous linear form on C0k .X / as a continuous linear form on L and then extend the latter as a continuous linear form on .C0 .X //m on account of the Dominated Extension Theorem, Theorem 8.11. But given a finite family of locally convex topological linear spaces .E1 ; : : : ; Em /, the product E10  0


 Em of the topological dual spaces is canonically isomorphic to the topological dual of the product, via the correspondence 0 E10 


 Em 3 .v1 ; : : : ; vm / ! P .E1 


 Em 3 .y1 ; : : : ; ym / 7! jmD1 vj .yj / 2 C/ 2 .E1 


 Em /0 :

Applying this to the product .C0 .X //m , we see that a continuous linear form on this product is an m-tuple of Radon measures .v˛ /j˛jk on X , operating in the following manner: X .v˛ /..˛ // D v˛ .˛ /: j˛jk

It suffices to assume that this linear form extends the linear form u transferred to L and to take ˛ D @˛ , to see that X uD . 1/j˛j @˛ v˛ : j˛jk

We must now verify the condition on the supports of the Radon measures u˛ . Take a function  equal to 1 in a neighborhood of supp u and to 0 outside some closed subset contained in U . Such a function exists in view of Corollary 2.16. First of all, u D  u since for all test functions , we have  u./ D u. / D u./, and supp .1 / is contained in the complement of supp u. Therefore we obtain X X uD . 1/j˛j  @˛ v˛ ; with  @˛ v˛ D cˇ @ˇ ..@˛ ˇ / v˛ / j˛jk

ˇ ˛

for suitable cˇ 2 R, as follows from Leibniz’s formula (2.8). Finally, observe that the supports of the Radon measures .@˛ ˇ / v˛ are contained in supp , which, in turn, is contained in U . 

8 Problems

81

Problems 8.1. Suppose that the singular support of u 2 D 0 .X / is a compact subset of X . Prove that u is of finite order.  8.2. Let .uj /j 2N be a sequence in E 0 .X / that converges in D 0 .X / to u 2 D 0 .X /. If there exists a compact subset K of X with supp uj  K for all j , then supp u  K and uj ! u in E 0 .X /, in the sense that uj ./ ! u./ for every  2 C 1 .X /. Prove this. Let .a.j //j 2N be a sequence in X with the property that d.a.j /; Rn n X / ! 0 or ka.j /k ! 1 as j ! 1. Prove that limj !1 ıa.j / D 0 in D 0 .X /, but not in E 0 .X /.
8.3. Let x .j / j 2N be a sequence in Rn that converges to x 2 Rn . In addition,

assume that all x .j / and x differ from each other. Finally, let .j /j 2N be a sequence in R>0 with the property X X j D 1; while also j kx .j / xk < 1: j 2N

j 2N

(i) Prove the existence of x .j / , x, and j with these properties, for j 2 N. Hint: let .n.k//k2N be a strictly increasing sequence satisfying kx .n.k// xk  k1 . Then take j D k1 if j D n.k/, and j D 0 otherwise. (ii) Verify that X
u./ D j .x .j / / .x/ . 2 C01 .Rn // j 2N

defines a distribution u of order  1. Prove supp u D f x .j / j j 2 N g [ fxg and show that this is a compact set. (iii) Verify that for every l 2 N there exists l 2 C01 .Rn / with l D 1 on a neighborhood of x .j / if 1  j  l, while l D 0 on a neighborhood of x .j / if j > l and l D 0 on a neighborhood of x. Prove that for every k 2 Z0 , kl kC k ; supp u D 1;

while

u .l / D

l X

j :

j D1

(iv) Demonstrate that (8.4) does not hold for any k 2 Z0 with K D supp u. 8.4. Prove that for every r > 0, ur ./ D

Z

2

.r cos t; r sin t/ dt 0

. 2 C01 .R2 //

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8 Distributions with Compact Support

defines a Radon measure ur on R2 with compact support and determine its support. Is ur a locally integrable function? Calculate ur .x 7! x ˛ / for all ˛ with j˛j  2. For what functions  does one have limr#0 .r/ ur D ı? And what condition has to be met by the functions ˛ and ˇ so that
lim ˛.r/ ur C ˇ.r/ ı D
ı‹ r#0

Prove that if ˛.r/ ur C ˇ.r/ ı converges to a distribution v of order  2, it follows that v must be a linear combination of ı and
ı.  8.5. Consider the function E on RnC1 ' Rn  R defined by E.x; t/ D ut .x/ for t > 0, with u t as in Problem 5.5, and E.x; t/ D 0 for t  0. Prove that E is locally integrable, so that E can be interpreted as a distribution on RnC1 . Determine sing supp E. Let v D @ t E
x E 2 D 0 .RnC1 /, where
x denotes the Laplace operator with respect to only the x variables. Determine supp v and estimate the order of v. Finally, calculate v using integration by parts, as in Problem 5.6. Note that in this case there is no requirement that n  3. See Problem 17.2 for another method.

Chapter 9

Multiplication by Functions

If X is an open subset of Rn , the function belongs to C 1 .X /, and f is locally integrable on X , one has, for every test function , .test. f //./ D .test f /. /: For an arbitrary u 2 D 0 .X / we now define . 2 C01 .X //:

. u/./ D u. /

This leads to a linear form u on C01 .X /. Using Leibniz’s formula (2.8) we see that the C k norm of  can be estimated by a constant times the C k norm of ; using Theorem 3.8 we therefore conclude that u 2 D 0 .X /. The definition is formulated such that test. f / D

.test f /

for every locally integrable function f ; thus we do not immediately run into notation difficulties if we omit the premodifier “test.” From formula (9.2) below it follows that the mapping “multiplication by ”: u 7! u is linear and continuous from D 0 .X / to D 0 .X /. One has u D u on U if .x/ D 1 for all x in the open subset U of X . Because  D 0 if  2 C01 .X / and supp \ supp  D ;, we also obtain supp . u/  supp \ supp u: Furthermore, . /u D . u/, for  2 C 1 .X /.

Example 9.1. Suppose X is an open set in Rn and let a 2 X and 2 C 1 .X /. Then ıa D .a/ ıa in D 0 .X /. Indeed, for every  2 C01 .X /, we have . ıa /./ D ıa . / D . /.a/ D

.a/ .a/ D

.a/ ıa ./:

In particular, x ı D 0 in D 0 .X / if 0 2 X . In turn, this implies x ı 0 D x ı 0 ./ D ı 0 .x / D ı.@x .x // D

ı./

ı, because

x ı. 0 / D ı./:

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_9, © Springer Science+Business Media, LLC 2010

83

84

9 Multiplication by Functions

This leads to x 2 ı 0 D 0 in D 0 .X /, on account of x 2 ı 0 ./ D x ı 0 .x / D

ı.x / D

˛

x ı./ D 0:

Example 9.2. We have x PV x1 D 1 in D 0 .R/. In fact, for  2 C01 .R/, Z Z 1 1 x PV ./ D PV .x / D lim .x/ dx D .x/ dx D 1./: x x #0 RnŒ ;   R Using Problem 1.3 and Example 9.1 we now obtain the following identities in D 0 .R/: 1 1 1 ˛ 1 D x PV D x ˙ i x ı D x : x x˙i0 x˙i0 A first application of multiplication by functions is the following: Lemma 9.3. Let .j /j 2N be a sequence in C01 .X / as in Lemma 8.2.(b). For every u 2 D 0 .X /, the sequence of terms uj WD j u 2 E 0 .X / then converges in D 0 .X / to u, as j ! 1. In other words, E 0 .X / is dense in D 0 .X /. Proof. If  2 C01 .X /, then j  !  in C01 .X / on account of Lemma 8.2.(b); therefore .j u/./ D u.j / ! u./ as j ! 1.  By Theorem 3.15 we can similarly define the product u if 2 C k .X / and the distribution u is of order  k; in that case, u is of order  k. Thus  is a Radon measure for every continuous function and Radon measure . The Leibniz rule (2.9), (2.8) holds if f D 2 C 1 .X / and g D u 2 D 0 .X /. 1 Indeed, for every 1  j  n and  2 C0 .X /, .@j . u//./ D . u/. @j / D u. D .@j

uC

@j / D u.@j



@j . //

@j u/./;

which leads to @j . u/ D @j

uC

In addition, the product this we mean that j j

!

@j u

.

2 C 1 .X /; u 2 D 0 .X //:

u is continuous as a function of

uj ! u in D 0 .X / if in C 1 .X / and uj ! u in

(9.1)

and u together. By

D 0 .X /:

(9.2)

For the proof we write . j uj /./ D uj . j /, for  2 C01 .X /. Next, we observe that j  !  in C01 .X / and we use Theorem 5.5.(ii).

9 Multiplication by Functions

85

The mapping u 7! u, of multiplication by 2 C 1 .X /, is also written by means of the notation . In quantum mechanics this is a familiar operation; multiplying by xj is said to be the jth position operator. It will be necessary to recall the definition from time to time, in order to avoid confusion in the notation. If 2 C01 .X /, then is a sequentially continuous linear mapping from D 0 .X / to 0 E .X /. It is not possible to extend the product u in a continuous manner to arbitrary distributions and u. For example, let u t 2 C 1 .Rn /, for t > 0, be as in Problem 5.5. Then u t ı D u t .0/ ı D .4 t/ n=2 ı according to Example 9.1, which does not converge in D 0 .Rn / as t # 0. On the other hand, u t does converge to ı in D 0 .Rn / as t # 0; this forms an obstacle to a well-behaved definition of the product ı ı as an element of D 0 .Rn /. The product that maps both . ; u/ 2 C 1 .X /  D 0 .X / and .u; / 2 D 0 .X /  1 1 1 C .X / to u 2 D 0 .X / is not associative. Indeed, xCi .x ı/ D xCi 0 D 0 0 0 1 by Example 9.1, while Example 9.2 implies . xCi 0 x/ ı D 1 ı D ı ¤ 0. On the other hand, the more elementary case of associativity . u/ D . / u, for and  2 C 1 .X / and u 2 D 0 .X /, does hold. Now that we have multiplication by functions available, we can study linear partial differential equations with variable coefficients, of the form X c˛ .x/ @˛ u D f; (9.3) P .x; @/u D j˛jm

for given distributions f and desired distributions u. Here the coefficients c˛ are required to be C 1 functions on the domain space X of the distributions. Variants involving C k coefficients with finite k are also possible, but the order of u must then be limited and one has to keep track of the degrees of differentiability and of the orders. By way of example we show how the theory of linear ordinary differential equations, the case of n D 1, is developed in the distributional context. Theorem 9.4. Let I be an open interval in R, cj 2 C 1 .I / for 0  j  m, and cm .x/ ¤ 0 for all x 2 I . Then there exists, for every f 2 D 0 .I /, a solution u 2 D 0 .I / of the equation m X P u WD cj .x/ u.j / D f: (9.4) j D0

The subset in D 0 .I / of all solutions consists of the u C h, with h denoting any solution of the homogeneous equation, given by f D 0. Every such h is a classical C 1 solution, and together they form an m-dimensional linear space over C. If f 2 C k .I /, then u 2 C mCk .I / and u is a solution in the classical sense. Proof. Because 1=cm 2 C 1 .I /, we can multiply the equation by this factor; thus we obtain an equivalent equation for u, with cm replaced by 1, the cj by cj =cm , and f by f =cm . Consequently, in what follows we may assume that cm D 1.

86

9 Multiplication by Functions

We now perform the usual reduction to an m-dimensional first-order system. To achieve this, we introduce vj WD u.j 1/ , for 1  j  m. Also, gj WD 0 if 0  j < m and gm WD f . Together these make (9.4) equivalent to a system of the form v 0 D L.x/ v C g, where the m  m matrix L.x/ can be obtained from the explicit form m X 0 vj0 D vj C1 .1  j < m/; vm D cj 1 .x/ vj C gm : j D1

Let x 7! ˚.x/ be a classical fundamental matrix for the homogeneous system v 0 D L.x/ v, that is, an m  m matrix whose columns form a basis for the linear space of solutions of v 0 D L.x/ v. Its existence is proved in the theory of ordinary differential equations; the coefficients are shown to be C 1 functions of x, and similarly those of the inverse matrix ˚.x/ 1 . Furthermore, ˚ satisfies the matrix differential equation ˚ 0 .x/ D L.x/ B ˚.x/. We now apply, in the distributional context, Lagrange’s method of variation of constants for solving the inhomogeneous equation. This consists in substituting v D ˚.x/ w and then deriving the differential equation that w must satisfy to make v a solution of the inhomogeneous equation v 0 D L.x/ v C g. Because a distribution may be multiplied by a C 1 function, the substitution v D ˚.x/ w yields a bijective relation between the vector distributions v and w on I . Using Leibniz’s rule we obtain for w, L.x/ B ˚.x/ w C g D @x .˚.x/ w/ D L.x/ B ˚.x/ w C ˚.x/ w 0 ; in other words, w 0 D ˚.x/ 1 g. The notation used is shorthand for a calculation by components. But for every component this is an equation like the one we have already studied in Theorem 4.3. This yields the existence of a solution w 2 .D 0 .I //n and therefore of a solution v D ˚.x/ w 2 .D 0 .I //n of v 0 D L.x/ v C g, and consequently, of a solution u D v1 2 D 0 .I / of (9.4). Owing to the linearity of the operator P , the solution space of the inhomogeneous equation consists of the u C h with h the solutions of the homogeneous equation. Now we have g D 0 if f D 0; this gives the equation w 0 D 0, the only solutions of which are of the form w D c 2 Rn according to the last assertion in Theorem 4.3. But this implies that v D ˚.x/ c, which leads to a classical solution of the homogeneous equation. Finally, if f 2 C k .I /, there exists a classical solution u0 2 C kCm .I /. Every solution u 2 D 0 .I / of the inhomogeneous equation is of the form u D u0 Ch with h a classical C 1 solution of the homogeneous equation. It follows that u 2 C kCm .I / is a classical solution.  The condition that the highest-order coefficient cm should have no zeros is essential. This is sufficiently demonstrated by the example of the equation x u D 0 in D 0 .R/, which has the nonclassical solution u D ı, as follows from Example 9.1. In the next theorem we formulate a converse assertion, considering Rn straight away.

9 Multiplication by Functions

87

Theorem 9.5. Let X be open in Rn and u 2 D 0 .X /. If 2 C 1 .X / and u D 0, then supp u is contained in the zero-set of . Consider real-valued j 2 C 1 .X /, for 1  j  n, and a 2 X . Further suppose that j .a/ D 0 and, additionally, that the total derivatives D j .a/ of the j at a are linearly independent, with 1  j  n. Then j u D 0, for all j , implies the existence of c 2 C and an open neighborhood U of a in X satisfying u D c ıa on the set U . Proof. Let C D f x 2 X j .x/ ¤ 0 g. Then C is an open subset of X and 1 2 C 1 .C /; therefore u D 1 . u/ D 1 0 D 0 on C . This proves that supp u  X n C , the zero-set of . Denote by W X ! Rn the C 1 mapping having the j as component functions. Since .a/ D 0, we may write, for x sufficiently close to a, .x/ D D

.x/ Z

Z

.a/ D

1

1 0

D .a C t.x

0

d .a C t.x dt

a// dt .x

a// dt

a/ DW .x/ .x

a/;

with an n  n matrix consisting of C 1 functions. In particular, .a/ D D .a/; and by assumption this matrix is invertible. By continuity (for instance, consider the determinant of .x/), there exists an open ball U about a in Rn such that .x/ is invertible, for all x 2 U . In view of Cramer’s rule for inverse matrices, the matrix coefficients of 1 belong to C 1 .U /. We now have a D .x/

x

1

.x/:

By combining this equality with the identity of vectors .x

a/ u D 0;

.xj

in other words,

.x/ u D 0 we obtain

aj / u D 0

.1  j  n; x 2 U /:

In view of the assertion proved above we may conclude that U \ supp u D fag; on U , therefore, we may consider u as an element of E 0 .U / on account of Theorem 8.8. Taylor expansion of an arbitrary  2 C 1 .U / about a leads to .x/ D .a/ C

n X

.xj

aj / j .x/

j D1

.x 2 U /;

for certain j 2 C 1 .U /. We obtain u./ D u..a// C

n X

j D1

u..xj

aj / j / D .a/ u.1/ C

D u.1/ ıa ./;

and from this we deduce u D u.1/ ıa on U .

n X

.xj

aj / u.j /

j D1



88

9 Multiplication by Functions

Note that Example 9.1 shows that the nondegeneracy condition in the theorem, requiring that the derivatives be linearly independent, in general cannot be omitted.

Problems 9.1. Suppose p 2 C 1 .  a; b Œ /, p.x/ > 0 for all x 2  a; b Œ and q and r continuous on  a; b Œ. Prove that every Radon measure u that is a solution of the variational equation (1.10) for all  2 C01 .  a; b Œ / is in fact C 2 and satisfies the Euler–Lagrange equation (1.8). Hint: use v2 D p.x/ u0 .

 9.2. Let 2 C 1 .Rn /. Describe the distribution as a linear combination of the @ˇ ı with ˇ  ˛.

@˛ ı, for every multi-index ˛,

 9.3. Determine all solutions u 2 D 0 .R/ of x k u D 0, for k 2 N.

 9.4. Determine all solutions u 2 D 0 .R/ of x u0 D 0. Show that these u form a linear space and determine its dimension.

 1 9.5. Observe that u D xCi is a solution of x u D 1, and prove that this implies 0 0 x u D u. Let u 2 D 0 .R/ be a solution of x k u D 1. Applying the differential operator x @x to this identity, show that v D k1 u0 is a solution of x kC1 v D 1. Finally, determine for every k 2 N, all solutions in D 0 .R/ of the equation x k u D 1.  9.6. Prove

1 D  i ı: (9.5) t !1 x Deduce (refer to Problem 14.13 for another proof); see Fig. 9.1, Z cos tx 1 t ix sin tx lim Dı and lim PV D 0: e d  D lim t !1 2 t !1 t !1 x x t lim e i tx PV

Π

50 -Π

-Π

Π

Π -10.86

Fig. 9.1 Graphs of x 7!

sin 50x , x

-Π

one with scales adjusted and one with equal scales, but truncated

9 Problems

89 i tx

e Next define u t; 2 C 1 .R/ by u t; .x/ D xCi  , where t and  2 R and  ¤ 0. Using Problem 1.3 prove the following equalities in D 0 .R/:

(i)

lim t !1 .lim#0 u t; / D 0; (ii) lim t !1 .lim"0 u t; / D 2 i ı; (iii) lim!0 .lim t !1 u t; / D 0:

In particular, limits in D 0 .Rn / cannot be freely interchanged.  9.7. Let u 2 D 0 .Rn / and xn u D 0. Prove that there exists a uniquely determined v 2 D 0 .Rn 1 / such that u./ D v. / for every  2 C01 .Rn /. Here . /.x1 ; : : : ; xn

1/

D .x1 ; : : : ; xn

1 ; 0/:

Hint: use  2 C01 .R/ with .x/ D 1 on a neighborhood of 0. For 2 C01 .Rn 1 /, define the function ˝  2 C01 .Rn / by . ˝ /.y; z/ D .y/ .z/ and deduce that v must be defined by v. / WD u. ˝/. See Problem 11.18 for a reformulation. 9.8. Prove that the differential equation .1

x 2 /2 u0

2x u D .1

x 2 /2

has no solution u 2 D 0 .I / if I is an open interval in R with Œ 1; 1   I .

9.9. Let a 2 C. Show that the solutions u 2 D 0 .R/ of the differential equation x u0 D a u form a linear space Ha . Demonstrate that multiplication by x, and the differentiation @x , define linear mappings from Ha to HaC1 and Ha 1 , respectively. Determine Ha . Hint: use Problem 7.6 and deal with the case a 2 Z 0, a resistor of resistance R  0, and a capacitor of capacitance C > 0 connected by ideal wires in which at time 0 a constant voltage of magnitude 1 is switched on. Prove that I is given by, for t 2 R (see Fig. 9.2), r 8 2 t 4L 4L R t ˆ ˆ ; if R2 > I q H.t/ e 2L sinh R2 ˆ ˆ 2L C C 4L ˆ 2 ˆ R ˆ C ˆ R2 4L ; in the second, it is also of exponential decay; in the third, it is C an exponentially damped sinusoidal oscillation. In this last case, under the additional condition ofq R D 0, the solution takes the form of an undamped sinusoidal oscillation I.t/ D

C L

H.t/ sin p t

LC

(see Problem 14.28 for a different approach).

Chapter 10

Transposition: Pullback and Pushforward

We now consider the behavior of distributions under coordinate transformations and, more generally, under suitable C 1 mappings. Suppose that .E; N / and .F; M/ are locally convex topological linear spaces and that A is a continuous linear mapping from E to F . (For the definitions of these notions, see Chap. 8.) For every v belonging to the topological dual F 0 of F , the mapping v B A, being a composition of continuous linear mappings A

v

.E; N / ! .F; M/ ! C, is a continuous linear form on E in view of a special case of Lemma 10.1 below. Therefore tA.v/ WD v B A is an element of the topological dual E 0 of E. Indeed, A

/F E@ @@ @@ v tA.v/DvBA @@  C

where

t

A.v/ W E 3 x 7! v.A.x// 2 C:

This defines a mapping tA W F 0 ! E 0 , said to be the transpose of A, on account of the defining formula t

A.v/.x/ WD .tA.v//.x/ WD v.A.x//

.v 2 F 0 ; x 2 E/:

The situation may be summarized by means of the following diagram: E E0 o

A tA

/F F0

Note that tA is automatically continuous, and therefore sequentially continuous, from F 0 to E 0 . Indeed, if x 2 E, one has, for every v 2 F 0 , that jtA.v/.x/j D jv.A.x//j, where the left-hand side is the x-seminorm of t A.v/ and the right-hand side is the A.x/-seminorm of v.

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_10, © Springer Science+Business Media, LLC 2010

91

92

10 Transposition: Pullback and Pushforward

Lemma 10.1. Let .E; N /, .F; M/, and .G; L/ be locally convex topological linear spaces. If A is a continuous linear mapping from .E; N / to .F; M/ and B a continuous linear mapping from .F; M/ to .G; L/, then B B A is a continuous linear mapping from .E; N / to .G; L/. For its transpose one has t .B B A/ D tA B tB. Proof. Let l 2 L. By virtue of the continuity of B and of formula (8.2), there exist a constant d > 0 and m 2 M with l.B.y//  d m.y/ for all y 2 F . On account of the continuity of A, there exist for this m a constant c > 0 and n 2 N such that m.A.x//  c n.x/ for all x 2 E. Combining these two estimates, we obtain l.B B A.x//  d m.A.x//  dc n.x/ for all x 2 E. Note that in the case of linear spaces endowed with a norm, this was the proof of the theorem that the operator norm of B B A is less than or equal to the product of the operator norms of B and of A. For the proof of the second assertion, consider w 2 G 0 and x 2 E. Then t

.B B A/.w/.x/ D w.B.A.x/// D tB.w/.A.x// D tA. tB.w//.x/:



We have, in fact, applied the principle of transposition several times before, when introducing operations involving distributions. For example, we defined @j W D 0 .X / ! D 0 .X / as the transpose of @j W C01 .X / ! C01 .X /; in other words, t @j D @j . Also, for 2 C 1 .X /, we introduced the multiplication by W 0 0 D .X / ! D .X / as the transpose of the multiplication by W C01 .X / ! C01 .X /. If, in addition, has compact support in X , multiplication by is even continuous and linear from C 1 .X / to C01 .X /, which then immediately implies that multiplication by , the transpose, defines a continuous linear mapping from D 0 .X / to E 0 .X /. Furthermore, the restriction mapping U V W D 0 .V / ! D 0 .U / from Chap. 7 is also a transpose, namely of the identical embedding V U W C01 .U / ! C01 .V /. We summarize the situation by the following diagram, where v 2 D 0 .V / and  2 C01 .U /: V U W C01 .U / D 0 .U /

! C01 .V / D 0 .V / W U V

with

.U V v/./ D v.V U /:

Remark 10.2. For general continuous linear A W C01 .U / ! C01 .V /, it is customary to denote tA W D 0 .V / ! D 0 .U / by B if B is a previously introduced operator acting on functions and B D tA on the subspace C01 .V / of D 0 .V /. This is what we have done in the case of differentiation and multiplication by functions, and this will also be our guiding principle in the subsequent naming of transpose operators. In Corollary 11.7 below we will see that C01 .V / is dense in D 0 .V /, which makes this procedure almost inevitable. Sometimes, however, the transpose operator turns out to be truly novel, for example when tA./ 2 D 0 .U / is not a function for some  2 C01 .V /. In later parts of this chapter we will encounter further examples of this, for instance in Proposition 10.21. ˛

10 Transposition: Pullback and Pushforward

93

Let X be an open subset of Rn , Y an open subset of Rp , and ˚ W X ! Y a C 1 mapping. Then B ˚ W x 7! .˚.x// belongs to C 1 .X /, for every 2 C 1 .Y /. The mapping ˚  W 7! B ˚ W C 1 .Y / ! C 1 .X / is linear (albeit between infinite-dimensional spaces) and by means of the chain rule for differentiation it can be shown to be also continuous. Note that .˚  /.x/ D .y/ if y D ˚.x/; thus, ˚  is something like “the change of variables y D ˚.x/ in functions of y.” The operator ˚  is said to be the pullback of functions on Y to functions on X , under the mapping ˚ W X ! Y . In fact, transposition is the pullback of continuous linear forms under continuous linear mappings, although the latter may have been defined on infinite-dimensional spaces. Furthermore, observe that for X D R and Y D f0g  R, we get .˚  /.x/ D .0/, for 2 C01 .Y / and x 2 R. In this case, therefore, we do not have ˚  W C01 .Y / ! C01 .X /. Example 10.3. Let ˚ W X ! Y and W Y ! Z be C 1 mappings, where X  Rn , Y  Rp , and Z  Rq are open sets. Then we have, on account of the chain rule p (6.1), X .@j B ˚  / i D @j . i B ˚/ D @j ˚k .˚  B @k / i ; kD1

for 1  j  n and 1  i  q. In particular, if q D 1, this equality is valid for all D 1 2 C 1 .Y /; therefore we obtain the following identity of continuous linear mappings, which describes composition of partial differentiation and pullback, for 1  j  n: p X  @j B ˚ D @j ˚k B ˚  B @k W C 1 .Y / ! C 1 .X /: (10.1) kD1

On the right-hand side of this equality @j ˚k denotes the linear operator of multiplication by @j ˚k acting in C 1 .X /. In the special case of a linear mapping ˚ D A W Rn ! Rn with matrix .aij /, n (10.1) takes the form X   @j B A D A B akj @k : kD1

Hence





t

@ B A D A B . A @ /;

where

@ D .@1 ; : : : ; @n /

(10.2)

is regarded as a column vector and tA W Rn ! Rn denotes the transpose linear mapping. ˛ The transpose of the continuous linear mapping ˚  W C 1 .Y / ! C 1 .X / is a continuous linear mapping t .˚  / W E 0 .X / ! E 0 .Y /. This is written as t .˚  / D ˚ and is said to be the pushforward of distributions on X with compact support to distributions on Y with compact support, under the mapping ˚ W X ! Y . The defining equation is .˚ u/. / D u.˚  / D u.

B ˚/

.u 2 E 0 .X /;

2 C 1 .Y //:

(10.3)

94

10 Transposition: Pullback and Pushforward

We describe the preceding results by means of the following diagram: ˚

X C 1 .X / o

/Y

˚

C 1 .Y /

˚

E 0 .X /

/ E 0 .Y /

In mathematics, ˚ is said to be covariant with respect to ˚, because ˚ acts in the same direction as ˚ itself, while ˚  is said to be contravariant with respect to ˚, seeing that the sense of the arrow is reversed. Distribution theory parallels geometry regarding these conventions. In the latter, the underlying space and the points it contains are the initial (covariant) objects, the smooth functions on the space are dual (contravariant) objects, while compactly supported distributions are objects dual to smooth functions, hence covariant again. This is corroborated by the fact that the Dirac measures, which correspond to the points, form a subset of the distributions. If is a C 1 mapping from Y to an open subset Z of Rq , then B ˚ is a C 1 mapping from X to Z. As in the proof of the second assertion in Lemma 10.1, we see that . B ˚/ D ˚  B  W C 1 .Z/ ! C 1 .X /: Transposing this in turn, we obtain . B ˚/ D  B ˚

E 0 .X / ! E 0 .Z/:

W

(10.4)

Example 10.4. For every x 2 X one has ˚ ıx D ı˚.x/ . Indeed, for every C 1 .Y / we obtain .˚ ıx /. / D ıx . B ˚/ D .˚.x// D ı˚.x/ . /.

2 ˛

Example 10.5. Set V D f x 2 R2 j kxk D 1 g and consider the distribution
ıV 2 E 0 .R2 / of order 2, where
denotes the Laplace operator as in Problem 4.7. Thus Z Z  .
ıV /./ D
.x/ dx D
.cos ˛; sin ˛/ d˛ . 2 C 1 .R2 //: 

V

Define ˚ W R2 ! R by ˚.x/ D kxk2 . Then we have ˚
ıV D 8.ı1 00

ı1 0 /

in E 0 .R/: 2 C 1 .R/,

Indeed, using the chain rule repeatedly we obtain, for @21 .

B ˚/.x/ D 4x12

00

.kxk2 / C 2

0

.kxk2 /;

and an analogous formula for differentiation with respect to x2 . It follows that .˚
ıV /. / D .
ıV /.˚  / D ıV .
. Z  ..cos2 ˛ C sin2 ˛/ D4 

B ˚// 00

.1/ C

0

.1// d˛ D 8.ı1 00

ı1 0 /. /:

10 Transposition: Pullback and Pushforward

95

See Problem 10.20.(vi) for another proof. 0

0

0

˛

We are not allowed to conclude from the density of E .X / in D .X / and of E .Y / in D 0 .Y / (see Lemma 9.3) that the continuous linear mapping ˚ W E 0 .X / ! E 0 .Y / extends to a continuous linear mapping ˚ W D 0 .X / ! D 0 .Y /. This is because the topology of E 0 .X / is finer than the topology that E 0 .X / inherits as a linear subspace of D 0 .X /. Nevertheless, in the following theorem ˚ is extended, subject to additional conditions on the mapping ˚, to distributions that may not have compact support. We need some preparation. For a closed subset A of X , we denote the space of the u 2 D 0 .X / with supp u  A by D 0 .X /A . We then say that uj ! u in D 0 .X /A if uj 2 D 0 .X /A for all j and uj ! u in D 0 .X / as j ! 1. This implies that u 2 D 0 .X /A . The mapping ˚ W X ! Y is said to be proper on A if for every compact subset L of Y , the set ˚ 1 .L/ \ A is a compact subset of X . An important property of a proper and continuous mapping ˚ W X ! Y is that it is closed, that is, the image under ˚ of every closed set in X is closed in Y ; see [7, Theorem 1.8.6]. Theorem 10.6. Let ˚ be a C 1 mapping from the open subset X of Rn to the open subset Y of Rp . Let A be a closed subset of X and suppose that ˚ is proper on A. Then there exists a uniquely determined extension of ˚ W E 0 .X / \ D 0 .X /A ! 0 E .Y / to a linear mapping ˚ W D 0 .X /A ! D 0 .Y / with the property that ˚ uj ! ˚ u in D 0 .Y / if uj ! u in D 0 .X /A . Furthermore, ˚.supp u/ is a closed subset of Y and supp .˚ u/  ˚.supp u/ .u 2 D 0 .X /A /: (10.5) In particular, if ˚ is proper from X to Y , then ˚ W E 0 .X / ! E 0 .Y / possesses an extension to a sequentially continuous linear mapping ˚ W D 0 .X / ! D 0 .Y /, and one has (10.5) for all u 2 D 0 .X /.

Proof. If u 2 D 0 .X /A , the sequence of terms uj D j u 2 E 0 .X / from Lemma 9.3 converges to u in D 0 .X /A . This leads to the uniqueness: if the extension exists, we have to take ˚ u D lim ˚ uj : j !1

Let L be an arbitrary compact subset of Y and  2 C01 .X / with  D 1 on a neighborhood of the compact set K D ˚ 1 .L/ \ A; see Corollary 2.16. For all  2 C01 .L/, we define v./ D .˚ . u//./ D u. ˚  /: Because the linear mapping ˚  is continuous from C01 .L/ to C 1 .X / and multiplication by  is continuous from C 1 .X / to C01 .X /, v is a continuous linear form on C01 .L/. Furthermore, if uj ! u in D 0 .X /A , then .˚ uj /./ D uj .˚  / D . uj /.˚  / D uj . ˚  / converges to v./ as j ! 1. From the resulting uniqueness it follows that this v has an extension to C01 .Y /; this is the ˚ u with the desired continuity properties.

96

10 Transposition: Pullback and Pushforward

With respect to (10.5) we observe that  2 C01 .Y / and supp  \ ˚.supp u/ D ; imply that supp .˚  /  ˚ 1 .supp / is disjunct from supp u, and so .˚ u/./ D lim .˚ uj /./ D lim uj .˚  / D 0: j !1

j !1

The last equality follows from supp uj D supp j u  supp u, which is disjoint from supp .˚  /. Thus, we may conclude that supp .˚ u/  ˚.supp u/ D ˚.supp u/, because the restriction of ˚ to A is a closed mapping.  Example 10.7. Denote by 1X 2 D 0 .X / the characteristic function of an open subset X of Rn . Write
W X ! X  X for the diagonal mapping1
.x/ D .x; x/, which is proper. In this case,
.X / D f .x; x/ 2 R2n j x 2 X g is called the diagonal in X  X . Next, consider the pushforward Radon measure Z
 1X 2 D 0 .X  X /I then
 1X . / D .x; x/ dx; (10.6) X

for 2 C01 .X  X /. Note that supp
 1X D
.X /, while
 1X D 2 where ı
.X / is the Radon measure on X  X as in Example 7.3.

n 2

ı
.X / , ˛

We will now discuss the pushforward of distributions under C 1 mappings having various additional properties. We begin by considering diffeomorphisms, and then, in Theorem 10.18 and Proposition 10.21, we study mappings between open sets of different dimensions. A C 1 diffeomorphism ˚ W X ! Y is a C 1 mapping from X to Y that is bijective and whose inverse mapping WD ˚ 1 is a C 1 mapping from Y to X . First, we recall some basic facts about diffeomorphisms. Applying the chain rule to B ˚.x/ D x and to ˚ B .y/ D y, we find, with the notation y D ˚.x/, that D .y/ B D˚.x/ D I and D˚.x/ B D .y/ D I . In other words, D˚.x/ W Rn ! Rp is a bijective linear mapping, with inverse D .y/; this also implies that n D p. Conversely, if ˚ is an injective C 1 mapping from an open subset X of Rn to Rn such that D˚.x/ is invertible for every x 2 X , the Inverse Function Theorem asserts that Y WD ˚.X / is an open subset of Rn and that ˚ is a C 1 diffeomorphism from X to Y . See, for example, [7, Theorem 3.2.8]. In the following theorem we use the notation j˚ .x/ D j det D˚.x/j

.x 2 X /

(10.7)

for a differentiable mapping ˚ from an open subset X of Rn to Rn . The matrix of derivatives D˚.x/ is also referred to as the Jacobi matrix of ˚ at the point x, and det D˚.x/ as the Jacobi determinant or Jacobian. Note that j˚ 2 C 1 .X / if ˚ W X ! Rn is a C 1 diffeomorphism. 1

In this book, one can distinguish between the diagonal mapping and the Laplace operator on the basis of the context in which we make use of them.

10 Transposition: Pullback and Pushforward

97

Theorem 10.8. Suppose that X and Y are open subsets of Rn and ˚ W X ! Y a C 1 diffeomorphism. Then ˚ is proper and the pushforward ˚ W D 0 .X / ! D 0 .Y / is a sequentially continuous linear mapping. In particular, for a locally integrable function f on X , ˚ f is the locally integrable function on Y given by ˚ f D j  f D 

f  ; j˚

where

WD ˚

1

:

(10.8)

In particular, we have j˚ ˚  B ˚ D I on C01 .X /. Proof. The properness of ˚ follows from the continuity of ; accordingly, the existence of the pushforward ˚ W D 0 .X / ! D 0 .Y / is a consequence of Theorem 10.6. For every 2 C01 .Y / we obtain, by means of the change of variables x D .y/, Z .˚ f /. / D f .˚  / D f . B ˚/ D f .x/ .˚.x// dx X Z (10.9) D f . .y// .y/ j .y/ dy D .j  f /. /: Y

See, for example, [7, Theorem 6.6.1] for the change of variables in an n-dimensional integral.  Example 10.9. Fix  > 0. The (isotropic) positive dilation  W Rn ! Rn given by x 7!  x is a diffeomorphism. Hence we obtain for a locally integrable function f on Rn and x 2 Rn , 1 1  x D f .x/:  f .x/ D  n . 1 / f .x/ D n f  

For the last function on the right-hand side we used the notation from Lemma 2.19. ˛ Remark 10.10. In the case of a diffeomorphism, pushforward leaves the space of compactly supported functions of class C 1 invariant, and therefore we can extend pullback of functions of class C 1 to a mapping acting on distributions. More precisely, as a consequence of Theorem 10.8, the restriction of ˚ to C01 .X / is a continuous linear mapping A from C01 .X / to C01 .Y /. Its transpose tA is a continuous linear mapping from D 0 .Y / to D 0 .X /. Let 2 C 1 .Y /. For every  2 C01 .X / one has .tA /./ D

.A/ D

.˚ / D .˚ /. / D .˚  / D .˚  /./;

and therefore tA D ˚  . In other words, tA W D 0 .Y / ! D 0 .X / is an extension of ˚  W C 1 .Y / ! C 1 .X /. This extension is also denoted by ˚  in order to conform to the convention in Remark 10.2; it is said to be the pullback under ˚ of distributions on Y to distributions on X . Because C01 .Y / is dense in D 0 .Y /, see Corollary 11.7 below, this continuous extension is uniquely determined and the naming is entirely natural. ˛

98

10 Transposition: Pullback and Pushforward

The following result summarizes the results above. Theorem 10.11. Under the conditions of Theorem 10.8 the pullback ˚  W D 0 .Y / ! D 0 .X / is a continuous linear mapping. For any v 2 D 0 .Y / and  2 C01 .X /, it satisfies     .˚  v/./ D j v.  / D v  ; where WD ˚ 1 : j˚ Example 10.12. It is a direct application of Theorem 10.8 that j˚ ˚  B ˚ D I on D 0 .X /. In particular, Example 10.4 therefore implies ˚  ı˚.x/ D

1 ıx j˚ .x/

˛

.x 2 X /:

From Theorem 10.8 it follows that for a diffeomorphism ˚, the pushforward under ˚ and the pullback under the inverse ˚ 1 of ˚ are closely related to each other. Indeed, we have ˚ B j˚ D .˚

1 

/

D 0 .X / ! D 0 .Y /:

W

(10.10)

In particular, ˚ and .˚ 1 / are identical if and only if j˚ equals 1, that is, if ˚ is volume-preserving. For u 2 D 0 .X /, the distribution ˚ u 2 D 0 .Y / is also said to be the transform of u under ˚ as a distributional density, while .˚ 1 / u is said to be the transform of u as a generalized function. Remark 10.13. The pullback of distributions under the diffeomorphisms that act as coordinate transformations enables one to define distributions on manifolds. See, for example, H¨ormander [12, Sect. 6.3]. ˛ Example 10.14. The reflection S W x 7! x in Rn about the origin satisfies S S and jS D 1, and therefore, in this case, S D .S

1

D

1 

/ D S :

This transformation of distributions on Rn is denoted by the same letter S ; for every u 2 D 0 .Rn /, S u is said to be the reflected distribution. Somewhat more generally, if A is an invertible linear transformation of Rn , then jA equals the constant j det Aj, so that (10.10) leads to the following identity of continuous linear mappings on D 0 .Rn /: j det Aj A D .A

1 

/ :

˛

10 Transposition: Pullback and Pushforward

99

Example 10.15. Let Th W x 7! x C h W Rn ! Rn be the translation in Rn by the vector h 2 Rn . This is evidently a diffeomorphism, whose Jacobi matrix equals the identity and whose Jacobi determinant therefore equals 1. Consequently, the mapping Th  W C 1 .Rn / ! C 1 .Rn / has an extension to a continuous linear mapping Th  D .Th

1

/ D .T

h /

W D 0 .Rn / ! D 0 .Rn /:

(10.11)

If e.j / denotes the jth basis vector in Rn , one has, for every  2 C01 .Rn /, T t e.j /  

Z 1
d .x C s te.j // ds  .x/ D .x C te.j // .x/ D 0 ds Z 1 @j .x C s te.j // ds: Dt 0

From this we obtain

1 T t e.j /  t !0 t lim


I  D @j 

in

C01 .Rn /:

(10.12)

In conjunction with (10.11), this implies that for u 2 D 0 .Rn / and  2 C01 .Rn /, 1 T t e.j /  u t


1 .T u ./ D t

1
T u ./ D u t

t e.j / / u

t e.j /

converges to u. @j / D @j u./ as t ! 0. In other words, lim

t !0

1 T t e.j /  t


I u D @j u

in

D 0 .Rn /:










(10.13)

For example, if u is a layer on a surface V , as introduced in Example 7.3, one may regard @j u as having been obtained by translating 1=t times the charge distribution on V by t e.j /, adding the opposite of this on V and then taking the limit as t ! 0. As long as one does not pass to the limit, this is similar to a capacitor from the theory of electricity. For this reason the derivative @j u of a layer u is also said to be a double layer and, more generally, the distribution @˛ u, with u a layer and j˛j > 0, is said to be a multiple layer. (The question whether a kth order derivative should be called a 2k -fold layer or a .k C 1/-fold layer is undecided.) In (1.13) translation of functions was written more concisely as Th .f / WD f B T h D .T h / .f /, in other words, Th D .T

h/



D .Th / :

(10.14)

For example, using this notation we obtain Th .ıa / D ıaCh . The disadvantage of this notation is that now @j u is the derivative of T t e.j / u with respect to t at t D 0. But that is the way it is: when the graph of an increasing function is translated to the right, it will lie lower. Summarizing, we have the following identities of linear mappings acting in D 0 .Rn /, for 1  j  n:

100

10 Transposition: Pullback and Pushforward

ˇ d ˇ T t e.j / ˇ D t D0 dt

@j

and

ˇ d ˇ T t e.j /  ˇ D @j : t D0 dt

(10.15)

In words, differentiation is the infinitesimal generator of translation; for more details, see [7, Sect. 5.9]. ˛ Equation (10.12) can be generalized as follows. Let I be an open interval in R and let As be a continuous linear mapping from .E; N / to .F; M/, for every s 2 I . This one-parameter family of linear mappings is said to be differentiable at the point a if there exists a continuous linear mapping B from .E; N / to .F; M/ such that for every x 2 E,
1 lim .As .x/ Aa .x/ D B.x/ in F: s!a s a

d In this case one writes B D ds As jsDa . Writing out the definitions, one immediately sees that this implies that the one-parameter family s 7! t.As /, of continuous linear mappings from F 0 to E 0 , is also differentiable at s D a, while

ˇ  ˇ t d d t ˇ ˇ As ˇ : .As /ˇ D sDa sDa ds ds

The generalization to families of mappings that depend on more than one parameter is obvious. Let X and Y be open subsets of Rn and Rp , respectively. Let I be an open interval in R and let ˚ W X  I ! Y be a C 1 mapping. Then ˚ t .x/ D ˚.x; t/ defines, for every t 2 I , a C 1 mapping from X to Y . Denote by ˚ t;k the kth component function of ˚ t , for 1  k  p. Applying (10.1) with j D n C 1 to the ˚ under consideration, one deduces that the one-parameter family t 7! ˚ t  , of continuous linear mappings from C 1 .Y / to C 1 .X /, is differentiable and that one has the identity of continuous linear mappings p X d d  ˚t D ˚ t;k B ˚ t  B @k dt dt kD1

W

C 1 .Y / ! C 1 .X /

(10.16)

for the derivative with respect to t of the pullback. On the right-hand side of this d d equality dt ˚ t;k denotes the linear operator of multiplication by dt ˚ t;k acting in 1 C .X /. In particular, if the ˚ t are C 1 diffeomorphisms and if we restrict the operators in this equality to C01 .Y /, we obtain an identity of continuous linear mappings sending C01 .Y / to C01 .X /. Transposition and Lemma 10.1 as well as the fact that the transpose of @k is @k then immediately lead to the formula d .˚ t / D dt

p X

kD1

@k B .˚ t / B

d ˚ t;k dt

W

D 0 .X / ! D 0 .Y /

10 Transposition: Pullback and Pushforward

101

d for the derivative with respect to t of the pushforward. In this case, dt ˚ t;k is the d 0 linear operator of multiplication by dt ˚ t;k acting in D .X /. If we restrict the operators in this formula to C01 .X / and transpose once more, we find that (10.16) also holds when we have the continuous linear mappings act on D 0 .Y /. These formulas become simpler if X D Y , 0 2 I , and ˚0 .x/ D x, for all x 2 X ; that is, ˚0 D I , the identity on X . In this case (compare with [7, Sect. 5.9]) ˇ d ˇ ˚ t .x/ˇ x 7! v.x/ D t D0 dt

is said to be the velocity vector field of ˚ at time t D 0. This defines a C 1 vector field on X with components vj 2 C 1 .X /; we now find for the pullback n ˇ X d ˇ ˚t ˇ D vj B @j t D0 dt

D 0 .X / ! D 0 .X /;

W

j D1

(10.17)

and for the pushforward

ˇ d ˇ .˚ t / ˇ D t D0 dt

n X

j D1

@j B vj

D 0 .X / ! D 0 .X /:

W

(10.18)

Conversely, for an arbitrary C 1 vector field v on X , let ˚.x; t/ be the solution x.t/ at time t of the system of differential equations x 0 .t/ D v.x.t//

with initial condition

x.0/ D x:

Suppose, for simplicity, that the solutions for all .x; t/ 2 X  R exist. This is equivalent to the condition that none of the solutions will leave every compact subset of X in a finite period of time. According to the theory of ordinary differential equations, ˚ in that case is a C 1 mapping from X  R to X and ˚ t W x 7! ˚ t .x/ D ˚.x; t/ is said to be the flow over time t with velocity vector field v. The latter satisfies the so-called group law ˚ t Cs D ˚ t B ˚s ;

˚ t Cs  D ˚s  B ˚ t 

which also implies

.t; s 2 R/:

In particular, all the ˚ t are C 1 diffeomorphisms. By differentiation with respect to t at t D 0 we find, in view of (10.17), that n

Transposition gives

X d ˚s  D ˚s  B vj B @j : ds

(10.19)

j D1

.˚ t Cs / D .˚ t / B .˚s / ;

from which we obtain, by differentiation with respect to s at s D 0, and on account of (10.18),

102

10 Transposition: Pullback and Pushforward

d .˚ t / D dt

.˚ t / B

n X

j D1

@j B vj :

(10.20)

The following theorem gives an application of this. The distribution u 2 D 0 .X / is said to be invariant under the diffeomorphism ˚ W X ! X as a generalized function if .˚ 1 / u D u. Applying ˚  , we see that this is equivalent to ˚  u D u. This u is said to be invariant under ˚ as a distributional density if ˚ u D u. Theorem 10.16. Let v be a C 1 vector field on an open set X in Rn whose solutions are defined for all .x; t/ 2 X  R. Then u 2 D 0 .X / is invariant as a generalized function, and as a distributional density, under the flow with velocity vector field v if and only if n X

j D1

vj B @j u D 0;

and

n X

j D1

@j B vj u D 0;

respectively:

Proof. Because ˚0 D I , the identity in X , we have ˚0  u D I  u D u. Therefore, ˚ t  u D u for all t 2 R is equivalent to the assertion that ˚ t  u is constant as a function of t. By testing with some  2 C01 .X / and applying the theorem that a real-valued function of a real variable is constant if and only if its derivative vanishes, we find that the first assertion follows from (10.19). Likewise, the second assertion follows from (10.20), on account of .˚0 / u D I u D u.  Homogeneity of functions and homogeneity of distributions can be discussed in analogous ways. For c > 0, we denote the mapping x 7! c x W Rn ! Rn by c. Then u 2 D 0 .Rn / is said to be homogeneous of degree a 2 C if for every c > 0, c  u D c a u;

or equivalently

c nCa c u D u:

Theorem 10.17. u 2 D 0 .Rn / is homogeneous of degree a 2 C if and only if it satisfies Euler’s differential equation: n X hx; grad ui D xj @j u D a u: j D1

Proof. x 7! e t x is the flow of the vector field v.x/ D x. Define A t .u/ D e at .e t / u. Then A t B As D A t Cs and we obtain that u is homogeneous of ded gree a () A t .u/ D u for all t 2 R () dt A t .u/j t D0 D 0. By PLeibniz’s rule and (10.17), the left-hand side in the latter equality equals a u C j xj @j u.  Thus, Problem 9.9 amounted to determining all homogeneous distributions on R.

As we have already established, pushforward under ˚ of distributions with compact support by application of formula (10.3) is possible for arbitrary C 1 mappings

10 Transposition: Pullback and Pushforward

103

˚ from an open subset X of Rn to an open subset Y of Rp , even if n > p or n < p. We conclude this chapter with a few remarks concerning these two cases under the assumption that ˚ is generic, i.e., that all of its corresponding tangent mappings are of maximal rank. If n > p, and under the further assumption that ˚ is a submersion, we can define pullback under ˚ of distributions; this result requires the preceding theory for diffeomorphisms. In the case n < p, and under the further assumption of ˚ being a proper immersion, we encounter a new phenomenon: the pushforward under ˚ of a function is not necessarily a function anymore. This is related to ˚.X / not being an open subset of Y . n > p in case of a submersion. We first derive an intermediate result. Write n D p C q with q > 0 and x D .y; z/ with y 2 Rp and z 2 Rq . Let ˘ W .y; z/ 7! y W Rn D Rp  Rq ! Rp be the projection onto the first p variables. This mapping is not proper, and therefore ˘ is defined only on E 0 .Rn /. For a continuous function f on Rn with compact support, (10.3) now gives Z Z Z  f .y; z/ .y/ d.y; z/ D f .y; z/ dz .y/ dy; .˘ f /. / D Rp Rq

Rp

Rq

for all 2 C 1 .Rp /. This means that the distribution ˘ f is given by the continuous function of y obtained from f .x/ D f .y; z/ by integration over the fiber of the mapping ˘ over the value y, that is, by “integrating out” the z-variable: Z .˘ f /.y/ D f .y; z/ dz: (10.21) Rq

On the strength of the Theorem about differentiation under the integral sign, we have ˘ f 2 C01 .Rp / if f 2 C01 .Rn /. We conclude that the restriction of ˘ to C01 .Rn / is a continuous linear mapping A from C01 .Rn / to C01 .Rp /. Exactly as in Remark 10.10, we now obtain that ˘  W C 1 .Rp / ! C 1 .Rn / has an extension to a continuous linear mapping ˘  W D 0 .Rp / ! D 0 .Rn /, defined as the transpose of the restriction to C01 .Rn / of ˘ . Recall that n > p and let ˚ be a C 1 mapping from an open X  Rn to an open Y  Rp , with the property that its derivative D˚.x/ 2 Lin.Rn ; Rp / at x is surjective, for every x 2 X . One says that such a mapping is a submersion; see for example [7, Definition 4.2.6]. Theorem 10.18. Let ˚ be a C 1 submersion from an open X  Rn to an open Y  Rp . Then the restriction of ˚ to C01 .X / is a continuous linear mapping from C01 .X / to C01 .Y /. Its transpose defines an extension of ˚  W C 1 .Y / ! C 1 .X / to a continuous linear mapping from D 0 .Y / to D 0 .X /; this is also denoted by ˚  and is said to be the pullback of distributions under ˚. Proof. By means of the Submersion Theorem (see [7, Theorem 4.5.2], for example) one can find, for every x 2 X , a neighborhood U D Ux of x in X and a C 1 diffeomorphism K from U onto an open subset V of Rn such that ˚ D ˘ B K on

104

10 Transposition: Pullback and Pushforward

U . (Since K and ˘ map open sets to open sets, it now follows that this holds for ˚ as well.) Applying this we obtain in E 0 .U /, ˚ D ˘ B K D ˘ B j B 

with

 WD K

1

:

(10.22)

Here we have used (10.4) and Theorem 10.8. In particular, it now follows that the restriction of ˚ to C01 .U / is a continuous linear mapping from C01 .U / to C01 .Y /. If for every compact subset K of X , we use a partition of unity over K subordinate to the cover by the Ux , for x 2 X , we conclude that the restriction of ˚ to C01 .K/ is a continuous linear mapping from C01 .K/ to C01 .Y /.  Remark 10.19. As we have seen, .˘ g/.y/ is obtained by integrating g over the linear manifold ˘ 1 .fyg/, for a continuous function g with compact support. If f is a continuous function with compact support contained in U , one can see, by application of (10.22), that .˚ f /.y/ is obtained by integrating g WD j .f B / over ˘ 1 .fyg/. But this amounts to an integration of f over the C 1 submanifold .˘

1

.fyg/ D K

1

.˘

1

.fyg/ D ˚

1

.fyg/  X;

the fiber of the mapping ˚ over the value y. If one chooses Euclidean .n p/-dimensional integration over the inverse image ˚ 1 .fyg/, then f must first be multiplied by a C 1 factor that does not depend on f but only on the mapping ˚. We state here (see Problem 10.21 and its solution for a proof, or Problem 10.22.(iv) in the special case of p D 1) that this factor equals gr ˚ (with gr associated with gradient and Gramian), where p .gr ˚/.x/ WD det.D˚.x/ B tD˚.x// .x 2 X /;

which is the p-dimensional Euclidean volume of the parallelepiped in Rn spanned by the vectors grad ˚j .x/, with 1  j  p. Here tD˚.x/ 2 Lin.Rp ; Rn / denotes the adjoint of D˚.x/, the matrix of which is given by the transpose of the matrix of D˚.x/. Observe that gr ˚ D k grad ˚k if p D 1. Summarizing, in the notation of Example 7.3 we have Z 1 f .x/ .˚ f /.y/ D dx D ı .f / .y 2 Y /: (10.23) 1 gr ˚ ˚ 1 .fyg/ ˚ .fyg/ .gr ˚/.x/ ˛ Pushforward under a submersion ˚ is not well-defined for all distributions, whereas pullback under ˚ of distributions is. The latter will play an important role in Chap. 13. Example 10.20. Define the difference mapping d W R n  R n ! Rn

by

d.x; y/ D x

y:

(10.24)

10 Transposition: Pullback and Pushforward

105

One directly verifies that d is a surjective linear mapping satisfying .gr d /.x; y/ D n 2 2 . (Note that K W Rn  Rn ! Rn  Rn with K.x; y/ D .x y; y/ is the diffeomorphism such that d D ˘ B K as in the proof of Theorem 10.18.) Now, in the notation of Example 10.7, d

1

.f0g/ D f .x; z/ 2 Rn  Rn j z D x g D
.Rn /;

the diagonal in Rn  Rn . More generally, for y 2 Rn , d

1

.fyg/ D f .x; z/ 2 Rn  Rn j z D x

y g D T.0;

y/ .
.R

n

//:

Let ı
.Rn / and
 1Rn be as in Example 10.7, while ı 2 D 0 .Rn / denotes the Dirac measure. Then (10.23) implies, for  2 C01 .Rn  Rn /, Z 1 .d /.y/ D n ı
.Rn / .T.0;y/ / D
 1Rn .T.0;y/ / D .x; x y/ dx; 22 Rn in particular, d  ı D
 1Rn 2 D 0 .Rn  Rn /: (10.25) See Problem 15.10 for another proof. Using Dirac notation, one writes d  ı.x; y/ D ı.x y/. Then the preceding identity leads to Z Z ˛ ı.x y/.x; y/ d.x; y/ D .x; x/ dx: Rn Rn

Rn

n < p in case of a proper embedding. Suppose, for convenience, that ˚ is proper. If, moreover, ˚ is injective and D˚.x/ W Rn ! Rp is injective for every x 2 X , then ˚ is a proper C 1 embedding of X onto the n-dimensional locally closed C 1 submanifold V D ˚.X / of Y . The n-dimensional Euclidean integral of a continuous function f with compact support in V then equals the integral of j˚ ˚  f over X , where p j˚ .x/ D det.tD˚.x/ B D˚.x// .x 2 X /

now denotes the Euclidean n-dimensional volume in Rp of the parallelepiped spanned by the D˚.x/.e.j // 2 Rp , 1  j  n. Here e.j / denotes the jth standard basis vector in Rn . See [7, Theorem 1.8.6, Exercise 4.20 and Sect. 7.3] among other references for more on these facts concerning proper embeddings and integration over submanifolds. We further recall the distribution g ıV of Euclidean integration over V with weight function g, as introduced in Example 7.3. Proposition 10.21. Let n  p, suppose X is open in Rn , and let ˚ be a proper C 1 embedding of X into the open subset Y of Rp , having as its image the manifold V D ˚.X /. Using the notation D ˚ 1 W V ! X we obtain ˚ f D 

f  j˚

ıV 2 D 0 .Y /;

(10.26)

106

10 Transposition: Pullback and Pushforward

for every continuous function f on X . Note that ıV is a distribution of order 0, multiplied by the continuous function  .f =j˚ / on V . Proof. This follows from a variant of (10.9). For 2 C01 .Y / one has R R .˚ f /. / D X f .x/ .˚.x// dx D X .f =j˚ /.x/ .˚.x// j˚ .x/ dx Z D .f =j˚ /. .y// .y/ dy D  .f =j˚ / ıV . /: V



Remark 10.22. Note the similarity in appearance between (10.26) and (10.8). Furthermore, observe that in this case of a proper embedding ˚, the pushforward ˚ of distributions is well-defined on account of ˚ being proper; however, ˚ does not map functions to functions. Owing to the latter phenomenon, it is not possible to define the pullback ˚  of distributions in a manner similar to the case of diffeomorphisms. ˛

Problems  10.1. (Composition of pushforward and partial differentiation.) Consider a C 1 mapping ˚ from an open subset X of Rn to an open subset Y of Rp . Prove, for every 1  j  n, the following identity of continuous linear mappings:

˚ B @j D

p X

kD1

@k B ˚ B @j ˚k

W

E 0 .X / ! E 0 .Y /:

10.2. Let X  Rn be an open subset,  a compactly supported Radon measure on X , and f W X ! R a C 1 function. Prove that f  is a Radon measure on R satisfying, in the notation of Theorem 20.34, Z Z .t/ f .dt/ D  B f .x/ .dx/ . 2 C01 .R//: R

X

In probability theory, f  is said to be the distribution of f under . In that context, X , , f , and  are usually more general than considered here. 10.3. Let u be a probability measure on X and ˚ W X ! Y a continuous mapping. Discuss how to define ˚ u so as to make it a probability measure on Y . 10.4. Let ˚ W X ! Y be a C 1 mapping that is not proper on the closed subset A of X . Demonstrate the existence of a sequence .x.j //j 2N in A with the following properties: (i) there exists y 2 Y such that limj !1 ˚.x.j // D y; (ii) for every compact subset K of X there exists a j0 with x.j / … K, whenever j  j0 .

10 Problems

107

Prove that ıx.j / ! 0 in D 0 .X /A , while ˚ ıx.j / ! ıy in D 0 .Y /, as j ! 1. Prove that the condition in Theorem 10.6, requiring ˚ to be proper on A, is necessary for the conclusion in Theorem 10.6. 10.5. Let ˚ be a C 1 mapping from the open subset X of Rn to the open subset Y of Rp . Prove supp ˚    ˚ 1 .supp /. For a counterexample to the reverse inclusion, consider X D Y D R, the function ˚.x/pD x 4 x 2 , pand 1   2 C0 .R/ such that supp  D Œ 0; 2 . Then supp ˚  D Œ 2; 1  [ Œ 1; 2  and ˚ 1 .supp / D f0g [ supp ˚   (see Fig. 10.1). 2

- 2

-1

1

2

Fig. 10.1 Illustration for Problem 10.5. Graph of function ˚

10.6. (Change of Variables Theorem.) Let W Y ! X be a C 1 diffeomorphism of open subsets Y and X of Rn . Show that 1 .Y / D  .j 1Y / in D 0 .X /.  10.7. (Composition of pullback and partial differentiation.) Consider a C 1 diffeomorphism W Y ! X of open subsets Y and X of Rn . Denote the inverse of the transpose of the Jacobi matrix of by . jk /1j;kn , that is, .tD .y// 1 D . jk .y//1j;kn , for y 2 Y . Derive the following identity of continuous linear mappings (compare with [7, Exercise 3.8]):

 B @j D

n X

kD1

jk

 B @k B 

W

D 0 .X / ! D 0 .Y /

.1  j  n/:

The change of variables x D .y/ relates the old variable x to the new variable y. The formula above expresses a partial derivative of a distribution with respect to the old variable in terms of partial derivatives of the transformed distribution with respect to the new variable.  10.8. (Interpretation of divergence.) Suppose ˚0 D I . Verify, by combining Theorem 10.8 with (10.17) and (10.18), that n ˇ X d ˇ det.D˚ t /ˇ D @j vj D div v: t D0 dt j D1

108

10 Transposition: Pullback and Pushforward

 10.9. Consider a C 1 vector field v on X with globally defined flow .˚ t / t 2R and fix x 2 X . Then prove the following: ıx is invariant under ˚ t , for all t 2 R, as a distributional density () ˚ t .x/ D x for all t 2 R () v.x/ D 0. And also: ıx is invariant under ˚ t , for all t 2 R, as a generalized function if and only if v.x/ D 0 and div v.x/ D 0.  10.10. The rotation in the plane R2 about the origin by the angle t is the linear mapping .x1 ; x2 / 7! .x1 cos t x2 sin t; x1 sin t C x2 cos t/. Demonstrate that u 2 D 0 .R2 / is invariant under all rotations if and only if x1 @2 u D x2 @1 u.

10.11. (Lorentz-invariant distributions.) Consider the quadratic form q W RnC1 ! R, of Lorentz type, as in (13.12) below and furthermore the collection Lo of all linear transformations ˚ W RnC1 ! RnC1 satisfying ˚  q D q. Prove that Lo is a group with composition as the binary operation. Lo is called the Lorentz group. (i) For all t 2 R and 1  j  n and 1  j < k  n, show that the linear j j;k transformations ˚ t and ˚ t in RnC1 that map y to .y1 ; : : : ; yj

1 ; yj

.y1 ; : : : ; yj

1 ; yj

cosh t C ynC1 sinh t; yj C1 ; : : : ; yj sinh t C ynC1 cosh t/;

cos t

yk sin t; yj C1 ; : : : ; yk

1 ; yj

sin t C yk cos t; : : : ; ynC1 /;

respectively, belong to Lo. (In the case of n D 3, the ˚ tj are boosts of rapidity t, which occur in special relativity.) Verify that the .˚ tj / t 2R and .˚ tj;k / t 2R actually are one-parameter subgroups in Lo consisting of C 1 diffeomorphisms of RnC1 . Prove that the velocity vector fields corresponding to these one-parameter subgroups are y 7! ynC1 e.j / C yj e.n C 1/

and

y 7!

yk e.j / C yj e.k/;

respectively, where .e.1/; : : : ; e.n C 1// denotes the standard basis in RnC1 . Furthermore, for any u 2 D 0 .RnC1 /, deduce ˇ ˇ d d ˇ ˇ .˚ tj / uˇ .˚ tj / uˇ ; D ynC1 @j u C yj @nC1 u D t D0 t D0 dt dt ˇ ˇ d d ˇ ˇ j;k j;k .˚ / uˇ .˚ / uˇ : D yj @k u yk @j u D t D0 t D0 dt t dt t

(ii) Prove that j det ˚j D 1, for ˚ 2 Lo, and deduce that invariance of u under Lo as generalized function and as distributional density amount to the same thing. In particular, if u is invariant under Lo, derive .ynC1 @j C yj @nC1 /u D 0 .yj @k yk @j /u D 0

.1  j  n/; .1  j < k  n/:

10.12. A distribution u in R is said to be periodic with period a > 0 if u is invariant under translation by a. Prove that every distributional Fourier series from Problem 5.10 is periodic with period a D 2 . !

10 Problems

109

 10.13. (Poisson summation formula.) For any  2 C, introduce e 2 C 1 .R/ by e .x/ D e x and verify that e D 1 @e . Fix ! > 0. Now prove that the distributional Fourier series (see Definition 16.5 for more about this notion) given by X uD ei n! n2Z

defines a distribution on R having the following property. For every compact subset K of R, there exists a constant c > 0 such that ju./j  c kkC 2 , for all  2 C01 .K/. Next show:

(i) ei ! u D u. (ii) u is periodic with period 2=!. (iii) There exists a constant c 2 C such that X X ei n! D c ık 2 : !

n2Z

k2Z

(iv) The order of u is 0.

For the determination of c D 2=!, see Example 16.8. This example and Problem 16.17 contain a different proof by means of Fourier transform. Furthermore, see Figs. 16.1 and 16.2.  10.14. Let 0 < s < 1. Prove the existence of c D c.n; s/ > 0 such that Z

Rn

je i hx; i 1j2 dx D kxknC2s

Z

Rn

4 sin2 hx;2i dx D c kk2s kxknC2s

. 2 Rn /:

See Problem 14.40 for a computation of c.1; s/.  10.15. (Distributions homogeneous of fixed degree.) Let Ha be the set of homogeneous distributions of degree a 2 C in Rn . Verify the following assertions.

(i) Ha is a linear subspace of D 0 .Rn /. Furthermore, Ha is a closed subspace of D 0 .Rn /, that is, if u 2 D 0 .Rn /, uj 2 Ha , and uj ! u in D 0 .Rn / as j ! 1, then u 2 Ha . (ii) If u 2 Ha , one has @j u 2 Ha 1 and xk u 2 HaC1 . (iii) If 2 Ha \ C 1 .Rn /, with ¤ 0, then a 2 Z0 and is a homogeneous
polynomial function of degree a. Prove dim.Ha \ C 1 .Rn // D aCna 1 . Hint: first prove that if the function t 7! t a W R>0 ! C possesses a C 1 extension to R, then a 2 Z0 , for example, by studying the behavior of the derivatives of this function as t # 0. Alternatively, a sufficiently high derivative would be homogeneous of negative degree and therefore not continuous at 0. (iv) Every continuous function f on the sphere kxk D 1 possesses a uniquely determined extension to a homogeneous function g of degree a on Rn n f0g. This g is continuous. If Re a > n, then g is locally integrable on Rn and g 2 Ha . (v) Ha is infinite-dimensional if n > 1 and a 2 C. Hint: use (iv) when Re a > n, and combine this with (ii) when Re a  n.

110

10 Transposition: Pullback and Pushforward

 10.16. For all ˛ 2 .Z0 /n , prove that @˛ ı 2 D 0 .Rn / is homogeneous of degree n j˛j.

10.17. Let ˚ W X ! Y be a C 1 submersion and let K W X ! X be a C 1 diffeomorphism such that ˚ B K D ˚. Prove that for every v 2 D 0 .Y / the distribution u D ˚  v 2 D 0 .X / is invariant under K (as a generalized function).  10.18. (Composition of partial differentiation and pullback under submersion.) Let X be open in Rn , Y open in Rp and ˚ W X ! Y a C 1 submersion. Prove that (10.1) can be extended to an identity of continuous linear mappings: D 0 .Y / ! D 0 .X /. 10.19. Let X D R2 n .R0  f0g/ and Y D R>0   ;  Œ and consider the C 1 diffeomorphism ˚ W X ! Y that assigns to x D r.cos ˛; sin ˛/ 2 X its polar coordinates .r; ˛/ in Y (see [7, Example 3.1.1]). For u 2 D 0 .X /, verify 2

@r .˚ u/ D

X xk 1 1 ˚ u C cos ˛ ˚ .@1 u/ C sin ˛ ˚ .@2 u/ D ˚ u C ˚ .@k u/: r r r kD1

 10.20. (Composition of second-order partial differential operator with pullback under associated quadratic form.) Consider the second-order linear partial differential operator with constant coefficients

P D

n X

Bij @i @j

i;j D1

in Rn , where .Bjk / is an invertible symmetric real n  n matrix. Let .Akl / be the inverse matrix and n X ˚ W x 7! Akl xk xl 2 R k;lD1

the corresponding quadratic form on Rn . Verify the following assertions.

(i) The restriction of ˚ to Rn n f0g is a submersion from Rn n f0g to R. (ii) One has the following identity of distributions in D 0 .Rn n f0g/: .P B ˚  /v D ˚  .4y @y2 C 2n @y /v

.v 2 D 0 .R//:

Here y D ˚.x/ denotes the variable in R. (iii) If v 2 D 0 .R/ is homogeneous of degree a, then ˚  v is homogeneous of degree 2a on Rn n f0g. (iv) If @y v is homogeneous of degree n2 , then u D ˚  v defines a solution on Rn n f0g of the partial differential equation P u D 0. When is u homogeneous? Of what degree? (v) From part (ii) deduce the following two identities of continuous linear mappings: P B ˚  D ˚  B .4y @y2 C 2n @y / ˚ B P D .4@y2 B y 2n @y / B ˚

W W

C 1 .R/ ! C 1 .Rn /; E 0 .Rn / ! E 0 .R/:

10 Problems

111

(vi) Apply the latter equality in part (v) to derive the result in Example 10.5.  10.21. Verify formula (10.23) using linear algebra to rewrite j , where j is as in (10.22). To this end, parametrize U \ ˚ 1 .fyg/ using  and show that the tangent space T of ˚ 1 .fyg/ at x D .x 0 / equals ker , where  D D˚.x/. Apply the chain rule to ˚ B  D ˘ and deduce that T is spanned by certain column vectors k that occur in the matrix of  WD D.x 0 /. Next, project the remaining column vectors in  along T onto T ? D im t and express .det /2 D det.t / in terms of these projections and the k spanning T . Finally, write the contribution coming from the projections in terms of the row vectors of .  10.22. (Pushforward of function under submersion to R as consequence of integration of a derivative.) Let X be an open subset of Rn and ˚ W X ! R a C 1 submersion. Denote by H the Heaviside function on R.

(i) Show, for all  2 C01 .X / and y 2 R, that .˚ /.y/ D .˚  ıy /./ D ˚  .@ Ty H /./ and apply Problem 10.17 to derive @j ˚ ˚  ıy D @j ˚ B ˚  B @.Ty H / D @j .˚  .Ty H //: (ii) By approximating H by C 1 functions, verify Z  .˚ .Ty H //./ D ˚

.x/ dx:

1 .  y; 1 Œ /

(iii) Apply the Theorem on Integration of a Total Derivative (see [7, Theorem 7.6.1]) to deduce Z @j ˚.x/ @j .˚  .Ty H //./ D dx: .x/ k grad ˚.x/k ˚ 1 .fyg/ Here @j ˚=k grad ˚k denotes the jth component of the normalized gradient vector field of ˚; the value of the vector field at x is the inner normal to ˚ 1 .fyg/ at x. Furthermore, dx denotes Euclidean .n 1/-dimensional integration over ˚ 1 .fyg/. (iv) Combine the results from parts (i) and (iii), replace  by  @j ˚=k grad ˚k, and sum over 1  j  n to obtain formula (10.23): Z .x/ .˚ /.y/ D dx .y 2 Y /: ˚ 1 .fyg/ k grad ˚.x/k  10.23. (Pullback of Dirac measure under submersion.) Let X be an open subset of n R and consider a C 1 submersion ˚ W X ! R.

112

10 Transposition: Pullback and Pushforward

(i) Let  2 C01 .X /. Prove .˚ /.y/ D

1 ı k grad ˚k ˚

1 .fyg/

./

.y 2 R/:

Verify the following identity in D 0 .X / (compare with Example 10.12): ˚  ıy D

1 ı k grad ˚k ˚

1 .fyg/

.y 2 R/:

(ii) On the basis of part (i) conclude (compare with [7, Exercise 7.36]) that 1

1; y Œ

.˚ / D 1˚

1. 

1; y Π/

and deduce from this, for all y 2 R, Z .˚ /.y/ D @y .x/ dx; ˚

1. 

1; y Π/

./

.y 2 R/I

˚  ıy D @y B .1˚

1. 

1; y Π/

/:

(iii) Give an independent proof of the second identity in part (ii) by means of successive application of the Fundamental Theorem of Integral Calculus on R (see [7, Theorem 2.10.1]), interchange of the order of integration, and integration by parts to .˚ /, where 2 C01 .R/.  10.24. (Integration of total derivative.) Let X be an open subset of Rn and ˚ W X ! R a C 1 submersion. Define the open set ˝ D ˚ 1 .R>0 /, and suppose that ˝ is nonempty and bounded. Note that the boundary @˝ of ˝ equals ˚ 1 .f0g/ and that this set is a C 1 submanifold in Rn of dimension n 1. Derive from Problem 10.23.(i)

˚ ı D

1 ı k grad ˚k @˝

in D 0 .X /:

With H the Heaviside function on R, prove on the basis of Problem 10.18 that @j ˚ ı : @j .˚  H / D k grad ˚k @˝ Deduce that for all  2 C01 .X /, Z Z @j .x/ dx D ˝

.y/ j .y/ dy; @˝

where .y/ denotes the outer normal to @˝ at y and dy denotes Euclidean .n 1/dimensional integration over @˝. Conclude that by this method we have proved the following Theorem on Integration of a Total Derivative; see [7, Theorem 7.6.1] and Z Z Problem 7.4: D.x/ dx D .y/ t.y/ dy: ˝

@˝

10 Problems

113

10.25. (Wave equation.) Denote the points in R2 by .x; t/ and consider the mappings plus W .x; t/ 7! x C t

minus W .x; t/ 7! x

and

t

from R2 to R. Verify that these are submersions. Prove, for every pair of distributions a 2 D 0 .R/ and b 2 D 0 .R/, that the distribution u WD .plus/ a C .minus/ b on R2 satisfies the wave equation @2t u D @2x u: This gives a distributional answer to the classical question whether .x; t/ 7! a.x C t/ C b.x

t/

is acceptable as a solution of the wave equation also when a and b are not C 2 functions, although continuous, for example. Describe .plus/ ı and .minus/ ı. 10.26. Let ˚ be a C 1 mapping from an open subset X of Rn to an open subset Y of Rp . Let C be the closure of ˚.X / in Y . Prove supp ˚ u  C

.u 2 E 0 .X //:

Now assume that the p-dimensional measure of C equals 0, so that ˚ u D 0 if ˚ u is locally integrable. Prove the existence of  2 C01 .X / such that ˚  is not locally integrable. Hint: approximate ıx by j 2 C01 .X /. 10.27. Let ˚ be a C 1 mapping from an open subset X of Rn to an open subset Y of Rp . Suppose that W Y ! X is a C 1 mapping such that B ˚.x/ D x, for all x 2 X . Prove that ˚ is a proper embedding. Write V WD ˚.X / and define E 0 .V / WD f v 2 E 0 .Y / j

v D 0 if

2 C 1 .Y / and

D 0 on V g:

Prove that ˚ is a bijective linear mapping from E 0 .X / to E 0 .V /, with the restriction of  to E 0 .V / as its inverse. Hint: use the fact that for every  2 C 1 .Y /, the function D   B ˚  ./ equals 0 on V . Finally, let f be a real-valued C 1 function on the open subset X of Rn . Denote the points of RnC1 by .x; y/, where x 2 Rn , y 2 R. Define ˚ W X ! X  R by ˚.x/ D .x; f .x//, for x 2 X . Let v 2 E 0 .X  R/. Prove that .y f .x// v D 0 if and only if there exists a u 2 E 0 .X / with v D ˚ u.

Chapter 11

Convolution of Distributions

Convolution involves translation; that makes it difficult to define the former operation for functions or distributions supported by arbitrary open subsets in Rn . Therefore we initially consider objects defined on all of Rn . In (3.2) we described the convolution .f  /.x/ of a continuous function f on Rn and  2 C01 .Rn / as the testing of f with the function Tx B S W y 7! .x y/. Here S and Tx are respectively the reflection, and the translation by the vector x, of functions, as described in Example 10.14 and Example 10.15. This suggests defining the convolution product u  , for u 2 D 0 .Rn / and  2 C01 .Rn /, as the function given by .u  /.x/ D u.Tx B S/ .x 2 Rn /: (11.1)

Note that the definition also makes sense if u 2 E 0 .Rn / and  2 C 1 .Rn /. In Theorem 11.2 we will show that in both cases u   belongs to C 1 .Rn /. Furthermore, with a suitable choice of  D  we can ensure that u   converges in D 0 .Rn / to u 2 D 0 .Rn / for  # 0; see Lemma 11.6. Example 11.1. For a 2 Rn we have

.ıa  /.x/ D .Tx B S/.a/ D .S/.a

x/ D .x

a/:

Consequently, ıa   D Ta  for every  2 C 1 .Rn /, and in particular ı D

. 2 C 1 .Rn //:

In other words, ı acts as a unit element for the operation of convolution. 0

n

1

n

(11.2) ˛

Theorem 11.2. If u 2 D .R / and  2 C .R / and at least one of the two sets supp u and supp  is compact, then u   2 C 1 .Rn /. One has Ta .u  / D .Ta u/   D u  .Ta /;

(11.3)

@˛ .u  / D .@˛ u/   D u  .@˛ /;

(11.4)

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_11, © Springer Science+Business Media, LLC 2010

115

for every a 2 Rn , and

116

11 Convolution of Distributions

for every multi-index ˛. Furthermore, supp .u  /  supp u C supp :

(11.5)

Proof. We suppose that supp  is compact; the other case is similar. Because x 7! Tx B S is continuous from Rn to C01 .Rn /, the function u   is continuous on Rn . Formula (11.3) follows when we compare .Ta .u  //.x/ D .u  /.x a/ D u.Tx a B S/; ..Ta u/  /.x/ D Ta u.Tx B S/ D u.T a B Tx B S/; .u  .Ta //.x/ D u.Tx B S B Ta / D u.Tx B T a B S/: Substituting a D

t e.j / in (11.3), we obtain

1 .u  /.x C t e.j // t



1 .u  /.x/ D T t e.j /  u u .Tx B S/ t

1 T t e.j /    : D u Tx B S B t

In view of (10.13) and (10.12) this converges as t ! 0. We see that u   is partially differentiable with respect to the jth variable, with derivative @j .u  / D .@j u/   D u  .@j /: Since the partial derivatives are continuous, the conclusion is that u   is of class C 1 . By mathematical induction on k one can then show that u   is of class C k and that (11.4) holds for all multi-indices ˛ with j˛j D k. Here, x … supp u C supp  means that supp u has an empty intersection with x C . supp / D x C supp S D supp .Tx B S/; which implies .u  /.x/ D u.Tx B S/ D 0. Because supp u is closed and supp  compact, C D supp u C supp  is a closed subset of Rn ; see the text following (2.2). Because u   D 0 on the open subset Rn n C of Rn , the conclusion is that supp .u  /  C .  A linear mapping U from functions on Rn to functions on Rn commutes with all translations if U B Ta D Ta B U, for all a 2 Rn . Convolution with u can be characterized in terms of such linear mappings: Theorem 11.3. Let u 2 D 0 .Rn /. Then u  W  7! u   is a continuous linear mapping from C01 .Rn / to C 1 .Rn / that commutes with all translations. Conversely, if U is a continuous linear mapping from C01 .Rn / to C 1 .Rn / that commutes with all translations, then there exists a uniquely determined u 2 D 0 .Rn / with U./ D

11 Convolution of Distributions

117

u  , for all  2 C01 .Rn /. More specifically, u D ı B U B S , with ı the Dirac measure on Rn . Proof. It is evident that u  is linear, and as (11.3) shows, u  commutes with all translations. Since we have to verify that the conditions in Example 8.6 are satisfied, the proof of the continuity of u  W C01 .Rn / ! C 1 .Rn / is more involved. Let A and B be compact subsets of Rn . Then K WD A C . B/ is compact and according to Theorem 3.8 there exist a constant c > 0 and k 2 Z0 such that ju. /j  c k kC k Applying this to

2 C01 .K//:

.

D Tx B S, with x 2 A and  2 C01 .B/, we get ku  kC 0 ; A  c kkC k

. 2 C01 .B//:

Because @˛ .u  / D u  .@˛ /, it now follows that for every m 2 Z0 there exists a constant c 0 > 0 with ku  kC m ; A  c 0 kkC kCm

. 2 C01 .B//:

Because this holds for all A, B, and m, the desired conclusion is obtained. Now consider U as described above. If U is of the form u , then .U /.0/ D u.S /; thus, u is determined by u. / D .U B S /.0/, for 2 C01 .Rn /. If we now interpret this as the definition of u, we see that u 2 D 0 .Rn / and .u  /.x/ D u.Tx B S/ D .U B S B Tx B S/.0/ D .U B T D .T

x

x /.0/

B U/.0/ D U./.x/;



for every x 2 Rn ; that is, u   D U./.

The preceding theorem implies that any translation acting in C01 .Rn / has to be a convolution operator, as has been explicitly demonstrated in Example 11.1. Furthermore, see Problem 15.11 for another proof of this theorem. In proving Theorem 11.5 we will use the following principle of integration under the distribution sign. Lemma 11.4. Let X be an open subset of Rn and suppose that the mapping A W Rm ! C01 .X / has the following properties:

(a) There exists a compact subset T of Rm such that A.t/ D 0 whenever t … T . (b) There is a compact subset K of X such that supp A.t/  K for all t 2 Rm . (c) For every k 2 Z0 and every  > 0 there exists ı > 0 such that ks tk < ı implies kA.s/ A.t/kC k < . Then

Z

Rm

Z  A.t/ dt .x/ D

A.t/.x/ dt Rm

118

11 Convolution of Distributions

R defines a function A.t/ dt belonging to C01 .X /. For every u 2 D 0 .X /, one has that u B A W Rm ! C is a continuous function with compact support, while  Z Z A.t/ dt D .u B A/.t/ dt: u Rm

Rm

Proof. It follows from the assumptions made that the (finite) Riemann sums X A.h t/ .h 2 R>0 / Sh WD hm t 2Zm

R converge in C01 .X / to A.t/ dt as h # 0. One begins by proving this in C00 .X / and then uses the Theorem about differentiation under the integral sign to obtain the result in C01 .X /. From the continuity of A and of u it follows that u B A is a continuous function. The complex number X X u.Sh / D hm u.A.h t// D hm .u B A/.h t/ t 2Zm

t 2Zm

is an approximating Riemann sum for the integral of u B A. Using the continuity of u once again, we now obtain Z  Z
A.t/ dt D u lim Sh D lim u.Sh / D .u B A/.t/ dt:  u h#0

Rm

h#0

Theorem 11.5. For u 2 D 0 .Rn / and  and

Rm

in C01 .Rn / one has

.u  /. / D u.S 

/:

(11.6)

Proof. The mapping A W x 7!

.x/ .Tx B S/

W

Rn ! C01 .Rn /

satisfies the conditions of Lemma 11.4. Applying (1.14), we obtain Z Z .u  /. / D .u  /.x/ .x/ dx D u.Tx B S/ .x/ dx Z Z  D u. .x/ .Tx B S// dx D u .x/ .Tx B S/ dx D u.S 

/:



Theorem 11.5 is useful in proving the following two results on approximation of distributions by functions. Lemma 11.6. Let  2 C01 .Rn / and 1./ D 1. Write  .x/ D  n . 1 x/, for  > 0. For every u 2 D 0 .Rn / we then have that u   2 C 1 .Rn / converges in D 0 .Rn / to u as  # 0.

11 Convolution of Distributions

119

Proof. Let 2 C01 .Rn /. Because 1.S/ D .S1/./ D 1./ D 1 and S. / D .S/ , we find, in view of Lemma 1.6, which can be extended to Rn in a straightforward manner, that S  ! in C01 .Rn /. In combination with Theorem 11.5 this implies .u   /. / D u.S  as  # 0. Since this holds for every D 0 .Rn /.

/ ! u. /

2 C01 .Rn /, we conclude that u   ! u in 

Next we extend a sharpening of this result to open subsets of Rn by “cutting and smoothing.” Corollary 11.7. Let X be an open subset of Rn . For every u 2 D 0 .X / there exists a sequence .uj /j 2N in C01 .X / with limj !1 uj D u in D 0 .X /. Proof. Choose an increasing sequence of compact sets Kj in X that absorbs X ; see Lemma 8.2.(a). Further, choose cut-off functions j 2 C01 .X / such that j D 1 on an open neighborhood of Kj . Finally, choose .j / > 0 such that limj !1 .j / D 0 and supp j C supp j  X for every j . Here j WD .j / with  as in Lemma 11.6. The third assumption is satisfied if .j / c < ıj , where c is the supremum of the kxk with x 2 supp  and where ıj > 0 has been chosen such that the ıj neighborhood of supp j is contained in X ; see Fig. 11.1. In addition, suppose that the ıj converge to 0 as j ! 1.

d j -nbhd of supp Χ j

supp Χ j

Kj

X

Fig. 11.1 Illustration of the proof of Corollary 11.7

Because j u 2 E 0 .X /, we can interpret it as an element of E 0 .Rn /  D 0 .Rn /; see Lemma 8.9. We now take uj WD .j u/  j 2 C 1 .Rn /: In view of (11.5), we find that supp uj is a compact subset of X , and so uj 2 C01 .X /.

120

11 Convolution of Distributions

Let 2 C01 .X /  C01 .Rn /. There exists a j0 such that for all j  j0 , the ıj neighborhood of supp is contained in Kj ; consequently, j .Sj  / D Sj  . Thus, on the strength of Theorem 11.5 we get, for j  j0 , uj . / D .j u/.Sj 

/ D u.j .Sj 

// D u.Sj 

/:

Because Sj  ! in C01 .X / we find that uj . / ! u. / as j ! 1 and conclude that uj ! u in D 0 .X /.  By Corollary 11.7, many identities that hold for test functions can be extended to distributions. The principle is that if A and B are sequentially continuous operators on D 0 .X / with A./ D B./ for all  2 C01 .X /, then A.u/ D B.u/ for all u 2 D 0 .X /. Indeed, with .uj /j 2N in C01 .X / as in Corollary 11.7 we obtain A.u/ D lim A.uj / D lim B.uj / D B.u/: j !1

j !1

(The principle as introduced here is the same as the “principle of uniqueness of continuous extensions to the closure.”) This was the basis for conjecturing (9.1), (11.3), (11.4), and (11.6), for instance, and even (11.5). The proof by transposition was then a matter of writing out the definitions. It is a very fruitful principle, if only for generating conjectures, but it should not be overlooked that its application presupposes that A and B have already been shown to be sequentially continuous extensions to D 0 .X / of A D B on C01 .X /. Because these extensions are often composed of operators defined by means of transposition, the method of transposition has not become redundant. The fact that C01 .X / is dense in D 0 .X / supports the concept of distributions as generalized functions. One can even go a step further, observing that, for example, the linear subspace in C 1 .X / consisting of the polynomial functions is dense in C 1 .X / with respect to uniform convergence on compacta. This is Weierstrass’s Approximation Theorem; see Example 14.31 below. Because C 1 .X / is dense in D 0 .X /, the polynomial functions are also dense in D 0 .X /. Our next aim is to define the convolution u  v of the most general distributions u and v in D 0 .Rn / possible. If u and v are continuous functions and if at least one of them has compact support, we obtain the following symmetric expression for testing u  v with  2 C01 .Rn /: “ “ .u  v/./ D u.x/ v.z x/ dx .z/ dz D u.x/ v.y/ .x C y/ dx dy:

(11.7) Actually, the left-hand side should read test.uv/, instead of uv. In the integration over z we have applied the change of variables z D x C y. In the right-hand side of (11.7), the function u ˝ v W .x; y/ 7! u.x/ v.y/

(11.8)

11 Convolution of Distributions

121

on Rn  Rn is tested against the function ˙   W .x; y/ 7! .x C y/:

(11.9)

The function u ˝ v is said to be the tensor product of u and v. In (11.9) we have used the sum mapping ˙ W .x; y/ 7! x C y

W

R n  R n ! Rn ;

(11.10)

by which  2 C 1 .Rn / is pulled back to a function in C 1 .Rn  Rn /. Using the transpose ˙ of ˙  in a formal way, see (10.3), we can write .u  v/./ D .u ˝ v/.˙  / D .˙ .u ˝ v//./

. 2 C01 .Rn //:

This suggests the following definition for the convolution product u  v of the distributions u and v in Rn : u  v D ˙ .u ˝ v/; (11.11) the tensor product of u and v pushed forward under the sum mapping. In explicit form this becomes .u  v/./ D u.x 7! v.y 7! .x C y///

. 2 C01 .Rn //:

(11.12)

To justify this, we first consider the extension of the definition of the tensor product u˝v to arbitrary distributions u and v. See Example 15.11 for another approach. Theorem 11.8. Let X and Y be open subsets of Rn and Rp , respectively, and let u 2 D 0 .X / and v 2 D 0 .Y /. Then there exists exactly one distribution on X  Y , denoted by u ˝ v and said to be the tensor product of u and v, with the property .u ˝ v/. ˝ for all  2 C01 .X / and

/ D u./ v. /;

(11.13)

2 C01 .Y /. One has, for  2 C01 .X  Y /,

.u ˝ v/./ D u.x 7! v.y 7! .x; y/// D v.y 7! u.x 7! .x; y///:

(11.14)

Equation (11.14) also applies if u 2 E 0 .X /, v 2 E 0 .Y /, and  2 C 1 .X  Y /. The mapping .u; v/ 7! u ˝ v

W

D 0 .X /  D 0 .Y / ! D 0 .X  Y /

(11.15)

is sequentially continuous with respect to each variable separately. Proof. We begin by verifying that the right-hand side in (11.14) is well-defined and is continuously dependent on the test function . Given x 2 X , introduce the partial function .x;
/ W y 7! .x; y/ in C01 .Y /; then x 7! .x;
/ is a continuous mapping: X ! C01 .Y /. Hence composition with the continuous v W C01 .Y / ! C implies that

122

11 Convolution of Distributions

.V /.x/ WD v B .x;
/ defines a continuous function V  on X with compact support. Using the linearity, and once again using the continuity of v, one can show that 1 .V /.x C t e.j // t !0 t lim


.V /.x/ D V .@j /.x/:

By mathematical induction on k we find that V  is of class C k and that @˛ .V / D V .@˛ /; for all multi-indices ˛ with j˛j  k. Thus we see that V  2 C01 .X /; moreover, making use of the estimates for v of the form (3.4), we conclude that V W  7! V 

W

C01 .X  Y / ! C01 .X /

defines a continuous linear mapping. Consequently, u ˝ v 2 D 0 .X  Y / if we set u˝v DuBV

W

C01 .X  Y / ! C:

(11.16)

Because .V . ˝

//.x/ D v..x/ / D .x/ v. /

.x 2 X /;

it follows that V . ˝ / D v. / . If we then apply u to this, we come to the conclusion that (11.13) holds. This proves the existence part of the theorem. Now suppose that w 2 D 0 .X  Y / and w. ˝ / D u./ v. / whenever  2 C01 .X / and 2 C01 .Y /. Then ! WD w .u ˝ v/ has the property that !. ˝ / D 0 for all  2 C01 .X / and 2 C01 .Y /. In Lemma 11.10 below, which is of independent interest, we will demonstrate that this implies ! D 0. In that case, w D u ˝ v, and thus we will also have proved the uniqueness. Furthermore, the right-hand side of the second identity in (11.14) defines a distribution that also satisfies (11.13); in view of the uniqueness just proved, this distribution also equals u ˝ v. To prove the last assertion we observe that on the strength of (11.16), it follows from limj !1 uj D 0 in D 0 .X / that for every  2 C01 .X  Y /, lim uj ˝ v./ D lim uj .V / D 0;

j !1

j !1

because V  2 C01 .X /. The analogous result holds for u ˝ vj .



