Function Spaces and Operators between them [1 ed.] 9783031416019, 9783031416026

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RSME Springer Series 11

José Bonet David Jornet Pablo Sevilla-Peris

Function Spaces and Operators between them

RSME Springer Series Volume 11

Editor-in-Chief Maria A. Hernández Cifre, Departamento de Matemáticas, Universidad de Murcia, Murcia, Spain Series Editors Nicolas Andruskiewitsch, FaMAF - CIEM (CONICET), Universidad Nacional de Córdoba, Córdoba, Argentina Francisco Marcellán, Departamento de Matemáticas, Universidad Carlos III de Madrid, Leganés, Madrid, Spain Pablo Mira, Departamento de Matemática Aplicada y Estadística, Universidad Politécnica de Cartagena, Cartagena, Spain Timothy G. Myers, Centre de Recerca Matemàtica, Barcelona, Spain Joaquín Pérez, Departamento de Geometría y Topología, Universidad de Granada, Granada, Spain Marta Sanz-Solé, Department of Mathematics and Computer Science, Barcelona Graduate School of Mathematics (BGSMath), Universitat de Barcelona, Barcelona, Spain Karl Schwede, Department of Mathematics, University of Utah, Salt Lake City, UT, USA

As of 2015, RSME - Real Sociedad Matemática Española - and Springer cooperate in order to publish works by authors and volume editors under the auspices of a co-branded series of publications including advanced textbooks, Lecture Notes, collections of surveys resulting from international workshops and Summer Schools, SpringerBriefs, monographs as well as contributed volumes and conference proceedings. The works in the series are written in English only, aiming to offer high level research results in the fields of pure and applied mathematics to a global readership of students, researchers, professionals, and policymakers.

José Bonet • David Jornet • Pablo Sevilla-Peris

Function Spaces and Operators between them

José Bonet Instituto Universitario de Matemática Pura y Aplicada Universitat Politècnica de València Valencia, Spain

David Jornet Instituto Universitario de Matemática Pura y Aplicada Universitat Politècnica de València Valencia, Spain

Pablo Sevilla-Peris Instituto Universitario de Matemática Pura y Aplicada Universitat Politècnica de València Valencia, Spain

ISSN 2509-8888 ISSN 2509-8896 (electronic) RSME Springer Series ISBN 978-3-031-41601-9 ISBN 978-3-031-41602-6 (eBook) https://doi.org/10.1007/978-3-031-41602-6 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 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 Paper in this product is recyclable.

Preface

The aim of this lecture notes is to present, in a unified and reasonably selfcontained way, certain aspects of functional analysis which are needed to treat function spaces whose topology is not derived from a single norm, their topological duals and continuous linear operators between those spaces. We treat spaces of continuous, analytic, and smooth functions as well as sequence spaces. Operators of differentiation, integration, composition, and multiplication as well as partial differential operators between those spaces are studied. A brief introduction to Laurent Schwartz’s theory of distributions and to Lars Hörmander’s approach to linear partial differential operators is presented. These two topics constituted certainly two of the cornerstones of Mathematics in the twentieth century. Our lecture notes had their origin in a course, entitled Operators between spaces of holomorphic and differentiable functions, which aimed at serving as an introduction to the topics mentioned above and related ones. Functional analysis is the proper language in which to formulate and understand these theories, and it is the guiding theme of the course. It has been taught several times over the last years as a part of the syllabus of the joint Masters Programme at Universitat Politècnica de València and Universitat de València. Having our notes as a starting point, we decided to polish, enrich, and expand them. Some topics have been enlarged; others that were not part of the original course have been added, and they could eventually become the content of different courses, with new points of view. The text is addressed to students at a master level, or even undergraduate at the last semesters, since only knowledge on real and complex analysis is assumed. We have intended to be as self-contained as possible, and wherever an external citation is needed, we try to be as precise as we can. Our aim is to be an introduction to topics in, or connected with, different aspects of functional analysis. Many of them are in some sense classical, but we tried to show a unified direct approach; some others are new. This is why parts of these lectures might be of some interest even for researchers in related areas of functional analysis or operator theory. The first three chapters of the text are devoted to establish the basic concepts and facts needed later. We begin with some elementary facts about pointwise and uniform convergence of sequences of functions. These are probably known to the v

vi

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reader but, since this is basic all along the text, we feel that it is good to dedicate some time to refresh these concepts. As motivating examples, we give a short look at uniform convergence for different situations: power series, Fourier series, and Dirichlet series. In Chaps. 2 and 3, we give an introduction to the basics of functional analysis that we are going to need later. We give some basic properties of Hausdorff locally convex spaces, looking also at properties like metrisability and normability. As first examples, we give the space of continuous functions endowed with the topology of uniform convergence on the compact subsets and Köthe echelon spaces. We then move to the study of duality theory and continuous linear operators. We define the dual of a locally convex space and give an introduction to the weak topology defined by a dual pair. We prove the fundamental theorems of the theory: Hahn-Banach, Alaouglu-Bourbaki, bipolar theorem, Mackey-Arens, and BanachSteinhaus. We also make a short introduction to the general theory of continuous linear operators (Banach-Schauder theorem, transpose, . . . ). All the notions and results in this part are absolutely classical, and can also be found in other references. We believe, however, that our approach and point of view is direct and relatively novel. We have no aim to be comprehensive or to develop the whole theory. There are already a number of excellent monographs that serve this purpose for the interested reader. Our aim here is to give a short, concise introduction to what will be needed later in the text. In each case, we try to get to the point as fast as possible. In the following two chapters, we introduce what are the objects of our main interest: spaces of functions and operators between them. We define and study the spaces of holomorphic complex functions and of infinitely differentiable real functions. We analyse their locally convex structure. For the space of holomorphic functions, we give a representation as a sequence space, and a description of the dual. Immediately after, we introduce the classical operators between spaces of functions which we are interested in: composition, multiplication, derivation, integration, and (mostly) partial differential operators. Once we have done this, we look at some dynamical properties of the operators. We give a brief introduction to the general theory of transitive and mean ergodic operators. Once again, we do not intend to be comprehensive, but to serve as a sort of gateway to this theory. Some readers might find a motivation in our exposition and get interested in the topic. Precise references are given to help with a more complete study of this theory. For some classical operators (multiples of the backward shift, composition, and partial differential operators), we classify when are they transitive/hipercyclic or power bounded/mean ergodic. This aspect is totally new and cannot be found in other monographs. In the last chapter of the text, we deal with the theory of distributions of Schwartz and linear partial differential operators. First, we give an introduction to the theory of distributions. We consider the spaces of test functions and of distributions and define the basic operations there. Then we consider the spaces of rapidly decreasing functions and of tempered distributions, paying special attention to the Fourier transform defined on these spaces. We finish by looking at partial differential operators defined on the spaces that we have been considering all along the text.

Preface

vii

We introduce the notion of fundamental solution of such an operator and prove the Malgrange–Ehrenpreis theorem on the existence of fundamental solutions. All in all the manuscript deals with different topics. Many of them are classical and can be found sparse in different texts. The novelty of our approach lies mainly on two facts. First of all, we show all these topics together in an accessible way, stressing the connection between them. Second, we keep it always at a level that is accessible to beginners and young researchers. A list of exercises with different levels of difficulty is included at the end of each chapter. As we have said, several parts of this text have already been tested in different courses at a masters level, and different combinations of chapters would lead to other possible courses, equally interesting. Our aim is not to build and describe a whole, complete theory, but to serve as an introduction to some aspects that we believe are interesting. We wish to guide any reader that wishes to enter in some of these topics in their first steps. Our hope is that they learn interesting aspects of functional analysis and become interested to broaden their knowledge about function and sequence spaces and operators between them. To this end, the bibliography collects several monographs of different levels, both classical and more modern, where the interested reader will find deep, comprehensive exposés of the topics we deal with here. At the end of each chapter, we give a list of references where the interested reader can delve into the subject. Acknowledgement The first and second authors were supported by the project PID2020-119457GB-100, and the third author by the project PID2021-122126NBC33. Both projects are funded by MCIN/AEI/10.13039/501100011033 and by ‘ERFD A way of making Europe’. València, Spain 18 June 2023

José Bonet David Jornet Pablo Sevilla-Peris

Contents

1

Convergence of Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Preliminaries and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Pointwise and Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Power Series in the Complex Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Dirichlet Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 2 9 10 11 24 34 37

2

Locally Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Topological Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Metric and Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Seminorms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Locally Convex Topology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Metrizable Locally Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Dual of a Locally Convex Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Examples of Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Space of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Köthe Echelon Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Normable Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Two Theorems on Spaces of Continuous Functions . . . . . . . . . . . . . . . . . 2.6.1 Stone–Weierstraß Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.2 Ascoli Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 A Short Introduction to Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 39 39 41 47 48 50 55 57 60 60 62 67 69 69 73 76 86 88

3

Duality and Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Hyperplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

91 91 ix

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3.2

The Hahn–Banach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Analytic Version . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Separation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Finite Dimensional Locally Convex Spaces . . . . . . . . . . . . . . . . . 3.2.4 Banach Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Weak Topologies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Bipolar Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 The Mackey–Arens Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 The Banach–Steinhaus Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 The Banach-Schauder Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Topologies on the Space of Continuous Linear Mappings . . . . . . . . . . 3.9 Transpose of an Operator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93 93 95 99 101 104 110 113 116 122 125 131 133 134

Spaces of Holomorphic and Differentiable Functions and Operators Between Them . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Space of Holomorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Locally Convex Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Representation as a Sequence Space . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Montel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Dual of the Space of Entire Functions. . . . . . . . . . . . . . . . . . . . . . . . 4.2 Spaces of Differentiable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Some Operators on Spaces of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 137 137 142 144 145 150 153 157 158

5

Transitive and Mean Ergodic Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Transitive Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Mean Ergodic Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 The Backward Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Composition Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Multiplication and Integration Operators . . . . . . . . . . . . . . . . . . . . 5.3.4 Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161 162 170 182 182 186 194 196 198 201

6

Schwartz Distributions and Linear Partial Differential Operators . . . . 6.1 Test Functions and Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Differentiation of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Multiplication of a Distribution by a C ∞ -Function . . . . . . . . . 6.1.4 Support of a Distribution and Distributions with Compact Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Space of Rapidly Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . .

203 203 203 212 214

4

217 219

Contents

Fourier Transform on S(RN ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Tempered Distributions and the Fourier Transform . . . . . . . . . . . . . . . . . . Linear Partial Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.1 Fundamental Solutions. The Malgrange–Ehrenpreis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.2 Solutions of Linear PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.3 6.4 6.5

xi

222 230 241 242 250 255 259

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

List of Symbols

·, · (x, y) .|x| . .χA .1 .f ∗ g .u ∗ ψ .A ˚ .A ⊥ .A ◦ .A  .β(E , E) .B(a, r) .B(a, r) .Bd (x, ε) .Bp (x, ε) .BX .C(z0 , r) .C[a, b] .C(K) .C() k .C () ∞ .C () .C(T) .c0 .Cϕ .δ, δa .D(z0 , r) .D .D(K) . .

Bilinear mapping defining a dual pair Scalar product Euclidean norm on .Rd Compact subset Characteristic function of A Constant sequence or function equal to 1 Convolution of functions Convolution of a distribution and a function Closure of A Interior of A Orthogonal of A Polar of A Strong topology on the dual of E Open ball of centre a and radius r in .RN Closed ball of centre a and radius r in .RN Ball of centre x and radius .ε defined by the distance d Ball of centre x and radius .ε defined by the seminrm p Open unit ball of a normed space X Circle of centre .z0 and radius r in .C Space of continuous functions on the interval .[a, b] Space of continuous functions on the compact set K Space of continuous functions on the open set . Space of .C k -functions on the open set . Space of .C ∞ -functions on the open set . Space of .2π -periodic continuous functions .f : R → C Space of null sequences Composition operator defined by .ϕ Dirac ‘delta’ functions Open disc of centre .z0 and radius r in .C Open unit disc of .C Space of .C ∞ -functions whose support is contained in K xiii

xiv

D() D  () .Dn .d(x, A) .en ∗ .E  .E  .E β .(E, F )  .E () .Exp (n) .f (w) .f .F . (A) .G(T ) .H() .H({0}) .H (x) .λp (A) .Im .ker .p .∞ .Lp (μ) .L∞ (μ) .L1,loc () .L(E, F ) .L(E) .Lβ (E, F ) . .

Lco (E, F )

.

Ls (E, F ) Lw (E, F ) .μ(E, F ) . μ .mN .Mg .N0 .Orb(x, T ) .P .pK .P (∂) . .

List of Symbols

Space of test functions on . Space of distributions on . Dirichlet kernel Distance between a point and a set Canonical vectors Algebric dual of E Topological dual of E Strong dual of E Dual pair Space of distributions with compact support on . Space of holomorphic functions of exponential type nth Fourier coefficient of f Fourier transform of a function Fourier transform operator Absolutely convex hull of A Graph of the mapping T Space of holomorphic functions on . Space of germs at 0 Heaviside function Köthe echelon space of order p Image of an operator Kernel of an operator Space of p-summable sequences Space of bounded sequences Space of p-integrable functions Space of esentially bounded functions Space of locally integrable functions on . Space of linear continuous operators from E to F Space of linear continuous operators from E to E Space .L(E, F ) with the topology of uniform convergence on bounded sets Space .L(E, F ) with the topology of uniform convergence on precompact sets Space .L(E, F ) with the strong operator topology Space .L(E, F ) with the weak operator topology Mackey topology on E defined by the dual pair .(E, F ) Fourier-Borel transform of a measure .μ Lebesgue measure on .RN Multiplication operator defined by g Set of non-negative integer numbers Orbit of x with respect to T Family of seminorms defining a topology Seminorm defined as the supremum on the compact set K Linear partial differential operator with constant coefficients

List of Symbols

P V x1 .σ (E, F ) .σ (F, E) s .supp N .S(R )  N .S (R ) .τ |Y .τd .τP .τco .τp .T  ∗ . T t . T n .T .Tn .Tf .T .U0 (E) .ω .

xv

Cauchy principal value of . x1 Weak topology on E defined by the dual pair .(E, F ) Weak topology on F defined by the dual pair .(E, F ) Space of rapidly decreasing sequences Support of a function or a distribution Space of rapidly decreasing functions (Schwartz) Space of tempered distributions Induced topology on Y Topology defined by the distance d Topology defined by the family of seminorms Compact-open topology (of uniform convergence on compact sets) Topology of pointwise convergence Norm of the operator T Transpose of the linear mapping T Transpose of the continuous linear operator T nth iterate of the operator T Cesàro mean Distribution defined by a function One-dimensional torus Set of absolutely convex neighbourhoods of 0 Space of all sequences, with pointwise convergence

Chapter 1

Convergence of Sequences of Functions

1.1 Preliminaries and Notation Throughout the text, .K will denote either the field of real numbers .R or  complex numbers .C. On .Rd for some .d ≥ 1, we consider the inner product .x·y = dn=1 xn yn for .x = (x1 , . . . , xd ) and .y = (y1 , . . . , yd ), and the Euclidean norm .|x| = (x ·x)1/2 . Then the open and closed balls centred at .a ∈ Rd with radius .r > 0 are defined as B(a, r) = {x ∈ Rd : |x − a| < r} and B(a, r) = {x ∈ Rd : |x − a| ≤ r} .

.

Given .A ⊆ Rd , we write .χA for the characteristic function, that is, .χA (x) is 1 if .x ∈ A and 0 otherwise. We identify .R2 with .C. We recall now some basic notations of complex numbers. Given .z ∈ C, we write .z = x + iy with .x = Re(z) (the real part) and .y = Im(z) (the imaginary part). The conjugate is defined as .z = x − iy and the modulus as 2 .|z| = z¯ z = x 2 + y 2 . The (open) disc centred at .z0 with radius .r > 0 is denoted by D(z0 , r) = {z ∈ C : |z − z0 | < r} .

.

Its boundary, the circle, is denoted by C(z0 , r) = {z ∈ C : |z − z0 | = r} ;

.

it will always be assumed to be oriented counterclockwise. We write .D = D(0, 1) and call it the open unit disc of .C. Throughout this text, we will write .N0 for the set of non-negative integers .N ∪ {0}.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_1

1

2

1 Convergence of Sequences of Functions

1.2 Pointwise and Uniform Convergence We are going to consider functions defined on some subset of .Rd and sequences .(fn )n of such functions. As we will see there are several possible ways to think of convergence to a certain f of such a sequence. The most elementary one is to ask that the values at each point converge to the value of f at that point. If . ⊆ Rd , a sequence of functions .fn :  → K (with .n ∈ N) converges pointwise to .f :  → K if .

lim fn (x) = f (x),

n→∞

for every .x ∈ . This means that for every .x ∈  and for every .ε > 0, there exists some .n0 = n0 (x, ε) ∈ N such that for every .n ≥ n0 , |fn (x) − f (x)| < ε .

.

(1.1)

Example 1.1 We take . = [0, 1] and define, for each .n ∈ N, the function .fn : [0, 1] −→ R by .fn (x) = x n . Note that .limn→∞ fn (x) = limn→∞ x n = 0 for all .0 ≤ x < 1 and .limx→∞ fn (1) = 1. This shows that .(fn )n converges pointwise to  f (x) =

.

See Fig. 1.1.

0 if x ∈ [0, 1[ 1 if x = 1.



Example 1.2 Take now . = R and define, for each .n ∈ N, the function .fn : R → R x 2n . It is easy to check that .(fn )n converges pointwise to by .fn (x) = 1 + x 2n ⎧ 1 ⎪ ⎪ ⎨ 2 if x = ±1 .f (x) = 0 if |x| < 1, ⎪ ⎪ ⎩1 if |x| > 1. See Fig. 1.2.



We need notions of convergence that preserve certain properties of the functions, such as continuity (that is, the limit of continuous functions is again continuous). Examples 1.1 and 1.2 show that pointwise convergence is not good enough for what we need. We need to go a little bit further. Recall that for a sequence to converge pointwise we have some .n0 (depending on .ε and x) for which (1.1) is satisfied. What we want now is to have some .n0 that depends only on .ε and works for every x.

1.2 Pointwise and Uniform Convergence

3

1

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

Fig. 1.1 Plot of .fn for .n = 2, 3, 4, 5, 10 1.5 1 0.5 0 -1

0

1

Fig. 1.2 Plot of .fn for .n = 1, 2, 3, 10, 20

Given . ⊆ Rd , a sequence of functions .fn :  → K converges uniformly to f :  → K if for every .ε > 0, there exists .n0 = n0 (ε) ∈ N such that

.

|fn (x) − f (x)| < ε ,

.

for every .x ∈  and every .n ≥ n0 . Remark 1.1 It is straightforward to check that a sequence .(fn )n converges uniformly to f if and only if for every .ε > 0 there exists some .n0 ∈ N such that .

sup |fn (x) − f (x)| < ε , x∈

4

1 Convergence of Sequences of Functions

for every .n ≥ n0 , or in other words, if and only if .



lim

sup |fn (x) − f (x)| = 0 .

(1.2)

n→∞ x∈

Clearly, if .(fn )n converges uniformly to f , then it also converges pointwise to the same function. However, Examples 1.1 and 1.2 show (just by checking (1.2)) that the converse does not hold. Example 1.3 We take the sequence .fn : [0, 1] → R given by .fn (x) = x n (1 − x). A simple computation shows that .

max |fn (x)| =

x∈[0,1]

nn . (n + 1)n+1

Since .

nn = 0, n→∞ (n + 1)n+1 lim

Remark 1.1 implies that .fn → 0 uniformly on .[0, 1] (see Fig. 1.3).



The Cauchy criterion for convergence of sequences of scalars transfers to the uniform convergence of functions.

0.25

0.2

0.15

0.1

0.05

0 0

0.2

Fig. 1.3 Plot of .fn for .n = 1, 2, 3, 5, 10

0.4

0.6

0.8

1

1.2 Pointwise and Uniform Convergence

5

Proposition 1.1 (Cauchy criterion) A sequence of functions .(fn )n defined on . ⊆ Rd converges uniformly if and only if for every .ε > 0 there is .n0 such that if .m, n ≥ n0 , then .

sup |fn (x) − fm (x)| < ε .

(1.3)

x∈

Proof Suppose first that the sequence converges uniformly on . to some f . Then, given .ε > 0, there is .n0 such that .supx∈ |fn (x) − f (x)| < 2ε . This immediately implies (1.3). Let us note that |fn (x0 ) − fm (x0 )| ≤ sup |fn (x) − fm (x)| ,

.

(1.4)

x∈

for every .x0 ∈  and all .n, m. Then, if (1.3) holds, this implies that the sequence (fn (x0 ))n is Cauchy in .K for each fixed .x0 ∈  and, therefore, converges. We define .f :  → K by doing .f (x) = limn fn (x). Taking the limit as .m → ∞ on the lefthand side of (1.4), we obtain .|fn (x) − f (x)| ≤ ε for all .n ≥ n0 and .x ∈ . This shows that .(fn )n converges uniformly to f .

.

As we announced, uniform convergence preserves continuity. Theorem 1.4 Let .(fn )n be a sequence of continuous functions on . ⊆ Rd . If .fn → f uniformly on ., then f is continuous on .. Proof Fix some .x0 ∈  and let us show that .limx→x0 f (x) = f (x0 ). To do that, we fix .ε > 0 and choose .n0 ∈ N such that .|fn (x) − f (x)| < 3ε for every .n ≥ n0 and .x ∈ . Now, since .fn0 is continuous at .x0 , there is some .δ > 0 such that for ε .x ∈  with .|x − x0 | < δ, we have .|fn0 (x) − fn0 (x0 )| < 3 . Then, if .x ∈  and .|x − x0 | < δ, we have |f (x) − f (x0 )| ≤ |f (x) − fn0 (x)| + |fn0 (x) − fn0 (x0 )| + |fn0 (x0 ) − f (x0 )| ≤ ε .

.

Remark 1.2 Suppose that .(fn )n is a sequence of bounded functions on . ⊆ Rd that converges uniformly. Let us see that the limit function f is also bounded. To see this, note in first place (recall (1.3)) that there exists .n0 ∈ N so that .

sup |f (x) − fn0 (x)| < 1 .

x∈

Being .fn0 bounded, we have .supx∈ |fn0 (x)| = K < ∞, and if .x0 ∈ , then |f (x0 )| ≤ |f (x) − fn0 (x)| + |fn0 (x)| < 1 + K .

.

Therefore f is bounded in ..

6

1 Convergence of Sequences of Functions

Proposition 1.2 Let .(fn )n be a sequence of continuous functions defined on . that converges uniformly to f and .(xn )n ⊆  such that .limn xn = x0 . Then .limn fn (xn ) = f (x0 ). Proof We fix .ε > 0 and choose .n1 ∈ N such that .|fn (x) − f (x)| < 2ε for every .x ∈  and .n ≥ n1 . On the other hand, since f is continuous at .x0 (because of Theorem 1.4), we can find .δ > 0 such that if .x ∈  and .|x − x0 | < δ we have ε .|f (x) − f (x0 )| < 2 . Now we take .n2 ∈ N such that .|xn − x0 | < δ for .n ≥ n2 and we have that if .n ≥ n0 = max(n1 , n2 ), then |fn (xn ) − f (x0 )| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x0 )| < ε .

.



This proves the claim. This becomes useful to check that a sequence does not converge uniformly.

Example 1.5 For each .n ∈ N, take the function .fn : [0, 1] → R given by .fn (x) = nx n (1 − x). First of all, it is a simple exercise to check that .limn nx n (1 − x) = 0 for every .x ∈ [0, 1]. Then the sequence .(fn )n converges pointwise to 0. Take now 1 .xn = 1 − , which tends to 1. We have n .

1 n 1 1− = = 0 = f (1) . n→∞ n e

lim fn (xn ) = lim

n→∞

By Proposition 1.2, the sequence .(fn )n does not converge uniformly (see Fig. 1.4).

As we have pointed out, uniform implies pointwise convergence, but the converse is not in general true. The following result shows that if both the set where the functions are defined and the functions themselves are nice enough, then pointwise also implies uniform convergence. Theorem 1.6 (Dini’s Theorem) Let .K ⊆ Rd be closed and bounded. Let .fn : K → R with .n ∈ N be continuous functions such that .fn+1 (x) ≤ fn (x) (or .fn+1 (x) ≥ fn (x)) for every .x ∈ K and .n ∈ N and so that .(fn )n converges pointwise to a continuous function .f : K → R. Then .(fn )n converges to f uniformly on K. Proof We proceed by contradiction. Assume that .fn does not converge uniformly to f . Then, there exist .ε0 > 0 and a sequence .m1 < m2 < · · · < mk < · · · such that for each .k ∈ N there is .xk ∈ K such that |fmk (xk ) − f (xk )| ≥ ε0 .

.

Since .(xk )k ⊆ K is a bounded sequence, by the Bolzano–Weierstraß theorem, there is a convergent subsequence (which for the sake of clarity we denote again by .(xk )k ), converging to some .x0 ∈ K.

1.2 Pointwise and Uniform Convergence

7

0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

0.2

0.4

0.6

0.8

1

Fig. 1.4 Plot of .fn for .n = 1, 2, 3, 5, 10

We fix now .mi and we have, for every .k > i, fmi (xk ) − f (xk ) ≥ fmk (xk ) − f (xk ) = |fmk (xk ) − f (xk )| ≥ ε0 .

.

Taking limit as k tends to .∞ on the left-hand side and using the fact that both .fmi and f are continuous at .x0 , we obtain fmi (x0 ) − f (x0 ) ≥ ε0

.

for every .i ∈ N. This contradicts the fact that .(fn )n converges pointwise to f (and, in particular, .limn fn (x0 ) = f (x0 )).

We have three hypotheses in this theorem: two regarding the set (being closed and bounded) and one on the sequence of functions (being monotone). The following three examples show that each one of them is necessary. Example 1.7 Consider the sequence .fn : [0, 1[→ R given by .fn (x) = x n . Note that it is decreasing for each fixed x and converges pointwise to .f : [0, 1[→ R given by .f (x) = 0. However, the sequence does not converge uniformly in .[0, 1[ to its pointwise limit since .

sup |fn (x) − f (x)| = 1 , x∈[0,1[

for every n. Here, the set, although being bounded, is not closed.



8

1 Convergence of Sequences of Functions

Example 1.8 We see now that the hypothesis of K being bounded is also needed. To do that, we consider the sequence .fn : [0, ∞[→ R (the set is closed but unbounded) defined as ⎧ ⎪ if x ∈ [0, n − 1], ⎪ ⎨0 .fn (x) = x − n + 1 if x ∈ [n − 1, n], ⎪ ⎪ ⎩1 if x ∈ [n, ∞[ . We clearly have .fn+1 (x) ≤ fn (x) for every .x ∈ [0, ∞[ and the sequence converges pointwise to .f : [0, ∞[→ R defined as .f (x) = 0 (i.e., the constant function 0). On the other hand, sup |fn (x) − f (x)| = 1,

.

x∈[0,∞[

for every n, and this shows that the convergence is not uniform (see Fig. 1.5).



Example 1.9 Now, we consider the sequence of continuous functions .fn : [0, 1/2] → R (defined on a closed, bounded set) given by ⎧ n+1 ⎪ if x ∈ [0, 2−(n+1) ], ⎪ ⎨2 x .fn (x) = −2n+1 x + 2 if x ∈]2−(n+1) , 2−n ], ⎪ ⎪ ⎩0 if x ∈ [2−n , 1/2] . This time .(fn )n is neither decreasing nor increasing. Again, the sequence converges pointwise to 0, but since .

sup

|fn (x) − f (x)| ≥ fn (2−(n+1) ) = 1

x∈[0,1/2]

for every n, it does not converge uniformly on .[0, 1/2] (see Fig. 1.6).



1

0 0

1

2

3

4

Fig. 1.5 Plot of .fn for .n = 2. For higher ns, the diagonal slides to the right

5

6

1.3 Series of Functions

9

1.2

1

0.8

0.6

0.4

0.2

0

-0.2 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Fig. 1.6 Plot of .fn for .n = 1, 2, 3

1.3 Series of Functions  Given a sequence of functions .fn :  ⊆ Rd → K, the series of functions . ∞ n=1 fn is (if it exists) the limit of the sequence .(Sn )n of partial sums of the sequence .(fn )n ,  given by .Sn = nk=1 fk . Depending on what kind of convergence we ask for the sequence of partial sums, we have a different convergence for the series. To be more precise, the series . ∞ n=1 fn converges pointwise  on . if .(Sn ) converges pointwise (that is, for each fixed .x ∈ , the scalar series . ∞ n=1 fn (x) converges in .K). The series converges uniformly if the sequence .(Sn ) converges uniformly  on .. Finally, we say that the series converges absolutely on . if the series . ∞ n=1 |fn | converges pointwise in .. The following criterion for uniform convergence of series is easily deduced from Proposition 1.1.  Proposition 1.3 (Cauchy criterion for series) The series . ∞ n=1 fn is uniformly convergent on . if and only if for all .ε > 0 there is .n0 ∈ N such that .

m sup fk (x) < ε

x∈ k=n

for every .m > n ≥ n0 . As a consequence, we obtain the following useful criterion.

10

1 Convergence of Sequences of Functions

Proposition 1.4 (Weierstraß M-test) ∞ Suppose that .(Mn )n is a sequence of nonnegative real numbers such that .  n=1 Mn < ∞. If for all .x ∈  and .n ∈ N we have .|fn (x)| ≤ Mn , then the series . ∞ n=1 fn converges uniformly on .. Proof Given .ε > 0, by Cauchy’s  criterion for scalar series, we can find .n0 ∈ N such that for .m > n ≥ n0 , then . m k=n Mn < ε. Then, for each .x ∈ , m m m

. fk (x) ≤ |fk (x)| ≤ Mn < ε . k=n

k=n

k=n



Then Proposition 1.3 gives the conclusion.

1.3.1 Power Series in the Complex Plane Our first example is a classical object in function theory of one complex variable: power series. We take a sequence .(an )n of complex numbers and consider the series ∞

.

an zn

n=0

 for .z ∈ C. Note that this can be seen as a series of functions . n fn , where .fn (z) = an zn for each n. We wonder now about the convergence of this series and find out that it converges at discs centred at 0. ∞ n Theorem 1.10 Let n )n ⊆ C and .z0 ∈ C such that . n=0 an z0 converges. Then .(a ∞ n the power series . n=0 an z converges absolutely at every .z ∈ C with .|z| < |z0 | and converges uniformly on .D(0, r) for every .0 < r < |z0 |. Proof If .z0 = 0, then the result is trivially satisfied. We assume, then, .z0 = 0. Since the series is convergent, .limn→∞ |an z0n | = 0, and we can find .M > 0 such that n .|an z | ≤ M. If .0 < r < |z0 | and .|z| ≤ r, then 0 |an zn | = |an ||z|n ≤ |an |r n = |an ||z0 |n

.

r n

r n ≤M . |z0 | |z0 |

n  r is a convergent geometric series, we immediately conclude Since . ∞ M n=1 |z0 | ∞ n that . n=1 |an z | < ∞ and that, by Proposition 1.4, the power series converges uniformly on .{z ∈ C : |z| ≤ r}.

If we write R = sup{|z| :

∞

.

n=0

an zn is convergent} ∈ [0, , +∞] ,

1.3 Series of Functions

11

then the power series converges (even absolutely) for every .|z| < R and diverges for every .|z| > R. Even more, the power series converges uniformly on .D(0, r) for every .r < R. The number R is called the radius of convergence of the power series. We have no information does the series do for zs with .|z| = R. For  on1 what n has the radius of convergence .R = 1, but for example, the power series . ∞ z n=1 n .z = 1, the series diverges, and for .z = −1 it converges. The radius of convergence of a power series can be computed using the Hadamard formula 1 . √ lim supn n |an |

R=

.

 n With all this in mind, the function .f : D(0, R) → C given by .f (z) = ∞ n=0 an z is well defined and, as we will see later (in Remark 4.1), holomorphic. We see, then, that the natural domains where power series are defined are open discs. Even more, if we consider the numbers Rc = sup{|z| :

∞

.

an zn is convergent}

n=0

Ru = sup{r :

∞

an zn converges uniformly on D(0, r)}

n=0 ∞

Ra = sup{|z| :

an zn is absolutely convergent},

n=0

then Rc = Ru = Ra .

.

(1.5)

1.3.2 Fourier Series We consider now periodic functions and ask to what point can these be represented by some series expansion. We start with the most elementary functions of this type. A trigonometric polynomial is a function .p : R → C of the form p(x) =

.

|j |≤n

cj eij x ,

12

1 Convergence of Sequences of Functions

where .cj ∈ C and .|cn | + |c−n | = 0. The number n is called the degree of p. Such a function p is clearly continuous and .2π -periodic, so that the polynomial is determined by its values in the interval .] − π, π ]. Since   π 2π if k = 0 ikx . e dx = 0 if k = 0, −π a straightforward computation yields cj =

.

1 2π



π −π

p(x)e−ij x dx

for every .|j | ≤ n. Starting from this idea, if .f : R → C is .2π -periodic and .|f | is integrable in .] − π, π ] (that is, f belongs to .L1 (−π, π )), then for each .j ∈ Z, the j th Fourier coefficient of f is defined as cj =

.

1 2π



π −π

f (x)e−ij x dx .

(1.6)

 We will also denote these by .f(j ). The series . j ∈Z cj eij x is called the Fourier   ij x (or .f ∼ ij x  series of f , and we write .f ∼ j ∈Z cj e j ∈Z f (j )e ). The two questions that arise naturally are 1. Is f determined by its Fourier coefficients? 2. If this is the case, to what extent does the Fourier series represent the function? The first question can be reformulated as whether or not the fact that .f(j ) =  g (j ) for every j implies .f = g or, equivalently, if .f(j ) = 0 for every j gives that necessarily .f = 0. Another way to look at the second question is to understand under which conditions do the trigonometric polynomials defined by the partial sums of the Fourier series

.sn (f, x) = (1.7) f(j )eij x for x ∈ R |j |≤n

converge (and in what sense) to f . So, our aim now is to give some answers to these questions. In many cases, we restrict our attention to .2π -periodic real functions. In this case, the description of the Fourier series with sine and cosine might be more convenient. Note that cj eij x + c−j e−ij x = (cj + c−j ) cos(j x) + i(cj − c−j ) sin(j x)

.

for every .j ∈ N. Then we can define .a0 = 2c0 and aj = cj + c−j and bj = i(cj − c−j )

.

(1.8)

1.3 Series of Functions

13

for .j ∈ N. Note that, since f is real-valued, we have .c−j = cj , and then .aj , bj ∈ R for every j . With all this, we can rewrite the Fourier series of f as f ∼

.

cj eij x =

j ∈Z

∞

∞

j =1

j =1

a0 aj cos(j x) + bj sin(j x) . + 2

These are called the exponential and trigonometric Fourier series of f (respectively). We have 1 .aj = π



π

1 f (x) cos(j x)dx and bj = π −π



π

−π

f (x) sin(j x)dx .

We can also recover the .cj coefficients from these by doing cj =

.

1 (aj − ibj ) 2

for .j ∈ Z (here we take .b0 = 0). From now on and for the rest of this section, we identify .L1 (−π, π ) with the space of .2π -periodic functions .f : R → R for which .|f | is integrable in .] − π, π ]. Also, if a function is defined in .] − π, π ], we assume that it is extended .2π periodically to .R. Example 1.11 Fix some .a > 0, and let us consider the function defined as f (x) =

 0

.

a

if − π < x < 0 if 0 ≤ x ≤ π

(extended periodically to .R). For each .j ∈ N, we have 1 .aj = π



π

a cos(j x)dx = 0 ,

0

and 1 .bj = π



π

 a sin(j x)dx =

0

0

if j is even

2a jπ

if j is odd .

Clearly .a0 = a, and therefore f ∼

.

∞  2a 1 a + sin (2k − 1)x . 2 π 2k − 1 k=1



14

1 Convergence of Sequences of Functions

Example 1.12 Consider the function .f (x) = ex for .x ∈] − π, π ] (extended periodically to the whole .R). For each .j ∈ Z, we have cj =

.

1 2π



π

−π

e(1−ij )x dx =

1 eπ − e−π (−1)j , 2π 1 − ij

so that eπ − e−π (−1)j ij x e . 2π 1 − ij

f ∼

.

j ∈Z

From (1.8), we immediately have aj =

.

eπ − e−π (−1)j eπ − e−π (−1)j +1 j and bj = 2 2 2 2 2π 2π j +1 j +1

for .j ∈ N, and .a0 = f ∼

.

eπ −e−π π

. This altogether yields



2(−1)j

2j (−1)j +1 eπ − e−π

cos(j x) + sin(j x) . 1+ 2π j2 + 1 j2 + 1 ∞

∞

j =1

j =1

Example 1.13 Take now the function given by .f (x) = x 2 for .x ∈] − π, π ] (again, 2 extended to .R periodically). Straightforward computations show that .a0 = 2π3 and aj =

.

1 π



π −π

x 2 cos(j x)dx = 4

(−1)j 1 and bj = 2 π j



π

−π

x 2 sin(j x)dx = 0 .

With this, ∞

f ∼

.

(−1)j π2 +4 cos(j x) . 3 j2 j =1

Example 1.14 Take now the function given by .f (x) = |x| for .x ∈] − π, π ] (once more, extended periodically to .R). We have a0 =

.

1 π



π −π

|x|dx =

2 π

 0

π

xdx = π ,

1.3 Series of Functions

15

and 1 .aj = π





π

2 |x| cos(j x)dx = π −π

π

 x cos(j x)dx =

0

0

for j even

− πj4 2

for j odd

as well as bj =

.

1 π



π

−π

1

π

|x| sin(j x)dx =



π

 x sin(j x)dx −

0

0

−π

 x sin(j x)dx = 0

for .j ∈ N. This yields f ∼

.

∞  1 π 4 cos (2k − 1)x . − 2 2 π (2k − 1)

(1.9)

k=1

Let us point out that if .f : R → R is .2π -periodic and • Even, then .bj = 0 and aj =

.

2 π



π

f (x) cos(j x)dx , 0

for every .j ∈ N0 . • Odd, then .aj = 0 and bj =

.

2 π



π

f (x) sin(j x)dx , 0

for every .j ∈ N0 .

 Let us give now a first look at the behaviour of the sequence . f(j ) j ∈Z of Fourier coefficients of a function f . We start with a result that we state without proof. There are a number of proofs available; one of them can be found in [11, Theorem 2.2]. Theorem 1.15 (Riemann–Lebesgue Lemma) Let .f ∈ L1 (R), then  .

lim

∞

|t|→∞ −∞

f (x)e−ixt dx = 0 .

As an immediate consequence, we have the following on the behaviour of the Fourier coefficients.

16

1 Convergence of Sequences of Functions

Corollary 1.1 Let .f ∈ L1 (−π, π ), then  π  1. .|f (j )| ≤ |f (x)|dx for every .j ∈ Z. −π

2.

.

lim |f(j )| = 0.

|j |→∞

Proof Statement 1 is a direct consequence of the definition (1.6), whereas 2 follows

from an application of Theorem 1.15 to .f χ]−π,π ] . 1.3.2.1

Dirichlet Kernel

For each .n ∈ N, we define the nth Dirichlet kernel .Dn as the function defined on .R by Dn (x) =

.

eij x ,

(1.10)

|j |≤n

which is clearly continuous and .2π -periodic. Moreover we have (see [11, Lemma 4.2]) Dn (x) = 1 + 2

n

.

 cos(j x) =

j =1

sin((n+1)/2)x) sin(x/2)

if 0 < |x| < π

2n + 1

if x = 0 .

(1.11)

In particular, .Dn is real-valued and even. It is a convenient tool to describe the partial sums of a Fourier series of a function. Proposition 1.5 Let .f ∈ L1 (−π, π ), then sn (f, x) =

.

1 2π



π

−π

 π 1 f (x − t)Dn (t)dt 2π −π  π 1 = (f (x + t) − f (x − t))Dn (t)dt , 2π 0

f (t)Dn (x − t)dt =

for every .x ∈] − π, π ] and .n ∈ N. In particular,  .

π

Dn (x)dx = π ,

0

for every .n ∈ N. Proof The result follows by direct computation, using the definition of the Fourier coefficients (1.6) and the periodicity of f .

We will see later (see (6.2)) that this is the convolution of f with the Dirichlet kernel. With this, we can already give a first result on the pointwise convergence of

1.3 Series of Functions

17

the Fourier series of a function (providing a first answer for our second question). Given .f : R → R and .a ∈ R, we define f (a + ) = lim f (a + h) and f (a − ) = lim f (a + h) .

.

h→0+

h→0−

(1.12)

Whenever these limits exist, we also define f+ (a) = lim

f (a + h) − f (a + ) h

f− (a) = lim

f (a + h) − f (a − ) . h

.

h→0+

and .

h→0−

Theorem 1.16 (Dini–Dirichlet) Let .f ∈ L1 (−π, π ) be continuous, except for a finite number of points, in which the limits in (1.12) exist. Let .c ∈ R so that both   .f+ (c) and .f− (c), then .

lim sn (f, c) =

n→∞

1 f (c+ ) + f (c− ) . 2

In particular, if f is differentiable in .R, then .

lim sn (f, x) = f (x) ,

n→∞

for every .x ∈ R Proof See [2, Theorem 11.9], [10, Theorem 8.45] or [11, Theorem 4.5].



Remark 1.3 All the functions in Examples 1.11–1.14 clearly satisfy the hypothesis of Theorem 1.16, so that the corresponding Fourier series converge (pointwise) at every .x ∈ R. Note that, as a consequence of the Weierstraß M-test (Proposition 1.4), the Fourier series of .ex , .x 2 , and .|x| converge uniformly on . [−π, π ]. We can use the 1 expansion as a Fourier series of .|x| to compute the value of . ∞ k=1 k 2 . To begin with, evaluating the function at .x = 0 and using that the series in (1.9) converges, we have 0=

.

∞ 4 1 π − , 2 π (2j − 1)2 j =1

from which we get ∞

.

j =1

1 π2 . = 8 (2j − 1)2

18

1 Convergence of Sequences of Functions

So, we have the sum of the odd terms in the series. Let us compute now the sum of the even terms. ∞

.

j =1

 1 1

1 1 1 1 1 1 π2 = = = + + , 2 2 2 2 2 4 4 4 32 (2j ) j (2j ) (2j − 1) (2j ) ∞

∞

∞

∞

j =1

j =1

j =1

j =1

and ∞

.

j =1

1 π2 = . 24 (2j )2

Joining these two finally yields ∞

π2 π2 1 π2 + = . = 2 8 24 6 k

.

k=1

Remark 1.4 Theorem 1.16 has two major drawbacks. First of all, the conditions on the function are rather restrictive and may not be accomplished even by continuous functions. As a matter of fact, we will see later (see Theorem 3.20) that there are continuous functions for which the Fourier series does not converge at 0. Second, even when the series converges, we can only ensure pointwise convergence, and we would like to have a more powerful (such as, for example, uniform) convergence. In order to overcome these two problems, we need a different approach.

1.3.2.2

Cesàro Means: Féjer Kernel

Given a function, the idea now is to take, instead of the partial sums, the arithmetic means of them. To be more precise, for .f ∈ L1 (−π, π ) and .n ∈ N0 , the nth Cesàro mean of f is defined as the arithmetic mean of the partial sums up to n, that is, 1 sk (f, x) , n+1 n

σn (f, x) =

.

k=0

for .x ∈ R. Note that, for each f and n, the Cesàro mean is again a trigonometric polynomial. Just as the Dirichlet kernel was a helpful tool when describing the partial sum of the Fourier series, we consider a new kernel that allows us to deal with the Cesàro means. For each .n ∈ N, the nth Féjer kernel is defined as the arithmetic mean of the first Dirichlet kernels, that is, 1 Dk (x) , n+1 n

Kn (x) =

.

k=0

1.3 Series of Functions

19

for .x ∈ R. As a straightforward consequence of Proposition 1.5, given .f ∈ L1 (−π, π ), we have 1 2π

σn (f, x) =

.



π

−π

f (t)Kn (x − t)dt =



1 2π

π

−π

f (x − t)Kn (t)dt ,

(1.13)

for every .n ∈ N0 and .x ∈ R. Once again, this is the convolution (see (6.2)) of f and .Kn . We establish now some basic properties of the Féjer kernel. A proof can be found, for example, in [11, Lemmas 4.3 and 4.4]. Proposition 1.6 For each .n ∈ N0 , we have 1. .Kn (0) = n + 1.   n+1 2 sin 2 x 1  2. .Kn (x) = for every .x ∈] − π, π ] \ {0}. n+1 sin x2 3. .K  nπ is non-negative and even. 4. . Kn (x)dx = 2π . −π  π 5. . lim Kn (x)dx = 0 for each fixed .0 < δ < π . n→∞ δ

With this, we can get what we aimed at uniform convergence, giving another answer to our second question. Theorem 1.17 (Féjer) Let .f : R → R be continuous and .2π -periodic. Then σn (f, x) → f uniformly on .R as .n → ∞.

.

Proof Since f is continuous and periodic, it is bounded, so that we can find .M > 0 with .|f (x)| ≤ M for every .x ∈ R. Then, the function is also uniformly continuousSo, for each fixed .ε > 0 there is .δ > 0 so that .|f (x + t) − f (x)| < ε for every .x ∈ R and .|t| < δ. Given any .x ∈ R and .n ∈ N, we have from (1.13) and Proposition 1.6 |σn (f, x) − f (x)| ≤

.

=

1 2π



−δ

−π

1 2π



π −π

|f (x + t) − f (x)|Kn (t)dt 

|f (x + t) − f (x)|Kn (t)dt +  +

δ

−δ π

|f (x + t) − f (x)|Kn (t)dt 

|f (x + t) − f (x)|Kn (t)dt .

δ

Let us bound now each one of these three integrals. For the first and third one,   2 2 let us observe in first place that . sin u2 ≥ πu 2 for every .u ∈ [0, π ], which, if .δ < |u| < π, implies   n+1 2  n+1 2 sin 2 u π2 1 π2 u sin u ≤ . .0 ≤ Kn (u) = ≤ 2 n+1 (n + 1)u2 (n + 1)δ 2 sin 2

20

1 Convergence of Sequences of Functions

Hence 

−δ

.

−π

|f (x + t) − f (x)|Kn (t)dt ≤ 2M

π3 , (n + 1)δ 2

|f (x + t) − f (x)|Kn (t)dt ≤ 2M

π3 . (n + 1)δ 2

and  .

π

δ

For the remaining integral, just observe that 

δ

.

−δ

 |f (x + t) − f (x)|Kn (t)dt ≤ ε

δ −δ

Kn (t)dt ≤ 2π ε .

Joining all these together, we obtain |σn (f, x) − f (x)| ≤ ε +

.

2Mπ 2 (n + 1)δ 2

for every .x ∈ R. Given .ε and .δ (recall that these are fixed), it is enough to choose δ2 n0 ∈ N with . n01+1 < ε 2Mπ 2 to get

.

|σn (f, x) − f (x)| ≤ 2ε

.

for every .x ∈ R and .n ≥ n0 . This completes the proof.



Remark 1.5 This result also answers our first question on whether or not the coefficients determine the function whenever it is continuous. Indeed, suppose that f is continuous and .f(j ) = 0 for every .j ∈ Z. Then .sn (f, x) = 0 for every n (and x) and so .σn (f, x) = 0. Theorem 1.17 implies that .f = 0. As a consequence of Féjer’s theorem, we obtain that the continuous functions can be uniformly approximated by polynomials. Corollary 1.2 Let .f : R → R be continuous and .2π -periodic. For each .ε > 0, there exists a trigonometric polynomial T so that .

sup |f (x) − T (x)| < ε . x∈R

Corollary 1.3 Let f be continuous in .[a, b]. For each .ε > 0, there exists a polynomial p so that .

sup |f (x) − p(x)| < ε . x∈[a,b]

1.3 Series of Functions

1.3.2.3

21

Poisson Kernel

Given a function .f :] − π, π ] → C (or a .2π -periodic function on .R), we can think of it as defined on .T = ∂D = {z ∈ C : |z| = 1} just by identifying .] − π, π ] with .T by means of .t  eit . The question then arises naturally of whether or not can this function be extended to the disc .D. The tool to achieve this is the Poisson kernel, defined as

.Pr (t) = r |j | eij t , (1.14) j ∈Z

for .t ∈] − π, π ] and .0 ≤ r < 1. Note that since .0 ≤ r < 1 the series converges (absolutely) for every t and r, and the function is well defined. The kth partial sum in (1.14) gives k

.

r |j | eij t =

j =−k

k

r j eij t +

j =0

k

r j e−ij t ,

j =1

which, summing up the geometric series, yields Pr (t) =

.

1 re−it 1 − r2 + = . it it 1 − re 1 − re 1 − 2r cos t + r 2

(1.15)

Proposition 1.7 For each .0 ≤ r < 1, the function .Pr is .2π -periodic, positive, and even. It satisfies 

π

.

−π

Pr (t)dt = 2π

and .

1−r 1+r ≤ Pr (t) ≤ 1+r 1−r

for every r and t. Proof The series in (1.14) converges uniformly on .[−π, π ]. This gives the statement on the integral. All the rest follows easily from (1.15).

By doing .z = reit (for .z ∈ D), we may look at it as a function defined on the disc P (z) = P (reit ) = Pr (t) =

.

j ∈Z

r |j | eij t =

1 − |z|2 . |1 − z|2

22

1 Convergence of Sequences of Functions

Proposition 1.8 Fix .0 < δ < π , and then for each .0 < r < 1, we have .Pr (t) ≤ Pr (δ) for every .δ < |t| < π. Also  .

lim

r→1−

 sup{Pr (t) : δ < |t| < π } = 0 .

Proof Let us define the function .g : [δ, π ] → R by .g(t) = Pr (t). Since .g  (t) < 0 for every .δ < t < π (this follows by direct computation from (1.15)), this gives the conclusion. Also, .

sup{Pr (t) : δ < |t| < π } = Pr (δ) =

1 − r2 , 1 − 2r cos δ + r 2

which clearly tends to 0 as .r → 1− .



This is the right tool to extend our functions from .T to the disc .D. Given .f ∈ L1 (−π, π ) (recall that we are identifying this with .2π -periodic functions on .R) for each .0 ≤ r < 1 and .t ∈ R, the integral P [f ](reit ) =

.

1 2π



π

−π

Pr (x − t)f (t)dt =

1 2π



π

−π

f (x − t)Pr (t)dt

is finite (note that, once again, this is a convolution). So, in this way, we may define a function .P [f ] : D → C (called the Poisson integral of f ). If f has Fourier coefficients . f(j ) j , then we can rewrite the previous as P [f ](reit ) =

.

r |j | f(j )eij t .

j ∈Z

So, starting from a function f defined on .T, we get a function .P [f ] defined on .D. Let us see now that (at least in some cases) this actually extends f . Theorem 1.18 Let .f : R → C be continuous and .2π -periodic. Let us define a function .F : D → C by F (eix ) = f (x) for x ∈ R

.

and F (reix ) = P [f ](reix ) for 0 ≤ r < 1 and x ∈ R .

.

Then F is continuous on .D.

1.3 Series of Functions

23

Proof The function F is clearly well defined. If .z = reix with .|z| < 1, we have F (z) =

.

r |j | f(j )eij t =

∞

f(j )zj +

∞

j =0

j ∈Z

f(−j )zj .

j =1

 Since . f(j ) j ∈Z is bounded (recall Corollary 1.1), each one of these two series converges absolutely for each .z ∈ D and, then, uniformly on each disc of radius strictly smaller than 1. Hence, F is continuous in .D. It is only left to see that F is continuous at each .eix ∈ T. To do this, we are going to see that .F (reix ) → F (eix ) uniformly on .[−π, π ] as .r → 1− , following a similar idea as in the proof of Theorem 1.17. To begin with, since f is continuous and .2π periodic, it is bounded (by, say, .M > 0) and uniformly continuous. Then, for a fixed .ε > 0, we can find .δ > 0 so that .|f (x − t) − f (x)| < ε for every .x ∈ R and .|t| < δ. Then, given .x ∈ R, we have |F (reix )−F (eix )| =

.



1 ≤ 2π 1 = 2π

π −π



|f (x − t) − f (x)|Pr (t)dt

−δ

−π



π

+

  π 1 π f (x − t)P (t)dt − f (x)P (t) r r 2π −π −π

 |f (x − t) − f (x)|Pr (t)dt +

δ −δ



|f (x − t) − f (x)|Pr (t)dt

|f (x − t) − f (x)|Pr (t)dt .

δ

We bound now each one of these three integrals. On the one hand, 

δ

.

−δ

 |f (x − t) − f (x)|Pr (t)dt ≤ ε

and, on the other hand,   π . |f (x − t) − f (x)|Pr (t)dt ≤ 2M δ

δ −δ

π

Pr (t)dt ≤ 2π ε ,

Pr (t)dt ≤ 2Mπ Pr (δ) .

δ

The same bound holds for the remaining integral. This altogether gives |F (reix ) − F (eix )| ≤ ε + 2MPr (δ) .

.

Recall that .δ > 0 is fixed and then, taking Proposition 1.8 into account, we can find ε 0 < r0 < 1 so that .Pr (δ) < 2M for every .r0 < r < 1. Then,

.

|F (reix ) − F (eix )| < ε

.

24

1 Convergence of Sequences of Functions

for every .x ∈ [−π, π ] and .r0 < r < 1. Since the choice of .r0 does not depend on x, this proves our claim and completes the proof.



1.3.3 Dirichlet Series Each sequence .(an )n of complex numbers defines a Dirichlet series ∞

.

an n−s ,

n=1

where the variable .s ∈ C (usually we would write z for a complex variable, but for historical reasons we write s). Note that, once again, this is a series of functions where .fn (s) = an n−s . We have seen in Theorem 1.10 that if a power series converges at some point, then it converges absolutely at every point with strictly smaller modulus. In this way, there is a maximal disc on which a power series converges, and it converges uniformly on every disc of strictly smaller radius (recall (1.5)). There are certain parallelism between Dirichlet and power series, replacing ‘modulus’ by ‘real part’ and ‘disc’ by half plane’. To be more precise, if a Dirichlet series converges at some point, then it converges at every point with strictly bigger real part (see Theorem 1.19). Then, the natural domains where Dirichlet series converge are halfplanes .[Re > σ ] = {s ∈ C : Re s > σ }. But there are also big differences: while the radius of convergence and the absolute convergence of a given power series are the same (recall (1.5)), in the case of Dirichlet series, the maximal half-planes of convergence and absolute convergence may not coincide (see (1.17)). We start by analysing the convergence of Dirichlet series. To begin with, we need a basic fact, known  as Abel summation: given .a1 , . . . , aN ∈ C and .b1 , . . . , bN ∈ C, we write .An = nk=1 ak for .2 ≤ n ≤ N, then N

.

an bn = AN bN +

n=1

N −1

An (bn − bn+1 ) .

(1.16)

n=1

If we observe that .A1 = a1 and .an = An − An−1 , we have N

.

an bn = A1 b1 +

n=1

N N −1 N

(An − An−1 )bn = AN bN + An bn − An−1 bn n=2

= AN bN +

N −1

n=1

An bn −

n=1 N −1

n=1

An bn+1 = AN bN +

n=2 N −1

n=1

An (bn − bn+1 ) .

1.3 Series of Functions

25

 Theorem 1.19 Let . n an n1s be a Dirichlet series that converges at .s0 = σ0 + it0 ∈ C, then it converges at s for all .Re s > σ0 . Even more, the series converges uniformly on every compact set in .[Re > σ0 ]. Proof Given .Re s > Re s0 , we write .u = s0 −s, and by (1.16), we have, for .M > N, M

.

n=N

an

M M M−1 n 



1 an 1 an  1 ak  1 1 . = = + − u s s u s u s u n n0n n0 M k0 n (n + 1) n=N

n=N

n=N

k=N

Take now some .K ⊆ [Re > σ0 ] and fix .ε > 0. Since .K1 = {s − s0 : s ∈ K} is again |u| compact, we may define .A = supu∈K1 Re u < ∞ (which clearly satisfies .A > 0).  Since the series . n an n1s0 is convergent, we can find .N0 so that M ε an < s 0 n A

.

n=N

for every .M > N ≥ N0 . Also 1  n+1 u  n+1 |u| 1 = dx dx u− ≤ n (n + 1)u x u+1 x Re u+1 n n  1 |u| 1 . − = Re u Re u Re u n (n + 1)

.

Taking all this into account, if .M > N ≥ N0 and .s ∈ K, we have M M−1 n M 

ak |u| 1 1 an 1 1 an s ≤ + − s s Re u Re u Re u n n0 M k 0 Re u n (n + 1) n=N n=N n=N k=N

.

≤

M−1 

1 1 ε 1 ε |u| 1 < A Re u < ε . + − Re u Re u Re u A Re u M A N n (n + 1) n=N

Proposition 1.3 gives the conclusion.



This shows that if a Dirichlet series converges at some point .s0 , then it converges on the open half-plane defined by the abscissa .Re s0 . In other words, the natural regions of convergence of Dirichlet series are half-planes. As we did for power series, we may consider

  σc = inf σ ∈ R : an n1s converges in [Re > σ ] ∈ [−∞, ∞] ,

.

the abscissa that defines the biggest half-plane on which a given Dirichlet series converges (this would be in some sense the analogue to the radius of convergence of

26

1 Convergence of Sequences of Functions

a power series). Then the Dirichlet series converges pointwise for every .Re s > σc and diverges for every .Re s < σc . In this way, the function .f : [Re s > σc ] → C given by f (s) =

∞

.

n=1

an

1 ns

is well defined. Moreover, the fact that the series converges uniformly on compact sets implies that the function defined by the Dirichlet series is holomorphic (check Theorem 4.4). We may also consider the abscissa of absolute convergence

  σa = inf σ ∈ R : an n1s converges absolutely in [Re > σ ] ∈ [−∞, ∞]

.

and wonder if, as in the case of power series (recall (1.5)), this equals .σc . Of course, for certain Dirichlet series, these two abscissas may coincide (just take a Dirichlet series with  positive coefficients), but this need not be the case. Take for example the series . n (−1)n n1s . This converges for every .Re s > 0 (just apply Leibniz criterion) but converges absolutely only on the plane .[Re > 1]. In other terms, σc = 0 and σa = 1 .

.

Now the question is ‘how big can it get?’ or ‘how far apart  can .σa and .σc be to each other?’ Actually not really too far away. Assume that . an n1s converges at some |an | .s0 = σ0 + it. Then the sequence .( σ0 )n is bounded by, say, K, and we have, for n every .ε > 0, ∞

. n=1

∞ ∞ ∞

|an | |an | 1 1 = ≤K < ∞. = nσ0 n1+ε ns0 +1+ε nσ0 +1+ε n1+ε

an

n=1

n=1

n=1

Then .σa ≤ σ0 + 1 + ε. This then shows that for every Dirichlet series .σa ≤ σc + 1 and, together with the previous example, we obtain .

  sup σa − σc : D Dirichlet series = 1 .

(1.17)

We focus now on a third abscissa: that of uniform convergence. Let ∞

  σu = inf σ ∈ R : an n1s converges uniformly on [Re > σ ] ∈ [−∞, ∞] .

.

n=1

Again, the question arises of whether or not this coincides with any of the other two abscissas. As a matter of fact, it does not. Taking the series . n (−1)n pn−s (where .pn stands for the nth prime number), it is easy to check using Leibniz criterion for

1.3 Series of Functions

27

alternate series that .σc = 0. On the other hand, it can be shown that .σu = 1 (this is less trivial, a proof can be found in [4, page 16]). Taking (1.17) into account (and that .σu ≤ σa ), this shows .

  sup σu − σc : D Dirichlet series = 1 .

A much more intricate problem is to separate .σa from .σu . It can be shown (see [4, Chapter 4]) that there is a Dirichlet series with .σa = 1 and .σu = 12 and that, in fact, .

  1 sup σa − σu : D Dirichlet series = . 2

The proofs of these facts exceed the scope of this text. We can, however, show an interesting and important fact about the abscissa of uniform convergence: it determines the biggest half-plane on which the holomorphic function defined by the series is bounded. Let us introduce two more abscissas. First of all, we define ∞

 σb = inf σ ∈ R : an n1s converges and defines a

.

n=1

 bounded holomorphic function on [Re > σ ] ∈ [−∞, ∞] .

 Second, if . an n1s is a somewhere convergent Dirichlet series (and defines a function f ), we define .σbext as the infimum of all .σ ∈ R so that f can be extended to .[Re > σ ] as a bounded holomorphic function. Obviously .σbext ≤ σb . On the other hand, it is easy to see that finite sums are bounded on half-planes and, then, Remark 1.2 gives .σb ≤ σu . A remarkable result, due to Harald Bohr (known today as Bohr’s theorem), shows that all three coincide. Theorem 1.20 (Bohr) For every Dirichlet series, we have .σbext = σb = σu . This follows from the following result.  Lemma 1.1 Let . an n1s be a Dirichlet series that converges absolutely in .[Re > a] (and then defines a holomorphic function .f : [Re > a] → C). Suppose that the function f can be extended as a bounded, holomorphic function to .[Re > b]. Then  the Dirichlet series . an n1s converges uniformly on .[Re > b + ε] for every .ε > 0. Let us note that as a straightforward consequence of this result .σu ≤ σbext , which gives Theorem 1.20. The proof relies heavily on a careful use of Cauchy’s theorem (see Theorem 4.1 ahead), Proof Let us denote .K = supRe s>b |f (s)| and fix .0 < ε < 1. Our aim is to show that there is a constant .C = C(ε, K) > 0 such that M

1 1 an s ≤ C ε log M f (s0 ) − 0 n M

.

n=0

(1.18)

28

1 Convergence of Sequences of Functions

Fig. 1.7 Contour of integration R

for every .M ≥ 2 and all .Re s0 > b + ε. This clearly will give our claim. Fix, then, .s0 with .Re s0 > b +ε and .M ≥ 2. We consider the rectangle of vertices a−b+2 , s + a − b ± iM a−b+2 (we denote it by R, see Fig. 1.7). Note .s0 − ε ± iM 0 that .Re s0 − ε > b and .Re s0 +a − b > a + ε. z−s0 The function .g(z) = f (z) M + 12 is holomorphic on a connected open set containing the rectangle (actually, on .[Re > b]). Then, by Cauchy’s Integral Formula (see Theorem 4.1),  .

