Theory of Distributions [2 ed.] 9783030812645, 9783030812652

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Svetlin G. Georgiev

Theory of Distributions Second Edition

Theory of Distributions

Svetlin G. Georgiev

Theory of Distributions

Second Edition

Svetlin G. Georgiev Department of Differential Equations Faculty of Mathematics and Informatics University of Sofia “St. Kliment Ohridski” Sofia, Bulgaria

ISBN 978-3-030-81264-5 ISBN 978-3-030-81265-2 (eBook) https://doi.org/10.1007/978-3-030-81265-2 1st edition: © Springer International Publishing 2015 2nd edition: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface to the Second Edition

In the 5 years since the first edition of this book was published, I have received a lot of messages and letters from readers commenting on the book and suggesting how it could be improved. With the aim of this information, I have revised the first edition of the book. The changes its second edition are as follows. In Chap. 1, a section titled “Lp Spaces” has been added. In this section, Lp spaces for p ≥ 1 are introduced. They are deduced from the Hölder, Young, Minkowski inequalities and an interpolation inequality. In this section, certain criteria for uniform integrability of some classes of functions are given. The RieszFischer theorem and the Lp dominated convergence theorem are studied. Also, the conception for separable and dual spaces is introduced and the Riesz representation theorem for bounded linear functionals on Lp spaces is proven. A new section titled "Change of Variables" has been added to Chap. 2. Here, the change of variables for some classes of distributions is investigated and a representation of the Dirac delta function is provided. In Chap. 3, a new section titled “The Local Structure of Distributions” has been added. In this section, a criterion for linear and continuous extension of a distribution with compact support is provided and that any distribution with compact support has a finite order is proven. A new section called “Notes and References” has been introduced in all chapters. In this section, some additional materials are provided for each chapter. Some problems are provided with detailed proofs. The book’s index has been updated. The aim of the second edition is to present a clear and well-organized treatment of the concept behind the development of mathematics and solution techniques. The material of this book is presented in a highly readable, mathematically solid format. Many practical problems are illustrated, displaying the scope of the theory of distributions. Paris, France February 2021

Svetlin G. Georgiev

v

Preface to the First Edition

The theory of partial differential equations is without a doubt one of the branches of analysis in which ideas and methods of different fields of mathematics manifest themselves and are interlaced—from functional and harmonic analysis to differential geometry and topology. Because of that, the study of this topic represents a constant endeavour and requires undertaking several challenges. The main aim of this book is to explain many of the fundamental ideas underlying the theory of distributions. The book consists of ten chapters. The first chapter deals with the well-known classical theory regarding the space C ∞ , the Schwartz space and the convolution of locally integrable functions. It may also serve as an introduction to typical questions related to cones in Rn . Chapter 2 collects the definitions of distributions, their order, sequences, support and singular support, and multiplication by C ∞ functions. In Chaps. 3 and 4 we introduce differentiation and homogeneous distributions. The notion of direct multiplication of distributions is developed in Chap. 5. The following two Chaps. 6 and 7, deal with specific problems about convolutions and tempered distributions. In Chaps. 8 and 9 we collected basic material and problems regarding integral transforms. Sobolev spaces are discussed in the tenth, and final, chapter. The volume is aimed at graduate students and mathematicians seeking an accessible introduction to some aspects of the theory of distributions, and is well suited for a one-semester lecture course. It is a pleasure to acknowledge the great help I received from Professor Mokhtar Kirane, University of La Rochelle, La Rochelle, France, who made valuable suggestions that have been incorporated in the text. I express my gratitude in advance to anybody who will inform me about mistakes, misprints, or express criticism or other comments, by writing to the e-mail addresses [email protected], [email protected]. Paris, France January 2015

Svetlin G. Georgiev

vii

Contents

1

Introduction .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 The Spaces C0∞ and S . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 The Lp Spaces.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.1 Definition .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.2 The Inequalities of Hölder and Minkowski .. . . . . . . . . . . . . . . . 1.2.3 Some Properties .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.4 The Riesz–Fischer Theorem .. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.5 Separability .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.6 Duality .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.7 General Lp Spaces .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 The Convolution of Locally Integrable Functions . . . . . . . . . . . . . . . . . . 1.4 Cones in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.6 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

1 1 12 12 14 18 20 27 29 44 48 55 56 76

2

Generalities on Distributions . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2 Order of a Distribution .. . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3 Change of Variables .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.4 Sequences and Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.5 Support.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.6 Singular Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.7 Measures .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.8 Multiplying Distributions by C ∞ Functions .. . .. . . . . . . . . . . . . . . . . . . . 2.9 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.10 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

77 77 84 85 88 92 98 100 102 103 112

3

Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1 Derivatives .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2 The Local Structure of Distributions . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3 The Primitive of a Distribution . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4 Simple and Double Layers on Surfaces . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

113 113 118 124 127 ix

x

Contents

3.5 3.6

Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 131 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 142

4

Homogeneous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

143 143 144 151 154

5

The Direct Product of Distributions . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

155 155 157 160 164

6

Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.3 Existence.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.4 The Convolution Algebras D  (Γ +) and D  (Γ ) . . . . . . . . . . . . . . . . . . . . 6.5 Regularization of Distributions . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.6 Fractional Differentiation and Integration.. . . . . .. . . . . . . . . . . . . . . . . . . . 6.7 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.8 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

165 165 167 169 170 171 172 176 194

7

Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2 Direct Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.3 Convolution .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.4 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.5 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

195 195 197 198 200 203

8

Integral Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.1 The Fourier Transform in S (Rn ) . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.2 The Fourier Transform in S  (Rn ) . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.3 Properties of the Fourier Transform in S  (Rn ) . . . . . . . . . . . . . . . . . . . . 8.4 The Fourier Transform of Distributions with Compact Support . . . 8.5 The Fourier Transform of Convolutions.. . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.6 The Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.6.1 Definition .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.6.2 Properties .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.7 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 8.8 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

205 205 206 208 209 210 211 211 212 214 218

9

Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.1 Definition and Properties.. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 9.2 Fundamental Solutions of Ordinary Differential Operators .. . . . . . . 9.3 Fundamental Solution of the Heat Operator . . . .. . . . . . . . . . . . . . . . . . . .

219 219 223 226

Contents

xi

9.4 9.5 9.6

Fundamental Solution of the Laplace Operator .. . . . . . . . . . . . . . . . . . . . 227 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 227 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 230

10 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2 Elementary Properties .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.3 Approximation by Smooth Functions . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.4 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.5 Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.6 Sobolev Inequalities .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.7 The Space H −s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.8 Advanced Practical Problems .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.9 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

231 231 232 236 241 244 247 255 256 258

References .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 259 Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 261

Chapter 1

Introduction

1.1 The Spaces C0∞ and S With Nn ∪ {0} we denote the space of multi-indices α = (α1 , α2 , . . . , αn ), αk ∈ N ∪ {0}, k = 1, 2, . . . , n. For α = (α1 , α2 , . . . , αn ), β = (β1 , β2 , . . . , βn ) ∈ Nn ∪ {0}, we will write α ≤ β if αk ≤ βk , k = 1, 2, . . . , n. Set D = (D1 , D2 , . . . , Dn ), ∂ ∂ |α| , k = 1, 2, . . . , n, D α = α1 α2 . Dk = ∂xk ∂x1 ∂x2 . . . ∂xαnn Let X ⊂ Rn be an open set. If K ⊂ X is a compact set we shall write K ⊂⊂ X. The following conventions will also be used throughout the book: U (x0 , R) is the open ball of radius R with centre at the point x0 , S(x0 , R) = ∂U (x0 , R) is the sphere of radius R with centre at x0 , U (x0 , R) = U (x0 , R) ∪ S(x0 , R) is the closed ball of radius R with centre at the point x0 and UR = U (0, R), SR = S(0, R), U R = U (0, R). If A and B are sets in Rn , by d(A, B) or dist(A, B) we shall denote the distance between the sets A and B, that is d(A, B) = dist(A, B) = infx∈A,y∈B |x − y|. We shall use A to denote the -neighbourhood of a set A, i.e., A = A + U . If A is an open set, then A will designate the set of points in A that are more than  away from the boundary ∂A, i.e., A = {x : x ∈ A, dist(x, ∂A) > }. We use intA to denote the set of interior points of the set A. With κA we will denote the characteristic function of A, i.e., κA (x) = 1 for x ∈ A and κA (x) = 0 for x ∈ A. Definition 1.1 The set A is called convex if for any points x and y in A the segment λx + (1 − λ)y,

λ ∈ [0, 1],

lies entirely in A. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_1

1

2

1 Introduction

We will write chA to denote the convex hull of a set A. Definition 1.2 We call space of basic functions the space C0∞ (X) of smooth functions with compact support defined on X. Example 1.1 The function ω(x) =

⎧ ⎨

Ce

−

1 1−|x|2

,

⎩0,

|x| ≤ 1, |x| > 1,

 where the constant C is chosen so that

ω(x)dx = 1, belongs in C0∞ (Rn ).

Rn

1 x  , x ∈ Rn . ω n  Definition 1.3 The function ω is called the “cap-shaped function”. Set ω (x) =

We have  ω (x)dx = Rn

1 n

 ω Rn

x  

 dx =

ω(x)dx = 1. Rn

∞ Definition 1.4 We say that the sequence {φk }∞ k=1 of elements of C0 (X) converges ∞ to the function φ ∈ C0 (X) if there exists a compact set K ⊂ X such that suppφk ⊂ K for every k ∈ N and lim D α φk (x) = D α φ(x) uniformly for every multi-index k→∞

α ∈ Nn ∪ {0}.

Example 1.2 Take φ ∈ C0∞ (R). The sequence C0∞ (R).



1 φ(x) k

Example 1.3 Take φ ∈ C0∞ (R). Then the sequence converge to 0 in C0∞ (R).

∞



converges to 0 in k=1

1 x  φ k k

∞ does not k=1

Lemma 1.1 For every set X1 ⊂ X and every  > 0 there exists a function φ ∈ C ∞ (Rn ) such that φ (x) = 1 when x ∈ X1 , φ (x) = 0 when x ∈ Rn \X13 and 0 ≤ φ (x) ≤ 1 when x ∈ Rn . Proof Let κX2 be the characteristic function of the set X12 , i.e., κX2 (x) = 1 for 1

1

x ∈ X12 and κX2 (x) = 0 for x ∈ / X12 . Define 1

 φ (x) =

 κX2 (y)ω (x − y)dy =

ω (x − y)dy,

1

Rn

X12

x ∈ Rn .

1.1 The Spaces C0∞ and S

3

We will prove that the function φ has the required properties. Since ω ∈ C ∞ (Rn ), 0 ≤ ω (x), x ∈ Rn , suppω = U  and  ω (x)dx = 1, Rn

we get φ ∈ C ∞ (Rn ) and 0 ≤ φ (x)  = ω (x − y)dy X12



ω (x − y)dy

≤ Rn



=

ω (y)dy Rn

= 1,

x ∈ Rn ,

and  φ (x) =

κX2 (y)ω (x − y)dy 1

Rn



=

κX2 (y)ω (x − y)dy 1

U (x,)

=

⎧  ⎪ ⎨

 ω (x − y)dy =

⎪ ⎩U(x,) 0, x ∈ X13 .

ω (y)dy = 1,

x ∈ X1 ,

U

This completes the proof. Lemma 1.2 (Expansion of the Unit Function) Let φ ∈ C0∞ (Rn ) and suppφ be covered by a finite number of neighbourhoods U (xk , rk ) for some xk ∈ Rn , rk > 0, k = 1, 2, . . . , m, and for some m ∈ N. Then there exist functions m  hk (x) = 1 for x in a φhk ∈ C0∞ (U (xk , rk )), k = 1, 2, . . . , m, such that k=1

neighbourhood of suppφ and supp(φhk ) ⊂ U (xk , rk ), k = 1, 2, . . . , m.  Proof For k = 1, 2, . . . , m, take 0 < rk < rk so that the union ∪m k=1 U (xk , rk ) cov∞ ers suppφ. Be Lemma 1.1, it follows that there exist functions ηk ∈ C0 (U (xk , rk )),

4

1 Introduction

k = 1, 2, . . . , m, so that ηk (x) = 1, x ∈ U (xk , rk ), suppηk ⊂ U (xk , rk ),  0 ≤ ηk (x) ≤ 1, x ∈ Rn , k = 1, 2, . . . , m. Then, for x ∈ ∪m k=1 U (xk , rk ), we set h(x) =

m 

ηk (x),

hk (x) =

k=1

ηk (x) , h(x)

k = 1, 2, . . . , m,



and for x ∈ Rn \ ∪m k=1 U (xk , rk ) we set h(x) = hk (x) = 0, k = 1, 2, . . . , m. Note that φhk ∈ C0∞ (U (xk , rk )), k = 1, 2, . . . , m, and h(x) ≥ 1,

m 

hk (x) =

k=1

m  ηk (x) k=1

h(x)

= 1,

 x ∈ ∪m k=1 U (xk , rk ).

This completes the proof. ∞ n n Definition 1.5 We say that the sequence {ηk }∞ k=1 in C0 (R ) converges to 1 in R if

1. for every α ∈ Nn ∪ {0} there exists a constant cα > 0 such that |D α ηk (x)| ≤ cα for every k ∈ N and every x ∈ Rn , 2. for every compact set K in Rn there exists N = N(K) ∈ N such that ηk (x) = 1 for every k > N and x ∈ K. ∞ n Example  x  1.4 Choose η ∈ C0 (R ) so that η(x) = 1 for |x| ≤ 1. Set ηk (x) = n , x ∈ Rn , k ∈ N. Then the sequence {ηk }∞ η k=1 tends to 1 in R . k Definition 1.6 With S (Rn ) we denote the space of C ∞ functions φ such that

sup |x|β |D α φ(x)| < ∞,

x∈Rn

∀α ∈ Nn ∪ {0},

β ∈ N ∪ {0}.

1

Here |x| = (x12 + x22 + · · · + xn2 ) 2 and x = (x1 , x2 , . . . , xn ). By || · ||S ,p , p ∈ N, we shall indicate the norm ||φ||S ,p =

sup x∈Rn ,|α|≤p

p

(1 + |x|2) 2 |D α φ(x)|,

φ ∈ S (Rn ).

(1.1)

Example 1.5 Let n = 1 and φ(x) = e−x , x ∈ R. Then φ ∈ S (R). 2

Example 1.6 Let n = 1, φ be as in Example 1.5 and p(x) = a0 x k + a1x k−1 + · · · + ak−1 x + ak , x ∈ R, where aj ∈ R, j = 0, 1, . . . , k, k ∈ N. Then φp ∈ S (R). Exercise 1.1 Prove that (1.1) satisfies all axioms for a norm. Theorem 1.1 The space C0∞ (Rn ) is a proper subset of the space S (Rn ).

1.1 The Spaces C0∞ and S

5

Proof Let φ ∈ C0∞ (Rn ), α ∈ Nn ∪ {0} and β ∈ N ∪ {0} be arbitrarily chosen. Set K = suppφ and ψ(x) = |x|β D α φ(x), x ∈ Rn . Then suppψ = K and there exists a constant K1 > 0 so that sup |ψ(x)| = max |ψ(x)| ≤ K1 < ∞. x∈K

x∈Rn

Because α ∈ Nn ∪ {0} and β ∈ N ∪ {0} were arbitrarily chosen, we conclude that φ ∈ S (Rn ). Now, using that φ ∈ C0∞ (Rn ) was arbitrarily chosen and we get that it is an element of the space S (Rn ), we obtain the inclusion C0∞ (Rn ) ⊆ S (Rn ). Let 2 ψ1 (x) = e−|x| , x ∈ Rn . Then ψ1 ∈ S (Rn ) and ψ1 ∈ C0∞ (Rn ). Thus, C0∞ (Rn ) ⊂ S (Rn ). This completes the proof. Theorem 1.2 Let φ, ψ ∈ S (Rn ). Then φψ ∈ S (Rn ). Proof Let α ∈ Nn ∪{0} and β ∈ N∪{0} be arbitrarily chosen. Since φ, ψ ∈ S (Rn ), we have that sup |x|β |D γ φ(x)| < ∞,

x∈Rn

sup |D α−γ ψ(x)| < ∞

x∈Rn

for any γ ∈ Nn ∪ {0}, γ ≤ α. Then, using Leibnitz’s rule, we find      α   sup |x|β |D α (φψ)(x)| = sup |x|β  D γ φ(x)D α−γ ψ(x) x∈Rn x∈Rn γ :γ ≤α γ     α ≤ sup |x|β |D γ φ(x)||D α−γ ψ(x)| γ x∈Rn γ :γ ≤α

   α  β γ α−γ ≤ sup |D ψ(x)| sup |x| |D φ(x)| γ x∈Rn x∈Rn γ :γ ≤α

< ∞, i.e., φψ ∈ S (Rn ). This completes the proof. n Definition 1.7 We say that the sequence {φk }∞ k=1 of elements of S (R ) converges n n to 0 in S (R ), if for every p ∈ N ∪ {0} and every α ∈ N ∪ {0}, we have p

lim (1 + |x|2) 2 |D α φk (x)| = 0

k→∞

uniformly. ∞ n Theorem 1.3 Let {φk }∞ k=1 be a sequence of elements of C0 (R ) such that ∞ n n φk →k→∞ 0 in C0 (R ). Then φk →k→∞ 0 in S (R ).

6

1 Introduction

Proof Since φk →k→∞ 0 in C0∞ (Rn ), there exists a compact set K ⊂ Rn so that suppφk ⊂ K for any k ∈ N and D α φk (x) →k→∞ 0 uniformly for any multi-index α ∈ Nn ∪ {0}. Let K1 > 0 be a constant so that 1 + |x|2 ≤ K1 for any x ∈ K. Then, for any p ∈ N ∪ {0}, we have 0 ≤ lim

k→∞

 p p 2 1 + |x|2 |D α φ(x)| ≤ K12 lim |D α φk (x)| = 0 k→∞

uniformly, i.e., φk →k→∞ 0 in S (Rn ). This completes the proof. Theorem 1.4 The space C0∞ (Rn ) is dense in the space S (Rn ). n ∞ n Proof Let φ ∈ S (R  x)and η ∈ C0 (R ) be chosen so that η(x) = 1 for |x| < 1. n , x ∈ R , k ∈ N. Then φk ∈ C0∞ (Rn ) and φk →k→∞ φ in Set φk (x) = φ(x)η k C0∞ (Rn ). Now, applying Theorem 1.3, we conclude that φk →k→∞ φ in S (Rn ) as well. This completes the proof.

Remark 1.1 Note that if a ∈ C ∞ (Rn ) and φ ∈ S (Rn ), it does not follow that aφ ∈ 2 2 S (Rn ). Take for instance a(x) = e|x| , x ∈ Rn , and φ(x) = e−|x| , x ∈ Rn . Then ∞ n n n a ∈ C (R ), φ ∈ S (R ) and a(x)φ(x) = 1, x ∈ R . Consequently aφ ∈ / S (Rn ). Definition 1.8 By ΘM we denote the space of functions a ∈ C ∞ (Rn ) for which there exist constants Ca > 0 and ma ∈ N such that |D α a(x)| ≤ Ca (1 + |x|)ma ,

x ∈ Rn ,

for every α ∈ Nn ∪ {0}. Such functions are called multipliers of S (Rn ). Theorem 1.5 Let a ∈ ΘM and φ ∈ S (Rn ). Then aφ ∈ S (Rn ). Proof Let β ∈ N ∪ {0} and α ∈ Nn ∪ {0} be arbitrarily chosen. Then     α   β α β γ α−γ sup |x| |D (aφ)(x)| = sup |x|  φ(x) D a(x)D γ n n x∈R x∈R γ ≤α  ⎛ ⎞  α  ≤ sup |x|β ⎝ |D γ a(x)||D α−γ φ(x)|⎠ γ n x∈R γ ≤α ⎛

≤ sup |x|β ⎝ x∈Rn

 α 

γ ≤α

γ

⎞

Ca (1 + |x|)ma |D α−γ φ(x)|⎠

 α 

= Ca sup |x|β (1 + |x|)ma |D α−γ φ(x)| γ n x∈R γ ≤α < ∞. Thus, aφ ∈ S (Rn ). This completes the proof.

1.1 The Spaces C0∞ and S

7

Theorem 1.6 Let a ∈ ΘM . Then the map φ → aφ from S (Rn ) to S (Rn ) is a continuous map. n Proof Let φ ∈ S (Rn ) and {φk }∞ k=1 be a sequence of elements of S (R ) that n n converges to φ in S (R ). Then, for any β ∈ N ∪ {0} and any α ∈ N ∪ {0}, we have

lim

k→∞

 β   2 1 + |x|2 D α φk (x) − D α φ(x) = 0

uniformly. Hence, for any β ∈ N ∪ {0} and α ∈ Nn ∪ {0}, we get  lim

k→∞

1 + |x|2

β   2 D α (aφk )(x) − D α (aφ)(x)

        β   α   α 2 = lim 1 + |x|2  D γ a(x)D α−γ φk (x) − D γ a(x)D α−γ φ(x) k→∞ γ  γ ≤α γ γ ≤α        β   α α−γ

 2 2  γ α−γ  = lim 1 + |x| a(x) D φ (x) − D φ(x) D k   k→∞  γ ≤α γ 

≤ lim

k→∞



1 + |x|

2

    α  γ D a(x) D α−γ φk (x) − D α−γ φ(x) γ γ ≤α

β  2

   β   α 2 Ca lim 1 + |x|2 (1 + |x|)ma D α−γ φk (x) − D α−γ φ(x) ≤ k→∞ γ γ ≤α 

=0

uniformly. Thus, aφk →k→∞ aφ in S (Rn ). This completes the proof. Exercise 1.2 Prove that the maps φ → D α φ, α ∈ Nn ∪{0}, and φ(x) → φ(Ax+b), where A is an n × n matrix with det A = 0, are linear and continuous maps from S (Rn ) to itself. Definition 1.9 For p ∈ N ∪ {0}, with Sp (Rn ) we will denote the completion of the space S (Rn ) with respect to  · S ,p . Definition 1.10 Let X and Y be two normed vector spaces with norms  · X and  · Y , respectively. We say that X is compactly embedded in Y , and we write X → Y or Y ← X, if 1. X is continuously embedded in Y . i.e., there is a constant C > 0 such that xY ≤ CxX for any x ∈ X, 2. any bounded set in X is totally bounded in Y , i.e., every sequence in such a bounded set has a subsequence that is a Cauchy sequence in the norm  · Y .

8

1 Introduction

Theorem 1.7 The spaces Sp (Rn ), p ∈ N ∪ {0}, are Banach spaces fitting in a chain of continuous and compact embedding S0 (Rn ) ← S1 (Rn ) ← S2 (Rn ) ← · · · .

(1.2)

Proof Let p ∈ N ∪ {0} be arbitrarily chosen. Take φ ∈ Sp+1 (Rn ) arbitrarily. Then φS ,p+1 = ≥

 sup x∈Rn ,|α|≤p

 sup x∈Rn ,|α|≤p

1 + |x|2 1 + |x|2

 p+1 2

p 2

|D α φ(x)|

|D α φ(x)|

= φS ,p . Hence, φ ∈ Sp (Rn ) and the embedding (1.2) are continuous. Now, let M be an infinitely bounded set in Sp+1 (Rn ). Then there exists a constant C > 0 such that ||φ||p+1 ≤ C for every φ ∈ M. Hence,     α D φ(x) ≤ C for every x ∈ Rn , α ∈ Nn ∪ {0}, |α| ≤ p, φ ∈ M. Therefore p 2

(1 + |x| ) |D φ(x)| = 2

α

(1 + |x|2)

p+1 2

|D α φ(x)|

(1 + |x|2 )

1 2

≤

C 1

(1 + |x|2 ) 2

→|x|→∞ 0.

Let {Rk }∞ k=1 be an increasing sequence of positive numbers such that p

(1 + |x|2 ) 2 |D α φ(x)| ≤

1 k

for |x| > Rk , |α| ≤ p.

By the Arzela–Ascoli theorem, it follows that there exists a sequence {φj }∞ j =1 of (1)

elements of M that converges in Sp (U R1 ). We may then find a sequence {φj }∞ j =1 (2)

n converging in Sp (U R2 ), and so on. The sequence {φk }∞ k=1 converges in Sp (R ). This completes the proof. (k)

Theorem 1.8 If φ ∈ C p (Rn ) and |x|p D α φ(x) →|x|→∞ 0 for α ∈ Nn ∪ {0}, |α| ≤ p, then φ ∈ Sp (Rn ). Proof To prove this assertion, we choose a sequence {ηk }∞ k=1 of elements in C0∞ (Rn ) such that ηk →k→∞ 1 in Rn . Fix  > 0. Since |x|p |D α φ(x)| →|x|→∞ 0 for any α ∈ Nn ∪ {0}, |α| ≤ p, it follows that there exists R = R() > 0 such that the inequality p

(1 + |x|2) 2 |D α φ(x)| < 

1.1 The Spaces C0∞ and S

9

holds for |x| > R. As ηk →k→∞ 1 in Rn , there exists N ∈ N such that ηk (x) = 1 for every k > N and |x| ≤ R + 1. Now, define  φ 1 (x) =

φ(y)ω 1 (x − y)dy.

k

k

Rn ∞ n Observe that {φ 1 ηk }∞ k=1 is a sequence in C0 (R ) and there exists N1 ∈ N so that k

 p     2   sup 1 + |x|2 D α φ − φ 1 ηk (x) <  k

x∈Rn |α|≤p

for k > N1 , |x| ≤ R + 1. Set N2 = max{N, N1 }. Then, for k > N2 , we have p

||φ − φ 1 ηk ||S ,p = sup (1 + |x|2 ) 2 |D α (φ − φ 1 ηk )(x)| k

k

x∈Rn |α|≤p

p

≤ sup (1 + |x|2 ) 2 |D α (φ − φ 1 ηk )(x)| k

|x|≤R+1 |α|≤p

p

+ sup (1 + |x|2 ) 2 |D α (φ − φ 1 ηk )(x)| k

|x|>R+1 |α|≤p

p

≤ sup (1 + |x|2 ) 2 |D α (φ − φ 1 ηk )(x)| k

|x|≤R+1 |α|≤p

+ sup (1 + |x| ) 2

p 2

|x|>R+1 |α|≤p



 α 

|D φ(x)| + α

β≤α

β

 |D β φ 1 (x)D α−β ηk (x)| k

p

<  + sup (1 + |x|2 ) 2 |D α φ(x)| |x|>R+1 |α|≤p

p

+ sup (1 + |x|2 ) 2 |x|>R+1 |α|≤p

 α  β≤α

β

|D β φ 1 (x)D α−β ηk (x)| k

p

< 2 + sup (1 + |x|2 ) 2 |D α φ 1 (x)| k

|x|>R+1 |α|≤p

p



≤ 2 + sup (1 + |x|2 ) 2 |x|>R+1 |α|≤p

= 2 + sup

 

|x|>R+1 |α|≤p Rn

|D α φ(x − y)|ω 1 (y)dy k

Rn

1 + |x − y + y|2

p   2 D α φ(x − y) ω 1 (y)dy k

10

1 Introduction

≤ 2 + sup

 

|x|>R+1 |α|≤p Rn

1 + (|x − y| + |y|)2

p   2 D α φ(x − y) ω 1 (y)dy k

  p   2 1 + |x − y|2 + |y|2 D α φ(x − y) ω 1 (y)dy

p

< 2 + 2 2 sup

k

|x|>R+1 |α|≤p Rn

≤ 2 + 2

 

p 2

sup |x|>R+1 |α|≤p Rn

|x|>R+1 |α|≤p Rn

= 2 + 2

k

 

≤ 2 + 2p sup

   p   2 D α φ(x − y) ω 1 (y)dy 1 + |x − y|2 + 1 + |y|2

1 + |x − y|2

p 2

 p    2 D α φ(x − y) ω 1 (y)dy + 1 + |y|2 k

  p   2 sup 1 + |x − y|2 D α φ(x − y) ω 1 (y)dy

p

k

|x|>R+1 |α|≤p Rn

+2p sup

  p   2 1 + |y|2 D α φ(x − y) ω 1 (y)dy k

|x|>R+1 |α|≤p Rn



 

ω 1 (y)dy + 2 

< 2 + 2  p

p

k

Rn 3p 2

2

ω 1 (y)dy k



ω 1 (y)dy + 2 

≤ 2 + 2 

p

Rn



p

1 + |y|2

ω 1 (y)dy

k

Rn

k

Rn

  3p =  2 + 2p + 2 2 .

Since  > 0 was arbitrarily chosen, we conclude that φ 1 ηk →k→∞ φ in Sp (Rn ). k Now, using the fact that Sp (Rn ) is a Banach space, we conclude φ ∈ Sp (Rn ). This completes the proof. Theorem 1.9 We have S (Rn ) =



Sp (Rn ).

(1.3)

p∈N∪{0}

Proof Let φ ∈ S (Rn ) be arbitrarily chosen. By the definition of the space S (Rn ), it follows that for any α ∈ Nn ∪ {0} and any p ∈ N ∪ {0}, we have   sup |x|p D α φ(x) < ∞.

x∈Rn

Hence,  p   2 sup 1 + |x|2 D α φ(x) < ∞

x∈Rn

(1.4)

1.1 The Spaces C0∞ and S

11

for any α ∈ Nn ∪ {0} and any p ∈ N ∪ {0}. In particular, by (1.4), we get sup x∈Rn ,|α|≤p

 p   2 1 + |x|2 D α φ(x) < ∞

for any p ∈ N ∪ {0}. From here, we obtain that φ ∈ Sp (Rn ) for any p ∈ N ∪ {0}  and so, φ ∈ Sp (Rn ). Because φ ∈ S (Rn ) was arbitrarily chosen and we p∈N∪{0}



get that it is an element of

Sp (Rn ), we find the following inclusion

p∈N∪{0}



S (Rn ) ⊆

Sp (Rn ).

(1.5)

p∈N∪{0}



Let now, ψ ∈

Sp (Rn ) be arbitrarily chosen. Then ψ ∈ Sp (Rn ) for any

p∈N∪{0}

p ∈ N ∪ {0}. From here, sup x∈Rn ,|α|≤p

 p   2 1 + |x|2 D α ψ(x) < ∞

(1.6)

for any p ∈ N ∪ {0}. Now, take β ∈ N ∪ {0} and α1 ∈ Nn ∪ {0} arbitrarily. Then there is a γ ∈ N ∪ {0} so that γ ≥ β and |α1 | ≤ γ . Hence,  γ     2 sup |x|β D α1 ψ(x) ≤ sup 1 + |x|2 D α1 ψ(x)

x∈Rn

x∈Rn

≤

 sup

x∈Rn ,|α1 |≤γ

1 + |x|2

γ   2 D α1 ψ(x)

< ∞, where we have used (1.6). Since β ∈ N ∪ {0} and α1 ∈ Nn ∪ {0}  were arbitrarily Sp (Rn ) was chosen, we conclude that ψ ∈ S (Rn ). Because ψ ∈ p∈N∪{0}

arbitrarily chosen and we find that it is an element of S (Rn ), we arrive at the inclusion  Sp (Rn ) ⊆ S (Rn ). p∈N∪{0}

By the last relation and by (1.5), we obtain (1.3). This completes the proof.

12

1 Introduction

1.2 The Lp Spaces 1.2.1 Definition Let E be a measurable set and F be the collection of all measurable extended real-valued functions on E that are finite a.e. on E. With m(E) we will denote the Lebesgue measure of E. Definition 1.11 Define two functions f and g in F to be equivalent, and we write f ∼ g, provided f (x) = g(x)

for almost all x ∈ E.

The relation ∼ is an equivalent relation, that is, it is reflexive, symmetric and transitive. Therefore it induces a partition of F into a disjoint collection of equivalence classes, which we denote by F / ∼. For given two functions f and g in F , their equivalence classes [f ] and [g] and real numbers α and β, we define α[f ] + β[g] to be the equivalence class of the functions in F that take the value αf (x)+βg(x) at points x ∈ E at which both f and g are finite. Note that these linear combinations are independent of the choice of the representatives of the equivalence classes. The zero element in F / ∼ is the equivalence class of functions that vanish a.e. in E. Thus, F / ∼ is a vector space. Definition 1.12 For 1 ≤ p < ∞, we define Lp (E) to be the collection of equivalence classes [f ] for which  |f |p < ∞. E

This is properly defined since if f ∼ g, then 

 |f |p = E

|g|p . E

Note that, if [f ], [g] ∈ Lp (E), then for any real constants α and β, we have 

⎛ ⎞   |αf + βg|p ≤ 2p ⎝ |αf |p + |βg|p ⎠

E

E

= |2α|



|f | + |2β|

p

p

E

i.e., α[f ] + β[g] ∈ Lp (E).

E



|g|p < ∞,

p E

1.2 The Lp Spaces

13 p

Definition 1.13 For 1 ≤ p < ∞, we define Lloc (E) to be the collection of equivalence classes [f ] for which  |f |p < ∞ K p

for any compact subsets K of E. The elements of Lloc (E) will be called locally p-integrable functions or p-locally integrable functions. If p = 1, then we will say locally integrable functions. Definition 1.14 We call a function f ∈ F essentially bounded provided there is some M ≥ 0, called an essential upper bound for f , for which |f (x)| ≤ M

for almost all x ∈ E.

Definition 1.15 We define L∞ (E) to be the collection of equivalence classes [f ] for which f is essentially bounded. Note that L∞ (E) is properly defined since if f ∼ g, then |f (x)| = |g(x)| ≤ M

for almost all x ∈ E.

Also, if [f ], [g] ∈ L∞ (E) and α, β ∈ R, there are nonnegative constants M1 and M2 such that |f (x)| ≤ M1 ,

|g(x)| ≤ M2

for almost all x ∈ E.

Hence, |αf (x) + βg(x)| ≤ |α| |f (x)| + |β| |g(x)| ≤ |α| M1 + |β| M2 for almost all x ∈ E. Therefore α[f ] + β[g] ∈ L∞ (E) and hence, L∞ (E) is a vector space. For simplicity and convenience, we refer to the equivalence classes in F / ∼ as functions and denote them by f rather than [f ]. Thus, f = g means f −g vanishes a.e. on E. Definition 1.16 For 1 ≤ p < ∞, in Lp (E) we define ⎞1 ⎛ p  f p = ⎝ |f |p ⎠ . E

For p = ∞, we define f ∞ to be the infimum of the essential upper bounds for f .

14

1 Introduction

1.2.2 The Inequalities of Hölder and Minkowski Definition 1.17 Let p ∈ [1, ∞]. The number q is said to be conjugate of p if 1 1 + = 1. p q Theorem 1.10 (Young’s Inequality) Let a, b > 0, p, q ∈ (1, ∞),

1 1 + = 1. p q

Then ab ≤

1 p 1 q a + b . p q

Proof Since the map x → ex is convex, we get 1

log a p + 1 log bq

q ab = elog a+log b = e p 1 1 1 1 p q ≤ elog a + elog b = a p + bq . p q p q

This completes the proof. Theorem 1.11 (Hölder’s Inequality) Let E be a measurable set, 1 ≤ p < ∞ and q be the conjugate of p. If f ∈ Lp (E) and g ∈ Lq (E), then the product fg is integrable over E and  |fg| ≤ f p gq .

(1.7)

E

Proof 1. Let p = 1. Then q = ∞ and 

 |fg| =

E

⎛ ⎞  |f ||g| ≤ ⎝ |f |⎠ g∞ = f 1 g∞ .

E

E

2. Let p ∈ (1, ∞). If f p = 0 or gq = 0, then the assertion is evident. Let f p = 0 and gq = 0. We set f1 =

f , f p

g1 =

g . gq

1.2 The Lp Spaces

15

Then f1 p = g1 q = 1. Now, using Young’s inequality, we have |f1 g1 | ≤

1 1 |f1 |p + |g1 |q . p q

Hence, 1 f p gq

 



 |fg| = E

|f1 g1 | ≤ E



=

1 1 |f1 |p + |g1 |q p q

E

1 |f1 |p + p

E



1 1 |g1 |q = q p

E



 |f1 |p + E

1 q

 |g1 |q E

1 1 1 1 p q = f1 p + g1 q = + = 1. p q p q From the last inequality, we get the inequality (1.7). This completes the proof. Remark 1.2 When p = q = 2, the Hölder inequality is known as the Cauchy– Schwartz inequality. Theorem 1.12 Let E be a measurable set, p > r, q > r, r > 0 be real numbers 1 1 1 such that + = . If f ∈ Lp (E), g ∈ Lq (E), then fg ∈ Lr (E) and p q r fgr ≤ f p gq . Proof Observe that q

1 p r

+

1 q r

(1.8) p

= 1. Set f1 = |f |r , g1 = |g|r . Then f1 ∈ L r (E) and

g1 ∈ L r (E). Now, applying the Hölder inequality, we arrive at  f1 g1 ≤ f1  p g1  q . r

r

E

Since 

 f1 g1 =

E

 |f |r |g|r =

E

|fg|r = fgrr E

(1.9)

16

1 Introduction

and ⎞r ⎛  p p f1  p = ⎝ f1r ⎠ = f rp , r

E

⎞r ⎛  q q g1  q = ⎝ g1r ⎠ = grq , r

E

by the inequality (1.9), we find that fgrr ≤ f rp grq , whereupon we get the inequality (1.8). This completes the proof. Theorem 1.13 Let E be a measurable set and pj ≥ 1, j = 1, 2, 3, be real numbers such that 1 1 1 + + = 1. p1 p2 p3 If fj ∈ Lpj (E), j = 1, 2, 3, then  |f1 f2 f3 | ≤ f1 p1 f2 p2 f3 p3 . E

Proof Let Lq (E) and

1 1 1 = + and g = f2 f3 . By Theorem 1.12, it follows that g ∈ q p2 p3 gq = f2 f3 q ≤ f2 p2 f3 p3 .

Now, we apply the Hölder inequality for f1 , g and p1 , q, and we find 

 |f1 f2 f3 | = E

|f1 g| ≤ f1 p1 gq ≤ f1 p1 f2 p2 f3 p3 . E

This completes the proof. Remark 1.3 Let E be a measurable set and pj ≥ 1, j = 1, 2, . . . , k, be real k  1 = 1 for some k ∈ N, k ≥ 2. If fj ∈ Lpj (E), numbers so that pj j =1

1.2 The Lp Spaces

17

j = 1, 2, . . . , k, using the principle of the mathematical induction, Theorem 1.12 and Theorem 1.13, one can get the following inequality       k  k   ≤ f fj pj . j   j =1  j =1

E

Lemma 1.3 (Interpolation Inequality) Let A be an open bounded set in Rn , 1 ≤ s ≤ r ≤ t ≤ ∞ and θ 1−θ 1 = + , r s t

0 ≤ θ ≤ 1.

The any u ∈ Ls (A) ∩ Lt (A) belongs in Lr (A) and . ||u||r ≤ ||u||θs ||u||1−θ t Proof We have 

 |u|r dx =

A

 |u|θr+(1−θ)r dx =

A

≤



|u|s dx

A

 rθ   s

|u|θr |u|(1−θ)r dx A

|u|t dx

 r(1−θ ) t

,

A

from where . ||u||r ≤ ||u||θs ||u||1−θ t This completes the proof. Theorem 1.14 (Minkowski’s Inequality) Let E be a measurable set, 1 ≤ p ≤ ∞. If f, g ∈ Lp (E), then f + gp ≤ f p + gp . Proof 1. Let p = 1. Then     f + g1 = |f + g| ≤ (|f | + |g|) = |f | + |g| = f 1 + g1 . E

E

E

E

18

1 Introduction

2. Let p = ∞. Then f + g∞ ≤ f ∞ + g∞ . 3. Let p ∈ (1, ∞) and q be its conjugate. If f + gp = 0, then the assertion is evident. Assume that f + gp = 0. Then, applying Hölder’s inequality, we get 

p



f + gp =

|f + g|p = E

≤

 

|f + g||f + g|p−1 E

   |f ||f + g|p−1 + |g||f + g|p−1 = |f ||f + g|p−1 + |g||f + g|p−1

E

E

E

⎛

⎞1 ⎛ ⎞1 ⎛ ⎞1 ⎛ ⎞1 p q p q     p⎠ ⎝ (p−1)q ⎠ p⎠ ⎝ (p−1)q ⎠ ⎝ ⎝ ≤ |f | |f + g| + |g| |f + g| E

E

E

p q

E

p q

= f + gp f p + f + gp gp .

Hence, p− pq

f + gp

≤ f p + gp

or f + gp ≤ f p + gp . This completes the proof.

1.2.3 Some Properties Theorem 1.15 Let E be a measurable set and 1 < p < ∞. Suppose that F is a family of functions in Lp (E) that is bounded in Lp (E) in the sense that there is a constant M > 0 such that f p ≤ M

for all f ∈ F .

Then the family F is uniformly integrable over E. Proof Let  > 0 be arbitrarily chosen. Suppose that A is a measurable subset of 1 1 E of finite measure. Let also, + = 1. Define g to be identically equal to 1 on p q A. Because m (A) < ∞, we have that g ∈ Lq (A). By Hölder’s inequality, for any

1.2 The Lp Spaces

19

f ∈ F , we have 

 |f | = A

⎞1 ⎛ ⎞1 ⎛ p q   p⎠ ⎝ q⎠ ⎝ |f |g ≤ |f | g .

A

A

A

On the other hand, ⎛ ⎝



⎞1

p

|f |p ⎠ ≤ M,

A

⎛ ⎞1 q  1 ⎝ |g|q ⎠ = (m (A)) q A

for any f ∈ F . Therefore 

1

|f | ≤ (m (A)) q M A

  q for any f ∈ F . Let δ = . Hence, if m (A) < δ, we have M  1 1  = |f | ≤ M (m (A)) q < Mδ q = M M A

for any f ∈ F . Therefore F is uniformly integrable over E. This completes the proof. Theorem 1.16 Let E be a measurable set of finite measure and 1 ≤ p1 < p2 ≤ ∞. Then Lp2 (E) ⊆ Lp1 (E) . Proof Let p2 < ∞ and f ∈ Lp2 (E) be arbitrarily chosen. Then p = take q > 1 such that

1 1 + = 1. Then p q ⎛ ⎞1 ⎞ p1 ⎛ p p2   ⎝ |f |p1 p ⎠ = ⎝ |f |p2 ⎠ < ∞, E

E

(1.10) p2 > 1. We p1

20

1 Introduction

i.e., |f |p1 ∈ Lp (E). Let g = κE . Since m (E) < ∞, then g ∈ Lq (E). Hence, using Hölder’s inequality, we get 

 |f |p1 = E

=

⎞1 ⎛ ⎞1 ⎛ p q   p1 p1 p ⎠ ⎝ q⎠ ⎝ |f | g ≤ |f | g

E

E

p f p12

1 q

E

(m (E)) < ∞.

Therefore f ∈ Lp1 (E). Because f ∈ Lp2 (E) was arbitrarily chosen and we get that it is an element of Lp1 (E), we obtain the relation (1.10). Let p2 = ∞ and f ∈ L∞ (E) be arbitrarily chosen. Then there is a positive constant M such that |f (x)| ≤ M for almost all x ∈ E. Hence,  |f |p1 ≤ M p1 m (E) . E

Then f ∈ Lp1 (E). Because f ∈ L∞ (E) was arbitrarily chosen and we get that it is an element of Lp1 (E), we obtain the relation (1.10). This completes the proof. p p Definition 1.18 Let {fn }∞ n=1 be a sequence of elements of L (E) and f ∈ L (E). ∞ When fn −f p → 0, as n → ∞, we will say that the sequence {fn }n=1 converges to f in Lp (E).

1.2.4 The Riesz–Fischer Theorem Definition 1.19 Let X be a normed vector space. A sequence {fn }∞ n=1 in X is said to be a rapidly Cauchy sequence provided that there is a convergent series of positive ∞  numbers k for which k=1

fk+1 − fk  ≤ k2 for all k ∈ N. Lemma 1.4 Let X be a normed vector space and {fn }∞ n=1 be a sequence in X such that fk+1 − fk  ≤ ak

1.2 The Lp Spaces

21

for all k ∈ N, where ak , k ∈ N, are nonnegative numbers. Then fn+k − fn  ≤

∞ 

(1.11)

al

l=n

for all k, n ∈ N. Proof For any n, k ∈ N, we have fn+k − fn = fn+k − fn+k−1 + fn+k−1 − · · · + fn+1 − fn =

n+k−1 



fj +1 − fj .

j =n

Hence, fn+k

   n+k−1 ∞        n+k−1

 n+k−1      − fn =  aj ≤ aj fj +1 − fj  ≤ fj +1 − fj ≤   j =n j =n j =n j =n

for any k, n ∈ N. This completes the proof. Theorem 1.17 Let X be a normed vector space. Then every rapidly Cauchy sequence in X is a Cauchy sequence in X. Furthermore, every Cauchy sequence in X has a rapidly Cauchy subsequence in X. Proof 1. Let {fn }∞ n=1 be a rapidly Cauchy sequence in X. Then there is a convergent series ∞  k of positive numbers such that k=1

fk+1 − fk  ≤ k2 for any k ∈ N. Hence and (1.11), we obtain fn+k − fk  ≤

∞ 

l2 → 0,

as n → ∞,

l=n

for any k ∈ N. Here we have used that the series Therefore {fn }∞ n=1 is a Cauchy sequence.

∞  k=1

k2 is a convergent series.

22

1 Introduction

2. Let {fn }∞ n=1 is a Cauchy sequence in X. Then there are M1 , M2 , M3 ∈ N such that   1   for any m11 , m12 ≥ M1 , fm1 − fm1  < 1 2 2   1   fm2 − fm2  < 2 for any m21 , m22 ≥ M2 , 1 2 2   1   fm3 − fm3  < 3 for any m31 , m32 ≥ M3 . 1 2 2 In particular, for m11 , m12 , m32 ≥ max{M1 , M2 , M3 }, we have   1   fm1 − fm1  < , 1 2 2   1   fm3 − fm1  < 2 , 1 2 2   1   fm3 − fm3  < 3 . 1 2 2 We set n1 = m12 ,

n2 = m11 ,

n3 = m32

for m11 , m12 , m32 ≥ max{M1 , M2 , M3 }. Then   fn − fn  < 1 , 2 1 2   fn − fn  < 1 , 3 2 22   1   fm3 − fn3  < 3 1 2 3 for this process, we obtain a subsequence  m1∞≥ max{M1 , M2 , M3 }. Continuing fnk k=1 of the sequence {fn }∞ n=1 such that

  fn − fn  < 1 k+1 k 2k for any k ∈ N. Since the series 

fnk

∞ k=1

∞  k=0

1 √ k is convergent, we conclude that 2

is a rapidly Cauchy sequence in X. This completes the proof.

1.2 The Lp Spaces

23

Theorem 1.18 Let E be a measurable set and 1 ≤ p ≤ ∞. Then every rapidly Cauchy sequence in Lp (E) converges both with respect to the Lp (E) norm and pointwise a.e. in E to a function f in Lp (E). Proof Let 1 ≤ p < ∞. We leave the case p = ∞ as an exercise. Let {fn }∞ n=1 be a ∞  rapidly convergent sequence in Lp (E). We choose k to be a convergent series k=1

of positive numbers such that fk+1 − fk p ≤ k2

(1.12)

for any k ∈ N. Let E k = {x ∈ E : |fk+1 (x) − fk (x)| ≥ k } ,

k ∈ N.

Then  p E k = x ∈ E : |fk+1 (x) − fk (x)|p ≥ k ,

k ∈ N.

Hence, 

  p |fk+1 (x) − fk (x)|p ≥ k m E k .

