Analysis on Polish Spaces and an Introduction to Optimal Transportation [1 ed.] 9781108421577, 1108421571

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London Mathematical Society Student Texts 89

Analysis on Polish Spaces and an Introduction to Optimal Transportation D. J. H. GARLING Emeritus Reader in Mathematical Analysis, University of Cambridge, and Fellow of St John’s College, Cambridge

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781108421577 DOI: 10.1017/9781108377362 © D. J. H. Garling 2018 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2018 Printed in the United Kingdom by Clays, St Ives plc A catalogue record for this publication is available from the British Library. Library of Congress Cataloging-in-Publication Data Names: Garling, D. J. H., author. Title: Analysis on Polish spaces and an introduction to optimal transportation / D.J.H. Garling (University of Cambridge). Other titles: London Mathematical Society student texts ; 89. Description: Cambridge, United Kingdom ; New York, NY : Cambridge University Press, 2018. | Series: London Mathematical Society student texts ; 89 | Includes bibliographical references and index. Identifiers: LCCN 2017028186 | ISBN 9781108421577 (hardback ; alk. paper) | ISBN 1108421571 (hardback ; alk. paper) | ISBN 9781108431767 (pbk. ; alk. paper) | ISBN 1108431763 (pbk. ; alk. paper) Subjects: LCSH: Polish spaces (Mathematics) | Mathematical analysis. | Transportation problems (Programming) | Topology. Classification: LCC QA611.28 .G36 2018 | DDC 514/.32–dc23 LC record available at https://lccn.loc.gov/2017028186 ISBN 978-1-108-42157-7 Hardback ISBN 978-1-108-43176-7 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Introduction PART ONE

1 TOPOLOGICAL PROPERTIES

7

1 1.1 1.2

General Topology Topological Spaces Compactness

9 9 15

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Metric Spaces Metric Spaces The Topology of Metric Spaces Completeness: Tietze’s Extension Theorem More on Completeness The Completion of a Metric Space Topologically Complete Spaces Baire’s Category Theorem Lipschitz Functions

18 18 21 24 27 29 31 33 35

3 3.1 3.2 3.3 3.4

Polish Spaces and Compactness Polish Spaces Totally Bounded Metric Spaces Compact Metrizable Spaces Locally Compact Polish Spaces

38 38 39 41 47

4 4.1 4.2 4.3 4.4

Semi-continuous Functions The Effective Domain and Proper Functions Semi-continuity The Br´ezis–Browder Lemma Ekeland’s Variational Principle

50 50 50 53 54

v

vi

Contents

5 5.1 5.2 5.3 5.4 5.5 5.6

Uniform Spaces and Topological Groups Uniform Spaces The Uniformity of a Compact Hausdorff Space Topological Groups The Uniformities of a Topological Group Group Actions Metrizable Topological Groups

56 56 59 61 64 66 67

6 6.1 6.2 6.3 6.4

C`adl`ag Functions C`adl`ag Functions The Space (D[0, 1], d∞ ) The Skorohod Topology The Metric dB

71 71 72 73 75

7 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Banach Spaces Normed Spaces and Banach Spaces The Space BL(X) of Bounded Lipschitz Functions Introduction to Convexity Convex Sets in a Normed Space Linear Operators Five Fundamental Theorems The Petal Theorem and Daneˇs’s Drop Theorem

79 79 82 83 86 88 91 95

8 8.1 8.2 8.3 8.4 8.5

Hilbert Spaces Inner-product Spaces Hilbert Space; Nearest Points Orthonormal Sequences; Gram–Schmidt Orthonormalization Orthonormal Bases The Fr´echet–Riesz Representation Theorem; Adjoints

97 97 101 104 107 108

9 9.1 9.2 9.3 9.4 9.5 9.6 9.7

The Hahn–Banach Theorem The Hahn–Banach Extension Theorem The Separation Theorem Weak Topologies Polarity Weak and Weak* Topologies for Normed Spaces Banach’s Theorem and the Banach–Alaoglu Theorem The Complex Hahn–Banach Theorem

112 112 116 118 119 120 124 125

Contents

vii

10 Convex Functions 10.1 Convex Envelopes 10.2 Continuous Convex Functions

128 128 130

11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

Subdifferentials and the Legendre Transform Differentials and Subdifferentials The Legendre Transform Some Examples of Legendre Transforms The Episum The Subdifferential of a Very Regular Convex Function Smoothness The Fenchel–Rockafeller Duality Theorem The Bishop–Phelps Theorem Monotone and Cyclically Monotone Sets

133 133 134 137 139 140 143 148 149 151

12 12.1 12.2 12.3

Compact Convex Polish Spaces Compact Polish Subsets of a Dual Pair Extreme Points Dentability

155 155 157 160

13 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8

Some Fixed Point Theorems The Contraction Mapping Theorem Fixed Point Theorems of Caristi and Clarke Simplices Sperner’s Lemma Brouwer’s Fixed Point Theorem Schauder’s Fixed Point Theorem Fixed Point Theorems of Markov and Kakutani The Ryll–Nardzewski Fixed Point Theorem

162 162 165 167 168 170 171 173 175

PART TWO MEASURES ON POLISH SPACES

177

Abstract Measure Theory Measurable Sets and Functions Measure Spaces Convergence of Measurable Functions Integration Integrable Functions

179 179 182 184 187 188

14 14.1 14.2 14.3 14.4 14.5

viii

15 15.1 15.2 15.3 15.4 15.5

Contents

Further Measure Theory Riesz Spaces Signed Measures M(X), L1 and L∞ The Radon–Nikodym Theorem Orlicz Spaces and Lp Spaces

191 191 194 196 199 203

16 Borel Measures 16.1 Borel Measures, Regularity and Tightness 16.2 Radon Measures 16.3 Borel Measures on Polish Spaces 16.4 Lusin’s Theorem 16.5 Measures on the Bernoulli Sequence Space (N) 16.6 The Riesz Representation Theorem 16.7 The Locally Compact Riesz Representation Theorem 16.8 The Stone–Weierstrass Theorem 16.9 Product Measures 16.10 Disintegration of Measures 16.11 The Gluing Lemma 16.12 Haar Measure on Compact Metrizable Groups 16.13 Haar Measure on Locally Compact Polish Topological Groups

210 210 214 215 216 218 222 225 226 228 231 234 236 238

17 17.1 17.2 17.3 17.4 17.5 17.6 17.7

Measures on Euclidean Space Borel Measures on R and Rd Functions of Bounded Variation Spherical Derivatives The Lebesgue Differentiation Theorem Differentiating Singular Measures Differentiating Functions in bv0 Rademacher’s Theorem

243 243 245 247 249 250 251 254

18 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9

Convergence of Measures The Norm .TV The Weak Topology w The Portmanteau Theorem Uniform Tightness The β Metric The Prokhorov Metric The Fourier Transform and the Central Limit Theorem Uniform Integrability Uniform Integrability in Orlicz Spaces

257 257 258 260 264 266 269 271 276 278

Contents

19 19.1 19.2 19.3 19.4 19.5 19.6 19.7

ix

Introduction to Choquet Theory Barycentres The Lower Convex Envelope Revisited Choquet’s Theorem Boundaries Peak Points The Choquet Ordering Dilations

280 280 282 284 285 289 291 293

PART THREE INTRODUCTION TO OPTIMAL TRANSPORTATION

297

20 20.1 20.2 20.3 20.4 20.5 20.6 20.7

Optimal Transportation The Monge Problem The Kantorovich Problem The Kantorovich–Rubinstein Theorem c-concavity c-cyclical Monotonicity Optimal Transport Plans Revisited Approximation

299 299 300 303 305 308 310 313

21 21.1 21.2 21.3 21.4 21.5 21.6

Wasserstein Metrics The Wasserstein Metrics Wp The Wasserstein Metric W1 W1 Compactness Wp Compactness Wp -Completeness The Mallows Distances

315 315 317 318 320 322 323

22 22.1 22.2 22.3 22.4 22.5 22.6

Some Examples Strictly Subadditive Metric Cost Functions The Real Line The Quadratic Cost Function The Monge Problem on Rd Strictly Convex Translation Invariant Costs on Rd Some Strictly Concave Translation–Invariant Costs on Rd

325 325 326 327 329 331 336

Further Reading Index

339 342

Introduction

Analysis is concerned with continuity and convergence. Investigation of these ideas led to the notions of topology and topological spaces. Once these had been introduced, they became subjects in their own right, which were investigated in fine detail to see how far the theory might lead (an excellent illustration of this is given by the fascinating book by Steen and Seebach [SS]). In practice, however, a great deal of analysis is concerned with what happens on a very restricted class of topological spaces, namely, the Polish spaces. A Polish space is a separable topological space whose topology is defined by a complete metric. Important examples include Euclidean space, pathwise-connected Riemannian manifolds, compact metric spaces and separable Banach spaces. The purpose of this book is to develop the study of analysis on Polish spaces. It consists of three parts. The first considers topological properties of Polish spaces, and the second deals with the theory of measures on Polish spaces. In the third part, we give an introduction to the theory of optimal transportation. This makes essential use of the results of the first two parts, or modifications of them. It was, in fact, study of optimal transportation that led to the realization of how much its study required properties of Polish spaces, and measures on them. There are three important advantages of restricting attention to Polish spaces. First, many of the curious complications of the general topological theory disappear. For example, a subspace of a separable topological space need not be separable, whereas a subspace of a separable metric space is always separable. Secondly, the proofs of standard results are frequently much easier in this restricted setting. For example, Urysohn’s lemma for normal topological spaces is quite delicate, whereas it is very easy for metric spaces. Thirdly, Polish spaces enjoy some very important properties. Thus it follows from Alexandroff’s theorem that a topological space is a Polish space if and only 1

2

Introduction

if it is homeomorphic to a Gδ subset of the Hilbert cube H = [0, 1]N , which is a compact metrizable space. From this, or directly, it follows that a Borel measure on a Polish space is tight (Ulam’s theorem: the measure of a Borel set can be approximated from below by the measures of compact sets contained in it). It also means that we can push forward a Borel measure on a Polish space X to a Borel measure on a compact metric space containing X. This greatly simplifies both the measure theory and also the construction of measures. In fact, I believe that almost all the probability measures that arise in practice are Borel measures on Polish spaces; one important exception, which we do not consider or need, is the theory of uniform central limit theorems. One major advantage of restricting attention to Polish spaces is that it is not necessary to appeal to the axiom of choice. Instead, we proceed by induction, using the axiom of dependent choice; we make an infinite sequence of decisions, each possibly dependent on what has gone before. In analysis, there are a few fundamental results which require the axiom of choice. The first is Tychonoff’s theorem, which states that an arbitrary product of compact topological spaces, with the product topology, is compact. We do not prove this, or use it. On the other hand, we do prove, and use, the fact that a countable product of compact metrizable spaces is compact and metrizable. Secondly, there are two fundamental results of linear analysis which need the axiom of choice, using Zorn’s lemma. The first of these is the Hahn–Banach theorem (together with the separation theorem). Using induction, we prove weak forms of these, for separable normed spaces; this is sufficient for our purposes. But for completeness’ sake we also give the classical results, using Zorn’s lemma; Here we first prove the separation theorem, showing that it essentially depends upon the connectedness of the unit circle T, and then derive the Hahn– Banach theorem from it. The other fundamental result which requires the axiom of choice is the Krein–Mil’man theorem, which states that every weakly compact convex subset K has an extreme point. Again, we only need, and use, the result in the case where K is metrizable, and we prove this without the axiom of choice. The fact that we avoid using the axiom of choice suggests that the proofs should, in some sense, be less abstract and more constructive. Unfortunately, this is not the case; the arguments that are used are frequently indirect (consider the collection of all sets with a particular property), so that for example a typical Borel subset of a Polish space does not have a simple description. Let us now describe the contents of the three parts of this book in more detail.

Introduction

3

Part I: Topological Properties Although it is assumed that the reader has some knowledge of general topology and metric spaces, the first two chapters give an account of these topics, including Tietze’s extension theorem, Baire’s category theorem and Lipschitz functions. This leads to the notion of a Polish space, a separable topological space whose topology is given by a complete metric. A fundamental example is given by a compact metrizable space, and Alexandroff’s theorem is used to show that a topological space is a Polish space if and only if it is homeomorphic to a Gδ subspace of a compact metric space, and in particular homeomorphic to a Gδ subspace of the Hilbert cube. We shall need to consider suprema of sets of real-valued continuous functions. Such functions are lower semi-continuous, and we consider such functions in Chapter 4. A lower semi-continuous function on a compact space attains its infimum, but this is not necessarily true for lower semi-continuous functions on a complete metric space. We establish its replacement, Ekeland’s variational principle, together with two of its corollaries, the petal theorem and Daneˇs’s drop theorem, and various other applications. Metric spaces have more structure than a topological one, and Chapter 5 contains an account of uniform spaces; uniformity is particularly important when we consider locally compact topological groups, in Part II. Chapter 6 is devoted to showing that the space of c`adl`ag functions is a Polish space under the Skorohod topology; many stochastic processes, and their underlying measures, lie on such spaces, and this helps justify the claim that almost all probability measures of interest lie on Polish spaces. Further examples are given by separable Banach spaces and Hilbert spaces; these are principally used to introduce the notion of convexity. The rest of Part I is concerned with convexity. The Hahn–Banach theorem is one of the key results here, and we give proofs of appropriate results, both without and with the axiom of choice. For us, the Hahn–Banach theorem is essentially a geometric theorem showing that two suitable convex sets can be separated by a hyperplane. It also leads onto the notion of weak topology. The Legendre transform provides an important duality theory for convex functions, and this leads naturally to the concept of subdifferentials and subdifferentiability. We prove the Bishop–Phelps theorem, and also introduce the notion of cyclic monotonicity. The rest of Part I is concerned with convex sets which are compact and metrizable in some suitable topology. We prove versions of the Krein–Mil’man theorem, Krein’s theorem and a swathe of fixed point theorems, many of which are used later.

4

Introduction

Part II: Measures on Polish Spaces We expect that the reader has some knowledge of abstract measure theory, but Chapter 14 contains a survey of the basic results. Chapter 15 contains some further results: we introduce the Banach space M(X) of finite measures on a Polish space X, its subspaces L1 (μ) and Orlicz spaces (with the use of Legendre duality). We give von Neumann’s Hilbert space proof of the Radon–Nikodym property and a proof of the strong law of large numbers (to be used later). In Chapter 16, we investigate Borel measures on Polish spaces. We prove regularity and tightness properties; we may not know what a typical Borel set looks like, but we can approximate the Borel measure of a Borel set from the outside by open sets, and on the inside by compact sets. This leads to Lusin’s theorem, which says that if μ is a Borel measure on a Polish space X then a Borel measurable function on X is continuous on a large compact subset. So far, all is theory, and no measures, other than trivial ones, have been shown to exist. We remedy this by showing how to construct Borel measures on the Bernoulli space (N), and then, pushing forward, constructing measures on compact metric spaces and Polish spaces. We prove the Riesz representation theorem, and use this to give a measure-theoretic proof of the Stone– Weierstrass theorem. We then show how Borel measures can be disintegrated, and establish the existence of Haar measure on compact and locally compact Polish spaces; we follow an account by Pedersen to show that this last result is relatively straightforward. In Chapter 17, we come down to earth and consider Borel measures on Euclidean space, where the point at issue is the differentiation of measures and of Borel measurable functions. We establish Lebesgue’s differentiation theorem and Rademacher’s theorem on the differentiability almost everywhere of Lipschitz functions. We now proceed to study one of the key points of this chapter, namely, the weak convergence of measures. We show that there are various metrics which define the weak topology w, and show that although the unit ball M1 (X) is generally not metrizable, the space of probability measures P(X) is a Polish space. Examples of weak convergence include the central limit theorem and the empirical law of large numbers. Finally, uniform integrability is investigated. Part II ends with an introduction to Choquet theory on a metrizable compact convex set. The theory is notoriously difficult for general weakly compact convex sets, but the difficulties disappear in the metrizable case. Parts I and II contain more than two hundred exercises. These are usually very straightforward, but most are an essential part of the text; do them.

Introduction

5

Part III: Introduction to Optimal Transportation The setting is this; μ and ν are Borel probability measures on Polish spaces X and Y, and c is a lower semi-continuous cost function on X × Y. We consider two problems. Kantorovich’s problem is to find a measure π on X × Y with marginals μ and ν with minimal cost X×Y c dπ. Monge’s problem is a special case of this; find a measurable mapping T : X → Y which pushes forward  μ to ν with minimal cost X c(x, T(x)) dμ(x). The results of Parts I and II are used, or modified, to tackle these problems. For example we can push forward μ and ν to measures on metrizable compactifications. We also consider the concepts of c-cyclic monotonicity and c-concavity. It is quite easy to show that Kantorovich’s problem has a solution, but with more care we introduce a ‘maximal Kantorovich potential’, which with its c-transform can give a great deal of information. When X = Y and c = dp , where d is a metric on X, we introduce and investigate the Wasserstein metric Wp , which is the minimal cost of transforming μ into ν. Similarly, we introduce the Mallows distance, which does the same for distributions of random variables. As an example, we prove a metric version of the central limit theorem. In the last chapter, we consider special cases. For example, we consider the case when X = Y = R, and the case where the cost is a quadratic function on a separable Hilbert space. Finally, following Gangbo and McCann [GMcC], we consider the cases when the cost on Rd is given by a strictly convex or strictly concave function. This only scratches the surface: for more, see the two large volumes by Villani, [V I] and [V II]. Although I have checked the proofs carefully, no doubt errors remain. Please consult my home page at www.dpmms.cam.ac.uk where a list of comments and corrections will be found, together with my email address, to which corrections should be sent.

PART ONE Topological Properties

1 General Topology

This chapter contains a brief account of topological spaces and their properties. It contains definitions and statements of fundamental results, and describes the notation that is used. Proofs are generally not given; they can be found in [G II] (and elsewhere).