R

g(z) dz = 2π ig(s0 ) = 2π if (s0 ) z − s0

and  M M

1 1 g(z) 1 = (s ) − a dz − a f 0 n s n s n0 2π i R z − s0 n0 n=0

.

n=0

 M

g(z) 1 1 g(z) dz − an s + dz n0 2π i γ1 z − s0 γ3 z − s0 n=0   1 1 g(z) g(z) dz + dz . + 2π i γ2 z − s0 2π i γ4 z − s0 (1.19)

 1 ≤ 2π i

1.3 Series of Functions

29

Now we have to bound each one of these integrals. In order to keep the notation as neat as possible, we set .c = a − b and .T = M a−b+2 . We begin by bounding the integral on the vertical left side. We have  T f (s0 − ε + it)(M + 1/2)−ε+it g(z) idt dz = −ε + it −T γ1 z − s0  T  T 1 |f (s0 − ε + it)|  1 1 −ε ≤ M+2 dt ≤ 2K dt . √ √ ε 2 2 2 (M + 1/2) 0 ε +t ε + t2 −T

 .

A change of variable .u = t/ε gives 

T

.

0



T /ε



1



T /ε 1 dt = du = du + du √ √ √ √ 2 2 2 2 1+u 1 + u2 ε +t 1+u 0 0 1  T /ε 1 ≤1+ du = 1 + log T − log ε = 1 + (a − b + 2) log M − log ε . u 1

1

1

1

Then, if .M > e/ε (so that .1 − log ε < log M), we have 

.

γ1

1 g(z) dz ≤ 2K (a − b + 1) log M , z − s0 (M + 1/2)ε

and hence there is .C1 > 0 so that  1 g(z) ≤ C1 1 log M , dz . 2π i Mε γ1 z − s0

(1.20)

for every .M ≥ 2. We now bound the integrals along the horizontal sides. We have 

.

γ2

 c t+iT g(z) f (s0 + t + iT )  M + 12 dz = dt z − s0 t + iT −ε  c t |f (s0 + t + iT )|  ≤ M + 12 dt √ 2 2 t +T −ε  c   t 1 K c 1 t M + 2 dt ≤ M + 12 dt ≤K √ 2 2 T t +T −ε −ε

 −ε   K 1 a−b = a−b+2 M + 12 − M + 12 log(M + 1/2) M K  ≤ a−b+2 2M)a−b . M

30

1 Convergence of Sequences of Functions

Then, there is .C2 > 0 so that  1 g(z) 1 . dz ≤ C2 2 , 2π i M γ2 z − s0

(1.21)

for every .M ≥ 2. Exactly the same argument shows that we can find .C4 > 0 with  1 . 2π i

γ4

g(z) 1 dz ≤ C4 2 , z − s0 M

(1.22)

for every .M ≥ 2. The remaining term in (1.19) requires some more work. Let us note first that the segment is within the half-plane of absolute convergence of the series. Hence the series converges uniformly on the segment, and we may replace f by the Dirichlet series and then interchange the sum and the integral to get  z−s0 g(z) f (z)  M + 12 dz = dz γ3 z − s0 γ3 z − s0 

∞  ∞

M + 1/2 z−s0 1 1 (M + 1/2)z−s0 an = an z dz = dz . s n z − s0 n 0 γ3 n z − s0 γ3

 .

n=1

n=1

Therefore,  M g(z) 1 1 an s − dz 0 n 2π i γ3 z − s0 n=1

.

M   

an 1 M + 1/2 z−s0 1 ≤ 1− dz s 0 n 2π i γ3 n z − s0 n=1

1 + 2π i

∞

n=M+1

an ns0

(1.23)



M + 1/2 z−s0 1 dz . n z − s0 γ3

So, to complete the proof, we have to bound each one of these terms, which correspond to sums with .1 ≤ n ≤ M and with .n ≥ M + 1. From now on, we write gn (z) =

.

M + 1/2 z−s0 1 . n z − s0

We begin by dealing with the case .n ≥ M + 1. For a given .σ > 0 (which we later will make tend to .∞), we consider a new rectangle (which we denote by .R1 , see Fig. 1.8) with vertices .s0 + c ± iT and .s0 + c + σ ± iT .

1.3 Series of Functions

31

Fig. 1.8 New contour of integration .R1

Note that the vertical left side here (.η1 ) is the same as the vertical right side of R (.γ3 ). Now, since .s0 ∈ R1 , the function  .gn is holomorphic on an open set containing .R1 and, by Cauchy’s Theorem 4.1, . R1 gn (z)dz = 0. Hence     . gn (z)dz ≤ gn (z)dz + gn (z)dz + gn (z)dz . γ3

η2

η3

(1.24)

η4

Once again, we bound each one of these integrals. For the horizontal sides, we have   gn (z)dz = η2

.



M + 1/2 t+iT 1 dt n t + iT c  c+σ

  1 M + 1/2 t 1 c+σ M + 1/2 t ≤ dt dt ≤ √ n T c n t2 + T 2 c  

M + 1/2 c+σ M + 1/2 c 1 1 = −  T log M+1/2 n n n 

 M + 1/2 c M + 1/2 c+σ 1 1 = −  M+1/2 T log n n c+σ

n

≤

M + 1/2 c 1 1 .  M+1/2 T log n n

(1.25)

32

1 Convergence of Sequences of Functions

  ≥ log M+1/2 > Note now that . log M+1/2 n M+1

1 2M+2

for every .n ≥ M + 1 (the

latter being true because .| log(1 − x)| > x for every .0 < x < 1, and . M+1/2 M+1 = 1 1 − 2M+2 ). Taking this into account, we have  . gn (z)dz η2

M + 1/2 a−b 2M + 1 a−b 1 1 (2M + 2) 2 a−b n 2M M M n 1 1 ≤ 3a−b 2 a−b . M n  The same estimate holds for . η4 gn (z)dz . We bound now the integral along the vertical right side. 1

≤

(2M + 2) a−b+2

 

M + 1/2 c+σ +it 1 = dt g (z)dz n n c + σ + it η3 η3 

M + 1/2 c+σ 1 ≤ dt n |c + σ + it| η3

M + 1/2 c+σ  M + 1/2 c+σ 1 = dt n n |c + σ + it| η3

M + 1/2 c+σ 2T . ≤ M +1 c+σ

.

Plugging all these in (1.24), we have 

M + 1/2 c+σ 2T 1 . ≤ 2 × 3a−b 1 + g (z)dz n M +1 c+σ M 2 na−b γ3 and, since this holds for every .σ > 0,  1 1 . gn (z)dz ≤ 2 × 3a−b 2 a−b , M n γ3 for every .n ≥ M + 1. Whit this, ∞ 

∞

an M + 1/2 z−s0 1 |an | 1 a−b . . dz ≤ 2 × 3 ns0 γ3 n z − s0 nRe s0 +a−b M n=M+1

n=M+1

1.3 Series of Functions

33

Fig. 1.9 Another contour of integration .R2

But this is the tail of a convergent series (because .Re s0 + a − b > a + ε) and, so, it is bounded. Then we can find .K1 > 0 so that ∞ 

an M + 1/2 z−s0 1 1 . dz ≤ K1 . s 0 n n z − s0 M γ3

(1.26)

n=M+1

We now consider the case .1 ≤ n ≤ M in (1.23). In this case, we have to estimate 

M + 1/2 z−s0 1 . 1 − 1 dz . 2π i γ3 n z − s0 Again, for each .σ > 0, we take the rectangle (which we now call .R2 , see Fig. 1.9) of vertices .s0 − σ ± iT and .s0 + c ± iT Note that now the vertical right sides of R (.γ3 ) and of .R2 (.ν3 ) coincide. Now, by the Cauchy Integral Formula (Theorem 4.2), 

 gn (z)dz =

.

R2

M + 1/2 z−s0 1 dz = 2π i , n z − s0 R2

and then       1 . 1 − 1 gn (z)dz ≤ gn (z)dz + gn (z)dz + gn (z)dz . 2π i γ3 2π ν1 ν2 ν4 Essentially the same computations as in (1.25) (using now that .log(x + 1) > x/2 for every .0 < x < 1) yield 

M+1/2 a−b

M+1/2 a−b 1 1 1 1 . ≤ ≤ 4 . g (z)dz  n M a−b+2 n M M na−b log M+1/2 ν2 M

34

1 Convergence of Sequences of Functions

The same estimate holds for .ν4 . Finally, integrating along the vertical left side, we have   T

M + 1/2 −σ +it i . dt gn (z)dz = n −σ + it ν1 −T  T

 −σ 2T

M + 1/2 −σ 1 ≤ . dt ≤ M + 12 √ n σ σ 2 + t2 −T Hence 

a−b+2  . 1 − 1 ≤ 1 8 × 2a−b 1 1 + 2M g (z)dz n 2π 2π i γ3 M na−b σ (M + 1/2)σ for every .σ > 0. Using this and proceeding as in (1.26), we find some .K2 > 0 so that M   

an M + 1/2 z−s0 1 1 ≤ K2 1 , dz 1 − s n0 2π i γ3 n z − s0 M

.

n=1

and this, together with (1.26) and (1.23), gives .C3 > 0 such that  M g(z) 1 1 1 . an s − dz ≤ C3 . 0 n 2π i γ3 z − s0 M

(1.27)

n=1

We have now at hand everything we need. Using (1.20), (1.21), (1.22), and (1.27) in (1.19), we get M

1 1 1 1 1 an s ≤ C1 ε log M + C4 2 + C3 + C4 2 . f (s0 ) − 0 n M M M M

.

n=0

1 ≤ Noting that . M12 ≤ M completes the proof.

1 Mε

log M for every .M ≥ 3 finally gives (1.18) and



1.4 Exercises 1.1 Calculate the pointwise limit of each sequence of functions fn :  → R nx for  = [0, ∞[. 1. fn (x) = 1 + nx 2 −nx 2. fn (x) = x e for  = [0, ∞[ 3. fn (x) = n2 x 2 e−nx for  = [0, ∞[.

1.4 Exercises

35

xn for  = [0, ∞[. 1 + xn 2n 5. fn (x) = cos π x for  = R.

4. fn (x) =

x . Show that (fn )n converges uniformly x+n on [0, a] for every a > 0 and pointwise but not uniformly on [0, ∞[. nx 1.3 Consider the sequence fn (x) = . Show that (fn )n converges 1 + n2 x 2 uniformly on [a, ∞[ for every a > 0 and pointwise but not uniformly on [0, ∞[.

1.2 Consider the sequence fn (x) =

1.4 Show that if fn (x) = x + 1/n and f (x) = x for x ∈ R, then (fn ) converges uniformly on R to f , but the sequence (fn2 ) does not converge uniformly on R. Thus the product of uniformly convergent sequences may not converge uniformly. 1.5 Let (fn ) and (gn ) be sequences of bounded functions on  that converge uniformly on  to f and g, respectively. Show that (fn gn ) converges uniformly on  to f g. 1.6 Let (fn ) be a sequence of functions that converges uniformly to f on  and that satisfies |fn (x)| ≤ M for all n ∈ N and all x ∈ . If g is continuous on the interval [−M, M], show that the sequence (g ◦ fn ) converges uniformly to g ◦ f on . 1 . Find the pointwise limit f 1.7 Let fn : [0, 1] → R be given by fn (x) = (1 + x)n of the sequence (fn ) on [0, 1]. Does (fn ) converge uniformly to f on [0, 1]?  1.8 Discuss the convergence and the uniform convergence of the series fn , where fn (x) is given in each case by 1 . + n2 1 fn (x) = for x = 0. 2 (nx) x fn (x) = sin n2 . 1 for x = 0. fn (x) = n x +1 n x for x ≥ 0. fn (x) = n x +1 n (−1) for x ≥ 0. fn (x) = n+x

1. fn (x) = 2. 3. 4. 5. 6.

x2

1.9 Determine the radius of convergence of the series case is given by 1 . nnα n 2. an = . n!

1. an =



an x n , where an in each

36

1 Convergence of Sequences of Functions

nn . n! 1 4. an = . log n (n!)2 . 5. an = (2n)! 1 6. an = √ . n n 1.10 For the following 2π -periodic functions, compute the Fourier series: 3. an =

1. 2. 3. 4. 5.

f (x) = 0, −π < x < 0, f (x) = a > 0, 0 < x < π . f (x) = 0, −π < x < 0, f (x) = x, 0 < x < π . f (x) = −1, −π < x < 0, f (x) = 1, 0 < x < π . f (x) = π − x, −π < x < π . f (x) = x cos 2x, −π < x < π .

1.11 Give the Fourier series and study its convergence of the following functions (extended periodically):  1 1. f (x) = x 4 − 2π 2 x 2 , |x| < π . Compute ∞ n=1 4 . ∞ (−1)n n 2. f (x) = | cos x|, |x| ≤ π . Compute n=1 4n2 −1 . 3. f (x) = x sin x, |x| < π .  1 4. f (x) = x+π, −π < x < 0, and f (x) = 0, 0 ≤ x ≤ π . Compute ∞ n=1 (2n−1)2 . 1.12 Show that the series ζ (s) =

∞

n−s converges uniformly in each semi-infinite

n=1

interval of the form [1 + h, +∞[ with h > 0.  1.13 Show that for every Dirichlet series an n1s , we have  N n=1 an .σc ≤ lim sup log N N and N σa ≤ lim sup

.

N

n=1 |an |

log N

.

Show also that, in each case, if the abscissa is not negative, then we have equality. Hint: Use Abel’s summation.  1 1.14 Let f (s) = ∞ n=1 an ns be a Dirichlet series which converges absolutely in [Re > σa ]. Prove that .

uniformly for t ∈ R.

lim f (σ + it) = f (1) = a1

σ →+∞

References

37

1.15 Deduce from the previous exercise that the function f (s) = s−1 s+1 , which is analytic and bounded in the half-plane [Re > 0], cannot be represented by a Dirichlet series on any half-plane [Re > σ ] (for any σ > 0).  1 1.16 The sum f (s) ∞ n=1 an ns of a Dirichlet series is holomorphic in the half-plane of convergence [Re > σc ] (recall Theorem 1.19). Show that the derivative f  (s) is represented in this half-plane by the series f  (s) = −

∞

.

an log n

n=1

1.17 Prove that if series

∞

1 n=1 an ns

1 . ns

converges on some half-plane [Re > σ0 ], then the

.

−

∞

an 1 log n ns n=2

also converges in the same half-plane. As a consequence, if a1 = 0, then the integral term by term of a Dirichlet series on a half-plane is represented by a Dirichlet series on the same half-plane. Further Reading [1–12]

References 1. Ahlfors, L.: Complex Analysis—An Introduction to the Theory of Analytic Functions of One Complex Variable, 3rd edn. AMS Chelsea Publishing, Providence, RI (2021). Reprint of the 1978 original 2. Apostol, T.M.: Mathematical Analysis, 2nd edn. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont (1974) 3. Conway, J.B.: Functions of one complex variable. In: Graduate Texts in Mathematics, vol. 11, 2nd edn. Springer, New York/Berlin (1978) 4. Defant, A., García, D., Maestre, M., Sevilla-Peris, P.: Dirichlet series and holomorphic functions in high dimensions. In: New Mathematical Monographs, vol. 37. Cambridge University Press, Cambridge (2019). https://doi.org/10.1017/9781108691611 5. Duoandikoetxea, J.: Fourier analysis. In: Graduate Studies in Mathematics, vol. 29. American Mathematical Society, Providence, RI (2001). Translated and revised from the 1995 Spanish original by David Cruz-Uribe 6. Gamelin, T.W.: Complex analysis. Undergraduate Texts in Mathematics. Springer, New York (2001). https://doi.org/10.1007/978-0-387-21607-2 7. Queffelec, H., Queffelec, M.: Diophantine approximation and Dirichlet series. In: Texts and Readings in Mathematics, vol. 80, 2nd edn. Hindustan Book Agency, New Delhi; Springer, Singapore (2020). https://doi.org/10.1007/978-981-15-9351-2 8. Rudin, W.: Principles of mathematical analysis, 3rd edn. In: International Series in Pure and Applied Mathematics. McGraw-Hill Book Co., New York-Auckland-Düsseldorf (1976)

38

1 Convergence of Sequences of Functions

9. Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Book Co., New York (1987) 10. Stromberg, K.R.: An Introduction to Classical Real Analysis. AMS Chelsea Publishing, Providence, RI (1981). https://doi.org/10.1090/chel/376. (2015, corrected reprint of the 1981 original) 11. Vretblad, A.: Fourier analysis and its applications. In: Graduate Texts in Mathematics, vol. 223. Springer, New York (2003). https://doi.org/10.1007/b97452 12. Werner, D.: Einführung in die höhere Analysis, corrected edn. Springer-Lehrbuch. [Springer Textbook]. Springer, Berlin (2009). https://doi.org/10.1007/978-3-540-79696-1

Chapter 2

Locally Convex Spaces

2.1 Topological Preliminaries 2.1.1 Basic Definitions We collect in this section some very basic definitions and facts on topology that will be used from now on. A topological space is a pair .(X, τ ), where X is a set and .τ (or .τX , when we need to stress the set) is a family of subsets of X (called open sets) satisfying (T1) (T2) (T3)

∅, X ∈ τ . If .G1 , G2 ∈ τ , then .G1 ∩ G2 ∈ τ . If  I is an (finite or infinite) indexing set and .Gi ∈ τ for every .i ∈ I , then . i∈I Gi ∈ τ . .

A set .C ⊆ X is closed if .X \ C is open. A set .U ⊆ X in a topological space .(X, τ ) is a neighbourhood of .x ∈ X if there exists an open set G such that .x ∈ G ⊆ U . A basis of neighbourhoods of x is a family of sets .(Ui )i∈I such that (BN1) .Ui is a neighbourhood of x for every .i ∈ I . (BN2) For every neighbourhood V of x, there is .i0 ∈ I such that .Ui0 ⊆ V . Remark 2.1 It is an easy exercise to see that a set in X is open if and only it is a neighbourhood of all its elements. If each point has a basis of neighbourhoods .(Ui (x))i∈I , then G is open if and only if for all .x ∈ G there exists .i ∈ I such that .Ui (x) ⊆ G. Remark 2.2 Let X be a set with two topologies .τ1 and .τ2 . If every .τ1 -open set is τ2 -open (that is, .τ1 ⊆ τ2 ), then .τ1 are said to be coarser (also smaller or weaker) than .τ2 and .τ2 is said to be finer (also bigger or stronger) than .τ1 .

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_2

39

40

2 Locally Convex Spaces

If each .x ∈ X has basis of neighbourhoods .(Ui (x))i∈I of .τ1 and .(Vj (x))j ∈J of .τ2 , then it is an easy exercise to show that .τ1 is coarser than .τ2 if and only if for every x, given .i ∈ I there is .j ∈ J such that .Vj (x) ⊆ Ui (x). A topological space .(X, τ ) is said to be Hausdorff if for every .x, y ∈ X, .x = y, there are open sets .G1 and .G2 with .x ∈ G1 and .y ∈ G2 , such that .G1 ∩ G2 = ∅. It is clear from what we have explained so far that the definition is equivalent if we replace ‘open sets’ by ‘neighbourhoods’. A function between two topological spaces .f : (X, τX ) → (Y, τY ) is continuous at .x ∈ X if for every .G ∈ τY with .f (x) ∈ G we have .f −1 (G) ∈ τX (that is, the pre-image of every open set in Y containing .f (x) is an open set in X). Remark 2.3 Suppose that X and Y are topological spaces, .f : X → Y , .(Ui )i∈I is a basis of neighbourhoods of x, and .(Vj )j ∈J is a basis of neighbourhoods of .f (x). Then f is continuous at x if and only if for every .j ∈ J there is .i ∈ I such that −1 (V ). .Ui ⊆ f j Also, f is continuous at x if and only if for every .j ∈ J there is .i ∈ I such that .f (Ui ) ⊆ Vj . Remark 2.4 A sequence .(xn )n in a topological space X converges to some .x ∈ X (we write .lim xn = x or .xn → x and call x the limit of the sequence) if for every neighbourhood U of x there exists .n0 ∈ N such that .xn ∈ U for all .n ≥ n0 . If .(Uj )j ∈J is a basis of neighbourhoods of x, then .xn → x if and only if for every .j ∈ J there exists .nj ∈ N such that .xn ∈ Uj for every .n ≥ nj . Remark 2.5 The notion of sequence is generalized by nets. Let us briefly recall the definition. First of all, a set I is partially ordered if there is a relationship .i ≤ j , defined for certain pairs of elements .i, j ∈ I , which is reflexive (.i ≤ i), transitive (.i ≤ j and .j ≤ k imply .i ≤ k), and antisymmetric (.i ≤ j and .j ≤ i imply .i = j ). A partially ordered set is totally ordered if for every .i, j ∈ I either .i ≤ j or .j ≤ i. A set I is said to be directed if it is partially ordered and for every pair .i, j ∈ I there is some .k ∈ I so that .i ≤ k and .j ≤ k. If X is a topological space, a net in X is any function .I → X, where I is a directed set, and is denoted by .(xi )i∈I (meaning that the index i is mapped to the element .xi ∈ X). Note that a sequence is just a net indexed on the natural numbers. A net .(xi )i∈I in a topological space X converges to some .x ∈ X (we write .xi → x) if for every neighbourhood U of x there exists .i0 ∈ I such that .xi ∈ U for all .i ≥ i0 . If .(Uj )j ∈J is a basis of neighbourhoods of x, then .xi → x if and only if for every .j ∈ J there exists .ij ∈ I such that .xi ∈ Uj for every .i ≥ ij . Take .A ⊆ X, and then .x ∈ X is a closure point of A if .A ∩ U = ∅ for every neighbourhood U of x. The set of all closure points of A is called the closure of A and is denoted by .A. It is the smallest closed set that includes A. The point .x ∈ X ˚ for the is interior to A if there is an open set G so that .x ∈ G ⊆ A. We denote .A set of all interior points of A (called the interior of A). This is the biggest open set contained in A.

2.1 Topological Preliminaries

41

A set .A ⊆ X is dense in X if .A = X. A straightforward computation shows that a set A is dense in X if and only if .A∩U = ∅ for every non-empty open set .U ⊆ X. A space X is separable if it has a countable dense subset. A set .K ⊆ X is compact (we usually denote this by .K  X) if whenever there  are open sets .(Gi )i∈I (being I some infinite indexing set) with .K ⊆ i∈I Gi there are .i1 , . . . , in ∈ I such that K ⊆ Gi1 ∪ . . . ∪ Gin .

.

A set is relatively compact if its closure is compact. If .(X, τ ) is a topological space, then each subset .Y ⊆ X inherits a topology, defined as τ |Y = {G ∩ Y : G ∈ τ } .

.

This is often called the induced topology on Y . The inclusion .i : (Y, τ |Y ) → (X, τ ) is clearly continuous. If .(X1 , τ1 ) and .(X2 , τ2 ) are topological spaces, then the set .X1 × X2 is a topological space with the topology defined by τ1 × τ2 = {G1 × G2 : G1 ∈ τ1 , G2 ∈ τ2 } .

.

This definition carries over in an obvious way for the product of finitely many topological spaces. More generally, if .(Xi , τi ) is an arbitrary  family of topological spaces, the product topology on the product space .X = i∈I Xi is defined in the following way. Given .x = (x i ) ∈ X, a set .U ⊆ X is a neighbourhood of x if (and only if) is of the form .U = i∈I Ui , in such a way that there is a finite .J ⊆ I so that .Ui = Xi for every .i ∈ I \ J and .Ui is a neighbourhood of .xi in .Xi for every .i ∈ J . This is the smallest topology that makes all projections .πi : X → Xi on the i-component continuous. Theorem 2.1 (Tikhonov) If .Ki for .i ∈ I is a family of compact spaces, then the  product space .K = i∈I Ki is also compact. Proof [19, Chapter 3], [20, Chapter 1, § 3], [35, Theorem 7.4.1]



2.1.2 Metric and Normed Spaces Given a non-empty set X, a metric (also distance) on X is any mapping .d : X×X → [0, +∞[ satisfying (D1) .d(x, y) ≥ 0, for all .x, y ∈ X, and if .d(x, y) = 0, then .x = y. (D2) .d(x, y) = d(y, x) for all .x, y ∈ X. (D3) .d(x, y) ≤ d(x, z) + d(z, y) for all .x, y, z ∈ X.

42

2 Locally Convex Spaces

The pair .(X, d) is called a metric space. Example 2.2 There are plenty of examples of such metrics. Probably the most classical one is the Euclidean distance on .X = Kn d2 (x, y) =

n 

.

|xi − yi |2

1 2

,

(2.1)

i=1

for .x = (x1 , . . . , xn ) and .y = (y1 , . . . , yn ) two vectors in .Kn . But there are other examples. Already on .Kn , we can consider dp (x, y) =

n 

.

|xi − yi |p

1

p

(2.2)

i=1

for .1 ≤ p < ∞ and d∞ (x, y) = max |xi − yi | .

.

1≤i≤n

(2.3)

Another example of a metric space is to consider .X =]0, ∞[ and   d(x, y) = log  yx  .

.

(2.4) 

Remark 2.6 A metric on a space defines in a natural way a topology (which we denote by .τd ) as follows: • .∅ is open. • .∅ =  G ⊆ X is open if for each .x ∈ G, there is .ε > 0 such that Bd (x, ε) = {y ∈ X : d(y, x) < ε} ⊆ G .

.

To begin with, (T1) is clearly satisfied. Take now two open sets .G1 and .G2 and fix x ∈ G1 ∩G2 . Then there are .ε1 , ε2 > 0 so that .Bd (x, εi ) ⊆ Gi (for .i = 1, 2). Taking .ε = min{ε1 , ε2 } gives .Bd (x, ε) ⊆ G1 ∩ G2 and shows that (T2) is also satisfied.  Finally, choose open sets .Gi with .i ∈ I and .x ∈ i∈I Gi . Then there exists  .i0 ∈ I such that .x ∈ Gi0 . So there is .ε0 > 0 such that .Bd (x, ε0 ) ⊆ Gi0 ⊆ i∈I Gi , and (T3) is also satisfied. Clearly every metric space is Haussdorff. The set .Bd (x, ε) is called the ball in the metric d, centred at x of radius .ε. It is fairly easy to check that for each .x ∈ X the families .

 Bd (x, ε) ε>0

.

and

 Bd (x, 1/n) n∈N

2.1 Topological Preliminaries

43

define bases of neighbourhoods of x for the topology .τd . Observe that in both cases they consist of open sets. In view of Remark 2.4 and with this at hand, a sequence .(xn )n in a metric space X converges to .x ∈ X if and only if for all .ε > 0 there exists .n0 ∈ N such that n .d(xn , x) < ε for all .n ≥ n0 or, in other words, if .d(xn , x) → 0 (as a sequence of real numbers). Being closed can be described in metric spaces through convergence of sequences. The proof is easy and left as an exercise to the reader. Proposition 2.1 Let C be a subset of a metric space X. The following are equivalent: 1. C is closed. 2. If .(xn )n ⊆ C converges to some x, then .x ∈ C. In a metric space X, the distance between a point .x ∈ X and a set .A ⊆ X is defined as d(x, A) = inf{d(x, a) : a ∈ A} .

.

We consider now a vector space X (over .R or .C). A norm on X is any mapping

· : X → [0, +∞[ which satisfies

.

(N1) . x ≥ 0, for all .x ∈ X and . x = 0 if and only if .x = 0. (N2) . αx = |α| x , for all .α ∈ K, and all .x ∈ X. (N3) . x + y ≤ x + y for all .x, y ∈ X. A pair .(X, · ) is called a normed space. A norm defines a metric (and then a topology) on X in a natural way by doing d(x, y) = x − y for x, y ∈ X .

.

(2.5)

The ball .BX = Bd (0, 1) is often called the open unit ball of X. Example 2.3 Given .x = (x1 , . . . , xn ) ∈ Kn , we can define

x ∞ = max |xk | ,

.

k=1,...,n

which clearly satisfies (N1)–(N3) and, then, defines a norm on .Kn (note that the distance defined in (2.3) is associated with this norm). For .1 ≤ p < ∞, we may define

x p =

n 

.

k=1

|xk |p

1

p

.

44

2 Locally Convex Spaces

This clearly satisfies (N1) and (N2). The inequality in (N3) is known as Minkowski inequality: n  .

|xk + yk |p

1

p

≤

k=1

n 

|xk |p

1

p

+

n 

k=1

|yk |p

1

p

,

(2.6)

k=1

a proof of which can be found in [13, (2.11.4)], [24, Proposition 7.1], or [9, Theorem 1.12]. So, . · p also defines a norm on .Kn , with which the distances defined on (2.1) and (2.2) are associated. For .x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Kn , we obviously have n  .

|xk yk | ≤ x 1 y ∞ .

k=1

This simple remark extends to other pairs of norms. If .1 < p < ∞ and . p1 + p1 = 1 (we say that p and .p are conjugate to each other, from now on we use the convention that 1 and .∞ are conjugated exponents), then n  .

|xk yk | ≤ x p y p .

(2.7)

k=1

This is known as Hölder’s inequality, and a proof can be found in [13, (2.8.3)], [24, Proposition 7.6] or [9, Theorem 1.12]. 

Example 2.4 There are metric spaces whose distance cannot be obtained from a norm as in (2.5). One example is the distance defined in (2.4), but in this case, the metric space is .]0, +∞[, which is not a vector space (hence no norm can be defined on it). There are, however, vector spaces on which a metric can be defined that cannot be obtained from a norm as in (2.5). For .x, y ∈ R, consider d(x, y) =

.

|x − y| . 1 + |x − y|

It is not difficult to see that this defines a metric on .R. To begin with, (D1) and (D2) are obviously satisfied. To see that (D3) also holds, note that the function .f (s) = s 1+s is increasing and .|x − y| ≤ |x − z| + |z − y| for all .x, y, z. Then   d(x, y) = f |x − y| ≤ f |x − z| + |z − y| =

.

=

|x − z| + |z − y| 1 + |x − z| + |z − y|

|z − y| |x − z| + 1 + |x − z| + |z − y| 1 + |x − z| + |z − y| ≤

|z − y| |x − z| + = d(x, z) + d(z, y) . 1 + |x − z| 1 + |z − y|

2.1 Topological Preliminaries

45

This shows that .(R, d) is a metric space (and .R is obviously a vector space). If there were a norm satisfying (2.5), we would have . x = d(x, 0) for every .x ∈ R. Taking 2 1 

.x0 = 1, we have . x0 = , but . 2x0 = 2 3 = 2 x0 , and (N2) is not satisfied. Example 2.5 The norms defined for finite tuples in Example 2.3 can be easily N extended to sequences. Given a sequence .x = (xn )∞ n=1 ∈ K , we can consider ∞ 

|xn |p for 1 ≤ p < ∞ and sup |xn | .

.

n∈N

n=1

Note that these may be finite or infinite. For .1 ≤ p < ∞, we consider the set ∞ 

 N p = x = (xn )∞ ∈ K : |xn |p < +∞ for 1 ≤ p < ∞ , n=1

.

n=1

and for .x ∈ p , we define

x p =

∞ 

.

|xn |p

1

p

.

(2.8)

n=1

Also, for .p = ∞, we define the space

 ∞ = x ∈ KN : x ∞ = sup |xi | < +∞ .

.

i∈N

Note that for .1 ≤ p ≤ ∞, the mapping . · p clearly satisfies (N1) and (N2). Minkowski’s inequality (2.6) extends to series, so that if .x, y ∈ p then .x + y ∈ p and

x + y p ≤ x p + y p .

.

Hence, they also satisfy (N3), and .p is a normed space for every .1 ≤ p ≤ ∞. Hölder’s inequality also extends to sequences in the following way: if .x ∈ p and  .y ∈ p  (where p and .p are conjugated exponents), then .(xn yn )n ∈ 1 and

(xn yn )n 1 ≤ x p y p .

.

Another important example is the space of null sequences

 c0 = x ∈ KN : lim xn = 0 .

.

n→∞

(2.9)

This is clearly a vector subspace of .∞ , and the topology defined by . · ∞ on .c0 is the induced topology.

46

2 Locally Convex Spaces

For each .n ∈ N, we define the sequence .en , which has 1 in the nth coordinate and 0 elsewhere. All these clearly belong to .p for every .1 ≤ p ≤ ∞, and . en p = 1 for every n and p. These sequences are commonly known as the canonical vectors.

 Example 2.6 The spaces defined in the previous example can be seen as a particular case of a wider class of spaces of functions. Let . be an arbitrary measure space and .μ a positive measure. Let us briefly recall that in this context we say that a given property is satisfied for almost every (a.e. for short) .x ∈  if there is .1 ⊆  with .μ(1 ) = 0 such that the property holds for every .x ∈  \ 1 . For .1 ≤ p < ∞, we write .Lp (μ) for the space of all measurable functions .f :  → K (where .K can be either .R or .C) so that . |f |p dμ < ∞ 

and define, for each such function,



f p =

1 p |f | dμ . p

.



For .p = ∞, we define .L∞ (μ) as the space of measurable functions f for which

f ∞ = inf{M : |f (x)| ≤ M a.e. x ∈ } < ∞ .

.

Again we have a version of Minkowski’s inequality (2.6): if .1 ≤ p ≤ ∞ and f, g ∈ Lp (μ), then .f + g ∈ Lp (μ) and

.

f + g p ≤ f p + g p ,

.

as well as Hölder’s inequality (2.7): if .f ∈ Lp (μ) and .g ∈ Lp (μ) (where p and .p are conjugated), then .fg ∈ L1 (μ) and

fg 1 ≤ f p g p .

.

(2.10)

We clearly have that . f p = 0 for .f = 0 and . αf p = |α| f p . The only feature left to be a norm is to have that . f p = 0 implies .f = 0, but this is not true. To overcome this, we define a relation of equivalence in .Lp (μ) by saying that .f ∼ g if  .μ {x ∈  : f (x) = g(x)} = 0 and defining .Lp (μ) as the space of all equivalence classes. If f and g are in the same class of equivalence, then . f p = g p . So, on .Lp (μ), we may define . · p as . f p for any representative of the equivalence class. With this, .Lp (μ) is a normed space for every .1 ≤ p ≤ ∞. A complete account on these spaces can be found, for example, in [28, Chapter 3]. In particular, Minkowski’s and Hölder’s inequalities are [28, Theorems 3.8 and 3.9]. These can also be found in [24, Theorem 13.4 and Corollary 13.3] or [9, Theorems 1.19 and 1.20].

2.2 Seminorms

47

Let us note that the spaces defined in Example 2.5 are a particular case of this construction, taking as measure space .N with the counting measure. When . is some subset of .Rd and .μ = md is the Lebesgue measure, we will write .Lp () instead of .Lp (md ). 

Example 2.7 Another classical example (within our setting) of normed space is that of continuous functions on a closed, bounded interval. Let .−∞ < a < b < ∞ and consider C[a, b] = {f : [a, b] → K : such that f is continuous} .

.

This is a vector space and

f = sup |f (x)|

.

x∈[a,b]

(that as a matter of fact, since .[a, b] is compact, can be taken as a maximum) defines a norm on .C[a, b]. Let us note that a sequence .(fn )n ⊆ C[a, b] converges to .f ∈ C[a, b] if .

lim fn − f = 0 .

n→∞

In other words, the convergence in the norm is the uniform convergence (see Remark 1.1). 

A sequence in a normed space is called Cauchy if for all .ε > 0 there exists n0 ∈ N such that . xn − xm < ε for every .n, m ≥ n0 . A normed space is said to be complete if every Cauchy sequence is convergent.

.

Theorem 2.8 The space .(C[a, b], · ) is complete. Proof Proposition 1.1 implies that every Cauchy sequence .(fn )n ⊆ C[a, b] converges uniformly to some function that, by Theorem 1.4, is continuous because the .fn s are so. 

With exactly the same idea, one can show that if .K ⊆ Rd is compact, then .C(K) (the space of continuous functions on K) with the norm . f = supx∈K |f (x)| is a complete normed space. Complete normed spaces are known as Banach spaces. All normed spaces given in Examples 2.3, 2.5, and 2.6 are Banach spaces.

2.2 Seminorms We have seen in Theorem 2.8 that .C[a, b] (or, more generally .C(K) for some compact .K ⊆ Rd ) with the supremum norm is a Banach space. Our next step is

48

2 Locally Convex Spaces

to give a structure to the space of continuous functions on an open set . ⊆ Rd . Here we cannot define a norm, and we need to develop different tools. If E is a vector space, a seminorm is a mapping .p : E → [0, +∞[ satisfying (S1) .p(x) ≥ 0 for all .x ∈ E. (S2) .p(λx) = |λ|p(x) for all .λ ∈ K and all .x ∈ E. (S3) .p(x + y) ≤ p(x) + p(y) for all .x, y ∈ E.

2.2.1 Locally Convex Topology With a similar idea as that for a metric (recall Remark 2.6), if we have a family of seminorms that behaves well enough, we can define a topology on E. To begin with, we need a family of seminorms on E (which we call .P ) satisfying ∀p1 , p2 ∈ P, ∃p ∈ P : max{p1 (x), p2 (x)} ≤ p(x) ∀x ∈ E .

.

(2.11)

This family induces a topology on E (which we denote by .τP ) by doing • .∅ is open. • .∅ =  G ⊆ X is open if for every .x ∈ G there is some .p ∈ P and .ε > 0 so that Bp (x, ε) = {y ∈ X : p(y − x) < ε} ⊆ G .

.

It is easy to see that .τP is a topology and that, for each .x ∈ E, {Bp (x, ε) : p ∈ P, ε > 0}

.

(2.12)

is a basis of neighbourhoods of x consisting of open sets. As a matter of fact, the basis on each point can be expressed in terms of a basis at 0 since Bp (x, ε) = x + Bp (0, ε)

.

(2.13)

for all .p ∈ P and all .ε > 0. A set A in some vector space E is called convex if {λx + (1 − λ)y : 0 ≤ λ ≤ 1} ,

.

(2.14)

for every .x, y ∈ A and balanced if .λx ∈ A for every .x ∈ A and .λ ∈ K with .|λ| ≤ 1. Sets that are balanced and convex are called absolutely convex. As a straightforward consequence of (S2) and (S3), the set .Bp (x0 , ε) is convex for every p, .x0 , and .ε, and .Bp (0, ε) is absolutely convex for every p and .ε. We denote by .U0 (E) the set of absolutely convex neighbourhoods of 0.

2.2 Seminorms

49

The property of being absolutely convex is clearly preserved by taking intersections, and, for a given set A we can define the absolutely convex hull as (A) =

.

 {B ⊆ E : B absolutely convex and A ⊆ B} .

It is a simple exercise to show that a set B is absolutely convex if and only if .λx + μy ∈ B for every .x, y ∈ B and .λ, μ ∈ K with .|λ| + |μ| ≤ 1. Then the absolutely convex hull of a set A can be reformulated as (A) =

n 

.

λi xi : x1 , . . . , xn ∈ A, λ1 , . . . , λn ∈ K,

i=1

n 

 |λi | ≤ 1, n ∈ N .

i=1

(2.15) A set A in a vector space E is absorbing if for every .x ∈ E there is .λ > 0 such that .x ∈ μA for every .μ ∈ K with .|μ| ≥ λ. If A is balanced, then this is equivalent to   . λA = {λx : x ∈ A} = E . λ>0

λ>0

Given a seminorm p and .x ∈ E so that .p(x) > 0, defining .λ = p(x) + 1, we have that .z = λ1 x ∈ Bp (0, 1) and .x = λz. Hence .Bp (0, 1) is absorbing. Proposition 2.2 Let E be a vector space with a family of seminorms .P satisfying (2.11). Then .(E, τP ) is Hausdorff if and only if for all .x ∈ E, .x = 0 there is .p ∈ P such that .p(x) > 0. Proof We suppose first that .(E, τP ) is Hausdorff and choose .x ∈ E with .x = 0. Then we can find .G1 , G2 ∈ τ with .0 ∈ G1 , .x ∈ G2 , and .G1 ∩ G2 = ∅. By the definition of the topology, there is some .ε0 > 0 such that .Bp (0, ε0 ) ⊆ G1 . This implies that .x ∈ Bp (0, ε0 ) and then .p(x) ≥ ε0 > 0. This gives one implication. To prove the converse, take .x, y ∈ E with .x = y. Then .y − x = 0 and we can find .p ∈ P such that .p(y − x) = ε > 0. We define the open sets .G1 = Bp (x, ε/2) and .G2 = Bp (y, ε/2). Obviously .G1 ∩ G2 = ∅, x ∈ G1 , and .y ∈ G2 . Hence .(E, τP ) is Hausdorff. 

With all this at hand, we call (Hausdorff) locally convex space (in short, l.c.s.) to every vector space E provided with a family of seminorms .P satisfying the following properties: (LC1) For every .p1 , p2 ∈ P, there exists .p ∈ P such that .max{p1 , p2 } ≤ p. (LC2) For all .x ∈ E, .x = 0, there is .p ∈ P such that .p(x) > 0. Note that by condition (LC1) the family of seminorms defines a topology .τP . On the other hand, condition (LC2) and Proposition 2.2 give that .(E, τP ) (which we also denote by .(E, P)) is Hausdorff. So from now on and for convenience, we assume that every locally convex space is already a Hausdorff topological space without

50

2 Locally Convex Spaces

mentioning it explicitly. Note also that, by (2.12) and (2.14), every point in a locally convex space has a basis of neighbourhoods consisting of convex sets. Also, 0 has a basis of neighbourhoods consisting of absolutely convex sets. Note also that every point .x0 in a locally convex space E has a basis of closed neighbourhoods, since, for example, .Bp (x0 , ε/2) ⊆ {x ∈ E : p(x − x0 ) ≤ ε/2} ⊆ Bp (x0 , ε) for every seminorm p and .ε > 0. Example 2.9 Let us consider the space .KN of all sequences .x = (xn )n∈N in .K. For each .k ∈ N and .x ∈ KN , we set pk (x) = max{|x1 |, . . . , |xk |} .

.

Each one of these is a seminorm, and they clearly satisfy (LC1) and (LC2) (for the latter just take k with .xk = 0). The space .KN endowed with the locally convex Hausdorff topology defined by this family of seminorms is denoted by .ω. 

Remark 2.7 A straightforward application of Remark 2.4 allows to reformulate the convergence of sequences in terms of seminorms. Then, a sequence .(xk )k ⊆ E in a locally convex space .(E, P) converges to .x0 if and only if for all .p ∈ P and .ε > 0 there is .k0 such that .p(xk − x0 ) < ε for every .k ≥ k0 . A subset A of an l.c.s. .(E, P) is said to be bounded if .

sup p(x) < +∞ x∈A

for every .p ∈ P or, equivalently, if for each .U ∈ U0 (E) there is .λ > 0 so that A ⊆ λU (that is, .x/λ ∈ U for every .x ∈ A). A straightforward use of the definition yields the following.

.

Proposition 2.3 Let E be a locally convex space and .A, B ⊆ E. Then 1. If both are bounded, then so also is . (A ∪ B). 2. If A is bounded and .B ⊆ A, then B is bounded. 3. If A is finite, then it is bounded. Example 2.10 If .(E, · ) is a normed space, then clearly .A ⊆ E is bounded if and only if there is .M > 0 such that . x ≤ M for every .x ∈ A.



2.2.2 Continuity We see now how some of the basic properties on continuity that we formulated in terms of neighbourhoods can be stated using the seminorms. If .(E, P) and .(F, Q) are two locally convex spaces, a straightforward computation from Remark 2.3 shows that a mapping .f : E → F is continuous at .x ∈ E if and only if for every .q ∈ Q and .ε > 0 there are .p ∈ P and .δ > 0 such that, if .p(x − y) < δ,

2.2 Seminorms

51

then .q(f (x) − f (y)) < ε. An important property is that the sum and the product by scalars are continuous. Proposition 2.4 Let .(E, P) be a locally convex space, then the mappings 1. .S : (E, P) × (E, P) → (E, P) given by .S(x, y) = x + y 2. .P : K × (E, P) → (E, P) given by .P (λ, x) = λx are continuous. Proof First of all, given .x0 , y0 ∈ E, .ε > 0, and .p ∈ P, it is easy to check that if p(x − x0 ) < ε/2 and .p(y − y0 ) < ε/2, then .p((x + y) − (x0 − y0 )) < ε. This gives the continuity of S. On the other hand, given .x0 ∈ E, .λ0 ∈ K, .ε > 0, and .p ∈ P, if

.

|λ − λ0 |
0. Then Bp (0, δ) = {x : p(x) < δ} ⊆ Bq (0, ε) = {x : q(x) < ε}

.

if and only if .q ≤ εδ p. Proof Let us suppose first that .Bp (0, δ) ⊆ Bq (0, ε) and choose .x ∈ E. First of all, if .p(x) = 0, then .p(λx) = λp(x) = 0 < δ for all .λ > 0, and this gives .λq(x) = q(λx) < ε for every .λ > 0. Hence .q(x) = 0. On the other hand, if .p(x) > 0, we fix .0 < r < 1 to have

rδx .p p(x)  Then .q

rδx p(x)

 = rδ < δ .

 < ε and this gives q(x)
0 we can find .δ > 0 so that Bp (x, δ) ⊆ Bq (x, ε) .

.

But by Remark 2.2, this implies that .τQ is coarser that .τP (that is, .τQ ⊆ τP ). Proposition 2.5 Let .(E, P) be a locally convex space. A seminorm q defined on E is continuous if and only if there exist .p ∈ P and .λ > 0 so that q(x) ≤ λp(x) for all x ∈ E .

.

(2.16)

Proof The result is a straightforward consequence of Remark 2.3 and Lemma 2.1. 

If the mapping is linear, then the situation is nice: if it is continuous at 0, then it is continuous everywhere, and moreover this is much easier to check. Theorem 2.11 Let .(E, P) and .(F, Q) be locally convex spaces and .T : E → F linear. The following are equivalent: 1. T is continuous. 2. T is continuous at 0. 3. For every .q ∈ Q, there exist .p ∈ P and .C > 0 so that .q(T x) ≤ Cp(x) for all .x ∈ E. 4. If .V ∈ U0 (F ), then .T −1 (V ) ∈ U0 (E). 5. For each .V ∈ U0 (F ), there exists .U ∈ U0 (E) so that .T (U ) ⊆ V . Proof Note first that 1 obviously implies 2. Suppose now that T is continuous at 0 and take some .q ∈ Q. We can find .δ > 0 and .p ∈ P so that .q(T x) < 1 whenever .p(x) < δ. We want to show that q(T x) ≤

.

2 p(x) δ

for every x. We consider two cases. If .p(x) = 0, then .p(λx) = 0 for every .λ > 0, and then .λq(T x) = q(T (λx)) < 1. Since this holds for every .λ > 0, we have .q(T x) = 0 and the claim holds.   δx δx Now, if .p(x) > 0, then .p 2p(x) = 2δ < δ. Therefore, .q T ( 2p(x) ) < 1, and we conclude .q(T x) < 2δ p(x), and this finally yields the claim. Let us assume that 3 holds, and let us see that T is continuous at an arbitrary .x0 ∈ X. We take, then, .q ∈ Q and .ε > 0. Since 3 holds, we can find a seminorm p and some .C > 0 satisfying the inequality. If .p(x − x0 ) < ε/C, then q(T x − T x0 ) = q(T (x − x0 )) ≤ Cp(x − x0 ) < ε ,

.

and T is continuous at .x0 .

2.2 Seminorms

53

This shows the equivalence between 1, 2, and 3. The fact that 4 and 5 are equivalent to 2 is a simple exercise using the definition of continuity and the linearity of T .

 A linear map .T : E → F is an isomorphism if it is one to one, surjective, and continuous and .T −1 : F → E is also continuous. In this case, X and Y are called isomorphic and we write .E ∼ = F. Remark 2.9 A particular case of Theorem 2.11 is when we have normed spaces. Then the system of seminorms reduced to one in each space, and the condition in Theorem 2.11–3 is simply to have a constant .C > 0 so that

T x F ≤ C x E

.

(2.17)

for every .x ∈ E. It is an easy exercise to check that, in this case,

T = inf{C : satisfying (2.17)}

.

defines a norm for the continuous linear operators between E and F , which can also be computed as . T = sup x E ≤1 T x F . If E and F are locally convex spaces, we define the space L(E, F ) = {T : E → F | linear and continuous} .

.

(2.18)

If .E = F , we simply write .L(E) = L(E, E). Note that if .A ⊆ E is bounded and T : E → F is linear and continuous, then .T (A) is bounded in F .

.

Example 2.12 Given a sequence .(αi )i∈N ∈ KN , we define .Tα : 1 −→ KN by .Tα ((xi )i ) = (αi xi )i . If we consider .Tα : 1 → ω (recall Example 2.9), then it is obviously well defined and continuous (since the topology is given by the pointwise convergence) for every .(αi )i . If we take a different space as target, then the situation changes. The following statements are equivalent: 1. .T : 1 → 1 is well defined and continuous. 2. .T 1 ⊆ 1 . 3. .α ∈ ∞ . That 1 implies 2 is trivial. To see that 2 implies 3, take some .α ∈ ∞ . For each k ∈ N, we can find .nk so that .|αnk | ≥ k. With this in mind, we define the sequence

.

 xn =

.

1 k2

if n = nk

0

otherwise .

54

2 Locally Convex Spaces

Clearly .x = (xn )n ∈ 1 but .Tα (x) ∈ 1 , and this completes the argument. Finally, if α ∈ ∞ , then

.

T x 1 =

∞ 

.

∞   |αn xn | ≤ sup |αk | |xn | = α ∞ x 1 k∈N

n=1

n=1

for every .x ∈ 1 , which shows that .Tα is well defined and continuous.



Suppose that H is a family of continuous functions .f : X → F (where X is a Hausdorff topological space and .(F, Q) is an l.c.s.). Given .x0 ∈ X, .q ∈ Q, and .ε > 0, the continuity of each .f ∈ H implies that there is a neighbourhood .V = Vf so that .q(f (x) − f (x0 )) < ε for every .x ∈ Vf . Of course, the neighbourhood depends not only on f but also on q and .ε, but we want to stress here the dependence on f . Now, H is said to be equicontinuous at .x0 if the neighbourhood can be taken independent of f (that is, the same V works for all f s in H ); more precisely, if for every .q ∈ Q and .ε > 0, there is a neighbourhood V of x such that q(f (x) − f (x0 )) < ε

.

(2.19)

for every .x ∈ V and every .f ∈ H . The family H is said to be equicontinuous if it is equicontinuous at .x0 for every .x0 ∈ X. The following result, dealing with equicontinuous families of linear mappings between locally convex spaces, follows with essentially the same proofs as its analogues for single linear mappings (Theorem 2.11). Proposition 2.6 Let .(E, P) and .(F, Q) be two l.c.s. and .H ⊆ L(E, F ). The following conditions are equivalent: 1. H is equicontinuous at 0. 2. H is equicontinuous at every .x ∈ E. 3. For each .q ∈ Q, there is .p ∈ P and a constant .C > 0 such that .q(T (x)) ≤ C p(x) for every .x ∈E and every .T ∈ H . 4. If .V ∈ U0 (F ) then . T ∈H T −1 (V ) ∈ U0 (E). 5. For each .V ∈ U0 (F ) there exists .U ∈ U0 (E) so that .T (U ) ⊆ V for every .T ∈ H . Remark 2.10 Given a sequence .(Tn )n ⊆ L(E, F ) such that .limn Tn x exists (in F ) for every .x ∈ E, then the mapping .T : E → F defined as .T x = limn Tn x is clearly linear. If .(Tn )n is equicontinuous, then it is an easy exercise to show that T is also continuous. Compare this with Proposition 2.16. Remark 2.11 Suppose that .(Tn )n ⊆ L(E, F ) is equicontinuous. If B is dense in A and .Tn y → 0 for every .y ∈ B, then it is easy to see that .Tn x → 0 for every .x ∈ A. In order to see this, take some continuous seminorm q and fix .ε > 0. Being .(Tn )n equicontinuous, we can find a seminorm p so that .p(Tn z − z) ≤ p(z) for every .z ∈ E. Now, by density, there is .y ∈ B with .p(y − x) < ε/2. Finally, since .Tn y → 0 , there is .n0 so that .p(Tn y) < ε/2 for every .n ≥ n0 . With all this, if

2.2 Seminorms

55

n ≥ n0 , then

.

q(Tn x) ≤ q(Tn (x − y)) + q(Tn y) ≤ p(x − y) + q(Tn y) < ε ,

.

which yields our claim.

2.2.3 Metrizable Locally Convex Spaces An l.c.s. E is metrizable if there is a metric on E that defines the same topology. Our next step is to find conditions that ensure that a given l.c.s. is metrizable. Theorem 2.13 A locally convex space .(E, P) is metrizable if and only if there exists a sequence .(pn )n of continuous seminorms on .(E, P) such that 1. .pn ≤ pn+1 for all .n ∈ N. 2. For each .p ∈ P, there is m such that .p ≤ pm . In this case, the metric can be taken invariant by translations. Before we go to the proof, let us note that if every .pn is continuous on .(E, P), then by Remark 2.8, .τ(pn ) ⊆ τP . On the other hand, if 2 is satisfied, then again Remark 2.8 gives .τP ⊆ τ(pn ) . All in all, if the family .(pn )n satisfies the conditions of the theorem, then the topologies .τ(pn ) and .τP coincide. Proof Suppose that E is metrizable. Then, there exists a distance d such that .τd and τP coincide. Then, by Remark 2.2, for each n, we can find .q˜n ∈ P and .ε˜ n so that .Bq˜n (0, ε ˜ n ) ⊆ Bd (0, 1/n). Now we define .

qn (x) = max{q˜1 (x), . . . , q˜n (x)}

.

for x ∈ X

εn = min{˜ε1 , . . . , ε˜ n } . We obviously have Bqn (0, εn ) ⊆ Bq˜n (0, εn ) ⊆ Bd (0, 1/n)

.

(2.20)

for all .n ∈ N. Moreover, .Bqn+1 (0, εn+1 ) ⊆ Bqn (0, εn ), and this, by Lemma 2.1, εn qn+1 for every n. gives .qn ≤ εn+1 Now, given .p ∈ P, there is n such that .Bd (0, 1/n) ⊆ Bp (0, 1). This, together with (2.20) and Lemma 2.1, gives .p ≤ ε1n qn . Defining .pn = ε1n qn , we have the desired sequence of seminorms. Conversely let us assume that .(pn )n is a sequence of continuous seminorms satisfying the two conditions in the statement. We define .d : E × E −→ [0, ∞[ by d(x, y) =

.

∞  1 min{1, pn (x − y)} , 2n n=1

56

2 Locally Convex Spaces

which is a translation-invariant metric since it satisfies the following conditions: 0 ≤ d(x, y) ≤ 1 d(x, y) = d(y, x) .d(x, y) ≤ d(x, z) + d(z, y) (because .pn (x − y) ≤ pn (x − z) + pn (z − y)) .d(x + z, y + z) = d(x, y)

1. 2. 3. 4.

. .

Now, we see that the topology .τd given by the metric and the topology .τ(pn ) (which, as we have pointed out, is the same as .τP ) coincide. We begin by seeing that .τ(pn ) ⊆ τd using Remark 2.2. Note first that for each .x ∈ E we can write Bd (x, ε) = x + Bd (0, ε) .

.

With this and (2.13) in mind, we can deal only with neighbourhoods of 0. Take n n (x)} and .1 > ε > 0. If .d(x, 0) < ε/2n , then . min{1,p < 2εn and this gives .pn (x) < ε. 2n Hence Bd (0, ε/2n ) ⊆ Bpn (0, ε) ,

.

which gives the conclusion. On the other hand, to see that .τd ⊆ τ(pn ) , we fix .δ > 0 and take .m ∈ N such that ∞ n . n=m+1 1/2 < δ/2, and for .pm (x) < δ/2, we have (recall that since the sequence of seminorms is increasing, we have .pk (x) < δ/2 for .k = 1, . . . , m) d(0, x) =

m 

.

2−k min{1, pk (x)} +

k=1

∞ 

2−k min{1, pk (x)}

k=m+1


0 there is .k0 such that .pn (xk − xl ) < ε for every .l, k ≥ k0 . Remark 2.12 If E is a locally convex space (not necessarily complete), a sequence (xn )n in E is said to be Cauchy if for every U , convex neighbourhood of 0, there is .n0 ∈ N such that .xn − xm ∈ U for all .n, m ≥ n0 or, equivalently, if for all .p ∈ P and .ε > 0 there is .k0 such that .p(xk − xl ) < ε for every .l, k ≥ k0 . Then E is said to be sequentially complete if every Cauchy sequence in E is convergent. If E is metrizable, we will omit ‘sequentially’ and say that the space is complete. Notice that this coincides with the notion of completeness that we introduced above for normed spaces. .

A metrizable and complete locally convex space is called a Fréchet space. We have some immediate examples of Fréchet spaces. Example 2.14 1. Every Banach space is a Fréchet space. 2. Every closed subspace of a Fréchet space is also a Fréchet space. 3. The family of seminorms .(pk )k defining the topology in .ω (recall Example 2.9) clearly satisfies that .pk ≤ pk+1 for all .k ∈ N. Then, .ω is a Fréchet space. 

More examples are given in Sect. 2.4.

2.3 The Dual of a Locally Convex Space We have dealt so far with linear operators between locally convex spaces. A particularly interesting case is when the target space is the scalar field. Let us look at this with some detail. In general, if E is a vector space, then the algebraic dual (denoted by .E ∗ ) is the space of all linear mappings .u : E → K. If E has more structure, then we may also consider the topological dual (or simply dual space, denoted by .E  ), consisting of all .u : E → K that are linear and continuous. If E is a locally convex space, then using the notation that introduced above we have   .E = L(E, K). The elements of .E are called functionals. As a direct consequence of Theorem 2.11, we have the following.

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2 Locally Convex Spaces

Proposition 2.8 Let .(E, P) be a locally convex space and .u ∈ E ∗ . Then the following are equivalent: 1. .u ∈ E  . 2. u is continuous at 0. 3. There exist .p ∈ P and .C > 0 so that |u(x)| ≤ Cp(x)

.

(2.22)

for every .x ∈ E. 4. There exists a 0-neighbourhood U in E such that .supx∈U |u(x)| < +∞. Remark 2.13 As a consequence of Proposition 2.6, if .(E, P) is an l.c.s., then .B ⊆ E  is equicontinuous if and only if there are .p ∈ P and .C > 0 such that .|u(x)| ≤ Cp(x) for all .x ∈ E and .u ∈ B. Let us describe now the duals of some of the classical spaces that we introduced before. Example 2.15 Let us see that the dual of .1 can be identified with .∞ . Note in first place that if .u ∈ 1 , then by (2.22) there is some .C > 0 so that |u(x)| ≤ C x 1 ,

.

 for every .x ∈ 1 . This in particular means that the sequence . u(en ) n is bounded,  and the mapping .T : 1 → ∞ given by .T (u) = u(en ) n is well defined. It is clearly linear. Let us see that it is also surjective. Given .a = (ai )∞ i=1 ∈ ∞ , we define .ua : 1 → K by ua (x) =

∞ 

.

ai xi ,

(2.23)

i=1

for .x = (xi )i ∈ 1 . Note that, for each .n ∈ N, we have n  .

|ai xi | ≤



sup |ai | 1≤i≤n

i=1

n 

|xi | ≤ a ∞ x 1 .

i=1

Then the series in (2.23) converges, .ua is well defined, and .|ua | ≤ a ∞ x 1 for every .x ∈ 1 . Since it is linear, this shows that it is also continuous, and .ua ∈ 1 . It clearly satisfies .ua (en ) = an for every n, which shows that T is onto. Moreover, this gives for each .u ∈ 1 that u(x) =

∞ 

.

i=1

u(ei )xi ,

2.3 The Dual of a Locally Convex Space

59

for every .x = (xi )i ∈ 1 . Finally, if .T (u) = 0, then .u(ei ) = 0 for every i and u(x) = 0 for every x. This shows that T is injective and yields .1 = ∞ . 

.

Example 2.16 We describe now the dual of .p for .1 < p < ∞. Fix some .u ∈ p , and, for each .i ∈ N take .ωi ∈ K with .|ωi | = 1 so that .u(ei )ωi = |u(ei )|. Now we consider .p so that . p1 + p1 = 1, and, for each .n ∈ N, we take n      x n = |u(e1 )|p −1 ω1 , . . . , |u(en )|p −1 ωn , 0, 0, . . . = |u(ei )|p −1 ωi .