Ek

Now, using (1.12), we get    1 1 p |fk+1 − fk |p = p fk+1 − fk p m Ek ≤ p k k E

≤

Since p ≥ 1, the series

∞ 

1 2p p p k = k . k p

k is convergent and

k=1 ∞  k=1

∞    p m Ek ≤ k < ∞. k=1

Hence and the Borel–Cantelli Lemma, it follows that there is E0 ⊂ E such that m (E0 ) = 0 and for each x ∈ E\E0 there is an index K(x) such that |fk+1 (x) − fk (x)| < k

24

1 Introduction

for any k ≥ K(x). Let x ∈ E\E0 . Then |fn+k (x) − fn (x)| ≤

n+k−1 

∞    fj +1 (x) − fj (x) ≤ j

j =n

j =n

for all n ≥ K(x) and for any k ∈ N. Therefore the sequence of real numbers {fn (x)}∞ n=1 is a Cauchy sequence in R for any x ∈ E\E0 . Consequently it is convergent for any x ∈ E\E0 and let lim fn (x) = f (x) for any x ∈ E\E0 . n→∞

By (1.12), we obtain  |fn+k

⎛ ⎞p ∞  − fn |p ≤ ⎝ j2 ⎠

(1.13)

j =n

E

for all n, k ∈ N. Since fn → f pointwise a.e. on E, we take the limit as k → ∞ in (1.13) and using Fatou’s Lemma, we get 

⎛ ⎞p ∞  |f − fn |p ≤ ⎝ j2 ⎠ j =n

E

for all n ∈ N. Hence, f ∈ Lp (E) and fn → f , as n → ∞, in Lp (E). This completes the proof. Theorem 1.19 (The Riesz–Fischer Theorem) Let E be a measurable set and 1 ≤ p ≤ ∞. Then Lp (E) is a Banach space. Moreover, if fn → f , as n → ∞, in Lp (E), a subsequence of {fn }∞ n=1 converges pointwise a.e. on E to f . in Lp (E). By Theorem 1.17, it follows Proof Let {fn }∞ n=1 be a Cauchy ∞  sequence a rapidly that there is a subsequence fnk k=1 of the sequence {fn }∞ n=1 that is  ∞ p Cauchy sequence in L (E). Hence and Theorem 1.18, it follows that fnk k=1 converges both with respect to the Lp (E) norm and pointwise a.e. on E. Let fnk → f , as k → ∞, in Lp (E). We take  > 0 arbitrarily. Then there exists K ∈ N such that   fn − f  <  k p 2

and

fl − fm p
0 arbitrarily. Then there is a finite disjoint collection {Ik }nk=1 of open intervals such n  that if U = Ik , then k=1

m (A\U ) + m (U \A) <  p .

28

1 Introduction

Let f = κU . Then ⎛ κU − κA p = ⎝

⎞1



p

|κU − κA |

p⎠

1

= (m (A\U ) + m (U \A)) p < .

E

Therefore the step functions are dense in the simple functions with respect to the Lp norm. Hence and Theorem 1.23, it follows that the step functions on [a, b] are dense in Lp ([a, b]). This completes the proof. Theorem 1.25 Let E be a measurable set and 1 ≤ p < ∞. Then Lp (E) is separable. Proof Let [a, b] be a closed bounded interval and S ([a, b]) be the collection of the step functions on [a, b]. Let also, S 1 ([a, b]) be the subcollection of the collection S ([a, b]) consisting of the step functions ψ on [a, b] that take rational values and for which there is a partition P = {x0 , . . . , xn } of [a, b] so that ψ is a rational constant on (xk−1 , xk ), 1 ≤ k ≤ n, and xk , 1 ≤ k ≤ n − 1, are rational numbers. Using the density of the rational numbers in the real numbers, we have that S 1 ([a, b]) is dense in S ([a, b]) with respect to the Lp (E) norm. Because the set of the rational numbers is countable, we have that S 1 ([a, b]) is a countable set. We have that S 1 ([a, b]) ⊆ S ([a, b]) ⊆ Lp ([a, b]) . Since S 1 ([a, b]) is dense in S ([a, b]), using Theorem 1.24, it follows that S 1 ([a, b]) is dense in Lp ([a, b]). For each n ∈ N, we define by Fn to be the collection of the functions that vanishes outside [−n, n] and whose restrictions to [−n, n] belong to S 1 ([−n, n]). Let F =

∞ 

Fn .

n=1

Also, we have 

 lim

n→∞ [−n,n]

|f |p =

|f |p R

for all f ∈ Lp (R). Using the definition of F , we conclude that F is a countable collection of functions that is dense in Lp (R). Hence, the collection of the restrictions on E of the functions in F is a countable dense set in Lp (E). Consequently Lp (E) is separable. This completes the proof.

1.2 The Lp Spaces

29

1.2.6 Duality Definition 1.21 For a normed vector space X, a linear functional T on X is said to be bounded if there is a constant M ≥ 0 such that |T (f )| ≤ Mf 

(1.15)

for all f ∈ X. The infimum of all such M will be called the norm of T and will be denoted by T . Let T be a bounded linear functional on the normed vector space X. Then, for any f, g ∈ X, we have |T (f ) − T (g)| ≤ f − g.

(1.16)

Hence, if fn → f , as n → ∞, in X, i.e., fn , f ∈ X, n ∈ N, fn − f  → 0,

as n → ∞,

using (1.16), we get T (fn ) → T (f ) ,

as

n → ∞.

Proposition 1.1 Let T be a bounded linear functional on the normed vector space X. Then T = sup |T (f )| . f ≤1

(1.17)

Proof By the inequality (1.15), we get |T (f )| ≤ T f  for any f ∈ X. Hence, sup |T (f )| ≤ T .

f ≤1

Let  > 0 be arbitrarily chosen. Then there exists g ∈ X, g = 0, such that |T (g)| ≥ (T − ) g. Hence,      T g  ≥ T −   g 

(1.18)

30

1 Introduction

and sup |T (f )| ≥ T − .

f ≤1

Because  > 0 was arbitrarily chosen, from the last inequality, we obtain sup |T (f )| ≥ T .

f ≤1

Hence and (1.18), we get (1.17). This completes the proof. Theorem 1.26 Let E be a measurable set, 1 ≤ p < ∞, gq = 0. Define the functional T on Lp (E) by

1 1 + = 1, g ∈ Lq (E), p q

 T (f ) =

gf E

for all f ∈ Lp (E). Then T is a bounded linear functional on Lp (E) and T = gq . Proof Let f1 , f2 ∈ Lp (E) and α, β ∈ F , where F = R or F = C. Then  T (αf1 + βf2 ) =

 g (αf1 + βf2 ) =

E



=α

gf1 + β E

(αgf1 + βgf2 ) = E





 αgf1 + E

βgf2 E

gf2 = αT (f1 ) + βT (f2 ) . E

Therefore T is a linear functional on Lp (E). Also, using Hölder’s inequality, we have         |T (f )| =  gf  ≤ |g||f |   E

E

⎛

⎞1 ⎛ ⎞1 q p   q⎠ ⎝ p⎠ ⎝ ≤ |g| |f | = gq f p E

E

for all f ∈ Lp (E). Consequently T is a bounded linear functional on Lp (E). By the last inequality, we get T ≤ gq .

(1.19)

1.2 The Lp Spaces

31

Let 1−q

g1 = gq

sign(g)|g|q−1 .

We have 

 |g1 |p = E

p(1−q)

gq

−q



|g|p(q−1) = gq

E

|g|q = 1, E

i.e., g1 ∈ Lp (E) and g1 p = 1. Next, 

 gg1 =

T (g1 ) = E

1−q

ggq

1−q



sign(g)|g|q−1 = gq

E

|g|q = gq . E

Therefore T ≥ gq . From the last inequality and from (1.19), we obtain T = gq . This completes the proof. Theorem 1.27 Let T and S be bounded linear functionals on a normed vector space X. If T = S on a dense subset X0 of X, then T = S on X. Proof Let g ∈ X be arbitrarily chosen. Then there exists a sequence {gn }∞ n=1 of elements of X0 such that gn → g, as n → ∞, in X. Hence, T (gn ) → T(g),

S (gn ) → S(g),

as

n → ∞,

and T (gn ) = S (gn ) for any n ∈ N. Therefore T (g) = S (g). Because g ∈ X was arbitrarily chosen, we obtain T = S on X. This completes the proof. Theorem 1.28 Let E be a measurable set and 1 ≤ p < ∞, integrable over E and there is an M ≥ 0 such that

1 1 + = 1, g is p q

       gf  ≤ Mf p     E

for any simple function f in Lp (E). Then g ∈ Lq (E) and gq ≤ M.

(1.20)

32

1 Introduction

Proof Since g is integrable over E, then it is finite a.e. on E. By excising a set of measure zero from E, we can assume that g is finite on all of E. 1. Let p > 1. Because |g| is a nonnegative measurable function on E, there exists a sequence {φn }∞ on E that converges pointwise n=1 of measurable simple functions  q ∞ on E to |g| and 0 ≤ φn ≤ |g| on E. Hence, φn n=1 is a sequence of nonnegative simple functions on E such that q

0 ≤ φn ≤ |g|q

on E

q

and φn → |g|q , as n → ∞, pointwise on E. Hence and Fatou’s Lemma, 

 |g| ≤ q

E

q

(1.21)

φn E

for every n ∈ N. Let n ∈ N be arbitrarily chosen. Then q

q−1

φn = φn φn

q−1

≤ |g|φn

q−1

= gsign(g)φn

on E.

Let q−1

fn = sign(g)φn

on

E.

Then fn is a simple function. Because g is integrable over E, we have that φn is q integrable over E. Then φn is integrable over E and 

 |fn |p = E

q

φn . E

Therefore fn ∈ Lp (E). Next, by (1.20), we obtain  E

q φn

 =

q−1 φn φn

 ≤

E

q−1 |g|φn

 =

E

E

⎞1 ⎛ p   q ⎠ ⎝ = gfn ≤ M fn p = M φn . E

E

From here, ⎛ ⎞1 q  ⎝ φnq ⎠ ≤ M, E

q−1

gsign(g)φn

1.2 The Lp Spaces

33

i.e., φn q ≤ M. Hence and (1.21), we obtain gq ≤ M. 2. Let p = 1. Suppose that M is not an essential upper bound for g. Then there is an  > 0 such that the set E = {x ∈ E : |g(x)| > M + } has finite positive measure. Let f = sign(g)κE . Then, by (1.20),               fg  =  gsign(g) = |g| > (M + ) m (E ) .         E E E

(1.22)

On the other hand, by (1.20),           fg  ≤ Mf 1 = M sign(g)κE  = Mm (E ) ,      E

E

which contradicts with (1.22). Therefore M is an essential upper bound for g. This completes the proof. 1 1 + = p q p 1. Suppose that T is a bounded linear functional on L ([a, b]). Then there is a function g ∈ Lq ([a, b]) such that

Theorem 1.29 Let [a, b] be a closed bounded interval, 1 ≤ p < ∞,

b T(f ) =

gf a

for all f ∈ Lp ([a, b]). Proof Let p > 1. We leave the case p = 1 for an exercise. For x ∈ [a, b], we define

Φ(x) = T κ[a,x) .

34

1 Introduction

For [c, d] ⊆ [a, b], we have κ[c,d) = κ[a,d) − κ[a,c) . Then







Φ(d) − Φ(c) = T κ[a,d) − T κ[a,c) = T κ[a,d) − κ[a,c) = T κ[c,d) . If {(ak , bk )}nk=1 is a finite disjoint collection of intervals in (a, b), then n 

|Φ (bk ) − Φ (ak )| =

k=1

n 

sign (Φ (bk ) − Φ (ak )) (Φ (bk ) − Φ (ak ))

k=1

=

n 

sign (Φ (bk ) − Φ (ak )) T κ[ak ,bk )

k=1

=

n 

T sign (Φ (bk ) − Φ (ak )) κ[ak ,bk ) k=1

=T

 n 

 sign (Φ (bk ) − Φ (ak )) κ[ak ,bk ) .

k=1

Consider the simple function f =

n 

sign (Φ (bk ) − Φ (ak )) κ[ak ,bk ) .

k=1

Then  |T(f )| ≤ T f p = T

n 

1

p

(bk − ak )

.

k=1

Consequently n 

 |Φ (bk ) − Φ (ak )| ≤ T

k=1

n 

1

p

(bk − ak )

k=1

and Φ is absolutely continuous on [a, b]. Therefore Φ is differentiable almost everywhere on [a, b], and if g = Φ  , then g is integrable over [a, b] and x Φ(x) =

g a

1.2 The Lp Spaces

35

for all x ∈ [a, b]. Consequently, for each [c, d] ⊆ (a, b), b



T κ[c,d) = Φ(d) − Φ(c) =

gκ[c,d) . a

b Since the functional T and the functional f →

gf are linear on the vector space a

of step functions in Lp ([a, b]), it follows that b T(f ) =

gf a

for all step functions f in Lp ([a, b]). There is a sequence {φn }∞ n=1 of step functions that converges to f in Lp ([a, b]) and also it is uniformly pointwise bounded on [a, b]. Hence and (1.16), we get lim T (φn ) = T(f ).

n→∞

On the other hand, by the Lebesgue Dominated Convergence Theorem, we obtain b

b gφn =

lim

n→∞ a

gf. a

Therefore b T(f ) =

gf a

for all simple functions f in Lp ([a, b]). Since T is bounded,  b       gf  = |T(f )| ≤ T f p     a

for all simple functions f in Lp ([a, b]). From here and from Theorem 1.28, b q gf is we have that g ∈ L ([a, b]). By Theorem 1.26, the functional f → a

bounded on Lp ([a, b]). This functional agrees with T on the simple functions

36

1 Introduction

in Lp ([a, b]). Because the set of the simple functions in Lp ([a, b]) is dense in Lp ([a, b])(see Theorem 1.23), using Theorem 1.27, these two functionals agree on all of Lp ([a, b]). This completes the proof. Theorem 1.30 (The Riesz Representation Theorem) Let E be a measurable set, 1 1 1 ≤ p < ∞, + = 1. For each g ∈ Lp (E), define the bounded linear functional p q Tg on Lp (E) by  Tg (f ) =

gf E

for all f ∈ Lp (E). Then for each bounded linear functional T on Lp (E), there is unique function g ∈ Lq (E) for which Tg = T

and

T = gq .

Proof By Theorem 1.26, it follows that Tg is a bounded linear functional on Lp (E) and Tg  = gq for each g ∈ Lp (E). Also, if g1 , g2 ∈ Lq (E), then 

 (g1 − g2 ) f =

Tg1 −g2 (f ) = E

 (g1 f − g2 f ) =

E

 g1 f −

E

g2 f = Tg1 (f ) − Tg2 (f ) E

for any f ∈ Lp (E). Therefore, if Tg1 = Tg2 , then Tg1 −g2 = 0 and hence, g1 − g2 q = 0, so that g1 = g2 . Therefore, for a bounded linear functional T on Lp (E) there is at most one g ∈ Lp (E) for which Tg = T. 1. Suppose that E = R. Let T be a bounded linear functional on Lp (R). For any n ∈ N, we define the linear functional Tn on Lp ([−n, n]) by   Tn (f ) = T fˆ for all f ∈ Lp ([−n, n]), where fˆ is the extension of f to all of R that vanishes outside [−n, n]. Then f p = fˆp and      |Tn (f )| = T fˆ  ≤ T fˆp = T f p for any f ∈ Lp ([−n, n]), n ∈ N. Hence, Tn  ≤ T ,

n ∈ N.

1.2 The Lp Spaces

37

By Theorem 1.29, it follows that there is a function gn ∈ Lq ([−n, n]), n ∈ N, for which  Tn (f ) = gn f, n ∈ N, E

for all f ∈ Lp ([−n, n]), n ∈ N, and gn q = Tn  ≤ T .

(1.23)

Note that the restriction of gn+1 to [−n, n] agree with gn a.e. on [−n, n], n ∈ N. Define g to be a measurable function on R that agrees with gn a.e. on [−n, n] for each n ∈ N. Hence, for all f ∈ Lp (R) that vanishes outside a bounded set,  T(f ) = gf. R

By (1.23), n

q

|g|q ≤ T ,

n ∈ N.

−n

Because the set of all functions of Lp (E) that vanishes outside a bounded set is dense in Lp (R), using Theorem 1.27, we conclude that Tg agrees with T on all Lp (R). 2. Let E be a measurable set and T be a bounded linear functional on Lp (E). Define the linear functional Tˆ on Lp (E) by    ˆ ) = T f  , T(f E

f ∈ Lp (R).

Then Tˆ is a bounded linear functional on Lp (R). Hence, there is a function gˆ ∈ Lq (R) for which ˆ )= T(f

 gf ˆ R

  for any f ∈ Lp (R). Define g = gˆ  . Then T = Tg . This completes the proof. E

Definition 1.22 Let 1 ≤ p ≤ ∞ and q is its conjugate. The space Lq (·) is called the dual space of the space Lp (·).

38

1 Introduction

Definition 1.23 Let X be a normed vector space. A sequence {fn }∞ n=1 in X is said to converge weakly in X to f ∈ X if lim T (fn ) = T(f )

n→∞

for any linear functional T on X. We will write fn  f

in X.

Remark 1.4 If X is a normed vector space, we will write fn → f in X if fn − f  → 0,

as n → ∞.

In this case, we will say that the sequence {fn }∞ n=1 converges strongly to f in X. Theorem 1.31 Let X be a normed vector space, {fn }∞ n=1 be a sequence in X, f ∈ X. If fn → f in X, then fn  f in X. Proof Since fn → f in X, we have fn − f  → 0,

as n → ∞.

Let T be arbitrarily chosen linear functional on X. Then |T (fn ) − T(f )| ≤ T fn − f  → 0,

as n → ∞.

Because T was arbitrarily chosen linear functional on X, we conclude that fn  f in X. This completes the proof. Theorem 1.32 Let E be a measurable set, 1 ≤ p < ∞ and q is its conjugate. Then fn  f in Lp (E) if and only if 

 gfn =

lim

n→∞ E

gf E

for all g ∈ Lq (E). Proof 1. Let fn  f in Lp (E). Then for every linear functional T we have lim T (fn ) = T(f ).

n→∞

(1.24)

1.2 The Lp Spaces

39

 Hence, using that h →

gh, h ∈ Lp (E), is a linear functional for each g ∈ E

Lq (E), we get (1.24). 2. Assume that (1.24) holds. Let T be arbitrarily chosen linear functional on Lp (E). By the Riesz representation theorem, it follows that for any h ∈ Lp (E) there is a unique g ∈ Lq (E) such that  T(h) =

gh. E

Hence and (1.24), we obtain lim T (fn ) = T(f ).

n→∞

Because T was arbitrarily chosen linear functional on Lp (E), we conclude that fn  f in Lp (E). This completes the proof. Theorem 1.33 Let E be a measurable set, 1 ≤ p < ∞. Then a sequence in Lp (E) can converge weakly to at most one function in Lp (E). p Proof Let q be the conjugate of p. Let also, {fn }∞ n=1 be a sequence in L (E) that p p converges weakly to f1 , f2 ∈ L (E). Then f1 − f2 ∈ L (E) and

   q   1−p 1−p |f1 − f2 |q(p−1) f1 − f2 p sign (f1 − f2 ) |f1 − f2 |p−1  ≤ f1 − f2 p E

E 1−p

 |f1 − f2 |p = f1 − f2 p ,

= f1 − f2 p

E

1−p

i.e., f1 − f2 p we get

sign (f1 − f2 ) |f1 − f2 |p−1 ∈ Lq (E). Hence and Theorem 1.32,



1−p

f1 − f2 p E

sign (f1 − f2 ) |f1 − f2 |p−1 f1

 = lim  =

1−p

f1 − f2 p

n→∞ E

1−p

f1 − f2 p E

sign (f1 − f2 ) |f1 − f2 |p−1 fn

sign (f1 − f2 ) |f1 − f2 |p−1 f2 .

40

1 Introduction

Therefore  0=

1−p

f1 − f2 p E

=

1−p f1 − f2 p

sign (f1 − f2 ) |f1 − f2 |p−1 (f1 − f2 )

 |f1 − f2 |p = f1 − f2 p . E

Consequently f1 = f2 . This completes the proof. Definition 1.24 Let E be a measurable set and 1 ≤ p < ∞. p p p 1. Let {fn }∞ n=1 be a sequence in L (E), f ∈ L (E) and fn  f in L (E). The function f will be called the weak sequential limit. 2. Let f ∈ Lp (E). The function 1−p

f  = f p

sign(f )|f |p−1

will be called the conjugate function of f . Note that f  ∈ Lq (E), where q is the conjugate of p. Exercise 1.3 Let E be a measurable set, 1 ≤ p < ∞, f ∈ Lp (E) and f  be the conjugate function of f . Prove that f ∗ q = 1, where q is the conjugate of p. Remark 1.5 Let X be a linear normed space and fn  f in X. Suppose that the sequence {fn }∞ n=1 is unbounded. By taking a subsequence and relabeling, we can suppose that fn  ≥ n3n , n ∈ N. By taking a further subsequence and relabeling, we can suppose that fn  → α ∈ [1, ∞), n3n

as n → ∞.

Let T be arbitrarily chosen linear functional. Then there is a constant M > 0 such that |T (fn )| ≤ M,

|T(f )| ≤ M

1.2 The Lp Spaces

41

for any n ∈ N. Hence,   n      T n3 fn − T 1 f    fn  α   n          n3 1 1 1  fn − T fn + T fn − T f  = T fn  α α α   n       n3 1 1 = T − fn + T (fn − f )  fn  α α  n   n3 1 1 ≤  −  |T (fn )| + |T (fn ) − T(f )| fn  α α   n  n3 1  1  −  M + |T (fn ) − T(f )| → 0, as n → ∞. ≤ fn  α α Therefore n3n 1 fn  f fn  α

in

X.

Theorem 1.34 Let E be a measurable set and 1 ≤ p < ∞. Suppose that fn  f p in Lp (E). Then {fn }∞ n=1 is bounded in L (E) and f p ≤ lim inf fn p . n→∞



Proof Let q be the conjugate of p and f be the conjugate function of f . Then f  ∈ Lq (E) and using Hölder’s inequality, we get  f  fn ≤ f  q fn p = fn p E

for any n ∈ N. Since fn  f in Lp (E), by Theorem 1.32, we get  f p =

 f  f = lim

f  fn ≤ lim inf fn p .

n→∞

E

n→∞

E

Now, we assume that the sequence {fn }∞ n=1 is unbounded. Using Remark 1.5 and by taking scalar multiples, we can suppose that fn p = n3n , n ∈ N. Let fk be the 1 conjugate functions of fk , k ∈ N. Let also, 1 = and we define 3 1 k = k 3

if

  k−1 E l=1

l fl fk ≥ 0

42

1 Introduction

and 1 k = − k 3

  k−1

if

l fl fk ≤ 0

E l=1

for k ∈ N, k ≥ 2. Then       n     n      1−p  k fk fn  =  k fk p sign (fk ) |fk |p−1 fn       k=1 k=1 E E        n−1  1−p 1−p p−1 p  |f = k fk p sign (fk ) k | fn + n fn p |fn |   k=1  E E       n−1  1 1−p p−1  = k fk p sign (fk ) |fk | fn + n fn p  ≥ n fn p = n, 3  

n ∈ N.

E k=1

Also, k fk q =

Since the series

1 , 3k

k ∈ N.

∞ ∞   1 is a convergent series, we have that k fk is a convergent 3k k=1

k=1

series in Lq (E) and let g=

∞ 

k fk .

k=1

For any n ∈ N, we have           ∞       gfn  =  k fk fn         k=1 E E           n ∞   k fk fn + k fk fn  =    k=1 k=n+1 E E         ∞     n      k fk fn  −  k fk fn  ≥      k=1 k=n+1 E

E

1.2 The Lp Spaces

43

 ∞       ≥ n− k fk  fn p   k=n+1

q

∞ ∞     1 n 1 k f   fn p = n − fn p = , ≥ n− fn p = n − k q 3k 2 (3n ) 2 k=n+1

k=n+1

which is a contradiction, because fn  f in Lp (E) and g ∈ Lq (E). This completes the proof. Theorem 1.35 Let E be a measurable set, 1 ≤ p < ∞ and q is its conjugate. Suppose that fn  f in Lp (E) and gn → g in Lq (E). Then 

 gn fn =

lim

n→∞ E

(1.25)

gf. E

Proof We have 





gn fn − E

gf = E

 (gn − g) fn +

E

g (fn − f ) .

(1.26)

E

By Theorem 1.34, it follows that there is a constant M > 0 such that fn p ≤ M for any n ∈ N. Then, using Hölder’s inequality, we obtain         (gn − g) fn  ≤ |gn − g| |fn | ≤ gn − gq fn p ≤ Mgn − gq → 0,     E

as

n → ∞.

E

(1.27) Since fn  f in Lp (E) and g ∈ Lq (E), using Theorem 1.32, we obtain  g (fn − f ) → 0,

as n → ∞.

E

Hence and (1.26), (1.27), we obtain (1.25). This completes the proof. Theorem 1.36 Let E be a measurable set, 1 ≤ p < ∞ and q be its conjugate. Let also, F ⊂ Lq (E) and its span is dense in Lq (E). Suppose that {fn }∞ n=1 is a bounded sequence in Lp (E) and f ∈ Lp (E). Then fn  f in Lp (E) if and only if 

 fn g =

lim

n→∞ E

for all g ∈ F .

fg E

(1.28)

44

1 Introduction

Proof 1. Let fn  f in Lp (E). Using Theorem 1.32, we conclude that (1.28) holds. 2. Suppose that (1.28) holds. Let g0 ∈ Lq (E) is arbitrarily chosen. For any g ∈ Lq (E) and for any n ∈ N, we have 

 (fn − f ) g0 = E

 (fn − f ) (g0 − g) +

E

(fn − f ) g E

and hence, using Hölder’s inequality, we obtain               (fn − f ) g0  =  (fn − f ) (g0 − g) + (fn − f ) g          E E E                ≤  (fn − f ) (g0 − g) +  (fn − f ) g      E  E      ≤ fn − f p g0 − gq +  (fn − f ) g  .  

(1.29)

E

p We take  > 0 arbitrarily. Since {fn }∞ n=1 is bounded in L (E) and the span of q F is dense in L (E), there is g ∈ F such that

fn − f p g − g0 q
0 arbitrarily. Then there exists an 0 > 0 so that from 0 <  < 0 and |z| ≤ 1, we have |f (x − z) − f (x)| < η for any x ∈ K. Therefore          |f (x) − f (x)| =  f (x − z)ω(z)dz − f (x) ω(z)dz   |z|≤1  |z|≤1         = (f (x − z) − f (x))ω(z)dz   |z|≤1   ≤ |f (x − z) − f (x)|ω(z)dz |z|≤1



0 arbitrarily. Then there exists a function g ∈ C0 (X) so that f − η gp < . By Theorem 1.49, it follows that there exists an 0 > 0 so that f − 3 η η gp < and f − g  < for any 0 <  < 0 . Now, applying the triangle 3 3 inequality, we arrive at f − f p = f − g + g − g + g − f p ≤ f − g p + g − gp + g − f p η η η < + + 3 3 3 = η, 0 <  < 0 . This completes the proof.

1.4 Cones in Rn

55

Corollary 1.1 Let X be an open set in Rn . Then the space C0∞ (X) is dense in Lp (X). Proof Let f ∈ Lp (X) be arbitrarily chosen. Take η > 0 arbitrarily. Let g ∈ C0 (X) η be chosen so that f − gp < . Since g ∈ C0∞ (X), there is an 0 > 0 so that 2 η g − gp < for any 0 <  < 0 . Hence, 2 f − g p = f − g + g − g p ≤ f − gp + g − g p η η < + 2 2 = η, 0 <  < 0 . This completes the proof.

1.4 Cones in Rn Definition 1.33 A cone in Rn with vertex at 0 is a set Γ with the property that if x ∈ Γ , then λx ∈ Γ for every λ > 0. The symbol prΓ will denote the intersection of Γ with the unit sphere centred at 0. A cone Γ  is called compact in the cone Γ if prΓ  ⊂ prΓ , in which case we write Γ  ⊂⊂ Γ . The cone Γ ∗ = {ξ ∈ Rn : (ξ, x) ≥ 0 ∀x ∈ Γ }, where (·, ·) is the standard inner product in Rn , is called the conjugate cone of Γ . Exercise 1.5 Prove that Γ ∗ is a closed, convex cone with vertex at 0. ∗ Exercise 1.6 Prove Γ ∗ = chΓ . Definition 1.34 A cone Γ is said to be acute if for any e ∈ pr intΓ ∗ the set {x : 0 ≤ (e, x) ≤ 1, x ∈ chΓ } is bounded in Rn . Example 1.8 Let {e1 , e2 , . . . , en } be a basis in Rn . Then Γ = {x ∈ Rn : (ek , x) > 0, k = 1, 2, . . . , n} is an acute cone. Exercise 1.7 Let Γ be an acute cone. Prove that chΓ does not contain an integral straight line. Exercise 1.8 Suppose chΓ does not contain a straight line. Prove that intΓ ∗ = Ø.

56

1 Introduction

Exercise 1.9 Let intΓ ∗ = Ø. Prove that for every C  ⊂⊂ intΓ ∗ there exists a constant σ > 0 such that (ξ, x) ≥ σ |ξ ||x| for every ξ ∈ C  and every x ∈ chΓ . Definition 1.35 The function μΓ (ξ ) = − inf (ξ, x) x∈prΓ

is called the indicator of the cone Γ . Exercise 1.10 Prove that μΓ (ξ ) is a convex function. Definition 1.36 Let Γ ⊂ Rn be a closed, convex, acute cone. Set C = intΓ ∗ . A smooth (n − 1)-dimensional surface S without an edge is said to be C-like if each straight line x = x0 + te, −∞ < t < ∞, e ∈ prΓ , intersects it in a unique point. Every C-like surface S cuts Rn in two unbounded regions S+ and S− such that 1. S+ lies “above” S, 2. S− lies “below” S, 3. Rn = S+ ∪ S ∪ S− . Exercise 1.11 Let Γ be a closed, convex, acute cone and suppose S is a C-like surface. Prove that S+ = S + Γ.

1.5 Advanced Practical Problems Problem 1.1 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets. Prove that for every φ ∈ ∞ ∞ C0∞ (X1 × X2 ) there exist sequences {φk }∞ k=1 ⊂ C0 (X1 × X2 ) and {Nk }k=1 ⊂ N ∪ {0} such that φk (x, y) =

Nk 

φik (x)ψik (y),

(x, y) ∈ X1 × X2 ,

i=1

and φk →k→∞ φ in C0∞ (X1 × X2 ). Solution Let suppφ ⊂⊂ X˜1 × X˜2 ⊂⊂ X1 × X2 ⊂⊂ X1 × X2 . By the Weierstrass theorem, it follows that there exists a sequence of polynomials Pk , k ∈ N, such that   1  α  D Pk (x, y) − D α φ(x, y) < , k

|α| ≤ k,

(x, y) ∈ X1 × X2 .

1.5 Advanced Practical Problems

57

Choose functions ξ ∈ C0∞ (X1 ) and η ∈ C0∞ (X2 ) so that ξ(x) = 1 for x ∈ X˜1 and η(y) = 1 for y ∈ X˜2 . Define φk (x, y) = ξ(x)η(y)Pk (x, y),

k ∈ N,

(x, y) ∈ X1 × X2 .

We have suppφk ⊂ X1 × X2 ⊂⊂ X1 × X2 , k ∈ N, and ⎧   ⎨1  α  D φ(x, y) − D α φk (x, y) ≤ kcα ⎩ k

if (x, y) ∈ X˜1 × X˜2 , if (x, y) ∈ X1 × X2

    for |α| ≤ k. Here the constants cα are obtained by using, for β ≤ α, sup D β ξ(x) x∈X 1     and sup D β η(y). Therefore φk →k→∞ φ in C0∞ (X1 × X2 ). y∈X2

Problem 1.2 Prove that for every function φ1 ∈ C0∞ (R) there exists a function φ2 ∈ C0∞ (R) such that φ1 (x) = φ2 (x), for every x ∈ R, if and only if ∞ φ1 (x)dx = 0. −∞

Solution 1. Let φ1 ∈

C0∞ (R)

∞ φ1 (x)dx = 0. We consider the function

and −∞

x φ2 (x) =

φ1 (s)ds,

x ∈ R.

−∞

Since φ1 ∈ C0∞ (R), it follows that φ2 ∈ C ∞ (R). If suppφ1 ⊂ [a, b] ⊂ R, a < b, then φ2 (x) = 0 for x < a. Therefore suppφ2 ⊂ [a, ∞). Since φ2 (∞) = ∞ φ1 (x)dx = 0, there exists a c > a such that suppφ2 ⊂ [a, c]. −∞

2. Let φ1 , φ2 ∈ C0∞ (R) and φ1 (x) = φ2 (x)

for x ∈ R.

Integrating from −∞ to x, we find x φ1 (s)ds = φ2 (x). −∞

(1.35)

58

1 Introduction

Since φ2 ∈ C0∞ (R), we have φ2 (∞) = 0. Hence, using (1.35), we obtain ∞ φ1 (s)ds = 0. −∞

Problem 1.3 Prove that for every φ ∈ C0∞ (R) there exists a function φ1 ∈ C0∞ (R) such that ∞ φ(x) = φ0 (x)

φ(s)ds + φ1 (x),

x ∈ R,

−∞

where φ0 ∈

C0∞ (R),

∞ φ0 (s)ds = 1.

if and only if −∞

Solution 1. Let φ0 ∈ C0∞ (R) and

∞ φ0 (s)ds = 1. Consider the function −∞

x

x

φ1 (x) =

φ(s)ds − −∞

∞ φ0 (s)ds

−∞

(1.36)

φ(s)ds.

−∞

Since φ, φ0 ∈ C0∞ (R), it follows φ1 ∈ C ∞ (R). Let suppφ, suppφ0 ⊂ [a, b] ⊂ R, a < b. Then φ1 (x) = 0 for x < a. Therefore suppφ1 ⊂ [a, ∞). From (1.36), for x = ∞, we have ∞ φ1 (∞) =

∞ φ(s)ds −

−∞

∞ φ0 (s)ds

−∞

∞ φ(s)ds =

−∞

∞ φ(s)ds −

−∞

φ(s)ds = 0.

−∞

Consequently there exists a constant c > 0 such that suppφ1 ⊂ [a, c], and therefore φ1 ∈ C0∞ (R). 2. Let φ, φ0 , φ1 ∈ C0∞ (R) and φ1 (x)

∞ = φ(x) − φ0 (x) −∞

φ(s)ds,

x ∈ R.

(1.37)

1.5 Advanced Practical Problems

59

We integrate the Eq. (1.37) from −∞ to ∞ and get ∞

φ1 (x)dx

∞ =

−∞

∞ φ(x)dx −

−∞

∞ φ0 (x)dx

−∞

φ(x)dx, −∞

that is, ∞ 0=

∞   φ(x)dx 1 − φ0 (x)dx

−∞

−∞

for every φ ∈ C0∞ (R). In particular, the last equation is valid for every φ ∈ ∞ ∞ C0 (R) for which φ(x)dx = 1. For such φ, we obtain −∞

∞ φ0 (x)dx = 1. −∞

Problem 1.4 Prove that for every φ ∈ C0∞ (R) there exists a function φ2 ∈ C0∞ (R) such that φ(x) =

φ1 (x)

∞ s

φ(τ )dτ ds + φ2 (x),

x ∈ R,

−∞ −∞

where φ1 ∈

C0∞ (R),

∞ φ1 (x)dx = 1.

if and only if −∞

Solution 1. Let φ, φ1 ∈

C0∞ (R)

∞ φ1 (x)dx = 1. Consider the function

and −∞

⎛ φ2 (x) = − ⎝

x

−∞

⎞⎛ φ1 (s)ds ⎠ ⎝

∞ s

−∞ −∞

⎞ φ(τ )dτ ds ⎠ +

x y

−∞ −∞

φ(s)dsdy,

x ∈ R.

60

1 Introduction

Since φ, φ1 ∈ C0∞ (R), we conclude that φ2 ∈ C ∞ (R). Let suppφ, suppφ1 ⊂ [a, b] ⊂ R, a < b. Then, for x < a, by the definition of the function φ2 , we find that φ2 (x) = 0. Thus, suppφ2 ⊂ [a, ∞). Next, ⎛ φ2 (∞) = − ⎝

⎞⎛

∞

φ1 (s)ds ⎠ ⎝

−∞

∞

∞ s

⎞ φ(τ )dτ ds ⎠ +

−∞ −∞

y

∞

=−

φ(s)dsdy

−∞ −∞

y

φ(s)dsdy +

−∞ −∞

∞ y

φ(s)dsdy

−∞ −∞

= 0. Therefore there exists a c > 0 so that suppφ2 ⊂ [a, c] and φ2 ∈ C0∞ (R). 2. Let φ, φ1 , φ2 ∈ C0∞ (R) and φ(x) =

φ1 (x)

∞ s

φ(τ )dτ ds + φ2 (x),

x ∈ R.

−∞ −∞

We integrate the last equation from −∞ to x and we get x −∞

⎛ φ(s)ds = ⎝

⎞⎛

x

φ1 (s)ds ⎠ ⎝

−∞

∞ s

⎞

−∞ −∞

∞ s = φ1 (x)

x

φ(τ )dτ ds ⎠ +

φ2 (s)ds

−∞

φ(τ )dτ ds + φ2 (x),

x ∈ R,

−∞ −∞

which we integrate from −∞ to x and we find x y −∞ −∞

⎛ φ(s)dsdy = ⎝

x

−∞

⎛ =⎝

x

−∞

⎞⎛ φ1 (s)ds ⎠ ⎝

∞ s

−∞ −∞

⎞⎛ φ1 (s)ds ⎠ ⎝

∞ s

−∞ −∞

⎞ φ(τ )dτ ds ⎠ + ⎞

x

φ2 (s)ds

−∞

φ(τ )dτ ds ⎠ + φ2 (x),

x ∈ Rn .

1.5 Advanced Practical Problems

61

In particular, for x = ∞, by the last equation and using that φ2 (∞) = 0, we get ∞ y

⎛ φ(s)dsdy = ⎝

−∞ −∞

∞

⎞⎛ φ1 (s)ds ⎠ ⎝

−∞

⎛ =⎝

∞

∞ s

⎞ φ(τ )dτ ds ⎠ + φ2 (∞)

−∞ −∞

⎞⎛ φ1 (s)ds ⎠ ⎝

−∞

∞ s

⎞ φ(τ )dτ ds ⎠

−∞ −∞

or ⎛ ⎝1 −

∞

⎞ φ1 (s)ds ⎠

−∞

∞ s φ(τ )dτ ds = 0.

−∞ −∞

Since the last equation holds for any φ ∈ C0∞ (R), it holds for those φ ∈ C0∞ (R) for which ∞ s φ(τ )dτ ds = 1. −∞ −∞

∞ φ1 (s)ds = 1.

For such φ, we obtain −∞

Problem 1.5 Prove that for every φ ∈ C0∞ (R) there exists a function φ2 ∈ C0∞ (R) such that ∞ s φ(x) = φ1 (x)

φ(τ )dτ ds + φ2 (x),

x ∈ R,

−∞ −∞

where φ1 ∈

C0∞ (R),

∞ s φ1 (τ )dτ ds = 1.

if and only if −∞ −∞

Hint For φ, φ1 ∈ x y φ2 (x) = −∞ −∞

C0∞ (R),

consider the function ⎛

φ(s)dsdy − ⎝

x y

−∞ −∞

⎞⎛ φ1 (s)dsdy ⎠ ⎝

∞ s

−∞ −∞

⎞ φ(τ )dτ ds ⎠ ,

x ∈ R.

62

1 Introduction

Problem 1.6 Prove that for every φ ∈ C0∞ (R) there exists φ3 ∈ C0∞ (R) such that ∞ s φ(x) = φ1 (x)

φ(τ )dτ ds

∞ s

+ φ2 (x)

−∞ −∞

φ(τ )dτ ds + φ3 (x),

x ∈ R,

−∞ −∞

where φ1 , φ2 ∈ C0∞ (R), if and only if ∞ s

∞ φ1 (τ )dτ ds +

−∞ −∞

φ2 (x)dx = 1.

−∞

Solution 1. Let φ, φ1 , φ2 ∈ C0∞ (R) and ∞ s

∞ φ1 (τ )dτ ds +

−∞ −∞

φ2 (x)dx = 1.

−∞

Define the function x y φ3 (x) =

⎛ φ(s)dsdy − ⎝

−∞ −∞

⎛

−⎝

x

⎞⎛ φ2 (s)ds ⎠ ⎝

−∞

x y

−∞ −∞

∞

s

⎞⎛ φ1 (s)dsdy ⎠ ⎝

∞ s

⎞ φ(τ )dτ ds ⎠

−∞ −∞

⎞

φ(τ )dτ ds ⎠ ,

x ∈ R.

−∞ −∞

Since φ, φ1 , φ2 ∈ C0∞ (R), by the definition of the function φ3 , it follows that φ3 ∈ C ∞ (R). Let suppφ, suppφ1 ,suppφ2 ⊂ [a, b]. Then, for x < a, by the definition of φ3 , we find φ3 (x) = 0. Consequently suppφ3 ⊂ [a, ∞). Note that ∞ y φ3 (∞) = −∞ −∞

⎛

−⎝

∞

⎛ φ(s)dsdy − ⎝ ⎞⎛ φ2 (s)ds ⎠ ⎝

−∞

φ(s)dsdy −∞ −∞

−∞ −∞

∞ s

−∞ −∞

∞ y =

∞ y

⎞⎛ φ1 (s)dsdy ⎠ ⎝ ⎞

φ(τ )dτ ds ⎠

∞ s

−∞ −∞

⎞ φ(τ )dτ ds ⎠

1.5 Advanced Practical Problems

∞ y −

63

⎛ φ(s)dsdy ⎝

−∞ −∞

∞

∞ φ1 (s)dsdy +

−∞ −∞

y

=

∞ y

∞

φ2 (s)ds ⎠

−∞

y

φ(s)dsdy − −∞ −∞

⎞

φ(s)dsdy

−∞ −∞

= 0. From the last relation, we conclude that there exists a c > 0 so that suppφ3 ⊂ [a, c] and φ3 ∈ C0∞ (R). 2. Let φ, φ1 , φ2 , φ3 ∈ C0∞ (R) be such that ∞ s φ(x) = φ1 (x)

φ(τ )dτ ds +

φ2 (x)

−∞ −∞

∞ s

φ(τ )dτ ds + φ3 (x),

x ∈ R.

−∞ −∞

We integrate the last equation from −∞ to x and we get x

⎛ φ(s)ds = ⎝

−∞

x

⎞⎛ φ1 (s)ds ⎠ ⎝

−∞

x +

=⎝

x

⎞

⎛

φ(τ )dτ ds ⎠ + ⎝

−∞ −∞

⎞⎛

x

φ2 (s)ds ⎠ ⎝

−∞

∞ s

−∞ −∞

φ3 (s)ds

−∞

⎛

∞ s

⎞⎛ φ1 (s)ds ⎠ ⎝

−∞

+φ3 (x),

∞ s

⎞ φ(τ )dτ ds ⎠ + φ2 (x)

−∞ −∞

∞ s φ(τ )dτ ds

−∞ −∞

x ∈ R,

which we integrate from −∞ to x and we find x y −∞ −∞

⎛ φ(s)dsdy = ⎝

x y

⎞⎛ φ1 (s)dsdy ⎠ ⎝

−∞ −∞

⎛

+⎝

x

φ2 (s)ds ⎠ ⎝

−∞

x + −∞

⎞⎛

φ3 (s)ds

∞ s

−∞ −∞

∞

s

−∞ −∞

⎞ φ(τ )dτ ds ⎠ ⎞

φ(τ )dτ ds ⎠

⎞ φ(τ )dτ ds ⎠

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1 Introduction

⎛ =⎝

x y

⎞⎛ φ1 (s)dsdy ⎠ ⎝

−∞ −∞

⎛

+⎝

x

∞ s

⎞⎛

φ2 (s)ds ⎠ ⎝

−∞

⎞ φ(τ )dτ ds ⎠

−∞ −∞

∞

s

⎞

φ(τ )dτ ds ⎠ + φ3 (x),

x ∈ R.

−∞ −∞

By the last equation, for x = ∞, we find ⎛ ⎝

∞ y

⎞⎛ φ(s)dsdy ⎠ ⎝1 −

−∞ −∞

Take φ ∈

∞ y

∞ φ1 (s)dsdy −

−∞ −∞

C0∞ (R)

⎞ φ2 (s)ds ⎠ = 0.

−∞

∞ y φ(s)dsdy = 1. For such φ, we arrive at

so that −∞ −∞

∞ y

∞ φ1 (s)dsdy +

−∞ −∞

φ2 (s)ds = 1.

−∞

Problem 1.7 Prove that for every φ ∈ C0∞ (R) there exists φ3 ∈ C0∞ (R) such that ∞ s φ(x) = φ1 (x)

φ(τ )dτ ds

+ φ2 (x)

−∞ −∞

∞ s

φ(τ )dτ ds + φ3 (x),

x ∈ R,

−∞ −∞

where φ1 , φ2 ∈ C0∞ (R), if and only if ∞ s

∞ φ1 (τ )dτ ds +

−∞ −∞

φ2 (x)dx = 1.

−∞

Hint Use the function x y φ3 (x) = −∞ −∞

⎛

−⎝

x

−∞

⎛ φ(s)dsdy − ⎝ ⎞⎛ φ2 (s)ds ⎠ ⎝

x y

−∞ −∞

∞

s

−∞ −∞

⎞⎛ φ1 (s)dsdy ⎠ ⎝ ⎞

φ(τ )dτ ds ⎠ ,

∞ s

−∞ −∞

x ∈ R.

⎞ φ(τ )dτ ds ⎠

1.5 Advanced Practical Problems

65

Problem 1.8 Prove that for every φ ∈ C0∞ (R) there exists φ4 ∈ C0∞ (R) such that φ(x) =

φ4 (x) +

∞ s1 s2



φ(τ )dτ ds2 ds1 φ1 (x) + φ2 (x) + φ3 (x) ,

x ∈ R,

−∞ −∞ −∞

where φ1 , φ2 , φ3 ∈ C0∞ (R), if and only if ∞ s1 s2

∞ s

∞

φ1 (τ )dτ ds2 ds1 + −∞ −∞ −∞

φ2 (τ )dτ ds +

−∞ −∞

φ3 (s)ds = 1.

−∞

Hint Use the function x x1 x2 φ4 (x) =

φ(s)dsdx2 dx1 −∞ −∞ −∞

⎛

−⎝

x x1 x2

x x1 φ1 (s)dsdx2 dx1 +

−∞ −∞ −∞

⎛ ×⎝

∞

s1

s2

φ(τ )dτ ds2 ds1 ⎠ ,

φ3 (s)ds ⎠

φ2 (s)dsdx1 +

−∞ −∞

⎞

⎞

x −∞

x ∈ R.

−∞ −∞ −∞

Problem 1.9 Prove that for every φ ∈ C0∞ (R) there exists φ3 ∈ C0∞ (R) such that φ(x) =

φ3 (x) +

∞ s1 s2



φ(τ )dτ ds2 ds1 φ1 (x) + φ2 (x) ,

−∞ −∞ −∞

where φ1 , φ2 ∈ C0∞ (R), if and only if ∞ s1 s2

∞ s φ1 (τ )dτ ds2 ds1 +

−∞ −∞ −∞

−∞ −∞

φ2 (τ )dτ ds = 1.

x ∈ R,

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1 Introduction

Hint Use the function x x1 x2 φ3 (x) =

φ(s)dsdx2 dx1 −∞ −∞ −∞

⎛

−⎝

x x1 x2

x x1 φ1 (s)dsdx2 dx1 +

−∞ −∞ −∞

⎛ ×⎝

∞

s1

s2

⎞ φ2 (s)dsdx1 ⎠

−∞ −∞

⎞ φ(τ )dτ ds2 ds1 ⎠ ,

x ∈ R.