1.1 Topological Spaces A topological space (X, τ ) is a set X together with a set τ of subsets of X, the topology, which satisfies (i) ∅ ∈ τ and X ∈ τ ; (ii) if F is a finite subset of τ , then ∩U∈F U ∈ τ ; and (iii) if G is any subset of τ , then ∪U∈G U ∈ τ . The elements of τ are called open sets. Here are some examples. The set P(X) of all subsets of X is a topology on X, the discrete topology, and the set {∅, X} is also a topology on X, the trivial topology. The usual topology on the real line R is defined by saying that U is open if whenever x ∈ U there exists δ > 0 such that (x − δ, x + δ) = {y : x − δ < y < x + δ} ⊆ U. Similarly, the usual topology on C is defined by saying that U is open if whenever z ∈ U there exists δ > 0 such that {w : |w − z| < δ} ⊆ U. Let R = [−∞, ∞]. The usual topology on R is defined by saying that U is open if

9

10

General Topology

(i) U ∩ R is open in R in the usual topology, (ii) if ∞ ∈ U there exists R ∈ R such that (R, ∞] ⊆ U, and (iii) if −∞ ∈ U there exists R ∈ R such that [−∞, R) ⊆ U. The right half-open topology τr on R is defined by saying that U ∈ τr is open if whenever x ∈ U there exists δ > 0 such that [x, x + δ) = {y : x ≤ y < x + δ} ⊆ U. Suppose that (X, τ ) is a topological space. A subset B of τ is a base for the topology if every U ∈ τ is the union of sets in B. Thus the open intervals (r, s) with r, s ∈ Q form a countable base for the usual topology on R. A subset C of X is closed if X \ C is open. If A ⊆ X, the interior Aint of A is the union of the open sets contained in A, the closure A of A is the intersection of the closed sets containing A, and the boundary, or frontier, ∂A of A is the set A \ Aint . Aint is the largest open set contained in A, and A is the smallest closed set containing A. Elements of Aint are called interior points of A, and elements of A are called closure points of A. A subset B of A is dense in A if A ⊆ B. A subset N of X is a neighbourhood of an element x of X if x ∈ N int . The set of neighbourhoods of x is denoted by Nx . A subset Bx of Nx is a base of neighbourhoods of x if every N ∈ Nx contains an element of Bx . A punctured neighbourhood of x is a set of the form N \ {x}, where N ∈ Nx . An element x of X is an accumulation point, or limit point, of A if N ∗ ∩ A = ∅, for each punctured neighbourhood N ∗ of x. A point of A is an isolated point of A if it is not an accumulation point of A. Here are two easy ways of constructing topological spaces. Suppose that Y is a subset of a topological space (X, τ ). The subspace topology on Y is defined by taking the set {U ∩ Y : U ∈ τ } of subsets of Y as the topology on Y. Suppose that f is a surjective mapping of X onto a set Z. The quotient topology on Z is defined by taking the sets {V : f −1 (V) ∈ τ } as the topology on Z. A most important construction is the construction of topological product spaces. Suppose that (Xα , τα )α∈A is a family of topological spaces. Let  X = α∈A Xα , and for each α ∈ A, let πα : X → Xα be the co-ordinate projection; πα (x) = xα . The product topology on X is defined by taking the collection {∩α∈F πα−1 (Uα ) : F a finite subset of A, Uα ∈ τα for α ∈ F} as base of open sets. One special case occurs when (Xα , τα ) = (X, τ ) for each  α ∈ A. In this case, α∈A Xα = X A , the space of mappings from A to X. In particular, X N (where N is the set {1, 2, 3, . . .} of natural numbers) is the space of sequences in X. A product of the form [0, 1]A , where [0, 1] is given its usual subspace topology, as a subspace of R, is called a hypercube. The space [0, 1]N is called the Hilbert cube.

1.1 Topological Spaces

11

Suppose that f is a mapping from a topological space (X, τ ) into a topological space (Y, σ ), and that x ∈ X. Then f is continuous at x if f −1 (N) ∈ Nx , for each N ∈ Nf (x) . Note that this agrees with the usual definition of continuity, when X = Y = R, with the usual topology. f is continuous on X if it is continuous at each point of X. Then f is continuous on X if and only if f −1 (V) ∈ τ , for each V ∈ σ , and if and only if f −1 (C) is closed in X for each closed subset C of Y. The composition of two continuous functions is continuous. If τ1 and τ2 are two topologies on X, then we say that τ1 is finer, or stronger, than τ2 , and that τ2 is coarser, or weaker, than τ1 , if the identity mapping i : (X, τ1 ) → (X, τ2 ) is continuous; that is, if τ2 ⊆ τ1 . A bijective mapping f : (X, τ ) → (Y, σ ) is a homeomorphism if f and f −1 are both continuous; that is, if f (τ ) = σ .  Suppose that (X, τ ) = α∈A (Xα , τα ), and that α ∈ A. The co-ordinate projection πα is a continuous mapping from (X, τ ) onto (Xα , τα ). A mapping f from a topological space (Y, σ ) into (X, τ ) is continuous if and only if πα ◦ f : (Y, σ ) → (Xα , τα ) is continuous, for each α ∈ A.  Suppose that (X, τ ) = α∈A (Xα , τα ), that α ∈ A and that x ∈ X. We define a mapping kx,α , the cross-section mapping, from Xα into X. If y ∈ Xα , let (kx,α (y))α = y, and let (kx,α (y))β = xβ if β = α. kx,α is a homeomorphism of (Xα , τα ) onto (kx,α (Xα ), τ ). A mapping f from a set X to a set Y is defined as a relation on X × Y which satisfies certain conditions. It is therefore natural to consider the corresponding graph mapping G( f ) from X to X×Y, by setting G( f )(x) = (x, f (x)), for x ∈ X. The set f = G( f )(X) is the graph of X. Exercise 1.1.1 Suppose that X and Y are topological spaces, and that f is a mapping from X into Y. Then f is continuous if and only if the graph mapping G( f ) is a homeomorphism of X onto f . Suppose that (xn )∞ n=1 is a sequence in a topological space (X, τ ) and that x ∈ X. Then xn converges to x if for each N ∈ Nx there exists n0 such that xn ∈ N for each n ≥ n0 ; if so, we write that xn → x as n → ∞. x is an accumulation point or limit point of the sequence if for each N ∈ Nx and each n ∈ N there exists m ≥ n such that xm ∈ N. If xn → x as n → ∞, then x is a limit point of the sequence. Suppose that f is a continuous mapping from a topological space (X, τ ) into a topological space (Y, σ ), and that x ∈ X. f is sequentially continuous at x if f (xn ) → f (x) as n → ∞ whenever xn → x as n → ∞, and is sequentially continuous on X if it is sequentially continuous at each point of X. A continuous mapping is sequentially continuous; as we shall see, the converse is generally not true.

12

General Topology

There are conditions that control the size of a topological space, and of topologies. A topological space is separable if there is a countable dense subset. It is first countable if every point has a countable base of neighbourhoods, and it is second countable if there is a countable base for the topology. These notions are related in the following way. Exercise 1.1.2 (i) A second countable space is first countable and separable. (ii) A subspace of a first countable space is first countable, and the product of countably many first countable spaces is first countable. (iii) A subspace of a second countable space is second countable, and the product of countably many second countable spaces is second countable. (iv) The product of countably many separable topological spaces is separable. (But see Proposition 1.1.3.) (v) A mapping from a first countable topological space into a topological space is continuous if and only if it is sequentially continuous. (vi) The topological space (R, τr ) is first countable and separable, but is not second countable. (Proofs can be found in [G II], Propositions 13.5.1 and 13.5.3.) Products of separable spaces behave remarkably well; this illustrates the fact that a product topology is a weak topology. Proposition 1.1.3 Suppose that (X, τ ) is a separable topological space. Then X (0,1] , with the product topology, is separable. Proof Let C be a countable dense subset of X, and let Y be the space of elements of X (0,1] which take constant values in C on each of the intervals (i/k, (i+1)/k] for 1 ≤ i ≤ k, for some k. Then Y is countable, and dense in X (0,1] . There are also conditions which ensure that points and closed sets can be distinguished topologically. A topological space (X, τ ) is • a T1 space if singleton sets are closed, so that finite sets are closed; • a T2 space, or Hausdorff space, if whenever x and y are distinct points of X there exist disjoint open sets U and V with x ∈ U and y ∈ V; • a T3 space if whenever A is a closed subset of X and x ∈ A there exist disjoint open sets U and V with x ∈ U and A ⊆ V; • a T4 space if whenever A and B are disjoint closed sets there exist disjoint open sets U and V with A ⊆ U and B ⊆ V. A topological space is a T3 space if and only if every point has a base of neighbourhoods consisting of closed sets. A Hausdorff T3 space is called a

1.1 Topological Spaces

13

regular space and a Hausdorff T4 space is called a normal space. A normal space is regular. Exercise 1.1.4 Show that a topological space (X, τ ) is Hausdorff if and only if the diagonal = {(x, x) : x ∈ X} is closed in X × X (with the product topology). If f is a continuous mapping from a topological space (X, τ ) into a Hausdorff topological space (Y, σ ) then the graph f is closed in X × Y. Theorem 1.1.5 A second countable regular space is normal. Proof Let B be a countable base for the topology. Suppose that C and D are disjoint closed sets. Let (Vi )∞ i=1 be an enumeration of {U ∈ B : U ∩C = ∅} and j ∞ let (Wj )j=1 be an enumeration of {U ∈ B : U ∩ D = ∅}. Let Pj = Wj \ (∪i=1 V i ) and let Qk = Vk \ (∪ki=1 W i ). Then Pj and Qk are disjoint open sets, for all j, k. ∞ Let P = ∪∞ j=1 Pj and Q = ∪k=1 Qk , so that P and Q are disjoint open sets. If x ∈ C then there exists Wj such that x ∈ Wj . But x ∈ V i for 1 ≤ i ≤ j. Thus x ∈ Pj ⊆ P, and so C ⊆ P. Similarly, D ⊆ Q, and so X is normal. The words ‘normal’ and ‘regular’ are sadly overused in a mathematical context. Later, we shall use the term ‘regular’, with a quite different meaning, in a measure-theoretic setting. In spite of their name, normal topological spaces can behave very badly. Exercise 1.1.6 (i) Show that the subspace L = {(x, y) : x + y = 0} of (X, τ ) = (R, τr ) × (R, τr ) has the discrete topology, and so is not separable. (ii) Show that every subset of L is closed in (X, τ ), and that (X, τ ) is not normal. In analysis, another property is important. A topological space (X, τ ) is completely regular if it is Hausdorff, and whenever A is a closed subset of X and x ∈ A there exists a continuous function f on X taking values in [0, 1], with f (x) = 0 and f (a) = 1 for each a ∈ A. A completely regular space is regular. Exercise 1.1.7 Show that a subspace of a T1 space (respectively Hausdorff space, regular space, completely regular space) is a T1 space (respectively Hausdorff space, regular space, completely regular space), and the product of T1 spaces (respectively Hausdorff spaces, regular spaces, completely regular spaces) is a T1 space (respectively Hausdorff space, regular space, completely regular space). Here is a more difficult result; we shall see that it is almost trivial when τ is given by a metric.

14

General Topology

Theorem 1.1.8 (Urysohn’s lemma) If A and B are disjoint closed subsets of a T4 space (X, τ ), there exists a continuous function f on X taking values in [0, 1], with f (a) = 0 for each a ∈ A and f (b) = 1 for each b ∈ B. Proof Let A0 = A. There exists a closed set A1 such that A0 ⊆ Aint 1 ⊆ A1 ⊆ X \ B. Let D be the set of dyadic fractions in [0, 1]. Arguing inductively, if r = p/2n , with p odd, define a closed set Ar such that int A(p−1)/2n ⊆ Aint r ⊆ Ar ⊆ A(p+1)/2n ⊆ A(p+1)/2n ;

(p − 1 and p + 1 are even, so that A(p−1)/2n and A(p+1)/2n have already been defined). Now if x ∈ X let f (x) = inf{r ∈ D : x ∈ Ar } (where inf(∅) = 1). Then f (x) = 0 for x ∈ A, f (x) = 1 for x ∈ B and 0 ≤ f (x) ≤ 1. It remains to show that f is continuous. But if 0 < s ≤ 1 then f (x) < s if and only if x ∈ Cs = ∪s t if and only if x ∈ Dt = ∪0≤r≤t Aint r . Then Cs and Dt are open sets, from which it follows that f is continuous. Consequently, a normal space is completely regular. Theorem 1.1.9 A topological space (X, τ ) is completely regular if and only if it is homeomorphic to a subspace of a hypercube. Proof The condition is certainly sufficient. Suppose that (X, τ ) is completely regular. Let A = {(x, F) : x ∈ X, F closed in X, x ∈ F}. If α = (x, F) ∈ A there exists a continuous function fα : X → [0, 1] such that fα (x) = 0 and f (y) = 1 for y ∈ F. Let f (x) = {fα (x) : α ∈ A}. If x = y then β = (x, {y}) ∈ A, and fβ (x) = 0 = 1 = fβ (y), and so f is injective. Since each fα is continuous, f : X → [0, 1]A is continuous. Conversely if x ∈ X and U is an open neighbourhood of x then γ = (x, X \U) ∈ A, V = {g ∈ f (X) : gγ < 1} is a neighbourhood of f (x) in f (X), and V ⊆ f (U), so that f −1 : f (X) → X is also continuous. Corollary 1.1.10 A second countable normal space is homeomorphic to a subspace of the Hilbert cube H. Proof Replace A by the countable set of pairs {(xn , Cm )}, where (Cm )∞ m=1 is a countable base of closed sets and (xn )∞ n=1 is a dense sequence, and xn ∈ Cm . Note that the theorem uses the axiom of choice, but the corollary does not.

1.2 Compactness

15

Let us end this section by observing that the definition of a topological space that we have given is short and easy, but is a little misleading. The important fact is that topology is a local phenomenon; it is therefore often appropriate to define a topology in terms of neighbourhoods. The neighbourhoods Nx of a point x form a filter; that is, (i) ∅ ∈ Nx ; (ii) if N ∈ Nx and N ⊆ M then M ∈ Nx ; (iii) if N1 , N2 ∈ Nx then N1 ∩ N2 ∈ Nx . Exercise 1.1.11 Suppose that (X, τ ) is a topological space and that {Nx }x∈X is the set of neighbourhood filters. Show that if N ∈ Nx then x ∈ N and there exists O ∈ Nx such that O ⊆ N and O ∈ Ny for each y ∈ O. Conversely, if {Nx }x∈X is a set of filters on a set X which has these properties, show that this defines a topology on X for which {Nx }x∈X is the set of neighbourhood filters.

1.2 Compactness Suppose that A is a subset of a topological space (X, τ ). A collection O of open sets is an open cover of A if A ⊆ ∪O∈O O. An open cover is finite if it has finitely many members. The set A is compact if every open cover of A has a finite subcover. Compact sets are a topological approximation to finite sets. By considering complements, it follows that (X, τ ) is compact if and only if it has the finite intersection property; if A is a collection of closed sets with the property that ∩A∈Af A is not empty for each finite collection Af of sets in A, then the total intersection ∩A∈A A is not empty. Here are some basic properties of compact sets. Exercise 1.2.1 (i) The union of a finite set of compact sets is compact. (ii) A compact Hausdorff space is normal. (iii) A subset of a compact Hausdorff space is compact if and only if it is closed. (iv) The continuous image of a compact set is compact. (v) A continuous bijection of a compact space onto a Hausdorff space is a homeomorphism. For details, see [G II], Propositions 15.1.3, 15.1.4, Corollary 15.1.7. Here is one of the fundamental results of general topology. It requires the axiom of choice, and indeed is equivalent to it; Tychonoff’s theorem is true if and only if the axiom of choice holds.

16

General Topology

Theorem 1.2.2 (Tychonoff’s theorem) The product of a set of compact topological spaces is compact. Proof See [G II], Appendix D. We shall only use this in Proposition 1.2.3. We shall however prove (and use) the fact that a countable product of compact metric spaces is compact. Corollary 1.2.3 A topological space is a compact Hausdorff space if and only if it is homeomorphic to a closed subset of a hypercube. Proof This follows from Theorem 1.1.9 and Exercise 1.2.1 (iii). Let us give an example of this. The Helly space H is the subset of functions h in [0, 1][0,1] which are non-decreasing; if 0 ≤ s < t ≤ 1 then h(s) ≤ h(t). H is closed in [0, 1][0,1] , and is therefore compact. It is first countable and separable, but is not second countable. For more details, see [SS: 107]. There are two related definitions. A is sequentially compact if every sequence in A has a convergent subsequence, convergent to an element of A, and A is countably compact if every sequence in A has a limit point in A. A compact set is countably compact, and a sequentially compact set is countably compact. In general, there are no further implications. Proposition 1.2.4 Let (N) = {0, 1}N and let X = {0, 1}(N) , with the product topology τ , when {0, 1} is given the discrete topology. (i) (X, τ ) is compact and separable, but not sequentially compact. (ii) Let X0 = {x ∈ X : {ω : x(ω) = 1 is countable}}. Then X0 is sequentially compact, but is not compact or separable. Proof (i) (X, τ ) is compact, by Tychonoff’s theorem, and is separable, by Proposition 1.1.3, since there is a bijection from (N) onto (0, 1]. If ω ∈ X, let en (ω) = ωn for n ∈ N. The sequence (en )∞ n=1 in X has no convergent subsequence; (X, τ ) is not sequentially compact. (ii) A diagonal argument shows that X0 is sequentially compact. It is a dense proper subspace of the Hausdorff space (X, τ ), and so it is not compact. If C is a countable subset of X0 , the set {ω : c(ω) = 1 for some c ∈ C} is countable, and so there exists ω such that c(ω ) = 0 for all c ∈ C. Then I{ω } is not in the closure of C, and so X0 is not separable. A topological space (X, τ ) is locally compact if every point has a base of neighbourhoods consisting of compact sets. Exercise 1.2.5 A Hausdorff topological space is locally compact if and only if each point has a compact neighbourhood.