.

i=1

Since u is continuous, there is some .C > 0 so that .|u(x)| ≤ C x p , which in particular implies n  .



|u(ei )|p = |u(x n )| ≤ C

n 

i=1



|u(ei )|(p −1)p

1/p

i=1

=C

n 

|u(ei )|p



1/p

,

i=1

and n  .

|u(ei )|p



1/p

≤C.

i=1

 Since this holds for every n, it implies that . u(ei ) i ∈ p , and the mapping .T :  p → p given by .T (u) = u(en ) n is well defined (and linear). Now, recall that for .a ∈ p and .x ∈ p , Hölder’s inequality (2.7) gives n  .

|ai xi | ≤ a p x p ,

i=1

for every n. Thus, the operator .ua defined on .p as in (2.23) belongs to .p . Proceeding as in Example 2.15, we conclude that T is a linear bijection that gives the identification .p = p . 

Example 2.17 With an up to some point similar idea, one has that for .1 ≤ p < ∞, the dual of .Lp () (recall Example 2.6) can be identified with .Lp (). A proof of this fact can be found in [9, Theorems 2.18 and 2.19]. 

Remark 2.14 It is an easy exercise to show that if E is finite dimensional, then every linear operator defined on E is continuous (and, in particular, the algebraic and the topological duals coincide). However E is infinite dimensional and this may not be the case. Let us show a couple of examples. Take E to be the space of all real polynomials of one variable and endow it with the norm

P = sup |P (x)| .

.

x∈[0,1]

60

2 Locally Convex Spaces

Consider now the operator .T : E → E given by .T (P ) = P  . This is clearly linear, but it is not continuous. To see this, consider, for each .n ∈ N, the polynomial n .Pn (x) = x . These satisfy . Pn = 1 but . T Pn = n, and this shows that (2.17) does not hold, and hence T is not continuous. Let us consider now E the space of functions .f : [−1, 1] → R that are continuous on .[−1, 1] and differentiable on .] − 1, 1[ and endow it with the norm

f =

.

sup |f (x)| . x∈[−1,1]

Define .ϕ : E → R as .ϕ(f ) = f  (0). Again this is linear but not continuous. Take sin(n2 x) . Clearly . fn ≤ n1 but .|ϕ(fn )n | = n. This shows that (2.22) does .fn (x) = n not hold and .ϕ is not continuous. Hence .E  and .E ∗ are in general not equal. What these two examples have in common is that the spaces are not complete. This allows to construct the non-continuous linear mappings. As a matter of fact in every infinite dimensional l.c.s. (also if the space is complete), there exist linear noncontinuous operators. This is more complicated and requires the Axiom of Choice.

2.4 Examples of Spaces 2.4.1 Space of Continuous Functions We have now at hand all tools that we need to define a topology on the space C() = {f :  → K : continuous} ,

.

where . ⊆ Rd is an open set. What we have to do is to define a family of seminorms, and we do it in the following way. For each compact set .K ⊆ , we define pK (f ) = sup |f (x)| .

.

(2.24)

x∈K

It is easy to check that each one of these is a seminorm. Now we consider the family P = {pK : K   compact} .

.

Since the union of two compact sets is again compact and .{x} is compact for every x, this family satisfies (LC1) and (LC2) and, then, defines a locally convex topology (denoted .τco ) on .C(). Our aim is to show that, in fact, it is a Fréchet space (metrizable and complete). But before we get into that, we see that the convergence in this topology has a very concrete meaning.

2.4 Examples of Spaces

61

As a straightforward application of Remark 2.7, we have that a sequence of functions .(fn )n ⊆ C() converges in .τco to some f if and only if .

 lim sup |fn (x) − f (x)| = 0 , n

x∈K τco

for every compact set .K ⊆ . In other words, .fn −→ f if and only if .(fn )n converges to f uniformly on every compact subset of K. For this reason, the topology .τco is known as the topology of uniform convergence over the compact sets or compact-open topology. As we have said, we want to prove that the space .(C(), τco ) is Fréchet. In view of Theorem 2.13, we have to find a countable family of increasing seminorms that generates the same topology. Given .  Rd open, for each .n ∈ N, we consider the set Kn = {x ∈  : x ≤ n, d(x, Rd \ ) ≥ 1/n} .

.

(2.25)

It is a simple exercise to check that the family .(Kn )n satisfies • Every .Kn ⊆  is compact. ˚n+1 for all .n ∈ N. • .Kn ⊆K ˚ • . = ∞ k=1 Kn . In particular, for every compact .K  , there exists .n = n(K) ∈ N such that K ⊆ Kn . A family of sets satisfying these three conditions is called a fundamental sequence of compact sets of .. A fundamental sequence of compact sets of .Rd is given by

.

Kn = {x ∈  : x ≤ n} .

.

Theorem 2.18 If . ⊆ Rd is an open set, then .(C(), τco ) is a Fréchet space. Proof As we already pointed out, the fact that the union of two compact sets is again compact gives (LC1). On the other hand, if .f = 0, then there exists .x0 ∈  for which .f (x0 ) = 0. Taking .K = {x0 }, we have .pK (f ) > 0, and the seminorms satisfy (LC2). We see now that the topology is metrizable. To do that, we take .(Kn )n a fundamental sequence of compact sets of . and denote, for the sake of clarity, .pn = pKn for .n ∈ N. Clearly these seminorms are all continuous and the sequence is increasing. Also, given any compact set .K  , since we have a fundamental sequence of compact sets, there is some n such that .pK ≤ pn . With all this, Theorem 2.13 gives that the topology is metrizable. In order to see that it is complete, take .(fn )n a Cauchy sequence in .(C(), τco ). By Remark 2.12, for each compact set K and each .ε > 0, there is .n0 such that .

sup |fn (x) − fm (x)| < ε for all m > n ≥ n0 . x∈K

(2.26)

62

2 Locally Convex Spaces

In particular, given any .x ∈  and taking .K = {x}, the sequence .(fn (x))n is Cauchy in .K and hence convergent. Let .f :  → K be the pointwise limit, i.e., .f (x) = limn fn (x) for each .x ∈ . To finish, we have to see that .f ∈ C() and .fn → f in .τco . Let .K   be a compact set. Then, from (2.26), we have that for all .ε > 0 there is some .n0 ∈ N such that |fn (x) − fm (x)| < ε for all m > n ≥ n0 and all x ∈ K .

.

Letting .m → ∞, this yields |fn (x) − f (x)| ≤ ε for all n ≥ n0 and all x ∈ K .

.

Hence .(fn )n converges uniformly on K for every compact set K of . or, in other words, .fn → f in .τco . Moreover, this implies that .f ∈ C(K) for every compact .K ⊆  and this, by Theorem 1.4, gives .f ∈ C(). 

Example 2.19 We consider .(C(), τco ). If .v :  → [0, +∞[ is continuous, then the set .Av = {f ∈ C() : |f | ≤ v} is bounded in .(C(), τco ). To see this, take a compact set K in . and denote .M = supx∈K v(x). Then, .

sup pK (f ) = sup

f ∈Av



f ∈Av

 sup |f (x)| ≤ sup v(x) = M . x∈K

x∈K



2.4.2 Köthe Echelon Spaces A Köthe matrix is a sequence .(an )n∈N of functions .an : N → [0, +∞[ such that (K1) .an (i) ≤ an+1 (i) for all .i ∈ N and .n ∈ N. (K2) For all .i ∈ N, there exists .n ∈ N such that .an (i) > 0. Example 2.20 Define .an (i) = ni . Then A can be represented by a matrix

n1 2 . 3 4

i 1 2 1 1 2 22 3 32 4 42 .. .. . .

3 1 23 33 43 .. .

4 1 24 34 44 .. .

··· ··· ··· ··· .. . 

2.4 Examples of Spaces

63

Example 2.21 Take .an (i) = (1 − n1 )i , and then A is

n1 2 . 3 4

i 1 2 3 4 0 0 0 0 2 3 1/2 1/2 1/2 1/24 2/3 (2/3)2 (2/3)3 (2/3)4 3/4 (3/4)2 (3/4)3 (3/4)4 .. .. .. .. . . . .

··· ··· ··· ··· .. . 

Example 2.22 Let .an (i) = i n . Then A is the matrix

n1 2 . 3 4

i 1 2 1 2 1 22 1 23 1 24 .. .. . .

3 3 32 33 34 .. .

4 4 42 43 44 .. .

··· ··· ··· ··· .. . 

We define now a scale of spaces associated with a Köthe matrix that, in some sense, are an extension of the .p -spaces that we introduced in Example 2.5. Given .1 ≤ p < ∞ and a Köthe matrix .A = (an )n , the Köthe echelon space of order p is defined as N λp (A) = {x = (xi )∞ i=1 ∈ K : qn (x) =

∞ 

.

an (i)|xi |p

1

p

< +∞, ∀ n ∈ N} ,

i=1

and, for .p = ∞, N λ∞ (A) = {x = (xi )∞ i=1 ∈ K : qn (x) = sup an (i)|xi | < +∞ ∀n ∈ N}.

.

(2.27)

i∈N

Proposition 2.9 For each .1 ≤ p ≤ ∞, the space .λp (A) with the topology generated by the corresponding family of seminorms is Fréchet. Proof Fix p and consider the system of seminorms .(qn )n corresponding to .λp (A). Note first that (K1) immediately implies that the system of seminorms is increasing. On the other hand, if .x = 0, then .xi = 0 for some i and, by (K2), there is some n

64

2 Locally Convex Spaces

for which .an (i) > 0. Hence .pn (x) > qn (x) > an (i)|xi | > 0 and the topology is Hausdorff (thus, by Theorem 2.13, metrizable). We only sketch the proof of the completeness (which finally gives the conclusion) for .p = 1. If .(x k )k ⊆ λ1 (A) is a Cauchy sequence in .λ1 (A), for all .n ∈ N and .ε > 0, there is .k0 ∈ N such that if .k, l ≥ k0 , then ∞  .

an (i)|xik − xil | = pn (x k − x l ) < ε .

i=1

By (K2), for each .i ∈ I , the limit .xi = limk xik exists. We denote .x = (xi )i . Then the proof will be completed by showing first that .x ∈ λ1 (A) (use that every Cauchy sequence is bounded) and then that .x k → x in the topology of .λ1 (A). The 

completeness of .λp (A) with .1 < p ≤ ∞ follows by a similar argument. As we said above, these Köthe spaces are some sort of generalization of the .p spaces. So, we can adapt some of the ideas in Example 2.15 to describe the dual of .λ1 (A) in terms of some sort of analogue of .∞ . Let A be a Köthe matrix so that .a1 (i) > 0 for every .i ∈ N (and then .an (i) > 0 for all .n, i). Define, for each n and i, the number vn (i) =

.

1 , an (i)

and consider the spaces

 N ∞ (vn ) = u = (ui )∞ i=1 ∈ K : sup vn (i)|ui | < ∞ ,

.

i

for each .n ∈ N and k∞ (A) =

∞ 

.

∞ (vn ) .

n=1

Proposition 2.10 Let A be a Köthe matrix so that .a1 (i) > 0 for every .i ∈ N. Then λ1 (A) can be identified with .k∞ (A), with the duality given by

.

x, u =

∞ 

.

xi ui ,

i=1

for .x = (xi )i ∈ λ1 (A) and .u = (ui )i ∈ k∞ (A). Proof To begin with, if .L ∈ λ1 (A) , then there exist .n0 ∈ N and .c > 0 so that |L(x)| ≤ c

∞ 

.

j =1

an0 (j )|xj |

2.4 Examples of Spaces

65

for every .x ∈ λ1 (A). This, in particular,  yields .|L(ei )| ≤ can0 (i) for every i ∈ N, and this immediately gives . L(ei ) i ∈ k∞ (A). Then the mapping .T :  λ1 (A) → k∞ (A) given by .ϕ(L) = L(ei ) i is well defined and clearly linear. Let us see that it is surjective. Given any ∞.u = (ui )i ∈ k∞ (A), let us consider .Lu : λ1 (A) → K given by .Lu (x) = i=1 xi ui for .x = (xi )i ∈ λ1 (A). Since .u ∈ k∞ (A), there is some .n0 ∈ N so that .supi vn0 (i)|ui | < ∞. Then, for .x ∈ λ1 (A), we have .

∞ ∞      |ψ(u)(x)| =  xi ui  ≤ |xi | |ui |

.

i=1

i=1

=

∞ 

∞   |xi |vn0 (i)an0 (i)|ui | ≤ sup |vn0 |ui | an0 (i)|xi | . i

i=1

i=1

This shows that .Lu is well defined and, since it is clearly linear, continuous. In other words, .Lu ∈ λ1 (A) . It is plain that .T (Lu ) = u, so that T is onto. Injectivity follows exactly as in Example 2.15, showing that T is a linear bijection and completing the proof. 

Example 2.23 Let .A = (an )n∈N be a Köthe matrix. We take .w = (wi )∞ i=1 ∈ λ∞ (A) (recall (2.27)) with .wi > 0, for .i ∈ N, and write ∞    |xi | N ≤ 1 . Bw = x = (xi )∞ ∈ K : i=1 wi

.

(2.28)

i=1

We denote .μn = supi an (i)wi < +∞ for every .n ∈ N. Finally, we deduce that for each .n ∈ N and .x = (xi )i ∈ Bw , qn (x) =

∞ 

.

an (i)|xi | =

i=1

n 

 |xi | |xi | ≤ μn ≤ μn . wi wi n

an (i)wi

i=1

This proves that .Bw is bounded in .λ1 (A).

i=1



In fact, starting from this example we can even characterize the bounded sets of λ1 (A).

.

Proposition 2.11 A subset .B ⊆ λ1 (A) is bounded if and only if there exists .w = (wi )∞ i=1 ∈ λ∞ (A) with .wi > 0 for every .i ∈ N such that .B ⊆ Bw . Proof Take some .B ⊆ λ1 (A) bounded. For each .n ∈ N, there is some .Mn > 0 so that .

sup

∞ 

x∈B i=1

an (i)|xi | ≤ Mn ,

66

2 Locally Convex Spaces

and we can consider  Cn = max Mn , n max (an (i), 1) .

.

1≤i≤n

Now, by (K2), for a given .i ∈ N, we can find .n(i) ∈ N so that .an(i) (i) > 0 (and, then, by (K1) also for every .n ≥ n(i)). Then 2n Cn 0 . a (i) n 1≤n≤max(i,n(i)) n∈N an (i)

wi ≥ inf

.

The sequence .w = (wi )∞ i=1 clearly belongs to .λ∞ (A). It is only left to see that B is contained in the set .Bw defined in (2.28). To see this, note first that for each i we have ∞

.

1 an (i)  an (i) = sup n ≤ , wi 2n Cn n∈N 2 Cn n=1

and the latter is finite because .Cn ≥ an (i) for every .n ≥ i. Now, if .x ∈ B, then .

∞  |xi | i=1

wi

≤

∞  i=1

|xi |

∞ ∞ ∞   an (i)   1  1  = a (i)|x | ≤ 1. n i 2n Cn 2n Cn n=1

n=1

i=1

This proves one implication. The converse one is a straightforward consequence of Example 2.23 and Proposition 2.3. 

If the Köthe matrix is nice enough, we can even see that bounded subsets are relatively compact. Proposition 2.12 Let A be a Köthe matrix so that for every .n ∈ N there exists .m > n such that . aamn ∈ c0 . Then, every bounded subset in .λ1 (A) is relatively compact. Proof By Proposition 2.11, it is enough to see that the sets .Bw defined as in (2.28) (with .w ∈ λ∞ (A) and .wi > 0 for every i) are relatively compact. To see this, choose some sequence .(x k )k ⊆ Bw , and let us see that it has a convergent subsequence. Note that .Bwis closed in .KN (with the topology of pointwise convergence) and is contained in . ∞ i=1 {x ∈ K : |x| ≤ wi }, which by Tikhonov’s theorem 2.1 is compact. Hence .Bw is compact, and we can find a subsequence .(x k(j ) )j that converges (pointwise) to some .x 0 = (xi0 )i ∈ Bw . Since .Bw ⊆ λ1 (A), we have that in fact 0 k(j ) → x 0 in .λ (A) as .j → ∞. To see this, fix .n ∈ N .x ∈ λ1 (A). Let us see that .x 1

2.5 Normable Spaces

67

and choose .m > n with . aamn ∈ c0 . Since .w ∈ λ∞ (A), we can find .M > 0 so that wi am (i) ≤ M

.

for every .i ∈ N. Now, given .ε > 0, there is .i0 ∈ N such that .

ε an (i) < am (i) 4M

for every .i ≥ i0 . Finally, since .(x k(j ) )j converges pointwise, we can find .j0 ∈ N so that if .1 ≤ i ≤ i0 − 1, then k(j )

|xi

.

− xi0 |
0 and some .p ∈ P so that q(x) ≤ λp(x)

.

for all .x ∈ E. We say that an l.c.s. .(E, P) is normable if there exists a norm q : X → [0, ∞[ such that the norm topology .τq coincides with the original one.

.

Remark 2.15 The fact that a space is normable basically requires two conditions to be satisfied.

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2 Locally Convex Spaces

1. The norm q is continuous. 2. The topology .τP is coarser than .τq (or .τq is finer than .τP ). These two conditions can be reformulated in different ways. First of all, in terms of neighbourhoods (see Remark 2.8), 1. There are .ε > 0 and .p0 ∈ P so that .Bp0 (0, ε) ⊆ Bq (0, 1). 2. For every .p ∈ P and all .ε > 0, there exists .δ > 0 such that .Bq (0, δ) ⊆ Bp (0, ε). In view of Lemma 2.1, this can also be reformulated by comparing the seminorms 1. There are .λ > 0 and .p0 ∈ P so that .q ≤ λp0 . 2. For every .p ∈ P, there is .μ > 0 such that .p ≤ μq. Proposition 2.13 The Fréchet space .ω (see Example 2.14–3) does not admit a continuous norm. In particular, it is not normable. Proof Suppose .q : ω → [0, ∞[ is a continuous norm. Then there exist n and .λ > 0 so that q(x) ≤ λ max{|x1 |, . . . , |xn |}

.

for every .x ∈ ω. Taking .x0 = (0, . n. ., 0, 1, 0, . . .) ∈ ω, this implies .q(x0 ) = 0, contradicting the fact that q is a norm.

 Proposition 2.14 The Fréchet space .(C(), τco ) does not admit a continuous norm. In particular, it is not normable. Proof Suppose .q : C() → [0, ∞[ is a continuous norm. Then there are some compact set .K   and some .λ > 0 so that q(f ) ≤ λ sup |f (x)|

.

x∈K

for all .f ∈ C(). Let .x0 ∈  \ K and choose .r > 0 such that .B(x0 , r) ⊆  \ K. Defining .f0 :  → K by  f0 (x) =

.

r− x−x0 r

if x − x0 ≤ r

0

otherwise,

we clearly have .f0 ∈ C(), .f0 (x0 ) = 1 and .f0 (x) = 0 for every .x ∈ K. Then 0 < q(f0 ) ≤ λ sup |f0 (x)| = 0 ,

.

x∈K

and this is a contradiction. We finish this section with a condition for a space to be normable.



2.6 Two Theorems on Spaces of Continuous Functions

69

Theorem 2.24 (Kolmogorov) A locally convex space E is normable if and only if there is a bounded 0-neighbourhood in X. Proof To begin with, if E is normable with a norm q, then the set B(0, 1) = {x ∈ X : q(x) < 1}

.

is a bounded 0-neighbourhood. Suppose now that U is a bounded 0-neighbourhood in E. We can find .ε > 0 and .p0 ∈ P so that .Bp0 (0, ε) ⊆ U and, then, .Bp0 (0, ε) is also bounded. Therefore, for each .p ∈ P, there exists .λp > 0 such that sup

.

p(x) < λp ,

x∈Bp0 (0,ε)

and this implies Bp0 (0, ε) ⊆ Bp (0, λp ) .

.

But now Lemma 2.1 yields p(x) ≤

.

λp p0 (x) ε

for all .x ∈ E. Since this holds for every .p ∈ P, it gives .p0 (x) = 0 for every .x =  0 and .p0 is in fact a norm, which (see Remark 2.15) defines the topology of E.



2.6 Two Theorems on Spaces of Continuous Functions 2.6.1 Stone–Weierstraß Theorem Weierstraß proved in 1885 his famous approximation theorem: every continuous function on a closed, bounded interval can be uniformly approximated by polynomials. Later, in 1937, Stone extended this result in the context of unital subalgebras of the space of continuous functions, a result that now is known as the Stone– Weierstraß theorem (see Theorem 2.27 below for a precise statement). We deduce it here from a result by Bishop. The proof is based on a result by Machado, following an approach from [26]. Let us first of all fix our setting. We take a compact Hausdorff space K, and .C(K) = C(K, C) the Banach space of continuous, complex-valued functions (recall Example 2.7, we restrict ourselves for a moment to functions taking values

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2 Locally Convex Spaces

in .C), and A a unital subalgebra of .C(K). For a given .f ∈ C(K) and .F ⊆ K we define df (F ) = inf sup |f (x) − g(x)| .

(2.29)

.

g∈A x∈F

Clearly, if .F1 ⊆ F2 , then .df (F1 ) ≤ df (F2 ). A non-empty .S ⊆ K is said to be A-antisymmetric if for every .h ∈ A with .h(x) ∈ R for every .x ∈ S we have that .h|S is constant. Theorem 2.25 (Machado) Let .f ∈ C(K), then there exists a non-empty, closed, A-antisymmetric .S ⊆ K such that .df (S) = df (K) . Proof Let .F denote the class of all non-empty, closed subsets .F ⊆ K so that df (F ) = df (K). We order it by inclusion and our aim is to see (using Zorn’s lemma, see e.g. [19, Chapter 0]) that it has a minimal element. In order to apply Zorn’s lemma, we consider C, a totally ordered subclass, and let us show that it has a lower bound. Take  .E = {F : F ∈ C} ,

.

which is not empty and closed. Let us see that .df (E) = df (K). To this purpose, pick some .g ∈ A and, for each .F ∈ C, consider the set K(F, g) = {x ∈ F : |f (x) − g(x)| ≥ df (K)} .

.

Since .df (F ) = df (K), the very definition (2.29) gives that .K(F, g) = ∅. It is clearly closed and then, being a subset of a compact Hausdorff space, is compact. Clearly, if .F1 ⊆ F2 belongs to C, then .K(F1 , g) ⊆ K(F2 , g). This altogether yields  .

K(F, g) = ∅ ,

F ∈C

and there is some .x0 ∈ E so that .|f (x0 ) − g(x0 )| ≥ df (K); hence .supx∈E |f (x) − g(x)| ≥ df (K), and (since g was arbitrary) .df (E) ≥ df (K). Then both are equal, .E ∈ F, and it is a lower bound of C. By Zorn’s lemma, there is a minimal element of .F, which we denote by S. It is not empty and closed and satisfies that .df (S) = df (K). So, to complete the proof, it is only left to see that S is also A-antisymmetric. Suppose it is not, then there is .h ∈ A such that .h(x) ∈ R for every .x ∈ S and that is not constant on S. Without loss of generality (taking if necessary a convenient linear combination of h and the constant function 1), we may assume .minx∈S h(x) = 0 and .maxx∈S h(x) = 1. Consider now the sets 



Y = x ∈ S : 0 ≤ h(x) ≤ 23 and Z = x ∈ S :

.

1 3

 ≤ h(x) ≤ 1 ,

2.6 Two Theorems on Spaces of Continuous Functions

71

both not empty, closed, and proper subsets of S. By the minimality of S, neither of these two belongs to .F, and then .df (Y ) < df (K) and .df (Z) < df (K). With this, there are .gY , gZ ∈ A so that .

sup |f (x) − gY (x)| < df (K)

(2.30)

x∈Y

and .

sup |f (x) − gZ (x)| < df (K) .

(2.31)

x∈Z

Now, for each .n ∈ N, we define n

hn = (1 − hn )2 and gn = hn gY + (1 − hn )gZ .

.

Since A is an algebra, .hn , gn ∈ A for every n. If we manage to find .n0 ∈ N so that .

sup |f (x) − gn0 (x)| < df (K) ,

(2.32)

x∈S

this will give .df (S) < df (K) and lead to a contradiction. Before we proceed, let us note that .0 ≤ hn ≤ 1 on S for every n and write .M = supx∈K |gY (x) − gZ (x)|. Take some .x0 ∈ S, and let us consider three possible cases. If .x0 ∈ Y ∩ Z, then, using (2.30) and (2.31), we have, for each n, |f (x0 )−gn (x0 )| ≤ hn (x0 )|f (x0 )−gY (x0 )|+(1−hn (x0 ))|f (x0 )−gZ (x0 )| < df (K) .

.

(2.33) Suppose now that .x0 ∈ Y \ Z (hence .0 ≤ h(x) < 13 ), a simple computation shows that .(1 − t)N ≥ 1 − Nt for every .N ∈ N and .0 ≤ t ≤ 1 and, then, hn (x0 ) ≥ 1 − 2n h(x0 ) ≥ 1 −

 2 n

.

3

.

With this, we have |gn (x0 ) − gY (x0 )| = (1 − hn (x0 ))|gZ (x0 ) − gY (x0 )|
0 and consider the following .τco -neighbourhood of .f0 in H , VK,ε = {f ∈ H : sup |f (x) − f0 (x)| < ε} .

.

x∈K

Now, since H is equicontinuous, we can find a neighbourhood .Vx of x (in X) such that ε for all f ∈ H, and all y ∈ Vx . 4  Being K compact, there are .x1 , . . . , xn ∈ K such that .K ⊆ ni=1 Vxi . Then we consider |f (y) − f (x)|
0. Since H is equicontinuous, there is a neighbourhood Vx of x in X such that

.

|f (y) − f (x)|
0. We fix K, a compact neighbourhood of x and, since H is relatively compact, choose .f1 , . . . , fn ∈ H such that

.

H ⊆ BpK (f1 , ε/3) ∪ . . . ∪ BpK (fn , ε/3) .

.

That is, for each .f ∈ H , there is .j = 1, . . . , n so that |f (x) − fj (x)|
k, then .(sk , uk ) = ck . Therefore   ck − (x, uk ) = |(sn − x, uk )| ≤ sn − x uk = sn − x .

.

Letting .n → ∞ yields .(x, uk ) = ck and completes the proof.



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2 Locally Convex Spaces

Theorem 2.36 Let H be a Hilbert space and .{uk : k ∈ N} an orthonormal sequence. The following statements are equivalent: 1. If .x ∈ H is such that .(x, uk ) = 0 for every k, then .x = 0. 2. For each .x ∈ H , we have

x 2 =

∞ 

.

|(x, uk )|2 .

k=1

3. For each .x ∈ H , we have x=

∞ 

.

(x, uk )uk .

k=1

Proof The statements in 2 and 3 are equivalent by Theorem 2.34. Since 2 clearly implies 1, itis only left to see that this implies 3. Given .x ∈ H , we know from 2 2 . Within the proof of Theorem 2.35, we have (2.41) that . ∞ k=1 |(x, uk )| ≤ x  shown that there is .y ∈ H with .y = ∞ k=1 (x, uk )uk and such that .(x, uk ) = (y, uk ) for every k. Then .(x − y, uk ) = 0 for every k. Since we are assuming that 1 holds, 

this implies that .x = y and completes the proof. An orthonormal sequence in a Hilbert space satisfying any of the equivalent conditions 1–2–3 in Theorem 2.36 is called complete. Corollary 2.3 Let H be a Hilbert space and .{uk : k ∈ N} an orthonormal sequence. If .M = span{uk : k ∈ N} is dense in H , then the sequence is complete. Proof Suppose that .x ∈ H is such that .(x, uk ) = 0 for every k. Then .(x, y) = 0 for every .y ∈ M and, by density .(x, z) = 0 for every .z ∈ H . In particular . x 2 = (x, x) = 0, and, then .x = 0. So, the sequence .(uk )k is complete. 

Proposition 2.21 Let H be a Hilbert space and .{uk : k ∈ N} a complete orthonormal sequence. If .x, y ∈ H , then (x, y) =

∞ 

.

(x, uk )(y, uk ) .

k=1

Proof Since the sequence is complete, we have ∞ ∞   .x = (x, uk )uk and y = (y, uj )uj . k=1

j =1

2.7 A Short Introduction to Hilbert Spaces

85

Then, by the continuity of the scalar product .(·, ·) and the orthonormality of the sequence .(uk )k , we have (x, y) =

.

∞ ∞  ∞   (x, uk )(y, uk ) . (x, uk )(y, uj )(uk , uj ) = k=1 j =1

k=1



We can look again with a new point of view at the problem of the convergence of Fourier series that we already tackled in Sect. 1.3.2. For a given function f , recall the definition of the Fourier coefficient .f(n) (with .n ∈ Z) given in (1.6). Theorem 2.37 The orthonormal sequence .{ψn (x) = einx : n ∈ Z} is complete in .L2 (−π, π ). In particular, for each .f ∈ L2 (−π, π ), we have ∞  .

|f(n)|2 =

n=−∞

1 2π



π −π

|f (x)|2 dx

(2.43)

(this is known as Parseval identity) and

.

n      lim f (x) − f(k)eikx 

n→∞

k=−n

L2 (−π,π )

= 0.

(2.44)

Proof In view of Corollary 2.3, in order to show that the sequence is complete, it is enough to see that the span of .{ψn : n ∈ Z} (that is, the space of trigonometric polynomials) is dense in .L2 (−π, π ). To begin with, the space .C(−π, π ) is dense in .L2 (−π, π ) (see, e.g., [28, Theorem 8.14]) and moreover . g L2 ≤ g = supx∈],π,π ] |g(x)| for every .g ∈ C(−π, π ). Therefore, it is enough to show that the trigonometric polynomials are dense in .C(−π, π ). But this is an immediate consequence of Féjer’s theorem (see Corollary 1.2) or of the Stone–Weierstraß theorem 2.27, once we identify .C(−π, π ) with .C(T), the space of continuous functions on the unit circle .T (recall the paragraph right before (1.14)). Hence, .{ψn : n ∈ Z} is a complete orthogonal sequence. Once we have this, (2.43) follows from Theorem 2.36–2 and (2.44) from Theorem 2.36–3. 

Corollary 2.4 The map .T : L2 (−π, π ) → 2 (Z) given by .T (f ) = (f(n))n∈Z is an isometric isomorphism. Proof The map T is clearly linear and, by (2.43), satisfies . T (f ) 2 = f for every .f ∈ L2 (−π, π ). This implies, in particular, that it is injective. Finally, the 

Riesz–Fischer theorem 2.35 shows that T is also surjective.

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2 Locally Convex Spaces

2.8 Exercises 2.1 Let (xn )n be a sequence in a Hausdorff locally convex space such that limn xn = x. Prove that .

1 (x1 + x2 + · · · + xn ) n

also converges to x as n → ∞. 2.2 Show that a set A in a vector space is absolutely convex if and only if λx +μy ∈ B for every x, y ∈ B and λ, μ ∈ K with |λ| + |μ| ≤ 1. Deduce that (2.15) holds. 2.3 Let X be a metric space, x ∈ X, and A ⊆ X. Show that x ∈ A if and only if d(x, A) = 0. 2.4 Let (E, P) be a metrizable locally convex space. Show that for each sequence (An ) n of bounded subsets in E there is a sequence of positive numbers (λn )n such that ∞ n=1 λn An is bounded. 2.5 A sequence (Bn )n of bounded sets in a locally convex space E is called a fundamental sequence of bounded sets if Bn ⊆ Bn+1 for every n and for each bounded set B in E there is n such that B ⊆ Bn . Show that a metrizable locally convex space with a fundamental sequence of bounded sets is normable. 2.6 Show that if 1 ≤ p < q ≤ ∞, then p ⊆ q with continuous inclusion, but they do not coincide. Show that the following inclusions are strict:  .

q ⊆ p ⊆

1≤q 0 such that for each y ∈ Y there is x ∈ X with T x = y and

x ≤ M y . 2.12 Show that a sequence converges in ω if and only if converges coordinatewise; that is, a sequence (x (n) )n (where x (n) = (x (n) )k for every n) converges to x = (xk )k (n) in ω if and only if xk → xk as n → ∞ (in C) for every k. Deduce that the space ω is complete. 2.13 Show that (c0 ) = 1 (recall the definitions from Example 2.5). 2.14 Let c be the Banach space of convergent sequences in C (with the · ∞ norm). 1. Prove that u : c → C given by u(x) = limn xn is a continuous linear form. 2. Show that T : c → c0 given by T (x) = (u(x), x1 − u(x), x2 − u(x), . . .) is an onto isomorphism such that T = T −1 = 2. 2.15 Consider the space ϕ = {x ∈ KN : card{i : xi = 0} < ∞} .

.

Show that ω = ϕ. ∞ 2.16 Let A = (an )n be a Köthe matrix, an = (an (i))∞ i=1 and α = (αn )n=1 a sequence of scalars.

1. Show that the diagonal operator Tα : λ1 (A) → λ1 (A) given by Tα (x) = (αi xi )i is continuous if and only if for every n there is m ≥ n such that .

sup i

an (i)|αi | < +∞ . am (i)

2. Characterise those αs for which the operator Sα a : λ1 (A) → λ1 (A) given by Sα (x1 , . . . , xn , . . .) = (α2 x2 , α3 x3 , . . . , αn xn , α)

.

is continuous. 2.17 Characterise the bounded subsets of λp (A) for 1 < p < ∞. 2.18 Let A = (an )n be a Köthe matrix, an = (an (i))∞ i=1 . 1. Show that λ1 (A) admits a continuous norm if and only if there exists n0 ∈ N such that an0 (i) > 0 for every i. 2. Show that λ1 (A) is normable if and only if there exists n0 ∈ N such that for every n ≥ n0 there is Cn > 0 with an (i) ≤ Cn an0 (i) for every i. 2.19 Give an example of a Köthe matrix A for which λ1 (A) is not normable. 2.20 Given a Köthe matrix with positive elements, define, for 1 < q < ∞ the space kq (A), and prove that λp (A) = kp (A).

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2 Locally Convex Spaces

2.21 Show that the space C() of continuous functions in the open set  ⊂ Rn contains bounded sets that for the compact-open topology are not compact. Give explicit examples. 2.22 With the notation from Example 2.19, prove that if A is bounded in (C(), τco ), then there is a continuous function v :  → [0, +∞[ continuous such that A ⊆ Av . Hint: Use partitions of unity. Further Reading [1–12, 14–25, 27, 29–36]

References 1. Bachman, G., Narici, L.: Functional Analysis. Dover Publications, Inc., Mineola, NY (2000). Reprint of the 1966 original 2. Banach, S.: Théorie des Opérations Linéaires. Éditions Jacques Gabay, Sceaux (1993). Reprint of the 1932 original 3. Bogachev, V.I., Smolyanov, O.G.: Topological Vector Spaces and Their Applications. Springer Monographs in Mathematics. Springer, Cham (2017). https://doi.org/10.1007/978-3-31957117-1 4. Brezis, H.: Functional analysis. In: Sobolev Spaces and Partial Differential Equations. Universitext. Springer, New York (2011) 5. Cerdà, J.: Linear functional analysis. In: Graduate Studies in Mathematics, vol. 116. American Mathematical Society, Providence, RI; Real Sociedad Matemática Española, Madrid (2010). https://doi.org/10.1090/gsm/116 6. Conway, J.B.: A course in functional analysis. In: Graduate Texts in Mathematics, vol. 96. Springer, New York (1985). https://doi.org/10.1007/978-1-4757-3828-5 7. Diestel, J.: Sequences and series in Banach spaces. In: Graduate Texts in Mathematics, vol. 92. Springer, New York (1984). https://doi.org/10.1007/978-1-4612-5200-9 8. Edwards, R.E.: Functional Analysis. Dover Publications, Inc., New York (1995). Theory and applications, Corrected reprint of the 1965 original 9. Fabian, M., Habala, P., Hájek, P., Montesinos, V., Zizler, V.: Banach Space Theory. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Springer, New York (2011). https:// doi.org/10.1007/978-1-4419-7515-7 10. Fabian, M., Habala, P., Hájek, P., Montesinos Santalucía, V., Pelant, J., Zizler, V.: Functional analysis and infinite-dimensional geometry. In: CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC, vol. 8. Springer, New York (2001). https://doi.org/10.1007/978-14757-3480-5 11. Floret, K., Wloka, J.: Einführung in die Theorie der lokalkonvexen Räume. Lecture Notes in Mathematics, No. 56. Springer, Berlin/New York (1968) 12. Grothendieck, A.: Topological Vector Spaces. In: Notes on Mathematics and Its Applications. Gordon and Breach Science Publishers, New York/London/Paris (1973). Translated from the French by Orlando Chaljub 13. Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge Mathematical Library. Cambridge University Press, Cambridge (1988). Reprint of the 1952 edition 14. Heuser, H.G.: Functional Analysis. A Wiley-Interscience Publication. John Wiley & Sons, Ltd., Chichester (1982). Translated from the German by John Horváth 15. Horváth, J.: Topological Vector Spaces and Distributions, vol. I. Addison-Wesley Publishing Co., Reading, Mass./London-Don Mills, Ont. (1966)

References

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16. Jameson, G.J.O.: Topology and Normed Spaces. Chapman and Hall, London; Halsted Press, New York (1974) 17. Jarchow, H.: Locally Convex Spaces. Mathematische Leitfäden. B. G. Teubner, Stuttgart (1981) 18. Kadets, V.: A Course in Functional Analysis and Measure Theory. Universitext. Springer, Cham (2018). https://doi.org/10.1007/978-3-319-92004-7. Translated from the 2006 Russian edition 19. Kelley, J.L.: General topology. In: Graduate Texts in Mathematics, No. 27. Springer, New York/Berlin (1975). Reprint of the 1955 edition 20. Köthe, G.: Topological vector spaces. I. Translated from the German by D. J. H. Garling. In: Die Grundlehren der Mathematischen Wissenschaften, Band 159. Springer, New York Inc., New York (1969) 21. Köthe, G.: Topological vector spaces. II. In: Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 237. Springer, New York/Berlin (1979) 22. Lindenstrauss, J., Tzafriri, L.: Classical Banach spaces. I. In: Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 92. Springer, Berlin/New York (1977). Sequence spaces 23. Lindenstrauss, J., Tzafriri, L.: Classical Banach spaces. II. In: Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], vol. 97. Springer, Berlin/New York (1979). Function spaces 24. Meise, R., Vogt, D.: Introduction to functional analysis. In: Oxford Graduate Texts in Mathematics, vol. 2. The Clarendon Press, Oxford University Press, New York (1997). Translated from the German by M. S. Ramanujan and revised by the authors 25. Pérez Carreras, P., Bonet, J.: Barrelled locally convex spaces. In: North-Holland Mathematics Studies, vol. 131. North-Holland Publishing Co., Amsterdam (1987). Notas de Matemática, 113 26. Ransford, T.J.: A short elementary proof of the Bishop-Stone-Weierstrass theorem. Math. Proc. Camb. Philos. Soc. 96(2), 309–311 (1984). https://doi.org/10.1017/S0305004100062204 27. Robertson, A.P., Robertson, W.: Topological vector spaces. In: Cambridge Tracts in Mathematics, vol. 53, 2nd edn. Cambridge University Press, Cambridge/New York (1980) 28. Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Book Co., New York (1987) 29. Rudin, W.: Functional Analysis, 2nd edn. In: International Series in Pure and Applied Mathematics. McGraw-Hill, Inc., New York (1991) 30. Schaefer, H.H., Wolff, M.P.: Topological vector spaces. In: Graduate Texts in Mathematics, vol. 3, 2nd edn. Springer, New York (1999). https://doi.org/10.1007/978-1-4612-1468-7 31. Taylor, A.E., Lay, D.C.: Introduction to Functional Analysis, 2nd edn. Robert E. Krieger Publishing Co., Inc., Melbourne, FL (1986) 32. Trèves, F.: Topological Vector Spaces, Distributions and Kernels. Dover Publications, Inc., Mineola, NY (2006). Unabridged republication of the 1967 original 33. Valdivia, M.: Topics in locally convex spaces. In: Notas de Matemática [Mathematical Notes], vol. 85. North-Holland Publishing Co., Amsterdam/New York (1982) 34. Wilansky, A.: Modern Methods in Topological Vector Spaces. McGraw-Hill International Book Co., New York (1978) 35. Wilansky, A.: Topology for Analysis. Robert E. Krieger Publishing Co., Inc., Melbourne, FL (1983). Reprint of the 1970 edition 36. Yosida, K.: Functional analysis. In: Classics in Mathematics. Springer, Berlin (1995). https:// doi.org/10.1007/978-3-642-61859-8. Reprint of the sixth (1980) edition

Chapter 3

Duality and Linear Operators

3.1 Hyperplanes We defined the dual of a locally convex space as .L(E, K), the space of continuous, linear scalar-valued mappings (called functionals). We already pointed out in Remark 2.14 that this is different from .E ∗ , the space of linear scalar-valued mappings on E (the algebraic dual of E). Given a linear mapping .u : E → K, we consider its kernel .ker(u) = {x ∈ E : u(x) = 0}. A subspace H of E is called a hyperplane if it is proper and maximal (i.e., if F is a subspace so that .H ⊆ F ⊆ E, then either .H = F or .F = E). Lemma 3.1 Let E be a vector space and H a subspace. Then H is a hyperplane if and only if there is .0 = u ∈ X∗ so that .ker(u) = H . Proof Suppose first that H is a hyperplane and choose .x0 ∈ H . Then H ⊕ [x0 ] = {x + λx0 : x ∈ H, λ ∈ K} = E ,

.

and we can define .u : E → K by doing .u(x + λx0 ) = λ. This clearly satisfies u ∈ E ∗ and .ker(u) = H . On the other hand, if there is .0 = u ∈ E ∗ so that .ker(u) = H , choose .x0 ∈ E u(x) with .u(x0 ) = 0. Given .x ∈ E, we denote .y = x − u(x x0 ∈ ker(u), and .x = 0)

.

y+

u(x) u(x0 ) x0 ,

showing that .E = H ⊕ [x0 ] and that H is a hyperplane.



Lemma 3.2 Let .u, u1 , u2 , . . . , un ∈ E ∗ . The following are equivalent: 1.

n  .

ker(ui ) ⊆ ker(u).

i=1

2. There exist .λ1 , λ2 , . . . , λn ∈ K so that .u = λ1 u1 + λ2 u2 + · · · + λn un . Proof Let us suppose that 1 holds, and let us see by induction on n that so also does 2. We begin with the case .n = 1 and suppose that .ker(u1 ) ⊆ ker(u). If .u1 = 0, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_3

91

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3 Duality and Linear Operators

then .ker(u) = ker(u1 ) = X and .u = 0. Suppose then .u1 = 0 and choose .x0 ∈ E such that .u1 (x0 ) = 0. Then, by Lemma 3.1, .ker u1 is a hyperplane, and for every .x ∈ E, there exist .λ ∈ K and .y ∈ ker(u1 ) ⊆ ker(u) so that .x = λx0 + y. Therefore u(x) = λu(x0 ) = λ

.

u(x0 ) u(x0 ) u1 (x0 ) = u1 (λx0 + y) = λ1 u1 (x) , u1 (x0 ) u1 (x0 )

where .λ1 = uu(x) . Assume now that the statement holds for .n − 1, and let us see 1 (x)  that it holds for n. Suppose, then, that . ni=1 ker(ui ) ⊆ ker(u) and .un = 0. Denote n−1 .H = ker(un ), then . i=1 ker(ui |H ) ⊆ ker(u|H ), and, by the induction hypothesis, we can find .λ1 , . . . , λn−1 ∈ K so that u|H =

n−1 

.

λi ui |H .

i=1

For each .x ∈ H , we have .u(x) −

n

i=1 λi ui (x)

= 0 and

  n−1  . ker u − λi ui ⊇ H = ker(un ) . i=1

But  the hypothesis holds for .n = 1 and this gives some .λn ∈ K so that u − n−1 i=1 λi ui = λn un . This shows that 2 holds. Since the converse implication  trivially holds, this completes the proof.

.

Proposition 3.1 Let .(E, P) be a locally convex space and .u ∈ E ∗ . Then .u ∈ E  if and only if .ker(u) is closed in E. Proof If .u ∈ E  , then .ker(u) = u−1 ({0}) is closed. To prove the converse, we assume .u = 0 and that .ker(u) = ker(u)  E. Then, there is .x ∈ E \ ker(u). So, there exist .ε > 0 and .p ∈ P such that  x + Bp (0, ε) ∩ ker(u) = ∅ .

.

(3.1)

Our next step is to see that .u(Bp (0, ε)) is balanced. Take some .λ ∈ K with .|λ| ≤ 1 and .t ∈ u(Bp (0, ε)). We can find .y ∈ Bp (0, ε) so that .t = u(y) and then (recall that .Bp (0, ε) is balanced, see (2.14)) λt = λu(y) = u(λy) ∈ u(Bp (0, ε)) .

.

Suppose now that .u(Bp (0, ε)) is not bounded. Since, as we have just seen, it is balanced, .u(Bp (0, ε)) = K, and there is some .y ∈ Bp (0, ε) such that .u(y) = −u(x). But this implies .u(x + y) = 0 and then .x + y ∈ ker(u) with .y ∈ Bp (0, ε). This contradicts (3.1), so .u(Bp (0, ε)) is bounded and we can find .M > 0 such that  .|u(y)| ≤ M for every .y ∈ Bp (0, ε), which proves that .u ∈ E . 

3.2 The Hahn–Banach Theorem

93

3.2 The Hahn–Banach Theorem 3.2.1 Analytic Version To formulate the theorem (and later to apply it), we need a weaker concept than that of seminorm. A mapping .q : E → R defined on a real vector space is called subadditive if .q(x + y) ≤ q(x) + q(y) for every .x, y ∈ E and positively homogeneous if .q(λx) = λq(x) for every .x ∈ E and .λ ≥ 0. Obviously every seminorm is subadditive and positively homogeneous. Theorem 3.1 (Hahn–Banach) Let E be a real vector space, F a proper subspace, q : E → R subadditive and positively homogeneous, and .v : F → R linear such that

.

v(y) ≤ q(y)

.

for every .y ∈ F . Then there exists a linear .u : E → R so that .u|F = v and u(x) ≤ q(x)

.

for every .x ∈ E. Proof We begin by checking that we can extend v to the subspace F ⊕ [x0 ] = {y + λx0 : y ∈ F, λ ∈ R} ,

.

where .x0 ∈ F . If we take any .s ∈ R and define .u : F ⊕ [x0 ] → R by .u(y + λx0 ) = v(y) + λs, we clearly have that u is linear and extends v. We only have to choose s in such a way that u is also dominated by q. To begin with, note that if .y ∈ F , then .v(y + y) ≤ q(y + x0 ) + q(y − x0 ) and we can choose .

  sup v(z) − q(z − x0 ≤ s ≤ inf q(z + x0 ) − v(z) . z∈F

z∈F

With this choice, it is straightforward to check that .u(z + x0 ) ≤ q(z + x0 ) and u(z − x0 ) ≤ q(z − x0 ) for every .z ∈ F . Now, if .λ > 0, then

.

   u(y + λx0 ) = v(y) + λs = λ v λ1 y + x0 ≤ λq λ1 y + x0 = q(y + λx0 ) ,

.

and if .λ < 0, then 1  1 y − x0 ≤ |λ|q |λ| y − x0 = q(y + λx0 ) . u(y + λx0 ) = v(y) + λs = |λ| v |λ|

.

This altogether shows that u is dominated by q on .F ⊕ [x0 ].

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3 Duality and Linear Operators

Consider now the family . of all subspaces H that contain F and all extensions u of v to H that are dominated by q on H . We establish the following order .(H1 , u1 ) ≤ (H2 , u2 ) if .H1 ⊆ H2 and .u2 |H1 = u1 . Now, if .(Hα , uα )α is a totally ordered set, we define .H = α Hα and .u : H → R by .u(y) = uα (y) if .y ∈ Hα (note that if .y ∈ Hβ , then the fact that the set is totally ordered gives .uα (y) = uβ (y)). The same argument shows that H is a vector subspace of E. Then .(H, u) is an upper ˜ in .. If bound for .(Hα , uα )α . By Zorn’s lemma, there is a maximal element .(H˜ , u) ˜ = E, we can find .x0 ∈ E \ H and then extend .u˜ to .H ⊕ [x0 ] as we did before. .H This contradicts the maximality of .(H˜ , u) ˜ and shows that, in fact, .H˜ = E (and .u˜ is the extension we were looking for).  We can now pass to complex vector spaces, but let us first make a short consideration on the linear functionals. If E is a complex vector space and .u : E → C is linear, then we can write .u(x) = u1 (x) + iu2 (x) for every .x ∈ E. Note that .u1 , u2 : E → R are linear and .R-linear (in the sense that, for .j = 1, 2, .uj (λx) = λuj (x) for every .x ∈ E and .λ ∈ R). We have iu1 (x) − u2 (x) = iu(x) = u(ix) = u1 (ix) + u2 (ix) ,

.

and, then, .u1 (ix) = −u2 (x) for every .x ∈ E. This implies that u is of the form u(x) = u1 (x) − iu1 (ix) .

.

(3.2)

On the other hand, if .v : E → R is .R-linear (recall that E is a complex vector space) and we define u(x) = v(x) − iv(ix),

.

(3.3)

it satisfies u(ix) = v(ix) − iv(−x) = v(ix) + iv(x) = i(−iv(ix) + v(x)) = iu(x)

.

and .u(rx) = ru(x) for every .r ∈ R. This gives that u is .C-linear. Theorem 3.2 (Hahn–Banach) Let E be a real or complex vector space and F a proper subspace. Let .p : X → [0, ∞[ be a seminorm and .v : F → K linear such that |v(x)| ≤ p(x)

.

for every .x ∈ F . Then there exists .u : E → K linear such that .u|F = v and |u(x)| ≤ p(x)

.

for every .x ∈ E.

3.2 The Hahn–Banach Theorem

95

Proof If E is real, the statement follows immediately from Theorem 3.1, which gives an extension u satisfying .u(x) ≤ p(x) and .−u(x) = u(−x) ≤ p(−x) = p(x) and, then, .|u(x)| ≤ p(x) for every .x ∈ E. If E is a complex vector space, we have seen in (3.2) that .v(x) = v1 (x)−iv1 (ix), where .v1 is an .R-linear form. Then .|v1 (x)| ≤ |v(x)| ≤ p(x) for every .x ∈ F and, by the previous case, it can be extended to some .R-linear .u1 : E → R. With this, .u : E → C given by .u(x) = u1 (x) − iu1 (ix) is a .C-linear form that extends u. Now, for each .x ∈ E, we can find .θx ∈ R such that .|u(x)| = eθx i u(x). Then |u(x)| = u(eθx i x) = u1 (eθx i x) ≤ p(eθx i x) = |eθx i |p(x) = p(x) .

.

 As an immediate consequence, we can extend any continuous linear functional to every superspace. Corollary 3.1 Let E be a locally convex space and F a subspace. For every .v ∈ F  , there exists .u ∈ E  that extends v. Proof Let .P be the family of seminorms defining the topology on E. Then the topology on F is defined by the restrictions of the seminorms in .P to F . Take  .v ∈ F . There is some .C > 0 such that .|v(x)| ≤ Cp(x) for every .x ∈ F . But ∗ .q = Cp is a seminorm on E, and we can find .u ∈ E extending v and such that  .|u(x)| ≤ Cp(x) for every .x ∈ E. Therefore .u ∈ E .  Corollary 3.2 Let E be a locally convex space. For every .0 = x ∈ E, there exists u ∈ E  such that .u(x) = 1.

.

Proof Since .x = 0, there is .p ∈ P, with .p(x) > 0. We consider .F = {λx : λ ∈ |λ| K} and define .v : F → K by .v(λx) = λ. Then, .|v(λx)| = |λ| = p(x) p(x) ≤ 1 p(x) p(λx),

and by the Hahn–Banach Theorem 3.2, there exists .u ∈ E ∗ such that

1 p(y) for every .y ∈ E. This implies that u is continuous u|H = v and .|u(y)| ≤ p(x)  and, moreover, .u(x) = v(x) = 1.

.

Remark 3.1 Let us point out that if E in Corollaries 3.1 and 3.2 is a normed space, then the family of seminorms .P reduces to the norm . · , and the proofs show that, in fact, u satisfies .u = v (in Corollary 3.1) and that .u(x) = 1 and .u = 1 (in Corollary 3.2).

3.2.2 Separation Theorems With help of the Hahn–Banach theorem, we can use functionals to separate sets. If the sets that we want to separate are ‘small’ (a point and a finite dimensional space, as in the following result), we do not need more than what we already have.

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Proposition 3.2 Let E be a locally convex space, M a finite dimensional subspace, and .x0 ∈ E \ M. Then there is .u ∈ E  such that .u|M = 0 and .u(x0 ) = 1. Proof Choose some basis .{x1 , . . . , xn } of M, consider .F = span{x0 , x1 , . . . , xn }, and define .v(λx0 + μ1 x1 + · · · + μn xn ) = λ. The argument now finishes as that in  Corollary 3.2. With essentially the same idea, we can separate closed subspaces and points. Proposition 3.3 Let E be a locally convex space, F a closed proper subspace, and x0 ∈ E \ F . Then there is .u ∈ E  such that .u|F = 0 and .u(x0 ) = 1.

.

Proof Consider the space .G = F + span{x0 } and endow it with the topology induced by E. For each .x ∈ F , there are .y ∈ F and .λ ∈ C so that .x = y + λx0 . By doing .v(x) = λ, we define a linear mapping .v : G → C that, since .ker v = F is closed, is continuous (recall Proposition 3.1). Corollary 3.1 gives a functional  .u ∈ E that extends v and satisfies the required conditions.  As a straightforward consequence, we have the following. Corollary 3.3 Let E be a locally convex space and F a subspace. The following are equivalent: 1. If .u ∈ E  is such that .u(x) = 0 for every .x ∈ F , then .u = 0. 2. F is dense in E. For ‘bigger’ sets, we need some more work. To begin with, for each convex, absorbing set A of some vector space E, we define the Minkowski gauge .pA : E → [0, ∞[ as pA (x) = inf{t > 0 : x ∈ tA} .

.

(3.4)

Clearly {x ∈ E : pA (x) < 1} ⊆ A ⊆ {x ∈ E : pA (x) ≤ 1} .

.

Lemma 3.3 If A is convex and absorbing, then .pA is subadditive and positively homogeneous. If A is absolutely convex, then .pA is a seminorm. Proof Note first that .pA (0) = 0, then for .λ = 0 we have .p(λ0) = λpA (x). On the other hand, if .λ > 0, then pA (λx) = inf{t > 0 : λx ∈ tA} = inf{t > 0 : x ∈ λt A} = λpA (x) .

.

This shows that it is positively homogeneous. To see that it is subadditive, take x y t , s ∈ A, and then, since the set is convex, we have

.

.

t x s y x+y = + ∈ A, t +s t +s t t +s s

3.2 The Hahn–Banach Theorem

97

and this implies .pA (x + y) ≤ t + s. Taking the infimum first with respect to t and then to s, we get .pA (x + y) ≤ pA (x) + pA (y). If A is balanced, then so also is tA for every .t > 0 and .λx ∈ tA if and only if .|λ| ∈ tA which gives (by the positive homogeneity) .pA (λx)pA (|λ|x) = |λ|pA (x).  Lemma 3.4 Let E be a real locally convex space, .u : E → R, non-zero and linear, and A a subset of E. If A is convex, balanced, or open, then so also is .u(A). Proof The statements about convexity and balancedness are straightforward consequences of the linearity of u. The fact that if A is open, then .u(A) is open requires slightly more work. It is enough to show that for every .p ∈ P and all .ε > 0 there is some .δ > 0 so that .] − δ, δ[⊆ u(Bp (0, ε)). Choose .x0 with .u(x0 ) > 0, take .q ∈ P so that .q(x0 ) > 0, and consider .r ∈ P 0) such that .max{p, q} ≤ r. Define .δ = u(x r(x0 ) ε. If .|t| < δ, then t =t

.

 tx0 u(x0 ) = u u(x . 0) u(x0 )

But  tx0 r(x0 ) < ε. = |t| u u(x 0) u(x0 )

.

Hence .] − δ, δ[⊆ u(Br (0, ε)) ⊆ u(Bp (0, ε)), and this completes the proof.



Theorem 3.3 Let E be a real locally convex space and A and B two non-empty disjoint convex sets in E, being A open. Then there exist .u ∈ E  and .γ ∈ R so that .u(a) < γ ≤ u(b) for every .a ∈ A and .b ∈ B. Proof Choose .a0 ∈ A and .b0 ∈ B. Both .A − a0 and .B − b0 are convex and then so also is .K = A − B + b0 − a0 . On the other

hand, for each .y ∈ B, the set .A − y + b0 − a0 is open and, then, since .K = y∈B A − y + b0 − a0 , it is also open. Note also that, since the sets are disjoint, .x0 = b0 − a0 ∈ K. Clearly .0 ∈ K, and therefore K is an open neighbourhood of 0 and, then absorbing. We consider its Minkowski gauge that, by Lemma 3.3, is subadditive and positively homogeneous. Note that, since .x0 ∈ K, we have .pK (x0 ) ≥ 1 and, since K is open, .pK (x) < 1 for every .x ∈ K. Consider now .F = {λx0 : λ ∈ R} and define .v(λx0 ) = λ. For .λ ≥ 0, we have .λ ≤ λpK (x0 ) = pK (λx0 ). Since the inequality is obvious for .λ < 0, we get that v is dominated by .pK on F , and, by Theorem 3.1, it can be extended to .u ∈ E ∗ satisfying .u(x) ≤ pK (x) for every .x ∈ E. If .x ∈ K, then .u(−x) ≤ pK (−x) = pK (x) < 1. We define .U = K ∩ (−K), which is an open neighbourhood of 0 and satisfies .|u(x)| < 1 for every .x ∈ U . This shows that u is also continuous, that is, .u ∈ E  . Now, if .a ∈ A and .b ∈ B, we have .a − b + x0 ∈ K and, then, .u(a − b + x0 ) < 1 = u(x0 ). This implies .u(a) < u(b). Now, Lemma 3.4 gives that both .u(A) and

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u(B) are convex (that is, intervals in .R) and .u(A) is open. Also, .u(A) ∩ u(B) = ∅. Taking .γ = infb∈B u(b) yields the conclusion. 

.

Remark 3.2 If E is a complex l.c.s. and A and B are as in Theorem 3.3, we can find a continuous, .R-linear .u1 : E → R separating A and B. Then, defining u as in (3.3) (which is obviously continuous), we get .u ∈ E  and .γ ∈ R such that .Re(u(a)) < γ ≤ Re(u(b)) for all .a ∈ A and .b ∈ B. Theorem 3.4 Let E be a real locally convex space and A and B two non-empty disjoint convex sets in E, being A closed and B compact. Then there exists .u ∈ E  such that sup u(a) < inf u(b) .

.

b∈B

a∈A

Before we proceed with the proof, let us note that this result does not hold if we only ask A and B to be convex and closed. To see this, just take .A = {(x, 0) : x ∈ R} and .B = {(x, y) : x, y ≤ 0, xy ≥ 1} as subsets of .R2 . Proof Since A is closed and B is compact, for each .b ∈ B, we can find .pb ∈ P and εb > 0 so that .Bpb (b, 2εb )∩A = ∅. Also, being B compact, there are .b1 , . . . , bN ∈ B so that (we denote .pn and .εn for .pbn and .εbn )

.

B⊆

N 

.

Bpn (bn , εn ) .

n=1

Define .ε = minn=1,...,N

εn and consider .p ∈ P such that .p ≥ maxn=1,...,N pn . Consider now .U = b∈B Bp (b, ε). Then N 

U = B + Bp (0, ε) ⊆

.

Bpn (bn , εn ) + Bp (0, ε)

n=1

⊆

N 

bn + Bpn (0, εn ) + Bpn (0, εn ) ⊆

n=1

N 

Bpn (bn , 2εn ) ,

n=1

and thus U and A are disjoint. Since U is convex and open, we can use Theorem 3.3 to find some .u ∈ E  so that .u(U ) and .u(A) are two disjoint intervals in .R. On the other hand, .u(B) is a compact interval contained in the open interval .u(U ). This gives the conclusion.  Corollary 3.4 Let E be a real or complex locally convex space and A and B two non-empty disjoint sets in E, being A closed and absolutely convex and B compact and convex. Then there exists .u ∈ E  such that .

sup |u(a)| < inf |u(b)| . a∈A

b∈B

3.2 The Hahn–Banach Theorem

99

Proof From Theorem 3.4, we can find some .u : E → R continuous and .R-linear so that .

sup u(a) = μ < γ = inf u(b) . a∈A

b∈B

Suppose first that E is real. Then for .b ∈ B, we have .γ ≤ u(b) ≤ |u(b)|. On the other hand, for each .a ∈ A, there is .λa ∈ R with .|λa | = 1 so that .λa u(a) = |u(a)|. Since A is absolutely convex .λa a ∈ A and, then, .|u(a)| = u(λa a) ≤ μ. This gives the conclusion. Suppose now that E is complex and define .v ∈ E  by .v(x) = u(x) − iu(ix). Again, .γ ≤ u(b) ≤ |v(b)| for every .b ∈ B. The same argument as before, finding for each .a ∈ A some .λa ∈ C with .|λa | = 1 and .|v(a)| = λa v(a) yields  |v(a)| = v(λa a) = Re v(λa a) = u(λa a) ≤ μ .

.

This completes the argument.