Example 11.9. Let the notation be as in Example 10.20. Now (10.25) takes the form, for  and 2 C01 .Rn / and y 2 Rn , Z d .˝ /.y/ D .x/ S .y x/ dx D S .y/; so d .˝ / D S : Rn

11 Convolution of Distributions

123

Theorem 11.5 then implies the following result, which will be of use in Example 15.11 below: d  u. ˝

/ D u.d . ˝

// D u.S

 / D u 

./

.u 2 D 0 .Rn //: ˛

The previously defined tensor product of continuous functions u and v satisfies (11.13), and in view of the uniqueness it therefore equals the distributional tensor product of u and v. In other words, the tensor product of distributions is an extension of the tensor product of continuous functions. The mapping in (11.15) is actually sequentially continuous with respect to all variables; this can be proved using Theorem 5.5. Next we come to the lemma that was invoked in the proof of Theorem 11.8. Lemma 11.10. Let X and Y be open subsets of Rn and Rp , respectively, and ! 2 D 0 .X  Y /. If !. ˝ / D 0, for all  2 C01 .X / and 2 C01 .Y /, then ! D 0. Proof. Let .x0 ; y0 / 2 X  Y . We can find a corresponding open neighborhood U of .x0 ; y0 / whose closure is a compact subset K of X  Y . If  2 C01 .X  Y / equals 1 on an open neighborhood of K, one has ! 2 E 0 .X  Y /  D 0 .Rn  Rp /. Then we get in the notation of Lemma 11.6 that, if  > 0 is sufficiently small, ..!/  . ˝

 //.x; y/

D .!/..Tx S / ˝ .Ty S

 //

D !..Tx S / ˝ .Ty S

 //

equals 0, for all .x; y/ 2 K, in view of supp Tz S D zC.  supp /. In particular, we have .!/  . ˝  / D 0 on U . On account of Lemma 11.6, the limit of the left-hand side, as  # 0, equals !; thus we obtain ! D 0 on U , and so ! D 0 on U . By applying Theorem 7.1 we see that this implies ! D 0.  This lemma has a useful application, as will be shown in Remark 11.12. In preparation we treat the following: Example 11.11. Write C01 .X / ˝ C01 .Y / for the linear subspace of C01 .X  Y / consisting of the finite linear combinations of the functions  ˝ , where  2 C01 .X / and 2 C01 .Y /. This linear subspace is a proper subset of C01 .X  Y /. In order to verify this, note that the following assertions are equivalent, for any  2 C01 .X  Y /:

(i)  2 C01 .X / ˝ C01 .Y /. (ii) The linear subspace of C01 .X / spanned by the set of functions f .
; y/ 2 C01 .X / j y 2 Y g is of finite dimension; here .
; y/.x/ D .x; y/. (iii) The linear subspace of C01 .Y / spanned by the set of functions f .x;
/ 2 C01 .Y / j x 2 X g is of finite dimension.

For an example of  2 C01 .R  R/ that does not satisfy condition (ii), consider (compare with Lemma 2.7)

124

11 Convolution of Distributions

.x; y/ D

(

e 0

1 k.x;y/k2

1

if if

k.x; y/k < 1; k.x; y/k  1:

Indeed, select k 2 Z0 and 1 > y1 >


> yk > 0 arbitrarily. Then q if jxj ≶ 1 yj2 : .x; yj / ≧ 0

(11.17)

P Suppose there exist cj 2 C such that 1j k cj .x; yj / D 0, for all x 2 R. q q Then consider x 2 R satisfying 1 yk2 1 < jxj < 1 yk2 and apply (11.17) to find that ck D 0. Continuing in this manner, we see that cj D 0 for all j . This demonstrates the linear independence of the functions .
; yj /, for 1  j  k. ˛ Remark 11.12. Notwithstanding that C01 .X / ˝ C01 .Y / is a proper linear subspace of C01 .X  Y /, it is a dense subset of the latter space. Indeed, using the Hahn– Banach Theorem, Theorem 8.12, one directly deduces the claim from Lemma 11.10. Phrased differently, test functions of both the variables x and y can be approximated by linear combinations of products of test functions of x and y, respectively. We will give a direct proof in Theorem 16.17 below and another one in Theorem 15.4. ˛ Example 11.13. If we denote the variable in X or Y by x or y, respectively, then u 2 D 0 .X / and v 2 D 0 .Y / can also be denoted by the function symbols u.x/ and v.y/, respectively. In that case it is usual to use the function notation .u ˝ v/.x; y/ D u.x/ v.y/ also for distributions u 2 D 0 .X / and v 2 D 0 .Y /. If a 2 X and b 2 Y are given points, then .ıa ˝ ıb / . ˝

/ D ıa ./ ıb . / D .a/ .b/ D ı.a; b/ . ˝

/;

from which we see that ıa ˝ ıb D ı.a; b/ . In function notation: ıa .x/ ıb .y/ D ı.a; b/ .x; y/. For example, if X D Rn and Y D Rp , then ı.x/ ı.y/ D ı.x; y/, a formula popular in the physics literature. Now let X D Y D Rn . Then ı.x C y/ ı.x y/ D 21 ı.x; y/ in D 0 .Rn  Rn /. Indeed, ı.xCy/ ı.x y/ corresponds to ˚  .ı˝ı/ 2 D 0 .Rn Rn /, with ˚.x; y/ D .x C y; x y/. Now we have, for  2 C01 .Rn  Rn /, 1 1 ˚  .ı ˝ ı/./ D .ı ˝ ı/.˚ / D j˚ 1 .ı ˝ ı/..˚ 1 / / D .0; 0/ D ı./; 2 2 where in order to establish the second identity, we have used Theorem 10.8, and where we have observed that ˚ 1 D 12 ˚ and j˚ 1 D 12 for the third. ˛ We now want to define the convolution uv of the distributions u and v belonging to D 0 .Rn / as the tensor product, pushed forward under the sum mapping, as in (11.11). In the pushing forward we intend to use Theorem 10.6. We are, however, faced with the problem that the sum mapping is not a proper mapping from Rn  Rn to Rn , unless n D 0. Indeed, the inverse image of z 2 Rn consists of the set of all .x; z x/ with x 2 Rn . In the following lemma we use sets of sums as defined in (2.2).

11 Convolution of Distributions

125

Lemma 11.14. Let A and B be closed subsets of Rn . Then the following assertions are equivalent: (a) The sum mapping (11.10) is proper on A  B. (b) For every compact subset L of Rn , A \ .L C . B// is bounded in Rn . (c) For every compact subset L of Rn , B \ .L C . A// is bounded in Rn . Assertions (a)–(c) obtain if either A or B is compact. Proof. (a) means that for every compact subset L of Rn the set V of the .x; y/ 2 A  B, with x C y 2 L, is compact. Because this set is closed, we may replace the word “compact” by “bounded.” Furthermore, the projection of V onto the first component equals A \ .L C . B// and the projection onto the second component is B \ .L C . A//. This proves (a) ) (b) and (a) ) (c). On the other hand,



˙ 1 .L/  A \ .L C . B//  L C A \ .L C . B// : This implies (b) ) (a) and by an analogous argument one shows that (c) ) (a). 

Definition 11.15. For every u and v in D 0 .Rn / such that the sum mapping (11.10) is proper on supp u  supp v, (11.11) defines a distribution u  v 2 D 0 .Rn /, said to be the convolution of u and v. In particular, u  v is defined if at least one of the two distributions u and v has compact support. ˛ Example 11.16. By substituting a Dirac measure into (11.12) we obtain u  ıa D ıa  u D Ta u

.u 2 D 0 .Rn /; a 2 Rn /:

(11.18)

In particular (compare with (11.2)), uı DıuDu

.u 2 D 0 .Rn //:

(11.19) ˛

Let A be a closed subset of Rn . In the following theorem, as in Theorem 10.6, we use the notation D 0 .Rn /A for the set of u 2 D 0 .Rn / such that supp u  A. We say that uj ! u in D 0 .Rn /A if uj 2 D 0 .Rn /A for all j and uj ! u in D 0 .Rn /. In that case, it automatically follows that u 2 D 0 .Rn /A . For the properness of the sum mapping, see Lemma 11.14. Theorem 11.17. Let A and B be closed subsets of Rn such that the sum mapping (11.10) is proper on A  B. The convolution product .u; v/ 7! u  v is the uniquely determined extension of the convolution product .u; / 7! u  

W

D 0 .Rn /A  C01 .Rn /B ! C 1 .Rn /;

126

11 Convolution of Distributions

as introduced in (11.1), to a mapping from D 0 .Rn /A  D 0 .Rn /B to D 0 .Rn / that is sequentially continuous with respect to each variable separately. The convolution of distributions satisfies the following computational rules: supp .u  v/  supp u C supp v;

(11.20)

uv D vu (commutative rule); .u  v/  w D u  .v  w/ (associative rule);

.u  v/./ D u.S v  / . 2 C01 .Rn //; .a 2 Rn /; Ta .u  v/ D .Ta u/  v D u  .Ta v/ @˛ .u  v/ D .@˛ u/  v D u  .@˛ v/

.˛ 2 .Z0 /n /:

(11.21) (11.22) (11.23) (11.24) (11.25)

Rule (11.22) holds if the sum mapping ˙ W .x; y; z/ 7! x C y C z is proper on supp u  supp v  supp w. This certainly is the case if at least two out of the three distributions u, v, and w have compact support. The right-hand side in (11.23) has to be understood as u..S v//, where  2 D 0 .Rn / and  D 1 on a neighborhood of A \ .L C . B// with L a compact subset of Rn containing supp . Proof. Regarding the extension of the convolution product to D 0 .Rn /A  D 0 .Rn /B , we first note the following. For P and Q closed subsets of Rn with P \ Q compact, there exists  2 C01 .Rn / with  D 1 on a neighborhood of P \ Q. Given any u 2 D 0 .Rn /P and 2 C 1 .Rn /Q , then u. / is well-defined and independent of the choice of . In this manner, we extend u to a continuous linear functional on C 1 .Rn /Q and the extension is uniquely determined, because C01 .Rn /Q is dense in the latter space. Now consider v 2 D 0 .Rn /B . Then S v 2 D 0 .Rn / B , so we obtain, for any compact L  Rn and  2 C01 .Rn /L , S v   2 C 1 .Rn /LC.

B/ :

Next apply the first result in the proof choosing P D A and Q D L C . B/. Then P \ Q is compact by Lemma 11.14.(b), and accordingly we may define, for u 2 D 0 .Rn /A , z v/./ D u.S v  /: .u 

If v 2 C01 .Rn /, then (11.6) implies that the right-hand side is equal to .u  v/./. z v D u  v if v 2 This being the case for every  2 C01 .Rn /, we conclude that u  C01 .Rn /: the convolution product of distributions is an extension of the convolution product of a distribution and a test function as introduced in (11.1). Furthermore, z with . In addition, we have for u 2 D 0 .Rn / (11.23) now follows on identifying  1 n and v and  2 C0 .R / that u.S v  / D ˙ .u ˝ v/./. This may be proved by observing that both sides depend continuously on u 2 D 0 .Rn / and that E 0 .Rn / is dense in D 0 .Rn /. For u 2 E 0 .Rn / we obtain ˙ .u ˝ v/./ D .u ˝ v/. B ˙/ D u.x 7! v..x C
/// D u.S v  /: Using (10.5), we see that

11 Convolution of Distributions

127

supp .u  v/ D supp ˙ .u ˝ v/  ˙.supp .u ˝ v//

D ˙.supp u  supp v/ D supp u C supp v:

Let  W .x; y/ 7! .y; x/ be the interchanging of variables. The second identity in (11.14) can be written as .u ˝ v/./ D .v ˝ u/.  .// for every test function , that is, u ˝ v D  .v ˝ u/:

The commutativity of addition means that ˙ B  D ˙. Application of (10.4) in the situation at hand is admissible, because it can be shown that (10.4) remains valid for ˚ a diffeomorphism and equal to ˚ as in Theorem 10.5. Since  is a diffeomorphism, this leads to u  v D ˙ B  .v ˝ u/ D .˙ B / .v ˝ u/ D ˙ .v ˝ u/ D v  u: Thus, the commutativity of convolution is seen to be a direct consequence of the commutativity of addition. For arbitrary u, v, and w 2 E 0 .Rn / we find by writing out the definitions that .u  .v  w//./ equals u.x 7! v.y 7! w.z 7! .x C .y C z/////; while ..u  v/  w/./ is equal to u.x 7! v.y 7! w.z 7! ..x C y/ C z////: In this case it follows that (11.22) is a direct consequence of the associativity of addition. For u 2 D 0 .Rn /A , v 2 D 0 .Rn /B , and w 2 D 0 .Rn /C with ˙ proper on A  B  C , we get associativity because it already holds on a dense subset. By virtue of Theorem 11.8 we conclude that u 7! u  v D ˙ .u ˝ v/, being a composition of two such mappings, is sequentially continuous; using the commutativity of convolution, we find that v 7! u  v is also sequentially continuous. Finally, (11.24) and (11.25) can be proved by writing out the definitions and applying transposition. These rules can also be derived from the corresponding formulas for v 2 C01 .Rn /, approximating v 2 D 0 .Rn / by a sequence of test functions and using the sequential continuity of the convolution product.  Example 11.18. Let U D u  as in Theorem 11.3. Then U has an extension to a sequentially continuous linear mapping U W E 0 .Rn / ! D 0 .Rn / satisfying U.v/ D u  v, for all v 2 E 0 .Rn /. In particular, u D U.ı/. ˛ Remark 11.19. The associativity of convolution implies that there is no need to use parentheses in the case of multiple convolution. The convolution u1  u2 


 uk of k distributions u1 ; u2 ; : : : ; uk is well-defined if the sum mapping ˙ W .x .1/ ; x .2/ ; : : : ; x .k/ / 7!

k X

j D1

x .j /

128

11 Convolution of Distributions

is proper on supp u1  supp u2 


 supp uk . One then has u1  u2 


 uk D ˙ .u1 ˝ u2 ˝


˝ uk /:

(11.26)

The notation for the repeated tensor product in the right-hand side of this formula suggests that the tensor product is associative as well. In fact, for test functions this is evident. Using Corollary 11.7 and the continuity of the tensor product, we then also obtain the associativity of the tensor product for arbitrary distributions. The proof of (11.26) can be given by mathematical induction on k. ˛ Remark 11.20. Assume that u and v 2 D 0 .Rn / and that the sum mapping is proper on supp u  supp v. If in this situation v 2 C 1 .Rn /, then also u  v 2 C 1 .Rn /, while .u  v/.x/ D u.Tx B S v/ .x 2 Rn /: (11.27) Although this is the same formula as (11.1), we first of all need to justify the right-hand side in (11.27), because we have not assumed u or v to have compact support. Consider a compact subset L  Rn . On account of Lemma 11.14.(b) there exists a compact K  Rn with the property that supp u \ .L C . supp v//  K. Next, write A D LC. supp v/. Then A is a closed subset of Rn , while supp u\A is compact. Therefore there exists a uniquely determined sequentially continuous extension of u to the space of  2 C 1 .Rn / with supp   A. This extension is also denoted by u. To verify its existence, we write u./ WD u. /, where  2 C01 .Rn / and  D 1 on a neighborhood of K (see Corollary 2.16). Now consider any x 2 L and write  D Tx B S v. Then supp  D fxg C . supp v/  A, and it follows that u.Tx B S v/ D u./ is well-defined. We will now show that u  v 2 C 1 .Rn / and that (11.27) holds. Let U be an arbitrary bounded open subset of Rn with compact closure L. According to the assumption made, K WD ˙ 1 .L/ \ .supp u  supp v/ is a compact subset of Rn  Rn . Let  2 C01 .Rn  Rn / with  D 1 on a neighborhood of K. The definition of distributions pushed forward, given in the proof of Theorem 10.6, leads to the formula .u  v/./ D .u ˝ v/.˙  .//; for every  2 C01 .U /. In addition, we can take  D ˛˝ˇ, with ˛ and ˇ 2 C01 .Rn / equal to 1 on a large bounded subset of Rn . But this means that on U , u  v D .˛u/  .ˇv/:

(11.28)

Here ˇv 2 C01 .Rn /. This identity is derived from .u ˝ v/..˛ ˝ ˇ/˙  .// D .˛u˝ˇv/.˙  .//, which follows from . u/./ D . u/.1/ D u. / D u. /, for 2 C01 , u 2 D 0 , and  2 C 1 . In Theorem 11.2 we have concluded that the right-hand side of (11.28) is a C 1 function given by ..˛u/  .ˇv//.x/ D .˛u/.Tx B S.ˇv//:

11 Convolution of Distributions

129

For x 2 U this is equal to the right-hand side of (11.27), with the interpretation given above. More precisely, we claim that u.˛ ˇ.x


/v.x


// D u.˛ v.x


//

.x 2 U /

and that ˛ D 1 on a neighborhood of supp u \ .L C . supp v//. Equivalently we may write the former identity as
/

u.˛.ˇ.x

1/v.x


// D 0

.x 2 U /:

(11.29)

For the proof of (11.29), fix x 2 U , which implies x 2 L. By the definition of K, we find for y 2 Rn the following three possibilities: y … supp u or x y … supp v or .y; x y/ 2 K. But K  U1  U2 , where U1 and U2 are open subsets of Rn on which ˛ and ˇ, respectively, are equal to 1. Hence, if .y; x y/ 2 K, then y 2 U1 \ .x C . U2 //. Thus supp u is covered by the following open sets: the complement in Rn of x C . supp v/ and the set U1 \ .x C . U2 //. For y belonging to either one of these sets we have ˛.y/.ˇ.x y/ 1/v.x y/ D 0; this leads to (11.29). Finally, we show that ˛ D 1 on a neighborhood of supp u \ .L C . supp v//. Since  D 1 on an open neighborhood of K, we see that ˛ D 1 on an open neighborhood of the image of K under the projection Rn  Rn ! Rn with .y; z/ 7! y. By the definition of K this image equals supp u \ .L C . supp v//. ˛ Remark 11.21. For application in (12.10) below we derive a result that is of interest in its own right. Suppose f and g are locally integrable functions on Rn and ˙ is proper on supp f  supp g. Then f  g 2 D 0 .Rn / as in the definition in (11.12) is actually a locally integrable function on Rn too. More precisely, let L  Rn be compact and set K D f .y; z/ 2 supp f  supp g j y C z 2 L g  Rn  Rn ; which is also compact on account of ˙ being proper. Then we obtain, for almost all x 2 L, Z .f  g/.x/ D

f .y/ g.x

y/ dy:

1 .K/\.LC. 2 .K///

Here i W Rn  Rn ! Rn denotes the projection onto the ith factor, for 1  i  2. Indeed, consider  2 C01 .Rn / with supp   L. Select  2 C01 .Rn  Rn / such that  equals 1 on a neighborhood of K. Then we obtain .test f  test g/./ D .test f ˝ test g/.. B ˙// D .test f /.y 7! .test g/.z 7! .y; z/ .y C z/// Z Z D f .y/ g.z/ .y; z/ .y C z/ dz dy n n ZR ZR f .y/ g.x y/ .y; x y/ .x/ dx dy D Rn

Rn

130

11 Convolution of Distributions

D D D

Z

n

Z

ZR Z

f .y/ g.x

y/ .y; x

L

1 .K/\.LC. 2 .K///

L

1 .K/\.LC. 2 .K///

Z Z

y/ dy .x/ dx

Rn

f .y/ g.x

y/ .y; x

y/ dy .x/ dx

f .y/ g.x

y/ dy .x/ dx:

In obtaining the last equality, we need the extra assumption that  equals 1 on a neighborhood of the set


1 .K/  L C L C . 2 .K// ;

which contains K. In the case of compact supp f , we may also write Z .f  g/.x/ D f .y/ g.x

y/ dy:

(11.30)

supp f

If  is a Radon measure on Rn of compact support and g is locally integrable on R , then the proof above can be modified to give, in the notation of Chap. 20, Z g.x y/ .dy/: (11.31) .  g/.x/ D n

supp 

˛ In general, we have the following estimate for the singular support of the convolution of two distributions. Theorem 11.22. Consider u and v 2 D 0 .Rn / and suppose that the sum mapping .x; y/ 7! x C y is proper on supp u  supp v. Then sing supp .u  v/  sing supp u C sing supp v:

(11.32)

Proof. Let A D sing supp u and B D sing supp v. For every ı > 0 we can find a cut-off function ˛ 2 C 1 .Rn / with ˛ D 1 on an open neighborhood of A and such that supp ˛ is contained in the ı-neighborhood Aı of A. For this purpose we can choose the convolution of the characteristic function of the 12 ı-neighborhood of A and  , for sufficiently small  > 0; see Lemma 2.19. Let ˇ be an analogous cut-off function, having the same properties with respect to B instead of A. Write u1 WD ˛ u and u2 WD .1 ˛/ u, and analogously v1 WD ˇ v and v2 WD .1 ˇ/ v. Then u D u1 C u2 with supp u1  Aı , u2 2 C 1 .Rn /. Furthermore, v D v1 C v2 with supp v1  Bı and v2 2 C 1 .Rn /. Out of the four terms in u  v D u1  v1 C u1  v2 C u2  v1 C u2  v2 ;

11 Convolution of Distributions

131

the last three are of class C 1 on account of Remark 11.20. This implies sing supp uv D sing supp u1 v1  supp u1 v1  supp u1 Csupp v1  Aı CBı : If we can demonstrate that for every x … A C B there exists ı > 0 with x … Aı C Bı , that is, x … sing supp .u  v/, then we will have proved that sing supp .u  v/  A C B: The mapping ˙ being proper on supp usupp v, it follows from [7, Theorem 1.8.6] that A C B is a closed subset of Rn . Such a set equals the intersection of all of its ı-neighborhoods; accordingly, there exists ı > 0 such that x … .A C B/2ı . The conclusion now follows because Aı C Bı  .A C B/2ı .  Remark 11.23. It may also happen that the convolution u  v can be formed without making any assumptions about the supports of u and of v. The space of all integrable functions on Rn is denoted by L1 .Rn /, or simply L1 , and provided with the integral Z norm kvkL1 WD jv.x/j dx: Rn

One of the main results of Lebesgue integration, see Theorem 20.40, asserts that L1 is a Banach space, that is, a normed linear space that is complete. Let E be a linear subspace of D 0 .Rn / endowed with a norm u 7! kuk, with the following four properties: (i) For every a 2 Rn and u 2 E, one has Ta u 2 E and kTa uk D kuk. (ii) E is complete. (iii) C01 .Rn /  E and convergence in C01 .Rn / implies convergence in E. (iv) C01 .Rn / is dense in E.

An example of such a space is E D L1 . In that case, y 7! .y/ Ty  is continuous from Rn to E, for every  and 2 1 C0 .Rn /. In view of (1.14), the integral of this function is equal to   . Applying the triangle inequality to the approximating Riemann sums, we obtain the inequality k 

k  k kL1 kk:

Now let u 2 E and v 2 L1 . Then there exist sequences .uj /j 2N and .vj /j 2N in C01 .Rn / with ku uj k ! 0 and kv vj kL1 ! 0 as j ! 1. The sequence .uj  vj /j 2N is a Cauchy sequence in E. The limit, denoted by u  v, is independent of the choice of the sequences .uj / and .vj / and satisfies ku  vk  kvkL1 kuk: The mapping .u; v/ 7! u  v thus defined is continuous from E  L1 to E and is the only continuous extension to E  L1 of the convolution product on C01 .Rn / 

132

11 Convolution of Distributions

C01 .Rn /. It is left to the reader as a problem to prove these assertions concerning u  v. Applying this for E D L1 , one can prove by means of the theory of Lebesgue integration on product spaces that for every u and v 2 L1 the integral Z u.x y/ v.y/ dy Rn

converges to .u  v/.x/ for almost all x; see also Problem 11.22. The example u D v D jxj a 1Œ 1; 1  in L1 .R/, for 12  a < 1, shows that the integral need not converge for all x. ˛

Problems  11.1. For each of the following cases, find a combination of a distribution u and a test function  on R that solves the equation:

(i) u   (ii) u   (iii) u   (iv) u  

D 0. D 1. D x. D sin.

11.2. If limj !1 uj D u in D 0 .Rn / and limj !1 j D  in C01 .Rn /, then lim uj  j D u  

j !1

in

C 1 .Rn /:

Prove this using the principle of uniform boundedness from Chap. 5.  11.3. Given a mapping A W C01 .Rn / ! C 1 .Rn /, verify that the following assertions are equivalent. (i) A is a continuous linear mapping and commutes with the partial differentiations @j , for all 1  j  n. (ii) There exists a u 2 D 0 .Rn / such that A./ D u  , for every  2 C01 .Rn /.

Hint: differentiate a 7! Ta B A B T

a.

11.4. If t 7! ut is a continuous mapping from Rm to DR0 .Rn / with u t D 0 for all t outside a bounded subset of Rm , we define the integral u t dt 2 D 0 .Rn / by Z Z  u t dt ./ WD u t ./ dt . 2 C01 .Rn //: Rm

Rm

Now prove that for u 2 D 0 .Rn / and  2 C01 .Rn /, Z u D .x/ Tx u dx: Rn

11 Problems

133

 11.5. Let u 2 D 0 .Rn / with @j u D 0, for all 1  j  n. Prove the existence of a constant c 2 C such that u D c. See Problem 12.9 for a generalization.  11.6. Fix u 2 D 0 .Rn / and set

L D f P .@/ u 2 D 0 .Rn / j P polynomial function on Rn g: Prove that the linear subspace L of D 0 .Rn / is of finite dimension if and only if u is P an exponential polynomial on Rn , that is, a function of the form u D lkD1 Pk eak , where Pk is a polynomial function on Rn and ak 2 Cn , while eak is defined in (14.1), for 1  k  l.

 11.7. Let f be a polynomial function on R of degree  m and T 2 E 0 .R/. Show that f  T is a polynomial function of degree  m.  11.8. Calculate ıa  ıb , for a and b 2 Rn .

11.9. Suppose that f and g 2 C.Rn / and that f has compact support. Verify that .test f /  .test g/ D test.f  g/.

 11.10. Let P be a linear partial differential operator in Rn with constant coefficients. Prove that P u D .P ı/  u for every u 2 D 0 .Rn /. Also prove that P .u  v/ D .P u/  v D u  .P v/ if u and v 2 D 0 .Rn / and if the sum mapping is proper on supp u  supp v.

 11.11. Let v 2 E 0 .Rn /. Prove that the convolution mapping S v sends C01 .Rn / 1 to C0 .Rn / and is continuous linear. Consider the transpose t

.S v/ W D 0 .Rn / ! D 0 .Rn /:

Prove that t .S v/u D u  v, for all u 2 D 0 .Rn /. See Example 15.11 for another proof. 11.12. Let E be the linear subspace of D 0 .Rn / consisting of the finite linear combinations of the ıa with a 2 Rn . Prove that for every continuous function f in Rn there exists a sequence .uj /j 2N in E such that limj !1 uj D f in D 0 .Rn /. Hint: see Problem 5.12. Now go on to show that for every u 2 D 0 .Rn / there exists a sequence .uj / in E with limj !1 uj D u in D 0 .Rn /. Given that the convolution of distributions is continuous in each of the variables, use this to prove, once again, the properties (11.24), (11.25), (11.23), (11.22), and (11.21). 11.13. Define D 0 .R/C as the union of the D 0 .R/Œ l; 1 Œ over all l 2 R. We say that uj ! u in D 0 .R/C if there exists an l 2 R with uj ! u in D 0 .R/Œ l; 1 Œ . Prove that for every u and v 2 D 0 .R/C the sum mapping is proper on supp u  supp v and additionally, that the convolution product .u; v/ 7! u  v

W

D 0 .R/C  D 0 .R/C ! D 0 .R/C

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11 Convolution of Distributions

satisfies all computational rules for the convolution product. In assertions concerning sequential continuity, one only needs to replace convergence in D 0 .R/ by convergence in D 0 .R/C .  11.14. Define, by mathematical induction on k, kC by 1C D H , the Heaviside function, and kC D H  kC 1 if k > 1. Calculate all kC and all derivatives .kC /.l/ . Prove that for every k 2 N and f 2 D 0 .R/C , there exists exactly one u 2 D 0 .R/C with u.k/ D f . For  2 C k .R/, verify the formula

. H /.k/ D

k X1

j D0

 .k

j 1/

.0/ ı .j / C  .k/ H:

Write out the formula resulting from the convolution of the left-hand side and the right-hand side, respectively, and kC . Do you recognize the result?

 11.15. Calculate .1  ı 0 /  H and 1  .ı 0  H /.

11.16. Prove that the tensor product of two probability measures is also a probability measure. Verify that for two probability measures  and  on Rn the convolution    is well-defined and defines a probability measure on Rn . The convolution    is said to be the independent sum of the probability measures  and . Do you recognize this from probability theory? Finally, calculate the probability measure    


 , the independent sum of N copies of , if  D p ı1 C .1 p/ ı0 and 0  p  1. 11.17. Calculate the distribution ı 0 .x C y/ ı.x y/ on R2 . Express the result in terms of the Dirac measure at 0 2 R2 and its derivatives. 11.18. Let u 2 D 0 .Rn / and xn u D 0. Show that the result from Problem 9.7 can be rephrased as the existence of a uniquely determined v 2 D 0 .Rn 1 / such that u D v ˝ı R in D 0 .Rn /, where ı R denotes the Dirac measure on R. (See Problem 15.7.(ii) for a different proof.)  11.19. For t 2 R, consider the translation T t in Rn given by T t W x D .x 0 ; xn / 7! 0 .x ; xn C t/ 2 Rn  R. Write T D .T t / t 2R for the corresponding one-parameter group of translations. Suppose that u 2 D 0 .Rn / is invariant under the action induced by T; in other words, it satisfies .T t / u D u, for all t 2 R.

(i) Show that this is the case if and only if @n u D 0. (ii) Let  W Rn ! Rn 1 be the orthogonal projection  W .x 0 ; xn / 7! x 0 , which corresponds to the decomposition Rn D Rn 1  R. Prove the existence of a v 2 D 0 .Rn 1 / such that u D   v D v ˝ 1R . (See Problem 15.7.(i) for a different proof.)

Background. The geometry underlying this result is as follows. Every orbit in Rn under the action of T is of the form f T t .x 0 ; 0/ D .x 0 ; t/ 2 Rn j t 2 R g for some

11 Problems

135

x 0 2 Rn 1 , and this set equals the level set  1 .fx 0 g/; geometrically, it is a line perpendicular to the hyperplane Rn 1  f0g. In other words, the set of orbits under T coincides exactly with the set of level sets in Rn of the surjective submersion . Therefore, distributions on Rn that are invariant under T factorize through , that is, they are pullbacks under  of distributions on Rn 1 .  11.20. (Distribution supported by linear subspace.) Corresponding to the decomposition Rn D Rp Rq for p and q > 0, write x D .y; z/. Suppose u 2 E 0 .Rn / satisfies supp u  Rp  f0g  Rn . Set A D f0g  .Z0 /q  .Z0 /n , let  W Rp ! Rn q be the natural embedding with .y/ D .y; 0/, and ı R the Dirac measure supported q 0 p at 0 2 R . Prove the existence of u˛ 2 E .R /, with ˛ 2 A, such that X X X q uD .@˛ B  /u˛ D .I ˝ @˛ / u˛ D u˛ ˝ @˛ ı R ; ˛2.Z0 /q

˛2A

˛2.Z0 /q

where the sum is actually finite. (See Problem 15.8 for another proof.) In other words, u is a finite sum of transversal derivatives of compactly supported distributions on Rp . Any linear subspace in Rn of dimension p can be transformed into Rp  f0g by means of a rotation about the origin. 11.21. In this problem we use the notation from Example 14.22. Let X be an open subset of Rn and consider p and p 0  1 satisfying 1=p C 1=p 0 D 1, on the understanding that p 0 D 1 if p D 1 and p 0 D 1 if p D 1. For all f 2 Lp .X / and 0 g 2 Lp .X /, prove H¨older’s inequality Z jf .x/g.x/j dx  kf kLp .X/ kgkLp0 .X/ : X

Conclude that fg 2 L1 .X /. Hint: see [7, Exercise 6.73.(i)]. 11.22. (Young’s inequality.) In this problem we use the notation from Example 14.22. Let p, q, and r  1 and 1=p C 1=q C 1=r D 2. Then we have Young’s inequality, which asserts, for all f , g, and h 2 C0 .Rn /, ˇ ˇZ ˇ ˇ f .x/ .g  h/.x/ dx ˇ  kf kLp kgkLq khkLr : ˇ Rn

Indeed, we are free to assume that f , g, and h are real and nonnegative. Introduce p 0  1 by 1=p C 1=p 0 D 1, and similarly q 0 and r 0 . Write the integral on the lefthand side as Z Z Z Z I D f .x/ g.x y/ h.y/ dy dx D a.x; y/ b.x; y/ c.x; y/ dx dy Rn

with

Rn

Rn

0

a.x; y/ D f .x/p=r g.x b.x; y/ D g.x

c.x; y/ D f .x/

Rn

0

y/q=r ;

0

0

y/q=p h.y/r=p ;

p=q 0

h.y/

r=q 0

:

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11 Convolution of Distributions

Noting that 1=p 0 C 1=q 0 C 1=r 0 D 1, we can use H¨older’s inequality from Problem 11.21 for three functions to obtain jI j  kakLr 0 kbkLp0 kckLq0 . But Z Z 1=r 0 p=r 0 q=r 0 f .x/p g.x y/q dx dy D kf kL kakLr 0 D p kgkLq ; Rn

Rn

and similarly for b and c. The second equality above is a consequence of changing variables from y to x y and integrating first with respect to y. This leads to the desired inequality. Now prove the following. For f and g 2 C0 .Rn / and p, q and r  1 satisfying 1 C 1=p D 1=q C 1=r, one has the inequality kf  gkLp  kf kLq kgkLr ;

in particular

kf  gkLp  kf kLp kgkL1 : 0

In order to obtain this estimate, apply Young’s inequality with f D s jg  hjp =p , where s is the function defined by s .g  h/ D jg  hj. This implies kg  hkLp0  kgkLq khkLr ;

where

1C

1 1 1 D C : 0 p q r

Then replace p 0 by p. Finally, prove the validity of the estimates if the functions belong to the appropriate spaces of type Lp .Rn / .

Chapter 12

Fundamental Solutions

P ˛ Definition 12.1. Let P D P .@/ D j˛jm c˛ @ be a linear partial differential n operator in R with constant coefficients, as introduced in (7.5). A fundamental solution of P is a distribution E 2 D 0 .Rn / such that PE D ı, the Dirac measure at the origin. ˛ Every linear partial differential operator with constant coefficients (not all of them equal to 0) has a fundamental solution. For more on this, see below: Theorem 17.11, Remarks 17.12 and 18.9 for special cases and Theorem 18.4 for the general result. For linear partial differential operators with variable coefficients the existence of fundamental solutions need not be the case. The importance of fundamental solutions lies in the following: Theorem 12.2. Suppose E is a fundamental solution of P . Then we have P .E  f / D f u D E  Pu

.f 2 E 0 .Rn //; .u 2 E 0 .Rn //:

(12.1) (12.2)

Proof. Using (11.25) and (11.19), we obtain P .E  f / D .PE/  f D ı  f D f , and therefore (12.1). Combination of (11.25) and (12.1) yields that E  .P u/ D P .E  u/ D u, which implies (12.2).  For every distribution f on Rn with compact support, formula (12.1) implies the existence of a distributional solution u D E  f 2 D 0 .Rn / of the inhomogeneous linear partial differential equation P u D f . Additionally, under the assumption that the solution u of P u D f has compact support, (12.2) means that the solution is uniquely determined and given by u D E  f . A word of warning: it may seem as if E  is a two-sided inverse of P , which would imply that P is bijective. But that is not actually the case, because the domain spaces do not correspond: generally speaking, we do not know more about E  than that it is a mapping from E 0 .Rn / to D 0 .Rn /. The differential operator P maps D 0 .Rn / to D 0 .Rn / and E 0 .Rn / to E 0 .Rn /, but not D 0 .Rn / to E 0 .Rn /. In general, J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_12, © Springer Science+Business Media, LLC 2010

137

138

12 Fundamental Solutions

partial differential operators are far from injective, as we will see in the next example and in Problem 14.3 below. Another consequence of this is that fundamental solutions are not uniquely determined. If E is a fundamental solution of P , then Ez is a fundamental solution of P if and only if Ez D E C u, with u 2 D 0 .Rn / a solution of P u D 0. P Example 12.3. Let P D
D jnD1 @j2 be the Laplace operator in Rn . A C 2 function u is said to be harmonic on the open subset U of Rn if
u D 0 on U ; this terminology is carried over to distributions u 2 D 0 .U /. In other words, the kernel of
consists of the harmonic functions, or distributions, respectively. If n > 1, then the harmonic polynomials on Rn form a linear space of infinite dimension. Indeed, the functions x 7! .x1 C i x2 /k on Rn are harmonic, for all k 2 Z0 . (If n D 1 and U is an interval, then u 2 D 0 .U / and
u D 0 if and only if u on U is equal to a polynomial function of degree  1.) In Problem 4.7 a fundamental solution of the Laplace operator was found to be the locally integrable function E on Rn that on Rn n f0g is given by E.x/ D

‚

1 .2 n/ cn kxkn 1 log kxk 2

2

if if

n ¤ 2;

(12.3)

n D 2:

For cn , see also (13.37) below. For f 2 E 0 .Rn /, the distribution u D E  f is said to be the potential of the distribution f , a terminology that has its origin in the situation in which n D 3 and f denotes a mass or charge density. We conclude that the potential u of f satisfies Poisson’s equation
u D f

in

D 0 .Rn /I

and in particular, the potential u is harmonic on the complement of supp f , the largest open set on which f D 0. Furthermore, if u is a solution of Poisson’s equation that has compact support, then u necessarily equals the potential of f . The general fundamental solution of the Laplace operator is equal to the sum of the ı-potential and a harmonic function on Rn . If n  3, then E can be characterized as the fundamental solution of
that converges to 0 as kxk ! 1. This can be proved by means of Fourier transform; see Problem 17.4 below. ˛ As far as the study of the singular supports (see Definition 7.8) of solutions is concerned, addition of C 1 functions is irrelevant. This can also be expressed by saying that calculations are performed modulo C 1 when summands of class C 1 are neglected. A distribution E on Rn is said to be a parametrix of P if there exists a 2 C 1 .Rn / such that PE D ı C ; (12.4) in other words, if E modulo C 1 satisfies the equation for a fundamental solution. For certain P a parametrix can be obtained by iterative methods.

12 Fundamental Solutions

139

Theorem 12.4. Suppose that P possesses a parametrix E with sing supp E D f0g. Then, for every open subset X of Rn , sing supp u D sing supp P u

.u 2 D 0 .X //:

(12.5)

Proof. We see immediately that sing supp P u  sing supp u, which is the case for every linear partial differential operator P in X with C 1 coefficients. We will now prove the converse inclusion. If u has compact support, we can interpret u as an element of E 0 .Rn / according to Lemma 8.9. We have E  P u D .PE/  u D ı  u C Since ı  u D u and

 u:

 u 2 C 1 .Rn / by Theorem 11.2, we obtain, using (11.32),

sing supp u D sing supp E  P u  sing supp E C sing supp P u D f0g C sing supp P u D sing supp P u: Now let u 2 D 0 .X / be arbitrary and x 2 X nsing supp P u. Choose  2 C01 .X / with  D 1 on an open neighborhood U of x. Then P . u/ D P u on U , and therefore x … sing supp P . u/. The above yields sing supp P . u/ D sing supp . u/, because  u has compact support. Hence x … sing supp . u/, which in turn implies that x … sing supp u, because  D 1 on a neighborhood of x.  The theorem means that if u 2 D 0 .X / is a solution of the partial differential equation P u D f , then u 2 C 1 on every open subset U of X where f 2 C 1 . A linear partial differential operator P with C 1 coefficients is said to be hypoelliptic if P has this property, that is, if (12.5) holds. If E is a parametrix of a hypoelliptic operator P with constant coefficients, then necessarily sing supp E D sing supp PE D sing supp ı D f0g. The term “hypoelliptic” conveys the fact that this condition is weaker than the condition that P is elliptic; see Theorem 17.6. For the definition of the term “elliptic,” see Definition 17.2. Remark 12.5. If all derivatives can be estimated such that convergence of each of the Taylor series is guaranteed, the proof of Theorem 12.4 can be modified so as to yield results on the real-analyticity of solutions. Thus one has that u is real-analytic wherever P u D 0, if there exist E 2 D 0 .Rn / and an open neighborhood U of 0 in Rn with the following properties: (a) E is real-analytic on U n f0g, (b) PE ı is real-analytic on U . Estimation of all derivatives can be circumvented by means of complex analysis; see Remark 12.13. This is then followed by the proof of this assertion. ˛

140

12 Fundamental Solutions

Remark 12.6. For linear partial differential operators P D P .x; @/ with variable coefficients there is a well-developed theory in which the convolution operator E , Z defined by .E  f /.x/ D E.x y/ f .y/ dy; is replaced by a singular integral operator K of the form Z .Kf /.x/ D k.x; y/ f .y/ dy:

Here the integral kernel k.x; y/ is a distribution on X  X if the open set X in Rn denotes the domain space of the functions f and Kf ; see Chap. 15 below. The operator K is then said to be a parametrix of P if P BK D I CR and KBP D I CS , where R and S are integral operators with a C 1 kernel on X  X . It is not difficult to prove that P is hypoelliptic when P has a parametrix K with sing supp k  f .x; y/ 2 X  X j x D y g: However, the construction of parametrices, for sufficiently general operators with variable coefficients, involves too much work to be dealt with in this text. ˛ Example 12.7. Every ordinary differential operator (the case n D 1) is hypoelliptic in the open set where the coefficient of the highest-order term does not vanish. See Theorem 9.4. ˛ Example 12.8. The fundamental solution E of the Laplace operator given in (12.3) satisfies sing supp E D f0g; consequently, the Laplace operator is hypoelliptic. In particular, every harmonic distribution is a harmonic C 1 function. The solution E is not only C 1 , but even real-analytic on the complement of the origin. Remark 12.5 therefore implies that u is real-analytic wherever
u D 0. In particular, every harmonic function and distribution is real-analytic. ˛ A note on history: R Weyl [24, Lemma 2] proved that every square-integrable function u such that u.x/
.x/ dx D 0 for every  2 C02 is in fact C 1 . This assertion is known as Weyl’s Lemma and has become the prototype of Regularity Theorems like Theorem 12.4. Weyl’s proof follows the same lines as the proof of Theorem 12.4 given here. Example 12.9. In Problem 8.5 a fundamental solution E was found for the operator P D @ t
x in .nC1/-dimensional .x; t/-space. This, too, satisfies sing supp E D f0g; consequently, the heat operator, too, is hypoelliptic. If the distribution u is a solution of the heat equation P u D 0 on an open subset U of RnC1 , then u is a C 1 function on U . However, in the present case E is not analytic on the entire plane of the .x; t/ with t D 0, and here it is not true that u is analytic wherever P u D 0. P is called the n-dimensional heat operator, although this operator is defined in RnC1 . This derives from the interpretation of x D .x1 ; : : : ; xn / as the position coordinates and t as the time coordinate, which are often treated as playing different roles. ˛

12 Fundamental Solutions

141

Example 12.10. Problems 10.25 and 12.7 describe u 2 D 0 .R2 / that satisfy the equation .@21 @22 /u D 0 in R2 but are far from C 1 functions. This shows that the wave operator @21 @22 is not hypoelliptic. Indeed, for every n  1 the n-dimensional wave operator  WD @2t
x , with x 2 Rn , is not hypoelliptic; see Theorem 13.3. ˛ Example 12.11. We now consider complex analysis from a distributional point of view. A function f defined on an open subset V of C is called complex-differentiable at the point z 2 V when f .zCh/h f .z/ converges as h ! 0 in C. The limit is said to be the complex derivative f 0 .z/ 2 C of f at z. If we apply the identification C ' R2 , by writing z 2 C as z D x C iy in the usual way, with x and y 2 R, then this condition is stronger than the condition that f is differentiable as a complex-valued function of two real variables .x; y/. Indeed, when f depends on the combination z D x C iy, its partial derivatives satisfy @x f .z/ D f 0 .z/ @x z D f 0 .z/

@y f .z/ D f 0 .z/ @y z D i f 0 .z/:

and

This means that f is complex-differentiable if and only if f is real-differentiable and if, moreover, @y f D i @x f

(Cauchy–Riemann equation):

(12.6)

It is then natural to call a distribution u 2 D 0 .V / complex-differentiable if u satisfies (12.6); in other words, if P u D 0, with P equal to the first-order linear partial differential operator with constant coefficients P D i @x

@y D i.@x C i @y /:

The formula for integration by parts in Rn reads, with 1  j  n, Z Z Z f .x/ @j g.x/ dx D g.x/ @j f .x/ dx C f .y/ g.y/ j .y/ dEucl y: U

U

@U

(12.7) Here U is an open subset of Rn with C 1 boundary @U and with the property that U lies at one side of @U . Furthermore, f and g 2 C 1 .U / and .supp f / \ U , for example, is compact. Finally, j .y/ is the jth component of the outer normal to @U at y. See, for example, [7, Corollary 7.6.2]. In R2 ' C we can write i 1 2 D i.1 C i 2 / D i , where the vector .y/ is now interpreted as a complex number. The vector i .y/ is equal to the tangent vector to @U at y of length 1, oriented to have U to the left. If @U is locally parametrized by a C 1 curve W Œ a; b  ! C, where 0 .t/ has the same orientation as i . .t//, one has, for a continuous function f with support in .Œ a; b /, Z

@U

f .z/ i .z/ dEucl z D

Z

b a

f . .t// 0 .t/ dt:

142

12 Fundamental Solutions

Here the vector 0 .t/ is also interpreted as a complex number. This is said to be the complex line integral of f over @U with respect to the orientation of @U described above, and is denoted by Z f .z/ dz with dz D i .z/ dEucl z: @U

Thus we obtain for the operator P the following transposition formula: Z Z Z f .z/ P g.z/ dz D g.z/ Pf .z/ dz C f .z/ g.z/ dz: U

U

(12.8)

@U

In this formula, U , f , and g are as above, with n D 2. The integrals over U are Euclidean 2-dimensional integrals, while that over @U is a complex line integral. For a first application of (12.8) we assume that U is bounded, that f is also a complex-differentiable function on U , and that g equals the constant function 1. We then immediately get the following version of Cauchy’s Integral Theorem: Z f .z/ dz D 0: (12.9) @U

In order to find a fundamental solution of P , we consider the function z 7! z1 on C n f0g. This function is locally integrable on C D R2 (use polar coordinates) and can therefore be interpreted as a distribution. From (12.6) it is evident that P z1 D 0 on R2 n f0g. For  2 C01 .R2 / we obtain, applying (12.8), Z Z  1 1 P .z/ .z/ P ./ D .P / D lim dz D lim dz z z #0 jzj> z #0 jzjD z Z 2 . e i t / D lim  e i t . i / dt D 2 i .0/: #0 0  e it Here we use that the outer normal to f z 2 C j jzj D  g points toward the origin; therefore this circle has to be traversed clockwise, which is the case under the mapping t 7!  e i t . We conclude that P z1 D 2 i ı, in other words, E.z/ D 21i z defines a fundamental solution of P . Because this fundamental solution is C 1 , and even complex-analytic, outside the origin, we conclude on the strength of Theorem 12.4 that P is hypoelliptic. It follows that in particular, every complex-differentiable distribution on an open set V equals a C 1 function on V and as such is complex-differentiable in the classical sense. Now let U be an open subset of V with C 1 boundary @U . Assume that U is a compact subset of V , f 2 C 1 .V /, and g 2 C01 .V /. Interpreting the function g in (12.8) as a test function and writing the left-hand side as .f 1U /.P g/ D P .f 1U /.g/, we can rewrite (12.8) as the identity P .f 1U / D .Pf / 1U

compl f ı@U

12 Fundamental Solutions

143

compl in the space of distributions on V of order  1 with compact support, where ı@U denotes the complex line integration over the boundary. If we now apply the convolution operator E  to this identity, and recall (12.2), the left-hand side becomes equal to f 1U ; and thus we obtain Pompeiu’s integral formula Z Z 1 1 Pf .z/ f .z/ dz C dz . 2 U /: (12.10) f ./ D 2 i U  z 2 i @U z 

Here we applied (11.30) and (11.31) for obtaining the first and second terms on the right-hand side, respectively. In particular, if f is complex-differentiable on V , that is, Pf D 0 and f 2 C 1 .V /, one obtains the well-known Cauchy integral formula Z 1 f .z/ dz . 2 U /: (12.11) f ./ D 2 i @U z  Thus, f is expressed on U in terms of the restriction f j@U of f to the boundary @U . Not every analytic function g on @U is of the form g D f j@U for a complexanalytic function f on U ; see Problem 16.13. Cauchy’s integral formula gives an arbitrary complex-differentiable function f as a “continuous linear combination” of the very simple complex-differentiable functions z 7! z 1  , where the variable z runs over the boundary @U . The singularity at z D  need not bother us here, provided we ensure that  2 U steers clear of @U . In particular, for a given point a 2 U and with  in a sufficiently small neighborhood U.a/ of a, we can substitute the power series 1 z



D

1 .z

a/

.

a/

D

1 z

1 a 1

 a z a

D

X

k2Z0

. a/k .z a/kC1

into (12.11). This implies that the complex-differentiable f can be expressed by a convergent complex power series in a neighborhood of a. More precisely,  X  1 Z f .z/ f ./ D dz . a/k . 2 U.a//: 2 i @U .z a/kC1 k2Z0

Conversely, it is known that every convergent complex power series is complexdifferentiable and even infinitely differentiable, and the power series is equal to the Taylor series. See any textbook about analysis in one variable. It follows that the conditions “complex-differentiable distribution” and “is locally equal to a convergent complex power series” are equivalent. In this case, f is said to be complexanalytic. Instead of working with the operator P D i @x @y , it is more usual in the literature to use 1 1 1 @z WD .@x i @y / P: (12.12) and @zN WD .@x C i @y / D 2 2 2i

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12 Fundamental Solutions

That makes (12.6) equivalent to @zN f D 0. If this is the case, then df D @z f . dz The operator @zN is said to be the Cauchy–Riemann operator; it has the fundamental 1 solution z , and in the literature (12.10) is mostly written in the form f ./ D

1 

Z

U

1 @zN f .z/ dz C z  2 i

Z

@U

f .z/ dz z 

. 2 U /:

(12.13)

In several variables one has an analogous theory: a function z 7! f .z/ on Cn is complex-differentiable if f is complex-differentiable as a function of each of the variables zj , that is, if @zNj f D 0, for 1  j  n. These are referred to as the Cauchy–Riemann equations for f . It can be proved that a distributional solution of these equations is of class C 1 and can locally be developed into power series, where the proof makes use of a higher-dimensional version of Cauchy’s integral formula. These functions are said to be complex-analytic in several variables. ˛ Remark 12.12. The notation becomes clear when we write f .x; y/ D F .z; z/, N with z D x C iy and zN D x iy. This yields @x f D @z F C @zN F

and

@y f D i @z F

i @zN F:

If we now solve this system for @z F and @zN F in terms of @x f and @y f , we obtain formulas that we recognize as the definitions of @z and @zN . ˛ Remark 12.13. In Lemma 2.6, a function f on an open subset U of Rn was said to be analytic on U if f has local power series representations at all points of U . To distinguish it from the above, such f is also said to be real-analytic on U . By substituting complex values for the variables into the power series, we see that f has a complex-analytic extension to an open neighborhood V of U in Cn ' R2n . Conversely, the restriction to U WD V \ Rn of a complex-analytic function on V is a real-analytic function on U . It follows that a function f is real-analytic if f can be extended to a function on a complex neighborhood that satisfies the Cauchy–Riemann equations on that neighborhood. In many cases, this involves less effort than working out the estimates for all derivatives required to establish the convergence of the Taylor series. ˛ Finally, we come back to the proof of the assertion in Remark 12.5. We observe that the analytic singular support sing supp anal .u/ may be defined in the obvious manner, for any u 2 D 0 .Rn /. Lemma 12.14. Let u 2 D 0 .Rn / and v 2 E 0 .Rn /. Then we have sing supp anal .u  v/  sing supp anal u C supp v:

Proof. Set U0 D Rn n sing supp anal u and denote by u0 the restriction of u to U0 . The function u0 has a complex-analytic extension to an open subset U  Cn

12 Fundamental Solutions

145

satisfying U \ Rn D U0 . Selecting a 2 Rn with a … sing supp anal u C supp v, we have a C . supp v/  U0 . Let V be an open neighborhood of a in Cn such that V C . supp v/  U . For z 2 V , write Z u0  v.z/ D v.u0 .z
// D u0 .z x/ v.x/ dx: Rn

This equality should be interpreted in the sense of distributions, and it defines a complex-analytic function on V . We note that on V \ Rn , the function u0  v coincides with the usual convolution product u  v as in (11.27). Thus, the latter is analytic on the open neighborhood V \ Rn of a in Rn ; this implies a … sing supp anal .u  v/.  Theorem 12.15. Let P be a linear partial differential operator in Rn with constant coefficients. Suppose there exist E 2 D 0 .Rn / and an open neighborhood U of 0 in Rn with the following properties: (a) E is analytic on U n f0g, (b) F WD PE ı is analytic on U .

We then have, for every open subset X  Rn and u 2 D 0 .X /, that u is analytic on X if P u D 0 on X .

Proof. We fix a point x 2 X and will prove that u is analytic on a neighborhood of x in X . To this end, we select  > 0 such that the open ball B.0I / is contained in U , while B.xI /  X . Furthermore, we consider  2 C01 .B.xI // satisfying  D 1 on an open neighborhood of x in X . Then we have P .u/ D f with f 2 E 0 .Rn / and V supp f  supp D  B.xI / WD B.xI / n fxg: (12.14) We obtain E  f D E  P .u/ D PE  .u/ D F  .u/ C u; which shows that it is sufficient to verify (i) x … sing supp anal .E  f /, (ii) x … sing supp anal .F  .u//. (i). On account of Lemma 12.14 and (12.14) we have V sing supp anal .E  f /  sing supp anal E C supp f  sing supp anal E C B.xI /: V / D ; on the strength of condition (a); therefore 0 … Now sing supp anal E \ B.0I V /, which implies x … sing supp anal E C B.xI V sing supp anal E C B.0I /. In other words, (i) is valid. (ii). We note that again on account of Lemma 12.14, sing supp anal .F  .u//  sing supp anal F C supp   sing supp anal F C B.xI /:

146

12 Fundamental Solutions

Furthermore, condition (b) gives sing supp anal F \ B.0I / D ;, which leads to 0 … sing supp anal F CB.0I /; hence x … sing supp anal F CB.xI / and we conclude that (ii) holds too. 

Problems 12.1. Determine all fundamental solutions Ek 2 D 0 .R/ of P D @k , for k 2 N. Which of these are homogeneous? Of what degree? P  12.2. (Dipoles.) The distribution j vj @j ıa is said to be a dipole at a point a with dipole vector v. Calculate the potential in R3 of the dipole at the origin, with dipole vector equal to the first basis vector. Sketch the corresponding level curves in the .x1 ; x2 /-plane. What do the level curves of the potential of ı look like? Also work out this problem for the potential in R2 . See Fig. 12.1.

Fig. 12.1 Illustration for Problem 12.2. Equipotential curves in the .x1 ; x2 /-plane in R3 and in R2 , respectively

 12.3. (Potential of a bar.) Calculate the potential u in R3 of the distribution f Z a defined by f ./ D .x1 ; 0; 0/ dx1 . 2 C01 .R3 //; a

a bar of length 2a. Indicate where the distribution u is harmonic. Is u a (locally integrable) function? Determine sing supp u. How does u behave as a ! 1?

 12.4. (Green’s formula and Dirichlet’s problem.) Suppose E 2 D 0 .Rn / is a fundamental solution of the Laplace operator
and consider a harmonic function u on the open subset X of Rn . Let U be an open and U a compact subset of X . Then prove that v WD
.u 1U / is a distribution on X with supp v  @U , the boundary of U . Also prove that v 2 E 0 .X / and u 1U D E  v.

12 Problems

147

Now assume that the boundary @U is of class C 1 and denote the outer normal to the boundary at the point y 2 @U by .y/. For a C 1 function f on a neighborhood of @U , the normal derivative @ f of f is defined by n X @ f .y/ WD j .y/ @j f .y/ .y 2 @U /: j D1

This is a continuous function on @U . Now prove Green’s formula Z
u.x/ D u.y/ @ .y 7! E.x y// E.x y/ @ u.y/ dy @U

.x 2 U /I

and furthermore, that the right-hand side vanishes for x 2 X n U . Observe that this formula expresses u in U in terms of u and @ u on the boundary @U . However, that is not the whole story, for it is known that u is completely determined on U by uj@U . This is the so-called Dirichlet problem, which is not discussed in the present book; see, however, Example 16.15 and Problems 14.38, 14.54, 16.14, and 18.8. 12.5. (Mean Value Theorem and maximum principle.) Let the notation be as in Problem R @u 12.4. Apply Green’s second identity with u and with v D 1 to obtain @U @ .y/ dy D 0. Next consider x 2 X and r > 0 such that B.xI r/  X and apply Green’s formula from Problem 12.4 with U D B.xI r/ and E as in (12.3) to obtain the following Mean Value Theorem for harmonic functions: Z 1 u.x/ D u.y/ dy: voln 1 [email protected] r// @B.xIr/ Z Deduce 1 u.x/ D u.y/ dy: voln .B.xI r// B.xIr/ Derive the maximum principle: if u is a harmonic function on X , then u cannot have a local maximum or minimum in X . More precisely, if u.x/ T u.y/ for all y 2 B.xI r/, where r > 0, then u is constant on the component of X containing x.  12.6. Let P be a harmonic polynomial function on Rn and x 2 Rn . Verify Z 1 n 2 P .x C y/ e 2 kyk dy D .2/ 2 P .x/: Rn

 12.7. (One-dimensional wave operator.) Define the open set V D f .x; t/ 2 R2 j jxj < t g, see Fig. 12.2. Compute a constant a 2 C such that E D a 1V is a fundamental solution of the wave operator  D @ t 2 @x 2 . Hint: for the computation one has to evaluate an integral. The evaluation of this integral can be simplified in several ways. One possibility is to apply the change of variables .x; t/ D .y/, where is the rotation in R2 about the origin by the angle =4. Alternatively, one can apply Green’s Integral Theorem (see [7, Theorem 8.3.5]). In this case, given a test function , determine a vector field v such that   D curl v D @x v2 @ t v1 .

148

12 Fundamental Solutions

t

V

x Fig. 12.2 Illustration for Problem 12.7. The set V

Determine supp E and sing supp E. Show that u D E  f is a well-defined distribution on R2 if the support of f 2 D 0 .R2 / is contained in a half-space of the form H D f .x; t/ j t  t0 g, for some t0 2 R. Prove that u is a solution of the inhomogeneous wave equation  u D f . 12.8. (Three-dimensional wave operator.) Denote the coordinates in R4 by .x; t/, where x 2 R3 and t 2 R, and the wave operator by  D @ t 2
x . Let u be the characteristic function of V D f .x; t/ j kxk < t g, the interior of the forward light cone. Further, let v D  u and w D  v. Write q.x; t/ D t 2 kxk2 and P D f .x; t/ j t > 0 g, the positive half-space. Verify that u D q  H on P . Use Problem 10.20 to determine v and w in P ; describe v. Prove that v and w are homogeneous distributions on Rn and determine their degrees. Prove that w D c ı for some constant c 2 C. Determine c by testing with the function t 7! e t , and justify this procedure. Calculate a fundamental solution E of  . If  W .x; t/ 7! .x; t/, then prove that   .E/ is also a fundamental solution of  and that U D E   .E/ satisfies the wave equation  U D 0. Determine supp U and sing supp U .  12.9. Prove the following generalization of Problem 11.5. Let X be a connected open subset of Rn and suppose u 2 D 0 .X / satisfies @j u D 0, for all 1  j  n. Prove the existence of a constant c 2 C such that u D c 1X .  12.10. (Distribution invariant under positive dilations and rotations.) Let u 2 D 0 .Rn / be invariant under positive dilations and rotations. Verify that u equals a constant function on Rn .

 12.11. Assume that P and Q are hypoelliptic operators and show in that case that the composition P B Q is also a hypoelliptic operator. Calculate the composition of @z and @zN . Use this to determine, from the fundamental solution (12.3) of
, a fundamental solution of @z , and also of @zN .  12.12. (Determining an antiderivative if integrability conditions are satisfied.) Let E be a fundamental solution of
in Rn and let gj , for 1  j  n, be distributions

12 Problems

149

with compact support in Rn , with the property that @j gk D @k gj for all 1  j; k  n. Prove that the distribution n X f D @j E  gj j D1

satisfies the system of differential equations @j f D gj

.1  j  n/:

(The problem with the usual method of obtaining f , namely by integration of the gj along lines, is that distributions on Rn cannot in all cases be restricted to curves.) Now prove that, for arbitrary distributions gj in Rn , there exists a distribution f in Rn with @j f D gj if and only if @j gk D @k gj , for all 1  j; k  n. Hint: in the formula above for f , replace the distributions gj by  gj , where  denotes a cut-off function.  12.13. (Helmholtz’s equation). Let k 2 R0 and define

E˙ 2 C 1 .R3 n f0g/

E˙ .x/ D

by

e ˙i kkxk : kxk

(i) Prove that E˙ 2 D 0 .R3 /. Begin by writing, for x 2 R3 n f0g, E˙ .x/ D cos kkxk

sin kkxk 1 1 ˙i DW g.x/ ˙ i h.x/; kxk kxk kxk

and show that both functions g and h belong to C 1 .R3 /. 1 E˙ is a fundamenNext one is asked to show, by two different methods, that 4 tal solution of Helmholtz’s differential equation with parameter k, or, differently phrased, that in D 0 .R3 / one has

.
C k 2 /E˙ D 4ı:

(12.15)

The first approach is a verification that the distribution E˙ does in fact satisfy (12.15); the second one is an actual construction of the solution. (ii) Verify the following equalities in D 0 .R3 /: grad g.x/ D

k h.x/ x;

grad

 1  .x/ D k
k


g D k.k g C 2h/:

1 x; kxk3

Prove that in addition, one has in D 0 .R3 /,  1  D  1 E  1  1 D g
C 2 grad g; grad C
g
g k
k k
k k
k k
k

150

12 Fundamental Solutions

D

k2 g

1 k
k

4 ı;

and use these results to prove (12.15). Now the second method. Suppose f 2 C 1 .R3 n f0g/ to be a radial function, that is, there exists f0 2 C 1 .R>0 / having the property f .x/ D f0 .kxk/. (iii) Evaluate
f in terms of derivatives of f0 . (iv) Further assume that .
C k 2 /f D 0. Determine the differential equation satisfied by fz0 2 C 1 .R>0 /, where fz0 .r/ D rf0 .r/, and use it to prove, for a and b 2 C, f .x/ D a

sin kkxk cos kkxk Cb kxk kxk

.x 2 R3 n f0g/:

(v) Use Green’s second identity (see for instance [7, Example 7.9.6]) to show that in D 0 .R3 /, .
C k 2 /f D 4a ı: (vi) Conclude that every rotation-invariant fundamental solution of Helmholtz’s equation is given by X X c˙ E˙ ; where c˙ 2 C; c˙ D 1: ˙

˙

 12.14. (Boundary values of complex-analytic functions.) In this problem we con1 struct generalizations of the distributions x˙i 2 D 0 .R/ from (1.5) and (1.6). We 0 need some notation. If U is an open subset of C, denote by O.U / the linear space of complexdifferentiable functions on U . Write HC D f z D x C iy 2 C j y > 0 g. For N 2 Z0 , a < b, and 0 < h, introduce a seminorm on O.HC / and linear subspaces, respectively, by

nN;a;b;h .f / D supf y N jf .x C iy/j j a  x  b; 0 < y  h g;

ON .HC / D f f 2 O.HC / j nN;a;b;h .f / < 1; for all a < b; 0 < h g; [ O .HC / D ON .HC /: N 2Z0

Furthermore, let G D f z 2 C j z ¤ 0 and  < arg z <  g and define the (principal) branch log W G ! C of the logarithm by putting log z D log jzjCi arg z. Denote the restriction of log to HC by logC 2 O.HC /. (i) Prove that both logC and z 7!

1 z

belong to O1 .HC /.

For every  2 C 1 .R/ and N 2 Z0 , introduce zN 2 C 1 .C/ and R;N 2 C 1 .R/, respectively, by

12 Problems

151

zN .x C iy/ D

N X  .k/ .x/ .iy/k kŠ

and

kD0

R;N .x/ D

 .N C1/ .x/ : 2NŠ

Here zN is the N th-order Taylor polynomial in y, which an analytic extension of  to HC would have if it did exist.

(ii) Demonstrate that @zN zN .z/ D R;N .x/ .iy/N , for all z 2 C. (iii) Show that for every 2 C 1 .HC / and f 2 O.HC / we have @zN . f / D .@zN /f . (iv) Select a < b and c < d such that R D Œ a; b   Œ c; d   HC . Verify, for all g 2 C 1 .HC /, Z Z g.z/ dz D 2i @zN g.x C iy/ dx dy: @R

R

From now on, suppose that f 2 ON .HC / and write f  .z/ WD f .z C i /, for  > 0. (v) Let a < b and 0 < h. Prove that for all  2 C01 .R/ with supp    a; b Œ , Z

b a

.x/ f  .x/ dx D

Z

b a

zN .x C ih/ f  .x C ih/ dx C2i

(vi) Demonstrate that for every  2

Z

h 0

Z

b a

C01 .R/

ˇC .f /./ WD lim #0

Z

R

R;N .x/ .iy/N f  .x C iy/ dx dy: the limit

.x/ f .x C i / dx

exists. In fact, prove that for any h > 0, Z N X .ih/k ˇC .f /./ D  .k/ .x/ f .x C ih/ dx kŠ R kD0 Z Z 1 .ih/N C1 C  .N C1/ .x/ t N f .x C i th/ dt dx: NŠ 0 R Deduce that ˇC .f / is an element of D 0 .R/ of order at most N C 1.

The preceding results lead to a linear operator ˇC W O .HC / ! D 0 .R/, which 1 is called the boundary value map. Note that ˇC . z1 / D xCi 0 . In analogy with this result we also write f .x C i 0/ instead of ˇC .f /. (vii) Verify

1 xCi 0

D PV x1

 i ı.

By means of Cauchy’s integral formula it is not difficult to prove that the complex differentiation @z maps ON .HC / into ON C1 .HC /. As a consequence, @z defines a linear operator from O .HC / into itself.

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12 Fundamental Solutions

(viii) Establish the identity of operators ˇC B @z D @x B ˇC on O .HC /.

Similarly as above, define H D f z D x C iy 2 C j y < 0 g, the corresponding boundary value map ˇ W O .H / ! D 0 .R/, as well as log . (ix) Demonstrate ˇC .logC / ˇ .log / D 2 i SH , where H denotes the Heaviside function and deduce 1 .x C i 0/k

1 . 1/k .k ı D 2 i .k 1/Š .x i 0/k

1/

.k 2 N/:

Here we have introduced, for x C iy 2 G and a 2 C, .x C iy/a D e a.log jxCiyjCi arg.xCiy// : For every a 2 C we thus obtain a complex-analytic function on G. Note that we have obtained the Plemelj–Sokhotsky jump relations (see Example 14.30 or Problem 1.3 for other proofs in the case of k D 1). Furthermore, observe that the jump relations do not depend on the choice of the argument of the complex logarithm. (x) Prove, for x 2 R and a 2 C, .x ˙ i 0/a D lim .x C iy/a D x a H.x/ C e ˙i a . x/a H. x/: ˙y#0

(xi) Define p W RH !C

by

p.r; z/ D r 2

z2 :

Demonstrate that p takes its values in G. Introduce, for every r 2 R and a 2 C, fr;a W H ! C

satisfying

fr;a .z/ D p.r; z/

a 2

:

For Re a  0 and x 2 R, show that fr;a .x i 0/ WD lim#0 fr;a .x i / exists and defines a distribution on R. In the case of Re a > 0, prove that fr;a .x

i 0/ WD ˇ .fr;a / 2 D 0 .R/

is well-defined. Give an estimate for the order of this boundary value and verify that its restriction to R n f˙rg equals the real-analytic function ( a jxj < jrjI .r 2 x 2 / 2 ; fr;a .x i 0/ D  a e i 2 a .x 2 r 2 / 2 ; ˙x > jrj:

Chapter 13

Fractional Integration and Differentiation

d In this chapter we deal with “complex powers of the operator dx ;” a concept already found in the posthumous works of Riemann and elaborated by Marcel Riesz in the 1930s and 1940s. The relevant article [18] is lengthy, but with the help of a little distribution theory and complex analysis, all results can readily be proved. We will also deal with Riesz’s treatment of the wave operator  D @2t
x in arbitrary dimension; thus we will obtain, among other things, a fundamental solution of  .

13.1 The Case of Dimension One Let D 0 .R/C be the space of distributions u 2 D 0 .R/ such that there exists l 2 R with supp u  Œ l; 1 Œ. For u and v 2 D 0 .R/C the convolution product u  v 2 D 0 .R/C is well-defined; see Problem 11.13 or Theorem 11.17. Every u 2 D 0 .R/C possesses a uniquely determined kth-order antiderivative in D 0 .R/C , that is, v 2 D 0 .R/C with @k v D u, for k 2 N. This is obtained via the definition (compare with [7, Exercise 2.75]), with H denoting the Heaviside function, v D kC  u;

where

kC .x/ D

xk 1 H.x/ .k 1/Š

.x 2 R/:

Indeed, by mathematical induction on k 2 N one gets @k kC D ı, since on account of the Leibniz rule and Examples 4.2 and 9.1,  xk  xk 1 xk xk 1 H D H C ı D H D kC : @kC1 D @ C kŠ .k 1/Š kŠ .k 1/Š In other words, finding kth-order antiderivatives is equal to convolution with the function kC . Conversely, kth-order differentiation can also be regarded as a convolution operator, namely, as convolution with the distribution ı .k/ . It was Riesz who discovered that these operators can be embedded in an analytic a way into a family of convolution operators IC D aC  depending on a 2 C. Here J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_13, © Springer Science+Business Media, LLC 2010

153

154

13 Fractional Integration and Differentiation

the aC have the following properties, for all a and b 2 C: aC 2 D 0 .R/ kC .x/ D 

bC

k 1

x H.x/ .k 1/Š

Ck D ı .k/ aC

and

D

supp aC  R0 ; .k 2 N/;

.k 2 Z0 /;

(13.1) (13.2) (13.3)

aCb C :

(13.4)

a By (13.1), the operator IC of convolution with aC is a continuous linear operator k is equal to finding a kthfrom D 0 .R/C to D 0 .R/C . On the strength of (13.2), IC 0 order antiderivative in D .R/C , while according to (13.3), IC k is equal to kth-order aCb a b differentiation. Finally, (13.4) implies the validity of the group law IC B IC D IC . 0 Because 0C D ı, one has IC D I , the identity in D 0 .R/C . On account of the group a law with b D a we also deduce from this that IC is bijective from D 0 .R/C to a D 0 .R/C , with inverse IC a . If one describes IC as “finding antiderivatives of order a 2 C,” then IC a could be called “differentiation of order a.” The family .aC /a2C forms a complex-analytic family of distributions, in the sense that a 7! aC ./ is a complex-analytic function on C, for every  2 C01 .R/. In this case one also uses the term distribution-valued complex-analytic function. This property will be used to deduce the validity for all a 2 C of numerous identities involving aC from the validity on an arbitrary nonempty open subset U of C. Thus, we are free to choose U such that the desired identity can be verified on U by direct calculation. This procedure involves the following principle. If f and g are analytic functions on a connected open subset V of C, one has f D g on V whenever f D g on a nonempty open subset U of V . This follows by applying Lemma 2.6 to  D f g. The functions f and g may be taken as the left-hand side and the right-hand side, or vice versa, of the identity involving aC , after testing with an arbitrary test function. This is called the principle of analytic continuation of identities.

We now give the definition of the family aC , followed by a summary of the properties. The starting point is the definition aC .x/ D

xa 1 H.x/ .a/

.x 2 R/:

(13.5)

Here x c D e c log x for x > 0 and c 2 C. The function aC .x/ is locally integrable on R if and only if Re a > 0; for these values of a we interpret aC as an element of D 0 .R/. The factor .a/ in the denominator is in terms of Euler’s Gamma function, see the appendix in Sect. 13.3 below and in particular, Corollary 13.6. This is a complexanalytic function on C n Z0 without zeros. For every k 2 Z0 , has a simple pole at k, with residue . 1/k =kŠ. Consequently, 1= possesses an extension to a complex-analytic function on C, with zeros only at the points k, with k 2 Z0 ,

13.1 The Case of Dimension One

155

and derivative . 1/k kŠ at those points. Further, .k/ D .k 1/Š for k 2 N, which means that (13.2) is satisfied. Integration by parts and the formula .a C 1/ D a .a/ imply the following identity in D 0 .R/: @aC1 D aC .Re a > 0/: (13.6) C This enables us to define aC 2 D 0 .R/ for every a 2 C, by means of aC D @k aCk C

.k 2 Z0 ; Re .a C k/ > 0/:

(13.7)

The right-hand side does not depend on the choice of k, while for Re a > 0 the definition is identical to the aC that we have defined above. This also implies that (13.6) holds for all a 2 C. Furthermore, for every  2 C01 .R/, the following function is complex-analytic on C: .k/ a 7! aC ./ D . 1/k aCk /; C .

as follows by interchanging the Cauchy–Riemann operator from (12.12) and integration. For every  2 C01 .R/ with supp   R 0; it follows by analytic continuation that this identity holds for all a 2 C; this proves (13.1). In the same way, analytic continuation yields the validity of (13.5) for all a 2 C. From the analytic continuation of (13.6) and Example 4.2 we obtain 0C D @1C D @H D ı:

(13.8)

By repeated differentiation of this we now get (13.3). From (13.5) we see that sing supp aC D f0g

(13.9)

if a … Z0 . For a 2 Z0 , (13.5) implies that supp aC  f0g; in this case we conclude again that (13.9) is valid, using (13.3). Because aC has order 0 if and only if Re a > 0, we see that the order of aC equals k if k < Re a  k C 1, with k 2 N. This can also be formulated by saying that the singularity of aC at 0 becomes worse and worse as Re a ! 1. If Re a > 0 and Re b > 0, we have, for x > 0, Z x .a/ .b/ .aC  bC /.x/ D y a 1 .x y/b 1 dy D x aCb

1

Z

0

1

ta 0

1

.1

t/b

1

dt D

.a C b/ B.a; b/ aCb C .x/:

Here B is Euler’s Beta function; see (13.34) below. In view of (13.33), this implies that (13.4) holds if Re a > 0 and Re b > 0. The identity (13.4) now follows by analytic continuation (after testing) for all a 2 C, for given b 2 C with Re b > 0,

156

13 Fractional Integration and Differentiation

and then also, by analytic continuation with respect to the variable b, for all a 2 C and b 2 C. By analytic continuation one also proves the following identity in D 0 .R/: x aC D a aC1 C

.a 2 C/:

Combining this with (13.6), we now obtain x @aC D .a

1/ aC :

(13.10)

In other words, for every a 2 C we conclude, on the strength of Theorem 10.17, that aC is a distribution homogeneous of degree a 1. a Marcel Riesz called IC .u/ D aC  u a Riemann–Liouville integral of u; he had encountered it in his work on so-called Ces`aro means. For suitable functions  and a in an appropriate subset of C, one writes Z M.a/ WD .a/ aC ./ D x a 1 .x/ dx; R>0

where M W  7! M is said to be the Mellin transform; see Problem 15.6 below 1 for more details. For the Mellin transform of x 7! e x , sin and cos, and x 7! xC1 , a see (13.30), Problem 13.13, and (16.32), respectively. In view of this, the C are sometimes called Mellin distributions or Mellin kernels. Fundamental solutions are a useful tool in proving results concerning the structure of distributions. Theorem 13.1. Let X be an open subset of Rn . For every u 2 D 0 .X / and bounded open subset U of X , there exists f 2 C.X / such that the restriction of u to U equals a derivative of finite order of f . Proof. Indeed, for k 2 Z0 , define Ek D kC ˝


˝ kC 2 D 0 .Rn / and the partial differential operator z @k WD @.k;:::;k/ in Rn . Then (13.6) and (13.8) lead to z @k Ek D ı 2 D 0 .Rn /;

while

Ek 2 C k

2

.Rn /

.k  2/:

@k . Next, select  2 C01 .X / In other words, Ek is a fundamental solution of z satisfying  D 1 on a neighborhood of U . Then u D u on U . Because u 2 E 0 .X /  E 0 .Rn /, it is of finite order, say k 2 Z0 , according to Theorem 8.8. From Theorem 11.17 it follows that u D u  @zkC2 EkC2 D @zkC2 .u  EkC2 / 2 D 0 .Rn /: It remains to prove that u  EkC2 is a continuous function. Define  as in Lemma 2.19 and apply Theorem 11.2 to get f WD .u  EkC2 /   2 C 1 .Rn /:

13.2 Wave Family

157

On account of (11.22), the associative law applies, and so application of (11.1) with u 2 E 0 .Rn / and EkC2   2 C 1 .Rn / gives f .x/ D u  .EkC2   /.x/ D u.Tx B S.EkC2   //:

(13.11)

As in (2.6) one shows that the EkC2   converge to EkC2 in C k .Rn / as  # 0. Since u is of order k, application of (8.3) to the right-hand side of (13.11) implies that the continuous functions f converge, uniformly on compact sets, to the function f defined by x 7! u.Tx B SEkC2 / as  # 0. Hence f is also continuous on X . On the other hand, by Lemma 11.6 the f converge to uEkC2 in D 0 .X / as  # 0. It is clear that the limits are the same in this case, so that one has uEkC2 D f in C.X /. See Example 18.2 for another proof. 

13.2 Wave Family We now describe the version for the wave operator  D @2t


x D @2nC1

n X

@j2

j D1

in RnC1 , whose points we denote by y D .x; t/. Here .x1 ; : : : ; xn / 2 Rn are the position coordinates and t 2 R is the time coordinate. (For generalization to multidimensional time, see Kolk–Varadarajan [15].) All definitions will be given in terms of the corresponding quadratic form q on RnC1 , of Lorentz type, defined by q.y/ D q.x; t/ D t 2

kxk2 D t 2

n X

j D1

xj2 D t .x; t/J.x; t/;

(13.12)

where t denotes the transpose and J is the .n C 1/  .n C 1/ diagonal matrix with coefficients . 1; : : : ; 1; 1/. Further, we denote the interiors CC and C of the (solid) forward and the backward cones by ˚ C˙ D y D .x; t/ 2 RnC1 j q.y/ > 0; t ≷ 0 ; (13.13)

respectively (see Fig. 13.1). Then the boundary @CC of CC is the nappe of the cone @CC D f .x; t/ 2 RnC1 j kxk D t g:

(13.14)

a The point of departure this time is the function RC on an open dense subset of R , defined by ( nC1

a RC .y/

D

c.a/ q.y/ 0

a n 1 2

if

if

y 2 CC ;

y 2 RnC1 n CC :

(13.15)

158

13 Fractional Integration and Differentiation

Fig. 13.1 Illustration for (13.13). CC and C are bounded by the upper and the lower nappes of the cone, respectively

Here c.a/ D cn .a/ is a constant to be determined; see Lemma 13.2 below. Note that a for Re a < n C 1 the function RC is unbounded along @CC . It is locally integrable on RnC1 if and only if Re a > n 1; if that condition is satisfied, we obtain an element of D 0 .RnC1 /. Lemma 13.2. If we choose (for the second equality, see (13.39) below) c.a/ D

/ . aC1 2 

n 2

.a/

we obtain, for Re a > n

. a nC1 / 2

D

1 2a

1 and Re b > n

1



n 1 2

. a2 / . a

nC1 / 2

;

(13.16)

1,

a b aCb RC  RC D RC :

(13.17)

a b Proof. For the calculation of the function R WD RC  RC at y D .x; t/ we have to integrate a b RC .x ; t / RC .; /

over  D .; / 2 CC with y  2 CC . The latter two relations imply that kk <  < t, which means that this set of integration is uniformly bounded whenever t varies over a bounded set; thus, there are no problems with the convergence of the integral. Next, on account of y  2 CC and  2 CC , we see that the sum y belongs to CC , and therefore supp R  CC . And finally, we observe that a change of variables  D  , with  > 0, results in R. y/ D 2m R.y/, where

13.2 Wave Family

2m D .a

159

n

1/ C .b

n

1/ C .n C 1/ D a C b

n

1:

That is, R is homogeneous of degree a C b n 1. The Lorentz group Lo is the collection of all linear transformations A in RnC1 satisfying A q D q; it is a group with composition as the binary operation. For the further determination of R we use the connected component Loı of Lo. This is the subgroup of all A 2 Lo with det A D 1 and A.CC / D CC (see [7, Exercise 5.70]) a and is called the proper orthochronous Lorentz group. It is clear that RC is invariant ı under all A 2 Lo . Conversely: if f is a function on CC that is invariant under Loı and homogeneous of degree 2m, then there exists a constant c 2 C with f D c q m on CC . Indeed, f is constant on all orbits f Ay j A 2 Loı g of points y 2 CC under the action of Loı . It is known that the orbits in CC under the action of Loı are equal to the level surfaces of q in CC (see Fig. 13.2), f y 2 CC j q.y/ D constant g:

Fig. 13.2 Illustration of (13.18). Surfaces of the levels 4, 1, 0, and

(13.18)

1 of q on R3

This implies that there exists a function g on R>0 such that f .y/ D g.q.y// for all y 2 CC . From the homogeneity of f we deduce that g must be homogeneous of degree m on R>0 . For every A 2 Loı , the change of variables  D A  in the integral for R D a b RC  RC yields that R.A y/ D R.y/; in other words, R is invariant under all ı A 2 Lo . Applying the preceding characterization, we find a constant c.a; b/ 2 C such that

160

13 Fractional Integration and Differentiation a b aCb RC  RC D c.a; b/ RC :

(13.19) 0

nC1

/ with For the calculation of c.a; b/ we test both sides of this identity in D .R the function .x; t/ 7! e t in C 1 .RnC1 /. On CC this function decreases fast enough to ensure convergence of the integrals. If we now write Z Z a T .a/ D e t RC .x; t/ dx dt; Rn

R

and note that e t D e t  e  , we find that testing of the left-hand side of (13.19) equals T .a/ T .b/; therefore, c.a; b/ is given by T .a/ T .b/ D c.a; b/ T .a C b/:

p On the other hand, with the changes of variables kxk D r D t s and the notation cn for the Euclidean .n 1/-dimensional volume of the unit sphere in Rn we obtain Z Z a n 1 T .a/ t D e .t 2 kxk2 / 2 dx dt c.a/ R>0 kxk0

cn 2 cn D 2 cn D 2 D

Z

e

t a 1

t

dt

Z

1

.1

s/

a n 1 2

s

n 2 2

ds

0

R>0

n a n C 1 ; 2 2  n   a n C 1 .  a C 1  .a/ : 2 2 2 .a/ B

Here B is Euler’s Beta function, see (13.34) and (13.33). Substituting formula (13.37) for cn , we see that the choice (13.16) means that T .a/ D T .b/ D T .a C b/ D 1; therefore, c.a; b/ D 1.  Next we will define RaC 2 D 0 .RnC1 /, for all a 2 C. To this end, we compute a  RC and apply a method similar to the one in (13.7). In fact, note that q is a submersion from RnC1 n f0g to R; according to Theorem 10.18, therefore, the distribution q  v is well-defined on RnC1 n f0g for every v 2 D 0 .R/. More precisely, comparing (13.5) and (13.15), we obtain, in view of a n2 1 D a nC1 1, 2 a nC1 2

a RC D d.a/ q  C

where

d.a/ D






. aC1 2 / n 2 .a/

on

RnC1 n C ; .Re a > n

1/:

(13.20) (13.21)

The specification “on RnC1 nC ” is necessary because q  cC behaves identically on C and CC . Indeed, the reflection y 7! y leaves q  cC invariant and carries CC a into C . However, RC D 0 everywhere on RnC1 n CC , which contains C n f0g. On

13.2 Wave Family

161

q 1 .  1; 0 Œ /, that is, the complement of CC [ C , equation (13.20) is correct, since both sides vanish there. Furthermore, if aC1 D k with k 2 Z0 , then 2 a D 1 2k 2 Z n C 1/:

2/, the end (13.22)

Initially this equality is valid in D 0 .RnC1 nf0g/, but because the partial derivatives of a a 2 RC vanish along @CC for Re a > n C 1, as does RC , it is also valid in D 0 .RnC1 /. This now enables us to define, for all a 2 C, as we did for the aC , the distribua tions RC by aC2k a RC D  k RC

.k 2 Z0 ; Re a C 2k > n

1/:

a The right-hand side is independent of k, and a 7! RC ./ is a complex-analytic 1 nC1 function on C, for every  2 C0 .R /. By analytic continuation, the identities (13.15), (13.17), (13.20), and (13.22) are now seen to be valid for all a 2 C, in a the distributional sense. Sometimes the RC are called Riesz distributions or Riesz a kernels. With respect to the distribution RC , (13.15) here means a  CC ; supp RC

(13.23)

a while on the open set CC the distribution RC equals the real-analytic function a n 1 c.a/ q 2 . This also implies in the notation of (13.14), a sing supp RC  @CC :

We further mention the equation

(13.24)

162

13 Fractional Integration and Differentiation nC1 X

j D1

a yj @j RC D .a

n

a 1/ RC ;

(13.25)

a which says that for every a 2 C, the distribution RC is homogeneous of degree a n 1. Furthermore, by analytic continuation of identities this is seen to imply a that for every a 2 C, the distribution RC is invariant under the proper orthochronous ı Lorentz group Lo and satisfies a q RC D a .a

aC2 n C 1/ RC :

(13.26)

0 2 Theorem 13.3. RC D ı and EC WD RC is a fundamental solution of  . For n D 1 or n even, one has supp EC D CC , while supp EC D @CC if n > 1 is odd. In all cases, sing supp EC D @CC .

Proof. In (13.21) we have 1.a/ D 0 and . aC1 2 / ¤ 0 at a D 0. Therefore it 0 D 0 on RnC1 n C . follows by analytic continuation of (13.20) to a D 0 that RC 0 By combining this with (13.23), we can conclude that supp RC  f0g. But on the strength of Theorem 8.10 this leads to X 0 RC D c˛ @˛ ı; j˛jm

0 is homogeneous of for some m 2 Z0 and c˛ 2 C. Now (13.25) implies that RC ˛ degree n 1, while @ ı is homogeneous of degree n 1 j˛j according to Problem 10.16. As a consequence we obtain

0 D .n C 1 C

nC1 X

j D1

0 yj @j / RC D

X

j˛jm

c˛ j˛j @˛ ı:

(13.27)

Further, testing with y 7! y ˇ shows that the distributions @˛ ı are linearly independent in D 0 .RnC1 /. This implies that (13.27) can be true only if c˛ D 0, for all 0 ˛ ¤ 0. This now proves that RC D c ı, for some c 2 C. But then (13.17) yields 0Ca a 0 a a a RC D RC D RC  RC D c ı  RC D c RC ; a which implies that c D 1, because there certainly exists an a 2 C such that RC ¤ 0. This proves the first assertion; the second one follows from (13.22) for a D 2. As regards the description of supp EC and sing supp EC , we observe that (13.20) leads to 3 n
1 EC D q  C2 on RnC1 n C : n 1 2 2

13.2 Wave Family

163

If n D 1, we have that 1C D H and therefore EC D 12 1CC , a fundamental solution that we have encountered before, in Problem 12.7. For n D 3 we obtain 1 EC D 2 q  ı on R4 n C ; compare this with Problems 12.8, 13.10, and 13.11. More generally we get, for n D 2k C 3 with k 2 Z0 , 1 EC D q  ı .k/ on RnC1 n C : 2  kC1 From this we can see that supp EC D @CC . If, however, n is even, then we have n/=2 D R0 , and so supp EC D CC . supp .3 C In (13.24) we already established that sing supp EC  @CC . For n odd, the equality of these sets follows from the foregoing, while if n is even, the equality follows from the observation that the 3 2 n th power of q is not differentiable coming from within CC at points y satisfying q.y/ D 0, that is, at points belonging to @CC .  Hadamard referred to the assertion supp EC D @CC for n > 1 odd as the Huygens principle. In view of what Huygens himself wrote about wave propagation, a case can also be made for calling the assertion sing supp EC D @CC , which holds in every dimension, the Huygens principle. Remark 13.4. The fact that supp EC  CC enables us to form EC  u for every u 2 D 0 .RnC1 / such that supp u is contained in the half-space f .x; t/ 2 RnC1 j t  0 g; see the first part of the proof of Lemma 13.2. If, moreover, the distribution u satisfies the wave equation  u D 0, then u D 0. Indeed, u D ı  u D .  EC /  u D  .EC  u/ D EC   u D 0: This implies that EC is the only fundamental solution E with support in the halfspace t  0. Indeed, in that case u WD EC E also has support in t  0, while  u D  EC  E D ı ı D 0, and therefore EC E D u D 0. By elaborating this further we can also obtain the uniqueness of solutions u D u.x; t/ for t > 0 of the Cauchy problem  u D f .x; t/;

lim u.x; t/ D a.x/; t #0

lim @ t u.x; t/ D b.x/; t #0

(13.28)

for a given inhomogeneous term f and given initial functions a and b. To prove the uniqueness, we have to demonstrate that u.x; t/ D 0 for t > 0, in the case that  u D 0 and u.x; t/ and @t u.x; t/ both converge to 0 as t # 0. On the strength of the foregoing we can indeed draw that conclusion if we can show that extension of u by u.x; t/ D 0 for t < 0 yields a distribution u on RnC1 , while additionally, this distribution satisfies  u D 0 on RnC1 . We now consider this Cauchy problem for the partial differential operator  as an d2 initial-value problem for the ordinary differential operator dt
x in the variable 2 t, with integral curves of the form t 7! u t 2 D 0 .Rn /. If we write ut .x/ D u.x; t/, then u t is a function on Rn for every t > 0 and we obtain, for every test function  with support in t > 0,

164

13 Fractional Integration and Differentiation

u./ D

Z

u t . t / dt:

(13.29)

R>0

This leads to the idea of defining a distributional solution of the Cauchy problem (13.28) as a family .ut / t 2R>0 of distributions on Rn with the following properties: (a) t 7! u t . / is of class C 2 , for every 2 C01 .Rn /, d2 (b) dt
x u t D f t , for every t > 0, 2 ut d (c) u t ! a and dt u t ! b in D 0 .Rn /, as t # 0.