−∞ −∞ −∞

Problem 1.10 Prove that for every φ ∈ C0∞ (R) there exists φ3 ∈ C0∞ (R) such that φ(x) =

φ3 (x) +

∞ s1 s2

φ(τ )dτ ds2 ds1 φ1 (x) + φ2 (x) ,

x ∈ R,

−∞ −∞ −∞

where φ1 , φ2 ∈ C0∞ (R), if and only if ∞ s1 s2

∞ φ1 (τ )dτ ds2 ds1 +

−∞ −∞ −∞

φ2 (τ )dτ = 1.

−∞

Hint Use the function x x1 x2 φ3 (x) =

⎛ φ(s)dsdx2 dx1 − ⎝

−∞ −∞ −∞

⎛

×⎝

∞

s1

x x1 x2

x φ1 (s)dsdx2 dx1 +

−∞ −∞ −∞

s2

⎞

φ(τ )dτ ds2 ds1 ⎠ ,

⎞ φ2 (s)ds ⎠

−∞

x ∈ R.

−∞ −∞ −∞

Problem 1.11 Prove that for every φ ∈ C0∞ (R) there exists φ3 ∈ C0∞ (R) such that φ(x) =

φ3 (x) +

∞ s1 s2

−∞ −∞ −∞



φ(τ )dτ ds2 ds1 φ1 (x) + φ2 (x) ,

x ∈ R,

1.5 Advanced Practical Problems

67

where φ1 , φ2 ∈ C0∞ (R), if and only if ∞ s1

∞ φ1 (τ )dτ ds1 +

−∞ −∞

φ2 (τ )dτ = 1.

−∞

Hint Use the function x x1 x2 φ3 (x) =

⎛ φ(s)dsdx2 dx1 − ⎝

−∞ −∞ −∞

⎛

×⎝

x x1

x φ1 (s)dsdx1 +

−∞ −∞

∞ s1 s2

φ2 (s)ds ⎠

−∞

⎞

φ(τ )dτ ds2 ds1 ⎠ ,

⎞

x ∈ R.

−∞ −∞ −∞

Problem 1.12 Prove that for every φ ∈ C0∞ (R) there exists φ2 ∈ C0∞ (R) such that φ(x) =

φ2 (x) + φ1 (x)

∞ s1 s2 φ(τ )dτ ds2 ds1 ,

x ∈ R,

−∞ −∞ −∞

where φ1 ∈

C0∞ (R),

∞ s1 s2 φ1 (τ )dτ ds2 ds1 = 1.

if and only if −∞ −∞ −∞

Hint Use the function x x1 x2 φ2 (x) =

⎛ φ(s)dsdx2 dx1 − ⎝

−∞ −∞ −∞

⎛

×⎝

∞

x1

x x1 x2

⎞ φ1 (s)dsdx2 dx1 ⎠

−∞ −∞ −∞

x2

⎞

φ(s)dsdx2 dx1 ⎠ ,

x ∈ R.

−∞ −∞ −∞

Problem 1.13 Prove that for every φ ∈ C0∞ (R) there exists φ2 ∈ C0∞ (R) such that φ(x) =

φ2 (x) + φ1 (x)

∞ s1 s2 φ(τ )dτ ds2 ds1 ,

−∞ −∞ −∞

where φ1 ∈

C0∞ (R),

∞ s1 φ1 (τ )dτ ds1 = 1.

if and only if −∞ −∞

x ∈ R,

68

1 Introduction

Hint Use the function x x1 x2 φ2 (x) =

⎛ φ(s)dsdx2 dx1 − ⎝

−∞ −∞ −∞

⎛

×⎝

∞

x x1

⎞ φ1 (s)dsdx1 ⎠

−∞ −∞

x1

x2

⎞

φ(s)dsdx2 dx1 ⎠ ,

x ∈ R.

−∞ −∞ −∞

Problem 1.14 Prove that for every φ ∈ C0∞ (R) there exists φ2 ∈ C0∞ (R) such that φ(x) =

φ2 (x) + φ1 (x)

∞ s1 s2 φ(τ )dτ ds2 ds1 ,

x ∈ R,

−∞ −∞ −∞

where φ1 ∈

C0∞ (R),

∞ φ1 (τ )dτ = 1.

if and only if −∞

Hint Use the function x x1 x2 φ2 (x) =

⎛ φ(s)dsdx2 dx1 − ⎝

−∞ −∞ −∞

x

⎞⎛ φ1 (s)ds ⎠ ⎝

−∞

∞ x1 x2

⎞ φ(s)dsdx2 dx1 ⎠ ,

−∞ −∞ −∞

x ∈ R. Problem 1.15 Let I be an open interval in R, V is a Banach space with a norm || · ||, f : I → V is a smooth map. Prove 1. f (y) − f (x) ≤ |y − x| sup f  (x + t (y − x)), x, y ∈ I , t ∈[0,1]

2. f (y) − f (x) − v(y − x) ≤ |y − x| sup f  (x + t (y − x)) − v, v ∈ V . t ∈[0,1]



1. Solution Let M = sup ||f (x + t (y − x))||. We define the set t ∈[0,1]

 E = t : 0 ≤ t ≤ 1,

 ||f (x + t (y − x)) − f (x)|| ≤ Mt|y − x| .

Since f is a continuous function, E is a closed subset of the interval [0, 1]. On the other hand, ||f (x + 0 · (y − x)) − f (x)|| = ||f (x) − f (x)|| ≤ M · 0 · |y − x|, so 0 belongs to E. From this, we conclude that E is compact and it has a maximal element s. We suppose that s < 1. Then we can find t > s such that t − s is

1.5 Advanced Practical Problems

69

sufficiently small. Hence, f (x + t (y − x)) − f (x) = f (x + t (y − x)) − f (x + s(y − x)) + f (x + s(y − x)) − f (x) ≤ f (x + t (y − x)) − f (x + s(y − x)) + f (x + s(y − x)) − f (x) = (t − s)(y − x)f  (x + ξ(y − x)) + f (x + s(y − x)) − f (x) ≤ M(t − s)|y − x| + Ms|y − x| = Mt|y − x|, where ξ is between t and s. This contradicts with the assumption that s is maximal. Therefore s = 1. For t = 1, we obtain f (y) − f (x) ≤ sup f  (x + t (y − x))|y − x|. t ∈[0,1]

2. Hint Use the function g(x) = f (x) − xv and part 1. Problem 1.16 Let I be an open interval in R, V is a Banach space with a norm || · ||, f : I → V is a continuous map that is differentiable on I \F , where F is a closed subset of I and f ≡ 0 on F . Prove that if x ∈ F and f  (y) →y→x 0, y ∈ I \F , then f  (x) exists and it is zero for every x ∈ I . Solution Let y ∈ F . Then f (y) − f (x) = 0. From this, f  (x) exists for every x ∈ F and f  (x) = 0 for every x ∈ F . Now, take y ∈ / F , x ≤ y, and let z be the point in F ∩ [x, y] closest to y. From the previous problem, we have ||f (y) − f (x)|| = ||f (y) − f (z) + f (z) − f (x)|| = ||f (y) − f (z)|| and ||f (y) − f (z)|| ≤ |y − z| sup ||f  (z + t (y − z))||. t ∈[0,1]

The last inequality implies ||f (y) − f (x)|| = o(|y − x|) when y → x. Therefore  f (x + h) − f (x)    lim   = 0, h→0 h

∀y ∈ I \F.

Consequently f  (x) exists for every x ∈ I and actually f  (x) = 0 for every x ∈ I .

70

1 Introduction

Problem 1.17 Let P be a polynomial and define ⎧   1 −1 ⎨ P e x for x > 0, f (x) = x ⎩ 0 for x ≤ 0. Prove that 1. f is a differentiable function on R and f  (0) = 0, 2. f ∈ C ∞ (R). Problem 1.18 Prove that there exists a function φ ∈ C0∞ (Rn ) for which φ(0) > 0. Hint Use the function 

−

1

1−|x|2 for |x| < 1, φ(x) = e 0 for |x| ≥ 1

and the previous problem. Problem 1.19 Let R > 0 and f ∈ C k (UR ), k ≥ 1. Prove  xj fj (x), fj ∈ C k−1 (UR ), x ∈ UR , 1. f (x) − f (0) = j

1 for every multi-index α such that |α| ≤ k, 2. ∂ fj (0) = ∂ ∂j f (0) 1 +    |α|     3. sup ∂ α fj  ≤ sup ∂ α ∂j f , |α| ≤ k. α

UR

α

UR

Solution We will prove the assertions for k = 1, as for k > 1 one can use induction. 1 1. Setting fj (x) =

∂ f (tx)dt, x ∈ UR , we get ∂xj

0

 j

xj fj (x) =

1  0

j

∂ xj f (tx)dt = ∂xj

1 df (tx) = f (x) − f (0), 0

We note that fj ∈ C (UR ). 2. From the definition of the functions fj , it follows that fj (0) =

∂ fj (0), ∂xj

∂ 1 ∂ ∂ fj (0) = fj (0). ∂xi 2 ∂xi ∂xj

x ∈ UR .

1.5 Advanced Practical Problems

71

3. The definition of the functions fj implies 1  1     ∂  ∂     ∂   f (tx)dt ≤ sup  f (tx)dt ≤ sup  f (x). |fj (x)| ≤  ∂xj ∂x ∂x j j |x| 0. Since h is continuous on X, there exists a neighbourhood U (a) ⊂ X of the point a such that h(x) > 0 for every x ∈ U (a). We may choose the function φ ∈ C0∞ (X) such that φ > 0 on U (a). Then 

 h(x)φ(x)dx ≥

X

h(x)φ(x)dx > 0, U (a)

which contradicts with (1.38). Consequently f ≡ g on X. Problem 1.21 Let X ⊂ Rn , U1 ⊆ X, f, g ∈ L1loc (X) with 

 f φdx =

X

gφdx X

for every φ ∈ C0∞ (X). Prove that f ≡ g almost everywhere on X. Solution Let h(x) = f (x) − g(x), x ∈ X. We have  h(x)φ(x)dx = 0 X

(1.39)

72

1 Introduction

for every φ ∈ C0∞ (X). Now, we choose a function φ ∈ C0∞ (X) so that suppφ = U1  φ(x)dx = 1. Then

and X



1 tn

φ

x − y  t

dy = 1,

x ∈ X.

X

Therefore h(x) = h(x) 1 = n t



1 tn

 φ

x − y  dy t

X

(h(x) − h(y)) φ

 x − y  x − y  1 dy + n h(y)φ dy, t t t

X

x ∈ X.

X

x − y    We take t small enough so that   < 1, x, y ∈ X. For this t, using (1.39), we t have  x − y  dy = 0, x ∈ X. h(y)φ t X

Consequently, h(x) =

1 tn

 (h(x) − h(y)) φ |x−y| 1.

Prove that f ∈ Lp ((1, ∞)) if and only if p = 2.   1 , x ∈ (0, 1], 1 ≤ p < ∞. Prove that f ∈ Problem 1.25 Let f (x) = log x Lp ((0, 1]) and f ∈ / L∞ ((0, 1]). Problem 1.26 Let E be a measurable set, 1 ≤ p < ∞ and q is the conjugate of p, f ∈ Lp (E). Prove that f ≡ 0 if and only if  fg = 0 E

for any g ∈ Lq (E). Problem 1.27 Let E be a measurable set of finite measure, 1 ≤ p1 < p2 ≤ ∞. Prove that if fn → f strongly in Lp2 (E), then fn → f strongly in Lp1 (E). Problem 1.28 Let E be a measurable set, 1 ≤ p <  ∞, q is the conjugate of p, S is dense in Lq (E). Prove that if g ∈ Lp (E) and fg = 0 for any f ∈ S, then E

g = 0.

Problem 1.29 Let E be a measurable set, 1 ≤ p < ∞. Prove that the functions in Lp (E) that vanish outside a bounded set are dense in Lp (E). Problem 1.30 Let [a, b] be a closed bounded interval and fn  f in C ([a, b]). Prove that {fn }n∈N converges pointwise on [a, b] to f . Problem 1.31 Let [a, b] be a closed bounded interval and fn  f in L∞ ([a, b]). Prove that x fn =

lim

n→∞ a

for any x ∈ [a, b].

x f a

74

1 Introduction

For any f ∈ S (Rn ), α, β ∈ Nn ∪ {0} and l, k, m ∈ N ∪ {0}, we set     qα,β (f ) = sup x α D β f (x), x∈Rn

    ql,β (f ) = sup sup x α D β f (x), |α|≤l x∈Rn

     ∗ qα,β (f ) = x α D β f (x)dx, Rn

  k   |f |k,m = sup sup  1 + |x|2 D β f (x). |β|≤m x∈Rn

Problem 1.32 Prove that the following assertions are equivalent 1. 2. 3. 4. 5.

f ∈ S (Rn ), ∗ qα,β (f ) < ∞ for any α, β ∈ Nn ∪ {0}, |f |k,m < ∞ for any k, m ∈ N ∪ {0}, ql,β (f ) < ∞ for any l ∈ N ∪ {0}, ∀β ∈ Nn ∪ {0}, qα,β (f ) < ∞ for any α, β ∈ Nn ∪ {0}.

Problem 1.33 Prove that S (Rn ) embeds continuously in every space Lp (Rn ), p ≥ 1. Solution Let f ∈ S (Rn ). Then 

 |f (x)|p dx = Rn

Rn

(1 + |x|2 )np |f (x)|p p dx ≤ |f |n,0 (1 + |x|2)np

 Rn

dx p ≤ c1 |f |n,0 , (1 + |x|2 )np

n where c1 is a constant. Let {fm }∞ m=1 be a sequence in S (R ) such that

|fm − f |n,0 →m→∞ 0. We obtain ||fm − f ||Lp ≤ C|fm − f |n,0 →m→∞ 0, Consequently fm →m→∞ f in Lp (Rn ). j

Problem 1.34 Let u ∈ C0 (Rn ). Prove 1. u ∗ v ∈ C j (Rn ) for v ∈ L1loc (Rn ), 2. u ∗ v ∈ C j +k (Rn ) for v ∈ C k (Rn ).

C = const.

1.5 Advanced Practical Problems

75

j

1. Solution Since u ∈ C0 (Rn ), there exists a compact set K ⊂ Rn such that suppu ⊂ K. On the other hand, 

 u(y)v(x − y)dy =

(u ∗ v)(x) = Rn

u(x − y)v(y)dy,

x ∈ Rn .

x−y∈K

From here, using that v ∈ L1loc (Rn ), we get  |(u ∗ v)(x)| ≤

 |u(x − y)||v(y)|dy ≤ c1

x−y∈K

|v(y)|dy ≤ C,

c1 , C = const,

x−y∈K

x ∈ Rn . Consequently the convolution u ∗ v exists. Since D l u ∈ C0 (Rn ) and D l (u ∗ v) = D l u ∗ v, as above, we conclude that D l (u ∗ v) exists for every 1 n l = 0, 1, 2, . . . , j . If {vn }∞ n=1 is a sequence of elements of Lloc (R ) converging 1 n to v in Lloc (R ), then      l    D (u ∗ vn )(x) − D l (u ∗ v)(x) = D l u ∗ (vn − v)(x) ≤ c2 ||vn − v||L1

n loc (R )

→n→∞ 0,

x ∈ Rn , l = 0, 1, 2, . . . , j , where c2 > 0 is a constant. So, D l (u ∗ vn ) →n→∞ D l (u ∗ v),

l = 0, 1, 2, . . . , j.

Consequently u ∗ v ∈ C j (Rn ). 2. Hint Use that D l+m (u ∗ v) = D l u ∗ D m v for l = 0, 1, . . . , j , m = 0, 1, . . . , k. Problem 1.35 Let Γ be a cone in Rn and σ > 0 a constant such that (ξ, x) ≥ σ |ξ ||x| ∀ξ ∈ C  , ∀x ∈ chΓ for every C  ⊂⊂ intΓ ∗ . Prove that Γ is an acute cone. Problem 1.36 Let Γ be a cone in Rn for which there exists a plane of support for chΓ that has a unique common point 0 with chΓ . Prove that Γ is an acute cone. Problem 1.37 Let Γ be a convex cone. Prove Γ = Γ + Γ . Problem 1.38 Let Γ be a cone in Rn . Show μΓ (ξ ) ≤ μchΓ (ξ ). Problem 1.39 Let Γ be a convex cone in Rn . Prove that for every a ≥ 0 {ξ : μΓ (ξ ) ≤ a} = Γ ∗ + Ua .

76

1 Introduction

Problem 1.40 Let Γ be a closed, convex, acute cone and S a C-like surface. Prove that for every R > 0 there exists a constant R  (R) > 0 such that TR = {(x, y) : x ∈ S, y ∈ Γ, |x + y| ≤ R} ⊂ UR  ⊂ R2n .

1.6 Notes and References In this chapter we introduce the spaces C0∞ , S , Lp , 1 ≤ p ≤ ∞, and they are deducted some of their properties. They are formulated and proved some variants of the Hölder and Minkowski inequalities. In the chapter is defined the convolution of locally integrable functions and they are proved some of its basic properties. They are introduced some basic facts for cones in Rn . Additional materials can be found in [3–11, 15, 17–19, 21–28, 31–36] and references therein.

Chapter 2

Generalities on Distributions

2.1 Definitions Let X be an open set in Rn , n ∈ N a fixed integer. Definition 2.1 Every linear continuous map u : C0∞ (X) → C is called a distribution or generalized function. In other words, a distribution is a linear map u : C0∞ (X) → C such that u(φn ) →n→∞ u(φ) for every sequence {φn }∞ n=1 in C0∞ (X) converging to φ ∈ C0∞ (X), as n → ∞. The space of distributions on X will be denoted by D  (X). We will write u(φ) or (u, φ) for the value of the functional (generalized function, distribution) u ∈ D  (X) on the element φ ∈ C0∞ (X). Example 2.1 Suppose 0 ∈ X and take the map u : C0∞ (X) → C defined as follows u(φ) = φ(0)

for φ ∈ C0∞ (X).

Let φ1 , φ2 ∈ C0∞ (X) and α1 , α2 ∈ C. As u(φ1 ) = φ1 (0), u(φ2 ) = φ2 (0), u(α1 φ1 + α2 φ2 ) = (α1 φ1 + α2 φ2 )(0) = α1 φ1 (0) + α2 φ2 (0) = α1 u(φ1 ) + α2 u(φ2 ), ∞ i.e., u : C0∞ (X) → C is linear. Let {φn }∞ n=1 be a sequence in C0 (X) for ∞ which φn →n→∞ φ in C0 (X). Then there exists a compact set K ⊂ X such that suppφn ⊂ K for every n ∈ N and D α φn → D α φ uniformly in X for every multi-index α ∈ N ∪ {0}. In particular, φn (0) →n→∞ φ(0), and therefore u(φn ) →n→∞ u(φ). Consequently the linear map u : C0∞ (X) → C is continuous. In other words, it is a distribution on X.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_2

77

78

2 Generalities on Distributions

Exercise 2.1 Let 0 ∈ X. For each multi-index α prove that the map u : C0∞ (X) → C, defined by u(φ) = D α φ(0)

for φ ∈ C0∞ (X),

is a distribution. Exercise 2.2 Denote by δa or δ(x − a), a ∈ Cn , Dirac’s “delta” function at the point a: δa (φ) = φ(a)

for φ ∈ C0∞ (X).

Prove that δa is a distribution. Exercise 2.3 Prove that the map 1 : C0∞ (X) → C, defined by  1(φ) =

φ(x)dx

for φ ∈ C0∞ (X),

X

is a distribution. Exercise 2.4 For u ∈ Lloc (X), p ≥ 1, we define u : C0∞ (X) → C by p

 u(φ) =

u(x)φ(x)dx. X

Prove that u is a distribution. 1 Exercise 2.5 Let P : C0∞ (X) → C be the map defined by x  1 φ(x) − φ(0) P (φ) = P .V . dx for φ ∈ C0∞ (X). x x X

Prove that P

1 ∈ D  (X). x

Definition 2.2 The distributions u, v ∈ D  (X) are said to be equal if u(φ) = v(φ) for any φ ∈ C0∞ (X).

2.1 Definitions

79

Definition 2.3 The linear combination λu + μv of the distributions u, v ∈ D  (X) is the functional acting by the rule (λu + μv)(φ) = λu(φ) + μv(φ),

φ ∈ C0∞ (X).

This makes the set D  (X) a vector space. Definition 2.4 Let u ∈ D  (X). We define a distribution u ∈ D  (X), called the complex conjugate of u, by u(φ) = u(φ),

φ ∈ C0∞ (X).

The distributions Re(u) =

u+u , 2

Im(u) =

u−u 2i

are called the real and imaginary parts of u, respectively. Equivalently, u = Re(u) + iIm(u),

u = Re(u) − iIm(u).

If Im(u) = 0, then the distribution u is said to be a real distribution. Exercise 2.6 Prove that the delta function is a real distribution. Here are elementary properties of the distributions. If u1 , u2 ∈ D  (X), then 1. u1 ± u2 ∈ D  (X), 2. αu1 ∈ D  (X) for ∀α ∈ C. These properties follow from the definition, so their proof is omitted. Theorem 2.1 u ∈ D  (X) if and only if for every compact subset K of X there exist constants C and k so that the inequality |u(φ)| ≤ C



    sup D α φ(x)

(2.1)

|α|≤k x∈K

holds for every φ ∈ C0∞ (K). Proof Let u ∈ D  (X). We will prove that the inequality (2.1) holds. Actually, we suppose that there exists a compact set K in X so that |u(φn )| > n



    sup D α φn (x)

α∈Nn ∪{0} x∈K

(2.2)

80

2 Generalities on Distributions

holds for φn ∈ C0∞ (K). We set φn (x) ,    sup D α φn (x)

ψn (x) = n

x ∈ K.

α∈Nn ∪{0} x∈K

From (2.2), we obtain |u(ψn )| > 1.

(2.3)

By the definition of ψn , it follows that ψn →n→∞ 0 in C0∞ (X). Since u : C0∞ (X) → C is continuous, we have u(ψn ) →n→∞ 0, which contradicts (2.3). Let now, u : C0∞ (X) → C is a linear map such that for every compact set K in X there exist constants C > 0 and k ∈ N ∪ {0} for which (2.1) holds. We will prove that u is a distribution on X. To show this, we will prove that u : C0∞ (X) → C ∞ is continuous at 0. Let {φn }∞ n=1 be a sequence in C0 (X) with φn →n→∞ 0 in ∞ C0 (X). Then     sup D α φn (x) →n→∞ 0

x∈K

for every |α| ≤ k. Hence with (2.1), we conclude u(φn ) →n→∞ 0. This completes the proof. Exercise 2.7 The function H defined by  H (x) =

1 for x ≥ 0, 0 for x < 0,

x ∈ R,

is called the Heaviside function. We define ∞ H (φ) =

H (x)φ(x)dx, −∞

φ ∈ C0∞ (R). Using the inequality (2.1), prove that H ∈ D  (R). Theorem 2.2 A linear map u : C0∞ (X) → C is a distribution if and only if there exist functions ρα ∈ C (X) such that |u(φ)| ≤

 α

    sup ρα D α φ , K

∀φ ∈ C0∞ (K),

(2.4)

for every compact set K ⊂ X, and only a finite number of ρα vanish identically.

2.1 Definitions

81

Proof 1. Let u be a linear map from C0∞ (X) to C and ρα ∈ C (X) be such that the inequality (2.4) holds for every φ ∈ C0∞ (X) and every compact set K. Since ρα ∈ C (X), there exists a constant C such that sup |ρα | ≤ C. K

From this and (2.4), it follows that |u(φ)| ≤ C

 α

    sup D α φ . K

As only finitely many ρα vanish identically, there is a constant k such that |u(φ)| ≤ C



    sup D α φ ,

|α|≤k K

i.e., u ∈ D  (X). 2. Let u ∈ D  (X) and {Kj } be compact subsets of X such that any compact subset of X is contained in some Kj . Take maps χj ∈ C0∞ (X) with χj ≡ 1 on Kj and define ψj = χj − χj −1 , ψ1 = χ1 .

j > 1,

Any φ ∈ C0∞ (X) satisfies φ=

∞ 

(2.5)

ψj φ.

j =1

Moreover, ψj = 0 on Kj \Kj −1 ψ1 = 0 on K1 .

for j > 1,

Consequently supp(ψj φ) ⊂ suppψj . As ψj φ has compact support, for every compact set K there are constants C and kj such that |u(ψj φ)| ≤ C



    sup D α (ψj φ).

|α|≤kj K

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2 Generalities on Distributions

From this and (2.5), we obtain ∞     u(ψj φ) |u(φ)| =  j =1

≤

∞ 

|u(ψj φ)|

j =1

≤C

∞  

    sup D α (ψj φ)

j =1 |α|≤kj K

≤C

∞         α     sup D β ψj  sup D α−β φ . β K K j =1 |α|≤kj β≤α

If we set ∞     α ρβ = D β ψj , β j =1 |α|≤kj

we obtain |u(φ)| ≤ C



    sup ρβ D α−β φ .

β≤α K

This completes the proof. n Theorem 2.3 Let {Xi }i∈I be an open  sets of R , X = ∪i∈I Xi and suppose that  ui ∈ D (Xi ) satisfy ui = uj on Xi Xj . Then there exists a unique distribution  u ∈ D (X) such that u|Xi = ui for every i ∈ I .  φi and Proof Take φi ∈ C0∞ (Xi ). Define φ = i

u(φ) =



(2.6)

ui (φi ).

i

We claim that the definition (2.6) does not depend on the choice of the sequence    φi = 0 implies u φi = 0. {φi }. For this purpose, it is enough to prove that Set K =



i

i

suppφi . Then K is a compact set. There exist functions ψk ∈ C0∞ (Xk )

i

such that 0 ≤ ψk ≤ 1 and

 k

ψk = 1 on K. Note that only a finite number

2.1 Definitions

83

 of the above summands are different from zero. Moreover, ψk φi ∈ C0∞ (Xk Xi ) and uk (ψk φi ) = ui (ψk φi ). Therefore       ui (φi ) = ui ψk φi = ui (ψk φi ) = uk (ψk φi ) i

i

=

 k

k

i



uk (ψk φi ) =

i

k



uk ψk



k



φi =

i

i



k

uk (0) = 0.

k

Consequently the definition (2.6) is consistent. Let φ ∈ C0∞ (K). Then φ =



φψk

k

and

            ui (ψi φ) ≤ ui (φψi ) u(φ) =  i

≤

 i

Ci



i

          !i sup D α (φψi ) ≤ sup D α φ  C

|α|≤k K

i

     ≤C sup D α φ ,

|α|≤k K

|α|≤k K

!i and C are nonnegative constants. We also showing u is a distribution. Here Ci , C have u = ui on Xi . Now, we will prove the uniqueness of u. Suppose there are two distributions u and u˜ with the previous properties. We conclude u|Xi = ui , u˜ |Xi = ui , so (u − u) ˜ |Xi = 0 for any i. Since X is open in Rn , it follows that u ≡ u˜ on X, proving uniqueness. This completes the proof. Example 2.2 Let f (x) = e x , x ∈ R\{0}. Suppose that f ∈ D  (R\{0}). Pick φ0 ∈ C0∞ (R\{0}) such that φ0 (x) ≥ 0 for every x = 0, φ0 (x) = 0 for x < 1 and x > 2, and 1

∞ φ0 (x)dx = 1. −∞

 ∞ Define the sequence φk

k=1

by φk (x) = e− 2 kφ0 (kx), x ∈ R\{0}, k ∈ N. It k

satisfies φk →k→∞ 0 in C0∞ (R\{0}), so f (φk ) →k→∞ 0. On the other hand, ∞ f (φk ) =

1

e x φk (x)dx −∞

2 =



e 1

k

1 1 y −2



φ0 (y)dy

84

2 Generalities on Distributions 3

2 ≥



e

k

1 1 y −2



φ0 (y)dy

1 3

≥e

k 6

2

φ0 (y)dy. 1

By this and the definition of φ0 , we conclude lim f (φk ) = ∞, which is a k→∞

contradiction. Therefore f ∈ D  (R\{0}).

2.2 Order of a Distribution Definition 2.5 If the inequality (2.1) holds for some integer k independent of the compact set K ⊂ X, the distribution u is said to be of finite order. The smallest such k is called the order of the distribution u. The space of distributions on X of finite order is denoted by DF (X), and the space k of distributions of order ≤ k is denoted by D  (X). Then DF (X) =



D  (X). k

k

Example 2.3 Dirac’s “delta” function is a distribution of order 0. Example 2.4 Let u ∈ D  (X) is defined by u(φ) =

∞ 

φ (k) (k),

φ ∈ C0∞ (X).

k=1

Then the order of u is infinite. 1 Exercise 2.8 Prove that P has order 1 on R. x 1 Exercise 2.9 Prove that P is of order 0 on R\{0}. x Theorem 2.4 Any u ∈ D  (X) can be extended in a unique way to a linear map on C0k (X) so that the inequality (2.1) holds for every φ ∈ C0k (X). k

2.3 Change of Variables

85

Proof Since the space C0∞ (X) is everywhere dense in C0k (X), for every φ ∈ C0k (X)  ∞ there exists a sequence φn in C0∞ (X) for which φn →n→∞ φ in C0k (X). n=1 Hence,      |u(φn ) − u(φl )| ≤ C sup D α φn − D α φl  →n,l→∞ 0. |α|≤k K

 ∞ Therefore u(φn ) is a Cauchy sequence in R, and as such it converges to, say, n=1

u(φ) = lim u(φn ).

(2.7)

n→∞

 ∞  ∞ , ψn be two sequences The claim is that (2.7) is consistent. In fact, let φn n=1

in C0∞ (X) for which

n=1

lim φn = lim ψn = φ

n→∞

n→∞

in C0k (X). Then u(φ) = lim u(γn ) = lim u(φn ) = lim u(ψn ), where n→∞ n→∞  ∞  ∞  n→∞ ∞ ∞ γn = φn ∪ ψn . For the sequence {γn }n=1 we have n=1

n=1

n=1

       sup |D α γn , u(γn ) ≤ C |α|≤k K

so          sup D α φ  u(φ) ≤ C |α|≤k K

when n → ∞. This completes the proof.

2.3 Change of Variables Let X1 be an open set in Rn . Take f ∈ L1loc (X1 ) and suppose that x = Ay + b is a nonsingular transformation from X to X1 , i.e., A is a n × n matrix with det A = 0 and b ∈ Rn . Then, for any φ ∈ C0∞ (X), we have     1 f (Ay + b)φ(y)dy = f (x)φ A−1 (x − b) dx. | det A| X

X1

This equality motivates us to give the following definition.

86

2 Generalities on Distributions

Definition 2.6 For any u ∈ D  (X), define

 φ A−1 (x − b) , (u(Ay + b), φ(y)) = u(x), | det A| 

x ∈ X,

y ∈ X1 .

(2.8)

  Note that the map φ(x) → φ A−1 (x − b) , x ∈ X, is a linear and continuous map from C0∞ (X) to C0∞ (X1 ). Therefore the functional u(Ay + b), y ∈ X1 , defined by (2.8), is well defined and it belongs to D  (X1 ). In particular, for u ∈ D  (X) and a ∈ Cn , |a| = 0, b ∈ C, b = 0, we have following distributions 1. u(φ)(x + a) = u(φ(x − a))(x), φ ∈ C0∞ (X), x ∈ X, 1   x  (x), φ ∈ C0∞ (X), x ∈ X. 2. u(φ)(bx) = u φ |b|n b Example 2.5 For φ ∈ C0∞ (R), we have δ(φ)(−x) = δ(φ)(x), δ(φ)(x + 1 − 2i) = δ(φ(x − 1 + 2i))(x) = φ(−1 + 2i), 1 1   x  (x) = φ(0), x ∈ R. δ(φ)(2ix) = δ φ 2 2i 2 Exercise 2.10 Compute δ(φ)(2x + 3i), x ∈ R, for φ ∈ C0∞ (R). 1  3i  Answer φ − . 2 2 Definition 2.7 For a ∈ C 1 (R), define δ(a(x)) = lim ω (a(x)), →0

x ∈ R.

Theorem 2.5 If a ∈ C 1 (R) has isolated simple zeros x1 , x2 , . . ., then δ(a(x)) =

 δ(x − xk ) k

|a  (xk )|

,

x ∈ R.

Proof It is enough to prove the assertion on a neighbourhood of the simple zero xk . Since xk is an isolated simple zero of a, there exists k > 0 such that a(x) = 0 for

2.3 Change of Variables

87

every x ∈ (xk − k , xk + k ), x = xk , a(xk ) = 0. As x k +k   δ(a(x)), φ(x) = δ(a(x))φ(x)dx xk −k x k +k

= lim

→0 xk −k

ω (a(x))φ(x)dx

a(xk +k )

= lim

→0 a(xk −k )

ω (y)

a(xk +k )

= lim

→0 a(xk −k )

ω (y)

φ(a −1 (y)) dy |a  (a −1 (y))| φ(a −1 (a(x))) dy |a  (a −1 (a(x)))|

φ(xk ) |a  (xk )|   δ(x − x ) k , φ(x) , = |a  (xk )|

=

x ∈ (xk − k , xk + k ),

for φ ∈ C0∞ (xk − k , xk + k ). Next, let (c1 , d1 ) ⊂ R does not contain a single zero of xk and φ ∈ C0∞ ((c1 , d1 )). Then  (δ(a(x)), φ(x)) = lim

d1

→0 c1

= 0,

ω (a(x))φ(x)dx

x ∈ (c1 , d1 ).

This completes the proof. Example 2.6 Let us consider δ(cos x), x ∈ R. Here a(x) = cos x, x ∈ R, and its (2k + 1)π , k ∈ Z. We notice that |a  (xk )| = 1 for k ∈ Z. isolated zeros are xk = 2 So, δ(cos x) =

  (2k + 1)π  , δ x− 2 k

Exercise 2.11 Compute δ(x 4 − 1), x ∈ R. Answer

δ(x − 1) + δ(x + 1) , x ∈ R. 4

x ∈ R.

88

2 Generalities on Distributions

2.4 Sequences and Series  ∞ Definition 2.8 We say that the sequence un of elements of D  (X) tends to n=1

the distribution u ∈ D  (X) if lim un (φ) = u(φ)

n→∞

for any φ ∈ C0∞ (X).

If so we write lim un = u

or

n→∞

un →n→∞ u.

∞ Theorem 2.6 If {un }∞ n=1 and {vn }n=1 are two sequences of distributions on X that converge to the distributions u and v, respectively, then {αun + βvn }∞ n=1 converges to αu + βv on X. Here α, β ∈ C.

Proof Indeed, let φ ∈ C0∞ (X) be arbitrary. Then un (φ) →n→∞ u(φ),

vn (φ) →n→∞ v(φ).

Hence, (αun + βvn )(φ) = (αun )(φ) + (βvn )(φ) = αun (φ) + βvn (φ) → αu(φ) + βv(φ),

as n → ∞.

This completes the proof. Example 2.7 Let x ∈ R and ⎧ ⎨ 1 for |x| ≤ , f (x) = 2 ⎩ 0 for |x| > . We will compute lim f (x), x ∈ R, in D  (R). Let φ ∈ C0∞ (R) be arbitrary. Then →0+

 lim f (φ)(x) = lim

→0+

→0+ |x|≤

=

1 lim 2 →0+

1 φ(x)dx 2



|y|≤1

φ(y)dy

2.4 Sequences and Series

89

= φ(0) = δ(φ)(x),

x ∈ R.

Consequently lim f (x) = δ(x), x ∈ R, in D  (R). →0+

Example 2.8 We will find lim

→0+ π(x 2

chosen. Then ∞ lim

→0+ −∞

 , x ∈ R. Let φ ∈ C0∞ (R) be arbitrarily +  2)

 φ(x)dx = lim →0+ x2 + 2

∞

−∞

∞ = φ(0) −∞

1 φ(y)dy 1 + y2 1 dy 1 + y2

y=∞ = φ(0) arctan y y=−∞ π π + = φ(0) 2 2 = πφ(0) = πδ(φ). Exercise 2.12 Prove that lim

→0+

 x sin2 = δ(x), πx 2 

x ∈ R.

 Theorem 2.7 Let {fn }∞ n=1 be a sequence in D (X) such that |fn (φ)| ≤ cφ for every ∞ ∞ ∞ φ ∈ C0 (X), and {φn }n=1 ⊂ C0 (X) a sequence converging to 0 in C0∞ (X) as n → ∞. Then fn (φn ) → 0, n → ∞.

Proof We suppose the contrary. Then there exists a constant c > 0 such that |fn (φn )| ≥ c > 0, for n large enough. Since φn → 0 in C0∞ (X) as n → ∞, there exists a compact set X such that suppφn ⊂ X for every n ∈ N and D α φn →n→∞ 0, for every α ∈ Nn ∪ {0}. Hence, |D α φn (x)| ≤

1 , 4n

|α| ≤ n = 0, 1, 2, . . . ,

90

2 Generalities on Distributions

for n large enough and every x ∈ X . We set ψn = 2n φn , n ∈ N. We have suppψn ⊂ X and 1 , 2n

|D α ψn (x)| ≤

|α| ≤ n = 0, 1, 2, . . . ,

(2.9)

|fn (ψn )| = 2n |fn (φn )| ≥ 2n c →n→∞ ∞.

(2.10)

∞ ∞ ∞ Let us take subsequences {fkν }∞ ν=1 of {fn }n=1 and {ψkν }ν=1 of {ψn }n=1 so that ν ∞ |fkν (ψkν )| ≥ 2 for ν = 1, 2, . . .. As ψk →k→∞ 0 in C0 (X), we have fkj (ψk ) →k→∞ 0 for j = 1, 2, . . . , ν − 1. Therefore there exists N ∈ N such that for every k ≥ N

|fkj (ψk )| ≤

1 2ν−j

j = 1, 2, . . . , ν − 1.

,

(2.11)

We note that |fk (ψkj )| ≤ ckj , j = 1, 2, . . . , ν − 1. From (2.10), we can choose kν ≥ N so that 

|fkν (ψkν )| ≥

ckj + ν + 1.

(2.12)

j = 1, 2, . . . , ν − 1,

(2.13)

|fkν (ψkj )| + ν + 1.

(2.14)

1≤j ≤ν−1

From (2.11) and (2.12), we have |fkj (ψkν )| ≤ |fkν (ψkν )| ≥

1 , 2ν−j  1≤j ≤ν−1

We set ψ = C0∞ (X) and



ψkj . From (2.9), it follows that ψ is a convergent series, ψ ∈

j ≥1

fkν (ψ) = fkν (ψkν ) +



fkν (ψkj ).

j ≥1,j =ν

Therefore 

|fkν (ψ)| ≥ |fkν (ψkν )| −

|fkν (ψkj )| −

1≤j ≤ν−1

≥ ν+1−

 j ≥ν+1

1 2j −ν

= ν,

 j ≥ν+1

|fkν (ψkj )|

2.4 Sequences and Series

91

and then fkν (ψ) →ν→∞ ∞, which contradicts with |fkν (ψ)| ≤ cψ . This completes the proof.  ∞ Theorem 2.8 Let {fn }∞ n=1 be a sequence in D (X) such that {fn (φ)}n=1 converges ∞ for every φ ∈ C0 (X). Then the functional

φ ∈ C0∞ (X),

f (φ) = lim fn (φ), n→∞

is an element of D  (X). Proof Let α1 , α2 ∈ C and φ1 , φ2 ∈ C0∞ (X). Then f (α1 φ1 + α2 φ2 ) = lim fn (α1 φ1 + α2 φ2 ) = lim (α1 fn (φ1 ) + α2 fn (φ2 )) n→∞

n→∞

= α1 lim fn (φ1 ) + α2 lim fn (φ2 ) = α1 f (φ1 ) + α2 f (φ2 ). n→∞

n→∞

Therefore f is a linear map on C0∞ (X). Now, we will prove that f is a continuous ∞ functional on C0∞ (X). Let {φn }∞ n=1 be a sequence in C0 (X) such that φn →n→∞ ∞ 0 in C0 (X). We claim f (φn ) →n→∞ 0. Suppose the contrary. There exists a constant a > 0 such that |f (φν )| ≥ a for every ν ∈ N. Since f (φν ) = lim fk (φν ), k→∞

there is kν ∈ N such that |fkν (φν )| ≥ a for every ν ∈ N, which is in contradiction with Theorem 2.7. Consequently f (φn ) →n→∞ 0 and f ∈ D  (X). This completes the proof. Definition 2.9 Let uj ∈ D  (X), j ∈ N. An expression like u1 + u2 + · · · or ∞ 

(2.15)

uj

j =1

is called infinite series of distributions. mth partial sum of the series (2.15) is defined by Sm =

m 

uj ,

m ∈ N.

j =1  If the sequence {Sm }∞ m=1 is convergent in D (X), then we say that the series (2.15)  is convergent in D (X). Otherwise, we say that the series (2.15) is divergent.

92

2 Generalities on Distributions

2.5 Support Definition 2.10 A distribution u ∈ D  (X) is said to vanish on an open set X1 ⊂ X if its restriction to X1 is the zero functional in D  (X1 ), i.e., u(φ) = 0 for all φ ∈ C0∞ (X1 ). This is written u(x) = 0, x ∈ X1 . Theorem 2.9 A distribution u ∈ D  (X) vanishes on X if and only if it vanishes on a neighbourhood of every point of X. Proof Suppose a distribution u ∈ D  (X) vanishes on X. Then it vanishes on a neighbourhood of every point in X. Conversely, let u ∈ D  (X) vanishes on a neighbourhood U (x) ⊂ X of every point x in X. Consider the cover {U (x), x ∈ X} of X. We will construct a locally finite cover {Xk } such that Xk is contained in some U (x). Let X11 ⊂⊂ X21 ⊂⊂ . . . ,



Xk1 = X.

k≥1 1

By the Heine–Borel lemma, the compact set X1 is covered by a finite number of neighbourhoods U (x), say U (x1 ), U (x2 ), . . ., U (xN1 ). Similarly, the compact set 1

X2 \X11 is covered by a finite number of neighbourhoods U (xN1 +1 ), . . ., U (xN1 +N2 ), and so on. We set  Xk = U (xk ) X1 , k = 1, 2, . . . , N1 ,  11 1 Xk = U (xk ) (X 2 \X1 ), k = N1 + 1, . . . , N1 + N2 , and so forth. In this way, we obtain the required cover {Xk }. Let {ek } be the partition of unity corresponding to the cover {Xk } of X. Then supp(φek ) = 0 for every φ ∈ C0∞ (X). This implies u(φ) = u



  φek = u(φek ) = 0.

k≥1

k≥1

Consequently the distribution u vanishes on the whole X. This completes the proof. Definition 2.11 The union of all neighbourhoods where a distribution u ∈ D  (X) vanishes forms an open set Xu , called the zero set of the distribution u. Therefore u = 0 on Xu and Xu is the largest open set where u vanishes. Definition 2.12 The support of a distribution u ∈ D  (X) is the complement suppu = X\Xu of Xu in X. Note that suppu is a closed subset in X.

2.5 Support

93

Definition 2.13 The distribution u ∈ D  (X) is said to have compact support if suppu ⊂⊂ X. Example 2.9 suppH = [0, ∞). Exercise 2.13 Find supp1. Let A be a closed set in X. With D  (X, A) we denote the subset of distributions on X whose supports are contained in A, endowed with the following notion of convergence: uk → 0 in D  (X, A), as k → ∞, if uk → 0 in D  (X), as k → ∞, and suppuk ⊂ A for every k = 1, 2, . . .. For simplicity, D  (A) will denote D  (Rn , A). Theorem 2.10 Suppose that for every point y ∈ X there is a neighbourhood U (y) ⊂⊂ X on which a given distribution uy is defined. Assume further that uy1 (x) = uy2 (x) if x ∈ U (y1 ) ∩ U (y2 ) = Ø. Then there exists a unique distribution u ∈ D  (X) so that u = uy in U (y) for every y ∈ X. Proof To see this we construct, starting as previously with the cover {U (y), y ∈ X}, the locally finite cover {Xk }, Xk ⊂ U (yk ), and the corresponding partition of unity {ek }. We also set u(φ) =



uyk (φek ),

φ ∈ C0∞ (X).

(2.16)

k≥1

The number of summands in the right-hand side of (2.16) is finite and does not depend on φ ∈ C0∞ (X ) for any X ⊂⊂ X. By the definition (2.16), it follows that u is linear and continuous on C0∞ (X), i.e., u ∈ D  (X). Furthermore, if φ ∈ C0∞ (U (y)), then φek ∈ C0∞ (U (yk )). From (2.16), we get    u(φ) = uy φ ek = uy (φ), k≥1

i.e., u = uy on U (y). If we suppose that there are two distributions u and u˜ such that u = uy and u˜ = uy on U (y) for every y ∈ X, then u − u˜ = 0 on U (y) for every y ∈ X. Therefore u − u˜ = 0 in X, showing that the distribution u is unique. This completes the proof. The set of distributions with compact support in X will be denoted by E  (X), and we set E k (X) = E  (X) ∩ D k (X).  Theorem 2.11 Let φ ∈ C0∞ (X) and suppu suppφ = Ø. Then u(φ) = 0.  Proof Since suppu suppφ = Ø, we have φ ∈ C0∞ (X\suppu). If x ∈ suppu, then φ(x) = 0, so u(φ) = 0. If x ∈ X\suppu, then u(φ)(x) = 0. This completes the proof.

94

2 Generalities on Distributions

Theorem 2.12 Let u ∈ D  (X) and let F be a relatively closed subset of X with suppu ⊂ F . Then there exists a unique linear map u˜ on   φ : φ ∈ C ∞ (X), F ∩ suppφ ⊂⊂ X such that 1. u(φ) ˜ = u(φ) for φ ∈ C0∞ (X), 2. u(φ) ˜ = 0 for φ ∈ C ∞ (X), F ∩ suppφ = Ø. Proof 1. (uniqueness) Let φ ∈ C ∞ (X) and F ∩ suppφ = K. As K is compact, there exists ψ ∈ C0∞ (X) such that ψ ≡ 1 on a neighbourhood of K. Let φ0 = ψφ,

φ1 = (1 − ψ)φ.

Then φ = φ0 + φ1 .

(2.17)

Therefore ˜ 1 ). u(φ) ˜ = u(φ ˜ 0 ) + u(φ Note that u(φ ˜ 1 ) = 0. So, u(φ) ˜ = u(φ ˜ 0 ) = u(φ0 ). ˜˜ Then Now, suppose that there are two such distributions u, ˜ u. u(φ) ˜ = u(φ ˜ 0 ),

˜˜ ˜˜ 0 ). u(φ) = u(φ

˜˜ Consequently u(φ) ˜ = u(φ) for every φ ∈ C ∞ (X) so that F ∩ suppφ = K. ˜˜ Therefore u˜ = u. 2. (existence) Let φ = φ0 + φ1 be another decomposition of kind (2.17). Define χ = φ0 − φ0 . Then χ ∈ C0∞ (X),

F ∩ suppχ = F ∩ supp(φ1 − φ1 ) = Ø

2.5 Support

95

and so, u(χ) = u(φ0 ) − u(φ0 ) = 0. Define u(φ) ˜ by u(φ) ˜ = u(φ0 ). This completes the proof. Theorem 2.13 The set of distributions on X with compact support coincides with the dual space of C ∞ (X) with the topology φ →



    sup D α φ ,

|α|≤k K

where K is a compact set in X. Proof Let u be a distribution with compact support and take φ ∈ C ∞ (X) and ψ ∈ C0∞ (X), ψ ≡ 1 on a neighbourhood of suppu. Then φ = ψφ + (1 − ψ)φ and u(φ) = u(ψφ + (1 − ψ)φ) = u(ψφ) + u((1 − ψ)φ) = u(ψφ). Define u on C ∞ (X) via u(φ) = u(ψφ) for φ ∈ C ∞ (X). Since u is a distribution and ψφ ∈ C0∞ (X), we have |u(φ)| = |u(ψφ)| ≤ C



        sup D α (φψ) ≤ C1 D α φ .

|α|≤k K

|α|≤k

Now, we suppose that v is a linear operator on C ∞ (X) for which |v(φ)| ≤ C



    sup D α φ 

|α|≤k K

for φ ∈ C ∞ (X) and K a compact set. Then v(φ) = 0 when suppφ ∩ K = Ø. If φ ∈ C0∞ (X) ⊂ C ∞ (X), v is a distribution. Therefore there exists a unique distribution u ∈ D  (X) such that u(φ) = v(φ) for every φ ∈ C ∞ (X). This completes the proof. Theorem 2.14 Let u be a distribution with a compact support of order ≤ k, φ a C k map with D α φ(x) = 0 for |α| ≤ k, x ∈ suppφ. Then u(φ) = 0.