1.2 Compactness

17

A compactification of a topological space (X, τ ) is a compact space (X ∗ , τ ∗ ), together with a homeomorphism j of X onto a dense subspace j(X) of X ∗ . There are usually many compactifications of a topological space (X, τ ); the one-point compactification is perhaps the simplest. If (X, τ ) is not compact, the one-point compactification of (X, τ ) is defined by taking X ∗ = X ∪ {∞}, where ∞ is a point not in X, and defining a subset U to be open if U ∩ X ∈ τ , and, if ∞ ∈ U, then X ∗ \ U is a compact subset of X. The one-point compactification of (X, τ ) is Hausdorff if and only if (X, τ ) is Hausdorff and locally compact. Thus a topological space (X, τ ) is Hausdorff and locally compact if and only if it is homeomorphic to an open subset of a compact Hausdorff space. A topological space (X, τ ) is σ -compact if it is the union of a sequence of compact subsets. If so, then by considering finite unions, X is the union of an increasing sequence of compact subsets. Proposition 1.2.6 A locally compact space (X, τ ) is a σ -compact if and only if there is an increasing sequence (Kn )∞ n=1 of compact subsets such that int , for each n ∈ N. ∪n∈N Kn = X and Kn ⊆ Kn+1 Proof The condition is certainly sufficient. Suppose that (X, τ ) is σ -compact and locally compact. Let (Ln )∞ n=1 be an increasing sequence of compact subsets of X whose union is X. Let K1 = L1 . Suppose that we have found K1 , . . . , Kn int for 1 ≤ j < n. For each such that Lj ⊆ Kj for 1 ≤ j ≤ n and that Kj ⊆ Kj+1 x ∈ Kn there exists an open neighbourhood Nx of x such that Nx is compact. The sets {Nx : x ∈ Kn } are an open cover of the compact set Kn , and so there is int a finite subcover {Nx : x ∈ F}. Let Kn+1 = (∪x∈F Nx ) ∪ Ln+1 . Then Kn ⊆ Kn+1 and Ln+1 ⊆ Kn+1 , so that X = ∪n∈N Kn . A separable locally compact topological space need not be σ -compact. Let {0, 1} be given the discrete topology, and let X = {0, 1}[0,1] be given the product topology τ . Then (X, τ ) is compact, by Tychonoff’s theorem, and is separable, but is not first countable. Let x ∈ X and let Y = X \ {x}. Then Y is locally compact and separable, but is not σ -compact, since x does not have a countable base of neighbourhoods in (X, τ ).

2 Metric Spaces

In this chapter, we introduce the idea of a metric space. Metric spaces have a natural topology, which we describe in the first two sections of the chapter, but they have more structure than that. They also have a uniform structure, which we shall consider in detail in Chapter 5, and which is the setting for the notion of completeness, which we discuss in Sections 2.3 to 2.7, a Lipschitz structure, which we consider in Section 2.8, and a geometric structure.

2.1 Metric Spaces A metric space is a set X, together with a function d : X ×X → R+ , the metric, which satisfies (i) d(x, y) = d(y, x) for all x, y ∈ X (symmetry); (ii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X (the triangle inequality); and (iii) d(x, y) = 0 if and only if x = y. The following inequality is a useful consequence of axioms (i) and (ii): (iv) |d(x1 , y1 ) − d(x2 , y2 )| ≤ d(x1 , x2 ) + d(y1 , y2 ) for all x1 , x2 , y1 , y2 (the quadrilateral inequality). A function p : X × X → R+ which satisfies (i) and (ii) is called a pseudometric. If p is a pseudometric, set x ∼ y if p(x, y) = 0. Then ∼ is an equivalence relation on X. Let X/ ∼ be the quotient space, and let q : X → X/ ∼ be the quotient mapping. If x ∼ x and y ∼ y then it follows from the quadrilateral inequality that p(x, y) = p(x , y ). Thus the function d(q(x), q(y)) = p(x, y) is well-defined, and is easily seen to be a metric on E/ ∼, the associated metric. 18

2.1 Metric Spaces

19

A metric defines a topology in a natural way. Suppose that (X, d) is a metric space. If x ∈ X and  > 0, the open -neighbourhood N (x) is defined as {y ∈ X : d(x, y) < }, and the closed -neighbourhood M (x) is defined as {y ∈ X : d(x, y) ≤ }. A subset U of X is open if whenever x ∈ U there exists  > 0 such that N (x) ⊆ U. The collection of open sets is then a topology τ on X, the metric topology. Open -neighbourhoods are open in the topology, and closed -neighbourhoods are closed. The sets {N (x) :  > 0} form a base of neighbourhoods of x, as does the collection {M (x) :  < 0}. Since the sets {N1/n (x) : n ∈ N} form a base of neighbourhoods, the metric topology is first countable. Thus if f is a mapping of (X, d) into a topological space (Y, σ ), f is continuous at x if and only if it is sequentially continuous; that is, f (xn ) → f (x) whenever d(xn , x) → 0 as n → ∞. If (X, d) and (Y, ρ) are metric spaces, then a mapping f :(X, d) → (Y, ρ) is an isometry if ρ(f (x), f (y)) = d(x, y) for all x, y ∈ X. An isometry is clearly a homeomorphism of (X, d) onto (f (X), ρ). Let us give some examples. (i) The function d(x, y) = |x−y| is the usual metric on R and the usual metric 1  on C. Similarly the function d(x, y) = ( nj=1 |xj − yj |2 ) 2 is the usual metric, or Euclidean metric, on Rn and the usual metric, or Hermitian metric, on Cn . (ii) If X is any set, the function defined as d(x, y) = 1 if x = y and d(x, y) = 0 if x = y is the discrete metric on X. (iii) If Y is a subset of a metric space (X, d) then the restriction of d to Y × Y is a metric on Y, the subspace metric. (iv) If A is a non-empty subset of a metric space, the diameter diam(A) is defined to be sup{d(a, b) : a, b ∈ A}. Note that it follows from the quadrilateral inequality that diam(A) = diam(A). A set A is bounded if A is empty or diam(A) < ∞. Suppose that S is a non-empty set, that (Y, ρ) is a metric space and that f is a mapping from S to Y. We define the oscillation ωS (f ) of f on S to be diam(f (S)) = sup{ρ(f (s), f (t)) : s, t ∈ S}. f is bounded on S if ωS (f ) < ∞, and B(S, Y) denotes the set of bounded mappings from S to Y. The function d∞ (f , g) = sups∈S ρ(f (s), g(s)) is then a metric on B(S, Y), the uniform metric on B(S, Y). A sequence (fn )∞ n=1 converges to f in this metric if and only if fn (s) → f (s) uniformly in s; sups∈S |fn (s) − f (s)| → 0 as n → ∞; thus convergence in d∞ is called uniform convergence. We denote B(S, R) by B(S), and we denote (B(N, Y) by l∞ (Y); thus l∞ (Y) is the space of bounded sequences in Y.

20

Metric Spaces

Exercise 2.1.1 (Dini’s theorem) Suppose (fn )∞ n=1 is a sequence of continuous real-valued functions on a compact Hausdorff space (X, τ ) which decreases pointwise to 0. Show that fn converges uniformly to 0 as n → ∞. We also introduce the notion of local oscillation. We define the local oscillation f (x) of f at x to be f (x) = inf{ωU (f ) : U a neighbourhood of x}. A function from a topological space (X, τ ) into [−∞, ∞] is upper semicontinuous at x if given  > 0 there exists a neighbourhood U of x such that if y ∈ U then f (y) < f (x) + , and it is upper semi-continuous on X if it is upper semi-continuous at every point of X. Proposition 2.1.2 Suppose that (X, τ ) is a topological space, that (Y, ρ) is a metric space, that f is a mapping from X to Y and that x ∈ X. Then f is a non-negative upper semi-continuous function on X. f is continuous at x if and only if f (x) = 0. Proof Given  > 0, there exists an open neighbourhood U of x such that if y and z belong to U then ρ(f (y), f (z)) < f (x) + ; but U is a neighbourhood of y, and so f (y) < f (x) +  for y ∈ U. If f (x) = 0 then if  > 0 there exists an open neighbourhood U  of x such that ρ(f (y), f (x)) <  for y ∈ U  ; that is, f is continuous at x. The converse is just as easy. We consider semi-continuity further in Chapter 4. Suppose that (X, d) and (Y, ρ) are metric spaces. Then Cb (X, Y) is the set of bounded continuous mappings from X to Y. Proposition 2.1.3 Suppose that (X, d) and (Y, ρ) are metric spaces. Then Cb (X, Y) is a closed subset of (B(X, Y), d∞ ). Proof Suppose that f ∈ Cb (X, Y), that x ∈ X and that  > 0. Then there exists g ∈ Cb (X, Y) such that d∞ (f , g) < /3. Since g is continuous, there exists δ > 0 such that if d(x, y) < δ then ρ(g(x), g(y)) < /3. If d(x, y) < δ then ρ(f (x), f (y)) ≤ ρ(f (x), g(x)) + ρ(g(x), g(y)) + ρ(g(y), f (y)) < , so that f is continuous at x. Exercise 2.1.4 Let L(Rn ) be the set of linear mappings from Rn to itself. If T ∈ L(Rn ) show that T is continuous. Let T = sup{d(T(x), 0) : d(x, 0) ≤ 1}.

2.2 The Topology of Metric Spaces

21

If S, T ∈ L(Rn ), let dOp (S, T) = S − T. Show that dOp is a metric on L(Rn ), the operator metric. A topological space (X, τ ) is metrizable if there is a metric d on X such that τ is the corresponding metric topology. In general, if (X, τ ) is metrizable, there are many metrics with metric topology τ . Two metrics are said to be equivalent if they have the same metric topology. Exercise 2.1.5 Suppose that (X, d) is a metric space, that c > 0 and that (d ∧ c)(x, y) = min(d(x, y), c). Show that d ∧ c is a bounded metric which is equivalent to d. Exercise 2.1.6 A metric ρ on a set X is an ultrametric if ρ(x, z) ≤ max(ρ(x, y), ρ(y, z)) for x, y, z ∈ X. Show that if (X, ρ) is an ultrametric space, then any open -neighbourhood is closed. Exercise 2.1.7 Suppose that p is a prime. If x, y ∈ Q, let dp (x, y) = 0 if x = y and let dp (x, y) = p−r if |x − y| = apr /b = 0, where p does not divide the integers a and b, and r ∈ Z. Show that dp is a metric (the p-adic metric) on Q. Show that dp is an ultrametric.

2.2 The Topology of Metric Spaces Let us now consider some topological properties of metric spaces. As we have seen, a metric space is first countable. Theorem 2.2.1 A metric space (X, d) is separable if and only if it is second countable. Proof Suppose first that (X, d) is separable. Let (xn )∞ n=1 be a dense sequence in (X, d). Then the countable collection {N1/j (xn ) : j, n ∈ N} of open sets is a basis for the topology. For if U is an open subset of X and x ∈ U there exists  > 0 such that N (x) ⊆ U. There exists j such that 1/j < , and there exists n such that d(x, xn ) < 1/4j. Then x ∈ N1/2j (xn ) ⊆ U, so that U = ∪{N1/j (xn ) : N1/j (xn ) ⊆ U}. Conversely, if (X, d) is second countable, and B is a countable base for the topology, choose a point xB from each non-empty B in B. Then the countable set {xB : B ∈ B, B = ∅} is dense in X. Corollary 2.2.2 A subspace of a separable metric space is separable. Proof For the subspace is second countable.

22

Metric Spaces

We have seen that the Helly space is separable and first countable, but not second countable, and so it is not metrizable – the topology is not given by a metric. We have also seen that a subspace of a separable topological space need not be separable. Theorem 2.2.3 Suppose that S is a dense subset of a metric space (X, d). Then there is an isometry of (X, d) onto a subset of (B(S), d∞ ). Proof Pick a point x0 ∈ X. If x ∈ X and s ∈ S let fx (s) = d(x, s) − d(x0 , s). By the quadrilateral inequality, |d(x, s) − d(x0 , s)| ≤ d(x, x0 ), so that fx ∈ B(S). If x, x ∈ X, |fx (y) − fx (y)| = |d(x, y) − d(x , y)| ≤ d(x, x ), so that d∞ (fx , fx ) ≤ d(x, x ). On the other hand if  > 0 there exists s ∈ S with d(x , s) < /2. Then |d(x, s) − d(x, x )| < /2, so that |(fx (s) − fx (s)) − d(x, x )| = |d(x, s) − d(x , s) − d(x, x )| < , from which it follows that d∞ (fx , fx ) ≥ d(x, x ). Corollary 2.2.4 If (X, d) is a separable metric space, then there exists an isometry of (X, d) onto a subset of (l∞ , d∞ ). On the other hand, (l∞ , d∞ ) is not separable. If A ⊆ N, let IA be the indicator function of A; then IA ∈ l∞ . If A = B, then d∞ (IA , IB ) = 1. Thus if C is a countable subset of l∞ and f ∈ C, then there is at most one A ∈ P(N) for which d(f , IA ) < 1/3. Since P(N) is uncountable, there exists A ∈ P(N) such that d(f , IA ) ≥ 1/3 for all f ∈ C, and so C is not dense in l∞ . If A is a non-empty closed subset of (X, d), let d(x, A) = infa∈A d(x, a) be the distance of x from A. Then d(x, A) = 0 if and only if x ∈ A. If y ∈ X and a ∈ A, then d(y, A) ≤ d(y, a) ≤ d(x, y) + d(x, a), so that d(y, A) ≤ d(x, y) + d(x, A). Similarly, d(x, A) ≤ d(x, y) + d(y, A), so that |d(x, A) − d(y, A)| ≤ d(x, y). Thus the function x → d(x, A) is continuous on X. Suppose now that A and B are disjoint non-empty closed subsets of X. Let fA,B (x) =

d(x, A) . d(x, A) + d(x, B)

Then fA,B is a continuous function on X taking values in [0, 1], fA,B = 0 if and only if x ∈ A and fA,B = 1 if and only if x ∈ B. This is so much easier than Urysohn’s lemma! Thus a metric space is normal, and completely regular. If A is a non-empty subset of (X, d), and r > 0 let Nr (A) = {y ∈ X : d(y, A) < r}. Then Nr (A) is an open subset of X, the open r-neighbourhood of A. If A is closed, A = ∩n∈N N1/n (A), A is a Gδ set: the intersection of a decreasing sequence of open sets. As we shall see, this has important implications for measure theory.

2.2 The Topology of Metric Spaces

23

Theorem 2.2.5 The countable topological product (X, τ ) of a sequence (Xn , dn ) of metric spaces is metrizable. ∞ Proof If x = (xn )∞ n=1 and y = (yn )n=1 are elements of X, let

d(x, y) =

∞ 

(dn ∧ 2−n )(xn , yn ).

n=1

Then the series converges, and it is easily verified that d is a metric on X. Let τ  be the corresponding metric topology. Let πn be the co-ordinate map from X to Xn . Since the identity mapping (Xn , dn ∧ 2−n ) → (Xn , dn ) is continuous, the identity mapping (X, τ  ) → (X, τ ) is continuous. On the other hand, suppose that N (x) is an open -neighbourhood of x for the metric d. There exists n0 such that 2−n0 < /2. Then N = {y ∈ X : dn (xn , yn ) < /2n for 1 ≤ n ≤ n0 } is a τ -neighbourhood, and if y ∈ N then d(x, y) < , and so the identity mapping (X, τ ) → (X, τ  ) is also continuous. Thus τ = τ  . There are many other metrics that can be used, such as ∞ 1/2   −n  −n 2 sup(dn (xn , yn ) ∧ 2 ) or dn (xn , yn ) ∧ 2 . n∈N

n=1

For finite products, there is no need to impose boundedness conditions on the metrics dn . There are two product spaces which will play an important role in what follows. First, there is the Hilbert cube, the product space [0, 1]N , which is denoted by H. Secondly, the product space {0, 1}N , where {0, 1} is given the discrete topology, is called the Bernoulli sequence space, and is denoted by (N). Thus the points of (N) are infinite sequences of 0s and 1s. Let us introduce some notions concerning (N). If n ∈ N and either η ∈ (N) or η ∈ {0, 1}m for some m ≥ n, we set Cη,n = {ω ∈ (N) : ωj = ηj for 1 ≤ j ≤ n}; Cη,n is a cylinder set, of rank n. Exercise 2.2.6 Show that there are 2n cylinder sets of rank n, which form a partition of (N). Show that cylinder sets are open and closed, and that the cylinder sets form a base for the topology of (N). Show that the countable collection of sets {Cη,n : n ∈ N} is a base of neighbourhoods of η. The Hilbert cube H and the Bernoulli sequence space (N) are metrizable. There are other metrics on them which define the product topology, and may be more natural to consider. For example, suppose that (an )∞ n=1 is a

24

Metric Spaces

 2 sequence of positive numbers for which ∞ j=1 aj < ∞. If x, y ∈ H let d(x, y) = ∞ 2 ( n=1 an (xn − yn )2 )1/2 . Then d is a metric on H which defines the product topology.   n If ω, ω ∈ (N), let d(ω, ω ) = 2 ∞ n=1 |ωn − ωn |/3 . Then d is a metric on (N) which defines the product topology. If ω ∈ (N), let f (ω) =  n 2 ∞ n=1 ωn /3 . Then f is an isometry of ((N), d) onto the Cantor set C, and so we call d the Cantor metric on (N). Note that the open and closed neighbourhood M1/3n (ω) of ω is the cylinder set Cω,n . When (X, τ ) is a separable topological space, there are necessary and sufficient conditions for (X, τ ) to be metrizable. Theorem 2.2.7 (Urysohn’s metrization theorem) Suppose that (X, τ ) is a separable topological space, the following are equivalent: (i) (X, τ ) is regular and second countable. (ii) (X, τ ) is homeomorphic to a subspace of the Hilbert cube H. (iii) (X, τ ) is metrizable. Proof Suppose that (i) holds. Then (X, τ ) is normal (Theorem 1.1.5), and so (X, τ ) is homeomorphic to a subspace of the Hilbert cube, by Corollary 1.1.10, and so (ii) is true. Condition (ii) certainly implies (iii). Finally suppose that (X, τ ) is the topology defined by a metric d. Then τ is normal, and (X, τ ) is second countable, by Theorem 2.2.1, so that (i) holds.