Here we assume that .f t / t 2R>0 is a continuous family of distributions on Rn , and that a and b belong to D 0 .Rn /. We obtain uniqueness for this Cauchy problem if f t  0 and a D b D 0 imply that u t  0. In the first instance one may think of C 2 functions u, instead of distributions. The identity (13.29), now for all  2 C01 .RnC1 / and without the restriction t > 0 for the support of , defines a distribution u on RnC1 , with support in the half-space t  0. This can be demonstrated by means of the principle of uniform boundedness, while also using the continuity of t 7! ut W R0 ! D 0 .Rn /. All we have to do is show that  u D 0. But this follows from Z 1  2  d . u/./ D u. / D lim ut  t
 t dt 2 s#0 s dt Z 1 2  d    d d s C us .s / C . u
u / dt D 0: D lim us t t t s#0 ds ds dt 2 s For C 2 functions this is clear; for the justification in the distributional context, the principle of uniform boundedness is invoked once more. An existence theorem for the Cauchy problem is given in Example 18.7 below, by means of Fourier transform. ˛

13.3 Appendix: Euler’s Gamma Function Euler’s Gamma function is defined by Z .a/ WD

e

t a 1

t

dt:

(13.30)

R>0

This is an absolutely convergent integral if a 2 C and Re a > 0. Differentiation with respect to a under the integral sign leads to the conclusion that satisfies the Cauchy–Riemann equation in the complex right half-plane, and therefore defines a complex-analytic function on that open set. d By writing e t D dt e t and integrating by parts, we obtain .a/ D .a

1/ .a

1/

.Re a > 1/:

(13.31)

13.3 Appendix: Euler’s Gamma Function

Because

165

.1/ D 1, it follows by mathematical induction on k that .k/ D .k

.k 2 N/:

1/Š

Euler’s brilliant idea was now to define, for every a 2 C with a … Z0 , .a/ WD

.a C k/ a.a C 1/


.a C k

.k 2 N; Re a C k > 0/:

1/

(13.32)

Using (13.31), one sees that the right-hand side does not depend on the choice of k, while in the complex right half-plane the definition is identical to the .a/ introduced before. In this way we obtain a complex-analytic extension of the Gamma function to C n Z0 ; see Fig. 13.3. Equation (13.31) holds for all a 2 C where the left-hand side and the right-hand side are both defined. For Re a > 0 and Re b > 0 we can write Z Z .a/ .b/ D e .t Cu/ t a 1 ub 1 dt du D D

Z

Z

R>0

R>0

Z e

R>0 s

s a 1

e

t

t/b

.s

dt ds

0 s

s aCb

1

Z

ds

R>0

1

ra

.a C b/ B.a; b/

1

.1

r/b

1

dr:

0

This has been obtained by first substituting u D s words, .a/ .b/ D

1

t and then t D r s. In other

.Re a > 0; Re b > 0/;

(13.33)

where B is Euler’s Beta function, defined by B.a; b/ D The change of variables t D B.a; b/ D

Z

R>0

u uC1

Z

1

ta

1

.1

 , 2

B.a; b/ D 2 In particular, taking a D b D leads to

1 2

1

dt:

(13.34)

gives

ua 1 du D 2 .1 C u/aCb

With t D cos2 ˛, for 0 < ˛
0

v 2a 1 dv: .1 C v 2 /aCb

(13.35)

we obtain Z

 2

cos2a

1

˛ sin2b

1

˛ d˛:

(13.36)

0

in either one of the preceding formulas immediately

166

13 Fractional Integration and Differentiation

Fig. 13.3 Illustration of (13.32). Graph of j j

 1 2 2

D2

Z

0

 2

d˛ D ;

that is,

This reconfirms (2.14), because of Z 1 Z 1 t 2 D e t dt D 2 e 2 R>0 R>0

x2

1

dx D

2

D

Z

e

p :

x2

dx:

R

A formula for the Euclidean .n 1/-dimensional volume cn of the unit sphere f x 2 Rn j kxk D 1 g in Rn now follows from Z Z n 2 2 e kxk dx D cn e r r n 1 dr 2 D Rn

R>0

13.3 Appendix: Euler’s Gamma Function

D cn

Z

e

t

167

t

n 1 2

R>0

1 t 2

1 2

dt D cn

n : 2

1 2

This yields (see [7, Example 7.9.1 or Exercise 7.21.(viii)] for alternative proofs) n

cn D

22 : . n2 /

(13.37)

With n D 2m even, one has . n2 / D .m/ D .m 1/Š. With n D 2m C 1 odd, one obtains m  1  n 1 1 Y D Cm D 2 : j 2 2 2 j D1

Note that sin 2˛ D ˛/ and apply the change of variables sin2 2˛ D t. This leads to 4 sin ˛ cos ˛ d˛ D p1 dt, and thus we obtain from (13.36) sin 2. 2

2 1 t

Z

 4

.cos ˛ sin ˛/2a 0  1 D 21 2a B a; : 2

B.a; a/ D 4

1

d˛ D 21

2a

Z

1

ta

1

.1

t/

1 2

dt

0

(13.38)

The following duplication formula of Legendre is an immediate consequence (see [7, Exercise 6.53] for a different proof): 1  1 .2a/ D 22a 1 .a/ (13.39) aC : 2 2 By analytic continuation, the formula is valid for a 2 C n 12 Z0 .

Lemma 13.5. For 0 < Re a < 1, one has the reflection formula for the Gamma function  .a/ .1 a/ D B.a; 1 a/ D : sin a

Proof. The first identity follows from .a C b/ D .1/ D 1 if a C b D 1. For 0 < Re a < 1, we obtain from (13.36) Z  2 .a/ .1 a/ D B.1 a; a/ D 2 tan2a 1 ˛ d˛: 0

Set tan ˛ D x. Then .1 C tan2 ˛/ d˛ D dx, and so we obtain from (16.32) below, Z 1 2a 1  x : (13.40) .a/ .1 a/ D 2 dx D 2 1 C x sin a 0 See (14.45) in Problem 14.39 or [7, Exercises 6.58 or 6.59] for other proofs.



168

13 Fractional Integration and Differentiation

Corollary 13.6. We have .a/ .1 a/ D sina , for all a 2 C n Z. For every a 2 C with a … Z0 we have .a/ ¤ 0. If k 2 Z0 , then lima! k .a C k/ .a/ D . 1/k =kŠ. The function 1 possesses an extension to an entire analytic function on C, with zeros only at k, for k 2 Z0 , and with derivative at those points equal to . 1/k kŠ; see Fig. 13.4. 4.076

1.129

-4

-3

-2

-1

Fig. 13.4 Illustration of Corollary 13.6. Graph of the restriction of

1

to a segment in R

Proof. The first assertion follows by analytic continuation. The assertion that has simple poles at the points k for k  0, with residue . 1/k =kŠ at those points, follows from this formula, or directly from (13.32).  Alternatively, for obtaining the analytic continuation and the properties of the poles one may use Z Z 1 Z 1 .a/ D e t t a 1 dt D e t t a 1 dt C e t t a 1 dt 0

R>0

D

X

k2Z0

k

. 1/ C kŠ .a C k/

Z

1

1

e

t a 1

t

dt;

1

where the second integral on the right-hand side defines a complex-analytic function.

Problems  13.1. (Abel’s integral equation.) Given 0 < a < 1 and  2 C01 .R>0 /, show that the solution u 2 D 0 .R/C to the integral equation Z x u.y/ .x/ D dy .x y/a 0

13 Problems

169

is given by

sin a u.x/ D 

Z

x 0

 0 .y/ .x y/1

and belongs to C 1 .R>0 /. The special case of a D tion.

1 2

a

dy

is called Abel’s integral equa-

 13.2. For a 2 C with Re a > 0, define a 2 D 0 .R/ by

a ./ D

Z

R 0, .x ˙ i 0/a D aC1 C e ˙i a aC1 : C .a C 1/

Derive this equality for every a 2 C and conclude that each of a 7! .x ˙ i 0/a is a complex-analytic family of distributions on R.  13.4. For a 1 x ..x/

1 < Re a < 0 and  2 C01 .R/, prove that the function x 7! .0// is integrable on R0 and that aC ./ D

Z

R>0

xa 1 ..x/ .a/

.0// dx: a 1

In particular, for these values of a it is not true that aC D test x .a/ H . More generally, show that for k 2 Z0 and k 1 < Re a < k, aC ./

D

Z

D

R>0

xa 1  .x/ .a/

 1 lim .a/ #0

Z

1 

xa

k X xj

j D0 1

jŠ

  .j / .0/ dx

.x/ dx C

k X  .j / .0/ aCj   : j Š .a C j /

j D0

Conclude that upon restriction to the linear subspace L  C01 .R/ consisting of the  2 C01 .R/ with 0 … supp , one actually has aC ./ D

Z

R>0

xa 1 .x/ dx; .a/

that is,

 ˇ x a 1 ˇˇ aC ˇL D test H ˇ : L .a/

170

13 Fractional Integration and Differentiation

Verify that the last identity holds for all a 2 C.

a 2 D 0 .R/, for k 13.5. Similarly to Problem 13.4, introduce x˙ with k 2 Z0 , by setting a x˙ ./

D

Z

 x .˙x/

R>0

  .j / .0/ .˙x/j dx jŠ

k X1

a

j D0

1 < Re a
0

X

.˙x/

2

k X1

j D0

˙

 .2j / .0/ 2j  x dx .2j /Š

. 2 C01 .R//:

In particular, for k 2 Z0 even, deduce that jxj

k

./ D

Z

x R>0

k

X

k

.˙x/

2

2 1 X  .2j / .0/

j D0

˙

 13.6. Prove @x PV x1 D jxj

2

 x 2j dx

.2j /Š

. 2 C01 .R//:

in D 0 .R/ in the notation of Problem 13.5.

13.7. Let a 2 C, k 2 Z, and 0 < Re a < k. Prove that for every f 2 C k .R/ with supp f  Πl; 1 Πthere exists exactly one continuous function u in R such that supp u  Πl; 1 Πand Z x .x y/a 1 u.y/ dy D f .x/ .x  0/: l

Derive a formula for u.x/ in terms of f , for every x  0. How does u D uj behave as j ! 1, if f D fj converges to the Heaviside function in D 0 .R/C ? 13.8. Prove

RC2k D  k ı

.k 2 Z0 /:

a  @CC ? For what a 2 C does one have that supp RC

 13.9. Consider the distribution

 WD Prove EC D

1 .1

n/

q

Is  homogeneous?

if

n > 1;

d a ˇˇ R ˇ : da C aD0 and

nC1 X

j D1

yj @j  D

.n C 1/  C ı:

13 Problems

171

 13.10. (Wave operator.) In this problem, we directly compute a fundamental solution in D 0 .R4 / of the wave operator  D @2t
x , where .x; t/ 2 R3  R ' R4 . As usual, we suppose that q W R4 ! R is given by q.x; t/ D t 2 kxk2 ; write X D R4 n f0g.

(i) Let  2 C01 .X / and prove by means of Problem 10.23.(ii) that ˇ Z Z pkxk2 Cy d ˇˇ  .q ı/./ D .x; t/ dt dx p dy ˇyD0 R3 kxk2 Cy D

X1Z ˙

2

R3

X .x; ˙kxk/ dx DW ı˙ ./: kxk ˙

(ii) Verify that the reflection in the hyperplane f .x; 0/ 2 R4 j x 2 R3 g transforms each of the distributions ı˙ 2 D 0 .X / into the other. Use positivity to show that ıC is a Radon measure on X having the forward cone @CC as its support, and prove that ıC can be extended to a Radon measure on R4 .

The next goal is a verification that  ı˙ D 2 ı in D 0 .R4 /.

(iii) Write q for the variable in R and in particular, v 0 D @q v for v 2 D 0 .R/. Show that we have the following identity in D 0 .X / (compare with Problem 10.20.(ii)): .q  v/ D 4q  ..q @q C 2/v 0 /: Conclude by means of the homogeneity of ı 2 D 0 .R/ that  .q  ı/ D 0 on X . The supports of ıC and ı being disjoint, we obtain  ı˙ D 0 on X . (iv) Next prove the existence of k 2 Z0 and c˛ 2 C such that in D 0 .R4 /, X c˛ @˛ ı:  ıC D j˛jk

Show that ıC is homogeneous of degree 2 and conclude that  ıC is homogeneous of degree 4. Verify that @˛ ı is homogeneous of degree 4 j˛j. Deduce the existence of c 2 C satisfying  ıC D c ı. (v) Verify that the definition of ıC ./ also applies with  2 C 1 .R4 / for which the support intersects @CC in a compact set. Hence, choose  of the form .x; t/ D .t/ with 2 C01 .R/. Now use spherical coordinates in R3 in order to prove c D 2. Conclude that Z 1 .x; kxk/  EC D ı dx: if EC ./ D 4 R3 kxk (vi) Given f 2 C01 .R4 /, verify that a solution u 2 C 1 .R4 / of the inhomogeneous wave equation  u D f is provided by the retarded potential Z 1 f .y; t kx yk/ u.x; t/ D dy: 4 R3 kx yk

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13 Fractional Integration and Differentiation

13.11. In the case of n D 3 and n D 2, according to (13.24) a fundamental solution EC for the wave operator is given on RnC1 n C by, respectively, EC D

1  q ı 2

and

1 1 2 EC D p q  C : 2 

We will derive more explicit forms of these fundamental solutions. Actually, with n D 3 and the sphere S.t/ D f x 2 R3 j kxk D t g we have, for all  2 C01 .R4 /, Z Z 1 EC ./ D .x; t/ dx dt: (13.41) R>0 4 t S.t / With n D 2 and the solid disk B.t/ D f x 2 R2 j kxk < t g we have, for all  2 C01 .R3 /, Z Z 1 .x; t/ EC ./ D p dx dt: (13.42) 2 R>0 B.t / t 2 kxk2

First we consider the case of n D 3. (i) Prove

1X q .0/ D 2 ˙

Z

R3

4

X .x; ˙kxk/ dx DW ı˙ ./: kxk ˙

Hint: use that q W R n f0g is a submersion and note that q while @C˙ D graph.h˙ / with h˙ W R3 ! R

defined by

1

.f0g/ D

S

˙

@C˙ ,

h˙ .x/ D ˙kxk:

Next p apply [7, Special case in Sect. 7.4.III]. Finally, note that k grad q.x; t/k D 2 2 kxk, for .x; t/ 2 q 1 .f0g/. (ii) Show that ı˙ 2 D 0 .R4 / is homogeneous of degree 2 by verification of the definition, by application of a result in the theory and in a problem, respectively. Prove that ı˙ is a Radon measure invariant under Loı and supported by @C˙ . (iii) Verify (13.41) by introducing spherical coordinates in R3 . Next we derive (13.42) from (13.41) by means of Hadamard’s method of descent. In this method solutions of a partial differential equation are obtained by considering them as special solutions of another equation that involves more variables and can be solved. Accordingly, consider  2 C01 .R3 / and define z 2 C01 .R4 / by z x3 ; t/ D .x; t/ .x;

.x 2 R2 ; x3 2 R; t 2 R/:

(iv) The sphere S.t/ minus its equator is the union of the graphs of the functions p given by h˙ .x/ D ˙ t 2 kxk2 : h˙ W B.t/ ! R Imitate the method from [7, Example 7.4.10] to prove

13 Problems

173

Z

S.t /

z t/ dx D 2t .x;

Z

B.t /

Now prove (13.42).

.x; t/ p dx t 2 kxk2

.t 2 R>0 /:

Formula (13.42) can also be proved directly as follows. (v) For p > 0, show that F .p/ WD q

1

.fp 2 g/ \ CC D f .x; t/ 2 R3 j t 2

kxk2 D p 2 ; t > 0 g;

which is one sheet of a two-sheeted hyperboloid. Deduce F .p/ D graph.h.p;
//, where p h W R  R2 ! R is given by h.p; x/ D p 2 C kxk2 : Derive, for  2 C01 .R3 n C /,

q .p 2 / D

1 2

Z

R2

.x; h.p; x// dx: h.p; x/

(vi) Using a change of variables in R>0 , verify that for 2 C01 .R/, Z 1 1 1 2 p C . /D .p 2 / dp:  R>0 2  (vii) Combining parts (v) and (vi), deduce, for  2 C01 .R3 n C /, Z Z 1
1 .x; h.p; x// 1  2 dp dx: p q C ./ D 2 2 h.p; x/ 2  R R>0

(viii) p Next introduce the change of variables h.p; x/ D t in R>0 , that is, p D t 2 kxk2 . Then p > 0 implies t > kxk. Deduce (see Fig. 13.5) Z Z 1 1 .x; t/ EC ./ D p dt dx: 2 2 R kxk t 2 kxk2 Furthermore, interchange the order of integration to obtain (13.42).

Finally we give an application of the results obtained. (ix) Given f 2 C01 .R4 /, verify that a solution u 2 C 1 .R4 / of the inhomogeneous wave equation  u D f is provided by the retarded potential Z 1 f .y; t kx yk/ u.x; t/ D dy: 4 R3 kx yk 13.12. (Lorentz-invariant distribution on RnC1 nf0g is a pullback under the quadratic form q.) Let q W RnC1 ! R be the quadratic form from (13.12) and define

174

13 Fractional Integration and Differentiation ÈÈ x ÈÈ

t

Fig. 13.5 Domain of integration in Problem 13.11.(viii)

˚˙ W Rn  R≷0 ! RnC1 WD f z 2 RnC1 j C

Pn

2 j D1 zj

C znC1 > 0 g

˚˙ .y/ D .y1 ; : : : ; yn ; q.y// D z: qP n 2 Prove that z 7! .z1 ; : : : ; zn ; ˙ j D1 zj C znC1 / are the inverses of ˚˙ and that the ˚˙ are C 1 diffeomorphisms. Show that 0 y1 1 1 0 ::: 0 ynC1 C B B: :: C B :: : C B C t 1 B C: y n .D˚˙ .y/ / D B C B 0 0 : : : 1 ynC1 C B C @ 1 A 0 0 ::: 0 2ynC1 by

Given u 2 D 0 .RnC1 nf0g/, consider its restrictions u˙ to the open subsets Rn R≷0 . Write v˙ D .˚˙1 / u˙ 2 D 0 .RnC1 C /. Derive from Problem 10.7 the following identity, for 1  j  n: 1  ˚˙ .@j v˙ / D .ynC1 @j C yj @nC1 /u˙ in D 0 .Rn  R≷0 /: ynC1 In the particular case of u being invariant under the Lorentz group Lo, deduce from Problem 10.11.(ii) that @j v˙ D 0, for 1  j  n. Next apply Problem 11.19 repeatedly to obtain w˙ 2 D 0 .R/ such that v˙ D 1Rn ˝ w˙ . Denote by ˘ W RnC1 ! R the projection onto the last factor. Then ˘ B ˚˙ D q. Deduce v˙ D 1Rn ˝ w˙ D ˘  w˙ D .˚˙1 / q  w˙ ;

that is,

u˙ D q  w˙ (13.43)

on Rn  R≷0 . In the arguments above, the coordinate ynC1 had a special role. Mutatis mutandis, by means of the equations from Problem 10.11.(ii) the conclusion in (13.43) can be drawn for every coordinate yj , with 1  j  n, instead of ynC1 . Furthermore, use

13 Problems

175

the Lorentz-invariance of u and the fact that the Lorentz group acts transitively on the level sets of q to conclude that w is independent of the choices made. Deduce that there exists a unique w 2 D 0 .R/ such u D q  w on nC1 [

j D1

f y 2 RnC1 j yj ¤ 0 g D RnC1 n f0g:

The mapping q is not submersive on all of RnC1 . As a consequence, a treatment of Lorentz-invariant distributions on RnC1 requires a more elaborate study. 13.13. We shall prove (see Problem 14.39 or [7, Exercise 6.60.(iii)] for another demonstration), for a 2 C, ( ( Z . 1 < Re a < 1/I sin    a 1 sin (13.44) x dx D .a/ a x .0 < Re a < 1/: cos cos 2 R>0 Let t > 0 and apply Cauchy’s Integral Theorem (12.9) to the function f .z/ D e z z a 1 and the open set U  C that equals the quarter-circle of radius t with vertices 0, t, and i t, of which the vertex 0, however, is cut off by a small quartercircle of radius  with 0 <  < t. Show that, for 0 < Re a < 1, 0D

Z

e



For 0    Z

t

x Z Ci

 2,

 2

e 0

x a 1

dx C i

0  2

e

Z

 2

e

e i  a i a

d 

t e

0

 e

we have cos   1

t cos 

t e i  a i a

 1 I 2 t

d DW 2 

4 X

d C i

Z



e

iy

.iy/a

1

1

e j Im aj 2 :

dy

t

Ij :

j D1

and therefore

hence

jI2 j 

 Re a t 2



Furthermore,

 Re a j Im aj  2 :  e 2 Conclude by taking limits for  # 0 and t ! 1, for 0 < Re a < 1, that Z Z  e x x a 1 dx D e i a 2 e iy y a 1 dy: .a/ D jI4 j 

R>0

R>0

If we carry out the same reasoning as above but using the quarter-circle of radius t of vertices 0, t, and i t, the quarter-circle again being indented at 0 by a quarter-circle of radius , we obtain Z  .a/ D e i a 2 e iy y a 1 dy: R>0

176

13 Fractional Integration and Differentiation

Deduce the formulas in (13.44) for 0 < Re a < 1 and verify that the first equation in (13.44) is valid for 1 < Re a < 1.

Chapter 14

Fourier Transform

Let  2 Cn . The function e W x 7! e hx; i D e x1 1 C


Cxn n

(14.1)

in C 1 .Rn / has the remarkable property that @˛ e D ˛ e

.˛ 2 .Z0 /n /:

As a consequence, for every linear partial differential operator with constant coefficients X P D P .@/ D c˛ @˛ W C 1 .Rn / ! C 1 .Rn / (14.2) one has

j˛jm

P .@/ e D P ./ e

with

P ./ D

X

c˛ ˛ :

j˛jm

In other words, e is an eigenvector of all of the operators P .@/ simultaneously. We note in passing that e is also an eigenvector of all translations Ta , for a 2 Rn , with eigenvalue e . a/. The idea behind the Fourier transform is that the operators P can better be understood if the vectors on which they act can be written as linear combinations of eigenvectors; indeed, P then acts as a scalar multiplication on each summand, the scalar being the eigenvalue corresponding to the eigenvector in question. A particularity is that here the eigenvalues do not form a discrete set but an entire continuum, namely Cn . For this reason it is natural to replace the linear combinations by integrals over , where a weight function plays the role of the coefficients in the integration. The approximating Riemann sums (in the -space) then become finite linear combinations. As will appear later on, for a large subspace S.Rn / of C 1 .Rn /, every function in S.Rn / can be written as an integral of this type over the set i Rn D f i  2 Cn j  2 Rn g. The function ei  is constant on the planes f x 2 Rn j hx; i D c g, for c 2 R, and behaves like a harmonic oscillation transversely to them; accordingly, J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_14, © Springer Science+Business Media, LLC 2010

177

178

14 Fourier Transform

this is called a plane harmonic wave. These are precisely the e , with  2 Cn , that are bounded on Rn , a great help in obtaining estimates. For a Fourier series on R of the form X .x/ D cn e i n!x ; n2Z

the Fourier coefficients cn are given in terms of the function  by cn D

! 2

Z

2=!

e

i n!x

.x/ dx;

0

the average of e i n!  over one period of ; compare with Chap. 16. If we cast off the shackles of periodic functions, the numerical factor and the interval of integration are no longer determined. For an integrable function u on Rn one now defines the Fourier transform F u (often written as u y in the literature) as Z F u./ D e i hx; i u.x/ dx . 2 Rn /: (14.3) Rn

Several other conventions are current in the literature. This definition is fairly standard in the theory of linear partial differential equations and leads to simple formulas, but is has the disadvantage of not being unitary on square-integrable functions (see Theorem 14.32 below). Some authors use the C instead of the sign, others add a factor 2 to the exponential or put .2/ n=2 in front of the integral. In our convention, the factor .2/ n occurs only in the inversion formula for the Fourier transform; see Theorem 14.13 below. Example 14.1. If 1Œ a; b  denotes the characteristic function of the interval Œ a; b , then e i b e i a F 1Œ a; b  ./ D i : (14.4)  In particular, if a D b D 1, the right-hand side takes the form ei 

e i

i

D

2 sin  DW 2 sinc ; 

where the notation sinc originates from sinus cardinalis; see Fig. 14.1. 1

-20 Π

Fig. 14.1 Example 14.1. Graph of sinc

-0.217

20 Π

˛

14 Fourier Transform

179

The following result is known as the Riemann–Lebesgue Theorem. In our development of the theory of distributions, it is the first instance in which certain properties of Lebesgue integrable functions have to be studied in some depth. Theorem 14.2. If u is a Lebesgue integrable function on Rn , then F u is uniformly continuous on Rn and F u./ converges to 0 if  2 Rn and kk ! 1. In particular, F u is bounded on Rn and satisfies Z ju.x/j dx . 2 Rn /: (14.5) jF u./j  Rn

Proof. Since je i hx; i j D 1, the estimate (14.5) follows from ˇZ ˇ Z ˇ Z ˇ ˇ ˇ ˇ i hx; i ˇ i hx; i ˇ ˇ e u.x/ dx ˇ  u.x/ˇ dx D ˇe ˇ Rn

Rn

Rn

ju.x/j dx:

For 0 ¤  2 Rn , we have ˝ ˛ Z Z i xC  2 ;  i hx; i kk F u./ D u.x/ dx e u.x/ dx D e Rn Rn Z    D e i hx; i u x  dx: kk2 Rn Hence  1  F u D F I T  2  u: (14.6) kk 2 Lemma 20.44 with p D 1 now implies that F u./ ! 0 as kk ! 1. It remains to show that F u is uniformly continuous. For arbitrary  and  2 Rn , we obtain Z ˇZ ˇ ˇ ˇ jF u. C / F u./j D ˇ e i hx; i .e i hx; i 1/ u.x/ dx ˇ  2 ju.x/j dx: Rn

Rn

1/ D 0. Therefore we see on acFor x 2 Rn , we have lim!0 .e i hx; i count of Lebesgue’s Dominated Convergence Theorem, see Theorem 20.26.(iv), that lim!0 F u. C / D F u./ uniformly in . 

Remark 14.3. An alternative proof of Theorem 14.2 can be based on Example 14.1 and runs as follows. Let C.0/ be the space of continuous functions f on Rn with the property that f ./ ! 0 as kk ! 1. If v is the characteristic function of an n-dimensional rectangle ˚ x 2 Rn j aj  xj  bj ; 1  j  n ;

we see from (14.4) that F v 2 C.0/. If v is a finite linear combination of characteristic functions of rectangles, then again F v 2 C.0/.

180

14 Fourier Transform

Furthermore, it is known from the theory of Lebesgue integration that for every Lebesgue integrable function u on Rn and every  > 0 there R exists a finite linear combination v of characteristic functions of rectangles with Rn ju.x/ v.x/j dx <  . On account of (14.5), with u replaced by u v, this implies that 2 sup jF u./

F v./j
0 with the property jF v./j < 2 whenever kk > R. From this it follows that   jF u./j  jF u./ F v./j C jF v./j < C D  .kk > R/: 2 2 Thus we have shown that F u./ ! 0 as kk ! 1. The limit of a uniformly convergent sequence of continuous functions is continuous. Because F u is the uniform limit of the above continuous functions F v, we conclude that F u is continuous. This completes the proof that F u 2 C.0/. ˛ If u has compact support, the definition of the Fourier transform can be extended to distributions and the Fourier transform is a complex-analytic function: Lemma 14.4. If u is an integrable function with compact support, then F u./ D u.e i  /. For every u 2 E 0 .Rn / the function  7! F u./ WD u.e i / is complexanalytic on Cn . Proof. The first assertion is evident. Let u 2 E 0 .Rn /. Then, by the continuity of  7! e i W Cn ! C 1 .Rn /, it follows that F u is continuous on Cn . The difference quotient 1 .e i.Ct e.j // e i / .t 2 C/ t converges to the function x 7! ixj e i .x/ in C 1 .Rn / as t ! 0. From this we see that F u is complex-differentiable with respect to every variable, with partial derivative equal to the continuous function @j F u./ D u. ixj e

i /

. 2 Cn /:

(14.7)

On the strength of Cauchy’s integral formula (12.11), applied to each variable, it now follows that v D F u is a complex-analytic function on Cn . That is, for every z 2 Cn the Taylor series of v, X @˛ v.z/ . z/˛ ; ˛Š ˛ converges to v./, for  in a suitable neighborhood in Cn of the point z.



Example 14.5. The function sinc is complex-analytic on C. This can also be seen from the power series expansion

14 Fourier Transform

181

sinc z D

X

k2Z0

. 1/k 2k z .2k C 1/Š

.z 2 C/:

˛

For  2 Rn the function u.e  / D F u. i/ is said to be the Laplace transform of u. The classical case is that in which n D 1, where u is a probability measure with support in R0 , and  > 0. For this reason the complex-analytic function F u on Cn , for u 2 E 0 .Rn /, is said to be the Fourier–Laplace transform of u. In Theorem 18.1 below we will prove that the restriction of F u to Rn is a function that is at most of polynomial growth. The analog of the Completeness Theorem for Fourier series is that “arbitrary” functions u can be written as continuous superpositions of the ei  with weight factors c F u./, where c is a constant to be determined. Here we run into a problem, because F u is not necessarily integrable on Rn for every integrable function u. For example, if 1Œ a; b  denotes the characteristic function of the interval Œ a; b , then, by (14.4), F 1Œ a; b  is a real-analytic function on R that is not absolutely integrable on R. The following space S of test functions, which lies between C01 .Rn / and C 1 .Rn /, proves to be very useful for constructing the theory. Definition 14.6. A function  on Rn is said to be rapidly decreasing if for every multi-index ˇ, the function x 7! x ˇ .x/ is bounded on Rn . One defines S D S.Rn / as the space of all  2 C 1 .Rn / such that @˛  is rapidly decreasing for every multi-index ˛. If .j /j 2N is a sequence in S and  2 S, then j is said to converge to  in S, notation limj !1 j D  in S, if for all multi-indices ˛ and ˇ, the sequence of functions .x ˇ @˛ j /j 2N converges uniformly on all of Rn to x ˇ @˛ . ˛ Two alternative formulations can be given: a function  on Rn
is rapidly decreasing if and only if for every N > 0 one has .x/ D O kxk N as kxk ! 1. We have  2 S if and only if  2 C 1 and P .x; @/ is bounded on Rn , for every linear partial differential operator P .x; @/ with polynomial coefficients. S D S.Rn / is a linear subspace of C 1 D C 1 .Rn /. When provided with the norms  7! kkS.k;N / WD sup jx ˇ @˛ .x/j (14.8) j˛jk; jˇ jN; x2Rn

it becomes a locally convex topological linear space in which convergence of sequences corresponds to that in Definition 14.6. Because it would be sufficient to consider the countable nondecreasing sequence of norms that is obtained by taking N D k, sequential continuity of linear operators on S is equivalent to continuity with respect to these norms; see Lemma 8.7. We have C01  S  C 1 . Furthermore, Definitions 2.13, 14.6, and 8.4 imply j !  in C01 ) j !  in S ) j !  in C 1 . This is tantamount to the assertion that the identity, interpreted as a mapping from C01 to S, is continuous, and the same applies to the identity from S to C 1 . Phrased differently, we have continuous inclusions in the following:

182

14 Fourier Transform

C01 .Rn /  S.Rn /  C 1 .Rn /:

(14.9)

By Lemma 8.2.(b), C01 .Rn / is dense in C 1 .Rn / and therefore, a fortiori, S.Rn / is dense in C 1 .Rn /. This implies that a continuous linear operator on C 1 .Rn / is uniquely determined by its restriction to S.Rn /. The following lemma shows that likewise, every continuous linear operator on S.Rn / is uniquely determined by its restriction to C01 .Rn /. Lemma 14.7. Let  2 C01 .Rn / with .x/ D 1 for kxk < 1. Write .  /.x/ D . x/. For every  2 S.Rn / and  > 0, one has that  WD .  /  belongs to C01 .Rn / and converges in S.Rn / to  as  # 0. Proof. By application of Leibniz’s formula we find that the value x ˇ @˛ . /.x/ equals .. x/ 1/ x ˇ @˛ .x/ plus terms that can be estimated by a constant times  j j jx ˇ @˛ .x/j, for j j ¤ 0 and  ˛. But the first term can be estimated by the supremum over the x with kxk  1= of jx ˇ @˛ .x/j. From this we see that lim#0  !  in S.Rn /.  Theorem 14.8. If P is a linear partial differential operator with polynomial coefficients, then P is a continuous linear mapping from S to S. Proof. For the norms defined in (14.8) and  2 S we have the estimates k@j kS.k;N /  kkS.kC1;N / ;

kxj kS.k;N /  kkS.k;N C1/ Ck kkS.k

where we have used that
@˛ xj  D xj @˛  C ˛j @˛

ıj



1;N / ;

.1  j  n/:

Here ıj is the multi-index having a 1 in the jth position and zeros everywhere else. From this follow the assertions for P D xj and for P D @j . The general assertion then is obtained by mathematical induction on the order of the operator and the degree of the polynomials in the coefficients.  For the next lemma the following notation proves useful: Dj WD

1 @j i

.1  j  n/:

The operator ~ Dj D ~i @j is known in quantum mechanics as the jth momentum h operator. In this expression ~ D 2 , where h is Planck’s constant, a quantity describing the atomic scale. Lemma 14.9. The Fourier transform F W u 7! F u defines a continuous linear mapping from S to S. For every 1  j  n,  2 S, and  and a 2 Rn , one has

14 Fourier Transform

183

F .Dj /./ D

F .xj /./ D F .Ta  /./ D

j .F /./;

(14.10)

Dj .F /./; .ei a F /./;

(14.11) (14.12)

F .ei a /./ D .T

a



F /./:

(14.13)

Here Ta  is as in Example 10.15. In compact notation, we have the following identities of continuous linear mappings from S to S: F B Dj D j B F F B Ta  D ei a B F

and and

F B xj D Dj B F ; F B ei a D T a  B F :

Proof. Let  2 S. Then the estimate j.x/j  c .1 C kxk/ N for all x 2 Rn and some c > 0 and N > n implies that  is integrable on Rn . On account of the Riemann–Lebesgue Theorem, Theorem 14.2, it follows that F  is a bounded continuous function on Rn . But xj  is also integrable and Dj D 1i @j acting on the integrand in (14.3) yields e i hx; i xj .x/. Therefore, the theorem on differentiation under the integral sign implies that F  is differentiable with respect to j , with continuous derivative given by (14.11). By mathematical induction on k we find that F  2 C k .Rn /, for all k 2 Z0 , and that all derivatives of F  are bounded. We obtain (14.10) when we apply integration by parts to the approximating integral over a large rectangle. For every pair of multi-indices ˛ and ˇ the function  ˇ D ˛ .F / D  ˇ F .. x/˛ / D F .D ˇ .. x/˛ // DW F

(14.14)

is bounded, because 2 S according to Theorem 14.8. Here we have used the following self-explanatory notation: Dˇ D

n Y

j D1

Dj ˇj D . i /jˇ j @ˇ :

We conclude that F .S/  S. For (14.12), use the substitution of variables x D y a to obtain Z Z  i hx; i i ha; i .F B Ta /./ D e .x C a/ dx D e e i hy; i .y/ dy Rn

Rn

D .ei a B F /./:

The remaining identity (14.13) has a similar proof. Finally we prove the continuity of F on the basis of (8.2). In particular, it is sufficient to show, for all k and N 2 Z0 , the existence of c D c.k; N / > 0 such that we have kF kS.k;N /  c kkS.N;kCnC1/ . 2 S/: To this end we first apply (14.5) to

in (14.14), then use the estimate

184

14 Fourier Transform

Z

Rn

j .x/j dx D

Z

Rn

.1 C kxk/

.nC1/

.1 C kxk/nC1 j .x/j dx

 c 0 sup .1 C kxk/nC1 j .x/j; x2Rn

R

where c 0 D Rn .1 C kxk/ to the desired estimate.

.nC1/

dx < 1. Application of Leibniz’s rule then leads 

Remark 14.10. Observe that (14.12) and (14.13) taken in conjunction with (10.15) imply (14.10) and (14.11), respectively. The reverse implications may be proved as in Problem 11.3. ˛ Example 14.11. For every a 2 C with Re a > 0, ua .x/ D e

a x 2 =2

.x 2 R/

defines a function ua 2 S.R/; the proof is analogous to that of Lemma 2.7. Therefore, F ua 2 S. Now Dua D i ax ua , and we obtain by Fourier transformation of both sides, also taking into account (14.10) and (14.11), 1 DF ua D i  F ua : a This yields F ua D c.a/ u 1

c.a/ D .F ua /.0/ D

with

a

Z

e

ax 2 =2

dx:

R

For a > 0, the change of variables x D .2=a/1=2 y leads to r 2 ; (14.15) c.a/ D a R p 2 where we have also used R e x dx D ; see (2.14). Now a 7! c.a/ is complex-analytic on H D f a 2 C j Re a > 0 g; differentiation under the integral sign shows, for example, that c satisfies the Cauchy–Riemann equation. With the definition a

1 2

D jaj

1 2

e

1 2i

arg a

;

where

  < arg a < ; 2 2

the right-hand side of (14.15) is a complex-analytic function on H as well. The difference v between the two sides equals 0 on the positive real axis. Therefore, the Taylor series of v vanishes at every point on that axis, which implies that v D 0 on an open neighborhood in C of R>0 . On account of Lemma 2.6 we conclude that v D 0 on H . In other words, (14.15) holds for all a 2 C with Re a > 0. The preceding results can be generalized to Rn by ua .x/ D e

Pn

j D1 aj

xj 2 =2

.a 2 Cn ; Re aj > 0; x 2 Rn /:

14 Fourier Transform

185

Since e

i hx; i

D

we conclude that

n Y

e

ixj j

j D1

ua .x/ D

and n

F ua D .2/ 2

n Y

j D1

1 p u a1 : aj j

n Y

uaj .xj /;

j D1

˛

The following lemma prepares the theorem that every  2 S can be written as a continuous linear combination of the ei  , with  2 Rn and c F ./ as weight function, giving the “Fourier coefficients.” Lemma 14.12. Define G D S B F B F W S.Rn / ! S.Rn /. Then G is a continuous linear mapping that commutes with the position and momentum operators and also with all translation operators. More specifically, for x 2 Rn and 1  j  n , .i/

G Bxj D xj BG ;

.ii/

G BDj D Dj BG ;

.iii/

G BTx  D Tx  BG : (14.16)

Proof. From (14.11) and (14.10) we deduce F B F B xj D F B . Dj / B F D xj B F B F ; F B F B Dj D F B xj B F D Dj B F B F : For the reflection S.x/ D . x/ we have S B . xj / D xj B S

and

S B . Dj / D Dj B S I

thus we conclude that (14.16).(i) and (ii) hold. Furthermore, (14.12) and (14.13) imply (14.16).(iii), for all x 2 Rn .  Theorem 14.13. The Fourier transform F W S.Rn / ! S.Rn / is bijective, with inverse F 1 D .2/ n S B F D .2/ n F B S . This is expressed by the formula Z 1 e i hx; i F ./ d  . 2 S.Rn /; x 2 Rn /: (14.17) .x/ D .2/n Rn Proof. With ı the Dirac measure on Rn and G as in Lemma 14.12, consider u WD ı B G 2 D 0 .Rn /. In view of (14.16).(i) we conclude xj u D u B xj D ı B G B xj D ı B xj B G D 0

.1  j  n/:

Theorem 9.5 then implies the existence of c 2 C such that u D c ı. Furthermore, for  2 C01 .Rn / and x 2 Rn , we now obtain, on account of (14.16).(iii),

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14 Fourier Transform

G .x/ D Tx  .G /.0/ D G .Tx  /.0/ D u.Tx  / D c ı.Tx  / D c .x/: Hence G D c I on C01 .Rn / and Lemma 14.7 then implies that G D c I on S.Rn /. 2 To determine c, we can apply G to  D e k
k =2 , for example. Then we see from Example 14.11, with all aj D 1, that F  D .2/n=2 . Therefore G  D .2/n , that is, c D .2/n . We can now conclude that ..2/ n S B F / B F D I . This is formula (14.17); in other words, F has left-inverse .2/ n S B F . It then follows that F is injective. A more symmetric form is obtained if we left-multiply by .2/n S : F B F D .2/n S: If we now right-multiply this in turn by .2/ n S , we see that .2/ n F B S is a right-inverse of F ; its existence implies that F is surjective. The end result is that F is bijective, with inverse as in the theorem.  Identity (14.17) is known as the Fourier inversion formula. Remark 14.18, Example 14.26, and Problems 14.7 and 15.9 below contain different proofs. In the case n D 1 the constant in the formula can also be determined by means of Fourier series; see Example 16.8 below. Theorem 14.24 below implies that G ı is a well-defined distribution. On account of this fact as well as Example 11.18, the proof of (14.17) can be simplified as follows. The example immediately leads to G D .G ı/ ;

while

xj .G ı/ D G .xj ı/ D G 0 D 0I

so

G D c I:

Remark 14.14. The plane waves are characterized by the fact that they are exactly the bounded eigenfunctions of the differential operator D D .D1 ; : : : ; Dn /, that is, if 2 C 1 .Rn / satisfies the eigenvalue equation D D  with  2 Cn , and if is a bounded function, then  D  with  2 Rn , and .x/ D .0/ e i h x;  i ; see Problem 14.1. The arbitrary function in S.Rn / can therefore be written as a superposition of bounded eigenfunctions of the differential operator D acting on C 1 .Rn /. The formula .F B D B F

1

/ D  

. 2 S.Rn //

shows that the differential operator D acting on S.Rn / can be diagonalized by conjugation with the Fourier transform F , and that the action in S.Rn / of the operator obtained by conjugation is that of multiplication by the coordinate . Apparently, F is the unitary (modulo a factor) linear coordinate transform in S.Rn / that diagonalizes the Hermitian linear operator D (see Problem 14.4) acting in the infinite-dimensional linear space S.Rn /. Note that this is a strong analog of the Spectral Theorem from linear algebra for Hermitian linear operators in a finitedimensional linear space over C provided with an inner product. See Theorem 14.32 below for the sense in which F modulo a factor is unitary. In fact, this case provides a model for the generalization of the theorem to the infinite-dimensional setting. ˛

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187

Example 14.15. Combining Theorem 14.13 and Example 14.11 gives immediately (compare with Problem 5.5), for t > 0, k
k2 1 2 4t : ˛ F 1 e t k
k D n e .4 t/ 2

In the following theorem we use the notation Z .x/ .x/ dx; h; i D Rn Z .x/ .x/ dx: .; / D h; i D

(14.18)

Rn

Here the overline denotes complex conjugation; thus, (14.18) defines a Hermitian inner product in the space of functions. These expressions are meaningful if  is integrable. Theorem 14.16. For all ;



2 S.Rn / we have

hF ;

i D h; F

F .

/ D .2/

i;

(14.19)

FF :

(14.22)

(Parseval’s formula), (14.20) .; / D .2/ n .F ; F / 2 S and F .  / D F  F ; (14.21) n

Proof. Both sides in (14.19) are equal to the double integral Z Z .x/ e i hx;i ./ dx d ; Rn

Rn

where we also use hx; i D h; xi. The proof of (14.20) begins with the observation that the complex conjugate of e i hx; i equals e i hx; i , which implies that F D S B F . Combining (14.19) and Theorem 14.13, we then conclude hF ; F

i D hF ; S B F

i D h; F B S B F

i D .2/n h;

i:

The convolution   is bounded; this merely requires one of the factors to be integrable and the other one to be bounded. Using ! X ˇ x ˇ D .x y C y/ˇ D .x y/ˇ y

ˇ ! we see that X ˇ ˇ x .  / D .x ˇ /  .x /:

ˇ

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14 Fourier Transform

Combining this with (11.4), we find that x ˇ @˛ .  / is bounded for all multiindices ˛ and ˇ, which implies   2 S. The formula F .  / D F  F is obtained by substituting e i hx; i D e i hx y; i e i hy;i in the double integral that represents the left-hand side. For (14.22) we replace  and in (14.21) by F  and F , respectively. Further, we apply F 1 to both sides and use F 2 D .2/n S . Thus we get FF

1

DF

.F 2  F 2 / D .2/n F B S.S S / D .2/n F .

/: 

Remark 14.17. According to (14.19) the restriction to S.Rn / of the transpose of F is F again. Using transposition and tDj D Dj , we see that the identities (14.10) and (14.11) actually are equivalent. Similarly, (10.11) leads to the equivalence of (14.12) and (14.13). From (1.14) and (14.12) we obtain, for  2 S.Rn /, the following identity of linear mappings from S to S: Z Z F B . / D F B .a/ T a  da D .a/ F B T a  da n n R R Z Z  .a/ e i a B F da D e i a .a/ da B F D .F / B F : D Rn

Rn

Note that this provides an alternative proof of the equality in (14.21). In view of Remark 14.10 and the proof of Theorem 14.13 it is obvious by now that all functorial properties of the Fourier transform are a consequence of the equalities (14.10) and t F D F . In fact, in Problem 15.13 below it will be shown that these two equalities determine the Fourier transform up to a scalar factor. In Problem 15.5 we show how to obtain these functorial properties by one uniform method. ˛ Remark 14.18. The identity (14.19) leads to an elegant proof of (14.17). To this end, apply (14.19) with replaced by   , for  > 0. Use a change of variables twice or Problem 14.30.(ii) as well as the Dominated Convergence Theorem of Arzel`a, see [7, Theorem 6.12.3], or that of Lebesgue, see Theorem 20.26.(iv), when passing to the limit as  # 0. This yields Z Z .0/ F ./ d  D .0/ F ./ d : (14.23) Rn

Rn

2

Next, take D e k
k =2 and apply Example 14.11 and Problem 2.7 to obtain (14.17) for x D 0. Finally, treat the case of arbitrary x 2 Rn by invoking (14.12). ˛ Definition 14.19. A tempered distribution on Rn is a (sequentially) continuous linear form on S D S.Rn /. The space of tempered distributions is denoted by S 0 or S 0 .Rn /. The term originates from the French “distribution temp´er´ee.”

14 Fourier Transform

189

One says that for a sequence .uj /j 2N and u in S 0 , lim uj D u

j !1

in S 0

if limj !1 uj ./ D u./ in C, for every  2 S (compare with Definitions 5.1 and 8.1). ˛ Restriction of u 2 E 0 D E 0 .Rn / to S yields a  u 2 S 0 ; restriction of u 2 S 0 to C01 gives a  u 2 D 0 D D 0 .Rn /. The restriction mappings  from E 0 to S 0 and from S 0 to D 0 are continuous and linear. These mappings are also injective, on account of Lemmas 8.2.(c) and 14.7, respectively. In both cases  u is identified with u; this leads to the continuous inclusions (compare with (14.9)) E 0 .Rn /  S 0 .Rn /  D 0 .Rn /:

(14.24)

We further note that Lemma 14.7 implies that for every u 2 S 0 the .  / u 2 E 0 converge in S 0 to u as  # 0. Hence, E 0 is dense in S 0 . Furthermore, C01 is dense in E 0 , see Corollary 11.7, and C01 is dense even in S 0 , in the sense that for every u 2 S 0 there exists a sequence .uj /j 2N in C01 such that uj ! u in S 0 as j ! 1. For example, take

uj D      u ;

with  as in Lemma 11.6,  as in Lemma 14.7, and  D 1=j , and apply (11.20). Indeed, on account of Theorem 11.5 this boils down to uj . / D u..  /.S  // tending to u. /, for every 2 S. Thus we have to prove that .  /.S  / tends to in S. The details of the argument are the same as in the initial part of the proof of Theorem 14.33 below. Consequently, every continuous operator on S 0 is uniquely determined by its restriction to C01 . For u 2 D 0 , one has that u 2 S 0 if and only if there exist a constant c > 0 and an S norm n such that ju./j  c n./

. 2 C01 /:

(14.25)

The continuous extension of u to S is obtained by taking u./ D lim#0 u. /, for  2 S and  as in Lemma 14.7. Example 14.20. Define u./ D

Z

.x/ e x cos e x dx R

. 2 S.R//:

In the given form, the integral is not absolutely convergent and has to be interRb preted as limb!1 1 .x/ e x cos e x dx. However, using integration by parts and limRx!˙1 j.x/j D 0, it can be rewritten as the absolutely convergent integral 0 x R  .x/ sin e dx. Hence Z Z 1 ju./j  j 0 .x/j dx  dx sup .1 C x 2 /j .j / .x/j: 2 0j 1; x2R R R 1Cx

190

14 Fourier Transform

This estimate establishes that u 2 S 0 .R/. Observe that there exists no polynomial p on R such that je x cos e x j  jp.x/j, for all x 2 R. ˛ The following is a direct consequence of Theorem 14.8. Theorem 14.21. If P is a linear partial differential operator with polynomial coefficients, then P is a continuous linear mapping from S 0 .Rn / to S 0 .Rn /. Example 14.22. For u integrable on Rn , we are going to prove that u 2 S 0 . Denoting the space of Lebesgue integrable functions on Rn by L1 D L1 .Rn /, see Theorem 20.40, we have the continuous inclusions C01 .Rn /  S.Rn /  L1 .Rn /  S 0 .Rn /: More generally: if 1  p < 1, then Lp is defined as the space of the locally integrable functions u on Rn , modulo functions that are equal to zero almost everywhere, such that jujp is integrable; see Theorem 20.41 for more details. In Lp , Z 1=p kukLp WD ju.x/jp dx Rn

defines a norm; Lp is complete with respect to this norm. L1 denotes the space of the essentially bounded locally integrable functions on Rn (modulo functions that are equal to zero almost everywhere); this is a Banach space with respect to the norm kukL1 WD ess sup juj:

Here ess sup f , the essential supremum of a function f , is defined as the infimum of all sup g as g runs through the set of all functions that are equal to f almost everywhere. Let q be the real number such that p1 C q1 D 1, on the understanding that q D 1 if p D 1 and q D 1 if p D 1. H¨older’s inequality from Problem 11.21 then implies, for u 2 Lp and  2 S, ju./j  kukLp kkLq : Next, choose N 2 Z0 such that N q > n, with N D 0 if q D 1, that is, p D 1. If we then write .x/ D .1 C kxk/ N .1 C kxk/N .x/

and use the integrability of .1 C kxk/ qN , we conclude that there exists a constant cN > 0 with the property that for every u 2 Lp and  2 S, ju./j  kukLp cN kkS.0;N / :

(14.26)

We also obtain that kkLq  cN kkS.0;N / . Furthermore, (14.26) proves that we have the following continuous inclusions: C01 .Rn /  S.Rn /  Lp .Rn /  S 0 .Rn /:

14 Fourier Transform

191

In other words, convergence with respect to the Lp norm implies convergence in S 0 ; see Remark 20.43 for more details. Combining this with Theorem 14.21, we find that the space of tempered distributions is really quite large: for every 1  p  1 we can start with u 2 Lp , then apply to it an arbitrary linear partial differential operator P with polynomial coefficients, and conclude that the result P u is a tempered distribution. ˛ Definition 14.23. If u 2 S 0 .Rn /, its Fourier transform F u 2 S 0 .Rn / is defined by F u./ D u.F /

. 2 S.Rn //:

˛

In view of (14.19) the new and the old definition of F  coincide for  2 S.Rn /  S 0 .Rn /. The slightly frivolous notation (3.3) is continued to Z F u./ D e i hx; i u.x/ dx; (14.27) Rn

where we are not allowed to give  a value, since the expression is meaningful only after integration over , with a test function as weight function. Theorem 14.24. For every u 2 S 0 .Rn / we have F u 2 S 0 .Rn /. The mapping F W u 7! F u is a continuous linear mapping from S 0 .Rn / to S 0 .Rn / that is a common extension of the Fourier transform on S.Rn /, on the space L1 .Rn / of Lebesgue integrable functions, or on the space E 0 .Rn / of distributions with compact support, respectively. If u 2 E 0 .Rn /, then F u 2 S 0 .Rn / corresponds to the analytic function F u from Lemma 14.4, and therefore certainly F u 2 C 1 .Rn /. For every u 2 S 0 .Rn / and 1  j  n we have F .Dj u/ D j F u

and

F .xj u/ D

Dj F u:

Finally, F is bijective from S 0 .Rn / to S 0 .Rn /, with inverse equal to .2/ .2/ n F B S .

(14.28) n

S BF D

Proof. These assertions follow from Lemma 14.9 and Theorem 14.13, where we recognize F W S 0 ! S 0 as the transpose of F W S ! S. The assertion that F is an extension of the Fourier transform on S follows from (14.19). On L1 , F corresponds to the Fourier transform, because the Fourier transform on L1 is also a continuous extension of that on S and (14.19) holds for  2 L1 and 2 S. Now let u 2 E 0 .Rn /. As we have seen above, E 0 .Rn /  S 0 .Rn /; therefore u 2 0 S .Rn /, and thus F u 2 S 0 .Rn / is defined. In order to prove that this distributional Fourier transform F u is equal to the analytic function u y D F u from Lemma 14.4, we write, for every  2 C01 .Rn /,  Z Z e i  ./ d  D u.e i  / ./ d  F u./ D u.F / D u Rn

Rn

192

14 Fourier Transform

D

Z

Rn

u y./ ./ d  D .test u y/./;

where for the third equality we have used Lemma 11.4 with A./ D ./ e i  as well as the linearity of u. So as to satisfy condition (b) from that lemma, i.e., that the support of A./ is contained in a compact set for all  2 Rn , we have, in fact, to perform the calculation above with u replaced by  u, where  2 C01 .Rn / is a test function that equals 1 on a neighborhood of supp u. Thus, we have shown that F u D test u y. The formulas in (14.28) follow from those in S by continuous extension or by transposition. The same holds for the formula F B F D .2/n S in S 0 .  Example 14.25. We have, for all a 2 Rn , Ta B F D F B e i a

and

F B Ta D e

ia

BF

in

S 0 .Rn /:

(14.29) ˛

Indeed, the two identities follow by transposition from (14.12) and (14.13).

Example 14.26. Because F ı./ D ı.F / D F .0/ D 1./, for every  2 S.Rn /, we see that F ı D 1 in S 0 .Rn /: (14.30) Since S1 D 1, we find on account of the Fourier inversion formula from Theorem 14.24 that F 1 D .2/n ı: (14.31)

This identity is equivalent to the inversion formula in the form (14.17). In order to see this, apply both sides of (14.30) to Tx  , with x 2 Rn , and use (14.29) to conclude .2/n .x/ D F 1.Tx  / D 1.F B Tx  /./ D 1.eix B F /./ Z D e i hx; i F ./ d : Rn

Significantly enough, (14.31) also may be obtained independently of the Fourier inversion formula (14.17) as follows. Observe that 0 D F 0 D F .Dj 1/ D j F 1

.1  j  n/: 2

Therefore F 1 D c ı on account of Theorem 9.5, and so c D 1.F e k
k =2 / D .2/n . Probably, this is the most transparent proof of (14.17). Furthermore, this proof too is based on (14.10) (actually, its global form (14.12)) and t F D F . See the natural extension of Problem 9.6 to Rn or Problems 14.31 and 14.32 for related proofs. In the Dirac notation (14.27) the equality (14.31) looks rather spectacular: Z 1 ı.x/ D e i hx;i d : (14.32) .2/n Rn

14 Fourier Transform

193

See (16.7) below for the analog in the periodic case With d equal to the difference mapping from (10.24), formula (14.31) implies .2/n d  ı D d  F 1; which in Dirac notation is known as the Fourier–Gel’fand formula Z 1 ı.x y/ D e i hx y; i d : .2/n Rn In fact, it leads to yet another proof of (14.17), in view of Z Z Z 1 .y/ ı.x y/ dy D e i hx .x/ D .2/n Rn Rn Rn

y; i

.y/ dy d :

If P is a polynomial function in n variables, then P D P 1, and so (14.28) implies F .P .D/ ı/ D P

and

F P D .2/n P . D/ ı:

(14.33)

In particular, if
denotes the Laplace operator, then F .P .
/ ı/ D . k
k2 / P:

(14.34)

Furthermore, application of (14.29) implies, for every a 2 Rn , F ıa D F B Ta ı D e

ia

BF ı De

i aI

(14.35)

in turn, this entails F .P ei a / D Ta .F P / D .2/n Ta .P .i @/ ı/ D .2/n P .i @/ ıa :

(14.36)

On account of Theorem 8.10 we obtain that a distribution on Rn has finite support if and only if it equals the Fourier transform of an exponential polynomial, which is defined in Problem 11.6, on Rn that is a tempered distribution, or more precisely, P the Fourier transform of a function of the form u D lkD1 Pk ei ak , where l 2 Z0 , Pk is a polynomial function on Rn , and ak 2 Rn , for 1  k  l (compare with Problem 14.24 below). Note that the linear subspace L D f P .@/ u 2 S 0 .Rn / j P a polynomial function on Rn g is of finite dimension and see Problem 11.6 for the converse assertion. Furthermore, similar to the problem of multiple eigenvalues in the theory of the indicial equation for nth-order ordinary differential equations, the occurrence of polynomials is related to confluence, but now of Dirac measures. ˛ Remark 14.27. By now, we have collected abundant evidence for the fact that the smoother a tempered distribution u, the faster the decay at infinity of its Fourier transform F u. Phrased differently, in the decomposition of a distribution into plane harmonic waves, many oscillations of high frequency are needed to account for

194

14 Fourier Transform

the small-scale variations of a “rough” distribution, while fewer are needed for a “smooth” distribution. Indeed: – If u is a derivative of a Dirac measure, then F u is an exponential polynomial that is tempered; see (14.36). – If u is of class L1 , then F u is continuous and vanishes at infinity; see the Riemann–Lebesgue Theorem, Theorem 14.2. – If u is an integrable function that is H¨older continuous of order 0 < ˛ < 1, then F u./ D O..1 C kk/ ˛ / as kk ! 1; see Problem 14.11 below. – If u is a function of class C k with derivatives of sufficient decay, then F u./ D O..1 C kk/ k / as kk ! 1; use integration by parts. – If u is a nontrivial polynomial, then supp F u D f0g; see (14.33). ˛ Remark 14.28. Define f W R ! R by f .x/ D e x . One might be tempted to argue that  X xk  X ik Ff DF D 2 ı .k/ : kŠ kŠ k2Z0

k2Z0

The last expression, however, does not define an element in D 0 .R/: its support is f0g but it is not of finite order. This means that f does not belong to S 0 .R/, whereas the partial sums of the series for f do. This shows that convergence of a sequence of functions uniformly on compact sets in R need not imply convergence of that sequence in S 0 .R/. ˛ Example 14.29. According to (1.3) we get, for PV x1 2 D 0 .R/ and any  2 C01 .R/, Z log jxj 1 ./ D .1 C x 2 / 0 .x/ dx; x 1 C x2 R ˇ ˇ Z log jxj 1 ˇ ˇ dx sup .1 C x 2 /j 0 .x/j: ˇ PV ./ˇ  2 x x2R R 1Cx PV

and so

This estimate proves that PV x1 2 S 0 .R/. On account of Example 9.2 we have x PV x1 D 1 in D 0 .R/ and so in S 0 .R/; hence (14.28) implies   1 1 i @x F PV D F x PV D F 1 D 2 ı x x

in view of the preceding example. Example 4.2 and Theorem 4.3 then lead to F PV x1 D 2 i H C c, for some c 2 C, where H denotes the Heaviside function, which is a tempered distribution. Furthermore, PV x1 is an odd distribution and so is F PV x1 in view of S F D F S ; this implies 2 i C c D c, which leads to c D  i . Therefore, with sgn denoting the sign function on R, F PV

1 D x

 i sgnI

and so

because 2 PV x1 D F S F PV x1 D

F sgn D

1 2i PV ; 

 i F S sgn D  i F sgn.

(14.37)

14 Fourier Transform

195

For any  2 S.R/, the latter identity in (14.37) entails Z Z Z .x/ dx: .x/ sin x dx d  D lim #0 R>0 R RnŒ ;   x Indeed, Z Z F sgn./ D sgn.F / D sgn./ e ix .x/ dx d  R R Z Z Z Z i x e .x/ dx d  C e i x .x/ dx d  D R>0 R R>0 R Z Z D 2i .x/ sin x dx d : Finally, note that H D

R>0

R

1 2

1 2

C

sgn. Applying (14.37), we now see that

FH Dı

1 i PV : 

(14.38) ˛

Example 14.30. According to Example 4.2 one has ı D @ H , and on account of (14.30) this implies 1 D F ı D F .iDH / D i  F H:

But this does not uniquely determine F H , because for every constant c 2 C, the equation above is also satisfied by F H C c ı. (All solutions are described in Problem 9.5.) For every  > 0, however, H .x/ D e  x H.x/ is a Lebesgue integrable function on R, and therefore F H is a continuous function. Furthermore, @ H D  H C ı, and so 1 ; (14.39) .i  C / F H D 1I that is, F H ./ D i C  because F H is continuous. Since lim#0 H D H in S 0 we now conclude, on the strength of Theorem 14.24, that F H D lim #0

1 1 1 D i C  i  i0

(14.40)

with convergence in S 0 .R/. A combination of (14.38) and (14.40) now leads to the following Plemelj–Sokhotsky jump relations (compare with Problems 1.3 and 12.14): 1 1 ˙  i ı D PV : (14.41) x˙i0 x ˛ Example 14.31. Consider u 2 E 0 .Rn / satisfying u.x 7! x ˛ / D 0, for every multiindex ˛. From (14.7) we derive @˛ F u./ D . i /j˛j u.x 7! x ˛ e i /. Since F u is complex-analytic on Cn on account of Lemma 14.4, we obtain, by power series expansion about 0,

196

14 Fourier Transform

X . i /j˛j u.x 7! x ˛ /  ˛ D 0 F u./ D ˛Š ˛

. 2 Cn /:

In other words, F u D 0, but then Theorem 14.24 implies u D 0. Using the Hahn–Banach Theorem, Theorem 8.12, one now deduces that the linear subspace in C 1 .Rn / consisting of the polynomial functions is dense in C 1 .Rn /. In other words, given a function  2 C 1 .Rn /, there exists a sequence .pj /j 2N of polynomial functions on Rn such that for every multi-index ˛ and compact set K in Rn , the sequence .@˛ pj /j 2N converges to @˛  uniformly on K as j ! 1. This is Weierstrass’s Approximation Theorem (see [7, Exercise 6.103] for a proof along classical lines and the corollary to Theorem 16.20 for still another); conversely, the vanishing of u is a direct consequence of this theorem. ˛ n Next, we generalize Parseval’s formula (14.20). A function f on to be R R is said square-integrable if f is a locally integrable function such that Rn jf .x/j2 dx < 1. The space of square-integrable functions is denoted by L2 D L2 .Rn /. According to the Cauchy–Schwarz inequality, the so-called L2 inner product .; / from (14.18) is well-defined for all  and 2 L2 . The corresponding norm Z 1=2 1=2 kf k D kf kL2 WD .f; f / D jf .x/j2 dx Rn

is said to be the L2 norm. The space L2 .Rn / is with respect to this norm. In other words: L2 .Rn / is a Hilbert space, an inner product space that is complete with respect to the corresponding norm. Additionally, the space of continuous functions with compact support is dense in L2 .Rn / by Theorem 20.41 and the definition preceding it. From Example 14.22 we obtain the continuous inclusion L2  S 0 . For a general Hilbert space H , a linear mapping U W H ! H is said to be a unitary isomorphism if U.H / D H and .Uf; Ug/ D .f; g/

.f; g 2 H /:

This implies that U is bijective and that U 1 from H to H is unitary as well. (Observe that in an infinite-dimensional Hilbert space an injective linear mapping is not necessarily surjective.) Theorem 14.32. If u belongs to L2 .Rn /, this is also true of its Fourier transform F u. Parseval’s formula (14.20) applies to all  and 2 L2 .Rn /. It follows that e WD .2/ n2 F to L2 .Rn / defines a unitary isomorphism from the restriction of F L2 .Rn / onto L2 .Rn /. Proof. Applying Remark 20.43 below, we obtain a sequence .j /j 2N in C01 .Rn / with limj !1 ku j k D 0. On account of (14.20) this leads to kFej

ek k D kj F

k k  kj

uk C ku

k k ! 0;

ej /j 2N is a Cauchy sequence in L2 , and in view for j; k ! 1. Therefore .F 2 of the completeness of L , see Theorem 20.41, there exists v 2 L2 satisfying

14 Fourier Transform

197

ej k D 0. This implies that F ej ! v in S 0 . Because, in addilimj !1 kv F 0 e e tion, j ! u in S , we obtain F j ! F u in S 0 , and on account of the uniqueness eu 2 L2 . of limits we derive that Feu D v. Because v 2 L2 , the conclusion is that F Combining the estimates euk  kF eu kF

ej k C kFej k F

and

and taking the limit as j ! 1, we also deduce

ej k D kj k  kuk C ku kF

j k

euk  kuk: kF

e is continuous with respect to the L2 norm. It follows that the This implies that F right-hand side in (14.20) is continuous with respect to both variables  and in L2 , as is the left-hand side. If we now approximate  and by j 2 C01 and 1 2 e e j 2 C0 , respectively, relative to the L norm and use .j ; j / D .F j ; F j /, 2 we obtain (14.20) for  and in L . e. According to Theorem 14.13, we have Finally we prove the surjectivity of F 2 e is e F .S/ D S, while S is dense in L . It follows that the image of L2 under F 2 e dense in L , and because F is unitary the image is closed; therefore, the image is euj ! v in L2 , then kF euj F euk k D kuj uk k equal to L2 .Rn /. Indeed, if F 2 shows that .uj /j 2N is a Cauchy sequence in L . Hence there exists u 2 L2 with euj ! F eu in L2 . But this leads to v D F eu. uj ! u in L2 , and therefore F  We give an extension of (14.21) to distributions. Note that the product F u F v is well-defined as an element of D 0 .Rn / if u 2 S 0 .Rn / and v 2 E 0 .Rn /, because then F v 2 C 1 .Rn / according to Theorem 14.24. Theorem 14.33. If u 2 S 0 .Rn / and v 2 E 0 .Rn /, then u  v 2 S 0 .Rn /, while F .u  v/ D F u F v. In particular, F u F v 2 S 0 .Rn /. Proof. On the strength of (11.23), we recognize that to prove u  v 2 S 0 we only have to demonstrate that  7! S v   W C01 ! C 1 possesses an extension to a continuous mapping from S to S. This means that for every pair of multi-indices ˛ and ˇ, we have to estimate the number x ˇ @˛ .S v  /.x/ D x ˇ .S v  @˛ /.x/ uniformly in x by an S norm of ; this requires only a uniform estimate by an S norm of D @˛ . In view of (11.1), .S v 

/.x/ D S v.Tx B S / D v.S B Tx B S / D v.T

x

/:

Using the continuity of v on C 1 .Rn /, we deduce from (8.4) that it is sufficient to show that for every multi-index and compact K  Rn , the number
x ˇ @y . .y C x// D x ˇ @ .y C x/

198

14 Fourier Transform

can be estimated uniformly in x 2 Rn and in y 2 K by an S norm of . The change of variables x D z y and the uniform estimate j.z y/ˇ j  c .1 C kzk/jˇ j , for all z 2 Rn and y 2 K, then yield the desired result. We now prove F .u  v/ D F u F v by continuous extension of this identity for u and v 2 C01 . We write, for u 2 S 0 .Rn /, v 2 E 0 .Rn /, and  2 C01 , F .u  v/./ D .u  v/.F / D u.S v  F /: Because F  2 S, we obtain from the above that S v  F  2 S. Following Definition 14.19 we observed that for every u 2 S 0 there exists a sequence .uj / in C01 with the property that uj ! u in S 0 as j ! 1. From this we now conclude that F .uj  v/ ! F .u  v/ in D 0 . But we also have F v  2 C01 , and therefore we deduce from .F u F v/./ D F u.F v / D u.F .F v //

that one also has F uj F v ! F u F v in D 0 . Thus we conclude that F .u  v/ D F u F v for all u 2 S 0 , if it holds for all u 2 C01 , for given v 2 E 0 . The commutativity of the convolution in turn implies that F .u  v/ D F u F v for all v 2 E 0 , because it holds for all v 2 C01 , for given u 2 C01 . 

A result related to the theorem above is that u   2 C 1 .Rn / and F .u  / D F u F  if u 2 S 0 .Rn / and  2 S.Rn /. In specific examples of homogeneous distributions it is often straightforward that the distributions are tempered. More generally one has H¨ormander’s Theorem [12, Thm. 7.1.18]: Theorem 14.34. If u 2 D 0 .Rn / and the restriction of u to Rn n f0g is homogeneous (of arbitrary degree a 2 C), then u 2 S 0 .Rn /. Proof. We begin with a  2 C01 .R>0 / such that   0 and  is not identically zero. Through multiplication by a suitable positive number we can ensure that R .1=s/ s 1 ds D 1. Let r > 0. By the change of variables s D t=r we deduce Z  r  dt D1 .r > 0/:  t t R>0 Z 1  Now define kxk  dt .x/ WD 1 :  t t 1

If x is bounded, the interval of integration can be replaced by a bounded interval, and the theorem on differentiation under the integral sign gives 2 C 1 .Rn /. On the other hand, if kxk is large, the interval of integration can be replaced by R>0 , which implies that .x/ D 0. In conclusion, 2 C01 .Rn /. Using the notation .x/ WD .kxk/ and t W x 7! t x, we now conclude, for every  2 C01 .Rn /,

14 Problems

199

Z 1  u./ D u. / C u .1=t/  t 1 dt  D u. / 1 Z 1 Z 1
1
 C u .1=t/   t dt D u. / C u  t   t aCn 1

1

dt;

1

where we have used the linearity of u, the identity .1=t/   D .1=t/ . t  /, and .1=t/ D t n t  (see Theorem 10.8), and t  u D t a u on Rn n f0g. Because both u and  u are distributions with compact support, we deduce the existence of a constant c1 > 0, a k 2 Z0 , and compact subsets K of Rn and L of Rn n f0g such that Z 1 ju./j  c1 kkC k ; K C c1 kt  kC k ; L t Re aCn 1 dtI 1

see (8.4). Now

kt  kC k ; L D t k kkC k ; t L and for every N there exists a constant c2 > 0 such that kkC k ; t L  c2 t

N

kkS.k;N / ;

(14.42)

with the notation of (14.8). Indeed, it is sufficient to prove the estimate for k D 0. Because 0 … L, there is a constant m > 0 such that m  max1j n jxj j for all x 2 L. Consequently, for x 2 t L there exists a j such that xj  t m, and therefore j.x/j D jxjN jj.x/jjxj j

N

 jxjN jj.x/j.tm/

N

m

N

t

N

kkS.0;N / :

Choosing N > k CRe aCn in (14.42), we obtain a constant c > 0 with the property ju./j  c kkS.k;N /

. 2 C01 .Rn //:

This implies that u can be extended to a continuous linear form on S.Rn /; on account of Lemma 14.7 this extension is uniquely determined. 

Problems 14.1. (Characterization of exponentials.) Suppose u 2 D 0 .Rn / and @j u D j u, for some  2 Cn and all 1  j  n. Demonstrate the existence of a constant c 2 C such that u D c e . 14.2. Prove e  B P .@/ B e D P .@ C /, for  2 Cn and P .@/ as in (14.2).  14.3. Suppose that n > 1 and let P be a polynomial function on Cn of degree at least one. Show that the zero-set of P is infinite. Deduce that the space of complexanalytic solutions u on Cn of the linear partial differential equation with constant coefficients P .@/u D 0 is of infinite dimension.

200

14 Fourier Transform

14.4. Prove that Dj is a Hermitian linear operator acting in S.Rn / when this linear space is provided with the Hermitian inner product from (14.18), that is, .Dj ; / D .; Dj /, for all 1  j  n and  and 2 S.Rn /.

 14.5. Assume that 1 ; : : : ; n are integrable functions on R. Using the notation (11.8) prove that  D ˝jnD1 j defines an integrable function on Rn with the property F  D ˝jnD1 F j .

14.6. Suppose that u and v WD F u are Lebesgue integrable on Rn . Prove that u equals the continuous function .2/ n S B F v almost everywhere on Rn .

 14.7. Let L W S ! S be a linear mapping that commutes with the position and momentum operators. Show that there exists a constant c 2 C such that L D c I , with I the identity on S. Deduce the Fourier inversion formula (14.17).  14.8. Let  2 S.Rn /. Prove one of the following assertions and then derive the other: P (i) If .0/ D 0, then we may write  D jnD1 xj j with j 2 S.Rn /. R P (ii) If Rn .x/ dx D 0, then we may write  D jnD1 @j j with j 2 S.Rn /.

 14.9. (Eigenvalues of Fourier transform.) What are the possible eigenvalues of the Fourier transform acting on S.R/? And when acting on L2 .R/? For three eigenvalues, try to find corresponding eigenfunctions. See Problem 14.42 for a different approach.  14.10. Let P be a harmonic polynomial function on Cn and define . i / P .z/ D P . i z/, for all z 2 Cn . Verify Hecke’s formula

F .e

1 2 2 k
k

n

P / D .2/ 2 e

1 2 2 k
k

. i / P:

 14.11. Suppose f 2 C0 .Rn / is H¨older continuous of order 0 < ˛  1 (for this notion, see Definition 19.4 below). Prove the existence of c > 0 such that

jF f ./j 

c .1 C kk/˛

. 2 Rn /:

 14.12. (L1 Sobolev inequality in special case.) For f 2 S.Rn /, write kf kL1 and kf kL1 as in Example 14.22. Show that there exists a constant c D cn > 0 such that for all f 2 S.Rn /, X kf kL1  cn kD ˛ f kL1 : j˛jnC1

14 Problems

201

 14.13. Prove, for every t > 0 and  2 S.R/, that Z t Z sin tx dx F ./ d  D 2 .x/ x t R

and deduce (compare with Problem 9.6) sin tx Dı t !1 x lim

S 0 .R/:

in

 14.14. Consider the following functions on R:

a.x/ D e

x 2 C2x

;

b.x/ D e

x

c.x/ D e

H.x/;

jxj

;

d.x/ D

1 : 1 C x2

(i) For each of these, sketch its graph. (ii) Verify that these functions belong to L1 D L1 .R/. Which of them belong to S D S.R/ and which to L2 D L2 .R/? (iii) Calculate the Fourier transforms of these functions. Deduce Laplace’s integral (compare with Problem 18.8.(iii) or [7, Exercise 2.85]) Z  cos x  dx D e jj . 2 R/: 2 2 R>0 1 C x (iv) Sketch the graphs of the Fourier transforms of the functions b, c, and d ; make separate sketches of the real and imaginary parts. (v) For each of the Fourier transforms found, determine whether it is a function in S, in L1 , or in L2 .  14.15. Define f D e j
j on R. Differentiate f twice and use the result to prove that 1 2 2 f is a fundamental solution of the differential operator D C I in R; in addition,

calculate F f . Derive F arctan D

 i

14.16. Suppose a > 0 and prove 1 F .1Œ a; a  / F .e 4

jj

PV e  .

j
j

/./ D

sin a DW g./: .1 C  2 /

Show that g 2 S 0 .R/ and that g 2 L1 .R/ actually. Verify that the solution f 2 S 0 .R/ of the equation F f D g is given by ( 1 e a cosh x .jxj  a/; 2f .x/ D jxj sinh a .jxj  a/: e Deduce that f 2 C.R/ \ L1 .R/ and verify that

R

R>0

g./ cos x d  D f .x/.

202

14 Fourier Transform

 14.17. Use the function d from Problem 14.14 to prove (compare with Problem 19.8.(viii)) x F f0 ./ D  i sgn./e jj if f0 .x/ D : 1 C x2

Next, suppose f 2 C.R/ satisfies f .x/ D x1 C O. x12 / as jxj ! 1. Show that F f is a function that is continuous except at the origin, where left and right limits exist. Verify that the jump F f .0C / F f .0 / equals 2 i .

 14.18. (Constant in Fourier inversion formula.) Prove that in the context of Problem 14.5, one has  2 S.Rn / if j 2 S.R/, for every j D 1; : : : ; n. Use this to prove that cnR D .c1 /n if cn is the constant in (14.17). Compute c1 by calculating G ./.0/ D F ./ d  for  D e j
j . A problem with this method is that  does not belong to S.R/. Argue that this can be overcome through approximating  by a suitable sequence of functions in S.R/.

14.19. Use Parseval’s formula to verify Z Z  1 x2 dx D dx D : 2 2 2 /2 .1 C x / .1 C x 2 R R 14.20. Suppose a and b > 0. Prove Z   sin ax sin bx dx D min.a; b/ D .a C b 2 x 2 4 R>0

ja

bj/:

14.21. Prove that (14.23) comes down to the equality .F 1/ ˝ ı D ı ˝ F 1 in S 0 .Rn  Rn /, which in turn is equivalent to ı ˝ ı D ı ˝ ı. 14.22. Let u be a locally integrable function on Rn such that there exists N 2 R with u.x/ D O.kxkN / as kxk ! 1. Prove that u 2 S 0 . 14.23. Prove that every tempered distribution is of finite order.  14.24. (Characterization of tempered exponentials.) Suppose u 2 S 0 .Rn / and  2 n S.R /. Prove that there exist constants C > 0 and N 2 Z0 such that

ju.Ta /j  C .1 C kak/N

.a 2 Rn /:

Now let  2 Cn . Prove that the function ei  defines a tempered distribution on Rn if and only if  2 Rn . p

14.25. For what combinations of p and q 2 N does x 7! e x e i e pered distribution on R?

xq

define a tem-

14.26. Let a > 0 be a constant. Calculate the Fourier transforms of the following distributions on R:

14 Problems

203

(i) x 7! e ax H.x/. (ii) e aj
j . 1 (iii) x 7! x 2 Ca 2.

Discuss possible complex-analytic extensions of the Fourier transforms to open subsets of C. Furthermore, try to establish whether the functions above converge in S 0 .R/ as a # 0. If so, determine the Fourier transforms of their limits. Calculate F 1H .

 14.27. Determine the Fourier transforms of the functions: R ! R given by the following formulas:

(i) cos, sin, x 7! x sin x, cos2 , cosk for k 2 N. (ii) sinc. (iii) x 7! x H.x/ (use Problem 13.6), j
j, sin j
j. Note that only in case (ii) is the Fourier transform a function. In the notation of Problem 13.5, deduce from part (iii) that F j
j 2 D  j
j.

 14.28. Derive the results from Problem 9.14 by means of the Fourier transform. The solutions of (systems of) linear ordinary differential equations with constant coefficients are at most of exponential growth at infinity, and for that reason might be nontempered. Therefore, Laplace transform is often used as a tool in situations like the present one; but it may be replaced by a combination of Fourier transform and analytic continuation. Q 14.29. Let P .D/ D jmD1 .D j / be a differential operator in R, where j 2 C for 1  j  m. Prove that all fundamental solutions of P .D/ are tempered distributions if and only if j 2 R for all 1  j  m. Now let P .D/ be a linear partial differential operator with constant coefficients on Rn , with n > 1. Prove that there are always solutions u 2 D 0 .Rn / of P .D/u D 0 such that u … S 0 .Rn /. Conclude that not all fundamental solutions of P .D/ are tempered distributions (compare with Problem 14.3).  14.30. (Fourier transform, pullback, and pushforward.) Let A be an invertible linear transformation of Rn . Write the transpose of the inverse of A as B; then B D .tA/ 1 .

(i) Demonstrate the following identity of continuous linear operators in S 0 .Rn /: F B A D .tA/ B F D j det Bj B  B F :

See Problem 15.5 for another proof. (ii) For c > 0, denote the mapping x 7! c x W Rn ! Rn by c. Deduce F B .c

1 

/ D cn c B F

or equivalently,

F B c D c  B F :

Prove that F u is homogeneous of degree n a if u 2 S 0 .Rn / is homogeneous of degree a 2 C.

204

14 Fourier Transform

(iii) Verify that u 2 S 0 .Rn / is invariant under all orthogonal transformations if and only if F u has that property. (iv) Let A be an invertible linear transformation of Rn . Prove that A commutes with F if and only if A is orthogonal. 14.31. Here we describe another derivation of (14.31) that is independent of (14.17) (see the natural extension of Problem 9.6 to Rn or Problem 14.32 for related proofs). 2 Set  D e k
k =2 . From a slight modification of Lemma 14.7 it follows that lim#0    D 1Rn in S 0 .Rn /. Hence, successively using the continuity of F , Problems 15.5 and 5.2, as well as the notation from Example 10.9, we obtain the following identities in S 0 .Rn /: F 1Rn D lim.F B   / D lim. B F / D lim.F / D c ı; #0

#0

where

#0

c D 1Rn .F / D .2/n :

 14.32. In the notation of Example 14.11 consider  D .2/ n=2 u.1;:::;1/ 2 S.Rn / 2 2 and  , for  > 0, as in Example 10.9. Prove that   F  D e  k
k =2 . Verify the equivalence of the following two assertions:

(i) lim#0   F  D 1 in S 0 .Rn /, (ii) lim#0   D ı in S 0 .Rn / (compare with Problem 5.5). Prove one of the two statements.  14.33. Let P be a polynomial function on Rn and suppose that u D P .D/ ı 2 0 n D .R / is invariant under all orthogonal transformations in Rn . Prove the existence of a polynomial function P0 on R such that u D P0 .
/ ı, where
is the Laplacian.

14.34. Let P .D/ be a linear partial differential operator in Rn with variable coefficients. Prove that P .D/ B A D A B P .D/ for all translations and orthogonal transformations A in Rn if and only if there exists a polynomial function P0 on R such that P .D/ D P0 .
/, where
is the Laplacian.  14.35. Define GL.n; R/ as the group of invertible real nn matrices and SL.n; R/ as the subgroup of GL.n; R/ consisting of the matrices with determinant 1. Suppose that n > 1 and that u 2 D 0 .Rn / is invariant under SL.n; R/ as a generalized function. Prove the existence of c1 and c2 2 C such that u D c1 1Rn C c2 ı. If, in addition, u is invariant under GL.n; R/, show that u D c1 1Rn .  14.36. (Bochner’s Theorem.) Consider p 2 C.Rn /. Prove that the following conditions are equivalent:

(i) The convolution operator  7!   p is positive on C01 .Rn /, that is, we have .p  ; /  0 for all  2 C01 .Rn /, where .
;
/ denotes the Hermitian inner product from (14.18).

14 Problems

205

P (ii) p is positive definite, that is, kl;mD1 cl cm p.xl xm /  0, for all c1 ; : : : ; ck 2 C and x1 ; : : : ; xk 2 Rn , where k 2 N. In other words, .p.xl xm //1l;mk is a positive semidefinite matrix. (iii) p equals F , where  is a positive Radon measure on Rn of finite mass .1/ D p.0/; see Remark 3.21. 14.37. Prove that the following functions on Rn are positive definite: ei a , for a 2 2 Rn ; e ak
k and e ak
k , both for a > 0 (for the latter; see Problem 18.8.(i)). Why is cos positive definite on R while sin is not?  14.38. (Dirichlet problem on Œ 0; 1  and Green’s function.) Set J D  0; 1 Œ and consider the pushforward Radon measure
 1J on R2 as in Example 10.7. Let I be the identity mapping acting in the space of locally integrable functions. The function g W R2 ! C is said to be Green’s function associated with the differential 2 operator D 2 on J if g is continuous and vanishes on R2 n J , while

.D 2 ˝ I /g D
 1J

g.0; / D g.1; / D 0

and

. 2 J /:

(14.43)

Here the differential equation is an equality in D 0 .R2 /. Further, suppose that f is a locally integrable function on R. Show that u 2 D 0 .R/ given by u.x/ D

Z

1 0

g.x; / f ./ d  D

Z

g.x; / f ./ d  R

.x 2 J /

is a solution to the Dirichlet problem D 2 u D f in D 0 .J / and u.0/ D u.1/ D 0. Consider the particular case of f 2 C01 .R/ and furthermore the continuous linear mapping ˝f W C01 .R/ ! C01 .R2 / satisfying ˝f ./ D  ˝ f . Then derive that u D t .˝f /g 2 E 0 .R/. Prove that E D 21 j
j is a fundamental solution of the differential operator D 2 on R and deduce that T E satisfies the inhomogeneous differential equation in (14.43). Note that x 7! ax C b is a solution of the homogeneous differential equation and determine a and b 2 C such that g.x; / D T E.x/ C ax C b. Verify that g is uniquely determined and that ( .1 x/; 0    x  1; g.x; / D x.1 /; 0  x    1: Now let p > 0 and apply the same method to the differential operator D 2 C p 2 I on J . Show that in this case, sinh p  sinh p .1 x/ ; 0    x  1; p sinh p e pj
j 1 and g.x; / D sinh p x sinh p .1 / ED 2p ; 0  x    1: p sinh p

€

Observe that Taylor expansion of E and g with respect to the variable p shows that the corresponding distributions obtained before arise in the limit as p # 0.

206

14 Fourier Transform

Finally, note that in [7, Exercises 7.69 and 7.70] this technique is applied in higherdimensional situations in a classical setting. 14.39. (Fourier transform of a˙ .) Let a be as in Problem 13.2. Prove that a˙ 2 S 0 .R/ for every a 2 C, and that for every  2 S.R/, a 7! a˙ ./ is a complexanalytic function on C. Deduce that a 7! a˙ is a complex-analytic family of tempered distributions and that a 7! F a˙ also has this property. Suppose 0 < a < 1. Show that aC C a and aC a are distributions on R that are homogeneous of degree a 1, and furthermore even and odd, respectively. Derive the existence of c˙ .a/ 2 C such that F .aC C a / D cC .a/.1C

a

C 1

a

F .aC

/;

a / D c .a/.1C

a

1

a

/:

By means of testing these identities against the Schwartz functions x 7! .x/ D 2 e x =2 and x 7! x.x/, respectively, and using (13.39) and Corollary 13.6, compute   cC .a/ D 2 .1 a/ cos a and c .a/ D 2i .1 a/ sin a: 2 2 In addition, deduce (see Problem 13.13 or [7, Exercise 6.60.(iii)] for different proofs)     Z cos cos   a : (14.44) x dx D .a/ xa 1 sin sin 2 R>0 Conversely, one may use (14.44) and Theorem 14.24 to prove cC .a/ cC .1

a/ D 2;

.a/ .1

that is,

a/ D

 ; sin a

(14.45)

for a 2 C n Z, the reflection formula from Corollary 13.6. Furthermore, in the notation of Problem 13.3, show that 

F a˙ D e

i 2a

.1

a/.1˙

a

C e i a 1 a / D

e i 2 a : .  i 0/a

Prove the validity of these formulas for all a 2 C using analytic continuation. Deduce, in the notation of Problem 13.5, for a 2 C not a positive even integer and not a negative odd integer, that F jja D F j
ja D

 a jj1 a ; 2  2 .1 C a/ sin a j
j 1 a ; 2 2 .1

a/ cos

for a 2 C not a nonnegative even integer and not a negative odd integer. Note that the case of a D 1 occurs in Problem 14.27.(iii), in the form F j
j D 2j
j 2 . Use the reflection formula from Lemma 13.5 to conclude, for a 2 C not a nonpositive even integer and not a positive odd integer, that

14 Problems

207

cos

 aFj
j 2

a

D

 j
ja .a/

1

:

(14.46)

Note that the case of a D 2 occurs in Problem 14.27, in the form F j
j

2

D

j
j.