Corollary 3.5 Let E be a real or complex locally convex space and .∅ = A ⊆ E be closed and absolutely convex. For each .x ∈ A, there exists .u ∈ E  such that .|u(a)| ≤ 1 for every .a ∈ A and .|u(x)| > 1. Proof Using Corollary 3.4, we can find .v ∈ E  and .γ ∈ R so that .

sup |v(a)| < γ < |v(x)| . a∈A

Defining .u =

1 γv

∈ E  yields the claim.



Remark 3.3 Let us note that if in Proposition 3.3 or Theorem 3.3 E is a normed space, then the operator v defined in the proofs clearly satisfies .v = 1, and therefore, by Remark 3.1, u can be taken with .u = 1.

3.2.3 Finite Dimensional Locally Convex Spaces The Hahn–Banach theorem allows us to characterize those locally convex spaces with finite dimension. Remark 3.4 First of all note that if .M = span{x1 , . . . , xn } is a finite dimensional subspace of an l.c.s. E, then we can choose some .x0 ∈ M, and, by Proposition 3.2, we can find .u ∈ E  such that .u(a) = 1 and .u(x) = 0 for all .x ∈ M. Since u is continuous, this gives .a ∈ M and shows that M is closed.

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A subset A of a locally convex space .(E, P) is said to be precompact if for each ε > 0 and each .p ∈ P, there exist a finite number of points .x1 , . . . , xN in A (or in E) so that

.

N 

A⊆

.

Bp (xn , ε) .

n=1

The fact that every open covering of a compact set has a finite subcovering immediately gives that every relatively compact set is precompact. On the other hand, if A is precompact in an l.c.s. E and p is some seminorm in .P, then there are .x1 , . . . , xN in A so that N 

A⊆

.

Bp (xn , 1) .

n=1

If .x ∈ A, then .x ∈ Bp (xn0 , 1) for some .n0 and .p(x) ≤ p(x − an0 ) + p(an0 ) = 1 + p(an0 ). This gives .

sup p(x) ≤ 1 + max p(an ) n=1,...,N

x∈A

and shows that A is bounded. Theorem 3.5 (Riesz) Let .(E, P) be an l.c.s. with a precompact 0-neighbourhood. Then, .dim(E) < ∞. Proof Let U be a precompact 0-neighbourhood. Since precompact sets are bounded, Kolmogorov’s Theorem 2.24 implies that E is normable. Since its open unit ball (which we denote by .BE ) is bounded, there is .λ > 0 such that .BE ⊆ λU , and this implies that .BE itself is precompact. Then we can find .x1 , . . . , xN ∈ BE such that N   1 xn + BE . .BE ⊆ 2 n=1

Define .M = span{x1 , . . . , xN }, then 1 1 1 1 BE ⊆ M + BE ⊆ M + M + BE ⊆ M + 2 BE . 2 2 2 2

.

Proceeding by induction, we get .BE ⊆ M + BE ⊆

.

1 2 n BE

for every .n ∈ N, and this gives

∞   1 M + n BE = M . 2

n=1

3.2 The Hahn–Banach Theorem

101

Since by Lemma 3.4 M is closed, .M = M and .E = span(BE ) ⊆ M. This implies dim(E) ≤ N. 

.

With this, we can finally show that for each n there is essentially only one ndimensional locally convex space. Theorem 3.6 A locally convex space .(E, P) has dimension .n ∈ N if and only if it is isomorphic to .Kn endowed with the sup-norm. Proof Let us assume first that .dim(E) = n and choose a basis .x1 , . . . , xn . We define φ : Kn → E by .φ(α) = ni=1 αi xi for .α = (α1 , . . . , αn ) ∈ Kn . Clearly, .φ is linear and bijective. Also, for each .p ∈ P, we have

.

p(φ(α)) = p

.

n n 

 ≤ max α x |α | p(x ) ≤ p(x ) α∞ , i i i i i i=1

 n

i=1,...,n

i=1

i=1

and .φ is continuous. It remains to show that .φ −1 : (E, τP ) → Kn is also continuous. First of all, since .x1 , . . . , xn ∈ E are linearly independent, by  such that .u (x ) = δ for .i, j = 1, . . . , n. Proposition 3.2 there are .u1 , . . . , un ∈ E i j ij Now, for each .x ∈ E, we can write .x = ni=1 ui (x)xi , and the inverse of .φ is given by φ −1 (x) = (u1 (x), . . . , un (x))

.

for each .x ∈ E. To see that this is continuous, let us note that, since .ui ∈ E  for .i = 1, . . . , n, for each .p ∈ P, there is .C > 0 such that .|ui (x)| ≤ Cp(x) for all .x ∈ E and .i = 1, . . . , n. Hence, φ −1 (x)∞ = max |ui (x)| ≤ Cp(x)

.

i=1,...,n

for each .x ∈ E. This shows that .φ −1 is continuous and, since every space that is isomorphic to .(Kn ,  · ∞ ) is obviously an n-dimensional normed (hence locally convex) space, concludes the proof.  Note that Theorem 3.6 in particular tells us that if E is a finite dimensional space, then there is only one possible Haussdorf locally convex topology on E.

3.2.4 Banach Limits Our aim now is to extend the notion of ‘limit’ to bounded (not necessarily convergent) sequences. Let us be more precise, and denote by c the space of all convergent sequences, endowed with the topology defined by the norm .·∞ (recall Example 2.5). Then .L(x) = limn xn defines a linear mapping .L : c → K that,

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since .|L(x)| ≤ supn |xn | for every .x = (xn )n ∈ c, is continuous (and .L ≤ 1). Moreover, denoting .1 = (1, 1, 1, . . .) (the constant sequence), we have .L(1) = 1, so that .L = 1. Additionally, .

lim xn = lim xn+1 n

n

(3.5)

for every .(xn )n ∈ c, and if .xn ≥ 0, then .

lim xn ≥ 0 n

(that is, the operator L is positive). So, our aim is to define a continuous operator L : ∞ → K that when restricted to c gives the limit and satisfies these two last properties. Let us have a look at (3.5): what we want here is that, when evaluated at .(xn )n and at .(xn+1 )n , the operator L gives the same result. To handle this, we  consider the backward shift operator .B : KN → KN , defined as .B (xn )∞ n=1 = (xn+1 )∞ , that is, n=1

.

B(x1 , x2 , x3 , . . .) = (x2 , x3 , . . .)

.

(3.6)

(this operator will be studied in more detail in Sect. 5.3.1). Then (3.5) can be reformulated as .lim x = lim B(x) for every .x ∈ c, and what we want is that .L = LB. We show the existence of such an operator (called Banach limit) in the following result. Theorem 3.7 There exists an operator .L : ∞ → K such that 1. 2. 3. 4.

L = 1. L(x) = limn xn for every .x = (xn )n ∈ c. If .x ∈ ∞ is such that .xn ≥ 0 for every n, then .L(x) ≥ 0. .L = LB on . ∞ . . .

Proof Let us handle first the real case, that is, if .K = R. Consider the subspace M = Im(I − B) = {x − Bx : x ∈ ∞ }. We proceed now in several steps. Let us see first that

.

d(1, M) = 1 .

.

(3.7)

To see this, note first that, since .0 ∈ M, we have .d(1, M) ≤ 0 − 1 = 1. Let us see now that .1 − (x − Bx) ≥ 1 for every .x ∈ ∞ . Suppose first that x is a bounded, decreasing sequence (i.e., .xn+1 ≤ xn for every n). Then it is convergent and .limn (xn − xn+1 ) = 0; hence 1 − (x − Bx) = sup |1 − (xn − xn+1 )| = 1 .

.

n

3.2 The Hahn–Banach Theorem

103

Now, if x is not decreasing, then there is some .m ∈ N such that .xm+1 ≥ xm , i.e., xm − xm+1 ≤ 0, and

.

1 − (x − Bx) ≥ |1 − (xm − xm+1 )| ≥ 1 .

.

This altogether shows (3.7), which clearly implies that .1 ∈ M. By Proposition 3.3 (see also Remark 3.3), there is a continuous linear operator .L : ∞ → R with .L = 1 such that .L(1) = 1 and .L(y) = 0 for every .y ∈ M. Note that the latter implies that .L(x−Bx) = 0 for every .x ∈ ∞ . We have, then, obtained L satisfying 1 and 4. To see that the other two properties hold, we need first to prove the following c0 ⊆ ker L

.

(recall from Example 2.5 that .c0 is the space of null sequences). Pick, then, some x ∈ c0 and set .x (0) = x and .x (n+1) = Bx (n) for .n ∈ N. Then,

.

x − x (n) =

n−1 

.

j =0

x (j ) − x (j +1) =

n−1 

x (j ) − Bx (j ) ∈ M

j =0

for every n, and therefore .L(x) = L(x (n) ). Now, given .ε > 0, there exists .n0 ∈ N such that .|xm | < ε for every .m ≥ n0 . Then, if .n ≥ n0 , we have |L(x)| = |L(x (n) )| ≤ x (n)  = sup |xm | < ε .

.

m≥n

This gives .L(x) = 0 and shows our claim. With this, we can now prove that the remaining properties are satisfied. Let us see first that 2 holds. If .x = (xn )n is convergent and we write .α = limn xn , then .x − 1α ∈ c0 , and then .L(x − α1) = 0. Hence L(x) = L(α1) = αL(1) = lim xn .

.

n

To finish the proof in the case .K = R, we take .x ∈ ∞ such that .xn ≥ 0 and suppose x that .L(x) < 0. Taking .z = x , we have .0 ≤ zn ≤ 1. Then .1 − z ≤ 1 and L(1 − z) = L(1) − L(z) = 1 −

.

L(x) > 1. x

This contradicts the fact that .L = 1 and shows that 3 holds. We address now the case .K = C. We write .LR for the operator defined on the real space of bounded real sequences. If .x = (xn )n ∈ ∞ (now the complex space of complex sequences), we write .Re x = (Re xn )n and .Im x = (Im xn )n and define L(x) = LR (Re x) + iLR (Im x) .

.

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3 Duality and Linear Operators

This gives .L : ∞ → C that is clearly .C-linear, continuous (with .1 ≤ L ≤ 2 and L(1) = 1) and satisfies 2–4. It remains only to see that .L ≤ 1. Choose .x ∈ ∞ with .x ≤ 1, and let us see that .|L(x)| ≤ 1. Let us suppose first that x has only a finite number of different values (we denote by . ∞,0 the space of such sequences), then there are .E1 , . . . , En ⊆ N, pairwise disjoint such that .N = E1 ∪ · · · ∪ En , and .α1 , . . . , αm ∈ C with .|αj | ≤ 1 for .j = 1, . . . , m so that .

x=

m 

.

αj 1Ej ,

j =1

where .1Ej is the sequence whose nth coordinate is 1 if .n ∈ Ej and  0 otherwise. Note that .L(1Ej ) = LR (1Ej ) ≥ 0 for every .j = 1, . . . , m and that . m j =1 1Ej = 1. Then m m m       |L(x)| =  αj L(1Ej ) ≤ |αj | |L(1Ej )| ≤ L(1Ej ) = L(1) = 1 .

.

j =1

j =1

j =1

To complete the proof, we need to see that the set . ∞,0 is dense in . ∞ . Choose, then,

m .x ∈ ∞ and .ε > 0. We can find .y1 , . . . , ym ∈ D(0, x) so that .D(0, x) ⊆ j =1 D(yj , ε). We define now a sequence .z = (zn )n in the following way: given .n ∈ N, there is some .1 ≤ jn ≤ m so that .|yjn − xn | < ε; then we set .zn = yjn . Clearly .z − x < ε and, since there are only a finite number of ys, .z ∈ ∞,0 . If .x ∈ ∞ with .x ≤ 1 is arbitrary, we may take a sequence .(x j )j in . ∞,0 converging in . ∞ to x. Note that by our argument above we may assume .x j  ≤ x ≤ 1. Then .|L(x j )| ≤ 1 for every j , and this gives .|L(x)| ≤ 1, completing the proof. 

3.3 Weak Topologies Our aim now is to see how the duality relationship can be used to define a topology. We are going to work in a fairly general setting, but since our main interest will be locally convex spaces E with their duals .E  , we start by looking at this situation. First of all, if .(E, P) is a locally convex space and .E  is its dual, then we have • For every .u ∈ E  , .u = 0, there exists .x ∈ E such that .u(x) = 0. • For every .x ∈ E, .x = 0, there exists .u ∈ E  such that .u(x) = 0. The first property is trivially satisfied, whereas the second one follows from the Hahn–Banach theorem. On the other hand, we can define a bilinear mapping .

·, · : E × E  −→ K (x, u) −→ x, u = u(x).

(3.8)

3.3 Weak Topologies

105

These are the properties that relate the two vector spaces and that we isolate. A dual pair consists of a pair of vector spaces .(E, F ) together with a bilinear form .·, · : E × F −→ K satisfying (DP1) For every .y ∈ F , .y = 0, there exists .x ∈ E such that .x, y = 0. (DP2) For every .x ∈ E, .x = 0, there exists .y ∈ F such that .x, y = 0. Example 3.8 If E is a locally convex space and .E  is its dual, then .(E, E  ) is a dual pair with the bilinear mapping described in (3.8). In particular (recall Example 2.16), .(c0 , 1 ), .( 1 , ∞ ), and .( p , p  ) are dual pairs, where the bilinear mapping is in each case defined by (xi ), (yi ) =

∞ 

.

xi yi .

i=1

 Example 3.9 Let us consider the Fréchet space .ω (recall Example 2.14) and ϕ = {y ∈ ω : there is i(x) ∈ N so that yi = 0 for all i ≥ i(x)} .

.

We define . ,  : ω × ϕ → K by (xi ), (yi ) =

∞ 

.

xi yi .

i=1

Note that this is in fact a finite sum and, then, . ,  is well defined and bilinear. Given 0 = x ∈ ω, there is i so that .xi = 0. Then .x, ei  = xi = 0 and (DP2) holds. The same argument shows that (DP1) holds, and thus .(ω, ϕ) is a dual pair. 

.

Example 3.10 If .L1 ([0, 1]) is the space of measurable functions .f : [0, 1] → K 1 such that . 0 |f (x)|dx < ∞, then .(L1 ([0, 1]), C([0, 1])) is a dual pair with the bilinear mapping defined by  f, g =

1

f (x)g(x)dx .

.

0

 Example 3.11 If .F ⊆ E ∗ , then a bilinear form can be defined as in (3.8). Condition (DP1) is obviously satisfied, and condition (DP2) is satisfied whenever F separates points. So, if this is the case, then .(E, F ) is a dual pair.  Remark 3.5 In fact, all dual pairs can be seen as a particular case of Example 3.11. Take some dual pair .(E, F ), fix .x ∈ E, and define .ux : F → K by .ux (y) = x, y.

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3 Duality and Linear Operators

This is clearly linear and then belongs to .F ∗ . On the other hand, given .λ, μ ∈ K, and .x1 , x2 ∈ E, by the bilinearity, we have uλx1 +μx2 (y) = λx1 + μx2 , y = λx1 , y + μx2 , y = λux1 (y) + μux2 (y)

.

for every .y ∈ F . On the other hand, if .ux = 0, then .0 = ux (y) = x, y for every y ∈ F , and by (DP1), .x = 0. This altogether shows that the mapping defined as .x  ux is a linear, bijective mapping, and then we can identify E with a subset of ∗ .F that separates points. With the same argument, F can be identified with a subset of .E ∗ that separates points. We will from now on always bear these identifications in mind without further notice. .

What we want to do now is to see how the structure of a dual pair defines a locally convex topology on each of the two spaces. To do that we begin by defining seminorms. If .J ⊆ F is finite, then we define, for .x ∈ E, pJ (x) = max{|x, u| : u ∈ J } .

.

This is a seminorm on E. With the same idea, given .I ⊆ E finite, we define, for u ∈ F,

.

pI (u) = max{|x, u| : x ∈ I } ,

.

which is a seminorm on F . Proposition 3.4 1. The family of seminorms .{pJ : J ⊆ F finite} defines a (Hausdorff) locally convex topology, denoted by .σ (E, F ), on E. 2. The family of seminorms .{pI : I ⊆ E finite} defines a (Hausdorff) locally convex topology, denoted by .σ (F, E), on F . Proof First of all, if.J1 and .J2 are finite subsets of F , then .K = J1 ∪ J2 is also finite and obviously .max pJ1 (x), pJ2 (x) ≤ pK (x) for all .x ∈ E. On the other hand, given .x = 0, there is .u ∈ F such that .x, u = 0, and then .p{u} (x) = |x, u| > 0, and this finally shows that .σ (E, F ) is a locally convex topology on E. The proof for .σ (F, E) is basically the same.  Once we have the topology, we describe the neighbourhoods in each of the two topologies. First, for .y1 , . . . , yn ∈ F and .ε > 0, then Uy1 ,...,yn ,ε = {x ∈ E : |x, yi | < ε , i = 1, . . . n}

.

is a neighbourhood of 0, and the family U = {Vy1 ,...,yn ,ε : n ∈ N, y1 , . . . , yn ∈ F, ε > 0}

.

defines a basis of neighbourhoods of 0 for .σ (E, F ).

3.3 Weak Topologies

107

In the same way, if .x1 , . . . , xn ∈ E and .ε > 0, then Vx1 ,...,xn ,ε = {y ∈ F : |xi , y| < ε , i = 1, . . . n}

.

is a neighbourhood of 0, and the family V = {Ux1 ,...,xn ,ε : n ∈ N, x1 , . . . , xn ∈ E, ε > 0}

.

defines a basis of neighbourhoods of 0 for .σ (F, E). Let us note that we have defined a topology, both on E and on F , that comes only from the linear structure, plus the duality assumption. There is no requirement whatsoever of a previous topological structure on any of the two spaces. Note also that, in general, the family of seminorms (or of neighbourhoods of 0) that define the topology is far from being countable; hence we cannot expect that it is always metrizable. Proposition 3.5 σ (E,F )

1. Let .(xi )i be a net in E. Then .xi −→ x if and only if .xi , u → x, u for all .u ∈ F . σ (F,E) 2. Let .(ui )i be a net in F . Then .ui −→ u if and only if .x, ui  → x, u for all .x ∈ E. σ (E,F )

Proof For the first statement, if .xi −→ x, then, given .u ∈ F , we have .p{u} (xi − x) → 0, and this immediately gives the claim. Suppose now that .xi , u → x, u for every .u ∈ F and fix some 0-neighbourhood .Vu1 ,...,un ,ε in .(E, σ (E, F )). Given .ε > 0, for each .j = 1, . . . , n, there is .ij ∈ I so that |xi − x, uj | = |xi , uj  − −x, uk | < ε

.

for every .i > ij . Choose now .i0 ∈ I so that .i0 > ij for every .j = 1, . . . , n and max |xi − x, uj | < ε

.

j =1,...,n

for every .i > i0 . Hence .xi − x ∈ Vu1 ,...,un ,ε and this gives the conclusion. The second statement follows in the same way.  Proposition 3.6 .(E, σ (E, F )) = F and .(F, σ (F, E)) = E. Proof We prove only the first statement, being the proof of the second one analogous. We start by taking some .u ∈ (E, σ (E, F )) (that is, a linear .u : E → K that is continuous with respect to .σ (E, F )). Then there are .u1 , . . . , un ∈ F and .c > 0 such that |u(x)| ≤ c max |ui (x)| .

.

i=1,...n

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3 Duality and Linear Operators

n Hence .  i=1 ker(ui ) ⊆ ker(u), and, by Lemma 3.2, there exist .λ1 , . . . , λn ∈ K such that .u = ni=1 λi ui ∈ F . Conversely, we fix .u ∈ F and, as we saw in Remark 3.5, .u(x) = x, u defines a linear form. To see that it is continuous, it is enough to take the seminorm .p{u} .  We see now that if E is a locally convex space, then the original topology is stronger than .σ (E, E  ). For this reason, this topology is often known as the weak topology. Also, the topology .σ (E  , E) on .E  is called the weak.∗ topology. Proposition 3.7 Let .(E, P) be a locally convex space. Then .σ (E, E  ) ⊆ τP . In τP σ (E,E  ) particular, for every sequence .(xn )n in E, if .xn −→ x, then .xn −→ x. Proof Take .J = {u1 , . . . , un } ⊆ E  , and then for each k, there are .ck > 0 and .pk ∈ P such that .|uk (x)| ≤ ck pk (x) for every .x ∈ E. Taking .c = max(c1 , . . . , cn ) and .p ∈ P with .max(p1 , . . . , pn ) ≤ p, we have pJ (x) ≤ cp(x) for all x ∈ E .

.

This, by Remark 2.8, gives the claim. The statement about the convergence of sequences follows immediately from this inequality. It also follows from Proposition 3.5 and the fact that all u are continuous.  Proposition 3.8 Let E be a vector space and .F ⊆ E ∗ that separates points. If .τ is a locally convex topology on E so that every .v ∈ F is .τ -continuous, then .σ (E, F ) ⊆ τ . Proof Fix some net .(xi )i ⊆ E converging in .τ to some .x ∈ E, and let us see that it also converges to x in .σ (E, F ). To do that, take any .v ∈ F . If .P is a family of seminorms defining .τ , by Theorem 2.11, there is .C > 0 so that .|v(y)| ≤ Cp(y) for every .y ∈ E. For each .ε > 0, since the net converges, we can find .i0 such that ε .p(xi − x) < C for all .i ≥ i0 . Then, for such i, we have |v(xi ) − v(x)| = |v(xi − x)| ≤ Cp(xi − x) < ε .

.

This shows that .v(xi ) → v(x) and, by Proposition 3.5, completes the proof.



Proposition 3.9 Let .(E, F ) be a dual pair and G a vector subspace of E. Then (G, F ) is a dual pair if and only if G is .σ (E, F )-dense in E.

.

σ (E,F )

. Since .(E, F ) is a dual pair, Proof Let us suppose that there exists  .x0 ∈ E \G  by Proposition 3.6, we have that . E, σ (E, F ) = F , and then by Proposition 3.3 σ (E,F )

(note that .G is a proper closed subspace), there is .u ∈ F such that .u(x0 ) = 0 (hence .u = 0) and .u(x) = 0 for every .x ∈ G. Then (DP1) is not satisfied and .(G, F ) is not a dual pair. Assume, conversely, that G is .σ (E, F )-dense in E, and let us see that .(G, F ) is a dual pair. Take some .0 = u ∈ F and suppose that .u(x) = 0 for every .x ∈ G. Given any .x ∈ E, we can find a net .(xi )i ⊆ G that .σ (E, F ) converges to x. This implies

3.3 Weak Topologies

109

(see Proposition 3.5) that .limi u(xj ) = u(x) and, since .u(xj ) = 0 for every j , we have that .u(x) = 0 and then .u = 0. This shows that condition (DP1) is satisfied. Since (DP2) is trivially satisfied (because .G ⊆ E), this completes the proof.  Proposition 3.10 Let .(E, P) be an l.c.s. and .A ⊆ E be convex. Then, .A = σ (E,E  )

A

.

Proof First of all, since the original topology .τP is finer than .σ (E, E  ), we have σ (E,E  )

σ (E,E  )

is closed in E and, then, .A ⊆ A . To see the converse inclusion, that .A take some .x0 ∈ A and use Theorem 3.4 to find .u ∈ E  and .γ ∈ R so that .

Re u(x0 ) < γ < inf Re u(x) . x∈A

By Proposition 3.6, .u : (E, σ (E, E  )) → K is continuous and, therefore, so also is  .Re u. Then the set .{x ∈ E : Re u(x) < γ } is a .σ (E, E )-neighbourhood of .x0 that does not intersect A. This shows that .x0 ∈ A

σ (E,E  )

and completes the proof.



Then .E 

Let us suppose now that E is a Banach space. is again a Banach space and .(E, E  ) is a dual pair. In this case, the weak topology .σ (E, E  ) on E is often denoted by w, while the weak.∗ topology .σ (E  , E) on .E  is denoted by .w ∗ . Example 3.12 The weak topology on .c0 is .w = σ (c0 , 1 ) (recall Example 2.16). w Take .en = (δin )∞ i=1 ∈ c0 for .n ∈ N, and let us see that .en → 0. Indeed, if .x ∈ 1 , then en , x =

∞ 

.

xi δin = xn → 0 = 0, x as n → ∞ .

i=1

But obviously .en does not converge to 0 in . · ∞ since .en ∞ = 1 for all n.



We can go one step further. Note that the dual space .E  is itself a Banach space, and we can consider its dual .E  and the dual pair .(E  , E  ). This defines a weak topology .σ (E  , E  ). Then, on .E  , we can consider two weak topologies: the .σ (E  , E) = w ∗ topology that comes from looking at it as dual of E and the weak topology .σ (E  , E  ) = w that comes from considering it as predual of .E  . Since E can be in a natural way looked at as a subspace of .E  by means of the identification .J : E → E  given by .J (x) : X → K with .J (x)(u) = u(x), then    .σ (E , E) ⊆ σ (E , E ). We see now that they may be different. Example 3.13 The .w ∗ topology on . 1 is the one defined by the dual pair .(c0 , 1 ). In the same way as in Example 3.12, we see that the sequence .(en )n in . 1 weak.∗ converges to 0. On the other hand, . 1 = ∞ , and the weak topology is .w = w

σ ( 1 , ∞ ). Then .en → 0 if and only if .en , u → 0, u for every .u ∈ ∞ . But if we take .u = (1, 1, 1, 1, 1, . . .) ∈ ∞ , we have .en , u = 1 for all n, and this shows that .(en )n cannot w-converge to 0. 

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3 Duality and Linear Operators

3.4 The Bipolar Theorem Given a dual pair .(E, F ), each subset A of E defines a set in F (known as polar set of A) in a natural way A◦ = {u ∈ F : |x, u| ≤ 1 for every x ∈ A} ⊆ F .

.

Proposition 3.11 Let .(E, F ) be a dual pair and A a subset of E. 1. 2. 3. 4. 5. 6. 7. 8.

A◦ is absolutely convex and .σ (F, E)-closed. If .B ⊆ A, then .A◦ ⊆ B ◦ . if .λ ∈ K, .λ = 0, then .(λA)◦ = λ−1 A◦ . ◦◦ .A ⊆ A . ◦ ◦◦◦ . .A = A  ◦ . A = A◦ . 

◦  . = i∈I A◦i . i∈I Ai If M is a vector subspace of E, then .M ◦ = {u ∈ F : x, u = 0 for all x ∈ M}. .

Proof For a fixed .x ∈ E, the set .{u ∈ F : |x, u| ≤ 1} is absolutely convex and, since x is .σ (E, E  )-continuous (see Proposition 3.6), closed. Then 1 follows once we have observed that  ◦ .A = {u ∈ F : |x, u| ≤ 1} . x∈A

Statements 2–4 follow immediately from the definition. As a straightforward consequence, we have  ◦◦  ◦◦ ◦ A◦ ⊆ A◦ = A ⊆ A◦ .

.

to x. Since For 6, take .u ∈ A◦ and .x ∈ A. We can find a net .(xj ) ⊆ A converging  ◦ .|xj , u| ≤ 1 for every j and .xj , u → x, u, this yields .u ∈ A . The converse inclusion follows from 2. ◦ 

In order to see 7, note that .u ∈ if and only if .|x, u| ≤ 1 for every i∈I Ai .x ∈ Ai and all .i ∈ I . This gives the conclusion. Suppose now that M is a vector subspace, choose some .u ∈ M ◦ and fix .x ∈ M. Since .λx ∈ M for every .λ > 0, we have .|λx, u| ≤ 1 and, then, .|x, u| ≤ 1/λ for every .λ > 0. Therefore .x, u = 0. This shows 8 since the converse inclusion is  obvious. Proposition 3.12 Let E be an l.c.s and consider the dual pair .(E, E  ). Then .H ⊆ E  is equicontinuous if and only if there is a 0-neighbourhood .U ⊆ E so that ◦ .H ⊆ U .

3.4 The Bipolar Theorem

111

Proof By Proposition 2.6, H is equicontinuous if and only if we can find .p ∈ P and .C > 0 so that .|x, u| ≤ Cp(x) for every .x ∈ X and .u ∈ H . Just taking .U = Bp (0, 1/C) gives the conclusion.  Proposition 3.13 Let E be an l.c.s. and consider the dual pair .(E, E ∗ ). If .U is any basis of 0-neighbourhoods, then .E  = U ∈U U ◦ . Proof Let us note first of all that in this case U ◦ = {u ∈ E ∗ : |x, u| ≤ 1 for every x ∈ U }

.

for each .U ∈ U . To begin with, if .u ∈ U ◦ , then u is bounded in U and, hence, is continuous. This shows . U ∈U U ◦ ⊆ E  . Conversely, if .u ∈ E  , then there is .U ∈ U such that .|x, u ≤ 1 for every .x ∈ U . This completes the proof.  This procedure of taking polars can be iterated. If .(E, E  ) is a dual pair and F is a vector space with .E ⊆ F ⊆ (E  )∗ , then .(E  , F ) is a dual pair. With this, given  .A ⊆ E, we take first the polar with respect to the dual pair .(E, E ) (we denote it ◦   .A EE for a moment). This is a subset of .E , and then we can consider its polar with the dual pair .(E  , F ), that is,  ◦  ◦E F ⊆F. A EE

.

This set is called the bipolar of A. A simple calculation from the definition shows that   A◦EE ◦E F = z ∈ F : |z, u| ≤ sup |x, u| for every u ∈ E  .

.

(3.9)

x∈A

Theorem 3.14 (Bipolar Theorem) Let .(E, E  ) be a dual pair and F a vector space with .E ⊆ F ⊆ (E  )∗ . Then A◦EE ◦E F = (A)

σ (F,E  )

.

for every .A ⊆ E. Proof In order to keep the notation as clean as possible, we write just .A◦◦ for the bipolar. By Proposition 3.11, .A◦◦ is absolutely convex and .σ (F, E  )-closed. This implies .(A) F \ (A)

σ (F,E  )

σ (F,E  )

⊆ A◦◦ . To see the converse inclusion, choose .x0 ∈

. We can use Corollary 3.5 (and Proposition 3.6) to find .u ∈ E  σ (F,E  )

(in particular, for so that .|x0 , u| > 1 and .|x, u| ≤ 1 for every .x ∈ (A)  every .x ∈ A). This implies .u ∈ A◦EE and, then, .x0 ∈ A◦EE ◦E F .

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3 Duality and Linear Operators

An interesting particular case is when we take .F = E. In this case, since .(A) is convex, Proposition 3.10 gives σ (E,E  )

A◦◦ = (A)

.

= (A)

τ

for every .A ⊆ E. Another interesting case is that of Banach spaces. The polar of the open unit ball of a Banach space X is easily found ◦ BX = {u ∈ X : |u(x)| ≤ 1, ∀x ∈ BX } = {u ∈ X : u ≤ 1} = B X

.

(3.10)

and, analogously, .(BX )◦ = B X . Then Theorem 3.14 yields BX = (BX )◦◦ = (BX )



σ (X ,X )

.

= BX



σ (X ,X )

.

As a matter of fact, this is a classical fact in Banach space theory, known as  Goldstine theorem: for each .x ∈ X with .x ≤ 1, there is a net .(xi ) ⊆ X with   .xi  ≤ 1 that .σ (X , X )-converges to x. Theorem 3.15 (Alaoglu-Bourbaki) If U is a 0-neighbourhood of an l.c.s. E, then U ◦ is .σ (E  , E)-compact.

.

ProofLet us denote .Kx = K for each .x ∈ E and consider the product space P = x∈E Kx . Define now the mapping .φ : (E  , σ (E  , E)) −→ P by .φ(u) = (u(x))x∈E . It is easy to check that .φ is a linear isomorphism onto the image .φ(E  ) when the latter is endowed with the induced topology. So, to get the conclusion, it is enough to see that .φ(U ◦ ) is compact in .P. First of all, note that, since U is a 0-neighbourhood, it is absorbing, and for each −1 −1 .x ∈ E, we can find .λx > 0 so that .λx x ∈ U . Hence .|u(λx x)| ≤ 1 and, therefore, ◦ .|u(x)| ≤ λx for every .u ∈ U . Thus .

φ(U ◦ ) ⊆



.

BK (0, λx ) ,

x∈E

which, by Tikhonov’s theorem 2.1, is compact. This shows that .φ(U ◦ ) is relatively compact; to prove that it is closed. To see this, take some net .(uj ) ⊆ U ◦  it remains so that . φ(ui ) i is convergent in .P to some .y = (yx )x∈E , and let us see that .y ∈ φ(U ◦ ). We define .u : E → K by .u(x) = yx . By the definition of .φ, we have that .uj (x) → yx (in .K) for every .x ∈ E and u(α1 x1 + α2 x2 ) = yα1 x1 +α2 x2 = lim uj (α1 x1 + α2 x2 )

.

j

= lim α1 uj (x1 ) + lim α2 uj (x2 ) = α1 u(x1 ) + α2 u(x2 ) . j

j

3.5 The Mackey–Arens Theorem

113

This shows that u is linear. Since .uj ∈ U ◦ for every j , we have .|uj (x)| ≤ 1 for every .x ∈ U . Taking limits yields .|u(x)| ≤ 1 for every .x ∈ U . Then u is continuous (hence .u ∈ E  ) and .u ∈ U ◦ . Since obviously .φ(u) = y, the set .φ(U ◦ ) is closed and  this completes the proof. A straightforward consequence of this result and (3.10) is the following. Corollary 3.6 Let X be a normed (or Banach) space. Then .B X is .σ (X , X)compact. In particular, every bounded subset of .X is .σ (X , X)-relatively compact.

3.5 The Mackey–Arens Theorem Given a dual pair .(E, F ), we say that a locally convex topology .τ on E is of the dual pair if .(E, τ ) = F . Clearly .σ (E, F ) is a topology of the dual pair (recall Proposition 3.6) and, by Proposition 3.8, if .τ is a topology of the dual pair .(E, F ), then .σ (E, F ) ⊆ τ . If .(E, P) is a locally convex space, then .τP is an obvious example of a topology of the dual pair .(E, E  ). Our aim in this section is to characterize the topologies of a dual pair, showing that these correspond to topologies of uniform convergence on certain families of compact sets (see Theorem 3.16 for a precise statement). Let .(E, P) be locally convex space and fix some equicontinuous .S ⊆ E  . Then there is some .U ∈ U0 (E) so that .S ⊆ U ◦ , and, for each .x ∈ E, we can find .λ > 0 so that .x ∈ λU (that is, .x/λ ∈ U ). Then .

      1 sup u xλ  ≤ sup u xλ  ≤ . λ u∈U ◦ u∈S

As a consequence, qS (x) = sup |u(x)|

.

(3.11)

u∈S

is finite for every .x ∈ E and, therefore, defines a seminorm on E. We consider the family Q = {qS : S ⊆ E  equicontinuous}

.

and the topology defined by it. Proposition 3.14 For every locally convex space, we have .τP = τQ . Proof We begin by showing that every .p ∈ P belongs to .Q. Given some p, consider .Up = {x ∈ E : p(x) ≤ 1} and its polar .Up◦ ⊆ E  that (see Proposition 3.12) is equicontinuous. We take the associated seminorm and observe that .qUp◦ (x) ≤ 1 if and only if .|u(x)| ≤ 1 for every .u ∈ Up◦ , and this happens if and only if

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3 Duality and Linear Operators

x ∈ Up◦◦ . Note that .Up is absolutely convex and closed and, then, by the Bipolar Theorem 3.14, .Up = Up◦◦ . Summarizing, .qUp◦ (x) ≤ 1 if and only if .p(x) ≤ 1. Hence, by Lemma 2.1, .p = qUp and (recall Remark 2.8) .τP ⊆ τQ . Fix now some equicontinuous .S ⊆ E  and (using Proposition 3.12) take U , an absolutely convex neighbourhood of 0 such that .S ⊆ U ◦ . Choose .ε > 0 and .p ∈ P so that .{x ∈ E : p(x) ≤ ε} ⊆ U . Now, if .x ∈ E is such that .p(x) ≤ ε, then .|u(x)| ≤ 1 for every .u ∈ S and, .qS (x) ≤ 1. Hence .{x ∈ E : p(x) ≤ ε} ⊆ {x ∈ E : qS (x) ≤ 1}. Using again Lemma 2.1, this gives .qS ≤ 1ε p and, by Remark 2.8, .τQ ⊆ τP and completes the proof.  .

Let us now observe that if .τ is a locally convex topology on E of the dual pair (E, F ) (that is, so that .F = (E, τ ) ) and .U, V ⊆ E are .τ -closed, absolutely convex neighbourhoods, then, by the Bipolar Theorem 3.14,

.

(U ◦ ∩ V ◦ )

.

σ (F,E)

 ◦◦  ◦ ◦ = (U ◦ ∪ V ◦ ) = (U ◦ ∪ V ◦ ◦   ◦ = U ◦◦ ∩ V ◦◦ = U ∩ V .

(3.12)

Theorem 3.16 (Mackey–Arens) Let .(E, F ) be a dual pair and .τ a Hausdorff locally convex topology on E. The following statements are equivalent: 1. .τ is a topology for the dual pair. 2. There is a family .A of .σ (F, E)-relatively compact subsets of F satisfying  .

S = F.

(3.13)

S∈A

(S ∪ T )

σ (F,E)

∈ A for every S, T ∈ A

(3.14)

and such that τ = τQA ,

(3.15)

.

where .QA = {qS : S ∈ A } and the seminorms are defined in (3.11). Proof We begin by assuming that .τ is a topology for the dual pair .(E, F ). We consider the family A = {S ⊆ F : (E, τ )-equicontinuous in F } .

.

Let us note first of all that, by the Alaoglu-Bourbaki Theorem 3.15, every .(E, τ )equicontinuous set is .(E, τ )-relatively compact. Since every point in F is .(E, τ )equicontinuous, we immediately have (3.13). On the other hand, if .S, T ∈ A , we can find .U, V ∈ U0 (E) so that .S ⊆ U ◦ and .T ⊆ V ◦ (recall Proposition 3.12). Using (3.12), we have σ (F,E)

S ∪ T ⊆ (U ∩ V )◦ = (U ◦ ∩ V ◦ )

.

,

3.5 The Mackey–Arens Theorem

115

σ (F,E)

∈ A . Finally, (3.15) follows immediately from and this implies .(S ∪ T ) Proposition 3.14. Assume now that 2 holds and that there is a family .A satisfying (3.13) and (3.14). Our aim is to see that .τQA is a topology for the dual pair, that is, .(E, τQA ) = F . Observe first of all that (3.13) implies .σ (E, F ) ⊆ τQA and then   F = E, σ (E, F ) ⊆ (E, τQA ) .

.

This gives one of the inclusions. To see the converse, one takes some .v ∈ (E, τQA ) ⊆ E ∗ and finds some .S ∈ A and .λ > 0 so that .|v(x)| ≤ λqS (x) for every .x ∈ E. This gives  1    v(x) ≤ sup |u(x)| λ u∈S

.

for every .x ∈ E, and (see (3.9) and the Bipolar Theorem 3.14) .

On the other hand, .(S) σ (F, E). Hence

1 σ (E ∗ ,E) . v ∈ S ◦F E ◦EE∗ = (S) λ

σ (F,E)

is by (3.14) .σ (F, E)-compact and .σ (E ∗ , E)|F =

S ◦F E ◦EE∗ = (S)

σ (F,E)

.

,

from which we deduce . λ1 v ∈ F and .v ∈ F . This shows that .(E, τQA ) ⊆ F and completes the proof.  Notice that, with this point of view, the family A = {S ⊆ F : subset of the absolutely convex hull of finite subsets of F } ,

.

which clearly consists of .σ (F, E)-compact sets, defines the topology .σ (E, F ). We consider now the family M = {S ⊆ F : subset of an absolutely convex σ (F, E)-compact set} .

.

(3.16)

This satisfies both (3.13) and (3.14), and then .τQA is a topology of the dual pair. We call it the Mackey topology and denote it by .μ(E, F ). Corollary 3.7 Given a dual pair .(E, F ), the Mackey topology is the finest locally convex topology of the dual pair. That is, a locally convex topology .τ is of the dual pair if and only if .σ (E, F ) ⊆ τ ⊆ μ(E, F ).

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3 Duality and Linear Operators

Proof Let .τ be a topology of the dual pair .(E, F ). Clearly .τ is finer than the topology .σ (E, F ) (recall Proposition 3.8). On the other hand, by Theorem 3.16, there is a family .A consisting of .σ (F, E)-relatively compact subsets of F satisfying (3.13) and (3.14) such that the topology .τ is the topology .τQA . Since .A is contained in the family .M from (3.16), it readily follows that .τ is coarser than .μ(E, F ). 

3.6 The Banach–Steinhaus Theorem Theorem 3.17 (Banach–Steinhaus) Let E be a Fréchet space and F a locally convex space. If H ⊆ L(E, F ) is such that H (x) = {T x : T ∈ H }

.

is bounded in F for every x ∈ E, then H is equicontinuous. Proof Let (pn )n be an increasing countable family of seminorms defining the topology of E. We proceed by contradiction, and suppose that H is not equicontinuous. Then, there exists a continuous seminorm q on F such that for every n and every c > 0 we may find Tn,c ∈ H and xn,c ∈ E satisfying q(Tn,c xn,c ) > c and pn (xn,c )
1 and p1 (x1 )
1 = n(1) so that q(T1 x) ≤ pn(2) (x) for every x ∈ E. Then, denoting M1 = supT ∈H q(T x1 ) < ∞, we can take x2 ∈ E and T2 ∈ H in such a way that q(T2 x2 ) > 2 + M1 + 1 and pn(2) (x2 )
j and l ∈ N, we get, using (3.19) and (3.17), pm

i+l 

.

k=i

i+l i+l i+l

   1 1 < i−1 . xk ≤ pn(j ) (xk ) ≤ pn(k) (xk ) < 2j 2 k=i

k=i

k=i



 n

This shows that the sequence k=1 xk n is Cauchy and, hence, convergent. Denote  x= ∞ x . Given j ∈ N, we have (using now (3.18), (3.19) and (3.17)) k=1 k q(Tj x) ≥ q(Tj xj ) −

j −1 

.

j −1 

Mi −

i=1

=j +1+

q(Tj xk )

k=j +1

k=1

>j +1+

∞ 

q(Tj xk ) −

j −1 

q(Tj xk ) −

∞ 

pn(k) (xk )

k=j +1

k=1

j −1 ∞    Mi − q(Tj xk ) − pn(k) (xk ) k=j +1

i=1

>j +1−

∞ 

pn(k) (xk ) ≥ j + 1 −

k=j +1

∞  1 =j. 2k k=1

This implies that H (x) is not bounded and completes the proof.



In particular, we have the following version for Banach and normed spaces. Theorem 3.18 Let X be a Banach space and Y a normed space. If H ⊆ L(X, Y ) is such that H (x) is bounded in Y for every x ∈ X, then .

sup T  < ∞ .

T ∈H

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3 Duality and Linear Operators

This means that if a set is pointwise bounded, then it is also uniformly bounded. For this reason this theorem is also known as the uniform boundedness principle. As a consequence of Theorem 3.17, we show that the bounded sets on every topology of a dual pair are the same (see Theorem 3.19). We need some preliminary work. Given an absolutely convex, bounded set B of a locally convex space E, we define  .EB = nB = span(B) . n∈N

The set B is clearly absorbing in EB , and we may consider on it the Minkowski gauge pB (recall the definition in (3.4)). Proposition 3.15 Let (E, P) be a locally convex space and B ⊆ E bounded and absolutely convex. Then (EB , pB ) is a normed space so that (EB , pB ) → (E, P) is continuous. Proof By Lemma 3.3, pB is a seminorm. To see that it is a norm, we pick x ∈ EB so that pB (x) = 0 (note that this implies that x ∈ εB for every ε > 0) and we want to see that x = 0. Take U ∈ U0 (E) and find λ > 0 so that B ⊆ λU . Then x ∈ (1/λ)B ⊆ U , and this implies that x ∈ ∩{U : U ∈ U0 (E)} = {0} (because E is Hausdorff). Therefore x = 0 and pB is a norm. If U ∈ U0 (E), it is easy to see (since B is bounded) that there is ε > 0 such that pB (x) < ε implies x ∈ U . This gives the continuity of the inclusion.  Proposition 3.16 Let E be a locally convex space and B ⊆ E bounded, absolutely convex and closed. If either E is sequentially complete or B is compact, then EB is a Banach space. Proof Let (xn )n be a Cauchy sequence in EB . This means that for every ε > 0 there is n0 so that xn − xm ∈ εB for every n, m ≥ n0 .

.

(3.20)

Suppose in first place that E is sequentially complete. Since EB → E is continuous, the sequence (xn )n is also Cauchy in E and, then, τP -converges to some x0 ∈ E. Now, the sequence, being Cauchy, is bounded in EB , and we can find some λ > 0 such that xn ∈ λB for every n. Then, since B is τP -closed, x0 ∈ λB ⊆ EB . Taking limits (with respect to τP ) as m → ∞ in (3.20), we have that for every ε > 0 there is n0 so that xn − x0 ∈ εB for every n ≥ n0 , and xn → x0 in EB . Suppose now that B is τP -compact. As before, there is some λ > 0 such that xn ∈ λB for every n. Then (by compactness) (xn )n has a τP -cluster point x0 ∈ EB . Since (xn )n is Cauchy in E, this implies that the sequence converges (in τP ) to x0 . The proof goes on as in the previous case. 

3.6 The Banach–Steinhaus Theorem

119

Theorem 3.19 (Banach–Mackey) Let (E, F ) be a dual pair and τ a topology (on E) of the dual pair. For each A ⊆ E, the following statements are equivalent: 1. A is μ(E, F )-bounded. 2. A is τ -bounded. 3. A is σ (E, F )-bounded. Proof As a straightforward consequence of Corollary 3.7, we have 1 ⇒ 2 ⇒ 3. To complete the proof, assume that A is σ (E, F )-bounded, and let us see that it is μ(E, F )-bounded. To do this, it suffices to show that for every σ (F, E)-compact σ (E,F ) , we have absolutely convex set B ⊆ F (recall (3.16)) and M = (A) .

sup sup |u(x)| < ∞ .

(3.21)

x∈M u∈B

So, we fix any such B and consider FB that by Proposition 3.16 (applied to (F, σ (F, E)) and B) is a Banach space. Since each x ∈ M ⊆ E is σ (F, E) continuous (Proposition 3.6) and FB → F, σ (F, E) is continuous (Proposition 3.15), the restriction x|FB is again continuous and H = {x|FB : x ∈ M} ⊆ (FB ) .

.

The set M is σ (E, F )-bounded, and this easily implies that H (u) = {x(u) : x ∈ M} is bounded in K for every u ∈ F . The Banach–Steinhaus theorem (in its version of Theorem 3.18) yields (note that B is the unit ball of FB ) .

sup sup |x(u)| < ∞ . x∈M u∈B



This gives (3.21) and completes the proof.

As a consequence of the Banach–Steinhaus theorem, we can conclude (as we already announced in Sect. 1.3.2) that there are continuous functions whose Fourier series does not converge at 0. We define C(T) as the space of 2π -periodic continuous functions f : R → C which, with the norm f ∞ =

.

sup

t∈[−π,π ]

|f (t)| ,

is a Banach space. Note that f ∞ = supt∈R |f (t)| for every f ∈ C(T). For each n ∈ N, we define un : C(T) → C as (recall (1.7) for the definition of sn ) un (f ) = sn (f, 0) =



.

|j |≤n

f(j ) .

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3 Duality and Linear Operators

This is clearly well defined and linear. Taking the Dirichlet kernel (recall (1.10)) and using Proposition 1.5, we can reformulate this as un (f ) =

.

1 2π



π −π

(3.22)

f (−t)Dn (t)dt .

From this, we have |un (f )| ≤

.

1 2π



π

−π

|f (−t)| |Dn (t)|dt ≤ f ∞

1 2π



π −π

|Dn (t)|dt

(3.23)

for every f ∈ C(T). Hence un is continuous and un ∈ C(T) for every n ∈ N. Lemma 3.5 For each n ∈ N, we have 

1 .un  = 2π

π

|Dn (t)|dt .

−π

Proof We fix n and define f : [−π, π ] → C as  f (t) =

.

1

if Dn (t) ≥ 0

−1

if Dn (t) < 0.

Since this is real-valued and discontinuous only on a finite number of points, there is a sequence (fj )j ⊆ C(T) so that −1 ≤ fj (t) ≤ 1 for every j and t and limj fj (t) = f (t) for every t ∈ [−π, π ]. From (3.22) and the theorem of the dominated convergence, we get 1 . lim un (fj ) = lim j j 2π



π

−π

fj (−t)Dn (t)dt =

1 2π



π

−π

f (−t)Dn (t)dt =

1 2π



π

−π

|Dn (t)|dt .

Hence un  ≥ sup |un (fj )| ≥

.

j ∈N

1 2π



π −π

|Dn (t)|dt .

The converse inequality follows immediately from (3.23), and this completes the proof.  Our next step is to estimate the behaviour of these norms as n grows.

3.6 The Banach–Steinhaus Theorem

121

Lemma 3.6 There exist c1 , c2 > 0 so that c1

2n 

.

k=0

1 1 ≤ k+1 2π



π

−π

|Dn (t)|dt ≤ c2

2n  k=0

1 , k+1

for every n ≥ 2. Proof We begin by noting that from the fact that Dn is even and (1.11), we have 1 . 2π



π

1 |Dn (t)|dt = π −π



  sin((n + 1)/2)t)  2 π/2  sin((2n + 1)u)      dt = du . sin(t/2) π 0 sin u (3.24)

π 0

Since .

2 u ≤ sin u ≤ u π

(3.25)

for every 0 ≤ u ≤ π/2, then .

1 2π



π

−π



| sin((2n + 1)u)| du u 0  2n  2  (k+1)π/2 | sin w| 2 (2n+1)π/2 | sin w| dw = dw = π 0 w π w kπ/2

|Dn (t)|dt ≥

2 π

π/2

k=0

≥

 (k+1)π/2 2n 2n 4  1 4  1 . | sin w|dw = k + 1 kπ/2 k+1 π2 π2 k=0

k=0

On the other hand, using again (3.25) after (3.24), we have .

1 2π





π

−π

=

0 2n  

(k+1)π/2

k=0 kπ/2

 (2n+1)π/2 | sin((2n + 1)u)| | sin w| du = dw u w 0 2n  π  (k+1)π/2 | sin w| | sin w| dw ≤ + dw w 2 w kπ/2

π/2

|Dn (t)|dt ≤

k=1

≤

 1 π 1 + ≤ c2 , 2 k k+1 2n

2n

k=1

k=0

for some c2 ≥ 1. We have now everything at hand to state and prove our claim.



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3 Duality and Linear Operators

Theorem 3.20 There exists a function in C(T) whose Fourier series does not converge at 0. Proof Let us suppose that this is not the case, and limn un (f ) = limn sn (f, 0) = f (0) for every f ∈ C(T). This implies that the set {un (f ) : n ∈ N} is bounded and, by Theorem 3.18, .

sup un  < ∞ , n∈N



and this contradicts Lemmas 3.5 and 3.6.

3.7 The Banach-Schauder Theorem If E and F are locally convex spaces, then the graph of a mapping .T : E → F is defined as the set G(T ) = {(x, y) ∈ E × F : T x = y} .

.

Suppose that T is continuous. If a net .(xi )i converges to some .x ∈ E, then the net .(Ti x)i converges to some .y ∈ F and .y = T x. Let us note that this then implies that if .(xi , T xi ) converges in .E × F to .(x, y), and then .y = T x and .G(T ) is closed. Our aim in this section is to show that in some cases (if E and F are Fréchet spaces and T is linear) this implication can be reversed, and the fact that the graph is closed gives the continuity of T . In this way, we obtain another characterization of the continuity of linear operators (recall that we already have one in Theorem 2.11) between Fréchet spaces. This result, known as the closed graph theorem (see Theorem 3.23), follows from another important result, known as the Banach-Schauder or open mapping theorem (see Theorem 3.22), which tells that every continuous linear mapping is open. The proof of this relies on the following result, known as the Baire category principle. Let M be a complete metric space. If .{Gn }n∈N is a ∞  Gn is dense in M. sequence of dense open subsets of M, then .

Theorem 3.21 (Baire)

∞

n=1

Proof If we write .A = n=1 Gn , what we have to show is that .A ∩ G = ∅ for every open, non-empty G. First of all, since .G1 is dense, we have .G1 ∩ G = ∅ and, being the intersection open, we can find .x1 ∈ M and .0 < r1 < 1 so that B(x1 , r1 ) = {x ∈ M : d(x, x1 ) ≤ r1 } ⊆ G1 ∩ G .

.

With the same idea, using the denseness of .G2 , we may find .x2 ∈ M and .0 < r2 < 1/2 so that .B(x2 , r2 ) ⊆ G2 ∩ B(x1 , r1 ). Proceeding in this way, by induction, we

3.7 The Banach-Schauder Theorem

123

can produce two sequences .(xn )n ⊆ M and .(rn )n ⊆ R so that B(xn , rn ) ⊆ Gn ∩ B(xn−1 , rn−1 ) and rn
0, consider the sets Vr = {x ∈ E : dE (x, 0) < r} and Wr = {y ∈ F : dF (y, 0) < r} ,

.

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3 Duality and Linear Operators

where .dE and .dF are the metric defining the topologies of E and F , respectively. Then, for every .r > 0, there is .ρ > 0 such that .Wρ ⊆ T (Vr ). In particular, .T (Vr ) is a neighbourhood of 0 for each .r > 0.  Proof Fix .r > 0 and take .x ∈ E. The sequence . n1 n tends to 0 as .n → ∞, and

1 . x ∈ Vr/2 for some n. This shows that .E = n∈N nVr/2 and, since T is surjective, n F = T (E) =



.

   nT Vr/2 = nT Vr/2 .

n∈N

n∈N

 As a consequence of Baire’s Theorem 3.21 (see Remark 3.6), .n0 T Vr/2 has non empty interior for some .n0 , which gives that .T Vr/2 has non-empty interior. Then  there are .y0 ∈ F and .ρ > 0 so that .y0 + Wρ ⊆ T Vr/2 , and if .y ∈ Wρ , there is   .w ∈ T Vr/2 with .y = w − y0 . Choose sequences .(xn )n and .(xn )n in .Vr/2 such that   .T xn → w and .T xn → y0 , and define .yn = xn −xn for each n. Simple computations show that .(yn )n ∈ Vr and .T (yn ) → y. Hence .y ∈ T (Vr ), and this completes the proof.  Theorem 3.22 (Banach-Schauder/open mapping) Let .E, F be Fréchet spaces and .T : E → F be linear, continuous and such that .T (E) = F . Then T is open. In particular, if T is bijective, linear and continuous, then .T −1 is continuous as well. Proof Let us denote by .dE and .dF the metrics defining the topology on E and F , respectively. Let .U ⊆ E be an absolutely convex neighbourhood of 0. Since T is linear, .T (U ) is again absolutely convex. So, in view of Lemma 3.7, if we show that .T (U ) is a neighbourhood of 0 in F , we are done. To begin with, there is .r > 0 so that V = {x ∈ E : dE (x, 0) < r} ⊆ V .

.

Now, for each .n ∈ N, consider the set  Vn = x ∈ E : dE (x, 0)
n, we have, using (2.21), dE

n 

.

xi ,

i=1

m m m m



   r xi = dE xi , 0 ≤ dE (xi , 0) < . 2i i=1

i=n+1

i=n+1

i=n+1



n This shows that the sequence . is Cauchy and, then (since E is i=1 xi n complete), converges to some .x ∈ E. Moreover dE (x, 0) = dE

∞ 

.

i=1

∞ ∞

  r xi , 0 ≤ dE (xi , 0) < =r, 2i i=1

i=1

and .x ∈ V ⊆ U . On the other hand, for each .n ∈ N, we have T

.

n n n

   xi = T xi = (yi − yi+1 ) = y1 − yn+1 . i=1

i=1

i=1

The fact that .dF (yn , 0) < δn implies .limn yn+1 = 0. Hence, the continuity of T yields n

 T x = T lim xi = y1 − lim yn+1 = y1 ,

.

n

i=1

n

and .y1 = T x ∈ T (U ). This gives our claim and completes the proof.



Theorem 3.23 (Closed Graph Theorem) Let E and F be Fréchet spaces. A linear map .T : E → F is continuous if and only if .G(T ) is closed in .E × F . Proof As we already pointed out, if T is continuous, then .G(T ) is closed. So, it remains only to prove the converse implication. We know (see 2.8) that .E × F is a Fréchet space. Now, if .G(T ) is closed in .E × F , it is a Fréchet space itself (with the restriction of the product topology), and the restrictions of the projections .π1 : G(T ) → E and .π2 : G(T ) → F given by .π1 (x, T x) = x and .π2 (x, T x) = T x are continuous linear maps. Note that .π1 is bijective, and, by Theorem 3.22 −1 −1 .π 1 : E → G(T ) given by .x  (x, T x) is continuous. Thus .T = π2 ◦ π1 , being the composition of two continuous maps, is continuous. 

3.8 Topologies on the Space of Continuous Linear Mappings If E and F are locally convex spaces, then the set .L(E, F ) of continuous linear mappings (recall (2.18)) is clearly a vector space. We can consider different topologies in this space.

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3 Duality and Linear Operators

In the first place, we consider the topology of uniform convergence on bounded sets is given by the family of seminorms defined for .T ∈ L(E, F ) as pB,q (T ) = sup q(T x) ,

.

x∈B

where .B ⊆ E is bounded, and q is a continuous seminorm on F . Note that this supremum is finite because, being T continuous and B bounded, the set .T (B) is bounded in F . We denote by .Lβ (E, F ) the space .L(E, F ) endowed with this topology. If .V is a basis of neighbourhoods of F , then for each .V ∈ V and .B ⊆ E bounded, the set W (B, V ) = {T ∈ L(E, F ) : T (B) ⊆ V }

.

is a neighbourhood of .0 ∈ L(E, F ) for this topology, and {W (B, V ) : V ∈ V, B ⊆ E bounded}

.

is a basis of neighbourhoods of .0 ∈ Lβ (E, F ). If .F = K, then .L(E, K) is .E  (the dual of E). In this case, the topology of uniform convergence on bounded sets is called the strong topology of .E  and is denoted by .β(E  , E). Let us note that the seminorms defining the topology in this case are of the form qB (u) = sup |u(x)| ,

.

(3.26)

x∈B

where B ranges over all bounded subsets of E, and the basis of neighbourhoods is given by W (B) = {u ∈ E  : sup |u(x)| < 1} .

.

x∈B

It is easy to check that .σ (E  , E) ⊆ μ(E  , E) ⊆ β(E  , E). We write .Eβ = (E  , β(E  , E)) and call it the strong dual of E. If both E and F are normed spaces, then .Lβ (E, F ) is the space .L(E, F ) endowed with the operator norm defined as T  = sup T xF ,

.

xE ≤1

which we briefly introduced in Remark 2.9. In this case, if .F = K, then .Eβ is the dual with the topology defined by the norm u = sup |u(x)| .

.

xE ≤1

3.8 Topologies on the Space of Continuous Linear Mappings

127

The second topology that we are going to consider is the topology of uniform convergence on precompact sets, given by the family of seminorms defined by pK,q (T ) = sup q(T x) ,

.

(3.27)

x∈K

for .T ∈ L(E, F ), where .K ⊆ E is precompact, and q is a continuous seminorm on F . We write .Lco (E, F ) for the space .L(E, F ) endowed with this topology. Given a continuous seminorm q on F and .x1 , . . . , xn ∈ E, then p{x1 ,...,xn },q (T ) = max q(T xj ) ,

.

i=1,...,n

(3.28)

for .T ∈ L(E, F ) defines a seminorm on .L(E, F ). The strong operator topology (also known as simple topology) is the one defined by these seminorms as .{x1 , . . . , xn } varies on the finite subsets of E and q on the continuous seminorms on F . We write .Ls (E, F ) for the space .L(E, F ) endowed with this topology. It is not difficult to check that a net .(Ti )i converges to some T in the strong operator topology if and only if .Ti x → T x (in F ) for every .x ∈ E. If .F = K, then we recover the weak topology on .E  , that is, .Ls (E, K) = (E  , σ (E  , E)). The weak operator topology is defined as the simple topology on .L(E, (F, σ (F, F  ))) and is given by the seminorms p{x1 ,...,xn },{u1 ,...,um } (T ) = max{|uj (T xi ) : i = 1, . . . , n , j = 1, . . . , m} ,

.

where .{x1 , . . . , xn } runs over all finite subsets of E and .{u1 , . . . , um } over all finite subsets of .F  . We write .Lw (E, F ) for .L(E, F ) endowed with the weak operator topology. The following inclusions are clearly continuous: Lβ (E, F ) → Lco (E, F ) → Ls (E, F ) → Lw (E, F ) ;

.

that is, each one of the topologies on the left hand side of an arrow is finer than the topology on the right-hand side of the arrow. We see now how equicontinuous sets behave with these topologies. Proposition 3.17 Every equicontinuous subset of .L(E, F ) is bounded in Lβ (E, F ).

.

Proof Fix some equicontinuous .H ⊆ L(E, F ) and take .B ⊆ E bounded and q a continuous seminorm on F . By Proposition 2.6, we can find some seminorm on E (call it p) and some .C > 0, so that q(T x) ≤ Cp(x) ,

.