2.3 Completeness: Tietze’s Extension Theorem For topological spaces, continuity is generally a local phenomenon, and it is not possible to compare what happens at one point with another. Metric spaces have other properties than topological ones, and here we consider uniform properties; we shall discuss uniform spaces in Chapter 5. A sequence (xn )∞ n=1 in a metric space (X, d) is a Cauchy sequence if whenever  > 0 there exists N ∈ N such that d(xm , xn ) < , for m, n ≥ N. A convergent sequence is a Cauchy sequence; conversely, if every Cauchy sequence converges, then (X, d) is said to be complete. The real line R, with its usual metric, is complete (the general principle of convergence): this property lies behind the construction of the real number system. Proposition 2.3.1 If S is a non-empty set and (Y, ρ) is a metric space then the metric space (B(S, Y), d∞ ) is complete if and only if (Y, ρ) is complete. Proof Suppose that (Y, ρ) is complete and that (fn )∞ n=1 is a Cauchy sequence in B(S). If s ∈ S then ρ(fm (s), fn (s)| ≤ d∞ (fm , fn ), and so (fn (s))∞ n=1 is a Cauchy

2.3 Completeness: Tietze’s Extension Theorem

25

sequence in Y, which converges, to f (s), say, as n → ∞. There exists N such that d∞ (fm , fn ) < 1 for m, n ≥ N, and so ρ(f (s), fN (s)) ≤ 1; thus f ∈ B(S). Finally, d∞ (f , fn ) ≤ supm≥n d∞ (fm , fn ) → 0 as n → ∞. Conversely, if (Y, ρ) is not complete then there is a Cauchy sequence (yn )∞ n=1 in Y which does not converge. Let fn (s) = yn , for s ∈ S and n ∈ N. Then (fn )∞ n=1 is a Cauchy sequence in B(S, Y) which does not converge. Corollary 2.3.2 If S is a non-empty set then (B(S), d∞ ) is complete. Proof For R, with its usual metric, is complete. Proposition 2.3.3 A subset Y of a complete metric space (X, d) is complete in the subspace metric if and only if it is closed. Proof If Y is closed, and if (yn )∞ n=1 is a Cauchy sequence in Y, then it is a Cauchy sequence in X, and so converges to an element x of X. Since Y is closed, x ∈ Y, and so Y is complete. Conversely, if Y is complete and x ∈ Y then there exists a sequence (yn )∞ n=1 in Y which converges to x. But then (yn )∞ is a Cauchy sequence in Y, and so n=1 converges to an element y of Y, since Y is complete. But limits are unique, and so x = y ∈ Y. Thus Y = Y, and so Y is closed. Corollary 2.3.4 If (X, τ ) is a topological space and (Y, ρ) is a metric space then the metric space (Cb (X, Y), d∞ ) is complete if and only if (Y, ρ) is complete. Proof For (Cb (X, Y), d∞ ) is closed in (B(X, Y), d∞ ). We use this, and Urysohn’s lemma, to prove an important extension theorem. Theorem 2.3.5 (Tietze’s extension theorem) Suppose that (X, τ ) is a normal topological space, that Y is a non-empty closed subset of X and that f ∈ Cb (Y). Let  = supy∈Y f (y) and λ = infy∈Y f (y). There exists g ∈ Cb (X) such that f (y) = g(y) for y ∈ Y, supx∈X g(x) =  and infx∈X g(x) = λ. If f (y) <  for all y ∈ Y then the extension g can be chosen so that g(x) <  for all x ∈ X. Similarly, for λ. Proof The result is trivially true if f is constant. Otherwise, by adding a constant and scaling we can suppose that  = 1 and that λ = −1. We shall show by induction that there exists a sequence (gn )∞ n=0 in Cb (X) such that (i) sup |f (y) − gn (y)| ≤ (2/3)n for n ∈ Z+ y∈Y

and (ii) sup |gn−1 (x) − gn (x)| ≤ 12 (2/3)n for n ∈ N. x∈X

26

Metric Spaces

We set g0 = 0. Suppose that we have found g0 , . . . , gn satisfying the conditions. Let An = {y ∈ Y : f (y) − gn (y) ≥ 2n /3n+1 , Bn = {y ∈ Y : f (y) − gn (y) ≤ −2n /3n+1 . Then An and Bn are disjoint closed subsets of X, and so by Urysohn’s lemma there exists kn ∈ Cb (X) such that kn (a) = 2n /3n+1 for a ∈ An , kn (b) = −2n /3n+1 for b ∈ Bn , and |kn (x)| ≤ 2n /3n+1 for x ∈ X. Let gn+1 = gn − kn . Then |f (y) − gn+1 (y)| ≤ (2/3)n+1 for y ∈ Y, and |gn (x) − gn+1 (x)| ≤ 12 (2/3)n+1 for x ∈ X; the induction is established. If m < n and x ∈ X then n n   j m 1 |gm (x) − gn (x)| ≤ |gj−1 (x) − gj (x)| ≤ 2 (2/3) < (2/3) , j=m+1

j=m+1

so that d∞ (gm , gn ) < (2/3)m . Thus (gn )∞ n=0 is a Cauchy sequence in (Cb (X), d∞ ). Since (Cb (X), d∞ ) is complete, there exists g ∈ Cb (X) such that gn → g as n → ∞. By (i), gn (y) → f (y) as n → ∞, for y ∈ Y, and so f (y) = g(y) for y ∈ Y. Since |gn (x)| = |gn (x)−g0 (x)| < 1 for x ∈ X and n ∈ N, it follows that |g(x)| ≤ 1 for x ∈ X, and so supx∈X g(x) = − infx∈X g(x) = 1. Suppose now that f (y) <  for all y ∈ Y. If g is an extension, as shown earlier, and Z = {x ∈ X : g(x) = } is not empty, then Y and Z are disjoint closed subsets of X, and so, by Urysohn’s lemma there exists h ∈ Cb (X) with h(y) = 1 for y ∈ Y, h(z) = 0 for z ∈ Z and 0 ≤ h ≤ 1. Then gh is an extension with the required property. Similarly for λ. If (X, τ ) is a topological space, we denote the space of continuous realvalued functions on X by C(X). Corollary 2.3.6 Suppose that (X, τ ) is a normal topological space, that Y is a non-empty closed subset of X and that f ∈ C(Y). Then there exists g ∈ C(X) such that g(y) = f (y) for y ∈ Y. Proof Let F(y) = tan−1 f (y), for y ∈ Y. Then F ∈ Cb (Y) and F(Y) ⊆ (−π/2, π/2). There exists an extension G with G(X) ⊆ (−π/2, π/2). Let g(x) = tan G(x), for x ∈ X. Then g is a continuous extension of f . Exercise 2.3.7 Suppose that p is an odd prime, that a is not a square and that x02 = a(mod p). Show by induction that there is a sequence (xn )∞ n=0 such that

 xn = xn−1 (mod pn ) and xn2 = a mod pn+1 .

2.4 More on Completeness

27

Show that (xn )∞ n=0 is a Cauchy sequence in the p-adic metric which does not converge in (Q, dp ); (Q, dp ) is not complete.

2.4 More on Completeness A mapping f from a metric space (X, d) to a metric space (Y, ρ) is uniformly continuous if whenever  > 0 there exists δ > 0 such that if d(x, y) < δ then ρ(f (x), f (y)) < . A bijection j : (X, d) → (Y, ρ) is a uniform homeomorhism if f and f −1 are uniformly continuous, and two metrics d and d are uniformly equivalent if the identity mapping i : (X, d) → (X, d ) is a uniform homeomorphism. As the names suggest, these ideas relate to the uniform structures of X and Y. It is often convenient to replace a metric with a uniformly equivalent metric. The following exercise suggests how this can be done. Exercise 2.4.1 Suppose that φ is an increasing real-valued function on [0, ∞), that φ(t) = 0 if and only if t = 0 and that φ is continuous at 0. Consider the following statements. (i) φ is concave: if a, b ≥ 0 and 0 ≤ λ ≤ 1 then φ((1 − λ)a + λb) ≥ (1 − λ)φ(a) + λφ(b). (ii) Let ψ(t) = φ(t)/t, for t > 0. Then ψ is a decreasing function on (0, ∞). (iii) If a, b ≥ 0 then φ(a + b) ≤ φ(a) + φ(b). (iv) If (X, d) is a metric space, and ρ(x, y) = φ(d(x, y)) then ρ is a metric on X which is uniformly equivalent to d. Show that (i) implies (ii), (ii) implies (iii) and (iii) implies (iv). Show that there are no converse implications. Thus if (X, d) is a metric space, and if φ(t) = t ∧ c, where c > 0, then d ∧ c is a bounded metric uniformly equivalent to d; another bounded uniformly equivalent metric is obtained by taking φ(t) = t/(1 + t). If 0 < p < 1 then the function φ(t) = tp is concave on [0, ∞), and so if dp (x, y) = d(x, y)p then dp is a metric on X uniformly equivalent to d. Note that dp is strictly subadditive: if x, y, z are three distinct elements of X then dp (x, z) < dp (x, y) + dp (y, z). If f : (X, d) → (Y, σ ) is a uniformly continuous mapping and (xn )∞ n=1 is a is a Cauchy sequence in Y. In particular, Cauchy sequence in X, then (f (xn ))∞ n=1 if d and d are uniformly equivalent metrics on X, then a sequence (xn )∞ n=1 in X is a d-Cauchy sequence if and only if it is a d -Cauchy sequence; consequently (X, d) is complete if and only if (X, d ) is complete. These properties are not topological ones. Let d be the usual metric on [1, ∞) and let d (x, y) = |1/x −  1/y|. Then d and d are equivalent metrics. The sequence (n)∞ n=1 is a d -Cauchy

28

Metric Spaces

sequence,but not a d-Cauchy sequence, and [1, ∞) is d-complete, but is not d complete. In the previous section, we defined a product metric on the product of a  sequence of metric spaces. If d is a product metric on X = ∞ i=1 (Xi , di ), d is a uniform product metric if each cross-section mapping kx,i is a uniform homeomorphism and each co-ordinate projection πi is uniformly continuous. The product metrics introduced in the previous section are all uniform product metrics, and we shall only consider uniform product metrics. Theorem 2.4.2 Suppose that ((Xn , dn ))∞ of complete metric n=1 is a sequence  spaces and that d is a uniform product metric on X = ∞ n=1 Xn . Then (X, d) is complete. Proof Suppose that (x(j) )∞ j=1 is a d-Cauchy sequence in X. If n ∈ N then (xn )∞ j=1 is a dn -Cauchy sequence in (Xn , dn ); since (Xn , dn ) is complete, it converges to an element xn of Xn . But then x(j) → x = (xn )∞ n=1 as j → ∞ in (X, d). (j)

Completeness can be characterized in terms of decreasing sequences of closed sets. Proposition 2.4.3 Suppose that (X, d) is a metric space. the following are equivalent. (i) (X, d) is complete. (ii) If (An )∞ n=1 is a decreasing sequence of non-empty closed subsets of X, and diam(An ) → 0 as n → ∞, then ∩n∈N An is non-empty. If so, then ∩n∈N An is a singleton {a}, and if an ∈ An for each n then an → a as n → ∞. Proof Suppose first that (ii) holds. Suppose that (xn )∞ n=1 is a Cauchy sequence is a decreasing sequence of nonin X. Let An = {xj : j ≥ n}. Then (An )∞ n=1 empty closed subsets of X, and diam(An ) → 0 as n → ∞. Thus ∩n∈N An is non-empty. If a ∈ ∩n∈N An , then xn → a as n → ∞, and so (X, d) is complete. Suppose conversely that (X, d) is complete and that (An )∞ n=1 is a decreasing sequence of non-empty closed subsets of X. Pick an ∈ An for each n ∈ N. If n ≥ m ≥ N then d(an , am ) ≤ diam(AN ) so that (an ) is a Cauchy sequence. Since (X, d) is complete, there exists a ∈ X such that an → a as n → ∞. Since each An is closed, a ∈ ∩n∈N An , so that ∩n∈N An = ∅. Finally, since diam(∩n∈N An ) = 0, ∩n∈N An = {a}. We now prove a fundamental extension result.

2.5 The Completion of a Metric Space

29

Theorem 2.4.4 Suppose that A is a dense subset of a metric space (X, d) and that f is a uniformly continuous mapping from A into a complete metric space (Y, ρ). Then there exists a uniformly continuous mapping f˜ : X → Y which extends f : f˜(a) = f (a) for a ∈ A. The extension is unique; indeed, if g is a continuous extension of f , then g = f˜. Proof Suppose that x ∈ X. If n ∈ N, let An (x) = f (N1/n (x) ∩ A). If  > 0, there exists N ∈ N such that if a, b ∈ A and d(a, b) < 2/N then ρ(f (a), f (b)) < /3. Thus if n ≥ N then diam(An (x)) ≤ /3. By Proposition 2.4.3, the set ∩n∈N An (x) is a singleton {a(x)}: we set f˜(x) = a(x). If x ∈ A, then f (x) ∈ ∩n∈N An , and so f˜(a) = f (a): f˜ extends f . Next we show that f˜ is uniformly continuous. Suppose that  and N are as above, and that d(x, y) < 1/N. Choose aN ∈ N1/2N (x) ∩ A and bN ∈ N1/2N (y) ∩ A. Then f (aN ) ∈ AN (x), so that ρ(f˜(x), f (aN )) ≤ /3. Similarly, ρ(f˜(y), f (bN )) < /3. Further, d(aN , bN ) < 2/N, so that ρ(f (aN ), f (bN )) < /3. Consequently ρ(f˜(x), f˜(y)) < : f˜ is uniformly continuous. Finally, if g is a continuous extension, then {x ∈ X : f˜(x) = g(x)} is closed, and contains A, and so is the whole of X. Here is a useful test for completeness. Theorem 2.4.5 Suppose that (X, d) is a complete metric space, that Y ⊆ X and that ρ is a metric on Y with the following properties. (i) If (yn )∞ n=1 is a ρ-Cauchy sequence then it is a d-Cauchy sequence. (ii) There exists r > 0 such that if y0 ∈ Y and 0 < η < r then the set Mη (y0 ) = {y ∈ Y : ρ(y, y0 ) ≤ η} is d-closed. Then (Y, ρ) is complete. Proof Suppose that (yn )∞ n=1 is a ρ-Cauchy sequence in Y. Then it is a d-Cauchy sequence in X, and so converges in (X, d) to an element x ∈ X. If 0 <  < r, there exists N such that ρ(ym , yn ) < /2 for m, n ≥ N. Thus yn ∈ M/2 (yN ) for n ≥ N. Since M/2 (yN ) is d-closed, x ∈ M/2 (yN ) ⊆ Y, and ρ(x, yn ) ≤ ρ(x, yN ) + ρ(yN , yn ) < , for n ≥ N. Thus yn → x in (Y, ρ) as n → ∞.

2.5 The Completion of a Metric Space ˆ j), where ˆ d), If (X, d) is a metric space, a completion of (X, d) is a pair ((X, ˆ is a complete metric space, and j is an isometry of (X, d) onto a dense ˆ d) (X, ˆ ˆ d). subspace of (X,

30

Metric Spaces

Theorem 2.5.1 Every metric space (X, d) has a completion, which is essentially unique: if ((Xˆ 1 , dˆ1 ), j1 ) and ((Xˆ 2 , dˆ2 ), j2 ) are two completions, there exists a unique isometry k of (Xˆ 1 , dˆ1 ) onto (Xˆ 2 , dˆ2 ) such that j2 = k ◦ j1 . Proof By Theorem 2.2.3, there exists an isometry j of (X, d) into (B(X), d∞ ), and so ((j(X), d∞ ), j) is a completion of (X, d). Suppose that ((Xˆ 1 , dˆ1 ), j1 ) and ((Xˆ 2 , dˆ2 ), j2 ) are two completions of (X, d). It follows from Theorem 2.4.4 that there exists a unique uniform homeomorphism k : (Xˆ 1 , dˆ1 ) → (Xˆ 2 , dˆ2 ) such that j2 = k ◦ j1 . The set ˆ k(y)) ˆ = dˆ1 (x, ˆ y)} ˆ {(x, ˆ y) ˆ ∈ Xˆ 1 × Xˆ 1 : dˆ2 (k(x), is closed in Xˆ 1 × Xˆ 1 , and contains the dense subset j1 (X) × j1 (X) of Xˆ 1 × Xˆ 1 , and so k is an isometry. ˆ of (X, d), and consider (X, d) ˆ d) As a result, we talk about the completion (X, ˆ ˆ as a subspace of (X, d). This construction of a completion is short, but artificial. It is useful to give a more natural construction of the completion of a metric space. We need a preliminary result. Proposition 2.5.2 Suppose that Y is a dense subspace of a metric space (X, d), and that every Cauchy sequence in Y converges to an element of X. Then (X, d) is complete. Proof Suppose that (xn )∞ n=1 is a Cauchy sequence in X. For each n ∈ N there exists yn ∈ Y with d(yn , xn ) < 1/n. Then d(ym , yn ) ≤ d(xm , xn ) + 1/m + 1/n, and so (yn )∞ n=1 is a Cauchy sequence in Y. By hypothesis, it converges to an element x of X. But d(x, xn ) ≤ d(x, yn ) + d(yn , xn ) < d(x, yn ) + 1/n, and so xn → x as n → ∞. Suppose now that (X, d) is a metric space. Let Ca(X) be the set of Cauchy sequences in X. If x, y ∈ Ca(X), then |d(xm , ym ) − d(xn , yn )| ≤ d(xm , xn ) + d(ym , yn ), by the quadrilateral inequality, and so (d(xn , yn ))∞ n=1 is a real Cauchy sequence, which converges to φ(x, y), say. Then it is easy to see that φ is a pseudometric on Ca(X). Let Xˆ be the corresponding quotient space, q the quotient mapping ˆ If x ∈ X, let c(x) be the constant sequence and dˆ the associated metric on X. with c(x)n = x, for all n. Then c(x) is a Cauchy sequence in X; let j(x) be the ˆ corresponding element of X.

2.6 Topologically Complete Spaces

31

ˆ j) is a completion of (X, d). ˆ d), Proposition 2.5.3 ((X, ˆ Proof First, it is clear that φ(c(x), c(y)) = d(x, y), so that d(j(x), j(y)) = ˆ Secondly, j(X) is dense in X. ˆ d). ˆ d(x, y): j is an isometry of (X, d) into (X, For if x = (xn )∞ ∈ Ca(X), then φ(x, c(x )) = lim d(x , x ) → 0 n m→∞ m n n=1 ∞ ∞ as n → ∞. Finally, if (j(xn ))n=1 is a Cauchy sequence in j(X), (xn )n=1 is a Cauchy sequence in X, and ˆ n ), q(x)) = lim d(xm , xn ) → 0 as n → ∞, d(j(x m→∞

ˆ is complete, by the preceding ˆ d) so that j(xn ) → q(x) as n → ∞. Thus (X, proposition.