14.40. For 0 < s < 1, consider the constant c.1; s/ from Problem 10.14. Use integration by parts to show that c.1; s/ D 4 . 2s/ cos s > 0. In particular, show that c.1; 12 / D 2 and lims#0 c.1; s/ D lims"1 c.1; s/ D C1. 14.41. For symmetric A 2 GL.n; R/, denote by sgn A the signature van A, i.e., the number of positive minus the number of negative eigenvalues of A, all counted with multiplicities. Prove by means of analytic continuation or (14.44) that n

Fe

i 2 hA
;
i

.2/ 2 Dp ei j det Aj

 4

sgn A

1 1
;
i 2 hA




:

14.42. (Metaplectic representation.) Let SL.2; R/ be as in Problem 14.35 and write sl.2; R/ for the corresponding Lie algebra; see [7, Sect. 5.10]. Then the following matrices are a basis for sl.2; R/: hD

1 0

0 ; 1

eC D

0 1 ; 00

e D

0 0 : 10

For a and b 2 sl.2; R/, define the commutator Œ a; b  as ab also belongs to sl.2; R/, and in addition, that Œ h; e ˙  D ˙2 e ˙ ;

ba. Show that Œ a; b 

Œ e C ; e  D h:

Next define the linear mapping ! from sl.2; R/ to the linear space of differential operators acting on S.R/ by 1 !.h/ D x @x C ; 2

!.e C / D

i 2 x ; 2

!.e / D

i 2 @ : 2 x

Verify that ! is a homomorphism of Lie algebras; more precisely, prove that ! maps into the linear space of skew-Hermitian operators with respect to the Hermitian inner product (14.18), and that ! preserves commutators. Furthermore, demonstrate that F B !.e / D !.e C /. Define the following mappings from R to the group GL.S.R// of invertible linear operators from S.R/ into itself: t

t 7! !.exp.th// z WD e 2 .e t / ;

i

2

C t 7! !.exp.te z // WD e 2 tx ;
 x2 1 t 7! !.exp.te z // WD p e i 4 C 2t  : 2 t

Here the last expression defines a convolution operator. Prove that we obtain oneparameter subgroups of GL.S.R//. Show that these subgroups have !.h/, !.e C /,

208

14 Fourier Transform

and !.e / as infinitesimal generators, respectively. Hint: in the third case, apply Problem 14.41 and compare with Problem 17.7. Verify that all the operators in these subgroups are unitary with respect to the Hermitian inner product on L2 .R/. The Theorem of Shale–Weil asserts that there exists a homomorphism of groups ! z f R/, the double covering group of SL.2; R/, to the group of unitary linear from SL.2; operators acting in L2 .R/ having ! as its tangent mapping. In particular, therefore, ! z B exp D exp B ! on sl.2; R/. The three one-parameter groups constructed above f R/. generate a group G of unitary operators in L2 .R/ that is isomorphic to SL.2; Now consider the following differential operators on S.R/: a D x ˙ @x :

Note that these operators are each others adjoints. Then we have the commutator relation Œ a ; aC  D 2I

Œ a ; .aC /j  D 2j .aC /j

hence

1

.j 2 Z>0 /:

Write 0 .x/ D e

1 2 2x

and

j D .aC /j 0

.j 2 Z>0 /:

(14.47)

Then a 0 D 0, and therefore, for j 2 Z>0 , a j D a .aC /j 0 D .Œ a ; .aC /j  C .aC /j a /0 D 2j .aC /j

1

0 D 2j j

1:

Consequently, aC is called a creation operator and similarly a an annihilation operator. In the notation (14.18), we now compute the inner products .j ; l / D .aC j

l / D 2l.j p D 2 lŠ ıjl .0 ; 0 / D 2 lŠ  ıjl : 1 ; l /

l

D .j

1; a

1 ; l 1 /

l

p 1 In other words, ..2j j Š / 2 j /j 2Z0 forms an orthonormal system in L2 .R/. By means of mathematical induction it is easy to prove the existence of polynomial functions Hj of degree j such that j D Hj 0 . The Hj are the Hermite polynomials. For any  2 S.R/, verify aC .x/ D

1

2

e 2 x @x .e

1 2 2x

/I

2

Hj .x/ D . 1/j e x @jx e

deduce

which is the Rodrigues formula for Hj . The Hermite functions j .x/

1 WD p p Hj .x/e j 2 jŠ 

1 2 2x

j

x2

;

with

.j 2 Z0 /

form an orthonormal basis for L2 .R/. For a proof of the completeness, suppose that f 2 L2 .R/ is orthogonal to all of the j . Then define F W C ! C by F ./ D R ix x 2 =2 f .x/ dx. Show that F is a complex-analytic function on C satisfying Re

14 Problems

209

@j F .0/ D . i /j

Z

xj e

1 2 2x

R

f .x/ dx D 0

.j 2 Z0 /:

Conclude that F must vanish identically on C. For arbitrary  2 R, this yields Z 1 2 e ix e 2 x f .x/ dxI 0 D F ./ D R

2

and hence x 7! e x =2 f .x/ D 0 in L2 .R/ on account of Theorem 14.32. Set k D i.e e C / and extend ! by linearity to sl.2; C/. The differential operator 2!.k/ D x 2 @2x is known as the Hermite operator. The Hermite functions are eigenfunctions of the Hermite operator. In fact, the identities x D 21 .a C aC / and @x D 12 .a aC / imply that i 1 !.e ˙ / D .a ˙ aC /2 ; !.k/ D .a aC C aC a /: 8 4 This entails  1 1 1 !.k/ j D .a j C1 C 2j aC j 1 / D .2j C 2 C 2j / j D j C j: 4 4 2

Note that the Fourier transform commutes with the Hermite operator and that the eigenspaces of the Hermite operator are of dimension 1. Deduce that the Hermite functions are eigenfunctions of the Fourier transform (compare with Problem 14.9). More specifically, the identities F B aC D i aC B F and Fe 0 D 0 lead to e F

j

e B .aC /j DF

0

e D . i /j .aC /j F

0

D . i /j

j

.j 2 Z0 /: (14.48)

It is remarkable that the Fourier transform also belongs to the group G. Indeed, we have  i  i 1 1 1 e jI exp !.k/ j D e 2 .j C 2 / j D . i /j C 2 j D . i / 2 F 2 hence   i   1 k : Fe D ! z i 2 exp 2 In turn, this explains why the Fourier transform and the Hermite operator commute. Once it is known that the collection of Hermite functions . /j 2Z0 is an orthonormal basis for L2 .R/, then one may define the normalized Fourier transform e as the linear mapping W L2 .R/ ! L2 .R/ satisfying (14.48); see Problem 15.14. F The construction above is essentially the “algebraic” treatment of the harmonic oscillator in quantum mechanics. For this reason the homomorphism ! z (or !) is called the metaplectic representation, Segal–Shale–Weil representation or oscillator f R/; see [9, Chap. 4]. representation of SL.2;

210

14 Fourier Transform

It is straightforward to generalize the treatment above to S.Rn / by considering tensor products of Hermite functions, starting with the differential operators ! n .h/ D

n X

j D1

xj @j C

n ; 2

! n .e C / D

i k
k2 ; 2

! n .e / D

i
: 2

 a 14.43. (Fourier transform of RC .) Let n  2. Consider the complex-analytic fama ily of distributions .RC /a2C from Chap. 13 and let C˙ be as in (13.13). Show that a a RC 2 S 0 .RnC1 /, for all a 2 C. Prove that .F RC /a2C is a complex-analytic family of tempered distributions. For a 2 C, apply Problem 12.14.(xi) to demonstrate the following identity in S 0 .RnC1 /: a F RC D .kk2

.

i 0/2 /

a 2

WD lim..; / 7! kk2 #0

a to RnC1 n Verify that the restriction of F RC satisfying

.; / 7!

€e

i  2a

. 2 .kk2

kk2 /  2/

a 2 a 2

S

˙

.

i /2 /

a 2

:

@C˙ is the real-analytic function

;

.; / 2 C˙ I

;

.; / 2 RnC1 n

[ ˙

C˙ :

a For Re a < 0, show that F RC is given by a locally integrable function on RnC1 . In a the case of Re a  0, prove that F RC is a distribution of order at most N C 1 if N 2 Z0 satisfies N  Re a.  14.44. Let a > 0. Compute F sinc and show that Z Z  sinc a x dx D sinc2 a x dx D : a R R

Determine F sinc2 and obtain the value of the second integral once more. See Problem 16.19 or 16.21 and [7, Example 2.10.14 or Exercises 0.14, 6.60 or 8.19] for different methods. 2

14.45. Consider ua .x/ D e ax =2 as in Example 14.11, but now for Re a  0. Let t 2 R. Prove that ui t 2 S 0 and that lima!i t; Re a>0 ua D ui t in S 0 . Calculate F ui t . 14.46. For what  2 C does the difference equation T1 u D  u have a solution u 2 S 0 .R/ with u ¤ 0? Describe the solution space for every such . The same questions for the difference equation 12 .T1 u C T 1 u/ D  u.

14.47. Let u 2 S 0 .Rn / with u ¤ 0 be a solution of the differential equation
u D  u, with  2 C. Prove that   0 and that u has an extension to an entire analytic function on Cn . Show that u is a polynomial if  D 0. Discuss the rotationally symmetric solutions u, for every   0.

14 Problems

211

14.48. Let  be a probability measure on Rn . Prove that  2 S 0 and that F  is a continuous function on Rn , with jF ./j  1 for all  2 Rn and F .0/ D 1. Hint: take r 2 C01 , 0  r  1 and r .x/ D 1 whenever kxk  r. Verify that F .r / converges uniformly on Rn as r ! 1. Also prove that F .  / D F  F  if  and  are probability measures.  14.49. (Hilbert transform.) The Hilbert transform H  2 C 1 .R/ of  2 C01 .R/ is defined as Z 1 1 1 .x y/ H .x/ D   PV .x/ D lim dy .x 2 R/:  y #0  y jyj> (i) Show that H  2 S 0 .R/ and that we have the following identity of linear operators on C01 .R/: F B H D i sgn B F : The question of natural domain and range spaces for H is subtle. For instance, extension of the definition of H to all of S 0 .R/ by means of the formula above is impossible on account of the discontinuity of the sign function. Furthermore, if  2 C01 .R/ and F .0/ ¤ 0, then H  … L1 .R/ because the Fourier transform of the latter function is discontinuous at 0. Nevertheless, there are positive results. (ii) Prove that kH k D kk, where k
k denotes the L2 norm. Verify that H possesses an extension to a unitary isomorphism H from L2 .R/ onto L2 .R/ and show that H 2 D I . 1 x (iii) Set f .x/ D 1Cx 2 and g.x/ D 1Cx 2 . Prove H f D g and verify that the identity kf k D kgk is corroborated by Problem 14.19. (iv) Demonstrate that H cos D sin and H sin D cos in two different ways. Deduce that H e˙i D i e˙i . (v) For a > 0, consider u D 1Œ a; a  2 L2 .R/. Verify, for x 2 R, ˇx C aˇ sin a
1 ˇ ˇ .F B H /u D 2i ; H u.x/ D log ˇ ˇ;  x a j
j Z Z ˇx C aˇ  ˇ ˇ log2 ˇ sinc2 a dx D : ˇ dx D a  2 ; x a a R>0 R

(vi) Deduce, for 0 < b ¤ a, that

ˇ aˇ  sin a
 ˇ ˇ ./ D i log ˇ F ˇ j
j  Ca

(vii) Verify H . sin
a
/ D

1 cos a



and and H . 1

Z

R>0

cos a
/


ˇa C b ˇ 1 sin ax sin bx ˇ ˇ dx D log ˇ ˇ: x 2 a b D

sin a
.


 14.50. Consider a and b 2 C with a2 C b 2 ¤ 0 and  2 C01 .R/. Show that 2 C 1 .R/ satisfies Cauchy’s integral equation Z 1 b .y/ dy D .x/ if D 2 a .x/ C PV .a I b H /  x y a C b2 R

212

14 Fourier Transform

in the notation of Problem 14.49. 14.51. (Poisson kernel, associated Poisson kernel, and Hilbert transform.) (See Problem 18.8 for a generalization in higher dimensions.) Let the notation be as in Problem 14.49. Define the Poisson kernel .P t / t >0 and the associated Poisson kernel .Q t / t >0 , respectively, by setting, with z D x C i t 2 C, P t .x/ WD P .x; t/ D Q t .x/ WD Q.x; t/ D

1 1 1 t D .t D z  x2 C t 2 t 1 1 1 x D .t Re D 2 2 z  x Ct t

Im

1 ; 1 C x2 x 1  / : 1 C x2

1 

/

(i) Prove that P and Q are harmonic functions on f .x; t/ 2 R2 j t > 0 g and that P t 2 L1 .R/ and Q t 2 L2 .R/, for all t > 0. (ii) Show that F P t D e t j
j and F Q t D i sgn e t j
j , for t > 0. (iii) Conclude (compare with Problem 5.3) lim P t D ı t #0

and

lim Q t D t #0

1 1 PV  x

in

S 0 .R/:

In particular, verify that we have the following identity of functions, for  2 S.R/: lim P t   D  and lim Q t   D H : t #0

t #0

(iv) Demonstrate H P t D Q t , kP t k D kQ t k and Q t  D P t H , for  2 S.R/. (v) Furthermore, deduce, for t and t 0 > 0, that 1.P t / D 1;

P t  P t 0 D P t Ct 0 ;

Q t  P t 0 D Q t Ct 0 :

As a consequence, .P t / is said to have the semigroup property. Young’s inequality from Problem 11.22 may be used to show that the convolution involving Q t is well-defined.  14.52. (Characterization of Hilbert transform.) According to Problem 14.49.(i) and (iv) the Hilbert transform H W C01 .R/ ! S 0 .R/ satisfies

F B H D i sgn B F ;

and as a consequence,

H cos D sin :

Without explicit computation of Fourier transforms we will show that these properties imply that H is given by H  D   1 PV x1 . (For a different proof, see Problem 15.15.) To this end, show that H commutes with translations and deduce that there exists a uniquely determined u 2 D 0 .R/ such that H D u . Next, prove that H commutes with positive dilations and deduce that u is homogeneous of degree 1. Conclude the existence of a1 and a2 2 C such that u D a1 PV x1 C a2 ı and finally prove that H anticommutes with reflection and deduce that u D 1 PV x1 .

14 Problems

213

14.53. (Complex-differentiable functions and Hilbert transform.) Let f W C ! C be complex-differentiable and U.r/  C the upper half of the disk of radius r > 0 with center 0. Suppose that f satisfies decay properties for jzj ! 1 with  < arg z < 0 that entail, for every w with Im.w/ > 0, Z Z f .z/ f .t/ lim dz D dt: r!1 @U.r/ z w t w R Application of Cauchy’s integral formula (12.11) leads to, for x 2 R, y > 0, and r sufficiently large, Z Z 1 1 f .z/ f .t/ f .x C iy/ D dz D dt: 2 i @U.r/ z x iy 2 i R t x iy Consider x temporarily as a constant and t 2 R as a variable. Then (14.41) implies 1 1 D PV C  i ıx : t x i0 t x As a consequence verify, denoting the restriction of f to R by fR , for x 2 R, 1 1 1 1 1 .fR / D PV .fR / C fR .x/ fR .x/ D lim f .x C iy/ D y#0 2 i t x i 0 2 i t x 2 Z 1 1 i 1 fR .t/ PV dt C fR .x/ D H fR .x/ C fR .x/; D 2 i t x 2 2 2 R with H the Hilbert transform from Problem 14.49. By taking real and imaginary parts of the resulting identity deduce the following identities of functions on R: H Re fR D Im fR

and

H Im fR D

Re fR :

Derive that H 2 D I on functions of the form fR . Verify the computations of the Hilbert transforms in Problem 14.49.(iv) and (vii) by applying the preceding i az identities in the case of the complex-analytic functions z 7! e iz and z 7! 1 ez , respectively. As a result, for a suitable class of complex-differentiable functions f , the real part of fR determines the imaginary part of fR via the Hilbert transform, and vice versa.  14.54. (Functions on R as “jump” of functions on CnR.) We use the notation from Problems 14.49 and 14.51. Set H˙ D f z D x C i t 2 C j ˙t > 0 g and define the function H iz , for z 2 HC , as in Example 14.30.

(i) Prove, for  2 S.R/ and z 2 HC , that Z Z 1 .y/ dy D F H iz ./ D e iz F ./ d : i Ry z R>0

(ii) More generally, show that

214

14 Fourier Transform

˚.z/ WD

1 i

Z

R

.y/ dy D ˙ y z

Z

e iz F ./ d  DW ˚˙ .z/

R≷0

.z 2 H˙ /:

Deduce that ˚ is complex-differentiable on C n R and that we may write ˚ D P ˚ ˙ ˙ with supp ˚˙  H˙ and ˚˙ complex-differentiable on H˙ .

Define u and v W HC ! C respectively, for z 2 HC , by u .z/ D

1 .˚C .z/ 2

˚ .z//

v .x; t/ D

and

(iii) Verify, for x C i t 2 HC , u .x C i t/ D P t  .x/ D

1 2

1 2 and prove in two different manners that v .x C i t/ D Q t  .x/ D

lim u .
C i t/ D  t #0

and

Z Z

i .˚C .z/ C ˚ .z//: 2

e

t jj

e ix F ./ d ;

e

t jj

e ix F .H /./ d ;

R

R

lim v .
C i t/ D H : t #0

(iv) Suppose that  is real-valued. Demonstrate the identities of functions u D 1 Re ˚C and v D 1 Im ˚C on HC . Deduce that u and v are harmonic  functions on HC . (v) Generalize the results above to the case of f 2 L2 .R/ instead of  2 S.R/. Define arg W C n R ! R by  < arg z < . Verify, for a > 0, f D 1Œ a; a  , and z D a C i t 2 HC , 1 aCx a x 1 arctan C arctan ; uf .z/ D .arg.z a/ arg.z C a// D   t t ˇ ˇ 1 ˇz C aˇ vf .z/ D log ˇ and lim uf .x C i t/ D 1Œ a; a  .x/; ˇI  z a t #0

for x 2 R and t > 0. Note the agreement with the results in Problem 14.49.(v). Furthermore, show that Z 1 aCx a x sin a d D arctan C arctan : e t  cos x  2 t t R>0

The complex-differentiable functions ˚˙ on H˙ have extensions to continuous functions on H˙ . In part (iii) it is asserted that lim t #0 .˚C .x C i t/ ˚ .x i t// D 2 .x/, for x 2 R. In general therefore, these extensions do not coincide on R. This means that it is impossible to “glue” the ˚˙ to define a single complexdifferentiable function on all of C. This point of view is of importance in the theory of hyperfunctions. On the positive side, u is a solution to the Dirichlet problem
u D 0 on HC and ujR D , for  2 S.R/; see Problem 12.4.

14 Problems

215

14.55. If u 2 L1 , v 2 E 0 , and u  v D 0, then prove that u D 0 or v D 0. Calculate 1  @j ı, for 1  j  n. 14.56. Prove the following assertions, in which the convolution product is as defined in Remark 11.23: (i) If u; v 2 L1 then u  v 2 L1 and F .u  v/ D F u F v. (ii) If u 2 L1 and u  u D u, then u D 0. (iii) If u 2 L2 and v 2 L1 , then u  v 2 L2 and F .u  v/ D F u F v. (iv) If v 2 L1 , there exists f 2 L2 such that the equation u  v D f has no solution u 2 L2 . 14.57. Let u0 2 S 0 .Rn /. Prove that there exists exactly one differentiable family t 7! u t W R>0 ! S 0 .Rn / such that

d u dt t

D
x u t and lim t #0 u t D u0 in S 0 .Rn /. Calculate F u t and u t .

 14.58. (Homogeneous fundamental solution of Laplace operator.) Let n  3. Prove the following assertions:

(i) v./ D kk 2 , for  2 Rn n f0g, defines a locally integrable function on Rn , and therefore a distribution on Rn , which we denote by v as well. (ii) v is homogeneous of degree 2 and v 2 S 0 .Rn /. (iii) Define E WD F 1 v. Then
E D ı, E 2 S 0 .Rn /, and E is homogeneous of degree 2 n. (iv) Every homogeneous fundamental solution of the Laplace operator is equal to E. (v) E is C 1 on Rn n f0g and invariant under all rotations. (vi) There exists a constant c such that E is equal to the locally integrable function x 7! c kxk2 n . (vii) c D .2 1n/cn with cn as in (13.37).  14.59. (Cauchy–Riemann operator.) In this problem the notation is as in Example 12.11. Our goal is to obtain a fundamental solution E of the Cauchy–Riemann @ operator @z by means of the Fourier transform. Differently phrased, we use that transform to solve the following partial differential equation for E on R2 :

@E .x; y/ @x

1 @E .x; y/ D 2ı.x; y/ D 2ı.x/ı.y/: i @y

To this end, suppose that x 7! E.x;
/ is a C 1 family in S 0 .R/ and deduce on the basis of the Fourier transform with respect to the other variable that under this assumption, dF E .x; / F E.x; / D 2ı.x/: dx Here F E denotes the partial Fourier transform. Write H for the Heaviside function on R and prove that, for a function  7! c./ still to be determined,

216

14 Fourier Transform

F E.x; / D 2.c./ C H.x//e x

..x; / 2 R2 /:

Observe that  7! e x does not define a tempered distribution on R if x > 0. Obviate this problem by means of the choice c./ D H./ and show that in this case F E.x; / D 2 sgn./ H. x sgn.// e x ..x; / 2 R2 /: Now prove (compare with the assertion preceding (12.13)) E.x; y/ D

1 1 ;  x C iy

E.z/ D

in other words,

1 ; z

and show that E is indeed a tempered distribution on C. Finally, derive (14.39) from the preceding formulas.  14.60. (Laplace operator.) In this problem we construct a fundamental solution of the Laplace operator
on Rn by a method analogous to the one in Chap. 13. To this end we define, for a 2 C with Re a > 0, the function Ra W Rn ! C by . n2 / 2 D Ra .x/ D c.a/ kxka n ; where c.a/ D : n cn . a2 /  2 . a2 /

(i) Prove that Ra is a locally integrable function on Rn and in consequence defines a distribution in D 0 .Rn /. Prove that in fact Ra 2 S 0 .Rn / and verify Ra .e

k
k2

/ D 1:

(ii) Let q W Rn ! R be the quadratic form x 7! kxk2 ; then verify that q W Rn nf0g ! R is a C 1 submersion. Now prove that on Rn n f0g and for Re a > 0 we have  a Ra D d.a/ q  C

nC2 2



with

d.a/ D

2 . a nC2 / 2 D a cn . 2 /

. n2 / . a n

2

nC2 / 2 : a .2/

Show that a 7! c.a/ admits an extension to a complex-analytic function on C and recall that a 7! aC is a family of distributions on R that is complex-analytic on C. Use this to verify that the formula above is valid on Rn nf0g for all a 2 C. (iii) Prove in two different ways that on Rn n f0g and for Re a > 0,
Ra D 2.a

n/ Ra

2

:

(14.49)

Now suppose n  3. Show that the definition of Ra 2 S 0 .Rn / may be extended to all a 2 C by setting, for k 2 Z0 and 0 < Re a C 2k  2, Ra D

2k

1 .a C 2j 1j k

Q

n/


k RaC2k :

(14.50)

(iv) Next prove that R0 D 0 on Rn n f0g and use positivity to deduce R0 D ı. Conclude that

14 Problems

217

 k
k2 n  D ı:
.2 n/cn

This is the way in which the fundamental solution of
from Problem 4.7 was obtained. More generally, demonstrate that R

2k

D

2k

. 1/k
k ı 0j 0. Also show that Z ˇ @nt 1 R.!; t/ˇ t D0 d!: A n ./ D 2 Sn

1

(v) Prove that the function Aa is homogeneous of degree a and is invariant under rotations acting on Rn , for all Re a > 0. Deduce the existence of d.a/ 2 C

14 Problems

219

such that Aa D d.a/ RaCn 2 D 0 .Rn / for all a 2 C, where .Ra /a2C is the complex-analytic family from Problem 14.60. Prove, for a 2 C n . 2N/, that d.a/ D

21 a  n . n2 / . a2 C 1/

RaCn D

and conclude that

1 Aa : d.a/

Note that it is sufficient to evaluate Aa .en / for Re a > 0 owing to the invariance under rotations, where en denotes the nth standard basis vector in Rn . Furthermore, use Legendre’s duplication formula (13.39). (vi) Now assume that n 2 N is odd. Apply Problem 14.60.(iv), part (ii), and the reflection formula from Lemma 13.5 to obtain, for x 2 Rn and  2 C01 .Rn //, n 1

. 1/ 2 .x/ D 2.2/n

1

Z

Sn

1

@nt

1

ˇ R.!; t/ˇ t Dh !; x i d!:

For n odd, the formula above is called Radon’s inversion formula, that is, it recovers  from its Radon transform R. Differently phrased:  is known when its integrals over all hyperplanes in Rn are known. Note that f N.!; h !; x i/ j ! 2 S n 1 g is the collection of all hyperplanes in Rn passing through x. Hence, this inversion formula is local, in the sense that only the integrals over hyperplanes close to x are needed. If n is even, the constant 1=d. n/ is undefined, while the definition of A n involves the constant 1= .1 n/ D 0. In this case, proceed as follows. (vii) Assume n to be even. Apply Legendre’s duplication formula twice, then apply the reflection formula in order to show that n

. 1/ 2 .n 1/Š D : n/ .2/n

1 d. n/ .1

In the notation of Problem 13.5, deduce n

. 1/ 2 .n 1/Š ıD .2/n

Z

Sn

that is, for  2 C01 .Rn / and x 2 Rn , n

.x/ D

. 1/ 2 .n 1/Š .2/n

Z

Sn

1

Z

t R>0 n

2

2 1 X

kD0

n

1

˚!  .j
j

X ˙

n

/ d!I

R.!; h !; x i ˙ t/ ˇ

ˇ @2k s R.!; s/ sDh !; x i

 t 2k dt d!: .2k/Š

Note that in the case that n is even, the inversion formula is not local. Consider the special case of n D 2 and define, for x 2 R2 and t  0,

220

14 Fourier Transform

R.x; t/ D

1 4

Z

X

S1 ˙

R.!; h !; x i ˙ t/ d!:

Then R.x; t/ is the average of R over the set of all lines in R2 at a distance t from the point x (note that N.!; h !; x i C t/ D N. !; h !; x i t/). The inversion formula now takes the form Z Z 1 1 R.x; t/ R.x; 0/ @ t R.x; t/ .x/ D dt dt D 2  R>0 t  R>0 t Z 1 d Rx .t/ D :  R>0 t Here the second identity is obtained by means of a formal integration by parts. The last expression is a Riemann–Stieltjes integral and equals, apart from differences in notation, the formula given by Radon in his 1917 paper.

Chapter 15

Distribution Kernels

A very important result in distribution theory is the Schwartz Kernel Theorem, which can be efficiently proved by means of the Fourier transform. Suppose X  Rn and Y  Rm are open subsets and k is a continuous function on X  Y . Then Z K W 7! K ; where K .x/ D k.x; y/ .y/ dy Y

defines a linear mapping K from the space of compactly supported continuous functions on Y to the space of continuous functions on X . The mapping K is said to be the linear integral operator or integral transform defined by the integral kernel1 k and is often denoted by kop . Integral transforms act in spaces of functions as changes of variables, which can be useful making problems more tractable. In the historical development of functional analysis, integral operators played an important role. Further assume 2 C01 .Y / and  2 C01 .X /. Integrating the function K against the function  gives Z k.x; y/ .x/ .y/ d.x; y/; XY

which implies

K ./ D k. ˝

/:

(15.1)

Here the tensor product  ˝ is defined similarly to (11.8). Since  ˝ belongs to C01 .X  Y /, the function k in (15.1) may be replaced by an arbitrary element of D 0 .X  Y /. We also observe that  ˝ depends continuously on  when is fixed and depends continuously on when  is fixed. As a consequence, (15.1) defines a continuous linear operator K W C01 .Y / ! D 0 .X /;

(15.2)

which is denoted by K D kop . These observations lead to a generalization of the concept of linear integral operator with far-reaching implications. 1

This notion of kernel is not to be confused with the one from linear algebra.

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_15, © Springer Science+Business Media, LLC 2010

221

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15 Distribution Kernels

Furthermore, the Kernel Theorem, Theorem 15.2, below asserts that for every K as in (15.2) there exists a unique k 2 D 0 .X  Y / such that K D kop , that is, (15.1) applies. On account of its uniqueness, we may call k, or kK , the distribution kernel, or Schwartz kernel, of K, while kop is said to be the operator generated by k. Some authors use the same symbol for the kernel and for the mapping generated by it. The Kernel Theorem enables the investigation of mappings from test functions to distributions by means of methods from distribution theory itself, through a study of the corresponding distribution kernels. Thus it is possible to obtain numerous properties of integral operators. In particular, this is the case for the Fourier transform: its intertwining properties with many operations on functions are easily obtained in this manner; see Problem 15.5 below. Kernels are important in the theory of linear partial differential equations; in particular, the Kernel Theorem unites differential and integral operators into a common framework. The verification of the theorem begins by deriving a formula for k in terms of K. This formula then implies the uniqueness of k but also provides a starting point for proving its existence. Principal ingredients in the proof are the basic property of Fourier decomposition, i.e., that a function can be represented as a continuous superposition of exponential functions, plus the fact that a multidimensional exponential function already equals a tensor product. After this introduction, we now discuss the notion of kernel in greater detail. Consider open sets X  Rn and Y  Rm , and let k 2 D 0 .X  Y /. Then .; / 7! k. ˝

. 2 C01 .X /;

/

2 C01 .Y //

(15.3)

is a bilinear form on C01 .X /  C01 .Y /. For fixed it becomes a linear form on C01 .X / and for fixed  it becomes a linear form on C01 .Y /. Both of these are distributions. Indeed, given compact sets K  X and K 0  Y , Theorem 3.8 implies the existence of a constant c > 0 and an order of differentiation N 2 Z0 such that jk. ˝ /j  c k ˝ kC N , for  2 C01 .K/ and 2 C01 .K 0 /. It is helpful to replace this by the equivalent estimate jk. ˝ Write K

/j  c kkC N k kC N

. 2 C01 .K/;

for the linear form  7! k. ˝ K ./ D k. ˝

/

2 C01 .K 0 //:

(15.4)

/ on C01 .X /, that is, . 2 C01 .X //:

(15.5)

The inequality (15.4) then gives an estimate as in Theorem 3.8 for K when fixed, so that K 2 D 0 .X /. Moreover, (15.4) shows that the linear mapping K W C01 .Y / ! D 0 .X /

given by

7! K

is sequentially continuous: if j ! 0 in C01 .Y / as j ! 1, then K D 0 .X /. It is also obvious from (15.4) that the linear mapping C01 .X / ! D 0 .Y /

given by

 7! .

is

7! k. ˝

//

j

! 0 in

15 Distribution Kernels

223

is sequentially continuous. To write this mapping in a similar way, one can first define the transposed kernel  k 2 D 0 .Y  X / of k, as follows. The isomorphism  W X  Y ! Y  X given by .x; y/ D .y; x/ induces isomorphisms of pullback   and pushforward  satisfying  k./ D k.  /, for every  2 C01 .Y X /. In particular,   . ˝/ D  ˝ . It follows from (15.5) that the mapping K k W C01 .X / ! D 0 .Y / generated by  k has the property K k . / D  k.

˝ / D k. ˝

/

.

2 C01 .Y //:

This notation thus makes it possible to discuss the two mappings associated with the bilinear form (15.3) on the same footing. It is useful to observe that the restriction to C01 .X / of the transpose mapping t K, and the mapping K k coincide, that is, t

K D K k :

(15.6)

In fact, one obtains tK. / D K ./ D k. ˝ / D K k . /, for  2 C01 .X / and 2 C01 .Y /. Furthermore, each of the mappings K and tK determines the other one. Let us consider two simple examples. If k 2 C 1 .X  Y /, then (15.5) yields the classical integral operator Z K .x/ D k.x; y/ .y/ dy . 2 C01 .Y //; Y

and the transpose operator is Z t K.y/ D k.x; y/.x/ dx

. 2 C01 .X //:

X

If k D u ˝ v, where u 2 D 0 .X / and v 2 D 0 .Y /, then the ranges of both mappings are one-dimensional. Indeed, K

D v. /u

and

t

K D u./v:

More generally, in the notation of Example 11.11, we note that the following assertions are equivalent, for any k 2 C01 .X  Y /:

(i) k 2 C01 .X / ˝ C01 .Y /. (ii) The integral operator kop W C01 .Y / ! C 1 .X / is of finite rank, that is, its range space is of finite dimension. (iii) The transpose of kop is of finite rank.

Finally, consider the special case of a sequentially continuous linear mapping K W C01 .Y / ! C 1 .X /. In order to derive a formula for at least one distribution kernel, say k, of K, suppose that  D  ˝ 2 C01 .X / ˝ C01 .Y / and note that .x/

D˝

.x;
/ D .x;
/ 2 C01 .Y /

and

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15 Distribution Kernels

K .x/ .x/ D K..x/ /.x/ D K..x;
//.x/; with .x;
/.y/ D .x; y/. Hence, (15.1) takes the form Z k./ D K..x;
//.x/ dx:

(15.7)

X

In fact, C01 .X / ˝ C01 .Y / is dense in C01 .X  Y / according to Remark 11.12 or Theorem 15.4 below; therefore (15.7) is valid for all  2 C01 .X  Y / and determines the unique kernel k of K. See Theorem 15.5 below for a more precise formulation. Example 15.1. Let I W C01 .X / ! C01 .X / be the identity mapping and denote the kernel as given in (15.7) by kI 2 D 0 .X  X /. We then obtain, in the notation of Example 10.7 and in particular with the natural embedding
W X ! X  X , Z kI ./ D .x; x/ dx D
 1X ./ . 2 C01 .X  X //: X

Furthermore, consider the special case of X D Rn and denote by d the difference mapping from (10.24) and by ı the Dirac measure on Rn . Using (10.25) one obtains kI D
 1Rn D d  ı 2 D 0 .Rn  Rn /:

(15.8)

This example shows that operators with very good regularity properties, meaning that they preserve C01 .X /, may have distribution kernels that are quite singular distributions. It also demonstrates that not every continuous linear operator from L2 .Y / to L2 .X / is necessarily generated by a kernel belonging to L2 .X  Y /. ˛ Suppose now that one has an arbitrary sequentially continuous linear mapping C01 .Y / ! D 0 .X /. Then it is natural to ask whether there exists a kernel in D 0 .X  Y / that generates this mapping. The main assertion of the following Kernel Theorem, which is due to Schwartz, is that this is indeed the case. Theorem 15.2. Consider open sets X  Rn and Y  Rm . A linear mapping K W C01 .Y / ! D 0 .X / is sequentially continuous if and only if it is generated by a distribution kernel k 2 D 0 .X  Y /; phrased differently, K ./ D k. ˝

/

. 2 C01 .X /;

2 C01 .Y //:

(15.9)

Moreover, the kernel k is uniquely determined by the mapping K. Proof. We begin by establishing uniqueness of k. This is the easy part of the proof; moreover, it suggests a way of verifying the existence of k. To this end, suppose that k 2 D 0 .X  Y / is a distribution kernel for K. Furthermore, let K be an arbitrary compact subset of X  Y and denote by K1 and K2 its orthogonal projections onto Rn and Rm , respectively. Then K1 and K2 are compact subsets of X and Y and consequently there exist  2 C01 .X / and  2 C01 .Y /

15 Distribution Kernels

225

such that  D 1 on K1 and  D 1 on K2 , respectively. Thus  ˝  2 C01 .X  Y /, while  ˝  D 1 on K  K1  K2 . Next, set l D . ˝ /k 2 E 0 .X  Y /. Then l and k coincide on C01 .K/, while l 2 E 0 .Rn  Rm /  S 0 .Rn  Rm / on account of Lemma 8.9 and (14.24). In view of (15.9) we now have, for .; / 2 Rn  Rm , l.ei  ˝ ei / D k.. ˝ /.ei  ˝ ei // D k.ei  ˝ ei / D K.ei /.ei  /: We recall that l is continuous and linear. Therefore, considering the integral as a limit of Riemann sums and using (14.17), we obtain, for every  2 C01 .K/  C01 .Rn  Rm /,   Z 1 k./ D l./ D l F .; / e ˝ e d.; / i i .2/nCm Rn Rm Z 1 F .; / l.ei  ˝ ei / d.; / D (15.10) .2/nCm Rn Rm Z 1 F .; / K.ei /.ei  / d.; /: D .2/nCm Rn Rm The identity (15.10) constitutes an explicit formula for the restriction of k to C01 .K/ in terms of K, which proves that this restriction is uniquely determined by K. Since K is an arbitrary compact subset of X Y , this implies that k is uniquely determined by K. Conversely, to prove the existence of k, let us assume that we are given a sequentially continuous linear mapping K W C01 .Y / ! D 0 .X /. Define the bilinear form B W C01 .X /  C01 .Y / ! C

B.; / D K ./:

by setting

(15.11)

Our next step is to derive an estimate similar to (15.4). Let K  X and K 0  Y be compact sets. Since K 2 D 0 .X /, there exists, on account of Theorem 3.8, an estimate jB.; /j D jK ./j  c kk

C

. 2 C01 .K//:

N

Here both c > 0 and N 2 Z0 depend on . Furthermore, since K is sequentially continuous and linear, 7! B.; / is an element of D 0 .Y /, for each  2 C01 .X /, and again on the basis of Theorem 3.8 we obtain jB.; /j  c0 k k

C

M

.

2 C01 .K 0 //:

These two inequalities show that the restriction of the bilinear form B to the linear subspace C01 .K/  C01 .K 0 / is separately continuous, which in turn yields that it is also jointly continuous. The verification of this fact is deferred to Proposition 15.3 below, so as to not interrupt the main argument of the proof of the theorem. In the present case the joint continuity implies that there exist constants c > 0 and N 2 Z0 such that

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15 Distribution Kernels

jB.; /j  c kkC N k kC N :

(15.12)

Note that this is, in effect, the estimate (15.4), which has thus been recovered. Let U and V be relatively compact open subsets of X and Y , and select  2 C01 .X / and  2 C01 .Y / such that  D 1 on U and  D 1 on V , respectively. Guided by (15.10), we introduce k;  2 D 0 .X  Y / by Z 1 k;  ./ D F .; / B.ei  ; ei / d.; /; (15.13) .2/nCm Rn Rm for all  2 C01 .X  Y /. Then k;  is well-defined. Indeed, (15.12) implies the existence of constants c > 0 and N 2 Z0 such that jB.ei  ; ei /j  c .1 C kk/N .1 C kk/N

..; / 2 Rn  Rm /:

Combination of this estimate with the fact that F  2 S.Rn  Rm / yields the convergence of the integral in (15.13). Let  2 C01 .U / and 2 C01 .V /. Then F . ˝ /.; / D F ./ F ./. Hence application of (15.13) with  D  ˝ implies Z 1 k;  . ˝ / D F ./ F ./ K.ei /.ei  / d.; / .2/nCm Rn Rm    Z Z 1 1 DK  F ./ e d  F ./ e d  i i .2/m Rm .2/n Rn D K. /./ D K ./: Phrased differently, t X Y U V B k;  2 D 0 .U  V / is the distribution kernel of the sequentially continuous linear mapping t X U B K B Y V W C01 .V / ! D 0 .U /, which is independent of  and . Here X U W C01 .U / ! C01 .X / is the linear injection defined in (7.1) (note that its transpose equals UX W D 0 .X / ! D 0 .U /, as in (7.2)). Since we have already established the uniqueness of distribution kernels, it follows that k;  is independent of the choices of  and , but it still depends on the choices of U and V . Thus, we may write k U;V instead of k;  . Invoking the uniqueness 0 0 once more, we see that k U;V D k U ;V on .U \ U 0 /  .V \ V 0 /. On account of Theorem 7.6, the k U;V have a common extension to k 2 D 0 .X  Y / that satisfies (15.9). This verifies the existence part of the theorem.  Next, we come to the proposition announced in the proof above. Proposition 15.3. Let the notation be as in the preceding proof. The restriction of B to C01 .K/  C01 .K 0 / is a jointly continuous bilinear form. Proof. Consider arbitrary sequences .j /j 2N in C01 .X / and . j /j 2N in C01 .Y / such that j ! 0 in C01 .X / and j ! 0 in C01 .Y /, respectively, as j ! 1. Define the linear functional uj on C01 .X / by uj ./ D B.; j /. Then uj is continuous, and so uj 2 D 0 .X /. Furthermore, limj !1 uj ./ D limj !1 B.; j / D 0,

15 Distribution Kernels

227

for all  2 C01 .X /; hence uj converges in D 0 .X / to 0. Accordingly, Theorem 5.5.(ii) implies B.j ; j / D uj . j / ! 0 as j ! 1. In other words, B is jointly sequentially continuous, and a fortiori this is also the case for the restriction of B to C01 .K/  C01 .K 0 /. Now both factors in this product can be provided with a countable separating collection of seminorms. But then Lemma 8.7 implies that the restriction is jointly continuous.  The preceding theorem immediately leads to the following result; Remark 11.12 and Theorem 16.17 contain different proofs. Theorem 15.4. Let the notation be as in Example 11.11. Then C01 .X / ˝ C01 .Y / is a dense linear subspace of C01 .X  Y /. Proof. Suppose that k 2 D 0 .X  Y / vanishes on C01 .X / ˝ C01 .Y /. If K W C01 .Y / ! D 0 .X / denotes the mapping generated by k, then (15.9) implies that K D 0 and this in turn gives k D 0, by the uniqueness of distribution kernels. Using the Hahn–Banach Theorem, Theorem 8.12, we now obtain the desired conclusion.  A continuous linear mapping K W C01 .Y / ! C 1 .X / is evidently also a continuous linear mapping C01 .Y / ! D 0 .X /. By Theorem 15.2 it is therefore generated by a distribution kernel; a rule for its computation is provided by the next theorem. The formulation of the result requires some notation. Define I ˝K W C01 .X Y / ! C 1 .X X /

by

.I ˝K/.x; x 0 / D K..x;
//.x 0 /;

for  2 C01 .X Y / and x and x 0 2 X . Then .x;
/ 2 C01 .Y /, so that K..x;
// 2 C 1 .X /. The notation for I ˝ K derives its justification from the fact that .I ˝ K/. ˝

/.x; x 0 / D . ˝ K /.x; x 0 / D .x/ K .x 0 / D K..x/ /.x 0 / D K.. ˝ /.x;
//.x 0 /;

(15.14)

taken in conjunction with Theorem 15.4. Theorem 15.5. The distribution kernel k 2 D 0 .X  Y / of a continuous linear mapping K W C01 .Y / ! C 1 .X / satisfies, in the notation discussed in Example 15.1, k D t .I ˝ K/
 1X D .I ˝ tK/
 1X ; or more explicitly k./ D

Z

X

K..x;
//.x/ dx

(15.15)

. 2 C01 .X  Y //:

Proof. On account of Theorem 15.4 it is sufficient to verify formula (15.15) for all  of the form  ˝ , where  2 C01 .X / and 2 C01 .Y /. It is not self-evident

228

15 Distribution Kernels

that
 1X 2 D 0 .X  X / may be applied to  ˝ K 2 C01 .X / ˝ C 1 .X /. For showing that this is the case, select  2 C01 .X / such that  D 1 on a neighborhood of supp . Then ˝K ˝ K D ˝.1 /K is a function in C 1 .X X / with support disjoint from the diagonal
.X /, which is supp
 1X . It follows that we may define
 1X . ˝ K / as
 1X . ˝  K / and that this definition is independent of the choice of . Thus, we have Z Z k. ˝ / D K ./ D .x/ K .x/ dx D .I ˝ K/. ˝ /.x; x/ dx X

X

D
 1X ..I ˝ K/. ˝

// D t .I ˝ K/
 1X . ˝

/:

Next, applying Theorem 15.4 once again and using Corollary 11.7, we see that the linear subspace C01 .X / ˝ C01 .Y / is dense in D 0 .X  Y /. Hence the identity t .I ˝K/ D I ˝ tK is valid on D 0 .X Y / as soon as it holds on C01 .X /˝C01 .Y /. Now observe, for  and  2 C01 .X / and  and 2 C01 .Y /, that t

.I ˝ K/. ˝ /. ˝

/ D . ˝ /. ˝ K / D ./ .K / D ./ tK. / D . ˝ tK/. ˝ / D .I ˝ tK/. ˝ /. ˝ /: 

This leads to the desired formula.

Example 15.6. Take X open in Rn , and consider a linear partial differential operator in X of order m with C 1 coefficients X P .@/ D a˛ @˛ I (15.16) j˛jm

here a˛ 2 C 1 .X /, for j˛j  m. Then we actually have a continuous linear mapping P .@/ W C01 .X / ! C01 .X /. Recall that tP .@/, the transpose operator of P .@/, is given by X t P .@/ D . @/˛ B a˛ : j˛jm

Applying (15.15), we obtain for the distribution kernel associated to P .@/ .I ˝ tP .@//
 1X 2 D 0 .X  X /:

(15.17)

Note that this assertion agrees with the following one, derived from (15.9): .I ˝ tP .@//
 1X . ˝

/ D
 1X . ˝ P .@/ / D P .@/ ./;

for  and 2 C01 .X /. For instance, if m D 0 and P .@/ D 2 C 1 .X /, then P .@/ reduces to the multiplication map  7! , with kernel .I ˝ /
 1X . In particular, the distribution
 1X , which satisfies supp
 1X D sing supp
 1X D
.X /, is the distribution

15 Distribution Kernels

229

kernel of the identity mapping; see Example 15.1. Furthermore, if P .@/ D @j , for 1  j  n, then its kernel is .I ˝ @j /
 1X . ˛ In applications, one often encounters (semi)regular kernels; these have images in spaces consisting of smooth functions instead of distributions, as we saw in the preceding example. Definition 15.7. Let k and K be as in the Kernel Theorem, Theorem 15.2. Then the distribution kernel k is said to be left semiregular if K sends C01 .Y / to C 1 .X / and if this mapping is continuous. Similarly, k is said to be right semiregular if the preceding condition holds with X and Y as well as K and tK interchanged. ˛ Theorem 15.8. Suppose that the distribution kernel k 2 D 0 .X  Y / is right semiregular; in other words, the mapping tK generated by  k is a continuous mapping C01 .X / ! C 1 .Y /. Then the mapping K extends to a linear mapping E 0 .Y / ! D 0 .X / that is sequentially continuous in the following sense: if a sequence .uj /j 2N converges in D 0 .Y / to u 2 E 0 .Y / and the supports of the uj are contained in a fixed compact set, then limj !1 Kuj D Ku in D 0 .X /. Proof. Define e W E 0 .Y / ! D 0 .X / K

by

e D u B tK Ku

.u 2 E 0 .Y //:

(15.18)

Being the composition of the continuous linear mappings tK W C01 .X / ! C 1 .Y / e is a distribution. The sequential continuity and u W C 1 .Y / ! C, the mapping Ku 0 0 e of K W E .Y / ! D .X / is also immediate from (15.18). e extends K W C 1 .Y / ! D 0 .X /, consider u D To show that K 2 C01 .Y /. 0 We then have, from (15.18), e ./ D K

.tK/ D K ./

. 2 C01 .X //;

which proves the claim. We can now obviously drop the tilde, and write K for the map defined by (15.18).  Definition 15.9. The case most frequently encountered in applications is that of a kernel that is both left and right semiregular. We then speak of a regular distribution kernel. ˛ For instance, kernels belonging to C01 .X  Y / are regular. Theorem 15.8 has the following consequence. Corollary 15.10. If k 2 D 0 .X  Y / is a regular distribution kernel, then the maps K and tK extend to sequentially continuous maps E 0 .Y / ! D 0 .X / and E 0 .X / ! D 0 .Y /, respectively.

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15 Distribution Kernels

Example 15.11. Consider u 2 D 0 .Rn /. On account of Theorem 11.2 we then have the linear mapping U WD u  W C01 .Rn / ! C 1 .Rn /

U .x/ D u.Tx B S /;

with

while the continuity of U is established in Theorem 11.3. The distribution kernel of U equals d  u 2 D 0 .Rn  Rn /, where d is the difference mapping from (10.24). Indeed, on the strength of Example 11.9 we conclude, for  and 2 C01 .Rn /, d  u. ˝ / D u 

./:

In particular, with u D ı one immediately obtains (15.8). In order to determine t U, it is sufficient to compute  .d  u/, in view of (15.6). It is straightforward to verify that  1 D  and d  D S d W Rn  Rn ! Rn  Rn . Theorem 10.8 and Corollary 11.7 then imply  D   on D 0 .Rn  Rn / and so  d  u D   d  u D .d / u D .S d / u D d  .S u/: Accordingly, t U D S u  (compare with Problem 11.11), which implies that t U W C01 .Rn / ! C 1 .Rn / is continuous too. It follows that k is a regular distribution kernel, and Corollary 15.10 then tells us that U can be extended to a sequentially continuous mapping E 0 .Rn / ! D 0 .Rn /. Thus we have recovered a major case of the existence part of Theorem 11.17. ˛ Example 15.12. As already pointed out in Chap. 12, fundamental solutions of differential operators with constant coefficients play an important part in the theory of such operators. In general, a similar part is played by fundamental kernels. Let P .@/ be as in (15.16). A distribution E 2 D 0 .X  X / is said to be a right fundamental kernel of P .@/ if the mapping E W C01 .X / ! D 0 .X / generated by it is a right inverse of P .@/, that is, P .@/E D I:

(15.19)

Likewise, E 0 2 D 0 .X  X / is said to be a left fundamental kernel of P .@/ if the corresponding mapping is a left inverse of P .@/, that is, E 0 P .@/ D I: Right fundamental kernels lead to existence theorems; left fundamental kernels immediately lead to uniqueness theorems. Let us consider (15.19) in more detail; in view of (15.9) and (10.6) it implies .P .@/ ˝ I /E. ˝ / D E.tP .@/ ˝ / D E .tP .@// D P .@/E ./ D ./ D
 1X . ˝ / .; 2 C01 .X //: Therefore, on account of Theorem 15.4, .P .@/ ˝ I /E D
 1X :

(15.20)

15 Distribution Kernels

231

Conversely, this identity leads to (15.19). Therefore (15.20) is an alternative characterization of right fundamental kernels of P .@/. A case frequently encountered is that of a right fundamental kernel that is regular. Suppose that E satisfies (15.19) and that both E and tE are continuous mappings C01 .X / ! C 1 .X /. Then, given arbitrary v 2 E 0 .X /, one can obtain a solution u 2 D 0 .X / of the inhomogeneous equation P .@/u D v. Indeed, one has only to set u D v B tE. This defines a distribution, and on account of Corollary 15.10 one may conclude P .@/u./ D u. tP .@// D v. tE tP .@// D v./

. 2 C01 .X //:

˛

Remark 15.13. In linear algebra, when considering linear spaces of finite dimension, one studies the null space and the image space of a linear operator, and one determines its inverse if the operator is bijective. In that theory, the operator and its inverse are quite similar in nature. In analysis, however, objects of major interest are differential operators, which act in linear spaces of infinite dimension. Under suitable conditions these operators are inverted by integral operators, which are given by classical kernels. It is only in the framework of distribution theory that the differential operators are also given by (distribution) kernels. As a consequence, in distribution theory both types of operators can be studied from a unified point of view. It should be noted, however, that the class of operators defined by distribution kernels is very large. For many aspects of the theory of differential operators, it is advantageous to retain important properties of the latter class, for instance the fact that they preserve the class of infinitely differentiable functions. In addition, a differential operator is a local continuous linear mapping, that is, it can only decrease the support of a function or distribution to which it is applied. The integral operators of interest do not have this property, but they are pseudo-local; this means that they decrease singular supports. Definition 15.14. A sequentially continuous linear mapping K W C01 .X / ! D 0 .X / is said to be pseudo-local if it has a sequentially continuous linear extension K W E 0 .X / ! D 0 .X / (recall that according to Theorem 15.8 this is the case if the kernel k 2 D 0 .X X / of K is right semiregular) and sing supp Ku  sing supp u, for all u 2 E 0 .X /. ˛

A sequentially continuous linear mapping K W C01 .X / ! D 0 .X / having a sequentially continuous linear extension K W E 0 .X / ! D 0 .X / is pseudo-local if and only if sing supp k 
.X / (for the analogous characterization of locality, see Problem 15.17). In fact, suppose that K is pseudo-local and select .x; y/ 2 X  X n
.X /. Then there exist disjoint open neighborhoods U of x and V of y in X such that .U  V / \
.X / D ;. Let X U W E 0 .U / ! E 0 .X / be the natural injection and V X W D 0 .X / ! D 0 .V / the restriction mapping. If u 2 E 0 .U /, then sing supp .V X B K B X U /u  U \ V D ;; in other words, we have the continuous linear mapping V X BK BX U W E 0 .U / ! C 1 .V /. This is equivalent to the assertion that the restriction of k to U  V belongs to C 1 .U  V /. The reverse implication is left to the reader.

232

15 Distribution Kernels

In this vein one obtains pseudo-differential operators (see Remarks 17.8 and 18.10) and their generalizations, the Fourier integral operators; see [6] or [14, Sects. VII.6 and VIII.6]. A starting point for these constructions is the formula, valid for P .D/ as in (15.16) with @ replaced by D and for  2 S.Rn /, Z 1 e i hx y; i P .x; /.y/ d.y; /: P .D/.x/ D .2/n Rn Rn In order to obtain the generalizations of the differential operators P .D/, one considers larger classes of functions P and more general phase functions in the second integral. Under some conditions, pseudo-differential operators can be composed; these operators then form an algebra. One of the fundamental aspects of their theory is that it is possible, in a canonical fashion, to associate with such an operator a certain function, called the symbol (for example, see (17.2)), with the property that algebraic operations on the operator get translated into differential-algebraic operations on its symbol. This is a sort of converse of the operation of quantization. There the problem is to associate to suitable functions f on the phase space R2n linear mappings Kf acting in L2 .Rn / such that the coordinate functions correspond to the momentum and position operators Dj and xj , for 1  j  n, and such that the properties of the functions f (algebra structure, positivity, boundedness, etc.) are reflected in some sense in the properties of the mappings Kf . For more details, see for instance [9]. ˛

Problems 15.1. (Generalization of Example 15.1.) Consider u 2 D 0 .X / and K W C01 .X / ! D 0 .X / given by K D  u. Prove that
 u 2 D 0 .X  X / is the distribution kernel of K. 15.2. (Kernel of a composition.) Let X  Rn , Y  Rm , and Z  Rp be open subsets, and let k 2 C01 .X  Y / and kz 2 C01 .Y  Z/. Prove that k and kz define sequentially continuous linear mappings K W C01 .Y / ! C01 .X / and e W C 1 .Z/ ! C 1 .X / is a sequene W C 1 .Z/ ! C 1 .Y /. Show that K B K K 0 0 0 0 tially continuous linear mapping with kernel Z kKBK kK .
; y/ kK fD f.y;
/ dy: Y

The right-hand side is said to be the Volterra composition product of kK f and kK . In this situation, define the diagonal mapping
and the projection  by
W X Y Z !X Y Y Z

and

 W X  Y  Z ! X  Z;

15 Problems

where

233


.x; y; z/ 7! .x; y; y; z/

and

.x; y; z/ 7! .x; z/;

 respectively. Verify that kKBK f D 
.kK ˝ kK f/.

 15.3. (Kernel and mappings of test functions.) Consider K as in the Kernel Theorem, Theorem 15.2.

(i) Let ˚ W C01 .X / ! C01 .X / and W C01 .Y / ! C01 .Y / be continuous linear mappings. Prove that t ˚ K W C01 .Y / ! D 0 .X / is a continuous linear mapping satisfying k t ˚ K D .t ˚ ˝ t /kK :

(ii) Now suppose that ˚ W X ! X and W Y ! Y are C 1 diffeomorphisms. Show that ˚ K  W C01 .Y / ! D 0 .X / is a continuous linear mapping satisfying k˚ K  D .˚ ˝  /kK : (iii) In particular, suppose that X D Y and ˚ D . Then demonstrate, in the notation (10.7), that k˚  K .˚

1 /

D .j˚ ˝ 1/.˚  ˝ ˚  /kK :

(iv) Verify that K commutes with ˚  if and only if kK D .j˚ ˝ 1/.˚  ˝ ˚  /kK : (v) Finally, let .˚ t / t 2R be a one-parameter family of volume-preserving C 1 diffeomorphisms with velocity vector field v and assume that K commutes with all ˚ t  , for t 2 R. Then prove, in the notation of (10.17), that n X

.vj @j ˝ I C I ˝ vj @j /kK D 0:

j D1

15.4. Apply Problem 15.3.(i) in order to derive (15.17) from Example 15.1.  15.5. (Kernel of Fourier transform.) Verify that the Fourier transform F W C01 .Rn / ! C 1 .Rn / is a continuous linear mapping. Prove that the distribution kernel kF of F is given by e i h
;
i 2 C 1 .Rn  Rn /. Show that the symmetry  kF D kF implies that F D tF and derive (14.19) and (14.20). Furthermore, obtain the identities from (14.10), (14.11), (14.29), and Problem 14.30.(i) by studying kF . (We note that (14.21) may also be obtained in this manner, but that Remark 14.17 contains an argument that is both shorter and more appropriate in this context.) Solve Problem 14.30.(iv). Finally, show that F extends to a sequentially continuous mapping F W E 0 .Rn / ! D 0 .Rn / satisfying F ı D 1.

15.6. (Mellin transform.) The linear integral operator M W C01 .R/ ! C 1 .R>0 / defined by the kernel k.a; x/ D H.x/ x a 1 is said to be the Mellin transform. (The identity k.a;
/ D .a/ C a and (13.7) can be used to extend this definition.) Prove

234

15 Distribution Kernels

that M B . x @x / D a B M, which expresses the fact that the Mellin transform diagonalizes the Euler operator x @x . In addition, show, for c > 0, that M B x c D Tc  B M;

M B log x D @ B M;

M B c D c

a

B M:

Finally, derive the second identity from the first by means of (10.15). See [7, Exercise 6.100] for more information about the Mellin transform.  15.7. Let X  Rn and Y  Rm be open subsets and consider u 2 D 0 .X  Y /.

(i) Suppose that Y is connected and that u satisfies .I ˝ @j /u D 0, for 1  j  m. Prove the existence of v 2 D 0 .X / such that u D v ˝ 1Y . (ii) Suppose that 0 2 Y and that u satisfies .I ˝ xj /u D 0, for 1  j  m. Prove the existence of v 2 D 0 .X / satisfying u D v ˝ ı Y , where ı Y denotes the Dirac measure on Y supported at 0.

For different proofs in special cases, see Problems 11.19 and 11.18.  15.8. (Distribution supported by linear submanifold.) Let X  Rn and Y  Rm be open subsets and assume that 0 2 Y . Suppose u 2 E 0 .X  Y / satisfies supp u  X  f0g. Let ı Y be the Dirac measure on Y supported at 0 and  W X ! X  Y the natural embedding with .x/ D .x; 0/. Prove the existence of u˛ 2 E 0 .X /, with ˛ 2 A WD f0g  .Z0 /m , such that X X X u˛ ˝ @˛ ı Y D .I ˝ @˛ / u˛ D @˛ . u˛ /; uD ˛2.Z0 /m

˛2.Z0 /m

˛2A

where the sum is actually finite. See Problem 11.20 for another proof in a special case. Observe that for k 2 N, any C k submanifold in Rp of dimension 0  n < p is locally C k diffeomorphic to a set of the form X  f0g, with X open in Rn ; see [7, Theorem 4.7.1.(iv)].  15.9. (Another proof of the Fourier inversion formula.) Let I and kI be as in Example 15.1 and k D kG , with G as in Lemma 14.12. Show that .Dj ˝ I C I ˝ Dj /k D 0

and

.xj ˝ I

I ˝ xj /k D 0

.1  j  n/:

Define the automorphism ˚ W Rn  R n ! Rn  R n

by

˚.x; y/ D .x; d.x; y// D .x; x

y/

and set kz D ˚  k 2 D 0 .Rn  Rn /. For 1  j  n, deduce that .Dj ˝ I /kz D ˚  .Dj ˝ I C I ˝ Dj /k D 0

and

.I ˝ xj /kz D 0:

Conclude that kz D 1Rn ˝ c ı, for some c 2 C and ı the Dirac measure on Rn and prove that k D ˚ ˚  k D ˚ .1Rn ˝ c ı/ D c d  ı D c kI . Finally, show that G D .2/n I .

15 Problems

235

Observe that an adaptation of the preceding arguments leads to another solution of Problem 14.7. The present approach is quite similar to the one in Problem 13.12.  15.10. Verify the formula for d  in (10.25) in a manner different from that in Example 10.20.  15.11. (Another proof of Theorem 11.3.) Let U be as in Theorem 11.3 and prove the characterization of U as in the theorem by computing kU .  15.12. Let u 2 D 0 .Rn / and define

U W C01 .Rn / ! D 0 .Rn /

by

kU D d  u:

For any 2 C01 .Rn /, prove that U D u  as in (11.1). Furthermore, verify that Ta and @j commute with U, for a 2 Rn and 1  j  n.

 15.13. (Characterization of Fourier transform.) Let K W C01 .Rn / ! D 0 .Rn / be a sequentially continuous linear mapping. Consider the following list of six conditions (compare with (14.10), (14.11), (14.19), (14.12), and Problem 14.30):

.i/ K Dj D j K; 

.iv/ K Ta D ei a K;

.ii/ .v/

K xj D 

Dj K;

K c D c K;

.iii/

t

K D K;

.vi/ K A D A K;

for every 1  j  n, a 2 Rn , c > 0, and every rotation A in Rn . (a) Prove that K equals the Fourier transform F if K satisfies only condition (i) but can be extended to a continuous linear mapping K W E 0 .Rn / ! D 0 .Rn / that satisfies K.ı/ D 1. Verify that K equals a multiple of F if K satisfies any of the following: (b) conditions (i) and (ii); (c) conditions (i) and (iii); (d) conditions (iv)–(vi); (e) conditions (iv) and (v), if, in addition, the range space of K consists of locally integrable functions that are continuous at 0.  15.14. In Problem 14.42 it is proved that the collection of functions ./j 2Z0 as in (14.47) is a basis in L2 .R/. Therefore one may define a continuous linear mape W L2 .R/ ! L2 .R/ by means of F ej D . 1/j j , for j 2 Z0 . Prove ping F e that this new definition of F coincides with the familiar one of normalized Fourier transform.  15.15. (Characterization of Hilbert transform.) Let K W C01 .R/ ! D 0 .R/ be a sequentially continuous linear mapping. Suppose that K commutes with all translations and positive dilations in R and anticommutes with reflection. Prove that K is a multiple of the Hilbert transform H from Problem 14.49. Compare with Problem 14.52.

236

15 Distribution Kernels

 15.16. Suppose that X is an open subset of Rn and that K W C01 .X / ! D 0 .X / is given by X X KD u˛ B @˛ ; that is, K D .@˛ / u˛ ; ˛2.Z0 /n

˛2.Z0 /n

where u˛ 2 D 0 .X / and the sum is locally finite. (We may think of K as a linear partial differential operator in X of locally constant order, with distributions as coefficients.) Show that K is a local operator. Prove that the distribution kernel k of K is given by X kD .I ˝ . @/˛ /
 u˛ and that supp k 
.X /: ˛2.Z0 /n

Conversely, suppose that K W C01 .X / ! D 0 .X / has a distribution kernel k that is supported by
.X /. Then prove that k, and therefore K, are of the form given above.  15.17. Consider a sequentially continuous linear mapping K W C01 .X / ! D 0 .X / and denote its distribution kernel by k 2 D 0 .X  X /. Show that K is local if and only if supp k 
.X /.

15.18. (Weak form of Peetre’s Theorem.) Let K W C01 .X / ! D 0 .X / be a local and sequentially continuous linear mapping. Prove that K is of the form given in Problem 15.16. In particular, if K maps into C 1 .X /, then K is locally a linear partial differential operator in X with C 1 coefficients. Hint: in order to demonstrate that the coefficients are of class C 1 , use ˛;V 2 C01 .X / that coincide with monomials x 7! x ˛ when restricted to a relatively compact open subset V of X . Peetre [17] proved that the latter conclusion holds even for local linear mappings; the continuity follows from the restriction on the supports. Let K W C01 .X / ! D 0 .X / be a local linear mapping. For this case, Peetre showed that for every x 2 X there exists a neighborhood U of x such that the restriction of K to U is a sequentially continuous linear mapping C01 .U / ! D 0 .U /, but that this is not true for x belonging to a locally finite set   X . Next, applying results from Problems 15.17 and 15.16, he obtained the existence of u˛ 2 D 0 .X / that are uniquely determined on X n  such that K is as in Problem 15.16 plus some distribution supported by , for any 2 C01 .X /.

Chapter 16

Fourier Series

The theory of Fourier series is a far-reaching analog of writing x 2 Rn as xD

n X

h x; ej iej ;

j D1

the decomposition of a vector into a sum of projections along the vectors ej , which form a complete system of orthonormal vectors .e1 ; : : : ; en / in Rn . Indeed, the theory of Fourier series asserts that a sufficiently smooth function u (of class C 2 is sufficient, see Problem 16.18) on R that is periodic with period a > 0 can be written as a uniformly convergent series X u.x/ D cn e i n!x : (16.1) n2Z

Here ! D

2 a

> 0 and cn , the nth Fourier coefficient of u, is given by 1 cn D a

Z

sCa

e

i n!x

.n 2 Z; s 2 R/:

u.x/ dx

s

For a locally integrable periodic function f with period a, the integral 1 M.f / WD a

Z

sCa

f .x/ dx

(16.2)

s

is said to be the mean of f . Note that the periodicity of f implies that f .x C na/ D f .x/, for all n 2 Z, and consequently M.f / D

1 na

Z

sCna

f .x/ dx s

.n 2 N/:

From this it follows in turn that

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_16, © Springer Science+Business Media, LLC 2010

237

238

M.f / D

lim

1

t s!1

t

s

Z

16 Fourier Series t

f .x/ dx: s

This result can be verified by choosing an n 2 N such that t  s C n a < t C a, then writing Z t 1 M.f / f .x/ dx t s s Z sCa Z sCn a  1 1  1 n f .x/ dx C f .x/ dx; D na t s t s t s n j D j ta .ts ns/a j  t 1 s . This further shows that (16.2) is and then using j a1 t s independent of the choice of s, as one also verifies directly. Our aim now is to reconstruct the theory of Fourier series from the theory of the Fourier transform of Chap. 14. In doing so, we will obtain (16.1), with convergence in the sense of distributions, even for arbitrary periodic distributions with period a. We begin by formulating an alternative definition of the mean M.u/ of a periodic distribution u. This definition is required because the formula

Z

sCa

u.x/ dx D u.1Œ s; sCa  /

s

cannot be generalized: the characteristic function 1Œ s; sCa  of the interval Œ s; s C a  is not a C 1 function. P Lemma 16.1. For every a > 0 there exists  2 C01 .R/ such that n2Z Tna  D 1. If u 2 D 0 .R/ and Ta u D u, then u./ D u./ if in addition  2 C01 .R/ and P n2Z Tna  D 1. If u is locally integrable, then u./ D a M.u/.

Proof. Begin byP choosing  2 C01 .R/ such that   0 and  > 0 on Œ s; s C a Œ . Then .x/ WD na/ defines a C 1 function on R. Indeed, note that n2Z .x for every bounded open interval I in R there exists a finiteP subset E of Z with .x na/ D 0 whenever x 2 I and n … E. This gives na/ D n2Z .x P .x na/, for x 2 I , and from this immediately follow both the convergence n2E and the fact that  2 C 1 .R/. Furthermore,  is periodic with period a. For every x 2 R there exists an n 2 Z with x na 2 Œ s; s C a Œ , and therefore .x/ D .x na/  .x na/ > 0. This gives  WD = 2 C01 .R/ and X n2Z

Tna  D

X n2Z

Tna

 

D

X Tna  n2Z

Tna 

D

X Tna  n2Z



D

 D 1: 

Using the periodicity of u we obtain, by a summation that is in fact finite,  X  X X u./ D u  Tna  D T na u. T na / D u. T na / D u./: n2Z

n2Z

n2Z

(16.3)

16 Fourier Series

239

To obtain the latter assertion we apply the foregoing to  D 1Œ s; sCa  , which is permitted if u is locally integrable. Naturally, in this case u./ D a M.u/.  Definition 16.2. Let a > 0. For an arbitrary u 2 D 0 .R/ satisfying Ta u D u, the number M.u/ WD a1 u./, with  as in Lemma 16.1, is said to be the mean of u. ˛ Theorem 16.3. Every periodic distribution on R is tempered. Proof. For u 2 D 0 .R/ we have an estimate of the form (3.4), with K D supp  and  as in Lemma 16.1. Formula (16.3), with  instead of , implies, for every  2 C01 .R/, ˇ ˇ X ˇX ˇ ju./j D ˇˇ u. T na /ˇˇ  c sup j@˛ .x C na/j: n2Z j˛jk; x2K

n2Z

There exists a constant b > 0 with 1 C jnjP b .1 C jx C naj/, for all n 2 Z and x 2 K. Now choose d > 1 such that s WD n2Z .1 C jnj/ d < 1. This yields ju./j  c b d s

sup

.1 C jx C naj/d j@˛ .x C na/j  c 0 kkS.k;d / ;

j˛jk; x2K; n2Z



for a constant c 0 > 0; see (14.8).

A sequence .cn /n2Z in C is said to be of moderate growth if positive constants c and N exist such that jcn j  c jnjN

.n 2 Z n f0g/:

(16.4)

Lemma 16.4. Let ! > 0. The distribution X vD cn ın! n2Z

on R is tempered if and only if the sequence .cn /n2Z of coefficients in C is of moderate growth. In that case k X lim cn ın! D v j;k!1

and

lim

j;k!1

both with convergence in S 0 .

k X

nD j

nD j

cn ei n! D u WD S B F v;

(16.5)

Proof. We have v 2 S 0 if and only if there exist c > 0, k 2 Z0 , and N 2 Z0 such that jv./j  c kkS.k;N / , for all  2 C01 .R/. See (14.8), Definition 14.19, and (14.25).

240

16 Fourier Series

First, suppose that v 2 S 0 . Let  2 C01 .R/, supp    !; ! Œ, and .0/ D 1. For n 2 Z and  D Tn!  we obtain v./ D cn , while on the other hand, kkS.k;N /  d .1 C jnj/N , where d is a constant that is independent of n but contains the C k norm of . This implies that .cn /n2Z is of moderate growth if v 2 S 0. Conversely, if the sequence .cn /n2Z satisfies (16.4), we obtain, for every b > 1, ˇ ˇ X ˇX ˇ ˇ jv./j D ˇ cn .n!/ˇˇ  c .1 C jnj/N j.n!/j n2Z

 X .1 C jnj/ c

n2Z

b

sup.1 C jnj/N Cb j.n!/j; n2Z

n2Z

for all  2 C01 .R/, which implies that v 2 S 0 . The convergence in S 0 follows from analogous estimates, with v replaced by the difference of v and the finite sum. The latter assertion follows from (14.35) and the continuity of S B F from S 0 to S 0 ; see Theorem 14.24.  Definition 16.5. When (16.5) holds, u is said to be the distributional Fourier series with coefficients .cn /n2Z . One uses the notation (see also Problem 5.10) X ˛ cn ei n! in S 0 : uD n2Z

Lemma 16.6. The derivative of a distributional Fourier series can be obtained by termwise differentiation; more exactly, if the sequence .cn /n2Z in C is of moderate growth, then X X @ cn ei n! D i n! cn ei n! in S 0 : n2Z

n2Z

Proof. Application of Lemma 5.9 yields @

X n2Z

cn ei n! D lim @ j;k!1

k X

cn ei n!

in S 0 :

nD j

But from Lemma 16.4 we deduce, considering that the sequence .i n! cn /n2Z is of moderate growth, k X X lim @ cn ei n! D i n! cn ei n! in S 0 :  j;k!1

nD j

n2Z

Theorem 16.7. Let a, ! > 0 and a ! D 2. The Pmapping S B F then is a linear bijection from the linear space of distributions n2Z cn ın! , where the sequence

16 Fourier Series

241

.cn /n2Z in C is of moderate growth, onto the linear space of periodic distributions with period a. If X uD cn ei n! in S 0 n2Z

then

cn D M.e

i n!

u/

.n 2 Z/:

(16.6)

Proof. If u 2 D 0 .R/ and Ta u D u, then u 2 S 0 on the strength of Theorem 16.3. 1 We obtain v WD 2 F u 2 S 0 and according to (14.29) the equation Ta u D u is equivalent to e i a vPD v, that is, .ei a 1/ v D 0. On account of Theorem 9.5 we conclude that v D n2Z cn ın! , for a sequence .cn / in C that can be of moderate growth at most growth, because v 2 S 0 . In reaching this conclusion we P have also seen that conversely, u WD S B F v is periodic with period a if v D n2Z cn ın! 2 S 0 , where we have used Theorem 14.24. The assertion in (16.5) concerning the convergence follows from Lemma 16.4 and the fact that F is continuous from S 0 to S 0 ; see Theorem 14.24. For (16.6) we write  X X M.e i n! u/ D M ck ei k! e i n! D ck M.ei.k n/! / D cn : k2Z

k2Z

The convergence does not pose a problem, because M is a continuous linear form on S 0 , as one straightforwardly reads from the definition of the mean.  Example 16.8. Let a and ! be positive numbers satisfying a ! D 2 and consider X ın! 2 S 0 .R/: u! D n2Z

P

Then F u! D S B F u! D n2Z ei n! on the basis of (14.35). A substitution of the index of summation implies T! u! D u! ; hence (14.29) leads to .ei ! 1/ F u! D 0. As in the proof of Theorem 16.7, weP deduce the existence of a sequence .cn / in C of moderate growth such that F u! D n2Z cn ına . But ei a ın! D ın! , for all n 2 Z, implies ei a u! D u! . Applying (14.29) once again, we obtain that F u! is a Radon measure invariant Punder Ta ; therefore P F u! has the same mass, say c, at every point. In other words, n2Z ei n! D c n2Z ına . See Problem 10.13 for another, strongly related, proof. We now conclude from (16.6) that X  c c c 1 D M.c ua / D ua ./ D ı Tka  D ; a a a k2Z

and therefore c D a D 2=!. In explicit form, the formula becomes (see Fig. 16.1), for  2 S.R/, X X X X ei n! D a ıka ; that is, F .n !/ D a .k a/: (16.7) n2Z

k2Z

n2Z

k2Z

242

16 Fourier Series 71

-Π

Fig. 16.1 Illustration of (16.7). Graph of

Π

P35

nD 35

ei n

This result is known as Poisson’s summation formula (see Example 1.4 for a heuristic discussion, Exercise 16.17 for another proof, and (14.32) for the analogous formula in the aperiodic case). Next, apply the summation formula with Tx   instead of  and note that F .Tx  / D eix F  on account of (14.12); this leads to X X Tk a  D F .n !/ ei n! . 2 S.R//: (16.8) a k2Z

n2Z

Observe that the formula gives the Fourier series of the periodization with period a of a Schwartz function on R. The identity of distributions in (16.7) is the quintessential Fourier series expansion, because from it one can obtain other such expansions, by testing the identity; this is what is expressed by (16.8) and (16.9) below. It is quite remarkable that many expansions of commonly encountered distributions can be directly derived from (16.7) by algebraic operations, translation, differentiation, or integration; see Example16.24 and Problems 16.2, 16.8, 16.17, and 16.22. One might say that the Fourier expansion of a distribution is determined by its singularities, similarly to Riemann’s key theme of the determination of a (complex-differentiable) function from its singularities. Furthermore, the identity (16.7) relates geometry of the circle R=a Z, specifically, the lengths (belonging to a Z) of its closed geodesics, to analysis on R=a Z, more precisely, the (multiplicities of) the eigenvalues n2 ! 2 and eigenfunctions ei n! of the Laplace operator @2 associated to the circle. The identity (16.8) is also valid for a much wider class of functions. This may be proved by convolving  by a suitable function f and then sending  to ı in the usual way. Alternatively, the next proof applies to continuous functions f satisfying
f .x/ D O x12 ; x ! 1:

16 Fourier Series

 M e

i n!

D D

a

243

X

k2Z XZ a

Tk a f

e

k2Z Z

e

R

0

i n!x



i n!x

Da

X

M.e

i n!

k2Z

f .x C k a/ dx D

f .x/ dx D F f .n !/:

T

k af

XZ k2Z

/

.kC1/a

e

i n!x

f .x/ dx

(16.9)

ka

The distribution u! can also be used to determine the constant in the Fourier inversion formula (14.17) in an alternative way. In the proof of (16.6), and therefore also of (16.7), formula (14.17) is not used. We now obtain, with a ! D 2, F B F u! D a F ua D a ! u! I thus, S B F B F D c I is seen to act on u! as the scalar multiplication by a ! D 2. Because u! ¤ 0, it follows that c D 2. ˛ P A distribution of the form cn ın! reminds one of the functions on a lattice ˝ WD f n! 2 R j n 2 Z g

used in numerical mathematics. By taking the limit as ! # 0 one can use these to approximate arbitrary distributions on R; compare Problem 5.12. Conversely, by application of the convergence in S one derives the Fourier transform in R from the theory of Fourier series. There also exists a Fourier series variant of Parseval’s formula. To formulate this on the natural maximal domain space, we introduce the following Hermitian inner product in the space of periodic functions with period a: .f; g/R=aZ D M.f g/: The corresponding L2 norm is kf k D kf kL2 .R=aZ/ D .f; f /R=aZ


1=2

:

The linear space of locally integrable periodic functions f on R with period a such that kf k < 1 is denoted by L2 .R=aZ/. This space is complete (a Hilbert space) and contains the space C 1 .R=aZ/ of C 1 periodic functions with period a as a dense linear subspace. In the space of sequences c D .cn /n2Z in C one defines the Hermitian inner product X .c; d / WD cn dn : The corresponding norm is

kck D kckl 2 D

n2Z

X n2Z

jcn j

2

1=2

;

244

16 Fourier Series

and the space of sequences c with kck < 1 is denoted by l 2 D l 2 .Z/. This, too, is a Hilbert space; the sequences .cn / of which only finitely many cn differ from zero (the sequences with compact support) lie dense in this space. Theorem 16.9. Consider a and ! > 0 with a ! D 2. The mapping F! that asP signs to .cn /n2Z 2 l 2 .Z/ the distributional Fourier series u D n2Z cn ei n! is a linear bijection from l 2 .Z/ to L2 .R=aZ/. We have Parseval’s formula X 2 jcn j2 D M.juj2 / D kukL k.cn /n2Z k2l2 .Z/ D 2 .R=aZ/ : n2Z

Proof. For a finite Fourier series, Parseval’s formula holds on account of X X cm cm M.ei.n m/! / D cn cn : M.juj2 / D M.u u/ D n;m

n

By continuous extension we immediately see that F! is a linear isometry from l 2 .Z/ to L2 .R=aZ/; in particular, it is injective. Furthermore, this implies that the image is complete with respect to the norm on L2 .R=aZ/, and therefore constitutes a closed subset of L2 .R=aZ/. For every  2 C 1 .R=aZ/ and every N 2 Z0 , repeated integration by parts yields cn D M.e

i n! /

D O..1 C jnj/

N

/

as jnj ! 1:

From this we deduce c 2 l 2 .Z/, and so  2 F! .l 2 .Z//. Approximating an arbitrary u 2 L2 .R=aZ/ with respect to the L2 norm by means of such , and using the fact that F! .l 2 .Z// is closed, we see that u 2 F! .l 2 .Z//. The conclusion is that the mapping F! is also surjective from l 2 .Z/ to L2 .R=aZ/.  Example 16.10. For n 2 Z0 , we have the (finite) Fourier series ! ! n 2n X X n 2n ei k e2i.n and 4n cos2n D .1 C ei /n D k k kD0

k/

kD0

of functions in L2 .R=2Z/. Since j1 C ei .x/j2 D 4 cos2 x2 , Parseval’s formula now leads to the Wallis integrals of even order: Z

 2n

cos 0

x dx D

Z



sin 0

2n

n  X n x dx D n 4 k kD0

!2

!  2n D n 4 n

.2n 1/ŠŠ 1
3


.2n 1/ DW  : D 2
4


2n .2n/ŠŠ For an evaluation of the integrals by means of Euler’s Beta function, use (13.38). Furthermore, note that we have obtained a binomial identity. ˛

16 Fourier Series

245

Remark 16.11. The extension of the theory of Fourier series to the n-dimensional variant below does not involve any real difficulties; our only reason for starting with the one-dimensional theory was to avoid being confronted with too many different aspects at the same time. Let a.j /, for 1  j  n, be a basis for Rn . A distribution u 2 D 0 .Rn / is said to be n-fold periodic with periods a.j / if Ta.j / u D u, for all 1  j  n. From this one immediately derives by mathematical induction that Ta u D u, for all a 2 A, n where nX o A WD kj a.j / j k 2 Zn j D1

n

is the lattice in R generated by the vectors a.j /. The distribution u is also said to 0 n be A-invariant and the space of A-invariant distributions is denoted P by D .R =A/. 1 n As in Lemma 16.1, one finds aP 2 C0 .R / such that a2A Ta  D 1. If, furthermore,  2 C01 .Rn / with a2A Ta  D 1, then u./ D u./, for every u 2 D 0 .Rn =A/. The mean of u 2 D 0 .Rn =A/ is defined as M.u/ D

1 u./; jA

where jA equals the n-dimensional volume of the parallelepiped spanned by the vectors a.j /, a fundamental domain for the lattice A. This corresponds to the usual mean if u is locally integrable. As in Theorem 16.3, we can use this function  to show that every A-invariant distribution is tempered. To study the Fourier transforms we need the dual lattice ˝ in Rn , generated by vectors !.k/, for 1  k  n, that are determined by the equations ha.j /; !.k/i D 2 ıjk : 1 The !.k/ also form a basis for Rn ; the 2 !.k/ form the so-called dual basis of the a.j /; see Problem 9.11. A multisequence .ck /k2Zn in C is said to be of moderate growth if positive constants c and N exist such that jck j  c .1 C kkk/N , for all k 2 Zn . As in Lemma 16.4, we obtain X v WD ck ı!.k/ 2 S 0 .Rn / k2Zn

if and only if the multisequence .ck /k2Zn of coefficients in C is of moderate growth; n the X !.k/ WD kj !.j / j D1

run through the points of the lattice ˝. The n-dimensional version of Theorem 16.7 asserts that the mapping S B F is a bijection from the space of these v 2 S 0 .Rn / to the space of A-invariant distributions in Rn . The relevant coefficients ck in the distributional Fourier series X ck ei !.k/ uD k2Zn

246

16 Fourier Series

are given by the formula ck D M.e

i !.k/ u/

.k 2 Zn /:

(16.10)

The n-dimensional Poisson summation formula takes the following form: X X if ı˝ D ı! and ıA WD ıa : F .ı˝ / D jA ıA !2˝

a2A

This also gives F 2 .ı˝ / D jA j˝ ı˝ D .2/n ı˝ , an alternative calculation of the factor in the Fourier inversion formula. Finally, we have Parseval’s formula. It asserts that the mapping that assigns the corresponding distributional Fourier series to a multisequence in l 2 .Zn /, is a bijective isometry from l 2 .Zn / to L2 .Rn =A/. ˛ Remark 16.12. The notation D 0 .Rn =A/ and L2 .Rn =A/ derives from the following. The quotient space Rn =A is defined as the space of cosets x C A with x 2 Rn . A function f is A-invariant if and only if f is constant on every coset relative to A, that is, if f .x/ D g.x C A/, for all x 2 Rn , where g denotes a uniquely determined function on Rn =A. To use yet another formulation, if  W x 7! xCA is the canonical projection from Rn to Rn =A, the mapping   is bijective from the space of functions on Rn =A to the space of A-invariant functions on Rn . It is customary to identify f and g with each other. Rn =A can be equipped with the structure of an n-dimensional C 1 manifold in such a way that   creates a correspondence between the C 1 functions on this manifold and the A-invariant C 1 functions on Rn , which explains the notation C 1 .Rn =A/ for the space of these functions. This manifold can in fact be obtained as a submanifold of R2n . Let C WD f z 2 C j jzj D 1 g

(16.11)

be the unit circle in the complex plane C ' R2 ; this is a compact, C 1 , onedimensional submanifold of R2 . The n-fold Cartesian product T WD C  C 


 C is a compact, C 1 , n-dimensional submanifold of R2n , known as an n-dimensional torus. With respect to the complex multiplication we can treat C , and therefore also T , as a group. This implies that
˚ W x 7! e i hx; !.1/i ; : : : ; e i hx; !.n/i

is a homomorphism from the additive group Rn to the multiplicative group T , with kernel equal to the period lattice A. This means that ˚ induces a bijective mapping from Rn =A to T , by which we can transfer the manifold structure of T to Rn =A. On the basis of this identification, Rn =A may also be referred to as an n-dimensional torus.

16 Fourier Series

247

It is possible on manifolds to integrate with respect to densities; considering, on T D Rn =A, the density that locally corresponds to the standard density on Rn , we Z Z have 1 u./ D u.x/ dx and M.u/ D u.x/ dx: jA T T In Remark 10.13 we have noted that it is possible to define distributions on arbitrary manifolds. Thus we now see that distributions on T correspond to the Ainvariant distributions on Rn . The point of Remark 16.11 is that there also exists a Fourier transform defined for distributions on the torus that leads to the space of functions on Zn , the multisequences. In the case of compact manifolds X , like the torus, we have the simplification that every distribution on X automatically has compact support; in this case, therefore, C 1 .X / D C01 .X / and D 0 .X / D E 0 .X /. By way of example, for the torus this yields a convolution product uv of arbitrary u and v 2 D 0 .T / such that F .uv/ D F u F v. For Lie groups G, groups that also possess a manifold structure, there exists a convolution product, which, however, is not commutative when G is not commutative. For compact Lie groups there is a theory of Fourier series that generalizes the theory outlined above for tori. For noncompact, noncommutative Lie groups, Fourier analysis is much more difficult. Here, Harish-Chandra, see [10], has done groundbreaking work. ˛ Until now we have considered convergence of Fourier series only in S 0 ; for the sake of completeness, we discuss some results on the pointwise convergence of Fourier series on Rn . In order to see that the former convergence is really weaker than the latter, derive from (16.8), for x 2 R, X X ux WD 2 ıxC2 n D e ix n e i n : n2Z

n2Z

The distribution ux has fxg C 2Z as its support (and therefore equals a continuous function on the complement of a countable discrete set), while its distributional Fourier series is nowhere on R pointwise convergent. We will study some subclasses in S 0 .Rn / possessing better behavior, to wit, pointwise convergence of the (weighted) Fourier series of their elements. WeP need some preparation. For 0 < r < 1, define the function Pr W R ! C by Pr D n2Z r jnj ei n . Summing two infinite geometric series, we obtain, for x 2 R, Pr .x/ D

X

n2N

r ne

i nx

C

X

n2Z0

r n e i nx D

1

1 C r2

r2 : 2r cos x

(16.12)

The family .Pr /0 0 arbitrarily, one therefore obtains the existence of g 2 C.Rn =2Zn / such that kg f kLp < . In turn, this leads to kAr f f kLp  kAr .f g/kLp CkAr g gkLp Ckg f kLp  2Cc kAr g gkL1 ; for a suitable constant c > 0. By (16.16) the middle term is less than enough to 1.

 c

if r is close 

Corollary 16.14. A continuous periodic function on Rn is uniquely determined by its Fourier series. Example 16.15. Consider the case of n D 1 and identify R=2Z with C as in (16.11) via x $ e ix . Similarly, write C.R=2Z/ 3 fz $ f 2 C.C / if P fz.x/ D f .e ix /. Suppose fz 2 C.R=2Z/ has Fourier series k2Z ck ei k and define the two following series: X X gC .z/ D ck z k and g .z/ D c kz k .z 2 C/: k2Z0

k2N

Write D˙ D f z 2 C j jzj ≶ 1 g. Then C D @D˙ . Observe that the bound jck j  M.jfzj/, for all k 2 Z, implies that g˙ define complex-analytic functions on D˙ ; and lim g .z/ D 0. jzj!1

(16.18)

We also obtain Ar fz.x/ D

X ˙

g˙ .r ˙1 e ix /

.0 < r < 1; x 2 R/:

Therefore it is a direct consequence of Theorem 16.13 that X g˙ .r ˙1 e ix / uniformly for x 2 R. f .e ix / D fz.x/ D lim r"1

˙

(16.19)

P In other words, the functions z 7! ˙ g˙ .r ˙1 z/ on C converge uniformly on C to f as r " 1. Furthermore, the function f on C is the uniform boundary value of the complex-analytic function gC on DC if and only if ck D 0, for every k < 0. In addition, note that z 2 C if and only if z1 D zN . Define the functions h and u W DC ! C by means of

250

16 Fourier Series

h .z/ D

X

c

k

k2N

zN k

u D gC C h :

and

Then it is obvious that @zN gC .z/ D 0 and @z h .z/ D 0 on DC , in the notation of (12.12). On account of Problem 12.11 we know that the Laplacian
equals 4@z @zN D 4@zN @z , which implies that
u D 0. In other words, u is harmonic on DC . Moreover, we verify directly, for 0  r < 1 and x 2 R, 1 u.re ix / D Ar fz.x/ D

r2 2

Now note that jre ix e iy j2 D jr e i.y

x/ 2

j D .r e i.y

Z

 

x/

1 C r2

/.r e

f .e iy / 2r cos.x

i.y x/

y/

dy:

/ D 1Cr 2 2r cos.x y/:

Hence, writing x 2 DC instead of re ix and y 2 @DC instead of e iy , we obtain Z f .y/ 1 jxj2 dy .x 2 DC /; (16.20) u.x/ D 2 jx yj2 @DC where dy denotes integration with respect to the Euclidean density on the circle @DC . Phrased differently, for every f 2 C.C /, Poisson’s integral formula (16.20) defines a harmonic function u on DC having f as its uniform boundary value, or u solves the Dirichlet problem for
on the disk DC given the boundary value f (see Problem 12.4). For the natural generalization of Poisson’s integral formula to the open unit ball in Rn , see Problem 16.14. ˛ n l n n Fourier Theorem P 16.16. If f 2 C .R =2Z / with l > 2 , then the distributional series k2Zn ck ei k of f is absolutely and uniformly convergent on Rn =2Zn . In addition, the Fourier inversion formula from Theorem 16.7 holds pointwise for f .