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3 Duality and Linear Operators

for every .x ∈ E and .T ∈ H . On the other hand, being B bounded, .supx∈B p(x) = λ < ∞. With this, we have pB,q (T ) = sup q(T x) ≤ C sup p(x) = Cλ

.

x∈B

x∈B

for every .T ∈ H . Since this holds for every bounded B and every seminorm q, the claim follows.  Proposition 3.18 Let .H ⊆ L(E, F ) be equicontinuous, and .N ⊆ E such that span N is dense in E. Then the strong operator topology on .L(E, F ) and the topology defined by the family of seminorms

.

  p{x1 ,...,xn },q : {x1 , . . . , xn } ⊆ N finite, q continous seminorm on F

.

(that is, the topology of simple convergence on N) coincide on H . In particular, a net .(Ti )i ⊆ H converges to T in the strong operator topology if and only if .Ti x → T x for every .x ∈ N . Proof Since the property of being equicontinous is maintained by translation, we may assume that .0 ∈ H . We fix .{x1 , . . . , xs } ⊆ E and a continuous seminorm q on F . Considering the seminorm .p{x1 ,...,xs },q defined in (3.28), we have that U = {T ∈ H :

.

max q(T xi ) ≤ 1}

i=1,...,s

(3.29)

is a neighbourhood of 0 in H for the strong operator topology. We want to find a neighbourhood V of 0 for the topology of simple convergence on N contained in U . Since H is equicontinuous, we can find a seminorm on E (say p) and a constant .C > 0 so that .q(T x) ≤ Cp(x) for every .x ∈ E and .T ∈ H . By density, for each .i = 1, . . . , s, we can choose .zi ∈ span N so that p(xi − zi )
max

n(i) 

.

i=1,...,s

|αi,k | ,

k=1

and define  V = T ∈ H : sup q(T w)
0 such that .

sup{|xn | : x = (xn )n ∈ A} ≤ mn .

7. Let A = {en : n ∈ N}, then a. σ ( 1 , c00 ) − limn en = 0.

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3 Duality and Linear Operators

b. A ∪ {0} is σ ( 1 , c00 )-compact. c. A is not β( 1 , c00 )-bounded. 3.6 Let (E, τ ) be a metrizable locally  convex space. Show that if there is a sequence (An )n of bounded sets such that E = An , then E is normable. 3.7 Let ϕ be the (vector) space of all sequences which are 0 except for a finite number of coordinates. Consider the dual pair (ϕ, ϕ) with the bilinear form given by (xn )n , (yn )n  =

∞ 

.

xn yn .

n=1

Prove that M ⊆ ϕ is 1. σ (ϕ, ϕ)-bounded if and only if supx∈M |xn | < ∞ for each n ∈ N. 2. β(ϕ, ϕ)-bounded if and only if it is σ (ϕ, ϕ)-bounded and there is n0 ∈ N such that xn = 0 for every n ≥ n0 and all x ∈ M. 3.8 Let (E, P) be a locally convex metrizable space and (F, Q) a locally convex space. Show that a linear map T : E −→ F is continuous if and only if for each bounded set B in E, T (B) is bounded in F . 3.9 Show that every open linear map is surjective. Show that if E is a Fréchet space and T : E → F is open (thus surjective), then F is also Fréchet. 3.10 Show that every separable Banach space is isometric to a quotient of 1 . Hint: Use the open mapping theorem. Further Reading [1–31]

References 1. Bachman, G., Narici, L.: Functional Analysis. Dover Publications, Inc., Mineola, NY (2000). Reprint of the 1966 original 2. Banach, S.: Théorie des Opérations Linéaires. Éditions Jacques Gabay, Sceaux (1993). Reprint of the 1932 original 3. Bogachev, V.I., Smolyanov, O.G.: Topological Vector Spaces and Their Applications. Springer Monographs in Mathematics. Springer, Cham (2017). https://doi.org/10.1007/978-3-31957117-1 4. Brezis, H.: Functional analysis. In: Sobolev Spaces and Partial Differential Equations. Universitext. Springer, New York (2011) 5. Cerdà, J.: Linear functional analysis. In: Graduate Studies in Mathematics, vol. 116. American Mathematical Society, Providence, RI; Real Sociedad Matemática Española, Madrid (2010). https://doi.org/10.1090/gsm/116 6. Conway, J.B.: A course in functional analysis. In: Graduate Texts in Mathematics, vol. 96. Springer, New York (1985). https://doi.org/10.1007/978-1-4757-3828-5

References

135

7. Dunford, N., Schwartz, J.T.: Linear Operators. Part I. Wiley Classics Library. John Wiley & Sons, Inc., New York (1988). Reprint of the 1958 original. 8. Edwards, R.E.: Functional Analysis. Dover Publications, Inc., New York (1995). Theory and applications, Corrected reprint of the 1965 original 9. Fabian, M., Habala, P., Hájek, P., Montesinos, V., Zizler, V.: Banach Space Theory. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Springer, New York (2011). https:// doi.org/10.1007/978-1-4419-7515-7 10. Fabian, M., Habala, P., Hájek, P., Montesinos Santalucía, V., Pelant, J., Zizler, V.: Functional analysis and infinite-dimensional geometry. In: CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC, vol. 8. Springer, New York (2001). https://doi.org/10.1007/978-14757-3480-5 11. Floret, K., Wloka, J.: Einführung in die Theorie der lokalkonvexen Räume. Lecture Notes in Mathematics, No. 56. Springer, Berlin/New York (1968) 12. Grothendieck, A.: Topological vector spaces. In: Notes on Mathematics and its Applications. Gordon and Breach Science Publishers, New York/London/Paris (1973). Translated from the French by Orlando Chaljub 13. Heuser, H.G.: Functional Analysis. A Wiley-Interscience Publication. John Wiley & Sons, Ltd., Chichester (1982). Translated from the German by John Horváth 14. Horváth, J.: Topological Vector Spaces and Distributions, vol. I. Addison-Wesley Publishing Co., Reading, Mass./London-Don Mills, Ont. (1966) 15. Jameson, G.J.O.: Topology and Normed Spaces. Chapman and Hall, London; Halsted Press, New York (1974) 16. Jarchow, H.: Locally Convex Spaces. Mathematische Leitfäden. B. G. Teubner, Stuttgart (1981) 17. Kadets, V.: A Course in Functional Analysis and Measure Theory. Universitext. Springer, Cham (2018). https://doi.org/10.1007/978-3-319-92004-7. Translated from the 2006 Russian edition 18. Kelley, J.L.: General topology. In: Graduate Texts in Mathematics, No. 27. Springer, New York/Berlin (1975). Reprint of the 1955 edition 19. Köthe, G.: Topological Vector Spaces. I. Translated from the German by D. J. H. Garling. Die Grundlehren der mathematischen Wissenschaften, Band 159. Springer, New York Inc., New York (1969) 20. Köthe, G.: Topological vector spaces. II. In: Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 237. Springer, New York/Berlin (1979) 21. Lindenstrauss, J., Tzafriri, L.: Classical Banach spaces. I. In: Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 92. Springer, Berlin/New York (1977). Sequence spaces 22. Lindenstrauss, J., Tzafriri, L.: Classical Banach spaces. II. In: Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], vol. 97. Springer, Berlin/New York (1979). Function spaces 23. Meise, R., Vogt, D.: Introduction to functional analysis. In: Oxford Graduate Texts in Mathematics, vol. 2. The Clarendon Press, Oxford University Press, New York (1997). Translated from the German by M. S. Ramanujan and revised by the authors 24. Pérez Carreras, P., Bonet, J.: Barrelled locally convex spaces. In: North-Holland Mathematics Studies, vol. 131. North-Holland Publishing Co., Amsterdam (1987). Notas de Matemática, 113 25. Robertson, A.P., Robertson, W.: Topological vector spaces. In: Cambridge Tracts in Mathematics, vol. 53, 2nd edn. Cambridge University Press, Cambridge/New York (1980) 26. Rudin, W.: Functional analysis, 2nd edn. In: International Series in Pure and Applied Mathematics. McGraw-Hill, Inc., New York (1991) 27. Schaefer, H.H., Wolff, M.P.: Topological vector spaces. In: Graduate Texts in Mathematics, vol. 3, 2nd edn. Springer, New York (1999). https://doi.org/10.1007/978-1-4612-1468-7 28. Taylor, A.E., Lay, D.C.: Introduction to Functional Analysis, 2nd edn. Robert E. Krieger Publishing Co., Inc., Melbourne, FL (1986)

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29. Valdivia, M.: Topics in locally convex spaces. In: Notas de Matemática [Mathematical Notes], vol. 85. North-Holland Publishing Co., Amsterdam/New York (1982) 30. Wilansky, A.: Topology for Analysis. Robert E. Krieger Publishing Co., Inc., Melbourne, FL (1983). Reprint of the 1970 edition 31. Yosida, K.: Functional analysis. In: Classics in Mathematics. Springer, Berlin (1995). https:// doi.org/10.1007/978-3-642-61859-8. Reprint of the sixth (1980) edition

Chapter 4

Spaces of Holomorphic and Differentiable Functions and Operators Between Them

So far, the only examples of spaces of functions we have dealt with are those consisting of continuous functions, either on a compact set (.C(K)) the spaces which are Banach spaces (recall Theorem 2.8) or on an open set (.C()) the spaces which we know that are Fréchet (Theorem 2.18) not normable (Proposition 2.14) spaces. Our aim now is to extend this, finding other classes of functions that appear naturally in the classical theory, looking at them from the point of view of the general theory that we presented in Chap. 2.

4.1 Space of Holomorphic Functions 4.1.1 Locally Convex Structure We begin by looking at holomorphic functions of one complex variable. We choose some open and connected set in the complex plane (that, for what topology is concerned, we identify with .R2 ) . ⊆ C. A function .f :  → C is holomorphic if for every .z0 ∈  the following limit exists f  (z0 ) = lim

.

z→z0

f (z) − f (z0 ) . z − z0

We define the space H() = {f :  → C holomorphic on } .

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_4

137

138

4 Holomorphic and Differentiable Functions and Operators

Every holomorphic function is continuous; hence, this is a subspace of .(C(), τco ), and on it, we consider the induced topology. That is, on .H(), we consider the topology of uniform convergence on compact sets, defined by the seminorms pK (f ) = sup |f (z)|, for K   .

.

z∈K

Obviously, in this way, .(H(), τco ) is a metrizable l.c.s. Our aim now is to see that it is in fact a Fréchet space. We address this question by seeing that it is a closed subspace of .C(). We have to see that the limit point of every convergent sequence in .(H(), τco ) also belongs to .H(). We recall (without proof) some classical results, the proof of which can be found, for example, in [1, 3–5, 8] (or in any other text on basic function theory of one complex variable). Theorem 4.1 (Cauchy Theorem) Let . ⊆ C be open and simply connected and f ∈ H(). Then

.

 f (z)dz = 0,

.

γ

for every closed piecewise smooth curve .γ ⊆ . Theorem 4.2 (Cauchy Integral Formula) Let .f ∈ H() and .D(z0 , r) ⊆ ; then, for every .z ∈ D(z0 , r) and all .n ∈ N ∪ {0}, we have f (n) (z) =

.

n! 2π i

 C(z0 ,r)

f (w) dw , (w − z)n+1

where .C(z0 , r) is the circle centred at .z0 of radius r with positive orientation. In particular, |f (n) (z0 )| ≤

.

n! rn

|f (z)| ,

sup

(4.1)

z∈D(z0 ,r)

for every .n ∈ N ∪ {0}. Theorem 4.3 (Morera’s Theorem) If f , a continuous function on .D(z0 , r) for some .z0 ∈ C and .r > 0, is such that for each triangle .T ⊆ D(z0 , r), we have  f (z)dz = 0,

.

T

then f is holomorphic on .D(z0 , r). τco

Theorem 4.4 Let .(fn )n ⊆ H() such that .fn → f ; then .f ∈ H(). That is, .(H(), τco ) is closed in .(C(), τco ). In particular, .(H(), τco ) is a Fréchet space.

4.1 Space of Holomorphic Functions

139

Proof Let .z0 ∈ , and we choose an open disc .D(z0 , r) ⊂ . Obviously (since the sequence converges uniformly on the compact sets), .f ∈ C(D(z0 , r)). Let T be a triangle in .D(z0 , r). Since T is compact, the sequence .(fn )n converges to f  f (z)dz. On the other hand, uniformly on T , and this implies that . T fn (z)dz →  T since each .fn is holomorphic, Theorem 4.1 gives . f (z)dz = 0 for every .n ∈ N. n T  Hence, . T f (z)dz = 0, and by Theorem 4.3, .f ∈ H(D(z0 , r)). In particular, f is holomorphic at .z0 , and since this holds for every .z0 ∈ , the conclusion follows.

But we have more. If the sequence converges uniformly on compact sets, then so also do all the derivatives. τco

Theorem 4.5 (Weierstraß Theorem) Let .(fn )n ⊆ H() be such that .fn → f . (k) Then .fn → f (k) uniformly on compact sets for every .k ∈ N0 . That is, if .fn → f (k) in .(H(), τco ), then .fn → f (k) in .(C(), τco ) for every k. Proof Fix .k ∈ N. Since each compact set can be covered by a finite family of closed discs, it is enough to prove that .fj(k) −→ f (k) uniformly on any closed disc. So, fix .z0 ∈  and choose .r > 0 such that .D(z0 , r) ⊆ . We take .R > r such that .D(z0 , R) ⊂ . By Theorem 4.2, for each .|z − z0 | ≤ r, we have fn(k) (z) − f (k) (z) =

.

k! 2π i

 C(z0 ,R)

fn (w) − f (w) dw . (w − z)k+1

This gives |fn(k) (z) − f (k) (z)| ≤

.

max|w−z0 |=R |fn (w) − f (w)| k! 2π R 2π (R − r)k+1 =

k!R (R − r)k+1

max

|w−z0 |=R

|fn (w) − f (w)| ,

and .

max |fn(k) (z) − f (k) (z)| ≤

z∈D(z0 ,r)

k!R (R − r)k+1

max

|w−z0 |=R

|fn (w) − f (w)| .

Since .C(z0 , R) is a compact set, the last term tends to 0 as .n → ∞ and this completes the proof.

Remark 4.1 Since every compact subset of .D(0, R) is contained in .D(0, r) for some .0 < r < R, Theorem 1.10 gives that the sequence of partial sums .SN (z) =  N n uniformly over the compact sets of .D(0, R) n=0 an z of a power series  converges n to the function .f (z) = ∞ n=0 an z . Each .SN is a polynomial (hence, holomorphic) and Theorem 4.4 implies that f is holomorphic on .D(0, R). Also, Theorem 4.5 immediately gives .f (n) (0) = n!an for every n.

140

4 Holomorphic and Differentiable Functions and Operators

So, the function defined by a convergent power series is always holomorphic. As a matter of fact, the converse also holds, and every holomorphic function can be represented by a power series (called the Taylor series of the function). Theorem 4.6 Suppose that f is holomorphic on .D(z0 , r) for some .r > 0. Then there exist .R ≥ r such that f (z) =

∞ 

.

an (z − z0 )n

n=0 (n)

for every .z ∈ D(z0 , R), where .an = f n!(z0 ) for every n. Moreover, the series converges uniformly on .D(z0 , s) for every .0 < s < R. Proposition 4.1 The space .(H(C), τco ) is separable. Proof Fix .0 < R and .ε > 0. By Theorem 4.6, there is some N such that

.

N      sup f (z) − an zn  < ε .

|z|≤R

n=0

Now, for each .j = 1, . . . , N, we can find .qj , rj ∈ Q so that .| Re aj − qj | < ε and | Im aj − rj | < ε. With this, if .|z| < R, we have

.

N N      an zn − (qj + irj )zn  < ε(R N + · · · + R + 1) . 

.

n=0

n=0

Joining these two facts, we conclude that for every compact .K ⊆ C and .ε > 0, we can find a polynomial .PN whose coefficients have rational real and imaginary parts so that .

sup |f (z) − PN (z)| < ε . z∈K

Hence, the set of such polynomials (which is obviously countable) is dense in .H(C).

We have seen that, like .C(), the space of holomorphic functions is a Fréchet space. Our aim now is to show that it is not normable, but, unlike the case for the space of continuous functions, it does admit continuous norms. We again need a classical result from the elementary theory of holomorphic functions in one complex variable. Theorem 4.7 (Identity Theorem) Let . be an open and connected set. Let .(zn )n be a sequence with a limit point in .. If .f (zn ) = 0 for all .n ∈ N, then .f (z) = 0 for every .z ∈ .

4.1 Space of Holomorphic Functions

141

Proposition 4.2 Let . be an open and connected set. The space .(H(), τco ) admits continuous norms but is not normable. Proof Let us see first that it admits continuous norms. We choose an infinite compact set K in .. We let .pK (f ) = supz∈K |f (z)|. Then .pK is obviously a continuous seminorm on .(H(), τco ). Since every sequence in K has a convergent subsequence, Theorem 4.7 easily gives that q is a norm. Let us suppose now that .(H(), τco ) is normable, and let us call q the norm that induces the topology. By Remark 2.15, we can find a compact .K0   and .λ > 0 so that q(f ) ≤ λ sup |f (z)|

.

z∈K0

for every .f ∈ H(). On the other hand (also by Remark 2.15), for every compact K  , there is .μ > 0 such that

.

.

sup |f (z)| ≤ μq(f ) z∈K

for all .f ∈ H(). Combining these two facts, we have that for every K there is cK > 0 such that

.

.

sup |f (z)| ≤ cK sup |f (z)| . z∈K

(4.2)

z∈K0

We want to see now that this is not possible. We consider two cases. First of all, if . = C, then we choose .R > 0 such that .K0 ⊆ D(0, R) and consider .K = D(0, R + 1). By (4.2), there exists .cR > 0 such that .

sup |f (z)| ≤ cR sup |f (z)|

|z|≤R+1

|z|≤R

for every .f ∈ H(C). Take now .fk (z) = zk for .k ∈ N, and we have .(R +1)k ≤ cR R k  k k for every .k ∈ N. But . R+1 −→ +∞, and this is a contradiction. R 1 Suppose now . = C. We choose .ξ ∈ ∂ and define .g :  → C by .g(z) = z−ξ . Obviously, .g ∈ H() and .limz→ξ |g(z)| = ∞. We set now .M = maxz∈K0 |g(z)|. Then there exists .z0 ∈  such that .|g(z0 )| > 2M. If .h(z) = g(z) M , we have .h ∈ H() and .|h(z)| ≤ 1 for every .z ∈ K0 , which gives .|h(z)k | ≤ 1 for all .k ∈ N and .z ∈ K0 . Taking .K = {z0 }, (4.2) gives some .cK > 0 such that .|f (z0 )| ≤ cK supz∈K0 |f (z)| for every .f ∈ H(). If we evaluate in .f = hk , we obtain    g(z0 ) k  ≤ cK sup |h(z)|k ≤ cK 2k <  M  z∈K0

.

for all .k ∈ N. This gives a contradiction and completes the proof.



142

4 Holomorphic and Differentiable Functions and Operators

4.1.2 Representation as a Sequence Space Our aim now is to see that the spaces of holomorphic functions on open discs can be represented as (that is, are isomorphic to) Köthe echelon spaces for a convenient choice of Köthe matrix. Given .0 < R ≤ ∞, we consider the set .DR = {z ∈ C : |z| < R} and the space .H(DR ) of holomorphic functions on .DR . Note that with this notation .H(D1 ) is the space of holomorphic functions on the open unit disc .H(D), and .H(D∞ ) is the space of entire functions .H(C). Given .(rn )n , an increasing sequence of positive numbers such that .0 < rn  R, the sequence of seminorms pn (f ) = sup |f (z)|

.

|z|≤rn

for .n ∈ N, defines a fundamental sequence of seminorms for the compact-open topology on .H(DR ). Theorem 4.8 Let .0 < R ≤ ∞ and .0 < rn  R. Define .an (k) = rnk for each .n ∈ N and .k ∈ N0 and the Köthe matrix .A = (an )n . Then the spaces .H(DR ) and .λ1 (A) are isomorphic. Proof We define the operator .

T :

H(D ) −→ λ1 (A) ∞ R k , k=0 ck z −→ (ck )k∈N0

f (k) (0) where .ck = . That is, for a function f , we define Tf as the sequence of the k! coefficients of the Taylor series expansion around 0. We have to see that T defines an isomorphism. We divide the proof into several steps. First of all, we have to see that the operator is well defined. Given .f ∈ H(DR ) k with Taylor expansion . ∞ k=0 ck z , we have, by the Cauchy Integral Formula, 1 .ck = 2π i

 C(0,rn+1 )

f (w) dw w k+1

for each .k ≥ 0. This gives |ck | ≤

.

1 k rn+1

max |f (w)| ≤

|w|=rn+1

1 k rn+1

pn+1 (f ) .

Therefore, qn ((ck )k ) =

∞ 

.

k=0

an (k)|ck | ≤ pn+1 (f )

∞  rn k ≤ cpn+1 (f ) . rn+1 k=0

(4.3)

4.1 Space of Holomorphic Functions

143

This shows that .(ck )k ∈ λ1 (A), and then T is well defined. Let us note that (4.3) also shows that T is continuous. If .Tf = 0, then .f (k) (0) = 0 for every .k ∈ N and .f = 0 in .D(0, R). This gives .ker T = {0}, and therefore, T is injective. We see now that it is also surjective and that .T −1 is continuous. To do that, we take some .(ck )k ∈ λ1 (A) and define f (z) =

∞ 

.

ck zk .

k=0

Now, given .z ∈ DR , we take n so that .|z| ≤ rn , and we have ∞  .

|ck zk | ≤

k=0

∞ 

|ck |rnk = qn ((ck )k ) .

k=0

Then the series converges absolutely on .DR and then uniformly on the compact sets. By Weierstraß theorem, .f ∈ H(DR ) and clearly .Tf = (ck )k . Hence, T is surjective. Finally, for each n, we have pn (T

.

−1

(ck )k ) = pn

∞ 

ck z

k



≤

k=0

∞ 

|ck |rnk = qn ((ck )k )

k=0

and .T −1 is continuous:



Corollary 4.1 1. .H(D) ∼ = λ1 (B), with .B = (bn )n defined by .bn (k) = (1 − n1 )k for .k = 0, 1, 2, . . .. 2. .H(C) ∼ = λ1 (C), with .C = (cn )n defined by .cn (k) = nk for .k = 0, 1, 2, . . .. Remark 4.2 Let us make a short comment on the seminorms defining the topology of the space .H(C) of entire functions. First of all, since .{z ∈ C : |z| ≤ n} is a fundamental sequence of compact sets in .C, the .τco topology is generated by the sequence of seminorms defined as pn (f ) = sup |f (z)|

.

|z|≤n

for .f ∈ H(C). But  Corollary 4.1 provides us with an alternative family of n seminorms: if .f (z) = ∞ n=0 an z , then the family f k =

∞ 

.

n=0

|an |k n ,

144

4 Holomorphic and Differentiable Functions and Operators

for .k ∈ N, also defines the .τco topology. We can give even a third family defining the topology. Letting |f |k = sup |an |nk ,

.

n∈N0

for .k ∈ N, it is straightforward to check that .|f |k ≤ f k ≤ .f ∈ H(C), so that the family .(| · |k )k defines the same topology.

π2 6 |f |2k

for every

4.1.3 Montel Theorem In Theorem 2.28, we gave a clear picture of how the compact sets of continuous functions are. We see now what do compact sets of holomorphic functions look like. Lemma 4.1 Every bounded subset H of .(H(), τco ) is equicontinuous. Proof For each compact .K ⊆ , there is .λK > 0 (see Remark 2.16) such that .

sup sup |f (z)| ≤ λK .

f ∈H z∈K

We fix now .a ∈ , and we want to see that H is equicontinuous at a. We fix .r > 0 and .δ > 0 so that .K = D(a, r) ⊆ L = D(a, r + δ) ⊆ . Note that, for each .z ∈ K, the circle with centre z and radius .δ is contained in L. Then for each .f ∈ H , we have     1  λL f (ω)  ≤ , .|f (z)| =  dω  2π i  2 δ C(z,δ) (ω − z) and .|f (z) − f (a)| ≤ |z − a| supw∈K |f  (w)| ≤ λδL |z − a|. Then, given .ε > 0 and z such that .|z − a| < λεδL , we have .|f (z) − f (a)| ≤ ε. This shows that H is equicontinuous at a.

As an immediate consequence of this lemma and Theorem 2.28, we have a description of the relatively compact sets of holomorphic functions. Theorem 4.9 (Montel) Every bounded subset H of .(H(), τco ) is relatively compact. In particular, if .(fn )n is .τco -bounded in .H(), there are a subsequence .(fnj )j and .f ∈ H() such that .fnj −→ f for .τco . We finish this section with an extension of this result. We need a short topological lemma.

4.1 Space of Holomorphic Functions

145

Lemma 4.2 Let .(K, τ ) be a Hausdorff compact topological space. If .σ is a Hausdorff topology on K and .σ ⊆ τ , then .τ = σ . Proof We just have to see that every .τ -closed set is .σ -closed. Take then F .τ closed. Since .(K, τ ) is compact, F is .τ -compact. Now, since every .σ -open set is also .τ -open, every covering by .σ -open sets admits a finite subcovering and F is .σ -compact. Being .σ Hausdorff, this implies that F is .σ -closed.

Theorem 4.10 (Vitali) Let . ⊂ C be an open connected set. Set .(fn )n ⊆ H(), τco -bounded. If there exists .A ⊂  with an accumulation point such that for each .x ∈ A the pointwise limit .limn→∞ fn (x) exists, then there is .f ∈ H() such that .fn −→ f in .τco . .

Proof The set .H = {fn : n ∈ N}

τco

is .τco -bounded. Then, by Theorem 4.9, there j →∞

are .f ∈ H() and .(fnj )j such that .fnj −→ f in (.H(), τco ). What we want to do now is to see that, in fact, .fn −→ f in .τco . To do that we consider on .H() the topology of pointwise convergence on A, that we denote by .τA . This is defined by the family of seminorms .pF (f ) = maxz∈F |f (z)| for .F ⊆ A finite. Let us note that a sequence .(gn )n converges to 0 in .τA if and only if .gn (x) → 0 for every .x ∈ A. We clearly have that .τA ⊆ τco . Let us see that it is also Hausdorff. If .g ∈ H() is such that .g(z) = 0 for every .z ∈ A, then, by the identity principle of holomorphic functions (Theorem 4.7), the fact that A has an accumulation point gives .g(z) = 0 for every .z ∈ . Proposition 2.2 shows then that the topology .τA is Hausdorff. Therefore, by Lemma  4.2,.τA |H = τco |H . Now, if .z ∈ A, we have on the one hand that . fn (z) n is convergent and on the τA

other hand that .limj fkj (x) = f (x). Hence, .limn fn (z) = f (z) and .fn −→ f . But,

since both topologies coincide on H , this implies that .fn converges to f on .τco . Remark 4.3 Montel’s theorem is a consequence of Vitali’s theorem. Let . ⊂ C be open and connected, and let .H ⊆ (H(), τco ) be bounded. Fix .(fn )n ⊆ H . Select a sequence .(zj )j ⊆ , .zj = zk if .j = k and .zj −→ z0 ∈ . Set .A = (zj )j . By a diagonal procedure, we can select a subsequence .(fnk )k such that .(fnk (zj ))k is convergent for each .j = 1, .. By Vitali’s theorem, there is .f ∈ H() such that .fnk −→ f in .(H(), τco ). This implies that H is relatively compact in .(H(), τco ).

4.1.4 Dual of the Space of Entire Functions Our aim now is to give a representation of the dual of .H(C). Let us note in the first place that Corollary 4.1 and Proposition 2.10 already give a representation in terms of a space of sequences. But we are interested in a representation in terms of holomorphic functions. In fact, we give two such representations: in terms of germs of holomorphic functions and in terms of holomorphic functions of exponential type by means of the Fourier–Borel transform.

146

4 Holomorphic and Differentiable Functions and Operators

Germs of holomorphic functions Let us consider the class of complex-valued functions f that are holomorphic at some open disc .D(0, ε) for .ε > 0 and, on it, define a relation .f ∼ g if there is some .ε > 0 so that .f (z) = g(z) for every .z ∈ D(0, ε). This clearly defines an equivalence relation, and we define .H({0}) to be the space of equivalence classes, which we call germs at 0. We identify each class of equivalence (germ) with any of its representatives. If .f ∈ H({0}), then it is holomorphic on some disc and, then, ∞ n for every .|z| < ε. Now, for .g ∼ f , we have that .f (z) = g(z) .f (z) = λ z n n=0  n in some .D(0, δ) and, hence, .g(z) = ∞ n=0 λn z for every .|z| < δ. In other words, the sequence of coefficients .(λ of the choice of representative of n )n is independent n . This power series converges on some disc the germ, and we write .f = ∞ λ z n n=0 .D(0, ε), and therefore, sup |λn |δ n < ∞ ,

.

(4.4)

n∈N0

 n for .0 < δ < ε. Note also that if the power series . ∞ n=0 λn z converges at some point, then it automatically defines a germ in .H({0}).  n n Theorem 4.11 If .L ∈ H(C) , then . ∞ n=0 L(z )z defines a germ  in .H({0}).  n n Moreover, the mapping .T : H(C) → H({0}) defined by .T (L) = ∞ n=0 L(z )z is a linear bijection. Proof Given any .L ∈ H(C) , there are .k ∈ N and .C > 0 so that (recall the seminorms defined in Remark 4.2) |L(f ) ≤ Cf k = C

∞ 

.

|an |k n ,

n=0

for every .f (z) =

∞

n=0 an z

n

∈ H(C). Taking .f = zn , we have

1

≤C kn  n n for every .n ∈ N0 . So, the power series . ∞ n=0 L(z )z converges (absolutely) on  1 .D 0, 2k and, then, defines a germ in .H({0}). This also shows that the mapping T is well defined. It is clearly linear, so it is only left to see that it is injective and surjective. For the injectivity, suppose that .T (L) = 0 for some .L ∈ H(C) . This means that .L(zn ) = 0 for every .n ∈ N0 , and since the polynomials are dense in .H(C), we have that .L(f ) = 0 for every entire function f . Thus .L = 0 and T is injective. |L(zn )|

.

4.1 Space of Holomorphic Functions

147

In order to see that T is surjective, pick any .g ∈ H({0}), so that .g(z) = ∞ n n=0 λn z for every .|z| < ε, and define .Lg : H(C) → C by Lg (f ) =

∞ 

.

an λn ,

n=0

 1 n for .f (z) = ∞ n=0 an z ∈ H(C). Choose some .k ∈ N such that . k < ε and denote (recall (4.4)) .

sup |λn | n∈N0

1 = S < ∞. kn

Then |Lg (f )| ≤

∞ 

.

|an λn | ≤ S

n=0

∞ 

|an |k n = Sf k .

n=0

This shows that .Lg is well defined and continuous. Since it is clearly linear, we get .Lg ∈ H(C) . Finally, since one easily has .Lg (zn ) = λn , we immediately get .T (Lg ) = g, which shows that T is surjective and completes the proof.

Remark 4.4 We can also give an integral of the duality between representation n that is holomorphic at some H(C) and .H({0}). Given a germ .g(z) = ∞ λ z n=0 n 1 .D(0, ε), we define for .|ζ | > , ε .

∞

G(ζ ) =

.

1 1 1  g ζ = λn n+1 , ζ ζ n=0

which is a holomorphic function on .C \ D(0, 1/ε). Now we fix some .0 < δ < ε and define (denoting by .γδ the circle centred at 0 and with radius .1/δ) .LG : H(C) → C by LG (f ) =

.

1 2π i

 f (z)G(z)dz . γδ

Note that |LG (f )| ≤

.

1 sup |g(ζ )| sup |f (z)| . δ |ζ |=δ |z|≤1/δ

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4 Holomorphic and Differentiable Functions and Operators

This shows that .LG is well defined Since it is clearly linear, we get  and continuous. n , then for each .n = 0, 1, 2, . . ., we have LG ∈ H(C) . Finally, if .g(z) = ∞ λ z n n=0

.

LG (zn ) =

.

1 2π i

 zn G(z)dz = γδ

 ∞ zn 1  λk dz = λn . k+1 2π i γδ z k=0

Accordingly, .LG coincides with the functional .Lg defined in the proof of Theorem 4.11. Let us finish by noting that the representation given in Theorem 4.11 is a particular case of the Köthe–da Silva Dias theorem on the duality of .H() for an open set . ⊆ C. The interested reader is referred to [6, Chapter Six]. Functions of exponential type An entire function f is of exponential type if there exist .k ∈ N and .C > 0 so that |f (z)| ≤ Cek|z|

.

for every .z ∈ C. We write .Exp(C) for the space of all .f ∈ H(C) of exponential type. Our aim now is to show that .H(C) can be identified with .Exp(C). Functionals in  .H(C) are now denoted by .μ and, if necessary, we will use a subscript to emphasize the variable in the function on which .μ is acting. To be more precise, if f is a function that depends on several variables, we will write .μw (f ) = μ([w  f (w)]). We define the Fourier–Borel transform of .μ ∈ H(C) as the function . μ:C→C given by  μ(z) = μw (ezw ) .

.

Theorem 4.12 The Fourier–Borel transform .F : H(C) → Exp(C) given by .F(μ) =  μ is a linear bijection. Proof Let us check in first place that it is well defined, and we begin by seeing that  μ is entire for every .μ ∈ H(C) . Take one such .μ and note that, by (2.22) (recall also Remark 4.2), we can find .k ∈ N and .C > 0 so that

.

|μ(f )| ≤ C sup |f (z)| ,

.

|z|≤k

(4.5)

for every .f ∈ H(C). Applying this to the monomial .f (w) = w n , we get |μw (w n )| ≤ Ck n ,

.

(4.6)

 λn n for every .n ∈ N0 . Note now that for each .λ ∈ C, the series . ∞ n=0 n! ζ converges in λζ . .H(C) (that is, absolutely and uniformly on compact sets, as a function on .ζ ) to .e

4.1 Space of Holomorphic Functions

149

Using this and the fact that from (4.6), we have .

kn |μw (w n )| |z|n ≤ C |zn |, n! n!

we obtain that the series .

∞  μw (w n ) n=0

n!

zn

(4.7)

converges in .H(C) to some entire function g. On the other hand, we have  μ(z) = μw (ewz ) = μw

.



∞ zn n n=0 n! w



=

∞  μw (w n ) n=0

n!

zn = g(z) ,

for each .z ∈ C. That is, . μ = g, and it is an entire function (note also that its Taylor expansion at 0 is given by (4.7)). Let us see now that it is of exponential type. This is a straightforward consequence of (4.5), from which we have | μ(z)| = |μw (ezw )| ≤ C sup |ezw | ≤ Cek|z| ,

.

|z|≤k

for every .z ∈ C. This finally shows that .μ ∈ Exp(C), and therefore, .F is well defined. It is clearly linear. If . μ = 0, all Taylor coefficients are 0 and, from (4.7), .μw (w n ) = 0 for every n. Since the polynomials are dense in .H(C), this implies that .μ(f ) = 0 for every .f ∈ H(C) and shows that .F is injective. It is only left to complete the proof to see that the .F is onto. To this end, fix some ∞ n .g ∈ Exp(C) with Taylor expansion . n=0 gn z and fix .k ∈ N and .C > 0 so that −k|z| .|g(z)|e ≤ C. With this and the Cauchy inequalities, we have |gn | ≤

.

1 Cekr sup |g(z)| ≤ , r n |z|=r rn

for every .r > 0 and .n ∈ N. In particular, |gn | ≤

.

Cekn nn

k for every n. Fix now .m ∈ N such ∞that .mn ≥ 2e , and observe that, if .f ∈ H(C) has Taylor expansion .f (z) = n=0 fn z , then as a consequence of the Cauchy inequalities, we have

|fn | ≤

.

1 sup |f (z)| . mn |z|=m

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4 Holomorphic and Differentiable Functions and Operators

Thus ∞  .

|fn |n!|gn | ≤ C

n=0



sup |f (z)|

|z|=m

∞ ∞ kn

  1

e n! sup |f (z)| , ≤ 2C n n m n 2n |z|≤m n=0

n=0

 and the mapping .μg : H(C) → C defined by .μg (f ) = ∞ n=0 n!fn gn is well defined and, since it is obviously linear, also continuous. In other words, .μg ∈ H(C) . Recall n that the Taylor coefficients of .ewz (as a function of w) are . zn! , and this implies .

∞ ∞     zn gn zn = g(z) . μ g (z) = μg w (ewz ) = n! gn = n! n=0

n=0

Hence, .F(μg ) = g, and this completes the proof.



4.2 Spaces of Differentiable Functions We move on and consider (real) differentiable functions of several (real) variables. We are going to consider different spaces of such functions. We start by fixing some basic notation. Given .d ∈ N, a multi-index is a vector .α = (α1 , . . . , αd ) ∈ Nd0 , and its order is defined to be .|α| = |α1 | + · · · + |αd |. With this, we write the partial derivatives of a function f defined on some open . ⊆ Rd as ∂αf =

.

∂ |α| f . ∂ α1 x1 · · · ∂ αd xd

For each .k ∈ N, we define .C k () as the space of all functions .f :  → K such that all partial derivatives up to order k exist and are continuous on .; that is,

 C k ()= f : → K : ∂ α f exist and are continuous functions on  for all |α|≤k .

.

We choose now a fundamental sequence of compact sets .(Kn )∞ n=1 of . and define the seminorms pn (f ) = sup sup |∂ α f (x)| .

.

|α|≤k x∈Kn

The space .C 0 () is just the space of continuous functions. By the definition, we clearly have C k+1 () ⊆ C k () ⊆ · · · ⊆ C 1 () ⊆ C()

.

4.2 Spaces of Differentiable Functions

151

for every k. We define now the space C ∞ () =



.

C k () ,

k∈N

and we endow it with the locally convex topology given by the system of seminorms pn (f ) = sup sup |∂ α f (x)| , for n ∈ N .

.

|α|≤n x∈Kn

(4.8)

Proposition 4.3 A sequence .(fk )k converges to f in .τ(pn ) if and only if ∂ α fk → ∂ α f in .τco for every .α.

.

k

Proof Suppose that .fk → f in .τ(pn ) , then .pn (fk − f ) → 0 for every n. Fix now some .α and some compact .K  . There is some .n0 so that .|α| ≤ n0 and .K ⊆ Kn0 . Hence .

sup |∂ α fk (x) − ∂ α f (x)| ≤ sup

sup |∂ β fk (x) − ∂ β f (x)| ≤ pn0 (fk − f ) → 0 .

|β|≤n0 x∈Kn0

x∈K

This shows that .∂ α fk → ∂ α f uniformly on compact sets. Suppose now that .∂ α fk converges to .∂ α f in .τco for every .α. Fix .n ∈ N0 and .ε > 0. Since .Kn is compact, for each .α with .|α| ≤ n, we can find .kα so that .

sup |∂ α fk (x) − ∂ α f (x)| < ε x∈Kn

for every .k ≥ kα . But the set .{α ∈ Nd0 : |α| ≤ n} is finite, and we may take .k0 the maximum of all .kα s. Then, for .k ≥ k0 , .

sup sup |∂ α fk (x) − ∂ α f (x)| < ε .

|α|≤n x∈Kn

Theorem 4.13

∞ .(C (), (pn )n )

is a Fréchet space.

Proof From Theorem 2.13, it follows immediately that the space is metrizable. In order to see that it is also complete, we choose a Cauchy sequence .(fk )k . Proceeding as in Proposition 4.3, for each fixed .α, the sequence .(∂ α fk )k is .τco -Cauchy. Then, by Theorem 2.18, there is .f ∈ C() so that .fk → f uniformly on compact sets, and for each .α, there is .fα ∈ C() so that .∂ α fk → fα uniformly on compact sets. What we have to see now is that .f ∈ C ∞ () and .∂ α f = fα for every .α. Take .α = (1, 0, . . . , 0), and let us see that .∂ α f exists and .∂ α f (a) = fα (a) for every .a ∈ . Fix, then, .a = (a1 , . . . , ad ) ∈  and choose .r > 0 such that K = {(a1 + t, a2 , . . . , ad ) : t ∈ [−r, r]} ⊆  .

.

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4 Holomorphic and Differentiable Functions and Operators

Now we define .g : [−r, r] → K by .g(t) = f (a1 + t, a2 , . . . , ad ) and, in an obvious way (replacing f by .fk or .fα ), .gk and .gα for .k ∈ N. Note that all these functions are continuous and that every .gk is differentiable on .] − r, r[ (and .gk = ∂ α fk ). Then, by the Fundamental Theorem of Calculus,  gk (t) − gk (−r) =

t

.

−r

gk (s)ds ,

for .t ∈ [−r, r]. On the other hand, the set K is compact, and the fact that .(∂ α fk )k converges to .fα uniformly on compact sets gives that .gk → gα uniformly on .[−r, r]. Therefore,  .

lim k

t −r

gk (s)ds

 =

t

lim gk (s)ds −r k

 =

t

−r

gα (s)ds .

This altogether gives  g(t) − g(−r) =

t

.

−r

gα (s)ds

and this implies that g is differentiable and .g  = gα for every .t ∈ [−r, r]. In particular, g is differentiable at 0, and .g  (0) = gα (0). This gives that .∂ α f (a) exists and .∂ α f (a) = fα (a). Iterating this procedure a finite number of times, differentiating each time once with respect to one variable, we finally get the conclusion.

With a similar proof, we also have the following. Proposition 4.4 .(C k (), (pn )n ) is a Fréchet space. Theorem 4.14 Every bounded subset of .C ∞ () is relatively compact. Proof Let .H ⊆ C ∞ () be bounded. We want to see that every sequence in H has a convergent subsequence. We need to see (recall Proposition 4.3) that the subsequence converges uniformly on compact sets to some function f , and that all the partial derivatives converge (again in .τco ) to the partial derivatives of f . To do so, we are going to use several times Ascoli’s Theorem 2.28. Our first step is to show that, for each fixed multi-index .α ∈ Nd0 , the set {∂ α f : f ∈ H }

.

is equicontinuous at every .x ∈ . Note that, since H is bounded, for each compact K ⊆  and such an .α, there is .μα,K such that

.

|∂ α f (x)| ≤ μα,K

.

(4.9)

4.3 Some Operators on Spaces of Functions

153

for every .x ∈ K and .f ∈ H . Fix now some .α ∈ Nd0 and .x0 ∈ . There is .δ0 > 0 such that .K = B(x0 , δ0 ) ⊆ . Given .y ∈ K, the segment .[x0 , y] is contained in K, and for each .f ∈ H the function .∂ α f is continuous. We may then apply the mean value theorem for several variables to have   |∂ α f (x0 ) − ∂ α f (y)| ≤ sup ∇∂ α f (c) |x0 − y|

.

c∈[x0 ,y]

= sup c∈[x0 ,y]

d   1    ∂i (∂ α f )(c)2 2 |x0 − y| ≤ d max μβ,K |x0 − y| . |β|≤|α|+1

i=1

This gives that .{∂ α f : f ∈ H } is equicontinuous (recall Remark 2.16), given .ε > 0 just take .|x0 − y| < δ < min(δ0 , d maxεμβ,K ). On the other hand, (4.9) gives that the set .{∂ α f (x) : f ∈ H } is bounded in .K for every fixed .x ∈ . With this, Ascoli’s Theorem 2.28 gives that .{∂ α f : f ∈ H } is d .τco -relatively compact in .C() for every .α ∈ N . 0 Take now a sequence .(fn )n ⊆ H . By what we have just shown (using Ascoli’s theorem), there is a subsequence .(n(k, 0))k and a continuous function .f (0) so that (0) in .τ as .k → ∞. Let us now enumerate the set of multi-indices .fn(k,0) → f co d .N = {α : j ∈ N j 0} in such a way that .α0 = (0, 0, . . . , 0). We take now the 0  sequence . ∂ α1 fn(k,0) k , and using again the compactness of the set, we find a subsequence .(n(k, 1))k and a continuous function .f (1) so that .fn(k,1) → f (1) in .τco as .k → ∞. Repeating this procedure, for each .j ∈ N0 , we define by induction a sequence .(n(k, j ))k and a continuous function .f (j ) so that each .(n(k, j ))k is a subsequence of .(n(k, j − 1))k and .∂ αj fn(k,j ) → f (j ) in .τco as .k → ∞. Now, for each .k ∈ N, define .nk = n(k, k). Then .(fnk )k is a subsequence of .(fn )n and ∂ αj f → f (j )

.

in .τco as .k → ∞ for every j . The same argument as in Theorem 4.13 shows that in fact .f = f (0) ∈ C ∞ () and .∂ αj f = f (j ) for every j . Thus .(fnk )k converges to f in .C ∞ (), and this completes the proof.



4.3 Some Operators on Spaces of Functions We give now some examples of continuous, linear operators defined on the spaces of functions that we have considered so far. These will appear several times in the rest of this text.

154

4 Holomorphic and Differentiable Functions and Operators

Example 4.15 (Multiplication operators) Let . be an open set in .Rd and fix a continuous function .g :  → K. With this, we define the associated multiplication operator Mg : C() → C(), by Mg (f ) := f · g .

.

This is clearly well defined and linear. Let us see that it is also continuous (recall that on .C(), we consider the compact-open topology). Given a compact set .K ⊆ , we have (taking .C = supx∈K |g(x)|) pK (Mg (f )) = sup |f (x)g(x)| ≤ sup |g(x)| pK (f ) = C pK (f )

.

x∈K

x∈K

and, by Theorem 2.11, .Mg is continuous. Note that, with the same argument, if . ⊆ C and g is holomorphic, then .Mg : H() → H() is also continuous.

Example 4.16 (Composition operators) We take now an open set . ⊆ Rd and a continuous function .ϕ :  → . We define the composition operator .Cϕ with symbol .ϕ as Cϕ : C() → C(), by Cϕ (f ) = f ◦ ϕ ,

.

that is clearly well defined and linear. In order to see that it is continuous, choose some compact .K ⊆ . Since .ϕ is continuous, .K2 = ϕ(K) is again compact, and we have pK (Cϕ (f )) = sup |(f ◦ ϕ)(z)| = sup |f (ϕ(z))| = sup |f (u)| = pK2 (f ) .

.

z∈K

z∈K

z∈K2

This shows that the operator is continuous. As in the previous example, if . ⊆ C is open and simply connected and .ϕ :  →  is holomorphic, the composition operator .Cϕ : H() → H() is well defined and continuous.

Example 4.17 (The differentiation operator) We consider now the open unit disc of the complex plane and define the differentiation operator D : H(D) → H(D) by D(f ) = f  .

.

This is clearly well defined and linear. We want to see now that it is continuous. To do that, let us briefly recall (see (2.24) and (2.25)) that the topology is generated by the family of seminorms .pn = pKn , where the fundamental sequence of compact sets is given by 

Kn = z ∈ D : |z| ≤ 1 − n1 ,

.

4.3 Some Operators on Spaces of Functions

155

for .n ∈ N. If .z ∈ Kn , then we have, by the Cauchy Integral Formula for the derivatives (Theorem 4.2), f  (z) =

.

1 2π i



Let us note that, for each .|w| = 1 −

1 C(0,1− n+1 )

1 n+1 ,

f (w) dw . (w − z)2

we have

|w − z| ≥ |w| − |z| = rn+1 − |z| ≥ rn+1 − rn =

.

1 , n(n + 1)

and this gives |f  (z)| ≤

.

1 2π

sup 1 |w|=1− n+1

|f (w)| 2π rn+1 |w − z|2 ≤ n3 (n + 1)

sup

|f (ω)| ≤ n3 (n + 1)pn+1 (f ) .

1 |w|=1− n+1

Hence, .pn (D(f )) = pn (f  ) ≤ n3 (n + 1)pn+1 (f ) for every f and D is continuous. A similar argument shows that the differentiation operator .D : H(C) → H(C) is continuous.

Example 4.18 (Integration operator) We now consider . = C. For .z ∈ C, we take .[0, z] = {λz : 0 ≤ λ ≤ 1}, the closed segment from 0 to z, and define the integration operator 

z

J : H(C) → H(C), as (Jf )(z) :=

f (w)dw .

.

0

Once again, it is clear that it is well defined and linear, and it is only left to see that it is continuous. To that purpose, we consider the fundamental sequence of compact sets given by .Kn = {z ∈ C : |z| ≤ n} for .n ∈ N and recall that the sequence of seminorms .pn = pKn defines the compact-open topology. If .z ∈ Kn , then clearly .[0, z] ⊆ Kn , and we have   .|Jf (z)| = 

0

z

  f (w)dw  ≤ |z| max |f (w)| ≤ n max |f (w)| . w∈[0,z]

w∈Kn

This gives .pn (J (f )) ≤ npn (f ) for every f and J is continuous. A similar argument shows that the integration operator .J : H(D) → H(D) is continuous.

Moreover, it is easy to see that .DJf = f and .J Df (z) = f (z) − f (0).

156

4 Holomorphic and Differentiable Functions and Operators

Example 4.19 For .1 ≤ j ≤ d, we write .∂j =

∂ ∂xj

(the partial derivative with α

α

∂ j αj ∂xj

respect to the j th variable) and, for .αj ∈ N0 , we write .∂j j = α = (α1 , . . . , αd ) ∈ Nd0 , we write .∂ α = ∂1α1 · · · ∂dαd =

.

α

∂α

α

∂x1 1 ···∂xd d

. Finally, given

.

If f is a function of d variables, at least m times differentiable, and .aα ∈ C for α ∈ Nd0 with .|α| ≤ m (with at least one .aα = 0 if .|α| = m, we write

.

P (∂)f =



aα ∂ α f .

.

|α|≤m

This defines an operator, called linear partial differential operator of order m with constant coefficients, P (∂) : C ∞ () → C ∞ ()

.

(here . ⊆ Rd is some open set). This is clearly well defined and linear. If .(Kn )n is a fundamental system of compact subsets of . and .pn = pKn , then      β    α  .pn (P (∂)f ) = sup sup |∂ aα ∂ f (x)  P (∂)f (x) | = sup sup ∂ β

|β|≤n x∈Kn

|β|≤n x∈Kn

     α+β  = sup  aα ∂ f (x) ≤ x∈Kn |α|≤m |β|≤n

≤

sup

|∂ f (x)| γ

x∈Kn+m |γ |≤n+m

|∂ f (x)| γ

sup x∈Kn |γ |≤n+m

 

|α|≤m



|aα | =

|α|≤m

 



|aα |

|α|≤m

 

 |aα | · pn+m (f ) ,

|α|≤m

and the operator is continuous. Let us note that if the operator is of order m, we may consider it as .P (∂) : C k+m () → C k () for each k and, with the same argument, it is continuous. Each linear partial differential operator with constant coefficients has an associated polynomial of N variables, called the symbol of the operator, given by P (z) =



.

aα zα ,

|α|≤m

where we denote .zα = z1α1 · · · zdαd for .z = (z1 , . . . , zd ) ∈ Cd . The classical theory of partial differential equations is full of such operators. We recall now just a few of the most classical ones.

4.4 Exercises

157

The Laplace operator P (∂) =  =

.

∂2 ∂2 + · · · + with symbol P (z) = z12 + · · · + zd2 . ∂x12 ∂xd2

The heat operator P (∂)=

.

∂  ∂2 ∂2  ∂ −x = − 2 +· · ·+ 2 with symbol P (z, t) = t −(z12 +· · ·+zd2 ) . ∂t ∂t ∂x1 ∂xd

The wave operator P (∂)=

.

∂2  ∂2 ∂2  ∂2 −x = 2 − 2 +· · ·+ 2 with symbol P (z, t)=t 2 −(z12 +· · ·+zd2 ) . 2 ∂t ∂t ∂x1 ∂xd



4.4 Exercises 4.1 Let  ⊆ C be an open, connected set and A ⊆ H() bounded. Show: 1. The set A = {f  : f ∈ A} is bounded in H(). 2. If A is bounded in H() and there is z0 ∈  such that {f (z0 ) : f ∈ A} is bounded in C, then A is bounded in H(). 4.2 Let  ⊆ C be an open and connected set. Let (fn )n be a sequence in H() of injective functions such that fn → f in (H(), τco ). Show that f is either injective or constant. 4.3 Let  ⊆ C be an open and connected set. Let (fn )n ⊆ H() be a sequence such that fn → f in (H(), τco ) and f is not identically zero. Suppose that B(a, r) ⊆  for some a ∈  and r > 0. If f (z) = 0 for every z ∈ C such that |z − a| = r, then there exists n0 ∈ N such that for any n ≥ n0 , f and fn have the same zeros in B(a, r). Hint: Recall Rouche’s Theorem. 4.4 For each f ∈ H(D), denote by an (f ) (for n = 0, 1, 2, . . .) the nth Taylor coefficient. Show that A ⊆ H(D) is bounded if and only if the following two conditions hold: 1. sup{|an (f )| : f ∈ A} is bounded in C for every n.  1/n 2. lim supn sup{|an (f )| : f ∈ A} ≤ 1. 4.5 Obtain a characterization of the bounded subsets of H(C), similar to the one in Exercise 4.4.

158

4 Holomorphic and Differentiable Functions and Operators

4.6 For each 0 < r < 1, let γr : [0, 1] → C be given by γr (t) = re2π it . Show that (fn )n ⊆ H(D) converges to f ∈ H(D) if and only if  |f (z) − fn (z)|dz −→ 0 as n → ∞

.

γr

for every r ∈]0, 1[. 4.7 Let f ∈ H(D) be non-constant and such that supz∈D |f (z)| ≤ 1. Prove that the function g : D → C defined by g(z) =

∞ 

.

f (z)n

n=0

is holomorphic on D. Is g(z) bounded?

 4.8 Let A ⊆ H() be a bounded set such that f ∈A f () is contained in an open subset D ⊆ C. Let g ∈ H(D) be bounded on bounded sets. Show that {g ◦ f : f ∈ A} is bounded in H(D). 4.9 Show that C 1 (R) contains bounded sets that are not relatively compact. 4.10 Let B() be the subset of C ∞ () consisting of functions f so that all the partial derivatives of f are bounded in , endowed with the locally convex topology defined by the seminorms   pn (f ) = sup sup ∂ α f (x) ,

.

|α|≤n x∈

for n = 0, 1, 2, . . .. Show that B() is a Fréchet space. Further Reading [1–5, 7–9]

References 1. Ahlfors, L.: Complex Analysis—An Introduction to the Theory of Analytic Functions of One Complex Variable, 3rd edn. AMS Chelsea Publishing, Providence (2021). Reprint of the 1978 original 2. Apostol, T.M.: Mathematical Analysis, 2nd edn. Addison-Wesley Publishing Co., Reading/London/Don Mills (1974) 3. Conway, J.B.: Functions of One Complex Variable. Graduate Texts in Mathematics, vol. 11, 2nd edn. Springer, New York/Berlin (1978) 4. Gamelin, T.W.: Complex Analysis. Undergraduate Texts in Mathematics. Springer, New York (2001). https://doi.org/10.1007/978-0-387-21607-2 5. Jameson, G.J.O.: A First Course on Complex Functions. Chapman and Hall, Ltd., London (1970)

References

159

6. Köthe, G.: Topological Vector Spaces. I. Translated from the German by D.J.H. Garling. Die Grundlehren der mathematischen Wissenschaften, Band 159. Springer, New York (1969) 7. Rudin, W.: Principles of Mathematical Analysis. International Series in Pure and Applied Mathematics, 3rd edn. McGraw-Hill Book Co., New York/Auckland/Düsseldorf (1976) 8. Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Book Co., New York (1987) 9. Shapiro, J.H.: Composition Operators and Classical Function Theory. Universitext: Tracts in Mathematics. Springer, New York (1993). https://doi.org/10.1007/978-1-4612-0887-7

Chapter 5

Transitive and Mean Ergodic Operators

If E is a complex locally convex (Hausdorff) space and .T : E → E is an operator (in order to make reading as easy as possible, all along this chapter, we assume without further notice that all operators are linear and continuous), we may compose it with itself. Our aim in this chapter is to analyse what happens when we iterate this process, that is, if we compose T with itself n times and let n tend to .∞. We begin by fixing some notation. First of all, we write T 0 =I (the identity on E)

.

T 1 =T T 2 =T ◦ T T 3 =T ◦ T ◦ T .. . Tn =T · · ◦ T = T n−1 ◦ T for n ∈ N .  ◦ · n times

Also, given some .x ∈ E, we define its orbit with respect to T as Orb(x, T ) = {x, T x, T 2 x, T 3 x, . . .} = {T n x : n = 0, 1, 2, . . .} .

(5.1)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_5

161

.

162

5 Transitive and Mean Ergodic Operators

5.1 Transitive Operators An operator .T ∈ L(E) is called transitive if for every pair of non-empty open sets U, V ⊆ E there is some .n ∈ N so that

.

T n (U ) ∩ V = ∅ .

.

Let us start by seeing some immediate consequences of this definition. Proposition 5.1 Let .T ∈ L(E) be a transitive operator. Then the following statements hold: 1. .T (E) is dense in E. 2. .T n (E) is dense in E for each .n ∈ N. 3. Given .U, V ⊆ E non-empty and open, for every .m ∈ N, there is some .n > m so that .T n (U ) ∩ V = ∅. 4. Let .U, V be neighbourhoods of 0 in E so that .V = E. Then for every .x ∈ E, there are .n ∈ N and .y ∈ E satisfying .x − y ∈ U and .T n x − T n y ∈ V (i.e., T has sensitive dependence on the initial conditions). Proof Fix .U = E and take .∅ = V ⊆ E any open set. By the transitivity of T , we can find n so that .∅ = T n (E) ∩ V ⊆ T (E) ∩ V , and .T (E) is dense. To check that 2 holds, let us note that if .S : E → F is any operator with dense range and .A ⊆ E is dense, then .S(E) = S(A) ⊆ S(A) and therefore .F = S(E) ⊆ S(A). With this, an argument with induction from 1 yields our second claim. Fix now m and some open and non-empty .U, V ⊆ E. By 2, .T m (E) is dense,  −1 and then .T m (E) ∩ V = ∅, and . T m (V ) ⊆ E is open and non-empty. Being  −1 T transitive, we can find .k ∈ N and .x ∈ U so that .T k x ∈ T m (V ). Hence, m+k x ∈ V and 3 holds. .T We finish by seeing that 4 holds. Take U and V as in the statement, pick .U ⊆ U , an absolutely convex, open neighbourhood of 0, and consider .G = E \ V (which is open and non-empty). By the transitivity of T , there is some n so that .T n (U )∩G = ∅. Given .x ∈ E, take .u ∈ U and define .y = x −u. Then .T n x −T n y = T n (u) ∈ G. This yields the claim and completes the proof. 

Our aim now is to find conditions that ensure that a given operator is transitive. Lemma 5.1 Let .T ∈ L(E) be such that there exist .E0 , F0 ⊆ E dense so that for every .x ∈ E0 and .y ∈ F0 , and for every absolutely convex, open 0-neighbourhood U , there are .m ∈ N and .u ∈ U with .T m x ∈ U and .T m u − y ∈ U . Then T is transitive. Proof Let .G1 , G2 be any non-empty, open subsets of E. By density, we can find x ∈ E0 and .y ∈ F0 , and some U , absolutely convex, open 0-neighbourhood so that

.

x + U ⊆ G1 and y + 2U ⊆ G2 .

.

5.1 Transitive Operators

163

Now, by hypothesis, there are m and .u ∈ U so that T m u ∈ y + U and T m x ∈ U .

.

This altogether gives .x + u ∈ x + U ⊆ G1 and .T m (x + u) ∈ y + 2U ⊆ G2 . This implies that .T m (G1 ) ∩ G2 = ∅, and T is transitive. 

Proposition 5.2 Let .T ∈ L(E) be such that there exist .E0 , F0 ⊆ E dense so that for every .x ∈ E0 and .y ∈ F0 , and for every absolutely convex, open 0-neighbourhood U , there are .m ∈ N and .u ∈ U with .T m (x + u) ∈ y + U . Then T is transitive. Proof Let .G1 , G2 be any non-empty, open subsets of E. By density, we can find x ∈ E0 and .y ∈ F0 , and some U , absolutely convex, open 0-neighbourhood so that

.

x + U ⊆ G1 and y + U ⊆ G2 .

.

By hypothesis, there are m and .u ∈ U so that .T m (x + u) ∈ y + U . This clearly gives the conclusion. 

Theorem 5.1 (Bès–Peris) Let .T ∈ L(E). Suppose that there exist .E0 , F0 ⊆ E dense, .(nk )k ⊆ N strictly increasing, and operators .Snk : F0 → E so that T nk x −→ 0 for every x ∈ E0.

(5.2)

Snk y −→ 0 for every y ∈ F0.

(5.3)

T nk Snk y −→ y for every y ∈ F0 .

(5.4)

.

k

k

k

Then T is transitive. Proof Take any .x ∈ E0 , .y ∈ F0 , and any absolutely convex open 0-neighbourhood U . Using (5.2), (5.3), and (5.4), we can find some .k0 ∈ N such that T nk x ∈

.

1 1 U, Snk y ∈ U, and y − T nk Snk y ∈ U , 2 2

for every .k ≥ k0 . Let now .m = nk0 and .u = Sm y ∈ U . We can now finish our argument in two ways. On the one hand, we have T mx ∈

.

1 1 U ⊆ U, and T m u − y ∈ U ⊆ U , 2 2

164

5 Transitive and Mean Ergodic Operators

and the conclusion follows from Lemma 5.1. On the other hand, we have 1 1 T m (x + u) ∈ y + U + U ⊆ y + U , 2 2

.

and Proposition 5.2 yields the conclusion.