2.6 Topologically Complete Spaces A topological space is topologically complete if it is homeomorphic to a complete metric space. The countable product of topologically complete spaces is topologically complete. A closed subspace of a topologically complete space is topologically complete. A subset A of a topological space is a Gδ -set (Fσ -set) if it is the countable intersection of open sets (countable union of closed sets). Theorem 2.6.1 A Gδ -subset of a topologically complete space (X, τ ) is topologically complete. Proof Let d be a complete metric on X which defines the topology on X. First suppose that A is open. If A = X, then A is topologically complete. Otherwise, let C(A) = X \ A and give X × R a uniform product metric σ . Then (X × R, σ ) is complete. Consider the injective mapping f from A to X × R defined by f (a) = (a, 1/d(a, C(A)). Since d(a, C(A)) > 0 for a ∈ A, this is well-defined. Since the mapping a → 1/d(a, C(A)) is continuous on A, f is continuous. We show that f (A) is closed in X × R. Suppose that f (an ) → y = (x, λ) ∈ X × R as n → ∞, so that an → x and 1/d(an , C(A)) → λ as n → ∞. Since |d(am , C(A)) − d(an , C(A))| ≤ d(am , an ), the sequence (d(an , C(A)))∞ n=1 is bounded, so that 1/d(an , C(A)) does not tend to 0 as n → ∞. Thus λ = 0, and d(an , C(A)) → 1/λ. Since d(an , C(A)) → d(x, C(A)), it follows that d(x, C(A)) = 1/λ, and so x ∈ A. Consequently f (an ) → f (x) as n → ∞, so that f (A) is closed in X × R, and (f (A), σ ) is complete. The mapping f : (A, d) → (X × R, σ ) is continuous, and the

32

Metric Spaces

inverse mapping f −1 : f (A) → A is continuous, since d(a, a ) ≤ σ (f (a), f (a )), so that f is a homeomorphism of (A, d) onto (f (A), σ ). Thus if we define ρ(a, a ) = σ (f (a), f (a )) then ρ is a complete metric on A equivalent to d. Next, suppose that A = ∩∞ j=1 Uj is a Gδ set. For each j there is a metric σj on Uj , equivalent to the restriction of d to Uj , under which Uj is complete. Let  U = ∞ j=1 (Uj , σj ), and let σ be a uniform product metric on U. Then (U, σ ) is complete. If a ∈ A, let ij : A → Uj be the inclusion map and let i : A → U be defined as i(a) = (ij (a))∞ j=1 . Then i is a continuous injective map of (A, d) into (U, σ ). We show that i(A) is closed in U. Suppose that i(an ) → u as n → ∞. Then, for each j ∈ N, ij (an ) → uj as n → ∞. Since the metric σj on Uj is equivalent to the metric d, d(an , uj ) → 0 as n → ∞. Since this is true for each j, there exists l in X such that uj = l for each j ∈ N. Thus l ∈ ∩∞ n=1 Uj = A, and u = i(l). Thus i(A) is closed in (U, σ ), and so (i(A), σ ) is complete. If i(an ) → i(a) in (i(A), σ ), then i1 (an ) → i1 (a) in (U1 , d1 ) as n → ∞. But d and d1 are equivalent on U1 , and so an → a in (A, d). Thus i−1 : (i(A), σ ) → (A, d) is continuous, and i is a homeomorphism of (A, d) onto (i(A), σ ). Thus if we define ρ(a, a ) = σ (i(a), i(a )) then ρ is a complete metric on A equivalent to d. There is also a converse result. Theorem 2.6.2 (Alexandroff) Suppose that Y is a topologically complete subspace of a metric space (X, d). Then Y is a Gδ subset of X. Proof Let ρ be a complete metric on Y which defines the topology of Y. If y ∈ Y and α > 0, let Nα (y) = {x ∈ X, d(x, y) < α} be the open α-neighbourhood of y in X. Since ρ and d agree on Y, for each y in Y there exists 0 < δn (y) < 1/n such that if z ∈ Y and d(y, z) < δn (y) then ρ(y, z) < 1/n. Let Un (y) = {x ∈ X : d(x, y) < δn (y)/2}, and let On = ∪y∈Y Un (y). Then On is an open subset of X containing Y. Let D = ∩n∈N On . Then D is a Gδ -subset of X which contains Y. We show that D = Y. Suppose that z ∈ D. For each n ∈ N, z ∈ On , and so there exists yn ∈ Y such that d(z, yn ) < δn (yn )/2 ≤ 1/2n. There exists kn ∈ N such that 1/kn < δn (yn ). If k ≥ kn then d(yn , yk ) ≤ d(yn , z) + d(z, yk ) < δn (yn )/2 + δk (yk )/2 ≤ δn (yn )/2 + 1/2k ≤ δn (yn ), and so ρ(yn , yk ) ≤ 1/n. Thus if k, l ≥ kn then ρ(yk , yl ) ≤ ρ(yk , yn ) + ρ(yn , yl ) ≤ 2/n. Thus (yn )∞ n=1 is a ρ-Cauchy sequence in Y. Since (Y, ρ) is complete, there exists y∞ ∈ Y such that ρ(yn , y∞ ) → 0 as n → ∞. Since the metrics d and ρ

2.7 Baire’s Category Theorem

33

are equivalent metrics on Y, d(y∞ , yn ) → 0 as n → ∞. But d(z, yn ) → 0 as n → ∞, so that z = y∞ ∈ Y.

2.7 Baire’s Category Theorem We now come to one of the most useful theorems of analysis. A topological space (X, τ ) is a Baire space if whenever (Un )∞ n=1 is a sequence of dense open subsets of X then ∩n∈N Un is dense in X. Theorem 2.7.1 (Baire’s category theorem) (i) A topologically space (X, τ ) is a Baire space. (ii) A locally compact Hausdorff space (X, τ ) is a Baire space.

complete

Proof Suppose that (Un )∞ n=1 is a sequence of dense open subsets of X, and that V is a non-empty open subset of X. We must show that V ∩ (∩∞ n=1 Un ) is not empty. (i) Let d be a complete metric on X which defines the topology τ . Since U1 is dense in X, there exists c1 ∈ V ∩ U1 . Since V ∩ U1 is open, there exists 0 < 1 ≤ 1/2 such that N1 (c1 ) ⊆ M1 (c1 ) ⊆ V ∩ U1 . We now iterate the argument; for each n ∈ N there exist cn ∈ Nn−1 (cn−1 ) ∩ Un and 0 < n < 1/2n such that Nn (cn ) ⊆ Mn (cn ) ⊆ Nn−1 (cn−1 ) ∩ Un . The sequence (Nn (cn ))∞ n=1 is decreasing, so that if m, p ≥ n then cm ∈ Mn (cn ) and cp ∈ Mn (cn ), so that d(cm , cp ) ≤ d(cm , cn ) + d(cn , cp ) < 2/2n ; thus (cn )∞ n=1 is a Cauchy sequence in (X, d). Since (X, d) is complete, it converges to an element c of X. Suppose that n ∈ N. Since cm ∈ Mn (cn ) for m ≥ n and since Mn (cn ) is closed, c ∈ Mn (cn ) ⊆ Un . Thus c ∈ ∩∞ n=1 Un . Further, c ∈ M1 (c1 ) ⊆ V, and so c ∈ V. (ii) The proof is similar, but easier. V ∩ U1 is non-empty; let x1 ∈ V ∩ U1 . There exists a compact neighbourhood K1 of x contained in V ∩ U1 . Let W1 = K1int . Now iterate the argument. For each n ∈ N there exist xn ∈ Wn−1 ∩Un , and a compact neighbourhood Kn of xn contained in Wn−1 ∩ Un . Let Wn = Knint . Then (Kn )∞ n=1 is a decreasing sequence of compact subsets of V, which has a non-empty intersection. If x ∈ ∩n∈N Kn then x ∈ V ∩ (∩n∈N Un ).

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The following corollary is particularly useful. Corollary 2.7.2 Suppose that (Cn )∞ n=1 is a sequence of closed subsets of a Baire space (X, τ ) whose union is X. Then there exists n such that Cn has a non-empty interior. ∞ Proof Let Un = X \ Cn . Then (Un )∞ n=1 is a sequence of open sets and ∩n=1 Un is empty, and so is certainly not dense in X. Thus there exists Un which is not dense in X; that is Cn has a non-empty interior.

It is sometimes useful to have a local version of this corollary. Corollary 2.7.3 Suppose that (X, τ ) is topologically complete or locally compact and Hausdorff. Suppose that (Cn )∞ n=1 is a sequence of closed subsets of X whose union contains a non-empty open set W. Then there exists n such that Cn ∩ W has a non-empty interior. Proof The space (W, τ ) is also topologically complete, or locally compact and Hausdorff. The sets Cn ∩ W are closed subsets of W whose union is W, and so there exists n and a non-empty open subset V of W such that V ⊆ Cn ∩ W. Since W is open in X, it follows that V is open in X. Why is Baire’s category theorem a ‘category’ theorem? There is a collection of terminologies related to it. A subset A of a topological space (X, τ ) is said to be nowhere dense if A has an empty interior, and is meagre, or of the first category if it is the union of a sequence of nowhere dense sets. Otherwise, A is of the second category. Thus the Baire category theorem says that a complete metric space is of the second category in itself. Here is an application of Baire’s category theorem. Proposition 2.7.4 Suppose that (X, τ ) is a topologically complete σ -compact topological space. Then there exists a dense open subset Y such that Y, with the subspace topology, is locally compact. Proof Let Y = {y ∈ X : y has a compact neighbourhood in X}. Then Y is open and locally compact. We must show that Y is dense in X. Let x ∈ X and let V be an open neighbourhood of x. Then V is topologically complete. Let (Kn )∞ n=1 be an increasing sequence of compact subsets of X whose union is X, and let Ln = V ∩ Kn , for each n. Each Ln is closed in V, and so, by Baire’s category theorem, there exists n such that Ln has a non-empty interior in V. Let y ∈ Vnint . Then y ∈ Y, and so Y is dense in X. A topological space (X, τ ) is locally homogeneous if whenever x, y ∈ X there exist open neighbourhoods Ux and Uy of x and y, and a homeomorphism φ : Uy → Ux with φ(y) = x.

2.8 Lipschitz Functions

35

Corollary 2.7.5 A locally homogeneous σ -compact topologically complete space is locally compact. Proof For there exists a dense open locally compact subset Y. Let y ∈ Y and x ∈ X, and let φ be a homeomorphism from an open neighbourhood Uy of y onto an open neighbourhood Ux of x. But Uy contains a compact neighbourhood of y, and so Ux contains a compact neighbourhood of x. On the other hand, let X = {z ∈ C : |z| < 1} ∪ {1}. Then X, with its usual topology, is σ -compact. It is topologically complete, by Theorem 2.6.1, and it is not locally compact. Exercise 2.7.6 Suppose that (X, τ ) is a countable locally homogeneous Baire space. Show that τ is the discrete topology. Give another proof that (Q, dp ) is not complete. Exercise 2.7.7 Suppose that f is a continuous function on [0, ∞) for which f (nx) → 0 as n → ∞ for each x > 0. Show that f (x)) → 0 as x → ∞.

2.8 Lipschitz Functions Suppose that f : X → Y, where (X, d) and (Y, ρ) are metric spaces, and that L ≥ 0. f is an L-Lipschitz function if ρ(f (x) − f (y)) ≤ Ld(x, y) for all x, y ∈ X. A Lipschitz function is a function which is an L-Lipschitz function, for some L > 0. A Lipschitz function is uniformly continuous. The space L(X) of realvalued Lipschitz functions on X is a linear subspace of C(X). If f is a Lipschitz function, there is a least L ≥ 0 such that f is an L-Lipschitz function: this is denoted by pL (f ); pL (f ) = 0 if and only if f is constant. Proposition 2.8.1 Suppose that A is a non-empty subset of a metric space (X, d). Let d(x, A) = inf{d(x, a) : a ∈ A} be the distance from x to A. Then d(·, A) is a 1-Lipschitz function on X, and d(x, A) = d(x, A). Proof If y ∈ X and a ∈ A then d(x, A) ≤ d(x, a) ≤ d(x, y) + d(y, a). Taking the infimum over A, we see that d(·, A) is a 1-Lipschitz function. Clearly, d(x, A) ≤ d(x, A). Given  > 0 there exist a ∈ A and a ∈ A such that d(x, A) ≤ d(x, a )+ /2 and d(a, a ) < /2. Then d(x, A) ≤ d(x, a) ≤ d(x, a ) + /2 ≤ d(x, A) + , so that d(x, A) ≤ d(x, A). Suppose now that  > 0. Recall that N (A) = {x ∈ X : d(x, A) < }: N (A) is the -neighbourhood of A. As usual, we write N (x) for N ({x}). We set H(A,) (x) = (1 − d(x, A)/)+ .

36

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Then H(A,) is a 1/-Lipschitz function on X, which takes the value 1 on A, the value 0 on X \ N (A) and values in (0, 1) otherwise. H(A,) converges pointwise to IA , the indicator function of A, as  → 0. We write H(x,) for H({x},) . The set L(X) of real-valued Lipschitz functions on a metric space (X, d) is a linear subspace of the vector space C(X) of continuous functions on X. Proposition 2.8.2 Suppose that F is a set of real-valued L-Lipschitz functions on a metric space (X, d), and that inf{f (x0 ) : f ∈ F} > −∞ for some x0 ∈ X. Let g(x) = infF f (x). Then g(x) > −∞ for all x ∈ X, and g is an L-Lipschitz function. Similarly for supF f . Proof If x ∈ X and f ∈ F then f (x) ≥ f (x0 ) − Ld(x, x0 ) ≥ g(x0 ) − Ld(x, x0 ), so that g(x) ≥ g(x0 ) − Ld(x, x0 ) > −∞. If x ∈ X then g(x) ≤ f (x) ≤ Ld(x, x ) + f (x ), so that g(x) ≤ Ld(x, x ) + g(x ), and similarly g(x ) ≤ Ld(x, x )+g(x), so that g is an L-Lipschitz function. Corollary 2.8.3 If (fn )∞ n=1 is a sequence of real-valued L-Lipschitz functions on (X, d) which converges pointwise to f , then f is an L-Lipschitz function. Proof For f = limn→∞ (infm≥n fm ). Suppose that f is a real-valued function on a metric space (X, d) and that L ≥ 0. Let UL (f ) = {h : h L-Lipschitz, h ≥ f }. If UL (f ) is not empty, we set f d,L = inf{h : h ∈ UL }. It follows from Proposition 2.8.2 that f d,L is a real-valued L-Lipschitz function, and is the smallest L-Lipschitz function greater than or equal to f . f d,L is the upper L-Lipschitz envelope of f . The lower L-Lipschitz envelope fd,L is defined in a similar way. We can characterize the upper L-Lipschitz envelope in the following way. Theorem 2.8.4 Suppose that f is a real-valued function on a metric space (X, d), that L ≥ 0 and that UL (f ) = ∅. If x ∈ X let h(x) = supy∈X (f (y) − Ld(x, y)). Then h = f d,L . Proof Setting y = x, it follows that h ≥ f . If x, y ∈ X then f d,L (x) ≥ f (y) − Ld(x, y), so that f d,L (x) ≥ h(x). But h is an L-Lipschitz function, and so h = f d,L . Corollary 2.8.5 If there exists a real-valued Lipschitz function less than or equal to f , then fd,L (x) = infy∈X (f (y) + Ld(x, y)). In the same vein, we have the following.

2.8 Lipschitz Functions

37

Theorem 2.8.6 (The McShane–Whitney extension theorem) Suppose that f is a real-valued L-Lipschitz function on a non-empty subset A of X. Then there exists an L-Lipschitz function g on X which extends f . Proof Let g(x) = inf{f (y) + Ld(x, y) : y ∈ A}. If x, x ∈ X and a ∈ A then g(x) ≤ f (a) + Ld(x, a) ≤ f (a) + Ld(x , a) + Ld(x, x ) ≤ g(x ) + Ld(x, x ). Thus g is an L-Lipschitz function. If a ∈ A then g(a) = inf (f (b) + Ld(b, a)) ≥ f (a) = f (a) + Ld(a, a) ≥ g(a), b∈A

so that g(a) = f (a). We denote by BL(X) the vector space of bounded real-valued Lipschitz functions on X. Corollary 2.8.7 If A is a non-empty subset of X and f ∈ BL(A), then there exists g ∈ BL(X) with pL (g) = pL (f ), supx∈X g(x) = supa∈A f (a) and infx∈X g(x) = infa∈A f (a) such that f = g|A . Proof By the McShane–Whitney theorem, there exists h ∈ L(X) with pL (h) = pL (f ), such that h|A = f . Let g = (h ∧ sup f ) ∨ (inf f ). Note that if X = N, with the discrete metric, then BL(N) = Cb (N) = l∞ .

3 Polish Spaces and Compactness

3.1 Polish Spaces A topological space (X, τ ) is a Polish space if it is separable and topologically complete; that is, it is separable, and there is a complete metric d on X which defines the topology τ . In particular, a complete separable metric space is called a Polish metric space. Let us bring earlier results together. Theorem 3.1.1 Suppose that (X, τ ) is a topological space. The following are equivalent. (i) (ii) (iii) (iv) (v)

(X, τ ) is a Polish space. (X, τ ) is homeomorphic to a Gδ subspace of the Hilbert cube H. (X, τ ) is homeomorphic to a Gδ subspace of a Polish space. (X, τ ) is homeomorphic to a closed subspace of a Polish space. (X, τ ) is homeomorphic to a closed separable subspace of (l∞ , d∞ ).

Further, a Polish space is second countable, and the product of a sequence of Polish spaces, with the product topology, is a Polish space. Here are some important examples. (i) (ii) (iii) (iv)

Euclidean space, and Hermitian space. The Hilbert cube H. The Bernoulli sequence space (N). The set I of irrational numbers, with the subspace topology.