Proof. With ck D M.f e i k /, we have on account of the Cauchy–Schwarz inequality and with suitable constants c and c 0 > 0, X

k2Zn

jck j

2

D 

X

k2Zn

X

c

k2Zn

X

k2Zn

.1 C jkj/

l

.1 C jkj/l jck j

.1 C jkj/

2l

 X

k2Zn

.1 C jkj/2l jck j2

.1 C jkj/ jck j  c 0 2l

2

2

X X

j˛jl k2Zn



jk ˛ ck j2 :

Parseval’s formula from Theorem 16.9 asserts that the right-hand side can be dominated by X c0 kD ˛ f kL2  c 00 kf kC l ;

for suitable c 00 > 0.

j˛jl

16 Fourier Series

251

Because of the uniform convergence of the Fourier series, its sum defines a function fz 2 C.Rn =2Zn /. In such a case, one a fortiori has X r jkj ck e i hx; ki D fz.x/ .x 2 Rn =2Zn /: lim Ar f .x/ D lim r"1

r"1

k2Zn

On the other hand, it follows from Theorem 16.13 that limr"1 Ar f .x/ D f .x/.  As an application, we give an elementary proof of the result discussed in Remark 11.12; for another argument see Theorem 15.4 below. We use the tensor product notation from (11.8). Theorem 16.17. Let X  Rn and Y  Rm be open sets and write C01 .X / ˝ C01 .Y / for the linear subspace of C01 .X  Y / consisting of finite linear combinations of functions  ˝ , where  2 C01 .X / and 2 C01 .Y /. Then 1 1 1 C0 .X / ˝ C0 .Y / is dense in C0 .X  Y /. Proof. It has to be shown that if  2 C01 .X  Y /, then there exists a sequence .m /m2N of functions in C01 .X / ˝ C01 .Y / that converges in C01 .X  Y / to  as m ! 1. By a partition of unity, the proof of this can be reduced to the case that supp  is contained in a cube. This, in turn, follows from the case that supp  is a subset of the unit cube in Rn  Rm . But then it is clear that the theorem is implied by the following lemma.  Lemma 16.18. Define I D  0; 1 Œ  R, let n 2 N, and denote by I n the open unit cube in Rn . Suppose that  2 C01 .I n /. Then one can find functions O m 2 C01 .I / .m 2 N/ (16.21) 1j n

such that the sequence .m /m2N converges in C01 .I n / to . Proof. Extend  to all of Rn as a C 1 function z that is periodic with respect to Zn . One can expand z as a distributional Fourier series, see (16.10), X y y e2 i k z z D with .k/ D M.e 2 i k /: .k/ k2Zn

z and thus in According to Theorem 16.16 this series converges in C 1 .Rn / to , 1 n C .I / to . Since  is supported in the open cube I n , there exists ı > 0 such that supp  is contained in the closed cube Œ 2ı; 1 2ı n . Choose  2 C01 .I / such that  D 1 on  ı; 1 ı Œ . Furthermore, note that   Y e2 i k .x/ D .k 2 Zn ; x 2 Rn /: e2 i kj .xj / D ˝ e2 i kj .x/ 1j n

1j n

252

16 Fourier Series

Now define m D

X

f k2Zn jjkj jm .1j n/ g

y .k/

˝

e2 i kj :

1j n

Then it is clear from the convergence in C 1 .I n / of the Fourier series to  that these functions m , which are of the desired form (16.21), converge in C01 .I n / to ; and so the assertion is proved.  In applications, the pointwise convergence of Fourier series on R under conditions weaker than the one in Theorem 16.16 is an important issue. Define the following functions on R: n n 1 X 1X Dn D ei k and Fn D Dk : (16.22) n kD n

kD0

The sequence .Dn /n2Z0 is called the Dirichlet kernel and .Fn /n2N the Fej´er kernel. We evaluate these sums explicitly. To this end, recall the summation formula Pn 1 z nC1 n for the geometric series, valid for all z 2 C. Accordingly, nD0 z D 1 z n X

kD n

e i kx D e

i nx

1

1

e i.nC 2 /x e i.2nC1/x D x 1 e ix ei 2

e e

i.nC 1 2 /x i

x 2

Furthermore, on account of the identity 2 sin p sin q D cos.p obtain from the preceding equality n Fn .x/ sin2

D q/

n 1 n 1  X x 1X 1 x D x sin D sin k C .cos kx 2 2 2 2 kD0

sin.n C 12 /x : sin x2

cos.p C q/ we

cos.k C 1/x/

kD0

1 nx .1 cos nx/ D sin2 : 2 2 Thus (see Figs. 16.2 and 16.3) D

Dn .x/ D

sin.n C 12 /x sin x2

and

Note the following properties: Z   1 Dn .x/ dx D 1I (i) 2  Fn (iii)

0  Fn .x/ 

2 n x2

.0 < jxj  /I

Fn .x/ D

(ii) (iv)

sin2 n

nx 2 2 x sin 2

:

Fn .x/  0

(16.23)

.x 2 R/I

lim Fn .x/ D 0

n!1

(16.24) uniformly on f x 2 R j ı  jxj   g, for every 0 < ı < . In property (i) we used (16.22), in (iii) that sin x2  x on Œ 0;  , and in (iv) the existence of a constant c > 0 such that

16 Fourier Series

253

2

-2 Π

-Π

Π

2Π

-2

Fig. 16.2 Illustration of (16.23). Truncated graphs of D20 and of x 7!

˙1 sin x 2

8

-2 Π

-Π

Π

2Π

Fig. 16.3 Illustration of (16.23). Graph of F8 .

jFn .x/j 

1 n sin2

x 2



c n ı2

.ı  jxj  /:

Furthermore, observe that the partial sums Dn nowhere have a pointwise limit as n ! 1, because their graphs all oscillate with increasing frequency between the graphs of two fixed functions. We begin by proving that the distributional Fourier series of a periodic distribution of the form test f is summable in the sense of Ces`aro at points where the function f satisfies a mild extra condition. Theorem 16.19. Suppose f is a locally integrable periodic function on R of period 2 and consider x 2 R. Introduce, in the notation of Definition 16.2, Z  n X 1 ik x Sn f .x/ D M.f e i k /e D f .x y/ Dn .y/ dy; 2  kD n

254

16 Fourier Series

Z n 1 1X 1 n f .x/ D Sn f .x/ D n 2 kD0



f .x

y/ Fn .y/ dy:



Suppose x 2 R and f .x / WD limy"x f .y/ and f .xC / WD limy#x f .y/ exist. Then lim n f .x/ D

n!1

1 .f .x / C f .xC // DW s.x/: 2

(16.25)

Proof. On account of (16.24).(i) and the evenness of Fn we have Z  2.n f .x/ s.x// D .f .x y/ s.x// Fn .y/ dy;  Z  Z  D .f .x y/ C f .x C y/ 2s.x// Fn .y/ dy DW g.x; y/ Fn .y/ dy; 0

0

which implies, according to (16.24).(ii), jn f .x/

1 s.x/j  2

Z

0



jg.x; y/j Fn .y/ dy:

(16.26)

Let  > 0 be arbitrary. By assumption there exists 0 < ı <  such that jg.x; y/j <  whenever 0 < y < ı. Next, choose N 2 N such that 2 2 2  .M.jf j/ C js.x/j/ < : 2 Nı 2 It follows directly that 1 2

Z

ı 0

1 jg.x; y/j Fn .y/ dy  2

Z

ı 0

  Fn .y/ dy  : 2 2

Furthermore, in view of (16.24).(iii), Z  Z  2 jg.x; y/j Fn .y/ dy  jg.x; y/j dy n y2 ı ı Z  2 .jf .x y/j C jf .x C y/j C 2js.x/j/ dy  n ı2 ı 2 4 3  .2M.jf j/ C 2M.jf j/ C 2js.x/j/  .M.jf j/ C js.x/j/ n ı2 n ı2 <  : We conclude that we have jn f .x/

s.x/j < , for all n  N .



The following result is known as Fej´er’s Theorem. Its proof closely parallels that of Theorem 16.13; see, however, Problem 16.15.

16 Fourier Series

255

Theorem 16.20. Suppose that f is a continuous periodic function on R of period 2. In the notation of Theorem 16.19 we then have that limn!1 n f D f uniformly on R. We mention an immediate corollary. A continuous periodic function can be uniformly Pn approximated by trigonometric polynomials, that is, by functions of the form kD n ck ei nk . In turn, this result leads via uniform convergence of the exponential series to the approximation of a continuous function by polynomials uniformly on compacta; compare with Weierstrass’s Approximation Theorem from Example 14.31. Now we come to results that imply pointwise convergence of the distributional Fourier series of a periodic distribution test f . Theorem 16.21. Let f be as in Theorem 16.19. At points x 2 R such that y 7! f .y/y xf .x/ is integrable in a neighborhood of x, in particular, at points x of differentiability of f , we have pointwise convergence, that is, lim

m;n!1

n X

M.f e

ik/ e

i kx

kD m

D f .x/:

Proof. Indeed, as usual we may suppose that x D 0 and f .x/ D 0; just shift the origin and subtract a constant from f . By assumption, the function g W x 7! efix.x/1 is integrable near 0. Furthermore, we have f .x/ D .e ix

1/g.x/;

so

M.f e

ik/

D M.g e

i.k 1/ /

M.g e

i k /:

Therefore the Fourier series of f at 0 is a telescoping series. Indeed, n X

M.f e

kD m

ik/

D M.g ei.mC1//

M.g e

i n /;

and this tends to 0 D f .0/ by the Riemann–Lebesgue Theorem, Theorem 14.2, applied to g 1Œ ;   , which is integrable on R.  We can also deal with a jump discontinuity. Theorem 16.22. Let f be as in Theorem 16.19. Suppose that f has left-hand and right-hand limits at x and, in addition, that it has one-sided slopes at x, that is, f .x˙h/ f .x˙ / h 7! converge to limits as h # 0. Then the symmetric partial sums h converge; more precisely, lim Sn f .x/ D

n!1

1 .f .x / C f .xC //: 2

256

16 Fourier Series

Proof. Using a translation we may suppose that x D 0. Subtract a constant such that f .0 / D f .0C /. We now must show that the Sn f .0/ converge to 0, where Z  1 Sn f .0/ D f .y/ Dn .y/ dy: 2  But we need only observe that Dn is an even function. Hence Z  1 1 .f . y/ C f .y// Dn .y/ dy: Sn f .0/ D 2  2 Now we simply apply Theorem 16.21 to the function x 7! 12 .f . x/ C f .x//.



Note that near a jump discontinuity the convergence of the symmetric partial sums cannot be uniform. Otherwise, the function would be continuous at that point. Example 16.23. Consider arbitrary be the periodic func z 2 C and let fz W R !2C 1 1 iz x tion of period 1 that on ; .x/ D e . A straightforward is given by f z 2 2 computation (based, for instance, on the generalization of (16.8)) gives the distributional Fourier series sin z X . 1/n fz D e2 i n ; (16.27)  n2Z z n

where a term with n D z should be read as e2 i n . In particular, application of the preceding theorem with x D 0 yields the partial-fraction decomposition of the cosecant (see Problem 16.21.(iii) for another proof) X . 1/n 1  D C 2z .z 2 C n Z/: (16.28) sin z z z 2 n2 n2N Furthermore, using the preceding theorem with x D 12 leads to the partial-fraction decomposition of the cotangent (see [7, Exercises 0.13.(i) and 0.21.(ii)] for other proofs) X 1 1  cot z D C 2z .z 2 C n Z/: (16.29) 2 z z n2 n2N In particular, we obtain the sum of Euler’s series 1 2 D lim z!0 6

X 1 z cot z D : 2z 2 n2 n2N

Parseval’s formula from Theorem 16.9 applied to fz or differentiation of (16.29) with respect to z implies (compare with Problem 16.10 and [7, Exercise 0.12.(ii)]) X 1 2 D .z 2 C n Z/: (16.30) 2 .z n/2 sin z n2Z See [7, Exercise 0.12] for further consequences of the identity, for instance, values of Riemann’s zeta function at positive even points (see also (16.35) below) and

16 Fourier Series

257

MacLaurin’s series of the tangent. Finally, one may rewrite (16.30) in the form (see Problems 16.10 or 18.9 for another proof) X 1D T n sinc2 on C; (16.31) n2Z

R which according to Problem 16.19 leads to R sinc x dx D  (compare with Problems 14.44 and 16.21). We give an application of the theory above. Let 0 < a < 1. Using the change of variables x D y1 , we get Z

R>0

xa 1 dx D 1Cx

Z

1 0

xa 1 dx C 1Cx

Z

1 0

y .1 a/ 1 dy: 1Cy

Furthermore, the following series is uniformly convergent for x 2 Œ ; 1  0 , where  and  0 > 0 are arbitrary: X xa 1 D . 1/k x aCk 1 : 1Cx k2Z0

Finally, apply termwise integration and (16.28) to obtain (see [7, Exercise 6.58.(v)] for another proof) Z  xa 1 dx D .0 < a < 1/: (16.32) sin a R>0 x C 1 ˛ Example 16.24. (Fourier series of Bernoulli functions.) The sequence of polynomial functions .bn /n2Z0 on R is uniquely determined by the following conditions: b0 D 1;

bn0 D nbn

1;

Z

1 0

bn .x/ dx D 0

.n 2 N/:

The bn are known as the Bernoulli polynomials and the Bn WD bn .0/ 2 R as the Bernoulli numbers. It follows that bn .0/ D bn .1/, for n  2. The bn .x/, for 1  n  4, are respectively given by x

1 ; 2

x2

1 xC ; 6

x3

3 2 1 x C x; 2 2

x4

2x 3 C x 2

1 : 30

Introduce the Bernoulli functions b n W R ! R by means of b n .x/ D bn .x Œx/; see Fig. 16.4. Then b n 2 S 0 .R/, for all n 2 N; in fact, b n 2 C n 2 .R/, for n  2. Observe that b 1 equals s 12 , where s is the sawtooth function from Example 4.2. Using the formula for s 0 from that example and Poisson’s summation formula, we P see that the distributional Fourier series of s 0 is given by k2Znf0g e2 i k . This leads to the following identity in S 0 .R/:

258

16 Fourier Series 1

1

2

6

-1

0.0481

-1

1 -1

1 -

-

2

1

1

1

-0.0481 12

7

1 0.0244

240

-1

42

1

-1

1

-1

1 31

1 -

-

-0.0244

1344

30

Fig. 16.4 Example 16.24. Graphs of Bernoulli functions b n , for 1  n  6, with different vertical scales and ordered in rows

b1 D s

1 D 2

X e2 i k : 2 i k

(16.33)

k2Znf0g

Here the mean has been used to fix the constant of integration; in fact, M.s/ D R1 1 0 x dx D 2 . It is also easy to verify (16.33) by direct computation of the distributional Fourier series of s. On account of Theorem 16.21 we even obtain the pointwise convergence of the series on R n Z; hence (compare with [7, Exercise 0.18.(i)] and Fig. 16.5) X sin 2kx 1 D .x 2 R n Z/: (16.34) b 1 .x/ D x Œx 2 k k2N

0.5895

-2

-1

1

2

-0.5895

Fig. 16.5 Illustration of (16.34). Graph of x 7!

P100

kD1

sin.2kx/ k

We would like to integrate (16.34) termwise in order to obtain the Fourier series of b 2 , but the series in (16.34) is uniformly convergent only on intervals of the form

16 Fourier Series

259

P Œ ı; 1 ı , for every 0 < ı < 12 and not on Œ 0; 1 . On the other hand, nkD1 sinkkx is uniformly bounded for n 2 N and x 2 Œ ;  . Indeed, in the notation of (16.22), n X 1 sin kx D k 2i

kD1

Now

Dn .t/ D

X

01 / nŠ .2 i k/n k2Znf0g

of functions on R. In particular, b 2n .x/ D 2. 1/n .2n/Š b 2nC1 .x/ D 2. 1/n .2n C 1/Š .2n/ WD

X 1 D . 1/n k 2n

k2N

1

X cos 2kx

k2N 1

.2k/2n

X sin 2kx .2k/2nC1

k2N 11

2

.2/2n

B2n .2n/Š

.n 2 N; x 2 R/; .n 2 N; x 2 R/; .n 2 N/:

(16.35)

260

16 Fourier Series

Here  P W f s 2 C j Re s > 1 g ! C denotes Riemann’s zeta function defined by .s/ D k2N k1s . Finally, note that B2nC1 D 0, for n 2 N. ˛

Problems  16.1. Without computing Fourier coefficients by means of integration, determine the Fourier series of the 2-periodic even continuous function arccos B cos on R from Problem 1.7; that is, show that

Deduce that

2X 1 ei.2n  n2Z .2n 1/2

 2

arccos B cos D P

1 n2N .2n 1/2 P 1 n2N .2n 1/4

D

1/

D

4 X cos.2n 1/
:  n2N .2n 1/2

 2

2 8 and Euler’s series .2/ P 1 4 and .4/ D n2N n4 96

D

P

furthermore D D (16.35) and [7, Exercise 0.12.(iv)].) In addition, derive for f D arcsin B sin (see Fig. 16.6), f D

4 X . 1/n 1 sin.2n  n2N .2n 1/2

Deduce Leibniz’s series

P

n2N

1/


f0 D

;

. 1/n 1 2n 1

D

1 2 n2N n2 D 6 , and 4 . (Compare with 90

4 X . 1/n 1 cos.2n  n2N 2n 1

1/


:

 4.

Π 2 -3 Π

-2 Π

-Π -

Π

Π

2Π

3Π

Π

2Π

3Π

2 1 -3 Π

Fig. 16.6 P100 . 4 

nD1

-2 Π

-Π

-1

Illustration of Problem 16.1. Graphs of 1/n

1 cos.2n

1/


2n 1

4 

P30

nD1

. 1/n 1 sin.2n 1/
.2n 1/2

and

 16.2. In the preceding problem new Fourier series were obtained from known ones by means of differentiation. The converse method of integration is effective as well. Generalize Poisson’s summation formula (16.7) as follows: X X a ıhCka D e i n!h ei n! .a; !; h 2 R; a ! D 2/: k2Z

n2Z

16 Problems

261

P Next deduce .arcsin B sin/00 D 4 n2N . 1/n sin.2n 1/
in S 0 .R/ (see Fig. 16.6) and integrate termwise. P 16.3. Show that k2Z ıp2 k is an eigendistribution for the Fourier transform with p eigenvalue 2; compare with Problem 14.9. 16.4. Demonstrate that Poisson’s summation formula may also be applied to  W 1 x 7! 1Cx 2 . Calculate X 1 Rh D h ; 1 C h2 n2 n2Z

a Riemann sum approximation of the integral of . Determine the limit, as h # 0, of e 2= h .Rh /. P 16.5. Show that n2N . 1/n 1 is summable in the sense of Abel and of Ces`aro to 1 2 in both cases. 16.6. For x 2 R n 2Z prove the following identities on summation of series in the sense of Abel: x  X X 1 1 1 cot ; ; .n C 1/ cos nx D .n C 1/ sin nx D 2 4 sin2 . x2 / 2 2 n2Z0 n2N X X 1 ; n cos nx D n sin nx D 0: 2 x 4 sin . 2 / n2N n2N

P For x D  the third series takes the form n2N . 1/n n. Show that its corresponding sequence of Ces`aro means has two limit points: 0 and 12 . In other words, the series is summable in the sense of Abel to 14 (the average of these limit points), but is not summable in the sense of PCes`aro. Conversely, suppose the series n2N cn of complex numbers is summable in the sense of Ces`aro toP s and prove that it is summable in the sense of Abel to s. Hint: P apply the identity n2N cn r n D .1 r/2 n2N nn r n and assume s D 0.  16.7. With n 2 N and Fn as in (16.22), demonstrate that

Fn D

X  1

jkjn 1

jkj  ei k n

and

F Fn D 2

X 

jkjn 1

1

jkj  ık : n

P PN  16.8. Calculate N nD0 cos n
and nD1 sin n
, and discuss the convergence in S 0 .R/ as N ! 1. Prove that in S 0 .R/ we have the following identity (compare with Example 1.4): X X 1 cos n
D C  ı2 n : 2 n2Z n2Z 0

In Problem 16.17.(v) we will use the Fourier transform to show

262

16 Fourier Series

u WD

X

n2N

sin n
D

 
1X ˙ i0 : cot 4 2 ˙

In this problem, compute only F u. Remark: it is also possible to prove, without usingPthe Fourier transform (see, for example, [7, Exercise 0.18]) the formula above for n2Z0 cos n
and furthermore x that the restriction of u to R n 2Z equals the analytic function x 7! 2.1sincos x/ D 1 x cot 2 ; see Fig. 16.7 and compare with Problem 16.6. For x near 0 we may write the 2 cos

x

last expression as sinc 2x x1 , where the first factor is a C 1 function. On account of the 2 2-periodicity of u, it has similar, not absolutely integrable behavior at all integer multiples of 2. For a test function  with support in the interval  2; 2 Œ one has Z cos x2  1 1 x u./ D ./ D ..x/ . x// cot dx x PV sinc 2 x 2 R>0 2 Z Z 1 ..x/ . x// sin x dx D D log.1 cos x/  0 .x/ dx; 4.1 cos x/ 2 R R where all integrals are absolutely convergent. Also, u equals the distributional derivative of the locally integrable function 12 log.1 cos/. Using Lemma 16.6

10

-2 Π -Π -2 Π

-Π

Π

Π

2Π

2Π

-5

-10

Fig. 16.7 Illustration of Problem 16.8. Graphs of

1 2

cot


2

and

1 2

log.1

cos/

and Problem 13.6 one obtains the following identity in D 0 . 2; 2 Œ/ (with some abuse of notation): X cos. x2 / 1 x sin x 1 2 u0 .x/ D n cos nx D PV x jxj : 2 x 4 x sinc. / sin 2 2 n2N Note that the restriction of u0 to R n 2Z equals the function x 7! pare with Problem 16.6

1 ; 4 sin2 . x 2/

com-

16 Problems

263

16.9. Let u0 be a periodic distribution on R with period a > 0. Prove that there exists exactly one differentiable family t 7! u t W R>0 ! D 0 .R/ with the following properties (compare with Problem 17.1): (i) For every t 2 R>0 , the distribution u t is periodic with period a, d (ii) dt u t D @2 u t , (iii) lim t #0 u t D u0 in D 0 .R/. Calculate F u t and u t .  Compute F sinc2 , and using Poisson’s summation formula deduce that 16.10. P 2 k2Z T k sinc D 1 on R. Compare with (16.31) and [7, Exercise 0.14]. Derive (16.30).

 16.11. (Relation between Fej´er kernel for line and unit circle.) For t > 0 define t F t .x/ D 2 sinc2 tx . Verify that F F t ./ D maxf0; 1 jj g, for every  2 R. 2 t Conclude that lim t !1 F t D ı in S 0 .R/. Furthermore, for every f 2 S.R/ and x 2 R, deduce that Z t  jj  1 F t  f .x/ D F f ./ d : e ix 1 2 t t

In this problem, write F n for the Fej´er kernel Fn from (16.23), for n 2 N. Prove in two different ways the following identity of functions on R: X 1 Fn D T2 k Fn 2

and

k2Z

The passage from Fn to

1 2

lim F n D 2

n!1

X

ı2 k :

k2Z

F n is said to be the periodization of Fn .

16.12. (Relation between Poisson kernel for half-space and unit disk.) Define the Poisson kernel for the half-space P t W R ! R, for t > 0, as in Problem 14.51, and the Poisson kernel for the unit disk P r W R=2Z ! R by (16.12) (here the bar is used temporarily in order to avoid confusion). Using that F P t D e t j
j , demonstrate that X P r D 2 T2k P t ; where r D e t: k2Z

Deduce that limr"1 P r D 2

P

k2Z ı2k

in two different ways.

16.13. Define C and D˙ as in Example 16.15. Let gC be complex-analytic on DC and write ur .x/ WD gC .r e ix /, for 0  r < 1 and x 2 R. Verify that as r " 1, .n/ .0/=nŠ, for ur converges in D 0 .R/ if and only if the sequence of terms cn WD gC n 2 Z0 , is of moderate growth. If this is the case, the limit f is said to be the distributional boundary value of gC , via the parametrization x 7! e ix W R ! C . Prove that f is periodic with period 2. Does every complex-analytic function on DC have a distributional boundary value? Does every 2-periodic distribution on R equal the distributional boundary value of an analytic function on DC ?

264

16 Fourier Series

Let g be complex-analytic on D . It is known that g .z/ ! 0 as jzj ! 1 if and only if g .z/ D hC . 1z / for an analytic function hC on DC with hC .0/ D 0. Write vr .x/ WD g .r e ix /, for r > 1 and x 2 R. Prove that as r # 1, vr converges in D 0 .R/ if and only if hC possesses a distributional boundary value h. If this is the case, the limit g is said to be the distributional boundary value of g . Prove that g D S h. Show that every 2-periodic distribution on R can be written uniquely as f C g, with f the distributional boundary value of a complex-analytic function gC on DC and g the distributional boundary value of a complex-analytic function g on D P with g .1/ D 0. Determine f and g for u D n2Z ı2 n .  16.14. (Poisson integral formula for open unit ball in Rn .) Write B n D f x 2 Rn j kxk < 1 g and denote, as usual, by cn the .n 1/-dimensional volume of the unit sphere S n 1 D @B n from (13.37). Consider f 2 C.S n 1 / and prove that Z f .y/ 1 kxk2 dy .x 2 B n / u.x/ D n 1 cn kx ykn S

defines a harmonic function u on B n that satisfies limr"0 u.r x/ D f .x/, uniformly for all x 2 S n 1 .  16.15. Give a direct proof of Theorem 16.20. P  16.16. Consider  2 S.R/. Prove that k2Z jT2 k F j2 D 1 on R, if and only if

.Tl ; Tm / D ılm

.l; m 2 Z/;

where ılm denotes the Kronecker delta. Equation (16.31) and Example 14.1 show that the equivalence also may hold in the case of functions belonging to L2 .R/.  16.17. (Poisson summation formula.) Define u 2 D 0 .R/ by

uD

X 1 ıC ı2k : 2 k2N

(i) Verify that u 2 S 0 .R/. Deduce the following equality of distributions in S 0 .R/: FuD (ii) Prove that u D lim e #0

x

1 X C e 2 n2N

2 i n :

u in S 0 .R/ and furthermore that

F u D lim #0



1 X C e 2 n2N

2 n

e

2 i n



Show by means of summation of a geometric series that

:

16 Problems

265

F u D lim #0

1 cot .
2i

i / DW

1 cot .
2i

i 0/:

Now use addition of the resulting identities to obtain X 2i e2 i n D cot .
i 0/ cot .
C i 0/: n2Z

(iii) In addition, show that the complex-analytic function z 7!  cot z on C has a simple pole of residue 1 at every point k 2 Z. Conclude by means of the 1 Plemelj–Sokhotsky jump relation  1i 0 Ci 0 D 2 i ı (see Problem 1.3) that one has X X X cot .
i 0/ cot .
C i 0/ D 2i ık I deduce e2 i n D ık ; n2Z

k2Z

k2Z

where the latter identity of tempered distributions is the Poisson summation formula known from (16.7) and Problem 10.13. (iv) Prove that (compare with Problem 16.6) X X 4 n sin 2 n
D ık0 : n2N

k2Z

(v) Also derive from part (ii) the following identity in S 0 .R/ (compare with Problem 16.8):  
X 1X ˙ i0 : sin n
D cot 4 2 n2N ˙

In particular, show for x near 0 that X cos x2  1 : PV sin n x D x sinc 2 x n2N (vi) Compute F cot.


i 0/ and F cot2 .


Now define s.x/ D (vii) Show that s.x/ D

i 0/.

X cos nx 1 C n2 n2N

1  cosh.x /  2 sinh 

.x 2 R/: 1



.0  x  2/:

To obtain this result, note that Poisson’s summation formula implies the following identities in S 0 .R/: X X 1 C s s 00 D cos n
D ı2k : 2 n2N k2Z

Now solve the differential equation for s using the fact that s is even and periodic. Conclude that (alternatively, apply (16.27) with z D ˙i and add, or

266

16 Fourier Series

compare with [7, Exercise 6.91.(ii) and (iii)]) cosh x D in particular

2 sinh  X . 1/n cos nx sinh  C .   x  /;   1 C n2 n2N X 1 1 D x . 1 C  coth / D 1:076 674 047 468


: 2 1 C n 2 n2N

 16.18. Suppose that f is a C 2 periodic function on R of period 2. Show that the Fourier series of f converges absolutely and uniformly on R to f , and that in fact Sn f f D O. n1 / uniformly as n ! 1.

16.19. Integrate (16.31) over Œ 0;   to obtain D

XZ n2Z

.nC1/ n

sinc2 x dx D

Z

sinc2 x dx: R

Integration by parts yields, for any a > 0, Z

a 0

sinc2 x dx D

sin2 a C a

Z

2a

sinc x dx: 0

R DeduceR the convergence of the improper integral R sinc x dx as well as the evaluation R sinc x dx D  (see Problem 14.44 or 16.21 and [7, Example 2.10.14 or Exercises 6.60 and 8.19] for other proofs).  16.20. Successively prove the following formulas:

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

X . 1/n 1  ; p D 2C4 16n2 1 2 n2N  z 1 X  2n 1 n D 2 2 C n Z ; . 1/ 2 cos 2 z z 2 .2n 1/2 2 n2N X . 1/n 1  D ; 4 2n 1 n2N  z 1 X 2 1 D4 2CnZ ; 2 2 cos z .2z 2n 1/ 2 n2Z  z 1 X 1 2 C n Z ;  tan z D 8z .2n 1/2 4z 2 2 n2N d 1 log sinc z D  cot z .z 2 C n Z/; dz z 2 Y z sin z D z 1 .z 2 C/; 2 n n2N 1  Y 1 1 D 1 ; 2 n2 nD2

16 Problems

(ix) (x)

267

Y 2 D 1  n2N Y 1 cos z D n2N

1  ; or 4n2 4z 2  .2n 1/2

.2n nŠ/2 p D n!1 .2n/Š 2n C 1 lim

r

 ; 2

.z 2 C/:

Note Leibniz’s series in (iii) and Wallis’s product in (ix).  16.21. Consider arbitrary z 2 C n Z and let fz W R ! C be the periodic function of period 2 that on  0; 2 Œ is given by fz .x/ D e iz x . Interpreting, if necessary, the sums over Z as limits of the symmetric partial sums, prove the following identity of functions on R n 2Z:

fz D e

i z

sin z X ei n I  n2Z z C n

or

X e i nx  e iz. x/ D ; sin z zCn n2Z

for 0 < x < 2. For 0 < x < 2, deduce that (i) (ii) (iii) (iv)

X e ˙i.zCn/x  e ˙i z D ; sin z zCn n2Z X cos.z C n/x ;  cot z D zCn n2Z X sin.z C n/x ; D zCn n2Z X 1D sinc.z C  n/ n2Z

(v) (vi)

.z 2 C/;

X cos nx ; z 2 n2 n2N X n sin nx sin z. x/ D 2z : sinc z z 2 n2 n2N

cos z. x/ sinc z

1 D 2z 2

Observe that the identities (i) and (iii) are not valid for x D 0. On the other hand, (ii) with x D 0 leads to the partial-fraction decomposition of the cotangent in (16.29); hence, (ii) is true in this case. Setting x D  in (iii) gives the partial-fraction decomposition of the cosecant inR (16.28). Use identity (iv) to prove R sinc x dx D . To this end, use the method from Problem 16.19 and estimates similar to those in Example 16.24 to justify the interchange of integration and summation. (See Problem 14.44 or 16.19 and [7, Example 2.10.14 or Exercises 6.60 and 8.19] for other proofs.)  16.22. Consider arbitrary z 2 C and let gz W R ! C be the periodic function of period 2 that on Œ ;   is given by gz .x/ D cos z x.

(i) Prove the following identity in S 0 .R/:

268

16 Fourier Series

gz00 C z 2 gz D 2z sin z

X

ı.2kC1/ :

k2Z

Verify that we have the following distributional Fourier series, where a term with n D z should be read as 12 ei n : gz D

z sin z X . 1/n ei n I  z 2 n2 n2Z

2

show

X

k2Z

ı.2kC1/ D

X

. 1/n ei n :

n2Z

Derive the latter identity also directly from Poisson’s summation formula. (ii) Define f W R ! R by f .x/ D arctan B tan. x2 / if x 2 R n . C 2Z/ and f .x/ D 0, for the remaining x. Verify that f is an odd sawtooth function, more 1 ix precisely, that f .x/ D x2  Œ xC 2  D 2i log e , for all x 2 R, if log is defined on C by its principal value (compare with Example 4.2). (Note the slight abuse of notation: the equality is not one of functions but of distributions, because the functions differ on  C 2Z.) Furthermore, show that f0 D

1 2



X

k2Z

ı.2kC1/ D

1 2

X 1X . 1/n ei n D . 1/nC1 cos n
: 2 n2Z n2N

P nC1 Verify by integration of the preceding identity that f D n2N . 1/n sin n
. (iii) As another application of the Fourier series of gz , deduce by taking z D 12 that j cos j D

4 X . 1/n 1 cos 2n
2 C ;   n2N 4n2 1

Derive X . 1/n 1  D 2 4n 1 4 n2N j cos j D

4 

1 ; 2

1C

X

n2N

1 4n2

1

j sin j D

D

1 ; 2

8 X . 1/n 1 cos2 n
;  n2N 4n2 1

4 X cos 2n
:  n2N 4n2 1

2 

X

n2N

.4n

j sin j D

1 2/2

1

D

 ; 8

8 X sin2 n
:  n2N 4n2 1

1 Since 4n21 1 D 12 . 2n1 1 2nC1 /, one recognizes the first and third numerical series as simple modifications of Leibniz’s series from Problem 16.1 and sees that the second one is telescoping. Finally, prove that X
8 X n sin 2n
; Tk 1 0; Œ cos D  n2N 4n2 1 k2Z

where the left-hand side is defined to attain the value 0 on  Z. In other words, the Fourier series of the even periodic extension to R of 1Œ 0;  sin and the odd extension of 1 0; Œ cos have been obtained; see Fig. 16.8. 16.23. (Gamma distribution, Lipschitz formula, and Eisenstein series.) Let ˛,  > 0. In statistics, the function f˛;  W R ! R with

16 Problems

269 1

-2 Π

-Π

Π

2Π

Π

2Π

1

-2 Π

-Π -1

Fig. 16.8 Illustration of Problem 16.22.(iii). Graphs of the even extension of sin and the odd extension of cos

f˛;  .x/ D ˛ ˛C .x/ e

x

is said to be the probability density of the Gamma distribution of order ˛ and parameter . In particular, for n 2 N, the function f n ; 1 .x/ D 2

2

1 2

n 2

n

. n2 /

x2

1

e

1 2x

H.x/

is called the probability density of Pearson’s 2 distribution with n degrees of freedom. (i) Using Problem 14.39 or by expansion of e ix into a power series and using [7, Exercise 0.11.(iii)] for the binomial series of .1 C i / ˛ , prove that F f˛;  ./ D



  C i

˛

. 2 R/:

For k 2 Z0 , introduce the shifted factorial, or Pochhammer symbol, .˛/k D

.˛ C k/ D ˛.˛ C 1/


.˛ C k .˛/

1/:

k (ii) Show that x k f˛;  .x/ D .˛/ f .x/. Use this identity or (14.28) in order k ˛Ck;  to deduce from part (i) that Z .˛/k ˛ .˛/k k F .x 7! x f˛;  .x//./ D ; x k f˛;  .x/ dx D k : . C i /˛Ck  R>0

The latter integral is called the kth moment of the function f˛;  . Conclude that Z Z ˛ ˛  2 WD  WD x f˛;  .x/ dx D ; .x /2 f˛;  .x/ dx D 2 :   R>0 R>0 In statistical terms,  is the expectation and  2 the variance of the Gamma distribution.

270

16 Fourier Series

(iii) Prove that f˛1 ;   f˛2 ;  D f˛1 C˛2 ;  , for ˛1 , ˛2 ,  > 0 on the basis of part (i) and also by a different method. Set H D f z 2 C j Im z > 0 g.

(iv) For z 2 H , deduce from part (i) that the Fourier transform of x 7! .x equals . i /˛ 1 iz  7! 2 i e H. /: .˛/

z/

˛

(v) Now use Poisson’s summation formula to derive from part (iv) Lipschitz’s formula: X . 2 i /k X k 1 2 i nz 1 D n e .z 2 H; k 2 Z2 /: .k 1/Š n2N .z C n/k n2Z (vi) On account of (16.29) we have the following identity, where 0 denotes symmetrical summation: X0 1 e 2 iz D  cot z D  i 2 i .z 2 H /: zCn 1 e 2 iz n2Z Lipschitz’s formula also can be obtained by expanding the right-hand side of the identity as a geometric series in e 2 iz (this is where the condition z 2 H is necessary) and differentiating k 1 times with respect to z. Conversely, the partial-fraction decomposition of the cotangent can be derived from Lipschitz’s formula. Let .2k/, for k 2 N, be as in (16.35). The Eisenstein series Gk of index k > 1 is the function Gk W H ! C given by Gk .z/ D

X

.0;0/¤.m;n/2ZZ

XX 1 1 D 2 .2k/ C 2 : 2k .mz C n/ .mz C n/2k m2N n2Z

(vii) Apply Lipschitz’s formula with z replaced by mz to get the so-called Fourier expansion of Gk at infinity 2. 2 i /2k X X 2k 1 2 i mnz Gk .z/ D 2 .2k/ C n e .2k 1/Š m2N n2N 2.2 i /2k X D 2 .2k/ C 2k 1 .j /e 2 ijz .2k 1/Š j 2N   4k X D 2 .2k/ 1 2k 1 .j /e 2 ijz : B2k j 2N

Here the coefficient 2k positive divisors of j .

1 .j /

denotes the sum of the .2k

1/th powers of

Chapter 17

Fundamental Solutions and Fourier Transform

The Fourier transform in S 0 .Rn / is a very useful tool in the analysis of linear partial differential operators X P .D/ D c˛ D ˛ j˛jm

in R with constant coefficients c˛ 2 C. If u 2 S 0 .Rn /, then (14.10) implies n

F .P .D/u/ D P ./ F u: Here P ./ D

X

c˛  ˛

(17.1)

(17.2)

j˛jm

is the polynomial in  obtained by substituting j for Dj everywhere. P ./ is said to be the symbol of P .D/. In other words, on the Fourier transform side the differential operator P .D/ changes over into multiplication by the polynomial P ./. Clearly, this is the expansion in eigenvectors that we discussed in the introduction to Chap. 14. Remark 17.1. A distribution u 2 S 0 .Rn / is a solution of the equation P .D/u D 0 if and only if F u 2 S 0 .Rn / satisfies P ./ F u D 0. If P ./ ¤ 0, for all  2 Rn , this leads to F u D 0 and therefore u D 0. In general, one has supp F u  N WD f  2 Rn j P ./ D 0 g: If the total derivative DP ./ is nonzero for all  2 N , then N is a C 1 (even analytic) submanifold in Rn of dimension n 1 according to the Submersion Theorem. In this case F u is given by a “distribution on N ”; but for a proper formulation we have to know what distributions on submanifolds are. This therefore represents a natural occasion to introduce these. An example of an entirely different nature was given by the distributions on the torus, in connection with Fourier series; see Remark 16.12. ˛ J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_17, © Springer Science+Business Media, LLC 2010

271

272

17 Fundamental Solutions and Fourier Transform

Suppose that E 2 D 0 .Rn / is a fundamental solution of P .D/ and, in addition, that E 2 S 0 .Rn /. Using (17.1), we conclude from (14.30) that 1 D F .P .D/E/ D P ./ F E:

(17.3)

Conversely, if Q 2 S 0 is a solution of the equation P ./ Q D 1, then E WD F 1 Q 2 S 0 is a fundamental solution of P .D/. Indeed, in that case one has F .P .D/E/ D P ./ Q D 1; therefore P .D/E D ı. It goes without saying that this result makes use of Theorem 14.24. Equation (17.3) implies that on the complement C D f  2 Rn j P ./ ¤ 0 g in n R of the zero-set N of the polynomial function P , the distribution Q D F E equals the C 1 function 1=P . In particular, a possible tempered fundamental solution of P .D/ is uniquely determined if P does not have real zeros. Even in this case it is not obvious, a priori, whether 1=P is a tempered distribution. This might lead to problems in cases in which P ./ were to converge to 0 too fast in too large subsets of Rn , as kk ! 1. At the present stage we do not have sufficient background on polynomials in n variables to reach general conclusions on this point. Instead, we now concentrate on a subclass of operators. Definition 17.2. The operator P .D/ of order m is said to be elliptic if the homogeneous part X Pm ./ WD c˛  ˛ j˛jDm

of degree m of the symbol P ./ of P .D/ does not have any real zeros  except  D 0. That is,  2 Rn

and  ¤ 0

H)

Pm ./ ¤ 0:

Pm ./ is said to be the principal symbol of P .D/. Example 17.3. The Laplace operator
is elliptic. The heat operator @ t the wave operator @2t
x are not elliptic.

˛
x and ˛

Note that the condition of ellipticity applies only to the highest-order part Pm .D/ of P .D/. Thus, if P .D/ is elliptic, the operator P .D/ C Q.D/ is elliptic for every Q.D/ of order lower than m. The condition of ellipticity becomes effective with the estimate in the following lemma. Lemma 17.4. The differential operator P .D/ in Rn is elliptic of order m if and only if there exist constants c > 0 and R  0 such that jP ./j  c kkm

. 2 Rn ; kk  R/:

(17.4)

Proof. Let P .D/ be elliptic. The continuous function jPm j attains its minimum on the compact subset S D f  2 Rn j kk D 1 g of Rn . That is, there exists an

17 Fundamental Solutions and Fourier Transform

273

 2 S such that jPm ./j  jPm ./j, for all  2 S . Now  ¤ 0, and therefore  WD jPm ./j > 0. For arbitrary  2 Rn n f0g one has kk 1  2 S , and therefore   jPm

 1   j D kk kk

m

jPm ./j:

Furthermore, Q./ D P ./ Pm ./ is a polynomial of degree  m implies the existence of a constant d > 0 such that jQ./j  d kkm

1

.kk  1/:

Combining the estimates, we obtain, for kk  1, jP ./j  jPm ./j

jQ./j   kkm

1, which

d kkm

1

 D 

d  kkm : kk

This yields the desired estimate, with c D  d=R, whenever R > d= and R  1. Conversely, if P .D/ is not elliptic, there exists  2 Rn such that  ¤ 0 and Pm ./ D 0. This implies Pm .t / D t m Pm ./ D 0, and consequently P .t / D Pm .t / C Q.t / D Q.t / is a polynomial in t of degree < m. This is in contradiction to (17.4).  The estimate (17.4) implies that for any c 2 R0 , the set of the  2 Rn with jP ./j  c is bounded. Lemma 17.5. If P .D/ is elliptic, it has a parametrix E (see (12.4)) satisfying sing supp E  f0g. Proof. Let R and c > 0, as in (17.4). Choose  2 C01 .Rn / with ./ D 1 for kk  R. The function 1=P is C 1 on a suitable open neighborhood U of supp .1 /. Therefore v 2 C 1 .Rn / if v./ WD

˚1 0

./ P ./

if

 2 U;

on the interior of the set where  D 1:

Furthermore, v./ D 1=P ./ if  … supp , that is, if kk is sufficiently large. On account of (17.4), one there has jv./j  1c kk m , if P .D/ is assumed to be of order m. It follows that v is certainly bounded and defines a tempered distribution on Rn . If we now choose E WD F 1 v, then F .P .D/E/ D P ./F E D P ./v D 1 . Applying F 1 , we obtain P .D/E D ı F 1 . Now F 1  D .2/ n S B F  is a C 1 function on Rn , even with a complex-analytic extension to Cn ; it follows that E is a parametrix of P . To prove that sing supp E  f0g, we begin by deriving an auxiliary estimate, guided by the knowledge that the way in which v decreases at infinity improves under differentiation. We observe that by mathematical induction on l,

274

17 Fundamental Solutions and Fourier Transform

Dj l

1 P

./ D

Ql ./ ; P ./lC1

if P ./ ¤ 0. Here Q0 D 1, and Ql D P Dj Ql function of degree  l.m 1/. Because l.m 1/ in combination with (17.4),
 ˛ Dj l v D O kkj˛j m l ;

1

l Ql 1 Dj P is a polynomial .1 C l/m D l m, this yields, kk ! 1:

For given k 2 N we choose l > nCk m, or k m l < n; then  ˛ Dj l v 2 L1 , for all multi-indices ˛ with j˛j  k. But this means that we may differentiate Z 1 l l 1 . xj / E.x/ D F .Dj v/.x/ D e i hx; i .Dj l v/./ d  .2/n Rn

to order k under the integral sign. In other words, xj l E is of class C k , that is, E 2 C k on the complement Cj of the hyperplane f x 2 Rn j xj D 0 g. Because this holds for all k, we have E 2 C 1 on S Cj , and because this in turn is true for all 1  j  n, it follows that E 2 C 1 on jnD1 Cj D Rn n f0g. 

From the proof it will be clear that we could have chosen E D F 1 .1=P / if P had no zeros in Rn . This applies, for example, if P .D/ D
 I , with  2 C and not   0. In that case, E also is a fundamental solution of P .D/, with the additional properties that E 2 S 0 and sing supp E  f0g. We obtain the following theorem by combining Lemma 17.5 and Theorem 12.4. Theorem 17.6. Every elliptic operator P .D/ is hypoelliptic.

This theorem is a classical regularity result; as a consequence, operators having the regularity property were called hypoelliptic. Remark 17.7. For the parametrix E in the proof of Lemma 17.5, P .D/E ı equals the Fourier transform of a function with compact support, and is therefore analytic. By replacing the integration over Rn in the formula for E by an integral over a suitable submanifold in Cn , which, on the strength of Cauchy’s Integral Theorem (see (12.9)), leaves the result unchanged, one also proves that E is analytic on Rn n f0g. Thus, it follows from Theorem 12.15 that for every elliptic operator P .D/, the distribution u is analytic wherever P .D/u D 0. ˛ Remark 17.8. An elliptic operator P .D/ with variable (C 1 ) coefficients has a parametrix K as discussed in Remark 12.6. Its construction is a generalization of the proof of Lemma 17.5, and the result is a so-called pseudo-differential operator K. This has a kernel k (see Theorem 15.2 below) that is smooth off the diagonal, and the conclusion is that elliptic operators with C 1 coefficients, too, are hypoelliptic. This theory can be found in H¨ormander [11, Chap. 18], for example. ˛ Example 17.9. We will prove the principle that every derivative can be majorized by a suitable power of the Laplacian. More specifically, suppose that  belongs to

17 Fundamental Solutions and Fourier Transform

275

C 1 .Rn / and
j  is of polynomial growth at infinity, for 0  j  k, with 2k > n. Then @˛  also is at most of polynomial growth at infinity, for all multi-indices ˛ with j˛j  l WD 2k n 1 2 Z0 . Indeed, define fk D .1 C k
k2 / k 2 C 1 .Rn /. Then fk 2 L1 .Rn /. According to the Riemann–Lebesgue Theorem, Theorem 14.2, the fundamental solution Ek WD F

1

fk D

1 F fk .2/n

(17.5)

of the differential operator .I
/k is a continuous function vanishing at infinity. Furthermore, Theorem 17.6 below and Theorem 12.4 imply that sing supp Ek D f0g. j˛j  l leads to j˛j 2k  n 1, which gives the convergence of R Now ˛ j f ./j d . In turn, this implies that Ek is of class C l and that n k R Z 1 ˛ D Ek .x/ D e i hx; i . /˛ fk ./ d : .2/n Rn Furthermore, D ˛ Ek is rapidly decreasing, as one sees using integration by parts, because all derivatives of  7!  ˛ fk ./ are linear combinations of functions  7!  ˇ D fk ./ with jˇj C j j  j˛j. As in the proof of Lemma 17.5 one obtains that the latter functions are O.kkj˛j 2k / D O.kk n 1 / as kk ! 1. Finally, @˛  D .I


/k Ek  @˛  D @˛ Ek  .I


/k ;

where the convolution on the right-hand side is given by the usual integral. Furthermore, this convolution is at most of polynomial growth. In the (minimal) case of 2k D n C 1 (assuming for the moment n to be odd), Problem 18.8 below shows that Ek D c e k
k , for some constant c > 0. This means that this Ek is actually of exponential decay at infinity. By applying the Fourier transform to the following identity, one sees that, for 2k > n, .I


/

2k n 1 2

Ek D c e

k
k

:

This makes it plausible that all such Ek decay exponentially fast; for a proof see Sect. 17.1 below. ˛ As another application of the preceding technique, in Example 17.9 we give a global characterization of tempered distributions; compare with Theorem 13.1 and Example 18.2. Theorem 17.10. A distribution in D 0 .Rn / belongs to S 0 .Rn / if and only if it is a finite sum of derivatives of a function in C.Rn / that is at most of polynomial growth at infinity. Proof. Suppose that u 2 S 0 .Rn /. Then there exist c > 0 and k, N 2 Z0 such that for all  2 S.Rn /,

276

17 Fundamental Solutions and Fourier Transform

ju./j  c kkS.k;N / D c

sup j˛jk; jˇ jN; x2Rn

jx ˇ @˛ .x/j:

(17.6)

Select l > n2 C k and consider the fundamental solution El 2 S 0 .Rn / of .I
/l as in (17.5). Then El belongs to C k .Rn /, with all of its derivatives rapidly decreasing on the strength of Example 17.9. We may define the function f on Rn by (compare with (11.1)) f D u  El

u  El .x/ D u.Tx El /;

with

(17.7)

because u is of order  k and El possesses the properties established above. Since x 7! Tx El is a continuous linear mapping with respect to the S.k; N / norm, the function f belongs to C.Rn /. Furthermore, to prove that f is at most of polynomial growth, apply (17.7) and (17.6). In addition, use that all derivatives of El of order at most k are rapidly decreasing. As a consequence, f 2 S 0 .Rn /. On the other hand, let  2 S.Rn / be arbitrary. Since El is absolutely integrable on Rn , the proof of Lemma 2.18 can be modified to give El   2 C 1 .Rn /. Furthermore, as in the proof of Theorem 14.16, we see that El   is rapidly decreasing. It follows that El   2 S.Rn /. (This conclusion is also suggested by the fact that F El F  D .1 C k
k2 / l F  belongs to S.Rn /. Actually, for every l 2 Z, the mapping  7! .1 C k
k2 / l  defines a linear isomorphism from S.Rn / onto itself, because .1 C k
k2 / l 2 C 1 .Rn /, while all of its derivatives are at most of polynomial growth.) Hence u.El  / is well-defined. A variation of the proof of Theorem 14.33 shows that the convolution in (17.7) is a continuous extension of convolution of distributions belonging to E 0 .Rn / and S 0 .Rn /. With    as in Lemma 14.7, we have    u 2 E 0 .Rn / and thus we obtain from Lemma 14.7 and (11.23), for all  2 C01 .Rn /, u.El  / D lim    u.El  / D lim.   u  El /./ D u  El ./ D f ./: #0

#0

In view of (11.4) this implies u./ D u.ı  / D u..I D .u  El /..I

l


/l El  / D u.El  .I


/ / D .I


/l /

l


/ f ./:

On account of Lemma 14.7 we now obtain the identity u D .I


/l f in S 0 .Rn /. 

We return to the properties of elliptic operators. Theorem 17.11. Every elliptic operator P .D/ in Rn has a fundamental solution E in D 0 .Rn /. Proof. Suppose P .D/ is of order m and let R > 0 be as in Lemma 17.4. Choose  2 Cn such that Pm ./ ¤ 0; for example,  2 Rn n f0g. For z 2 C we write P . C z/ as a polynomial in z:

17 Fundamental Solutions and Fourier Transform

277

P . C z/ D Pm ./ z m C

m X1

Rl ./ z l ;

lD0

where the coefficients Rl ./ D Rl .; / are polynomials in . Introducing cl D supkkR jRl ./j, we obtain the estimate  jP . C z/j  r m jPm ./j

m X1 lD0

cl rm

l



.jzj D r; kk  R/:

Because the quantity within parentheses converges to jPm ./j > 0 as r ! 1, we conclude that there exist  > 0 and r > 0 such that jP . C z/j  

.z 2 C; jzj D r;  2 Rn ; kk  R/:

Write B D f  2 Rn j kk  R g

and

C D Rn n B:

We now define E D F C G W Rn ! C, with F D F 1 .1C =P / and G the analytic function given by Z Z 1 1 e i hx; Czi 1 G.x/ D dz d : .2/n B 2 i jzjDr P . C z/ z Here the integral over z is the complex line integral, whose value can be regarded as the sum of all residues in the disk f z 2 C j jzj < r g of the meromorphic function z 7! e i hx;Czi =.z P . C z//. This yields P .D/F D F 1 1C , while Z Z 1 1 1 P .D/G.x/ D e i hx; Czi dz d  n .2/ B 2 i jzjDr z Z 1 D e i hx; i d  D F 1 1B .x/; .2/n B where in deriving the latter identity we have used Cauchy’s integral formula (12.11). The conclusion is P .D/E D F 1 1C C F 1 1B D F 1 1 D ı.  Note that sing supp E  f0g for every fundamental solution of an elliptic operator; this follows from Theorem 17.6. Remark 17.12. The Theorem of Ehrenpreis–Malgrange asserts the existence of a fundamental solution in D 0 .Rn / for an arbitrary linear partial differential operator P .D/ ¤ 0 with constant coefficients. The construction given above is a simple variant of H¨ormander’s proof [12, Thm. 7.3.10] of that theorem. The simplification is made possible by the ellipticity of P .D/, which ensures that no problems arise due to P ./ becoming small for large kk. By combining the Fourier transform with deep results from algebraic geometry, Bernstein and Gel’fand [2] and Atiyah [1] were able to prove that every P .D/

278

17 Fundamental Solutions and Fourier Transform

as above has a fundamental solution belonging to S 0 .Rn /. However, in the nonelliptic case these solutions often have local properties less favorable than those of H¨ormander. Remarkably enough, we now are prepared enough to establish the Theorem of Ehrenpreis–Malgrange in full generality. They proved the result in [8] and [16], respectively, by an approach different from the one below; to be more specific, by application of the Hahn–Banach Theorem, Theorem 8.11. The proof presented here is extremely efficient as well as explicit, but at this stage of the theory it might be difficult to see how to find it yourself. For a more “natural” verification, see the proof of Theorem 18.4 below. ˛ Theorem 17.13. Every linear partial differential operator P .D/ in Rn with constant coefficients (not all of them equal to 0) possesses a fundamental solution in D 0 .Rn /. More specifically, if P .D/ is of order m and the polynomial function Pm is the principal symbol as in Definition 17.2, there exists  2 Rn such that Pm ./ ¤ 0. Then E 2 D 0 .Rn / defined by   Z 1 P .
C z/ .x/ dz E.x/ D z m 1 e z hx; i F 1 P .
C z/ 2 i Pm ./ C is a distribution having the desired property. Here C denotes the unit circle in C, and the integration with respect to z is the complex line integral over C . Proof. The set of zeros of the nontrivial polynomial function P on Rn is an algebraic hypersurface in Rn and therefore a set of measure 0; this can be deduced from the structure theory in Whitney [25] or apply [13, Problem VI.10]. Given z 2 C , this .Cz/ implies that  7! P is an essentially bounded function on Rn and therefore P .Cz/ 0 n defines an element of S .R /. Furthermore, the distribution-valued mapping C ! S 0 .Rn /

z 7!

given by

P .
C z/ P .
C z/

is continuous on account of Lebesgue’s Dominated Convergence Theorem, Theorem 20.26.(iv), applied to Rn . A continuous distribution-valued function may be integrated over a compact set; hence E is well-defined. Successively using Problem 14.2, the identity P .D/ B F 1 D F 1 B P ./, and (14.33), one obtains with D D Dx , P .D/E.x/ D D D

1 2 i Pm ./ 1

Z Z

zm C

2 i Pm ./ C Z 1 2 i Pm ./

C

zm zm

  P .
C z/ .x/ dz P .
C z/   P .
C z/ 1 z hx; i .x/ dz e P .D C z/ F 1 P .
C z/ 1

 P .D/ e z hx; i F

1 z hx; i

e

F

1

1



P .
C z/.x/ dz

17.1 Appendix: Fundamental Solution of .I

D

1 2 i Pm ./

Z

zm


/k

1 z hx; i

e

C

279

P .D C z/ ı dz;

with ı 2 D 0 .Rn / the Dirac measure. Next, expansion of the polynomial P .D C z/ in the variable z and the equality z D z1 for z 2 C lead to P .D C z/ D

m X1 1 1 P ./ C Qj .D; /; m zm zj j D0

for certain polynomial functions Qj . The integral of the leading term on the righthand side yields ı on account of Cauchy’s integral formula (12.11), while the integral of the sum vanishes by Cauchy’s Integral Theorem in (12.9). 

17.1 Appendix: Fundamental Solution of .I


/k

The fundamental solution Ek from (17.5), for k > n2 , deserves closer study on account of its useful properties; for instance, see Remark 18.8 below. The method used here is a vast generalization of the one employed in Problem 14.15 and uses a differential equation satisfied by Ek . By means of a change of variables one sees that Ek is invariant under all rotations about the origin in Rn . Special functions with this property often satisfy a differential equation of Bessel type; see (17.9) below. In order to derive this equation, first rewrite Ek as a Fourier transform of a function of one variable. Note that Ek .x/ D Ek .kxk e1 /, where e1 is the first standard basis vector in Rn . Now q one has, writing  D .1 ;  0 / 2 R  Rn 1 and using the change of variables  0 D

1 C 12  in Rn

1

as well as [7, Exercise 7.23],

Z

e i kxk1 d 2 0 2 k Rn .1 C 1 C k k / Z Z 1 e i kxk1 1 D d  d 1 n 1 n 2 k .2/ R .1 C  2 /k 2 Rn 1 .1 C kk / 1 Z .k n 2 1 / e i kxk D d : nC1 n 1 2n  2 .k/ R .1 C  2 /k 2

1 Ek .x/ D .2/n

It is a direct verification that for x and  2 R, @2 B F D xB@BF D Next introduce

F B  2; F B @ B ;

x 2 B @2 B F D x2 B F D

F B @2 B  2 ; F B @2 :

280

17 Fundamental Solutions and Fourier Transform

n

Dk

1

and

2

F D F a

a./ D

with

1 : .1 C  2 /

Now try to determine constants p, q, and r 2 C such that .x 2 @2 C p x @ C q C r x 2 /F D 0; F .@ . 2 a/ p @.a/ C q a r @2 a/ D 0I @2 .. 2 r/a/ p @.a/ C q a D 0

i.e., hence by Fourier inversion:

2

The left-hand side of the last equation is a rational function in the variable  with .1 C  2 /C2 in the denominator and c4  4 C c2  2 C c0 in the numerator, where c4 D .2

c2 D 2..

1/p C q

1/p C q C 4 2

.2 2 C /r C 2

5/;

6 C 2; c0 D

p C q C 2 r C 2:

Equate the ci to 0 and note that this inhomogeneous linear system for p, q, and r is nonsingular, because  > 0. Its solution is p D 2 2, q D 0, and r D 1; hence one obtains the following ordinary differential equation for the function F : .x @2 C 2.1

/@

x/F D 0:

(17.8)

In fact, on account of the theory of the indicial equation, see [13, p. 223], one may predict the vanishing of q, because Ek is nonsingular at 0. In that case, the equations c4 D c0 D 0 take the simple form p C 2 2 D 0 and p C 2 r C 2 D 0. Now it is a straightforward to compute that for any  2 C and C 2 function u defined on R, x



.x 2 @2 .x  u/ C x @.x  u/

2 .x  u// D x 2 @2 u C .2 C 1/x @u:

Therefore select  such that 2 C 1 D 2

2 D 2

2k C n

1;

that is,

D

n 2

k:

It follows that n

v.x/ WD x 2

k

Ek .x/

satisfies

x 2 v 00 C x v

.x 2 C 2 /v D 0:

(17.9)

This is a differential equation for Bessel functions of purely imaginary argument, which differs from Bessel’s equation x 2 u00 C x u0 C .x 2 2 / u D 0 only in the coefficient of u; see Watson [23, ÷ 3.7] for more information. As we will see in a moment, any two linearly independent solutions to the former equation have the property that one is exponentially increasing at infinity and the other exponentially decreasing. Hence, this differential equation satisfies the following principle: a solution that is a tempered distribution is of exponential decay at infinity. A tempered solution is given by the Macdonald function K of order  > 0 that is normalized by the following condition, which is a natural one in the theory of Bessel functions:


/k

17.1 Appendix: Fundamental Solution of .I

lim

x#0

 x  2

281

./ : 2

K .x/ D

K tends exponentially to 0 at infinity through positive values. On account of (13.35) one obtains n kxkk 2 Ek .x/ D n Ck 1 n .x 2 Rn /: (17.10) Kk n2 .kxk/ 22  2 .k/ For this reason the .Ek / are said to be the Bessel kernel. As a byproduct of this identity and [7, Exercise 7.30.(iii)], which reduces the Fourier transform of a function on Rn that is invariant under rotations to an integral over R>0 involving ordinary Bessel functions, we mention (compare with [23, ÷ 13.6 (2)], which, however, is not correct) Z

n

R>0

r2 Jn .1 C r 2 /k 2

1 .x

r/ dr D

xk 2k

1

1

.k/

Kk

n 2

.x/

.x 2 R/:

Conversely, this equality can be applied to derive (17.10). Finally, we derive formulas for Ek that are more explicit than the one in (17.10). It is straightforward to verify that Z 1 Z x F1 .x/ D e  . 2 x 2 / 1 d  and F2 .x/ D e  .x 2  2 / 1 d  x

x

(17.11) define functions on R annihilated by the differential operator in (17.8). These integrals are related to Schl¨afli’s integral; see [23, ÷ 6.15 (4)]. Furthermore, F1 is bounded on R, whereas F2 is unbounded. Hence, there exists a constant c 2 C such that F D c F1 . In this case, one again computes the constant by taking the limit as x # 0. Using Legendre’s duplication formula (13.39) one derives c D 4 1 ./2 . Combination of the preceding results leads to, for x 2 Rn , Z 1 nC1 1 Ek .x/ D e  . 2 kxk2 /k 2 d : n 1 n 1 2k 1 2  2 .k / .k/ kxk 2 In particular (compare with (18.18) and Problem 18.8.(i), and furthermore [23, ÷ 3.71 (13)]), e kxk E nC1 .x/ D .x 2 Rn /: (17.12) n 1 nC1 2 n 2 2  . 2 / Using the change of variables  D x C t in (17.11) one obtains Z x 1 F .x/ D e e t .2xt C t 2 / 1 dt R>0 Z t  1  1 x / e t t  1 .1 C dt D .2x/ e 2x R>0   1  ; x ! C1:  ./.2x/ 1 e x 1 C O x

(17.13)

282

17 Fundamental Solutions and Fourier Transform

Hence (note the agreement with [23, ÷ 7.23 (1)]) Ek .x/ 

nC1 2

kxkk

n 1 2 Ck

2



e

  1  1CO ; kxk .k/

kxk

n 1 2

kxk ! 1:

In (17.13) we restrict ourselves to deriving a crude asymptotic estimate. The t  1 binomial series expansion of .1C 2x / is convergent only as long as 0 < t < 2jxj. Therefore interchange of integration and summation is not admissible. In fact, a formal interchange leads to the divergent series .2x/

1

x

e

1 X

j D0

./ . C j / 1 : j Š . j / .2x/j

Instead, one uses a finite number of terms plus a remainder in a correct analysis. In the special case of  2 N, observe that the same change of variables and application of the Binomial Theorem to .2x C t/ 1 imply (the comparable final formula in [23, ÷ 6.33] is erroneous) F1 .x/ D . Therefore, if k

nC1 2

Ek .x/ D

x

1/Š e

 1 X .2 2 j /Š .2x/j j Š . 1 j /Š

j D0

.x 2 R/:

2 Z0 and x 2 Rn (see [23, ÷ 3.71 (12)]), then

22k

e

kxk

1

n 1 2

k

nC1

X2 .2k .n C 1/ j /Š .2kxk/j : nC1 j /Š .k/ j D0 j Š .k 2

Problems 17.1. (Heat equation on R and R=2Z). For t > 0, define u t 2 S.R/ as in Prob2 lem 5.5. Prove that F u t ./ D e t  , for  2 R, both by direct computation and by solving the following initial value problem on R: d u t D
u t dt

.t > 0/

lim u t D ı:

and

t #0

Show that the solution of the following initial value problem on R: d v t D
v t dt

.t > 0/

is given by the periodic function

and

lim v t D t #0

X n2Z

ı2 n ;

17 Problems

283

vt D Verify that v t .x/ D

X n2Z

1 X e 2 n2Z

t n2

ei n :

u t .x C 2 n/

.x 2 R/:

(Compare with Problem 16.9.) In other words, the solution to the initial value problem on the circle R=2Z equals the periodization (in the sense of Problem 16.12) of the solution to the initial value problem on R. 17.2. (Fundamental solution of heat operator.) Let a 2 R. Prove that a fundamental solution of the differential operator @ C a I in R is given by e a
H ; see Example 14.30. Next, use the partial Fourier transform with respect to the variable x 2 Rn to show that a fundamental solution E 2 S 0 .Rn  R/ of the heat operator @ t
x , for .x; t/ 2 Rn  R, is given by the locally integrable function (see Problem 8.5 for another proof) E W .x; t/ 7! u t .x/ H.t/ D

H.t/ .4 t/

n 2

e

kxk2 4t

:

Here u t W Rn ! R, for t > 0, is as in Problem 5.5.

 17.3. For what linear partial differential operators P .D/ in Rn , with constant coefficients and of order > 0, does the equation P .D/u D 0 possess a solution u ¤ 0 in (i) D 0 .), (ii) S 0 , (iii) E 0 .), (iv) L1 , (v) C01 , and (vi) S? (Compare with Problem 14.3.)

 17.4. Prove that every harmonic and tempered distribution u on Rn is a polynomial. Prove that u is constant if u is a bounded harmonic function. Prove that for the Laplace operator
in Rn , with n  3, the potential of ı as in (12.3) is the only fundamental solution E with the property that E.x/ ! 0 as kxk ! 1.

17.5. (Liouville’s Theorem.) Suppose that u is a tempered distribution on C ' R2 and is complex-differentiable in the sense that @u=@Nz D 0; then prove that u is a complex polynomial. Verify Liouville’s Theorem, which asserts that any bounded complex-analytic function u on C is constant. Generalize this result as follows. Let P .D/ be a partial differential operator in Rn with constant coefficients satisfying P ./ ¤ 0 for all  2 Rn n f0g. Show that u is a polynomial function on Rn if u 2 S 0 .Rn / and P .D/u D 0. Conclude that u is constant if it is a bounded function. Verify that the heat operator @ t
x in RnC1 is an example of an operator as discussed here.  17.6. Let P .D/ be an elliptic operator of order m. Prove the following assertions.

(i) If P .D/ possesses a homogeneous fundamental solution E, then P .D/ D Pm .D/, that is, P is homogeneous of degree m. Furthermore, E is homogeneous of degree m n.

284

17 Fundamental Solutions and Fourier Transform

(ii) Suppose that P is homogeneous and that E is a homogeneous fundamental solution E of P .D/. On account of Theorem 14.34, E is tempered. Prove that Ez is a tempered fundamental solution of P if and only if there exists a polynomial function f such that Ez D E C f and P f D 0. Furthermore, if m  n, then Ez is a homogeneous fundamental solution for P if and only if there exists a homogeneous polynomial function f of degree m n such that Ez D E C f and P f D 0. If m < n and if Ez is a homogeneous fundamental solution of P , then Ez D E. (iii) If P is homogeneous and m < n, then 1=P is locally integrable on Rn and P has exactly one homogeneous fundamental solution. (iv) The Laplace operator in R2 has no homogeneous fundamental solution. 17.7. (Fundamental solution of Schr¨odinger operator.) Prove that E W Rn  R ! C is a fundamental solution of the Schr¨odinger operator on Rn  R
kxk2 1 i .2 n/  4 C 4t if E.x; t/ D H.t/ D t
x : n e .4 t/ 2 Hint: apply Problem 14.41.  17.8. Let  2 C and  … R. Show that P .D/ D D  has exactly one tempered fundamental solution E0 . Calculate the fundamental solution E from the proof of Theorem 17.11. Show that E is not a tempered distribution on R whenever r > .jj C R/=jj. 17.9. Fill in the details of the following arguments. In the notation of Sect. 17.1 we have on account of (17.10), for k > n2 , Z   1 1 Ek D ; and so F Ek .x/ dx D 1I .2/n 1 C k
k2 /k Rn Z  n n . > 0/I hence kxk K .kxk/ dx D 2Cn 1  2 C 2 Rn see [23, ÷ 13.21 (8)]. Using the Fourier transform prove that Ek El D EkCl and .I
/Ek D Ek 1 , for admissible values of k and l. Investigate whether these properties enable the construction of a complex-analytic family .Ek /k2C of distributions; see Fig. 17.1. Derive from Definition 13.30 and Example 14.11, respectively, Z k
k2 1 1 n n t t kk2 k 1 t k
k2 2e 4t t 2 : D e e t dt; F e D  2 k .k/ .1 C kk / R>0 Deduce that

Z

Ek .x/ D e R>0

t

r2 4t

t

1 n 2

Z

e

t

kxk2 4t

tk

n 2

1

2n  .k/ R>0  r  1 dt D 2 K .r/ .r > 0;  2 R/: 2

dt

.x 2 Rn /;

17 Problems

285

For the last identity, compare with [23, ÷ 6.22 (15)] and [7, Exercise 2.87.(v)] for the elementary case of  D 12 . In addition, the penultimate identity allows one to define Ek explicitly, for 0 < k  n2 . Show by direct computation that in this case one also R has Rn Ek .x/ dx D 1. In the last identity above involving Ek .x/ introduce a new variable s by means 2 of t D kxk . Next, suppose 0 < k < n2 and use dominated convergence to show s z2k .x/.1 .x//, where R z2k is as in (14.51) and limx!0 .x/ D 0. that Ek .x/ D R z2k , as is to be This shows that there is an intimate connection between Ek and R expected. f .t/ WD e

t

kxk2 4t

attains its maximum e 1 .t C 4t /

kxk

on R>0 at t D

kxk . 2

Further, if

. Combine the two estimates in order to obtain kxk  1, then f .t/  e kxk kxk 1 1 0  f .t/  e 2 e 2 .t C 4t / and thus prove Ek .x/ D O.e 2 /; x ! 1.

2

5 64

10 -1

1

Fig. 17.1 Graphs of E k , for 1  k  7, and of E4 5

Chapter 18

Supports and Fourier Transform

In Lemma 14.4 we have proved that the Fourier transform F u of a distribution with compact support can be extended to a complex-analytic function U on Cn , called the Fourier–Laplace transform of u. In this chapter we deal with questions like, What complex-analytic functions U occur as F u for some u 2 E 0 ? and How can U provide information on the support of u? For a compact subset K of Rn we define sK W Rn ! R

by

sK ./ D sup hx; i:

(18.1)

x2K

Theorem 18.1. Let K  Rn be compact and u 2 E 0 .Rn / with supp u  K. If u is of order k 2 Z0 , there exists a constant c > 0 such that jF u./j  c .1 C kk/k e sK .Im /

. 2 Cn /:

(18.2)

If, in addition, u is of class C k with k 2 Z0 , there exists a constant ck > 0 such that jF u./j  ck .1 C kk/ k e sK .Im / . 2 Cn /: (18.3)

Proof. We are going to apply Theorem 8.8; this requires some preparation. For given  > 0 we can, on the strength of Corollary 2.4, cover the compact set K by a finite number of open balls whose union U is contained in the =3neighborhood K=3 . Applying Lemma 2.19 with this set U and with  and ı replaced by =3 and 2=3, respectively, we conclude that it is possible to find  2 C01 .Rn / with  D 1 on an open neighborhood of K, while supp   K . Then there exists, for every multi-index ˇ, a constant cˇ > 0 such that sup j@ˇ  .x/j  cˇ  x

jˇ j

:

Writing  D  C i with  and  2 Rn , we have, for every multi-index , J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_18, © Springer Science+Business Media, LLC 2010

287

288

18 Supports and Fourier Transform

j@x e

i hx; i

j  j j e hx; i :

On account of hy; i  kyk kk, the Cauchy–Schwarz inequality, we get e hx; i  e sK ./C kk

.x 2 K /:

If we now apply Theorem 8.8 with  D  , and work out an explicit form of the C k norm of  e i by means of Leibniz’s rule, we obtain an estimate of the form jF u./j D ju.e

i /j

 c0

sup jˇ jCj jk



jˇ j

j j e sK ./C kk :

With the choice  D 1=kk we now deduce (18.2), where we have used 

jˇ j

j j D kkjˇ j j j  .1 C kk/jˇ jCj j ;

while e  kk D e may be included with the constant. As regards (18.3), we observe that ˇZ ˇ Z ˇ ˇ jF u./j D ˇ e i hx; i u.x/ dx ˇ  e hx; i ju.x/j dx  e sK ./ kukL1 ; supp u

whenever u 2 E 0 .Rn / is integrable. On account of (14.28) this yields, for all multiindices ˛ with j˛j  k and for all u of class C k , j ˛ F u./j  e sK ./ kD ˛ ukL1 : 

This proves (18.3).

For u 2 C01 .Rn / we thus find that an estimate of the form (18.3) obtains for every k 2 Z0 . To account for the fact that the constant c will depend on k, it is written with subscript k. Example 18.2. Let u 2 E 0 .Rn /. Then there exist a continuous function f on Rn and N 2 Z0 such that u D .I
/N f . Indeed, every distribution with compact support is of finite order k, and Theorem 18.1 therefore yields an estimate of the form jF u./j  c .1 C kk/k . 2 Rn /: Now choose N sufficiently large to ensure that 2N k > n. One then has g 2 L1 .Rn / if
N g./ WD 1 C kk2 F u./ . 2 Rn /:

On account of the Riemann–Lebesgue Theorem, Theorem 14.2, f D F continuous function on Rn with f .x/ ! 0 as kxk ! 1. Furthermore, F .I It follows that .I





N
N
f ./ D 1 C kk2 g./ D F u./;


/N f D u.

1

g is a

18 Supports and Fourier Transform

289

Using a smooth cut-off function we can deduce, for every distribution u on an open subset X of Rn and for every compact subset K of X , the existence of a continuous function f on X and an N 2 Z0 such that u D .I
/N f on a neighborhood of K. This implies that every distribution can locally be written as a linear combination of derivatives of a continuous function. In Theorem 13.1 we gave a different proof. ˛ Now we come to another proof of the Ehrenpreis–Malgrange Theorem, Theorem 17.13; it is based on Theorem 18.1. We begin by considering a special case. Proposition 18.3. Let c ¤ 0 and P .D/ D c

Dnm

C

m X1

Pj .D 0 / Dnj

D 0 D .D1 ; : : : ; Dn

with

j D0

1/

be a linear partial differential operator in Rn of order m 2 N with constant coefficients. Then P .D/ has a fundamental solution in D 0 .Rn /. Proof. The symbol of P .D/ is P ./ D c nm C

m X1

Pj . 0 / nj ;

where

j D0

 0 D .1 ; : : : ; n

1/

2 Rn

1

and the Pj are polynomial functions on Rn 1 . For 0 2 Rn 1 , the polynomial function n 7! P .0 ; n / on C has m zeros counted with multiplicities. Denote by D.0 / the set of those zeros with repetitions if they have positive multiplicities. Then there exists .0 / 2 R such that j.0 /j < m C 1

and

j.0 /

Im .0 /j > 1

..0 / 2 D.0 //: (18.4)

The roots of a polynomial with constant highest coefficient depend continuously on the remaining coefficients (see Rouch´e’s Theorem from complex function theory or [7, Example 8.11.11], both of which are concerned with winding numbers, that is, degrees of mappings). Therefore one can find an open neighborhood U.0 / of 0 in Rn 1 such that (18.4) also is valid with 0 replaced by arbitrary  0 2 U.0 /. One may take U.0 / to be an open cube in Rn 1 parallel to the coordinate axes and centered at 0 . These cubes form an open cover of Rn 1 . The Heine–Borel Theorem, Theorem 2.2.(c), allows one to extract a countable and locally finite subcover of Rn 1 , whose elements we denote by Uk D U.0k /, for k 2 N. Next, one can define sets Vk  Uk by k[1 V1 D U 1 ; Vk D U k n Vj .k 2 Z>1 /: j D1

Then the Vk are mutually disjoint, their union equals Rn 1 , and any compact subset of Rn 1 meets only a finite number of them. Furthermore, define the (real) n-

290

18 Supports and Fourier Transform

dimensional subsets Zk and Z of Rn Zk D Vk  .R C i .0k //

1

 C by

.k 2 N/

Obviously, for  D . 0 ; n / 2 Rn 1  C, Y jP ./j D jcj jn . 0 /j  jcj . 0 /2D. 0 /

and Y

. 0 /2D. 0 /

ZD

[

Zk :

(18.5)

k2N

j Im n

Im . 0 /j:

By construction, (18.4) holds for each  0 belonging to some Uk . Therefore it is clear from (18.5) that jP ./j  jcj . 2 Z/: (18.6) Now define E 2 D 0 .Rn / by setting, for  2 C01 .Rn /, Z Z 1 1 X F ./ F ./ E./ D d  D d : n n .2/ Z P . / .2/ Zk P . /

(18.7)

k2N

This sum converges on account of (18.6) (with P replaced by SP ) as well as (18.3), which leads to the existence of constants c > 0 and d > 0 such that, for  D . 0 ; n C i .0k // 2 Zk with arbitrary k, jF ./j  c .1 C k. 0 ; n /k/

n 1

0

e d j.k /j  c .1 C k. 0 ; n /k/

n 1

e d.mC1/ :

Note that the estimate on the right-hand side is independent of k and thus applies to all  2 Z. Furthermore, c and d can be taken to be the same for all  2 C01 .Rn / supported in a fixed compact set. Therefore E 2 D 0 .Rn /. Finally, first applying Cauchy’s Integral Theorem from (12.9) to the integration with respect to the variable n and then Fourier inversion, one obtains Z 1 P . / F ./ d P .D/E./ D E.P . D// D n .2/ Z P . / Z Z 1 1 F ./ d  D F ./ d  D .0/ D ı./: D .2/n Z .2/n Rn  Theorem 18.4. Every linear partial differential operator in Rn with constant coefficients (not all of them equal to 0) possesses a fundamental solution in D 0 .Rn /.

Proof. Denote the differential operator of order m 2 N by P .D/ and its principal symbol by Pm ./; see Definition 17.2. Let A W Rn ! Rn be a linear isomorphism and write An for the last column of the matrix of A. A straightforward computation shows that the coefficient of nm in the symbol P .A / of the differential operator P .A D/ equals Pm .An /. One can clearly choose A as above such that Pm .An / ¤ 0. The preceding proposition then yields the existence of F 2 D 0 .Rn / such that P .A D/F D ı. Hence, application of (10.2) with A replaced by B WD tA and of

18 Supports and Fourier Transform

291

Example 10.12 leads to P .D/.B  F / D B  B P .A D/F D B  ı D j det Bj

1

ı:

Phrased differently, j det Bj B  F 2 D 0 .Rn / is a fundamental solution of P .D/.  Next, we present a converse of Theorem 18.1. For a real-valued function s on a subset Y of Rn n f0g we define the subset Ks of Rn by \ f x 2 Rn j hx; i  s./ g: (18.8) Ks WD 2Y

Before proceeding further, we give a characterization of the sets of the form Ks . Theorem 18.5. For every Y  Rn n f0g and s W Y ! R [ f1g, the subset Ks of Rn as defined by (18.8) is closed and convex. Conversely, if K is a closed and convex subset of Rn , then K D Ks if s D sK is defined by (18.1). Proof. A subset K of a linear space is said to be convex if x and y 2 K implies that the line segment f x C t .y x/ 2 Rn j 0  t  1 g from x to y is contained in K. For  2 Rn n f0g and c 2 R, it is evident that the half-space H.; c/ WD f x 2 Rn j hx; i  c g is convex. Because the intersection of a collection of convex sets is also convex, we deduce the convexity of Ks , the intersection of all H.; s.//, where  2 Y . Furthermore, Ks is closed, being the intersection of the closed subsets H.; s.// of Rn . With respect to the converse assertion we begin by observing that K  Ks , for every K  Rn , if s D sK ; this immediately follows from the definitions. What remains to be shown, therefore, is that if K is convex and closed, and if y … K, it follows that y … Ks . That is, there exists  2 Rn such that sK ./ < hy; i. Because K is closed, there is a 2 K satisfying ky

ak D d.y; K/ WD inf ky x2K

xk:

Indeed, there exists a sequence .xj /j 2N in K such that ky xj k ! d.x; K/. This sequence is bounded, and hence it contains a subsequence that converges to some a; then ky ak D d.y; K/. Because K is closed, a 2 K. Now let x 2 K. Because K is convex, we see, for every 0  t  1, that a C t .x a/ 2 K; as a result, ky

ak2  ky

.a C t .x

a//k2 D ky

ak2

2t hy

a; x

ai C t 2 kx

ak2 :

This implies that the derivative of the right-hand side with respect to t cannot be negative at t D 0, that is,

292

18 Supports and Fourier Transform

hx; y

ai  ha; y

ai

.x 2 K/:

In other words, sK .y a/ D ha; y ai < hy; y ai, where the inequality is a consequence of the assumption that y … K; therefore, y ¤ a. This is the desired inequality, with  D y a.  Theorem 18.6. Let Y  Rn n f0g and s W Y ! R such that Ks is a bounded (and therefore compact) subset of Rn . Let N 2 R. Suppose that U is a complexanalytic function on Cn with the property that for every  2 Y there exists a constant c./ > 0 such that (compare with (18.2)) jU. C i /j  c./ .1 C kk/N e  s./

. 2 Rn ;   0/:

(18.9)

Then U D F u for some u 2 E 0 .Rn / satisfying supp u  Ks . If, in addition, N < n k, then u is of class C k . Proof. Denote the restriction of U to Rn by v. First assume that N < n k. In that case, the functions  ˛ v are integrable on Rn if j˛j  k, and consequently, u WD F 1 v is a bounded C k function on Rn . Now let  2 Y . Because  ¤ 0, there exists a linear mapping A W Rn 1 ! Rn such that B W .; / 7!   C A is a bijective linear mapping from Rn to Rn ; the latter has a complex-linear extension B W Cn ! Cn . The change of variables  D B.; / yields Z Z u.x/ D j det Bj F B B.; / d d; Rn

where

1

F ./ D

R

1 e i hx; i U./: .2/n

On account of Cauchy’s Integral Theorem (see (12.9)), applied to the complexanalytic function F B B.; / of  D  C i  2 C, the integration over the real -axis may be replaced by integration over  7!  C i . Here the estimate (18.9) and the condition on N guarantee that the contributions over t 7!  C i t, for t 2 Œ 0; 1 , converge to 0 as jj ! 1. Returning to the old variables, we get Z 1 u.x/ D e i hx; Ci  i U. C i  / d : .2/n Rn Since je i hx; Ci  i j D e

 hx; i

, substitution of the estimate (18.9) gives

ju.x/j  c e

 hx; i

e  s./ ;

with a constant c D c./ > 0 that is independent of   0. If hx; i > s./, the right-hand side converges to 0 as  ! 1; in that case, therefore, ju.x/j  0, implying u.x/ D 0. The conclusion is that hx; i  s./, for all x 2 supp u. Because this is true for all  2 Y , we have proved supp u  Ks .