96

2 Generalities on Distributions

Proof Let χ ∈ C0∞ (X), χ ≡ 1 on a neighbourhood U of suppu, while χ ≡ 0 on X\U . Define the set M ,  > 0, by   M = y : |x − y| ≤ , x ∈ suppu , making M an -neighbourhood of suppu. Moreover,    α  D χ  ≤ C |α| , |α| ≤ k, for some positive constant C. Since suppu ∩ supp(1 − χ )φ = Ø, we have u(φ) = u(φχ ) + u((1 − χ )φ) = u(φχ ),       sup D α (φχ )  |u(φ)| ≤ C   |α|≤k        ≤ C1 sup D α φ D β χ  |α|+|β|≤k

≤ C2



    sup D α φ  k−|α| →→0 0,

|α| ≤ k.

|α|+|β|≤k

Consequently u(φ) = 0. This completes the proof. Theorem 2.15 Let u be a distribution of order k with support {y}. Then u(φ) =  α aα D φ(y), φ ∈ C k . |α|≤k

Proof Fix x ∈ Rn . For φ ∈ C k (Rn ), we have φ(x) =



D α φ(y)

|α|≤k

(x − y)α + ψ(x), α!

where D α ψ(y) = 0

for |α| ≤ k. Hence, u(ψ) = 0. Therefore   (x − y)α + ψ(x) u(φ(x)) = u D α φ(y) α! |α|≤k

=u

 |α|≤k

D α φ(y)

(x − y)α  + u(ψ(x)) α!

  (x − y)α  = D α φ(y). u α! |α|≤k

2.5 Support

97

Let aα = u

 (x − y)α  α!

.

Then u(φ) =



aα D α φ(y).

|α|≤k

This completes the proof. Theorem 2.16 Write x = (x  , x  ) ∈ Rn . Then for every distribution u ∈ D  (Rn ) of order k with compact support contained in the plane x  = 0, we have u(φ) =



(2.18)

uα (φα ),

|α|≤k

where α = (α  , 0), uα is a distribution in the variables x  of order k − |α| with compact support and φα (x  ) = D α φ(x  , x  )|x  =0 . Proof For φ ∈ C ∞ (Rn ), we have 

φ(x) =

D α φ(0, x  )

|α  |≤k,α  =0

x α + Φ(x), α!

where D α Φ(x)|x  =0 = 0 for |α| ≤ k. This implies u(Φ) = 0. Since u is a distribution, u(φ) =

 |α  |≤k,α  =0

 x α  . u D α φ(0, x  ) α!

Now, let  x α  uα (φ) = u D α φ(0, x  ) . α! We want to show that uα is a distribution of order k − |α|. Set ψ(x) = D α φ(0, x  )

x α + O(|x  |k+1 ) α!

for x  → 0.

Then u(ψ) = uα (φ)

for ψ ∈ C ∞ (Rn )

(2.19)

98

2 Generalities on Distributions

and     sup D γ φ  ≤ C

 |γ |≤k

    sup D β ψ ,

 |β|≤k−|α|

so     sup D α φ  ≤ C

    sup D β ψ .

 |β|≤k−|α|

Consequently uα (ψ) ≤ C 



    sup D β ψ 

|β|≤k−|α|

for every ψ ∈ C0∞ (Rn ), proving uα is a distribution of order k − |α| in the variable x  . From (2.19), it follows that uα has compact support. This completes the proof.

2.6 Singular Support Definition 2.14 The set of points of X not admitting neighbourhoods where u ∈ D  (X) coincides with a C ∞ function is called the singular support of u, written singsuppu. Hence, u coincides with a C ∞ function on X\singsuppu. Example 2.10 Let f ∈ C ∞ (X). We define the functional u in the following manner  u(φ) =

f (x)φ(x)dx,

φ ∈ C0∞ (X).

X

For φ1 , φ2 ∈ C0∞ (X) and α1 , α2 ∈ C, we have  u(α1 φ1 + α2 φ2 ) =

f (x)(α1 φ1 (x) + α2 φ2 (x))dx X



=

(α1 f (x)φ1 (x) + α2 f (x)φ2 (x))dx X



 f (x)φ1 (x)dx + α2

= α1 X

= α1 u(φ1 ) + α2 u(φ2 ).

f (x)φ2 (x)dx X

2.6 Singular Support

99

Therefore u is a linear functional on C0∞ (X). For φ ∈ C0∞ (X), there exists a compact subset K of X such that suppφ ⊂ K and         |u(φ)| =  f (x)φ(x)dx  =  f (x)φ(x)dx  X

K

 ≤

 |f (x)||φ(x)|dx ≤

K

|f (x)|dx sup |φ(x)| < ∞. x∈K

K

Consequently the linear functional u : C0∞ (X) → C is well defined. Let {φn }∞ n=1 be a sequence in C0∞ (X) such that φn → φ, n → ∞, φ ∈ C0∞ (X), in C0∞ (X). Then  u(φn ) =

 f (x)φn (x)dx →n→∞ u(φ) =

X

f (x)φ(x)dx. X

Therefore u : C0∞ (X) → C is a linear continuous functional, i.e., u ∈ D  (X). Note that u ≡ f ∈ C ∞ (X) and therefore singsuppu = Ø. 1 for x ∈ R\{0}. x 1 Exercise 2.15 Find singsuppP for x ∈ R. x 1 Exercise 2.16 Find singsuppP 2 for x ∈ R\{0}. x 1 Exercise 2.17 Find singsuppP 2 for x ∈ R. x

Exercise 2.14 Find singsuppP

Definition 2.15 The distribution u ∈ D  (X) is called regular if there exists f ∈ L1loc (X) such that  u(φ) =

f (x)φ(x)dx

for ∀φ ∈ C0∞ (X).

X

In this case, we will write u = uf . If no such f exists, u is called singular. 1 , x ∈ R. The map u : C0∞ (X) → C, 1 + x2  u(φ) = f (x)φ(x)dx, φ ∈ C0∞ (R),

Example 2.11 Let f =

R

is a regular distribution.

100

2 Generalities on Distributions

Example 2.12 Consider δ(x), x ∈ R, and suppose that δ is a regular distribution. Then there exists f ∈ L1loc (R) such that uf = δ. Choose ρ ∈ C0∞ (R) for which supp(ρ) ⊂ U1 , ρ(0) = 1. Define the sequence {ρn }∞ n=1 by ρn (x) = ρ(nx),

x ∈ R,

n = 1, 2, . . . .

Then supp(ρn ) ⊂ U 1 and ρn (0) = 1. In addition, n

δ(ρn ) = ρn (0) = 1 and      1 = |δ(ρn )| =  f (x)ρ(nx)dx  ≤ |f (x)||ρ(nx)|dx U1

 ≤ sup |ρ(x)|

U1

n

n

|f (x)|dx →n→∞ 0,

x∈R U1

n

which is a contradiction. Therefore δ ∈ D  (R) is a singular distribution. Exercise 2.18 Let u1 , u2 ∈ D  (X) be regular distributions. Prove that α1 u1 + α2 u2 is a regular distribution for every α1 , α2 ∈ C.

2.7 Measures Definition 2.16 A measure on a Borel set A is a complex-valued additive function  μ(E) =

μ(dx), E

that is finite (|μ(E)| < ∞) on any bounded Borel subset E of A. The measure μ of A can be represented in a unique way in terms of four nonnegative measures μi ≥ 0, i = 1, 2, 3, 4, on A in the following way μ = (μ1 − μ2 ) + i(μ3 − μ4 ) and 

 μ(dx) =

E

 μ1 (dx) −

E

 μ2 (dx) + i

E

 μ3 (dx) − i

E

μ4 (dx) E

2.7 Measures

101

for any bounded Borel subset E of A. The measure μ on the open set X determines a distribution μ on X as follows  μ(φ) = φ(x)μ(dx), φ ∈ C0∞ (X), X

 where

is the Lebesgue–Stieltjes integral. From the integral properties, it follows

that μ ∈ D  (X). Every measure μ of X for which μ(dx) = f (x)dx, f ∈ L1loc (X), defines a regular distribution. Theorem 2.17 A distribution u ∈ D  (X) defines a measure μ on X if and only if u ∈ D 0 (X). Proof Let u ∈ D  (X) defines a measure μ of X. Then      |u(φ)| =  φ(x)μ(dx) ≤ μ(dx) sup |φ(x)| X1

X1

x∈X1

for every X1 ⊂⊂ X and every φ ∈ C0∞ (X1 ). Hence, u ∈ D 0 (X). Now, we suppose u ∈ D 0 (X), i.e., for every X1 ⊂⊂ X |u(φ)| ≤ C(X1 ) sup |φ(x)|, x∈X1

where C(X1 ) is a constant which depends on X1 . Let {Xk }∞ k=1 be a family of sets such that Xk ⊂⊂ Xk+1 , ∪k Xk = X. Since C0∞ (Xk ) is dense in C0 (Xk ), the RieszRadon theorem implies that there exists a measure μk of Xk such that  u(φ) =

φ(x)μk (dx),

φ ∈ C0 (Xk ),

k = 1, 2, . . . .

Xk

Therefore the measures μk and μk+1 coincide on Xk . From this, we conclude that there is a measure μ on X which coincides with μk on Xk and with the distribution u on X. This completes the proof. Definition 2.17 The distribution u ∈ D  (X) is called nonnegative on X if u(φ) ≥ 0 for every φ ∈ C0∞ (X), φ(x) ≥ 0, x ∈ X. Example 2.13 The distribution 1 is nonnegative. Exercise 2.19 Prove that the distribution H is nonnegative. Exercise 2.20 Prove that the distribution 1 is a measure.

102

2 Generalities on Distributions

2.8 Multiplying Distributions by C ∞ Functions Definition 2.18 The product of a distribution u ∈ D  (X) by a function b ∈ C ∞ (X) is defined by bu(φ) = u(bφ) for φ ∈ C0∞ (X). Note that the map φ → bφ is a linear and continuous map from C0∞ (X) to C0∞ (X). We have bu(α1 φ1 + α2 φ2 ) = u(b(α1 φ1 + α2 φ2 )) = u(α1 bφ1 + α2 bφ2 ) = α1 u(bφ1) + α2 u(bφ2 ) = α1 bu(φ1) + α2 bu(φ2 ) for α1 , α2 ∈ C, φ1 , φ2 ∈ C0∞ (X), i.e., bu is a linear map on C0∞ (X). Let {φn }∞ n=1 be a sequence in C0∞ (X) such that φn →n→∞ φ, φ ∈ C0∞ (X), in C0∞ (X). Then bφn →n→∞ bφ in C0∞ (X). Since u ∈ D  (X), we have u(bφn ) →n→∞ u(bφ), so bu(φn ) →n→∞ bu(φ). Consequently bu is a continuous functional on C0∞ (X) and bu ∈ D  (X). Example 2.14 Take x 2 δ. Then x 2 δ(φ) = δ(x 2 φ) = 02 φ(0) = 0 for φ ∈ C0∞ (X). Therefore x 2 δ = 0. Exercise 2.21 Compute (x 2 + 1)δ. Answer δ. Let α1 , α2 ∈ C, b1 , b2 ∈ C ∞ (X) and u1 , u2 ∈ D  (X). Then 1. (α1 b1 (x) + α2 b2 (x))u1 = α1 b1 (x)u1 + α2 b2 (x)u1 , 2. b1 (x)(α1 u1 + α2 u2 ) = α1 b1 (x)u1 + α2 b1 (x)u2 , x ∈ X.

2.9 Advanced Practical Problems

103

Let us prove that this multiplication is neither associative nor commutative. Suppose the contrary, so xδ(φ) = δ(xφ) = 0φ(0) = 0(φ),  1 1 xP (φ) = P (xφ) = P .V . φ(x)dx = 1(φ), x x

x ∈ R,

R

for φ ∈ C0∞ (R). Hence, 0 = 0P

1 1 1 1 = (xδ(x))P = (δ(x)x)P = δ(x)(xP ) = δ(x)1 = δ(x), x x x x

x ∈ R,

a contradiction.

2.9 Advanced Practical Problems Problem 2.1 Let P

1 be defined on C0∞ (R) by x2 1 P 2 (φ) = P .V . x

∞

−∞

φ(x) − φ(0) dx. x2

1 ∈ D  (R). x2 Problem 2.2 Define u by Prove that P

 u(φ) =

φ(x)dx

∀φ ∈ C0∞ (Rn ).

|x|≤1

Prove that u ∈ D  (Rn ). Problem 2.3 Define  u(φ) =

D α φ(x)dx

∀φ ∈ C0∞ (Rn ),

|x|≤1

where α is a multi-index. Show that u ∈ D  (Rn ). Problem 2.4 Let α be a multi-index and set u(φ) = D α φ(x0 ), φ ∈ C0∞ (X), for a given x0 ∈ X. Prove that u is a distribution of order |α|.

104

2 Generalities on Distributions

Solution Let φ1 , φ2 ∈ C0∞ (X) and a, b ∈ C. Then u(aφ1 +bφ2) = D α (aφ1 +bφ2 )(x0 ) = aD α φ1 (x0 )+bD α φ2 (x0 ) = au(φ1 )+bu(φ2). Consequently u is a linear map on C0∞ (X). Let K be a compact subset of X and φ ∈ C0∞ (K). Since supp φ ⊂ K we have to consider two cases: x0 ∈ K and x0 ∈ / K. If x0 ∈ K,      (2.20) |u(φ)| ≤ C sup D β φ(x) |β|≤|α| K

for C ≥ 1. If x0 ∈ / K, then u(φ) = 0. Therefore the inequality (2.20) holds, and then u ∈ D  (X). Using the definition of u and (2.20), we conclude that u has order |α|. Problem 2.5 Take f ∈ C (Rn ) and a multi-index α. Let D α f be defined on C0∞ (Rn ) as follows: D α f (φ) = (−1)|α|

 f (x)D α φ(x)dx. Rn

Prove that D α f is a distribution of order |α|. Problem 2.6 Prove that H ∈ D  (R).  aα (x)D α , where q ∈ N ∪ {0} is fixed, and a ∈ Problem 2.7 Let P (x, D) = 0

|α|≤q

C (Rn ). Let u be defined on C0∞ (Rn ) by  u(φ) =

u(x)P (x, D)φ(x)dx. Rn

Prove that u ∈ D  (Rn ). q

Problem 2.8 Let u ∈ D  (X) and suppose u(φ) ≥ 0 for every nonnegative function φ ∈ C0∞ (X). Prove that u is a measure, i.e., a distribution of order 0. Solution Let K ⊂ X be a compact set. Then there exists a function χ ∈ C0∞ (X) such that 0 ≤ χ ≤ 1 on X and χ = 1 on K. Then χ sup |φ| ± φ ≥ 0 K

for every φ ∈ C0∞ (K), and therefore u(χ sup |φ| ± φ) ≥ 0. K

(2.21)

2.9 Advanced Practical Problems

105

On the other hand, u(χ sup |φ| ± φ) = sup |φ|u(χ) ± u(φ). K

K

Consequently, using (2.21), ±u(φ) ≤ u(χ) sup |φ|. K

Therefore u ∈ D  (X), i.e., u is a measure. 0

Problem 2.9 Take φ ∈ C ∞ (X ×Y ), where Y is an open set in Rm , m ≥ 1. Suppose that there is a compact set K ⊂ X such that φ(x, y) = 0 for every x ∈ / K and for every y ∈ Y . Prove that the map y → u(φ(·, y)) is a C ∞ function for every u ∈ D  (X) and Dyα u(φ(·, y)) = u(Dyα φ(·, y)) for every multi-index α. Solution Since u ∈ D  (X) and φ ∈ C0∞ (X × Y ), we have that u(φ) is continuous with respect to the second variable. We will prove  ∂  ∂ u(φ(x, y)) = u φ(x, y) for x ∈ K, ∂yj ∂yj

y ∈ Y,

and j = 1, 2, . . . , m. For y ∈ Y given, φ(x, y + h) = φ(x, y) +

m 

hj

j =1

∂φ (x, y) + o(|h|2 ) ∂yj

for φ ∈ C0∞ (K × Y ). Let h = (0, . . . , 0, hj , 0, . . . , 0). Then ∂φ 1 φ(x, y + h) − φ(x, y) = (x, y) + o(h2j ). hj ∂yj hj Since u is linear, we have u

 φ(x, y + h) − φ(x, y)  hj

=u

 ∂φ

  1  (x, y) + u o(h2j ) . ∂yj hj

106

2 Generalities on Distributions

From this equality, we obtain u

 ∂  ∂ φ(x, y) = u(φ(x, y)), ∂yj ∂yj

as hj → 0. By induction, we get   u Dyα φ(x, y) = Dyα u(φ(x, y)),

x ∈ K,

y ∈ Y.

Problem 2.10 Let un ∈ D  (X), un (φ) ≥ 0 for every nonnegative φ ∈ C0∞ (X) and un →n→∞ u in D  (X). Prove that u ≥ 0 and un (φ) →n→∞ u(φ) for every φ ∈ C00 (X). Problem 2.11 Prove that the functions 1

1. f = e x 2 , 1 2. f = e x m , m ∈ N do not define distributions, i.e. f ∈ D  (R\{0}) in all cases. 1. Hint Use k2

φk (x) = e− 4 kφ0 (kx). 2. Hint Use φk (x) = e

 m − k2

kφ0 (kx).

Problem 2.12 Given constants m ∈ N, ai , i = 1, 2, . . . , m, prove that 1

1

1

f = a1 e x + a2 e x 2 + · · · + am e x m ∈ D  (R\{0}). Hint Use the previous problem. Problem 2.13 (Sochozki Formulas) Show that   φ(x) φ(x) dx = iπφ(0) + P .V . dx, φ ∈ C0∞ (R), 1. lim →0 x − i x R R  φ(x) φ(x) 2. lim dx = −iπφ(0) + P .V . dx, φ ∈ C0∞ (R). →0 x + i x R

R

2.9 Advanced Practical Problems

107

1. Solution Take φ ∈ C0∞ (R) with supp φ ⊂ [−R, R]. Then  R

φ(x) dx = x − i

R −R

R = −R

(x + i)φ(x) dx x2 + 2 (x + i)(φ(x) − φ(0)) dx + x2 + 2

R

−R

(x + i)φ(0) dx. x2 + 2

From this, R lim

→0 −R

(x + i)(φ(x) − φ(0)) dx = P .V . x2 + 2

∞

−∞

φ(x) dx. x

What is more, R lim

→0 −R

R (x + i)φ(0) dx = 2iφ(0) lim arctan = iπφ(0) = iπδ(φ). →0 x2 + 2 

2. Hint Use the solution of part 1. Problem 2.14 (Sochozki Formulas) Prove that 1 1 = iπδ + P , x − i0 x

1 1 = −iπδ + P . x + i0 x

Hint Use the previous problem. Problem 2.15 Prove that x 1 1 1. lim √ sin = δ(x), = δ(x), lim 2 x →0+ →0+ πx  2 πe− 4 ixt −ixt e e 2. lim = 2πiδ(x), lim = 0, t →∞ x − i0 t →∞ x − i0 ixt −ixt e e = 0, lim = −2πiδ(x), 3. lim t →∞ x + i0 t →∞ x + i0  cos(tx)  = 0, 4. lim t m eixt = 0, m ≥ 0, lim P t →∞ t →∞ x 1 x  2n3 x 2 = δ(x), lim 5. lim ω = δ(x), n→∞ π(1 + n2 x 2 )2 →0+   n 1 sin2 (nx) 6. lim = δ(x), = δ(x), lim n→∞ π(1 + n2 x 2 ) n→∞ nπ x2

108

7. 8. 9. 10.

2 Generalities on Distributions

⎧ 1 ⎨n for |x| ≤ lim fn (x) = δ(x), where fn (x) = 2 n n→∞ ⎩ 0 otherwise; 2 2 n x n sin(nx) lim √ e− 2 = δ(x), lim = δ(x), n→∞ 2π n→∞ πx n 1 1 lim ne−n|x| = δ(x), lim = δ(x), n→∞ 2 n→∞ π e nx + e −nx " n −nx 2 n e lim = δ(x), lim 2 2 = δ(x)π. n→∞ π n→∞ n x + 1

1. Solution Take φ ∈ C0∞ (R). Then there exists R > 0 such that suppφ ⊂ [−R, R]. Now, 2  R e− x4 2 1 − x4 , φ(x) = √ e √ φ(x)dx 2 π 2 π  2 −R  2 x R √ −  R − 2√x  2  # $ e e φ(0) 1 √ φ(x) − φ(0) dx + √ √ dx = √ π 2  π 2  −R −R  2  2 x R √  − 2  R  x  x − 2√ 1 φ(0) e φ(x) − φ(0)  = √ dx + √ e d √ x √ x π π 2  2  −R −R  2 R √ R − 2√x  2  φ(0) e φ(x) − φ(0) 1 2 dx + √ e−y dy. √ x = √ x π 2  π



−R

R − 2√ 

Therefore  lim

→0+

 x2 1 √ e− 4 , φ(x) 2 π  

1 = lim √ →0+ π φ(0) = √ π

∞

R

−R

2

e

−

x √ 2 

√ 2 

x

φ(0) φ(x) − φ(0) dx + √ lim x π →0+

R √ 

2

R − 2√ 

e−y dy = φ(0) = δ(φ). 2

−∞

Problem 2.16 Prove that suppδ = {0}.

e−y dy 2

2.9 Advanced Practical Problems

109

Problem 2.17 Let K be a compact set in Rn which cannot be written as union of finitely many compact connected domains. Prove that there exists a distribution u ∈ E  (K) of order 1 that does not satisfy 

u(φ) ≤ C

    sup ∂ α φ ,

φ ∈ C ∞ (X),

|α|≤k K

for any constants C and k. Problem 2.18 Let K be a compact set in Rn and uα , |α| ≤ k, be continuous functions on K. For |α| ≤ k, we set   Uα (x, y) = uα (x) −



uα+β (y)

|β|≤k−|α|

(x − y)β  |x − y||α|−k , β!

for x, y ∈ K, x = y, and Uα (x, x) = 0 for x ∈ K. Suppose that every function Uα , |α| ≤ k, is continuous on K × K. Prove that there exists v ∈ C k (Rn ) such that ∂ α v(x) = uα (x) for x ∈ K, |α| ≤ k. Then prove that v can be chosen so that 

       sup ∂ α v  ≤ C sup Uα + sup uα , |α|≤k K×K

|α|≤k

|α|≤k K

where C is a constant depending on K only. Problem 2.19 Prove that |u(φ)| ≤ C



+

sup

|α|≤k x,y∈K,x =y



  α ∂ φ(x) −

    sup ∂ α φ  ,

 |β|≤k−|α|

∂ α+β φ(y)

(x − y)β  |x − y||α|−k β!

φ ∈ C ∞ (Rn ),

|α|≤k K

for every distribution u of order k with compact support K ⊂ Rn . Problem 2.20 Let K be a compact set in Rn with finitely many connected components such that every two points x and y in the same component can be joined by a rectifiable curve in K of length ≤ C|x − y|. Prove that for every distribution u of order k with suppu ⊂ K the estimate |u(φ)| ≤ C



    sup ∂ α φ ,

|α|≤k K

holds.

φ ∈ C k (Rn ),

110

2 Generalities on Distributions

Problem 2.21 Let fn , f ∈ L1loc (X) and  |fn (x) − f (x)|dx →n→∞ 0 K

for every compact subset K of X. Prove that fn →n→∞ f in D  (X). Problem 2.22 Prove that $ 1# δ(x − a) + δ(x + a) , a = 0, 1. δ(x 2 − a 2 ) = 2a ∞  2. δ(sin x) = δ(x − kπ). k=−∞

Problem 2.23 Prove that δ(x), x ∈ R, is a measure. Problem 2.24 Prove that H (x), x ∈ R, is a measure. Problem 2.25 Let u ∈ D  (X) and b ∈ C ∞ (X) be such that b ≡ 1 on a neighbourhood of suppu. Show that u = bu. Solution For the function 1 − b we have that 1 − b ≡ 0 on suppu. Then for φ ∈ C0∞ (X), we have 0 = u((1 − b)φ) = u(φ − bφ) = u(φ) − u(bφ) = u(φ) − bu(φ), so u(φ) = bu(φ) for every φ ∈ C0∞ (X). Therefore u = bu. Problem 2.26 Compute 1 (x 4 + x 2 + 3)δ(x) + xP , x Answer 3δ + 1. Problem 2.27 Let b ∈ C ∞ (R). Compute b(x)δ(x),

x ∈ R.

x ∈ R.

2.9 Advanced Practical Problems

111

Answer b(0)δ. Problem 2.28 Let a ∈ C ∞ (X), u ∈ D  (X). Prove that supp(au) ⊂ suppa ∩suppu. Problem 2.29 Let f , u ∈ D  (X) and singsuppu ∩ singsuppf = Ø. Prove that f ◦ u ∈ D  (X). Problem 2.30 Let f ∈ C ∞ (X), u ∈ D  (X) and suppu ∩ suppf ⊂⊂ X. Prove that u(f ) can be defined by u(f ) = (f u)(1). Problem 2.31 Let f ∈ C k (X), u ∈ D  (X). Prove that f u ∈ D  (X). k

k

Problem 2.32 Solve the equation (x − 3)u = 0 in D  (X). Solution Let φ ∈ C0∞ (R). Then we have (x − 3)u(φ) = 0

or u((x − 3)φ) = 0.

(2.22)

Let now ψ ∈ C0∞ (R). We choose η ∈ C0∞ (R) so that η ≡ 1 on [3 − , 3 + ] and ψ(x) − η(x)ψ(3) , η ≡ 0 on R\[3−, 3+], for a small enough  > 0. Let g(x) = x−3 ∞ x ∈ R. Then g ∈ C0 (R). From this and (2.22), we have that  ψ(x) − η(x)ψ(3)  = 0. u (x − 3) x−3 Hence,   ψ(x) − η(x)ψ(3) u(ψ) = u (x − 3) + η(x)ψ(3) x−3  ψ(x) − η(x)ψ(3)  + u(η(x)ψ(3)) = u (x − 3) x−3 = ψ(3)u(η) = Cψ(3) = Cδ(x − 3)(ψ). Here C = u(η) = const. Since ψ ∈ C0∞ (R) was arbitrarily chosen, we conclude that u = Cδ(x − 3). Problem 2.33 Solve the equation (x − 3)u = P in D  (R).

1 x−3

112

2 Generalities on Distributions

Solution By using the previous problem, the corresponding homogeneous equation (x − 3)u = 0 is solved by u = Cδ(x − 3), C = const, and a particular solution is 1 P . Therefore (x − 3)2 u = Cδ(x − 3) + P

1 . (x − 3)2

Problem 2.34 Solve the equations 1. (x − 1)(x − 2)u = 0, 2. (sin x)u = 0. Answer 1. u = C1 δ(x − 1) + c2 δ(x − 2), C1 , C2 = const, ∞  Ck δ(x − kπ), Ck = const. 2. k=−∞

2.10 Notes and References In this chapter are collected the definitions for distributions, their order, sequences and series, support and singular support, and multiplication by C ∞ functions. They are deducted some of their basic properties. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 3

Differentiation

3.1 Derivatives Let X be an open set in Rn . Definition 3.1 For u ∈ D  (X) and α ∈ Nn ∪ {0}, we define D α u as follows D α u(φ) = (−1)|α| u(D α φ)

(3.1)

for every φ ∈ C0∞ (X). Since the operation φ → D α φ is linear and continuous on C0∞ (X), the functional D α u, determined by (3.1), is linear and continuous, i.e., D α u ∈ D  (X). Below, we will list some of the properties of the derivatives of the distributions. Theorem 3.1 The operation u → D α u : D  (X) → D  (X) is linear and continuous. Proof We start by showing linearity. Let α1 , α2 ∈ C, u1 , u2 ∈ D  (X) and φ ∈ C0∞ (X). Then (α1 u1 + α2 u2 )(φ) → D α (α1 u1 + α2 u2 )(φ) = (−1)|α| (α1 u1 + α2 u2 )(D α φ) = (−1)|α| α1 u1 (D α φ) + (−1)|α| α2 u2 (D α φ) = α1 D α u1 (φ) + α2 D α u2 (φ).  Now, we will prove the continuity. Let {un }∞ n=1 be a sequence in D (X) such that  un →n→∞ 0 in D (X). Then

D α un (φ) = (−1)|α| un (D α φ) →n→∞ 0 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_3

113

114

3 Differentiation

for φ ∈ C0∞ (X). Consequently D α un →n→∞ 0 in D  (X). This completes the proof. Example 3.1 Let us consider the Heaviside function H (x), x ∈ R. For its derivatives, we have D α H (φ) = (−)α H (D α φ) ∞ α = (−1) H (x)D α φ(x)dx −∞ ∞

= (−1)α

D α φ(x)dx 0

= (−1)α D α−1 φ(x)|x=∞ x=0 = (−1)α+1D α−1 φ(0) for φ ∈ C0∞ (R) and α ∈ N. Example 3.2 Let us compute lim D α ω

→0

in D  (R),

for α ∈ N.

Let φ ∈ C0∞ (R). Then lim D α ω (φ) = (−1)α lim ω (D α φ)

→0

→0

∞ = (−) lim α

→0 −∞

ω (x)D α φ(x)dx



= (−1) lim α

→0 |x|≤

= (−1)α D α φ(0) = (−1)α δ(D α φ) = D α δ(φ). Therefore lim D α ω = D α δ in D  (X). →0

Exercise 3.1 Find δ (k) (x), x ∈ R, k ∈ N. Answer (−1)k φ (k) (0), φ ∈ C0∞ (R).

e

−

2  2 −|x|2

D α φ(x)dx

3.1 Derivatives

115

By Theorem 3.1, it follows that if the series 

uk (x) = S(x),

uk ∈ L1loc (X),

k≥1

is uniformly convergent on every compact subset K of X, it may be differentiated term by term any number of times, and the resulting series will converge in D  (X). We have  D α uk (x) = D α S(x). k≥1

Theorem 3.2 Every distribution u ∈ D  (X) is differentiable infinitely many times. Proof Since u(D α φ) exists for every α ∈ Nn ∪ {0} and every φ ∈ C0∞ (X), we conclude that u is differentiable infinitely many times. This completes the proof. Theorem 3.3 We have D α+β u = D α (D β u) for every α, β ∈ Nn ∪ {0} and every u ∈ D  (X). Proof Let in fact φ ∈ C0∞ (X) be arbitrary. Then D α+β u(φ) = (−1)|α+β| u(D α+β φ) = (−1)|α| (−1)|β| u(D β (D α φ)) = (−1)|α| D β u(D α φ) = D α (D β u)(φ). This completes the proof. Theorem 3.4 We have D α (α1 u1 + α2 u2 ) = α1 D α u1 + α2 D α u2 for u1 , u2 ∈ D  (X), α1 , α2 ∈ C, α ∈ Nn ∪ {0}. Proof Let φ ∈ C0∞ (X) be arbitrarily chosen. Then D α (α1 u1 + α2 u2 )(φ) = (−1)|α| (α1 u1 + α2 u2 )(D α φ) = (−1)|α| α1 u1 (D α φ) + (−1)|α| α2 u2 (D α φ) = α1 D α u1 (φ) + α2 D α u2 (φ)

= α1 D α u1 + α2 D α u2 (φ). This completes the proof.

116

3 Differentiation

Theorem 3.5 If u ∈ D  (X) , a ∈ C ∞ (X), we have D (au) = α

 α β≤α

β

D β aD α−β u,

for every α ∈ Nn ∪ {0}. Proof We will prove the simple case ∂ ∂u ∂a (au) = a + u ∂xi ∂xi ∂xi for some i ∈ {1, 2, . . . , n}. Using induction, the reader can deduce the general case. If φ ∈ C0∞ (X), then  ∂  ∂ (au)(φ) = −au φ ∂xi ∂xi  ∂  = −u a φ ∂xi  ∂ ∂a  (aφ) − φ = −u ∂xi ∂xi   ∂a   ∂ (aφ) + u φ = −u ∂xi ∂xi =

∂u ∂a (aφ) + u(φ) ∂xi ∂xi

∂a ∂u (φ) + u(φ) ∂xi ∂xi  ∂u ∂a  + u (φ), = a ∂xi ∂xi =a

so

∂ ∂a ∂u (au) = u+a in D  (X). This completes the proof. ∂xi ∂xi ∂xi

Theorem 3.6 For every u ∈ D  (X) and every α ∈ Nn ∪ {0}, suppD α u ⊂ suppu. Proof To see this, let u ∈ D  (X), φ ∈ C0∞ (X) and suppφ ∩ suppu = Ø. Then D α φ ∈ C0∞ (X) and suppD α φ ∩ suppu = Ø. Hence, D α u(φ) = (−1)|α| u(D α φ) = 0, and the assertion follows.

3.1 Derivatives

117

Theorem 3.7 Let f be a function defined on (a, b) that is piecewise-differentiable with continuity. Call {xk } the points in (a, b) where f or its derivative have jump discontinuities. Write [f ]xk = f (xk + 0) − f (xk − 0), and denote by {f  } the classical derivative of f at x ∈ (a, b). Then f  = {f  } +



[f ]xk δ(x − xk ).

k

Proof For φ ∈ C0∞ (a, b), we have f  (φ) = −f (φ  ) xk+1   =− f (x)φ  (x)dx k

=

 k

b =

xk

xk+1

#

{f  }(x)φ(x)dx −

f (xk+1 − 0)φ(xk+1 ) − f (xk + 0)φ(xk )

k

xk

{f  }(x)φ(x)dx +

#

$ f (xk + 0) − f (xk − 0) φ(xk )

k

a

 [f ]xk δ(x − xk )(φ). = {f  }(φ) + k

This completes the proof. Theorem 3.8 The trigonometric series k=∞ 

ak eikx ,

x ∈ R,

k=−∞

where ak ∈ C, k ∈ Z, satisfy |ak | ≤ A(1 + |k|)m ,

k ∈ Z,

for some m ∈ N ∪ {0}, is convergent in D  (R). Proof Since the series a0 x 2m+2  ak + eikx , (m + 2)! (ik)m+2 k =0

x ∈ R,

$

118

3 Differentiation

is uniformly convergent in R, the series which is a derivative of it of order m + 2 is convergent in D  (R). This completes the proof. Theorem 3.9 We have 1  ikx  e = δ(x − 2kπ). 2π k

k

Proof The function f0 (x) =

x2 x − , 2 4π

0 ≤ x < 2π,

has Fourier series f0 (x) =

1  eikx π − , 6 2π k2

x ∈ R.

k =0

Using Theorem 3.7, we have f0 (x) = −

1 x i  eikx = − 2π k 2 2π k =0

and f0 (x) =

 1  ikx 1 + e =− δ(x − 2kπ). 2π 2π k =0

k

From the last equation, we obtain 1  ikx  e = δ(x − 2kπ). 2π k

k

This completes the proof.

3.2 The Local Structure of Distributions Theorem 3.10 Let u ∈ D  (X) and X1 ⊂ X be a bounded set. Then there exist a function f ∈ L∞ (X1 ) and an integer m ≥ 0 such that u(x) = D1m . . . Dnm f (x),

x ∈ X1 .

3.2 The Local Structure of Distributions

119

Proof Since u ∈ D  (X) and X1 ⊂ X, there exist constants C and k so that 

|u(φ)| ≤ C

sup |D α φ(x)|,

|α|≤k x∈X1

φ ∈ C0∞ (X1 ).

With d we will denote the diameter of X1 . Applying the inequality  xj      ∂φ |φ(x)| =  (x1 , . . . , xj −1 , t, xj +1 , . . . , xn )dt  ∂xj   −∞

 xj   ∂φ    ≤  ∂x (x1 , . . . , xj −1 , t, .xj +1 , . . . , xn ) dt j −∞

   ∂φ   (x) , ≤ d max  x∈X1 ∂xj

x ∈ X1 ,

j ∈ {1, . . . , n},

a sufficiently number of times, we obtain |u(φ)| ≤ c1 max |D1k . . . Dnk φ(x)|, x∈X1

φ ∈ C0∞ (X1 ),

for some constant c1 > 0. Note that, for any ψ ∈ C0∞ (X1 ), we have x1 ψ(x) =

xn ...

−∞

−∞

∂ nψ (y)dyn . . . dy1 , ∂y1 . . . ∂yn

x ∈ X1 ,

and then  |ψ(x)| ≤

|D1 . . . Dn ψ(y)|dy,

x ∈ X1 .

X1

From here and from (3.2), for m = k + 1, we find  |u(φ)| ≤ c2 |D1m . . . Dnm φ(x)|dx,

φ ∈ C0∞ (X1 ),

X1

for some constant c2 > 0. For φ ∈ C0∞ (X1 ), denote χ(x) = (−1)mn D1m . . . Dnm φ(x),

x ∈ X1 ,

(3.2)

120

3 Differentiation

and define the functional u∗ (χ) = u(φ). By the Hahn–Banach theorem, it follows that there exist an extension ! u of u∗ on 1 L (X1 ) such that |! u(φ)| = |u(φ)| ≤ c3 χ1 for some constant c3 > 0. By a F. Riesz theorem, we find that there is a function f ∈ L∞ (X1 ) so that f ∞ ≤ c4 and  ! u(χ) = f (x)χ(x)dx. X1

Hence and the definitions of χ, u∗ and ! u, we arrive at  u(φ) = (−1)mn f (x)D1m . . . Dnm φ(x)dx  =

X1

D1m . . . Dnm f (x)φ(x)dx,

φ ∈ C0∞ (X1 ).

X1

This completes the proof. ∞ Definition 3.2 We say that the sequence {φk }∞ k=1 ⊂ C (X) converges to 0 in ∞ α C (X) if D φk →k→∞ 0 uniformly on any X1 ⊂ X for any α ∈ Nn ∪ {0}.

Note that the convergence in C0∞ (X) implies the convergence in C ∞ (X), but not vice versa. Theorem 3.11 For a distribution u ∈ D  (X) to have a compact support in X, it is necessary and sufficient that it admits of a linear and continuous extension onto C ∞ (X). Proof 1. Suppose that u ∈ D  (X) has a compact support in X, i.e., suppu = K ⊂⊂ X. Take η ∈ C0∞ (X) such that η ≡ 1 in a neighbourhood of K. Define the functional ! u(φ) = u(ηφ),

φ ∈ C ∞ (X).

Let φ1 , φ2 ∈ C ∞ (X) and α1 , α2 ∈ C. Then ! u(α1 φ1 + α2 φ2 ) = u (η(α1 φ1 + α2 φ2 )) = u(α1 ηφ1 + α2 ηφ2 )

(3.3)

3.2 The Local Structure of Distributions

121

= α1 u(ηφ1 ) + α2 u(ηφ2 ) u(φ1 ) + α2! u(φ2 ), = α1! ı.e., ! u is a linear functional on C ∞ (X). Since the map φ → ηφ is a continuous map from C ∞ (X) into C0∞ (X), we conclude that ! u is a continuous functional on C ∞ (X). Next, for φ ∈ C0∞ (X), we have ! u(φ) = u(ηφ) = (ηu)(φ) = u(φ). Thus, ! u is an extension of u from C0∞ (X) onto C ∞ (X). Now, we will prove that this extension is unique. Suppose that ! ! u is another extension of u onto C ∞ (X). Let Xk , k ∈ N, be a sequence of bounded subsets of X such that X = ∪∞ k=1 Xk ∞ and X1 ⊂ X2 ⊂ . . .. Let {ηk }∞ be a sequence of elements of C (X) such k=1 0 that ηk ≡ 1 on Xk , k ∈ N, and ηk →k→∞ 1 in C ∞ (X). Therefore, for any φ ∈ C ∞ (X), we have ηk φ →k→∞ φ in C ∞ (X). From here, we get 

 ! u(φ) = ! u

lim (ηk φ)

k→∞

= lim ! u(ηk φ) k→∞

= lim u(ηk φ) k→∞

= lim ! ! u(ηk φ) k→∞

=! ! u





lim (ηk φ)

k→∞

=! ! u(φ),

φ ∈ C ∞ (X).

So, ! u =! u. u onto C ∞ (X). 2. Let u ∈ D  (X) admits of a linear and continuous extension ! Suppose that suppu is not a compact set. Then there exists a sequence {φk }∞ k=1 of elements of C0∞ (X) such that suppφk ⊂ X\Xk , X1 ⊂ X2 ⊂ . . ., ∪∞ X k=1 k = X, and u(φk ) = 1, k ∈ N. Note that φk →k→∞ 0 in C ∞ (X) and then ! u(φk ) →k→∞ 0, which is a contradiction because ! u(φk ) = u(φk ) = 1, This completes the proof.

k ∈ N.

122

3 Differentiation

Theorem 3.12 Any distribution u ∈ D  (X) with compact support has a finite order in X. Proof Since u ∈ D  (X) has a compact support, it admits of a linear and continuous extension ! u onto C ∞ (X) and (3.3) holds. Because η ∈ C0∞ (X) and suppη ⊂ X, there exists an X1 ⊂ X so that η ∈ C0∞ (X1 ). Therefore ηφ ∈ C0∞ (X1 ) for any φ ∈ C ∞ (X). From the last conclusion, it follows that there exist constants K = K(X1 ) and m = m(X1 ) so that |! u(φ)| = |u(ηφ)| ≤ K1



sup |D α (ηφ)(x)|,

|α|≤m x∈X1

φ ∈ C0∞ (X). Hence, |! u(φ)| ≤ C1



φ ∈ C0∞ (X),

sup |D α φ(x)|,

|α|≤m x∈X1

for some constant C1 > 0. This completes the proof. Theorem 3.13 Let 0 ∈ X, u ∈ D  (X) and suppu = {0}. Then u is uniquely representable in the following form u(x) =



cα D α δ(x),

x ∈ X,

(3.4)

|α|≤k

where k is the order of u and cα ∈ C, |α| ≤ k. Proof Let η ∈ C0∞ (U1 ) be such that η ≡ 1 in U 1 . For φ ∈ C0∞ (X), define 2

φN (x) =

 Dα φ (0)x α , α!

x ∈ X.

|α|≤N

Then, for any  > 0, we have u = η

x  

u. Hence,

 x   u (φ) u(φ) = η   x    x   =u η (φ − φN ) + u η φN .  

(3.5)

3.2 The Local Structure of Distributions

Note that η

x  

123

(φ − φN ) ∈ C0∞ (U ). Then

   x   x        (φ − φN )  ≤ C max D α η (φ(x) − φN (x))  u η   |x| ≤  |α| ≤ N  x    α    ≤ C max D α−β (φ(x) − φN (x)) D β η β  |x| ≤  β≤α |α| ≤ N  ≤ C1 max  |β|  N−|α−β|  |α≤N

β≤α

≤ C1 max  N−|α|+1 |α|≤N

≤ C2 , u is an extension of u onto C ∞ (X). Now, for some positive constants C, C1 , C2 . Let ! employing (3.5), we arrive at u(φ) = ! u(φ) =

 Dα φ (0)! u xα . α!

|α|≤k

Set cα =

(−1)|α| α ! u x . α!

Then u(φ) =

 |α|≤k

=



(−1)|α| cα D α φ(0)

(−1)|α| cα δ D α φ

|α|≤k

=



cα D α δ(φ).

|α|≤k

Now, we will prove the uniqueness of the representation (3.4). Assume that there is another such representation u(x) =

 |α|≤k

cα D α δ(x),

x ∈ X.

124

3 Differentiation

Hence, 

cα − cα D α δ(x),

0=

x ∈ X,

|α|≤k

and then 

cα − cα D α δ(x m )

0=

|α|≤k

 

 cα − cα (−1)|α| D α x m 

=

x=0

|α|≤k

 = (−1)m m! cm − cm ,

x ∈ X,

 i.e., cm = cm . This completes the proof.

3.3 The Primitive of a Distribution Let (a, b) ⊂ R, u ∈ D  (a, b), φ ∈ C0∞ (a, b) and x0 ∈ (a, b) be arbitrary but fixed. We also fix  > 0 such that  < min{x0 − a, b − x0 }. The function φ can be represented as ∞



φ(x) = ψ (x) + ω (x − x0 )

x ∈ (a, b),

φ(ξ )dξ,

(3.6)

−∞

where ψ is determined by the equality x  ∞  ψ(x) = φ(s) − ω (s − x0 ) φ(ξ )dξ ds, −∞

x ∈ (a, b).

(3.7)

−∞

Suppose suppφ ⊂ [a  , b  ] ⊂ (a, b). Since φ ∈ C0∞ (a, b) and ω ∈ C ∞ (a, b), we have that ψ ∈ C ∞ (a, b). Moreover, ψ(x) ≡ 0 if x < a  = min{a  , x0 − }. As  < {x0 −a, b−x0}, it follows  < x0 −a and a < x0 −, and since [a  , b  ] ⊂ (a, b), we get a < a  and a  > a. For x > b = max{b , x0 + } < b, using (3.7), we have ∞  ∞  φ(ξ )dξ ds φ(s) − ω (s − x0 ) ψ(x) = −∞

−∞

∞ =

∞ φ(s)ds −

−∞

−∞

∞ ω (s − x0 ) −∞

φ(ξ )dξ ds

3.3 The Primitive of a Distribution

125

∞ =

∞ φ(s)ds −

ω (s − x0 )ds

−∞

−∞

∞

∞

=

φ(s)ds − −∞

∞ φ(ξ )dξ

−∞

φ(ξ )dξ = 0.

−∞

Therefore suppψ ⊂ [a  , b  ] and ψ ∈ C0∞ (a, b). Definition 3.3 A distribution u(−1) ∈ D  (a, b) is said to be a primitive of the distribution u if   u(−1) = u in D  (a, b). We assume that the primitive u(−1) of the distribution u ∈ D  (a, b) exists in D  (a, b). Then we have the representation

u

(−1)

(φ) = u

(−1)

∞    ψ + ω (x − x0 ) φ(ξ )dξ −∞

=u

(−1)



(ψ ) + u

(−1)



∞ ω (x − x0 )

 φ(ξ )dξ

−∞

∞   (−1) (−1) (ψ) + u (ω (x − x0 )) φ(ξ )dξ =− u −∞

where φ ∈ C0∞ (a, b) and ψ ∈ C0∞ (a, b) satisfy (3.7). Setting C = u(−1) (ω (x − x0 )) = const, we obtain ∞ u

(−1)

(φ) = −u(ψ) + C

φ(ξ )dξ.

(3.8)

−∞

Now, we will show that if the functional u(−1) satisfies (3.8) for an arbitrary constant C, then it is first of all a distribution in D  (a, b), and also a primitive for u ∈ D  (a, b).

126

3 Differentiation

Let α1 , α2 ∈ C, φ1 , φ2 ∈ C0∞ (a, b). Take ψ1 , ψ2 ∈ C0∞ (a, b) such that ∞ u

(−1)

(φ1 ) = −u(ψ1 ) + C

φ1 (ξ )dξ,

−∞

∞ u

(−1)

(φ2 ) = −u(ψ2 ) + C

φ2 (ξ )dξ,

−∞

x ψ1 (x) =

∞   φ1 (s) − ω (s − x0 ) φ1 (ξ )dξ ds,

−∞

−∞

x  ∞  ψ2 (x) = φ2 (ξ )dξ ds, φ2 (s) − ω (s − x0 ) −∞

x ∈ (a, b).