As a straightforward consequence, we have the following reformulation. Corollary 5.1 Suppose that .T ∈ L(E) satisfies any of the following two conditions: (1) (Godefroy–Shapiro) There exist .E0 , F0 ⊆ E dense, .(nk )k ⊆ N strictly increasing and operators .Snk : F0 → E so that T nk ◦ Snk = IF0 for every k ∈ N

.

T nk x −−−→ 0 for every x ∈ E0 k

Snk y −→ 0 for every y ∈ F0 . k

(2) (Kitai) There exist .E0 , F0 ⊆ E dense, an operator .S : F0 → E with .T ◦S = IF0 and .(nk )k ⊆ N strictly increasing so that T nk x −−−→ 0 for every x ∈ E0

.

k

S y −→ 0 for every y ∈ F0 . nk

k

Then T is transitive. A notion that is closely related with transitivity is that of hypercyclicity. An operator .T : E → E is said to be hypercyclic if there is a vector .x ∈ E (called hypercyclic vector) such that .Orb(x, T ) (recall the definition in (5.1)) is dense in E. Observe that if E is a locally convex Hausdorff space on which there is a hypercyclic operator, then necessarily E has to be separable. Remark 5.1 Suppose that .T ∈ L(E) is hypercyclic, with hypercyclic vector x, and fix two non-empty open sets .U, V ⊆ E. Since the orbit of x is dense in E, we can find some m so that .T m x ∈ U . The set .{T m+1 x, T m+2 x, . . .} is again dense in E, and we can find k such that .T m+k x ∈ V . Therefore, .T k (U ) ∩ V = ∅, and this shows that T is transitive. In fact, if the space is Fréchet (and separable), both concepts are equivalent. Theorem 5.2 (Birkhoff) Let E be a separable Fréchet space. Then .T ∈ L(E) is hypercyclic if and only if it is transitive. Proof Let us suppose that T is transitive and see that it is also hypercyclic (in view of Remark 5.1, this yields the result). Since E is separable, we can choose a dense

5.1 Transitive Operators

165

countable set .(xj )j ∈N in E and consider the family .{B(xj , 1/m) : j, m ∈ N} that, being countable, we can write as .{Vn : n ∈ N}. Then, for every open set .G ⊆ E, there is some n with .Vn ⊆ G. Now, for each .n ∈ N, we define the set Gn = {x ∈ E : T m x ∈ Vn for some m} .

.

On the one hand, .T m (E) is dense in E for every m (because T is transitive, see Proposition 5.1) and then .T m (E) ∩ Vn = ∅ for every n. This shows that, for each n, the set .Gn is not empty. Now, .Vn is open, and from this, we deduce that for each m .x ∈ Gn , we can find m and U an open neighbourhood of x such that .T U ⊆ Vn . Hence, .U ⊆ Gn and .Gn is open. Our next step is to see that, moreover, .Gn is dense in E. Let .V ⊆ E be open and non-empty and choose (by the transitivity of T ) some m with .T m V ∩ Vn = ∅. Then there is some .y ∈ V with .T m y ∈ Vn (that is, .y ∈ Gn ). Hence, .Gn ∩ V = ∅, and since the open set V was arbitrary, this shows that .Gn is dense in E for every n. By  Baire’s Theorem 3.21, .G = n Gn is dense in E. We are going to finish the proof by seeing that every vector in G is hypercyclic. Take .z ∈ G and .U ⊆ E open, and find k so that .Vk ⊆ U . Then (since .z ∈ Gk ) there is some m such that .T m z ∈ Vk ⊆ U , and this shows that .Orb(z, T ) ∩ U = ∅. Hence, .Orb(z, T ) is dense, and this completes the proof. 

Let us note that in the proof we have seen that a hypercyclic operator on a separable Fréchet space not only has a hypercyclic vector, but the set of such vectors is dense and, moreover, is the intersection of a countable family of open sets. We can give a sort of version of Corollary 5.1 for hypercyclic operators. Proposition 5.3 (Godefroy–Shapiro) Let E be a complex, separable locally convex Hausdorff space and .T ∈ L(E). Suppose that the spaces E0 = span{x ∈ E : T x = λx for some λ ∈ C, |λ| < 1}

.

F0 = span{y ∈ E : T y = λy for some λ ∈ C, |λ| > 1} are both dense in E. Then T is hypercyclic. Proof First of all, note that .T n x → 0 as .n → ∞ for every .x ∈ E0 . On the other hand, if .y ∈ E is such that there is .λ ∈ C with .|λ| > 1 so that .T y = λy, we define .Sy = λ1 y and extend it by linearity to .F0 . It is straightforward to check that n .T Sy = y and .S y → 0 for every .y ∈ F0 . Corollary 5.1–(1) then gives that T is transitive, and being E separable, Theorem 5.2 gives that it is hypercyclic. 

So far we have given some sufficient conditions for an operator to be transitive. We change now the point of view, looking for necessary conditions. The transpose of the operator (recall Sect. 3.9) plays a fundamental rôle. Lemma 5.2 Let .T ∈ L(E) be transitive. Then for every .0 = u ∈ E and every t non-empty open set .G ⊆ E, the set . n∈N ( T )n (u)(G) is dense in .C.

166

5 Transitive and Mean Ergodic Operators

Proof Fix some non-empty open set .G ⊆ E. If .∅ = A ⊆ C is open, then .F = u−1 (A) is open (and not empty) in E, and we can  find

t.n ∈n N and .x ∈ G so that n n .T x ∈ F (because T is transitive). Then .u T x = T (u) (x) ∈ A, and this implies that . n∈N (t T )n (u)(G) ∩ A = ∅. Since the open set A was arbitrary, this yields the claim. 

t n Proposition 5.4 Let .T ∈ L(E) be transitive. Then .{ T (u) : n ∈ N} is not equicontinuous in .E for every .0 = u ∈ E . t n Proof Let us suppose that there is some .0 = u ∈ E for which .{ T (u) : n ∈ N} is t n equicontinuous. Then there is some open 0-neighbourhood U so that . T (u) ∈ U ◦ t n for n. Therefore, . T (u)(x) ≤ 1 for every n and all .x ∈ U , and every t n . n∈N ( T ) (u)(G) cannot be dense in .C. By Lemma 5.2, the operator is not transitive. 

Corollary 5.2 Let .T ∈ L(E) be transitive. Then .{T n : n ∈ N} is not equicontinuous in .L(E). Proof Suppose that .{T n : n ∈ N} is equicontinuous. Then for each .0 = v ∈ E there is some 0-neighbourhood U so that .|v(T n x)| ≤ 1 for every .x ∈ U . Then t n .{( T ) (v) : n ∈ N} is equicontinuous and, by Proposition 5.4, T is not transitive. 

Proposition 5.5 Let .T ∈ L(E) be transitive. Then .t T does not have eigenvalues. In particular, .t T is injective. Proof Suppose that .t T has an eigenvalue, that is, there are .0 = u ∈ E and .λ ∈ C so that .t T u = λu. Our aim is to show that we can find some open non-empty .G ⊆ E for which

t n . T (u)(G) n∈N

is not dense in .C. This, in view of Lemma 5.2, will yield the conclusion. In first place, note that for each .n ∈ N and .x ∈ E, we have |(t T )n (u)(x)| = |u(T n x)| = |λ|n |u(x)| .

.

Now we distinguish two cases. If .|λ| ≤ 1, we consider the open set G1 = {x ∈ E : |u(x)| < 1} ,

.

and .|(t T )n (u)(x)| < 1 for every n and .x ∈ G1 . On the other hand, if .|λ| > 1, we take G2 = {x ∈ E : |u(x)| > 1} ,

.

5.1 Transitive Operators

167

and .|(t T )n (u)(x)| > 1 for every n and .x ∈ G2 . Hence, for eigenvalue eachpossible t n .λ, we can find some open, non-empty set G for which . T (u)(G 1 ) is not n∈N 

dense. This completes the proof. Corollary 5.3 If E is a finite dimensional space, then no operator .T ∈ L(E) is transitive. Proof If E is finite dimensional, then .t T ∈ L(E ) has at least one eigenvalue.



Corollary 5.4 Let .x ∈ E be a hypercyclic vector of an operator .T ∈ L(E). Then Orb(x, T ) is linearly independent.

.

Proof Suppose that there is some .n ∈ N, and .α0 , . . . , αn ∈ C so that T n+1 x =

n 

.

αj T j x .

j =0

Then .Orb(x, T ) ⊆ span{x, T x, . . . , T n x}, which is a finite dimensional subspace of E. Since .Orb(x, T ) is dense in E, we deduce that .E = span{x, T x, . . . , T n x}. This contradicts Corollary 5.3. 

Theorem 5.3 (Bourdon) Let E be a separable locally convex space and .T ∈ L(E) hypercyclic. Then, there exists a dense subspace F of E such that every nonzero .x ∈ F is hypercyclic for T . Proof Choose some .x ∈ E so that .Orb(x, T ) = E, and define .F = span(Orb(x, T )), which is obviously  dense and satisfies .T (F ) ⊆ F . What we have to see is that, if .0 = y = nj=0 αj T j x ∈ F , then .Orb(y, T ) = E. To do n j this, we consider .P (z) = j =0 αj z , so that .y = P (T )(x), and observe that k k .T (P (T ))x = P (T )(T x) for every k. Then, if .P (T ) has dense range, Orb(y, T ) = P (T ) Orb(x, T ) = P (T )Orb(x, T ) = P (T )(E) = E ,

.

and we are done. It suffices, then, to see that .P (T ) has dense range. By the Fundamental Theorem of Algebra, we can find .am = 0 and .c1 , . . . , cm ∈ C so that P (T ) = am (T − c1 I ) · · · (T − cm I ) .

.

So, if we see that .T −λI has dense range for each .λ ∈ C, then we are done. Suppose that this is not the case and there is some .λ ∈ C with .(T − λI )(E) = E. Then, by the Hahn–Banach theorem, we can find some non-zero .u ∈ E such that .u(z) = 0 for every .z ∈ (T − λI )(E). With this, for each .w ∈ E, we have   (t T u − λu)(w) = u(T w) − u(λw) = u (T − λI )(w) = 0 ,

.

168

5 Transitive and Mean Ergodic Operators

from which .t T u = λu, and .λ is an eigenvalue of .t T . By Proposition 5.5, T cannot be transitive and, hence, neither hypercyclic. This is a contradiction and completes 

the proof. We finish this section by introducing another notion concerning the asymptotic behaviour of an operator. A point .x ∈ E is said to be n-periodic for the operator n .T ∈ L(E) if .T x = x, and periodic if it is n-periodic for some n. With this, an operator in .L(E) is chaotic (in the sense of Devaney) if it is transitive and the set of periodic points is dense in E. The set of periodic points of an operator has an interesting description. Proposition 5.6 Let E be a complex locally convex space, and .T ∈ L(E). Then the set of periodic points of T is exactly .

span{x ∈ E : there exists λ ∈ C with λn = 1 for some n ∈ N so that T x = λx} . (5.5)

 Proof Let us denote the set in (5.5) by .G0 . If .x ∈ G0 , then .x = m k=1 αk xk , where, for each k, there are .λk ∈ C and .nk ∈ N with .λnk k = 1 and .T xk = λk xk . Taking n to be the least common multiple of .{n1 , . . . , nm }, it is straightforward to see that each .xk is n-periodic and then so also is x. This gives one inclusion. Let us see now that every periodic point lies in .G0 . Choose, then, some .x0 ∈ E so that .T n x0 = x0 for some .n ∈ N, and let .λ1 , . . . , λn be the nth roots of 1. For each .i = 1, . . . , n, we define  .pi (z) = (z − λj ), j =i

and since all .λk s are different, the system .{p1 (z), . . . , pn (z)} is a basis of the space of polynomials of degree strictly less than n. Then, we can find .α1 , . . . , αn ∈ C so that 1=

n 

.

αi pi (z) ,

i=1

 k for every .z ∈ C. In other words, if we write the left hand side as . n−1 k=0 ak z , we have that .a0 = 1 and .ak = 0 for .k = 1, . . . , n − 1; hence, I=

n 

.

αi pi (T ) .

i=1

Then, writing .yi = pi (T )(x0 ), we have x0 =

n 

.

i=1

αi yi .

(5.6)

5.1 Transitive Operators

169

On the other hand, note that .(z − λi )pi (z) = zn − 1 for every .i = 1, . . . , n, which yields (T − λi I )(yi ) = (T − λi I )pi (T )(x0 ) = (T n − I )(x0 ) = T n x0 − x0 = 0,

.

and then .T yi = λi yi . Since .λni = 1, this and (5.6) show that .x0 is in .G0 and 

completes the proof. As a straightforward consequence of Propositions 5.3 and 5.6, we have the following. Proposition 5.7 (Godefroy–Shapiro) Let E be a complex, separable locally convex Hausdorff space and .T ∈ L(E). Suppose that the spaces E0 = span{x ∈ E : T x = λx for some λ ∈ C, |λ| < 1}

.

F0 = span{y ∈ E : T y = λy for some λ ∈ C, |λ| > 1} G0 = span{x ∈ E : there exists λ ∈ C with λn = 1 for some n∈N so that T x=λx} are dense in E. Then T is chaotic. Given two locally convex spaces E and F , the operators .T ∈ L(E) and .S ∈ L(F ) are said to be quasi-conjugate if there is a continuous operator .R : F → E with dense range such that .R ◦ S = T ◦ R. Proposition 5.8 Let .T ∈ L(E) and .S ∈ L(F ) be quasi-conjugate operators. If S is transitive (resp., hypercyclic or chaotic), then so also is T . Proof We only give the details for the transitivity, the other cases being left to the reader. Take .U, V ⊆ E open and not empty. Since R is continuous and has dense range, .R −1 (U ) and .R −1 (V ) are non-empty open subsets of F . Then (because S is transitive), there are n and .x ∈ R −1 (U ) so that .S n x ∈ R −1 (V ). Therefore, .Rx ∈ U and .RS n x ∈ V . One can easily check that .R ◦ S n = T n ◦ R. This gives the conclusion. 

Corollary 5.5 Let .T ∈ L(E) be such that there exist a dense subspace .F ⊆ E, invariant for T (i.e., .T F ⊆ F ), and a locally convex topology .τ on F , finer than the one induced by E and so that the restriction .T |F : (F, τ ) → (F, τ ) is continuous and transitive (resp., hypercyclic or chaotic). Then, so also is .T : E → E. Proof It is enough to take the inclusion .(F, τ ) → E as R in Proposition 5.8.



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5 Transitive and Mean Ergodic Operators

5.2 Mean Ergodic Operators Given an operator .T ∈ L(E), we define its Cesàro means as 1 k T n n

Tn =

.

k=1

for .n ∈ N (we set .T0 = I ). Clearly, each .Tn ∈ L(E) (i.e., is linear and continuous). An operator .T ∈ L(E) is said to be: • Power bounded if .(T n )n is equicontinuous in .L(E). • Cesàro bounded if .(Tn )n is equicontinuous in .L(E). • Mean ergodic if there is .P ∈ L(E) such that .Tn → P in .Ls (E) (i.e., .Tn x → P x as .n → ∞ for every .x ∈ E). • Uniformly mean ergodic if there is .P ∈ L(E) such that .Tn → P in .Lβ (E) (i.e., uniformly on the bounded subsets of E). Our aim is to see how these properties relate to each other. Clearly, every uniformly mean ergodic operator is mean ergodic. Also, if T is power bounded, then .(T n )n is equicontinuous, and given a continuous seminorm q, there is some continuous seminorm p so that q(T n x) ≤ p(x) for every x ∈ E and n ∈ N .

.

This has two consequences. On the one hand, .q holds for every q (and every n), we have .

1

nT

nx



≤

1 n p(x),

1 n T x → 0 for every x ∈ E . n

and since this

(5.7)

But we can say even more. If .B ⊆ E is bounded, then .supx∈B p(x) < ∞. This immediately gives that .

1 n T x → 0 uniformly on the bounded subsets of E . n

(5.8)

On the other hand, q(Tn x) ≤

.

1 1 q(T k x) ≤ p(x) = p(x) n n n

n

k=1

k=1

(5.9)

for every x and n. This shows that .(Tn )n is equicontinuous in .L(E). So, every power bounded operator is Cesàro bounded. For this reason, from now on, we focus on the relation between Cesàro boundedness and mean ergodicity. We begin with two simple observations that follow by direct computation from the definition. For every

5.2 Mean Ergodic Operators

171

T ∈ L(E) and all .n ∈ N, we have

.

.

1 n n−1 T = Tn − Tn−1 n n

(5.10)

and (I − T )Tn = Tn (I − T ) =

.

1 (T − T n+1 ) . n

(5.11)

As a straightforward consequence of the definitions and (5.10), we have the following. Proposition 5.9 Let .T ∈ L(E): 1. If T is mean ergodic, then .

1 n T → 0 in Ls (E) n

(that is, . n1 T n x → 0 for every .x ∈ E). 2. If T is uniformly mean ergodic, then .

1 n T → 0 in Lβ (E) n

(that is, . n1 T n → 0 uniformly on the bounded subsets of E). Remark 5.2 Let us note that from (5.11) we have, for each .x ∈ E, Tn (x − T x) =

.

 n + 1 1 1 1 (T x − T n+1 x) = T x − T n+1 x . n n n n+1

So, if x satisfies . n1 T n x → 0, then .Tn (x − T x) → 0. The same computations show that, if . n1 T n → 0 uniformly on the bounded sets of E, then .Tn (I − T ) → 0 uniformly on the bounded sets. Remark 5.3 Suppose that E is a complex locally convex Hausdorff space and .T ∈ L(E) is such that there are .0 = x0 ∈ E and .λ ∈ C with .|λ| > 1 so that .T x0 = λx0 . Then . n1 T n x0 = n1 λn x0 does not tend to 0 and T cannot be neither power bounded nor mean ergodic (recall (5.7) and Proposition 5.9). Theorem 5.4 (Eberlein’s mean ergodic theorem) bounded and .x ∈ E such that .

1 lim T n x = 0 . n n

Let .T ∈ L(E) be Cesàro

(5.12)

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5 Transitive and Mean Ergodic Operators

Then for each .y ∈ E, the following statements are equivalent: 1. .T y = y and y belongs to the closed convex hull of .Orb(x, T ). 2. .y = lim Tn x. n

3. .y = σ (E, E ) − lim Tn x. n

4. y is a .σ (E , E)-cluster point of .(Tn x)n (i.e., for every .σ (E , E)-neighbourhood U of 0 and every m, there is some .n > m such that .y − Tn x ∈ U ). Proof The implications 2 .⇒ 3 .⇒ 4 are obvious. In order to see 1 .⇒ 2, let us take some y in the closed convex hull of .Orb(x, T ) and such that .y = T y. We fix some seminorm q on E and .ε > 0. Being T Cesàro bounded, .(Tn )n is equicontinuous, and we can find a seminorm p so that .q(Tn z) ≤ p(z) forevery .z ∈ E and every m .n ∈ N. By the choice of y, there are .α0 , . . . , αm ≥ 0 with . i=0 αi = 1 satisfying m   ε  p y− αi T i x < . 2

(5.13)

.

i=0

If .k ∈ N is fixed, a straightforward computation shows that Tn T k x − Tn x =

.

1 j 1  n+j T x− T x. n n k

k

j =1

j =1

We clearly have .limn n1 T j x = 0 for each .j = 1, . . . , k, and by (5.12), .

 n + j  1  1 T n+j x = 0 . lim T n+j x = lim n n n n n+j

Therefore, for each fixed k, we have .Tn T k x − Tn x → 0 as .n → ∞, and we can find .n(k) ∈ N so that q(Tn T k x − Tn x)
0 and a seminorm q on E. Being (Tn )n equicontinuous (because T is Cesàro bounded), we can find a seminorm p so that .q(Tn x) ≤ p(x) for every x and n. First, .z ∈ Im(I − T ); hence, there exists some .y ∈ E with .p(z − (I − T )y) < ε and then

.

  q Tn (z − (I − T )y) < ε .

.

Second, z is in .ker(I − T ), and this implies that .Tn z = z for every n. Third, by Remark 5.2 and (5.16), we have .Tn (y − T y) → 0 as .n → ∞, and there is some m for which   q Tm (y − T y) < ε .

.

Joining these three facts together, we have     q(z) = q(Tm z) ≤ q Tm (z − (I − T )y) + q Tm (y − T y) < 2ε .

.

Since this holds for arbitrary .ε and q, we have .z = 0. As a second step, we show that Eme (T ) ⊆ ker(I − T ) + Im(I − T ) .

.

To do this, we pick some .x ∈ Eme (T ) and define .P x = limn Tn x. By Eberlein’s Theorem 5.4, .T P x = P x, which gives .P x ∈ ker(I − T ). This also implies that

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5 Transitive and Mean Ergodic Operators

Tn P x = P x for every n, and letting .z = x − P x, we have

.

T n z = T n x − T n P x = Tn x − P x ,

.

and this clearly tends to 0 as .n → ∞. Lemma 5.3 shows that .z ∈ Im(I − T ) and yields our claim. We see now that .

ker(I − T ) + Im(I − T ) ⊆ Eme (T ) ,

which will complete the proof of (5.17). Suppose that .z = a +b with .a = T a (hence Tn a = a for every n) and .b ∈ Im(I − T ) (and, then, .limn Tn b = 0, by Lemma 5.3). Thus .limn Tn z = a and .z ∈ Eme (T ). This completes the proof of (5.17). To finish the proof, just observe that we have just seen that Eberlein’s theorem implies that .T P = P , and then .P x ∈ ker(I − T ) for every .x ∈ Eme (T ). Also, .Tn P x = P x for every n and x. Then .

P 2 x = lim Tn P x = P x

.

n

for every .x ∈ Eme (T ). Finally, since .Tn x − Tn T x → 0 as .n → ∞ and .Eme (T ) is invariant by T , we immediately have .P x − P T x = 0 for every .x ∈ Eme (T ). 

Corollary 5.8 Let .T ∈ L(E) be Cesàro bounded satisfying .

1 n T x → 0 as n → ∞ for every x ∈ E . n

Then T is mean ergodic if and only if .ker(I − T ) separates points of .ker(I − t T ). Proof Let us note in first place that, if .u ∈ ker(I − t T ), then .u(x) = u(T x) for every .x ∈ E, and a straightforward computation shows that .u(x) = u(Tn x) for every .x ∈ E and .n ∈ N. If T is mean ergodic, this implies that .u(P x) = u(x) for every x. Choose now .u1 , u2 ∈ ker(I − t T ) with .u1 = u2 . Then there is some .x0 ∈ E so that .u1 (x0 ) = u2 (x0 ). By what we have just seen, .u1 (P x0 ) = u2 (P x0 ), and by Yosida’s Theorem 5.5, .P x0 ∈ ker(I − T ). This shows that indeed .ker(I − T ) separates points of .ker(I − t T ). Let us suppose now that T is not mean ergodic and take .y ∈ ker(I − T ) ⊕ Im(I − T ) (which we know that exists by Yosida’s theorem). By the Hahn– Banach theorem, there is .u ∈ E such that .u(y) = 1 and .u(x) = 0 for every .x ∈ ker(I − T ) ⊕ Im(I − T ). This implies that .u((I − T )x) = 0 for every .x ∈ E and all .u ∈ ker(I − t T ). By assumption .ker(I − T ) separates u and 0, so there is some .x0 ∈ ker(I − T ) with .u(x0 ) = 0. This gives a contradiction and completes the 

proof. As a straightforward consequence, we have the following.

5.2 Mean Ergodic Operators

177

Corollary 5.9 Let .T ∈ L(E) be Cesàro bounded satisfying .

1 n T x → 0 as n → ∞ for every x ∈ E , n

and .ker(I − T ) = {0}. Then T is mean ergodic if and only if .ker(I − t T ) = {0}. An operator .T : E → F between (Hausdorff) locally convex spaces is called Montel if it maps bounded subsets of E into relatively compact subsets of F . A space E is called semi-Montel if the identity is a Montel operator (that is, every bounded subset of E is relatively compact). Let us point out that every semi-Montel space is semi-reflexive. Proposition 5.10 Let .T ∈ L(E) be a Cesàro bounded Montel operator such that .

1 n T → 0 as n → ∞ uniformly on the bounded subsets of E . n

(5.18)

Then T is uniformly mean ergodic. Proof By Lemma 5.7, the operator T is mean ergodic, and there is .P ∈ L(E) such that .Tn x → P x for every .x ∈ E. Moreover, Theorem 5.5 shows that .P = P 2 = P T = T P , from which we immediately have .P = P Tn = Tn P for every n. We fix now some continuous seminorm p on E, some .ε > 0, and take a bounded set .B ⊆ E. Since T is Cesàro bounded, the sequence .(Tn )n is equicontinuous, and we can find a continuous seminorm q and a constant .C > 0 so that p((Tn − P )x) ≤ Cq(x)

.

for every .x ∈ E and .n ∈ N. Now, the set .T (B) is relatively compact (because T is Montel); hence, there exist .z1 , . . . , zm ∈ T (B) so that for each .y ∈ T (B) there is some .i = 1, . . . , m with q(y − zi )
0 was arbitrary yield .

lim sup p(Tn T x − P x) = 0 , n x∈B

and .Tn T → P as .n → ∞ uniformly on the bounded subsets of E. Now we have Tn − P = Tn (I − T ) + (Tn T − P ) .

.

We have just seen that the second term of the sum tends to 0 uniformly on the bounded subsets of E and, by Remark 5.2, so also does the first term. This altogether shows that .Tn → P uniformly on bounded sets, and T is uniformly mean ergodic. 

Note that every operator on a semi-Montel space is Montel. Recall also that every power bounded operator is Cesàro bounded (see (5.9)) and satisfies (5.8). Then we can apply Proposition 5.10 to have the following result. Proposition 5.11 Every power bounded operator on a semi-Montel locally convex space is uniformly mean ergodic. Theorem 5.6 (Lin) Let E be a Fréchet space and .T ∈ L(E) such that .

1 n T → 0 as n → ∞ uniformly on the bounded subsets of E , n

(5.20)

and .ker(I − T ) = {0}. Consider the following statements: 1. .I − Tn is surjective for some .n ∈ N. 2. .I − T is surjective. 3. .Tn → 0 in .Lβ (E) as .n → ∞. Then 1 .⇒ 2 .⇒ 3. If E is Banach, then also 3 .⇒ 1. Proof It is easy to check that I − Tn = (I − T )

.

k n−1  1 

n

Tj



k=1 j =0

for every n. Then, if .I − Tn is surjective for some n, so also is .I − T . Let us suppose now that .I −T ∈ L(E) is surjective. Since, by assumption, it is also injective, by the Open Mapping Theorem 3.22, it is a surjective linear isomorphism. Given a bounded set .B ⊆ E, the set .C = (I − T )−1 (B) is also bounded and .(I − T )(C) = B. Take now any continuous seminorm q in E. We have qB (Tn ) = sup q(Tn z) = sup q((I − T )Tn z) = qC ((I − T )Tn ) ,

.

z∈B

z∈C

5.2 Mean Ergodic Operators

179

and this tends to 0 as .n → ∞, since (5.20) holds (recall Remark 5.2). This shows that .Tn → 0 in .Lβ (E), and 2 .⇒ 3. Let us finally suppose that E is a Banach space and that 3 holds. Then there is some .n ∈ N such that .Tn  < 1. A straightforward computation shows that S=

.

∞  (Tn )j j =0

is a continuous linear inverse of .I − Tn , and then 1 holds.



Let us see that the implication 3 .⇒ 1 does not hold in general for non-normable Fréchet spaces. To this end, consider the Köthe echelon space .λ1 (A) defined by the matrix .an (i) = i n , that is, ∞    λ1 (A) = x = (xi )i ∈ CN : qn (x) = i n |xi | < ∞ for every n ∈ N ,

.

i=1

known as the space of rapidly decreasing sequences, and often denoted by s. Example 5.7 The operator .T ∈ L(s) given by T (x) =

.

 1−

1 2i

 i

for x ∈ s

satisfies: 1. .ker(I − T ) = {0}. 2. . n1 T n → 0 uniformly on bounded sets. 3. .Tn → 0 in .Lβ (s). 4. .I − T is not surjective. Note that the first two ones are the hypothesis of Proposition 5.6, and the remaining two are the proof of our claim that the implication 3 .⇒ 1 does not hold in general. It is straightforward that .ker(I − T ) = {0}. On the other hand, for each m, we have  m  m .T x = 1 − 21i xi for every .x ∈ s. Then, ∞     qn T m x = in 1 −

.

i=1

 1 m |xi | 2i

≤

∞ 

i n |xi | = qn (x) .

i=1

From this, we deduce that . n1 T n → 0 uniformly on bounded sets (that is, 2 holds) and that T is power bounded. By Proposition 2.12, the space s is Montel and then

180

5 Transitive and Mean Ergodic Operators

Proposition 5.11 gives that T is uniformly mean ergodic. Then there is an operator P ∈ L(s) so that .Tk → P in .Lβ (s) as .k → ∞. Let us see that, in fact, .P = 0. We begin with the canonical basis vectors. Given .j, k ∈ N, we have

.

 1 j T j ek = 1 − i ek , 2

.

and this tends to 0 (for each fixed k) as .j → ∞. Then 1 j T ek −→ 0 as n → ∞, n n

.

j =1

and therefore, .P ek = 0 for each fixed k. As .span{ek : k ∈ N} is dense in s (since the finite truncations of each sequence .x ∈ s converge to x) and .P ∈ L(s), we conclude that .P = 0, giving our claim. Finally, a simple computation shows that −i ) ∈ Im(I − T ), which gives that .I − T is not surjective. .(2 

i Theorem 5.8 (Lin) Let E be a Banach space and .T ∈ L(E) such that .

1 n T  → 0 as n → ∞ . n

(5.21)

The following are equivalent: 1. T is uniformly mean ergodic. 2. .Im(I − T ) is closed and .E = ker(I − T ) ⊕ Im(I − T ). 3. .Im(I − T ) is closed. Proof In order to keep the notation as clear as possible, all through the proof we write .(I − T )(E) = Im(I − T ), .F = Im(I − T ), and .S = T |F (the restriction of T to F ). Suppose that T is uniformly mean ergodic. By Yosida’s Theorem 5.5, .E = ker(I − T ) ⊕ F . We have     T (F ) = T (I − T )(E) ⊆ T (I − T )(E) ⊆ F = F ;

.

thus .S(F ) ⊆ F and .S ∈ L(F ). There is some .P ∈ L(E) so that .limn Tn − P  = 0. We know that (recall again Yosida’s theorem) .Im P = ker(I − T ) and then .F = ker P , from which we have Sn  = sup Sn x ≤ sup Sn x − P x + sup P x = Sn − P  ,

.

x∈F x≤1

x∈F x≤1

x∈F x≤1

5.2 Mean Ergodic Operators

181

and this clearly implies that .limn Sn  = 0 and then . n1 Sn  → 0. We also have .ker(I − S) = {0}, so that S fulfils all conditions in Proposition 5.6–3 and .I − S is surjective. Then (I − T )(E) = F = (I − S)(F ) = (I − T )(F ) ⊆ (I − T )(E) ⊆ F = (I − T )(E) ,

.

and .(I − T )(E) is closed. This shows that 1 implies 3. Let us suppose that .Im(I − T ) is closed (hence Banach). Note that .Im(I − T ) = F , and the operator .I − T : E → F is continuous and surjective so, by the Open Mapping Theorem 3.22, .I − T is open and there is .K > 0 such that for every .y ∈ E there is .z ∈ E so that .(I − T )z = y and .z ≤ Ky. We fix some .y ∈ E and select z as above to get (recall (5.11)) Tn y = Tn (I − T )z ≤

.

 K 1 T  + T n+1  y . T − T n+1  z ≤ n n

The restriction .S = T |F : F → F is well defined, linear, and continuous. Then, taking the supremum over .y ≤ 1 in the previous expression, and using (5.21), we have .limn Sn  = 0, and then . n1 Sn  → 0. On the other hand, proceeding as in the proof of Yosida’s Theorem 5.5, we get .E = ker(I − T ) ∩ Im(I − T ) = {0}, from which .ker(I − S) = {0} follows. Then, by Proposition 5.6, .I − S : F → F is surjective (hence, an isomorphism onto). Denote by .J : F → E the canonical inclusion and define P = I − J (I − S)−1 (I − T ) ∈ L(E) .

.

Let us finally see that .limn Tn − P  = 0, which will give that T is uniformly mean ergodic and complete the proof. For .x ∈ E, let .y = (I − S)−1 (I − T )x, and note that .(I − T )y = (I − S)y = (I − T )x, which gives .T (x − y) = x − y, and hence, .Tn (x − y) = x − y for every n. Note also that .P x = x − y. With this, we have (recall again (5.11)) T. n x−P x = Tn x−(x−y) = Tn x−Tn (x−y)=Tn y=Tn (I −S)−1 (I −T )x ≤ (I − S)−1  Tn (I − T )x ≤

  1 (I − S)−1  T  + T n+1  x . n

Then Tn − S ≤

.

  1 (I − S)−1  T  + T n+1  n

for every .n ∈ N, and since the latter tends to 0 as .n → ∞, this yields the claim and completes the proof. 

Remark 5.4 We finish this section by noting that, as a straightforward consequence of the Banach–Steinhaus theorem, an operator .T ∈ L(E) on a Fréchet space is

182

5 Transitive and Mean Ergodic Operators

power bounded if and only if the orbit .Orb(x, T ) of each .x ∈ E is bounded. Then Corollary 5.2 gives that a power bounded operator cannot be hypercyclic. Also, if T is an operator on a Banach space, then it is power bounded if and only if n .supn∈N T  < ∞. Hence, a hypercyclic operator T on a Banach space must satisfy .T  > 1.

5.3 Examples We look now at some concrete operators and study when they share the properties that we have been looking at.

5.3.1 The Backward Shift We start by considering operators defined on sequence spaces. For the moment, let X denote either . p with .1 ≤ p < ∞ (recall Example 2.5) or .c0 (recall (2.9)).   We consider the backward shift operator .B : X → X, defined as .B (xn )∞ n=1 = (xn+1 )∞ n=1 , that is, B(x1 , x2 , x3 , . . .) = (x2 , x3 , . . .)

.

(recall (3.6)). Clearly, this defines a continuous linear operator (i.e., .B ∈ L(X)), and moreover, .B ≤ 1 (in fact, .B = 1, since .B(e2 ) = e1 ). Our aim is to study the properties (transitivity and mean ergodicity) of the operator T = λB for λ ∈ C

.

in terms of .λ. For .λ = 0, there is not much to say, so we assume, from now on, λ = 0. We collect all the information in the following result.

.

Theorem 5.9 Let .T = λB with .λ = 0: 1. If .|λ| > 1, then T is transitive, hypercyclic, and chaotic. 2. If .|λ| ≤ 1, then T is power bounded, and: a. If .|λ| < 1, then T is uniformly mean ergodic. b. If .|λ| = 1, then T is mean ergodic but not uniformly mean ergodic. Proof Let us suppose in first place that .|λ| > 1. We use Corollary 5.1–(2) to see that the operator is transitive. Let .X0 be the subspace of X consisting of those sequences with only finitely many non-zero coordinates, which is clearly dense in X. Define an operator .S : X0 → X by S(y1 , y2 , y3 , . . .) =

.

1 (0, y1 , y2 , y3 , . . .) . λ

5.3 Examples

183

Obviously, .(T ◦ S)(y) = y for every .y ∈ X0 . Also, given .y ∈ X0 , we have limn T n y = 0 (because y has only a finite number of non-zero coordinates). Finally, if .y ∈ X0 , then

.

S n y =

.

1 y λn

for each n, and .limn S n y = 0. Then Corollary 5.1–(2) gives that T is transitive. The space X is separable, and then by Birkhoff’s Theorem 5.2, T is also hypercyclic. Let us see that it is also chaotic. First of all, it is easy to see that .x ∈ X is periodic if and only if it is of the form   x = x1 , . . . , xN , λ−N x1 , . . . , λ−N xN , λ−2N x1 , . . . , λ−2N xN , . . . .

.

Since .X0 is dense in X, in order to see that the set of periodic points is dense in X, it suffices to see that every .y ∈ X0 can be approximated by periodic points. Take, then, some .y = (y1 , . . . , ym , 0, 0, . . .) in .X0 . Now, for each .N ≥ m, we define .xN as   xN = y1 , . . . , ym , 0, . . . , 0, λ−N y1 , . . . , λ−N ym , 0, . . . , 0, . . . ,      

.

N

N

each one periodic. If .X = c0 , then 1 N →∞ y −−−−→ 0 . N λ

xN − y∞ =

.

On the other hand, if .X = p for .1 ≤ p < ∞, we have xN − yp =

m ∞  

.

|λ−Nj |p |yk |p

1

p

=

j =1 k=1

∞ 

|λ−Nj |p yp

1

p

j =1

≤

∞ 

|λ−Nj |y =

j =1

|λ|−N N →∞ −−−−→ 0 . −N 1 − |λ|

So, in each case, we have a sequence of periodic points converging to y. This gives our claim and completes the proof of 1. The proof of 2 begins by noting that, since .B = 1, we have .T  = |λ| ≤ 1, and T is power bounded. Suppose now that .|λ| < 1, then .

n n n  1 1 j 1 1 n→∞  j 1  j j ≤ T λ B  ≤ |λ| B ≤ −−−→ 0 .   n n n n 1 − |λ| j =1

j =1

j =1

184

5 Transitive and Mean Ergodic Operators

This shows that the Cesàro means .Tn converge (in the operator norm) to the operator P = 0. Hence, T is uniformly mean ergodic (showing 2a). We complete the proof by showing that if .|λ| = 1, then T is mean ergodic but not uniformly mean ergodic. Since .B j x → 0 for every .x ∈ X, we get .T j x → 0 and then

.

1 j T x −→ 0 n n

.

j =1

for every .x ∈ X. This yields that T is mean ergodic. Observe that, if .X = p with 1 < p < ∞, the space is reflexive, and the result follows also from Corollary 5.7. Let us see now that T is not uniformly mean ergodic. We consider two cases. Assume first that .X = p with .1 ≤ p < ∞. Fix .n ∈ N and observe that

.

n n 1   1   jp  p1   T j (en+1 ) = |λ| = 1.  p n n

.

j =1

j =1

Since the canonical vectors have norm 1, this implies that

.

n  1  j T  → 0 ,  p n j =1

and T is not uniformly mean ergodic. The case of .X = c0 is slightly more involved. To begin with, from the fact that 1 n .B = 1 and .|λ| = 1, we have .limn T  = 0, and by Lin’s Theorem 5.6, to see n that T is not uniformly mean ergodic, it suffices to see that .

Im(I − T ) = (I − λB)(c0 )

is not closed in .c0 . We see in first place that .I − T is injective. If this were not the case, there would be some non-zero .x ∈ c0 so that .x = λBx. This forces ∞ x to be of the form .x = x1 1, 1/λ, 1/λ2 , 1/λ3 . . . , but the sequence . 1/λj j =0 does not belong to .c0 . This gives a contradiction and yields our claim. In the second place, we observe that (I − λB)(e1 ) = e1 , and (I − λB)(en ) = en − λen−1 for n ≥ 2 ,

.

which clearly implies X0 = span{en : n ∈ N} ⊆ Im(I − T ) .

.

5.3 Examples

185

This gives that .Im(I − T ) is dense in .c0 (because .X0 is so). Therefore, if .Im(I − T ) were closed, we would have .

Im(I − T ) = c0

and, by the Open Mapping Theorem 3.22, .I − T : c0 → c0 would be an isomorphism. Then its bi-transpose .

tt

(I − T ) : ∞ −→ ∞

would be also an isomorphism. It is easy to see that . tt (I − T ) = (I − λB) : ∞ → ∞ , where B is the backward shift defined by (3.6) on . ∞ . However,   (I − λB) 1, λ1 , λ12 , . . . = 0 ,

.

  and . 1, λ1 , λ12 , . . . ∈ ∞ (since .|λ| = 1). This shows that .I − λB is not injective on . ∞ , and henceforth not an isomorphism. Then .Im I − T is not closed in .c0 and this completes the proof. 

Remark 5.5 The case of .T = λB : ∞ → ∞ is more delicate. To begin with, it is easy to see that .T n  = |λ|n and, then, .λB is power bounded if and only if .|λ| ≤ 1. On the other hand, if .|λ| < 1, the same argument as in Theorem 5.9 shows that .λB ∈ L( ∞ ) is uniformly mean ergodic. Suppose that .|λ| > 1. Since the space is not separable, .λB cannot be hypercyclic in any case (actually, no operator on . ∞ can be hypercyclic). Let us see that neither can it be transitive. It is enough to see that .t T ∈ L( ∞ ) has eigenvalues (recall Proposition 5.5). Let .L : ∞ → C be a Banach limit (recall Sect. 3.2.4). By Theorem 3.7, .L ∈ ∞ and .L(Bx) = L(x) for every .x ∈ ∞ , which implies t  T (L)(x) = t (λB)(L)(x) = L(λBx) = λL(x)

.

for every .x ∈ ∞ , that is, .t T (L) = λL, and L is an eigenvector of .t T . This shows that T is not transitive. As a matter of fact, Bermúdez and Kalton showed in [2, Theorem 3.4] that there cannot be any transitive operator on . ∞ . It remains to study the behaviour of the operator for .|λ| = 1. In this case, the operator .λB is not mean ergodic (hence, obviously, neither is it uniformly mean ergodic). This relies on some known results. We just indicate how to use them to get the conclusion. First of all, by [8, Theorem 5], an operator on . ∞ is mean ergodic if and only if it is uniformly mean ergodic, so it suffices to see that the latter fails for .λB. Now, by [7, Chapter 2, Theorem 2.7, p.90], if S is any operator on a Banach space and .I − S is invertible, then S is uniformly mean ergodic. But, on the other hand, from [4, Chapter VII, Corollary 6.6, p. 214], we get that for .|λ| = 1, the operator .I − λB is not invertible. Hence, .λB is not uniformly mean ergodic, and this gives our claim.

186

5 Transitive and Mean Ergodic Operators

5.3.2 Composition Operators We look now at composition operators defined on .H(D). Let us recall (see Example 4.16) that the composition operator .Cϕ : H(D) → H(D) is always well defined and continuous for every holomorphic self-map .ϕ : D → D. Our aim now is to study when these operators are power bounded, mean ergodic, or hypercyclic. We look for conditions on .ϕ that ensure that the composition operator .Cϕ has these properties. We answer these questions in Theorems 5.12 and 5.13. Given a self-map .ϕ : D → D and .n ∈ N, we denote, as usual, .ϕ n for the n-iterate of .ϕ. That is, we denote .ϕ 0 = idD (the identity mapping on .D), and .ϕ n = ϕ n−1 ◦ ϕ for .n ≥ 1. We clearly have that .(Cϕ )n (f ) = f ◦ ϕ n for every .f ∈ H(D). In other words, .Cϕn = Cϕ n for every n. Hence, if we want to study the behaviour of .Cϕn as n .n → ∞, then the behaviour of the sequence of iterates .(ϕ )n has to appear in one way or another. We begin by looking at power boundedness and mean ergodicity. Let us point out in first place that by Montel’s Theorem 4.9 and Proposition 5.11, every power bounded operator on .H(D) is uniformly mean ergodic. Whether or not .Cϕ has some of these properties depends on the shape of symbol .ϕ. We use some very classical results on complex analysis. We state them without proof. The first one is the Schwarz lemma, the proof of which can be found in [9, Theorem 12.2]. Theorem 5.10 (Schwarz Lemma) Let .ϕ : D → D be holomorphic such that ϕ(0) = 0. Then:

.

1. .|ϕ(z)| ≤ |z| for every .z ∈ D. If .|ϕ(z0 )| = |z0 | for some .0 = z0 ∈ D, then there exists .λ ∈ C with .|λ| = 1 such that .ϕ(z) = λz for every .z ∈ D. 2. .|ϕ (0)| ≤ 1. Moreover, .|ϕ (0)| = 1 if and only if there exists .λ ∈ C with .|λ| = 1 such that .ϕ(z) = λz for every .z ∈ D. We begin by looking at a particular class of symbols. A self-map .T : D → D is an automorphism of .D if it is holomorphic and bijective (note that, by the open mapping theorem, .T −1 is also holomorphic). The set of all automorphisms of .D clearly is a group and is denoted by .Aut(D). Automorphisms can be described in a very precise way. For .a ∈ D, we define .Ta : D → D by Ta (z) =

.

a−z , 1 − az

which is clearly holomorphic. Moreover, it is easy to see that .Ta (D) = D and that it is invertible and .Ta−1 = Ta . So, .Ta ∈ Aut(D) and (this is also straightforward) √ .Ta (0) = a and .Ta (a) = 0. Also, some simple computations show 1−

1−|a|2

is the only fixed point of .Ta in .D. Every automorphism is the that .z0 = a |a|2 composition of one of this type and a rotation. This is the content of the following result (see [9, Theorem 12.6]).

5.3 Examples

187

Proposition 5.12 Let .T ∈ Aut(D) with .T (a) = 0. Then there exists .λ ∈ C with |λ| = 1 so that .T (z) = λTa (z) for every .z ∈ D.

.

As a consequence, every automorphism of .D can be continuously extended to the closed disc .D. From Proposition 5.12, we can study the fixed points of automorphisms. The proof of the following result is not difficult and can be found in [3, Lemma 1.8.1]. Lemma 5.4 Let .T ∈ Aut(D) be different from the identity. Then T has at least one fixed point in .D. Moreover, if T does not have fixed points in .D, then it has at most two fixed points .τ, σ ∈ ∂D (that may be equal) that satisfy .T (σ )T (τ ) = 1. Taking this into account, we say that an automorphism T of .D is: • Elliptic if it has a (then unique) fixed point in .D. • Parabolic if it has a unique fixed point in .∂D. • Hyperbolic if it has two fixed points in .∂D. For example, .T (z) = iz is elliptic (with .τ = 0 as unique fixed point), .T (z) = z+1/2 (1+i)z−i iz+1−i is parabolic (with fixed point .τ = 1), and .T (z) = 1+z/2 is hyperbolic (with fixed points .τ = 1 and .σ = −1). We begin by studying composition operators defined by elliptic automorphisms. As we said before, the iterates of the symbol play an important rôle. In this case (thanks to Proposition 5.12), they have a very convenient shape. Lemma 5.5 Let T be an elliptic automorphism with fixed point .τ ∈ D. Then there is .λ ∈ C with .|λ| = 1 such that   T n (z) = Tτ λn Tτ (z)

.

(5.22)

for every .z ∈ D and .n ∈ N. Proof Consider .S = Tτ ◦ T ◦ Tτ ∈ Aut(D), which satisfies .S(0) = 0. By Proposition 5.12, there is .λ with .|λ| = 1 so that .S(z) = λz for every .z ∈ D. Now, using the fact that .Tτ ◦ Tτ = idD , we have n  λn z = S n (z) = Tτ ◦ T ◦ Tτ (z) = Tτ ◦ T n ◦ Tτ (z)

.

for every n and z. This yields the conclusion.



Proposition 5.13 Let T be an elliptic automorphism. Then the composition operator .CT : H(D) → H(D) is power bounded and uniformly mean ergodic. Proof We just have to show that the composition operator is power bounded since, as we have already noticed, this (by Theorem 4.9 and Proposition 5.11) automatically gives that the operator is also uniformly mean ergodic. Let .τ ∈ D be the fixed point of T and take .λ as in Lemma 5.5. Now, .Tτ is continuous, and then, for each .0 < ρ < 1, there is some .0 < R(ρ) < 1 so that .|Tτ (z)| < R(ρ) for every .|z| < ρ. Fix now .0 < r < 1 and write .s = R(r) and .R = R(s). Then, for

188

5 Transitive and Mean Ergodic Operators

every .f ∈ H(D) and all .|z| < r and .n ∈ N, we have, using (5.22),     |f (T n z)| = f Tτ (λn Tτ z) ≤ sup f Tτ (w) ≤ sup |f (ζ )| .

.

|w|≤s

|ζ |≤R

This shows that .CT is power bounded and completes the proof.



This describes completely the composition operators defined by elliptic automorphisms, showing that they are power bounded and, then, uniformly mean ergodic. What can be said for composition operators defined by arbitrary holomorphic selfmaps? When are they power bounded? As we have seen, having a fixed point plays some rôle. A holomorphic self-map of the disc may (or may not) have a fixed point in .D, but if it does have one, then it has only one (unless the mapping is the identity). Proposition 5.14 Let .ϕ : D → D be holomorphic and different from the identity. Then there exists at most one point .z0 ∈ D such that .ϕ(z0 ) = z0 . Proof Suppose that there are .a = b so that .ϕ(a) = a and .ϕ(b) = b. We define ψ = Ta ◦ ϕ ◦ Ta ∈ H(D) that satisfies .ψ(0) = 0 and take .w ∈ D so that .Ta (w) = b. Then

.

ψ(w) = Ta ◦ ϕ(b) = Ta (b) = w .

.

(5.23)

By Theorem 5.10, there exists some .λ ∈ C with .|λ| = 1 such that .ψ(z) = λz for every .z ∈ D. But (5.23) forces .λ = 1, and, in fact, .ψ = idD . Finally, the fact that .Ta ◦ Ta = idD yields that .ϕ is the identity. 

Our next step is to show that, if a holomorphic self-map (not an automorphism) has a fixed point .τ , then the iterates converge uniformly on the compact subsets of .D to the constant function given by this point. We denote this function by .cτ ; that is, .cτ (z) = τ for every .z ∈ D. Proposition 5.15 Let .ϕ : D → D be holomorphic and not an automorphism such that there is .τ ∈ D with .ϕ(τ ) = τ . Then the sequence of iterates .(ϕ n )n converges to .cτ uniformly on the compact subsets of .D. Proof Let us suppose first that .τ = 0 (i.e., .ϕ(0) = 0). Since .ϕ is not an automorphism, from Schwarz’s Lemma (Theorem 5.10), we have |ϕ(z)| < |z|

(5.24)

.

for every .0 = z ∈ D. For each .0 < r < 1, we define M(ϕ, r) = max |ϕ(z)| and δ(r) =

.

|z|≤r

M(ϕ, r) . r

5.3 Examples

189

From (5.24), we have .δ(r) < 1 for every .0 < r < 1. Fix one such r and consider the function given by ψr (z) =

.

ϕ(rz) M(ϕ, r)

for .z ∈ D. Clearly, it is holomorphic and .ψr (D) ⊆ D. Even more, it can be extended to a continuous function on .D and .ψr (0) = 0. Then, Schwarz’s Lemma again gives |ψr (z)| ≤ |z|

.

for every .z ∈ D. With this, for .|z| ≤ r, we have z   |ϕ(z)| = M(ϕ, r) ψr z/r ≤ M(ϕ, r) = δ(r)|z| < δ(r) . r

.

Iterating this, we get |ϕ n (z)| ≤ δ(r)n

.

for every .z ∈ D(0, r) and all n. Since .δ(r) < 1, this shows that .(ϕ n ) converges to 0 uniformly on the compact sets, completing the proof if .τ = 0. Now, if .τ = 0, we define .ψ = Tτ ◦ ϕ ◦ Tτ . This is a holomorphic self-map that fixes 0 and that is not an automorphism. By the previous case, the iterates .(ψ n )n converge to 0 uniformly on compact sets. But, since clearly .ϕ n = Tτ ◦ ψ n ◦ Tτ for every n, we have the conclusion. 

We are now in the position to give an analogue of Proposition 5.13 for holomorphic self-maps with a fixed point. Proposition 5.16 Let .ϕ : D → D be holomorphic, not an automorphism and with a fixed point .τ ∈ D. Then the composition operator .Cϕ : H(D) → H(D) is power bounded and uniformly mean ergodic. Proof As in Proposition 5.13, it suffices to see that .Cϕ is power bounded. We know from Proposition 5.15 that the iterates of .ϕ converge to the constant function .cτ uniformly on the compact subsets of .D. Our first step in the proof is to see that for every compact .K ⊆ D there is some compact .L ⊆ D such that ϕ n (K) ⊆ L for every n ∈ N .

.

Fix some compact set .K ⊆ C and, for .ε =

1−|τ | 2

> 0, choose .n0 ∈ N so that

|ϕ n (z) − τ | < ε ,

.

(5.25)

190

5 Transitive and Mean Ergodic Operators

for every .z ∈ K and all .n ≥ n0 we have. Then, for every such z and n, we have |ϕ n (z)|
r for every .n ≥ n0 and .|z| ≤ r (in particular for .z ∈ K). This gives (5.26) and completes the proof. 

As a straightforward consequence, we have the following: Corollary 5.10 Let .ϕ : D → D be holomorphic. Then .ϕ has stable orbits if and only if it has a fixed point in .D. We have finally everything at hand to characterize mean ergodicity for composition operators. Theorem 5.12 Let .ϕ : D → D be holomorphic. The following statements are equivalent: 1. 2. 3. 4. 5. 6.

ϕ has stable orbits. ϕ has a fixed point in .D. .Cϕ : H(D) → H(D) is power bounded. .Cϕ : H(D) → H(D) is uniformly mean ergodic. .Cϕ : H(D) → H(D) is mean ergodic.   1 n . (Cϕ ) → 0 in .Ls H(D) . n . .

Proof The equivalence between 1 and 2 has been already established in Corollary 5.10. That 2 implies 3 follows from Proposition 5.13 (if .ϕ is an automorphism) and Proposition 5.16 (if it is not). The implication 3.⇒ 4 holds for any arbitrary operator on .H(D) (recall Proposition 5.11). Obviously, 4 implies 5, and this implies 6 (by Proposition 5.9). So, in order to complete the proof, only the  implication 6 .⇒ 1 is left. Suppose, then that . n1 (Cϕ )n → 0 in .Ls H(D) . This implies   that . n1 Cϕ n n is equicontinuous and, in particular, for each compact set .K ⊆ D there are .0 < s < 1 and .M > 0 so that .

 1  sup f ϕ n (z) ≤ M sup |f (w)| , |w|≤s z∈K n

for every .n ∈ N and .f ∈ H(D). Our aim now is to see that ϕ n (K) ⊆ D(0, s)

.

(5.27)

192

5 Transitive and Mean Ergodic Operators

for every n, which clearly will give the conclusion. Suppose that this does not hold; then there are .a ∈ K and .m ∈ N so that, if .α = ϕ m (a), then .|α| > s. Take |α| .1 < R < s and define the function .g : D → C by g(z) =

.

Rαz , |α|2

which is clearly holomorphic. Then .g(α) = R and, for .|w| ≤ s, we have |g(w)| =

.

R|αw| Rs < 1, ≤ 2 |α| |α|

  from which we deduce that .g k ϕ m (a) = g(α)k = R k and .sup|w|≤s |g k (w)| < 1 for every k. Now, using (5.27) with .f = g k and .n = m, we have .

  1 k 1  1  R = g k ϕ m (a) ≤ sup g k ϕ m (z) ≤ M sup |g k (w)| < M . m m |w|≤s z∈K m

This gives .R k ≤ Mm for every .k ∈ N, but this contradicts the fact that .R > 1.



We have clarified completely when a composition operator is mean ergodic. Let us see now when is it hypercyclic. Given a holomorphic self-map, .ϕ : D → D, we say that the sequence of iterates .(ϕ n )n runs away if for every compact .K ⊆ D there is some .n ∈ N so that ϕ n (K) ∩ K = ∅ .

.

Proposition 5.18 Let .ϕ : D → D be holomorphic. Then the iterates run away if and only if .ϕ does not have fixed points in .D. Proof Let us suppose that .ϕ does has a fixed point. By Corollary 5.10, it has stable orbits and, given the compact set .K = {0}, there exists .L ⊆ D compact such that n .ϕ (0) ⊆ L for every n. Using again that .ϕ has stable orbits, we can find another compact .M ⊆ D such that .ϕ n (L) ⊆ M for every n. Now, the set L˜ = L ∪

∞ 

.

 ϕ n (L) ⊆ L ∪ M

n=1

˜ ⊆ L˜ for every .n ∈ N. Hence, .(ϕ n )n do not run away. is compact and satisfies .ϕ n (L) This gives one implication. The converse one follows from Proposition 5.17. 

With this, we can describe the hypercyclic composition operators.

5.3 Examples

193

Theorem 5.13 Let .ϕ : D → D be holomorphic. The following statements are equivalent: 1. .ϕ is injective and .(ϕ n )n runs away. 2. .Cϕ : H(D) → H(D) is hypercyclic. 3. .Cϕ : H(D) → H(D) is transitive. Proof The equivalence between 2 and 3 follows from Birkhoff’s Theorem 5.2, since H(D) is Fréchet (Theorem 4.4) and separable (Proposition 4.1). Assume now that .Cϕ is hypercyclic, and let us see that 1 holds. Suppose in first place that .ϕ is not injective. Then we can find .a, b ∈ D with .a = b so that .ϕ(a) = ϕ(b). Choose some .g ∈ H(D) such that .g(a) = g(b) and a hypercyclic vector f of .Cϕ . Since the orbit of f is .τco -dense, there is a subsequence .(nj )j for which .(f ◦ ϕ nj )j converges to g uniformly on compact subsets of .D and, in particular, on every compact set that contains a and b. Since .f (ϕ n (a)) = f (ϕ n (b)) for every n, we have .g(a) = g(b). This is a contradiction and shows that .ϕ is injective. Suppose now that .(ϕ n )n does not run away. Then there is a compact .K ⊆ D and a sequence .(zn )n ⊆ K so that .ϕ n (zn ) ⊆ K. Now, for each .f ∈ H(D), set .M = supz∈K |f (z)| and note that, for every .n ∈ N, we have .

.

  n  inf Cϕ (f )(z) ≤ f ϕ n (zn ) ≤ M ,

z∈K

And this shows that .(f ◦ ϕ n )n cannot approximate the constant function .M + 1 uniformly on K. Hence, .Cϕ cannot have hypercyclic vectors. This completes the proof of 2 .⇒ 1. The most difficult part of the proof is the remaining one: if 1 holds, then .Cϕ ∈ L(H(D)) is transitive. Assume that .ϕ is injective and .(ϕ n )n runs away, and fix two open non-empty sets .U, V ⊆ H(D). Choose .K ⊆ D compact, .f, g ∈ H(D) and .ε > 0 so that  .

   h∈H(D) : sup |h(z)−f (z)| 0 so that .|ak | ≤ RcRk for every k. Now, given some compact set .K ⊆ C, let .r > 0 be such that .K ⊆ D(0, r), and fix .R > r. Note that, given .z ∈ K, the power series converges uniformly on .[0, z] and, then,  Jf (z) =

∞ 

.

[0,z] k=0

ak w k dw =

∞ 

 ak

k=0

[0,z]

=

w k dw ∞ ∞   ak k+1 ak k  z z =z k+1 k+1 k=0

and, iterating this, J n f (z) = zn

∞ 

.

k=0

 ak zk , (k + 1) · · · (k + n)

k=0

196

5 Transitive and Mean Ergodic Operators

for every .n ∈ N. Then |J n f (z)| ≤ |zn |

∞ 

.

k=0

∞

|ak | |zn |  |ak ||zk | |zk | ≤ (k + 1) · · · (k + n) n! k=0

≤

∞ rn 

n!

k=0

cR k r n   r k r = cR . k R n! R ∞

k=0

The latter tends to 0 as .n → ∞ and does not depend on z. This yields the claim and 

completes the proof. With a similar argument, it can be shown that the integration operator .J : H(D) → H(D) is power bounded, uniformly mean ergodic but not transitive.

5.3.4 Differential Operators Let . ⊆ C be open. Let us recall that the differentiation operator .D : H() → H() is defined by .D(f ) = f . An obvious modification of Example 4.17 ≥ 1 and a polynomial mshowskthat this operator is continuous. Given .m  m k .P (z) = a z , we consider the operator . P (D) = k k=0 k=0 ak D . That is, .P (D) : H() → H() is given by P (D)(f ) =

m 

.

ak f (k) .

k=0

Our aim now is to study whether or not is this operator hypercyclic, chaotic, power bounded, or uniformly mean ergodic. For the remainder of this section, we establish the following notation: for .λ ∈ C, we consider the function .eλ ∈ H(C) given by λz .eλ (z) = e . Note that, for each such function (and every operator .P (D)), we have P (D)(eλ ) =

m 

.

k=0

(k) ak eλ

=

m 

ak λk eλ = P (λ)eλ .

(5.30)

k=0

Lemma 5.6 If . ⊆ C has an accumulation point, then .span{eλ : λ ∈ } is dense in .H(C). Proof Let us recall from Theorem 4.12 that, if .μ ∈ H(C) , then its Fourier–Borel transform . μ given by . μ(z) = μw (ezw ) = μ(ez ) is a holomorphic function of exponential type. Suppose now that .μ ∈ H(C) is such that .μ(ez ) = 0 for every .λ ∈ . Then . μ(λ) = 0 for every .λ ∈ , and since . μ is entire and . has an μ = 0 and, accumulation point, the identity principle (Theorem 4.7) yields that . therefore, .μ = 0. Corollary 3.3 gives the claim. 