For I = R \ Q is a Gδ subset of the Polish space R. (v) The product NN , with the product topology. Exercise 3.1.2 Use continued fraction expansions to show that I and NN are homeomorphic. 38

3.2 Totally Bounded Metric Spaces

39

Exercise 3.1.3 Show that it follows from Baire’s category theorem that the set Q of rational numbers, with the subspace topology, is not a Polish space.

3.2 Totally Bounded Metric Spaces We need some more definitions. Suppose that  > 0. If B is a subset of a metric space (X, d), then a subset A of B is an -net in B if B ⊆ ∪a∈A N (a). Thus A is an -net if and only if each element of B is within  of a member of A. A subset B of X is totally bounded if there is a finite -net in B for each  > 0. Total boundedness is not a topological property: (0, 1] is totally bounded under the usual metric, and is homeomorphic to [1, ∞), which is not totally bounded under its usual metric. On the other hand, if f is a uniformly continuous mapping of a metric space (X, d) into a metric space (Y, ρ) and B is a totally bounded subset of X, then f (B) is a totally bounded subset of Y. In particular, if f is a uniform homeomorphism then B is totally bounded if and only if f (B) is. Proposition 3.2.1 A totally bounded subset B of a metric space is separable, and is therefore second countable. Proof For each n ∈ N there is a finite 1/n-net An in B. Then ∪n∈N An is a countable dense subset of B. Here is a characterization of sets that are not totally bounded. Proposition 3.2.2 A subset C of a metric space (X, d) is not totally bounded if and only if there exist  > 0 and a sequence (cn )∞ n=1 in C such that d(cm , cn ) ≥  for m = n. Proof Suppose that C is not totally bounded. Thus there exists  > 0 such that C has no finite -net. We use an inductive argument to find a sequence (cn )∞ n=1 ∈ C such that d(cm , nn ) ≥  for m = n. If we have found c1 , . . . , cn satisfying this condition, then {cj : 1 ≤ j ≤ n} is not an -net in C, and so there exists cn+1 ∈ C such that d(cj , cn+1 ) ≥  for 1 ≤ j ≤ n. Conversely, if the condition is satisfied, and if c ∈ C then N/2 (c) can contain at most one term of the sequence, and so C contains no finite /2-net. Totally bounded sets can be characterized in terms of Cauchy sequences. Theorem 3.2.3 A subset B of a metric space (X, d) is totally bounded if and only if every sequence in B has a Cauchy subsequence.

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Proof Suppose first that B is totally bounded, and that (xn )∞ n=1 is a sequence in B. We use a diagonal argument. For each n ∈ N, let An be a finite 1/n-net in B. First, there exists a1 ∈ A1 such that xn ∈ N1 (a1 ) for infinitely many n; that is, there exists a subsequence (x1n )∞ n=1 such that x1n ∈ N1 (a1 ) for n ∈ N. Arguing inductively, for each j ∈ N there exist aj ∈ Aj and a subsequence (xjn )∞ n=1 of ∞ such that x ∈ N (a ) for n ∈ N. Let x = x . Then (x ) (xj−1,n )∞ jn 1/j j kj jj kn n=1 is n=1 a subsequence of (xn )∞ n=1 , and d(xkm , xkp ) < 2/n for m, p ≥ n, and so it is a Cauchy sequence. Conversely, if B is not totally bounded, there exist  > 0 and a sequence (bn )∞ n=1 in B such that d(bm , bn ) ≥  for m = n. This sequence clearly has no convergent subsequence. Let us establish a result about -nets that we shall need later, when we prove the existence of Haar measure on a compact metric group. If (X, d) is a totally bounded metric space, and  > 0, then there is a finite -net in X. A finite -net with as few elements as possible is called a minimal -net. (Beware! An -net which contains no proper -subnet need not be minimal.) The members of two minimal -nets can be paired so that members of a pair are close together. For this we need Hall’s marriage theorem. Theorem 3.2.4 (Hall’s marriage theorem) Suppose that A and B are finite sets and that H ⊂ A×B. If C ⊂ A, let h(C) = {b ∈ B : (a, b) ∈ H for some a ∈ C}. Then there exists an injective mapping ψ : A → B such that (a, ψ(a)) ∈ H for all a ∈ A if and only if |h(C)| ≥ |C| for all C ⊆ A. Proof The condition is certainly necessary. We prove sufficiency by induction on |A|. The result is true for |A| = 1. Suppose that |A| = n and that the result is true for 1 ≤ k < n. We consider two cases. First, suppose that |h(C)| > |C| for each proper subset of A. Pick a1 ∈ A, and choose ψ(a1 ) ∈ h({a1 }). Let A1 = A \ {a1 }, and let H1 = {(a, b) : a ∈ A1 , (a, b) ∈ H}. Then A1 , B1 = B \ {ψ(a1 )} and H1 satisfy the conditions of the theorem, and |A1 | < n, so that we can define an injective mapping ψ : A1 → B1 so that (a, ψ(a)) ∈ H1 for each a ∈ A1 . Then ψ : A → B satisfies the theorem. Secondly, suppose that there exists a proper subset C of A such that |h(C)| = |C|. Then by the inductive hypothesis we can define an injective mapping ψ : C → h(C) so that (c, ψ(c)) ∈ H for c ∈ C. Let D = A \ C, E = B \ h(C) and HD = {(d, e) : (d, e) ∈ D × E, (d, e) ∈ H}. Then |D| < n, and D, E, HD satisfy the conditions, so that we can define an injective mapping ψ : D → E so that (d, ψ(d)) ∈ HD for each d ∈ D. Then ψ : A → B satisfies the theorem.

3.3 Compact Metrizable Spaces

41

Theorem 3.2.5 Suppose that (X, d) is a totally bounded metric space, that  > 0 and that m and n are minimal -nets in X. Then there exists a bijective mapping χ : m → n such that d(m, χ (m)) < 2 for all m ∈ m . Proof Define a relation H on m × n by setting H(m, n) if and only if d(m, n) < 2. If m ∈ m , let h(m) = {n ∈ n : H(m, n)}, and if A ⊆ m let h(A) = ∪m∈A h(m). We shall show that |h(A)| ≥ |A|, for each A ⊆ m . If not, let L = (m \ A) ∪ h(A). Then |L| < |N|. We shall show that L is an -net, giving a contradiction. If x ∈ X, then d(x, m) <  for some m ∈ m . Suppose that m ∈ A. Then there exists n ∈ n such that d(x, n) < . But then d(m, n) ≤ d(m, x) + d(x, m) < 2 and so n ∈ h(A). If m ∈ A then x ∈ ∪m∈mx \A N (m). Thus X = (∪m∈mx \A N (m)) ∪ (∪n∈h(A) N (n)) = ∪m∈L N (m), so that L is an -net. We now apply Hall’s marriage theorem, which ensures that a suitable mapping χ exists.

3.3 Compact Metrizable Spaces Compact metrizable spaces form one of the most fundamental and important classes of Polish spaces. When is a metric space compact? Theorem 3.3.1 Suppose that (X, d) is a metric space. The following are equivalent. (i) (ii) (iii) (iv)

(X, d) is compact. (X, d) is countably compact. (X, d) is sequentially compact. (X, d) is complete and totally bounded.

Proof A compact topological space is countably compact, so that (i) implies (ii). Since (X, d) is first countable, (ii) and (iii) are equivalent. Suppose that (X, d) is sequentially compact. Then every sequence in X has a convergent sequence, and this is a Cauchy sequence, so that (X, d) is totally bounded. Further, a Cauchy sequence in X has a convergent subsequence, and is therefore convergent, so that (X, d) is complete. Thus (iii) implies (iv). Conversely, if (X, d) is totally bounded and complete, then every sequence in X has a Cauchy subsequence (by total boundedness), and this converges (by completeness), and so (iv) implies (iii). Thus (ii), (iii) and (iv) are equivalent.

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Finally we show that (iii) and (iv) together imply (i). Suppose that (iii) and (iv) hold. Let O be an open cover of X. First we show that there exists δ > 0 such that if x ∈ X then there exists O ∈ O such that Nδ (x) ⊆ O. Suppose not. Then for each n ∈ N there exists xn ∈ X such that N1/n (xn ) ⊆ O, for all O ∈ O. Since (X, d) is sequentially compact, there is a convergent subsequence (xnk )∞ k=1 , convergent to x, say. Then there exists O ∈ O and N ∈ N such that N1/N (x) ⊆ O. There exists K ∈ N, with nK > 2N, such that d(xnk , x) < 1/2N for k ≥ K. But then N1/2nK (xnK ) ⊆ O, giving a contradiction. Since (X, d) is totally bounded, there exists a finite δ-net Aδ in X. For each a ∈ Aδ there exists Oa ∈ O such that Nδ (a) ⊆ Oa . Then X = ∪a∈A Nδ (a) ⊆ ∪a∈A Oa , so that {Oa : a ∈ A} is a finite subcover of X; (X, d) is compact. Corollary 3.3.2 A compact metrizable space is a Polish space. It is a remarkable fact that neither completeness nor total boundedness is a topological property, but that together they are equivalent to compactness, which is a topological property. ˆ of a totally bounded metric space is ˆ d) Corollary 3.3.3 The completion (X, compact. ˆ Proof For if 0 < δ <  then a δ-net in X is an -net in X. For this reason, a totally bounded set is sometimes called a precompact set. Exercise 3.3.4 Suppose that (Un )N n=1 is a finite open cover of a compact metric space. Show that there is a corresponding partition of unity (fn )N n=1 ; each fn is a continuous real-valued function on X, with 0 ≤ fn ≤ 1 and is zero outside  Un , and N n=1 fn = 1. Theorem 3.3.5 A continuous mapping f from a compact metric space (X, d) into a metric space (Y, ρ) is uniformly continuous. Proof Suppose not. Then there exists  > 0 such that for each n ∈ N there exist xn , yn ∈ X with d(xn , yn ) < 1/n and ρ(f (xn ), f (yn )) ≥ . Since (X, d) is sequentially compact, there exists a subsequence (xnk )∞ k=1 which converges to a point x of X. But then ynk → x as k → ∞. Since f is continuous at x, ρ(f (xnk ), f (x)) → 0 and ρ(f (ynk ), f (x)) → 0 as k → ∞, and so ρ(f (xnk ), f (ynk )) → 0 as k → ∞, giving a contradiction. Theorem 3.3.6 The topological product (X, d) of a sequence (Xn , dn ) of compact metrizable spaces is compact and metrizable. Proof (X, d) is metrizable (Theorem 2.2.5), and so it is sufficient to show that (j) ∞ ∞ it is countably compact. Suppose that (x(j) )∞ j=1 = ((xn )n=1 )j=1 is a sequence

3.3 Compact Metrizable Spaces

43

in X. For each n there is a limit point xn of the sequence (xn )∞ j=1 . Then x = ∞ ∞ (j) (xn )n=1 is a limit point of the sequence (x )j=1 . (j)

Of course, this is a special case of Tychonoff’s theorem. But the proof does not use the axiom of choice. Corollary 3.3.7 The Hilbert cube H and the Bernoulli sequence space (N) are compact metrizable spaces. A topological space (X, τ ) is a compact metrizable space if and only if it is homeomorphic to a closed subset of the Hilbert cube. Proof A compact metrizable space is second countable, and so it is homeomorphic to a subspace Y of the Hilbert cube. Since a compact subspace of a Hausdorff space is closed, Y must be closed. Conversely, since the Hilbert cube is compact and metrizable, a topological space which is homeomorphic to a closed subspace of H must be compact and metrizable. Corollary 3.3.8 A topological space (X, τ ) is a Polish space if and only if it is ˜ If so, there is a metric d ˜ d). a dense Gδ subset of a compact metric space (X, on (X, τ ) which defines the topology τ such that (X, d) is totally bounded. Proof For we can take X˜ as the closure of X, when (X, τ ) is considered as a subspace of the Hilbert cube H, take d˜ to be a metric defining the subspace ˜ and take d to be the restriction of d˜ to X. topology on X, Proposition 3.3.9 The subspace ⎫ ⎧ n ⎬ ⎨ aj ICj : n ∈ N, aj ∈ R, Cj a cylinder set S() = ⎭ ⎩ j=1

of C() is dense in C(). Proof Let d be the Cantor metric on . If f ∈ C() then f is uniformly continuous, and so if  > 0 there exists n ∈ N such that if d(ω, ω ) < 1/3n then |f (ω) − f (ω )| < . For each cylinder set C of rank n pick ωC ∈ C, and let  g= {f (ωC )IC : C a cylinder set of rank n}. Then g ∈ S() and f − g∞ < . Functions in S() are called step functions. Suppose that A is a subset of a topological space (X, τ ). A continuous mapping r : X → A is a retraction if r(a) = a for a ∈ A. If there is a retraction of X onto A then A is a retract of X.

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Theorem 3.3.10 Suppose that K is a non-empty compact subset of a metric space (X, d) and that there is a mapping n : X → K such that d(x, n(x)) < d(x, k) for k = n(x). Then n is a retraction of X onto K. Proof We must show that n is continuous. Suppose that it is not continuous at x. Then there exists  > 0 and a sequence (xj )∞ j=1 in X such that xj → x as j → ∞, while d(n(x), n(xj )) ≥  for all j. Since K is compact, the sequence ∞ (n(xj ))∞ j=1 has a subsequence (n(xji ))i=1 which converges to a point y of K as i → ∞. Then d(n(x), y) ≥ . Now d(x, y) = lim d(xji , n(xji )) ≤ lim d(xji , n(x)) i→∞

n→∞

≤ lim d(xji , x) + d(x, n(x)) = d(x, n(x)). i→∞

But n(x) is the unique nearest point to x in K and y = n(x), giving a contradiction. We shall also need the following theorem. Theorem 3.3.11 A non-empty metric space X is compact if and only if there is a continuous surjective mapping of the Bernoulli sequence space (N) onto X. Proof The condition is sufficient, since (N) is compact, and the continuous image of a compact space is compact. Suppose that (X, d) is compact. Then it is totally bounded, and so it is the union of finitely many sets of diameter at most ; since the diameter of a set is the same as the diameter of its closure, we can suppose that the sets are closed. First we find a finite set A of closed non-empty subsets of X, each of diameter at most 1 such that X = ∪A∈A A. We choose k1 = l1 such that 2k1 ≥ |A|, and a surjective mapping φ0 of (k1 ) onto A. We set Aω = φ0 (ω). We now repeat the procedure for each Aω . For each ω ∈ (l1 ), there exists a finite set Aω of non-empty closed subsets of Aω of diameter at most 1/2 such that Aω = ∪A∈Aω A. We choose k2 such that 2k2 ≥ maxω∈(l1 ) |Aω | and for each ω ∈ (l1 ) define a surjective mapping φ1,ω of (k1 ) onto Aω . We set l2 = l1 + k2 . If ω ∈ (l2 ), we can write ω as (ω , ω ), with ω ∈ (l1 ) and ω ∈ (k2 ). We set Aω = φ1,ω (ω ). Thus Aω = ∪ω ∈(k2 ) A(ω ,ω ) . We now iterate. There exists a strictly increasing sequence (ln )∞ n=1 , with ln+1 = ln + kn , such that for each n ∈ N there exists a family {Aω : ω ∈ (ln )} of closed non-empty subsets of X, each of diameter at most 1/n, such that if ω ∈ (ln ) then Aω = ∪ω ∈(kn+1 ) (A(ω ,ω ) ). Suppose now that ω ∈ (N). For n ∈ N, let An (ω) = A(ω1 ,...ωln ) . Then (An (ω))∞ n=1 is a decreasing sequence of closed subsets of X, and diam(An (ω)) → 0 as n → ∞. Since (X, d) is complete, it follows from

3.3 Compact Metrizable Spaces

45

Proposition 2.4.3 that ∩n∈N An is a singleton, {ψ(ω)}, say. Thus ψ is a mapping of (N) into X. Next we show that ψ is continuous. If ω ∈ (N) and n ∈ N then the cylinder set Cω,ln is contained in An (ω), so that if ω ∈ Cω,ln then d(ψ(ω), ψ(ω )) ≤ 1/n. Finally, we show that ψ is surjective. Since ψ((N)) is compact, and is therefore closed in X, it is enough to show that ψ((N)) is dense in X. If x ∈ X and n ∈ N then x ∈ An (ω ) for some ω ∈ (ln ). There exists ω ∈ (N) such that ω = (ω1 , . . . , ωln ). Then ψ(ω) ∈ An (ω ), so that d(x, ψ(ω)) ≤ 1/n. We use this theorem to give an easy proof of the following result. Theorem 3.3.12 Suppose that (X, d) is a compact metric space and that (Y, ρ) is a Polish space. Then (C(X, Y), d∞ ) is a Polish space. Proof Certainly (C(X, Y), d∞ ) is complete. We must show that it is separable. Let ψ : (N) → X be a continuous surjection. If f ∈ C(X, Y), let Tψ (f ) = f ◦ ψ. Then Tψ is an isometry of (C(X, Y), d∞ ) into (C((N), Y), d∞ ), and so it is sufficient to show that (C((N), Y), d∞ ) is separable. Let D be a countable dense subset of Y. If ω ∈ (N) and n ∈ N, let Pn (ω) = (ω1 , . . . , ωn ), so that Pn maps (N) onto n . Let Fn be the (countable) set of mappings from n into D, let Gn = {g ◦ Pn : g ∈ Fn }, and let G = ∪∞ n=1 Gn . Then G is a countable subset of C((N), Y): we show that G is dense in (C((N), Y), d∞ ). Suppose that f ∈ C((N), Y) and that  > 0. Since f is uniformly continuous, there exists n ∈ N such that if Pn (ω) = Pn (ω ), ˜ ω) ˜ ∈ D such that then ρ(f (ω), f (ω )) < /2. Thus if ω˜ ∈ n there exists d( ˜ ˜ ˜ Let g = d ◦ Pn . Then g ∈ G and ρ(d(ω), ˜ f (ω)) <  whenever Pn (ω) = ω. d∞ (f , g) ≤ . Corollary 3.3.13 If (X, d) is a compact metric space, then (C(X), d∞ ) (=(C(X, R), d∞ )) is a Polish space. Corollary 3.3.13 characterizes those compact spaces which are metrizable. Proposition 3.3.14 Suppose that (X, τ ) is a compact topological space for which (C(X), d∞ ) is separable. Then X is metrizable. Proof Suppose that (fn )∞ n=1 is a dense sequence in (C(X), d∞ ). If x ∈ X, let ∞ F(x) = (fn (x))n=1 . Then F is a continuous mapping of X into (RN , d), where d is a metric on RN which defines the product topology, so that (RN , d) is a Polish space. Suppose that x = y. Let g(·) = d(·,y). There exists n ∈ N such that d∞ (fn , g) < d(x, y)/2. Then fn (x) = fn (y), and so the mapping F is injective. By Exercise 1.2.1(iii), F is then a homeomorphism of X onto F(X), and so (X, τ ) is metrizable.