−∞

Then we get ∞ (−1)

u

(α1 φ1 + α2 φ2 ) = −u(α1 ψ1 + α2 ψ2 ) + C

(α1 φ1 (ξ ) + α2 φ2 (ξ ))dξ −∞

∞ = −α1 u(ψ1 ) − α2 u(ψ2 ) + Cα1

∞ φ1 (ξ )dξ + Cα2

−∞

φ2 (ξ )dξ

−∞

= α1 u(−1) (φ1 ) + α2 u(−1) (φ2 ).

Consequently u(−1) is a linear functional on C0∞ (a, b). ∞ ∞ Let now {φn }∞ n=1 be a sequence in C0 (a, b) with φn → 0, n → ∞, in C0 (a, b). ∞ Choose ψn ∈ C0 (a, b) so that ∞ u

(−1)

(φn ) = −u(ψn ) + C

φn (ξ )dξ,

−∞

x  ∞  φn (ξ )dξ ds. φn (s) − ω (s − x0 ) ψn (x) = −∞

−∞

Then ψn → 0, n → ∞, in C0∞ (a, b). From here, u(−1)(φn ) →n→∞ 0.

3.4 Simple and Double Layers on Surfaces

127

Therefore u(−1) is a linear continuous functional on C0∞ (a, b), i.e., u(−1) ∈ D  (a, b). Now, we will show that u(−1) is a primitive of u. We replace φ by φ  in (3.8) and we get

u

(−1)

∞



(φ ) = −u(ψ) + C

φ  (ξ )dξ = −u(ψ),

(3.9)

−∞

where x  ∞ x    ψ(x) = φ (ξ )dξ ds = φ  (s)ds = φ(x), φ (s)−ω (s−x0 ) −∞

−∞

x ∈ (a, b).

−∞

  The last relation and (3.9) imply u(−1) (φ  ) = −u(φ), so − u(−1) (φ) = −u(φ),   i.e., u(−1) (φ) = u(φ). Since φ ∈ C0∞ (a, b) was arbitrary, we conclude   u(−1) = u. The solution to the equation u = f,

(3.10)

for u, f ∈ D  (a, b), can be represented in the form u = f (−1) + C,

(3.11)

where C is an arbitrary constant. If f ∈ C (a, b), (3.11) is a classical solution of the Eq. (3.10). Proceeding as above, we can define successive primitives u(−n) using the relation ship u(−n) = u(−n−1) .

3.4 Simple and Double Layers on Surfaces Definition 3.4 Let S be a piecewise smooth surface in Rn and let f be a continuous function on S. We introduce the generalized function f δS in the following manner  f δS (φ) = f (x)φ(x)dS, φ ∈ C0∞ (S). S

The generalized function f δS is called a simple layer on the surface S. By the definition, it follows that f δS ∈ D  (S), f δS (x) = 0 for x ∈ S and so, suppf δS ⊂ S.

128

3 Differentiation

Theorem 3.14 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S and write Ω = Rn \Ω1 . Consider f ∈ C 1 (Ω) ∩ C 1 (Ω1 ), [f ]S (x) = lim f (x  ) − lim f (x  ),   x →x x  ∈Ω1

x →x x  ∈Ω

x ∈ S.

 ∂f  (x), i = 1, 2, . . . , n, we will denote the classical derivatives of f at ∂xi ∂ x ∈ Rn , x ∈ / S, while , i = 1, 2, . . . , n, will denote derivatives in D  (Rn ). Then ∂xi for every i = 1, 2, . . . , n, With

 ∂f  ∂f = + [f ]S cos(n, xi )δS , ∂xi ∂xi

f ∈ D  (Rn ),

where n is the outer normal to S. Proof Let φ ∈ C0∞ (Rn ). The Gauss theorem tells us that  ∂φ  ∂f (φ) = −f ∂xi ∂xi  ∂φ = − f (x) (x)dx ∂xi Rn



=−

f (x) Ω



=−



∂φ (x)dx − ∂xi

f (x) Ω1

    ∂f  ∂f  (x)φ(x)dx + (x)φ(x)dx ∂xi ∂xi

∂φ f (x) (x)dx − ∂xi

Ω



−  =

∂φ (x)dx ∂xi

Ω

Ω

Ω1

Ω1

    ∂f  ∂f  (x)φ(x)dx + (x)φ(x)dx ∂xi ∂xi

∂φ f (x) (x)dx − ∂xi

Ω1

  ∂f  [f ]S cos(n, xi )φ(x)dS + (x)φ(x)dx ∂xi Rn

S

= [f ]S cos(n, xi )δS (φ) +

 ∂f  ∂xi

(φ).

As φ ∈ C0∞ (Rn ) was arbitrary, we conclude  ∂f  ∂f + [f ]S cos(n, xi )δS . = ∂xi ∂xi This completes the proof.

3.4 Simple and Double Layers on Surfaces

129

Theorem 3.15 Consider the plane R2 with complex coordinate z = x + iy, and the differential form dz = dx + idy annihilating the Cauchy–Riemann operator ∂ 1 ∂ ∂  = +i . ∂z 2 ∂x ∂y Let Ω be a bounded domain in R2 with piecewise-smooth boundary S. We take ∂f f ∈ C 1 (Ω) with f = 0 on R2 \Ω. If is the derivative of f in the sense of ∂z  ∂f  is the classical derivative of f at z, z ∈ / S, then distributions and ∂z  ∂f  1   ∂f = + f cos(nx) + i cos(ny) δS . ∂ z¯ ∂ z¯ 2 Proof As ∂f 1 ∂f i ∂f = + , ∂z 2 ∂x 2 ∂y applying Theorem 3.14 to

(3.12)

∂f ∂f and gives ∂x ∂y  ∂f  ∂f = + f cos(n, x)δS , ∂x ∂x   ∂f ∂f = + f cos(n, y)δS . ∂y ∂y

From here and (3.12), it follows 1  ∂f  i  ∂f  1 i ∂f = + + f cos(n, x)δS + f cos(n, y)δS ∂z 2 ∂x  2 ∂y 2 2  1 ∂f + f cos(n, x) + i cos(n, y) δS . = ∂z 2 This completes the proof. Definition 3.5 Let S be a piecewise smooth two-sided surface with normal n, and ∂ ν a continuous function on S. Define the functional − (f δS ) on C0∞ (X) in the ∂n following way ∂ − (f δS )(φ) = ∂n

 f (x) S

∂φ(x) dS, ∂n

φ ∈ C0∞ (X).

130

3 Differentiation

∂ (f δS ) ∈ D  (X). ∂n  ∂  Exercise 3.3 Prove that supp − (f δS ) ⊂ S. ∂n Definition 3.6 Let S be a piecewise smooth two-sided surface with normal n, f is ∂ a continuous function on S. The distribution − (f δS ) is called a double layer on ∂n the surface S. Exercise 3.2 Prove that −

Theorem 3.16 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S, Ω = Rn \Ω1 , f ∈ C 1 (Ω)∩C 1 (Ω1 ),

By

 ∂f  ∂xi

[f ]S (x) =

lim

x  →x,x  ∈Ω1

f (x  )−

lim

x  →x,x  ∈Ω

f (x  ),

x ∈ S.

(x), i = 1, 2, . . . , n, we denote the classical derivatives of f at the point

∂ , i = 1, 2, . . . , n, the derivatives in D  (Rn ). If n is the ∂xi # ∂f $ ∂ ∂f  is the normal derivative, (x ) − outer normal to S, (x) = lim   ∂n ∂n S x →x,x ∈Ω1 ∂n $   # ∂f  ∂f ∂f (x ) and (x  ) −  lim lim (x) =  lim  x →x,x ∈Ω ∂n ∂xi S x →x,x ∈Ω1 ∂xi x →x,x ∈Ω  ∂f   (x ), x ∈ S, i = 1, 2, . . . , n. Then ∂xi

/ S, and by x ∈ Rn , x ∈

n 

∂ ∂ [f ]S δS . [f ]S cos(n, xi )δS = ∂xi ∂n i=1

Proof Let φ ∈ C0∞ (Rn ) be arbitrarily chosen. Then   n  n   

∂φ ∂ [f ]S cos(n, xi )δS , [f ]S cos(n, xi )δS , φ = − ∂xi ∂xi i=1

i=1

=−

n  

[f ]S cos(n, xi )

i=1 S

=−

n   i=1

=−

S

[f ]S

∂φ dS ∂ni

 n   ∂φ [f ]S δS , ∂ni i=1

∂φ dS ∂xi

3.5 Advanced Practical Problems

131

 n  

∂ [f ]S δS , φ ∂ni i=1  n   ∂

= [f ]S δS , φ ∂ni i=1  

∂ = [f ]S δS , φ . ∂n

=

This completes the proof.

3.5 Advanced Practical Problems Problem 3.1 Compute d3 |x|, dx 3

x ∈ R.

Answer 2δ  (x). Problem 3.2 Compute x m δ (k) (x). Answer ⎧ 0 for k < m; ⎪ ⎪ ⎨ (−1)m m!δ for k = m;   x m δ (k) (x) = ⎪ k ⎪ ⎩ (−1)m m!δ (k−m) for k > m. m Problem 3.3 Prove that δ(x) + δ  (x − 1) + δ  (x − 2) + · · · converges in D  (X) and that it has a finite order.

132

3 Differentiation

Solution Let φ ∈ C0∞ (X). Then ∞ 

(j )

δj (φ) =

j =0

∞  (−1)j φ (j ) (j ), j =0

∞ ∞        (j )   (j ) δj (φ) ≤  φ (j ) < ∞. j =0

j =0

Since φ has compact support, only a finite number of terms in nonzero. Therefore δ(x) + δ  (x − 1) + · · · has a finite order.

∞     (j )  φ (j ) are j =0

Problem 3.4 Prove ∞ ∞  2 1. cos(2k + 1)x = (−1)k δ(x − kπ), π k=0

2. | sin x| + | sin x| = 2

k=−∞ ∞ 

δ(x − kπ).

k=−∞

Problem 3.5 Prove 1 d log |x| = P , 1. dx x d 1 1 2. P = −P 2 , dx x x d 1 1 3. = iπδ  (x) − P 2 , dx x − i0 x 1 1 d  = −iπδ (x) − P 2 . 4. dx x + i0 x Hint 3. Use 1 1 = P + iπδ. x − i.0 x 4. Use 1 1 = P − iπδ. x + i.0 x Problem 3.6 Compute the first and the second derivative of the following functions in D  (R)  sin x x ≥ 0, 1. u(x) = cos x − 1 x ≤ 0,

3.5 Advanced Practical Problems

133

⎧ ⎨ x − 1 x ≥ 0, 2. u(x) = −1 − 1 ≤ x ≤ 0, ⎩ 2 −x x ≤ −1,  4 x − 1 ≤ x ≤ 1, 3. u(x) = 0 |x| > 1,  2 x + x + 1 − 1 ≤ x ≤ 1, 4. u(x) = 0 |x| > 1. 1. Solution Let φ ∈ C0∞ (R). Then u (φ) = −u(φ  ) ∞ =−

u(x)φ  (x)dx

−∞

∞

0



sin xφ (x)dx −

=−

(cos x − 1)φ  (x)dx

−∞

0

x=∞ ∞ x=0   = − sin xφ(x) + cos xφ(x)dx − (cos x − 1)φ(x) x=0

x=−∞

0

0 −

sin xφ(x)dx

−∞

= H (x)(cos xφ) − H (−x)(sin xφ) = cos xH (x)(φ) − sin xH (−x)(φ).

Since φ ∈ C0∞ (R) is arbitrary, we conclude that u = cos xH (x) − H (−x) sin x. As H  (x) = δ(x), H  (−x) = δ(−x), cos xδ(x) = δ(x), sin xδ(x) = 0, the second derivative reads u = − sin xH (x) + cos xH  (x) − cos xH (−x) − sin xH  (−x) = − sin xH (x) + cos xδ(x) − cos xH (−x) + sin xδ(x) = − sin xH (x) − cos xH (−x) + δ(x). 2. Answer u = −2xH (−x − 1) + H (x), u = −2H (−x − 1) + 2δ(−x − 1) + δ(x),

134

3 Differentiation

3. Answer u = −δ(x − 1) + δ(x + 1) + 4x 3H (x + 1) − 4x 3 H (x − 1), u = −δ  (x − 1) + δ  (x + 1) + 12x 2H (x + 1) −12x 2H (x − 1) − 4δ(x + 1) − 4δ(x − 1), 4. Answer u = −3δ(x − 1) + δ(x + 1) + (2x + 1)H (x + 1) − (2x + 1)H (x − 1), u = −3δ  (x − 1) + δ  (x + 1) + 2H (x + 1) − 2H (x − 1) − δ(x + 1) − 3δ(x − 1).

Problem 3.7 Compute d3 H (x + 1), dx 3 d (xsignx), 2. dx   d 3. (sin x + cos x)H (x + 2) dx 1.

in the space D  (R). Answer 1. δ  (x + 1), 2. H (x) − H (−x), 3. (cos 2 − sin 2)δ(x + 2) + (cos x − sin x)H (x + 2). Problem 3.8 Compute the first, second and third derivatives of the function u(x) = |x| sin(2x) in the space D  (R). Answer 1. u = −(sin(2x) + 2x cos(2x))H (−x) + (sin(2x) + 2x cos(2x))H (x), 2. u = −4(cos(2x) − x sin(2x))H (−x) + 4(cos(2x) − x sin(2x))H (x), 3. u = 4(3 sin(2x)+2x cos(2x))H (−x)−4(3 sin(2x)+2x cos(2x))H (x)+8δ(x). Problem 3.9 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S, Ω = Rn \Ω1 , f ∈ C 2 (Ω)∩C 2 (Ω1 ),

[f ]S (x) =

lim

x  →x,x  ∈Ω1

f (x  )−

lim

x  →x,x  ∈Ω

f (x  ),

x ∈ S.

 ∂ 2f   ∂f  (x), (x), i, j = 1, 2, . . . , n, we denote the classical derivatives ∂xi ∂xi ∂xj  ∂f  # ∂f $ (x  )− of f at the point x ∈ Rn , x ∈ / S, (x) = lim lim ∂xi S x  →x,x  ∈Ω1 ∂xi x  →x,x  ∈Ω By

3.5 Advanced Practical Problems

135

 ∂f  ∂2 ∂ (x  ), x ∈ S, i = 1, 2, . . . , n, and by , , i, j = 1, 2, . . . , n, the ∂xi ∂xi ∂xj ∂xi corresponding derivatives in D  (Rn ). Prove that for every i, j ∈ {1, 2, . . . , n},  ∂ 2f   # ∂f $ ∂ 2f ∂  1. = + [f ]S cos(n, xj )δS + cos(n, xi )δS , ∂xi ∂xj ∂xi ∂xj ∂xi ∂xj S n n #      ∂  ∂f $ [f ]S cos(n, xi )δS + cos(n, xi )δS . 2. Δf = Δf + ∂xi ∂xi S i=1

i=1

Here n is the outer normal to S. Problem 3.10 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S, Ω = Rn \Ω1 , f ∈ C 1 (Ω)∩C 1 (Ω1 ),

[f ]S (x) =

lim

x  →x,x  ∈Ω1

f (x  )−

lim

x  →x,x  ∈Ω

f (x  ),

x ∈ S.

 ∂f  (x), i = 1, 2, . . . , n, we denote the classical derivatives of f at the point ∂xi ∂ x ∈ Rn , x ∈ / S, and by , i = 1, 2, . . . , n, the derivatives in D  (Rn ). If n is the ∂xi # ∂f $ ∂ ∂f  is the normal derivative, (x ) − outer normal to S, (x) =  lim ∂n ∂n S x →x,x ∈Ω1 ∂n # $   ∂f  ∂f ∂f lim (x ) and (x) = lim (x  ) − lim     ∂xi S x →x,x ∈Ω ∂n x →x,x ∈Ω1 ∂xi x  →x,x  ∈Ω  ∂f  (x  ), x ∈ S, i = 1, 2, . . . , n. Prove ∂xi By

n #  ∂f $ i=1

∂xi

S

cos(n, xi )δS =

# ∂f $ ∂n

S

δS .

Problem 3.11 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S, Ω = Rn \Ω1 , f ∈ C 2 (Ω)∩C 2 (Ω1 ),  ∂f 

[f ]S (x) =

lim

x  →x,x  ∈Ω1

f (x  )−

lim

x  →x,x  ∈Ω

f (x  ),

x ∈ S.

 ∂ 2f  (x), i, j = 1, 2, . . . , n, we will denote the classical ∂xi ∂xi ∂xj ∂2 ∂ / S, and by , , i, j = 1, 2, . . . , n, derivatives of f at the point x ∈ Rn , x ∈ ∂xi ∂xj ∂xi ∂ is the the corresponding derivatives in D  (Rn ). If n is the outer normal to S, ∂n # ∂f $ ∂f  ∂f  (x ) − (x ) for x ∈ S. normal derivative, (x) = lim lim ∂n S x  →x,x  ∈Ω1 ∂n x  →x,x  ∈Ω ∂n By

(x),

136

3 Differentiation

Prove    # ∂f $ ∂  [f ]S δS , Δf = Δf + δS + ∂n S ∂n

f ∈ D  (Rn ).

Problem 3.12 Let Ω ⊂ Rn be a bounded domain with boundary ∂Ω = S, let also, Ω = Rn \Ω1 , f ∈ C 2 (Ω)∩C 2 (Ω1 ),  ∂f 

[f ]S (x) =

lim

x  →x,x  ∈Ω1

f (x  )−

lim

x  →x,x  ∈Ω

f (x  ),

x ∈ S.

 ∂ 2f  (x), i, j = 1, 2, . . . , n, we will denote the classical ∂xi ∂xi ∂xj ∂2 ∂ / S, and by , , i, j = 1, 2, . . . , n, derivatives of f at the point x ∈ Rn , x ∈ ∂xi ∂xj ∂xi ∂ indicates the corresponding derivatives in D  (Rn ). If n is the outer normal to S, ∂n # ∂f $ ∂f  ∂f  the normal derivative, (x ) − (x ) for (x) = lim lim ∂n S x  →x,x  ∈Ω1 ∂n x  →x,x  ∈Ω ∂n x ∈ S. Assume f = 0 on Ω1 . Prove By

(x),

  ∂f ∂ δS − (f δS ). Δf = Δf − ∂n ∂n Problem 3.13 Let u ∈ D  (R) and u = 0 in the sense of the distributions. Prove that u is a constant. Solution Let φ ∈ C0∞ (R). Then u (φ) = 0, whereupon u(φ  ) = 0 for every φ ∈ C0∞ (R). Take ψ ∈ C0∞ (R). There exists a ψ1 ∈ C0∞ (R) such that ∞ ψ(x) = ψ0 (x)

ψ(s)ds + ψ1 (x),

x ∈ R,

−∞

for ψ0 ∈

C0∞ (R)

∞ such that

ψ0 (x)dx = 1. Then u(ψ1 ) = 0 and

−∞

  ∞ ψ(x)dx + ψ1 u(ψ) = u ψ0 −∞

∞ = −∞

ψ(x)dxu(ψ0 ) + u(ψ1 )

3.5 Advanced Practical Problems

137

∞ =

ψ(x)dxu(ψ0 ) −∞

∞ =C

ψ(x)dx

−∞

= C1(ψ) = C(ψ), where C = u(ψ0 ). Consequently u = C. If C ∈ C is arbitrary, then u = C solves u = 0. Problem 3.14 Take u ∈ D  (X), where X ⊂ R is an open set. Consider u + au = f for given f ∈ C (X), a ∈ C ∞ (X). Prove that u ∈ C 1 (X). Solution (1st case)

Let a ≡ 0, so u = f.

Since f ∈ C (X), f has a primitive v ∈ C 1 (X), so v  = f and (u − v) = u − v  = f − f = 0. Using the previous problem, we conclude that u − v = C = const, and u = v + C ∈ C 1 (X). (2nd case)

Suppose a is not identically zero and define E(x) = e

%

a(x)dx

.

Since a ∈ C ∞ (X), then E ∈ C ∞ (X). The product Eu is well defined and the chain rule says (Eu) = E  u + Eu = E(−au + f ) + Eau = Ef ∈ C (X), because f ∈ C (X). Therefore Eu ∈ C 1 (X). Since E ∈ C ∞ (X), we obtain that u ∈ C 1 (X).

138

3 Differentiation

Problem 3.15 Let X ⊂ R be an open set, and suppose u = (u1 , u2 , . . . , un ) ∈ D  (X)×D  (X)×. . .×D  (X), f = (f1 , f2 , . . . , fn ) ∈ C (X)×C (X)×. . .×C (X), a = {aij }ni,j =1 , aij ∈ C ∞ (X), satisfy u + au = f. Prove that u ∈ C 1 (X) × C 1 (X) × . . . × C 1 (X). Hint Use the previous problem. Problem 3.16 Let X ⊂ R be an open set where u ∈ D  (X), ai ∈ C ∞ (X), i = 0, 1, . . . , m − 1, f ∈ C (X) satisfy u(m) + am−1 u(m−1) + · · · + a1 u + a0 u = f. Prove that u ∈ C m (X). Solution Setting uj = u(j −1) ,

j = 1, 2, . . . , m,

we have uj = u(j ) = uj +1

for j = 1, 2, . . . , m − 1,

and u(m) + am−1 um + · · · + a1 u2 + a0 u1 = f. The previous problem tells us that uj ∈ C 1 (X) for j = 1, 2, . . . , m. Using u(m) = −am−1 um − · · · − a1 u2 − a0 u1 + f, we conclude that u(m) ∈ C (X), so u ∈ C (m) (X). Problem 3.17 Solve the equation u = 0 in D  (X). Solution Set u = v, so that v  = 0 and therefore v = C0 , C0 = const. Hence, u = C0 . The solution of the homogeneous equation u = 0 is u = C1 , C1 = const. A particular solution of u = C0 is u = C0 x. Therefore the general solution reads u = C0 + C1 x.

3.5 Advanced Practical Problems

139

Problem 3.18 Solve u(m) = 0,

m ≥ 3,

in D  (R). Answer u = Cm−1 x m−1 + · · ·+ C1 x + C0, where Ci = const, i = 0, 1, . . . , m − 1. Problem 3.19 Solve the following equations 1. xu = 1,

1 2. xu = P , x 3. x 2 u = 0, 4. xu = signx, 5. (sin x)u = 0, 6. (cos x)u = 0, 7. x n u(m) = 0, n > m, 8. u = δ(x), 9. x 2 u = 1. Answer 1. c1 + c2 H (x) + log |x|, ci = const, i = 1, 2, 1 2. c1 + c2 H (x) − P , ci = const, i = 1, 2, x 3. c1 + c2 H (x) + c3 δ(x), ci = const, i = 1, 2, 3, 1 , c = const, 4. cδ(x) + P |x|  5. ck δ(x − kπ), ck = const, k

6. 7.

 k m−1  k=0

 (2k + 1)π  , ck = const, ck δ x − 2 ck H (x)x m−1−k +

n−1  k=m

ck δ (k−m) (x) +

8. xH (x) + c1 x + c2 , ci = const, i = 1, 2, 1 9. P 2 + c1 δ(x), c1 = const. x

m−1 

dk x k , ck , dk = const,

k=0

Problem 3.20 Let u ∈ D  (R), u(x) = 0 when x < x0 for some given x0 in R. Prove that there exists a unique primitive U −1 of u for which U −1 = 0 when x < x0 .

140

3 Differentiation

  Problem 3.21 Let {fn }∞ n=1 ⊂ D (R) converges to f ∈ D (R). Prove

b

b fn (x + t)dt →n→∞

a

f (x + t)dt a

in D  , where a < b are arbitrary fixed constants. Problem 3.22 Let

∞ 

gn (x) be convergent. Prove

n=1

b  ∞

gn (x + t)dt =

a n=1

∞  

b

gn (x + t)dt,

n=1 a

where a < b are arbitrary fixed numbers. Problem 3.23 Prove that the functions D α δ(x), |α| = m, m = 0, 1, . . ., are linearly independent. Problem 3.24 Let Y be an open set in Rn−1 , I an open interval of R, and take ∂ u ∈ D  (Y × I ) with u = 0. Prove that ∂xn  u(φ) = u0 (φ(x1 , x2 , . . . , xn ))dxn , φ ∈ C0∞ (Y × I ), u0 ∈ D  (Y ). I

Solution Choose ψ0 ∈ C0∞ (I ) so that



ψ0 (x)dx = 1. For a given g ∈ C0∞ (Y )

I

we write g0 (x) = g(x1 , , . . . , xn−1 )ψ0 (xn ), u0 (g) = u(g0 ).

x = (x1 , x2 , . . . , xn ),

We have that u0 ∈ D  (Y ). Let φ ∈ C0∞ (Y × I ) and  I φ(x1 , . . . , xn−1 ) =

φ(x1 , . . . , xn−1 , xn )dxn . I

The function φ − (I φ)ψ0 has a primitive Φ with respect to the variable xn , i.e., φ(x) − I φ(x1 , . . . , xn−1 )ψ0 (xn ) =

∂ Φ(x1 , . . . , xn ). ∂xn

3.6 Notes and References

141

Therefore φ(x) − I φ0 (x) = so u(φ − I φ0 ) = u u0 (φ) = 0, i.e.,

∂ Φ(x), ∂xn

 ∂  ∂ Φ , whence u(φ) − u(I φ0 ) = − u(Φ) and then u(φ) − ∂xn ∂xn

⎛ ⎞   u(φ) = u0 ⎝ φ(x1 , x2 , . . . , xn−1 , xn )dxn ⎠ = u0 (φ(x1 , x2 , . . . , xn−1 , xn ))dxn . I

I

Problem 3.25 Let X ⊂ Rn be an open set, u, f ∈ C (X) with some j = 1, 2, . . . , n, in D  (X). Prove that

∂ u = f , for ∂xj

∂ u exists at every point x ∈ X and ∂xj

∂u = f , j = 1, 2, . . . , n. ∂xj Solution We will prove the assertion for j = n. Since f ∈ C (X), it has a primitive v with respect to the variable xn and ∂ v = f. ∂xn We consider u − v. Then ∂ ∂ ∂ (u − v) = u− v = f − f = 0. ∂xn ∂xn ∂xn Let X = Y × I , where Y is open in Rn−1 and I is a real open interval. Then  (u − v)(φ) =

u0 (φ(x1 , x2 , . . . , xn−1 , xn ))dxn ,

u0 ∈ D  (Y ).

I

As v and u0 are piecewise-differentiable in xn and is piecewise-differentiable in xn and

∂ u = f. ∂xn

∂ v = f , we conclude that u ∂xn

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3 Differentiation

3.6 Notes and References In this chapter we introduce derivatives of distributions and we deduct some of their properties. It is investigated the local structure of distributions. In the chapter is defined antiderivative of a distribution. They are defined simple and double layers on surfaces and they are given some of their applications. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 4

Homogeneous Distributions

4.1 Definition Definition 4.1 A distribution u ∈ D  (Rn \{0}) is said homogeneous of degree a if   u(φ(x)) = t a u t n φ(tx) , x ∈ Rn \{0}, t > 0, for every φ ∈ C0∞ (Rn \{0}). We introduce the notation φt (x) = t n φ(tx) for x ∈ Rn \{0}, φ ∈ C0∞ (Rn \{0}). Take φ ∈ C0∞ (R\{0}) and a ∈ C, Rea > −1. Define the function  a x x > 0, a x+ = 0 x ≤ 0, and the functional ∞ Ia (φ) =

a x+ (φ)

=

x a φ(x)dx. 0

a ∈ D  (R\{0}) for a ∈ C, Rea > −1, and We have x+

∞ a (φ) x+

=

∞ x φ(x)dx = a

0

t a y a tφ(ty)dy 0

∞ a a y a tφ(ty)dy = t a x+ (tφ(tx)) = t a x+ (φt )

= ta 0

a is a homogeneous distribution of degree a. for t > 0. Consequently x+

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_4

143

144

4 Homogeneous Distributions

a Exercise 4.1 Let a ∈ C with Rea > −1. Prove that the function x+ is a locally integrable function.

Exercise 4.2 Take a ∈ C with Re(a) > −1, and distinct points x1 , x2 ∈ R\{0}. Prove that x1 a+ = x2 a+ .

4.2 Properties In this section, we will deduct some of the properties of the homogeneous distributions. Theorem 4.1 Let Re(a) > 0. Then for every φ ∈ C0∞ (R\{0}) Ia (φ  ) = −aIa−1 (φ). Proof Let φ ∈ C0∞ (R\{0}). Then 

∞

Ia (φ ) =

∞ x dφ(x) = −a

x a−1 φ(x)dx = −aIa−1(φ).

a

0

0

This completes the proof. Theorem 4.2 Let Re(a) > −1 and k ∈ N. Then for every φ ∈ C0∞ (R\{0})   Ia (φ) = (−1)k Ia+k φ (k)

1 . (a + 1) . . . (a + k)

Proof Let φ ∈ C0∞ (R\{0}). From the previous theorem, we get Ia (φ) = −

Ia+1 (φ  ) a+1

and Ia+1 (φ  ) = −

Ia+2 (φ  ) . a+2

Therefore Ia (φ) = (−1)2

Ia+2 (φ  ) . (a + 1)(a + 2)

(4.1)

4.2 Properties

145

Using induction, we obtain Ia (φ) = (−1)k

Ia+k (φ (k) ) . (a + 1)(a + 2) . . . (a + k)

This completes the proof. By (4.1), it follows that if a ∈ C, Re(a) > −1, then Ia ∈ D  (R\{0}). k

Theorem 4.3 Given k ∈ N, we have a lim (a + k)x+ = (−1)k−1

a→−k

δ (k−1) (x) , (k − 1)!

x ∈ R\{0}.

(4.2)

Proof Let φ ∈ C0∞ (R\{0}). Then, using (4.1), we have   1 lim (a + k)Ia (φ) = lim (−1)k Ia+k φ (k) a→−k a→−k (a + 1) . . . (a + k − 1)     I0 φ (k) 1 =− I0 φ (k) = (−1)k (1 − k)(1 − k + 1) . . . (−1) (k − 1)! ∞ x=∞ 1 1  (k−1) =− φ (k) (x)dx = − (x) φ  x=0 (k − 1)! (k − 1) ! 0  (−1)k−1  φ (k−1)(0) 1 = = δ φ (k−1) = δ (k−1) (φ). (k − 1)! (k − 1)! (k − 1)! Since φ ∈ C0∞ (R\{0}) was chosen arbitrarily, we conclude that a lim (a + k)x+ =

a→−k

(−1)k−1 (k−1) δ (x), (k − 1)!

x ∈ R\{0}.

This completes the proof. Theorem 4.4 Let k ∈ N. Then ∞ k−1  φ (k−1) (0)  1 (log x)φ (k)(x) φ (k−1) (0)  =− dx + , lim Ia (φ) − →0 (k − 1)! (k − 1)! (k − 1)! j 0

(4.3)

j =1

x ∈ R\{0}, for φ ∈ C0∞ (R\{0}). Here a + k = . Proof Let φ ∈ C0∞ (R\{0}). Using (4.1), we can represent Ia (φ) in the following form   I φ (k) Ia (φ) = (−1)k . ( + 1 − k) . . . 

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4 Homogeneous Distributions

Then    I φ (k) φ (k−1) (0)  φ (k−1) (0)  = lim (−1)k lim Ia (φ) − − →0 →0 (k − 1)! ( + 1 − k) . . .  (k − 1)! ∞ ∞   (k) (k) x φ (x) φ (x) = lim (−1)k dx − (−1)k dx →0 ( + 1 − k) . . .  ( + 1 − k) . . .  0 1  0 1 1 +φ (k−1) (0) − (k − 1 − ) . . . (1 − ) (k − 1)!  ∞ x  − 1 φ (k) (x) 1   1 1 dx + φ (k−1) (0) − = lim (−1)k →0 ( + 1 − k) . . .  (k − 1 − ) . . . (1 − ) (k − 1)!  

0

1 =− (k − 1)!

∞ log xφ (k) (x)dx + φ (k−1) (0)

k−1  1 1 . j (k − 1)! j =1

0

This completes the proof. Definition 4.2 Define −k (φ) x+

1 =− (k − 1)!

∞ log xφ (k)(x)dx + φ (k−1)(0)

k−1  1 1 , j (k − 1)!

(4.4)

j =1

0

for k ∈ N and φ ∈ C0∞ (R\{0}). −k , defined by (4.4), is an element of D  (R\{0}). Note that, for k ∈ N, x+ k

Theorem 4.5 Let k ∈ N. Then  d  δ (k) (x) a−1 a x+ . + kx+ = (−1)k a→−k dx k! lim

Proof We have, using (4.2),  d    a−1 a−1 a−1 a x+ + kx+ + kx+ = lim ax+ a→−k dx a→−k lim

a−1 = = lim (a + k)x+ a→−k

a lim (a + k + 1)x+ = (−1)k

a→−k−1

δ (k) (x) . k!

This completes the proof. Theorem 4.6 Let k ∈ N. Then −k −k (φ) = t −k x+ (φt ) + log t x+

φ (k−1)(0) (k − 1)!

for φ ∈ C0∞ (R\{0}),

t > 0.

4.2 Properties

147

Proof Let φ ∈ C0∞ (R\{0}). Using (4.4), for t > 0, we have −k t −k x+ (φt )

=− =− =−

t −k =− (k − 1)!

1 (k − 1)! 1 (k − 1)! 1 (k − 1)!

+ log t

−k = x+ (φ) +

=

−k x+ (φ) −

 (k) (k−1) log x φt (x) dx + φt (0) x

k−1

k−1  1 1 (k − 1)! j j =1

0

∞ (log(tx) − log t)φ (k) (tx)d(tx) + φ (k−1) (0)

k  1 1 (k − 1)! j j =1

0

∞ log(tx)φ (k) (tx)d(tx) + φ (k−1) (0) 0

t −k  1 (k − 1)! j j =1

0

∞ (log x)tφ (k) (tx)dx + φ (k−1) (0)

1 (k − 1)!

1 =− (k − 1)!

∞

1 (k − 1)!

∞

k  j =1

1 j

φ (k) (tx)d(tx) 0

∞

log yφ 0

log t (k − 1)!

 k−1  1 1 log t (0) φ (k) (y)dy + (k − 1)! j (k − 1)! ∞

(k)

(y)dy + φ

(k−1)

j =1

∞

0

φ (k) (y)dy 0

log t φ (k−1) (0). (k − 1)!

This completes the proof. Theorem 4.7 Fix a ∈ / Z− and take the smallest k ∈ N so that k + Re(a) > −1. We define, for  > 0 and φ ∈ C0∞ (R\{0}), ∞ Ha, (φ) =

x a φ(x)dx. 

Then there exist unique constants C0 , Bj , j = 0, 1, . . . , k − 1, such that Ha, (φ) = C0 +

k−1  j =0

where λj = −(a + j + 1).

Bj  −λj + o(1) when  → 0,

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4 Homogeneous Distributions

Proof Let φ ∈ C0∞ (R\{0}). We integrate Ha, (φ) by parts k times, to the effect that ∞ Ha, (φ) =

x a φ(x)dx = 

1 a+1

∞ φ(x)dx a+1 

∞ x=∞ 1 1  x a+1 φ(x) = − x a+1 φ  (x)dx x= a+1 a+1 

1 1 =−  a+1 φ() − a+1 (a + 1)(a + 2)

∞

φ  (x)dx a+2



∞ x=∞ 1 1 1  a+1 a+2   φ() − x =− φ (x) + x a+2 φ  (x)dx x= a+1 (a + 1)(a + 2) (a + 1)(a + 2) 

1 1 1 =−  a+1 φ() +  a+2 φ  () + a+1 (a + 1)(a + 2) (a + 1)(a + 2)(a + 3) .. . =

(−1)k (a + 1)(a + 2) . . . (a + k) k−1 

∞

φ  (x)dx a+3



∞ x a+k φ (k) (x)dx 

(−1)j +1 φ (j ) ()  a+j +1 + (a + 1)(a + 2) . . . (a + j + 1) j =0 ∞ (−1)k = x a+k φ (k) (x)dx (a + 1)(a + 2) . . . (a + k) +

k−1  j =0

0

(−1)j +1 φ (j ) (0)  a+j +1 + o(1). (a + 1)(a + 2) . . . (a + j + 1)

Now, let (−1)k C0 = (a + 1)(a + 2) . . . (a + k)

∞ x a+k φ (k) (x)dx, 0

(−1)j +1 φ (j ) (0) Bj = , (a + 1)(a + 2) . . . (a + j + 1)

j = 0, 1, . . . , k − 1.

Using the above expression of Ha, (φ), we obtain Ha, (φ) = C0 +

k−1  j =0

Bj  −λj + o(1).

4.2 Properties

149

Suppose Ha, (φ) = D0 +

k−1 

Qj  −λj + o(1),

j =0

where D0 , Qj , j = 0, 1, 2, . . . , k − 1, are constants. Then k−1  C0 − D0 + (Bj − Qj ) −λj →→0 0. j =0

Since λi = λj for i = j , Reλj ≥ 0, i, j = 0, 1, . . . , k − 1, the above limit exists if and only if C0 − D0 = 0, Bj − Qj = 0, j = 0, 1, 2, . . . , k − 1. This completes the proof. Theorem 4.8 We have a e∓iπa x+ +

δ (k−1) (x) δ (k−1) −k ∓ iπ →a→−k (−1)k x+ . (k + a)(k − 1)! (k − 1)!

−k Proof By (4.2) and the definition of x+ , we have

 (−1)k−1 δ (k−1) (x)  −k a x+ = x+ − . a→−k (k − 1)!(a + k) lim

(4.5)

Using (4.5) and e∓πa = (−1)k (1 ∓ πi(a + k) + O(a + k)2 )

when a → −k,

we have  (−1)k−1 δ (k−1) (x)  −k a = (−1)k x+ lim e∓iπa x+ − . a→−k (k − 1)!(a + k) Now, using (4.6), we get (−1)k−1 δ (k−1) (x) (k − 1)!(a + k) (−1)k−1 δ (k−1)(x) ∓iπa a x+ − (−1)k =e (k − 1)!(a + k) k−1 δ (k−1) (x) (−1)k−1 δ (k−1) (x) k (−1) iπ(a + k) − O(a + k)2 . ±(−1) (k − 1)!(a + k) (k − 1)!(a + k) a − e∓iπa e∓iπa x+

(4.6)

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4 Homogeneous Distributions

Therefore 

(−1)k−1 δ (k−1) (x)  a e∓iπa x+ − e∓iπa a→−k (k − 1)!(a + k)  (−1)k−1 δ (k−1)(x) ∓iπa a x+ − (−1)k = lim e a→−k (k − 1)!(a + k)  k−1 δ (k−1) (x) (−1)k−1 δ (k−1) (x) k (−1) iπ(a + k) − O(a + k)2 ±(−1) (k − 1)!(a + k) (k − 1)!(a + k)  k−1 δ (k−1) (x) (−1)k−1 δ (k−1)(x)  ∓iπa a k (−1) ± (−1)k iπ x+ − (−1) = lim e a→−k (k − 1)!(a + k) (k − 1)! −k = (−1)k x+ . lim

This completes the proof. Theorem 4.9 We have −k −k (x ± i.0)−k = x+ + (−1)k x− ± iπ(−1)k

δ (k−1)(x) . (k − 1)!

Proof Since a a (x ± i.0)a = x+ + e±iπa x− ,

we have a a + x− . e∓iπa (x ± i.0)a = e∓iπa x+

(4.7)

e∓iπa (x ± i.0)a →a→−k (−1)k (x ± i.0)−k

(4.8)

Moreover,

and −k a a e∓iπa x+ + x− →a→−k (−1)k x+ ± iπ

δ (k−1) (x) −k + x− . (k − 1)!

From (4.7), (4.8) and (4.9), we then get −k −k + (−1)k x+ ± iπ (−1)k (x ± i.0)−k = x−

δ (k−1) (x) . (k − 1)!

This completes the proof. Theorem 4.10 We have (x + i.0)−k − (x − i.0)−k = 2iπ(−1)k

δ (k−1)(x) . (k − 1)!

(4.9)

4.3 Advanced Practical Problems

151

Proof From the previous theorem, we have δ (k−1) (x) , (k − 1)! (k−1) δ (x) −k −k , = x+ + (−1)k x− − (−1)k iπ (k − 1)!

−k −k (x + i.0)−k = x+ + (−1)k x− + (−1)k iπ

(x − i.0)−k and immediately

(x + i.0)−k − (x − i.0)−k = 2iπ(−1)k

δ (k−1)(x) . (k − 1)!

This completes the proof.

4.3 Advanced Practical Problems Problem 4.1 Prove that the function a → Ia (φ), φ ∈ C0∞ (R\{0}), is analytic when Re(a) > −1. Problem 4.2 Let a ∈ N ∪ {0}. Prove that Ia can alternatively be defined as the analytic continuation with respect to a. Hint Use (4.1). Problem 4.3 Let k ∈ N. Prove d −k δ (k) (x) −k−1 x+ = −kx+ + (−1)k dx k!

on R\{0}.

Problem 4.4 Let k ∈ N, k ≥ 2. Prove that there exist unique constants Aj , j = 0, 1, . . . , k − 2, such that 1 H−k, (φ) = − (k − 1)! +

k−2 

∞

j =0

j =1

0

Aj φ (j ) (0) j +1−k −

1 1 (k − 1)! j k−1

log xφ (k) (x)dx + φ (k−1) (0)

log  (k−1) φ (0) + o(1) when  → 0, (k − 1)!

for every φ ∈ C0∞ (R\{0}). Problem 4.5 Let a ∈ C, Re(a) > −1 and define a x−

=

 0 |x|a

x > 0, x < 0.

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4 Homogeneous Distributions

Prove a a ˘ x− (φ) = x+ (φ),

˘ φ(x) = φ(−x),

for every φ ∈ C0∞ (R\{0}). Problem 4.6 Let a ∈ C and a ∈ / Z− ∪ {0}. Prove that a a (x ± i.0)a = lim (x ± i.)a = x+ + e±iπa x− , →0

x ∈ R.

Problem 4.7 Prove d (x ± i.0)a = a(x ± i.0)a−1 . dx Problem 4.8 Define x −k =

(x + i.0)−k + (x − i.0)−k . 2

Prove that −k −k 1. x −k = x+ + (−1)k x− ,   d −k −k−1 = −kx , 2. x dx −k 1−k 3. xx = x .

Problem 4.9 Show that x −1 =

d log |x|. dx

Problem 4.10 Define the function χ+a as follows χ+a =

a x+ Γ (a + 1)

for a ∈ C, Re(a) > −1. Prove 1. χ+a (φ  ) = −χ+a−1 (φ), 2. χ+−k = δ (k−1)(x).

φ ∈ C0∞ (R),

Problem 4.11 Let u be a homogeneous distribution of degree a on Rn \{0} and λ=

 j

xj ∂j .

4.3 Advanced Practical Problems

153

Prove au − λu = 0. Hint Differentiate with respect to t the equality u(φ(x)) = t a u(t n φ(tx)) for φ ∈ C0∞ (Rn \{0}). Problem 4.12 Let ψ ∈ C ∞ (Rn \{0}) be a homogeneous function of degree b and u ∈ D  (Rn \{0}) a homogeneous distribution of degree a. Prove that ψu is a homogeneous distribution of degree a + b in Rn \{0}. Problem 4.13 Let u be a homogeneous distribution of degree a on Rn \{0}, ψ ∈ C0∞ (Rn \{0}) and ∞ r a+n−1 ψ(rx)dr = 0,

x = rw ∈ Rn \{0}.

0

Prove u(ψ) = 0. Hint Use au = λu. Deduce u((a + n)φ(x) + λφ(x)) = 0 for every φ ∈ C0∞ (Rn \{0}). Then rewrite the last equality in polar coordinates and multiply by r a+n−1 . ∂u , Problem 4.14 Let u ∈ D  (Rn \{0}) be homogeneous of degree a. Prove that ∂xj j = 1, 2, . . . , n, are homogeneous distributions of degree a − 1 on Rn \{0}. Problem 4.15 Let u ∈ D  (Rn \{0}) be homogeneous of degree a and α ∈ Nn . Prove that D α u is homogeneous of degree a − |α| on Rn \{0}. Problem 4.16 Let uj ∈ D  (Rn \{0}), j = 1, 2, be homogeneous of degree aj . Find conditions on a1 , a2 so that the combination α1 u1 + α2 u2 becomes homogeneous on Rn \{0}, for any α1 , α2 ∈ C. Answer a1 = a2 . Problem 4.17 Let u ∈ D  (Rn \{0}) be homogeneous of degree a. Prove that xj u is a homogeneous distribution of degree a + 1 on Rn \{0}), for any j = 1, 2, . . . , n.

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4 Homogeneous Distributions

4.4 Notes and References In this chapter we define homogeneous distributions and we deduct some of their properties. In the chapter are given some applications of homogeneous distributions. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 5

The Direct Product of Distributions

5.1 Definition Definition 5.1 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets. The direct product of the distributions u1 ∈ D  (X1 ), u2 ∈ D  (X2 ) is defined through u1 (x) × u2 (y)(φ) = u1 (x)(u2 (y)(φ(x, y))), u2 (y) × u1 (x)(φ) = u2 (y)(u1 (x)(φ(x, y))),

φ ∈ C0∞ (X1 × X2 ).

Take X ⊂⊂ X1 ×X2 and φ ∈ C0∞ (X ). Since suppφ ⊂ X ⊂⊂ X1 ×X2 is compact, there exist open sets X1 ⊂⊂ X1 , X2 ⊂⊂ X2 such that X ⊂⊂ X1 × X2 . Let x ∈ X1 \X1 . Then φ(x, y) = 0 for every y ∈ X2 . Hence, ψ(x) = u2 (y)(φ(x, y)) = 0, i.e., ψ ≡ 0 on X1 \X1 . We may choose an open set X˜1 so that X1 ⊂⊂ X˜1 ⊂⊂ X1 , and consequently suppψ ⊂ X˜1 . Take x ∈ X1 and let {xk }∞ k=1 be a sequence in X1 tending to x, as k → ∞. Then φ(xk , y) →k→∞ φ(x, y) in C0∞ (X2 ) for every φ ∈ C0∞ (X1 × X2 ). In fact, y∈X2

suppφ(xk , y) ⊂ X2 ⊂⊂ X2 and Dyα φ(xk , y) → Dyα φ(x, y), k → ∞, for every multi-index α ∈ Nm . Because u2 ∈ D  (X2 ), we have ψ(xk ) = u2 (y)(φ(xk , y)) →k→∞ u2 (y)(φ(x, y)) = ψ(x), i.e., ψ is continuous at x. Since x ∈ X1 was completely arbitrary, we conclude that ψ ∈ C (X1 ). Let now e1 = (1, 0, . . . , 0) and consider the function χh (y) =

1 (φ(x + he1 , y) − φ(x, y)) h

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_5

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5 The Direct Product of Distributions

for x ∈ X1 . For it we have suppχh ⊂ X2 ⊂⊂ X2 and y∈X2

D α χh (y) → Dyα

∂φ(x, y) , ∂x1

h → 0,

for every α ∈ Nm . Because u2 ∈ D  (X2 ), we have  ψ(x + he1 ) − ψ(x) 1 = u2 (y)(φ(x + he1 , y)) − u2 (y)(φ(x, y)) h h  φ(x + he , y) − φ(x, y)  1 = u2 (y) h  ∂φ  (x, y) . = u2 (y)(χh (y)) →h→0 u2 (y) ∂x1 By induction, we conclude that   D α ψ(x) = u2 (y) Dxα φ(x, y) for every α ∈ Nn ∪ {0} and φ ∈ C0∞ (X1 × X2 ). Therefore ψ ∈ C0∞ (X˜1 ) for φ ∈ C0∞ (X ). Let φ ∈ C0∞ (X ) and x ∈ X1 . Then Dxα φ(x, y) ∈ C0∞ (X2 ). Since u2 ∈ D  (X2 ), there exist constants C ≥ 0 and m ∈ N ∪ {0}, C = C(u2 ), m = m(u2 ), such that |D α ψ(x)||u2 (y)(Dxα φ(x, y))| ≤ C

max

y∈X2 ,|β|≤m

|Dyβ Dxα φ(x, y)|

for x ∈ X1 . Now, we consider the operation φ(x, y) → ψ(x) = u2 (y)(φ(x, y))

(5.1)

from C0∞ (X1 × X2 ) to C0∞ (X1 ). If φ1 , φ2 ∈ C0∞ (X1 × X2 ) and α1 , α2 ∈ C, then α1 φ1 + α2 φ2 → u2 (y)(α1 φ1 (x, y) + α2 φ2 (x, y)) = α1 u2 (y)(φ1 (x, y)) + α2 u2 (y)(φ2 (x, y)) = α1 ψ1 (x) + α2 ψ2 (x), i.e., the operation φ → u2 (y)(φ) from C0∞ (X1 × X2 ) to C0∞ (X1 ) is linear. ∞ Let now {φn }∞ n=1 be a sequence in C0 (X1 × X2 ) such that φn →n→∞ 0 in ∞ C0 (X1 × X2 ). Then there exists a compact set X3 ⊂ X1 × X2 such that suppφn ⊂ X3 for every n ∈ N and (x,y)

Dxα Dyβ φn (x, y) → n→∞ 0

5.2 Properties

157

for every α ∈ Nn , β ∈ Nm . From here, we conclude that there exists a compact set x X4 ⊂ X1 such that suppψn ⊂ X4 for every n ∈ N and D α ψn (x)→0, n → ∞, and ∞ ψn →n→∞ 0 in C0 (X1 ). Therefore (5.1) is a linear and continuous operation from C0∞ (X1 ×X2 ) to C0∞ (X1 ). It follows that u1 (x)(u2 (y)(·)) is a linear and continuous functional on C0∞ (X1 × X2 ), so u1 × u2 ∈ D  (X1 × X2 ). In a similar way it can be proved that u2 × u1 ∈ D  (X2 × X1 ). Example 5.1 Let us consider δ(x) × δ(y)(φ(x, y)) for φ ∈ C0∞ (R × R). We have δ(x) × δ(y)(φ(x, y)) = δ(x)(δ(y)(φ(x, y))) = δ(x)(φ(x, 0)) = φ(0, 0). Exercise 5.1 Compute H  (x) × δ(y)(φ(x, y)),

φ ∈ C0∞ (R × R).