18 Supports and Fourier Transform

293

If we drop the assumption N < n k, we still have v 2 S 0 ; so v D F u, for some u 2 S 0 . Choose  2 C01 .Rn / with .x/ D 0 if kxk > 1 and 1./ D 1. For  .x/ WD  n . 1 x/ we find that u WD u ! u in S 0 as  ! 0. Furthermore, one has, on the strength of Theorem 14.33, v ./ WD F .u /./ D F u./ F  ./ D U./ F . /: The right-hand side has an extension to a complex-analytic function on Cn , which we denote by U . Using the Cauchy–Schwarz inequality we obtain sK ./ D kk if K D fx 2 Rn j kxk  1g: Combining the estimate (18.3) for  with (18.9) we get, for  D  C i , jU ./j  c.; k/ .1 C kk/

k

e kk .1 C kk/N e  s./ :

Choosing k > N C n we may apply the theorem, with k as in the theorem set to 0, to conclude that supp u is contained in the half-space H ./ WD f x 2 Rn j hx; i  s./ C kk g: On account of supp uı  Hı ./  H ./, whenever 0 < ı   and u D limı#0 uı , this implies supp u  H ./. Because this is true for every  > 0, we conclude that supp u  H0 ./. Finally, since this applies for all  2 Y , it follows that supp u  Ks .  Theorems of this kind are called Paley–Wiener Theorems, and the estimates for the function U that they contain are the Paley–Wiener estimates. Theorems 18.1 and 18.6 in essentially this form are due to Schwartz; Paley and Wiener studied the Fourier transformation of integrable functions on R with support in a half-line. Example 18.7. It is a classical idea, due to Cauchy, to solve the wave equation d2 u t D
u t dt 2 in .x; t/-space by treating t as a parameter in the Fourier transform with respect to the variable x. We will use the notation from Remark 13.4. To initiate this treatment; we assume that t 7! u t is a C 2 family in S 0 .Rn /. Because F W S 0 ! S 0 is a continuous linear mapping, t 7! F u t in that case also is a C 2 family in S 0 and  d2  d2 F u D F u t D F .
u t / D t dt 2 dt 2

kk2 F u t ;

in view of Theorem 14.24. But this implies

F u t ./ D p./ cos t kk C q./ with

sin t kk ; kk

(18.10)

294

18 Supports and Fourier Transform

p D F u0

and

qD

ˇ d ˇ  d ˇ ˇ F ut ˇ ut ˇ ; DF t D0 t D0 dt dt

(18.11)

or p D F a and q D F b in the notation of Remark 13.4. This causes us to turn to the functions A t ./ WD cos t kk

B t ./ WD

and

sin t kk ; kk

(18.12)

the solutions for p D 1 and q D 0, and for p D 0 and q D 1, respectively. Both functions possess an extension to a complex-analytic function on Cn , owing to the fact that cos z D C.z 2 / and sinc z D S.z 2 /; where C and S are complex-analytic functions on C; this can be seen from the power series, for example. Accordingly, we may write A t ./ D C.t 2 kk2 /

B t ./ D t S.t 2 kk2 /I

and

here  7! kk2 has the complex-analytic extension  C i D  7!

n X

j D1

j 2 D kk2

kk2 C 2i h; i:

In order to estimate A t ./, we observe that j cos zj  e jyj if z D x C iy with x, y 2 R. Solving y 2 from x2

y 2 C 2ixy D z 2 D t 2 .kk2

kk2 C 2i h; i/;

we obtain y2 D

t2 .kk2 2

kk2 ˙

p .kk2

kk2 /2 C 4h; i2 /  t 2 kk2 :

Here we have used the Cauchy–Schwarz inequality. This leads to jA t . C i/j  e jt j kk

.;  2 Rn /:

Now hx; i  jtj kk, for all  2 Rn , if and only if kxk  jtj. Similar Paley– d Wiener estimates hold for B t if it is observed that dt sin tz D z cos tz, which R1 jyj implies j sin zj D jz 0 cos tz dtj  jzj e . On the basis of Theorem 18.6 we thus find that A t D F a t and B t D F b t , for distributions at and b t in Rn with supp at and supp b t both contained in the ball about 0 of radius jtj. Moreover, these are C 2 families in E 0 .Rn / with d2 a t D
a t dt 2 and initial conditions

and

d2 b t D
b t dt 2

18 Supports and Fourier Transform

a0 D ı;

d ˇˇ at ˇ D0 t D0 dt

295

and

b0 D 0;

d ˇˇ bt ˇ D ı: dt t D0

(18.13)

The fact that a t and b t are distributions with compact support enables us to convolve them with arbitrary distributions. The conclusion is that for every a and b 2 D 0 .Rn /, the distributions u t WD a t  a C b t  b

.t 2 R/

give a C 2 family t 7! u t in D 0 .Rn / that forms a solution of the Cauchy problem d2 u t D
u t dt 2

with

u0 D a

and

d ˇˇ ut ˇ D b: t D0 dt

(18.14)

Combining this with the uniqueness from Remark 13.4, we conclude that the Cauchy problem for the wave equation has a unique solution, even for arbitrary distributional initial values. Next, we consider the wave equation with an inhomogeneous term f 2 D 0 .R  n R / as in (13.28). Once again, we assume f toRbe associated with a family f t of distributions in D 0 .Rn / via the formula f ./ D R f t . t / dt, for all  2 C01 .RnC1 /, with the notation  t W x 7! .t; x/. Furthermore, we suppose that the f t 2 D 0 .Rn / depend continuously on the parameter t 2 R. Recall that this means that the function t 7! f t ./ is continuous, for every  2 C01 .Rn /. (The principle of uniform boundedness is required in showing that one really defines a distribution in this manner.) In order to get an idea about the solution, we initially consider the situation in which t 7! f t is continuous with values in S 0 .Rn /. In that case, t 7! F f t is also continuous with values in S 0 .Rn /. Upon Fourier transformation the inhomogeneous wave equation takes the form d2 F v t D kk2 F v t C F f t : dt 2 Moreover, let us suppose that F f t ./ depends continuously on t and . Under that assumption we may, for fixed , consider the equation as an inhomogeneous secondorder equation with constant coefficients and continuous inhomogeneous term. As in the proof of Theorem 9.4, we write it as a first-order inhomogeneous system d V t ./ D L./ V t ./ C F t ./; dt  0  Fv  1 t ; L./ D ; Vt D d 2 kk 0 dt F v t

where and

Ft D

 0  : F ft

Let ˚ t ./ be the fundamental matrix determined by A t ./ and B t ./ for the homogeneous system, more precisely,   Bt At ˚ t ./ D d D exp t L./ and ˚0 ./ D I: d dt A t dt B t

296

18 Supports and Fourier Transform

Next, we use Lagrange’s method of variation of constants to obtain a solution of the inhomogeneous system having the form ˚ t ./ W t ./. This leads to Z t Z t V t ./ D ˚ t ./ ˚ s ./ Fs ./ ds D ˚ t s ./ Fs ./ ds; t0

t0

for some arbitrary t0 2 R. Thus, F v t D .1; 0/ V t ./ D .1; 0/ D and hence

Z

t

F bt t0

Z

t

˚t t0

s F fs ds D F

vt D

Z

s

Z

Fs ds D

t

bt t0

s

Z

t

Bt

s

F fs ds

t0

  fs ds ;

t

bt t0

s

 fs ds:

(18.15)

The formula for v t is a version of Duhamel’s principle. Recall that supp b t s is contained in the closed ball around 0 of radius jt sj. Therefore (18.15) also makes sense for a continuous family t 7! f t of distributions in D 0 .Rn /. The family t 7! v t thus defined is a C 2 family in D 0 .Rn /. A direct verification that the family defined in (18.15) provides a solution to the inhomogeneous equation is also feasible. Indeed, Z t Z t d d d v t D b t s  fs jsDt C bs t  fs ds D bs t  fs ds dt dt dt t0 t0 in view of (18.13). Using these conditions once more, we obtain ˇ Z t 2 Z t ˇ d d2 d ˇ b t s  fs ˇ vt D C b  fs ds D ı  f t C
bs 2 s t dt 2 dt t0 dt t0 sDt D f t C
v t :

t

 fs ds

Furthermore, the initial values can be adjusted by adding to v t a solution of the Cauchy problem (18.14) with suitably chosen constants a and b. If f t is a family of distributions with f t D 0 for t < t0 , then the lower limit t0 may be replaced by 1. For the corresponding distribution v we thus obtain the following formula, where  2 C01 .RnC1 /: Z Z Z Z Z v./ D v t . t / dt D b t s fs . t / ds dt D b t s .Sfs  t / dt ds: R

R

st

R

nC1

t s

If we take f D ı R , the Dirac measure on RnC1 , then certainly t 7! f t is not a continuous family. Yet we have obtained a formula that makes sense, namely Z Z Rn v./ D b t 0 .Sı   t / dt D b t . t / dt: R0

R>0

18 Supports and Fourier Transform

297

This formula really defines a fundamental solution. In fact, Z Z  d2   v./ D v.  / D dt bt  b t .
 t / dt t dt 2 R>0 R>0 Z Z  d2  bt  t dt
b t . t / dt D dt 2 R>0 R>0 Z   d2  d2  bt  b . / dt: D t t t dt 2 dt 2 R>0 Using integration by parts we now deduce h d   v./ D b t t dt

i1 d n nC1 b t . t / D ı R .0 / D .0/ D ı R ./: 0 dt

Denoting the fundamental solution thus obtained by EC , we may write Z b t . t / dt . 2 C01 .RnC1 //: EC ./ D R>0

We note that supp b t is contained in the closed ball around 0 of radius jtj. Therefore EC D 0 on the union of the set in R  Rn where t < 0 and the set where kxk > t. In other words, supp EC is contained in the solid cone CC D f .t; x/ 2 R  Rn j kxk  t g: On account of the uniqueness result from Remark 13.4, this fundamental solution equals the one from Theorem 13.3. This implies that b t is analytic on f x 2 Rn j kxk < t g and even vanishes on that set for odd n > 1. More precisely, we obtain the formula bt D Here

1 2

n 1 2

 3 n q t  C2

q t D q B i t W x 7! t 2

kxk2

.t > 0/: W

Rn ! R;

in other words, the composition of q by the mapping i t W x 7! .x; t/ from Rn to RnC1 . Furthermore, q t is a submersion from Rn n f0g to R; there is not a real 3 n

problem at x D 0, however, since b t 2 C 1 at x D 0 and C2 2 C 1 at q D t. One hesitates to write b t D i t  .EC /, because i t is not a submersion. For b t with t < 0 one uses b t D bt ; while at can be expressed in terms of b t by means of the formula at D

d bt : dt

(18.16)

298

18 Supports and Fourier Transform

The analyticity of a t and b t in kxk < t can also be demonstrated directly if the paths of integration in the integral formulas for at D F 1 A t and b t D F 1 B t , respectively, are slightly rotated in suitably chosen complex directions. ˛ Remark 18.8. It is natural to try to derive explicit formulas for the distributions at and b t , whose respective Fourier transforms A t and B t are given by (18.12). In view of (18.16), only b t needs to be determined. A uniform formula for all dimensions n is given by
.n 1/=2 . nC1 / b t .x/ D pn lim Im kxk2 .t C i /2 : with pn D nC1 2 #0  2 .n 1/ (18.17) If n D 2, then b t equals the locally integrable function bt D

(

sgn t .2/

1

.t 2

kxk2 /

1=2

0

if

kxk < jtj;

if

kxk > jtj;

as can be seen from (18.17) for n D 2, for which some extra work is required, however. If n D 3, then b t ./ equals t times the mean of  over the sphere of radius jtj about the origin. For n D 3, the derivation of this classical description of b t from (18.17) involves further complications. For n  4 it is useful to observe that kxk2

.t C i /2




.n 1/=2

D .n 3/ 1 .t Ci /

1

d kxk2 dt

.t C i /2




.n 3/=2

;

which shows that the right-hand side of (18.17) can be identified with a constant d p 1 times the differential operator .t 1 dt / of order p 1 applied to the distribution lim Im kxk2 #0

.t C i /2




1=2

or

lim Im kxk2 #0

.t C i /2




1

;

depending on whether n D 2p or n D 2p C 1, respectively. To within a constant
1=2 factor, the latter distributions equal the integral of t 2 kxk2 .x/ over the open ball in Rn of radius jtj about 0, and t n 2 times the mean of the test function  over the sphere of radius jtj about 0, respectively. Equation (18.17) may be verified as follows. The function 1 v.x; t/ D p e t

x2 4t

.x 2 R; t > 0/

satisfies the heat equation @ t v D @2x v (see Problem 5.5), and furthermore, Z v.x; t/ dx D 1: R>0

Under the assumption x > 0, all derivatives @kt v.x; t/ converge to 0 as t # 0 and also as t ! 1. This enables us to conclude that the integral

18 Supports and Fourier Transform

299

I.x/ WD converges and that 00

I .x/ D

Z

t

e

@2x

Z

t

e

v.x; t/ dt

R>0

v.x; t/ dt D

R>0

D

Z

v.x; t/ @ t e

t

R>0

Z

e

t

@ t v.x; t/ dt

R>0

dt D I.x/:

Indeed, the first identity follows by differentiating twice under the integral sign, the second from the heat equation, and the third through integration by parts. By solving the differential equation we obtain the existence of a constant c 2 C such that I.x/ D c e x , because I.x/ ! 0 as x ! 1. As a result we have Z Z Z I.x/ dx D e t v.x; t/ dt dx cD D

Z

R>0

e

t

R>0

Z

R>0

R>0

R>0

v.x; t/ dx dt D

Z

e R>0

t

dt D 1:

This proves the subordination identity (see [7, Exercise 2.87.(v)] for another proof) Z 1 e t x2 x e Dp p e 4t dt .x > 0/:  R>0 t Replacing x by kk with  2 Rn , then multiplying by .2/ n e i hx; i , integrating over  2 Rn , changing the order of integration, and finally using Example 14.11, we obtain the following formula for the inverse Fourier transform of the function f ./ D e kk (see (17.12) for another proof): F

1

f .x/ D D

If we now define f ./ D e F

1

Z n 1 .4 t/ 2 e .2/n R>0 . t/ 12 . nC1 1 2 / 

 kk

nC1 2

.1 C kxk2 /

t .1Ckxk2 /

dt (18.18)

: nC1 2

, then

f .x/ D

. nC1 2 / 

nC1 2

 .kxk2

C  2/

nC1 2

:

If n > 1, which we assume from now on, then F ./ WD e  kk =kk defines a locally integrable function of  whose derivative with respect to  equals f ./. This leads to pn G .x/ WD F 1 F .x/ D ; (18.19) n 1 2 .kxk C  2 / 2

300

18 Supports and Fourier Transform

where we have used the fact that both sides converge to zero as  ! 1, and where the constant pn is as in (18.17). Now F is a complex-differentiable function of  with values in S 0 .Rn /, defined on the complex half-plane f  2 C j Re  > 0. Furthermore, this S 0 .Rn /-valued function is continuous on f  2 C j Re   0 g. Because F 1 is a continuous operator, we arrive at the same conclusions, with G D F 1 F instead of F , where formula (18.19) remains valid for Re  > 0. If we write  D  i t D i.t C i /, with t 2 R and  > 0, and let  # 0, this leads to F

1



 e i t k
k .x/ D pn lim kxk2 k
k #0

.t C i /2




.n 1/=2

:

(18.20)

Because F . / D F ./, the complex conjugate of F 1 F equals the inverse Fourier transform of the complex conjugate of F  ; therefore, we also obtain that Im F 1 F D F 1 .Im F /. On account of B t D Im Fi t , (18.17) now follows from (18.20). ˛ Remark 18.9. The theory of the wave operator  has a generalization to a wide class of linear partial differential operators P .D/ with constant coefficients. Let P ./ be the symbol of P and Pm ./ the homogeneous part of degree m of P ./, where m is the order of P . Let  2 Rn n f0g. The operator P is said to be hyperbolic with respect to  if for every  2 Rn , the polynomial  7! Pm . C / on C of degree m has only real zeros. For n > 1, this excludes the possibility that P is elliptic, because in that case  7! Pm . C / does not have any real zeros if  and  2 Rn and  is not a multiple of . Hyperbolicity is equivalent to the assertion that Pm . C i / ¤ 0 whenever  2 Rn ,  2 R, and  ¤ 0. Using the homogeneity of Pm , we obtain jPm . C i /j  c .kk C jj/m

. 2 Rn ;  2 R/;

where c is the infimum of the left-hand side for kkCjj D 1; that is, c > 0. Because the other terms in P ./ are of degree < m, there exist 0  0 and a constant c 0 > 0 such that c0 1  . 2 Rn ;  > 0 /: jP . C i /j .kk C jj/m This allows the definition of E 2 D 0 .Rn / by means of the formula Z F 1 . C i / d . 2 C01 .Rn //; E./ D P . C i / Rn

where we have chosen  > 0 . For F 1 ./ D .2/ n F . / we use Paley– Wiener estimates of the form (18.3). In view of the identity F 1 .P . D//./ D P ./ F 1 ./, translation of the path of integration leads to Z Z F 1 .Ci / d  D F 1 ./ d  D .0/: P .D/E./ D E.P . D// D Rn

Rn

18 Supports and Fourier Transform

301

That is, P .D/E D ı; in other words, E is a fundamental solution of P . In addition, by Cauchy’s Integral Theorem (see (12.9)), the definition of E is independent of the choice of   0 . Using the Paley–Wiener estimates for F . / we see that the limit as  ! 1 equals 0 if hx; i > 0 for all x 2 supp . That is, supp E  f x 2 Rn j hx; i  0 g: We obtain a fundamental solution with support in the opposite half-space by substituting  for . Moreover, the Cauchy problem with initial values on the hyperplane f x 2 Rn j hx; i D 0 g can be shown to have a unique solution. In this text, we will not elaborate the theory any further; the main purpose of the remarks above was to demonstrate that the ideas underlying the Paley–Wiener Theorems can find application in quite a few situations. ˛ Remark 18.10. The constructions of parametrices and solutions of the Cauchy problem of hyperbolic differential equations with variable coefficients find their natural context in the class of the so-called Fourier integral operators. These form an extension of the class of pseudo-differential operators mentioned in Remark 17.8. See, for example, H¨ormander [11, Chap. 25] or Duistermaat [6]. ˛ Suppose we want to reconstruct a function f 2 S.R/ from its values sampled at the points of the lattice !Z in R, for ! > 0. The process of sampling may be represented by X X fsam WD f .k!/ ık! D f .k!/ Tk! ı 2 S 0 .R/: k2Z

k2Z

Then, using (14.35) and applying (16.8) with x replaced by  and  by F 1 f D 1 S F f , we obtain 2 X a X F fsam ./ D f .k!/ e i k!  D F f . C ka/ . 2 R/: (18.21) 2 k2Z

k2Z

Here, as usual, a ! D 2. For a suitable function u that is still to be determined, we introduce the reconstructed function by frec WD fsam  u;

and therefore

F frec D F fsam F u:

(18.22)

Furthermore, assume f to be band-limited, more precisely that supp F 1 f  Œ ;  . Then  a D 2 and !D : (18.23)  Select also u such that, see Example 14.1, a F u D 1Œ 2

;  

2 E 0 .R/;

that is,

u.x/ D sinc x:

(18.24)

302

18 Supports and Fourier Transform

Under these assumptions, we obtain, by combination of (18.22), (18.21), and (18.24), X F frec ./ D 1Œ ;   ./ F f . C ka/ D F f ./: k2Z

Hence, by Fourier inversion we obtain the following cardinal series expansion of f , valid for all x 2 R: X X f .x/ D frec .x/ D f .k!/ u.x k!/ D f .k!/ sinc .x k!/ D

X

k2Z

k2Z

sin x X . 1/k f .k!/ : k/ D  x k!

f .k!/ sinc.x

k2Z

(18.25)

k2Z

The identity (18.25) is the Sampling Theorem, which is due to Whittaker, Kotel’nikov, and Shannon, in the case that g WD F

1

f 2 C01 .R/:

It asserts that perfect reconstruction of the band-limited function f D F g is possible. Of course, this fact is related to the special nature of f : it can be extended to C as a complex-analytic function satisfying estimates as in (18.3), for every k 2 Z0 . Remark 18.11. Note that (18.25) is not true for arbitrary distributions g satisfying supp g  Œ ;  . For instance, if f .x/ D sin x, then g D 2i .ı ı  /, but f .k!/ D sin k! D sin k D 0, for all k 2 Z. The inverse Fourier transform of a nonconstant polynomial even has support in f0g, while the cardinal series of such a polynomial is pointwise divergent. On the other hand, (16.27) implies that the Fourier series of the function ei t on  ;  Œ , for t 2 R, is given by e i tx D

sin  t X . 1/k i kx e  t k

.  < x < /:

(18.26)

k2Z

However, for fixed  < x < , the function f D eix on R satisfies g D ı x and thus supp g  Œ ;  . Therefore (18.26) may be recognized as the cardinal series of f . Here we have to interpret the series as a limit of symmetric partial sums. ˛ By expanding the periodization of g into a Fourier series, we now extend (18.25) to the more general case of g 2 E 0 .R/ satisfying the extra condition (18.28) below. In Remark 18.13 we will discuss properties of g or f that imply this condition. Theorem 18.12. Let  > 0 and g 2 E 0 .R/ with supp g  Œ ;   and define f D F g 2 C 1 .R/. Select  2 C01 .R/ with 1./ D 1 and define  as in Lemma 2.19. Put  D   1Œ ;   2 C01 .R/ (18.27) and suppose g D lim  .g #0

Ta g/

in

E 0 .R/:

(18.28)

18 Supports and Fourier Transform

303

Then we have f D

X k2Z

f .k!/ .Tk! B  / sinc

in S 0 .R/:

(18.29)

P Proof. We employ the usual notation as in (18.23). Consider h D k2Z Tka g in D 0 .R/. Then h is periodic of period a and so belongs to S 0 .R/ in view of Theorem 16.3. On account of its periodicity we may expand h into its Fourier series, while Lemma 16.4 implies X X hD ck ei k! D S F v; where vD ck ık! : k2Z

k2Z

By means of Theorem 16.6, Definition 16.2, and Lemma 16.1 we derive 1 1X h.e i k! / D Tna g.e a a n2Z   X 1 X 1  1 D g T na .e i k! / D g e i k! T na  D g.e a n2Z a a n2Z

ck D M.e

D

i k!

h/ D

1 e a

h./ D

i k!

i k! /

i k! /

1 1 F g.k!/ D f .k!/: a a

Next, Fourier inversion leads to   1 X F S F v D 2 v D ! f .k!/ ık! : 2 k2Z Furthermore, condition (18.28) implies F h D 2

g D lim  h in S 0 .R/;

hence

#0

f D F g D lim F . h/ #0

in S 0 .R/:

On the other hand, we have F . h/ D

1 1 F  F h D F . 1Π2 2

;  

/F h D

1 .F  F 1Π2

;  

Write D F , then F  D   in view of Problem 14.30.(ii), while F 1Πa  sinc as follows from (18.24). Accordingly

/F h: ;  

D

X a!  .. / . sinc//  f .k!/ ık! 2 k2Z X D f .k!/ Tk! ..  / . sinc//:

F . h/ D

k2Z

Now .0/ D 1./ D 1. On account of Theorem 18.1 we know that f is at most of polynomial growth and that is rapidly decreasing on R; therefore the series on the right-hand side converges absolutely and uniformly on R, which leads to

304

18 Supports and Fourier Transform

f D lim #0

X

k2Z

f .k!/ Tk! ..  / . sinc// D

X

k2Z

f .k!/ .Tk! B  / sinc 

in S 0 .R/. This finishes the proof.

Remark 18.13. We discuss some conditions on the function g that imply that (18.28) is satisfied. If supp g is contained in the interior of Œ ;  , then there exists 0 > 0 such that g D  .g Ta g/ as soon as 0 <   0 . Furthermore, (18.28) holds if there exist neighborhoods of the points ˙ on which g is a locally integrable function. Of course, we are also interested in conditions that guarantee the pointwise convergence of (18.29) and that are weaker than those for (18.25). Therefore, let us suppose in addition to (18.28) that X jf .k!/j < 1: (18.30) 1 C jkj k2Z

Then the series on the right-hand side in (18.29) converges absolutely and uniformly on R, which implies that its sum is a continuous function. On the other hand, (18.29) is an equality in the weak sense, but then pointwise too, because both sides in (18.29) are continuous. Because the cardinal series of f is uniformly convergent on R in this case, termwise integration of the series is admissible. Problem 14.44 asserts Z Z  D !I sinc.x k/ dx D sinc x dx D  R R in consequence, we obtain the following integration formula: Z X f .x/ dx D ! f .k!/: R

(18.31)

k2Z

Observe that (18.30) is certainly the case if g is sufficiently smooth; in fact, H¨older continuity is enough according to Problem 14.11. In addition, (18.30) is also satisfied if g 2 L2 .R/, or equivalently on account of Theorem 14.32 if f 2 L2 .R/. Indeed, g then is locally integrable, and considering it as a function that is periodic of period a, we have its Fourier series Z  X 1X gD f .k!/ ei k! I and so jf .k!/j2 D jg.x/j2 dx < 1 a  k2Z

k2Z

according to Parseval’s formula from Theorem 16.9. The Cauchy–Schwarz inequality in l 2 .Z/ then leads to   X X X 1 jf .k!/j 2 2 jf .k!/j2 < 1: jkj k2 k2Znf0g

k2Z

k2N

(18.32)

18 Supports and Fourier Transform

305

Finally, consider f 2 L1 .R/. On account of the Riemann–Lebesgue Theorem, Theorem 14.2, we have g 2 C.R/, which implies g 2 L2 .R/, and therefore f 2 L2 .R/. Example 18.14. Now we study the cardinal series (18.25) from the point of view of Hilbert space theory. Consider g 2 L2 .R/ satisfying supp g  Œ ;   and write f D F g. Then f is a complex-analytic function on C according to Lemma 14.4, while there exist constants c > 0 and k 2 Z0 such that f ./  c.1 C jj/k e j Im j

. 2 C/I

in addition,

f 2 L2 .R/:

Denote the linear space of all such functions f by PW2 .C/ and introduce an inner R product on PW2 .C/ by .f; h/PW D R f ./ h./ d . In fact, the complex analyticity of f guarantees that this is an inner product. Denote by k
kPW the corresponding norm on PW2 .C/. Now Theorems 14.32 and 18.6 imply that e F

1

e D p1 S F W PW2 .C/ ! L2 .Œ ;  /  L2 .R/ D SF  2

is a unitary isomorphism. It follows that PW2 .C/ is a Hilbert space, which is called the Paley–Wiener space of type . Define eiCk! D p1a ei k! 1Œ ;   2 E 0 .R/, for k 2 Z. Then .eiCk! /k2Z forms an orthonormal basis in L2 .Œ ;  /. Using Example 14.1 and (14.29) one easily computes, for z and  2 C, e F

1 C eiz!

D

s

z;

where

1 sz ./ D p sinc . !

z !/:

(18.33)

It follows that .sk /k2Z is an orthonormal basis in PW2 .C/. In particular, in the R case of  D , the identity ks0 kPW D 1 implies R sinc2 x dx D ; compare with Problem 14.44. Furthermore, one finds that each function f 2 PW2 .C/ has a unique expansion of the form X X f D ck sk with jck j2 < 1: (18.34) k2Z

k2Z

The convergence of the series on the left-hand side is understood with respect to the norm k
kPW . But convergence in PW2 .C/ implies uniform convergence on each horizontal strip in C. Indeed, on account of the Cauchy–Schwarz inequality we obtain, for  and  2 R, Z  ˇZ  ˇ p ˇ ˇ i.Ci/x jj jf . C i/j D ˇ e g.x/ dx ˇ  e jg.x/j dx  a e jj kgk 

e jj D p kf kPW : !



306

18 Supports and Fourier Transform

p If we evaluate the identity in (18.34) at k!, we see that ck D ! f .k!/; and as a consequence we obtain another proof of the cardinal series for f : X p X f .x/ D ! f .k!/ sk .x/ D f .k!/ sinc .x k!/ .x 2 R/: k2Z

k2Z

In particular, Parseval’s formula in PW2 .C/ leads to the following integration forZ mula: X jf .x/j2 dx D ! jf .k!/j2 ; (18.35) R

k2Z

which as in (18.32) corroborates the pointwise convergence of the cardinal series. e 1 is unitary, we have, denoting the complex conjugate of Furthermore, since F  2 C by  and using (18.33), Z  Z p  1 ix f ./ D e g.x/ dx D a g.x/ p e i ! !x 1Œ ;   .x/ dx a  R  p    p  1 e 1 g; F e 1 eC D a g; e C Sf; sinc .
C D a F D /  i!! i! ! PW ! PW Z 1 1 sin z : D f .x/ sinc . x/ dx D f  k./ with k.z/ D ! R  z In other words, f is a solution to Bateman’s integral equation for the unknown Z function g: 1 sin . x/ f ./ D dx: g.x/  R  x In addition, we obtain the integration formula Z X f .x/ sinc . x/ dx D ! f .k!/ sinc . R

k!/:

k2Z

Finally, by writing K.z; / D k.z

/ D

1 sin .z /  z 

..z; / 2 C  C/;

we obtain the reproducing kernel K for the Hilbert space PW2 .C/ satisfying f ./ D .f; K.
; //PW

. 2 C/:

˛

Problems  18.1. (Fast convergence of Riemann sums for test functions.) Let  2 C01 .R/ and let k 2 Z>1 be arbitrary. Prove that, for all N 2 N,

18 Problems

307

Z

R

.x/ dx D

 1  1 X n CO ;  N n2Z N Nk

N ! 1:

Here the implied constant may depend on  and k. 18.2. (Sequel to Problem 14.47.) Prove that there exist constants c > 0 and k 2 Z0 such that p ju.z/j  c .1 C kzk/k e  k Im zk .z 2 Cn /: 18.3. Suppose that Y  Rn consists P of a basis .j / of Rn , where 1  j  n, supplemented by a vector .n C 1/ D jnD1 cj .j /, with cj < 0 for all j . Further suppose that U is a complex-analytic function on Cn and that for every  > 0 there exist constants c > 0 and N 2 R with jU. C i  /j  c .1 C kk/N e  

. 2 Rn ;  2 Y;   0/:

Prove that U is a polynomial. 18.4. (Klein–Gordon operator.) Discuss fundamental solutions and the Cauchy problem for P D  C , where  2 C is a given constant. (For  D m2 > 0 this is the Klein–Gordon operator.) 18.5. Let u be an integrable function on R. Let H D f  2 C j Im  < 0 g and denote by H the closure of H in C. Prove that supp u  R0 if and only if F u has an extension to a bounded continuous function U on H that is complex-analytic on H.  18.6. (Distributions as boundary values of holomorphic functions.) We will prove a result converse to that in Problem 12.14. In doing so, we use the notation and results from this problem. Consider u 2 E 0 .R/. (i) Prove that f˙ W H˙ ! C are well-defined and belong to f˙ 2 O.H˙ / if Z 1 e iz  F u./ d  .z 2 H˙ /: f˙ .z/ D 2 R≷0

(ii) Show the existence of N 2 Z0 and c > 0 such that j Im zjN jf˙ .z/j  c, for j Im zj sufficiently Psmall. (iii) Verify that u D ˙ ˇ˙ .f˙ /.

18.7. (Conservation of energy for wave equation.) The energy E.t/ of a solution ut of the wave equation as in Example 18.7 is defined as Z ˇ ˇ2  1 ˇd ˇ E.t/ D ˇ u t .x/ˇ C k grad u t .x/k2 dx; 2 Rn dt

under the assumption of convergence of the integral for all t 2 R (see Problem 19.3.(ii)). The first term is kinetic energy and the second potential energy. Show that

308

18 Supports and Fourier Transform

E.t/ D

1 2.2/n

Z

ˇ2 ˇ d  ˇ ˇ ˇ F u t ./ˇ C kk2 jF u t ./j2 d  dt Rn

and use (18.10) to prove that Z  ˇ d ˇ ˇ2  1 ˇ ˇ ˇ 2 2 u kk jF u ./j C ./ E.t/ D ˇ ˇF ˇ d ; 0 t t D0 2.2/n Rn dt

which is independent of t.

18.8. (Fourier transform of x 7! e kxk and Poisson’s integral formula for halfspace – higher-dimensional generalization of Problem 14.51.) For t > 0, let f t W Rn ! R be given by f t .x/ D e t kxk . (i) Prove that f1 2 S 0 .R/ and also (compare with Problems 14.14 and 14.15) n C 1

1 nC1 2 .1 C kk2 / 2 Deduce, for all  2 Rn and t > 0, that F f1 ./ D 2n 

n 1 2

F f t ./ D 2n 

n 1 2

n C 1 2

. 2 Rn /:

t .kk2

C t 2/

nC1 2

:

D RnC1 n f0g. Define Poisson’s kernel P W RnC1 ! R by Let RnC1  
nC1 jtj 2 P .x; t/ D : nC1 nC1  2 k.x; t/k Note that P .x; t/ D P .x; t/. In statistics the function P .
; 1/ W Rn ! R is said to be the probability density of the Cauchy distribution. R (ii) Conclude that Rn P .x; t/ dx D 1, for all t ¤ 0. More generally, verify that Z

cosh ; x i

Rn

.kxk2

C

t 2/

nC1 2

dx D



nC1 2


nC1 2

e

t kk

t

. 2 Rn ; t > 0/;

and show that one therefore has the following, known as Laplace’s integrals: Z

 e t cos x .  0; t > 0/; (18.36) dx D x2 C t 2 2 t R Z >0  x sin x dx D e t  . > 0; t  0/: 2 2 2 R>0 x C t R Use the identity R>0 sinc x dx D 2 from, for instance, Problem 14.44 to obtain the validity of the latter formula for t D 0. (iii) Deduce from part (i) that we have the following semigroup property, for all t1 and t2 > 0: P .
; t1 C t2 / D P .
; t1 /  P .
; t2 /:

18 Problems

309

(iv) Let ı > 0 be arbitrary. Check that Z this to prove lim

R

f x2Rn j kxk>ı g

t !0 f x2Rn j kxk>ı g

kxk

n 1

dx < 1, and use

P .x; t/ dx D 0:

D f .x; t/ 2 RnC1 j ˙t > 0 g. Define RnC1 ˙ (v) From Example 12.3 we know that the functions .x; t/ 7! log k.x; t/k

.n D 1/I

.x; t/ 7!

1 k.x; t/kn

.n > 1/;

1

. Verify that, leaving scalars aside, P on RnC1 is the are harmonic on RnC1  ˙ partial derivative with respect toS t of the preceding function, and conclude that P is a harmonic function on ˙ RnC1 ˙ . S Let h 2 C.Rn / be bounded and define Poisson’s integral P h W ˙ RnC1 ! R of ˙ Z h by .P h/.x; t/ D .h  P .
; t//.x/ D h.x y/P .y; t/ dy: Rn

S (vi) Prove that P h is a well-defined harmonic function on ˙ RnC1 ˙ . Verify by means of parts (ii) and (iv) that, uniformly for x in compact sets in Rn , lim .P h/.x; t/ D h.x/:

˙t #0

Evidently, in the terminology of Problem 12.4, the function f D P h is a solution of the following Dirichlet problem on U D RnC1 or RnC1 : C
f D 0

with

f j@U D h:

(vii) Show that, with cn as in (13.37), Z 2 h.x C ty/ .P h/.x; t/ D dy cn Rn k.y; 1/knC1 (viii) Write h D 12 h . S nC1 with ˙ R˙

1 2 h/

..x; t/ 2 RnC1 ˙ /:

and prove the existence of a harmonic function u on

lim u.x; t/ t #0

lim u.x; t/ D h.x/ t "0

.x 2 Rn /:

Evidently, a bounded continuous function on Rn can be represented by means of the S jump along Rn  f0g made by a suitably chosen harmonic function on ˙ RnC1 ˙ . (ix) What can be said if h 2 C.Rn / above is replaced by u 2 E 0 .Rn /?

 18.9. Let t 2 R. For f W R ! R given by

f .x/ D sinc.x C t/

show that

Ff De

i t

1Œ

;  

:

310

18 Supports and Fourier Transform

Deduce that supp F f D Œ ;   and derive with (16.31). Similarly, verify that 2

F f ./ D

e

 it 

2



max 0; 2

 jj I 

P

k2Z sinc

deduce

2

.k C t/ D 1; compare

X

sinc2

k2Z

 2

 k C t D 2:

18.10. Verify that PW2 .C/ is closed under differentiation, for every  > 0. 18.11. For k 2 Z, define tk W C ! C by  1 sin2 2 . C k!/ ; tk ./ D p  ! 2 . C k!/

and show that

e F

1

. i sgn
eiCk! / D tk :

In particular, deduce that .tk /k2Z forms an orthonormal basis in PW2 .C/. Using Parseval’s formula and the notation and results from Problem 14.49, prove that, for f 2 PW2 .C/, Z

R

f .x/ tk .x/ dx D

p ! H f . k!/;

Hf D

X

f .k!/

k2Z

Verify the truth of the latter formula in the case of f D lem 14.49.(vii).

sin 



sin2 21 .
k/ 1 . 2


k/

:

by means of Prob-

Chapter 19

Sobolev Spaces

The degree of differentiability of a function u with compact support cannot easily be seen from the Fourier transform F u. For instance, there is a difference of more than n between the exponents in the estimates for F u in Theorems 18.1 and 18.6, which are necessary and sufficient, respectively, for u to be of class C k . The question how to decide by inspection whether a function is the Fourier transform of a bounded continuous function is very difficult to answer in general. In contrast, the situation for the space L2 .Rn / and the corresponding L2 -norm e D .2/ n2 F is a unitary isomorphism is much simpler, because the mapping F from L2 .Rn / to L2 .Rn / according to Theorem 14.32. Now F D ˛ u D  ˛ F u, see (14.28), implying that u is an L2 function whose derivatives of order  k all belong to L2 .Rn / if and only if .1 C kk/k F u 2 L2 .Rn /. Hence, in this case the degree of differentiability can be derived from the Fourier transform. The space of these functions u is denoted by H.k/ D H.k/ .Rn /. The following extension of this definition proves very flexible. Definition 19.1. For every real number s we define the Sobolev space H.s/ D H.s/ .Rn / of order s as the space of all u 2 S 0 .Rn / such that .1 C k
k2 /s=2 F u 2 L2 .Rn /. This space is provided with the norm Z 1=2 eu./j2 .1 C kk2 /s d  kuk.s/ WD jF : (19.1) Rn

˛

The norm has been chosen such that H.0/ D L2 . The convention to use the weight factor .1 C kk2 /1=2 , instead of 1 C kk, derives from the fact that 1 C kk2 is the symbol of I
, where
is the Laplace operator. The operators .I


/z WD F

1

B .1 C kk2 /z B F

can be defined for all z 2 C; they form a complex-analytic family of operators, analogous to the families of Riesz from Chap. 13 (see also Sect. 17.1). Thus, under this convention H.s/ can also be characterized as the space of u 2 S 0 such that J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_19, © Springer Science+Business Media, LLC 2010

311

312

19 Sobolev Spaces

v WD .I
/s=2 u 2 L2 , and the H.s/ norm of u as the L2 norm of v. For estimates the norm convention is hardly of importance, considering that 1 C kk2  .1 C kk/2  2 .1 C kk2 /: Part (a) of the following result is a version of Sobolev’s Embedding Theorem. Theorem 19.2. Let k 2 Z0 .

(a) If u 2 H.s/ and s > n2 Ck, then u is of class C k and @˛ u.x/ ! 0 as kxk ! 1, for every multi-index ˛ with j˛j  k. (b) For every u 2 E 0 , we have that u is of finite order, say k, and u 2 H.s/ if s < n2 k.

Proof. (a). s > n2 C k means 2.k s/ < n, that is, .1 C kk/k s 2 L2 . The Cauchy–Schwarz inequality for the integral inner product yields the integrability of the product of two L2 functions. Thus we see, for every multi-index ˛ with j˛j  k, that the function  7! .F .D ˛ u//./ D  ˛ .1 C kk/

s

.1 C kk/s F u./

is integrable. In view of the Riemann–Lebesgue Theorem, Theorem 14.2, this proves (a). (b). Let u 2 E 0 . From Theorem 8.8 we know that u is of finite order. From (18.2) we read that there exists c > 0 such that jF u./j  c .1 C kk/k

. 2 Rn /:

This implies that the square of .1 C kk/s jF u./j can be estimated by the function .1 C kk/2.sCk/ , which is integrable if 2.s C k/ < n, that is, s < n2 k.  Corollary:

\

s2R

H.s/  C 1

and

E0 

[

H.s/ ;

s2R

but the assertions in Theorem 19.2 are more detailed, of course. Example 19.3. In Example 1.5 we discussed functions v defined on an open interval  a; b Œ such that v and its derivative v 0 are square-integrable. If  belongs to C01 .  a; b Œ /, we find by means of Leibniz’s rule that  v 2 H.1/ ; therefore, on account of Theorem 19.2.(a), the function  v is continuous on R. Because this holds for every  2 C01 .  a; b Œ /, we conclude that v is continuous on  a; b Œ . Furthermore, one has the estimate (1.9), for every a < x < y < b. Indeed, let R  2 C01 .R/ with   0 and .z/ dz D 1. As usual, we introduce  .z/ D 1 . z / and define the function  in terms of  .z/ by (see Fig. 19.1)   .z/

D

Z

z

 . 1

x/

 .


y/ d :

(19.2)

19 Sobolev Spaces

313

If  is sufficiently small, one has



2 C01 .  a; b Œ /. Furthermore, v.x/

v.y/

1

-1

-0.5

Fig. 19.1 Illustration of (19.2). Graph of bD2

1



1.5

with  D 1=5, a D

2

1, x D

1=2, y D 1, and

equals the limit of the v. 0 / as  # 0, that is, of the v 0 .  /. On the strength of the Cauchy–Schwarz inequality, the absolute value of v 0 .  / is smaller than or equal to the product of the L2 norm of v 0 and the L2 norm of  . Now,  converges, in the L2 -sense, to the characteristic function of Œ x; y , and so the L2 norm of  converges to .y x/1=2 as  # 0. This proves (1.9). Applying the Cauchy criterion for convergence we now also see that v.x/ converges as x # a or x " b; write the corresponding limits as v.a/ and v.b/. Thus we have extended v to a continuous function on Œ a; b . It now makes sense to speak ˛;ˇ of the space H.1/ of all v 2 L2 .  a; b Œ / such that v 0 2 L2 . a; bŒ / and for which v.a/ D ˛ and v.b/ D ˇ. What is more, v is H¨older continuous of order 1=2 (see the following definition), because (1.9) now holds for all x, y 2 Œ a; b . ˛ Definition 19.4. Let X  Rn be open. The function f is said to be H¨older continuous on X of order 0 < ˛  1, with notation f 2 C ˛ .X /, if for every compact set K  X, jf .x/ f .y/j sup kx yk˛ x; y2K; x¤y is finite. ˛ In a certain sense, H¨older continuity may be viewed as fractional differentiability. The Sobolev spaces H.s/ .Rn /, for 0 < s < 1, can be characterized in terms of integrated H¨older conditions. Theorem 19.5. Consider 0 < s < 1. A function u 2 L2 .Rn / belongs to H.s/ .Rn / Z Z if and only if ju.x C y/ u.x/j2 dy dx < 1: kyknC2s Rn Rn Proof. Let y 2 Rn . In view of (14.29) we obtain F .Ty  u u/ D .eiy 1/F u, and thus Theorem 14.32 implies Z Z 1 ju.x C y/ u.x/j2 dx D je i hy; i 1j2 jF u./j2 d : .2/n Rn Rn

314

19 Sobolev Spaces

Using the nonnegativity of the integrands as well as Lebesgue’s Dominated Convergence Theorem, Theorem 20.26.(iv), we may interchange the order of integration twice to obtain Z Z ju.x C y/ u.x/j2 dy dx kyknC2s Rn Rn Z Z Z 1 je i hy; i 1j2 2 dy jF u./j d  D c kk2s jF u./j2 d ; D .2/n Rn Rn kyknC2s Rn on account of Problem 10.14, for some constant c > 0. Finally, it is easy to see that kuk.s/ < 1 if and only if F u and k
ks F u belong to L2 .Rn /.  Theorem 19.6. Let P .D/ be a linear partial differential operator in Rn of order m with constant coefficients. One then has, for every s 2 R, (a) P .D/ is a continuous linear mapping from H.s/ to H.s m/ . (b) If P .D/ is elliptic, u 2 E 0 , and P .D/u 2 H.s m/ , then u 2 H.s/ . Proof. (a). One only needs to observe that  7! .1 C kk/ m P ./ is bounded, so that the L2 norm of .1 C kk/s m P ./ Feu./ can be estimated as a constant times eu./. the L2 norm of .1 C kk/s F (b). To prove this we use (17.4). If kk  R we have U./ WD .1 C kk/s F u./ D

.1 C kk/m .1 C kk/s P ./

m

F .P .D/u/./:

Write B D f  2 Rn j kk < R g. The function jU./j2 is integrable on Rn n B, n because the factor .1 C kk/m =P ./ is bounded on R R n B2 and the other factor in n U./ is square-integrable on R . The conclusion B jU./j d  < 1 follows from the fact that U is continuous (and even analytic).  The definition of H.s/ .Rn / makes use of the global operation of the Fourier transform, which precludes a similar definition of a Sobolev space H.s/ .X /, for arbitrary open X  Rn . Still, there exists a localized version of the Sobolev spaces. Definition 19.7. Let X be an open subset of Rn and u 2 D 0 .X / a distribution on X . The distribution u is said locally to belong to H.s/ if  u 2 H.s/ .Rn / for every loc  2 C01 .X /. The space of all such u is denoted by H.s/ .X /. ˛ Theorem 19.8. Let k 2 Z0 and s 2 R.

loc loc .X /  C k .X /  H.k/ .X / for s > n2 C k. (a) H.s/ loc (b) If u is a distribution of order  k in X and s < n2 k, then u 2 H.s/ .X /. If u 2 H.lock/ .X /, it follows that u is a distribution of order k.

19 Sobolev Spaces

315

loc Proof. (a). Let u 2 H.s/ .X / and s > n2 C k. For every  2 C01 .X / one then has  u 2 H.s/ .Rn /, and therefore  u 2 C k .Rn / on account of Theorem 19.2.(a). This implies u D 1  u 2 C k , where .x/ ¤ 0. For every a 2 X , there exists a  2 C01 .X / with .x/ ¤ 0 for all x in an open neighborhood of a; this implies that u 2 C k .X /. The second inclusion follows from the fact that a continuous function with compact support is square-integrable. (b). As regards the first assertion, we note that  u 2 E 0 .X / is of order  k; hence Theorem 19.2.(b) implies  u 2 H.s/ .Rn /. Because this holds for every loc  2 C01 .X /, we conclude that u 2 H.s/ X . The proof of the second assertion is left to the reader; it may be seen in relation to the continuous inclusion of C k .X / in loc H.k/ .X /. 

Theorem 19.9. (a) Let X be an open subset of Rn and let P .x; D/ be a linear partial differenloc tial operator of order m with C 1 coefficients on X . If u 2 H.s/ .X /, one has loc P .x; D/u 2 H.s m/ .X /. (b) In addition, suppose that P .D/ is an elliptic operator with constant coefficients loc and let P .D/u 2 H.sloc m/ .X /. Then u 2 H.s/ .X /. Proof. (a) follows if the following two assertions are combined: loc .X /, one has Dj u 2 H.sloc 1/ .X /, for every 1  j  n. (i) If u 2 H.s/ loc loc (ii) If u 2 H.s/ .X / and 2 C 1 .X /, one has u 2 H.s/ .X /.

In the arguments below,  is an arbitrary element of C01 .X /. As regards (i) we note that  Dj u D Dj . u/ .Dj / u. Here  u 2 H.s/ .Rn /, and hence Dj . u/ 2 H.s 1/ .Rn / on the strength of Theorem 19.6.(a). Because the second term lies in H.s/  H.s 1/ , the conclusion is that  Dj u 2 H.s 1/ , for all  2 C01 .X /; that is, Dj u 2 H.sloc 1/ .X /. With respect to (ii) we note that  2 C01 .X /, and therefore  . u/ D . / u 2 H.s/ . (b). Let K be a compact subset of X . On account of Theorem 8.8, there exists a k 2 Z0 such that u is of order  k whenever 2 C01 .X / and supp  K. Choose l 2 Z such that s l < n2 k. In view of Theorem 19.2.(b) we then have u 2 H.s l/ . Now write X P .D/. u/ D  P .D/u C P.˛/ .D/.D ˛  u/; (19.3) ˛¤0; j˛jm

where P.˛/ .D/ are operators of order < m. This is true for every operator P .D/ of order  m. The proof proceeds by first considering the case of P .D/ D D ˛ , by mathematical induction on j˛j. Now suppose that  2 C01 .X / and supp   K. In that case, the support of WD D ˛  is contained in K, and so u 2 H.s l/ . Using Theorem 19.6.(a), we

316

19 Sobolev Spaces

find that the terms following the summation sign in (19.3) belong to H.s l mC1/ . Because it is given that  P .D/u 2 H.s m/ , we can conclude that P .D/. u/ 2 H.s l mC1/ , that is, so long as l  1. If we now apply Theorem 19.6.(b), we find that  u 2 H.s lC1/ . By descending mathematical induction on l we arrive at the result that  u 2 H.s/ , whenever  2 C01 .X / and supp   K. The fact that this loc holds for every compact subset K of X implies u 2 H.s/ .X /.  loc Remark 19.10. An important fact that justifies the name H.s/ is that loc H.s/ .Rn /  H.s/ .Rn /:

This means that for every  2 C01 .Rn /, the operator
/s=2 B  B .I

A WD .I


/

s=2

maps the space L2 .Rn / to itself. The main step in the argument is to show that  u 2 H. s/ .Rn / if  2 C01 .Rn / and u 2 H. s/ .Rn /. We treat the case of s  0; the other one is handled similarly. Indeed, for all  and  2 Rn , we have kk  1 C kk2 . Hence 1 C k

k2  1 C .kk C kk/2 D 1 C kk2 C 2kk kk C kk2  1 C kk2 C 2kk C 2kk kk2 C kk2 C kk2 kk2 D .1 C kk2 /.1 C kk/2 :

Hence .1 C kk2 / n

s 2

 .1 C kk/s .1 C k

s 2

k2 /

:

On account of F  2 S.R / we may extend the identity in (14.22) to the case of the product of  and u. Next, observe that s

f ./ WD .1 C kk2 / 2 jF   F u./j Z  .1 C kk/s jF ./j .1 C k Rn

k2 /

s 2

jF u.

/j d D g  h./

with

g D .1 C k
k/s jF j

and

h D .1 C k
k2 /

s 2

jF uj:

Using the rapid decrease of F  once more we get g 2 L1 .Rn /, while h 2 L2 .Rn / by the assumption on u. Therefore f 2 L2 .Rn /, because Problem 11.22 implies kf kL2  kg  hkL2  kgkL1 khkL2 . By means of pseudo-differential operators one proves that the assertion in Theorem 19.9.(b) remains valid when an elliptic operator P .x; D/ with C 1 coefficients on X is substituted for P .D/. ˛ Remark 19.11. The fact that Sobolev norms are better suited to partial differential operators than C k norms is also illustrated by the fact that there exist continuous

19 Problems

317

functions f with compact support in Rn whose potential is not C 2 , at least not if n > 1. For example, let f .x/ D

˚

0

.x/ x12 kxk2 log kxk

if 0 < kxk < 1; for x in all other cases.

Here  2 C01 .Rn /, .0/ > 0, and .x/ D 0 if kxk  r, where 0 < r < 1. Then f is continuous if we take f .0/ D 0. Furthermore, f has compact support and sing supp f  f0g. For the potential u D E  f of f we obtain
u D f , and therefore sing supp u  f0g. However, further computation yields lim

x!0; x¤0

@21 u.x/ D 1;

in other words, @21 u is certainly not continuous at 0. If, however, f is H¨older continuous of order ˛, the potential u of f is actually C 2 , and all second-order derivatives of u are also H¨older continuous of order ˛. Furthermore, Sobolev norms are meaningful for general distributions. However, the notion of “C k plus H¨older” is preferred if uniform estimates for function values are required. ˛

Problems 19.1. Let P .D/ be an elliptic operator of order m and E the parametrix of P .D/ constructed in Lemma 17.5. Prove that E  W u 7! E  u is a continuous linear mapping from H.s m/ to H.s/ . 19.2. Let E be the potential of ı in Rn , for n  3. Is E  u 2 H.s/ for every u 2 H.s 2/ ?

 19.3. Let a t and b t be the solutions of the Cauchy problem for the wave equation from Example 18.7. Prove, for arbitrary s 2 R, d (i) b t , a t  and dt a t  are continuous linear mappings from H.s/ to H.sC1/ , to H.s/ , and to H.s 1/ , respectively. (ii) If u t ˇ is a solution of the homogeneous wave equation with u0 2 H.s/ and d d uˇ 2 H.s 1/ , one has, for every t 2 R, that u t 2 H.s/ and dt ut 2 dt t t D0 H.s 1/ .

19.4. Examine the following functions on R and decide in which Sobolev spaces they are contained, and in which they are not (parts (iii) and (iv) are difficult): (i) the characteristic function of the interval  2; 5 , (ii) x 7! e x H.x/,

318

19 Sobolev Spaces

(iii) x ! 7 H.x/=.x C 1/, (iv) x ! 7 x H.x/, (v) x 7! 1 x 2 for jxj  1, and zero elsewhere.

 19.5. Let s > n2 and 0 < ˛ < 1. If s D n2 C ˛, then H.s/ .Rn /  C ˛ .Rn /. More generally, if s D n2 C k C ˛, where k 2 N, then H.s/ .Rn / is contained in

n

f 2 C k .Rn / j

o lim @ˇ f .x/ D 0; kˇj  kI @ˇ f 2 C ˛ .Rn /; jˇj D k :

kxk!1

 19.6. Suppose P .D/, s, and m are as in Theorem 19.6.(b). If u 2 L2 .Rn / and P .D/u 2 H.s m/ .Rn /, prove that u 2 H.s/ .Rn /. Furthermore, prove the existence a constant c > 0 such that for all such u,

kuk.s/  c .kuk.0/ C kP .D/uk.s

m/ /:

Observe that some condition on the growth of u is necessary in view of examples 2 2 such as u W R2 ! R with u.x/ D e x1 x2 cos 2x1 x2 . This u satisfies
u D 0, but it does not belong to H.s/ for any s 2 R. Furthermore, the assumption u 2 L2 .Rn / may be weakened to u 2 H. t / .Rn /, for some t > 0. 19.7. For all  and  2 R, prove that 0
0 or x < 0, and that the contour is followed in the clockwise direction in the latter case; or use the fact that the integral is an odd function of x. Next, conclude by means of Fourier theory that Z 3 d  D @2 h.x/ D  i e wjxj sgn.x/ cos wx: e ix 4 (19.5)  C1 R (Note that the integral is not absolutely convergent and should, in fact, be interpreted as the Fourier transform of a square-integrable function.) Conclude that u is given by (19.4). Background. The following is an example of the situation described in part (v). The function u from (19.4) belongs to C 3 .R/ but not to C 4 .R/ (that is, u is not a classical solution). Indeed (see Fig. 19.2 below), p u.0/ D 0; @u.0/ D 1 C 2; @2 u.0/ D 0; @3 u.0/ D 1; limx"0 @4 u.x/ D 2 ¤ 2 D limx#0 @4 u.x/:

Alternatively, the solution u in (19.4) may be obtained by requiring that the L2 solutions u˙ on R≷0 given by u˙ .x/ D ˙e

x

C a˙ e w.1Ci /x C b˙ e w.1

i /x

.x 2 R≷0 /;

have coinciding derivatives of order 3 or less at 0, and solving the resulting system of linear equations for the constants a˙ and b˙ . This amounts to a computation of the following integrals without complex analysis: Z   cos x d  D e wjxj sin wx; 4C1  2 R>0 Z   3 cos x d  D e wjxj sgn.x/ cos wx: 4 C1  2 R>0

Chapter 20

Appendix: Integration

As we have seen, in Theorems 3.15 and 3.18 above, continuous linear forms on C0 .X /, for X an open subset of Rn , arise naturally in the theory of distributions. The alternative name of Radon measure for such a form finds its origin in the existence of a bijective correspondence that associates a complex-valued measure on X to the linear form. More specifically, we have the following (Frigyes)1 Riesz Representation Theorem. Theorem 20.1. Suppose X  Rn is an open subset and u 2 D 0 .X /. Then u is of order zero if and only if there exist positive measures j on X such that every  2 C01 .X / is integrable with respect to all j , for 1  j  4, and Z Z Z Z u./ D .x/ 1 .dx/ .x/ 2 .dx/Ci .x/ 3 .dx/ i .x/ 4 .dx/: We do not assume the reader to be familiar with the notions of a positive measure and integrability with respect to it; the precise formulation of the theorem is included here for the benefit of those who are. In this appendix we will show that certain aspects of the theory of distributions are related to this result in a natural way. We also give a condensed but complete proof of the theorem, which will, however, be brought to a close only after Theorem 20.35 has been verified. It should be noted that the notation and nomenclature of measure theory, which are needed in this proof, are not completely standardized in the literature. Moreover, the usual treatments of measure theory start from functions acting on sets, rather than linear forms acting on continuous functions; see for example, Knapp [13] or Stroock [21]. This is another reason why we survey the theory of Lebesgue integration with respect to a measure from a perspective that corresponds as closely as possible to that of the theory of distributions. The next lemma is stated here in order to motivate the subsequent development. 1

Marcel Riesz, see Chap. 13, was his younger brother.

J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_20, © Springer Science+Business Media, LLC 2010

321

322

20 Appendix: Integration

Lemma 20.2. A continuous linear form u W C0 .X / ! C possesses the monotone convergence property, in the sense that for a sequence .k /k2N in C0 .X /, k # 0

H)

lim u.k / D 0:

k!1

(20.1)

Here k # 0 means that .k /k2N is a nonincreasing sequence of real-valued functions with pointwise limit 0 as k ! 1. Proof. We have supp k  K WD supp 1 , for every k 2 N, where K is a compact subset of X . On account of Dini’s Theorem (see [7, Theorem 1.8.19]) the convergence to 0 of the sequence .k / is uniform on K. Hence, the continuity of the linear form u on C0 .X / leads to limk!1 u.k / D 0; that is, u satisfies (20.1).  In measure theory, one usually begins by introducing measures of sets and subsequently defines an operation of integration with respect to these measures that acts on suitable functions, which are then said to be integrable. The introduction of the integral of step functions is often a first step in this construction. Integration is the counterpart in analysis of the arithmetic operation of computing a weighted average of a finite collection of numbers. Thus, integrating a function, which in general assumes countably or even uncountably many values, means averaging its values. In this process different weights can be assigned to different values: the weight assigned to a value can depend on properties such as location and size of the set of points where the value is attained. By and large, customary requirements in a theory of integration of functions are the following: (i) (ii) (iii) (iv)

Integrable functions are not necessarily compactly supported and bounded. On integrable functions integration acts as a linear form. The integral of a nonnegative integrable function is nonnegative. The class of integrable functions is closed with respect to taking specific kinds of limits. More precisely, pointwise limits of monotone sequences of integrable functions are integrable, too, and the same holds for arbitrary sequences of integrable functions that are all dominated by an integrable function. (v) The integral of such a limit is the limit of the integrals.

Conditions (ii) and (iii) relate to algebraic properties, while (i), (iv), and (v) refer to analytic ones. More specifically, (v) demands that integration be continuous on certain sequences of functions. Observe that (i) is a necessary consequence of (iv). The monotone convergence property as in Lemma 20.2 implies that u 2 D 0 .X / of order 0 acting on the space C0 .X / of functions satisfies conditions (ii) and (iv), mutatis mutandis. If one wants to consider u as an integral, one needs to show that u can be extended to a class of integrable functions, which properly contains C0 .X /, in such a way that (i)–(v) are valid. Preferably, the class of integrable functions should include the characteristic functions of a large class of subsets of X . Note that no characteristic function of a nonempty subset of X belongs to C0 .X / itself.

20 Appendix: Integration

323

The fact that we impose conditions (i) and (iv) disqualifies Riemann integration as the context for the extension of u. Indeed, Riemann integration primarily applies to bounded functions with compact support. Moreover, a pointwise limit of a sequence of Riemann integrable functions need not be Riemann integrable. For instance, the characteristic function 1Q of the set Q WD Q \ Œ 0; 1  of rational numbers in Œ 0; 1  is not Riemann integrable. On the other hand, if .qj /j 2N is an enumeration of the elements in Q, then 1Q is the pointwise limit of the nondecreas
P ing sequence 1j n 1fqj g n2N of functions, all of which have Riemann integral equal to 0. Accordingly, we will develop the more appropriate tool of Lebesgue integration with respect to a measure. Just as with the Riemann integral, the Lebesgue integral is defined by means of an approximation process. In the case of the Riemann integral, the process is to use upper sums and lower sums, which capture an approximate value of an integral by adding contributions influenced by proximity in the domain of the integrand. The process is qualitatively different for the Lebesgue integral, which captures an approximate value of an integral by adding contributions based on what happens in the image of the integrand. As a consequence, the Lebesgue integral is more suitable for handling functions of rapid variation: it leads to a theory of integration of wider applicability and to a simpler treatment of singular behavior of functions. The theories of Lebesgue and Riemann have similar permanence properties concerning algebraic operations, but in Lebesgue’s theory the properties concerning limits are superior. On the other hand, a complete treatment is technically more demanding for Lebesgue integration than for Riemann integration. In addition, exists a theory of conditionally convergent Riemann integrals, such as R sin there x dx D  (see Problem 14.44), which has no counterpart in Lebesgue inteR x gration. More explicitly, this integral is not absolutely convergent, which would be the case for a Lebesgue integrable function. Nevertheless, the distinction between the two theories rarely plays a role for the functions encountered in this book. More specifically, in practically all situations in which the interchange of a limit and integration is usually justified by an appeal to Lebesgue’s Dominated Convergence Theorem, Theorem 20.26.(iv) below, one may as well apply Arzel`a’s Dominated Convergence Theorem (see [7, Theorem 6.12.3] for a simple proof) for Riemann integration. This is because the Riemann integrability of the limit function is seldom an issue. However, the limit properties of Lebesgue integration established in the former theorem are indispensable for demonstrating the remarkable fact that completions of C0 .X / with respect to natural integral (semi)norms are again spaces of functions and do not consist of objects not encountered before. Furthermore, interchanging the order of integration in multiple integrals under minimal conditions on the integrands can better be handled in Lebesgue’s theory. In the remainder of this appendix we will develop Lebesgue’s theory by means of a method due to Daniell, in which functions, instead of measures, are the initial objects of investigation and (semi)norms play an important role. For these reasons, the method naturally suggests itself in the context of distribution theory. If one starts with a ring of subsets and a positive measure on it, and then applies Daniell’s exten-

324

20 Appendix: Integration

sion method to the corresponding step functions, the method equals the one used in the standard theory of Lebesgue integration, but it applies in more general settings. We consider the case of a linear form u W E ! R, where E will be specified in (20.8) below. The properties of E are abstracted from those of C0 .X /. Example 20.3. A most important example of a linear formR u on C0 .X / is provided by the Riemann integral 1X D test 1X , with 1X ./ D X .x/ dx. In this case, the Daniell extension produces the Lebesgue integral acting on the (large) space of functions integrable with respect to Lebesgue measure. ˛ For a characterization of Riemann integrable functions on Rn in the context of Lebesgue integration, see [13, Theorem 6.31]. We begin our treatment of integration theory by discussing some properties and definitions that are intimately connected with the ordering that R possesses but that is not present in C. Denote by E the real-linear subspace C0 .X; R/ of real-valued functions in C0 .X /. Since the real and imaginary parts of a function in C0 .X / belong to E, any complex-valued u as above is uniquely determined by its restriction to E. Accordingly, in the following it will be sufficient to study u W E ! C. Furthermore, we have u D v1 C i v2

with unique real-linear forms

v1 ; v2 W E ! R:

(20.2)

Note that E and C01 .X; R/ both are linear spaces over R. In contrast to C01 .X; R/, the space E possesses the Riesz property, in the sense that max .f; g/ D 1 .f C g ˙ jf min 2

gj/ 2 E

.f; g 2 E/:

(20.3)

The function max.f; g/ is the smallest element in E that is  f and g; likewise, min.f; g/ is the largest element in E that is  f and g. In particular, then, 0  f˙ WD max.˙f; 0/ 2 E;

jf j D fC C f 2 E: (20.4) Recall that a linear form u W E ! R is said to be positive if u.f /  0

f D fC

f ;

.f 2 E; f  0/:

(20.5)

According to Theorem 20.8 below, every real-valued distribution of order 0 can be written as the difference of two positive distributions. This result requires some preparation. Definition 20.4. The linear form u W E ! R is of bounded variation if juj./ WD supf ju. /j j

2 E; j j   g < 1

If this is the case, then the mapping

.0   2 E/:

(20.6)

20 Appendix: Integration

325

juj W E0 WD f 2 E j   0 g ! R0 ˛

is called the variation of u.

Note that a positive linear form u W E ! R is of bounded variation, and so is the difference of two positive linear forms. Therefore, the linear form u W E ! R being of bounded variation is a necessary condition for the validity of Theorem 20.8. Theorem 20.5. A linear form u W E ! R defines a distribution of order 0 if and only if u is of bounded variation. Proof. ). Consider 0   2 E and set K D supp . Applying (3.4) with k D 0, we find a constant c > 0 such that ju. /j  c k kC 0 , for all 2 C0 .K/. In particular, if 2 E with j j  , then supp  K; hence, ju. /j  c k kC 0  c kkC 0 . This implies juj./  c kkC 0 , which proves that u is of bounded variation. (. Suppose that u is not continuous. As in the proof of Theorem 3.8, we find a compact set K  X and a sequence .j / in E satisfying supp j  K

and kj kC 0
0. Next, supn2N gn is the pointwise supremum of the collection f n;k 2 E j n; k 2 N g. Lastly, observe that gn is the pointwise limit of the nondecreasing sequence .zn;k /k2N with zn;k D max n;j 2 E: 1j k

On account of the preceding argument, min.g1 ; g2 / D supk2N min.z1;k ; z2;k / belongs to E " . (ii). According to Lemma 20.9.(iii), the limit in (20.11) equals the supremum. The inequality u" .g/  sup u" .gn / n2N

is straightforward. For the reverse inequality, consider any r 2 R with r < u" .g/. There exist E 3   g with u./ > r and collections f n;k 2 E g as in (20.12). The sequence .zn / of functions   E 3 zn D min ; max m;k  gn 1m; kn

is pointwise nondecreasing to . From (20.1) it follows that limn!1 u.zn / > r. Since u" .gn /  u.zn /, we have supn2N u" .gn / > r and thus the desired inequality sup u" .gn /  u" .g/:

n2N

(iii). The nonnegative-homogeneity is obvious. Now let g and g 0 2 E " . There exist sequences .n / and .n0 / of functions in E that are pointwise nondecreasing to g and g 0 , respectively. Clearly .n C n0 / is pointwise nondecreasing to g C g0 . According to part (ii), u" .g C g 0 / D lim u.n C n0 / D lim u.n / C u.n0 / D u" .g/ C u" .g 0 /:  n!1

n!1

330

20 Appendix: Integration

Proposition 20.11. The function u W F ! R [ f 1 g as in (20.9) has the following properties: (i) u is monotone. (ii) The restriction of u to E " coincides with u" ; in particular, u agrees with u on E. (iii) u is nonnegative-homogeneous. (iv) u is countably subadditive; that is to say, for any sequence .fn /n2N of nonnegative functions in F X  X u fn  u.fn /: n2N

n2N

Proof. (i). Consider f and f 0 2 F with f  f 0 . If g 2 E " satisfies f 0  g, then f  g; hence u.f /  u" .g/. Taking the infimum over all such g, we obtain u.f /  u.f 0 /. (ii). If f 2 E " , then u" .f /  u" .g/ for all f  g 2 E " according to Lemma 20.9.(iii). Hence u" .f /  u.f /, whereas taking g D f we obtain that u.f /  u" .f /. (iii). If f 2 F , g 2 E " , and c > 0, then f  g if and only if c f  c g, while u" .c g/ P D c u" .g/ in view of Lemma 20.10.(iii). (iv). If P n2N u.fn / < r for some r 2 R, there exist functions fn  gn 2 E " with n2N u" .gn / < r. We now successively apply parts (i) and (ii) and Lemma 20.10.(ii) and (iii), to derive u

X

n2N

n n  X    X  X fn  u" gn D u" lim gk D lim u" gk

D lim

n2N n X

n!1

n!1

kD1

n!1

kD1

u" .gk / < r:

kD1

Since this holds for all admissible r, the desired inequality follows.



The countable subadditivity is the only one of these features of u that the Riemann upper integral does not share. All limit theorems of the integral constructed by means of the Daniell extension are consequences of this. Definition 20.12. On the set F we define the Daniell seminorm k
k D k
ku associated with the extension u of u to F by kf k D u.jf j/ 2 R [ f 1 g

.f 2 F /:

Denote the space of all functions f W X ! R by F . We then say that the sequence .fn /n2N in F converges with respect to the Daniell seminorm or in seminorm to f 2 F if limn!1 kf fn k D 0; compare with Chap. 8. The terminology is justified by the fact that k
k satisfies conditions (a)–(c) in Definition 8.5, as will be shown in the following theorem. ˛

20 Appendix: Integration

331

Theorem 20.13. The Daniell seminorm k
k W F ! R [ f 1 g has the following properties: (i) k
k is solid, that is, jf j  jgj implies kf k  kgk, for any f and g 2 F . Furthermore, k
k is absolute-homogeneous as well as countably subadditive; that is to say, for any sequence .fn /n2N of nonnegative functions in F ,

X X

fn  kfn k:

n2N

n2N

(ii) k
k is finite on all of E; in addition, it dominates j u j on E; that is, ju./j  kk, for all  2 E. (iii) For every sequence .n /n2N of nonnegative functions in E, m

X

H) lim kn k D 0: sup n < 1 m2N

n!1

nD1

Proof. (i). This is immediate from Proposition 20.11.(i), (iii), and (iv). (ii). Consequence of Proposition 20.11.(ii) and (i). P Pm (iii). If k m nD1 n k D nD1 u.n / admits a bound independent of m, then obviously limn!1 kn k D limn!1 u.n / D 0.  Definition 20.14. A function f 2 F is said to be negligible if kf k D 0. A subset A of X is negligible if its characteristic function 1A as defined in Definition 2.17 is negligible. A property of the points of X is said to hold almost everywhere, or a.e. for short, if the subset of points in X where it fails to hold is negligible. ˛ For f and g 2 F , we define the following subsets of X : .f D g/ D f x 2 X j f .x/ D g.x/ g; .f ¤ g/ D f x 2 X j f .x/ ¤ g.x/ g:

(20.13)

Proposition 20.15. Negligibility has the following permanence properties. (i) The sum of countably many nonnegative negligible functions is negligible. The union of countably many negligible sets is negligible. Any subset of a negligible set is negligible. The empty set ; is negligible. (ii) A function f 2 F is negligible if and only if it vanishes almost everywhere, that is to say, if and only if the set .f ¤ 0/ is negligible. (iii) If f and f 0 2 F agree almost everywhere, then kf k D kf 0 k. (iv) A function with finite seminorm assumes finite values almost everywhere. Proof. (i). If fn 2 F are nonnegative negligible functions, Theorem 20.13.(i) leads to

X X X

fn  kfn k D 0 D 0:

n2N

n2N

n2N

332

20 Appendix: Integration

The second claim is a particular instance of this, since characteristic functions of sets are nonnegative. The third claim is immediate from the solidity of k
k. Finally, 1; D 0 and u.0/ D 0 by Theorem 20.13.(i). P (ii). For the implication ) note that 1.f ¤0/  n2N jf j. Hence, if kf k D 0, then k1.f ¤0/ k 

Conversely, jf j 

P

n2N 1.f ¤0/ ;

kf k 

X

n2N

kf k D

X

n2N

0 D 0:

accordingly k1.f ¤0/ k D 0 implies

X

n2N

k1.f ¤0/ k D

X

n2N

0 D 0:

(iii). Write A and B for the sets .f D f 0 / and .f ¤ f 0 /, respectively. Theorem 20.13.(i) and part (ii) then imply kf k D k1A jf j C 1B jf j k  k1A f k C k1B f k D k1A f k C 0  kf k; so

kf k D k1A f k D k1A f 0 k D kf 0 k:

(iv). We have n 1.jf jD1/  jf j and so n k1.jf jD1/ k  kf k, for all n 2 N. If .jf j D 1/ is not negligible, kf k must be infinite.  The only functions of interest for the purpose at hand are, of course, those with finite seminorm. We should like to argue that the sum of any two of them also has finite seminorm, in view of the subadditivity of k
k. A technical difficulty appears: even if f and g have finite seminorm, there may be points x 2 X where f .x/ D 1 and g.x/ D 1 or vice versa; f .x/ C g.x/ is not defined for such x. A similar problem arises with 0 f .x/. The solution is to note that according to Proposition 20.15.(iv) such ambiguities can occur in a negligible set of points x at the most. We simply extend k
k to functions that are defined only almost everywhere, as follows. Definition 20.16. A function f 2 F is said to be defined almost everywhere on X if the complement in X of the domain of f is negligible. For such f , set kf k D kf 0 k, where f 0 is any function defined everywhere on X and coinciding with f almost everywhere at the points where f is defined. ˛ Proposition 20.15.(iii) entails that this definition is sound: it does not matter which function f 0 we choose to agree almost everywhere with f ; any two will differ negligibly and thus have the same seminorm. Given two functions f and g with finite seminorm that are defined almost everywhere, we declare their sum f Cg to equal f .x/ C g.x/ at points x where both f .x/ and g.x/ are finite. This function is defined almost everywhere, because the set of points where f and/or g are either infinite or not defined is negligible on account of Proposition 20.15.(iv). It is clear how to define the maximum, minimum, scalar multiples, etc. of functions that are defined almost everywhere, as well as how to read f  g almost everywhere etc. when f and g are defined almost everywhere.

20 Appendix: Integration

333

Definition 20.17. A function f 2 F defined almost everywhere is said to have finite seminorm, or to be finite in seminorm, if kf k < 1. Let F k k denote the collection of functions defined almost everywhere with finite seminorm. ˛ We now derive a crucial technical result. Theorem 20.18. (i) F k k is closed under taking finite linear combinations, and finite maxima and minima. In addition, k
k is a solid and countably subadditive seminorm on F k k . (ii) Every Cauchy sequence in F k k with respect to the seminorm has a subsequence that converges pointwise almost everywhere to a limit in F k k with respect to the seminorm. (iii) The space .F k k ; k
k/ is complete, that is, every Cauchy sequence in it is convergent to an element contained in the space. Proof. (i). For finite sums, apply jf C gj  jf j C jgj and Theorem 20.13.(i) to obtain kf C gk  kf k C kgk. Next, for the maxima and minima use (20.3) plus the fact that jf gj  jf j C jgj. Solidity is obvious and countable subadditivity is proved by applying similar arguments as for finite sums. (ii). Let .fn / be a Cauchy sequence in seminorm in F k k ; that is to say, kfn fm k ! 0 as n and m ! 1. For every n 2 N, there exists a function fn0 that is defined and finite everywhere and agrees with fn almost everywhere. Let Nn denote the negligible set of points where fn is not defined or does not agree with fn0 . There is a nondecreasing sequence .nk /k2N of indices such that kfn0 Clearly

kfn0kC1

fn0k k
. But the sequence .fnkC1 fnk /k consists of nonnegative functions in E satisfying m

X

.fnkC1 sup

m2N

kD1

fnk / D sup kfnm m2N

fn1 k  kfn1 k C sup kfn k < 1: n

By Lemma 20.22.(ii), though, .fnkC1 fnk /k must converge to 0 in seminorm. Now that we know that .fn / is a Cauchy sequence, we may employ Theorem 20.21.(ii). Thus we find a limit in seminorm f 0 2 E and a subsequence .fnk /k such that fnk .x/ ! f 0 .x/ for all x outside some negligible set N . For every x 2 X , however, fn .x/ ! f .x/. Therefore f .x/ D lim fn .x/ D lim fnk .x/ D f 0 .x/ n!1

k!1

.x … N /:

Hence f is equal almost everywhere to the limit in seminorm f 0 . Consequently, it is a limit in seminorm itself, on account of Theorem 20.21.(iii). If .fn / is nonincreasing rather than nondecreasing, then . fn / is nondecreasing pointwise, and therefore, by the above, nondecreasing in seminorm, to f ; again limn!1 fn D f in seminorm.  Next comes the Lebesgue Dominated Convergence Theorem. It is a central result in integration theory: many other results can be derived from it. Theorem 20.24. Let .fn /n2N be a sequence of functions in E and assume: (i) limn!1 fn DW f pointwise almost everywhere; (ii) there exists a function g with finite seminorm such that jfn j  g, for all n 2 N.

Then .fn /n2N converges to f in seminorm, and consequently f 2 E.

Proof. As in the proof of the preceding theorem, we begin by showing that .fn / has the Cauchy property. To this end we consider the nonnegative function gk D sup jfn kn; m

fm j D lim

max jfn

l!1 kn; ml

fm j  2g

.k 2 N/:

By Proposition 20.20 and Theorem 20.23, we get gk 2 E. Moreover, the sequence .gk .x//k2N converges nonincreasingly to 0 at all points x where .fn .x// converges, that is, almost everywhere. A simple modification of the preceding theorem then leads to limk!1 kgk k D 0. Now kfn fm k  kgk k, for m and n  k, so that .fn / is indeed a Cauchy sequence in seminorm. By Theorem 20.21.(ii), the sequence possesses a limit in seminorm f 0 and a subsequence .fnk /k that converges pointwise almost everywhere to f 0 . Since .fnk /k also converges to f almost everywhere, the functions f and f 0 agree almost everywhere, namely, at all points x where both .fn .x// and .fnk .x//k converge. Thus limn!1 kfn f k D limn!1 kfn f 0 k D 0. 

20 Appendix: Integration

337

Definition 20.25. We are now in a position to define the extension u W E ! R of the linear form u W E ! R. Namely, for any f 2 E, select an arbitrary sequence .n /n2N of functions in E converging in seminorm to f and set u.f / D limn!1 u.n /. ˛ The extension is well-defined. First, on the strength of Theorem 20.13.(ii) and (i), ju.n /

u.m /j D ju.n

m /j  kn

m k  kn

f k C kf

m k ! 0;

as n and m ! 1. Hence, .u.n // is a Cauchy sequence in R and does have a limit. Next, if .n0 / is another sequence of functions in E converging in seminorm to f , then ju.n /

u.n0 /j D ju.n

n0 /j  kn

n0 k  kn

f k C kf

n0 k ! 0;

as n ! 1. Phrased differently, .u.n // and .u.n0 // have the same limit. Next, we present the fundamental properties of the extension. Part (iv) below is also referred to as the Lebesgue Dominated Convergence Theorem. Theorem 20.26. (i) The extension u W E ! R is linear and monotone; that is to say, for any f and g 2 E and r and s 2 R, u.r f Cs g/ D r u.f /Cs u.g/;

and

f g

H)

u.f /  u.g/:

(ii) The extension is still majorized by the seminorm. More specifically, ju.f /j  u.jf j/ D kf k

.f 2 E/:

In particular, u assigns equal values to functions in E that are equal almost everywhere. (iii) If the sequence .fn /n2N in E converges in seminorm to f 2 F , then f 2 E and limn!1 u.fn / D u.f /. (iv) The conclusion in part (iii) holds if the sequence .fn /n2N in E converges almost everywhere to f and jfn j  g, for all n 2 N and for some function g 2 E. In particular, the choice g D supn2N jfn j is admissible if k sup jfn j k < 1: n2N

Proof. (i). Let .n / and . n / be sequences in E converging in seminorm to f and g 2 E, respectively. Then .r n C s n / converges to r f C s g in seminorm, because kr f C s g Thus

.r n C s

n /k

 jrj kf

n k C jsj kg

nk

!0

as n ! 1:

338

20 Appendix: Integration

u.r f Cs g/ D lim u.r n Cs n!1

n/

D lim .r u.n /Cs u. n!1

n //

D r u.f /Cs u.g/:

This shows that the extension is linear. On account of the linearity of the extension its monotonicity follows if f  0 implies u.f /  0, for all f 2 E. By Lemma 20.22.(i), there exist nonnegative functions n 2 E converging in seminorm to f . Then u.n /  0 implies u.f / D limn!1 u.n /  0. (ii). Suppose .n / in E satisfies limn!1 kf n k D 0. Then the fact that u.f / D limn!1 u.n / leads to ju.f /j D limn!1 ju.n /j. In addition, since j jf j jn j j  jf n j, it follows from Theorem 20.13.(i) that limn!1 k jf j jn j k D 0; hence u.jf j/ D lim u.jn j/ D lim kn k D kf k: n!1

n!1

The monotonicity of u on E leads to ju.n /j  u.jn j/. Combination of the latter two results yields the desired ju.f /j D lim ju.n /j  lim u.jn j/ D u.jf j/ D kf k: n!1

n!1

(iii). Consequence of part (ii). (iv). Evident from part (iii) and Lebesgue’s Dominated Convergence Theorem, Theorem 20.24.  Theorem 20.26 completes the construction of the Daniell closure as well as the description of its properties. Example 20.27. Consider the family of functions .Ar /0r1 from Example 1.4. We recall the argument at the end of the example that .Ar /0r1 does not admit an integrable majorant on Œ ;  ; see Fig. 1.4. ˛ We now introduce some concepts from measure theory. Definition 20.28. Let X be a set. A ring of subsets of X is a collection A of subsets of X such that ; 2 A and A [ B 2 A;

A \ B 2 A;

AnB 2A

.A; B 2 A/:

Note the redundancy in this definition; for instance, ; D A n A and A \ B D A n .A n B/. A positive measure on A, or on X , is a set function  W A ! R that is positive in the sense that .A/  0 .A 2 A/; (finitely) additive in the sense that A; B 2 A and A \ B D ;

H)

.A [ B/ D .A/ C .B/;

and has the monotone convergence property in the sense that

20 Appendix: Integration

339

.An /n2N nonincreasing in A with

1 \

nD1

An D ;

H)

lim .An / D 0:

n!1

(20.14) Given the ring A of subsets of X , a function f W X ! R is called A-elementary if f .X /  R is a finite subset and f 1 .fcg/ 2 A, for every c 2 R n f0g. Denote the set of all A-elementary functions by EA . For any f 2 EA , the integral of f with respect to the positive measure  on A is defined as Z X I .f / WD f .x/ .dx/ WD c .f 1 .fcg/: (20.15) c2f .X/nf0g

˛

For arbitrary A  X , recall the characteristic function 1A of A as defined in Definition 2.17. Definition 20.29. Given E and u W E ! R as in (20.8) with Daniell extension u W E ! R, introduce the collection A of subsets of X and the function  on A by A D Au D f A  X j 1A 2 E g;  D u W A ! R by .A/ D u.1A /

˛

.A 2 A/:

Lemma 20.30. In the notation as above, we have the following properties: (i) A is a ring of subsets of X . (ii)  is a positive measure on A. (iii) EA and I possess the properties of E and u listed in (20.8). In particular, Daniell’s extension method is applicable to I W EA ! R. Furthermore, EA  E and I equals the restriction of u to EA . Proof. (i). 1; D 0. For A and B 2 A, the following functions all belong to E in view of Proposition 20.20: 1A[B D max.1A ; 1B /;

1A\B D min.1A ; 1B /;

1AnB D 1A

1A\B :

(ii). We have .A/ D u.1A /  u.0/ D 0 because 1A  0. In addition, .A [ B/ D u.1A[B / D u.1A C 1B / D u.1A / C u.1B / D .A/ C .B/ if A and B 2 A and A \ B D ;. Finally, let .An / be a nondecreasing sequence in A such that ..An // is bounded from above, and let A be the union of all An . Then E 3 1An " 1A and .u.1An // is bounded from above. Therefore the Monotone Convergence Theorem, Theorem 20.23, implies that 1A 2 E and u.1A / D limn!1 u.1An /; hence, A 2 A and .A/ D limn!1 .An /. An analogous statement can be made for nonincreasing sequences in A, which implies (20.14).

340

20 Appendix: Integration

(iii). Provided with the pointwise addition and scalar multiplication of functions, the space F of all functions f W X ! R is a linear space over R, and EA is a linear subspace of F . Also, in F we have the partial ordering defined by f  g. The ring property of A implies that EA possesses the Riesz property (20.3). P For f 2 EA , we have f D c2f .X/nf0g c 1f 1 .fcg/ 2 E, because the summation is finite. On account of (20.15) and Definition 20.29 this leads to   X X I .f / D c .f 1 .fcg// D u c 1f 1 .fcg/ D u.f /: c2f .X/nf0g

c2f .X/nf0g

Consequently I is linear and positive. Any sequence of functions E  EA 3 fn # 0 is majorized by f1 2 E; therefore application of Theorem 20.26.(iv) entails lim I .fn / D lim u.fn / D u.0/ D 0:

n!1

n!1



This means that I has the monotone convergence property as in (20.1).

Example 20.31. Application of the preceding lemma to the positive linear form from Definition 20.3 gives rise to the Lebesgue measure on the ring of Lebesgue measurable sets in X  Rn . ˛ I

Definition 20.32. Let the notation be as above. Functions belonging to EA , the Daniell closure of EA with respect to the measure , are said to be integrable with respect to  or -integrable. ˛ Observe that EA is not contained in E. On the other hand, EA  E and I equals u on EA . When applying the extension method with u W E ! R replaced by I I W EA ! R, it is natural to ask whether E  EA and whether u equals the I restriction to E of the extension of I to EA . This is not always the case. Indeed, for arbitrary f 2 E, the value u.f / cannot always be recovered as the integral of f with respect to the measure . The following example, admittedly somewhat artificial, shows this convincingly. In other words, the theory of the Daniell closure is a true generalization of the theory of Lebesgue integration with respect to a measure. Example 20.33. Let X be a subset of R>0 having at least two elements. Let E be the set of all restrictions to X of linear functions on R. That is, f 2 E if and only if there exists c 2 R such that f .x/ D c x for every x 2 X . Define u.f / D c if f .x/ D c x for every x 2 X . Phrased differently, if we choose x0 2 X , then u.f / D f .x0 /=x0 for every f 2 E. It follows that E is a vector space of functions on X having the Riesz property (20.3) and that u is a linear form on E satisfying (20.5) and (20.1). And yet one has E " D E, E D E, Au D f ; g, EAu D f0g, u D 0, and Iu D 0. ˛ Under an additional assumption, however, one obtains a positive result.

20 Appendix: Integration

341

Theorem 20.34. Suppose that E is a linear space of real-valued functions on a set X having the Riesz property (20.3) and that u W E ! R is a linear form on E satisfying (20.5) and (20.1). Furthermore, assume that E has the Stone property, that is, min.; 1X / 2 E for every  2 E. Then Z I E D EA and u.f / D f .x/ .dx/ .f 2 E/: R I Proof. First we prove that  2 EA and u./ D .x/ .dx/, for any  2 E. We will do this by approximating  in a monotone way by means of functions in EA . In view of (20.4), with f replaced by , we may assume that   0. For any c > 0, we have min.; c 1X / D c min. 1c ; 1X / 2 E. Accordingly we also obtain c; c 0

WD

1 c

c0

.min.; c 1X /

min.; c 0 1X // 2 E

.0 < c 0 < c/:

Now c; c 0 .x/ takes the following values: 0 if .x/  c 0 ; ..x/ c 0 /=.c c 0 / 2 Œ 0; 1  if c 0  .x/  c; and 1 if c  .x/. Let .cn / be a sequence in R>0 such that 0 < cn < c for every n and cn " c as n ! 1. Then n WD min1mn c; cm 2 E and also n # 1.c/ , in the notation (20.13). The Monotone Convergence Theorem, Theorem 20.23, for E implies 1.c/ 2 E; this in turn leads to .  c/ 2 A

and

1.c/ 2 EA :

Let F  R>0 be nonempty and finite. Set cF0 D minf c 0 2 F j c 0 > c g for every c 2 F , where cF0 D 1 if c D max F . In view of the fact that .cF0 >   c/ D .  c/ n .  cF0 / 2 A, we obtain fF WD

X

c2F

c 1.c0

F

>c/

2 EA  E:

In view of   fF and .cF0 >   c/ D fF 1 .fcg/ and applying Lemma 20.30.(iii) Z we deduce u./  u.fF / D fF .x/ .dx/:

If .Fn / is a nondecreasing sequence of finite subsets of R>0 such that the union of all Fn is dense in R>0 , then fFn "  as n ! 1. Since the u.fFn / are bounded from above by u./, the Monotone Convergence Theorem for u W E ! R leads to Z fFn .x/ .dx/: u./ D lim u.fFn / D lim n!1

n!1

On the other hand, the Monotone Convergence Theorem for I implies that  2 I EA and

342

lim

n!1

Z

fFn .x/ .dx/ D

Z

20 Appendix: Integration

.x/ .dx/:

Combination of the two identities gives the desired equality u./ D I

R

.x/ .dx/.

Thus we have established that E  EA and ujE D I jE . Since the Daniell closure of I equals I , the Daniell closure of ujE D I jE is contained in the I Daniell closure of I , that is, E  EA and ujE D I jE . On the other hand, since EA  E and ujE is Daniell closed, we also have EA implies E D EA

I

and u D I on E.

I

u

 EA  E, which 

As a direct consequence we derive the following important result. Recall that on account of Theorem 3.18, a positive distribution is a distribution of order 0. Theorem 20.35. Let X  Rn be an open subset. Then u is a positive distribution on X if and only if there existsRa positive measure  on X such that  is integrable with respect to  and u./ D .x/ .dx/, for every  2 C01 .X /.

Proof. (. Lemma 20.30.(iii) implies that u D I is a linear form on C01 .X /. If K is an arbitrary compact subset of X we have, for all  2 C01 .K/, ˇZ ˇ Z ˇ ˇ ju./j D ˇ .x/ .dx/ˇ  j.x/j .dx/  sup j.x/j .K/ D .K/kkC 0 : x2X

In view of Theorem 3.8 this implies that u 2 D 0 .X /. The positivity of u is a consequence of the positivity of . ). From Theorem 3.18 and Lemma 20.2 it follows that u possesses the property in (20.1). We conclude that the space E D C0 .X; R/ and the linear form u W E ! R satisfy (20.3), the Stone property, (20.5), and (20.1). Since E contains C01 .X; R/, the desired conclusion now follows from Theorem 20.34.  We have now completed the preparations required for the proof of Riesz’s Representation Theorem, Theorem 20.1. Proof. (. This implication follows as in the proof of Theorem 20.35. ). Apply (20.2) and Theorems 20.8 and 20.35.



The condition imposed on the measure , namely that every  2 C01 .X / be integrable with respect to , is not very direct in terms of the measure  as a set function itself. This is remedied by the following proposition. Proposition 20.36. Let  be as in Theorem 20.35. Then the following assertions are equivalent: (i) Every  2 C01 .X / is -integrable. (ii) 1U is -integrable for every relatively compact (i.e., having compact closure) open subset U of X . (iii) 1K is -integrable for every compact subset K of X .

20 Appendix: Integration

343

(iv) Every  2 C0 .X / is -integrable. I

Proof. Write E D EA , E D EA and u D I . Then f 2 E if and only if f is -integrable. (i) ) (ii). Use Lemma 8.2 and Corollary 2.16 to choose  2 C01 .X / and an increasing sequence .n / in C01 .X / such that n " 1U  . Because u.n /  u./ for every n, it follows from the Monotone Convergence Theorem that 1U 2 E. (ii) ) (iii). For K  X compact, Corollary 2.4 implies the existence of a relatively compact open U  X with K  U . Then K D U n .U n K/ exhibits K as a difference of two relatively compact open subsets; hence K belongs to the ring A and so 1K 2 E. (ii) ( (iii). Application of the preceding argument with the roles of K and U interchanged leads to the conclusion 1U 2 E. (ii) ) (iv). In view of (20.4) it is sufficient to consider  2 C0 .X / satisfying   0. Then . > c/ is a relatively compact open subset of X for every c > 0; hence 1.>c/ 2 E. Since .  c/ is compact, (iii) implies that 1.c/ 2 E. The ring property of A leads to 1.a 0. There exists  2 E such that kf k < . For ı > 0 sufficiently small, we then obtain from (20.17) and (20.16), plus the subadditivity of j
j, 0  jj

j kf k

so

j kk

kı kC 0  1;

ju.ı f /j j  j kf k kk j C j kk ju.ı /j j Cj ju.ı /j ju.ı f /j j < 2 C kı kC 0 kf

k < 3: 

Corollary 20.39. Suppose the function f is locally integrable with respect to the Lebesgue measure on an open set X in Rn and f D 0 in D 0 .X /, that is, test f D 0. Then f D 0 almost everywhere on X . It is by now evident that two functions in E defined almost everywhere that differ only negligibly are the same for all practical purposes. From Definition 20.14 and Theorem 20.18.(i) it is clear that “equality” of two such functions is an equivalence relation. Hence, we may identify them by the formation of equivalence classes. For any function f defined almost everywhere with f 2 E almost everywhere, we denote by Œf  the equivalence class of all functions in E that agree with f almost everywhere, that is, Œf  D f f 0 2 E j f 0 D f almost everywhere g:

(20.18)

Proposition 20.15.(iv) implies that every class contains a function that is defined everywhere and assumes only finite values. The sum and scalar product of classes are defined in the obvious way: Œf CŒg is the class of f C g, and r Œf  is the class of r f . These classes do not depend on the choice of the representatives f 2 Œf  and g 2 Œg. For instance, if f D f 0 almost everywhere and g D g0 almost everywhere, then the class of f C g is clearly the same as the class of f 0 C g 0 . In addition, the Daniell seminorm of functions in E now leads to the norm of a class by means of k Œf  k D kf k; where f 2 Œf ; again, this number will not depend on the choice of f 2 Œf  according to Definition 20.16. Let us denote the collection of equivalence classes of functions in E defined almost everywhere by L1 .X /: Observe that L1 .X / is the quotient of E by its linear subspace of negligible functions, while k
k is the quotient norm. By application of Theorem 20.21.(ii) one immediately establishes the following result.

20 Appendix: Integration

345

Theorem 20.40. .L1 .X /; k
k/, the space of Lebesgue integrable functions on X , is a Banach space, i.e., a normed linear space that is complete. In addition, the linear form u is continuous with respect to the topology of L1 .X / defined by the norm. There is an additional structure on E, the order. The question arises whether it is inherited by the space L1 .X /. It is. Let us say that the class Œf  is smaller than the class Œg if a representative of Œf  is smaller than a representative of Œg almost everywhere. In other words, Œf   Œg if and only if f  g almost everywhere, for any, and then clearly all, representatives f 2 Œf  and g 2 Œg.

Finally, we define the spaces Lp .X /, for 1 < p < 1 and any open subset X  Rn . We denote by k
k D k
k1 the Daniell seminorm corresponding to the Lebesgue measure on X and introduce k
kp by Z 1=p kkp WD kkLp .X / WD k jjp k1=p D j.x/jp dx ; (20.19) X

for  2 C0 .X /. Minkowski’s inequality, see Remark 20.42 below or [7, Exercise 6.73.(iii)], k C kp  kkp C k kp .; 2 C0 .X //;

then expresses the subadditivity of k
kp on C0 .X /.

Theorem 20.41. Let the notation be as in (20.19). Then all assertions of Theorem 20.13 remain valid if the seminorm k
k acting on E D C0 .X; R/ is replaced by k
kp , but for the fact that k
kp dominates juj as in part (ii) of the theorem. Proof. (i). The solidity of k
kp follows from the fact that r 7! r p is strictly increasing on R0 and the solidity of k
k. The absolute homogeneity of k
kp is a consequence of the same property of k
k. If .k / is a nonincreasing sequence of functions with pointwise limit 0, then .jk jp / has the same property. Since k
k is monotone decreasing, we now obtain that this is also true of k
kp . The countable subadditivity follows easily from this and the subadditivity. Indeed, for any sequence .k / of nonnegative functions in E we obtain l l

X

X X X



k D lim k  lim kk kp D kk kp :

p

k2N

l!1

kD1

p

l!1

kD1

k2N

(ii). The finiteness of k
kp on E is implied by the corresponding property of k
k. (iii). Let .k / be a sequence of nonnegative functions in E and set

l

p / l l

D

l X

kD1

k 2 EI

then, by assumption

sup k l2N

l kp

< 1:

Now . is a nondecreasing sequence in E, while supl k lp k < 1. The Monotone Convergence Theorem, Theorem 20.23, then implies that . lp / is a Cauchy

346

20 Appendix: Integration

p sequence with respect to k
k. In particular, liml!1 k lp k D 0. Furtherl 1 p p p p more l  l , by virtue of the fact that 1 C t  .1 C t/p , for t  0. l 1 (The latter assertion follows from the observation that we have equality for t D 0, while the derivative with respect to t of the left-hand side is dominated by that of the right-hand side.) Thus we obtain liml!1 klp k D liml!1 kl kpp D 0 and this gives the desired conclusion. 