46

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Recall that the Helly space is compact, separable and first countable, but is not metrizable. Let us consider compactness in (C(X, Y), when X and Y are compact. First we consider total boundedness. We need some definitions. Suppose that (X, d) and (Y, ρ) are metric spaces, that x ∈ X and that F is a set of mappings from X to Y. Then F is equicontinuous at x if whenever  > 0 there exists δ > 0 such that if d(x, y) < δ then ρ(f (x), f (y)) <  for all f ∈ F. F is equicontinuous on X if it is equicontinuous at each point of X. F is uniformly equicontinuous if whenever  > 0 there exists δ > 0 such that if d(y, z) < δ then ρ(f (y), f (z)) <  for all f ∈ F. Proposition 3.3.15 Suppose that (X, d) and (Y, ρ) are metric spaces and that F is a set of mappings from X to Y which is equicontinuous on X. If (X, d) is compact, then F is uniformly equicontinuous. Proof Make obvious changes to the proof of Theorem 3.3.5. Proposition 3.3.16 Suppose that (X, d) and (Y, ρ) are totally bounded metric spaces and that F ⊆ C(X, Y). If F is a uniformly equicontinuous set of mappings from X to Y, then F is totally bounded in C(X, Y). Proof Suppose that  > 0. Then there exists δ > 0 such that if d(x, y) < δ then ρ(f (x), f (y)) <  for all f ∈ F. There is a finite partition S of X into non-empty sets of diameter at most δ, and a finite partition T of Y into non-empty sets of diameter at most . For each S ∈ S pick xS ∈ S. For each φ in the finite set T S , let Fφ = {f ∈ F : f (xS ) ∈ φ(S) for all S ∈ S}. Let  = {φ ∈ T S : Fφ = ∅}. Then F = ∪φ∈ Fφ is a finite partition of F. If f , g ∈ Fφ and x ∈ X, then x ∈ S for some S ∈ S, and so ρ(f (x), g(x)) < . Hence diam(Fφ ) < . Thus F is totally bounded in Cb (X, Y). Theorem 3.3.17 (The Arzel`a–Ascoli theorem) Suppose that (X, d) and (Y, ρ) are compact metric spaces and that F ⊆ C(X, Y). Then F is compact if and only if F is equicontinuous on X. Proof If F is equicontinuous on X then it is uniformly equicontinuous, and so F is totally bounded. Then F is totally bounded. But it is also complete, since (C(X, Y), d∞ ) is complete, and so F is compact. Conversely, suppose that F is not equicontinuous. Then there exists x ∈ X for which F is not equicontinuous at x, and so there exists  > 0 for which there is no suitable δ > 0. Take δ1 = 1. Then there exists x1 ∈ X such that d(x1 , x) < δ1 and f1 ∈ F such that ρ(f1 (x1 ), f1 (x)) ≥ . We show by induction that there exist a decreasing sequence (δn )∞ n=1 of positive numbers, a sequence ∞ in F such that ρ(f (x ), f (x)) ≥  and in X and a sequence (f ) (xn )∞ n n=1 n n n n=1

3.4 Locally Compact Polish Spaces

47

ρ(fn (y), fn (x)) < /3 if d(y, x) < δn+1 . Suppose that we have found δj , xj and fj for 1 ≤ j ≤ n which satisfy the conditions. Since fn is continuous at x there exists δn+1 such that ρ(fn (y), fn (x)) < /3 for d(y, x) < δn+1 . Since F is not equicontinuous at x, there exist xn+1 with d(xn+1 , x) < δn+1 and fn+1 ∈ F such that ρ(fn+1 (xn+1 , fn+1 (x))) ≥ . This establishes the induction. If m > n then either ρ(fm (x), fn (x)) ≥ /3 or ρ(fm (xm ), fn (xm )) ≥ ρ(fm (xm ), fm (x)) − ρ(fm (x), fn (x)) − ρ(fn (xm ), fn (x)) ≥ /3, so that d∞ (fm , fn ) ≥ /3. Thus F is not totally bounded, and F is not compact. Corollary 3.3.18 Suppose that (X, d) is a compact metric space and that F ⊆ C(X), the space of continuous real-valued functions on X. The following are equivalent. (i) F is totally bounded. (ii) F is equicontinuous, and there exists M > 0 such that |f (x)| ≤ M for each x ∈ X and f ∈ F. (iii) F is equicontinuous, and {f (x) : f ∈ F} is bounded, for each x ∈ X. Proof Suppose that F is totally bounded, so that F is equicontinuous. Suppose that N is a 1-net in F. Then |f (x)| ≤ maxg∈N |g(x)| + 1 ≤ maxg∈N g∞ + 1, and so (i) implies (ii). Conversely, suppose that (ii) holds. Then we can consider F as a subset of C(X, [−M, M], and so (i) holds. (ii) certainly implies (iii). Suppose that (iii) holds. For each x ∈ X, there exists an open neighbourhood Ox of x in X such that if y ∈ Ox then | f (y) − f (x)| ≤ 1 for all f ∈ F. Since X is compact, there exists a finite set X0 of X such that X = ∪x∈X0 Ox . Then sup sup |f (x)| ≤ max sup |f (x)| + 1, x∈X f ∈F

x∈X0 f ∈F

so that (iii) implies (ii).

3.4 Locally Compact Polish Spaces When is a metrizable locally compact space a Polish space? Theorem 3.4.1 Suppose that (X, τ ) is a metrizable locally compact space. The following are equivalent.

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(i) (ii) (iii) (iv) (v)

(X, τ ) is a Polish space. (X, τ ) is separable. (X, τ ) is second countable. (X, τ ) is σ -compact. There is an increasing sequence (Kn )∞ n=1 of compact subsets of X such int , for each n ∈ N. that ∪n∈N Kn = X and Kn ⊆ Kn+1 ˜ τ˜ ) of (X, τ ) is metrizable. (vi) The one-point compactification (X, (vii) (X, τ ) is homeomorphic to an open subset of a compact metric space. Proof (i) trivially implies (ii), and (ii) implies (iii), by Theorem 2.2.1. Suppose that (X, τ ) is second countable, and that B is a countable base for the topology. Let Bc = {B ∈ B : B is compact}. If x ∈ X, let N(x) be a compact neighbourhood of X. There exists B ∈ B such that x ∈ B ⊆ N(x). But then B ∈ Bc , and so X = ∪B∈Bc B, and X is σ -compact. If (iv) holds, then (v) holds by Proposition 1.2.6. If (v) holds then (X, d) is separable and therefore second ˜ n : n ∈ N} is a countable countable. If B is a countable base for τ , then B∪{X\K ˜ base for τ˜ . Thus (X, τ˜ ) is metrizable, by Urysohn’s metrization theorem. Since ˜ τ˜ ), (vi) implies (vii), and (vii) implies that (X, τ ) X is an open subset of (X, is a Polish space, by Theorem 2.6.1, since a compact metric space is a Polish space. Let X = {z ∈ C : |z| < 1} ∪ {1}. With its usual topology, X is an example of a σ -compact Polish space which is not locally compact. Let (X, τ ) be a locally compact Polish space, and suppose that (Kn )∞ n=1 is int , a sequence of compact subsets of X such that ∪n∈N Kn = X and Kn ⊆ Kn+1 for each n ∈ N. If f ∈ C(X), the space of continuous real-valued functions on from Tietze’s extension X, let fn = f|Kn and let j(f ) = (fn )∞ n=1 . Then it follows ∞ theorem that j maps C(X) onto a closed subset of n=1 C(Xn ). Thus if we give C(X) a metric such as d(f , g) =

∞ 

sup (|f (x) − g(x)| ∧ 2−n ),

n=1 x∈Kn

(C(X), d) becomes a complete Polish metric space. We also consider the space C0 (X). Suppose that f is a continuous real-valued function on a locally compact space (X, τ ), considered as a subspace of its one-point compactification (X ∗ = X ∪ {∞}, τ ∗ ). Then f ∈ C0 (X) if and only if f (x) → 0 as x → ∞; that is, given  > 0 there exists a compact subset K of X such that |f (x)| <  for x ∈ X \ K. We give C0 (X) the metric d∞ . Then (C0 (X), d∞ ) is homeomorphic to the closed hyperplane {f ∈ C(X ∗ ) : f (∞) = 0}, and so is complete.

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49

Proposition 3.4.2 If (X, τ ) is a locally compact space, then (C0 (X), d∞ ) is separable if and only if X is σ -finite and metrizable. Proof This follows immediately from Corollary 3.3.13 and Proposition 3.3.14. In particular, C0 (Rd ) is separable, for d ∈ N.

4 Semi-continuous Functions

4.1 The Effective Domain and Proper Functions We shall need to consider the suprema and infima of sets of real-valued functions defined on a set X. A supremum can however take the value +∞ and an infimum take the value −∞. It is therefore frequently convenient to consider functions taking values in the extended real line R = [−∞, ∞]. If f is such a function then we define the effective domain f , or dom(f ) to be the set {x ∈ X : f (x) ∈ R}, and say that f is a proper function if f is not empty.

4.2 Semi-continuity Suppose that (X, τ ) is a topological space, that f is a proper function on X taking values in the extended real line R = [−∞, ∞] and that x ∈ X. Then f is lower semi-continuous at x if whenever λ < f (x) there exists a neighbourhood N of x such that f (y) > λ for each y ∈ N. (Note that the condition is trivially satisfied if f (x) = −∞.) f is lower semi-continuous on X if it is lower semicontinuous at each point of X. Upper semi-continuity, which we have met earlier, is defined in a similar way. There are corresponding results for upper semi-continuous functions, but we shall concentrate for the most part on lower semi-continuous functions. Exercise 4.2.1 Let τ− be the one-sided topology {∅}∪{(a, +∞] : a < +∞} ∪ {R} on R. Show that an R-valued function on X is lower semi-continuous on X if and only if it is continuous from X to (R, τ− ). Suppose that f is a mapping from a set X into R. The epigraph Af is the set {(x, μ) ∈ X × R : f (x) ≤ μ}, and the strict epigraph Sf is the set {(x, μ) ∈ X × R : f (x) < μ}. (Note that we exclude the values ±∞ in the definitions of the epigraph and strict epigraph.) 50

4.2 Semi-continuity

51

Proposition 4.2.2 Suppose that f is a proper function on a topological space (X, τ ) taking values in [−∞, ∞]. f is lower semi-continuous if and only if the epigraph Af is closed in X × [−∞, ∞). Proof Suppose that f is lower semi-continuous. If x ∈ X and λ < f (x), let μ = 12 (λ + f (x)). Then there exists a neighbourhood N of x such that f (y) > μ for y ∈ N. Then N×(−∞, μ) is an open subset of X × R disjoint from Af . Thus Af is closed. Conversely, if Af is closed, and f (x) > λ then there is an open neighbourhood U of x and 0 <  < f (x) − λ such that U × (−∞, λ + ) is disjoint from Af , so that f (y) > λ for y ∈ U; thus f is lower semi-continuous. This proposition implies many basic results. Exercise 4.2.3 Suppose that F is a set of lower semi-continuous functions on X. Show the following. g = supf ∈F f is lower semi-continuous. If F is finite, then h = inff ∈F f is lower semi-continuous. If F is infinite, then inff ∈F f need not be lower semi-continuous. If f , g ∈ F, then f + g is lower semi-continuous. If f , g ∈ F and f and g are non-negative, then f .g is lower semicontinuous (when 0.∞ is defined as 0). (vi) If f ∈ F and f ≥ 0 then 1/f is upper semi-continuous. (vii) f is upper semi-continuous if and only if Sf is open, and f is continuous if and only if G(f ) is closed. (i) (ii) (iii) (iv) (v)

Proposition 4.2.4 Suppose that f is a proper lower semi-continuous function on a compact space (X, τ ), taking values in (−∞, ∞]. Then f attains its infimum. Proof Let c = inf{f (x) : x ∈ X}, let L = {λ : λ > c} and for each λ ∈ L let Cλ = {x : f (x) ≤ λ}. Then each Cλ is a closed non-empty set, by Proposition 4.2.2, and so since X is compact, C = ∩λ∈L Cλ is non-empty. If x ∈ C, then f (x) = c. Note that this implies that c > −∞. Suppose that f is a proper function on a topological space (X, τ ) taking values in (R, ∞]. We define f† (x) = sup{ inf f (y) : U a neighbourhood of x}. y∈U

Then f† ∈ R. Proposition 4.2.5 If f is a proper function on a topological space (X, τ ) taking values in (R, ∞], then f† ≤ f , f† is lower semi-continuous on f , and f† is the largest lower semi-continuous function which is less than or equal to f .

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Proof Since x ∈ U for each neighbourhood U of x, f† ≤ f . If x ∈ f and α < f† (x), choose α < α  < f (x). There exists an open neighbourhood U of x such that f (y) > α  for y ∈ U. If z ∈ U then U is a neighbourhood of z, and so f† (z) ≥ α  > α. Thus f† is lower semi-continuous at x. If f is lower semi-continuous at x and α < f (x), there exists an open neighbourhood U of x such that f (y) > α for y ∈ U. Thus f† (x) ≥ α. Since α is arbitrary, f† (x) ≥ f (x). But f† ≤ f , and so f† (x) = f (x). If g is lower semi-continuous and g ≤ f , then g = g† ≤ f† , so that f† is the largest lower semi-continuous function less than or equal to f . f† is called the lower semi-continuous envelope of f . The upper semicontinuous envelope f † is defined similarly. Corollary 4.2.6 (i) If f is lower semi-continuous at x and xn → x as n → ∞, then lim infn→∞ f (xn ) ≥ f (x). (ii) If (X, τ ) is first countable, then conversely f is lower semi-continuous at x if lim infn→∞ f (xn ) ≥ f (x) whenever xn → x as n → ∞. Proof (i) This follows, since lim infn→∞ f (xn ) ≥ lim infy→x f (y) ≥ f† (x) = f (x). (ii) Conversely, suppose that (X, τ ) is first countable, and that f is not lower semi-continuous at x. Let (Un )∞ n=1 be a base of neighbourhoods of x. There exists  > 0 such that inf{ f (y) : y ∈ Un } ≤ f (x) −  for each n ∈ N. Thus there exists xn ∈ Un such that f (xn ) < f (x) − /2 for n ∈ N. Then xn → x as n → ∞ and lim infn→∞ f (xn ) ≤ f (x) − /2. Suppose that f is a real-valued function on a topological space (X, τ ), and that x ∈ X. Then the local oscillation f of X is f † − f† , so that f is a nonnegative upper semi-continuous function. For clearly f † ≥ f† , and f † and −f† are upper semi-continuous. We can also use the epigraph of a function to construct its lower semicontinuous envelope. In fact, we do a bit more. Proposition 4.2.7 Suppose that Y is a subset of a topological space (X, τ ) and that f is a function from Y to R which is bounded below. Let Af = {(x, μ) ∈ X × R : f (x) ≤ μ}, and let Bf = Af ∪ (X × {∞}). (i) There is a function g : x → (−∞, ∞] such that Bf = Ag . (ii) g(y) = f† (y) for y ∈ Y. Proof (i) Let Z = {z ∈ X : there exists (z, t) ∈ Bf }, so that Y ⊆ Z ⊆ X. Suppose that (z, t) ∈ Bf and that s > t. First we show that (z, s) ∈ Bf . If U × I is an open neighbourhood of (x, s), then U × (I − (s − t)) is an open

4.3 The Br´ezis–Browder Lemma

53

neighbourhood of (x, t), and there exists (y, r) ∈ Af ∩ (U × (I − (s − t))). Then (y, r + s − t) ∈ Af ∩ (U × I), so that (x, s) ∈ Bf . Thus if we set g(z) = inf(t : (z, t) ∈ Bf ) for z ∈ Z and g(z) = ∞ otherwise, then Bf = Ag . (ii) Since Ag is closed, g is lower semi-continuous, by Proposition 4.2.2. Since Af ⊆ Ag , g ≤ f . If h is lower semi-continuous and h ≤ f , then Af ⊆ Ah , and so Ag = Af ⊆ Ah : h ≤ g. Theorem 4.2.8 (The extension theorem for lower semi-continuous functions) Suppose that Y is a subset of a topological space (X, τ ) and that f is a lower semi-continuous function on Y which is bounded below. There exists a lower semi-continuous function g on X such that g|Y = f . Proof Let g be the function of Proposition 4.2.7. Since Af is closed in Y × R, Bf ∩ (Y × R) = Af , and so g|Y = f . The extension of the function f (x) = 1/x on (0, 1] to [0, 1] shows that the extension of a real-valued lower semi-continuous function need not be realvalued. Here is a useful approximation result. If (X, d) is a metric space, let BL(X) be the space of bounded Lipschitz functions on X. Theorem 4.2.9 Suppose that f is a non-negative lower semi-continuous proper function on a non-empty subset A of a metric space (X, d). Then there exists an increasing sequence (fn )∞ n=1 of non-negative functions in BL(X) such that fn (a) → f (a) for each a ∈ A. Proof Let fn (x) = (infa∈A {f (a) + nd(x, a)}) ∧ n. Then fn ∈ BL(X), (fn ) is a pointwise increasing sequence of functions, and if a ∈ A then fn (a) ≤ f (a). Suppose that a ∈ A. Then fn (a) → f∞ (a) as n → ∞, for some f∞ (a) ≤ f (a). There exists an ∈ A such that f (an ) + nd(a, an ) ≤ fn (a) + 1/n. Thus d(a, an ) ≤ ( fn (a) + 1/n)/n ≤ ( f∞ (a) + 1/n)/n, so that an → a as n → ∞. Consequently f (a) ≤ lim infn→∞ f (an ) ≤ f∞ (a), and so f (a) = f∞ (a).