Answer φ(0, 0). Exercise 5.2 Compute δ(x − 2) × H  (y)(φ(x, y)),

φ ∈ C0∞ (R × R).

Answer φ(2, 0).

5.2 Properties Theorem 5.1 We have u1 (x) × u2 (y) = u2 (y) × u1 (x),

u1 ∈ D  (X1 ), u2 ∈ D  (X2 ).

Proof To prove the property we take φ ∈ C0∞ (X1 × X2 ). So, there exist sequences ∞ ∞ {ψk }∞ k=1 in C0 (X1 × X2 ) and {Nk }k=1 in N ∪ {0} such that φk (x, y) =

Nk 

φik (x)ψik (y)

i=1

and φk →k→∞ φ in C0∞ (X1 × X2 ). From here, for (x, y) ∈ X1 × X2 , we get u1 (x) × u2 (y)(φ(x, y)) = u1 (x)(u2 (y)(φ(x, y))) = lim u1 (x)(u2 (y)(φk (x, y))) = lim u1 (x)(u2 (y)( k→∞

k→∞

Nk  i=1

φik (x)ψik (y)))

158

5 The Direct Product of Distributions

= lim u1 (x)( k→∞

= lim

k→∞

= lim

k→∞

Nk 

φik (x)u2 (y)(ψik (y))) = lim

Nk 

u2 (y)(ψik (y))u1 (x)(φik (x)) = lim

k→∞

i=1 Nk 

u2 (y)(u1 (x)(ψik (y)φik (x))) = lim

k→∞

i=1

= lim u2 (y)(u1 (x)( k→∞

k→∞

i=1

Nk 

Nk 

u1 (x)(φik (x))u2 (y)(ψik (y))

i=1 Nk 

u2 (y)(ψik (y)u1 (x)(φik (x)))

i=1 Nk 

u2 (y)(u1 (x)(φik (x)ψik (y)))

i=1

φik (x)ψik (y))) = lim u2 (y)(u1 (x)(φk (x, y))) k→∞

i=1

= u2 (y)(u1 (x)(φ(x, y))) = u2 (y) × u1 (x)(φ(x, y)),

(x, y) ∈ X1 × X2 .

Since φ ∈ C0∞ (X1 × X2 ) was arbitrary, u1 (x) × u2 (y) = u2 (y) × u1 (x). This completes the proof. Theorem 5.2 We have (u1 (x) × u2 (y)) × u3 (z) = u1 (x) × (u2 (y) × u3 (z)) for u1 ∈ D  (X1 ), u2 ∈ D  (X2 ), u3 ∈ D  (X3 ), (x, y, z) ∈ X1 × X2 × X3 , where X3 ⊂ Rk is an open set. Proof Let φ ∈ C0∞ (X1 × X2 × X3 ). Then (u1 (x) × u2 (y)) × u3 (z)(φ(x, y, z)) = (u1 (x) × u2 (y))(u3 (z)(φ(x, y, z))) = u1 (x)(u2 (y)(u3 (z)(φ(x, y, z)))) = u1 (x)((u2 (y) × u3 (z))(φ(x, y, z))) = u1 (x) × (u2 (y) × u3 (z))(φ(x, y, z)), (x, y, z) ∈ X1 × X2 × X3 . Since φ ∈ C0∞ (X1 × X2 × X3 ) was arbitrary, (u1 (x) × u2 (y)) × u3 (z) = u1 (x) × (u2 (y) × u3 (z)) for (x, y, z) ∈ X1 × X2 × X3 . This completes the proof. Exercise 5.3 Let u1 ∈ D  (X1 ), u2 ∈ D  (X2 ). Prove that the operator u1 (x) → u1 (x) × u2 (y),

(x, y) ∈ X1 × X2 ,

defined from D  (X1 ) to D  (X1 × X2 ) is linear and continuous. Definition 5.2 We will say that the distribution u(x, y) ∈ D  (X1 × X2 ) does not depend on the variable y if there exists a distribution u1 (x) ∈ D  (X1 ) such that u(x, y) = u1 (x) × 1(y).

5.2 Properties

159

If this is the case, u ∈ D  (X1 × Rm ) and for φ ∈ C0∞ (X1 × Rm ) u(x, y)(φ(x, y)) = u1 (x) × 1(y)(φ(x,  y))

= u1 (x)(1(y)(φ(x, y))) = u1 (x) φ(x, y)dy 

Rm

= 1(y) × u1 (x)(φ(x, y)) = 1(y)(u1 (x)(φ(x, y))) =

u1 (x)(φ(x, y))dy, Rm

i.e.,  u1 (x)

  φ(x, y)dy = u1 (x)(φ(x, y))dy.

Rm

Rm

Exercise 5.4 Let (a, b) ⊂ R, a < b, and take u(x, y) ∈ D  (X1 × (a, b)) not depending on y. Prove that u(x, y + h) = u(x, y),

∀x ∈ X1 , ∀y, y + h ∈ (a, b).

Solution There exists a distribution u1 (x) ∈ D  (X1 ) such that u(x, y) = u1 (x) × 1(y). Since 1(y) = 1(y + h) for every y, y + h ∈ (a, b), we have u(x, y) = u1 (x) × 1(y + h) = u(x, y + h). For u ∈ D  (X1 × X2 ) and φ ∈ C0∞ (X1 ), we define the distribution uφ on by

C0∞ (X2 )

uφ (ψ) = u(φ(x)ψ(y))

for ψ ∈ C0∞ (X2 ).

Definition 5.3 The distribution u ∈ D  (X1 × X2 ) is said to be an element of C p (X2 ), p = 0, 1, 2, . . ., if for every φ ∈ C0∞ (X1 ) we have uφ ∈ C p (X2 ).   Exercise 5.5 Prove D α uφ = Dyα u . φ

Solution Choose ψ ∈

C0∞ (X2 )

arbitrarily. Then

D α uφ (ψ) = (−1)|α| uφ(D α ψ(y)) = (−1)|α| u(Dyα (φ(x)ψ(y)))  = Dyα u(φ(x)ψ(y)) = Dyα u

φ

(ψ).

160

5 The Direct Product of Distributions

5.3 Advanced Practical Problems Problem 5.1 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets and take u1 ∈ D  (X1 ), u2 ∈ D  (X2 ). Prove that supp(u1 × u2 ) = suppu1 × suppu2 . Solution Let (x0 , y0 ) ∈ suppu1 × suppu2 be arbitrary, and suppose U is a neighbourhood of the point (x0 , y0 ) contained in X1 × X2 . Let U1 ⊂ X1 be a neighbourhood of x0 , U2 ⊂ X2 a neighbourhood of y0 . As (x0 , y0 ) ∈ suppu1 × suppu2 , there exist φ1 ∈ C0∞ (U1 ), φ2 ∈ C0∞ (U2 ) such that u1 (φ1 ) = 0, u2 (φ2 ) = 0. Therefore, u1 × u2 (φ1 φ2 )(x0 , y0 ) = u1 (φ1 )(x0 )u2 (φ2 )(y0 ) = 0 by the definition of the direct product. Consequently (x0 , y0 ) ∈ supp(u1 × u2 ), so we conclude suppu1 × suppu2 ⊂ supp(u1 × u2 ).

(5.2)

Let now φ ∈ C0∞ (X1 × X2 ) be chosen so that suppφ ⊂ (X1 × X2 )\(suppu1 × suppu2 ). Then there exists a neighbourhood U3 of suppu1 such that suppφ(x, y) ⊂ X2 \suppu2 for every x ∈ U3 . Consequently, ψ(x) = u2 (y)(φ(x, y)) = 0 for x ∈ U3 . As suppψ ∩ suppu1 = Ø, (X1 × X2 )\(suppu1 × suppu2 ) ⊂ (X1 × X2 )\(supp(u1 × u2 )), from which supp(u1 × u2 ) ⊂ suppu1 × suppu2 . From here and (5.2), we get supp(u1 × u2 ) = suppu1 × suppu2 . Problem 5.2 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets, u1 ∈ D  (X1 ), u2 ∈ D  (X2 ). Prove 1. Dxα1 u1 (x1 ) × Dxβ2 u2 (x2 ) = Dxα1 Dxβ2 (u1 (x1 ) × u2 (x2 )) for any α ∈ Nn ∪ {0}, β ∈ Nm ∪ {0}. 2. a(x1 )b(x2)(u1 (x1 ) × u2 (x2 )) = (a1 (x1 )u1 (x1 )) × (b(x2 )u2 (x2 )), where a ∈ C ∞ (X1 ), b ∈ C ∞ (X2 ).

5.3 Advanced Practical Problems

161

Problem 5.3 Let X1 ⊂ Rn be an open set and (a, b) ⊂ R, a < b. Take u ∈ D  (X1 ×(a, b)) satisfying u(x, y) = u(x, y +h) for every x ∈ X1 , y, y +h ∈ (a, b). Prove ∂u (x, y) = 0, ∂y

(x, y) ∈ X1 × (a, b).

Solution Since for every (x, y), (x, y + h) ∈ X1 × (a, b), h = 0, we have u(x, y + h) − u(x, y) = 0. h→0 h lim

Then ∂u =0 ∂y

on X1 × (a, b).

Problem 5.4 Let X1 ⊂ Rn be an open set, (a, b) ⊂ R, u ∈ D  (X1 × (a, b)) and ∂u = 0 on X1 × (a, b). Prove that u does not depend on y. ∂y Solution Let φ ∈ C0∞ (X1 × (a, b)). From here, ∂u (φ) = 0 ∂y i.e. u

 ∂φ  ∂y

=0

(5.3)

for every φ ∈ C0∞ (X1 × (a, b)). Let ψ ∈ C0∞ (X1 × (a, b)). Then there exists ψ1 ∈ C0∞ (X1 × (a, b)) such that ψ(x, y) =

∂ψ1 (x, y) + ω (y − y0 ) ∂y

b ψ(x, ξ )dξ. a

We define the distribution u1 ∈ D  (X1 ) by u1 (ψ2 ) = u(ω (y − y0 )ψ2 (x)) for ψ2 ∈ C0∞ (X1 ).

162

5 The Direct Product of Distributions

Using (5.3), b   ∂ψ (x, y) 1 + ω (y − y0 ) ψ(x, ξ )dξ u(ψ) = u ∂y a

b  ∂ψ (x, y)    1 =u + u ω (y − y0 ) ψ(x, ξ )dξ ∂y = u1

a

 b

 ψ(x, ξ )dξ ,

a

i.e., u(x, y) = u1 (x) × 1(y). Problem 5.5 Let X1 ⊂ Rn be an open set and F ∈ D  (X1 × R). Prove that the distribution u ∈ D  (X1 × R), defined by u(φ) = F (ψ) + f (x) × δ(y)(φ),

φ ∈ C0∞ (X1 × R),

satisfies the equation yu(x, y) = F (x, y). Here f ∈ D  (X1 ), ψ(x, y) =

1 (φ(x, y) − η(y)φ(x, 0)), y

and η ∈ C0∞ (R) equals 1 on a neighbourhood of y = 0. Problem 5.6 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets, u ∈ C (X2 ) in y. Prove that for every y ∈ X2 there exists uy (x) ∈ D  (X1 ) such that uφ (y) = uy (φ),

φ ∈ C0∞ (X1 ).

Problem 5.7 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets, u ∈ D  (X1 × X2 ), u ∈ C (X2 ) in y. Prove that for every φ ∈ C0∞ (X1 ), every y ∈ X2 and every α ∈ Nm   D α (uy (φ)) = Dyα u (φ). y

5.3 Advanced Practical Problems

163

Problem 5.8 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets, u ∈ D  (X1 × X2 ), u ∈ C (X2 ) in y. Prove that the operation ψ → uy (ψ(x, y)) is linear and continuous from C0∞ (X1 × X2 ) to C0 (X2 ). Problem 5.9 Let X1 ⊂ Rn , X2 ⊂ Rm be open sets, u ∈ D  (X1 × X2 ), u ∈ C (X2 ). Prove that  u(ψ) = uy (ψ(x, y))dy X2

for every ψ ∈ C0∞ (X1 × X2 ). Problem 5.10 Compute 1. 2. 3. 4. 5.

∂ n H (x) , where H (x) = H (x1 ) · · · H (xn ), ∂x1 . . . ∂xn δ(x1 ) × · · · × δ(xn ), ∂2 H (x, t), where H (x, t) = H (x)H (t), ∂t 2 ν(x) × δ(t), where ν ∈ C (Rnx ), −ν(x) × δ  (t), where ν ∈ C (Rnx ).

Answer 1. (−1)n δx1 × δx2 × · · · × δxn , 2. δ(x), 3. H (x) × δ  (t), 4. ν(x)δt , d 5. −ν(x) δt . dt Problem 5.11 Let X1 ⊂ Rn , X2 ⊂ Rm , X3 ⊂ Rk be open sets, u1 ∈ D  (X1 ), u2 ∈ D  (X2 ), u3 ∈ D  (X3 ). Prove u1 × (u2 + u3 ) = u1 × u2 + u1 × u3 . Problem 5.12 Let u ∈ D  (Rn ) and α ∈ Nn . Determine Dyα (u(x) × 1(y)). Answer 0.

164

5 The Direct Product of Distributions

5.4 Notes and References In this chapter we define the direct product of two distributions and we prove that it is well defined. In the chapter are proved the commutativity and associativity of the direct product of distributions. They are considered some applications of the direct product of distributions. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 6

Convolutions

6.1 Definition ∞ 2n Consider u1 , u2 ∈ D  (Rn ) and a sequence {ηk (x, y)}∞ k=1 in C0 (R ) converging to 1 in R2n . Suppose that

lim u1 (x) × u2 (y)(ηk (x, y)φ(x + y)) = u1 (x) × u2 (y)(φ(x + y))

k→∞

exists for every φ ∈ C0∞ (Rn ) and does not depend on the choice of the sequence {nk (x, y)}∞ k=1 . Definition 6.1 The convolution of the distributions u1 and u2 is defined by u1 ∗ u2 (φ) = u1 (x) × u2 (y)(φ(x + y)) = lim u1 (x) × u2 (y)(ηk (x, y)φ(x + y)) k→∞

for any φ ∈ C0∞ (Rn ). For φ1 , φ2 ∈ C0∞ (Rn ) and α1 , α2 ∈ C, we have u1 ∗ u2 (α1 φ1 + α2 φ2 ) = lim u1 (x) × u2 (y)(ηk (x, y)(α1 φ1 (x + y) + α2 φ2 (x + y))) = lim

k→∞



k→∞

 α1 u1 (x) × u2 (y)(ηk (x, y)φ1 (x + y)) + α2 u1 (x) × u2 (y)(ηk (x, y)φ2 (x + y))

= α1 lim u1 (x) × u2 (y)(ηk (x, y)φ1 (x + y)) + α2 lim u1 (x) × u2 (y)(ηk (x, y)φ2 (x + y)) k→∞

k→∞

= α1 u1 ∗ u2 (φ1 ) + α2 u1 ∗ u2 (φ2 ),

proving the convolution ∗ is a linear functional on C0∞ (Rn ).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_6

165

166

6 Convolutions

∞ n ∞ n Let now {φn }∞ n=1 be a sequence in C0 (R ) which tends to 0 in C0 (R ). Then

lim u1 ∗ u2 (φn ) = lim lim u1 (x) × u2 (y)(ηk (x, y)φn (x + y))

n→0

n→∞ k→∞

= lim lim u1 (x) × u2 (y)(ηk (x, y)φn (x + y)) k→∞ n→∞

= lim u1 (x) × u2 (y)(0) = 0. k→∞

Consequently u1 ∗ u2 ∈ D  (Rn ). Exercise 6.1 Let u1 , u2 ∈ D  (Rn ) and assume u1 ∗ u2 exists. Prove that u1 → u1 ∗ u2

(6.1)

is a linear map from D  (Rn ) to itself. Example 6.1 The operation defined in (6.1) is not continuous on D  (Rn ), because δ(x − k) →k→∞ 0 but 1 ∗ δ(x − k) = 1. Exercise 6.2 Let u1 , u2 ∈ D  (Rn ) and assume u1 ∗ u2 exists. Prove supp(u1 ∗ u2 ) ⊂ suppu1 + suppu2 . Solution Pick φ ∈ C0∞ (Rn ) so that suppφ ∩ suppu1 + suppu2 = Ø. Then, since supp(u1 × u2 ) = suppu1 × suppu2 , supp(u1 × u2 ) ∩ supp(ηk (x, y)φ(x + y)) ⊂ (suppu1 × suppu2 ) ∩ {(x, y) ∈ R2n : x + y ∈ suppφ} = Ø. Consequently, supp(u1 ∗ u2 ) ⊂ suppu1 + suppu2 .

6.2 Properties

167

6.2 Properties Let u1 , u2 ∈ D  (Rn ) and suppose u1 ∗ u2 ∈ D  (Rn ) exists. 1. Commutativity. u1 ∗ u2 (φ) = lim u1 (x) × u2 (y)(ηk (x, y)φ(x + y)) k→∞

= lim u2 (y) × u1 (x)(ηk (x, y)φ(x + y)) k→∞

= u2 (y) × u1 (x)(φ(x + y)) = u2 ∗ u1 (φ),

φ ∈ C0∞ (Rn ).

2. Convolution with the δ function. u1 ∗ δ(φ) = u1 (x) × δ(y)(φ(x + y)) = lim u1 (x) × δ(y)(ηk (x, y)φ(x + y)) k→∞

= lim u1 (x)(δ(y)(ηk (x, y)φ(x + y))) k→∞

= lim u1 (x)(ηk (x, 0)φ(x)) k→∞

= u1 (φ),

φ ∈ C0∞ (Rn ),

and analogously, δ ∗ u1 = u1 . 3. Translation. Let h ∈ Rn . Then for φ ∈ C0∞ (Rn ), we have (u1 ∗ u2 )(x + h)(φ) = u1 ∗ u2 (φ(x − h)) = lim u1 (x) × u2 (y)(ηk (x − h, y)φ(x − h + y)) k→∞

= lim u1 (x + h) × u2 (y)(ηk (x, y)φ(x + y)) k→∞

= u1 (x + h) ∗ u2 (φ). Moreover, u1 (−x) ∗ u2 (−x)(φ) = lim u1 (−x) × u2 (−y)(ηk (x, y)φ(x + y)) k→∞

= lim u1 (x) × u2 (y)(ηk (−x, −y)φ(−x − y)) k→∞

= u1 ∗ u2 (φ(−x)) = (u1 ∗ u2 )(−x)(φ),

φ ∈ C0∞ (Rn ).

4. Let α ∈ Nn ∪ {0}. Then D α u1 ∗ u2 , u1 ∗ D α u2 exist and satisfy D α (u1 ∗ u2 ) = D α u1 ∗ u2 = u1 ∗ D α u2 .

168

6 Convolutions

In fact, for φ ∈ C0∞ (Rn ), we have D α (u1 ∗ u2 )(φ) = (−1)|α| u1 ∗ u2 (D α φ) = (−1)|α| lim u1 (x) × u2 (y)(ηk (x, y)Dxα φ(x + y)) k→∞  = (−1)|α| lim u1 (x) × u2 (y) Dxα (ηk (x, y)φ(x + y)) k→∞   α − Dxα−β ηk (x, y)Dxβ φ(x + y) β β 0, x ∈ R+ , fα (x) = Γ (α) ⎩ (n) fα+n (x) for α ≤ 0, x ∈ R+ , n ∈ N,

n + α > 0.

Note that fα ∈ D+ and H (x) fα ∗ fβ (x) = Γ (α)Γ (β)

x y β−1 (x − y)α−1 dy,

x ∈ R+ ,

α > 0,

β > 0.

0

(6.6) Theorem 6.3 We have fα ∗ fβ = fα+β .

6.6 Fractional Differentiation and Integration

173

Proof Case 1.

α > 0, β > 0. We take y = tx in (6.4) and we get

H (x)x α+β−1 fα ∗ fβ (x) = Γ (α)Γ (β)

1 t β−1 (1 − t)α−1 dt 0

= Case 2.

H (x)x α+β−1 H (x)x α+β−1 B(α, β) = = fα+β (x), Γ (α)Γ (β) Γ (α + β)

x ∈ R+ .

α ≤ 0, β > 0. Then (n)

fα (x) = fα+n (x),

α + n > 0,

n ∈ N,

x ∈ R+ ,

and we get (n)  (n) fα ∗ fβ (x) = fα+n ∗ fβ (x) = fα+n ∗ fβ (x)  (n)  H (x)x α+β+n−1 (n) = fα+β+n (x) = Γ (α + β + n) =

H (x)(α + β + n − 1) . . . (α + β)x α+β−1 Γ (α + β + n)

=

H (x)x α+β−1 = fα+β (x), Γ (α + β)

x ∈ R+ .

Case 3. α > 0, β ≤ 0. We omit the proof and leave it to the reader, since it merely reproduces that of case 2. Case 4. α ≤ 0, β ≤ 0. Let n1 , n2 ∈ N be fixed so that (n1 ) fα = fα+n , 1

(n2 ) fβ = fβ+n , 2

α + n1 > 0,

β + n2 > 0.

Then (n )

(n )

1 2 fα ∗ fβ (x) = fα+n ∗ fβ+n (x) 1 2 (n1 +n2 )  = fα+n1 ∗ fβ+n2 (x)

(n1 +n2 )  = fα+β+n1 +n2 (x) =

(n1 +n2 )  1 H (x)x α+β+n1+n2 −1 Γ (α + β + n1 + n2 )

174

6 Convolutions

=

H (x) (α + β + n1 + n2 − 1) . . . (α + β)x α+β−1 Γ (α + β + n1 + n2 )

=

H (x)x α+β−1 = fα+β (x), Γ (α + β)

x ∈ R+ .

This completes the proof. Example 6.3 Let us consider f0 . We have f0 (x) = f1 (x) =

H  (x) = δ(x). Γ (1)

Since D+ is a commutative convolution algebra, fα ∗ f−α (x) = f0 (x) = δ(x),

x ∈ R+ ,

for α ∈ R. Consequently fα−1 = f−α . For n ∈ Z− , we have (−n)

fn = f0

= δ (−n) .

So, fn ∗ u = δ (−n) ∗ u = δ ∗ u(−n) = u(−n) for u ∈ D+ . In the case when n ∈ Z+ , we have fn (x) =

H (x)x n−1 , x ∈ R+ . Γ (n)

Hence,   f1 ∗ u (x) = f1 ∗ u(x) =

H  (x) ∗ u(x) Γ (1)

= δ ∗ u(x) = u(x),

x ∈ R+ ,

for u ∈ D+ . Definition 6.6 When α < 0 the operator fα ∗ is called fractional differentiation of order α. When α > 0 it is known as fractional integration of order α. Let k ∈ (0, 1). Then     D k u = D D k−1 u = D f1−k ∗ u for u ∈ D+ .

(6.7)

6.6 Fractional Differentiation and Integration

175

Let φ ∈ C0∞ (R+ ), ξ, η ∈ D+ be chosen so that ξ ≡ 1 on (suppf1−k ) , η ≡ 1 on (suppu) , ξ ≡ 0 on R+ \(suppf1−k )2 , η ≡ 0 on R+ \(suppf1−k )2 . Then f1−k ∗ u(φ) = f1−k (x) × u(y)(ξ(x)η(y)φ(x + y)) =

H (x)x −k (ξ(x)u(y)(η(y)φ(x + y))) Γ (1 − k)

H (x)x −k  = ξ(x) Γ (1 − k)

∞ u(y)φ(x + y)dy



−∞

1 = Γ (1 − k) =

1 Γ (1 − k)

1 = Γ (1 − k) 1 = Γ (1 − k) 1 = Γ (1 − k) 1 = Γ (1 − k)

∞ x

−k

∞ u(y)φ(x + y)dydx −∞

0

∞ ∞

x −k φ(x + y)dxu(y)dy

(x + y = z)

−∞ 0

∞

∞ u(y) (z − y)−k φ(z)dzdy

−∞

y

∞ ∞

u(y)(x − y)−k φ(x)dxdy

−∞ y

∞ x

u(y)(x − y)−k dyφ(x)dx

−∞ 0

x

u(y)(x − y)−k dy(φ),

x ∈ R+ .

0

As φ ∈ C0∞ (R+ ) was chosen arbitrarily, we conclude f1−k

1 ∗ u(x) = Γ (1 − k)

x

u(y)(x − y)−k dy,

x ∈ R+ .

0

The latter representation and (6.7) imply d 1 D u(x) = Γ (1 − k) dx

x

k

0

u(y)(x − y)−k dy,

k ∈ (0, 1),

x ∈ R+ .

176

6 Convolutions

If l ∈ N, then D l+k u = D l (D k u) for u ∈ D+ . 1

Exercise 6.10 Compute D 2 H (x), x ∈ R+ . 1 H (x) Answer √ √ . π x 1

Exercise 6.11 Compute D 3 δ(x), x ∈ R+ .  1 d  − 31 . H (x)x Answer Γ ( 23 ) dx Exercise 6.12 Let k ∈ (0, 1) and u ∈ D+ , and call u−k the kth primitive of u. Prove that u−k

1 = fk ∗ u = Γ (k)

x (x − y)k−1 u(y)dy,

x ∈ R+ .

0

Exercise 6.13 Let k ∈ (0, 1), l ∈ N and u ∈ D+ . Prove that

u−l−k

1 = Γ (k)

x x1

xl−1xl 

(xl − y)k−1 u(y)dydxl . . . dx1 ,

... 0

0

0

0

6.7 Advanced Practical Problems Problem 6.1 Let u(x, t) ∈ D  (Rnx × Rt ). Find   D α δ(x) × δ (β) (t) ∗ u(x, t), where α ∈ Nn , β ∈ N. Answer β

Dxα Dt u(x, t). Problem 6.2 Compute 1. H (x) ∗ H (x)x 2, 2. H (x) ∗ H (x) sin x, 3. H (x) ∗ H (x)x 3,

x ∈ R+ .

6.7 Advanced Practical Problems

177

4. H (x) ∗ H (x)(x + cos x), 5. H (x) ∗ H (x)f (x), f ∈ C ∞ (R) in D  (R). 1. Solution Fix φ ∈ C0∞ (R) and choose ξ, η ∈ C0∞ (R) so that ξ ≡ 1 on (supp(H (x)x 2)) , η ≡ 1 on (suppH (x)) , ξ ≡ 0 on R\(supp(H (x)x 2))2 , η ≡ 0 on R\(suppH (x))2 . Then H (x) ∗ H (x)x 2(φ) = H (x) × H (y)y 2(ξ(x)η(y)φ(x + y)) ∞ =

∞ H (y)y 2φ(x + y)ξ(x)η(y)dydx

H (x) −∞

−∞

∞ ∞ =

H (x)H (y)y 2φ(x + y)dydx

(y + x = z)

−∞ −∞

∞ ∞ =

H (x)H (z − x)(z − x)2 φ(z)dzdx −∞ −∞

∞ =

∞ H (x)H (z − x)(z − x)2 dxdz

φ(z) −∞

−∞

∞

z

=

(z − x)2 dxφ(z)dz

H (z) −∞

0

∞ =

H (z) −∞

= H (x)

z3 φ(z)dz 3

x3 (φ). 3

Since φ ∈ C0∞ (R) was arbitrarily chosen, we conclude that H (x) ∗ H (x)x 2 = H (x) 2. Answer 2H (x) sin2 3. Answer H (x)

x4 , 4

x , 2

x3 . 3

178

6 Convolutions

4. Answer H (x)

 x2

 + sin x ,

2 x 5. Answer H (x) f (x − t)dt. 0

Problem 6.3 Compute 1. 2. 3. 4. 5.

H (x)x ∗ H (x)x 2, H (x)x ∗ H (x) sin x, H (x) cos x ∗ H (x)x 3, H (x)x ∗ H (x)e−x , H (x)f (x) ∗ H (x)g(x), f, g ∈ C ∞ (R)

in D  (R). 1. Solution Let φ ∈ C0∞ (R), ξ, η ∈ C0∞ (R) such that ξ ≡ 1 on (supp(H (x)x)) , η ≡ 1 on (supp(H (x)x 2)) , ξ ≡ 0 on R\(supp(H (x)x))2 , η ≡ 0 on R\(supp(H (x)x 2))2 . Then H (x)x ∗ H (x)x 2(φ) = H (x)x × H (y)y 2(ξ(x)η(y)φ(x + y)) ∞ =

∞ H (y)y 2ξ(x)η(y)φ(x + y)dydx

H (x)x −∞

−∞

∞

∞

=

H (y)y 2φ(x + y)dydx

H (x)x −∞

−∞

∞

∞

=

H (z − x)(z − x)2 φ(z)dzdx

H (x)x −∞

−∞

∞ =

∞ H (x)H (z − x)x(z − x)2 dxdz

φ(z) −∞

−∞

∞ =

z x(z − x)2 dxdz

φ(z)H (z) −∞

0

∞ =

H (z) −∞

= H (x)

z4 φ(z)dz 12

x4 (φ). 12

6.7 Advanced Practical Problems

179

Therefore H (x)x ∗ H (x)x 2 = H (x)

x4 . 4

2. Answer H (x)(x − sin x), 3. Answer H (x)(3x 2 + 6 cos x − 6),   4. Answer H (x) x − 1 + e−x , x f (y)g(x − y)dy.

5. Answer H (x) 0

Problem 6.4 Prove 1. δ(x − a) ∗ δ(x− b) = δ(x − a − b), x ∈ R, a, b = const, 2. δ (m) (x − a) ∗ δ (k) (x − b) ∗ u(x) = u(k+m) (x − a − b), x ∈ R, u ∈ D  (R), a, b = const, k, m ∈ N. Problem 6.5 In D  (R2 ) compute   1. H (at − |x|) ∗ H (t) × δ(x) , a > 0,   2. H (at − |x|) ∗ δ(t) × δ(x) , a > 0,   3. H (at − |x|) ∗ H (t) sin t × δ(x) , a > 0,   4. H (at − |x|) ∗ H (t)(t 2 + t + 1) × δ(x) , a > 0,   5. H (at − |x|) ∗ H (t)(1 + cos t) × δ(x) , a > 0,   6. H (at − |x|) ∗ f (t) × δ(x) , a > 0, where f ∈ C (t ≥ 0) and f ≡ 0 for t < 0. Answer |x|  , 1. H (t) t − a 2. H (at − |x|),

t |x|  3. H (at − |x|)2 sin2 − , 2 2a 1  |x| 3 1 |x| 2 |x|  4. H (t) t− t− , + +t − a  3 |x| a  2 |x| a + sin t − , 5. H (t) t − a a |x| t− a  6. H (at − |x|) f (τ )dτ . 0

180

6 Convolutions

Problem 6.6 Let X1 ⊂ Rn1 , X2 ⊂ Rn2 be open sets and K ∈ D  (X1 × X2 ). Define the map K : C0∞ (X2 ) → D  (X1 ) by  K (φ)(x1 ) =

φ ∈ C0∞ (X2 ).

K(x1 , x2 )φ(x2 )dx2 , X2

Prove that 1. K is continuous if and only if K φj →j →∞ 0 in D  (X1 ) as φj →j →∞ 0 in C0∞ (X2 ). 2. K φ(ψ) = K(ψ × φ),

φ ∈ C0∞ (X2 ), ψ ∈ C0∞ (X1 ).

Hint Use the definition of distribution and direct product by C0∞ functions. Definition 6.7 The distribution K(x1 , x2 ) is called the kernel of the map K . Problem 6.7 Let X1 ⊂ Rn1 , X2 ⊂ Rn2 be open sets, K ∈ C ∞ (X1 × X2 ), and define K : C ∞ (X2 ) → C ∞ (X1 ) by  K φ(x1 ) =

K(x1 , x2 )φ(x2 )dx2 ,

φ ∈ C0∞ (X2 ).

X2

Prove that K can be extended to a map from E  (X2 ) to C ∞ (X1 )   K u(x1 ) = u K(x1 , ·) ,

u ∈ E  (X2 ),

x1 ∈ X1 .

Problem 6.8 Let u1 ∈ D  (Rn ), u2 ∈ C0∞ (Rn ). Prove that u1 ∗ u2 defines a continuous function k

x → u1 (u2 (x − ·)). Problem 6.9 Take u1 , u2 ∈ D  (X), u2 with compact support. Prove that singsupp(u1 ∗ u2 ) ⊂ singsuppu1 + singsuppu2 . Solution Let ψ ∈ C0∞ (Rn ) be such that ψ ≡ 1 on a neighbourhood of singsuppu2 . Then u2 = (1 − ψ)u2 + ψu2 .

6.7 Advanced Practical Problems

181

By the definition of ψ it follows that (1 − ψ)u2 ∈ C0∞ (Rn ). Therefore u1 ∗ ((1 − ψ)u2 ) is a C ∞ function on 

 x : {x} − supp(ψu2 ) ⊂ singsuppu1 .

Consequently singsupp(u1 ∗ u2 ) = singsupp(u1 ∗ (ψu2 )) ⊂ singsuppu1 + singsupp(ψu2 ). We also have singsupp(ψu2 ) ⊂ singsuppu2 , and the claim follows. Problem 6.10 Let P be a differential operator with constant coefficients P =



aα D α .

α

Prove 1. P (u) = P (δ) ∗ u for u ∈ D  (Rn ), 2. P (u1 ∗ u2 ) = P (u1 ) ∗ u2 = u1 ∗ P (u2 ) for u1 , u2 ∈ D  (X), where u2 has compact support. Solution 1. We have        P (u) = aα D α δ ∗ u = aα D α δ ∗ u = aα D α δ ∗ u = P (δ) ∗ u. α

α

α

2. We have P (u1 ∗u2 ) =



aα D α (u1 ∗u2 ) =

α

 α

aα D α u1 ∗u2 =



 aα D α u1 ∗u2 = P (u1 )∗u2 .

α

At the same time P (u1 ∗u2 ) =

 α

aα D α (u1 ∗u2 ) =

 α

aα u1 ∗D α u2 = u1 ∗

 α

 aα D α u2 = u1 ∗P (u2 ).

182

6 Convolutions

Problem 6.11 Let u1 , u2 ∈ D  (Rn ), u2 with compact support. Suppose that for every y ∈ suppu2 we can find an integer j ≥ 0 and an open neighbourhood Vy of y for which 1. u1 ∈ D  ({x} − Vy ), 2. u2 ∈ C k+j (Vy ) or 1. u1 ∈ C k+j ({x} − Vy ), j 2. u2 ∈ D  (Vy ), where x  ∈ Rn . Prove that u1 ∗ u2 ∈ C k on a neighbourhood of x. Problem 6.12 Given f ∈ C (Rn ), compute f ∗ μδS . Problem 6.13 Let

∂ (νδS ) ∈ D  (Rn ), f ∈ C 1 (Rn ). Find ∂n f∗

∂ (νδS ), ∂n

where ν is a piecewise-continuous function. Problem 6.14 Let μ ∈ C (Rn ). Find 1. |x|2−n ∗ μδS , n ≥ 3, 2. log |x| ∗ μδS , n = 2. Problem 6.15 Let En (x) = |x|2−n , n ≥ 3 and g ∈ L1 (Rn ). Compute 1. Vn = En ∗ g, 2. Δn (En ∗ g). Problem 6.16 Compute 1. 2. 3. 4. 5.

|x|2 ∗ δSR , sin |x|2 ∗ δSR , 2 e|x| ∗ δSR , |x| ∗ δSR , f (|x|) ∗ δSR ,

where f (x) ∈ C ([0, ∞)). Problem 6.17 Let w(t) be a continuous function on t ≥ 0 and w(t) = 0 for t < 0. H (t) δS (x). Find E3 (t) ∗ w(t). Define E3 (x, t) = 4πt t

6.7 Advanced Practical Problems

183

Problem 6.18 Let 1 H (t − |x|), 2 H (t − |x|) & , E2 (x, t) = 2π t 2 − |x|2 H (t) δS(t )(x). E3 (x, t) = 4πt E1 (x, t) =

˜ i = 1, 2, 3. Also let u(x) ˜ ∈ C (Ri ), i = 1, 2, 3. Find Ei (x, t) ∗ u(x), Problem 6.19 Let f (x, t) ∈ D  (Rix × R1t ), i = 1, 2, 3, be a distribution for which suppf ⊂ {(x, t) : t ≥ 0}. Find Ei ∗ f , i = 1, 2, 3. Problem 6.20 Let f (λ) ∈ C 1 (λ ≥ 0), f  (0) = 0. Find −f (|x|) ∗

∂ δS . ∂n R

Problem 6.21 Let u1 , u2 , . . . , un ∈ C0 . Prove that |u1 ∗ u2 ∗ · · · ∗ un (0)| ≤ ||u1 ||p1 · · · ||un ||pn , where 1 1 1 + + ···+ = n − 1, p1 p2 pn

1 ≤ pj ≤ ∞,

j = 1, 2, · · · , n.

Hint Distinguish two cases: k = 2 and k > 2. Problem 6.22 Let 1 ≤ pj ≤ ∞, j = 1, 2, · · · , n and 1 1 1 + ···+ = n−1+ , p1 pn q

1 ≤ q ≤ ∞.

Take ui ∈ C0 , i = 1, 2, · · · , n, and prove ||u1 ∗ u2 ∗ · · · ∗ un ||q ≤ ||u1 ||p1 · · · ||un ||pn . n

Problem 6.23 Let ka (y) = |y|− a and u ∈ L∞ ∩ Lp . Prove

1 1 1 1 + = 1, +  = 1, 1 ≤ p < a  ,  a a p p p

1− p

||ka ∗ u||∞ ≤ Cp,a ||u||pa ||u||∞ a .

184

6 Convolutions

Solution Let R > 0 be fixed. Then         |ka ∗ u(x)| =  ka (x − y)u(y)dy  =  ka (y)u(x − y)dy   |y|

≤



Rn − an

|u(x − y)|dy =



Rn

|y|

− an

|u(x − y)|dy +

|y|≤R

Rn

n

|y|− a |u(x − y)|dy.

|y|≥R

For 

n

|y|− a |u(x − y)|dy,

|y|≤R

we have the following estimate  |y|

− an

R |u(x − y)|dy ≤ c1 ||u||∞

|y|≤R

ρ − a ρ n−1 dρ = ||u||∞ c2 R n− a . n

n

(6.8)

0

For 

|y|− a |u(x − y)|dy n

|y|≥R

we have the following estimate 

n

|y|− a |u(x − y)|dy ≤ ||u||p

|y|≥R

 

n



|y|− a p dy

 1 p

|y|≥R

 ∞  1 n n  −n p = ||u||p c3 ρ n−1− a p dρ = c4 ||u||p R p a . R

Combining (6.8) and (6.9),   n n −n |ka ∗ u(x)| ≤ Cp,a R n− a ||u||∞ + R p a ||u||p . We choose R so that n

R p = ||u||p

1 . ||u||∞

(6.9)

6.7 Advanced Practical Problems

185

Then p 

n

− p

R n− a = ||u||pa ||u||∞a , p 

n

1− ap

R n− a ||u||∞ = ||u||pa u∞ R

n −n p a

||u||p =

,

p a

1− p ||u||p ||u||∞ a .

Consequently p 

1− p

|ka ∗ u(x)| ≤ Cp,a ||u||pa ||u||∞ a . Here c1 , c2 , c3 , c4 and Cp,a are nonnegative constants. Problem 6.24 Let u ∈ L1 (Rn ) and s be a positive number. Prove that 1. u can be written as u=v+

∞ 

wk ,

k=1

where ∞ 

||v||1 +

||wk ||1 ≤ 3||u||1,

|v(x)| ≤ 2n s,

x ∈ Rn .

k=1

2. for every sequence of pairwise disjoint cubes Ik we have wk (x) = 0 for x ∈ Ik ,  ∞  wk (x)dx = 0, s μ(Ik ) ≤ ||u||1. k=1

Rn

Solution Let us subdivide Rn into cubes In such that μ(In ) >  |u(x)|dx =

sμ(In ) >



i.e., 1 μ(In )

 |u(x)|dx. Then Rn

 |u(x)|dx >

i I i

Rn

1 s

|u(x)|dx, In

 |u(x)|dx < s. In

Now, we divide I1 into 2n equal parts so that the average of |u| on each one is greater than or equal to s. Therefore  sμ(I1k ) ≤

 |u(x)|dx ≤

I1k

|u(x)|dx ≤ sμ(I1 ) = 2n sμ(I1k ). I1

186

6 Convolutions

Let v(x) =

1 μ(I1k )

 x ∈ I1k ,

u(y)dy,

(6.10)

I1k

and  w1k (x) =

u(x) − v(x) for x ∈ I1k , 0 for x ∈ / I1k .

(6.11)

Now, we divide I2 into 2n equal parts so that the average of |u| on each is greater than or equal to s. We extend the definitions (6.10) and (6.11) to these cubes. Continuing in this way we produce a sequence of functions wj k and a sequence of cubes Ij k . We complete the definition of v by setting v(x) = u(x) for x ∈ / O = ∪k Ik . Then u=v+

∞ 

wk .

k=1

We also have     |v(x)| + |wk (x)| dx ≤ 3 |u(x)|dx. Ik

Ik

Therefore ∞  

|v(x)|dx +

k=1 I



k

|v(x)|dx + Rn

∞ 

∞  

|wk (x)|dx ≤ 3

k=1 I

∞  

|u(x)|dx,

k=1 I

k

k

||wk ||1 ≤ 3||u||1.

k=1

Since Ik ∩ Il = Ø, k = l, we have wk (x) = 0 for x ∈ / Ik , v(x) = u(x) for x ∈ O and |v(x)| ≤ 2n s

for x ∈ O.

If x ∈ / O, there exist sufficiently small cubes containing x on each of which the average of |u| is less than s. Consequently |u(x)| ≤ s,

x∈ / O.

6.7 Advanced Practical Problems

187

From the inequality  sμ(Ik ) ≤

|u(x)|dx, Ik

we obtain s

∞ 

μ(Ik ) ≤ ||u||1.

k=1

Problem 6.25 Let I be a cube with centre at the origin,  I ∗ a cube with the same centre and twice the edge. Take w ∈ L1 (Rn ), suppw ⊂ I , w(x)dx = 0. Prove Rn



|ka ∗ w(x)|a dx

1 a

≤ Ca ||w||1 .

(6.12)

CI ∗

Here CI ∗ is the complement of I ∗ . Solution Let the side of the cube I is L. By the mean value theorem, we have    n   |ka ∗ w(x)| =  ka (x − y)w(y)dy  ≤ |(ka (x − y) − ka (x))||w(y)|dy ≤ CL|x|−1− a ||w||1 Rn

Rn

when x ∈ / I ∗ . Here C is a constant. Hence, 

|ka ∗ w(x)|a dx

1 a

⎛ ≤ CL||w||1 ⎝

CI ∗



⎞1 a

|x|−a−n dx ⎠ .

CI ∗

Note that ⎛ ⎜ ⎝



⎞1

⎛∞ ⎞1 a a  ⎟ −a−n −a−1 ⎠ ⎝ |x| dx ⎠ = C1 r dr

|x|≥L

L

=

C1 , L

where C1 is a constant. Therefore there exists a constant Ca so that (6.12) holds. Problem 6.26 Prove   μ x : |ka ∗ u(x)| > t t a ≤ Ca ||u||a1 , for t > 0, a > 0.

188

6 Convolutions

Solution Let us suppose ||u||1 = 1 (otherwise, we may consider be represented as u=v+

∞ 

u ). Then u can ||u||1

wk .

k=1

We also have (when p = 1) 1

1

|ka ∗ v(x)| ≤ c||v||1a ≤ c1 s a ,

x ∈ Rn .

Let s satisfies 1

c1 s a =

t . 2

Then |ka ∗ u(x)| > t,

x ∈ Rn ,

implies ∞ 

|ka ∗ wk (x)| >

k=1

t , 2

x ∈ Rn .

Let O = ∪k Ik∗ , where Ik∗ is the cube with twice the edge of Ik . We have μ(O)
≤ c1 t −a . ≤ +c 2 s 2 k=1

Problem 6.27 Let 1 < a < ∞, 1 < p < q < ∞,

1 1 1 + = 1 + . Prove p a q

||ka ∗ u||q ≤ Cp,a ||u||p for u ∈ C0 (Rn ).

6.7 Advanced Practical Problems

189

Solution For convenience we will suppose that ||u||p = 1. Let   μ(t) = μ x : |ka ∗ w(x)| > t . Then q ||ka ∗ u||q

 =

∞ |ka ∗ u(x)| dx = −

Rn

∞ t dμ(t) = q

q

q

0

0

Set u = v + w, where  u=

v for |u| ≤ s, w for |u| > s.

Then p

p

||ka ∗ v||∞ ≤ cs 1− a = cs q . Now, we choose s so that p

cs q =

t . 2

If |ka ∗ u(x)| > t,

x ∈ Rn ,

then we have |ka ∗ w(x)| >

t , 2

t q−1 μ(t)dt.

x ∈ Rn .