The arguments and definitions following Theorem 20.13 that underlie Theorem 20.21 can now be copied, and they lead to the linear space E consisting of the p-integrable functions, which is complete with respect to the seminorm k
kp . Analogous to Theorem 20.40 for L1 .X /, one obtains the Banach space Lp .X /. Remark 20.42. In order to give a direct proof of Minkowski’s inequality, for f and g belonging to Lp .X /, write a D kf kp and b D kgkp . To avoid trivialities, we may suppose that a > 0 and b > 0. By scaling, it suffices to consider the case that R R z p aCb D 1. Let fz D jfa j and gz D jgj z.x/p dx D 1. Now b . Then X f .x/ dx D X g jf C gj  jf j C jgj D afz C b gz. The second derivative of the function r 7! r p is nonnegative on R0 , which implies that this function is convex on R0 . Therefore R we have jf C gjp  afzp C b gzp , and so X jf C gjp .x/ dx  a C b D 1. This leads to kf C gkp  1 D kf kp C kgkp . ˛ Remark 20.43. We list some more properties of the spaces Lp .X /. C0 .X / is dense in Lp .X / with respect to k
kp . H¨older’s inequality from Problem 11.21 immediately implies that any f 2 Lp .X / is locally integrable. The inclusion mapping Lp .X / ! D 0 .X / given by u 7! test u is injective, as a consequence of Remark 3.7. On account of H¨older’s inequality we also obtain, for arbitrary u 2 Lp .X / and  2 C01 .X /, ˇZ ˇ ˇ ˇ ju./j D ˇ u.x/.x/ dx ˇ  kukLp kkLq : X

Here q is the real number such that

1 p

C

continuous injection

1 q

D 1. This entails that we have a Lp .X / ! D 0 .X /:

(20.20)

As an application, consider a sequence .uj /j 2N in C01 .X / that has the Cauchy property with respect to the Lp norm and satisfies limj !1 uj D 0 in D 0 .X /. Then we claim that limj !1 kuj kLp D 0. Indeed, in view of the completeness of Lp .X /, there exists u 2 Lp .X /  D 0 .X / such that limj !1 uj D u in Lp .X /. Then the continuity in (20.20) gives 0 D limj !1 uj D u in D 0 .X / and therefore u D 0 in Lp .X / by the injectivity in (20.20). This in turn leads to the desired conclusion. ˛ Finally we give a useful result on the continuity of pullback under translation acting in Lp .Rn /. Lemma 20.44. Let 1  p < 1 and f 2 Lp .Rn /. Then

20 Appendix: Integration

347

lim k.Th 

h!0

I /f kp D 0:

Proof. Let  > 0 be arbitrary. Applying Theorem 20.41, choose  2 C0 .Rn / such that kf kp < 3 . Then it is clear from the invariance under translation of the Lebesgue measure that kTh  f Th  kp D kf kp < 3 , for all h 2 Rn . Furthermore, there exists R > 0 such that .x/ D 0, for kxk  R; write B D f x 2 Rn j kxk  R C 1 g. Since  is uniformly continuous on Rn , we may select 0 < ı < 1 such that j.x C h/ Hence Z j.x C h/ Rn

p

.x/j dx D

which is to say that kTh   kTh  f

.x/j < Z

B

 p1  1 3 voln .B/

j.x C h/

.khk < ı/:

.x/jp dx
0. We introduce some notation. For   0 and x 2 R, set .x/ .0/ and Ia D  a; a Œ .a 2 R/: x C i Note that ˚ is continuous on R (see [7, Proposition 2.2.1.(ii)]), for all   0. Now we get, writing .x/ D .0/ C ..x/ .0//, Z Z Z Z .x/ .x/ x i dx D dx D .0/ dx C ˚ .x/ dx: 2 2 R x C i Im x C i  Im x C  Im ˚ .x/ D

Since x 7! x=.x 2 C  2 / is an odd function on R and Im is symmetric about the origin, we obtain, by means of the change of variables x D  y, Z Z Z x i  1 dy dx D i dx D i 2 2 2 2 2 Im x C  Im x C  Im y C 1 

D

m 2i arctan ! 

Z

˚ .x/ dx D

Furthermore, we have lim

#0 Im

Z

i

as  # 0:

˚0 .x/ dx:

(21.1)

Im

Postponing for the moment the proof of this identity, we apply it to obtain lim #0

Z

Im

˚ .x/ dx D lim ı#0

Z

ı m

˚0 .x/ dx C

Z

m ı

  1 ./: ˚0 .x/ dx D PV x

With the notation ı./ D .0/, combination of these identities leads to the following Plemelj–Sokhotsky jump relations (see Example 14.30 or Problem 12.14.(ix) for other proofs): J.J. Duistermaat and J.A.C. Kolk, Distributions: Theory and Applications, Cornerstones, DOI 10.1007/978-0-8176-4675-2_21, © Springer Science+Business Media, LLC 2010

349

350

21 Solutions to Selected Problems

1 1 ˙  i ı D PV ; x˙i0 x

ıD

1  1 2 i x i 0

1  : xCi0

Formula (21.1) follows from the Dominated Convergence Theorem of Arzel`a, see [7, Theorem 6.12.3], or that of Lebesgue, see Theorem 20.26.(iv). Indeed, we have ˇ ˇ j˚ .x/j D ˇ

ˇ x jxj ˇ ˚0 .x/ˇ D p j˚0 .x/j  j˚0 .x/j: 2 x C i x C 2

A more elementary proof of (21.1) is as follows. The integrand on the left-hand side is a continuous function that admits a majorant that is independent of . Accordingly, for arbitrary  > 0 there exists m > ı > 0 such that for every   0, ˇZ ˇ  ˇ ˇ ˚ .x/ dx ˇ < : ˇ 3 Iı

Next, observe that for all  > 0 and x 2 Iı0 WD Œ m; m  n Iı , ˇ ˇ ˚0 .x/j D ˇ

j˚ .x/

1 x C i

1 ˇˇ ˇ j.x/ x

.0/j D 

j˚0 .x/j j˚0 .x/j  : jx C i j ı

In other words, the convergence of ˚ to ˚0 is uniform on Iı0 as  # 0. This implies the existence of 0 > 0 such that for all 0 <   0 , ˇZ
ˇˇ  ˇ ˚ .x/ ˚0 .x/ dx ˇ < : ˇ 3 Iı0 As a consequence we obtain, for all 0 <   0 , Z ˇ ˇZ ˇ ˇ ˚ .x/ dx ˚0 .x/ dx ˇ ˇ Im

Im

ˇ ˇZ ˇ ˇZ ˇZ ˇ ˇ ˇ ˇ ˇ ˚0 .x/ dx ˇ C ˇ  ˇ ˚ .x/ dx ˇCˇ Iı

Iı

Iı0

˚ .x/


ˇˇ  ˚0 .x/ dx ˇ < 3 D : 3

1 Finally note that for x D  D n1 with n 2 N, we have jxCi D pn ; this implies j 2 that the convergence of ˚ to ˚0 is not uniform on all of Im . A second proof can be based on integration by parts. For x 2 R and  > 0, we d x d   have dx log jx C i j D x 2 C , and therefore 2 and dx arctan x D x 2 C 2

x i 1 d log.x C i / D 2 : D 2 dx x C x C i Hence integration by parts leads to

21 Solutions to Selected Problems

Z

R

351

Z

d log.x C i / dx .x/ dx RZ  0 .x/ log.x C i / dx C Œ.x/ log.x C i /11 D R Z Z 0 D  .x/ log jx C i j dx i  0 .x/ arg.x C i / dx:

.x/ dx D x C i

R

R

Now lim#0 arg.x C i / equals 0 for x > 0, and  for x < 0. Furthermore, we have j arg.x C i /j  2. By dominated convergence we obtain lim #0

Z

R

0

 .x/ arg.x C i / dx D 

Z

0 1

 0 .x/ dx D  .0/ D  ı./:

Assuming  < 1 we see that

j log jx C i jj D

€

1 log.x 2 C 1/; 2 1 1  log D j log jxjj; log jx C i j jxj

log jx C i j 

jx C i j  1I jx C i j < 1:

So another application of dominated convergence leads to Z Z .x/ lim dx D  0 .x/ log jxj dx  i ı./: #0 R x C i  R x 6 . Then 1.4 Consider p.x/ z D .x 1/3 .x C 1/3 D 1 3x 2 C 3x 4 R1 35 2 3 z dx D 32=35 and therefore p.x/ D 32 .1 x / . That  is twice con1 p.x/ tinuously differentiable on some neighborhood of ˙1, respectively, is an easy verification. /00 D 2 . Indeed, Z jyj 00 .x y/ dy /00 .x/ D .j
j  00 /.x/ D R Z 1 Z 0 y 00 .x y/ dy C y 00 .x y/ dy D 1 0 Z 1 Z 0 0 0 .x y/ dy C .x y/ dy D 2 .x/: D

1.5 Integration by parts leads to .j
j  .j
j 

1

Now select g 0 .x/ D

0

D  . Then integration implies

1C2

Z

x=

p.t/ dt 1

and

g.x/ D

xC2

Z

x= 1

Z

=

p.t/ dt d : 1

R 1.6 (i). f  g equals 0 outside Œ 2; 2 . Indeed, R f .x y/ g.y/ dy ¤ 0 implies that there is y 2 Œ 1; 1  with 1  x y  1, which leads to 2  y 1  x  y C 1  2.

352

21 Solutions to Selected Problems

(ii). Take both f and g equal to the characteristic function 1Œ 1;1  . Then f  g.x/ D R1 1 1Œ 1;1  .x y/ dy, where the interval of integration may actually be restricted to Œ x 1; x C 1 . Hence, for 2  x  0 and 0  x  2 we find that f  g.x/ equals Z

Z

xC1 1

dy D 2 C x

and

1

dy D 2

x 1

x;

respectively:

It follows that f  g.x/ D maxf 0; 2 jxj g, for all x 2 R. (iii). The functions from part (ii) satisfy all demands in (iii). (iv). Following the hint, we find by means of the substitution y D x t that f  g.x/ D

Z

x

.x 0

˛

˛

y/ y dy D x

2˛C1

Z

1

.1

t/˛ t ˛ dt;

0

where the latter integral converges. f  g is discontinuous at 0 if 2˛ C 1 < 0, that is, if 1 < ˛ < 1=2. 2.1 There exists z 2 U such that d.y; U / D ky d.x; U /  kx which implies d.x; U / kx yk.

zk  kx

d.y; U /  kx

zk. Therefore we obtain yk C ky

zk;

yk. Similarly d.y; U /

d.x; U / 

2.2 Note that supp  .k/  supp , for all k 2 Z0 . (i). Suppose there exists a compact set K  R such that fj g C supp  D supp j  K, for all j 2 N. Then j 2 K C . supp /, which is a compact set; hence no convergence in C01 .R/. On the other hand, for fixed k and arbitrary j and x, jj.k/ .x/j D

1 .k/ j  .x j

j /j 

1 sup j .k/ .x/j; j x2R

which shows that limj !1 j.k/ D 0 converges uniformly on R. (ii).  cannot be a polynomial, owing to the compactness of supp ; consequently, there exists x0 2 R with  .pC1/ .x0 / ¤ 0. Now set xj D x0 =j . Then we have, as j ! 1, ˇ  x ˇ 0 ˇ ˇ .pC1/ jj .xj /j D j pC1 p ˇ .pC1/ j ˇ D j j .pC1/ .x0 /j ! 1: j .pC1/

This implies that .j /j 2N is not uniformly convergent on R, and therefore .j /j 2N is not convergent in C01 .R/. Now consider x 2 R fixed. Since 0 does not belong to the compact set supp , there is N D N.x/ such that jx … supp , for all j  N . Accordingly, for such j and fixed k 2 Z0 , we have j.k/ .x/ D j k p  .k/ .jx/ D 0 and thus limj !1 j.k/ .x/ D 0. (iii). Select a < 0 < b such that supp   Œ a; b . Then supp j  Œ a; b , for all j . Fix k arbitrarily. Then we have, for all j and x,

1 j

supp  

21 Solutions to Selected Problems

353

jj.k/ .x/j D j k e

j

j .k/ .jx/j  j k e

j

sup j .k/ .x/j;

x2R

whereas limj !1 j k e j D 0. It follows that limj !1 j.k/ D 0 uniformly on R, and this holds for every k. As a consequence, we have convergence to 0 in C01 .R/. 2.3 supp  is contained in the closed ball B.0I /, while supp is a compact subset of the open set X . According to Corollary 2.4, there exist 0 > 0 and a compact set K  X such that the -neighborhood of supp is contained in K, for 0 <   0 . Now Lemma 2.18 implies, for 0 <  < 0 ,   /  supp

supp .

C B.0I /  K  X;

which shows that condition (a) in Definition 2.13 is satisfied. According to the proof of Lemma 2.18 one has @˛ .   / D @˛   , for arbitrary multi-index ˛. On account of Lemma 1.6, it follows that @˛  converges uniformly to @˛ on K as  # 0; this leads to the uniform convergence of @˛ .   / to @˛ on X . Therefore condition (b) in Definition 2.13 is satisfied too. R x R1 2 2.5 We have y  .y/ D 2 0 .y/ and furthermore 1  .y/ dy D x  .y/ dy, because  is an even function. Hence, for any x 2 Rn , Z x Z Z 1 .j
j   /.x/ D .x y/  .y/ dy jx yj  .y/ dy D 1

R

x

 2

 .y/ dy C .Œ  x 1 Œ  1 Dx x / 2 1 x Z x= Z x x 2 Dx

 .y/ dy C  2  .x/ D p e y dy C  2  .x/;  x= x Z

Z

x

1

where we have obtained the last integral by a change of variables. In view of (2.14), lim  2  .x/ D 0 #0

1 lim p #0 

and

Z

x=

e

y2

x=

dy D



˙1; 0;

x ≷ 0I x D 0:

Accordingly lim.j
j   /.x/ D jxj: #0

A standard computation now gives 1 .j
j   /0 .x/ D p 

Z

x=

e

y2

dy

and

x=

.j
j   /00 .x/ D 2  .x/:

3.2 Suppose there exists a neighborhood U of 0 in R such that u is of order < k on U . Then for every compact subset K  U , there exists c > 0 such that j .k/ .0/j D ju./j  c kkC k

1

. 2 C01 .K//:

354

21 Solutions to Selected Problems

For arbitrary 2 C01 .R/ satisfying .0/ ¤ 0 and supp  U and t > 0, 1 consider  t 2 C0 .R/ given by  t .x/ D e i tx .x/. Then  t .k/ .0/ D .i t/k .0/ C O.t k 1 / on account of Leibniz’s formula (2.8), while k t kC k 1 D O.t k 1 / as t ! 1. This leads to a contradiction and therefore u is of order k on U .

3.5 (i). Both functions are locally integrable on R2 and therefore define distributions, of order 0. (ii).(a). A straightforward estimate yields that ju./j  2 kkC 1 , for all  2 C01 .R2 /; this implies that u is a distribution of order  1. The integrand in the definition of u./ is equal to @r .r cos t; r sin t/jrD1 . Therefore, if we take .x/ D .kxk/ for some 2 C01 .R>0 /, then u./ D  0 .1/, whereas on the other hand supx2R2 j.x/j D supr>0 j .r/j. Hence, if u were of order 0, there would exist c > 0 such that j 0 .1/j  c supr>0 j .r/j, for all 2 C01 .R>0 /; this is in contradiction to Problem 3.2 for k D 1. We conclude that u has order 1. d (ii).(b). Note that . sin t @1 C cos t @2 /.cos t; sin t/ D dt .cos t; sin t/, which implies that v is of order 0. 4.1 For any  2 C01 .R/, one finds by integration by parts that j
j0 ./ D D Similarly,

Z 0 Z 1 j
j. 0 / D x  0 .x/ dx x  0 .x/ dx 1 0 Z 1 Z Z 0 .x/ dx C .x/ dx D sgn.x/ .x/ dx D sgn./: 1

j
j00 ./ D sgn0 ./ D

0

sgn. 0 / D

D 2ı./:

R

Z

0 1

 0 .x/ dx

Z

1 0

 0 .x/ dx D 2.0/

4.2 This is a direct consequence of (1.3). 4.6 Without any restriction in generality we may assume that p D 0, i.e., 1 n 1 v.x/ D kxk in Rn via n x. Introduce spherical coordinates .r; y/ 2 R>0  S x D .r; y/ D r y; one then has j det D .r; y/j D r n 1 !.y/, where ! denotes the Euclidean .n 1/-dimensional density on S n 1 . The local integrability of the vj over Rn , for 1  j  n, then follows from Z Z  Z nC1Cn 1 jvj .x/j dx D r dr jyj j !.y/ dy 0 B.0I/ Sn 1 Z D jyj j !.y/ dy < 1 . > 0/: Sn

For any  2 C01 .Rn /, we have

1

21 Solutions to Selected Problems n X

.div v/./ D

vj .@j / D

j D1

Z

D

Rn

355

Z

n X

Rn j D1

vj .x/ @j .x/ dx

h v.x/; grad .x/ i dx D

Z

D.x/v.x/ dx:

Rn

Upon the introduction of spherical coordinates in Rn this leads to Z Z r n 1 D.r y/v.r y/ dr !.y/ dy .div v/./ D Z

D

Sn 1

Z

Sn

R>0

1

R>0

D.r y/y dr !.y/ dy D

D cn .0/ D cn ı./:

Z

Sn 1

Z

R>0

d .r y/ dr !.y/ dy dr

4.7 The local integrability of E over Rn follows as in the preceding problem. For any  2 C01 .Rn / and n ¤ 2 we obtain, through integration by parts, Z .@j E/./ D E.@j / D E.x/ @j .x/ dx Rn

Z

Z 1 kxk2 n @j .x/ dxj dxjy .n 2/cn Rn 1 R Z Z Z 1 1 xj xj .x/ dxj dxjy D D kxk1 n .x/ dx cn Rn 1 R kxk cn Rn kxkn 1 D vj ./: cn D

Finally,
E D div.grad E/ D in a similar way.

1 cn

div v D

1 cn

cn ı D ı. The case of n D 2 is treated

5.1 For any 0 ¤  2 C01 .Rn / and r > 0, we have, in view of condition (a), Z .test fj /./ ı./ D fj .x/ ..x/ .0// dx Rn

D

Z

fj .x/..x/ kxkr

.0// dx C

Z

fj .x/ ..x/ kxkr

.0// dx DW

2 X

Ii :

i D1

Now select  > 0 arbitrarily. In view of the continuity of  there exists r > 0 such that j.x/ .0/j < =2, for all kxk  r. Since fj  0, condition (a) implies Z Z   .j 2 N/: jI1 j  fj .x/j.x/ .0/j dx < fj .x/ dx  2 kxkr 2 kxkr Next, write m D supx2Rn j.x/j > 0. On account of condition (b) there exists N 2 N such that for all j  N ,

356

21 Solutions to Selected Problems

Z

 ; 0 fj .x/ dx < 4m kxkr

so

jI2 j  2m

Z

fj .x/ dx
0 satisfies conditions (a) and (b) from Problem 5.1. Indeed, Z Z r fı .x/ dx D f ı .x/ dx with lim D 1: r #0  kxkr kxk  Thus the conclusion follows from the same problem. Alternatively, one can proceed as follows. For any  2 C01 .Rn /, one has Z Z .test f /./ D f .x/.x/ dx D f .x/. x/ dx: Rn

Rn

Now jf .x/. x/j  .supx2Rn j.x/j/ jf .x/j, for all x 2 Rn and all  > 0, with a dominating function that is integrable on Rn . Furthermore, lim#0 f .x/. x/ D .0/ f .x/. As one can see by applying either the Dominated Convergence Theorem of Arzel`a (see [7, Theorem 6.12.3]) or that of Lebesgue (see Theorem 20.26.(iv)), R one has lim#0 .test f /./ D Rn f .x/ dx ı./. 5.5 Note that u t .x/ D fpt .x/ in the notation of Problem 5.2, where f .x/ D .4/

n 2

e

kxk2 4

and

Z

Rn

f .x/ dx D 1:

Application of that same problem therefore implies lim t #0 u t .x/ D ı in D 0 .Rn /. The following formulas show that u t satisfies the heat equation:  kxk2 n u t .x/ d u t .x/ D u t .x/; x; gradx u t .x/ D 2 dt 4t 2t 2t  kxk2 u t .x/ 1 n divx x h gradx u t .x/; x i D u t .x/;
u t .x/ D 2t 2t 4t 2 2t where we have used that
D div grad. Consequently, lim t #0 D 0 .Rn /. Indeed, for any  2 C01 .Rn /, lim t #0

d dt u t

D
ı in

d u t ./ D lim
u t ./ D lim u t .
/ D ı.
/ D
ı./: dt t #0 t #0 2

5.6 Here we evaluate only the integral. Use the change of variables t D kxk 4 u to obtain Z Z kxk2 1 kxk2 n n n 2 e 4t dt D E.x/ D t u 2 2 e u du: n n .4/ 2 R>0 4 2 R>0

21 Solutions to Selected Problems

357

Note that according to (13.30) and (13.31) the integral on the right-hand side equals n

 2 . n2 / 1 D I n 2

2

E.x/ D

so

.2

1 n/ cn kxkn

2

in view of (13.37). 6.2 We discuss two different methods to prove the result. Consider arbitrary  2 C01 .Rn / and define f W R>0 ! C by f .t/ D u t ./. From Problem 5.5 we obtain lim t #0 f .t/ D .0/, and more generally, f .j / .t/ D

dj u t ./ D
j u t ./ D u t .
j / !
j .0/ as t # 0: dt j

As a consequence, f 2 C 1 .R>0 / and the f .j / .t/ have limits as t # 0, for all j 2 Z0 . Therefore we may use Taylor expansion to derive f .t/ D

k X1 j

j D0

t f .j / .C0/ C jŠ

Z

t

.t s/k 1 .k/ f .s/ ds .k 1/Š

0

.t > 0/:

Hence, by means of the change of variables s D tu in the integral, we obtain, for k 2 Z0 ,   Z 1 k X1 t j 1 .1 u/k 1 .k/ ˇˇ j
ı ./ D lim f .s/ˇ lim k u t du sDt u t #0 t jŠ t #0 0 .k 1/Š j D0

D

1 1 .k/ f .C0/ D
k ı./: kŠ kŠ

Next, the second method. According to Problem 6.1 we may apply Proposition 6.3 to the function u t . Indeed, if 2t D  2 , then u t .x/ D 1n f . 1 x/ where 2 f .x/ D .2/ n=2 e kxk =2 . In the notation of the proposition and on account of [7, Exercise 6.51], we have, for all ˛ 2 .Z0 /n , c˛ D D

Z

€ 0; Rn

x ˛ f .x/ dx D



n 2

n 2 Y n

n

n

.2/ 2 2 2

n Z Y

1 n

.2/ 2

j D1 R

.˛j /Š

˛j j D1 . 2

/Š 2

˛

D

˛j 2

xj j e

dxj if there exists j with ˛j oddI

˛Š 2

1 2 2 xj

j˛j 2

. ˛2 /Š

;

for ˛ 2 .2Z0 /n :

Therefore we have, with j 2 Z0 , X c˛ @˛ ı D uj D ˛Š j˛jDj

€ 0;

1 j

22

X

j˛jDj; ˛2.2Z0 /n

j oddI 1 ˛ @ ı; ˛ . 2 /Š

j even.

358

21 Solutions to Selected Problems

Next apply Proposition 6.3 with k replaced by 2k. This leads to 0 D lim #0

1  2k

 f

k X



2j

u2j

j D0



 1 1 D k lim k u t 2 t #0 t

k X

t

j

j D0

 X 1 2˛ @ ı : ˛Š

j˛jDj

The desired formula now follows by application of the following Multinomial Theorem (see [7, Exercise 2.52.(ii)]): P X 1 X x 2˛ . niD1 xi2 /j 1 j D ; that is, @2˛ D
: ˛Š jŠ ˛Š jŠ j˛jDj

j˛jDj

6.3 (i). Introducing spherical coordinates in Rn , we see that Z Z n r2 e 4t r n 1 .r y/ dy dr u t ./ D .4 t/ 2 R>0 S Z n r2 D cn .4 t/ 2 e 4t r n 1 S .r/ dr: R>0

(ii). Expand S in its Taylor series at 0 and next apply the substitution of variables p r D 4ts as well as (13.30). Thus we get, for any k 2 Z0 , ut ./ D D

cn .4 t/

n 2

k cn X n

2

Z k X S.j / .0/ jŠ

j D0

2

j 1

j D0

e R>0

r2 4t

rn

1Cj

dr C O.t kC1 /;

j n C j  t2 .j / S .0/ C O.t kC1 /; 2 jŠ

t #0

t # 0:

On the other hand, according to Problem 6.2 we have, for any k 2 Z0 , u t ./ D

k X


j .0/

j D0

tj C O.t kC1 /; jŠ

t # 0: .j /

Equate the coefficients of like powers of t to find that S .0/ D 0 if j is odd. Therefore, suppose that j D 2k; then n

cn 22k 1 kŠ n  2 .2k/Š

2

 C k S.2k/ .0/ D
k .0/:

On account of (13.31) this leads to S2k .0/ D

2k

.2k/Š
k .0/: Qk 1 kŠ j D0 .n C 2j /

(iii). Multiply y 7! y ˛ by a suitable cut-off function and apply part (ii).

21 Solutions to Selected Problems

359

7.4 See [7, Formula (7.44)]. 7.5 See [7, Exercise 6.95]. 7.6 (i) and (ii). For these results, see the beginning of Chap. 13. a (ii). Consider k 2 Z such that Re a 1 < k < Re a. It is easy to show that xC a has order  k. We will show that xC cannot have order < k. If this were the case, there would exist c > 0 such that a jxC ./j  c kkC k

1

. 2 C01 .Œ 0; 1 //:

For  2 C01 .Œ 0; 1 / and 0 <   1, define  .x/ D . x /. Routine computation shows that a a xC . / D  aC1 xC ./:

a ./ ¤ 0. Such  exist, because otherwise Now select  2 C01 .Œ 0; 1 / such that xC a xC D 0 on  0; 1 Œ , which by Theorem 4.3 would imply that on  0; 1 Œ the distribution a xC equals some polynomial of degree  k 1, which cannot be the case. Therefore we have, for all 0 <   1, a a  Re aC1 jxC ./j D jxC . /j  c k kC k

1

Dc

sup l0 R>0 Z Z 1 1 x D @k .x/ log dx D @k .x/ log x dx C @k 1 .0/: log j R>0 j log j R>0 As a consequence, limj !1 . 1/k @k lC .j / D 1. On the other hand, for 0  l < k we obtain   1 k@l j kC 0 D O k l 1 D o.1/; j ! 1: j log j

It follows that @k lC is of order k on R, by arguments similar to those in the solution to Problem 3.2.

360

21 Solutions to Selected Problems

8.2 If C D X n K, then C open in Rn . For any x 2 C , there exists an open neighborhood U of x with K \ U D ;. Now consider  2 C01 .U /. Since supp uj \ supp   K \ U , formula (7.3) says that uj ./ D 0 and thus u./ D limj !1 uj ./ D 0. In other words, u D 0 on U and then Theorem 7.1 asserts that u D 0 on C ; that is, supp u  K. This implies u 2 E 0 .X /, in view of Theorem 8.8. Next, select  2 C01 .X / with  D 1 on a neighborhood of K and let 2 C 1 .X / be arbitrarily chosen. Then supp .  / \ K D ;, and so  /D0

uj .

.j 2 N/

u.

and

Because limj !1 uj D u in D 0 .X / and 

 / D 0:

2 C01 .X /, we obtain

lim uj . / D lim uj . / D u. / D u. /;

j !1

j !1

which implies limj !1 uj D u in E 0 .X /. The final assertion is a direct consequence of the preceding results. 8.5 The function R.x; t/ 7! E.x; t/ is locally integrable on RnC1 because it equals 0 if t  0, while Rn E.x; t/ dx D 1, if t > 0, which defines a locally integrable function on R. Further, sing supp E D f0g. In order to see this, we have only to consider a neighborhood of a point .x; 0/ with x ¤ 0. In fact, we have to show that all partial derivatives of E.x; t/ tend to 0 as t # 0. We have, for  2 C01 .RnC1 /, v./ D .@ t E
x E/./ D E.@ t  C
x / Z Z D E.x; t/.@ t  C
x /.x; t/ dt dx DW lim I C J ; Rn

with

I D

Z

Rn

Z

1

#0

R>0

.E @ t /.x; t/ dt dx;



Z

J D

1 

Z

Rn

.E
x /.x; t/ dx dt:

Integration by parts with respect to the variable t gives Z Z Z 1 I D .E /.x; / dx C . @ t E/.x; t/ dt dx: Rn

Rn



Apply Green’s second identity (see [7, Example 7.9.6]) to J . There is no boundary term, because  has compact support; hence we see that Z 1Z J D .
x E/.x; t/ dx dt: 

Rn

n Since E satisfies the heat equation pon R   ; 1 Œ , we therefore obtain, by means of the change of variables x D 2  y, Z Z p n 2 .E /.x; / dx D  2 e kyk .2  y; / dy: I C J D Rn

Rn

21 Solutions to Selected Problems

361

On account of the Mean Value Theorem and the compactness of supp  we can find c > 0 such that for all 0 <   1 and y 2 Rn , p p j.2  y; / .0/j  c  kyk: p p Hence the equality .2  y; / D .0/ C .2  y; / .0/ leads to Z n 2 e kyk dy C R ; I C J D .0/  2 where

jR j 

p  c

n 2

Z

Rn

Rn

kyk e

kyk2

p dy D O. /;

 # 0:

It follows that v D ı 2 D 0 .RnC1 /, in view of v./ D lim I C J D .0/ C lim R D .0/ D ı./: #0

#0

9.2 For every  2 C01 .Rn /, we have on account of Leibniz’s rule (2.8), . @˛ ı/./ D .@˛ ı/. / D . 1/j˛j @˛ . /.0/ ! ! X ˛ X ˛ j˛j ˛ ˇ ˇ @ @˛ .0/ @ .0/ D D . 1/ ˇ ˇ ˇ ˛ ˇ ˛ ! X ˛ ˛ ˇ @ . 1/j˛ ˇ j .0/ @ˇ ı : D ˇ

ˇ

.0/ . 1/j˛j

jˇ j

@ˇ ı 

ˇ ˛

9.3 Application of Problem 9.2 implies ! m X m k m x @ ıD . 1/m l .@m l x k /.0/ @l ı l lD0 ! m X m . 1/m l k.k 1/


.k m C l C 1/x k D l

€

lD0

0;

ˇ

mCl ˇ

xD0

@l ı; m

l  kI

m

l > k:

Therefore x k @m ı ¤ 0 only if there exists an l such that l  0 and k that is, l D m k  0. If this is the case, we obtain x k @m ı D . 1/k

.m

mŠ @m k/Š

k

m C l D 0,

ı:

Now consider u 2 D 0 .R/ satisfying x k u D 0. On account of Theorem 9.5 we have P supp u  f x 2 R j x k D 0 g D f0g, while Theorem 8.10 then implies u D m cm @m ı, with cm 2 C. Now

362

21 Solutions to Selected Problems

0 D xk u D

X m

X

cm x k @m ı D

cm . 1/k

mk

Accordingly cm D 0, for m  k, and so u D for x k u D 0.

mŠ @m .m k/Š

Pk

1 m mD0 cm @ ı

k

ı:

is the general solution

9.4 According to Problem 9.3 we have u0 D c ı with c 2 C arbitrary. In view of Example 4.2 the latter equation has the particular solution u D c H , while Theorem 4.3 says that u D c H C d is the general solution, where d 2 C.

9.5 Since x ı D 0, the desired formula may be easily deduced from the formula 1 1 xCi 0 D  i ı C PV x , which occurs in the solution of Problem 1.3. The problem k of solving x u D 1 can be reduced to finding all solutions u0 to the homogeneous problem x k u0 D 0, which is done in Problem 9.3, plus a particular solution to the inhomogeneous problem x k u D 1. In the case k D 1, we already know that PV x1 solves the latter problem. Next, suppose that uk satisfies x k uk D 1. This implies 0 D x .x k uk /0 D kx k uk C x kC1 u0k D k C x kC1 u0k ;  1  u0 D 1: so x kC1 ukC1 WD x kC1 k k

Therefore, the general solution to x k u D 1 is given by, see Problem 4.2, uD

. 1/k 1 k @ .k 1/Š

1

PV

k k X1 X1 1 . 1/k 1 C log.k/ jxj C cm @m ı D cm @m ı: x mD0 .k 1/Š mD0

9.6 Let  2 C01 .R/ and suppose .x/ D 0 if jxj  m > 0. We have  1 ./ D lim J ; e i tx PV x #0

where, by means of a change of variables, J WD

Z

jxjm

e i tx .x/ dx D x

Z



m  i tx

e

.x/

e x

i tx

. x/ 

dx:

Writing .x/ D .0/ C x .x/ for all x 2 R, with continuously differentiable on R, we get Z m Z m i tx e e i tx J D .0/ dx C .e i tx .x/ C e i tx . x// dx x   Z tm Z m sin x D 2i ı./ dx C .e i tx .x/ C e i tx . x// dx: x   Both integrands on the right-hand side are continuous, which implies

21 Solutions to Selected Problems

363

lim J D 2i ı./ #0

where Kt D

Z

Z

tm

0

sin x dx C K t ; x

m 0

.e i tx .x/ C e

i tx

. x// dx:

R Now we have the well-known evaluation R>0 sinxx dx D 2 , see Problems 14.44, 16.19, or 16.21 or [7, Example 2.10.14 or Exercises 0.14, 6.60, 8.19]. Furthermore, the identity e i tx D i1t @x e i tx and integration by parts lead to, for t > 0, ˇi Z m ˇ .e i tx jK t j D ˇ t 0

0

.x/ C e

i tx

ˇ 1Z ˇ . x// dx ˇ  t

m m

j

0

.x/j dx:

(21.2)

We now immediately obtain (9.5). The second identity follows by adding together (9.5) and the analogous identity obtained by replacing x by x. By combination of Problem 1.3 and (9.5) it follows that lim e i tx

t !1

 1 1 D lim  i e i tx ı C e i tx PV D  i ı C  i ı; t !1 x˙i0 x

and this proves (i) and (ii). Assertion (iii) follows from the fact that the inner limit is already equal to 0, as can be demonstrated in a way similar to the proof of (21.2). 9.7 Theorem 9.5 implies that supp u is contained in the hyperplane H D f x 2 Rn j xn D 0 g. Select  2 C01 .R/ with  D 1 on an open neighborhood of 0 and define z 2 C 1 .Rn / by .x z 1 ; : : : ; xn / D .xn /. Then 1 z D 0 on an open neighborhood of H ; hence u D z u. Now consider  2 C01 .Rn /. Then .x1 ; : : : ; xn / D .x1 ; : : : ; xn

1 ; 0/ C xn

D ..x// C xn .x/; where  W x 7! .x1 ; : : : ; xn

1 ; 0/

Z

1

@n .x1 ; : : : ; xn

1 ; txn / dt

0

and  2 C01 .Rn /. This leads to

u./ D u.z / D u.z . B // C u.xn z / D u.z . B //: Now .x/ z . B /.x/ D .x1 ; : : : ; xn 1 ; 0/ .xn / D   ˝ .x/, so u./ D  u.  ˝ /. Finally, we note that 7! ˝  defines a continuous linear mapping C01 .Rn 1 / ! C01 .Rn /. Hence, v W 7! u. ˝ / belongs to D 0 .Rn 1 / and satisfies u./ D v. /. 9.14 On account of the data, the solution I should be supported by R0 , that is, it has to contain the Heaviside function H as a factor. Note that H 0 D ı. Furthermore, the restriction of I to R>0 should satisfy the corresponding homogeneous differential equation L I 00 C R I 0 C C1 I D 0. Now the solutions of the latter equation are given by, for c˙ 2 C,

364

21 Solutions to Selected Problems

I0 .t/ D

X ˙

c˙ e t .

A˙B/

with

aD

R 2L

BD

and

1 2L

r

R2

4L C

under the assumption of B ¤ 0. Hence, consider I D I0 H ; then by Leibniz’s rule and Example 9.1, I 0 D I0 .0/ ı C I00 H

I 00 D I0 .0/ ı 0 C I00 .0/ ı C I000 H:

and

As a consequence, the condition L I 00 C R I 0 C

1 I D L.I0 .0/ ı 0 C I00 .0/ ı/ C R I0 .0/ ı C D L I0 .0/ ı 0 C .L I00 .0/ C R I0 .0//ı D ı

leads to I0 .0/ D 0 and L I00 .0/ D 1, whence c D cC D

1 1 D q 2LB R2

: 4L C

In the case of B D 0, one uses I0 .t/ D c e tA C cC t e equations as before. These imply c D 0 and L cC D 1.

tA

and derives similar

10.1 Taking the transpose of (10.1) and using Lemma 10.1 as well as the fact that the transpose of @k is @k leads to the desired identity. 10.7 Successively applying the chain rule and the transposition of a matrix, we obtain, for  2 C01 .X /, D. B / D .D/ B D ; H) D.  / D  .D/ D ; H) grad.  / D tD  .grad /; H)  B grad./ D .tD / 1 B grad B  ./: 10.8 The mapping t 7! det D˚ t is continuous and nonvanishing on the connected set R, while det D˚0 D det I D 1. Therefore, det D˚ t > 0, for all t 2 R. On account of (10.10) and the fact that .˚ t / 1 D ˚ t , one obtains, on D 0 .X /, ˇ d d ˇ .˚ t / ˇ ..˚ t / D t D0 dt dt

1 

/ .det D.˚ t //

According to (10.18) and (10.17), this implies n X

j D1

@j B vj D D

Hence,

1


ˇˇ ˇ

t D0

D

ˇ d ˇ .˚ t  det D˚ t /ˇ : t D0 dt

ˇ ˇ d d ˇ ˇ ˚t  ˇ det D˚ t ˇ det DI C I  t D0 t D0 dt dt

n X

j D1

vj B @j C

ˇ d ˇ det D˚ t ˇ : t D0 dt

21 Solutions to Selected Problems

365

n ˇ X d ˇ det D˚ t ˇ D .@j B vj t D0 dt j D1

vj B @j / D

n X

j D1

@j vj D div v:

In fact, one can prove that

d det D˚ t D .div v/ B ˚ t det D˚ t dt

.t 2 R/:

In view of det D˚0 D 1, solving the differential equation (compare with [7, Formula (5.32)]) yields det D˚ t .x/ D e

Rt

0

div v.˚ .x// d 

..t; x/ 2 R  X /:

10.9 ıx is invariant under .˚ t / t 2R as a distributional density iff .˚ t / ıx D ıx for all t 2 R, and in view of Example 10.4 this is the case iff ı˚ t .x/ D ıx , that is, iff ˚ t .x/ D x, which is equivalent to v.x/ D 0. Furthermore, assume that ıx is invariant under .˚ t / t 2R as a generalized function. On account of Example 10.12, we have 1 ˚ t  ıx D ı˚ t .x/ .t 2 R/: (21.3) j det D˚ t .˚ t .x//j

By considering the supports of the distributions occurring in (21.3), it follows that the assumption implies x D ˚ t .x/, or equivalently ˚ t .x/ D x, for all t 2 R. In turn, (21.3) then leads to det D˚ t .x/ D 1, for all t 2 R. Upon differentiation with respect to t at t D 0 and application of Problem 10.8 we obtain the desired conclusion. ˇ d 10.10 Writing ˚ t for the rotation by the angle t, we obtain v.x/ D dt ˚ t .x/ˇ t D0 D . x2 ; x1 /. On account of Theorem 10.16, the distributional density u 2 D 0 .R2 / is invariant under all of the ˚ t if and only if 2 X

j D1

@j .vj u/ D @1 .x2 u/ C @2 .x1 u/ D x2 @1 u C x1 @2 u D 0:

Furthermore, u 2 D 0 .R2 / is invariant as a generalized function if and only if P2 j D1 vj @j u D x2 @1 u C x1 @2 u D 0.

10.13 Using the identityPfor the exponential function, integration by parts twice, and the convergence of n2N n12 , one obtains the desired estimate. Theorem 3.8 then implies that u 2 D 0 .R/. Part (i) follows by a shift in the index of summation. For (ii), consider, with  2 C01 .R/, XZ 2 / dx .T 2 / u./ D u.T 2  / D n2Z ei n! .T 2  / D e i n! x .x C ! ! ! ! n2Z R XZ D e i n! x .x/ dx D u./: n2Z

R

366

21 Solutions to Selected Problems

Next, e i ! x 1 D 0 only if x 2 2 Z, while @.ei ! 1/ D i ! ¤ 0 at such ! x. Hence, Theorem 9.5 implies the existence of constants ck 2 C such that u D P . The periodicity of u and Example 10.4 now lead to ck D ckC1 , and k2Z ck ık 2 ! so ck D c, for all k 2 Z. 10.14 The integral is convergent at 0 because near 0 the numerator can be estimated by kxk2 while s < 1, and at 1 because the numerator is bounded by 4 and s > 0. The integral defines a function on Rn homogeneous of degree 2s, as follows by the change of variables x D 1t y. Finally, the integral is invariant under every rotation A about the origin on account of the substitution x D Ay. P 10.15 (i). According to Theorem 10.17, we have nkD1 xk @k uj D a uj , for all j . Since @k W D 0 .Rn / ! D 0 .Rn / is a continuous linear mapping, on account of (8.2), 0 n it follows Pnthat @k uj ! @k u in D .R /; by taking the limit as j ! 1, we therefore obtain kD1 xk @k u D a u, which implies u 2 Ha . (ii). Applying @j to the last identity in (i), we see that X X xk @j @k u C @j u C xj @j2 u D xk @k @j u C @j u D a @j uI k¤j

k

in other words, @j u 2 Ha 1 . The assertion for xk u follows by a similar argument. (iii). If a 2 R, then t 7! t a belongs to C 1 .R/ if and only if a 2 Z0 . For any a 2 C n R,
t a D t Re a e i.Im a/ log t D t Re a cos..Im a/ log t/ C i sin..Im a/ log t/ ;

where Im a ¤ 0. Hence, t 7! t a does not belong to C 1 .R/ in this case. Now suppose x 2 Rn and a 2 Z0 . Then .t x/ D t a .x/ is an identity of functions in C 1 .R/. Now differentiate this identity a times with respect to t at t D 0; the right-hand side then gives aŠ .x/, while on account of the chain rule the left-hand side leads to D a .tx/.x; : : : ; x/j t D0 D D a .0/.x; : : : ; x/ DW p.x/; with a copies of x, and where p is a polynomial function on Rn . Accordingly, 1 D aŠ p. According to [7, Exercise 0.11.(iv)], properties of the binomial series imply X

a2Z0

jf ˛ 2 .Z0 /n j j˛j D a gj t a D

X

n Y

˛2.Z0 /n j D1

D .1

t/

n

D

t ˛j D X

a2Z0

 X

a2Z0

aCn a

ta

n 1

!

t a:

1 (iv). Set g.x/ D kxka f . kxk x/. The local integrability of g can be proved using spherical coordinates. And g 2 Ha , because according to Theorem 10.8 we have,

21 Solutions to Selected Problems

367

for any  2 C01 Rn ,

Z c  g./ D g.c / D g.c n .c 1 / / D g.x/ .c Rn Z g.cx/ .x/ dx D c a g./: D

1

x/ c

n

dx

Rn

10.16 On the basis of Theorem 10.8 we see, for all c > 0 and  2 C01 .Rn /, that c  ı./ D ı.c / D ı.c

n

1 

/ / D c

.c

n

ı./:

This proves that ı is homogeneous of degree n. Using Problem 10.15.(ii) and mathematical induction over j˛j, we now find that @˛ ı is homogeneous of degree n j˛j. 10.18 Problem 10.1 implies that ˚ B @j D

p X

kD1

@k B ˚ B @j ˚k

W

C01 .X / ! C01 .Y /

is an identity of continuous linear mappings. On account of Theorem 10.18, the pullback ˚  W D 0 .Y / ! D 0 .X / is well-defined as the transpose of ˚ W C01 .X / ! C01 .Y /, the mapping ˚ being a C 1 submersion. Therefore the desired identity can be obtained by transposition. 10.20 (i). From ˚.x/ D h Ax; x i one obtains grad ˚.x/ D 2Ax. Because A is invertible, its kernel consists of 0 only. This proves the assertion. (ii). Apply Problem 10.18 with p D 1 to obtain, since ˚.x/ D y and y denotes the variable in R, .@j B ˚  /v D .@j ˚/ ˚  v 0

in

D 0 .Rn n f0g/:

Using this identity leads to ..@i @j / B ˚  /v D @i .@j B ˚  /v D .@i @j ˚/ ˚  v 0 C .@j ˚/.@i B ˚  /v 0 D .@i @j ˚/ ˚  v 0 C .@j ˚/.@i ˚/ ˚  v 00 : P Part (i) now implies @j ˚.x/ D 2 nkD1 Ajk xk and @i @j ˚ D 2Aj i . Hence, in D 0 .Rn n f0g/, n X .P B ˚  /v D Bij ..@i @j / B ˚  /v i;j D1

D

n X

i;j D1

D2

n X i D1

Bij 2Aj i ˚  v 0 C

n X

Bij 4Ajk Ai l xk xl ˚  v 00

i;j;k;lD1

ıi i ˚  v 0 C 4y ˚  v 00 D ˚  .2n v 0 C 4y v 00 /;

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21 Solutions to Selected Problems

because n X

i;j;k;lD1

Bij Ajk Ai l xk xl D

n X

i;k;lD1

ıi k Ai l xk xl D

n X

k;lD1

Akl xk xl D y:

(iii). Application of the first identity in part (ii) and of part (i) leads to n X

j D1

xj .@j B ˚  /v D

n X

j D1

xj .@j ˚/ ˚  v 0 D 2  0



0

n X

xj .Ax/j ˚  v 0

j D1

D 2y ˚ v D 2˚ .y v / D 2a ˚  v:

Thus, the claim is a consequence of Theorem 10.17. (iv). The first assertion also follows from Theorem 10.17. If v is homogeneous of degree 1 n2 , then u is homogeneous of degree 2 n. (v). The former identity follows upon restriction of the identity in part (ii) from D 0 .R/ to C 1 .R/, while the latter is obtained by means of transposition. (vi). In this case, the matrix B is the identity matrix on R2 and so is A. Hence ˚ as in this problem coincides with ˚ as in Example 10.5. Furthermore, Z  .˚ ıV /./ D .1/ d˛ D 2 ı1 ./: 

Since y ı1 D ı1 according to Example 9.1, one derives ˚
ıV D 4.@y2 B y

@y /2 ı1 D 8.ı1 00

ı1 0 /:

10.21 We use the notation of the proof of Theorem 10.18 and Remark 10.19. Given y 2 Y , write V .y/ D ˘

1

.fyg/ \ V D f z 2 Rq j .y; z/ 2 V g:

In view of formulas (10.21) and (10.22), we obtain Z j .y; z/ f ..y; z// dz .f 2 C0 .U /; y 2 Y /: .˚ f /.y/ D

(21.4)

V .y/

We will rewrite this integral in a more intrinsic manner, as an integral with respect to Euclidean integration over the fiber of ˚ over the value y 2 Y , which possesses the following parametrization: y W V .y/ ! U \ ˚

1

.fyg/

given by

y .z/ D .y; z/:

To this end, we derive the identity (21.9) below for j .y; z/. Identify x 0 2 Rn with .y; z/ 2 Rp  Rq and write x D .x 0 / D .y; z/, for .y; z/ 2 V . Suppose x D .x 0 / 2 U . Then ˚.x/ D ˘ B K B .x 0 / D ˘.x 0 / D y; in other words, x 2 ˚ 1 .fyg/. On account of the Submersion Theorem, near x the set ˚ 1 .fyg/ is a C 1 submanifold of Rn of dimension n p D q. According to [7,

21 Solutions to Selected Problems

369 1

.fyg/ at x is the following linear subspace

where

 D D˚.x/ 2 Lin.Rn ; Rp /: (21.5)

Theorem 5.1.2], the tangent space of ˚ of Rn of dimension q: T WD Tx ˚

1

.fyg/ D ker 

Application of the chain rule to the identity ˚ B  D ˘ on V gives the following identity in Lin.Rn ; Rp /: D˚.x/ B D.x 0 / D D˘.x 0 / D ˘

.x 0 2 V /:

For the moment we keep x and x 0 fixed and rewrite this equality more concisely as   D .` k / D . `  k / D ˘:

(21.6)

Here  D D.x 0 / 2 Aut.Rn /, while for 1  j  n, j D Dj .x 0 / 2 Rn ;

` D .1


p /;

k D .pC1


n / D Dy .z/:

As is customary, we identify a linear mapping and its matrix with respect to the standard basis vectors. It is a consequence of (21.6) and (21.5) that T is spanned by the vectors that occur in k ; these are the linearly independent vectors k , for p < k  n. Furthermore, (21.6) implies that  j D ej , for 1  j  p. In particular, the vectors that occur in ` are transversal to T in view of (21.5). Now consider the parallelepiped in Rn spanned by all j , for 1  j  n. Its volume equals J WD j det D.x 0 /j D j .y; z/ D j det.1


n /j D j det  j: Denote the orthocomplement in Rn of T by T ? . Then dim T ? D p. The value of J does not change if we replace j 2 Rn by its projection j 2 T ? along T , for all 1  j  p. Using the notation  D .1


p / this leads to J 2 D .det /2 D .det.1


n //2 D .det.1


p pC1


n //2 D .det. k //2 D det.t. k / . k //:

The vectors in  belong to T ? , while those in k belong to T ; hence . k / . k / D t  ˝ tk k ;

t

where the former factor on the right-hand side belongs to Aut.Rp / and the latter one to Aut.Rq /. This implies J 2 D det.t / det.tk k /:

(21.7)

Next, we evaluate det.t /. From linear algebra it is well-known that T ? may be described by

370

21 Solutions to Selected Problems

T ? D .ker /? D im t;

Rn D ker  ˚ im t:

in particular,

(21.8)

(Observe that (21.8) proves the identity rank t D rank ; see [7, Rank Lemma 4.2.7].) Indeed, for p 2 Rn , p 2 ker 

()

p D0

()

h p; qi D 0 for all q 2 Rp

hp; t  qi D 0 for all q 2 Rp

()

p 2 .im t/? ;

()

which shows that .ker /? D .im.t//?? D im.t/. Now (21.8) yields the existence of vectors 1 ; : : : ; p 2 Rp such that  D t . Since j j 2 T , for 1  j  p, it follows from (21.5) that .` / D 0. Hence, we obtain from (21.6), Ip D  ` D   D  t  D ;

where

D  t:

Note that 2 Aut.Rp / because it is Gram’s matrix associated with the linearly independent row vectors of . As a consequence,  D 1 , and thus  D t 1 . In turn, this leads to t

  D .t /

1

. t/

1

On account of (21.7) we therefore obtain 1 and det.t / D det

D

1



J2 D

1

D

1

:

det.tk k / : det. t/

More explicitly, this means that p p det.tDy .z/ B Dy .z// det.tDy .z/ B Dy .z// j .y; z/ D q :
D .gr ˚/..y; z// det D˚..y; z// B tD˚..y; z// (21.9) On account of the definition of Euclidean q-dimensional integration (see [7, Sect. 7.3]), one now deduces from (21.4), for f 2 C0 .U / and y 2 Y , that Z f .y .z// q .˚ f /.y/ D det.tDy .z/ B Dy .z// dz V .y/ .gr ˚/.y .z// (21.10) Z f .x/ dx: D U \˚ 1 .fyg/ .gr ˚/.x/ 10.22 (i). The first identity follows from .˚ /.y/ D ıy .˚ / D .˚  ıy /./ and ıy D @ Ty H , which is a consequence of Z 1 .y/ D @.x C y/ dx D H.T y @/ D @ Ty H./: 0

21 Solutions to Selected Problems

371

The second may be obtained from (10.1). (ii). Select  2 C 1 .R/ satisfying 0    1,  D 0 on  1;  1; 1 Œ . For t > 0, write .t  /.y/ D .ty/. Then, for x 2 X , ˚  .Ty t  /.x/ D .Ty t  /.˚.x// D .t.˚.x/

1 Π, and  D 1 on y//;

and so lim ˚  .Ty t  /.x/ D H.˚.x/

t !1

y/ D 1˚

1 .  y; 1 Œ /

.x/:

the Dominated Convergence Theorem of Arzel`a, see [7, Theorem 6.12.3], or that of Lebesgue, see Theorem 20.26.(iv), now leads to Z ˚  .Ty H /./ D lim ˚  .Ty t  /./ D lim ˚  .Ty t  /.x/ .x/ dx t !1 t !1 X Z Z 1˚ 1 .  y; 1 Œ / .x/ .x/ dx D .x/ dx: D X

˚

1 .  y; 1 Œ /

(iii). Part (ii) and the Theorem on Integration of a Total Derivative imply Z @j .x/ dx @j .˚  .Ty H //./ D .˚  .Ty H //.@j / D ˚ 1 .  y; 1 Œ / Z @j ˚.x/ dx: .x/ D k grad ˚.x/k ˚ 1 .fyg/ Here we have used that grad ˚.x/ points inward in ˚ (iv). Parts (i) and (iii) lead to

and so

1

.  y; 1 Œ /.

  @j ˚ .@j ˚/2   ˚ ı ./ D @ .˚ .T H //  y j y k grad ˚k2 k grad ˚k2 Z .@j ˚.x//2 .x/ dx; D k grad ˚.x/k3 ˚ 1 .fyg/ .˚ /.y/ D

Pn

j D1 .@j ˚/ k grad ˚k2

2 

˚ ıy ./ D

Z

˚

1 .fyg/

.x/ dx: k grad ˚.x/k

10.23 (i). The first identity is a direct consequence of (10.23). The second follows from the first by noting that .˚  ıy /./ D ıy .˚ / D .˚ /.y/. (ii). From part (i) we obtain, for any y 2 R, Z y Z y Z .x/ dx dt .˚ /.t/ dt D 1 k grad ˚.x/k 1 1 ˚ .ft g/ Z D .x/ dx: ˚

1. 

1; y Π/

In view of the Fundamental Theorem of Integral Calculus on R this implies

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21 Solutions to Selected Problems 

.˚ ıy /./ D .˚ /.y/ D @y (iii). We have

Z

Z

y 1

.˚ /.t/ dt D @y

1; y Π/

Z



.x/ .˚.x// dx .˚ / D .˚ /./ D Rn Z Z 1 Z Z D .x/ @ .y/ dy dx D @ .y/ ˚.x/ Rn R ˚ Z Z D .y/ @y .x/ dx dy: R

.x/ dx: 1. 

˚

˚

1. 

1 k grad ˚k

@j .˚  H / D @j ˚ ˚  .@H / D @j ˚ ˚  ı D For arbitrary  2 C01 .X / we now obtain Z  @j .˚ H /./ D .H B ˚/.@j / D

˚

@j ˚ ı ./ D k grad ˚k @˝

1; y Π/

1; y Π/

10.24 Combining of Problem 10.18 and ˚  ı D

while

.x/ dx dy 1. 

Z

1 .R

>0 /

ı@˝ leads to

@j ˚ ı : k grad ˚k @˝

@j .x/ dx D

@j ˚.y/ dy D .y/ k grad ˚.y/k @˝

Z

Z

@j .x/ dx; ˝

.y/ j .y/ dy: @˝

Here we have used that @j ˚.y/=k grad ˚.y/k equals the jth component of the normalized gradient vector of ˚ at y, which points inward in ˝, since the latter is the inverse image ˚ 1 .R>0 /; phrased differently, it equals j .y/, where j .y/ is the outer normal to @˝ at y. The desired equality is now obvious. 11.1 (i). u D  D 0. R (ii). Select u D 1 2 C 1 .R/ and  2 C01 .R/ satisfying R .y/ dy D 1. Then Z Z .u  /.x/ D u.x y/.y/ dy D .y/ dy D 1: R

R

(iii). Select u 2 C 1 .R/ with u.x/ D x and  as in part (ii) and, in addition, assume that  is an even function. Then Z Z Z .u  /.x/ D .x y/.y/ dy D x .y/ dy y .y/ dy D x: R

R

(iv). Select u.x/ D sin x and an even  2 C01 .R/ with Then

R

R

R

.y/ cos y dy D 1.

21 Solutions to Selected Problems

.u  /.x/ D

Z

373

sin.x y/ .y/ dy Z D sin x .y/ cos y dy R

cos x

R

Z

R

.y/ sin y dy D sin x:

11.3 The implication (ii) ) (i) follows from (11.4). For the reverse implication, note that for every  2 C01 .Rn / and x 2 Rn , @aj Ta .x/ D @aj .a 7! .x

a// D

a/ D

@j .x

Ta @j .x/ D

@j Ta .x/;

because @j commutes with Ta . Application of the chain rule to a 7! .a; a/ 7! Ta B A B T

a .x/

W

Rn ! Rn  Rn ! R;

taken in conjunction with assumption (i), implies @aj .Ta B A B T

a /

D .@aj Ta / B A B T a  C Ta B A B .@aj T a / D Ta B @j B A B T a  C Ta B A B @j B T a  D 0:

Apparently a 7! Ta B A B T a is a constant mapping, with value A, for a D 0; or Ta B A D A B Ta . The desired conclusion now follows from Theorem 11.3. 11.5 Consider  2 C01 .Rn / with 1./ D 1. On the basis of Problem 5.2 we then have lim#0  D ı. Accordingly, u   2 C 1 .Rn / satisfies @j .u   / D @j u   D 0, for 1  j  n. Thus there exists c 2 C with u   D c , and this leads to u D lim#0 u   D lim#0 c in D 0 .Rn /. This in turn implies c WD u./ D lim#0 c 1./ D lim#0 c in C and therefore u D c 2 C.

11.6 Set m D dim L and let u D u1 ; : : : ; um be some enumeration of all the @˛ u 2 D 0 .Rn /, for suitable multi-indices ˛, and denote by U the vector in .D 0 .Rn //m having these distributions as components. Then there exist m  m matrices Aj with coefficients in C P such that @j U D Aj U , where the differentiation n is componentwise; hence @j .e i D1 xi Ai U / D 0, for all 1  j  n (see [7, Example 2.4.10] for the exponential of a matrix). But then Problem 11.5 implies the Pn existence of some C 2 Cm such that U D e i D1 xi Ai C . The assertion follows by considering the first component of U . 11.7 f is characterized by the condition f .mC1/ D 0 on account of Example 4.4. 11.8 In view of (11.11) and Example 11.13 we obtain .ıa  ıb /./ D .ıa ˝ ıb /.˙  / D ı.a;b/ . B ˙/ D .a C b/ D ıaCb ./: 11.10 From (11.19) and (11.25) we deduce P v D ı  P v D .P ı/  v. The second assertion is a direct consequence of (11.25). 11.11 According to Theorem 11.2 the image under the mapping is C01 .Rn /, while Theorem 11.17 implies that it is continuous. The desired equality follows from (11.23), in view of t.S v/u./ D u.S v  / D .u  v/./, for all  2 C01 .Rn /.

374

21 Solutions to Selected Problems

11.14 A short proof of the formula for .H /.k/ may be given by mathematical induction on k. On the other hand, one might apply Leibniz’s rule (9.1) and obtain .H /

.k/

! k  .k i 1/ ı .i / C  .k/ H D i C1 i D0 ! i ! k X1 X k i j i  .k j . 1/ D i C1 j i D0 j D0 ! ! k 1k X X1 k i j i  .k . 1/ D j i C1 k 1 X

1/

.0/ ı .j / C  .k/ H

j 1/

j D0 i Dj

In the second equality we used that for all . .k

i 1/ .i /

ı /. / D ı .i / . .k

i 1/

.0/ ı .j / C  .k/ H:

2 C01 .R/, / D . 1/i . .k

i 1/

/.i / .0/;

as well as Leibniz’s rule once again. Comparison of the two right-hand sides implies that the inner sum equals 1. 11.15 On account of (11.25), 1  ı 0 D 10  ı D 0 and so .1  ı 0 /  H D 0. On the other hand, ı 0  H D ı  H 0 D ı  ı D ı and so 1  .ı 0  H / D 1. This result does not violate (11.22), because neither of the distributions 1 and H has compact support. 11.19 (i). This assertion follows by means of (10.15). (ii). Select  2 C01 .R/ such that 1R ./ D 1. Since u./ D u.T t  / D u.T t /, for all t 2 R and  2 C01 .Rn /, we see that Z Z Z  u./ D .t/ dt u./ D .t/ u.T t / dt D u .t/ T t  dt : (21.11) R

R

R

For any s 2 R, consider the embedding s W Rn 1 ! Rn satisfying x 0 7! .x 0 ; s/. Thus, we have Z Z Z  0 .t/ T t  dt .x/ D .t/ .x ; xn t/ dt D .x 0 ; s/ .xn s/ ds R R R Z Z  s  .x 0 / Ts .xn / ds D s   ˝ Ts  ds .x/ D R RZ   D Ts .s  ˝ / ds .x/: R

(21.12) Note that in the fourth and fifth expressions, Ts denotes a translation acting in R. Hence, combining (21.11) and (21.12) as well as using the invariance of u under T once more, we may write Z Z Z  u./ D u.Ts .s   ˝ // ds D u.s   ˝ / ds D u s   ds ˝  R

R

R

21 Solutions to Selected Problems

375

D u.  ˝ / D v. / D .  v/./; if we use (10.21) in the fourth equality and introduce v 2 D 0 .Rn 1 / by v. / D u. ˝ /, for 2 D 0 .Rn 1 /. This demonstrates that u D   v. Finally, according to (11.14), .v ˝ 1R /./ D v.x 0 7! 1R .xn 7! .x 0 ; xn /// D v. / D   v./: 11.20 The proof is a generalization of that of Theorem 8.10. We describe its new aspects. u is of finite order, say  k, according to Theorem 8.8. Next, consider the Taylor expansion of the function on Rq given by z 7! .y; z/ to order k at 0, with y 2 Rp as a parameter, X z˛ .y; z/ D @˛ .y; 0/ C R.y; z/: ˛Š ˛2A; j˛jk

z z/ z ˛ with z 2 C 1 .Rn / Here R.y; z/ is a finite sum of terms of the form .y; and ˛ 2 A satisfying j˛j D k C 1. This implies u.R/ D 0. Note that .y; 0/ D ..y// D  .y/, where   2 C01 .Rp /. Furthermore, p˛ is a polynomial func˛ tion on Rq if p˛ .z/ D z˛Š . Replacing A by its finite subset f ˛ 2 A j j˛j  k g, we may write X D  @˛  ˝ p˛ C R: ˛2A

Next define

u˛ 2 E 0 .Rp /

by

u˛ . / D . 1/j˛j u.

˝ p˛ /

.

2 C01 .Rp //:

Since . 1/j˛j u˛ . @˛ / D .@˛ B  /u˛ ./, we obtain X X .@˛ B  /u˛ ./ C u.R/ D .@˛ B  /u˛ ./: u./ D ˛2A

˛2A

12.2 Successively using (11.4), (11.18), and (11.3), we obtain, for the potential u 2 D 0 .R3 / of the dipole f 2 E 0 .R3 /, X  X  X uD vj @j ıa  E D vj Ta ı  @j E D Ta vj ı  @j E j

D Ta Now @j E.x/ D

j

X j

 vj @j E ;

xj ; 4kxk3

so

where

j

E.x/ D

1 1 : 4 kxk

 hv; xi  1 hv; x ai D u.x/ D Ta x 7! ; 3 4kxk 4 kx ak3

for x ¤ a. In particular, for a D 0, v D e1 , and x in the .x1 ; x2 /-plane provided with polar coordinates .r; ˛/, we see that

376

21 Solutions to Selected Problems

u.x/ D

1 1 cos ˛ x1 D : 4 .x12 C x22 /3=2 4 r 2

Note that the equipotential lines satisfy a polynomial equation of degree 6. In the case of n D 2, E.x/ D

1 log kxk; 2

@j E.x/ D

1 xj ; 2 kxk2

so

u.x/ D

1 x1 : 2 kxk2

The equipotential lines now become circles centered on the x1 -axis and tangent to the x2 -axis; they satisfy .x1 c/2 C x22 D c 2 , where c 2 R. 12.3 In order to obtain concise formulas, we introduce some notation. Denote by S  R2 the strip parallel to the main diagonal, given by S D f .x1 ; 1 / 2 R2 j x1 2 R; x1 a  1  x1 C a g D f .x1 ; 1 / 2 R2 j 1 2 R; 1 a  x1  1 C a gI furthermore, .x/ D k.0; x2 ; x3 /k and ˇ.a; x/ D

x1 C a .x/

.x 2 R3 /:

p R Moreover, recall the antiderivative p dt 2 D log.t C 1 C t 2 /. Then, for any 1Ct  2 C01 .R3 /, we obtain, by changing the order of integration over S , Z 4 ua ./ D 4 .E  f /./ D 4 E.x/f .y/ .x C y/ d.x; y/ R3 R3 Z a Z 1 D .x1 C y1 ; x2 ; x3 / dy1 dx 3 kxk a R Z x1 Ca Z 1 .1 ; x2 ; x3 / d 1 dx D R3 kxk x1 a Z Z 1 .1 ; x2 ; x3 / d.x1 ; 1 / d.x2 ; x3 / D R2 S kxk Z Z Z 1 Ca 1 q dx1 .1 ; x2 ; x3 / d 1 d.x2 ; x3 / D R2 R  1 a x12 C .x/2 Z p D Œlog.t C 1 C t 2 /ˇ.a;x/ .x/ dx: ˇ. a;x/ R3

Hence

p 1 C ˇ.a; x/2 exp. 4 ua .x// D p ˇ. a; x/ C 1 C ˇ. a; x/2 p 1 C x1 a 1 C .x/2 a 2 C .1 C x1 a 1 /2 D : p 1 C x1 a 1 C .x/2 a 2 C . 1 C x1 a 1 /2 ˇ.a; x/ C

21 Solutions to Selected Problems

377

For the purpose of studying the behavior of ua as a ! 1, we note that 2 1 C h2 h8 C O.h3 / as h ! 0. This implies, as a ! 1, p .x/2 a

2

C .˙1 C x1 a

1 /2

D

p 1 ˙ 2x1 a

p 1ChD

C kxk2 a 2 1 D 1 ˙ x1 a 1 C .x/2 a 2 C O.a 2 1

3

/:

Therefore, as a ! 1, exp. 4 ua .x// D

2 C 2x1 a 1 2 2 .x/ a

1

4a2 C O.a 2 / D .1 C x1 a 2 C O.a 3 / .x/2

1

2

C O.a

//:

The unnormalized potential ua becomes unbounded as a ! 1. To remedy this, consider va D ua ua .e3 /. Then we have, for x 2 R3 with .x/ > 0 and as a ! 1, 4a2 .1 C x1 a 1 C O.a 2 // exp. 4 ua .x// exp. 4 va .x// D D exp. 4 ua .e3 // 4a2 .x/2 .1 C O.a 1 // 1 D .1 C O.a 1 //: .x/2 This leads to va .x/ D

1 log.1 C O.a 4

1 log .x/ 2

1

// D

1 log .x/ C O.a 2

1

/;

and accordingly lim va .x/ D

a!1

1 1 log .x/ D log k.0; x2 ; x3 /k: 2 2

Observe that in this manner we have obtained the fundamental solution as in (12.3) of the Laplacian
acting on R2 . 12.4 The first assertions follow in a straightforward manner. Green’s second identity (see [7, Example 7.9.6]) asserts that for u 2 C 1 .X / and  2 C01 .Rn / and in the notation of Example 7.3, Z Z .1U u
 1U 
u/.x/ dx D .ı@U u @  ı@U  @ u/.x/ dx: Rn

Rn

This implies the following identity in E 0 .Rn /:
.u 1U /

.
u/ 1U D

@ .u ı@U /

.@ u/ ı@U :

The function u being harmonic on X , it satisfies
u D 0; therefore one obtains the identity
.u 1U / D @ .u ı@U / .@ u/ ı@U in E 0 .Rn /. Note that the fundamental solution E 2 D 0 .Rn / of
as in (12.3) is a locally integrable function on Rn with sing supp E D f0g. Convolution of the preceding identity with E is well-defined on account of (11.31); alternatively, one might apply Theorems 14.34 and 14.33.

378

21 Solutions to Selected Problems

Furthermore, U is disjoint with sing supp E  .u ı@U /. Therefore, one deduces u 1U D .
E/  .u 1U / D E 
.u 1U / D E  @ .u ı@U / D

Hence u.x/ D

@ E  .u ı@U /

Z

@U

 @ .y 7! E.x

E  .@ u/ ı@U :

y// u.y/

E  .@ u/ ı@U

 y/ @ u.y/ dy

E.x

.x 2 U /:

The vanishing of the integral on X n U can also be derived from the formula above.

12.6 According to Problem 12.5 the mean value of y 7! P .x C y/ over a sphere 1 2 centered at the origin equals P .x/; the function e 2 k
k is constant on such spheres, n while its total integral over Rn is .2/ 2 on the strength of Problem 2.7.

12.7 The direct approach to prove that a D 12 is as follows. Write, for arbitrary  2 C01 .R2 /, Z  1V ./ D 1V .  / D .@2t @2x / .x; t/ d.x; t/ V Z Z 1 Z t Z 2 D @ t .x; t/ dt dx @2x .x; t/ dx dt: R

jxj

R>0

t

Continuing this evaluation is straightforward but leads to tedious calculations. Instead, we discuss two alternative, more conceptual, approaches. First, note that 2 the rotation as described
in the problem satisfies V t D .R>01 / and is given by 1 1 1 p the matrix D D D .y/ D , 1 1 2 SO.2; R/. In particular, . D .y// 2

and so det D .y/ D 1, for all y 2 R2 . By application of Problem 10.7 we obtain, for 1  j  2, 1 1  .@j2 / D p .@1 C . 1/j @2 /  .@j / D .@1 C . 1/j @2 /2  ; 2 2
1  2 .@1 @2 /2   D 2 @1 @2 .  /: H) . / D .@1 C @2 / 2 Hence, the Change of Variables Theorem (see [7, Theorem 6.6.1]) implies Z Z  1V ./ D 1V . / D  .x/ dx D  .x/ dx D D

Z

V

R2 >0

2

Z

 . /.y/j .y/ dy D 2

R>0

Z

R>0

.R2 >0 /

Z

@1 .@2 .  //.y1 ; y2 / dy1 dy2 R>0

@2 .  /.0; y2 / dy2 D 2  .0/ D 2 . .0// D 2 .0/ D 2 ı./:

Next, we describe the second method. Consider the vector field

21 Solutions to Selected Problems

379

v D S grad  W R2 ! R2

with

SD

0 1 10

2 Mat.2; R/:

Introducing the notation J D 2 , the rotation in R2 by =2, and .g1 ; g2 / D .g1 ; g2 / for a vector field g, we have SD

0 1

1  1 0 0

0 DJ B 1

:

Furthermore, J 2 SO.2; R/ implies curl v D div.tJ v/ D

div.tJJ grad / D

div. grad  / D .@22

@21 / D  :

It is a straightforward verification that a positive parametrization y W R ! @V is given by y.s/ D .s; sgn.s/ s/, for s 2 R. Differentiation then leads to the following formula for Dy.s/, where we note that sgn is a locally constant function: Dy.s/ D



1  sgn.s/

and

Dy.s/ D sgn.s/S Dy.s/

.s 2 R n f0g/:

Next we obtain, on account of the chain rule and tS D S , for s 2 R n f0g, . B y/0 .s/ D D.y.s// Dy.s/ D sgn.s/hgrad .y.s//; SDy.s/ i

D sgn.s/h .S grad / B y.s/; Dy.s/ i D sgn. s/h v B y; Dy i.s/:

On the basis of Green’s Integral Theorem and the compact support of  we obtain Z Z Z Z  .x/ dx D curl v.x/ dx D h v.y/; d1 y i D h v B y; Dy i.s/ ds V V @V R Z 0 Z sgn. s/ . B y/0 .s/ ds D . B y/0 .s/ ds D R 1 Z 1 . B y/0 .s/ ds D .y.0// . .y.0/// D 2.0/: 0

It is straightforward that supp E D V and sing supp E D @V . For the last two assertions, we verify that the condition of Theorem 11.17 is satisfied with A D V and B D H or, differently phrased, that the sum mapping ˙ W V  H ! R2 is proper. Indeed, consider arbitrary .x; t/ 2 V and .x 0 ; t 0 / 2 H and suppose that .x C x 0 ; t C t 0 / belongs to a compact subset in R2 . Then x C x 0 and t C t 0 belong to compact subsets in R. From t  0 and t 0  t0 we get that both t and t 0 belong to a compact subset of R. But now jxj  t implies that x belongs to a compact subset of R, which finally implies that x 0 belongs to a compact subset of R. This proves the claim. The theorem says that u is a well-defined element of D 0 .R2 /, while (11.25) leads to  u D . E/  f D f .

380

21 Solutions to Selected Problems

12.9 Since
u D 0, it follows from Example 12.8 that u 2 C 1 .X /. But for smooth functions u the result is classical. 12.10 Since u is homogeneous of degree 0 as well as invariant under rotations, it follows from Theorem 10.17 and Problem 10.11.(ii) that u satisfies the system of partial differential equations n X

j D1

xj @j u D 0

.xj @k

and

xk @j /u D 0

.1  j < k  n/:

Hence, for all 1  j  k  n, we have xk xj @j u D xj2 @k u, which implies kxk2 @k u D 0:

Thus

@k u D 0

on Rn n f0g

.1  k  n/:

On account of Problem 12.9 and Theorem 8.10 we obtain c 2 C and a polynomial function p on Rn such that u D c C p.@/ ı. Problem 10.16 then leads to p D 0. 12.11 For the first assertion, iterate (12.5). Note that on D 0 .R2 /, @z @zN D

1 .@x 4

1 2 1 .@x C @y2 / D
: 4 4

i @y /.@x C i @y / D

Hence 1 4 1
.log k.x; y/k/ D @zN @z log.x 2 C y 2 / 2 2 2    1   1  1 1 1  x iy  @x i @y log.x 2 C y 2 / D @zN 2 D @ : D @zN D @ z N z  2  x C y2 z  zN

ıD

12.12 The idea underlying the solution is that, formally speaking, the equation grad f D g in .D 0 .Rn //n leads to
f D div g, which in turn implies that f D E 
f D E  div g 2 D 0 .Rn /. Select  2 C01 .Rn / such that  D 1 on an open neighborhood of the closed ball around 0 of radius r. Now introduce n X gzj D  gj and fz D @j E  gzj : j D1

Note that fz is well-defined, because the gzj have compact support. On account of the identity
E D ı as well as the integrability conditions satisfied by the gj , this leads to n n n X X X @k fz D @k @j E  gzj D @j2 E  gzk C .@k @j E  gzj @j2 E  gzk / j D1

D gzk C

n X

j D1

j D1

@j E  .@k gzj

j D1

@j gzk / D gzk C

n X

j D1

@j E  ..@k / gj

.@j / gk /

21 Solutions to Selected Problems

381

DW gzk C hk : Take E as in (12.3), note that supp ..@k / gj .@j / gk /  Rn n B.0I r/, and apply Theorem 11.32. One obtains sing supp hk  Rn n B.0I r/. Furthermore, in view of hk D @k fz gzk , one has on B.0I r/, for all j and k, @j hk

@k hj D @k gzj

@j gzk D @k gj

@j gk D 0:

Applying Poincar´e’s Lemma in the classical setting (see [7, Lemma 8.2.6]), one finds H 2 C 1 .B.0I r// such that in C 1 .B.0I r//, @k H D hk

.1  k  n/:

Any two such solutions H differ by an additive constant. Hence, one may assume that solutions H and H 0 associated with balls B.0I r/ and B.0I r 0 /, for r < r 0 , coincide on B.0; r/. It follows that f WD fz H defines a distribution on B.0I r/ satisfying @fk D gk on B.0I r/, for 1  k  n, while any two such solutions f agree on the intersection of open balls centered at 0. Therefore Theorem 7.6 implies the existence of f 2 D 0 .Rn / with the desired properties; indeed, it is of the form given in the problem. 12.13 (i). PBecause the power series expansions of g and h involve only powers of kxk2 D j3D1 xj2 , both g and h belong to C 1 .R3 /. According to Problem 4.7, the 1 function x 7! kxk on R3 n f0g defines an element of D 0 .R3 /, and therefore E˙ does too. (ii). By direct computation or by using a computer algebra system such as Mathematica one directly verifies that h is a solution of
the homogeneous equation and 1 furthermore that the identities for grad g, grad k
k on R3 n f0g, and
g are valid.
1 On the basis of Problem 4.7 we have
kxk D 4 ı. Furthermore, Leibniz’s rule (9.1) may now be used to obtain the second identity (compare with [7, Exercise 2.40.(i)]), which then leads to  1 
g .x/ D k
k

4 g.x/ ı C2k

h.x/ kxk

k2

g.x/ kxk

2k

h.x/ g.x/ D k2 kxk kxk

4 ı:

(iii). According to the well-known formula for the Laplacian in spherical coordinates in R3 (see [7, Exercise 3.8.(vi)]), one has 2 .
f /0 .r/ D f000 .r/ C f00 .r/: r (iv). The function fe0 W r 7! r f0 .r/ satisfies 0 fe0 .r/ D r f00 .r/ C f0 .r/;

and this implies

  2 00 fe0 .r/ D r f000 .r/ C f00 .r/ ; r

382

21 Solutions to Selected Problems 00 fe0 .r/ D r.
f /0 .r/ D

k 2 r f0 .r/ D k 2 fe0 .r/:

Accordingly, there exist constants a and b 2 C such that fe0 .r/ D r f0 .r/ D a cos kr C b sin kr;

so

f .x/ D a

g.x/ C b h.x/: kxk

(v). h 2 C 1 .R3 / is a classical solution of Helmholtz’ equation, as has been obg served in part (i). Therefore, one only needs to compute .
C k 2 / k
k in D 0 .R3 /. For arbitrary  2 C01 .R3 / one obtains Z  g g  g.x/ .
C k 2 / ./ D .
C k 2 /./ D .
C k 2 /.x/ dx k
k k
k R3 kxk Z g.x/ .
C k 2 /.x/ dx DW lim I : D lim #0 kxk> kxk #0
g Since .
C k 2 / k
k D 0 on the open set f x 2 R3 j kxk >  g, application of Green’s second identity leads to Z  g.y/ g.y/  I D @ .y/ .y/ @ d2 y: kyk kykD kyk For the evaluation of the integral, introduce spherical coordinates .r; !/ 2 R>0 S 2 for x 2 R3 and use dx D r 2 dr d!, where d! indicates two-dimensional integration over the unit sphere S 2 . On S 2 , considered as the boundary of the complement of the unit ball, the direction of the outer normal is the opposite of the radial direction; hence, ˇ kr sin kr C cos kr g.y/ ˇˇ cos kr @ D D @r : kyk ˇyDr! r r2 Conclude that Z Z Z I D  cos k @ .!/ d! k sin k .!/ d! cos k .!/ d! S2

DW

3 X

S2

S2

I.i / :

i D1

To estimate I.1/ , observe that for any y 2 R3 , j@ .y/j D jh grad .y/; .y/ ij  sup k grad .x/k DW m: x2R3

Hence jI.1/ j

 j cos kj m

Z

S2

d! ! 0;

for  # 0;

21 Solutions to Selected Problems

383

Z

jI.2/ j  j sin kj j sup j.x/j x2R3

I.3/ D

cos k .0/

Z

S2

d!

d! ! 0;

cos k

S2

Z

for  # 0;

..!/

.0// d!:

S2

On account of the Mean Value Theorem one has j.!/ which implies ˇZ ˇ ..!/ ˇ S2

.0/j   sup kD.x/k DW  M; x2R3

ˇ Z ˇ .0// d! ˇ 

S2

j.!/

Accordingly,  g  ./ D lim I D lim I.3/ D .
C k 2 / k
k #0 #0

.0/j d!   4M:

4.0/ D

4 ı./:

(vi). The results above imply that rotation-invariant fundamental solutions are obtained only by taking P a D 1 and b 2 C arbitrarily in part (iv). This amounts to c˙ D 1bi , so that ˙ c˙ D 1. 2

12.14 (i). Recall that limy#0 log y D 1 and limy#0 y j log yj D 0. Therefore the nontrivial case of the estimate occurs if x C iy 2 HC is near 0. If so, we have j log.x 2 C y 2 /j  2j log yj and thus ˇ ˇ y j logC .x C iy/j D y ˇ log jx C iyj C i arg.x C iy/ˇ y  j log.x 2 C y 2 /j C y  y j log yj C y: 2 Next, 2 jxyj  x 2 C y 2 leads to jyx iy 2 j 3 jxyj y2 1 y D 2  C  C1D : 2 2 2 2 2 jx C iyj x Cy x Cy x Cy 2 2 (ii). We have 2 @zN zN .z/ D .@x C i @y / D

N C1 X kD1

.k/

N X  .k/ .x/ .iy/k kŠ

kD0

 .x/ .iy/k .k 1/Š

1

N X  .k/ .x/ .iy/k .k 1/Š

kD1

1

D

 .N C1/ .x/ .iy/N : NŠ

(iii). This follows directly from Leibniz’s rule (9.1) and the fact that @zN f D 0, for all f 2 O.HC /, on account of (12.6). (iv). Identify g D g1 C ig2 with the following vector field on R2 :

384

21 Solutions to Selected Problems

gD

g  1

g2

then g D

;

 g  0 1 and J g WD 1 g2

1  g1   g2  D : 0 g2 g1

In addition, curl.g C iJ g/ D @x i.g1 C ig2 / @y .g1 C ig2 / D i.@x C i @y /.g1 C ig2 / D 2i @zN g: Application of Green’s Integral Theorem (see [7, formula (8.27) and Theorem 8.3.5]) now implies Z Z Z g.z/ dz D h.g C iJ g/.z/; d1 zi D 2i @zN g.x C iy/ dx dy: @R

@R

R

Here the second integral denotes the oriented line integral of a vector field on R2 . (v). The function z 7! zN .z/ f  .z/ is well-defined on R D Œ a; b   Œ 0; h  and continuous on R, while zN .x/ D .x/ and zN .z/ D 0 if z D a C iy or b C iy. Hence, successive application of parts (iv), (iii), and (ii) leads to Z

Z

b 

.x/ f .x/ dx

a

D 2i

Z

R

b a

zN .x C ih/ f  .x C ih/ dx D

@zN .zN f  /.z/ dz D 2i

Z

b a

Z

h

0

Z

@R

.zN f  /.z/ dz

R;N .x/ .iy/N f  .x C iy/ dx dy:

(vi). There exist a < b such that supp    a; b Œ . Because the functions x 7! zN .x C ih/ f  .x C ih/ are bounded on Œ a; b  uniformly in 0 <   1, Arzel`a’s or Lebesgue’s Dominated Convergence Theorem implies lim #0

Z

b a

zN .x C ih/ f  .x C ih/ dx D

Z N X .ih/k  .k/ .x/ f .x C ih/ dx: kŠ R

kD0

By definition of f 2 ON .HC /, we have that y C  7! .y C /N jf .x C iy C i /j is bounded, and so y 7! y N jf  .x C iy/j is uniformly bounded for 0 <   1. Accordingly, also in this case we obtain lim #0

Z

h 0

Z

b

a Z h

R;N .x/ .iy/N f  .x C iy/ dx dy

Z

b

R;N .x/ .iy/N f .x C iy/ dx dy 0 a Z Z 1 .ih/N C1 .N C1/  .x/ t N f .x C i th/ dt dx: D NŠ 0 R D

The assertion about the order of ˇC .f / is a straightforward estimate. (vii). Successively use part (vi) with N D 0, integration by parts, logC .x C i 0/ D log jxj C  i H. x/ for x 2 R, and (1.3) in order to obtain, for  2 C01 .R/,

21 Solutions to Selected Problems

1 ./ D x C i0

385

Z

Z

.x/ dx C  0 .x/.logC .x C ih/ logC .x C i 0// dx R x C ih R Z Z 0  .x/ logC .x C i 0/ dx D  0 .x/ log jxj dx D R R Z 0   1  i ı ./:  0 .x/ dx D PV i x 1

(viii). Because f 2 O.HC /, we have @z f D .@z C @zN /f D @x f on HC . Therefore Z Z ˇC .@z f /./ D lim .x/ @z f .x C i / D lim .x/ @x f .x C i / dx #0 R #0 R Z D lim  0 .x/ f .x C i / dx D ˇC .f /. 0 / D @x .ˇC .f //./: #0

R

(ix). ˇC .logC / and ˇ .log / agree along R>0 and differ by 2 i along R 0. In other words, p.r; z/ 2 G. Accordingly, for every r 2 R and a 2 C, the function fr;a is welldefined and complex-differentiable on H . For Re a  0 and x 2 R, we have that lim#0 fr;a .x i / exists and is a locally integrable function of the variable x; hence it defines a distribution on R. Next, suppose Re a > 0. Then jp.r; x C iy/j2 D p.r; x/2 C y 4 C 2.r 2 C x 2 /y 2

.x C iy 2 H /; (21.13)

which implies jp.r; x C iy/j  y 2 . Conclude from part (vi) that fr;a 2 ON .H / if N 2 Z0 satisfies N  Re a. In turn, this means that we have the boundary value fr;a .x i 0/ D ˇ .fr;a / 2 D 0 .R/, which is given by Z ˇ .fr;a /./ D lim .x/ fr;a .x i / dx . 2 C01 .R//: #0

R

In addition, it is a distribution of order at most N C 1.

13.1 Note that  D .1 a/ u  1C a . According to (13.4) and (13.3) we have aC 1 D C1  aC D ı 0  aC ; so another application of (13.4) and of (11.4) gives u D .u  1C a /  aC

1

D

1 .  ı 0 /  aC D .1 a/

1  0  aC : .1 a/

The reflection formula from Lemma 13.5 now leads to the desired formula. That u 2 C 1 .R/ follows from Theorem 11.2. 13.2 For Re a > 0 we have

386

21 Solutions to Selected Problems

a ./ D

Z

0

1

xa 1 . x/ dx D .a/

Z

1 0

xa 1 S.x/ dx D aC .S/: .a/

The identity now follows for all a 2 C by analytic continuation. In particular, it follows from (13.3) that  k ./ D Ck .S/ D ı .k/ .S/ D . 1/k ı .k/ ./.

2 C 1 .R/ such

13.4 Taylor expansion of  around 0 leads to the existence of that .x/ .0/ D x .x/. Hence xa

1

.0// D x a .x/:

..x/

This function is integrable on R>0 ; indeed, the right-hand side is locally integrable at 0 since 1 < Re a, and the left-hand side is locally integrable at 1 because d Re a 1 < 1. On account of (13.7) one has aC D dx aC1 C , where 0 < Re a C 1. Therefore, integration by parts gives Z xa 0 aC ./ D aC1  0 .x/ dx . / D C .a C 1/ R>0 i1 Z h xa 1 xa ..x/ .0// ..x/ .0// dx C D 0 .a C 1/ .a/ R>0 Z xa 1 ..x/ .0// dx: D .a/ R>0 More generally, for k 2 Z0 and 0 < Re.a C k C 1/ < 1, define kC1 D  .kC1/ Z x and also j .x/ D j C1 .t/ dt 0

by means of downward mathematical induction on 0  j  k. Using mathematical Rx induction and 0  .j / .t/ dt D  .j 1/ .x/  .j 1/ .0/ one proves j .x/ D 

Then x aCj j .x/ D

(

.j /

.x/

k Xj lD0

x aCj O.x k

j

x aCj O.x k

j C1

x l .j Cl/  .0/: lŠ

/ D O.x aCk / D o.1/;

x ! 1;

/ D O.x aCkC1 / D o.1/;

x # 0:

In the particular case of j D 0, the O-estimates also imply that x 7! x a 1 0 .x/ is integrable on R0 . On the basis of these estimates, .k C 1/-fold integration by parts leads to d kC1 aCkC1 aC ./ D  ./ D . 1/kC1 aCkC1 . .kC1/ / C dx kC1 C Z x aCk kC1 kC1 .x/ dx D . 1/ .a C k C 1/ R>0

21 Solutions to Selected Problems

D

k X

. 1/j C1

j D0

Z

D

h

i1 x aCj j .x/ C 0 .a C j C 1/

xa 1  .x/ .a/

R>0

 1 lim .a/ #0

D

387

Z

1

k X xj

j D0

xa

1



jŠ

Z

R>0

xa 1 0 .x/ dx .a/

  .j / .0/ dx

.x/ dx C

k X  .j / .0/ aCj   : j Š .a C j /

j D0

The last equality follows by meansˇof a simple integration. j ˇ Finally, (13.3) implies that C ˇ D 0 if j 2 Z0 , which is also the case for L ˇ j 1 .test x .j / H /ˇL because of the zero of 1 at j ; see Corollary 13.6.