5.3 Examples

197

We give our first step by describing the dynamical properties of the differential operators defined on the space of entire functions. m k Theorem 5.14 Let .P (z) = k=0 ak z be a non-constant polynomial of degree .m ≥ 1. The operator .P (D) : H(C) → H(C) is hypercyclic and chaotic, and neither power bounded, nor mean ergodic. Proof Let us note in first place that the set .{λ ∈ C : |P (λ)| < 1} is open and nonempty and hence has an accumulation point. By Lemma 5.6, .{eλ : |P (λ)| < 1} is dense in .H(C). With this and (5.30), we have that E0 = span{f ∈ H(C) : P (D)f = μf for some μ ∈ C, |μ| < 1}

.

is dense in .H(C). A similar argument shows that F0 = span{f ∈ H(C) : P (D)f = μf for some μ ∈ C, |μ| > 1}

.

is also dense and, by Proposition 5.3, the operator .P (D) is hypercyclic (hence not power bounded; see Remark 5.4). Let us see now that the operator is also chaotic. By Proposition 5.7, it is enough to show that G0 = span{f ∈ H(C) : there exists μ ∈ C with μn = 1

.

for some n ∈ N so that P (D)f = μf } is dense in .H(C). Note that, by (5.30), the set {eλ : P (λ)n = 1 for some n}

.

(5.31)

is contained in .G0 . Note now that for every .w ∈ C with .|w| = 1, there is some λ with .P (λ) = w. Also, the set .{λ ∈ C : P (λ)n = 1 for some n} consists of preimages through P of roots of unity, and since these are dense in .T (because .eqπ is a root of 1 for every .q ∈ Q and .Q is dense in .R), this set has an accumulation point. Lemma 5.6 gives that the set in (5.31) is dense in .H(C) and then so also is .G0 . This yields the claim. Finally, take .λ0 such that .P (λ0 ) = 2 to have, by (5.30), that .P (D)eλ0 = 2eλ0 . 

Remark 5.3 gives that .P (D) is not mean ergodic.

.

We move now to operators defined on spaces of holomorphic functions on some open set. Theorem 5.15 (Shapiro) Let . ⊆ C be open and connected, and .P (z) = m k be a non-constant polynomial of degree .m ≥ 1. The following statements a z k=0 k are equivalent: 1. .P (D) : H() → H() is chaotic. 2. .P (D) : H() → H() is hypercyclic. 3. . is simply connected.

198

5 Transitive and Mean Ergodic Operators

Proof If the operator is chaotic, it is transitive, and then, by Birkhoff’s Theorem 5.2, hypercyclic. Let us suppose now that . is not simply connected. Then we can find 1 .a ∈ C \  and a closed piecewise .C curve .γ contained in . so that  .

γ

1 dz = 0 . z−a

(5.32)

 We define now .u : H() → C by .u(f ) = γ f (z)dz. It is easy to check that .u ∈ H() and (5.32) gives .u = 0. Moreover, .u(f ) = 0 for every .f ∈ H(). We now write the polynomial as P (z) = P (0) +

m 

.

ak zk

k=1

and, for each .f ∈ H(), we have t P (D)u, f  = u, P (D)f  = u, P (0)f  +

m 

.

ak u, f (k)  = P (0)u(f ) .

k=1

Hence, .t P (D)u = P (0)u, and .u = 0 is an eigenvector of .t P (D), which, by Proposition 5.5, cannot be transitive. Remark 5.1 gives that it cannot be hypercyclic. Let us finally assume that . is simply connected. The polynomials are dense in .H() (see, e.g., [9, Theorem 13.11] for a proof of this fact) and, therefore, .H(C) is dense in .H(). From Theorem 5.14, we know that .P (D) : H(C) → H(C) is chaotic. Corollary 5.5 yields 1 and completes the proof. 

5.4 Exercises 5.1 Let T be an invertible transitive operator on a locally convex space. Show that T −1 ∈ L(E) is also transitive. 5.2 Let H ⊆ L(E) be equicontinuous. Show that there is a fundamental system of seminorms P defining the topology of E such that p(T x) ≤ p(x)

.

for every T ∈ H , every x ∈ E, and p ∈ P. Hint: Given a fundamental system of seminorms Q of E, we know that for each q ∈ Q there are q ∈ Q and C > 0 such that q(T x) ≤ Cq (x) for every T ∈ H and x ∈ E. Set, for a given q ∈ Q,   q(x) ˜ = max q(x), sup q(T x) .

.

T ∈H

5.4 Exercises

199

This defines a continuous seminorm on E that satisfies q ≤ q˜ ≤ Cq . Taking P = {q˜ : q ∈ Q} gives the conclusion. 5.3 The resolvent ρ(T ) of an operator T ∈ L(E) is defined as ρ(T ) = {λ ∈ C : λI − T is invertible} ,

.

  and the spectrum as σ (T ) = C \ ρ(T ). Prove that, if T is such that n1 T n n is equicontinuous in L(E), then σ (T ) ⊆ {λ ∈ C : |λ| ≤ 1}. Hint: By the previous exercise, there is a fundamental system of seminorms P such that p(T n x) ≤ np(x) for every x ∈ E and n ∈ N. If λ ∈ C satisfies |λ| > 1, then ∞ T nx the series n=0 λn converges in E and so defines a continuous operator R, since ∞ n Now (λI − T )R = R(λI − T ) = λI , as can be easily seen. The n=1  λn < ∞. ∞ Tn series n=0 λn is called the von Neumann series. 5.4 Prove that, if T ∈ L(E) is power bounded or mean ergodic on a Fréchet space, then σ (T ) ⊆ {λ ∈ C : |λ| ≤ 1}.   Hint: Recall that if E is a Fréchet space and n1 T n → 0 in Ls (E), then n1 T n n is equicontinuous. 5.5 Show that the operator T : c0 → c0 given by T (x1 , x2 , x3 , . . .) = (x1 , x1 , x2 , x3 , . . .) is power bounded but not mean ergodic. 5.6 Show that the operator T : 1 → 1 given by T (x1 , x2 , x3 , . . .) = (0, x1 , x2 , x3 , . . .) is power bounded but not mean ergodic. 5.7 Let E and F be Hausdorff locally convex spaces. Assume that every β(F , F )-bounded subset of F is equicontinuous. Show that the linear map φ : Lβ (E, F ) → Lβ (Fβ , Eβ ) defined as φ(T ) = t T is continuous. Prove that, as a consequence, if E satisfies that every bounded subset of Eβ is equicontinuous and Tn → T in Lβ (E), then t Tn → t T in Lβ (Eβ ). 5.8 Let E be a Hausdorff locally convex space and T ∈ L(E). Prove: 1. If T is uniformly mean ergodic, then t T ∈ L(Eβ ) is mean ergodic. 2. If every bounded subset of Eβ is equicontinuous and T is uniformly mean ergodic, then t T ∈ L(Eβ ) is uniformly mean ergodic. 5.9 A linear operator T : E → F between Hausdorff locally convex spaces is called compact if there is a neighbourhood U in E such that T (U ) is relatively compact in F : 1. Let E be a complex Banach space. Show that if T ∈ L(E) is compact, then T cannot be hypercyclic. Hint: The proof that we suggest relies on several results on spectral theory. The transpose t T : E → E is also compact. By Proposition 5.5, t T has no eigenvalues. By Riesz’s spectral theory for compact operators, the spectrum σ (t T ) reduces to {0} (see, e.g., [4, Theorem 7.7.1]). Then σ (T ) = σ (t T ) = {0}.

200

5 Transitive and Mean Ergodic Operators

By the spectral radius formula ([4, Proposition 7.3.8]), limn T n  = 0. This implies that T is power bounded, which contradicts Remark 5.4. 2. Show that no compact operator on a Hausdorff locally convex space is hypercyclic. Hint: Let T ∈ L(E) be compact and U be a closed absolutely convex set so that T (U ) is relatively compact. Let pU be the Minkowski functional of U and EU the normed space (E/ ker pU , p U ), where p U (x + ker U ) = pU (x). The map T˜ : EU → EU defined as T˜ (x + ker pU ) = T x + pU (x) is well defined and continuous. Denote also by T˜ the extension to the completion. Show that if T is hypercyclic, then so also is T˜ . 5.10 Let A be a Köthe matrix such that a1 (i) > 0 for every i ∈ N. Let 1 ≤ p ≤ ∞ or p = 0. For λ = (λ1 , λ2 , . . .), define the diagonal operator by Dλ x = (λ1 x1 , λ2 x2 , . . .): 1. 2. 3. 4. 5. 6. 7.

When is Dλ continuous on λp (A)? Show that Dλ is never hypercyclic on λp (A). Can Dλ have periodic points other than (0, 0, 0, . . .)? When is Dλ power bounded on λp (A)? When is Dλ mean ergodic on λp (A)? When is Dλ uniformly mean ergodic on λp (A)? When is Im(I − Dλ ) closed in λp (A)?

5.11 Consider the Banach space C[0, 1] of all complex continuous functions on [0, 1]. For each f ∈ C[0, 1], define the Volterra integration operator 

x

Vf (x) =

f (t)dt

.

0

for 0 ≤ x ≤ 1. 1. Show that V is continuous and V  ≤ 1. 2. Show by induction that 1 .V f (x) = (n − 1)!



n

x

(x − t)n−1 f (t)dt

0

for every n ∈ N (and 0 ≤ x ≤ 1). 3. Compute the norm of V n for each n ∈ N. 4. Is V power bounded, hypercyclic, or mean ergodic? 5. Show that σ (V ) = {0}. Further Reading [1, 3, 5–7, 10]

References

201

References 1. Bayart, F., Matheron, E.: Dynamics of Linear Operators. Cambridge Tracts in Mathematics, vol. 179. Cambridge University Press, Cambridge (2009). https://doi.org/10.1017/ CBO9780511581113 2. Bermúdez, T., Kalton, N.J.: The range of operators on von Neumann algebras. Proc. Am. Math. Soc. 130(5), 1447–1455 (2002). https://doi.org/10.1090/S0002-9939-01-06292-X 3. Bracci, F., Contreras, M.D., Díaz-Madrigal, S.: Continuous Semigroups of Holomorphic SelfMaps of the Unit Disc. Springer Monographs in Mathematics. Springer, Cham (2020). https:// doi.org/10.1007/978-3-030-36782-4 4. Conway, J.B.: A Course in Functional Analysis. Graduate Texts in Mathematics, vol. 96. Springer, New York (1985). https://doi.org/10.1007/978-1-4757-3828-5 5. Eisner, T., Farkas, B., Haase, M., Nagel, R.: Operator Theoretic Aspects of Ergodic Theory. Graduate Texts in Mathematics, vol. 272. Springer, Cham (2015). https://doi.org/10.1007/ 978-3-319-16898-2 6. Grosse-Erdmann, K.G., Peris Manguillot, A.: Linear Chaos. Universitext. Springer, London (2011). https://doi.org/10.1007/978-1-4471-2170-1 7. Krengel, U.: Ergodic Theorems. De Gruyter Studies in Mathematics, vol. 6. Walter de Gruyter & Co., Berlin (1985). https://doi.org/10.1515/9783110844641. With a supplement by Antoine Brunel 8. Lotz, H.P.: Uniform convergence of operators on L∞ and similar spaces. Math. Z. 190(2), 207–220 (1985). https://doi.org/10.1007/BF01160459 9. Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Book Co., New York (1987) 10. Shapiro, J.H.: Composition Operators and Classical Function Theory. Universitext: Tracts in Mathematics. Springer, New York (1993). https://doi.org/10.1007/978-1-4612-0887-7

Chapter 6

Schwartz Distributions and Linear Partial Differential Operators

6.1 Test Functions and Distributions 6.1.1 Definition and Examples If . is some open subset of .RN (with .N ∈ N), we want to deal with functions in ∞ .C () that are 0 except on a relatively compact subset of .. To achieve this, we define the support of an arbitrary function .f : RN → K as the set .

supp(f ) = {x ∈ RN : f (x) = 0} .

If .K ⊆ RN is a compact subset, we consider the space of .C ∞ -functions whose support is contained in K; that is, D(K) = {f ∈ C ∞ (RN ) : supp(f ) ⊆ K} .

.

If we endow it the induced topology from .C ∞ (RN ), this is clearly a closed subspace of .C ∞ (RN ). Then, in view of Theorem 4.13, .D(K) is a Fréchet space. Note that the family of seminorms (recall (4.8)), which are in fact norms, defining the topology of .D(K) are   pK,n (f ) = sup sup ∂ α f (x) ,

.

|α|≤n x∈K

for .n ∈ N0 . If K is a subset of the open set . ⊆ RN , then .D(K) is also a closed subspace of .C ∞ () for the induced topology.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6_6

203

204

6 Distributions and Partial Differential Operators

Example 6.1 Consider the function .g : R → R given by  g(x) =

e−1/x

if x > 0

0

if x ≤ 0

.

.

For each n, there is a polynomial .Pn of degree n so that .g (n) (x) = Pn (1/x)e−1/x for every .x > 0. Then it is easy to see that .g ∈ C ∞ (R) and, moreover, .g (n) (0) = 0 for every .n ∈ N. Define now a function .ρ : RN → R by .ρ(x) = Cg(1 − |x|2 ), where .C > 0 is chosen in such a way that . RN ρ(x)dx = 1. Then .ρ ∈ C ∞ (RN ) and .supp ρ = B(0, 1) (that is, .ρ ∈ D(B(0, 1))). Finally, for a given .ε > 0, we define .ρε ∈ C ∞ (RN ) by ρε (x) =

.

1 x  , ρ εN ε

(6.1)

N for  .x ∈ R . This clearly satisfies .supp ρε = B(0, ε), .0 ≤ ρε ≤ 1, and .  RN ρε (x)dx = 1 for each .ε > 0.

Given any two measurable functions .f, g : RN → K, their convolution is defined, for .x ∈ RN , as  .(f ∗ g)(x) = f (y)g(x − y)dy , (6.2) RN

whenever the integral exists. It satisfies the following properties. Proposition 6.1 Let .f, g ∈ L1 (RN ); then: 1. 2. 3. 4.

f ∗ g ∈ L1 (RN ) and .f ∗ gL1 (RN ) ≤ f L1 (RN ) gL1 (RN ) . .f ∗ g = g ∗ f almost everywhere. If .supp f is compact, then .supp(f ∗ g) ⊆ supp f + supp g. If .g ∈ C 1 (RN ) ∩ L1 (RN ), then .f ∗ g ∈ C 1 (RN ) ∩ L1 (RN ), and .

.

∂ ∂g (f ∗ g) = f ∗ , ∂xk ∂xk

for .k = 1, . . . , N . A detailed study of the convolution of functions (including the proof of these properties) can be found, e.g., in [2, pages 53–54] or [12]. If one of the functions is in .L1 , then we know that we can calculate its convolution with any other function in .L1 . However, if our function has ‘better’ properties (namely, it is continuous and has compact support), then we can pick functions in a larger space to convolve it with. Let us be more precise.

6.1 Test Functions and Distributions

205

For an open set . ⊆ RN , a measurable function .f :  → K is said to be locally integrable if .f ∈ L1 (K) for every compact set .K ⊆ . We write .L1,loc () for the space of all locally integrable functions on .. If .f ∈ L1,loc () and .K ⊆  is compact, then clearly .f χK ∈ L1 (). If .ϕ is continuous and has compact support (that is, .supp ϕ ⊆ RN is compact), and .f ∈ L1,loc (RN ), then we have  (ϕ ∗ f )(x) =



.

RN

ϕ(y)f (x − y)dy =

ϕ(y)f (x − y)dy . supp ϕ

In particular, if .ϕ ∈ D(K) for some compact K, then we can convolute it with any f ∈ L1,loc (RN ). Given a set .A ⊆ RN , it is convenient to have a function that takes value 1 on A and 0 elsewhere. Obviously, the characteristic function does this job, but it has the drawback that it is not continuous. If we go down to one variable, it is very easy to see that, given .A < B, one may construct a function that is 1 on .[−A, A] and 0 on N .R \ [−B, B]. It is slightly more involved (but still easy) to see that, if .K,  ⊆ R are, respectively, compact and open, and .K ⊆ , then a continuous function can be constructed taking value 1 on K and 0 outside .. Or next result shows that, in fact, the function can be taken to be .C ∞ . .

Proposition 6.2 Let .  RN be open, and .K ⊆  compact. Then there exists ∞ N .ψ ∈ C (R ) such that: 1. .supp ψ is a compact subset of .. 2. .0 ≤ ψ(x) ≤ 1 for every .x ∈ RN . 3. .ψ(x) = 1 for every .x ∈ K. Proof Select .δ > 0 such that .2δ < d(K, RN \ ) and set L = K + B(0, δ) = {x ∈ RN : d(x, K) ≤ δ} ⊆  ,

.

which is again compact. By Example 6.1, we can find .ϕ ∈ D(B(0, δ)) satisfying  0 ≤ ϕ ≤ 1 and . RN ϕ(x)dx = 1. We then define .ψ = ϕ ∗ χL . To begin with, since .ϕ ∈ C ∞ (RN ), Proposition 6.1 gives .ψ ∈ C ∞ (RN ) and (since .ϕ has compact support) .supp ψ ⊆ K + supp ϕ ⊆ L and, then, it is also compact. Finally, since .ϕ ≥ 0 and .0 ≤ χL ≤ 1, we have

.

 0 ≤ ψ(x) =



.

RN

ϕ(y)χL (x − y)dy ≤

RN

ϕ(y)dy ≤ 1 ,

206

6 Distributions and Partial Differential Operators

for every .x ∈ RN . On the other hand, if .x ∈ K, note that .x − y ∈ K + B(0, δ) ⊆ L for each .y ∈ supp ϕ. Therefore,  ψ(x) = (ϕ ∗ χL )(x) =

ϕ(y)χL (x − y)dy

.

RN



 ϕ(y)χL (x − y)dy =

= supp ϕ

ϕ(y)dy = 1 . supp ϕ

 Functions that are .C ∞ and have compact support are called test functions. If N is open and not empty, we denote by .D() the space of all test functions . ⊆ R belonging to .C ∞ () (that is, whose support is compact in .). Note that D() =

.

 {D(K) : K ⊆ , K compact in } .

Note that if .(Kn )n is a fundamental sequence of compact subsets in ., then D() =

∞ 

.

D(Kn ) .

n=1

Proposition 6.3 Let . ⊆ RN be open. Then the space .D() is dense in .C ∞ (). Proof Let .(Km )m a fundamental sequence of compact subsets of . such that ◦ ∞ .Km ⊆K m+1 for each .m ∈ N and . = m=1 Km . We apply Proposition 6.2 to find, for each m, a test function .ϕm ∈ D() such that .ϕm (x) = 1 for each .x ∈ Km , ◦

supp ϕm ⊆K m+1 , and .0 ≤ ϕm ≤ 1. Given .f ∈ C ∞ (), we consider the sequence .(ϕm f )m ⊆ D(). Let us see that this sequence converges to f in .C ∞ () (recall in (4.8) the seminorms defining the topology). Note that .ϕm f = f on .Km for every m. Now, for each fixed .n ∈ N, if

.

◦

m ≥ n + 1, we have .Kn ⊆K n+1 ⊆ Km , and then

.

pn (ϕm f − f ) = sup sup |∂ α (ϕm f − f )(x)| = 0 .

.

|α|≤n x∈Kn

This gives our claim and completes the proof. be open proper subsets of .RN

m



Proposition 6.4 Let .1 , . . . , m and .K ⊆ j =1 j be compact. Then there are .ϕ1 , . . . , ϕm with .ϕj ∈ D(j ) for .j = 1, . . . , m, such that: 1. .0 ≤ ϕj (x) ≤ 1 for every .x ∈ RN and .j = 1, . . . , m. m

2. . ϕj (x) = 1 for every .x ∈ K. j =1

6.1 Test Functions and Distributions

207

The family .{ϕ1 , . . . , ϕm } is often called a partition of unity on .1 , . . . , m . Proof Given .x ∈ K, there is some .j ∈ {1, . . . , m} such that .x ∈ j , and since this set is open, we can find .r(x) > 0 with .B(x, r(x)) ⊆ j . This builds an open cover of K, which is compact. Then there are .x1 , . . . , xs ∈ K such that K ⊆ B(x1 , r(x1 )) ∪ · · · ∪ B(xs , r(xs )) .

.

Now, for each .i = 1, . . . , s, there is .j = j (i) ∈ {1, . . . , m} such that .B(xi , r(xi )) ⊆ j (i) . Then we define Kj =



.

B(xi , r(xi )) : j (i) = j ,

which is clearly compact and .Kj ⊆ j . Note also that .K ⊆ m j =1 Kj . Now, for each j , we apply Proposition 6.2 to find some .ψj ∈ C ∞ (RN ) with compact support contained in .j (i.e., .ψj ∈ D(j )) with .0 ≤ ψj ≤ 1 on .RN and .ψj (x) = 1 for every .x ∈ Kj . Now we define ϕ1 =ψ1

.

ϕk =(1 − ψ1 ) · · · (1 − ψk−1 )ψk for k = 2, . . . , m . Clearly, .supp ϕj ⊆ supp ψj ⊆ Kj ⊆ j and .0 ≤ ϕj ≤ 1 for each .j = 1, . . . , m. Note also that ϕ1 + ϕ2 = ψ1 + (1 − ψ1 )ψ2 = ψ1 + ψ2 − ψ1 ψ2 = 1 − (1 − ψ1 )(1 − ψ2 )

.

and, with the same argument, ϕ1 + · · · + ϕk = 1 − (1 − ψ1 ) · · · (1 − ψk ) ,

.

for each .k = 2, . . . , m. If .x ∈ K, then .x ∈ Kj for some .1 ≤ j ≤ m and, so, ψj (x) = 1. Hence, .ϕ1 (x) + · · · + ϕk (x) = 1 for each .j ≤ k ≤ m. This gives 2 and completes the proof. 

.

A distribution on an open subset . of .RN is a linear map .T : D() → K such that for each compact .K ⊆  there are .c > 0 and .n ∈ N0 so that |T (ϕ)| ≤ c sup sup |∂ α ϕ(x)| = c pK,n (ϕ),

.

|α|≤n x∈K

(6.3)

for every .ϕ ∈ D(K). The space of all distributions on . is denoted by .D  (). Observe that, if T is linear, then .T ∈ D  () if and only if, for each .K ⊆  compact, the restriction .T |D (K) of T to .D(K) is continuous on .D(K).

208

6 Distributions and Partial Differential Operators

We give now some examples of distributions that will play a fundamental rôle for now on. Example 6.2 The Dirac ‘delta’ function .δ : D(RN ) → K, defined by .δ(ϕ) = ϕ(0), is clearly linear. Also, for each compact .K ⊆ RN and .ϕ ∈ D(K), we have |δ(ϕ)| = |ϕ(0)| ≤ sup |ϕ(x)| = pK,0 (ϕ) .

.

x∈K

Then .δ ∈ D  (RN ) (that is, .δ is a distribution). Note that .ϕ(0) = 0 whenever .0 ∈ K. With the same argument, for each fixed .a ∈ RN , the linear map .δa : D(RN ) → K, defined by .δa (ϕ) = ϕ(a), is a distribution.  Example 6.3 If . ⊆ RN is open, then each locally integrable function on . defines a distribution in the following way. Given .f ∈ L1,loc (), we define the map .Tf : D() → K as   .Tf (ϕ) = f (x)ϕ(x)dx = f (x)ϕ(x)dx . (6.4) RN

supp(ϕ)

This is linear and, for each compact subset .K ⊆  and every .ϕ ∈ D(K), we get         . Tf (ϕ) =  f (x)ϕ(x)dx  ≤ |f (x)|dx sup |ϕ(x)| . K

K

x∈K

Hence, .Tf ∈ D  ().



Example 6.4 The Heaviside step function, defined as  H (x) =

.

0 if x < 0 1 if x ≥ 0,

is a locally integrable function (which is not in .L1 (R)). We denote also by H the associated distribution .H ∈ D  (R), that is,  H (ϕ) =



.

R

ϕ(x)H (x)dx =

∞

ϕ(x)dx , 0

for each .ϕ ∈ D(R). For .a ∈ R, the translated Heaviside function is defined as Ha (x) = H (x − a) for .x ∈ R. It is locally integrable and, therefore, defines a distribution. 

.

Example 6.5 Define a mapping .T : D(R) −→ K by T (ϕ) =

.

∞ 

  ϕ  n12 − ϕ  (0) . n=1

6.1 Test Functions and Distributions

209

Let us see that it is well defined and that .T ∈ D  (R). Once it is well defined, it is straightforward to check that it is linear. Take .A > 0 and .ϕ ∈ D([−A, A]). For each .n ∈ N, there is .cn ∈]0, 1/n2 [ such that ϕ



.

1 n2



− ϕ  (0) =

1  ϕ (cn ). n2

This gives |T (ϕ)| ≤

∞

|ϕ  (1/n2 ) − ϕ  (0)| =

.

n=1

∞

1  |ϕ (cn )| n2 n=1

≤



∞

n=1

1 π2 sup |ϕ  (x)| ≤ 2 6 n x∈[−A,A]

sup

|ϕ  (x)| ,

x∈[−A,A]

which (recall (6.3)) shows that .T ∈ D  (R).



Example 6.6 We define the Cauchy principal value of D(R) → K given by  PV

.

1 x (ϕ)

= lim

ε→0+ |x|>ε

1 . x

as the mapping .P V

1 x

:

ϕ(x) dx . x

We show that .P V x1 is well defined and that it belongs to .D  (R). Fix .A > 0 and take some .ϕ ∈ D([−A, A]). Since .ϕ(A) = ϕ(−A) = 0 (because .supp ϕ ⊆ [−A, A]), we get, integrating by parts PV

.



  −ε ϕ(x) ϕ(x) dx + dx = lim ε→0+ ε 0, select .m ∈ N such that .A ≤ m. For each .ϕ ∈ D([−A, A]), we have |T (ϕ)| ≤

m

.

|ϕ (n) (n)| ≤ m sup

sup

|ϕ (k) (x)|,

k≤m x∈[−A,A]

n=1

and .T ∈ D  (R). However, this distribution is not of finite order. Otherwise, there is .k ∈ N such that for each compact set .K ⊆ R there is .c > 0 with |T (f )| ≤ c sup sup |f (n) (x)| ,

.

n≤k x∈K

for every .f ∈ D(K). Take .ψ ∈ D([−1, 1]) with .ψ (k+1) (0) = 1 and define, for each .ε > 0, the function .ϕε (x) = ψ( x−k−1 ). If .ε < 1, we have ε .

supp ϕε ⊆ [k + 1 − ε, k + 1 + ε] ⊆ [k, k + 2],

and T (ϕε ) =

∞

.

ϕε(n) (n) = ϕε(k+1) (k + 1) =

n=0

1 εk+1

.

On the other hand, for each .n = 1, . . . , k, we get .

sup x∈[k,k+2]

|ϕε(n) (x)| ≤

C1 1 sup |ψ (n) (y)| ≤ k , εn y∈R ε

for some .C1 > 0 independent of .ε ∈]0, 1[. Hence, .

1 CC1 = |T (ϕε )| ≤ C sup sup |ϕε(n) (x)| ≤ k , ε εk+1 n≤k x∈L

and .1/ε < CC1 for each .0 < ε < 1. This is a contradiction and yields our claim.



Remark 6.1 The pair .(D(), D  ()) is a dual pair (with the bilinear mapping .T , ϕ = T (ϕ)). Indeed, clearly, if .T ∈ D  () and .T = 0, then there is .ϕ ∈ D() with .T (ϕ) = 0. On the other hand, if .ϕ ∈ D() is non-zero, then   2 .Tϕ (ϕ) = RN ϕ(x)ϕ(x)dx = RN |ϕ(x)| dx > 0.

6.1.2 Differentiation of Distributions Our aim now is to define the derivative of a distribution. We would expect to do it in such a way that, when we consider the distribution defined by a differentiable

6.1 Test Functions and Distributions

213

function of one variable, its derivative coincides with the distribution defined by the derivative of the function. That is, if .f ∈ C 1 (R), we have that .f  ∈ C(R) ⊆  L1,loc (R), and what we want is to define a derivative in such a way that . Tf = Tf  . If .ϕ ∈ D(R) is so that .supp ϕ ⊆ [−A, A] (note that then .ϕ(A) = ϕ(−A) = 0), we can integrate by parts to have  Tf  ϕ =

.

f  (x)ϕ(x)dx =

 [−A,A]

f  (x)ϕ(x)dx



=−





[−A,A]

f (x)ϕ (x)dx = −

f (x)ϕ  (x)dx .

This suggests to define the derivative of .Tf as .(Tf ) (ϕ) = −Tf (ϕ  ) for .ϕ ∈ D(R). We generalize this to arbitrary distributions. If . ⊆ RN is an open set and .α ∈ NN 0 , then the .αth derivative of a distribution α .T ∈ D() is defined as the mapping .∂ T : D() −→ K given by ∂ α T (ϕ) = (−1)|α| T (∂ α ϕ) .

.

The first natural question we have to answer is if .∂ α T defines a distribution. It is clearly linear, so we only have to check that (6.3) holds or, equivalently, that for each compact .K ⊆  the restriction .∂ α T |D (K) : D(K) → K is continuous. We know that the operator .∂ α : C ∞ () → C ∞ () is continuous. If K is compact and .supp ϕ ⊆ K, then .supp ∂ α ϕ ⊆ K. This (since the topology in .D(K) is induced by .C ∞ ()) implies that the operator .∂ α : D(K) → D(K) is well defined and continuous. Then we can see .∂ α T |D (K) as the composition of two continuous operators (−1)|α| T

∂α

∂ α T |D (K) : D(K) −→ D(K) −−−−−→ K

.

and it is continuous. We compute now the derivative of some of the basic distributions. Example 6.8 1. The derivative of the Dirac function .δa : D(R) → K is easily computed: δa (ϕ) = −ϕ  (a) .

.

2. For the Heaviside function, we have, for each .ϕ ∈ D(R) with .supp ϕ ⊆ [−A, A], (TH ) (ϕ) = −

.



H (x)ϕ  (x)dx = −



A 0

ϕ  (x)dx = ϕ(0) = δ(ϕ) .

214

6 Distributions and Partial Differential Operators

That is, H = δ .

.

(6.6)

Similarly, we have .Ha = δa . 3. We look finally at the derivative of Cauchy’s principal value of . x1 . If .ϕ ∈ D(R), we have (recall (6.5)) 1 (Tlog |x| ) (ϕ) = −Tlog |x| (ϕ  ) = P V (ϕ) . x

.

Hence, .(Tlog |x| ) = P V x1 .



6.1.3 Multiplication of a Distribution by a C ∞ -Function What we want to do now is to define a product of distributions and .C ∞ -functions. As we did before, we check first how it should work for distributions defined by locally integrable functions of one variable. If .f ∈ L1,loc (R) and .g ∈ C ∞ (R), then .gf ∈ L1,loc (R), and what we want is to have that .gTf = Tfg . But this is  Tfg (ϕ) =

.

 f (x)g(x)ϕ(x)dx =

  f (x) gϕ (x)dx = Tf (gϕ) .

So we define .gTf (ϕ) = Tf (gϕ), and we want to follow this idea for arbitrary distributions. Given .g ∈ C ∞ (), note that for every .ϕ ∈ D(), the product is obviously .C ∞ and has compact support, that is, .gϕ ∈ D(). With this in mind, given .T ∈ D  (), we define the mapping .gT : D() → K by  .

 gT (ϕ) = T (gϕ) ,

(6.7)

which is clearly linear. To see that it is a distribution, we have to check that, for each compact set .K ⊆ , the restriction to .D(K) is continuous. To do that we have to see first that multiplying test functions and .C ∞ -functions is continuous. For multiindices .α, β ∈ Nn0 , we write .β ≤ α whenever .βk ≤ αk for .k = 1, . . . , n. Also α  α! .α! = α1 ! · · · αn ! and . β = β!(α−β)! . Lemma 6.2 Let .g ∈ C ∞ (); then the function .Mg : C ∞ () −→ C ∞ () given by .Mg (f ) = fg is continuous.

6.1 Test Functions and Distributions

215

Proof First of all, by Leibniz rule, we have .∂ α (f g) = for each .K ⊆  compact and each .n ∈ N, we have



α  β β≤α β ∂ f

·∂ α−β g. Then,

α  . sup sup |∂ (f g)(x)| ≤ sup sup |∂ β f ∂ α−β g| β x∈K |α|≤n x∈K |α|≤n α

 ≤

sup

α 

|α|≤n β≤α

β

β≤α



sup sup |∂ g(x)| γ

x∈K |γ |≤n

 sup sup |∂ f (x)| . η

x∈K |η|≤n

    supx∈K sup|γ |≤n |∂ γ g(x)| , this shows that .Mg Taking .c = sup|α|≤n β≤α βα is continuous.  We can now see that the linear mapping gT defined in (6.7) is actually a distribution. Note first of all that if K is compact and .supp ϕ ⊆ K, then .supp gϕ ⊆ K. So, by Lemma 6.2, the mapping .Mg |D (K) : D(K) −→ D(K) is continuous. Therefore, .(gT )|D (K) = T |D (K) ◦ Mg |D (K) is also continuous. Since this holds for every compact .K ⊆ , we have that gT is a distribution. Example 6.9 Let us see some examples of products of functions and distributions. To begin with, if .g ∈ C ∞ (R), then, for each .ϕ ∈ R, we have (gδ)(ϕ) = δ(gϕ) = g(0)ϕ(0) = g(0)δ(ϕ) ,

.

which gives .g(x)δ = g(0)δ. In particular, .xδ = 0. The same argument yields .gδa = g(a)δa . Let us multiply the function .g(x) = x with the distribution .δ  (the derivative of the Dirac function). If .ϕ ∈ R, then (xδ  )(ϕ) = δ  (xϕ) = −δ((xϕ) ) = −δ(ϕ + xϕ  ) = −δ(ϕ) − (xδ)(ϕ  ) = −δ(ϕ) .

.

Hence, .xδ  = −δ. The same argument shows that .δa = (a − x)δa for each fixed .a ∈ R. Finally, taking Cauchy’s principal value of . x1 , we have   1 1 )(xϕ) = lim . x(P V ) (ϕ) = (P V x ε→0+ x  = lim



1 x ϕ(x)dx x ε 0 so that |S(f )| ≤ C sup sup |∂ α f (x)| ,

.

|α|≤m x∈L

for every .f ∈ C ∞ (). If .ϕ ∈ D() is such that .supp ϕ ⊆ \L, we have .S(ϕ) = 0. This implies .supp T ⊆ L, and then T has compact support. Note also that, since .D() is dense in .C ∞ () (recall Proposition 6.3), .S ∈ ∞ C () is uniquely determined by its restriction to .D(). Suppose now that .T ∈ D  () has compact support .K = supp T . Take some relatively compact .V ⊆  such that .K ⊆ V ⊆ V ⊆ , and use Proposition 6.2 to find a function .ψ ∈ D() with .ψ(x) = 1 for each .x ∈ V . We define .S : C ∞ () → K by .S(f ) = T (ψf ). We first prove that .S ∈ C ∞ () . It is clearly linear, so it is only left to see that it is continuous. The set .L = supp ψ is compact in .; then (see (6.3)), there are .C > 0 and .m ∈ N with |T (ϕ)| ≤ C sup sup |∂ α ϕ(x)| ,

.

|α|≤m x∈L

for every .ϕ ∈ D(L). If .f ∈ C ∞ (), then clearly .ψf ∈ D(L), and |S(f )| ≤ sup sup |∂ α (ψf )(x)| .

.

|α|≤m x∈L

6.2 The Space of Rapidly Decreasing Functions

219

Now, using Leibniz formula, we can find .D > 0 such that |S(f )| ≤ D sup sup |∂ α f (x)|,

.

|α|≤m x∈L

for every .f ∈ C ∞ (). This shows that .S : C ∞ () → K is continuous. Finally, we show that T is the restriction of S to .D(). Given .ϕ ∈ D(), the function .ϕ − ψϕ ∈ D() vanishes on V . Thus .supp(ϕ − ψϕ) ⊆  \ V ⊆  \ K and, hence, .ϕ − ψϕ ∈ D( \ K). Since T vanishes in . \ K, .T (ϕ − ψϕ) = 0, which gives .S(ϕ) = T (ψϕ) = T (ϕ). 

6.2 The Space of Rapidly Decreasing Functions So far we have been working with the space .C ∞ (RN ) of all .C ∞ -functions, and its subspace .D(RN ) of functions with compact support. We are going to consider now functions that, although not having compact support (and maybe even being nonzero everywhere), decrease to 0 very fast as x grows. Let us give a precise meaning to this idea and say that a function .f ∈ C ∞ (RN ) is rapidly decreasing if pk (f ) = sup sup (1 + |x|2 )k |∂ α f (x)| < +∞ ,

.

|α|≤k x∈RN

(6.9)

for each .k ∈ N. The space of all rapidly decreasing functions (also known as the Schwartz space) is denoted by .S(RN ) and is endowed with the metrizable locally convex topology defined by the increasing system of seminorms .(pk )k given in (6.9). Let us point out that if .f ∈ S(RN ), then for each .k ∈ N there is .Ck > 0 such that 1 α for every x and all .|α| ≤ k. In particular, .|∂ f (x)| ≤ Ck (1+|x|2 )k/2 .

lim ∂ α f (x) = 0 ,

|x|→∞

(6.10)

for every .α. Remark 6.3 Every test function is clearly rapidly decreasing. In other words, we have .D(RN ) ⊆ S(RN ). This inclusion is strict, since the function .f ∈ C ∞ (RN ) 2 given by .f (x) = e−π |x| clearly belongs to .S(RN ) but does not have compact support. Remark 6.4 For each .α, β ∈ NN 0 , we consider qα,β (f ) = sup |x β ∂ α f (x)| .

.

x∈RN

220

6 Distributions and Partial Differential Operators

It is a simple exercise to show that the family of seminorms .(qα,β )α,β is equivalent to .(pk )k in (6.9) and, therefore, also defines the topology on .S(RN ). As a straightforward consequence, we have the following. Proposition 6.6 A sequence .(fj )j ⊆ S(RN ) converges to zero in .S(RN ) if and only if, for each pair of multi-indices .α, β ∈ NN 0 , we have .

lim x β ∂ α fj (x) = 0

j →∞

uniformly on .RN . Proposition 6.7 For each compact .K ⊆ RN , the inclusions .D(K) ⊆ S(RN ) ⊆ C ∞ (RN ) are continuous. Moreover, .D(RN ) is dense in .S(RN ), and .S(RN ) is dense in .C ∞ (RN ). Proof Fix some compact set .K ⊆ RN . For each .k ∈ N0 and every .f ∈ D(K), we have .

sup sup (1 + |x|2 )k |∂ α f (x)| ≤ max(1 + |x|2 )k sup sup |∂ α f (x)| . x∈K

|α|≤k x∈RN

|α|≤k x∈K

This shows that the inclusion .D(K) ⊆ S(RN ) is continuous. To show the continuity of the inclusion .S(RN ) ⊆ C ∞ (RN ), we fix .K ⊆ RN compact and .k ∈ N. Since .(1 + |x|2 )k ≥ 1 for each .x ∈ K, we have .

sup sup |∂ α f (x)| ≤ sup sup (1 + |x|2 )k |∂ α f (x)| ,

|α|≤k x∈K

|α|≤k x∈RN

for every .f ∈ S(RN ). The fact that .S(RN ) is dense in .C ∞ (RN ) is an immediate consequence of Remark 6.3 and Proposition 6.3. It is only left, then, to see that .D(RN ) is dense in .S(RN ). To this end, we choose .ϕ ∈ D(RN ) such that .ϕ(x) = 1 on an open neighbourhood of .B(0, 1). For each .j ∈ N, we consider the function N .ϕj (x) = ϕ(x/j ). Clearly, .ϕj ∈ D(R ) and .ϕj (x) = 1 for every x in some open neighbourhood of .B(0, j ). Given .f ∈ S(RN ), we set .fj = ϕj f ∈ D(RN ). Let us see that .(fj )j converges to f in .S(RN ). First of all, for each .x ∈ RN , using Leibniz formula, we have (1 + |x|2 )k |∂ α (fj − f )(x)|

α  1 |∂ β (ϕj (x) − 1)| sup (1 + |y|2 )k+1 |∂ α−β f (y)| . ≤ β 1 + |x|2 y∈RN

.

β≤α

Note now that .|∂ β (ϕ(x) − 1)| ≤ supy∈RN |∂ β ϕ(y)| for every .β = 0, and recall that .ϕj (x) = 1 in some open neighbourhood of .B(0, j ). With this, we can find .C > 0

6.2 The Space of Rapidly Decreasing Functions

221

such that .

sup (1 + |x|2 )k |∂ α (fj − f )(x)| x∈RN

≤ C sup

|x|≥j

α  1 sup (1 + |x|2 )k+1 |∂ α−β f (x)| . β x∈RN 1 + |x|2 β≤α

Since .f ∈ S(RN ), we conclude .

lim sup (1 + |x|2 )k |∂ α (fj − f )(x)| = 0,

j →∞ x∈RN

which proves that .D(RN ) is dense in .S(RN ).



Theorem 6.15 The space .S(RN ) is complete, and thus a Fréchet space. Proof Let .(fj )j be a Cauchy sequence in .S(RN ). Since, by Proposition 6.7, the inclusion if S in .C ∞ is continuous, the sequence is also Cauchy in .C ∞ (RN ), which is complete (see Theorem 4.13). Then there is some .f ∈ C ∞ (RN ) such that .(fj )j converges to f in .C ∞ (RN ). Let us see now that, in fact, .f ∈ S(RN ) and .(fj )j converges to f also in .S(RN ). Given .ε > 0, .k ∈ N0 , and .α ∈ NN 0 , there is .j0 ∈ N N such that, for .j, l ≥ j0 and every .x ∈ R , we have (1 + |x|2 )k |∂ α fj (x) − ∂ α fl (x)| < ε .

.

Fixing .l = j0 and letting .j → ∞, it follows that .f ∈ S(RN ). Now, for a fixed N .l ≥ j0 and letting .j → ∞, we conclude that .(fj )j converges to f in .S(R ).  Proposition 6.8 The space .S(RN ) is contained in .L1 (RN ) with continuous inclusion. Proof First note that  .

RN

1 dx = cN (1 + |x|2 )k



∞ 0

r N −1 dr < ∞ , (1 + r 2 )k

if .N − 1 − 2k < −1. Then, for .k > N/2 and .f ∈ S(RN ), we get  f L1 =

.

RN

|f (x)|dx ≤ Cpk (f ) ,

for some .C > 0. This completes the proof.

(6.11) 

The following two results are straightforward consequences of the definitions, and their proof is left to the reader.

222

6 Distributions and Partial Differential Operators

α β Proposition 6.9 Fix .f ∈ S(RN ) and .α, β ∈ NN 0 . Then .∂ f and .x f belong to N .S(R ).

Proposition 6.10 For .a ∈ RN and .λ > 0, the following operators Ta f (x) = f (x − a)

.

Ea f (x) = e2π ia·x f (x) N

Dλ f (x) = λ 2 f (λx) fq(x) = f (−x) are well defined and continuous from .S(RN ) into .S(RN ).

6.3 Fourier Transform on S(RN ) Given a function .f ∈ L1 (RN ), its Fourier transform is defined as (w) = .F(w) = f

 RN

e−2π ix·w f (x)dx ,

for .w ∈ RN . The integral is well defined because .|e−2π ix·w f (x)| = |f (x)| for all N .x, w ∈ R . Proposition 6.11 Let .f, g ∈ L1 (RN ), .t ∈ RN , and .ε > 0. Then: 1. The Fourier transform .f is a bounded continuous function. Moreover, .

sup |f(w)| ≤



w∈RN

RN

|f (x)|dx .

2. The following formula holds  .

RN

3. 4. 5.

f(w)g(w)dw =

 Rn

f (w) g (w)dw .

f ∗ g(w) = f(w) g (w) for every .w ∈ RN . | q= f (−w), that is, .f . .Ffq(w) = f N If .f (x) = g(x + t) for .x ∈ R , we have .

f(w) =  g (w)e2π it·w .

.

(6.12)

6.3 Fourier Transform on S(RN )

223

6. If .f (x) = g(εx) for .x ∈ RN , we have w . g f(w) = ε−N  ε

.

Proof To see 1, we note in first place that the continuity of .fis a direct consequence of the dominated convergence theorem. On the other hand, if .w ∈ RN , then 

|f(w)| ≤

.

RN

|f (x)||e−2π ix·w |dx ≤

 RN

|f (x)|dx .

For the proof of 2, define .h : RN × RN → K by .h(x, w) = f (x)e−2π ix·w g(w), which is measurable and satisfies    . |h(x, w)|dx dw < ∞ . RN

RN

Thus .h ∈ L1 (RN × RN ) and, by Fubini’s theorem, we get  .

RN

f(w)g(w)dw =



 RN

 f (x)e−2π ix·w dx g(w)dw

RN





=

h(x, w)dxdw =

RN ×RN

Rn

f (x) g (x)dx.

Fubini’s theorem and a simple change of variable give 3:  f ∗ g (w) =



.

RN

 =

e−2π ix·w (f ∗ g)(x)dx = 

RN

 =

RN

 =

RN

=

e−2π iy·w f (y) e−2π iy·w f (y)

RN

e

RN

e−2π ix·w

 RN

f (y)g(x − y)dydx

e−2π iy·w f (y)e−2π i(x−y)·w g(x − y)dydx

RN





−2π iy·w

 RN



RN

 e−2π i(x−y)·w g(x − y)dx dy  e−2π iu·w g(u)du dy

 

f (y)dy

RN

e

−2π iu·w

 g(u)du = f(w) g (w) .

Again a change of variable .−x = u yields 4 .

q(w) = f

 RN

e−2π ix·w fq(x)dx =  =

RN

e

2π iu·w

 RN

e−2π ix·w f (−x)dx

f (u)du =

 RN

e−2π iu·(−w) f (u)du = f(−w) .

224

6 Distributions and Partial Differential Operators

For 5, we apply the definition of Fourier transform to obtain f(w) =



g(x + t)e−2π ix·w dx =

.

RN

 RN

g(z)e−2π i(z−t)·w dz =  g (w)e2π it·w .

Finally, 6 again follows directly from the definition of Fourier transform by doing 

f(w) =

.

RN

1 εN

g(εx)e−2π ix·w dx =



w

RN

g(z)e−2π iz· ε dz =

1 w .  g εN ε 

The Fourier transform of a function in .L1 (RN ) may not belong to .L1 (RN ), as our next example shows. Example 6.16 Fix .a > 0 and take .f = χ[−a/2,a/2] , the characteristic function of the interval .[−a/2, a/2], of length .a > 0. Then (w) = .f

 R

f (x)e

−2π ix·w

 dx = =

a/2

−a/2

w

e−2π ix· ε dx

 sin(π aw) 1  iπ aw e . − e−iπ aw = 2π iw πw

Observe that .f is continuous and bounded, but .f ∈ / L1 (R).



Example 6.17 Let us compute the Fourier transform of the function .f (x) = for .x ∈ R. We have (w) = .f  =

+∞

 R

e

−2π ixw −|x|

e

e2π ixw e−x dx +

 dx =



0

+∞

0

−∞

e



−2π ixw x

+∞

e dx +

e−2π ixw e−x dx =

e−2π ixw e−x dx

0



0

e−|x|

+∞ 

 e2π ixw + e−2π ixw e−x dx

0



=

+∞

2 cos(2π ixw)e−x dx .

0

If .a = 0, elementary computations show that  .

cos(ax)e−x dx =

  1 e−x a sin(ax) − cos(ax) . 1 + a2

This altogether shows that f(w) =

.

for .w ∈ R.

2 , 1 + 4π 2 w 2 

6.3 Fourier Transform on S(RN )

225

The following basic properties that relate the Fourier transform with differentiation are a direct consequence of derivation under the integral sign and integration by parts. Proposition 6.12 Fix .k ∈ N: 1. If .f ∈ L1 (RN ) satisfies .x α f (x) ∈ L1 (RN ) for every .|α| ≤ k, then .f ∈ C k (RN ) and   α .∂ f (w) = (−2π i)|α| F x α f (x) (w) , for every .w ∈ RN . 2. If .f ∈ L1 (RN ) ∩ C k (RN ), and .∂ α f ∈ L1 (RN ) for every .|α| ≤ k, then   α f (w) = F ∂ α f (w) = (2π i)|α| w α f(w) , .∂ for every .w ∈ RN . Proposition 6.12 shows that the Fourier transform has a nice behaviour with derivatives. This suggests that it can be a useful tool when dealing with differential equations. We will later give an example of how to use it (see Example 6.19). The general idea will be to apply the Fourier transform to the equation, to get an algebraic equation, and then solve it to get an expression for the Fourier transform of the solution. We need then a way to ‘come back’, recovering the function from its Fourier transform. We show in Theorem 6.18 how to do this, but we need some preliminary work. Lemma 6.4 We have 

e−π t dt = 1 . 2

.

R

Proof Since  .

R

and .x  e

−x 2

e

−π t 2

1 dt = √ π

 R

e−x dx , 2

is an even function, it is enough to show that √  ∞ π −x 2 . . e dx = 2 0

Consider the function .f (x, y) = e−x e−y , for .(x, y) ∈ R2 , and the sets 2

2

S = [0, R] × [0, R] ,

.

DR = B(0, R) ∩ {(x, y) ∈ R2 : x, y ≥ 0} √ D√2R = B(0, 2R) ∩ {(x, y) ∈ R2 : x, y ≥ 0} .

(6.13)

226

6 Distributions and Partial Differential Operators

Clearly, .DR ⊆ S ⊆ D√2R ; hence, since .f (x, y) > 0 and it is continuous, we have 





f (x, y)dxdy ≤

.

f (x, y)dxdy ≤

DR

S

D√2R

f (x, y)dxdy .

On the other hand, Fubini’s theorem yields  f (x, y)dxdy =

.



R

e−x dx 2

2

,

0

S

and with the change to polar coordinates .x = r cos θ , .y = r sin θ , for .θ ∈ [0, π/2] and .r ∈ [0, R], we get .

1  π 1− 2 ≤ 4 eR



R

e−x dx 2

2

0

≤

1  π 1− . 2 4 e2R

Letting .R → ∞, we conclude

 .

∞

e−x dx 2

2

0

=

π , 4 

which yields (6.13) and gives the conclusion. Lemma 6.5 If .f (x) = e

−π |x|2

for .x ∈ RN , then .f(w) = e

−π |w|2

for all .w ∈ RN .

Proof Since .f (x) = e−π x1 · · · e−π xN for all .x ∈ RN , by Fubini’s theorem, it is 2 g = g for .g(x) = e−π x for .x ∈ R. To do this, we observe enough to show that . that g is a solution of the differential equation .2π xy + y  = 0 (this follows by straightforward computation). Suppose now that .h(x) is a differentiable function that is also a solution of the same differential equation. Then we have 2

2

(eπ x h) = (2π xh + h )eπ x = 0 . 2

2

.

So .eπ x h is constant, and .h(x) = Ce−π x for some .C ∈ R and all .x ∈ R. Now, since .g  ∈ L1 (R) and .xg(x) ∈ L1 (R), we can apply Proposition 6.12 to conclude that . g is also differentiable and 2

2

0 = F(2π xg + g  )(w) = 2π F(xg(x))(w) + F(g  )(w)

.

= −2π

   1   g (w) + 2π iw g (w) = i  g (w) + 2π iw g (w) . 2π i

6.3 Fourier Transform on S(RN )

227

This implies that . g is a solution of the same differential equation, and, by what we 2 g (w) = Ce−π w . On the other hand, using have just observed, there is .C ∈ R with . Lemma 6.4, we have  2 .C =  g (0) = e−π x dx = 1 . R

This shows that .g =  g and completes the proof.



Theorem 6.18 (Fourier inversion formula) Let f be a bounded and continuous function in .L1 (RN ). If .f ∈ L1 (RN ), then  f (x) =

.

RN

f(w)e2π ix·w dw ,

(6.14)

for every .x ∈ RN . Proof We set .g(x) = e−π |x| and .gε (x) = g(εx) for each .ε > 0. For .x ∈ RN fixed, we consider  .Iε = f(w)e2π ix·w gε (w)dw. 2

RN

The function .gε satisfies .|gε (w)| ≤ 1 for all .w ∈ R and .gε → 0 pointwise as ε → 0+, and then

.

|f(w)e2π ix·w gε (w)| ≤ |f(w)| ,

.

for all .w, x ∈ RN and .ε > 0. Since .f ∈ L1 (RN ), we can apply the dominated convergence theorem to get  .

lim Iε =

ε→0+

RN

f(w)e2π ix·w dw

for every .x ∈ RN . On the other hand, by Proposition 6.11–2 and 5,  Iε =

.

RN

f(w)e2π ix·w gε (w)dw =

 RN

f (x + w) gε (w)dw .

(6.15)

From Propositions 6.11–6 and Lemma 6.5, we have  gε (w) =

.

RN

gε (x)e−2π ix·w dx =

1   1 w = N g wε .  g εN ε ε

Joining (6.15) and (6.16), we obtain Iε =

.

1 εN

 RN

f (x + w)g

w ε dw =

 RN

f (x + εt)g(t)dt .

(6.16)

228

6 Distributions and Partial Differential Operators

We let .ε → 0+ and get that .f (x + εt)g(t) converges pointwise to .f (x)g(t), and |f (x + εt)g(t)| ≤ sup |f (z)||g(t)| ,

.

z∈RN

as f is continuous and bounded in .RN . We apply again the dominated convergence theorem, which is possible since .g ∈ L1 (RN ), to get  .

lim Iε = f (x)

ε→0+

 RN

g(t)dt = f (x)

e−π |t| dt = f (x) . 2

RN

Hence,  f (x) =

.

RN

f(w)e2π ix·w dw . 

Corollary 6.1 Let .f, g ∈ L1 f = g.

(RN )

be continuous and bounded. If .f =  g , then

.

Proof Define .h(x) = f (x) − g(x) for .x ∈ RN . Taking the Fourier transform, we have . h = f − g = f−  g = 0, which clearly belongs to .L1 (RN ). Then (6.14) gives N .h(x) = 0 for every .x ∈ R , which gives the conclusion.  Example 6.19 Take .f ∈ L1 (R) and consider the differential equation on .R given by y  − y = f (x) .

.

If we apply Fourier transform to the equation and use Proposition 6.12, it becomes y (w) −  y (w) = −(4π 2 w 2 + 1) y (w) , f = F(y  − y) = (2π i)2

.

where .f is continuous and bounded. Taking Example 6.17 and Proposition 6.11–3 into account, we have  y (w) = −

.

1 4π 2 w 2

+1

    f(w) = F − 12 e−|x| (w)F(f )(w) = F − 12 e−|x| ∗ f (w) .

Corollary 6.1 yields y(x) = −

.

1 2

 R

e−|x−y| f (y)dy . 

6.3 Fourier Transform on S(RN )

229

The action .f  f = F(f ) defines an operator (say .F) on .L1 (RN ). But taking .L1 (RN ) as the domain has certain flaws. First of all, as we have seen in Example 6.16, the Fourier transform of an .L1 -function may not belong to .L1 (RN ) (that is, we cannot define the operator .F : L1 (RN ) → L1 (RN )). Also, we need to be in a setting where we are able to use the inversion formula to recover a function from its Fourier transform. We need, then, to find a subspace of .L1 (RN ) that is closed under the Fourier transform, and where we can safely play this game of going forth and back. The Schwartz space is the one. We recall from the notation of Proposition 6.10 that, given any .f ∈ L1 (RN ), we denote .fq(x) = f (−x) for N .x ∈ R . Theorem 6.20 The Fourier transform .F : S(RN ) → S(RN ) is a topological isomorphism. Also, if .f ∈ S(RN ), then  .f (x) = f(w)e2π ix·w dw , RN

|  for every .f ∈ S(RN ). for every .x ∈ RN . In particular, .f = F f Proof Let us see in first place that it is well defined and continuous.   Take .f ∈ S(RN ), and recall that, by Proposition 6.9, the function .∂ β x α f (x) belongs to N N N N .S(R ) for all .α, β ∈ N . Since .S(R ) ⊆ L1 (R ) (recall Proposition 6.8), we can 0 apply Proposition 6.12 to get   w β ∂ α f(w) = w β (−2π i)|α| F x α f (x) (w)

.

   (−2π i)|α|  β α = (−2π i)|α| w β F x α f (x) (w) = F ∂ (x f (x)) (w) . |β| (2π i) Once we have this, (6.12) gives .

sup |w β ∂ α f(w)| ≤ w∈RN

(−2π i)|α| (2π i)|β|

 RN

|∂ β (x α f (x))|dx < ∞ .

Since this holds for every .α and .β, we have .f ∈ S(RN ) (recall Remark 6.4). This shows that .F : S(RN ) → S(RN ) is well defined, and (6.11) gives that it is continuous. We already saw in Corollary 6.1 that .F is injective. In order to see that it is surjective, take some .f ∈ S(RN ). Since .f ∈ S(RN ) ⊆ L1 (RN ), we can use the inversion formula in Theorem 6.18 and a change of variable .w = −t to get   2π ix·w  .f (x) = dw = f (w)e f(−t)e−2π ix·t dt RN

RN

for every .x ∈ RN . Hence, | , f =F f

.

(6.17)

230

6 Distributions and Partial Differential Operators

|  ∈ S(RN ), shows that .F is surjective. Also, (6.17) immediately gives which, since .f −1  that .F is continuous and completes the proof. Corollary 6.2 For each .f ∈ S(RN ), we have .f = fq. Proof We use the Fourier inversion formula to get   2π i(−x)·w q  .f (x) = f (−x) = dw = f (w)e f(w)e−2π ix·w dw = f (x) . RN

RN

 We know from Proposition 6.11 that the Fourier transform of an .L1 -function is continuous and bounded. We can now say more. Corollary 6.3 (Riemann–Lebesgue lemma) If .f ∈ L1 (RN ), then .f is a continuous function vanishing at infinity, that is, .

lim f(x) = 0 .

|x|→∞

Proof Fix .f ∈ L1 (RN ), and for each .n ∈ N, consider the function .ρ1/n defined in (6.1). Each one of these belongs to .C ∞ (RN ) (see Example 6.1) and, then, by Proposition 6.1, we have .ρ1/n ∗ f ∈ C ∞ (RN ) for every n, and this sequence converges to f in .L1 (RN ) (see [2, Theorem 6.2] for a proof of this fact). This shows that .C ∞ (RN ) is dense in .L1 (RN ), and this implies (see Proposition 6.3) that N N N .D(R ) is dense in .L1 (R ). So, we can find a sequence .(ϕn )n ⊆ D(R ) such that N .ϕn → f in .L1 (R ). Now, Proposition 6.11–1 shows that the Fourier transform defines a continuous mapping from .L1 (RN ) into the space of bounded continuous ϕn → f uniformly on .RN . By Theorem 6.20, each functions. This implies that . N . ϕn ∈ S(R ) and this (recall (6.10)) gives the conclusion.  Proposition 6.13 If .f, g ∈ S(RN ), then the convolution .f ∗ g ∈ S(RN ). Proof By Proposition 6.11–3, as .f, g ∈ L1 (RN ), we have .f ∗ g = f  g , and, by N  Theorem 6.20, .f ,  g ∈ S(R ). Leibniz formula easily implies that .f  g ∈ S(RN ).  An application of Theorem 6.20 again shows that .f ∗ g ∈ S(RN ).

6.4 Tempered Distributions and the Fourier Transform We look now at the dual of the space .S(RN ), denoted .S  (RN ). This consists, then, of the continuous linear maps .T : S(RN ) → K, which are called tempered distributions. Note that this means that there are .C > 0 and .m ∈ N such that |T (ϕ)| ≤ C sup sup (1 + |x|2 )m |∂ α ϕ(x)| ,

.

|α|≤m x∈RN

for all .ϕ ∈ S(RN ).

6.4 Tempered Distributions and the Fourier Transform

231

Remark 6.5 The first question that we have to address at this point is why do we call them ‘distributions’. The reason is easy: if .T ∈ S  (RN ), then its restriction N .T |D (RN ) defines a distribution, since for each compact set .K ⊆ R , the inclusion N .D(K) → S(R ) is continuous (recall Proposition 6.7). Even more, if .T |D (RN ) vanishes, the density of .D(RN ) in .S(RN ) (see again Proposition 6.8) implies that N  N .T (ϕ) = 0 for each .ϕ ∈ S(R ). Hence, .T ∈ S (R ) is uniquely determined by its N restriction to .D(R ) or, in other words, that each .T ∈ S  (RN ) defines, by restricting it to .D(RN ), a unique distribution in .D  (RN ). Summarizing, the restriction map  N  N .J : S (R ) → D (R ) given by .J (T ) = T |D (RN ) , is well defined and injective. Example 6.21 Not every distribution is a tempered distribution. Consider the .C ∞ function given by .f (x) = ex for .x ∈ R, and its associated distribution, defined as  .Tf (ϕ) = ex ϕ(x)dx , R

for .ϕ ∈ D(R). Choose some .φ ∈ C ∞ (R) such that .φ(x) = 1 for .x ≥ 0, and .φ(x) = 0 for .x ≤ −1. Consider now the function .ψ ∈ C ∞ (R) given by .ψ(x) = φ(x)e−x , which clearly satisfies .

lim x β |∂ α ψ(x)| = 0 ,

x→+∞

for all .β, α ∈ N0 . Hence, .ψ ∈ S(R). If .Tf ∈ S  (R), then  Tf (ψ) =

.

R

ex ψ(x)dx ∈ K.

But .ex ψ(x) = 1 for all .x > 0 and so .ex ψ(x) ∈ / L1 (R).