190

6 Convolutions

Consequently  t q ≤ c t −a ||w||1 , μ(t) ≤ μ x : |ka ∗ w(x)| > 2  ∞   a q  q−1−a t |u(x)|dx dt ||ka ∗ u||q ≤ c 0

≤ c



∞

0



|u(x)|>s

t q−1−a dt

1 a

|u(x)|dx

a .

|u(x)|>s

We note that  t q−1−a dt ≈ t q−a . s n, Dj u ∈ Lloc , j = 1, 2, · · · , n. Prove p

sup

x =y;x,y∈K

|u(x) − u(y)| < ∞, |x − y|γ

γ =1−

n . p

192

6 Convolutions

Problem 6.30 Let u ∈ D  (X), 1 < p < ∞, m ∈ N. Let D α u ∈ Lloc (X) for |α| = m. Prove that for |α| < m we have p

q

1. D α u ∈ Lloc (X) if q < ∞,

1 m − |α| 1 ≤ + , p q n

2. D α u is Hölder continuous of order γ , where 0 < γ < 1 and

1 1 ≤ (m−|α|−γ ) . p n

Problem 6.31 In D  (R 1 ) compute  d 1 4 (H (x) ∗ δ(x)), 1. dx  d 1 3 2. H (x). dx Answer H (x) 1.   1 , Γ 34 x 4 H (x) 2.   1 . Γ 23 x 3 Problem 6.32 In D  (R) compute 1. lim δ(x + k), k→∞

2. lim δ(x − k), k→∞

3. δ(x + k) ∗ δ(x − k), k ∈ R. Answer 1. 0, 2. 0, 3. δ(x). Problem 6.33 Let 2 1 −x fα (x) = √ e 2α2 , 2πα

x ∈ R,

α > 0.

Prove that fα ∈ D  (R) and fα ∗ fβ = f√α 2 −β 2 . Problem 6.34 Let fα (x) =

α 1 , π α2 + x 2

x ∈ R,

α > 0.

6.7 Advanced Practical Problems

193

Prove that fα ∈ D  (R) and fα ∗ fβ = fα+β . Problem 6.35 Prove that the function sin πα u(x) = π

x 0

g  (ξ ) dξ (x − ξ )1−α

solves x 0

u(ξ ) dξ = g(x), (x − ξ )α

g(0) = 0,

g ∈ C 1 (x ≥ 0),

0 < α < 1.

Problem 6.36 Prove eax f ∗ eax g = eax (f ∗ g),

f, g ∈ D+ .

Problem 6.37 Let f ∈ D  (Rn ). Prove that the convolution f ∗ 1 exists and is constant. Problem 6.38 Let u ∈ D  (Rn ), φ ∈ C0∞ (Rn ). Prove 1. u ∗ φ ∈ C ∞ (Rn ), 2. supp(u ∗ φ) ⊂ suppu + suppφ, 3. D α (u ∗ φ) = D α u ∗ φ = u ∗ D α φ for every α ∈ Nn . 2. Solution Let u ∗ φ(x) = 0. Then x − y ∈ suppφ, so x ∈ suppu + suppφ. Since x is arbitrary in supp(u ∗ φ), we conclude that supp(u ∗ φ) ⊂ suppu + suppφ. 3. Solution From the definition of D α u, it follows that D α (u ∗ φ) = D α u ∗ φ = u ∗ D α φ. Problem 6.39 Let u ∈ D  (Rn ), φ, ψ ∈ C0∞ (Rn ). Prove u ∗ (φ ∗ ψ) = (u ∗ φ) ∗ ψ.

194

6 Convolutions

Solution   φ(x − · − kh)hn ψ(kh) u ∗ (φ ∗ ψ)(x) = lim u h→0 k∈Zn  (u ∗ φ)(x − kh)ψ(kh)hn = lim h→0  k∈Zn = (u ∗ φ)(x − y)ψ(y)dy = (u ∗ φ) ∗ ψ. Rn

Problem 6.40 Let u, v ∈ D  (X), where X is a real open interval. Prove 1. u ≥ 0 if and only if u defines a nondecreasing function, 2. v  ≥ 0 if and only if v defines a convex function.

6.8 Notes and References In this chapter we define convolution of distributions. We deduct some of its basic properties. We prove that the convolution of distributions exists and it is well defined. We introduce a regularization of distributions. As applications of the convolution of distributions, we introduce fractional differentiation and integration. In the chapter are investigated the convolution algebras D  (Γ +) and D  (Γ ). Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 7

Tempered Distributions

7.1 Definition Definition 7.1 A linear continuous functional on S (Rn ) is called a tempered distribution. The space of tempered distributions is indicated by S  (Rn ).  n  n Definition 7.2 A sequence {un }∞ n=1 in S (R ) is said to converge in S (R ) to  n n u ∈ S (R ) if un (φ) →n→∞ u(φ) for every φ ∈ S (R ).

Note that the convergence in S  (Rn ) implies convergence in D  (Rn ). Definition 7.3 A set M  ⊂ S  (Rn ) is called weakly bounded if for every φ ∈ S (Rn ) there is a constant Cφ such that |u(φ)| ≤ Cφ for every u ∈ M  . Theorem 7.1 If M  ⊂ S  (Rn ) is a weakly bounded set, there exist constants K > 0 and m ∈ N such that |u(φ)| ≤ K||φ||S ,m ,

u ∈ M  , φ ∈ S (Rn ).

Proof In order to show this let us suppose that the assertion is false, i.e., there exist  ∞ n sequences {uk }∞ k=1 in M and {φk }k=1 in S (R ) such that |uk (φk )| > k||φk ||S ,k ,

k ∈ N.

1 φk (x) , ψk (x) = √ k ||φk ||S ,k

k ∈ N.

(7.1)

We define the functions

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_7

195

196

7 Tempered Distributions

So, ψk ∈ S (Rn ), k ∈ N, and 1 ||φk ||S ,p ||ψk ||S ,p = √ , k ||φk ||S ,k

k, p ∈ N.

Moreover, ||φk ||S ,p ≤ ||φk ||S ,k for every k ≥ p. Hence, 1 ||ψk ||S ,p ≤ √ →k→∞ 0. k  Since p ∈ N was arbitrary and S = Sp , we have ψk →k→∞ 0 in S (Rn ). p∈N∪{0}

Using techniques of the sort of (2.9)–(2.14), we conclude that uk (ψk ) →k→∞ 0. On the other hand, using (7.1), we have √ √ |uk (φk )| > k k||φk ||S ,k ,

(7.2)

k ∈ N,

from which |uk (ψk )| >

√ k →k→∞ ∞.

This contradicts with (7.2). This completes the proof. From this we also deduce that any tempered distribution u has finite order m. It can be extended to a linear continuous functional from the smallest dual space Sm , and |u(φ)| ≤ ||u||S ,−m ||φ||S ,m , where ||u||S ,−m is a norm in Sm . Example 7.1 Let u be defined on Rn and suppose  Rn

|u(x)| dx < ∞ (1 + |x|)m

for some m ≥ 0. Define the functional on S (Rn )  u(φ) = u(x)φ(x)dx, φ ∈ S (Rn ). Rn

(7.3)

7.2 Direct Product

197

This is well defined. In fact, let φ ∈ S (Rn ) and C be a positive constant such that sup ((1 + |x|)m |φ(x)|) ≤ C.

x∈Rn

Then      |u(φ)| =  u(x)φ(x)dx  ≤ |u(x)||φ(x)|dx 

Rn

= Rn

Rn

|u(x)| (1 + |x|)m |φ(x)|dx ≤ C (1 + |x|)m



|u(x)| dx < ∞. (1 + |x|)m

Rn

It is a linear and continuous functional on S (Rn ), so u ∈ S  (Rn ). / S  (R). Exercise 7.1 Prove that ex ∈   Exercise 7.2 Prove that cos ex belongs to S  (R) but not to S (R). Exercise 7.3 Show that S  (Rn ) is a C-vector space.

7.2 Direct Product We remark that the function ψ(x) = u1 (y)(φ(x, y)), where φ ∈ S (Rn+m ), u1 ∈ S  (Rm ), satisfies D α ψ(x) = u1 (y)(Dxα φ(x, y)) for every α ∈ Nn ∪ {0}. Since u1 ∈ S  (Rm ), there exist q ∈ N and a positive constant Cu1 such that |D α ψ(x)| ≤ Cu1

q

sup y∈Rm ,|β|≤q

(1 + |y|2 ) 2 |Dxα Dyβ φ(x, y)|.

Therefore ||ψ||S ,p = ≤ Cu1

p

sup x∈Rn ,|α|≤p

sup (x,y)∈Rn+m |α|≤p,|β|≤q

(1 + |x|2 ) 2 |D α ψ(x)| p

q

(1 + |x|2 ) 2 (1 + |y|2 ) 2 |Dxα Dyβ φ(x, y)|

≤ Cu1 ||φ||S ,p+q ,

φ ∈ S (Rn+m ),

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7 Tempered Distributions

for p, q ∈ N. Let u1 ∈ S  (Rn ), u2 ∈ S  (Rm ). Then the functional u1 (ψ) = u1 (x)(u2 (y)(φ(x, y))),

φ ∈ S (Rn+m ),

is linear and continuous on S (Rn+m ). Definition 7.4 The direct product of u1 and u2 is u1 (x) × u2 (y)(φ) = u1 (x)(u2 (y)(φ(x, y))),

φ ∈ S (Rn+m ).

Notice that u1 (x) × u2 (y) ∈ S  (Rn+m ). Since C0∞ (Rn ) is dense in S (Rn ), all properties of direct products in D  (Rn ) hold true for the space S  (Rn ). Exercise 7.4 Let u1 ∈ S  (Rn ), u2 ∈ S  (Rm ). Prove that the operation u1 (x) → u1 (x) × u2 (y) from S  (Rn ) to S  (Rn+m ) is linear and continuous.

7.3 Convolution Take u1 , u2 ∈ S  (Rn ) so that the convolution u1 ∗ u2 exists in D  (Rn ). 1. Let u1 ∈ S  (Rn ), u2 ∈ E  (Rn ). Since u1 ∗ u2 exists in D  (Rn ), we have u1 ∗ u2 (φ) = u1 (x) × u2 (y)(η(y)φ(x + y)),

φ ∈ S (Rn ),

where η ∈ C0∞ (Rn ) and η ≡ 1 on (suppu2 ) . As u1 ∈ S  (Rn ), u2 ∈ E  (Rn ), the direct product u1 (x) × u2 (y) exists in S  (R2n ). Then the convolution u1 ∗ u2 exists in S  (Rn ). We claim that φ → η(y)φ(x + y) is a continuous operation on S (Rn ). In fact, we have ||η(y)φ(x + y)||S ,p ≤ ≤ Cα

p

(1 + |x|2 + |y|2 ) 2 |D α (η(y)φ(x + y))|

sup (x,y)∈R2n,|α|≤p

p

(1 + |x|2 + |y|2 ) 2 |D α φ(x + y)|

sup (x,y)∈R2n,|α|≤p

= Cα ||φ||S ,p ,

Cα = const.

Therefore the map φ → η(y)φ(x + y) from S (Rn ) to itself is continuous, and u1 → u1 ∗ u2 is continuous from S  (Rn ) to S  (Rn ).

7.3 Convolution

199



Γ ∗ , S is a

2. Let Γ be a closed, convex, acute cone in Rn with vertex at 0, C =

strictly C-like surface. Now, we suppose u1 ∈ S  (Γ +) and u2 ∈ S  (S+ ). The convolution u1 ∗ u2 exists in S  (Rn ) and can be represented as u1 ∗ u2 (φ) = u1 (x) × u2 (y)(ξ(x)η(y)φ(x + y)),

φ ∈ S (Rn ),

where ξ, η ∈ C0∞ (Rn ), ξ ≡ 1 on (suppu1 ) , η ≡ 1 on (suppu2 ) and ξ ≡ 0 on Rn \(suppu1 )2 , η ≡ 0 on Rn \(suppu2 )2 . If K is a compact set in Rn and suppu1 ⊂ Γ + K, the map u1 → u1 ∗ u2 is continuous from S  (Γ + K) to S  (S+ + K). The set S  (Γ +) is a convolution subalgebra of D  (Γ +) and S  (Γ ) is a convolution subalgebra of S  (Γ +). 3. Let u ∈ S  (Rn ) and η ∈ S (Rn ). Then the convolution u1 ∗ η exists in ΘM . It can be represented in the form u ∗ η(φ) = u(η ∗ φ(−x)),

φ ∈ S (Rn ).

We note that there exists a natural number m such that m

|D α (u ∗ η)(x)| ≤ Cu (1 + |x|2 ) 2 ||η||S ,m+|α| ,

x ∈ Rn .

∞ 2n Here Cu = const. In fact, let {ηk (x, y)}∞ k=1 be a sequence in C0 (R ) such that 2n n ηk →k→∞ 1 in R and φ ∈ S (R ). Then   η(y)ηk (x, y)φ(x + y)dy →k→∞ η(y)φ(x + y)dy Rn

Rn

in S (Rn ). Since u ∗ η exists in D  (Rn ), we have u ∗ η(φ) = lim u(x) × η(y)(ηk (x, y)φ(x + y)) k→∞   = lim u(x) η(y)ηk (x, y)φ(x + y)dy k→∞

= u(x)



Rn



η(y)φ(x + y)dy = u(x)

Rn

φ(ξ )η(ξ − x)dξ

Rn

= u(x)(η ∗ φ(−x)). We note that φη ∈ S (Rn ) and





u ∗ η(φ) =

u(x)(η(ξ − x))φ(ξ )dξ. Rn

But φ ∈ S (Rn ) was chosen arbitrarily, so u ∗ η(x) = u(y)(η(x − y)).



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7 Tempered Distributions

If m is the order of u, then |D α (u ∗ η)(x)| ≤ Cu ||Dxα η(x − y)||S ,m m = Cu sup (1 + |y|2) 2 |Dxα Dyβ η(x − y)| y∈Rn ,|β|≤m

= Cu

m

(1 + |x − ξ |2 ) 2 |D α+β η(ξ )|

sup

ξ ∈Rn ,|β|≤m 2 m 2

≤ Cu (1 + |x| )

m

sup

m

(1 + |ξ |2 ) 2 |D α+β η(ξ )|

ξ ∈Rn ,|β|≤m

≤ Cu (1 + |x|2 ) 2 ||η||S ,m+|α| .

7.4 Advanced Practical Problems Problem 7.1 Prove that for every distribution u ∈ S  (Rn ) there exist constants K ≥ 0 and m ∈ N such that |u(φ)| ≤ K||φ||m ,

φ ∈ S (Rn ).

Problem 7.2 Prove that any tempered distribution has finite order. Problem 7.3 Prove 

S0 ⊂ S1 ⊂ · · · ⊂ Sm ⊂ · · · ⊂ S  =

Sp .

p∈N∪{0}  Problem 7.4 Prove that the embedding Sp ⊂ Sp+1 is continuous for any p ∈ N.

Problem 7.5 Prove that every weakly convergent sequence in Sp , p ∈ N,  converges in the norm of Sp+1 . Problem 7.6 Prove that Sp , p ∈ N, is a weakly complete space. Problem 7.7 Show that S  (Rn ) is a complete space. Problem 7.8 Let u ∈ E  (Rn ). Prove that u ∈ S  (Rn ) and u(φ) = u(ηφ) for every φ ∈ S (Rn ), where η ∈ C0∞ (Rn ) and η ≡ 1 on a neighbourhood of suppu. Definition 7.5 A measure μ on Rn is called tempered if  Rn

for some m ≥ 0.

(1 + |x|)−m μ(dx) < ∞

7.4 Advanced Practical Problems

201

Problem 7.9 Let μ be a tempered measure on Rn and define the functional  μ(φ) = φ(x)μ(dx), φ ∈ S (Rn ). Rn

Prove that μ ∈ S  (Rn ). Problem 7.10 Let u ∈ S  (Rn ). Prove 1. D α u ∈ S  (Rn ) for every α ∈ Nn ∪ {0}, 2. the map u → D α u is a linear continuous operation on S  (Rn ). Problem 7.11 Let u ∈ S  (Rn ), A is an invertible n × n matrix. Prove that u(Ax + b) ∈ S  (Rn ), where b = (b1 , b2 , . . . , bn ), bl = const, l = 1, 2, . . . , n, and the map u(x) → u(Ax + b) is a linear and continuous operation on S  (Rn ). Problem 7.12 Let u ∈ S  (Rn ), a ∈ ΘM . Prove that au ∈ S  (Rn ) and that u → au is a linear and continuous operation from S  (Rn ) to S  (Rn ). Problem 7.13 Let an ∈ C, |an | ≤ C(1 + |n|)N for some constants C > 0 and N ≥ 0, n = 1, 2, . . .. Prove that ∞ 

ak δ(x − k) ∈ S  (Rn ).

k=1

Problem 7.14 Let u ∈ S  (Rn ). Prove that there exists a tempered function g in Rn and a constant m ∈ N such that u(x) = D1m D2m . . . Dnm g(x),

x ∈ Rn .

Solution Since u ∈ S  (Rn ), there exist p ∈ N and a positive constant Cu such that p

sup (1 + |x|2 ) 2 |D α φ(x)| |u(φ)| ≤ Cu ||φ||p = Cu n ,|α|≤p x∈R     p   ≤ Cu max D1 D2 . . . Dn (1 + |x|2 ) 2 D α φ(x) dx, φ ∈ S (Rn ).

(7.4)

|α|≤p Rn

We define the functions

  p ψα (x) = D1 D2 . . . Dn (1 + |x|2) 2 D α φ(x) ,

φ ∈ S (Rn ).

In this) way we have a one-to-one mapping φ → {ψα } from S (Rn ) to the direct L1 (Rn ) equipped with the norm sum |α|≤p

||{fα }|| = max ||fα ||1 . |α|≤p

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7 Tempered Distributions

Call   M = {ψα }, φ ∈ S (Rn ) . Then M is a subset of

)

L1 (Rn ), on which we define the functional

|α|≤p

  u∗ {ψα } = u(φ),

{ψα } ∈ M.

Using (7.4), we get        p  ∗      u ({ψα }) = |u(φ)| ≤ Cu D1 D2 . . . Dn (1 + |x|2 ) 2 D α φ(x)  = Cu {ψα }. 1

We conclude that u∗ is continuous. We also recall that L∞ (Rn ) is the dual space to L1 (Rn ).) By the Hahn–Banach and Riesz theorems, there exists a vector-valued map L∞ (Rn ) such that {χα } ∈ |α|≤p

    u∗ {ψα } =

n |α|≤p R

χα (x)ψα (x)dx.

Hence, u(φ) =

 

  p χα (x)D1 D2 . . . Dn (1 + |x|2 ) 2 D α φ(x) dx

|α|≤pRn

for φ ∈ S (Rn ). Integrating by parts, we infer the existence of functions gα , |α| ≤ p + 2, gα ∈ S (Rn ), such that   p+2 p+2 p+2 gα (x)D1 D2 . . . Dn φ(x)dx. u(φ) = (−1)pn Rn |α|≤(p+2)n

Since φ ∈ S (Rn ), we conclude that u(x) =



p+2

D1

p+2

D2

p+2

. . . Dn

gα (x).

|α|≤(p+2)n

Problem 7.15 Let u ∈ S  (Rn ). Prove that there exists p ∈ N ∪ {0} such that for every positive  there are functions gα, ∈ S (Rn ), |α| ≤ p, such that gα, ≡ 0 on

7.5 Notes and References

203

Rn \(suppu) and u(x) =



D α gα, (x).

(7.5)

|α|≤p 

Solution Let  > 0 and η ∈ ΘM , η ≡ 1 on (suppu) 3 , η ≡ 0 on Rn \(suppu) . Using the previous problem, there exist m ∈ N ∪ {0} and g ∈ S (Rn ) such that u(x) = D1m D2m . . . Dnm g(x). Since u(x) = η(x)u(x), u(x) = η(x)D1m D2m . . . Dnm g(x) = D1m D2m . . . Dnm (η(x)g(x)) −



ηα (x)D α g(x),

|α|≤mn−1

where ηα ∈ ΘM , ηα (x) = 0 for x ∈ / (suppu) . The function ηα (x)D α g(x) we represent in the form ηα (x)D α g(x) = D α (ηα (x)g(x)) − F (x), and so forth. Note that we obtain (7.5) for p = mn and gα, = χα g, where χα ∈ ΘM and suppχα ⊂ (suppu) . Problem 7.16 Let u1 ∈ S  (Rm ). Prove that D α ψ(x) = u1 (y)(Dxα φ(x, y)) for every φ ∈ S (Rn+m ) and α ∈ Nn ∪ {0}. Problem 7.17 Prove that S (Rn ) is dense in S  (Rn ). Solution Let u ∈ S  (Rn ) and consider u = u∗ω . Then u ∈ ΘM and u →→0 u in S  (Rn ). Since the space S (Rn ) is dense in ΘM , because if a ∈ ΘM we have 2 2 ae−|x| ∈ S (Rn ),  > 0, and ae−|x| →→0 a in S  (Rn ). The claim follows.

7.5 Notes and References In this chapter we introduce tempered distributions. They are investigated the direct product of tempered distributions and convolution of tempered distributions. In the chapter are given some applications of tempered distributions. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 8

Integral Transforms

8.1 The Fourier Transform in S (Rn ) Definition 8.1 The Fourier transform of φ ∈ S (Rn ) is the integral  F (φ)(x) =

e−i(x,ξ )φ(ξ )dξ,

x ∈ Rn .

Rn

Note that F (φ) is bounded and continuous on Rn . Furthermore, F (φ) ∈ C ∞ (Rn ) and  D α F (φ)(x) = (−iξ )α e−i(x,ξ )φ(ξ )dξ = F ((−iξ )α φ)(x), Rn

F (D φ)(x) = α

D α φ(ξ )e−i(x,ξ ) dξ = (ix)α F (φ)(x),

x ∈ Rn ,

Rn

for every α ∈ Nn ∪ {0}, where (−iξ )α = (−i)|α| ξ1α1 ξ2α2 . . . ξnαn , (ix)α = i |α| x1α1 x2α2 . . . xnαn ,

ξ = (ξ1 , ξ2 , . . . , ξn ),

x = (x1 , x2 , . . . , xn ).

In particular, F (φ) is an integrable function on Rn . Observe that every function φ ∈ S (Rn ) can be represented by means of its Fourier transform F (φ) and inverse Fourier transform  1 −1 F (φ)(ξ ) = ei(x,ξ )φ(x)dx, ξ ∈ Rn , (2π)n Rn

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_8

205

206

8 Integral Transforms

as follows φ = F −1 (F (φ)) = F (F −1 (φ)). Explicitly, 1 φ(x) = (2π)n

 ei(x,ξ )F (φ)(ξ )dξ,

x ∈ Rn .

Rn

As p 2

#

|D#α F (φ)(ξ )| $

(1 + |ξ | )   ≤  (I − Δ) 2

p+1 2

Rn

≤ C sup (1 + |x| ) 2

x∈Rn

p+1 2

$

≤ (1 + |ξ | ) |D α F (φ)(ξ )|    (−ix)α φ(x) e−i(x,ξ )dx 

n+1 2

2

# $   p+1   2 (x α φ(x)), (I − Δ)

ξ ∈ Rn ,

for p ∈ N, it follows that ||F (φ)||S ,p ≤ Cp ||φ||S ,p+n+1 , where Cp is a constant independent of φ. By the last estimate, we conclude that φ → F (φ) is a linear and continuous map on S (Rn ). Every element φ ∈ S (Rn ) can be represented as the Fourier transform of the function ψ = F −1 (φ) ∈ S (Rn ), where φ = F (ψ). If F (φ) = 0, then φ = 0. Therefore the map φ → F (φ) is one-to-one on S (Rn ).   2 Exercise 8.1 Compute F e−ax , x ∈ R, a = const > 0. " π − ξ2 Answer e 4a , ξ ∈ R. a Exercise 8.2 Let A be a positive definite n × n matrix. Prove that 

F e

i(Ax,x)



n

πn i π2 −1 e−i 4 − 4 (A ξ,ξ ) , = √ detA

x, ξ ∈ Rn .

8.2 The Fourier Transform in S  (Rn ) Definition 8.2 The Fourier transform of a distribution u ∈ S  (Rn ) is defined as follows F (u)(φ) = u(F (φ))

for φ ∈ S (Rn ).

8.2 The Fourier Transform in S  (Rn )

207

Since the map φ → F (φ) : S (Rn ) → S (Rn ) is linear and continuous, the operation u → F (u) is linear and continuous from S  (Rn ) to itself. For u ∈ S  (Rn ), we define the operator F −1 in the following manner F −1 (u)(x) =

1 F (u)(−x). (2π)n

As φ → F (φ) is continuous and goes from S (Rn ) to S (Rn ), and S (Rn ) is dense in S  (Rn ), we conclude that F −1 (F (u)) = F (F −1 (u)) = u,

u ∈ S  (Rn ).

It follows that for every distribution u ∈ S  (Rn ) there exists a distribution v ∈ S  (Rn ) such that v = F −1 (u) and u = F (v). If F (u) = 0, then u = 0. Example 8.1 Let us compute F (δ). Take φ ∈ S (Rn ), so  F (δ)(φ) = δ(F (φ)) = F (φ)(0) =

φ(x)dx. Rn

This implies F (δ) = 1. Exercise 8.3 Compute F (H (x)e−x ), x ∈ R. Answer

1 . 1 + iξ

Let u(x, y) ∈ S  (Rn+m ), x ∈ Rn , y ∈ Rm . We introduce the Fourier transform Fx (u) with respect to the variable x = (x1 , x2 , . . . , xn ) by Fx (u)(φ) = u(Fξ (φ)),

φ ∈ S (Rn+m ),

where  Fξ (φ)(x, y) =

e−i(ξ,x)φ(ξ, y)dξ.

Rn

The map φ(ξ, y) → Fξ (φ) is an isomorphism from S (Rn+m ) to itself, and Fx (u) ∈ S  (Rn+m ) for u ∈ S  (Rn+m ). The inverse Fourier transform Fξ−1 is defined by Fξ−1 (u) =

1 Fξ (u(−ξ, y))(x, y). (2π)n

The map u → Fx (u) is an isomorphism of S  (Rn+m ) onto S  (Rn+m ).

208

8 Integral Transforms

Example 8.2 Let a = const ∈ R. Then  F (δ(x − a))(φ) = δ(x − a)(F (φ)) = F (φ)(a) =

e−iξ a φ(ξ )dξ,

φ ∈ S (R).

Rn

Since φ ∈ S (R) is arbitrary, F (δ(x − a)) = e−iξ a . Exercise 8.4 Prove that F

 δ(x − a) + δ(x + a)  2

= cos(aξ ),

a = const,

in S  (R).

8.3 Properties of the Fourier Transform in S  (Rn ) 1. For any u ∈ S  (Rn ), α ∈ Nn , we have D α F (u) = F ((−ix)α u). Example 8.3 Let φ ∈ S (Rn ). Then F (x α )(φ) = (i)|α| F ((−ix)α 1)(φ) = (i)|α| D α F (1)(φ), α ∈ Nn ∪ {0}. On the other hand, F (δ)(φ) = 1(φ). Using the inverse Fourier transform, we get δ(φ) = F −1 (1)(φ) =

1 F (1)(φ), (2π)n

i.e.,

F (1)(φ) = (2π)n δ(φ).

Therefore F (x α )(φ) = (2π)n (i)|α| D α δ(φ), in other words F (x α ) = (2π)n (i)|α| D α δ(ξ ). 2. For any u ∈ S  (Rn ), α ∈ Nn ∪ {0}, we have F (D α u) = (iξ )α F (u).

8.4 The Fourier Transform of Distributions with Compact Support

209

Example 8.4 Let us find F (δ  ) in S  (R). Take φ ∈ S (R) and compute F (δ  (φ))(ξ ) = (iξ )2 F (δ(φ)) = (iξ )2 δ(F (φ)) ∞  ∞  2 −ixξ 2 = (iξ ) δ e φ(x)dx = (iξ ) φ(x)dx, −∞

−∞

so F (δ  )(x) = (ix)2. 3. For any u ∈ S  (Rn ), we have F (u(x − x0 )) = e−i(ξ,x0 ) F (u). 4. For any u ∈ S  (Rn ), we have F (u)(ξ + ξ0 ) = F (e−i(ξ0 ,x) u)(ξ ). 5. For any nonsingular n × n matrix A, we have F (u(Ax))(ξ ) =

1 F (u)(A−1T ξ ), | det A|

u ∈ S  (Rn ).

6. For any u ∈ S  (Rn ), v ∈ S  (Rm ), we have F (u(x) × v(y))(ξ, η) = Fx (u(x) × F (v)(η)) = Fy (F (u)(ξ ) × v(y)) = F u(ξ ) × F (v)(η). 7. For any u ∈ S  (Rn+m ), α ∈ Nn ∪ {0}, β ∈ Nm ∪ {0}, we have Dxα Dyβ Fx (u) = Fx ((−ix)α Dyβ u), Fx (Dxα Dyβ u) = (iξ )α Dyβ Fx (u). The proofs of the above properties are left to the reader.

8.4 The Fourier Transform of Distributions with Compact Support If we take u ∈ E  (Rn ) we know already that u ∈ S  (Rn ), so it admits a Fourier transform in S  (Rn ). What is more, the Fourier transform exists in ΘM and can be represented in the form F (u)(ξ ) = u(x)(η(x)e−i(ξ,x)),

(8.1)

210

8 Integral Transforms

where η ∈ C0∞ (Rn ) and η ≡ 1 on a neighbourhood of suppu. We claim that there are constants Cα > 0 and m ∈ N such that m

|D α F (u)(ξ )| ≤ ||u||S ,−m Cα (1 + |ξ |2 ) 2 ,

ξ ∈ Rn , α ∈ Nn ∪ {0}.

Indeed, let φ ∈ S (Rn ) be arbitrary. So, D α F (u)(φ) = (−1)|α| F (u)(D α φ) = (−1)|α| u(F (D α φ))   = (−1)|α| u(η(x)(ix)α F (φ)) = u(x) η(x)(−ix)α φ(ξ )e−i(x,ξ ) dξ  =

R   α −i(x,ξ ) φ(ξ )dξ u(x) η(x)(−ix) e

n

Rn

and therefore D α F (u)(ξ ) = u(x)(η(x)(−ix)α e−i(x,ξ ) ),

α ∈ Nn ∪ {0},

(8.2)

From here, we obtain (8.1) for α = 0. But (8.2) implies |D α F (u)(ξ )| = |u(x)(η(x)(−ix)α e−i(x,ξ ))| ≤ ||u||S ,−m ||η(x)(−ix)α e−ixξ ||S ,m m = ||u||S ,−m sup (1 + |x|2 ) 2 |Dxβ (η(x)(−ix)α e−i(x,ξ ))| x ∈ Rn |β| ≤ m m ≤ ||u||S ,−m Cα (1 + |ξ |2 ) 2 . Therefore F (u) ∈ ΘM .

8.5 The Fourier Transform of Convolutions Let u ∈ S  (Rn ) and v ∈ E  (Rn ). So, the convolution u ∗ v is defined in S  (Rn ). Choose φ ∈ S (Rn ) and η ∈ C0∞ (Rn ) so that η ≡ 1 on a neighbourhood of suppv. Then u ∗ v(φ) = u(x) × v(y)(η(y)φ(x + y)),

8.6 The Laplace Transform

211

and F (u ∗ v)(φ) = u(x)(v(y)(η(y)F (φ(x + y))))     = u(x) v(y) η(y) φ(ξ )e−i((x+y),ξ )dξ = u(x) = u(x)

 n R

Rn

v(y)(η(y)e−i(y,ξ ))e−i(x,ξ )φ(ξ )dξ F (v)(ξ )e−i(x,ξ ) φ(ξ )dξ





Rn

= u(F (F (v)φ) = F (u)(F (v)φ) = F (u)F (v)(φ). Consequently F (u ∗ v) = F (u)F (v).

8.6 The Laplace Transform 8.6.1 Definition Definition 8.3 Let Γ be a closed, convex, acute cone in Rn with vertex at 0, and set C = intΓ ∗ , so C = Ø is an open, convex cone. Define T C = Rn + iC = {z = x + iy : x ∈ Rn , y ∈ C}. The Laplace transform of u ∈ S  (Γ +) is defined by L(u)(z) = F (u(ξ )e(y,ξ ) )(x).

(8.3)

This is well defined. Indeed, pick η ∈ C ∞ (Rn ) with the following properties: |D α η(ξ )| ≤ cα , η ≡ 1 on (suppu) and η ≡ 0 on Rn \(suppu)2 , and  > 0 arbitrary. Since η(ξ )e(y,ξ ) ∈ S (Rn ) for every y ∈ C, we have u(ξ )e(y,ξ ) = u(ξ )η(ξ )e(y,ξ ) ∈ S  (Γ +) for any y ∈ C. We conclude that (8.3) is well defined. It has the following representation L(u)(z) = u(ξ )(η(ξ )e−i(z,ξ ) ).

(8.4)

212

8 Integral Transforms

Observe that the Laplace transform does not depend on the choice of η. Indeed, let φ ∈ S (Γ +) be arbitrary. Then η(ξ )e(y,ξ )φ(x)e−i(x,ξ ) ∈ S (R2n ) and L(u)(z) = F (u(ξ )e(y,ξ ) )(φ) = u(ξ )e(y,ξ )F (φ)      = u(ξ )e(y,ξ ) φ(x)e−i(x,ξ )dx = u(ξ ) η(ξ )e(y,ξ ) φ(x)e−i(x,ξ )dx = u(ξ )



Rn

φ(x)e

−i(z,ξ )





η(ξ )dx =

Rn

R   −i(z,ξ ) u(ξ ) e η(ξ ) φ(x)dx. n

Rn

Equation (8.4) now follows. Example 8.5 Let us compute L(δ(ξ − ξ0 )). With φ ∈ S (Γ +), L(δ(ξ − ξ0 ))(φ) = F (δ(ξ − ξ0 )e(y,ξ ))(φ) = δ(ξ − ξ0 )(e(y,ξ )F (φ))     = δ(ξ − ξ0 ) e(y,ξ ) e−i(x,ξ )φ(x)dx = δ(ξ − ξ0 ) e−i(z,ξ ) φ(x)dx  =

Rn

e

−i(z,ξ0 )

Rn

φ(x)dx.

Rn

Since φ ∈ S (Γ +) is arbitrary, we find L(δ(ξ − ξ0 ))(z) = e−i(z,ξ0 ) .

8.6.2 Properties Let us write v(z) = L(u)(z). Since η(ξ )e−i(z,ξ ) is a continuous function in the variable z ∈ T C in S (Γ +), for z, z0 ∈ T C , we have η(ξ )e−i(z,ξ ) →z→z0 η(ξ )e−i(z0 ,ξ ) in S (Γ +). Hence, v(z) = u(ξ )(η(ξ )e−i(z,ξ ) ) →z→z0 u(ξ )(η(ξ )e−i(z0 ,ξ ) ) = v(z0 ) and v is continuous in T C . Take e1 = (1, 0, . . . , 0) and z ∈ T C and consider χh (ξ ) =

 1 η(ξ )e−i(z+he1 ,ξ ) − η(ξ )e−i(z,ξ ) →h→0 η(ξ )(−iξ1 )e−i(z,ξ ) h

8.6 The Laplace Transform

213

in S (Γ +). Then  1 v(z + he1 ) − v(z) = u(ξ )(η(ξ )e−i(z+he1 ,ξ ) ) − u(ξ )(η(ξ )e−i(z,ξ ) ) h h = u(ξ )(χh (ξ )) →h→0 u(ξ )(η(ξ )(−iξ1 )e−i(z,ξ )) = (−iξ1 )u(ξ )(η(ξ )e−i(z,ξ ) ), so ∂v = (−iξ1 )u(ξ )(η(ξ )e−i(z,ξ ) ) ∂z1 and finally D α L(u) = L((−iξ )α u),

∀α ∈ Nn ∪ {0}.

Definition 8.4 The distribution u ∈ S  (Γ +) for which v = L(u) is called a spectral function of v. If a spectral function u exists, then it must be unique, and we have a representation u(ξ ) = e−(y,ξ )Fx−1 (v(x + iy))(ξ ). Using the features of the Fourier transform one can easily deduce the following properties for the Laplace transform. 1. L(D α u) = (iz)α L(u)

for any u ∈ S  (Γ +), α ∈ Nn ∪ {0}.

Example 8.6 L(D α δ(ξ − ξ0 )) = (iz)α e−i(z,ξ0 ) . 2. L(u(ξ )e−i(a,ξ ) ) = L(u)(z + a) for any u ∈ S  (Γ +), Ima ∈ C. 3. L(u(ξ + ξ0 )) = ei(z,ξ0) L(u)(z). 1 TC L(u)(A−1T z) for z ∈ T A , where A is invertible of 4. L(u(Aξ )) = | det A| order n. 5. L(u1 × u2 )(z, ζ ) = L(u1 )(z)L(u2 )(ζ ) for any u1 ∈ S  (Γ1 +), u2 ∈ S  (Γ2 +), (z, ζ ) ∈ T C1 ×C2 . 6. L(u1 ∗ u2 ) = L(u1 )L(u2 ) for any u1 , u2 ∈ S  (Γ +).

214

8 Integral Transforms

Example 8.7 Let us compute L(H (ξ ) sin(ωξ )) in S  (Γ +), ω ∈ C, n = 1. Let φ ∈ S (Γ +). Then L(H (ξ ) sin(ωξ ))(φ) = F (H (ξ ) sin(ωξ )eyξ )(φ) ∞ yξ yξ e−ixξ φ(x)dx = H (ξ ) sin(ωξ )e (F (φ)) = H (ξ ) sin(ωξ )e ∞ ∞ =

∞

sin(ωξ )eyξ e−ixξ φ(x)dxdξ =

0 −∞ ∞

−∞ ∞

φ(x) −∞

sin(ωξ )e−izξ dξ dx

0

ω φ(x)dx. ω 2 − z2

= −∞

This proves that L(H (ξ ) sin(ωξ )) =

ω . ω 2 − z2

Exercise 8.5 Compute L(H (ξ ) cos(ωξ )) in S  (Γ +), n = 1, ω ∈ C. Answer

ω2

iz . − z2

Exercise 8.6 Compute L(H (ξ )eiωξ )) in S  (Γ +), n = 1, ω ∈ C. Answer

i . w−z

Exercise 8.7 Compute L(H (ξ )e−iωξ ) in S  (Γ +), n = 1, ω ∈ C. Answer −

i . w+z

8.7 Advanced Practical Problems Problem 8.1 Compute in S (R)   x2 F e− 4 cos(αx) ,

α = const.

Problem 8.2 Let u, v ∈ S (Rn ). Prove   F (u)(x)v(x)dx = u(x)F (v)(x)dx. Rn

Rn

8.7 Advanced Practical Problems

215

Solution We have      F (u)(x)v(x)dx = e−i(x,ξ ) u(ξ )dξ v(x)dx = e−i(x,ξ )u(ξ )v(x)dξ dx Rn



=

 u(ξ )

Rn



Rn Rn

e

−i(x,ξ )

v(x)dxdξ =

Rn

Rn Rn

u(ξ )F (v)(ξ )dξ. Rn

Problem 8.3 Prove that F F −1 = F −1 F = I

in

S (Rn ).

Problem 8.4 Show F (f ∗ g) = F (f )F (g) for f, g ∈ S (Rn ). Problem 8.5 Prove lim F (f )(λ) = 0

|λ|→∞

for f ∈ S (Rn ). Problem 8.6 Let f ∈ S (R). Prove 1. F (f (−x)) = F (f )(−ξ ), ξ  1 bξ , a, b, c = const, a > 0, 2. F (f (ax + b)) = ei a F (f ) a a iax 3. F (e f (x)) = F (f )(ξ − a), a = const = 0, ξ − a  1 c , a, b, c = const, b > 0. 4. F (eiax f (bx + c)) = ei b (ξ −a) F (f ) b b Problem 8.7 In S  (R) compute F (u) when u = H (1 − |x|), 2 u = e−4x , 2 u = eix , 2 u = e−ix , u = H (x)e−3x , u = H (−x)e4x , u = e−2|x| , 2 8. u = , 1 + x2 x α−1 . 9. u = H (x)e−2x Γ (α)

1. 2. 3. 4. 5. 6. 7.

216

8 Integral Transforms

1. Solution Fix an arbitrary φ ∈ S (R), so F (H (1 − |x|))(φ) = H (1 − |x|)(F (φ))  1 ∞  ∞ −ixξ e φ(ξ )dξ = e−ixξ φ(ξ )dξ dx = H (1 − |x|) ∞ =

1 φ(ξ )

−∞

−1

−∞

e−ixξ dxdξ =

−1 −∞

∞ 2 −∞

sin ξ sin ξ φ(ξ )dξ = 2 (φ). ξ ξ

Hence, F (H (1 − |x|)) = 2

sin ξ . ξ

Problem 8.8 In S  (R) compute F (u) when 1. u = H (x − a), a = const, 2. u = signx, 1 3. u = P , x 1 4. u = , x ± i0 5. u = |x|, 6. u = H (x)x k , k ∈ N, 7. u = |x|k , k ∈ N, k ≥ 2, 1 8. u = x k P , k ∈ N, x 1 9. u = P 2 , x 10. u = x k δ(x), k ∈ N, 11. u = x k δ (m) (x), k, m ∈ N, m ≥ k, 1 12. u = P 3 , x 13. u = H (1/2)(x), ∞  ak δ(x − k), ak = const, |ak | ≤ C(1 + |k|)m , C = const > 0, for 14. u = k=−∞

some m ≥ 2, 15. u = H (±x), 1 16. u = P . x Problem 8.9 Find  1  1. F P 2 , |x|

8.7 Advanced Practical Problems

2. F

217

 H (1 − |x|)  & 1 − |x|2

in S  (R2 ). Problem 8.10 Find F

 1  , 0 < k < n, in S  (Rn ). |x|k

Problem 8.11 Find 1. Fx (δ(x, t)), 2. Fx (H (t − |x|)), n = 1, in S  (Rn+1 (x, t)), (x, t) = (x1 , x2 , . . . , xn , t). Problem 8.12 Find   2 1. Fξ−1 H (t)e−ξ t ,  sin(ξ t)  2. Fξ−1 H (t) ξ in S  (Rn ).

 sin(|ξ |t)  in S  (R3 ). Problem 8.13 Find Fξ−1 H (t) |ξ | Problem 8.14 Find L(H (ξ )J0 (ξ )) in S  (Γ +), n = 1.

Problem 8.15 Using the Laplace transform, solve the following Cauchy problems in S  (Γ +), n = 1: 1. u (t) + 3u(t) = e−2t , u(0) = 0, 2. u (t) + 5u (t) + 6u(t) = 12, u(0) = 2, u (0) = 0, 3. ⎧  −t ⎨ u (t) + 5u(t) + 2v(t) = e , v  (t) + 2v(t) + 2u(t) = 0, ⎩ u(0) = 1, v(0) = 0. Problem 8.16 Using the Laplace transform solve the following equations in S  (Γ +), n = 1: 1. (H (t) sin t) ∗ u(t) = δ(t), 2. (H (t) cos t) ∗ u(t) = δ(t), 3. u(t) + 2(H (t) cos t) ∗ u(t) = δ(t), 4.  H (t) ∗ u1 (t) + δ  (t) ∗ u2 (t) = δ(t) δ(t) ∗ u1 (t) + δ  (t) ∗ u2 (t) = 0.

218

8 Integral Transforms

8.8 Notes and References In this chapter we introduce the Fourier transform for S functions and the Fourier transform for tempered distributions. They are deducted some of their basic properties. It is investigated the Fourier transform of convolution of tempered distributions. It is defined the Laplace transform for distributions and they are deduct some of its properties. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 9

Fundamental Solutions

9.1 Definition and Properties Let us write P (D) =



aα D α ,

aα = const,

|α|≤m



|aα | = 0.

|α|=m

Definition 9.1 Given P as above, the distribution u ∈ D  (Rn ) is called fundamental solution if P (D)u = δ. Consider the polynomial P (ξ ) =



aα ξ α .

|α|≤m

There exists a transformation ξ = Aξ  ,

with det A = 0,

A = (akj ),

under which P reads P (ξ  ) = aξ1 + m



Pk (ξ2 , . . . , ξn )ξ1 , k

a = const = 0.

(9.1)

0≤k≤m−1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_9

219

220

9 Fundamental Solutions

Note that there exists a constant κ = κ(m) such that for every point ξ ∈ Rn there is a k ∈ N ∪ {0} for which |P (ξ1 + iτ

k , ξ2 , . . . , ξn )| ≥ aκ, m

|τ | = 1.

(9.2)

Before to prove the classical Malgrange theorem, we will formulate the PaleyWiener-Schwartz theorem. For its proof we refer the reader to [35]. Theorem 9.1 (Paley-Wiener-Schwartz Theorem) For a function f to be integral and to satisfy the conditions of growth: for any  > 0 there is a constant M = M() > 0 such that α

|f (z)| ≤ M()e(a+)|y|(1 + |z|2 ) 2 ,

z ∈ C,

for a = a(f ) ≥ 0, α = α(f ) ≥ 0, z = x + iy, it is necessary and sufficient that its spectral function g ∈ E  (Ua ). Theorem 9.2 (Malgrange-Ehrenpreis Theorem) Every differential operator with constant coefficients has a fundamental solution in D  (Rn ). Proof Without loss of generality, we suppose P (ξ ) = aξ1m +

m−1 

Pk (ξ2 , . . . , ξn )ξ1k ,

a = const > 0.

k=0

We will prove the Malgrange-Ehrenpreis theorem for the polynomial P (iξ ). Let m  φ0 , φ1 , φ2 , . . . , φm ∈ C0∞ (Rn ) be chosen so that φk (ξ ) = 1, φk (ξ ) ≥ 0 for ξ ∈ Rn , and φk (ξ ) = 0 for those ξ ∈ Rn for which

k=0

     k min P iξ1 − τ , iξ2 , . . . , iξn  < aκ. |τ |=1 m If L(φ) denotes the Laplace transform of φ ∈ C0∞ (Rn ), we set u(φ) =

 m  1 1  φ (ξ ) k (2π)n 2πi k=0Rn

|τ |=1

L(φ)(ξ1 + iτ mk , ξ2 , . . . , ξn ) dτ dξ. P (ξ1 − τ mk , iξ2 , . . . , iξn ) τ

(9.3)

9.1 Definition and Properties

221

We fix φ ∈ C0∞ (Rn ) and choose R > 0 so that suppφ ⊂ UR . Since L(φ) is an entire function, by the Paley-Wiener-Schwartz theorem we have      L(φ) ξ1 + iτ k , ξ2 , . . . , ξn  ≤   m 2  −N    k k max e|Reτ | m 1 + ξ1 + iτ  + |ξ2 |2 + · · · + |ξn |2 m |τ |=1

|(1 − Δ)N φ(x)|dx

|x|

Rn

We note that      k  min P iξ1 − τ , iξ2 , . . . , iξn  ≥ aκ |τ |=1 m for ξ ∈ Rn with φk (ξ ) = 0 and for every k = 0, 1, 2, . . . , m. Then, using (9.4), max |L(φ)(ξ1 + iτ mk , ξ2 , . . . , ξn )| n  1  |τ |=1 |u(φ)| ≤ dξ φk (ξ ) (2π)n min |P (iξ1 − τ mk , iξ2 , . . . , iξn )| k=0Rn |τ |=1   −N   m   k 1 k 2 |Reτ | m 2 2  ≤ 1 + ξ1 + iτ  + ξ2 + · · · + ξn max e dξ (2π)n aκ |τ |=1 m k=0 n R  N × |(1 − Δ) φ(x)|dx. |x| 0.

9.5 Advanced Practical Problems

227

9.4 Fundamental Solution of the Laplace Operator Consider the differential operator P (D) =

n 

Dj2

j =1

and the equation P (D)E = δ. Take the Fourier transform and we get −|ξ |2 F (E)(ξ ) = 1(ξ ) or F (E)(ξ ) = −

1 . |ξ |2

Then E(x) = −F

−1



1 |ξ |2

 (x),

x ∈ Rn .

Hence, for n = 2, we have E(x) =

1 log |x|, 2π

x ∈ R2 ,

and for n ≥ 3, we have E(x) = −

1 1 , (n − 2)σn |x|n−2

x ∈ Rn ,

where σn is the “area” of the boundary of the n-dimensional unit ball.

9.5 Advanced Practical Problems Problem 9.1 Using (9.3) find a fundamental solution for the following operators 1.

d2 d +4 , 2 dx dx

228

9 Fundamental Solutions

d2 d + 1, −4 2 dx dx 2 d d + 2, 3. +3 2 dx dx 2 d d 4. + 5, −4 2 dx dx 3 d 5. − 1, dx 3 3 d2 d d − 3 +2 , 6. 3 2 dx dx dx d4 7. − 1, dx 4 4 d d2 8. − 2 + 1. dx 4 dx 2 Answer 2.