13.6 Indeed, for  2 C01 .R/, 1 ./ D x

@x PV

PV

1 0 . / D x

lim #0

Z

RnŒ ;  

 0 .x/ dx DW x

lim I : #0

Integration by parts gives h .x/ i

I D



C

h .x/ i1

C

Z

.x/ dx x2

1 x x  RnŒ ;   Z 1 1 X 1X .˙/ C .˙x/ dx:  x2 

D

˙

˙

0 2 Next, Taylor expansion of  about P 0 implies .x/ D .0/ PC x  .0/ C x .x/, 1 2 2 C .R/. Hence ˙ .˙/ D 2.0/ C  ˙ .˙/. In view of Rwhere 1 1 1 dx D , one obtains  x2 

I D

and therefore

Z

1



lim I D #0

 1 X .˙x/ x2 ˙

Z

x R>0

2

X ˙

 2.0/ dx

.˙x/



X

.˙/;

˙

 2.0/ dx D jxj

2

./:

13.9 Using (13.20) and (13.21), we obtain outside C , D

 a . aC1 d d a ˇˇ 2 /  RC ˇ C D q n aD0 da da  2 .a/

nC1 2

ˇ ˇ ˇ

aD0

:

Next, we use Leibniz’s rule and Corollary 13.6 to conclude that outside C ,

388

21 Solutions to Selected Problems

D

. 12 / d n  2 da

 nC1   nC1  1 n 1 ˇˇ D  2 q  C 2 : q  C 2 ˇ .a/ aD0

1 n n 1
In the particular case of n 1 even, (13.3) implies  D  2 q  ı . 2 / outside C , which is a distribution with support contained in @CC . In view of (13.26) we have d a ˇˇ 2 q D q R ˇ D . n C 1/ RC D .1 n/EC : da C aD0 Finally, the last equality is immediate from (13.25). This equality in turn implies that  is not homogeneous. p 13.10 (i). Note that q.x; t/ < y implies jtj < kxk2 C y. Using Problem 10.23.(ii) and changing the order of differentiation and integration over R3 , we therefore get ˇ Z d ˇˇ  q .ı/./ D .x; t/ d.x; t/ dy ˇyD0 f .x;t /2R4 j q.x;t / 0,

21 Solutions to Selected Problems

c4

2

ıC .c  f / D

c2 2

Z

R3

389

1 f .cx; ckxk/ dx D kxk 2

Z

R3

f .x; kxk/ dx D ıC .f /: kxk

Recalling Problem 10.15.(ii), we deduce the homogeneity of  ıC of degree 4, and thus the argument can be finished by copying the proof of Theorem 13.3. (v). Consider  as given. Using a suitable cut-off function we can construct a function 2 C01 .R4 / that coincides with  on an open neighborhood of the compact set supp  \ supp ıC . The definition ıC ./ WD ıC . / is now independent of the particular choice of . For  as specifically given, we have Z 00 1 .kxk/ c .0/ D ıC . / D dx: 3 2 R kxk Employing spherical coordinates in R3 we thus obtain Z Z 1 00 0 r .r/ dr D 2 .r/ dr D 2 c .0/ D 4 2 R>0 R>0

.0/ D 2 .0/:

(vi). u D EC  f is a solution of the inhomogeneous equation according to Theorem 12.2. In view of (11.1) and Theorem 11.2, Z .T.x;t / Sf /.y 0 ; ky 0 k/ 0 1 EC  f .x; t/ D EC .T.x;t / Sf / D dy 4 R3 ky 0 k Z Z 1 1 f .x y 0 ; t ky 0 k/ 0 f .y; t kx yk/ dy dy: D D 4 R3 ky 0 k 4 R3 kx yk 14.3 We may assume that P is a function actually depending on its first variable, that is, that 1 7! P .1 ; : : : ; n / is a polynomial of degree at least 1. For every .2 ; : : : ; n / 2 Cn 1 , this polynomial function has at least one zero on account of the Fundamental Theorem of Algebra (see [7, Exercise 8.13], for instance). This proves that the zero-set is infinite. It follows that there are infinitely many  2 Cn such that P .@/ e D P ./ e D 0, while the e are linearly independent complexanalytic functions on Cn . 14.5 Straightforward verification, by writing an integral over Rn as iterated onedimensional integrals. 14.7 We first show that L .a/ D 0 if there exist 1 ; : : : ; n 2 S such that .x/ D

n X

.xj

j D1

aj /

2 S, a 2 Rn , and

j .x/

.x 2 Rn /:

.a/ D 0. Note that

(21.14)

d .a C t .x a//, Indeed, by writing .x/ .a/ as the integral from 0 to 1 of dt 1 z we obtain (21.14), with j 2 C instead of the j 2 S. Now choose  2 C01 with  D 1 on a neighborhood of a. Then

390

21 Solutions to Selected Problems j .x/

WD .x/ zj .x/ C .1

.x// .x/

satisfy (21.14). This implies L

X n DL .xj

aj /

j

j D1



D

n X

L.xj

j/

j D1

xj kx

aj ak2

L.aj

j/

.1  j  n/

D

n X

.xj

aj / L

j;

j D1

from which we conclude that L .a/ D 0 if 2 S and .a/ D 0. Now let 2 S be chosen such that is nowhere zero on Rn . For arbitrary  2 S,

2 S; then has a zero at a, and so define D  .a/

.a/ .a/ L .a/:

.a/

0 D L .a/ D L.a/

Because this holds for all a 2 Rn , the conclusion is, for all  2 S, L D c 

with

c WD

L 2 C 1 .Rn /:

Finally, we see that c is constant, because application of this identity with  successively replaced by and Dj implies

Dj c C c Dj D Dj .c / D Dj .L / D L.Dj / D c Dj ; and so Dj c D 0; in other words, Dj c D 0, for every 1  j  n.

14.8 For assertion (i) see the initial part of the solution to Problem 14.7 with a D 0. The two assertions are just Fourier transforms of one another. 14.9 Suppose  2 S.R/ satisfies F  D , for some  2 C. According to Theorem 14.13 we p have .2/2  D F .F S /.S F /F  D F 4  D 4 . Thus p 4 D 2 j x 2 =2 and  D 2. .2/ , that is,  D 2 i , for 0  j  3. Next, set .x/ D e Then F  D  according to Example 14.11. Note that @ D x . Using this, we see that F .x / D DF  D i @./ D i .x /. Finally, F

 x2

1   D 2

DF .x /  D   x2 

1 1  D i @. i x /  2 2    1 1  D  x2 : 2 2

More generally, for j 2 Z0 , define the Hermite polynomials Hj and the Hermite functions j on R (the j are also called the quantum harmonic oscillator wave functions) by 1 2 2 Hj .x/ D . 1/j e x @jx e x and j D p p  Hj : j 2 jŠ 

21 Solutions to Selected Problems

391

By means of integration by parts and the identity @Hj D 2j Hj 1 one sees that . j /j 2Z0 forms an orthonormal system in L2 .R/. In fact, it is a complete system; for a proof, see Problem 14.42. In addition, every j is an eigenfunction of the Fourier transform; more specifp ically, F j D 2. i /j j , for j 2 Z0 . Indeed, using integration by parts and Example 14.11 one obtains Z Z 1 2 1 2 2 2 j 1 2 F . Hj /./ D e x @jx e ixC 2 x dx D i j e 2  e x @ e 2 .x i / dx R R Z p 1 2 1 2 2 j j 1   ix x 2 e dx D 2. i /j . Hj /./: D i e 2 @ e 2 R

14.10 By analytic continuation it follows that the identity from Problem 12.6 is valid for all x 2 Cn . In particular, for x D i  with  2 Rn , we obtain Z 1 n 2 e 2 kyk P .y i / dy D .2/ 2 . i / P ./: Rn

Next an n-fold application of Cauchy’s Integral Theorem as in (12.9) leads to the third equality in the following: Z 1 1 1 2 2 kk2 k
k2 2 2 e F .e P /./ D e 2 .kyk C2i hy;i kk / P .y/ dy n R Z Z 1 1 n 2 2 kyCi k e 2 P .y/ dy D e 2 kyk P .y i / dy D .2/ 2 . i / P ./: D Rn

Rn

1

2

Observe that the argument above gives a computation of F e 2 k
k different from the one in Example 14.11. In addition, in the case of n > 1 Hecke’s formula produces infinitely many eigenfunctions similarly as in Problem 14.9 where n D 1. 14.11 Applying the estimate in Definition 19.4 with K D supp f , we obtain a  constant c 0 > 0 such that for arbitrary x 2 supp f and  2 Rn nf0g with x kk 2 2 supp f , j.I T  2  /f .x/j  c 0 kk ˛ : kk

Application of (14.6) and (14.5) then leads to the desired estimate. 14.12 Prove .1 C kk2 /

nC1 2

and deduce .1 C kk2 /

nC1 2

 nC1 X X  1C jj j  cn0 j ˛ j 1j n

jF f ./j  cn0 

cn0

X

j˛jnC1

X

j˛jnC1

j˛jnC1

j ˛ F f ./j D cn0 ˛

kD f kL1 :

X

j˛jnC1

. 2 Rn /;

jF .D ˛ f /./j

392

21 Solutions to Selected Problems n

Verify kf kL1  .2/ kf kL1

kF f kL1 by means of Theorem 14.13, and conclude that Z X nC1 00  cn .1 C kk2 / 2 d  kD ˛ f kL1 : Rn

j˛jnC1

14.13 Interchanging the order of integration implies, for all t > 0, Z t Z tZ Z Z t ix F ./ d  D e .x/ dx d  D .x/ e t t R Zt R sin tx dx: D 2 .x/ x R

ix

d  dx

The desired equality for the limit now follows on account of Definition 14.19 and Theorem 14.13. 14.14 (ii). We have a 2 S. On the other hand, the functions b and c do not belong to S because they are not differentiable at 0, while d is not in S since x 7! x 3 d.x/ is unbounded on R. All functions belong to L2 . (iii). Since all functions belong to L1 .R/, their Fourier transforms are given by 2 (14.3). Note that a.x/ D e e .x 1/ . Hence Example 14.11 leads to Z Z 2 ix .x 1/2 e dx D e e i.xC1/ e x dx F a./ D e e R

D e1

i

Z

R

e

ix

e

2x 2 =2

R

dx D e 1

i

p 1 2 p e 2

2 4

D

p e

. 2 Ci /2

:

1 follows from Example 14.30 or by direct computation. Further, F b./ D i C1 observe that c D b C S b. Since F S D S F , this means that

F c./ D F d./ D  e

jj

1 2 1 C D : 1 C i 1 i 1 C 2

. Indeed, note that d D 12 F c; therefore, Fourier inversion gives

Fd D

  1 1 2 F cD S F F .Sc/ D  Sc D  c: 2 2

On the other hand, F d is given by (14.3). Combination of these two results immediately produces Laplace’s integral. (v). Part (ii) taken in conjunction with Theorem 14.32 gives that all Fourier transforms belong to L2 . Together with Theorem 14.13 it also implies that F a 2 S, whereas the remaining Fourier transforms do not. Finally, F b … L1 but the remaining do belong to L1 . 14.15 On account of Problem 4.1 we have @f .x/ D f .x/ @jxj D .f sgn/.x/. Hence, @2 f D @f sgn 2f ı D f sgn2 2f .0/ı D f 2ı:

21 Solutions to Selected Problems

393

So D 2 f C f D 2ı. Since F ı D 1, this means that . 2 C 1/F f ./ D 2, that is, 2 F f ./ D 1C 2 (compare with Problem 14.14.(iii)). Next, note that i D arctan x D 1 . 1Cx 2

Hence, i  F arctan  D e jj ; therefore Theorem 9.5 implies the existence of c 2 C such that  e jj F arctan D PV C c ı: i  Since arctan is odd, so is F arctan, which leads to c D 0.

14.17 We have f0 .x/ D x d.x/ and so F f0 D i @F d according to (14.28). Now F d./ D e jj , and so the solution to Problem 14.15 leads to the desired formula for F f0 . Next note that f f0 D g, where g 2 L1 .R/. The Riemann–Lebesgue Theorem, Theorem 14.2, then implies that F f F f0 D F g is continuous. Q 14.18  2 S.Rn / is a direct consequence of x ˇ @˛ .x/ D jnD1 x ˇj @˛j j .xj /, while the identity cn D .c1 /n results from a similar computation. Write f .x/ D e jxj . From Problem 14.14.(iii) or 14.15 we obtain F f ./ D 2=.1 C  2 /; in particular, F f 2 L1 .R/. Select  2 C01 .R/ satisfying   0 and 1./ D 1, and define  .x/ D  1 .x=/. Now F  ./ D F . / and jF j  1./ D 1 imply jF  j  1. Furthermore, Z jF  ./ 1j  je i x 1j .x/ dx R

implies F  ! 1 uniformly on compacta as  # 0. This in turn leads to lim kF  F f #0

F f kL1 D 0:

Indeed, split the integration into two parts: R one over a suitable compact set, the other part over its complement C such that C F f ./ d  is sufficiently small. Since  2 C01 .R/  E 0 .R/ and f 2 L1 .R/  S 0 .R/, we have F .  f / D F  F f on account of Theorem 14.33. But F  belongs to S.R/ and the properties of F f now imply F  F f 2 S.R/ and so   f 2 S.R/ according to Theorem 14.13. From the proof of that same theorem we get the existence of c1 > 0 such that G . / D c1 , for all 2 C01 .R/. In particular, we obtain, for all  > 0, Z F  ./ F f ./ d  D c1 .  f /.0/: R

Now we are sufficiently equipped to take the limit in this identity as  # 0; thus we Z Z obtain 2 2 D d  D F f ./ d  D c1 f .0/ D c1 : 2 R 1C R

14.24 By definition of S 0 .Rn / there exist, for every u 2 S 0 .Rn /, constants c > 0 and k, N 2 Z0 such that for all a 2 Rn and  2 S.Rn /, ju.Ta /j  c kTa kS.k;N / D c

sup j˛jk; jˇ jN; x2Rn

jx ˇ .@˛ /.x

a/j

394

21 Solutions to Selected Problems

Dc

sup j˛jk; jˇ jN; x2Rn

j.x C a/ˇ .@˛ /.x/j

! ˇX ˇ ˇ Dc sup x aˇ ˇ

j˛jk; jˇ jN; x2Rn 0

 c .1 C kak/

N

ˇ

sup

j˛jk; j jN; x2Rn

ˇ ˇ .@˛ /.x/ˇ

jx .@˛ /.x/j  d .1 C kak/N ;

for a suitable d > 0. Now let  2 Cn and suppose that ei  2 S 0 .Rn /. Then ei  .Ta / D .Ta  ei  /./ D e i ha; i ei  ./; hence we have jei  .Ta /j  d .1 C kak/N H) e i ha; i D O.kakN /;

kak ! 1 H)  2 Rn :

14.27 (i). Formula (14.36) immediately leads to the identity F ei a D 2 ıa in S 0 .R/, for all a 2 R. It follows that we have in S 0 .R/, e C e  i i D .ı 1 C ı1 /; F cos D F 2 e  e i i F sin D F D  i.ı 1 ı1 /; 2i F sin D i F D cos D  i .ı 1 C ı1 / D  i.ı 1 ı1 /; DF .sin/ D i @. i.ı 1 ı1 // D .ı 0 1  1 .F cos/  .F cos/ D .ı 1 C ı1 /  .ı F cos2 D 2 2  D .ı 2 C 2ı0 C ı2 /: 2

ı10 /;

F .x 7! x sin x/ D

1

C ı1 /

In the third identity we have applied (14.28) and Example 9.1, in the fourth (14.28) once again, and in the fifth (14.22) and Theorem 14.33. (ii). We have F sinc D  1Œ 1; 1 . Indeed, this follows by Fourier inversion from F .1Œ 1; 1 / D 2 sinc; see Example 14.1. Another way to prove this is as follows. On account of Example 4.2 we get i @ F sinc D F .x sinc/ D F .sin/ D i .ı

1

ı1 / D i @  1 Œ

1;1 

:

Therefore there exists c 2 C such R that F sinc D  1Œ 1; 1 C c, and the wellknown evaluation F sinc.0/ D R sinc x dx D  (see Problem 14.44 or 16.19 or [7, Example 2.10.14 or Exercise 0.14, 6.60, or 8.19]) then leads to c D 0. (iii). Applying (14.38) and Problem 13.6, we see that  1 F .x H / D i @ F H D i @  ı i PV D  i ı0 j
j 2:  Observe that jxj D x sgn x; hence (14.37) and Problem 13.6 lead to F j
j D i @ F sgn D i. 2i / @ PV

1 D 2j
j 

2

:

21 Solutions to Selected Problems

395

We have sin jxj D sgn.x/ sin x D

e ix

e 2i

ix

sgn.x/:

Next, apply the first identity in (14.29) to obtain 1 1 PV : F sin j
j D PV  C1  1 Note that 2 j
j D S F F j
j D F S. 2j
j

2

/ D 2F j
j

2

.

14.28 Suppose R ¤ 0. Then application of (14.28) leads to I DF

1

  7!

1 2 L .i / C R i  C

1 C

 :

The roots of the polynomial in the variable i  in the denominator are r R˙iQ 4L ; where QD R2 : 2L C

(21.15)

If these two roots are distinct, we apply partial-fraction decomposition to obtain   1 1 1 1 I D F  7! : Q Q R R iQ i  C 2L i 2L i  C 2L C i 2L Thus, analytic continuation of (14.39) to f  2 C j Re  > 0 g immediately implies r 2 2 Qt t 4L R R t t 2L 2L I.t/ D H.t/ e D q R2 : sin H.t/ e sin Q 2L 2L C 4L 2 R C

2 In the case of R2 < 4L C , this is the desired formula for I . If, however, R > rewrite this expression as r 2 t 4L R t 2L I.t/ D q : H.t/ e sinh R2 2L C R2 4L

4L C ,

we

C

The case of R D 0 is obtained by taking limits for R ! 0 in the preceding formulas. Finally, if the roots in (21.15) coincide, we have R2 D 4L > 0 and C 1 I D F L

1

  7!

1 .i  C

R 2 2L /

Hence (14.28) implies



1 D F L

1

  7! D

1 i C

R 2L

 :

1 R H.t/ t e 2L t : L 14.30 (i). In view of Example 10.14 we have, for any u 2 S 0 .Rn / and  2 S.Rn /, I.t/ D

396

21 Solutions to Selected Problems

F .A u/./ D .A u/.F / D u.A .F // D

1 u..A j det Aj

1 

/ F /:

By means of the change of variable x D B 1 y D tAy we obtain, for every  2 Rn , Z Z 1 .A 1 / F ./ D F .A 1 / D e i hx; A i .x/ dx D e i hBx; i .x/ dx Z D e i hy; i .tAy/j det Aj dy D j det Aj F ..tA/ /./: On the basis of Example 10.14 again, this leads to F .A u/./ D u.F ..tA/ // D .tA/ .F u/./ D D j det Bj B  .F u/./:

1 .tA j det tAj

1 

/ .F u/./

(ii). The identity is a direct consequence of part (i). For the homogeneity of F u, note that c  F u D c n F B .c 1 / u D c n F c a u D c n a F u. (iii). If A is orthogonal, then tAA D I ; so B D tA 1 D A and j det Aj D 1, which implies F B A D A B F . Using the injectivity of F we see that A u D u, for all orthogonal A, if and only if A .F u/ D F u, for all such A. (iv). Part (i) implies A D j det.tA/ 1 j .tA/ 1 , which gives j det Aj tAA D I . In turn this leads to j det Aj D 1 and thus tAA D I . 14.32 The equivalence of (i) and (ii) follows from Problems 14.30.(ii) and (14.30), while Arzel`a’s or Lebesgue’s Dominated Convergence Theorem may be used to prove (i). 14.33 F u D P is a polynomial function on Rn that is invariant under all orthogonal transformations in Rn according to (14.33) and Problem 14.30.(iii). Therefore it is the pullback under k
k2 of a polynomial function P0 on R, that is, P ./ D P0 .kk2 /, for all  2 Rn . Indeed, a polynomial function P on Rn invariant under orthogonal transformations is constant on spheres centered at the origin, and these spheres intersect L WD R  f0g  Rn in a vector and its opposite. The restriction of P to L may be identified with a polynomial P2 on R invariant under reflection. The latter condition is satisfied if and only if P2 is a linear combination of monomials of even degree, that is, P2 .x/ D P1 .x 2 /, for some polynomial P1 on R0 . One directly verifies that the mapping P 7! P0 WD SP1 is a linear bijection. 14.35 For 1  i; j  n with i ¤ j , define the vector field vij on Rn by vij .x/ D xj ei , where ei 2 Rn is the ith standard basis vector; that is, the linear mapping vij W Rn ! Rn has a matrix Eij with all entries 0 with the exception of 1 at the position .i; j /. Then the flow .˚ t / t 2R in Rn generated by vij satisfies d ˚ t D Eij ˚ t ; dt

and therefore

˚ t D e t Eij I

see [7, Example 2.4.10]. It follows that det ˚ t D 1, because of [7, Exercise 2.44.(v)]; in other words, ˚ t 2 SL.n; R/. Now apply Theorem 10.16 to obtain xj @i u D 0.

21 Solutions to Selected Problems

397

For every connected component U of the complement of the union of the hyperplanes f x 2 Rn j xj D 0 g for all 1  j  n, Problem 12.9 then implies the existence of cU 2 C such that u D cU 1U on U . The invariance of u under rotations in Rn now gives u D c1 1Rn on Rn n f0g, for some c1 2 C. Next, according to Theorem 8.10 there exists a polynomial function p on Rn such that u D c1 1Rn C p.D/ ı 2 S 0 .Rn /:

Hence

F u D c1 .2/n ı C p 1Rn ;

while F u is invariant under SL.n; R/, too, on account of Problem 14.30.(i). On the strength of the preceding arguments we therefore obtain F u D p 0 .D/ ı C c2 1Rn . Since ı and 1Rn are homogeneous of degree n and 0 according to Problem 10.16, respectively, this gives that p equals the constant function c2 . In other words, u is as desired. Under the final assumption, u is also invariant under dilations. Now 1Rn possesses this invariance and ı not. Observe that SL.1; R/ D f.1/g and that all factors occurring in the Iwasawa decomposition of GL.n; R/ (see [7, Exercise 5.49]) play a role in this solution. P 14.36 (i) , (ii). Approximate klD1 cl ıxl by functions  2 C01 .Rn / to show that (i) ) (ii), and approximate integrals by Riemann sums to prove the converse. (i) and (ii) ) (iii). Note that the choices c1 D 1, c2 D c, x1 D 0, and x2 D x in (ii) imply .1 C jcj2 / p.0/ C c p.x/ C c p. x/  0: (21.16)

The choice c D 0 gives p.0/  0. Selecting c D 1 and c D i leads to p.x/ C p. x/ 2 R and i.p.x/ p. x// 2 R, respectively, which give p. x/ D p.x/. Furthermore, suppose p.x/ ¤ 0 and take c D jp.x/j p.x/ . Then we obtain that p.0/ jp.x/j  0 from (21.16). Thus jp.x/j  p.0/, for all x 2 Rn . As a consequence, p 2 S 0 .Rn /, and thus we may write p D F , for some  2 S 0 .Rn /. Furthermore, on account of Theorems 11.2 and 11.5 and (14.21) we have, for all  2 C01 .Rn /, 0  .p  S; S/ D .p  S/.S/ D p.  S/ D F .  S/ D .F  F .S// D .F  F / D .jF j2 /:

(21.17)

By continuity this inequality holds for all  2 S.Rn /. Therefore we obtain .; /  0, for all  2 S.Rn / satisfying   0; to verify this, approximate  by the square 2 1 of . C  e k
k / 2 2 S.Rn /, for  > 0. The term containing  is required because 1  2 itself is nondifferentiable at zeros of . On account of Theorem 3.18 we find that  is a positive Radon measure. Furthermore, if  is an even function as in Lemma 2.19, then (21.17) and Problem 14.30.(ii) imply p.0/ D lim.p   ;  / D lim .j  F j2 /  .1/: #0

#0

398

21 Solutions to Selected Problems

Thus,  is of finite mass and equality holds. (iii) ) (i). In this case, the identities in (21.17) lead to .p; / D .jF Sj2 /  0, for any  2 C01 .Rn /. 14.38 On account of (14.43) we have, for  2 C01 .R/, Z Z D 2 u./ D u.D 2 / D g.x; / f ./ d  D 2 .x/ dx R R Z g.x; / .D 2 ˝ I /. ˝ f /.x; / d.x; / D .D 2 ˝ I /g. ˝ f / D RR Z f ./ ./ d  D f ./: D
 1J . ˝ f / D J

The properties of the distribution g allow us to extend its action to locally integrable functions. In the particular case that f 2 C01 .R/ we may write Z Z u./ D g.x; / f ./ d  .x/ dx D g. ˝ f / D g..˝f // D t.˝f /g./: R

R

Problems 4.1, 14.27.(iii) or 14.39 imply that E is the desired fundamental solution. The solution E  ı D T E of the inhomogeneous equation follows from (11.24). The uniqueness of g follows from Theorem 9.4. Solving the boundary conditions from (14.43) for a and b leads to 1 g.x; / D .x 2x C  jx j/: 2 Note that g D 0 along the boundary @J 2 , where x D 0 or 1 or  D 0 or 1. 1 Now F E./ D  2 Cp 2 . In the case of p D 1, the formula for E follows from Problem 14.15; the case of general p > 0 then follows from Problem 14.30.(ii). The general solution of the homogeneous equation is given by x 7! ae px C be px . For the remainder, use Mathematica. a 14.43 According to (13.25) the RC for a 2 C are homogeneous distributions. a a Theorem 14.34 then implies that all of the RC belong to S 0 .RnC1 /. Since .RC /a2C a is a complex-analytic family of distributions and F is continuous linear, .F RC /a2C is a similar family. For the computation of the Fourier transforms we use a variation of the method of Example 14.30. For Re a > n C 1, define Ra on RnC1 ' Rn  R by

Ra .x; t/ D e

t

a RC .x; t/

. > 0/:

Then Ra 2 L1 .Rn /. Hence, for .; / 2 RnC1 and c.a/ as in Lemma 13.2, the Fourier transform of this function is given by Z a n 1 a F R .; / D c.a/ e i hx;i .i C/t H.t kxk/.t 2 kxk2 / 2 d.x; t/: RnC1

21 Solutions to Selected Problems

399

On account of the invariance of the integrand under rotations in Rn  f0g we may assume that  D kke1 with e1 the first standard basis vector in Rn . Upon the change to spherical coordinates in Rn 1 and with B tn D f x 2 Rn j kxk  t g and cn 1 as in (13.37), the integral becomes Z Z a n 1 e i hx;i .i C/t .t 2 kxk2 / 2 dx dt R>0

D

Z

B tn

Z

R>0

D cn

Z

1

t

t

Z

n 1 Bp t 2 y2

R>0

Z

t t

e

Z pt 2

iykk .i C/t

R>0

D

Z

Z

Z

1 1 1



1

rn

s2/

.1 2

.1

a n 1 2

dz dy dt

2

iykk .i C/t

e

.t 2

y2

r 2/

a n 1 2

dr dy dt:

s 2 . Then the triple integral takes the form

a 2 2

n

0

n 2

kzk2 /

0

1

ta

y2

y2

p Next set y D ts and r D t 1 Z

.t 2

 /

a n 1 2

d

0

Z

2

2 /

.1

a n 1 2

1 2

.1

s /

a 2 2

1

e

Z

.i skkCi C/t

ta

1

d ds dt

.i skkCi C/t

e

dt ds:

R>0

In view of (13.34) and (13.30), respectively,

Z

Z

1

n

2

ta

1

2 /

.1

a n 1 2

0

e

d D

.i skkCi C/t

R>0

1 n 1 a B ; 2 2

n C 1 ; 2

dt D .i skk C i  C /

a

.a/:

Thus we obtain, for F Ra .; /, c.a/ cn

1

1 n 1 a B ; 2 2

n C 1 .a/ 2

Z

1

.1

s2/

a 2 2

1

.i skk C i  C /

a

ds:

The remaining integral equals B Here we used that Z 1 .1 1

s2/

a 1 ; .kk2 C .i  C /2 / 2 2

a 2 2

.i ˛s C ˇ/

a

ds D B

a 2

:

a 1 ; .˛ 2 C ˇ 2 / 2 2

a 2

;

which can be seen by expanding .1 . i ˇ˛ s// a in its binomial series in s, see [7, Exercise 0.11], interchanging integration and summation and summing the resulting binomial series. If necessary, we have to use analytic continuation in ˛ and/or ˇ. Combining the preceding results and Legendre’s duplication formula (13.39), we

400

21 Solutions to Selected Problems

obtain F Ra .; / D .kk2 C .i  C /2 /

a 2

:

Application of Problem 12.14.(xi) now leads to the desired formula. Now we give a more conceptual proof. The subgroup Loı of the Lorentz group Lo consisting of all proper orthochronous Lorentz transformations, i.e., all Lorentz transformations that preserve the (solid) forward cone CC , acts transitively on each of the connected components of the level sets of the quadratic form q with the exS ception of q 1 .f0g/ D ˙ @C˙ ; Loı does, however, act transitively on each of the three sets @C˙ n f0g and f0g. a In view of Problem 14.30.(i) and (ii) it follows that F RC is invariant under the ı ı action of Lo (because Lo is invariant under taking transposes) and homogeneous of degree .n C 1/ .a n 1/ D a, respectively. Now S suppose Re a < 0 and let X denote the set of connected components of RnC1 n ˙ @C˙ . It follows that a F RC 2 S 0 .RnC1 / has to be of the form X a a F RC D fX .a/jqj 2 1X C ua ; X2X

where fX .a/ 2 C and uaS2 S 0 .RnC1 / is Loı -invariant, homogeneous of degree a and satisfies supp ua  ˙ @C˙ D q 1 .f0g/. Define  W Rn  R ! Rn  R by .; / D .; /. Now C 2 S 0 .RnC1 n f0g/ that is defined as ua on Rn R>0 and   ua on Rn R 0, because ua is homogeneous of degree a; thus w must equal 0 on account of Problem 10.16. This implies C D 0. By a similar argument  2 S 0 .RnC1 n f0g/ that equals ua on Rn  R0 must be equal to 0. Therefore ua vanishes away from the origin. The homogeneity of ua now implies that ua D 0 in S 0 .RnC1 /. a Our next goal is the evaluation of the fX .a/. Note that F RC is a continuous a b function, since Re a < 0. Therefore the product F RC F RC is well-defined if Re a and Re b < 0. Although the conditions in Theorem 14.33 are not satisfied, its cona b a b clusion F .RC  RC / D F RC F RC nevertheless holds in this case, as we will now 1 nC1 / is dense in S 0 .RnC1 /, there exists a sequence . kb /k2N in prove. Since C0 .R b in S 0 .RnC1 / as k ! 1. Let  2 S.RnC1 /. Then C01 .RnC1 / such that kb ! RC a b a b a b a  RC /./ D .RC  RC /.F / D RC .SRC  F / D RC . lim S F .RC k!1

b k

 F /:

Here S denotes reflection in the origin. Let  be a C 1 function that is supported on a sufficiently small open neighborhood of CC and equals 1 on CC . Then we have b that  .S kb  F / converges to  .SRC  F / in S.RnC1 / as k ! 1. So   a b a a lim  .S kb  F / D lim  RC  RC /./ D RC .S kb  F / F .RC k!1

k!1

21 Solutions to Selected Problems a D lim RC .S k!1

b k

401

a  F / D lim .RC 

b k /.F

k!1

a / D lim F .RC 

b k /./:

k!1

Now Theorem 14.33 can be applied to obtain a b  RC /./ D lim .F F .RC k!1

b k/F

a a b RC ./ D F RC F RC ./:

For the last equality, Arzel`a’s Dominated Convergence Theorem has been used. As a consequence of (13.17) we now obtain aCb a b a b F RC D F .RC  RC / D F RC F RC I

fX .a C b/ D fX .a/fX .b/:

so

Note that fX has an analytic continuation to C; therefore, fX .a/ D fX .1/a on account of [7, Exercise 0.2 (iii)]. In addition F RC2 D F . ı/ D q F ı D q; therefore fX .1/ 2 D fX . 2/ D sgn qjX . This leads to fX .1/ 2 f˙i g if X SD C˙ and fX .1/ 2 f˙1g if XSequals one of the connected components of RnC1 n ˙ C˙ . Let X0 D RnC1 n ˙ C˙ . Observe that X0 is connected if and only if n > 1. If n D 1, then X0 consists of two connected components X1 and X2 . Due to the a Loı -invariance, the functions fX1 and fX2 coincide. Now suppose n > 1. Since RC 1 is not invariant under reflection of t, this precludes the invariance of F RC . Since X0 is invariant under reflection of t, it follows that fCC .1/ and fC .1/ are distinct; hence fCC .1/ D fC .1/. k
k2 From now on we will assume n > 1. Let  W RnC1 ! R be the function e 2 . Then   D .n 1 C q/. Therefore RC2 ./ D ı. / D n

1 > 0:

a a The function R 3 a 7! RC ./ is real-analytic and real-valued; hence RC ./ > 0 for a in a sufficiently small open neighborhood of 2 in R. In addition, since F  D n .2/ 2  and  is invariant under reflection in t, we find for a with Re a < 0 that n

a a a .2/ 2 RC ./ D RC .F / D F RC ./ D fX0 .a/. q/

a 2

1X0 ./:

a

Note that . q/ 2 1X0 ./ > 0 for a < 0. Therefore fX0 .a/ > 0 for a in a sufficiently small open neighborhood of 2 in R and thus fX0 .a/ D 1 first for these a and then for all a 2 C by analytic continuation. To determine fC˙ .1/ we consider some integrals. For Re a > n C 1, Z a n 1 a fC˙ .a/ D F RC .0; ˙1/ D lim c.a/ e .˙i C/t .t 2 kxk2 / 2 d.x; t/; #0

CC

which by the first part of the calculation in the preceding proof equals lim c.a/ cn #0

1

1 n 1 a B ; 2 2 

D .a/e i 2 a ;

n C 1 .a/ 2

Z

1

.1 1

s2/

a 2 2

.˙i C /

a

ds

402

21 Solutions to Selected Problems

where is a positive function. Because the absolute value of fC˙ .a/ equals 1,

must be equal to the constant function 1. By analytic continuation fC˙ .1/ D  e i 2 D i .

14.44 From Example 14.1 or Exercise 14.27.(ii) we obtain F sinc D  1Œ 1; 1 . R Evaluating this identity at 0 gives R sinc t dt D , while Parseval’s formula from R Theorem 14.32 implies R sinc2 t dt D , since sinc 2 L2 .R/. The substitution of variables t D ax leads to the desired formulas. An extension of (14.22) to the case at hand in combination with Exercise 1.6.(ii) implies F sinc2 ./ D 2 maxf 0; 2 jj g; hence, the integral now follows by setting  D 0. 14.49 (i). Note that H  2 C 1 .R/ on account of Theorem 11.2. From Example 14.29 it is known that PV y1 2 S 0 .R/. Therefore Theorem 14.33 implies H  2 S 0 .R/ as well as F B H ./ D F .H / D 1 F  F PV y1 D i sgn B F , in view of (14.37). (ii). Apply Parseval’s formula (14.20) in order to obtain kF .H /k D kF k D p 2 kk. This leads to F .H / 2 L2 .R/, and invoking Parseval’s formula once more, one gets the desired equality. For the extension of H to a continuous linear e mapping from L2 .R/ into itself, one copies the proof of Theorem 14.32 with F 2 replaced by H . Finally, H D I follows by means of Fourier inversion from F B H 2 D .F B H / B H D

i sgn B F B H D .sgn/2 B F D

F:

(iii). From Problem 14.15 we obtain F f D  e j
j , and Problem 14.17 then leads to F g D i sgn B F f D F H f . (iv). From Problem 14.27.(i) it is known that F cos D .ı 1 C ı1 / and F sin D  R i.ı 1 ı1 /; this implies F cos D i sgn B F sin. On the other hand, the identity R sinc y dy D  from Problem 14.44 leads to Z Z Z cos sin cos.
y/ cos y sin y 1 PV dy D PV dy C dy H cos D  y y   R R R y D sin : Finally, use H 2 D I . (v). The first formula follows from the definition of H u and Problem 1.1 and the second from part (i) and (14.4), the third from the identity kH uk2 D kuk2 in part (ii) and the fourth from Parseval’s formula and the third. (vi). The former equality follows by the inverse Fourier transform from the second and first in part (iv). Finally, for the latter, note that for  2 S.R/, Z Z Z ˇx aˇ  sin a
 sin a ˇ ˇ i ./ D .x/ log ˇ e ix .x/ dx d  ˇ dx D F xCa j
j R R jj R Z Z Z Z sin a sin a d  dx D i d  dx: D .x/ e ix .x/ sin x jj jj R R R R

21 Solutions to Selected Problems

403

Actually, one should insert a convergence factor in the variable  in order to justify the interchange in the order of integrations, and use a limit argument. The resulting integral is of Frullani type; see [7, Example 2.10.8]. (vii). From F . sin
a
/ D  1Œ a; a  we get H . sin
a
/ D F 1 .  i sgn 1Œ a; a  /.

14.50 The integral equation can be written as a C b H D , which leads to aH b D H  on account of Problem 14.49.(ii). Therefore .a2 C b 2 / D a  b H , and this proves that 2 C 1 .R/. 14.52 On account of (14.29) one gets, for every c 2 R, F B Ta B H D e i a B i sgn B F D D F B H B Ta :

i sgn B e

ia

BF D

i sgn B F B Ta

Theorem 11.3 now implies the existence and uniqueness of u. In view of Problem 14.30.(ii) one obtains, for all c 2 R, F B c B H D c D

1

.c

1 

/ BF BH Dc

i sgn B c

1

.c

1 

1

/ BF D

.c

1 

/ B i sgn B F

i sgn B F B c  D F B H B c  :

On the one hand, one gets by applying (11.1), for all  2 C01 .R/, .c  B H /.0/ D H .0/ D u  .0/ D u.S/: And on the other, by Theorem 10.8, .H B c  /.0/ D H .c  /.0/ D u.Sc  / D u.c  S/ D c u.S/ D c 1 .c 1 / u.S/: In other words, u D c 1 .c 1 / u, that is, c  u D c 1 u. A combination of Euler’s Theorem, Theorem 10.17, and Problem 9.5 now implies the desired formula for u. 14.54 (i). Replace  > 0 in Example 14.30 by iz, where z D x C i t 2 HC ; then Re iz D t > 0. Hence, for  and x 2 R, H

iz ./

D e iz H./

and

FH

iz .x/

D

1

i.x

z/

:

As a consequence, the desired formula comes down to F H iz ./ D H (ii). If z 2 H , replace  by S and z by z 2 HC in order to obtain Z Z 1 .y/ dy D e iz F ./ d : i Ry z R0 2 R0

e ix

D 2i Z

R

y x jy zj2

D 2i

x y . .x y/2 Ct 2

Therefore

x y .y/ dy D Q t  .x/: .x y/2 C t 2

Z i e t  e ix F ./ d  2 R 0;  < 0 < x; x < 0 < :

This particular solution of the differential equation defines a tempered distribution on R, as is required by assumption, and from its description one obtains, for .x; / 2 R2 , F E.x; / D 2 sgn./H. x/ e x D 2 sgn./ H. x sgn.// e x : Consider  2 S.R2 /. Then y 7! .x; y/ belongs to S.R/, for every x 2 R. Since  7! F E.x; / defines an element in S 0 .R/, Theorem 14.24 now implies, for all x 2 R, 1 1 E.x;
/..x;
// D .F B S B F E.x;
//..x;
// D .F E.x;
//.S F .x;
//: 2 2 F E being a locally integrable and rapidly decreasing function on R2 , one obtains F E 2 S 0 .R2 / and thus E 2 S 0 .R2 / too. Accordingly, the admissible interchange of the order of integration yields Z 1 1 E./ D .F E/.S F / D .F E/.x; /.S F /.x; / d.x; / 2 2 R2 Z Z Z 2 sgn./ H. x sgn.// e x e iy .x; y/ dy dx d D 2 R R R Z Z Z 1 sgn./ H. x sgn.// e .xCiy/ .x; y/ d dx dy D  R R R Z Z  Z 0 Z 1  1 H.x/ e .xCiy/ d H. x/ e .xCiy/ d .x; y/ dx dy D  R R 1 0 Z Z Z 0 1  1 .x; y/ .x; y/  dx C dx dy D  R 0 x C iy 1 x C iy Z 1 1 1 1 1 D .x; y/ d.x; y/ D ./ D ./:  R2 x C iy  x C iy z Furthermore, Theorem 14.34 confirms once more the fact that E belongs to S 0 .C/. 1 Finally, suppose  > 0 and consider E.; / D 1 i C . Then, in the notation of Example 14.30, F E.; x/ D 2 sgn.x/ H.  sgn.x// e x D 2H . x/ D 2SH .x/: Equation (14.39) now follows by Fourier inversion; indeed,  E.;
/ D

1 F S F E.;
/ D F S 2 H D F H : 2

21 Solutions to Selected Problems

407

14.60 (i). Ra is continuous on Rn n f0g, for all a 2 C, and therefore it is locally integrable on this set. To prove the local integrability of Ra at 0 if Re a > 0, we integrate this function over the unit ball B centered at 0, using spherical coordinates (note that . a2 / ¤ 0 if Re a > 0): Z

B

jRa .x/j dx D c.a/ D

Z

B

kxkRe a

2 . a2 /

Re a

n

lim.1 #0

dx D lim #0

2 . a2 /

Z

1

r Re a

1

dr



 Re a / < 1:

Obviously, Ra is homogeneous on Rn nf0g if Re a > 0; according to Theorem 14.34 this implies that the distribution Ra is temperate. We may therefore test it with 2 e k
k 2 S.Rn /, which yields Z Z 2 2 2 2 kxka n e kxk dx D r a 1 e r dr Ra .e k
k / D c.a/ . a2 / R>0 Rn Z Z 1 1 1 2 a 1 a 2 e s 2 ds D s s s 2 1 e s ds D 1; D a a . 2 / R>0 2 . 2 / R>0 in view of (13.30). (ii). It is straightforward that q is a C 1 function satisfying Dq.x/ D 2 tx W Rn ! R, and the latter is ¤ 0 if x ¤ 0. Hence Dq.x/ is surjective for all x 2 Rn n f0g; in other words, q W Rn n f0g ! R is a C 1 submersion. The complex-analyticity of c follows from Corollary 13.6. Comparison of (13.5) with the definition of Ra directly implies the following identity of functions on Rn n f0g and hence also of distributions belonging to D 0 .Rn n f0g/:  a nC2  Ra D d.a/ q  C 2 ;

initially for Re a > 0, but then by analytic continuation of identities for all a 2 C. (iii). We now come to the first proof of the formula for
Ra . For x 2 Rn n f0g we have
Ra .x/ D c.a/
.kxka n / D c.a/ div..a n/kxka n 2 x/ D c.a/.n.a n/kxka n 2 C .a n/.a n 2/kxka a n 2

D c.a/.a 2/.a n/kxk D 2.a n/Ra 2 .x/:

D 2.a

n/c.a

n 2

/

a 2 n

2/kxk

For the second proof we denote the variable in the range space R of q also by q. a nC2

a n

With the notation v D C 2 we get v 0 D C2 by (13.6); on account of (13.10) this implies q v 00 D a n2 2 v 0 . In turn, that leads to 2n v 0 C 4q v 00 D .2n C 2a

2n

4/ v 0 D 2.a

2/ v 0 D 2.a

a n

2/ C2 :

408

21 Solutions to Selected Problems

Now applying Problem 10.20 with P D
, B D A D I , and ˛ D q, we obtain, using part (ii) and (13.31),  a n d.a/
Ra D d.a/
.q  v/ D d.a/2.a 2/ q  C2 D 2.a 2/ Ra 2 d.a 2/ . a 2 n C 1/ . a2 1/ a 2 R D 2.a 2/ D 2.a n/ Ra 2 : . a2 / . a2n / Equation (14.50) follows by repeated application of (14.49), while (14.50) makes sense since Re.a C 2j n/  Re.a C 2k n/  2 n < 0; furthermore, it is a routine computation to see that the extension is actually independent of the choice of k. (iv). R0 D 0 on Rn n f0g, because a 7! .1a / has a zero at a D 0, according 2

to Corollary 13.6; thus, supp R0  f0g. Note that Ra ./  0 if  2 C01 .Rn / is positive, that is, the distributions Ra are positive. Theorems 8.10 and 3.18 then 2 imply R0 D c0 ı. Furthermore, 1 D R0 .e k
k / D c0 by analytic continuation of the identity in part (i); in other words, R0 D ı. The second identity now follows as a direct consequence of part (iii) and the substitution k j 7! j of the index in the product. (v). Both properties follow by means of analytic continuation of identities. (vi). Direct computation. (vii). F Ra is radial and homogeneous of degree .n a/ n on account of part (v) and Problem 14.30.(ii) and (iii). Hence there exists a constant p.a/ 2 C such that 2 F Ra D p.a/ Rn a . Now, the Fourier transform of the function e k
k on Rn equals 2  n=2 e k
k =4 according to Example 14.11. Furthermore, from the homogeneity of 2 Rn a of degree a and part (i) we obtain Rn a .e k
k =4 / D 2n a . Therefore part (i) and Parseval’s formula (14.20) imply 1 D Ra e

k
k2

D p.a/ 

n 2




2

D a

:

 n 1 a 2e F R .2/n

k
k2 4



D p.a/ 2

n



n 2

Rn

a

 e

k
k2 4



For the remainder, replace a by n a and next select a D n 2. Finally, the equivalence of (14.52) with (14.46) is shown using Legendre’s duplication formula (13.39) (with a replaced by a2 ) and the reflection formula from Lemma 13.5. (viii). Take a D 2k in part (vii) and apply (14.34) to the formulas in part (iv).

14.61 (i). k grad ˚! .x/k D k!k D 1, for all x 2 Rn , which proves that ˚! is submersive. From (10.23) one finds that the value at a point of the pushforward of a function under a submersion is obtained by integrating the function over the fiber over that point. (ii). One has Z Z R.Th  /.!; t/ D .x C h/ dn 1 x D .y/ dn 1 y N.!;t /

N.!; t Ch !; h i/

21 Solutions to Selected Problems

409

D R.!; t C h !; h i/: In view of (10.12) the remaining identities follow by taking limits in this formula. (iii). The pullback ˚!  W D 0 .R/ ! D 0 .Rn / is well-defined according to Theorem 10.18, while a combination of Problem 13.2 and part (i) yields ˚!  .jj

nC1

/./ D ˚!  .2ı .n

1/ ..˚! / / ˇ ˇ 1 n 1 @ t .˚! / .t/ˇ t D0 D 2@nt 1 R.!; t/ˇt D0 :

D 2. 1/n

1/

/./ D 2ı .n

For the last equality, use part (ii). (iv). .Aa /a2C is a complex-analytic family of distributions because .jja /a2C is one, and because ˚!  is a linear mapping. Changing the order of integration, one obtains Z Z ˚!  .jjaC1 /.x/.x/ dx d! Aa ./ D n 1 n ZR ZS jjaC1 .˚! .x//.x/ dx d! D S n 1 Rn Z Z jh !; x ija d! .x/ dx: D .a C 1/ Rn S n 1 The second identity is a direct consequence of part (iii). (v). The homogeneity is obvious from the definition and the invariance follows by an orthogonal substitution of variables. As a consequence, we obtain that Aa .x/ D d.a/ c.a C n/kxka D d.a/ RaCn .x/, initially for Re a > 0, but then for all a 2 C by means of analytic continuation. For the evaluation of d.a/, we note that d.a/ D Aa .en /=c.a C n/, with c.a/ as in Problem 14.60. Next, introduce spherical coordinates; in particular, in the notation of [7, Exercise 7.21.(vii)], set h !; en i D !n D sin n 2 and d! D dn 1 ! D dn 2 ! cosn 2 n 2 dn 2 . Thus we conclude, for Re a > 0, with the value (13.37) for cn and by means of (13.33), Z

a

Sn

1

jh !; x ij d! D D cn

1

Z

2

Sn

Z

 2

2

d!n

2

Z

sina  cosn

0

 2  2

j sin n

2j

a

cosn

2

n 1

2

 d D

2 2 . n2 1 /

In other words, application of Legendre’s duplication formula to d.a/ D D

2

n 1 2

.a C 1/ . aCn / 2 21

. n2 /

a

n

2 . aC1 2 / 

n : . a2 C 1/

. aCn 2 / D n .2/

2 n . n2 /2a 

1 2

1 2

n

2

dn

2

. aC1 / . n2 1 / 2 . aCn 2 /

:

.a C 1/ leads to . aC1 2 /

a . aC1 2 / . 2 C 1/

410

21 Solutions to Selected Problems

1 According to Corollary 13.6, the function a 7! d.a/ is well-defined, unless we have a . 2 C 1/ 2 Z0 . (vi). From part (v) and the reflection formula one deduces

1 d. n/

. n2 / .1 2nC1  n

D

n / 2

D

n 1

1 2nC1  n

1

sin n 2

. 1/ 2 : 4.2/n 1

D

The formula for .0/ follows by successive application of Theorem 13.3, the identity from part (v) with a D n, and part (iv). Furthermore, the first identity from part (ii) implies the formula for .x/, for arbitrary x 2 Rn . (vii). The duplication formula implies n 2

1

D

 2 .n/ n 2 1 . 1Cn 2 /

n 2/

.1 .1

and

n/

1

D

2n  2 : . 1 2n /

Hence, the reflection formula leads to b.n/ WD D

1 d. n/ .1

n/

D

.n/ 2n  n 1

. 1Cn / 2

. 1 2n /

n

1/Š sin.n C 1/ 2 . 1/ 2 .n 1/Š D : .2/n .2/n

.n

In this case we have 1 1 1 .0/ D ı./ D R0 ./ D b.n/ b.n/ b.n/ Z j
j n .R.!;
// d! D Sn

D

Z

Sn

Z

Sn 1

˚!  .j
j

n

/./ d!

1

1

Z

t R>0

n

X ˙

R.!; ˙t/

n

2

2 1 X

kD0

ˇ

ˇ @2k s R.!; s/ sD0

 t 2k dt d!: .2k/Š

Finally, apply the first identity from part (ii) in order to get .x/.

15.3 (i). In view of Theorem 15.4 it is sufficient to observe, for  2 C01 .X / and 2 C01 .Y /, that k t˚ K . ˝

/ D .t˚ K / ./ D K. /.˚/ D kK .˚ ˝ /

D kK ..˚ ˝ /. ˝

// D t.˚ ˝ /kK . ˝ / D .t˚ ˝ t /kK . ˝

/:

Here the last equality is obtained as in the proof of Theorem 15.5. (ii). Apply part (i) with the roles of ˚ and played by ˚  and  , respectively. (iii). In this case, (ii) implies k  K  D . ˝ / kK . According to Theorem 10.8 we have  D j˚ ˚  if ˚ now denotes 1 . Since j˚˝˚ D j˚ ˝ j˚ , the desired identity follows from

21 Solutions to Selected Problems

411

.1 ˝ j˚ /k˚  K .˚

1 /

D .j˚ ˝ j˚ /.˚ ˝ ˚/ kK :

(iv). Note that K ˚  D ˚  K is equivalent to K D ˚  K.˚ (v). The identity is a direct consequence of Theorem 10.16.

1 

/ .

15.5 By means of differentiation under the integral sign one sees that F maps n C01 .Rn / to C 1 .R ˛ 2 .Z0 /n and a compact K  Rn , the existence of R /. Given ˛ c > 0 such that j Rn x .e i  /.x/ dxj  c kkC 0 , for all  2 C01 .K/, leads to estimates as in Example 8.6 and, consequently, to the continuity of F . The formula for kF follows from the fact that F is an integral operator with integral kernel kF . Let I be as in Example 15.1. Since tDj D Dj and txj D xj and furthermore .I ˝ Dj /h; xi D i j and .I ˝ ei a /e i h; xi D e i h a; xi , Problem 15.3.(i) implies kF Dj D .I ˝ Dj /e kF xj D .I ˝ xj /e kF ei a D .I ˝ ei a /e

i h
;
i i h
;
i i h
;
i

D .j ˝ I /e D . Dj ˝ I /e D .Ta ˝ I /e

i h
;
i i h
;
i i h
;
i

D kj F ; D k Dj F ; D kTa F :

In the notation of Problem 14.30 we obtain B D t.A 1 / and so hB; Axi D h; xi, for all  and x 2 Rn ; this implies .B ˝ A/ kF D kF . In view of det.B ˝ A/ D .det B/.det A/ D 1, we have kB F A D .B ˝ A/ kF D kF on account of Problem 15.3.(ii). Hence B F A D F . Next, note that e i h; xi D j det Aj e i hA; Axi implies j det Aj D 1 and thus t AA D I . The last assertion follows from Theorem 15.8. 15.7 (i). For the moment, consider a fixed  2 C01 .X /. For every 2 C01 .Y /, we have that w W 7! u. ˝ / is a distribution on Y satisfying, for 1  j  m, @j w. / D u. ˝ @j / D

u..I ˝ @j /. ˝

// D .I ˝ @j /u. ˝

/ D 0I

that is, @j w D 0. Hence, on account of Problem 12.9, there exists v./ 2 C with u. ˝ / D w. / D v./ 1Y . /: Applying this identity with a satisfying 1Y . / D 1, we obtain that v belongs to D 0 .X /. In turn, this leads to u D v ˝ 1Y . As a consequence of Theorem 15.4 this identity now holds on C01 .X Y /, which implies that it is an equality in D 0 .X Y /. (ii). In this case we have xj w. / D u. ˝xj / D .I ˝xj /u. ˝ / D 0. Hence, on account of Theorem 9.5, there exists v./ 2 C with u. ˝

/ D w. / D v./ ı Y . / D v ˝ ı Y . ˝

/:

15.8 Since u is compactly supported, it is of order  k, for some k 2 Z0 . For the moment, consider a fixed  2 C01 .X /. For every 2 C01 .Y / with 0 … supp ,

412

21 Solutions to Selected Problems

we have supp . ˝ / \ .X  f0g/ D ;, and so u. ˝ / D 0. This shows that 7! u. ˝ / is a distribution on Y of order  k with support contained in f0g. Hence, for ˛ 2 .Z0 /m , there exist u˛ ./ 2 C such that this distribution is given by X 7! u. ˝ / D u˛ ./ @˛ ı Y . /: j˛jk

Applying this identity with D ˛ 2 C01 .Y / that equals y 7! y ˛ in a neighborhood of 0, we obtain a constant c˛ such that u˛ ./ D c˛ u. ˝

. 2 C01 .X //:

˛/

But this implies that u˛ W  7! u˛ ./ belongs to E 0 .X / and is of order  k. In other words, X uD u˛ ˝ @˛ ı Y on C01 .X / ˝ C01 .Y /: j˛jk

As a consequence of Theorem 15.4 this identity now holds on C01 .X  Y /, which implies that it is an equality in D 0 .X  Y /. For the remaining equalities we note that for .x; y/ 2 X  Y ,  .I ˝ . @/˛ /. ˝

so

˛

/.x/ D . ˝ . @/˛ /.x; 0/ D .x/ @˛ ı Y . /;

.I ˝ @ / u˛ . ˝

/ D u˛ ./ @˛ ı Y . / D u˛ ˝ @˛ ı Y . ˝

/:

15.9 On account of Problem 15.3.(i) and (14.16).(ii) and (i) we have, for 1  j  n, .Dj ˝ I C I ˝ Dj /k D kDj G C k G Dj D 0; D 0: .xj ˝ I I ˝ xj /k D kxj G kG xj According to Problem 10.18 we then obtain 2n X Dj ˚l B ˚  B Dl k .Dj ˝ I /kz D .Dj ˝ I /˚  k D lD1

(21.18)

D ˚  .Dj ˝ I C I ˝ Dj /k D 0

and analogously

.I ˝ xj /kz D 0:

(21.19)

In view of Problem 15.7.(i) we obtain from (21.18) the existence of u 2 D 0 .Rn / such that kz D 1Rn ˝u, while Problem 15.7.(ii) and (21.19) imply that kz D 1Rn ˝c ı, for some c 2 C and ı the Dirac measure on Rn . Application of Theorem 10.8 now leads to k D ˚ ˚  k D ˚ .1Rn ˝ c ı/ D c kI , see (10.25). Finally, testing against 2 e k
k =2 as in the proof of Theorem 14.13 gives c D .2/n ; hence G D .2/n I . Observe that the preceding arguments and Problem 14.7 are strongly related. 15.10 In view of Theorem 15.4 is sufficient to compute d . ˝ / with  and 2 C01 .Rn /. To this end, consider arbitrary u 2 D 0 .Rn / and successively apply Example 15.11 and Theorem 11.5 to get

21 Solutions to Selected Problems

413

u.d . ˝ // D d  u. ˝ with

/ D u  ./ D u.S  /; Z  .y/ D  ˝ .x; x y/ dx:

S

Rn

15.11 If U D u , then kU D d  u in view of Example 15.11. For h 2 Rn , we have d.Th ˝ Th / D d and so kU D .Th ˝ Th / kU . Now Problem 15.3.(iv) implies that U commutes with all translations in Rn . For the reverse implication, use results from the proof of Problem 15.9. 15.12 For  2 C01 .Rn /, we have, according to Example 11.9 and Theorem 11.5, / D d  u. ˝

U ./ D kU . ˝

/ D u.S

 / D u 

./:

That Ta commutes with U follows as in Problem 15.11; hence, (10.15) implies that @j commutes with U. 15.13 If k 2 D 0 .Rn  Rn / denotes the kernel of K, we will compute it to be (a multiple of) the function .; x/ 7! e i h; xi occurring in Problem 15.5. (a). Application of Problem 15.3.(i) implies .I ˝ Dj /k D .j ˝ I /k, for 1  j  z n. In order to solve this system of partial differential equations, write k D e i h
;
i l, 0 n n z z for some l 2 D .R  R /. Then .I ˝ @j /l D 0, for 1  j  n; see the solution to Problem 15.5. According to Problem 15.7.(i) this entails lz D l ˝ 1Rn , for some l 2 D 0 .Rn /; therefore k D e i h
;
i .l ˝ 1Rn /: (21.20) Thus we have, for K ./ D k. ˝

and  2 C01 .Rn /, /De

i h
;
i

D l. 7! 1Rn .xZ7! e  D l  7! ./ e Rn

D l..F

// D ..F

.l ˝ 1Rn /. ˝

i h; xi i h; xi

/ D .l ˝ 1Rn /.e

./ .x///  .x/ dx D l. 7! ./F

i h
;
i

. ˝

//

.//

/l/./:

Here (11.14) has been used in the fourth equality. Hence, K satisfies K

D .F

/l 2 D 0 .Rn /

.

2 C01 .Rn //:

(21.21)

In particular, 1 D K.ı/ D .F ı/l D l on account of (14.30), which proves the claim in (c). (b). Using the same method as in (a) we obtain c 2 C such that l D c 1Rn . (c). Conditions (i) and (iii) imply (ii). Now apply (b). (d). As in Remark 14.10, condition (i) follows from (iv) and so (21.20) holds. From (v) we obtain K D c K .c 1 / . Hence Problem 15.3.(ii) and .c ˝ c 1 / h
;
i D h
;
i imply e

i h
;
i

.l ˝ 1Rn / D k D kc K .c D .c ˝ c

1

/ .e

1 /

D .c ˝ .c

i h
;
i

1

/ /k

.l ˝ 1Rn // D e

i h
;
i

.c l ˝ 1Rn /:

414

21 Solutions to Selected Problems

This leads to l D c l, for every c > 0; in other words, l is invariant under positive dilations. Similarly, from (vi) and (10.8) we get K D A K A . So .A˝A/ h
;
i D h
;
i gives e

i h
;
i

.l ˝ 1Rn / D .A ˝ A/ .e

i h
;
i

.l ˝ 1Rn // D e

i h
;
i

.A l ˝ 1Rn /:

Hence, l D A l, for every rotation A; in other words, l is invariant under rotations. Finally, application of Problem 12.10 implies that l is a constant function on Rn . (e). Now it follows from (21.21) that l is a function continuous at 0. Since l is also homogeneous of degree 0, it has to be a constant function. 15.14 We prove that the conditions in Problem 15.13.(b) are satisfied. Since we e to have continuous inclusions C01 .R/  L2 .R/  D 0 .R/, the restriction of F C01 .R/ is a sequentially continuous linear mapping to D 0 .R/. Use the identities @x D 21 .a aC /, a j D 2j j 1 , aC j D j C1 , and x D 12 .a C aC / from Problem 14.42 to conclude that e B 1 @x /j D 1 .2j. i /j j .F i 2 e/j : D .x B F

1

1 . i /j C2 j C1 / D . i /j .2j j 2

1

C j C1 /

e Dx D x Fe of linear mappings in L2 .R/, and so in This leads to the identity F 1 e x D Dx F e. Finally, particular in C0 .R/. Similarly, one establishes the identity F e the identity F 0 D 0 determines the multiplicative constant.

15.15 Denote by k 2 D 0 .R  R/ the kernel of K. Since K commutes with translations, it follows from Problem 15.11 that k D d  u, for some u 2 D 0 .R/. Furthermore, for every c > 0, we have c  K D K c  ; hence Problem 15.3.(iv) implies that k D .c ˝ 1/.c  ˝ c  /k; in other words, .c  ˝ c  /k D c 1 k. Now d.c ˝ c/ D c d gives that .c  ˝ c  /d  D d  c  . Hence d  c  u D d  c 1 u, and the injectivity of d  leads to c  u D c 1 u; phrased differently, u is homogeneous of degree 1. According to Problem 15.3.(i) the equality K S D S K entails .I ˝ S /k D .S ˝ I /k; that is, .S ˝ S /k D k. Since d.S ˝ S / D S d , this gives d  S u D d  . u/, and so u is antisymmetric. Now Theorem 10.17 and Problem 9.5 imply that u is a multiple of PV x1 .

15.16 K is local because the support of the distribution K is contained in the support of the function 2 C01 .X /. For the claim about k, we verify that (15.9) is satisfied. Indeed, for  2 C01 .X / and x 2 X ,
 .I ˝ @˛ /. ˝ /.x/ D . ˝ @˛ /.x; x/ D .x/ @˛ .x/ D ..@˛ //.x/: Hence .I ˝ . @/˛ /
 u˛ . ˝ / D u˛ ..@˛ // D .@˛ / u˛ ./; X and so k. ˝ / D .@˛ / u˛ ./ D K ./: ˛2.Z0 /n

Now the proof of the converse statement. Suppose k 2 E 0 .Rn / satisfies supp k 
.Rn / and let ˚ be the automorphism of Rn  Rn as in Problem 15.9. Then

21 Solutions to Selected Problems

415

˚ k 2 E 0 .Rn  Rn / satisfies supp .˚ k/  ˚.supp k/  Rn  f0g in view of Theorem 10.6; hence, from Problem 15.8 we obtain X X ˚ k D .I ˝ @˛ / u˛ ; thus kD ˚ .I ˝ @˛ / u˛ ; ˛2.Z0 /n

˛2.Z0 /n

because of ˚ 2 D I . Now Problem 10.1 and ˚  D
on Rn imply X .I ˝ . @/˛ /
 u˛ ; kD ˛2.Z0 /n

which is the kernel of

P

˛2.Z0 /n

u˛ B@˛ according to the initial part of the problem.

15.17 Suppose that K is local. Select .x; y/ 2 .X  X / n
.X / arbitrarily. Then there exist disjoint open neighborhoods U of x and V of y in X such that we have .U  V / \
.X / D ;. Consider any 2 C01 .V / and  2 C01 .U /. Since supp K  supp  V and supp   U , it follows that 0 D K ./ D k. ˝ /. On the basis of Theorem 15.4 this implies that the restriction of k to U  V vanishes. Therefore .x; y/ … supp k, which proves supp k 
.X /. The reverse implication follows from Problem 15.16. 16.1 According to Problem 1.7 and (16.8) we have arccos B cos D

1 X F .n/ei n 2 n2Z

 D 1Œ 0;   1Œ 0;  :

with

As a consequence of (14.21) (use approximation) and (14.4),  e F ./ D .F 1Œ 0;  /2 D i In particular, F .n/ D

i 

‚ ; 

2

1

2

D  2e

.2k

1/2

sinc2

   2

. 2 R/:

n D 0I n D 2k ¤ 0I

0;

4

i 

;

n D 2k

1:

Hence the coefficients are as desired, while we have pointwise convergence on account of Theorem 16.21. So the first numerical identity follows by evaluation at 0 of the identity of functions. For the second, split the desired series into a series of even terms and one of odd terms and use the preceding identity. Application of Parseval’s formula to arccos B cos on Œ ;   leads to Z  2 2 8 X 2 1 D C 2 x 2 dx D : 3 2 0 4  n2N .2n 1/4 For the final three identities, use arccos B cos.
C 2 / D 2 C arcsin B sin. Note that the pointwise convergence of the Fourier series for f 0 follows from Theorem 16.22.

416

21 Solutions to Selected Problems

16.2 On account of Example 10.4 and (10.14), application of .Th / to the Poisson summation formula leads to the desired result. In turn, it implies X 0 X 00 .arcsin B sin/ D T2k .2 1Œ  ;   1Œ ;  / D 2 .ı 2 C2k ı 2 C2k / 2

k2Z

1 X in .e 2 e D  n2Z 2i X D .ei.4nC1/  n2Z

2

k2Z

  2i X ei n / ei n D sin n  n2Z 2 4 X ei.4nC3// D . 1/n sin.2n  n2N

in 2

1/
:

Now integrate termwise and determine the constants of integration using XZ 1 X . 1/n 1 D . x 2 /n 2n 1 n2N n2N 0

1

dx D

Z

1 0

 1 dx D 1 C x2 4

and the fact that arcsin B sin is an odd function. 16.7 By writing Fn as a double sum and interchanging the order of summation we see that n n k n 1 X X1 X X1 X n Fn D eij D eij D .n jj j/eij : kD0 j D k

j D .n 1/ kDjj j

jj jn 1

16.8 Denote by DN the function as in (16.22); then limN !1 DN exists on account of Lemma 16.4. In combination with Poisson’s summation formula (16.7) this implies X X lim DN D ei n D 2 ı2k : N !1

Furthermore, N X

nD0

cos n
D

which implies

n2Z

N 1X .ei n C e 2 nD0

X

n2Z0

k2Z

i n/ D

cos n
D

N 1 1 X 1 ei n C D .DN C 1/; 2 2 2 nD N

X 1 C ı2k : 2 k2Z

From the solution to Problem 14.27.(i) we get F .sin n
/ D  i.ı X  X F sin n
D  i .ı n ın /: n2N

We have, for x 2 R n 2Z, X cos nx

n2N

n

D

ˇ x ˇˇ ˇ log ˇ2 sin ˇ D 2

n

ın /, so

n2N

ˇ 1 x ˇˇ ˇ log ˇ4 sin2 ˇ D 2 2

1 log 2.1 2

cos x/:

21 Solutions to Selected Problems

417

On account of Lemma 5.9, termwise differentiation gives the following identity in S 0 .R n 2Z/: X cos 2

sin 1 cot D : sin n
D
D 2 sin 2 2 2 2.1 cos/ n2N Taylor expansion around 0 implies .x/ . x/ D 2x 0 .0/ C O.x 2 / as x ! 0. Thus, the following integral is absolutely convergent, while integration by parts yields Z Z sin x ..x/ . x// dx D log.1 cos x/. 0 .x/ C  0 . x// dx: cos x R 1 R Since the integrand in the latter integral is an even function, the right-hand side takes R the form 2 R log.1 cos x/  0 .x/ dx.

16.10 From Exercise 14.44 we obtain F sinc2 ./ D 2 maxf 0; 2 jj g. The desired formula now follows from (16.8) by taking  D sinc2 , a D , and thus ! D 2. For the final identity, replace x by x.

16.11 The formula for F F t follows from Problems 14.44 and 14.30.(ii). In view of the continuity of F we obtain F .lim t !1 F t / D lim t !1 F F t D 1 D F ı in 1 1 S 0 .R/. Next use that F t  f D 2 S F F .F t  f / D 2 S F .F F t F f /. Finally, on account of (14.29), (16.7), and Problem 16.7, X k2Z

n F .T2 k Fn / D max 0; 1 D

X 

jkjn 1

1

n j
jo X e 2 i k D max 0; 1 n k2Z jkj  1 ık D F F n: n 2

j
jo X ık n k2Z

Alternatively, for x 2 R, we have in view of (16.30), X

1 X 1 1 1 D D x 2 2 2 .x C 2 k/ 4 . k/ 4 sin2 k2Z k2Z 2

X

k2Z

Fn .x C 2 k/ D

x 2

;

hence

X 2 sin2 nx 1 sin2 nx 1 2 2 D : n .x C 2 k/2 2 n sin2 x2 k2Z

Note that in this manner one may obtain another verification of (16.30). 2

1 kxk n 16.14 For a given y 2 S n 1 set v.x/ D kx ykn . Then v is harmonic on R n fyg. It suffices to prove that w W x 7! v.x C y/ is harmonic on Rn n f0g. In view of 1 kx C yk2 D .2hx; yi C kxk2 /, we have

w.x/ D

2hx; yi 1 C kxkn kxkn

2

:

418

21 Solutions to Selected Problems

From (12.3) we know that the second term on the right-hand side is harmonic on Rn n f0g. Now, applying @j to a harmonic function on an open set in Rn gives another, so the following are harmonic on Rn n f0g: wj .x/ D @j

1 kxkn

2

D .2

In the case of n D 2, we take @j log kxk D

n/

xj kxk2

xj kxkn

.1  j  n/:

.1  j  2/:

This proves the claim about v. By means of differentiation under the integral sign it now follows that u is harmonic on B n . Define p W R>0  S n We claim that

Z

1

Sn

 Sn

1

1

!R

p.r; x; y/ D

by

p.r; x; y/ dy D 1

.r > 0; x 2 S n

1 r2 : cn kr x ykn 1

/:

By rotational invariance, this integral is clearly independent of x; hence we may denote its value by d.r/. We obtain Z Z Z Z 1 1 d.r/ D p.r; x; y/ dy dx D p.r; x; y/ dx dy: cn S n 1 S n 1 cn S n 1 S n 1 But x 7! p.r; x; y/ is harmonic, for kxk < 1r , so the Mean Value Theorem for harmonic functions from Problem 12.5 implies that the inner integral is independent of r. Evaluation at r D 0 gives its value as 1, which leads to d.r/ D 1. Furthermore, limr"1 p.r; x; y/ D 0 uniformly for y belonging to the complement in S n 1 of any open subset in S n 1 containing x. As in Problem 5.1 we therefore obtain limr"1 u.r x/ D f .x/, uniformly for all x 2 S n 1 . 16.15 From (16.24).(i) and (ii) it is obvious that Z  1 jf .x y/ jn f .x/ f .x/j  2 

f .x/j Fn .y/ dy:

The continuous periodic function f is uniformly continuous on R. As a result, given  > 0 arbitrarily, we can find ı > 0 such that for all x and all y 2 R with jyj < ı we have jf .x y/ f .x/j < 2 . Hence, for all n 2 N, Z

ı ı

jf .x

y/

f .x/j Fn .y/ dy 

 2

Z

ı ı

Fn .y/ dy 

 : 2

Next (16.24).(iv) implies that we can find N 2 N such that for all n  N ,

21 Solutions to Selected Problems

X ˙

This entails X Z ˙

˙

˙ı

˙

Z

˙

˙ ˙ı

419

Fn .y/ dy 

jf .x

y/

It follows that jn f .x/

 ; 4m

m D sup jf .x/j:

where

x2R

f .x/jFn .y/ dy  2m

X ˙

˙

Z

˙ ˙ı

Fn .y/ dy 

 : 2

f .x/j  , for all n  N and x 2 R.

16.16 As in (21.17), we have jF j2 D F .  S /;

so

F jF j2 D F 2 .  S / D 2 S.  S /:

Therefore (16.8) with a D 2 and  replaced by jF j2 leads to, for all  2 R, X X X jF . C 2 k/j2 D S.  S /.k/ e i k  D .  S /.k/ e i k  : k2Z

k2Z

Next observe, for all l 2 Z, that Z .x/.x .  S /.k/ D R

k2Z

k/ dx D

Z

.x

.k C l// dx:

l/.x

R

P 16.17 (i). We have u D k2Z 21 .1 C sgn.k// ı2k , which shows that u is a distribution as in Lemma 16.4, with coefficients forming a sequence of moderate growth. According to the lemma, u belongs to S 0 .R/ and satisfies S BFuD

1 X C e2 i n 2 n2N

and therefore

1 X C e 2 n2N

FuD

2 i n :

(ii). For any  > 0 and x 2 supp u D 2Z0 , we see that e x  1, which means 1 that we can find a function  2 S.R/ coinciding with x 7! e x on  2; 1 Œ . Thus we obtain, for any  2 S.R/, x

e

u./ D  u./ D u. / D u.e

x

/:

As a consequence, ju./

e



x

u./j D ju./

X1

k2N

u.e

x

/j 

X

k2N

j.1

e

2k

kkS.0;2/ X 1 e 2k sup jx 2 .x/j D 2 .2k/ 4 2 x2R k2N

/.2k/j e 2k : k2

The series at the right converges uniformly for all   0, which implies lim #0

X1

k2N

X e 2k 1 D lim 2 k #0 k2N

e 2k D 0; k2

420

21 Solutions to Selected Problems

lim ju./

and so

x

e

#0

u./j D 0:

This proves lim#0 e x u D u in S 0 .R/. According to Theorem 14.24 the mapping F W S 0 .R/ ! S 0 .R/ is continuous; consequently F u D lim#0 F .e x u/. Once more on the basis of Lemma 16.4 and summing a convergent geometric series, we obtain, for  2 R, F .e

x

u/./ D F

DS

1 2

1

X

C

ıC

2

e

X

2k

k2N

2n

n2N

e

 ı2k ./

 1 X e2 i n ./ D C e 2 n2N

D

e 2.Ci / 1 11Ce C D 2 21 e 1 e 2.Ci /

D

1 e .Ci / C e 2 e .Ci / e

.Ci / .Ci /

D

2.Ci /n

2.Ci / 2.Ci /

1 e i . 2 e i .

i /

Ce i / e

i . i / i . i /

D

1 cot . 2i

i /:

Combining the preceding results now leads to F u D lim #0

1 cot .
2i

i / D

1 cot .
2i

i 0/:

(21.22)

Furthermore, X 2i e 2 i n D 2i.F u C S F u/./ D 2i.F u./ C F u. // n2Z

D cot .

cot . C i 0/ DW c./:

i 0/

(iii). The properties of the function f .z/ D cot z follow directly, as well as the fact that f is complex-analytic on C n Z. As a consequence we have, for  belonging to the open subset R n Z, X c./ D lim.f . i / f . C i // D f ./ f ./ D 0 D 2i ık jRnZ : #0

k2Z

On the other hand, for z 2 Uk D f z 2 C j jz f .z/ D

1 .z

k/

kj < 1=2 g, we may write C g.z/;

where g is complex-analytic on Uk . The Plemelj–Sokhotsky jump relation then implies, for  2 R \ Uk , c./ D lim #0



.

1 k

i /

.

1 C g. k C i /

i /

g. C i /



21 Solutions to Selected Problems

D

.

1 k

421

i 0/

In view of R D .R n Z/ [

.

S

cot .


X 1 D 2i ık D 2i ıl jUk : k C i 0/

k2Z .R

i 0/

l2Z

\ Uk / and Theorem 7.1 we now obtain X cot .
C i 0/ D 2i ık : k2Z

(iv). Poisson’s summation formula from the preceding part and Lemma 16.6 imply X X X X 0 ık0 D e2 2 i n e2 i n D 4 n sin 2 n
: in D k2Z

n2Z

n2N

n2Znf0g

Alternatively, differentiate the first formula in Problem 16.8. (v). The identity from part (ii) and its transform under S imply   1 X 1 Fu D C e2 i n D ˙ cot .
 i 0/: SF u 2 n2N 2i X 1X 1 So sin 2 n
D .S F u F u/ D cot .
˙ i 0/; 2i 4 n2N ˙ X 1X
sin n
D cot. ˙ i 0/: and hence 4 2 n2N ˙

cos x 1 For x near 0 we have cot x D sinc , where the first factor is a C 1 function having xx value 1 at 0. So the Plemelj–Sokhotsky jump relation now leads to

cot.x ˙ i 0/ D

 1 cos x  1 cos x PV   i ı D PV   i ı: sinc x x sinc x x

In other words, for x near 0, X

n2N

sin n x D

cos x2 1 x PV : sinc 2 x

(vi). Define A W R ! R by Ax D  1 x. Then t A D A. According to Problem 14.30.(i) and Theorem 14.24, and Example 10.4, respectively, we have F B A B F D A B F 2 D 2A B S

and

Applying F B A to (21.22) in part (ii), we now obtain  X F cot.
i 0/ D 2 i ı C 2 ı

Next note that cot2 D .1 C cot0 /, and so

k2N

A ı

2k

 :

2k

Dı

2k :

422

21 Solutions to Selected Problems 2

F cot .


i 0/ D

F1

D

2ı

 X i 0/ D 2ı C 2 x ı C 2 ı

ix F cot.
X

8

kı

2k

k2N



2k :

k2N

(vii). On the basis of Lemma 16.6 and Problem 16.8 the desired formula follows from X X cos n
X n2 cos n
X 1 C s s 00 D C D cos n
D ı2k : 2 2 1Cn 1Cn 2 n2N n2N n2N k2Z

Restricting s to the open interval  0; 2 Œ , we obtain for it the differential equation s s 00 D 1=2, which on R has the general solution sQ .x/ D a e x C b e

1 2

x

a; b 2 C:

for arbitrary

From the definition of s we see that it is 2-periodic, even, and continuous on R, because the series is uniformly convergent on R. Consequently, the function s is completely determined on all of R by the choice of a and b. Periodicity implies s.0/ D s.2/, that is, aCb

1 D a e 2 C b e 2

1 ; 2

2

that is,

bDa

e 2 1 : 1 e 2

In view of the uniform convergence of the series on R we may write Z

2 0

s.x/ dx D

On the other hand, Z Z 2 s.x/ dx D 0

0

Z

2 0

2



X 1 Z 2 X cos nx dx D cos nx dx D 0: 1 C n2 1 C n2 0 n2N n2N

a ex C b e

x

1 dx D a e 2 2

be

2

aCb

hence aD

 2.e 2

1/

and

bD

2.1

 e

2 /

:

Inserting the values for a and b, we get (see Fig. 21.1)  1   ex e x   2 C 1 2 e 1 1 e 2  1   ex  1  cosh.x / e .x /     D 1 D C 2 e e  e e  2 sinh 

sQ .x/ D

 1 :

I

21 Solutions to Selected Problems

423

A less elegant, but slightly more illuminating, method for obtaining a second equation for a and b is as follows. Let 0 < 1 < 2 < 1 and 1 ;2 2 C01 .R/ be such that 1 ;2 .x/ D 1 for jxj < 1 , and 1 ;2 .x/ D 0 for jxj > 2 . Then we have  X 1 .1 ;2 / D ı2k .1 ;2 / D s s 00 C 2 k2Z Z 2  1 1 ;2 .x/ dx: s.x/ s 00 .x/ C D 2 2 This implies Z 2  Z 1 1 1 ;2 .x/ dx D lim  D lim s.x/ s 00 .x/ C s 00 .x/ dx 2 2 #0 1 #0 2 1 D lim .s 0 . 1 /

s 0 .1 // D sQ 0 .2/

1 #0

sQ 0 .0/ D a e 2

be

2

a C b:

Observe that this formula implies the jump relations (see Fig. 21.1) lim s 0 .x/

lim s 0 .x/ D

x#2k

x"2k

.k 2 Z/:



The choice x D 0 finally leads to the desired value of the numerical series. 1

2Π

4Π

Fig. 21.1 On Œ 0; 2  the graph of s is a catenary; therefore on R the graph looks like a suspension bridge. On account of the jump relations one computes the angles between the tangent lines at the points 2 n to be  2 arctan 2  65 ı

Note that the Fourier transform provides another method for solving the differential equation for s; it turns out to be somewhat more laborious. We describe the main steps. Application of Theorem 14.24 and Lemma 16.4 leads to F s./ D and therefore s.x/ D

ı./ C 

1 1X C 2 2 n2Z

Z

R

Xe n2Z

2 i n

1 C 2

;

e i.x 2n/ d : 1 C 2

On the basis of [7, Exercise 2.85 or 6.99.(iii)] or using the Residue Theorem from complex analysis, one obtains Z

R

e i.x 2n/ d  D e 1 C 2

.x 2n/ sgn.x 2n/

;

and so

424

21 Solutions to Selected Problems

s.x/ D

 1 C 2 2

X

ex

2n

f n2Z j x 2n 0 such that jgx0 .y/j  m, for all x and y 2 R. Now apply integration by parts. 16.20 Use the results from Example 16.23 and successively apply the following arguments. (i) Set z D 41 in (16.28). (ii) Evaluate the Fourier series of fz at x D 14 and take its imaginary part. (iii) Put z D 0 in (ii). (iv) Replace z in (16.30) by 1 z. (v) Differentiation of this identity gives the one in (iv), while the constant of 2 integration equals 0, as follows by taking z D 0. (vi) Direct verification. (vii) Use 2 d that dz log.1 nz 2 / D z 22zn2 . At the outset, the identity is obtained for 0 < jzj < 1, but on account of analytic continuation, the identity holds for all z 2 C. (viii) Take the limit for z ! 1 in 1  Y sin z sin  z2  D z.z C 1/ 1 : .z 1/ n2 nD2 (ix) Set z D 12 in (vii). (x) The identity sin z cos z D 12 sin 2z in combination with (vii) implies Y .2z/2  Y  .2z/2  sin z cos z D z 1 1 : .2n/2 n2N .2n 1/2 n2N 16.21 The distributional Fourier series of fz has coefficients

21 Solutions to Selected Problems

1 2

Z

2

e

i.zCn/x

0

dx D

425

1 e 2 iz De 2i .z C n/

 iz

sin z .z C n/

.n 2 Z/:

The pointwise convergence on R n 2Z follows directly from Theorem 16.21. Identity (i) with the C sign is a simple rewrite of the Fourier series, P while Pthe case of is obtained by replacing z by z. For (ii) and (iii), apply ˙ and ˙ ˙ to (i), respectively. Next, (iv) follows from (iii) by taking x D  and replacing z by z. For (v) and (vi) note that e i nx e C zCn z

i nx

n

D

2z cos nx z 2 n2

2n sin nx : z 2 n2

i

Next, apply this identity to the Fourier series, suppose z 2 R for the moment, split into real and imaginary parts, and finally use analytic continuation in z. Note that (vi) is the derivative of (v) with respect to x, under the assumption that termwise differentiation is admissible. 16.22 (i). We have, for arbitrary  2 S 0 .R/, XZ .gz00 C z 2 gz /./ D gz . 00 C z 2 / D n2Z

.2nC1/

.2n 1/

gz .x/. 00 .x/ C z 2 .x// dx:

Next, twice integrating by parts, we obtain Z

.2nC1/ .2n 1/

D

Z

. 00 .x/ C z 2 .x//gz .x/ dx .2nC1/

.2n 1/

D cos z

. 00 .x/ C z 2 .x// cos z.x

X ˙

˙ 0 ..2n ˙ 1// C z sin z

2n/ dx X ˙

..2n ˙ 1//:

The desired formula now follows upon summation, since the sum involving the terms  0 ..2n ˙ 1// is telescoping. In the notation of Example 16.23 we get, for x 2 R, 1 sin z X . 1/n  1 .fz .x/ C f z .x// D 2  n2Z 2 z n n X z sin z . 1/ D e 2 i n x ; 2 2  z n n2Z

1

gz .2z x/ D

z

n



e 2 i n x

which leads to the distributional Fourier series for gz . The final identity is deduced by means of repeated termwise differentiation of that series. Poisson’s summation formula (16.7) implies X X X X ei n D 2 ı2k and 2 e2i n D 2 ık : n2Z

k2Z

n2Z

k2Z

426

21 Solutions to Selected Problems

Subtracting the former identity from the latter, we get the desired result. (ii). This is established by straightforward computations. The pointwise convergence of the Fourier series of f is a consequence of Theorem 16.22. (iii). Replace x by 2x for j cosP j and then replace x by 2 x for j sin j. For the final identity, note that j sin j D k2Z Tk .1Œ 0;  sin/ and differentiate. The Fourier series for j cos j and j sin converge pointwise on account of Theorem 16.21, while the pointwise convergence of the last Fourier series follows from Theorem 16.22. 17.3 With respect to the cases (i), (ii), and (v), observe that P .D/ei  D P ./ei  , for all  2 Cn . Because the degree of P is positive, there exists  2 Cn with P ./ D 0; hence, u D ei  2 C 1 .Rn /  D 0 .Rn / satisfies P .D/u D 0. Note that according to Problem 14.24 we have ei  2 S 0 .Rn / if and only if  2 Rn ; therefore, in case (ii) it is necessary that the zero  belong to Rn . For the cases (iii), (iv), and (vi), note that P .D/u D 0 implies P ./ u y./ D 0 . 2 Rn /:

This equality holds if and only if at least one of P ./ and u y./ equals 0. Furthermore, the closed subset N D f  2 Rn j P ./ D 0 g is nowhere dense; therefore we have u y./ D 0, for  belonging to the dense open subset Rn n N . According to Lemma 14.4, Theorem 14.2, and Lemma 14.9, depending on the case respectively, u y is continuous on Rn . This yields u y D 0, and we see that Theorem 14.24 implies u D 0.

17.4 Consider u 2 S 0 .Rn / with
u D 0. The Fourier transform implies kk2 F u D 0, or supp F u  f0g; and so Theorem 8.10 leads to the existence of k 2 Z0 and c˛ 2 C such that on account of (14.33), FuD

X

j˛jk

c˛ @˛ ı D

X 1 . i /j˛j c˛ F .x ˛ /: .2/n j˛jk

Hence, the injectivity of F gives that u is a polynomial. If u is a bounded harmonic function, it defines a tempered distribution. The preceding argument then gives that u is a bounded polynomial, which is possible only if it is a constant. Next, suppose n  3. Denote by E the potential of ı as in (12.3). Now suppose F 2 D 0 .Rn / is a fundamental solution of
satisfying F .x/ ! 0 as kxk ! 1. In particular,
.E F / D 0, while
is hypoelliptic according to Example 12.8. Therefore E F is a bounded harmonic function, and the preceding argument gives that it is a constant. The condition at infinity implies that E F D 0.

17.6 (i). E 2 S 0 .Rn / according to Theorem 14.34, while Problem 14.30.(ii) implies that F E is homogeneous Pm of degree a n if E is homogeneous of degree a. Next, write P ./ D l homogeneous of degree l. Then the lD0 Pl ./ with PP Fourier transform of P .D/E D ı gives 1 D m lD0 Pl F E, where Pl F E is homogeneous of degree l a n and Pm F E ¤ 0. Accordingly, P .D/ D Pm .D/ and m a n D 0, that is, a D m n. (ii). It follows immediately that F E D F Ez D 1=P on Rn n f0g and thus z  f0g. As in Problem 17.4, this implies Ez E D f , with supp .F .E E//

21 Solutions to Selected Problems

427

f a polynomial function on Rn . If Ez is homogeneous, then f is homogeneous of degree m n according to part (i). Consequently, f D 0 if m < n, and thus E D Ez in this case. If m  n, then f is a homogeneous polynomial function of degree m n. (iii). In this case, 1=P is a homogeneous function on Rn n f0g of degree m > n, which implies its local integrability on Rn . From Theorem 14.34 we obtain 1=P 2 S 0 .Rn /, and therefore E WD F 1 .1=P / is a homogeneous fundamental solution, of degree . m/ n D m n. The uniqueness of E finally follows from part (ii). 1 (iv). According to (12.3), 2 log kxk is a fundamental solution of
, and as a consequence of the hypoellipticity of
any two of its fundamental solutions differ by a function belonging to C 1 .R2 /. Therefore any fundamental solution possesses a logarithmic singularity at 0, which is incompatible with homogeneity. 17.8 The fundamental solution E0 2 S 0 .R/ is given by F 1 .  1 /. Set ˝ D f z 2 C j jzj < r g. For the solution E from the proof of Theorem 17.11 we first compute, for jj  R and  2 Rn n f0g, 1 2 i

Z

@˝

e ix e i hx;Czi dz D z P . C z/ 2 i

ix

Z

e D 2 i

@˝

Z

e ixz dz z. C z /

@˝

e ix e ixz dz DW z.z ˛/ 2 i

Z

f .z/ dz; @˝

where we have written ˛ D   . For the moment, assume that r is large enough that the singular points of f .z/ within ˝ occur at 0 and ˛. Then, by residue calculus, Z 1 f .z/ dz D Res f .z/ C Res f .z/ D lim z f .z/ C lim .z ˛/ f .z/ z!˛ zD˛ z!0 zD0 2 i @˝ ix. / e 1 C : D ˛ ˛ We obtain Z e ix e ix. / 1 e ix e ix 1 e i hx;Czi dz D D : 2 i @˝ z P . C z/  ˛   Accordingly, Z 1 G.x/ D 2

R R

e ix 

1 e ix d D  2

Z

R R

e ix d  

e ix 2

Z

R R

1 



d ;

which is nontempered, since  2 C n R. For the point ˛ to belong to ˝ (while jj  R), we need j j < rjj. In particular, if  D R=jj, we obtain j j D jj.1 C R=jj/ D jj C R. Hence, rjj > jj C R is necessary. On the other hand, this condition is sufficient. Indeed, rjj > jj C R  jj C jj  j j. 18.1 Use Poisson’s summation formula 16.7, the fact that F .0/ D 1./, and (18.3) for estimating the remainder.

428

21 Solutions to Selected Problems

18.6 (i). We prove the results for fC ; the arguments for f are similar, mutatis mutandis. First note that F u 2 C 1 .R/ according to Lemma 14.4. Observe that Z 2 fC .z/ D e i Re z  e Im z  F u./ d  .z 2 HC /: (21.23) R>0

From Theorem 18.1 we obtain the existence of constants c 0 > 0 and N 0 2 Z0 0 such that jF u./j  c 0 .1 C jj/N , for all  2 R. Set N D N 0 C 3. Because p N e  N Š p , for p > 0, we obtain, for all sufficiently large  > 0 and a suitable constant d > 0, 0 d N d 1 e Im z  jF u./j  : 0 C3 D N N N .Im z/  .Im z/  3 This estimate proves that the integrand in fC is integrable on R>0 . In addition, it shows that fC may be differentiated under the integral sign with respect to x D Re z and y D Im z. That fC is complex-differentiable on HC follows from Z 1 .i  i /e ix  e y  F u./ d  D 0: 2@zN fC .z/ D .@x C i @y /fC .x C iy/ D 2 R>0 (ii). (21.23) implies that for a sufficiently large constant m > 0, Z m Z 1 2 jfC .z/j  e Im z  jF u./j d  C e Im z  jF u./j d  0 m Z m Z 1 d d 00 1 0  e Im z  jF u./j d  C d   d C : .Im z/N m  2 .Im z/N 0 00

For Im z sufficiently small, we have d 0  .Imdz/N , which leads to the desired estimate. (iii). On the basis of Problem 12.14.(vi) and part (ii) we obtain the existence of ˇ˙ .f˙ / 2 D 0 .R/. Indeed, we get, for  2 C01 .R/, Z Z 2 ˇC .fC /./ D lim e ix  e y  F u./ d  .x/ dx: y#0

R

R>0

For fixed y > 0, the function .; x/ 7! e y  jF u./j j.x/j is integrable on the product R  R>0 . Therefore, we may interchange the order of integration to obtain Z Z e y  F u./ e ix  .x/ dx d  2 ˇC .fC /./ D lim y#0 R>0 R Z y D lim e F u./ F . / d : y#0

R>0

Since F u is at most of polynomial growth,R we may take the limit under the integral sign. Thus we get 2 ˇC .fC /./ D R>0 F u./ F . / d  and similarly R 2 ˇ .f /./ D R