Proposition 6.14 1. .Lp (RN ) ⊆ S  (RN ) for every .1 ≤ p ≤ ∞. In particular, .S(RN ) ⊆ S  (RN ). 2. If .f ∈ L1,loc (RN ) satisfies .|f (x)| ≤ C(1 + |x|2 )n/2 for some .n ∈ N0 and  N .C > 0, we have .Tf ∈ S (R ). In particular, every polynomial defines a tempered distribution. Proof If .f ∈ Lp (RN ), then it belongs to .L1,loc (RN ), and we have by the definition of the associated distribution (6.4) and Hölder’s inequality  |Tf (ϕ)| ≤

.

|f (x)ϕ(x)|dx ≤ f p ϕp .

232

6 Distributions and Partial Differential Operators

Then, to show that .Tf is a tempered distribution, we just have to bound .ϕp by the action on .ϕ of some convenient element of the fundamental system of seminorms from (6.9). We distinguish two cases. First of all, if .p = ∞, then it is obvious since ϕ∞ = sup |ϕ(x)| = p0 (ϕ) .

.

x∈RN

Now, if .1 ≤ p < ∞, we choose .m > N/p and have ϕp =



.

p

|ϕ(x)| dx

 1 p

=

 (1 + |x|2 ) m2 (1 + |x|2 )

m 2



|ϕ(x)|p dx

≤ pm (ϕ)

 1 p



1 (1 + |x|2 )

mp 2

dx

 1 p

.

The last integral converges because of the choice of m. This completes the proof of the first statement. The argument for the second one is rather similar. Take some .ϕ ∈ S(RN ) and estimate  |Tf (ϕ)| ≤

.

≤C



 |f (x)ϕ(x)|dx ≤ C 

1 (1 + |x|2 )

N+1 2

dx

(1 + |x|2 )

N+1 2

(1 + |x|2 )

N+1 2

sup (1 + |x|2 )

n

(1 + |x|2 ) 2 |ϕ(x)|dx

N+1+n 2

x∈RN

|ϕ(x)| ≤ C1 pN +1+n (ϕ) . 

As we have just seen, every polynomial defines a tempered distribution. Nevertheless, there are tempered distributions defined by functions other than polynomials, as our next example shows. Example 6.22 Consider the function .f (x) = ex cos ex for .x ∈ R. If .ϕ ∈ S(R), then we have   .Tf (ϕ) = ex cos(ex )ϕ(x)dx = − sin(ex )ϕ  (x)dx , R

R

and  |Tf (ϕ)| ≤

.

R

| sin(ex )||ϕ  (x)|dx ≤ C sup (1 + x 2 )|ϕ  (x)| ,

which shows that .Tf ∈ S  (RN ).

x∈R

6.4 Tempered Distributions and the Fourier Transform

233

However, there is no polynomial .P (x) with .|f (x)| ≤ |P (x)| for every .x ∈ R. If this were the case, we would have | cos(ex )| ≤

.

|P (x)| , ex

for every .x ∈ R, which implies .

lim cos(ex ) = 0 .

x→+∞

But this is not true, since .cos(elog 2kπ ) = 1 for each .k ∈ N.



Proposition 6.15 Every distribution with compact support is tempered. That is, E  (RN ) ⊆ S  (RN ).

.

Proof If .T ∈ E  (RN ), then by Theorem 6.14 there is a unique continuous linear map .S : C ∞ (RN ) → K such that .T = S|D (RN ) . Now, by Theorem 6.7, the inclusion N ∞ N .S(R ) ⊆ C (R ) is continuous and has dense range. Then the restriction .S|S(RN ) is a uniquely determined distribution that coincides with T on .D(RN ).  We have already seen how some of the basic operations on functions (such as differentiation or multiplication by a .C ∞ -function) can be extended also to distributions. Our aim now is to extend the definition of the Fourier transform to tempered distributions. As in previous occasions, we look first at distributions defined by functions, in order to get some ideas. If .f ∈ L1 (RN ) is such that  ∈ L1 (RN ), we would expect that the Fourier transform of .Tf coincides with .f N .Tf. If .ϕ ∈ S(R ), we can use Propositions 6.11–2 to get  Tf(ϕ) =

.

RN

f(x)ϕ(x)dx =

 RN

f (x) ϕ (x)dx = Tf ( ϕ) .

 This suggests to define the Fourier transform of .Tf as .T ϕ ) for .ϕ ∈ f (ϕ) = Tf ( N S(R ). With this idea, we define the Fourier transform of a tempered distribution .u ∈ u : S(RN ) → K defined as S  (RN ) as the mapping .  u(ϕ) = u( ϕ) .

.

Let us note that . u = u◦F for every .u ∈ S  (RN ), where .F : S(RN ) → S(RN ). Since u ∈ S  (RN ) we know that .F is continuous (see Theorem 6.20), we immediately have .  N for every .u ∈ S (R ).

234

6 Distributions and Partial Differential Operators

Example 6.23 The Dirac function .δ is a tempered distribution, since .δ ∈ E  (RN ). Let us compute its Fourier transform. For .ϕ ∈ S(RN ), we have  ϕ) =  ϕ (0) = .δ (ϕ) = δ(



e−2π i(0·w) ϕ(w)dw =

 ϕ(w)dw = 1(ϕ) ,

which shows that  .δ is the constant function .1. With the same idea, for every .a ∈ RN , we have .δa ∈ S  (RN ), and  .δa (ϕ) = δa ( ϕ) =  ϕ (a) =

 RN

ϕ(x)e−2π i(a·x) dx ,

−2π i(a·x) in .S  (RN ). for every .ϕ ∈ S(RN ). So,  .δa = e

Proposition 6.16 The Fourier transform .F : is a linear isomorphism.

S  (RN )

 →

S  (RN ) given by .F(u)

= u

Proof It is easy to see that it is linear. Let us see in first place that it is injective. If .F(u) = 0, then .u( ϕ ) = 0 for every .ϕ ∈ S(RN ). Taking into account that .F : N N S(R ) → S(R ) is surjective (see Theorem 6.20), this implies that .u(ψ) = 0 for all .ψ ∈ S(RN ) and, then, .u = 0. This shows the injectivity. To see that .F is surjective, fix some .v ∈ S  (RN ) and define .u : S(RN ) → K by −1 .u(ϕ) = v(F ϕ), that is, .u = v ◦ F −1 , which (again by Theorem 6.20) belongs to  N .S (R ). Then  u(ϕ) = u( ϕ ) = v(F −1 Fϕ) = v(ϕ) ,

.

for all .ϕ ∈ S(RN ). This shows that .F : S  (RN ) → S  (RN ) is surjective and completes the proof.  Proposition 6.17 Let .u ∈ S  (RN ), and then: α u = (2π i)|α| w α 1. .∂ u.  2. .∂ α u = (−2π i)|α| F x α u .

Proof This follows directly from the definition of the Fourier transform for tempered distributions and Proposition 6.12. First, we have

   α u(ϕ) = ∂ α u( ϕ ) = u (−1)|α| ∂ α  ϕ = u (−1)|α| (−2π i)|α| F(x α ϕ) ∂   u((2π i)|α| x α ϕ) = (2π i)|α| (x α u)(ϕ) . = u F((2π i)|α| x α ϕ) = 

.

6.4 Tempered Distributions and the Fourier Transform

235

The second formula follows from the following computation:

   α u(ϕ) = x α u ( αϕ x ϕ ) = u(x α  ϕ ) = u (2π1i)|α| ∂

.

=

  1 1 1 αϕ = u ∂  u(∂ α ϕ) = ∂ α u(ϕ) . |α| |α| (2π i) (2π i) (−2π i)|α| 

Our aim is to use the Fourier transform on partial differential operators (recall its definition in Example 4.19) in order to solve partial differential equations. As a consequence of Proposition 6.17, we see how the Fourier transform acts on such an operator: Corollary 6.4 α δ = (2π i)|α| w α . 1. .∂



2. If .P (z) = aα zα and .P (∂) = aα ∂ α , we have |α|≤m

|α|≤m

  F P (∂)u = aα (2π i)|α| w α u = P (2π iw) u,

.

(6.18)

|α|lem

for all .u ∈ S  (RN ). We give now some consequences of the use for the inverse Fourier transform. We q for the distribution defined as fix some notation. If .u ∈ D  (RN ), then we write .u q(ϕ) = u(q .u ϕ ), for .ϕ ∈ D(RN ).  q and . u =u Proposition 6.18 If .u ∈ S  (RN ), then . uq = u. Proof Both formulas follow immediately from the definitions and Corollary 6.2. If ϕ ∈ S(RN ), then we have

.

  q q q ) = u(ϕ) ,  q(ϕ) , and  ϕ ) = u(q ϕ) = u .u(ϕ) = u( u (ϕ) =  u(q ϕ ) = u( ϕ which gives the claim.



 δ. But, for .ϕ ∈ S(RN ), we have .q Example 6.24 We have  .1 = δ = q δ(ϕ) = δ(q ϕ) = q .1 = δ.  ϕq(0) = ϕ(0), and hence, .δ = δ. This implies  An important result in the classical theory of functions of one variable is that a function is constant if and only if its derivative is 0. Our aim now is to see that this is also the case when we deal with distributions. We need first a result. Lemma 6.6 Let .ϕ ∈ S(R). Then there is .φ ∈ S(R) such that .φ  = ϕ if and only if  . R ϕ(x)dx = 0.

236

6 Distributions and Partial Differential Operators

Proof Suppose first that .φ ∈ S(R) is such that .φ  = ϕ. Then Barrow rule gives 



.

R

ϕ(x)dx = lim

A

A→+∞ −A

ϕ(x)dx = lim [φ(A) − φ(−A)] = 0 . A→+∞

 Assume conversely that . R ϕ(x)dx = 0 and define  φ(x) =

x

.

−∞



+∞

ϕ(y)dy = −

ϕ(y)dy . x

By the Fundamental Theorem of Calculus, we have .φ  = ϕ, which implies that ∞ (k) = ϕ (k−1) for every .k ∈ N. To complete the proof, we have to .φ ∈ C (R) and .φ see that .φ ∈ S(R). First of all, since .ϕ ∈ S(R), we have m

.

sup sup (1 + |x|2 ) 2 |φ (k) (x)| = 1≤k≤m x∈R

sup

m

sup (1 + |x|2 ) 2 |ϕ (k) (x)| < ∞ ,

0≤k≤m−1 x∈R

for every m. It is only left to see what happens with .k = 0. Take C = sup (1 + |x|2 )

.

m+2 2

|ϕ(x)| < ∞ .

x∈R

Then, for .x ≥ 0, we have m 2

(1 + |x| ) |φ(x)| ≤ (1 + |x| )

.

2

2

m



≤ (1 + |x|2 ) 2

x

∞

m 2



x −∞

1 (1 + |y|2 )

|ϕ(y)|dy 

Cdy ≤ C m+2 2

For .x < 0, we proceed similarly with .φ(x) = proof.

x

x

∞

1 dy = C1 . (1 + |y|2 )2

−∞ ϕ(y)dy,

and this completes the 

Proposition 6.19 Let .u ∈ S  (R). Then .u = 0 if and only if u is constant. Proof Let us suppose first that u is constant. This means that u is the distribution defined by a constant function and then .u (as a distribution) coincides with .u (as a function). Therefore, .u = 0. Let us conversely assume that .u ∈ S  (R)  is such that  .u = 0. Our aim is to see that .u = Tc for some c, that is, .u(ϕ) = R cϕ(x)dx for  every .ϕ. We begin by fixing .ψ0 ∈ D(R) ⊆ S(R) with . R ψ0 (x)dx = 1. Now, given .ϕ ∈ S(R) ⊆ L1 (R), we define  .Aϕ = ϕ(x)dx R

6.4 Tempered Distributions and the Fourier Transform

237

and consider the function .ϕ − Aϕ ψ0 ∈ S(R). Note that  .

R

(ϕ − Aϕ ψ0 )(x)dx = 0 ,

which, by Lemma 6.6, implies that there is .φ ∈ S(R) so that .φ  = ϕ − Aϕ ψ0 . We have .ϕ = φ  + Aϕ ψ0 and (recall that .u = 0) u(ϕ) = u(φ  ) + Aϕ u(ψ0 ) = −u (φ) + Aϕ u(ψ0 )   u(ψ0 )ϕ(x)dx . = Aϕ u(ψ0 ) = u(ψ0 ) ϕ(x)dx =

.

R

R

Denoting .c = u(ψ0 ) gives the claim and completes the proof.



Example 6.25 We calculate the Fourier transform of Cauchy’s principal value of 1 x and the Heaviside function that, as we know, are closely related. For the sake of clarity, let us denote .u = P V x1 . We know from Example 6.9 that .xu = 1 in the sense of (tempered) distributions. Then Example 6.24 and (6.6) give

.

x u = 1 = δ = H .

.

On the other hand, Propositions 6.17–2 give x u=−

.

1   u . 2π i

Therefore, . u = −2π iH  , and .( u + 2π iH ) = 0. By Proposition 6.19, there is a constant C with . u = C − 2π iH , that is,   u(ϕ) = C



.

R

ϕ(x)dx − 2π i

∞

(6.19)

ϕ(x)dx , 0

for every .ϕ ∈ S(R). A straightforward computation from the definition (recall Example 6.6) shows that .u(q ϕ ) = −u(ϕ) for every .ϕ ∈ S(R). Then, using Propositions 6.11–4, we have  q) = u(  u(q ϕ ) = u(ϕ ϕq) = −u( ϕ ) = − u(ϕ) .

(6.20)

.

Note that, with a change of variables, we have   u(q ϕ) = C

.

R



∞

ϕ(−x)dx − 2π i 0

 ϕ(−x)dx = C

R

 ϕ(t)dt − 2π i

0

−∞

ϕ(t)dt .

238

6 Distributions and Partial Differential Operators

This, combined with (6.20) and (6.19), gives .C = π i and, then, P V x1 = π i(1 − 2H ) .

.

q (recall u =u We obtain from here the Fourier transform of H . First of all, using that . Proposition 6.18), we have    π iF 1 − 2H = P V

.

1 w

  q= − PV

1 w



.

On the other hand, Example 6.24 gives        = π i δ − 2H  . π iF 1 − 2H = π i  1 − 2H

.

 = π iδ + P V Consequently, .2π i H

1 w,

which yields

 = δ − 1 PV H πi

1 w

.

. 

As we announced above, our aim is to use the Fourier transform to solve partial differential equations where the unknown is a tempered distribution (see Examples 6.27 and 6.28), in a similar way as we did in Example 6.19 in a more ‘classical’ setting. As we saw there, convolution is a fundamental tool. So, we need to extend the definition to have the convolution of a distribution and a function. As a first step (as many times so far), we check how this should work for distributions defined by functions. So, given .f ∈ L1 (RN ) and .ϕ, ψ ∈ S(RN ), we have by Fubini’s theorem    .Tf ∗ψ (ϕ) = (f ∗ ψ)(x)ϕ(x)dx = f (t)ψ(x − t)dt ϕ(x)dx  =

RN



RN

RN

RN



ψ(x − t)ϕ(x)dx f (t)dt =  =

RN

RN

RN

 RN

q − x)ϕ(x)dx f (t)dt ψ(t

q ∗ ϕ)(t)f (t)dt = Tf (ψ q ∗ ϕ) . (ψ

With this idea, the convolution of a tempered distribution .u ∈ S  (RN ) and a function N .ψ ∈ S(R ) is defined as  .

   q ∗ϕ , u ∗ ψ (ϕ) = u ψ

6.4 Tempered Distributions and the Fourier Transform

239

q ∗ ϕ ∈ S(RN ). for .ϕ ∈ S(RN ). Observe that, by Proposition 6.13, we have .ψ Moreover, it can be shown that .u ∗ ψ ∈ S  (RN ) and that it satisfies ∂ α (u ∗ ϕ) = ∂ α u ∗ ϕ = u ∗ ∂ α ϕ

.

u ∗ (ϕ ∗ ψ) = (u ∗ ϕ) ∗ ψ u ∗ϕ = u ϕ. Example 6.26 If .a ∈ R and .φ, ϕ ∈ S(R), we have    q q q .(δa ∗ φ)(ϕ) = δa (φ ∗ ϕ) = φ ∗ ϕ (a) = φ(u)ϕ(a − u)du RN

 =

RN



φ(−u)ϕ(a − u)du =

RN

φ(t − a)ϕ(t)dt = TTa φ (ϕ)

(recall the notation from Proposition 6.10). That is, .δa ∗ φ = φ(x − a), as distributions.  The reader is referred to [13, Chapter 7] for a more detailed study of the convolution of tempered distributions and functions in .S(RN ), in particular to [13, Theorem 7.19], where the proof of the aforementioned properties can be found. Example 6.27 (Laplacian) We consider the Laplacian, defined for .u ∈ S  (RN ) as u =

.

∂ 2u ∂ 2u + · · · + , ∂xn2 ∂x12

and we want to solve the equation (1 − )u = f ,

.

where .f ∈ S  (RN ) (or .S(RN )). Then our operator is .P (∂) = 1 − , whose symbol is .P (x) = 1 − x12 − x22 − · · · − xn2 , for .x ∈ RN . We apply the Fourier transform to the equation, taking (6.18) into account and noting that .P (2π iw) = 1 + 4π 2 |w|2 , to have (1 + 4π 2 |w|2 ) u = f.

.

This gives  u=

.

f 1 + 4π 2 |w|2

240

6 Distributions and Partial Differential Operators

and u = F −1

.

 f ∈ S  (RN ) or S(RN ) . 1 + 4π 2 |w|2

Taking, for instance, .N = 1 and .f = δ, from Examples 6.23 and 6.17, we have u = F −1

.

 1 1 = e−|x| . 2 2 2 1 + 4π |w| 

Example 6.28 (Wave equation) We consider the problem ⎫ − c2 u = 0 ⎪ ⎬ u(0, x) = φ(x) ⎪ ⎭ ∂u ∂t (0, x) = ψ(x) ∂2u ∂t 2

.

(6.21)

with .φ, ψ ∈ S(RN ) and .c > 0. We take the Fourier transform with respect to x (which we denote by .Fx ) and leave t fixed to get ⎫  u=0 ⎪ + c2 4π 2 |w|2  ⎬ (w) .  u(0, w) = φ ⎪ ⎭ ∂ u  ∂t (0, w) = ψ (w)  ∂2

∂t 2

.

This is a linear ordinary differential equation, whose solution is (w) +  u(t, w) = cos(c2π |w|t)φ

.

sin(c2π t|w|) (w) . ψ 2π c|w|

Therefore, 

sin(c2π t|w|) (w) (w) + ψ u(t, x) = Fw−1 cos(c2π |w|t)φ 2π c|w| 



sin(c2π t|w|) (w) , (w) + Fw−1 ψ = Fw−1 cos(c2π |w|t)φ 2π c|w|

.

and, by Proposition 6.11–3,



sin(c2π t|w|)  ∗x ψ . u(t, x) = Fw−1 cos(c2π |w|t) ∗x φ + Fw−1 2π c|w|

.

(6.22)

6.5 Linear Partial Differential Operators

241

From now on, we assume .N = 1. On the one hand, we have (recall Example 6.23)     Fw−1 cos(2π ct|w|) = Fw−1 cos(2π ctw)

.

=

 1 1  −1 2π icwt Fw (e ) + Fw−1 (e−2π icwt ) = (δ−ct + δct ) . 2 2

On the other hand, Example 6.16 yields Fw−1

sin(2π ct|w|) 

.

2π c|w|

= Fw−1

sin(2π ctw)  2π cw

=

1 χ[−ct,ct] . 2c

So we have the two inverse Fourier transforms appearing in (6.22), and u(t, x) =

.

1 1 (δ−ct + δct ) ∗x φ + χ[−ct,ct] ∗x ψ . 2 2c

We know the result of the first convolution from Example 6.26. We compute now the second one, having  (χ[−ct,ct] ∗ ψ)(x) =

ct

.

−ct

 ψ(x − s)ds = −

x−ct

 ψ(u)du =

x+ct

x+ct

ψ(u)du , x−ct

for .ψ ∈ S(R). With all this, we finally have d’Alembert’s formula for the solution of (6.21) u(t, x) =

.

1 1 (φ(x − ct) + φ(x + ct)) + 2 2c



x+ct

ψ(y)dy . x−ct



6.5 Linear Partial Differential Operators In this chapter, we present a few results about linear partial differential operators with constant coefficients (they were introduced in Example 4.19). Let us briefly recall that a polynomial of N variables of degree m is of the form P (z) =



.

α∈NN 0 |α|≤m

aα zα ,

242

6 Distributions and Partial Differential Operators

where we assume that there is at least one non-zero .aα with .|α| = m. The principal part of P is the m-homogeneous polynomial Pm (z) =



aα zα .

.

|α|=m

We recall that each polynomial has an associated partial differential operator (with constant coefficients) P (∂) =



.

aα ∂ α .

|α|≤m

We already saw (recall again Example 4.19) that, if . is some open set in .RN , then the operator .P (∂) : C ∞ () → C ∞ () is continuous. Observe that this implies that P (∂) : D(K) → D(K)

.

is continuous for each compact set .K ⊆ RN . Our main interest now is to find out when the partial differential equation defined by a given partial differential operator has solution. That is, we wonder now under what circumstances does the partial differential equation .P (∂)u = v have a solution. This can be seen under several points of view. First of all, if v is a function, we may ask to find another function u so that .P (∂)u = v. But this is often too restrictive, and the chances to find an answer are small. If we broaden the field where we look for solutions, our chances to find one will certainly increase. Then, we are going to interpret the equations in a distributional sense. That is, if v is a distribution, we want to find some distribution u so that .P (∂)u = v as distributions.

6.5.1 Fundamental Solutions. The Malgrange–Ehrenpreis Theorem We say that a distribution .E ∈ D  (RN ) is a fundamental solution of a partial differential operator .P (∂) if P (∂)E = δ.

.

Example 6.29 The Heaviside function .H (x) is a fundamental solution of the d in .D  (R) (recall (6.6)). Note that, in this case, the symbol operator .P (∂) = dx is given by .P (z) = z.  We give now fundamental solution for some classical partial differential operators (see Example 4.19). A complete and detailed justification can be found in [17, Chapter 1].

6.5 Linear Partial Differential Operators

243

Example 6.30 1. The Cauchy–Riemann operator ∂=

.

1 ∂ ∂  +i 2 ∂x ∂y

(6.23)

(the symbol of which is .P (x, y) = 12 (x + iy)) has a fundamental solution given by E(z) =

.

∂2 ∂x12

2. The Laplacian . =

+ ··· +

∂2 2 ∂xN

E(x) =

.

1 . πz

has a fundamental solution given by

1 log |x| if N = 2 , 2π

and E(x) =

.

1 1 if N ≥ 3 , (N − 2)ωN |x|N −2

2π N/2 is the volume of the Euclidean unit ball of .RN . (N/2) 3. The heat operator .P (∂) = ∂t∂ − x has a fundamental solution given by where .ωN =

E(x, t) = H (t)

.

4. The wave operator .P (∂) =

∂2 ∂t 2

−

∂2 ∂x 2

1  −|x|2 e 4t . √ 2 πt

has a fundamental solution given by

 1 1 H (t − x)H (t + x) = 2 .E(x, t) = 2 0

if t − |x| > 0, if t − |x| < 0. 

Our aim in this section is to show that these operators always have a fundamental solution. This is an important theorem, due to Malgrange [10] and Ehrenpreis [5]. Theorem 6.31 (Malgrange–Ehrenpreis) Every non-constant linear partial differential operator with constants coefficient has a fundamental solution. The original proof uses the Hahn–Banach theorem. Shortly after, Trèves gave in [16] a constructive proof, based on ‘Hörmander’s staircase’ and partitions of unity. Rosay in [11] gave an elementary proof, based on the .L2 theory. We follow here

244

6 Distributions and Partial Differential Operators

the proof of Wagner [19] that gives an explicit formula for the fundamental solution (see (6.26) later). Before getting there, we need several lemmas. Lemma 6.7 Let .P (z) be a non-constant polynomial of N variables. Then, the set VR (P ) = {x ∈ RN : P (x) = 0}

.

has null Lebesgue measure in .RN . Proof If .N = 1, the result is trivial, since it is a finite set. We proceed by induction, and assume that the result holds for polynomials of N variables. Take now .P (x1 . . . . , xN +1 ) a polynomial of .N + 1 variables, and write l 

P =

.

Qj ,

j =1

where each .Qj is irreducible. Note that VR (P ) =

l 

.

VR (Qj ) ,

j =1

so we may assume without loss of generality that P is irreducible. For each fixed .a ∈ R, we consider the polynomial of N variables given by .Pa (x1 , . . . , xN ) = P (x1 , . . . , xN , a). We want to see now that .Pa = 0 for every a. Suppose that this is not the case, and there is some a for which .P (x  , a) = 0 for every .x  = (x1 , . . . , xN ) ∈ RN . If P has degree m, taking the Taylor expansion of the function .λ ∈ R  P (x  , λ) around a, we get P (x  , xN +1 ) = P (x  , a) +

.

m

1 ∂j j ! ∂x j j =1

=

m

1 ∂j j ! ∂x j j =1

P (x  , a) (xN +1 − a)

N +1

P (x  , a) (xN +1 − a) = (xN +1 − a)S(x  , a) ,

N +1

contradicting the fact that P is irreducible. So, .Pa = 0 for every .a ∈ R, and we can use Fubini’s theorem and the induction hypothesis to get  mN +1 (VR (P )) =

∞

.

−∞

  mN {x  ∈RN : P (x  , a) = 0} da =



∞ −∞

mN (VR (Pa ))da = 0 . 

6.5 Linear Partial Differential Operators

245

Corollary 6.5 Let .P (z) be a polynomial of degree .m ≥ 1 of N variables. Then, for each .λ ∈ R and .η ∈ RN , the set {w ∈ RN : P (2π iw + λη) = 0}

.

has null Lebesgue measure in .RN . Proof Just observe that the mapping .w  P (2π iw + λη) = λη)α is a polynomial of N variables, and apply Lemma 6.7.



|α|≤m aα (2π iw

+ 

Lemma 6.8 The Fourier transform satisfies, for each .u ∈ S  (RN ):   1. .F −1 (u) = Fu q.   2. .P (∂)F −1 u = Fw−1 P (2π iw)u . Proof For the first assertion, we write .Fu =  u. From Proposition 6.18, we know  q that . u = u. Taking the Fourier transform in both sides of the equality yields the claim. For the second statement, applying (6.18) to the distribution .Fu ∈ S  (RN ), we have F(P (∂)F −1 u) = P (2π iw)u ,

.



which immediately gives our claim.

Lemma 6.9 Let .P (z) be a non-constant polynomial of N variables. For every .ζ ∈ CN and .T ∈ D  (RN ), we have   P (∂) eζ x T = eζ x P (ζ + ∂)(T ) .

.

Proof We proceed in two steps. In the first one, let us assume that .T = ϕ ∈ D(RN ). It is easy to check by induction that for every .j = 1, . . . , N and .αj ∈ N0 , we have α

∂j j (eζ x ϕ) = eζ x (ζj + ∂j )αj ϕ .

.

Then ∂ α (eζ x ϕ) = eζ x (ζ + ∂)α ϕ ,

.

for every .α ∈ NN 0 , and therefore,

 

P (∂) eζ x ϕ = aα ∂ α (eζ x ϕ) = aα eζ x (ζ + ∂)α ϕ = eζ x P (ζ + ∂)(ϕ) .

.

α

α

  For our second step, recall first that . D(RN ), D  (RN ) is a dual pair (see Remark 6.1). Also, since clearly .D(RN ) ⊆ L1,loc (), it can be seen as a subspace

246

6 Distributions and Partial Differential Operators

  of .D  (RN ). Then Proposition 3.9 gives that .D(RN ) is .σ D  (RN ), D(RN ) -dense in .D  (RN ). Take now an arbitrary .T ∈ D  (RN ), and choose a net .(ϕi )i∈I ⊆ D(RN ) such that .T (ψ) = limi ϕi (ψ) for every .ψ ∈ D(RN ) (we shall later see that in fact we can take a sequence, see Corollary 6.6, but for our purposes now this is enough). Before we proceed, let us note that .(−∂)α = (−1)|α| ∂ α and, then, for any distribution u, we have



  .P (∂)u, ψ = aα (∂ α u)ψ = aα (−1)|α| u(∂ α ψ) = u P (−∂)ψ . α

α

This, using what we have just obtained in the first step, gives P. (∂)(eζ x T ), ψ = eζ x T , P (−∂)ψ = T , eζ x P (−∂)ψ = limϕi , eζ x P (−∂)ψ i

= limP (∂)(eζ x ϕi ), ψ = limeζ x P (ζ + ∂)(ϕi ), ψ = limϕi , P (−∂ + ζ )(eζ x ψ) i

i

i

= T , P (−∂ + ζ )(eζ x ψ) = eζ x P (∂ + ζ )T , ψ . Since this holds for every .ψ ∈ D(RN ), this completes the proof.



Lemma 6.10 Let .λ0 , λ1 , . . . , λm ∈ C be pairwise different. Then, the system of linear equations m

.

 aj λkj =

j =0

0

for k = 0, . . . , m − 1

1

for k = m

(6.24)

has a unique solution given by aj =

m 

.

k=0 k=j

1 λj − λk

for .j = 0, 1, . . . , m . Proof Let us note in first place that the matrix of the system ⎛

1 ⎜ λ0 ⎜ .⎜ . ⎝ ..

1 ··· λ1 · · · .. .

⎞ 1 λm ⎟ ⎟ .. ⎟ . ⎠

m m λm 0 λ1 · · · λm

is a Vandermonde matrix that, being all .λi =  λj for .i = j , is nonsingular. Therefore, it has a unique solution .(a0 , a1 , . . . , am ) ∈ Cm+1 . In order to find it, we consider

6.5 Linear Partial Differential Operators

247

the polynomial p(z) =

m 

.

(z − λj ) ,

j =0

of degree m, and whose roots are exactly .λ0 , λ1 , . . . , λm . Fix .M > 2 maxj |λj |, take zk any .0 ≤ k ≤ m, and use Cauchy’s residue theorem with the function .f (z) = p(z) to get  .

|z|=M

m m



λkj zk dz = 2π i . Res(f, λj ) = 2π i p(z) p (λj ) j =0

j =0

This gives m

.

j =0

λkj p (λ

j)

=



1 2π i

2π

(Meit )k iMeit dt it − λ ) (Me j j =0

%m

0

1 M k+1 = 2π M m+1



2π 0

ei(k+1)t dt . it j =0 (e − λj /M)

%m

Since   |λj | 1  it λj  > e −  ≥ 1 − M M 2

.

for each .j = 0, . . . , m and every .t ∈ [0, 2π ], we have   . 

2π 0

  2π  ei(k+1)t 1  dt ≤ 2m+2 π ,  %m %m dt  ≤ it − λ /M)  it   (e j 0 j =0 j =0 (e − λj /M)

for every .k = 0, . . . , m. Now, if .0 ≤ k ≤ m − 1, this gives m 

k+1 λkj   m+1 M ≤ 2 .   p (z) M m+1

.

j =0

Since the right-hand side of the inequality tends to 0 as .M → ∞, we have m

.

j =0

λkj p (λj )

= 0 for k = 0, . . . , m − 1 .

(6.25)

248

6 Distributions and Partial Differential Operators

Now, for .k = m, we have m

.

j =0

λm j

1 = p (λj ) 2π



2π

ei(m+1)t dt . it j =0 (e − λj /M)

%m

0

A straightforward application of Lebesgue’s dominated convergence theorem then yields m

.

j =0



This and (6.25) show that Noting that

.

λm j p (λj )

= 1.

1 1 1 p (λ0 ) , p (λ1 ) , . . . , p (λm )

p (z) =

m m 

.

i=0



is the solution of (6.24).

(z − λk )

k=0 k=i



finally gives the conclusion.

We can now give Wagner’s formula for the fundamental solution of a linear partial differential operator with constant coefficients. This clearly proves Theorem 6.31. Theorem 6.32 Let P be a polynomial of degree m and N variables, and .η ∈ RN be such that .Pm (η) = 0. Take .λ0 , λ1 , . . . , λm ∈ R pairwise different and define aj =

m 

.

k=0 k=j

1 . λj − λk

Then E=

1

m

Pm (2η)

j =0

.

aj e

λj ηx

F−1 w



P (2π iw + λj η) P (2π iw + λj η)



is a fundamental solution of .P (∂). Proof Let us take for the moment an arbitrary .λ ∈ R and define S(w) =

.

P (2π iw + λη) . P (2π iw + λη)

(6.26)

6.5 Linear Partial Differential Operators

249

By Lemma 6.5, this is well defined for almost all .w ∈ RN (and then, clearly N  N .|S(w)| = 1). Then .S ∈ L∞ (R ) and, by Proposition 6.14, .S ∈ S (R ).  N N Take any .T ∈ S (R ), and .ζ ∈ C . Using Lemmas 6.8 and 6.9, we obtain       P (∂) eζ x F −1 (T ) = eζ x P (∂ + ζ )F −1 (T ) = eζ x F−1 w P (2π iw + ζ )T .

.

We apply this to .T = S(w) and .ζ = λη ∈ RN , to get  

ληx −1 P (2π iw+λη) .P (∂) e Fw P (2π iw+λη)     P (2π iw + λη) P (2π iw + λη) = eληx F−1 = eληx F−1 w w P (2π iw + λη) . P (2π iw + λη) If .P (z) =



|α|≤m aα z

α,

we write .P (z) =

P (2π iw + λη) =





|α|≤m aα z

α

and

aα (2π iw + λη)α

.

|α|≤m

=



aα (−2π iw + λη)α = P (−2π iw + λη) .

|α|≤m

Using again Lemmas 6.8 and 6.9 (note that .P is again a polynomial of N variables and degree m), we have     −1 F−1 w P (2π iw + λη) = Fw P (−2π iw + λη)   = P (−∂ + λη) F−1 w (1) = P (−∂ + λη)δ .

.

Recall that .ψδ = ψ(0)δ for every .ψ ∈ C ∞ (RN ), which immediately gives .eληx δ = δ. Taking all this into account, and using once again Lemma 6.9, we get 

 ληx −1 P (2π iw+λη) = eληx P (−∂ + λη)δ = P (−∂ + 2λη)(eληx δ) .P (∂) e Fw P (2π iw+λη) = P (−∂ + 2λη)δ = =

aα (−∂ + 2λη)α δ

|α|≤m



aα (−∂ + 2λη)α δ +

|α|=m

= λm





|α|=m



aα (−∂ + 2λη)α δ

|α|≤m−1 m−1 

aα (2η)α δ + λ k Tk , k=0

250

6 Distributions and Partial Differential Operators

where each .Tk ∈ E  (RN ) is a linear combination of derivatives .∂ α (where .1 ≤ |α| ≤ m) of .δ. Note also that . |α|=m aα (2η)α = Pm (2η). With this and Lemma 6.10, we finally have P (∂)E =

1

m

Pm (2η)

j =0

.

=

   P (2π iw + λj η) aj P (∂) eλj ηx F−1 w P (2π iw + λj η) 1

Pm (2η)

m

λm j aj Pm (2η)δ +

j =0

=

m

m−1

λkj Tk



k=0

λm j aj δ +

j =0

1

m−1

Pm (2η)

k=0

Tk

m

 λkj aj = δ ,

j =0



which completes the proof.

6.5.2 Solutions of Linear PDEs Our aim now is to find conditions under which a linear partial differential equation with constant coefficients has a solution. To be more precise, the problem that we address now is to find conditions that ensure that, if P is a polynomial of N variables and g is some .C ∞ -function, then the PDE P (∂)f = g

.

(6.27)

has a solution in some sense. This will come as a consequence of the Malgrange– Ehrenpreis Theorem 6.31. To begin with, we need some basic facts on convolution of distributions and functions that we briefly recall here. We refer the interested reader to [7], [13], or [18, Chapter 27] for a complete account and detailed proofs. If .ϕ ∈ C ∞ (RN ) and .T ∈ D  (RN ) are such that at least one of them has compact support, then the function defined for .x ∈ RN by (T ∗ ϕ)(x) = Ty , ϕ(x − y)

.

(meaning that, for a fixed x, the distribution T acts on the function .y  ϕ(x + y)) is a well defined function in .C ∞ (RN ) that satisfies .

supp(T ∗ ϕ) ⊆ supp ϕ + supp T .

Proposition 6.20 If .T ∈ E  (RN ) (has compact support), then the map .CT : C ∞ (RN ) → C ∞ (RN ) given by .CT (ϕ) = T ∗ ϕ is linear and continuous.

6.5 Linear Partial Differential Operators

251

Proposition 6.21 If .T ∈ D  (RN ) and .ϕ, ψ ∈ D(RN ), then .(T ∗ ϕ) ∗ ψ = T ∗ (ϕ ∗ ψ).  Proposition 6.22 Let .ϕ ∈ D(RN ) be such that .ϕ ≥ 0, . RN ϕ(x)dx = 1, and N .supp ϕ = {x ∈ R : |x| ≤ 1}. For each .ε > 0, we consider the function given by x  1  N .ϕε (x) = N ϕ ε . For every .T ∈ D (R ), we have: ε 1. .T ∗ ϕε ∈ C ∞ (RN ). 2. .supp(T ∗ ϕε ) ⊆ supp T + {x ∈ RN : |x| ≤ ε}. 3. .limε↓0+ (T ∗ ϕε )(ψ) = T (ψ) for every .ψ ∈ D(RN ). We denote .Tε = T ∗ ϕε and call them the regularizations of T . Just taking ε = 1/n, we obtain the following:

.

Corollary 6.6 Let . ⊆ RN be open, and .T ∈ D  (). Then there exists a sequence ∞ .(Sj )j ⊆ C () such that .limj →∞ Sj (ψ) = T (ψ) for every .ψ ∈ D(). We can now consider the convolution of two distributions. Theorem 6.33 Let .S ∈ E  (RN ) and .T ∈ D  (RN ). There is a well defined distribution, denoted .S ∗ T such that & '   S ∗ T (ϕ) = Sx , Ty , ϕ(x + y) = S ∗ (T ∗ ϕq) (0)

.

for every .ϕ ∈ D(RN ). Proposition 6.23 Let T be any distribution; then .δ ∗ T = T . Also, if T and S are two distributions, one of them having compact support, then: 1. 2. 3. 4.

T ∗S = S ∗T. (S ∗ T ) ∗ R = S ∗ (T ∗ R) for every .R ∈ E  (RN ). N α α α .∂ (S ∗ T ) = S ∗ (∂ T ) = (∂ S) ∗ T for every .α ∈ N . 0 .supp(S ∗ T ) ⊆ supp T + supp S. In particular, if T and S have compact support, so also has .T ∗ S. . .

We can now give our first result on the existence of solutions of linear partial differential equations with constant coefficients (6.27). Proposition 6.24 Let P be a non-zero polynomial of N variables: 1. For each .T ∈ E  (RN ), there is .S ∈ D  (RN ) such that P (∂)S = T .

.

2. For each .g ∈ D(RN ), there is .f ∈ C ∞ (RN ) such that P (∂)f = g .

.

3. Let .,  ⊆ RN be open and such that . is compact and contained in .. Then, for each .g ∈ C ∞ (), there is .f ∈ C ∞ () such that .(P (∂)f )| = g| .

252

6 Distributions and Partial Differential Operators

The last statement is known as the theorem of local solvability. Proof Let E be a fundamental solution of .P (∂) (which, by Theorem 6.31, exists). If .T ∈ E  (RN ), let .S = E ∗ T , and use Proposition 6.23 to get P (∂)S = P (∂)(E ∗ T ) = (P (∂)E) ∗ T = δ ∗ T = T ,

.

which yields 1. Exactly the same argument shows that 2 holds. In order to see 3, let us take .ϕ ∈ D(RN ) with .ϕ(x) = 1 for every x in some neighbourhood of . . Given .g ∈ C ∞ (RN ), the product .ϕg is in .D(RN ) and, by 2, we can find .f˜ ∈ C ∞ (RN ) with .P (∂)f˜ = ϕg. If we denote the restriction of .f˜ to  ., we have that .P (∂)f (x) = ϕ(x)g(x) = g(x) for every .x ∈  . This completes the proof.  Remark 6.6 Since .D() is dense in .C ∞ () for each open . ⊆ RN , Proposition 6.24 yields that the continuous linear operator .P (∂) : C ∞ () → C ∞ () has dense range. We are interested now in seeing if this operator is surjective. Note that, if this is the case, then the Eq. (6.27) has a solution in .C ∞ () for every .g ∈ C ∞ (). In order to study the surjectivity of the operator .P (∂) : C ∞ () → C ∞ (), we introduce the following notation: NP () = {f ∈ C ∞ () : P (∂)f = 0} .

.

This is just the kernel of the operator .P (∂), that is, the set of solutions of the homogeneous equation .P (∂) = 0. With this notation, .NP (RN ) is the space of global solutions of the homogeneous equation. It is easy to check that .P (∂)(eζ x ) = P (ζ )eζ x for every .ζ ∈ CN . Then, .ez0 x ∈ NP (RN ) for every .z0 ∈ CN with .P (z0 ) = 0. The following result was proved by Malgrange in 1953. We refer to [7, Theorem 7.3.6] for a proof of it. Theorem 6.34 Let . ⊆ RN be open and convex and P a non-zero polynomial of N variables. For each .f ∈ NP (), there is a sequence .(hk )k ⊆ NP (RN ) such that ∞ .limk→∞ hk = f in .C (). We use this result and a Mittag-Leffler method to show that, whenever the set . is convex, the operator .P (∂) : C ∞ () → C ∞ () is surjective (that is, the equation in (6.27) has a solution for every .g). Theorem 6.35 Let . ⊆ RN be open and convex and P a non-zero polynomial of N variables. For each .g ∈ C ∞ (), there is .f ∈ C ∞ () such that .P (∂)f = g. Proof Let .(Kn )n be a fundamental sequence of compact, convex subsets of . such that ˚2 ⊆ K2 ⊆ K ˚3 ⊆ K3 ⊆ · · · ⊆  . K1 ⊆ K

.

6.5 Linear Partial Differential Operators

253

˚j +1 and (recall For each .j ∈ N, we take some open set .Uj with .Kj ⊆ Uj ⊆ K Proposition 6.2) consider .ψj ∈ D() such that .ψj (x) = 1 for every .x ∈ Uj and .ψj (x) = 0 for .x ∈  \ Kj +1 . Define ϕ1 = ψ1 and ϕj = ψj − ψj −1 for j ≥ 2 .

.

Given any .x ∈ , there is some .n0 with .x ∈ Kn0 \ Kn0 −1 . Then .ψj (x) = 1 for every .j ≥ n0 , and .ϕj = 0 for every .j ≥ n0 + 1. Hence, ∞

.

n0

ϕj (x) =

j =1

ϕj (x) = ψ1 (x) +

j =1

n0

  ψj (x) − ψj −1 (x) = ψn0 (x) = 1 . j =2

 In other words, the family is locally finite and . ∞ j =1 ϕj = 1. For each .j ∈ N, we define .gj : RN → C by .gj (x) = g(x)ϕj (x) if .x ∈  and 0 otherwise. Clearly, .gj ∈ D(RN ) and, by Proposition 6.24–2, there is .fj ∈ C ∞ (RN ) with P (∂)fj = gϕj .

(6.28)

.

˚j (an open, convex subset Note that for each fixed .j ≥ 2 we have .P (∂)fj = 0 on .K ∞ of .), and then by Theorem 6.34, there exists .hj ∈ C (RN ) such that P (∂)hj = 0

(6.29)

.

and (recall the definition of the seminorms in (4.8)) .

  1 sup ∂ α (fj − hj ) < j . 2 x∈Kj −1 |α| m and note that .Km ⊆ Kk−1 for every .k > n. Then, for .p ∈ N, we have

.

n+p  n+p 

1  α  sup sup  ∂ (fj − hj )(x) < . 2j x∈Km |α|≤m j =n+1

j =n+1

254

6 Distributions and Partial Differential Operators

This implies our claim and then .f ∈ C ∞ (). Finally, since (6.28) and (6.29) hold for every j , we have P (∂)f =

.

∞ ∞ ∞



 

gϕj = g ϕj = g , P (∂)fj − P (∂)hj = j =1

j =1

j =1



which completes the proof.  .D ()

For a detailed study of partial differential operators .P (∂) on and on C ∞ (), we refer the reader to [8, Chapter X]. We finish by considering a notion introduced by Schwartz. A non-zero polynomial of N variables P and its associated partial differential operator .P (∂) are called hypoelliptic on an open set . ⊆ RN if for each open, non-empty .U ⊆  and each  ∞ ∞ .u ∈ D () such that .P (∂)u ∈ C (U ) we have .u ∈ C (U ). If the polynomial  P is hypoelliptic on ., then every distribution .u ∈ D () that is solution of the homogeneous equation .P (∂)u = 0 belongs to .C ∞ (). .

Theorem 6.36 A non-zero polynomial of N variables P is hypoelliptic on .RN if and only if .P (∂) has a fundamental solution whose restriction to .RN \ {0} belongs to .C ∞ (RN \ {0}). Proof Assume that P is hypoelliptic on .RN , and let E be a fundamental solution of N ∞ N .P (∂). Since .P (∂)E = δ and the restriction of .δ to .R \{0} belongs to .C (R \{0}), the hypoellicity of P gives the conclusion. To prove the converse, let E be a fundamental solution whose restriction to .RN \ {0} belongs to .C ∞ (RN \ {0}). Fix some open set .U ⊆ RN and take .u ∈ D  (RN ) such that .f = P (∂)u ∈ C ∞ (U ). We want to show that .u ∈ C ∞ (U ). To do this, take .x0 ∈ U and .g ∈ D(U ) such that .g(x) = 1 for every x in some neighbourhood .U0 of .x0 with .U0 ⊆ U . Then P (∂)(gu) =



.

aα ∂ α (gu) = gP (∂)u

|α|≤m

+

0 −1, we define the functions  λ .x+

=

xλ

for x > 0,

0

for x ≤ 0

and

λ x−

=

 |x|λ

for x < 0,

0

for x ≥ 0 .

λ with Show that they define distributions on R. Compute the derivative of T = x+ −1 < λ < 0.

6.8 Prove that Remark 6.4 holds true. 6.9 Compute the Fourier transform of: ⎧ ⎨1 − |x| if |x| ≤ R 1. f (x) = R ⎩0 otherwise .  2 1−x if |x| ≤ 1 2. f (x) = 0 otherwise .  e−ax if x ≥ 0 3. Given a > 0, f (x) = 0 otherwise . −a|x| 4. f (x) = xe with a > 0. 5. f (x) = e−a|x| cos(bx), with a > 0. 1 . 6. f (x) = 1 + x2 x . 7. f (x) = 1 + x2 8. f (x) = cos(ax). 9. f (x) = sin3 (x). sin(ax) . 10. f (x) = πx (k) 11. (δa ) for k = 0, 1, 2, 3, . . .. 12. x k for k = 0, 1, 2, 3, . . .. 6.10 Let f, g ∈ S(R) satisfy f ∗ g = 0. Does it follow that f = 0 or g = 0?

6.6 Exercises

257

6.11 Let f ∈ S(R2 ) and define g(x) = f (x, 0). Find the relation between f and  g. 6.12 Use the inversion formula to find the Fourier transform of f (x) =

.

2a . a 2 + 4π 2 x 2

This can also be done directly solving the integral using residues. 6.13 Let F (w) be the Fourier transform of f (x) ∈ L1 (R). Compute the Fourier transform of the following functions: 1. f (x) = f (x − 1) + f (−x − 1). 2. f (x) = f (3x − 6). 3. f3 (x) = f  (x + 1). 6.14 It is possible to improve the inversion formula for the Fourier transform for piecewise functions that are in L1 (R) and in L∞ (R) (see [4, pages 228–231]). Use this result, Plancherel’s formula and the Fourier transform of χ[−1,1] to compute the following integrals:  +∞ sin x dx. 1. x −∞ +∞ sin2 x dx. 2. x2 −∞ 6.15 Give a direct proof, using the definitions, of the following statements: d is hypoelliptic on R. dx ∂ is not hypoelliptic on R2 . 2. P (∂) = ∂x

1. P (∂) =

6.16 Show that if T ∈ D  (RN ) and supp T = {a} for some a ∈ RN , then there are k ∈ N0 and cα ∈ K for |α| ≤ k so that T =



.

cα ∂ α δa .

|α|≤k

6.17 Let T ∈ S  (RN ) be a tempered distribution such that T = 0 (where  is the Laplace operator). Prove that T is a polynomial. Hint: Use the Fourier transform and the previous exercise. 6.18 Let a ∈ R. Prove that a fundamental solution of the differential operator d + a is E = e−a H (where H is the Heaviside function). P (∂) = dx ∂ on R2 . 6.19 Find a fundamental solution of P (∂) = ∂x

258

6 Distributions and Partial Differential Operators

6.20 The Schrödinger partial differential operator is P (∂) =

.

1 ∂ − x , i ∂t

for (t, x) ∈ Rn+1 . A fundamental solution is given by   2  n E = H (t) exp − i(n − 2) π4 (4π t)− 2 exp |x| 4it

.

(see [17, p. 43–44]). Show that E is not a C ∞ -function in any neighbourhood of a point (0, x0 ) and the Schrödinger is not hypoelliptic. 6.21 Let P (∂) be a linear partial differential operator in R2 such that the equation ⎫ ⎧ ⎬ ⎨ P (∂)u = f on R2 . ∂u ⎩ u(x, 0) = 0 = (x, 0) ⎭ ∂t

(6.31)

has a unique solution u ∈ C ∞ (R2 ) for each f ∈ C ∞ (R2 ). This happens for the ∂2 ∂2 − , for example. Show that R : C ∞ (R2 ) → wave equation P (∂) = 2 ∂t ∂x 2 C ∞ (R2 ) given by Rf = u (the unique solution of (6.31)) defines a continuous linear operator and that P (∂)R is the identity operator on C ∞ (R2 ). Hint: Use the closed graph theorem and the uniqueness of the solution of (6.31). 6.22 Let f ∈ S(RN ), and P (∂) be a non-constant partial differential operator on RN with constant coefficients. Show that if P (∂)f = 0, then f = 0. Hint: Use the Fourier transform and that the set {z ∈ CN : P (z) = 0} is dense in CN . d2 d + b. Let +a dx dx 2   f, g be solutions of P (∂)u = 0 such that f (0) = g(0) and f (0) − g (0) = 1. Show that  f (x) if x ≤ 0 .G(x) = g(x) if x > 0

6.23 Consider the differential operator on R given by P (∂) =

is a fundamental solution of P (∂). 6.24 Use polar coordinates to show that f : R2 → C given by f (x, y) = satisfies ∂f = π δ (recall that ∂ is the Cauchy–Riemann operator, see (6.23)). Further Reading [1–3, 6–9, 13–15, 17, 18, 20]

1 x+iy

References

259

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Index

Symbols 105 .(x, y), 76 ◦ .A , 110 ⊥ .A , 78 .B(a, r), 1 .BX , 43 .Bd (x, ε), 42 .Bp (x, ε), 48 .C(K), 47 .C(T), 119 .C(), 60 .C(z0 , r), 1 .C[a, b], 47 ∞ .C (), 151 k .C (), 150 .Cϕ , 154 .Dn , 16  .E , 57 ∗ .E , 57  .Eβ , 126 .H (x), 208 .L(E), 53 .L(E, F ), 53 .L1,loc (), 205 .Lβ (E, F ), 126 .L∞ (μ), 46 .Lco (E, F ), 127 .Lp (μ), 46 .Ls (E, F ), 127 .Lw (E, F ), 127 .Mg , 154 .P (∂), 156 1 .P V , 209 x  N .S (R ), 230 .(E, F ),

N ),

219 161 .Tf , 208 .Tn , 170 .Exp, 148 ., 41 .T , 53  .β(E , E), 126 .χA , 1 .δ, δa , 208 . ∞ , 45 . p , 45 .Im, 132 .ker, 91, 132 .λp (A), 63 .·, · , 104 .1, 102, 215 .D, 1 .D(z0 , r), 1 .N0 , 1 .T, 21 .F, 148, 222, 234 .U0 (E), 48 ˚ 40 .A, .H(), 137 .H({0}), 146 .D (K), 203 .D (), 206  .D (), 207  .E (), 218 .P , 48 .G(T ), 122 .μ(E, F ), 115 .ω, 50 .Orb(x, T ), 161 .S(R .T

n,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 J. Bonet et al., Function Spaces and Operators between them, RSME Springer Series 11, https://doi.org/10.1007/978-3-031-41602-6

265

266 .A,

40

1 .σ (E, F ), 106 .σ (F, E), 106 .supp, 203, 217 .τ |Y , 41 .τP , 48 .τco , 60 .τd , 42 .τp , 73 .(A), 49 .|x|, 1 . μ, 148 (n), 12 .f (w), 222 .f ∗ . T , 131 t . T , 131 .c0 , 45 .d(x, A), 43 .en , 46 .f ∗ g, 204 .mN , 47 .pK , 60 s, 179 .u ∗ ψ, 238 .B(a, r),

A Abel summation, 24 Absolutely convex hull, 49 Absolutely convex set, 48 Absorbing set, 49 Alembert formula, 241 Automorphism, 186 elliptic, 187 hyperbolic, 187 parabollic, 187

B Balanced set, 48 Ball, 42, 43 Banach limit, 102 Bipolar of a set, 111 Bounded set, 50, 53, 62, 126, 127

C Canonical vectors, 46 Cauchy criterion, 4, 9 Cauchy principal value, 208, 214, 215, 237 Cauchy sequence, 47, 57 Cesàro mean, 18, 170

Index Closed set, 39, 43, 98 Closure, 40 Compact, 41, 98 locally, 73 operator, 86 relatively, 41, 66, 73, 75, 114, 144, 152 Continuous, 40, 50 linear mapping, 52 seminorm, 52 Convergence absolute, 9 of nets, 40, 107 pointwise, 2, 9 of sequences, 40, 43, 50, 57 uniform, 3, 5, 6, 9, 19, 47 and continuity, 5 Convex set, 48, 97 Convolution of distributions, 251 of distributions and functions, 238, 241, 250 of functions, 16, 19, 22, 204, 230

D Denjoy–Wolff point, 190 Dense, 41 Dirac function, 208, 213, 215, 218, 234 Dirichlet kernel, 16, 120 Dirichlet series, 24 abscissas of convergence, 25, 26 convergence on half-planes, 25 Distance, see Metric Distance between a point and a set, 43 Distribution, 203, 207 defined by a function, 208, 212, 214 derivative, 213 of finte order, 211 multiplication by a function, 214 tempered, 230, 233 vanish on a set, 217 Dual, 57, 81, 91, 105, 126 algebraic, 57 pair, 105, 131, 132, 212, 245 strong, 126

E Equicontinuous, 54, 58, 74, 75, 110, 113, 116, 127–130, 144, 166 for linear mappings, 54 Exponential type, 148

Index

267

F Féjer kernel, 18 Fourier–Borel transform, 148, 196 Fourier coefficient, 12, 15, 20, 22, 85 Fourier series, 12, 85, 119 exponential, trigonometric, 13 Fourier transform, 234 and differentiation, 225 of a distribution, 233, 237–240 of functions, 222 is an isomorphism, 229 Functional, 57, 91, 94 Fundamental sequence of compact sets, 61

M Metric, 41 Minkowski gauge, 96, 118 Multi-index, 150

G Germ of a holomorphic function, 146 Graph of a mapping, 122

O Open mapping, 123 Open set, 39 Operator backward shift, 102, 182 dynamical/ergodic properties, 182, 185 Cauchy–Riemann, 243, 255 Cesàro bounded, 170, 171, 173, 176 chaotic, 168 compact, 199 composition, 154, 186 dynamical/ergodic properties, 187, 189, 191, 193, 194 differential dynamical/ergodic properties, 197 differentiation, 154 heat, 157, 243 hypercyclic, 164, 165, 182, 193 is transitive, 164 integration, 155, 194, 195 Laplacian, 157, 239, 243, 255 linear partial differential, 155, 241 associated polynomial, 241 existence of solution, 250–252 fundamental solution, 242, 248, 254 hypoelliptic, 254 principal part, 242 symbol, 156 mean ergodic, 170, 173–176, 191, 194 multiplication, 154, 194 dynamical/ergodic properties, 194 never hypercyclic, 194 power bounded, 170, 182, 187, 189, 191, 194 is Cesàro bounded, 170 reflexive, 173 semi-reflexive, 173

H Hadamard formula, 11 Haviside function, 208 Heaviside function, 213, 237 Holomorphic, 137 Hypercyclic vector, 164, 167 Hyperplane, 91

I Inequality Bessel, 82 Hölder, 44–46, 59 Minkowski, 44–46 Schwarz, 77 Inner product, 76 Interior, 40 Isomorphism, 53

K Köthe echelon space, 63, 142, 179 bounded sets, 65 dual, 64 is Fréchet, 63 Köthe matrix, 62, 65, 142 Kernel of an operator, 91, 132, 252

L Limit of a sequence, 40 Locally integrable function, 205

N Neighbourhood, 39, 106 basis of, 39, 40, 43, 48, 50, 111, 126 Net, 40 Norm, 43, 47 of an inner product space, 76 operator, 126

268 transisitve, 164 transitive, 162, 163, 166, 193 uniformly mean ergodic, 170, 177, 178, 180, 187, 189, 191, 194 Volterra, 200 wave, 157, 240, 243, 255 Orbit, 161, 167, 172, 182 Order partial, 40 total, 40 Orthogonal projection, 80, 81 set, 78 vectors, 78 Orthonormal set, 81, 83, 84 complete, 84

P Parseval identity, 85 Partition of unity, 207 Periodic point, 168 Poisson integral, 22 Polar of a set, 112 Polar set, 110 Positively homogeneous, 93 Possion kernel, 21 Power series, 10, 139, 146 Precompact set, 100, 127

Q Quasi conjugate, 169

R Radius of convergence, 11 Rapidly decreasing function, 219 Riesz–Fisher, 83 Run away, 192

S Seminorm, 48 Sensitive dependence on the initial conditions, 162 Series of functions, 9 Space of all sequences, 50, 57, 68, 105 of bounded sequences, 45 of .C ∞ functions, 151 bounded sets, 152 is Fréchet, 151

Index of .C ∞ -functions with compact support, 203 is Fréchet, 203 of .C k functions, 150 is Fréchet, 152 complete, 47 of continuous functions, 47, 60 is complete, 47, 61 not normable, 68 of distributions with compact support, 218 contained in .S  , 233 Fréchet, 57, 138 Hilbert, 78 of holomorphic functions, 137 bounded sets, 144 dual, 146, 148 is Fréchet, 138 is not normable, 141 representation as a sequence space, 142 is separable, 140 inner product, 76 locally convex, 49 admits continuous norm, 67 finite dimensional, 101 metrizable, 55 normable, 67 sequentially complete, 57, 118 metric, 42 Montel, 177 normed, 43, 47 of null sequences, 45, 103 of p-integrable functions, 46 of rapidly decreasing functions, 219 is Fréchet, 221 of rapidly decreasing sequences, 179 of Schwartz, see rapidly decreasing functions semi-Montel, 177 separable, 41, 164 of tempred distributions, 230 of test functions, 206 is dense in .C ∞ , 206 topological, 39 Stable orbits, 190 Subadditive, 93 Support of a distribution, 217 of a function, 203

T Taylor series, 140, 142 Theorem Alaoglu-Bourbaki, 112

Index Ascoli, 75, 152 Bès-Peris, 163 Baire, 122, 165 Banach–Mackey, 119 Banach-Schauder, 122, 124, 178 Banach–Steinhaus, 116, 119, 130, 181 best approximation, 81 bipolar, 111, 114 Birkhoff, 164, 183, 193 Bohr, 27 Bourdon, 167 Cauchy, 27, 31, 138 Cauchy Integral Formula, 33, 138 closed graph, 122, 125 Denjoy–Wolff, 190 Dini, 6 Dini–Dirichlet, 17 Eberlein, 171, 173, 175 Féjer, 19, 85 Fourier inversion formula, 227 Godefroy–Shapiro, 165, 169 Goldstine, 112 Hahn–Banach, 133, 167, 173 analytic version, 93, 94 separation, 96–98 identity, 140 Kolmogorov, 69, 100 Lin, 178, 180, 184 of local solvability, 252 Mackey–Arens, 114 Malgrange–Ehrenpreis, 243 Montel, 144, 186 Morera, 138 open mapping, see Banach–Schauder Riemann–Lebesgue, 230 Riemann–Lebesgue Lemma, 15 Riesz, 100 Riesz representation, 80 Schwarz lemma, 186

269 Stone–Weierstraß, 69, 73, 85 Tikhonov, 41, 66, 75, 112 uniform boundedness principle, 118 Vitali, 145 Weierstraß, 73, 139 Weierstraß M-test, 10 Yosida, 175, 176, 180 Zorn’s lemma, 70 Topology of .C ∞ functions, 151 coarser, 39 compact-open, 61, 73, 138, 139, 142, 143 defined by a metric, 42 defined by a norm, 43 defined by seminorms, 48 of a dual pair, 113–115, 119 finer, 39 Hausdorff, 40, 49 induced, 41, 45 Mackey, 115 product, 41 simple, see strong operator strong, 126 strong operator, 127–129 of uniform convergence on bounded sets, 126 of uniform convergence on compact sets, see compact-open of uniform convergence on precompact sets, 127, 129 weak, 106, 108, 109, 127, 172 weak operator, 127 weak.∗ , 108, 109 Transpose of a mapping, 131, 165 Trigonometric polynomial, 11, 85 V von Neumann series, 199