1 − e−4x , 4 x 2. H (x)xe , 3. H (x)(e−x − e−2x ), 4. H (x)e2x sin x, √  √  √ H (x)  x 3 3 − x2 e −e x + 3 sin x , 5. cos 3 2 2 H (x) 6. (1 − ex )2 , 2 H (x) 7. (shx − sin x), 2 H (x) 8. (xchx − shx). 2 Problem 9.2 Prove that 1. H (x)

(x+t)2 H (t) u(x, t) = √ et − 4t 2 πt

is a fundamental solution for the operator ∂2 ∂ ∂ − 2− − 1. ∂t ∂x ∂x Problem 9.3 Prove that −H (t)H (−x)et +x

9.5 Advanced Practical Problems

229

is a fundamental solution for the operator ∂ ∂ ∂2 − − + 1. ∂x∂t ∂x ∂t Problem 9.4 Let n = 3. Prove that u(x) = −

e±ik|x| 4π|x|

satisfies the equation Δu + k 2 u = δ(x). Problem 9.5 Prove that u(x, t) =

1 H (at − |x|) 2a

satisfies the equation ut t − a 2 uxx = δ(x, t),

x ∈ R,

t ∈ R,

and u(x, t) →t →+0 0,

∂u(x, t) →t →+0 δ(x), ∂t

∂ 2 u(x, t) →t →+0 0 ∂t 2

in D  (R). Problem 9.6 Prove that u(x, y) =

1 π(x + iy)

satisfies the equation 1  ∂u ∂u  +i = δ(x, y). 2 ∂x ∂y Problem 9.7 Prove that u(x, t) =

iH (t) −i π n i |x|2 √ e 4 e 4t (2 πt)n

230

9 Fundamental Solutions

solves 1 ut − Δu = δ(x, t), i

x ∈ Rn ,

t ∈ R.

Problem 9.8 Define u(x, t) =

H (t) − |x|2 e 4t √ (2 πt)n

and suppose ∞ F (x) =

u(x, t)dt, 0

exists for any t ≥ 0 and almost every x ∈ Rn . Prove ΔF = −δ(x, t).

9.6 Notes and References In this chapter it is given a definition for a fundamental solution of a differential operator. It is proved the classical Malgrange-Eherenpreis theorem. As applications, they are deducted the fundamental solutions for ordinary differential operators, the heat operator and the Laplace operator. Additional materials can be found in [7, 16, 17, 20, 21, 24, 25] and references therein.

Chapter 10

Sobolev Spaces

10.1 Definitions Definition 10.1 Let A be an open set in Rn , m ∈ N and 1 ≤ p ≤ ∞. The Sobolev space W m,p (A) consists of functions in Lp (A) whose partial derivatives up to order m, in the sense of distributions, can be identified with functions in Lp (A). Equivalently, W m,p (A) = {u ∈ Lp (A) : D α u ∈ Lp (A) for any α ∈ Nn ∪ {0}, |α| ≤ m}. Notice that clearly W 0,q (A) = Lq (A). For p = 2, the symbol W m,2 (A) is generally replaced by H m (A), and in the case A = Rn , we can use the Fourier transform ξ → F (u)(ξ ) of u ∈ L2 (A) to give the following characterisation m

W m,2 (Rn ) = H m (Rn ) = {u ∈ L2 (Rn ) : ξ → (1 + |ξ |2 ) 2 F (u)(ξ ) ∈ L2 (Rn )}. Example 10.1 Let n = 2. We seek conditions on β > 0 so that the function u(x, y) = x(x 2 +y 2 )−β , x, y ∈ U1 , is, away from the origin, an element of H 1 (U1 ). In polar coordinates x = r cos φ, y = r sin φ, 0 < r < 1, φ ∈ [0, 2π], we have u(x, y) = x(x 2 + y 2 )−β = r 1−2β cos φ, |u(x, y)|2 = r 2−4β cos2 φ, ux (x, y) = (x 2 + y 2 )−β − 2βx 2(x 2 + y 2 )−β−1 = r −2β (1 − 2β cos2 φ), |ux (x, y)|2 = r −4β (1 − 2β cos2 φ)2 , uy (x, y) = −2βxy(x 2 + y 2 )−β−1 = −2βr −2β cos φ sin φ, |uy (x, y)|2 = 4β 2 r −4β cos2 φ sin2 φ,

x, y ∈ U1 ,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2_10

231

232

10 Sobolev Spaces

substituting which produces 1 2π

 |u(x, y)| dxdy =

r 3−4β cos2 φdφdr < ∞ ⇐⇒ β < 1,

2

0

U1



0

1 2π |ux (x, y)| dxdy =

r 1−4β (1 − 2β cos2 φ)2 dφdr < ∞ ⇐⇒ β
0 and A ⊂ Rn be open. We define u (x) = ω  u(x) for u ∈ W m,p (A) and x ∈ A . Theorem 10.5 If u ∈ W m,p (A), then u ∈ C ∞ (A ). Proof To see this, let α be an arbitrary multi-index. Then  D α u (x) = D α  =

ω (x − y)u(y)dy A

Dxα ω (x − y)u(y)dy A

= (−1)

|α|

 Dyα ω (x − y)u(y)dy,

x ∈ A,

A

from which  D α u (x) =

ω (x − y)D α u(y)dy = ω  D α u(x), A

x ∈ A,

10.3 Approximation by Smooth Functions

and

237

    |D α u (x)| =  Dxα ω (x − y)u(y)dy  A

 ≤

|Dxα ω (x − y)||u(y)|dy A

≤



|Dxα ω (x − y)|q dy

 1 

A

q

|u(y)|p dy

1

p

A

≤ C||u||

W m,p (A)

,

x ∈ A.

This completes the proof. m,p

Theorem 10.6 If u ∈ Wloc (A), then u →→0 u m,p

in Wloc (A). Proof Let α be given. From the previous property D α u exists, and if V ⊂⊂ A, then   1 p ||u − u||W m,p (V ) = |D α u − D α u|p dx |α|≤m V

≤2

1 p

1

= 2p

 

|D α u − D α u|p dx

1

p

(10.1)

|α|≤m V



||D α u − D α u||Lp (V ) .

|α|≤m

Since (D α u) = D α u and D α u ∈ Lp (V ), 0 ≤ |α| ≤ m, using the properties of the convolution, we see that ||D α u − D α u||Lp (V ) →→0 0. m,p

Hence, u →→0 u in Wloc (A). This completes the proof. Theorem 10.7 If A is a bounded set in Rn and u ∈ W m,p (A), there exists a ∞ m,p sequence {ul }∞ (A) such that l=1 in C (A) ∩ W ul →l→∞ u in W m,p (A). 1 }, i ∈ N. i 1 Let x ∈ A and x ∈ ∂A. Therefore there exists i ∈ N such that dist(x, ∂A) > , so i Proof We claim A = ∪∞ i=1 Ai , where Ai = {x ∈ A : dist(x, ∂A) >

238

10 Sobolev Spaces

∞ ∞ x ∈ Ai and x ∈ ∪∞ i=1 Ai . Consequently A ⊂ ∪i=1 Ai . Conversely, if x ∈ ∪i=1 Ai 1 there exists j ∈ N such that x ∈ Aj . Therefore x ∈ A and dist(x, ∂A) > , so we j conclude ∪∞ i=1 Ai ⊂ A. Let Vi = Ai+3 \Ai , i = 1, 2, . . ., and choose V0 ⊂⊂ A. ∞ Then, as above, A = ∪∞ i=0 Vi . Let {ζi }i=1 be a sequence of smooth functions such that

ζi ∈ C0∞ (Vi ),

0 ≤ ζi ≤ 1,

∞ 

ζi = 1.

i=0

We define i > 0 and δ > 0, ui = (ζi u)  ωi so that δ , 2i+1 i suppu ⊂ Wi = Ai+4 \Ai ⊃ Vi .

||ui − ζi u||W m,p (A)
γ (x1 , x2 , . . . , xn−1 )}. r Let V = A ∩ U (x 0 , ) and define 2 x  = x + λen ,

x ∈ V ,  > 0, λ > 0, en = (0, 0, . . . , n).

There exists a large enough λ > 0 and a small enough  > 0 such that U (x  , ) lies in A ∩ U (x 0 , r) for every x ∈ V . Let u (x) = u(x  ) and define v = ω  u . Take |α| ≤ m, so ||D α v − D α u||Lp (V ) = ||D α v − D α u + D α u − D α u||Lp (V ) ≤ ||D α v − D α u ||Lp (V ) + ||D α u − D α u||Lp (V ) .

(10.2)

We have |D α v − D α u ||Lp (V ) →→0 0

(10.3)

||D α u − D α u||Lp (V ) →→0 0.

(10.4)

and

240

10 Sobolev Spaces

From (10.2)–(10.4), it follows ||D α v − D α u||Lp (V ) →→0 0 for 0 ≤ |α| ≤ m. Consequently ||v − u||W m,p (V ) →→0 0. Since ∂A is compact, there exist finitely many points xi0 , radii ri and functions vi ∈ C ∞ (V i )(i = 1, 2, . . . , N) for which 0 ∂A ⊂ ∪N i=1 U (xi ,

ri ), 2

||vi − u||W m,p (Vi ) ≤ δ,

where Vi = A ∩ U (xi0 ,

ri ). 2

Let V0 ⊂⊂ U . Then A ⊂ ∪N i=0 Vi . There exist functions ζi , i = 0, 1, . . . , N, such that ζi ∈ C0∞ (Vi ), 0 ≤ ζi ≤ 1 and N  ζi = 1. Now, we define the function i=0

v=

N 

ζi vi .

i=0

Then u=

N 

ζi u

i=0

and ||D α v − D α u||Lp (A) = ||

N 

D α (ζi (vi − u))||Lp (A)

i=0

≤

N  i=0

||D α (ζi (vi − u))||Lp (A)

10.4 Extensions

241

=

N 

||D α vi − D α u||Lp (Vi )

i=0

≤

N 

||vi − u||W m,p (Vi )

i=0

≤ δ(N + 1) for 0 ≤ |α| ≤ m.

10.4 Extensions Theorem 10.8 (Extension Theorem) Let A ⊂⊂ V be bounded subsets of Rn and assume the boundary ∂A is C 1 . There exists a linear operator E : W 1,p (A) → W 1,p (Rn ) such that 1. Eu = u for any u ∈ W 1,p (A), 2. suppE ⊂ V , 3. ||Eu||W 1,p (Rn ) ≤ C||u||W 1,p (A) . Proof Fix x 0 ∈ ∂A. Case 1. u ∈ C ∞ (A). First of all, we suppose that locally, around x 0 , the boundary ∂A belongs to {xn = 0}. Since ∂A is C 1 , there is a ball U such that U + = U ∩ {xn ≥ 0} ⊂ A, U − = U ∩ {xn ≤ 0} ⊂ Rn \A. Now, define  u(x) =

u(x)

for

x ∈ U +,

−3u(x1 , x2 , . . . , xn−1 , −xn ) + 4u(x1 , x2 , . . . , xn−1 , −

xn ) 2

for

x ∈ U −.

We will show that u ∈ C 1 (U ). In fact, let us define u+ = u|xn ≥0,x∈U + ,

u− = u|xn ≤0,x∈U − .

We have u+ |x

n =0

u− |x

n =0

= u(x1 , x2 , . . . , xn−1 , 0), = −3u(x1 , x2 , . . . , xn−1 , 0) + 4u(x1 , x2 , . . . , xn−1 , 0) = u(x1 , x2 , . . . , xn−1 , 0).

242

10 Sobolev Spaces

Consequently − u+ |x =0 = u|x =0 , n

n

and then ∂u+ ∂u− = , ∂xi |xn =0 ∂xi |xn =0

i = 1, 2, . . . , n − 1.

On the other hand, ∂u+ ∂u = , ∂xn |xn =0 ∂xn |xn =0 ∂u− ∂u ∂u ∂u =3 −2 = , ∂xn |xn =0 ∂xn |xn =0 ∂xn |xn =0 ∂xn |xn =0 so overall, ∂u+ ∂u− = ∂xn |xn =0 ∂xn |xn =0 α + 1 and D α u− |xn =0 = D u|xn =0 is well defined for 0 ≤ |α| ≤ 1. Hence, u ∈ C (U ). Additionally,

||u||W 1,p (U ) ≤ C||u||W 1,p (U + ) ≤ C||u||W 1,p (A) . If ∂A does not belong on such hyperplane locally (in a neighbourhood of x 0 ), since the boundary is C 1 there exists a function Φ with inverse Ψ , say x = Φ(y), y = Ψ (x), that maps a neighbourhood of x 0 to a neighbourhood of Ψ (x 0 ) in such a way that, locally, Ψ (∂A) lies on {yn = 0}. Let u1 (y) = u(x) = u(Φ(y)). As above, we construct a function u1 such that ||u1 ||W 1,p (U ) ≤ C||u1 ||W 1,p (U + ) ≤ C||u1 ||W 1,p (A) , and u1 = u1 in U + . If W = Ψ (U ), then ||u||W 1,p (W ) ≤ C||u||W 1,p (A) . Now, we define the operator Eu = u. Since u is bounded in W 1,p (A), the map u → Eu is linear and bounded.

10.4 Extensions

243

0 Because ∂A is compact, there exist finitely many points x10 , x20 , . . . , xN , open sets Wi , extensions ui of u on Wi such that if we take W0 ⊂⊂ A, then we have

A ⊂ ∪N i=0 Wi .

∂A ⊂ ∪N i=1 Wi ,

Let {ζi }N i=0 be the partition of unity corresponding to the system W0 , W1 , . . . , WN . Now, we put u0 = u. Then u=

N 

ζ i ui

i=0

and N     ||u||W 1,p (Rn ) =  ζ i u i  i=0

≤

N 

W 1,p (Rn )

||ζi ui ||W 1,p (Rn )

i=0

=

N 

||ζi ui ||W 1,p (Wi )

i=0

=

N 

||ui ||W 1,p (Wi )

i=0

≤

N 

C||u||W 1,p (A)

i=0

= C(N + 1)||u||W 1,p (A) . ∞ Case 2. Let u ∈ W 1,p (A). Then there exists a sequence {ul }∞ l=1 in C (A) ∩ 1,p 1,p W (A) such that ul → u, l → ∞, in W (A). For any ul , l = 1, 2, . . ., we apply case 1. We also have

||Eum − Eul ||W 1,p (Rn ) = ||E(um − ul )||W 1,p (Rn ) ≤ C||um − ul ||W 1,p (A) →l,m→∞ 0.

Consequently {Eum }∞ m=1 is a fundamental sequence in the Banach space 1,p W (A), so it converges to some u ∈ W 1,p (A). But as Eum = u on A, we have u = u on A. This completes the proof. Definition 10.5 We call Eu an extension of u to Rn .

244

10 Sobolev Spaces

Exercise 10.10 Let A ⊂⊂ V ⊂ Rn be bounded sets with ∂A of class C 2 . Then there is a linear operator E : W 2,p (A) → W 2,p (Rn ) such that 1. Eu = u for u ∈ W 2,p (A), 2. suppE ⊂ V , 3. ||Eu||W 2,p (Rn ) ≤ C||u||W 2,p (A) . Exercise 10.11 Extend u ∈ W 1,p ((0, ∞)) onto (−∞, 0) by setting u(x) = u(−x). Prove that this extension u is an element of W 1,p (R).

10.5 Traces Theorem 10.9 (Trace Theorem) Let A be a bounded set in Rn with C 1 boundary ∂A. There exists a linear bounded operator T : W 1,p (A) → W 1,p (∂A) such that T u = u|∂A

if u ∈ W 1,p (A) ∩ C (A)

and ||T u||Lp (∂U ) ≤ C||u||W 1,p (A). Proof We assume u ∈ C 1 (A) and take x 0 ∈ ∂A. We also suppose intersected with some neighbourhood of x 0 lies on the plane {xn = r r > 0 be such that A ∩ U (x 0 , r) ⊂ {xn = 0}. We consider U (x 0 , ) 2 r Γ = ∂(A ∩ U (x 0 , )). We choose ζ ∈ C0∞ (U (x 0 , r)) so that 0 ≤ ζ 2 r 0 U (x , r), ζ ≡ 1 on U (x 0 , ). 2 By denoting

that ∂A 0}. Let and call ≤ 1 on

x  = (x1 , x2 , . . . , xn−1 ) ∈ Rn−1 = {xn = 0}, we have 





|u| dx ≤ p



Γ

=− U+

 ζ |u| dx = − p

{xn =0}



ζxn |u|p dx − p U+

(ζ |u|p )xn dx

U+

ζ |u|p−1 (signu)uxn dx.

(10.5)

10.5 Traces

245

Young’s inequality, with

1 1 + = 1, gives p q

|u|p−1 |uxn | ≤

|uxn |p (|u|p−1 )q + ≤ C(|u|p + |Du|p ). q p

From this and (10.5), we deduce   |u|p dx ≤ C (|u|p + |Du|p )dx, Γ

U+

so ||u||Lp (Γ ) ≤ C||u||W 1,p (U + ) ≤ C||u||W 1,p (A) .

(10.6)

If we cannot find a neighbourhood of x 0 the restriction of ∂A to which belongs in {xn = 0}, there exist a C 1 map Φ, with inverse Ψ , mapping a neighbourhood of x 0 to a neighbourhood of y 0 = Ψ (x 0 ) so that, locally, Ψ (∂A) lies in {yn = Ψ (xn ) = 0}. 0 Since ∂A is compact, there exists a finite number of points x10 , x20 , . . . , xN and 0 N balls U (xi , ri ) = Vi such that ∂A ⊂ ∪i=1 Vi , V0 ⊂⊂ A and A ⊂ ∪N i=0 Vi . From (10.6), ||u||Lp (Γi ) ≤ C||u||W 1,p (U ) . N Let {ζi }N i=0 be the partition of unity of the system {Vi }i=0 . Then N     ||u||Lp (∂A) =  ζ i u 

≤ =

i=0 N 

Lp (∂A)

||ζi u||Lp (∂A)

i=0 N 

||u||Lp (Γi )

i=0

≤ C||u||W 1,p (A) . Now, define the operator T : W 1,p (A) → W 1,p (∂A) by T u = u|∂A .

(10.7)

246

10 Sobolev Spaces

Using (10.7), we see that ||T u||Lp (∂A) ≤ C||u||W 1,p (A) . ∞ Let u ∈ W 1,p (A) ∩ C (A). There exists a sequence {um }∞ m=1 in C (A) such that um → u in W 1,p (A), as m → ∞. From (10.7), we have

||T um − T ul ||Lp (∂A) ≤ C||um − ul ||W 1,p (A) →m,l→∞ 0. p Therefore the sequence {T um }∞ m=1 is fundamental in the Banach space L (∂A), and p as such it converges in L (∂A):

lim T um = T u.

m→∞

As T um = u|∂A , we infer that T u = u|∂A and T is a bounded operator. Definition 10.6 We call T u the trace of u on ∂U . Exercise 10.12 Let A be a bounded set in Rn and assume ∂A is C 1 . Prove that T u 1,p vanishes on ∂A, provided u ∈ W0 (A). ∞ Hint Use the fact that there exists a sequence {um }∞ m=1 in C0 (A) such that um → 1,p u in W (A), as m → ∞. Since T um is zero on ∂A, we also have T u = 0 on ∂A.

Exercise 10.13 Let A be a bounded set in Rn and let ∂A be C 1 . Take u ∈ W 1,p (A) with T u = 0 on ∂A. Prove that 





|u(x , xn )| dx ≤ p

p−1 Cxn

Rn−1

xn 

|Du|p dx  dt

0 Rn−1

for a.e. xn > 0. Here x  = (x1 , x2 , . . . , xn−1 ). 1 Hint Use the extension theorem, then choose {um }∞ m=1 in C (R + ) such that um → n u in W 1,p (R + ), as m → ∞. Then use the identity n





xn

um (x , xn ) − um (x , 0) =

umxn (x  , s)ds.

0 n

Here R + is the closure of {x = (x1 , x2 , . . . , xn ) ∈ Rn : xn > 0}. Exercise 10.14 Let A be a bounded set in Rn with ∂A of class C 1 . Take u ∈ 1,p W 1,p (A) with T u = 0 on ∂A. Prove that u ∈ W0 (A).

10.6 Sobolev Inequalities

247

Hint Use the extension theorem. Consider the function ζ ∈ C ∞ (R) such that ζ ≡ 1 on [0, 1], 0 ≤ ζ ≤ 1 on R, ζ ≡ 0 on R\[0, 2], and the two sequences ζm (x) = ζ (mxn ), wm (x) = u(x)(1 − ζm (x)), x ∈ Rn+ . Prove that wm →m→∞ u in W 1,p (Rn+ ). Mollify wm to produce functions um ∈ C0∞ (Rn+ ) such that um →m→∞ u in W 1,p (Rn+ ).

10.6 Sobolev Inequalities Definition 10.7 Let 1 ≤ p < n. The Sobolev conjugate p∗ of p is defined by 1 1 1 = − . ∗ p p n Theorem 10.10 (Gagliardo-Nirenberg-Sobolev Inequality) Let u ∈ C01 (Rn ). Then ||u||Lp∗ (Rn ) ≤ C||Du||Lp (Rn ) for some constant C > 0, 1 ≤ p < n. Proof Case 1.

Let p = 1. Then p∗ = ||u||

n and we have to prove that n−1 n

L n−1 (Rn )

≤ C||Du||L1 (Rn ) .

Our first observation is  xi    |u(x)| =  uxi (x1 , . . . , xi−1 , yi , xi+1 , . . . , xn )dyi  −∞

xi ≤

|uxi (x1 , . . . , xi−1 , yi , xi+1 , . . . , xn )|dyi −∞ ∞

≤

|Du(yi )|dyi ,

i = 1, 2, . . . , n.

−∞

Then |u(x)|

1 n−1

≤

 ∞ −∞

|Du(yi )|dyi



1 n−1

,

i = 1, 2, . . . , n,

248

10 Sobolev Spaces

i.e., 1

|u(x)| n−1 ≤

 ∞

|Du(y1 )|dy1



1 n−1

,

−∞

|u(x)|

1 n−1

≤

 ∞

|Du(y2 )|dy2



1 n−1

,

−∞

... |u(x)|

1 n−1

≤

 ∞

|Du(yn )|dyn



1 n−1

.

−∞

We multiply the above inequalities and get |u(x)|

n n−1

≤

n  ∞ 

|Du(yi )|dyi



1 n−1

.

i=1 −∞

Now, we integrate in the variable x1 and obtain ∞ |u(x)|

n n−1

dx1 ≤

≤

|Du(yi )|dyi

−∞ i=1 −∞ n  ∞ ∞  1 

−∞

 ∞

∞  n  ∞

n−1

|Du(y1 )|dy1



1 n−1

dx1

|Du(yi )|dyi dx1



1 n−1

.

i=2 −∞ −∞

−∞

Integrating now in x2 gives ∞ ∞

n

|u(x)| n−1 dx1 dx2 −∞ −∞ ∞

 ∞

≤

|Du(y1 )|dy1



1 n−1

≤

|Du(y2 )|dy2 dx1

−∞ −∞

×

|Du(yi )|dyi dx1



i=2 −∞ −∞ ∞ ∞

−∞ −∞ ∞ ∞



n  ∞ ∞ 

n  ∞ ∞ ∞  i=3 −∞ −∞ −∞



1 n−1



|Du(y1 )|dy1dx2

−∞ −∞

|Du(yi )|dyi dx1 dx2



1 n−1

.

1 n−1



dx2

1 n−1

10.6 Sobolev Inequalities

249

Iterating,  |u(x)|

n n−1

dx ≤ C



Rn

|Du(x)|dx



n n−1

,

(10.8)

Rn

and hence, ||u||

n

L n−1 (Rn )

≤ C||Du||L1 (Rn ) .

Case 2. Let p > 1. We put v = |u|γ , where γ will be determined subsequently. We apply the inequality (10.8) to v and get 

γn

|u(x)| n−1 dx

 n−1 n

 ≤C

Rn

|u(x)|γ −1 |Du(x)|dx.

Rn

By Hölder’s inequality 

γn

|u(x)| n−1 dx

 n−1 n

≤C

Rn



|u(x)|

(γ −1)p p−1

 p−1 p

dx

||Du||Lp (Rn ) .

(10.9)

Rn

Now, we take γ > 0 such that (γ − 1)p γn = , n−1 p−1 so γ =

p(n − 1) , n−p

γn pn = = p∗ . n−1 n−p

From (10.9), we find 

∗

|u(x)|p dx

Rn

 n−1 n

≤C



∗

|u(x)|p dx

 p−1 p

||Du||Lp (Rn ) ,

Rn

and then ||u||Lp∗ (Rn ) ≤ C||Du||Lp (Rn ) . This completes the proof. Exercise 10.15 Let 1 ≤ p < n and consider A bounded, open in Rn with C 1 boundary. For u ∈ W 1,p (A) prove ||u||Lp∗ (A) ≤ ||u||W 1,p (A) .

250

10 Sobolev Spaces

Hint Apply the Extension theorem so to ensure the existence of the extension u ∈ ∞ n W 1,p (Rn ) of u. Then choose a sequence {um }∞ m=1 in C0 (R ) such that um → u 1,p n in W (R ), as m → ∞. Apply the Gagliardo-Nirenberg-Sobolev inequality to p∗ n um − ul and conclude that {um }∞ m=1 converges to u in L (R ). Eventually, the Gagliardo-Nirenberg-Sobolev inequality on um gives the desired result. 1,p

Exercise 10.16 Let A be a bounded set in Rn with ∂A of class C 1 , u ∈ W0 (A) for 1 ≤ p < n. Prove ||u||Lq (A) ≤ C||Du||Lp (A) for some constant C > 0 and for every q ∈ [1, p∗ ]. Hint Use approximation and the Gagliardo-Nirenberg-Sobolev inequality. Theorem 10.11 (Morrey Inequality) Let n < p ≤ ∞. Then for every u ∈ C 1 (Rn ) there exists a constant C = C(n, p) > 0 such that ||u||C 0,γ (Rn ) ≤ C||u||W 1,p (Rn ) , where γ = 1 −

n . p

Proof Let U (x, r) ⊂ Rn be arbitrary ball, r > 0, s ∈ (0, r]. Fix w ∈ ∂U (0, 1), so   s   s d     u(x + tw)dt  =  Du(x + tw) · wdt  |u(x + sw) − u(x)| =  dt 0

s ≤

|Du(x + tw) · w|dt ≤ 0

0

s

|Du(x + tw)|dt, 0

which we integrate on ∂U (0, 1) and get 



s

|u(x + sw) − u(x)|dS ≤ ∂U (0,1) s

∂U (0,1) 0 s 



=

|Du(x + tw)|dtdS

|Du(x + tw)|dSdt = 0 ∂U (0,1)

(x + tw = y) s  |Du(y)| ≤ 0 ∂U (x,t )

≤ U (x,r)

|Du(x + tw)|

t n−1 dSdt t n−1

0 ∂U (0,1)

1 dSy dt = |x − y|n−1

1 |Du(y)| dy. |x − y|n−1

 |Du(y)| U (x,s)

1 dy |x − y|n−1

10.6 Sobolev Inequalities

251

From here,  s

 |u(x + sw) − u(x)|dS ≤ s

n−1

|Du(y)|

n−1

∂U (0,1)

U (x,r)

1 dy. |x − y|n−1

Therefore r

 s

r |u(x + sw) − u(x)|dSds ≤

n−1

0

 s

0

∂U (0,1)

|Du(y)|

n−1 U (x,r)

1 dyds, |x − y|n−1

so r

 s

0



rn |u(x + sw) − u(x)|dSds ≤ n

n−1 ∂U (0,1)

|Du(y)| U (x,r)

1 dy, |x − y|n−1

and substituting x + sw = y finally r





rn n

|u(y) − u(x)|dSy ds ≤ 0 ∂U (x,s)

|Du(y)| U (x,r)

1 dy |x − y|n−1

and 

1 rn

|u(y) − u(x)|dy ≤ U (x,r)

|Du(y)| U (x,r)



1 We set n r



1 n

1 dy. |x − y|n−1

 (·)dy =

(·)dy. Then there exists a constant C > 0 such that U (x,r)

U (x,r)



 |u(y) − u(x)|dy ≤ C U (x,r)

|Du(y)|

U (x,r)

1 dy. |x − y|n−1

On the other hand, 



|u(x)| =

|u(y)|dy = U (x,1)

 ≤

|u(x) − u(y) + u(y)|dy U (x,1)



|u(x) − u(y)|dy + U (x,1)

|u(y)|dy U (x,1)

(10.10)

252

10 Sobolev Spaces

 ≤

|u(x) − u(y)|dy + C||u||Lp (U (x,1)) 

U (x,1)

≤

|u(x) − u(y)|dy + C||u||Lp (Rn ) (by (10.10)) 

U (x,1)

|Du(y)| dy + C||u||Lp (Rn ) |x − y|n−1 U (x,1)  |Du(y)| ≤C dy + C||u||Lp (Rn ) |x − y|n−1 U (x,1) ≤

(Hölder’s inequality)   1   p p ≤C |Du(y)| dy U (x,1)

≤ C(||u||

U (x,1)

Lp (Rn )

+ ||Du||

Lp (Rn )

 p−1

1 |x − y|

(n−1)p p−1

dy

p

+ C||u||Lp (Rn )

) ≤ C||u||W 1,p (Rn ) .

Therefore sup |u(x)| ≤ C||u||W 1,p (Rn ) .

x∈Rn

Let x, y ∈ Rn be arbitrary points and W = U (x, r) ∩ U (y, r). Then   |u(x) − u(y)| = |u(x) − u(y)|dz = |u(x) − u(z) + u(z) − u(y)|dz W W   |u(x) − u(z)|dz + |u(z) − u(y)|dz. ≤ W

W

(10.11) The inequality (10.10) allows us to estimate 





|u(x) − u(z)|dz ≤

|u(x) − u(z)|dz ≤ C

W

≤C

U (x,r)

  U (x,r)

|Du(z)| dz p

1   p

U (x,r)

U (x,r)

 p−1

1 |x − z|

|Du(z)| dz(Hölder’s inequality) |x − z|n−1

(n−1)p p−1

dz

p

 p−1  p n− (n−1)p 1− n p−1 ≤C r ||Du||Lp (U (x,r)) ≤ C|x − y| p ||Du||Lp (Rn ) = C|x − y|γ ||Du||Lp (Rn ) ≤ C|x − y|γ ||Du||W 1,p (Rn ) .

(10.12)

10.6 Sobolev Inequalities

253

As above, we deduce  W

|u(y) − u(z)|dz ≤ C|x − y|γ ||Du||W 1,p (Rn ) .

From the latter, (10.11) and (10.12), we get |u(x) − u(y)| ≤ C|x − y|γ ||u||W 1,p (Rn ) , so sup

x =y,x,y∈Rn

|u(x) − u(y)| ≤ C||u||W 1,p (Rn ) |x − y|γ

or ||u||C 0,γ (Rn ) ≤ C||u||W 1,p (Rn ) . This completes the proof. To prove the Poincaré inequality we have a need of the following RellichKondrachov compactness theorem. For its proof we refer the reader to [6]. Theorem 10.12 (Rellich-Kondrachov Compactness Theorem) bounded open set in Rn with C 1 boundary and 1 ≤ p < n. Then

Let A be a

W 1,p (A) → Lq (A) for every 1 ≤ q < p∗ , where p∗ is the Sobolev conjugate of p. Exercise 10.17 Let A be a bounded open set in Rn with C 1 boundary. Prove that W 1,p (A) → Lp (A), 1 ≤ p ≤ ∞. Definition 10.8 One calls (u)A =

1 |A|

 u(y)dy A

the average of u over A. Theorem 10.13 (Poincaré Inequality) Let A be a bounded, connected, open set in Rn , 1 ≤ p ≤ ∞. Then for every u ∈ W 1,p (A) there exists a constant C = C(u, p, A) such that ||u − (u)A ||Lp (A) ≤ C||Du||Lp (A) .

254

10 Sobolev Spaces

Proof Let us suppose that there exists a function uk ∈ W 1,p (A) such that ||uk − (uk )A ||Lp (A) > k||Duk ||Lp (A) .

(10.13)

Let vk =

uk − (uk )A . ||uk − (uk )A ||Lp (A)

Then 1 (vk )A = uk − (uk )A Lp (A)

 (uk (y) − (uk )A )dy A

 1 = uk − (uk )A Lp (A)

=





uk (y)dy − (uk )A A

dy A

1 ((uk )A − (uk )A ) = 0 uk − (uk )A Lp (A)

and   ||vk ||Lp (A) = 

 uk − (uk )A ||uk − (uk )A ||Lp (A)  = = 1.  ||uk − (uk )A ||Lp (A) Lp (A) ||uk − (uk )A ||Lp (A)

From (10.13), we infer ||Duk ||Lp (A) 1 > k ||uk − (uk )A ||Lp (A) ||D(uk − (uk )A )||Lp (A) = ||u − (uk )A ||Lp (A)   k u − (uk )A k   = D  ||uk − (uk )A ||Lp (A) Lp (A) = ||Dvk ||Lp (A) .

(10.14)

p 1,p (A). By the Consequently {vk }∞ k=1 is a bounded sequence in L (A) and W 1,p p Rellich-Kondrachov compactness theorem W (A) → L (A). Therefore there p is a subsequence {vkj }∞ j =1 that converges to some v ∈ L (A). From (10.14), we have

lim ||Dvkj ||Lp (A) = 0.

j →∞

Fix an arbitrary φ ∈ C0∞ (A). Then 

 vφxi dx = lim

j →∞

A

 vkj φxi dx = − lim

vkj xi φdx = 0

j →∞

A

A

10.7 The Space H −s

255

and Dv = 0. Consequently v ∈ W 1,p (A). Since A is connected and Dv = 0, it follows that v = const. Let v = l. Because vk Lp (A) = 1, we have ||v||Lp (A) = 1. From (vk )A = 0, we conclude that (v)A = 0, so l = 0, ||v||Lp (A) = 0, which is a contradiction. Exercise 10.18 (Poincaré Inequality on the Ball) Given 1 ≤ p ≤ ∞, there exists a constant C = C(u, p, A) such that ||u − (u)U (x,r)||Lp (U (x,r)) ≤ Cr||Du||Lp (U (x,r)). Hint Consider the function v(y) = u(x + ry), y ∈ U1 . Use the Poincaré inequality z−x . for U1 and change variables y = r

10.7 The Space H −s Definition 10.9 We denote by H −s (Rn ), for any 0 < s < ∞, the dual space to H0s (Rn ). In other words, f ∈ H −s (Rn ) is a bounded linear functional on H0s (Rn ). We also set H −∞ (Rn ) = ∪s∈R H s (Rn ). Theorem 10.14 (Characterization of H −s ) For any given 0 < s < ∞, any element u ∈ S  (Rn ) ∩ H −s (Rn ) can be written as u=



D α hα ,

with

hα ∈ L2 (Rn ).

|α|≤s

Proof We have that f = (1 + |ξ |2 )− 2 F (u) ∈ L2 (Rn ). s

Then n  (1 + |ξ |2 ) 2s f   s F (u) = (1 + |ξ |2 ) 2 f = 1 + |ξj |s n j =1 1+ |ξj |s



= 1+

n  j =1

n   |ξ |s   j |ξj |s g = g + ξjs g , ξjs j =1

j =1

256

10 Sobolev Spaces

where s

g=

(1 + |ξ |2 ) 2 f 1+

n j =1

∈ L2 (Rn ).

|ξj |s

We set gj =

|ξj |s g ∈ L2 (Rn ), (−iξj )s

h = F −1 (g),

hj = F −1 (gj ),

j = 1, 2, . . . , n.

Then h, hj ∈ L2 (Rn ), j = 1, 2, . . . , n, so F (u) = F (h) +

n 

(−iξj ) F (hj ) = F (h) + s

j =1

n 

F (Djs hj ),

j =1

and therefore u = h+

n 

Djs hj .

j =1

This completes the proof. Exercise 10.19 Prove that H −∞ (Rn ) ⊂ S  (Rn ).

10.8 Advanced Practical Problems Problem 10.1 Let k > 0 be given, and set U = {(x, y) ∈ R2 : 0 < x < 1, x k < y < 2x k }, u(x, y) = y α , (x, y) ∈ U . Find conditions on α so that u ∈ H m (U ), m = 1, 2, . . .. Problem 10.2 Let u(x) = |x|−α , x ∈ U1 , x = 0. Find conditions on α > 0, n and p so that u ∈ W 1,p (U1 ). Problem 10.3 Find conditions for s so that e−|x| ∈ H s (Rn ). Problem 10.4 Find conditions for s so that δ ∈ H s (Rn ).  1  belongs to W 1,p (U1 ). Problem 10.5 Let n > 1. Prove that u = log log 1 + |x|

10.8 Advanced Practical Problems

257

Problem 10.6 Prove the following interpolation inequality  |Du|2 dx ≤ A



|u|2 dx

 1  2

A

|D 2 u|2 dx

1 2

,

A

for every u ∈ C0∞ (A), A ⊂ Rn open and bounded. Using approximation, prove it for u ∈ H 1 (A) ∩ H01 (A). Problem 10.7 Prove the interpolation inequality  |Du|p dx ≤ A



|u|p dx

 1  2

A

|D 2 u|p dx

1 2

A 1,p

for 2 ≤ p < ∞ and every u ∈ W 2,p (A) ∩ W0 (A), where A ⊂ Rn is open and bounded. Problem 10.8 Prove that any u ∈ W 1,p ((a, b)) can be extended to W 1,p (R). Problem 10.9 Let A be an open bounded set in Rn with C 1 boundary, take u ∈ 1 m n 1 W m,p (A), m < . Prove that u ∈ Lq (A), where = − , and p q p n ||u||Lq (A) ≤ C||u||W m,p (A) . Hint Use the fact that D α u ∈ Lp (A) for any u ∈ W m,p (A) and every α such that |α| ≤ m. Applying the Gagliardo-Nirenberg-Sobolev inequality deduce ||D β u||Lp∗ (A) ≤ C||D α u||Lp (A) ≤ C||u||W m,p (A) 1 1 1 − . Conclude = p∗ p n ∗ u ∈ W m−1,p (A). Then using again the Gagliardo-Nirenberg-Sobolev inequality, prove that

for some constant C > 0, |α| = m and |β| = m − 1,

||D γ u||Lp∗∗ (A) ≤ C||D β u||Lp∗ (A) ≤ C||u||W m,p (A) . 1 2 1 1 1 = − , and so forth. = ∗ − ∗∗ p p n p n Eventually conclude that u ∈ W 0,q (A) and that the given inequality holds. ∗∗

Deduce that u ∈ W m−2,p , where

Problem 10.10 Let A be an open bounded set in Rn with C 1 boundary. Take u ∈ n W m,p (A), m > and prove p u∈C

*n+

m−

p

−1,γ

(A),

258

10 Sobolev Spaces

where ⎧* n + n ⎨ +1− , p p γ = ⎩any positive number < 1,

n is not an integer p otherwise.

if

Also show that u satisfies the inequality ||u||

* +

C

n −1,γ m− p (A)

≤ C||u||W m,p (A)

for some positive constant C. Hint Use the Extension theorem, Morrey’s inequality and the GagliardoNirenberg-Sobolev inequality. Problem 10.11 Prove H +∞ = ∩s∈R H s (Rn ). Problem 10.12 Show H +∞ (Rn ) ⊂ C ∞ (Rn ).

10.9 Notes and References In this chapter we introduce Sobolev spaces and we deduct some of their basic properties. They are investigated approximations by smooth functions and it is proved the main extension theorem. It is defined a trace of an element of a Sobolev space and it is proved the trace theorem. In the chapter are deducted the GagliardoNirenberg-Sobolev inequality, the Morrey inequality and the Poincaré inequality. They are introduced the dual spaces H −s of the spaces H s . Additional materials can be found in [1, 2, 6, 29, 30].

References

1. 2. 3. 4.

R. Adams, Sobolev Spaces (Academic Press, New York, 1975) R. Adams, J.J.F. Fournier, Sobolev Spaces, 2nd edn. (Elsevier, Oxford, 2003) N. Bourbaki, Elements of Mathematics. General Topology. Part 1 (Hermann, Paris, 1966) R. Courant, D. Hilbert, Methods of Mathematical Physics, vol. I (Interscience, New York, 1953) 5. R. Courant, D. Hilbert, Methods of Mathematical Physics. Partial Differential Equations, vol. II (Interscience, New York, 1962) 6. L. Evans, Partial Differential Equations, vol. 19 (AMS, Providence, 1997) 7. G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, New York, 1998) 8. B. Fuchssteiner, D. Laugwitz, Funktional Analysis (BI Wissenschaftsverlag, Zürich, 1974) 9. G.B. Folland, Introduction to Partial Differential Equations, 2nd edn. (Princeton University Press, Princeton, 1995) 10. G.B. Folland, Real Analysis (Wiley, New York, 1999) 11. O. Forster, Analysis 1, 8 Auflage. (Vieweg Verlag, Wiesbaden, 2006) 12. O. Forster, Analysis 2, 6 Auflage. (Vieweg Verlag, Wiesbaden, 2005) 13. O. Forster, Analysis 3, 3 Auflage. (Vieweg Verlag, Wiesbaden, 1984) 14. D. Gillbarg, N. Trudinger, Elliptic Partial Differential Equations of Second Order, 2nd edn. (Springer, Berlin, 1987) 15. M. Grosser, M. Kunzinger, M. Oberguggenberger, R. Steinbauer, Geometric Theory of Generalized Functions (Kluwer, Dordrecht, 2001) 16. D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Math. Society, Zürich, 2008) 17. L. Hörmander, The Analysis of Linear Partial Differential Operators, vol. I, 2nd edn. (Springer, Berlin, 1990) 18. G. Hörmander, Analysis (Lecture notes, University of Vienna). Available electronically at http://www.mat.univie.ac.at/gue/material.html, 2008–09 19. R.V. Kadison, J.R. Ringrose, Fundamentals of the Theory of Operator Algebras. Advanced Theory, vol. II (Academic, New York, 1986) 20. M. Kunzinger, Distributionen theorie II (Lecture notes, University of Vienna, spring term 1998). Available from the author, 1998 21. E.H. Lieb, M. Loss, Analysis. Graduate Studies in Mathematics, vol. 14, 2nd edn. (American Mathematical Society, Providence, 2001) 22. F. Riesz, B.Sz. Nagy, Vorlesungen über funktional analysis (VEB Deutscher Verlag der Wissenschaften, Berlin, 1982)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2

259

260

References

23. X. Saint Raymond, Elementary Introduction to the Theory of Pseudodifferential Operators (CRC Press, Boca Raton, 1991) 24. H.H. Schaefer, Topological Vector Spaces (Springer, New York, 1966) 25. W. Schempp, B. Dreseler, Einführung in die harmonische analyse (B. G. Teubner, Stuttgart, 1980) 26. L.A. Steen, J.A. Seebach, Counterexamples in Topology (Dover, Mineola, 1995). Reprint of the second edition (1978) 27. E. Stein, R. Shakarchi, Complex Analysis. Princeton Lectures in Analysis II (Princeton University Press, Princeton, 2003) 28. R. Steinbauer, Locally convex vector spaces (Lecture notes, University of Vienna, fall term 2008). Available from the author, 2009 29. L. Tartar, An Introduction to Sobolev Spaces and Interpolation Spaces (Springer, Berlin, 2007) 30. R. Temam, Navier-Stokes Equations (North Holland, Amsterdam, 1977) 31. H. Triebel, Theory of Function Spaces. Monogr. Math., vol. 78 (Birkhäuser, Basel, 1983) 32. H. Triebel, Theory of Function Spaces. Monogr. Math., vol. 84 (Birkhäuser II, Basel, 1992) 33. H. Triebel, Theory of Function Spaces. Monogr. Math., vol. 91 (Birkhäuser III, Basel, 2006) 34. V. Vladimirov, Elements of Mathematical Physics (Dekker, New York, 1971) 35. V. Vladimirov, Methods of the Theory of Generalized Functions (Dekker, New York, 1984) 36. D. Werner, Funktional Analysis, fünfte Auflage. (Springer, Berlin, 2005)

Index

Symbols (u, φ), 77 C-like surface, 56 D |prime (X), 77 H −s , 255 Ia , 143 ΘM , 6 C0∞ (X), 2 D  (A), 93 D  (X, A), 93 S (Rn ), 4 Sp (Rn ), 7 p-locally integrable function, 13 u(φ), 77 a , 143 x+ −k x+ , 146 Lp Dominated Convergence Theorem, 26 Lp (E) Spaces, 12 L∞ (E), 13 L∞ (X, μ), 45

A Acute cone, 55

B Bounded linear functional, 29

C Cap-shaped function, 2 Cauchy–Schwartz inequality, 15 Change of Variables, 86 Characterization of H −s , 255

Compactly embedded normed space, 7 Complex conjugate of distribution, 79 Cone, 55 Conjugate cone, 55 Conjugate function, 40 Continuously embedded normed space, 7 Convergence in C ∞ (X), 2 Convergence in C0∞ (X), 2 Convergence in D  (Γ +), 170 Convergence in S (Rn ), 5 Convergence in W m,p , 233 Convergent series of distributions, 91 Convex hull, 2 Convex set, 1 Convolution of distributions, 165 Convolution of locally integrable functions, 48

D Derivative of distribution, 113 Dirac delta function, 78 Direct product of distributions, 155 Distribution, 77 Distribution of finite order, 84 Divergent series of distributions, 91 Double layer of surface, 130

E Equal distributions, 78 Equivalent functions, 12, 44 Essentially bounded function, 13, 45 Essential upper bound, 13, 45 Expansion of the unit function, 3

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Theory of Distributions, https://doi.org/10.1007/978-3-030-81265-2

261

262 Extension, 243 Extension theorem, 241 F Fourier transform in S  (Rn ), 206 Fourier transform in S (Rn ), 205 Fractional differentiation, 174 Fundamental solution, 219 G Gagliardo-Nirenberg-Sobolev inequality, 247 Generalized function, 77 H Heaviside function, 80 Hölder’s inequality, 14 Homogeneous distribution, 143 I Imaginary part of distribution, 79 Indicator, 56 Infinite series of distributions, 91 Interpolation inequality, 17 L Laplace transform of distribution, 211 Leibnitz formula, 234 Linear combination of distributions, 79 Locally integrable function, 13 Locally p-integrable function, 13 M Malgrange-Ehrenpreis theorem, 220 Mean function of locally integrable function, 51 Measure, 100 Minkowski inequality, 17 Morrey inequality, 250 Multiplier, 6

Index Poincaré inequality, 253 Poincaré inequality on the ball, 255 Primitive of distribution, 125 Product of a distribution by a C ∞ function, 102 R Rapidly Cauchy sequence, 20 Real distribution, 79 Real part of distribution, 79 Regular distribution, 99 Regularization of distribution, 171 Regularization of locally integrable function, 51 Rellich-Kondrachov compactness theorem, 253 Riesz representation theorem, 36 Riesz representation theorem for the dual of Lp (X, μ), 47 S Sequence of distributions, 88 Simple layer on a surface, 127 Singular distribution, 99 Singular support of a distribution, 98 Sobolev space, 231 Sochozki formulas, 106, 107 Space of the basic functions, 2 Spectral function, 213 Support of distribution, 92 T Tempered distribution, 195 Tempered measure, 200 Trace, 246 Trace theorem, 244 V Vitalli Lp convergence theorem, 47

N Nonnegative distribution, 101 Norm of a bounded linear functional, 29

W Weakly bounded set, 195 Weakly convergent sequence, 38 Weak sequential limit, 40

O Order of distribution, 84

Y Young convolution inequality, 49

P Paley-Wiener-Schwartz theorem, 220

Z Zero set of a distribution, 92