4.3 The Br´ezis–Browder Lemma We now consider abstract partially ordered sets. A partially ordered set (X, ) is said to be countably inductive if whenever (xn )∞ n=1 is an increasing sequence in X then there exists an upper bound y ∈ X; xn  y for each n ∈ N.

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Theorem 4.3.1 (The Br´ezis–Browder lemma) Suppose that φ is a decreasing real-valued function on a countably inductive partially ordered space X, and that φ is bounded below. Then there exists y ∈ X such that if y  z then φ(y) = φ(z). Proof If x ∈ X, let S(x) = inf{φ(y) : x ≺ y}. Pick x1 arbitrarily in X. An inductive argument then shows that there is an increasing sequence (xn )∞ n=1 such that φ(xn+1 ) − S(xn ) ≤ 12 (φ(xn ) − S(xn )) for each n ∈ N. The sequence of real numbers (S(xn ))∞ n=1 is increasing and bounded above by φ(x1 ); let β = limn→∞ S(xn ). There exists an upper bound y to the sequence (xn )∞ n=1 . Suppose that y  z. Since φ(z) ≥ S(xn ), for each n, φ(xn+1 ) − φ(z) ≤ 12 (φ(xn ) − φ(z)) for each n ∈ N. Taking the limit as n → ∞, it follows that β ≤ φ(z). But β ≥ φ(y) ≥ φ(z), so that φ(y) = φ(z) = β. Note that this theorem does not use the axiom of choice, and can be thought of as a countable version of Zorn’s lemma. Exercise 4.3.2 Suppose that φ is a decreasing real-valued function on a countably inductive partially ordered space X for which, if (xn )∞ n=1 is an increasing sequence then there exists an upper bound y for which φ(xn ) → φ(y). Suppose also that if t < φ(x) there exists z > x for which t < φ(z) < φ(x). By considering X0 = {z ≥ x : φ(z) > t}, show that {φ(z) : z ≥ x} = (−∞, φ(x)). Exercise 4.3.3 Suppose that (X, τ , ≤) is a partially ordered topological space for which each increasing sequence converges, and for which each set {z : z ≥ x} is closed. Show that if there exists a strictly decreasing real-valued function φ on X (if x < y then φ(y) < φ(x)) then if x ∈ X there exists y ≥ x which is maximal.

4.4 Ekeland’s Variational Principle If f is a real-valued lower semi-continuous function on a compact Hausdorff space (X, τ ) then f is bounded below, and there exists x˜ ∈ X such that f (x) ˜ = infx∈X f (x). This result does not extend to more general spaces. Ekeland’s variational principle provides a powerful substitute for proper lower semi-continuous functions on a complete metric space. Theorem 4.4.1 (Ekeland’s variational principle) Suppose that f is a proper non-negative lower semi-continuous function on a complete metric space

4.4 Ekeland’s Variational Principle

55

(X, d) and that infx∈X f (x) = 0. Suppose that α > 0, that  > 0 and that f (x0 ) ≤ . Then there exists x˜ ∈ X such that (i) d(x, ˜ x0 ) ≤ /α, ˜ x0 ) ≤ , and (ii) 0 ≤ f (x) ˜ ≤ f (x0 ) − αd(x, (iii) f (x) ˜ < f (z) + αd(z, x) ˜ for z ∈ X \ {x}. ˜ Proof Set x ≺ y if f (y) < f (x)−αd(x, y). It follows from the triangle inequality that this is a partial order: let us show that it is countably inductive. If (xn )∞ n=1 is an increasing sequence, f (xn ) is a decreasing sequence of non-negative real numbers, which is therefore convergent, to l say. If m > n then αd(xm , xn ) ≤ f (xn ) − f (xm ), so that (xn )∞ n=1 is a Cauchy sequence, which therefore converges to a point y of X. By lower semi-continuity, l ≤ f (y). Thus l ≤ f (y) ≤ f (xm ) ≤ f (xn ) − αd(xm , xn ). Letting m → ∞, f (y) ≤ f (xn ) − αd(y, xn ), so that y  xn , and the partial order is countably inductive. We now apply the Br´ezis–Browder lemma to {x : x  x0 }; there exists ˜ Since x˜  x0 , (ii) is satisfied, and x˜  x0 such that if y  x˜ then f (y) = f (x). (i) follows from it. Suppose that y  x. ˜ Then f (x) ˜ = f (y) ≤ f (x) ˜ − αd(y, x), ˜ so that y = x. ˜ Thus if z = x˜ then f (y) < f (x) ˜ − αd(x, ˜ y), giving (iii). This theorem has many applications. We shall use it later to prove the petal theorem (Theorem 7.7.2), Daneˇs’s drop theorem (Theorem 7.7.3), the Bishop– Phelps theorem (Theorem 11.8.4), and the fixed point theorems of Caristi (Theorem 13.2.1) and Clarke (Theorem 13.2.3).

5 Uniform Spaces and Topological Groups

5.1 Uniform Spaces Metric spaces not only have a topological structure, but also have a (more restrictive) uniform structure. Here we introduce the notion of a uniform space, show how a metric space has a natural uniform structure and prove a fundamental metrization theorem. Suppose that X is a set. A relation A is a subset of the product X × X. It is reflexive if (x, x) ∈ A for all x ∈ X; that is A contains the diagonal = {(x, x) : x ∈ X}. The transposed relation AT is defined as {(x, y) : (y, x) ∈ A}. The relation A is symmetric if A = AT . If A and B are relations, then the product A.B is defined as the relation {(x, y) ∈ X × X : there exists z ∈ X with (x, z) ∈ A, (z, y) ∈ B}. A uniformity U on X is a set of relations on X satisfying the following conditions. (i) U is a filter; that is, ∅ ∈ U, the intersection of two elements of U is in U, and if U ∈ U and U ⊆ V, then V ∈ U. (ii) If U ∈ U , then U is reflexive and U T ∈ U . (iii) If U ∈ U, then there exists V ∈ U with V.V ⊆ U. The relations in U are called entourages, or vicinities. A uniform space is a pair (X, U ), where X is a set and U is a uniformity on X. A base B for the uniformity U is a subset of U such that if U ∈ U then there exists B ∈ B with B ⊆ U. For example, the symmetric elements of U form a base for U . If U has a countable base, then there is a base {Bn : n ∈ N} consisting of symmetric sets, for which Un+1 .Un+1 .Un+1 ⊆ Un , for all n ∈ N. For example, if (X, d) is a metric space and n ∈ N, let Bn = {(x, y) ∈ X × X : d(x, y) < 1/3n }, 56

5.1 Uniform Spaces

57

and let Ud = {V ∈ X × X : V ⊃ Bn for some n ∈ N}. Then Ud is a uniformity on X, the metric uniformity, and {Bn : n ∈ N} is a base satisfying the conditions of the previous paragraph. As a more special example, let d be the usual metric on R. The corresponding uniformity has a base (Cn )∞ n=1 of entourages, where Cn = {(x, y) : |x − y| < 3−n }. If we give R the equivalent metric d (x, y) = | tan−1 (x) − tan−1 (y)|, the corresponding uniformity has base (Cn )∞ n=1 , where Cn = {(x, y) : | tan−1 (x) − tan−1 (y)| < 3−n }, which is not the same as the uniformity defined by d: equivalent metrics can define different uniformities. A uniform space has a natural topology. If A is a relation on X and x ∈ X, we set A(x) = {y ∈ X : (x, y) ∈ A}. If U is a uniformity on X, we define a topology τU on X by saying that a subset O of X is open if whenever x ∈ O then there exists U ∈ U with U(x) ⊆ O. It follows immediately from the filter properties of U that τU is a topology on X, and that the sets {U(x); U ∈ U } form a base of neighbourhoods of x, for x ∈ X. If d is a metric in X, then the topology τUd is simply the topology defined by the metric d. Suppose that (X, U) and (Y, V) are uniform spaces, and that f is a mapping from X into Y. We define the mapping f˜ : X×X → Y ×Y by setting f˜(x1 , x2 ) = (f (x1 ), f (x2 )). Then f is said to be uniformly continuous if whenever V ∈ V then f˜−1 (V) ∈ U . In the case where U is defined by a metric d and V is defined by a metric ρ, then it is immediate that f is uniformly continuous in the uniform space sense if and only if it is uniformly continuous in the metric space sense. Theorem 5.1.1 Suppose that (X, U) and (Y, V) are uniform spaces, and that f is a uniformly continuous mapping from X into Y. Then f is a continuous mapping from the topological space (X, τU ) into the topological space (Y, τV ). Proof Suppose that x ∈ X and that O is a τV -open neighbourhood of f (x). Then there exists V ∈ V such that V(f (x)) ⊆ O. Then U = f˜−1 (V) ∈ U , and U(x) is a τU -neigbourhood of x. If x ∈ U(x) then (f (x ), f (x)) ∈ V, so that f (x ) ∈ V(f (x)) ⊆ O: f is continuous at x. A bijective mapping from a uniform space (X, U) onto a uniform space (Y, V) is a uniform homeomorphism if both f and f −1 are uniformly continuous.

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Proposition 5.1.2 If U is a uniformity on X then the τU × τU -open entourages form a base for U . Proof Suppose that U ∈ U. Then there exists a symmetric entourage V with V.V.V ⊆ U. If (x, y) ∈ V then V(x) × V(y) ⊆ U, so that V ⊆ U int , and so the sets U int form a base for the uniformity. Suppose that D is a family of pseudometrics on X. We can then define a uniformity U (D) on X by setting B(D) = {{(x, y) : d(x, y) < } : d ∈ D} :  > 0, D a finite subset of D}, and U (D) = {U : V ⊆ U for some V ∈ B(D)}. The sets in B(D) are then τU (D) open, and form a base for U (B). In fact, the converse is true; a uniformity can always be defined by a family of pseudometrics. Theorem 5.1.3 If U is a uniformity on X then there is a family D of continuous pseudometrics on X such that U = U(D). Proof Suppose that B is a base for U consisting of symmetric entourages. If U ∈ B, there exists a decreasing sequence (Un )∞ n=0 of symmetric entourages such that U0 = X × X, U1 = U and Un+1 .Un+1 .Un+1 ⊆ Un , for n ∈ N. If (x, y) ∈ X × X, let kU (x, y) = inf{n : (x, y) ∈ Un }, let ρU (x, y) = 1/2kU (x,y) and let ⎧ ⎫ j ⎨ ⎬ ρU (xi−1 , xi ) : j ∈ N, xi ∈ X, x0 = x, xj = y . dU (x, y) = inf ⎩ ⎭ i=1

We show that d is a continuous pseudometric on X, and that as U and  vary, the sets {(x, y) : d(x, y) < } form a base for the uniformity. Since each Un is symmetric, d(x, y) = d(y, x), and it follows from the definition that d(x, z) ≤ d(x, y) + d(y, z), so that d is a pseudometric on X. If (x, y) ∈ Un , then, taking j = 1, it follows that d(x, y) ≤ 1/2n , so that d is uniformly continuous on X × X and the sets {(x, y) : d(x, y) < } are entourages in U . j Suppose that x = x0 , x1 , . . . , xj = y and that i=1 d(xi−1 , xi ) < 1/2n . We prove by induction on j that (x, y) ∈ Un . If j = 1 then certainly (x, y) ∈ Un+1 ⊆ Un . Suppose that the result is true for j − 1. We consider two cases. First, suppose that ρ(x0 , x1 ) = 1/2n+1 . Then (x0 , x1 ) ∈ Un+1 and j n+1 . By the inductive hypothesis, (x , y) ∈ U 1 n+1 , and i=2 ρ(xi−1 , xi ) < 1/2 n+1 so (x, y) ∈ Un+1 .Un+1 ⊆ Un . Secondly, suppose that ρ(x0 , x1 ) < 1/2 . Let

5.2 The Uniformity of a Compact Hausdorff Space  k = sup l ∈ {1, . . . , j − 1} :

l 

59

 k+1

d(xi−1 , xi ) < 1/2

.

i=1

Then (x, xk ) ∈ Un+1 , by the inductive hypothesis. Further, ρ(xk , xk+1 ) ≤ 1/2n+1, so that (xk , xk+1 ) ∈ Uk+1 . Thus if k = j − 1 then (x, y) ∈ Un+1 .Un+1 ⊆ Un . j Otherwise, i=k+1 d(xi−1 , xi ) < 1/2k+1 , so that, by the inductive hypothesis, (xk , y) ∈ Un+1 , and (x, y) ∈ Un+1 .Un+1 .Un+1 ⊆ Un . Consequently, if d(x, y) < 1/2n then (x, y) ∈ Un . Thus, as U varies, the sets {(x, y) : d(x, y) < } form a base for the uniformity U . What about the topological properties of a uniform space? A uniformity U is Hausdorff if ∩U∈U U = . Proposition 5.1.4 If U is Hausdorff then (X, τU ) is completely regular. Proof Suppose that x ∈ X, that C is a closed subset of X and that x ∈ C. There exists a symmetric entourage U such that (x, c) ∈ U, for c ∈ C. Let d be a continuous pseudometric as defined earlier and let f (y) = d(x, y). Then f is a continuous function on X taking values in [0, 1], f (x) = 0 and f (c) = 1 for c ∈ C. Exercise 5.1.5 Show that a Hausdorff uniformity U has a base of τU × τU closed entourages. We have the following fundamental metrization theorem. Theorem 5.1.6 If U is a Hausdorff uniformity on X with a countable base, then there is a metric D on X such that U is the corresponding metric uniformity. Proof We can find a sequence (Un )∞ n=0 of symmetric entourages which satisfy the conditions of Theorem 5.1.3 and which form a base for the uniformity. Let d be the corresponding pseudometric. If x = y, there exists n such that (x, y) ∈ Un , since the uniformity is Hausdorff, and so d(x, y) ≥ 1/2n . Thus d is a metric, which defines the uniformity U.

5.2 The Uniformity of a Compact Hausdorff Space Theorem 5.2.1 Suppose that (X, τ ) is a compact Hausdorff space. The collection O of symmetric open subsets of X × X which contain is a base for a uniformity U on X with τU = τ . Proof O is closed under finite intersections, and OT ∈ O if and only if O ∈ O. Suppose that O ∈ O. We show that there exists P ∈ O with P.P ⊆ O. Suppose

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not. Let C = (X × X) \ O. Then for each P ∈ O, the set S(P) = P.P ∩ C is non-empty. The sets {S(P) : P ∈ O} have the finite intersection property, and so S = ∩P∈O S(P) is not empty. Suppose that (x, y) ∈ S. Then x = y. Since (X, τ ) is normal, there exist open sets Ux , Vx , Uy and Vy in X such that x ∈ Ux ⊆ U x ⊆ Vx , y ∈ Uy ⊆ U y ⊆ Vy and Vx ∩ Vy = ∅. Let W = X \ (U x ∪ U y ): W is open in X. Now let P = (Vx × Vx ) ∪ (Vy × Vy ) ∪ (W × W). Then P ∈ O. Suppose if possible that (x , y ) ∈ (Ux × Uy ) ∩ P.P, so that there exists z ∈ X such that (x , z) ∈ P and (y , z) ∈ P. Since (x , z) ∈ P, z ∈ Vx , and similarly z ∈ Vy . But Vx ∩ Vy = ∅, giving a contradiction. Thus (x, y) ∈ P.P, giving the required contradiction. Consequently, O is a base for a uniformity U on X, and clearly τU = τ . Theorem 5.2.2 If (X, τ ) is a compact Hausdorff space, then the uniformity U of the preceding theorem is the unique uniformity on X which defines the topology τ . Proof Suppose that V is a uniformity on X for which τV = τ . Since V has a base of open sets, V ⊆ U. Suppose if possible that O ∈ O \ V. Let F be the set of closed symmetric elements in V. Then the sets {F \ O : F ∈ F} have the finite intersection property, and so there exists (x, y) ∈ (∩F∈F F) \ O. But then x = y, contradicting the fact that V is Hausdorff. When is a compact Hausdorff space metrizable? Theorem 5.2.3 A compact Hausdorff space (X, τ ) is metrizable if and only if the diagonal is a Gδ subset of (X, τ ) × (X, τ ). Proof If (X, τ ) is metrizable then so is (X, τ ) × (X, τ ), and so the closed set is a Gδ subset of X × X. Conversely, suppose that is a Gδ subset of X × X. There therefore exists a decreasing sequence (Un )∞ n=1 of open subsets of X × X, with ∩n∈N Un = . By considering Un ∩ UnT , we can suppose that each Un is symmetric. Let O1 = U1 . Since the symmetric reflexive open subsets of X × X form a base for the uniformity of X, there exists a symmetric reflexive open set V1 such that V1 .V1 ⊆ O1 . Let O2 = U2 ∩ V1 , so that O2 .O2 ⊆ O1 . Iterating this procedure, there exists a decreasing sequence (On )∞ n=1 of open symmetric reflexive sets, with On+1 .On+1 ⊆ On ⊆ Un , for n ∈ N. Consequently, ∩n∈N On = . Thus if we use the sequence (On )∞ n=1 to define a pseudometric d on X, then d is in fact a metric on X. The metric d is τ -continuous on X, and the identity mapping (X, τ ) → (X, d) is continuous. But (X, τ ) is compact and (X, d) is Hausdorff, and so the identity mapping (X, τ ) → (X, d) is a homeomorphism.

5.3 Topological Groups

61

5.3 Topological Groups A topological group (G, τ ) is a group G, with identity element e, together with a topology τ on G which satisfies: (a) the mapping (g, h) → gh : (G, τ ) × (G, τ ) → (G, τ ) is continuous; (b) the mapping g → g−1 : (G, τ ) → (G, τ ) is continuous. Exercise 5.3.1 Suppose that (G, τ ) is a topological group, and that g ∈ G. Establish the following, which follow immediately from the definition. (i) Let lg (h) = gh and rg = hg. Then lg and rg are homeomorphisms of (G, τ ). (ii) Let i(h) = h−1 . Then i is a homeomorphism of (G, τ ). (iii) If U is a neighbourhood of e then there exists a neighbourhood V of e such that V 2 = {gh : g, h ∈ V} ⊆ U. A subset A of G is symmetric if g−1 ∈ A whenever g ∈ A. Exercise 5.3.2 Suppose that (G, τ ) is a topological group. Establish the following. (i) (ii) (iii) (iv)

N (g) = {gN : N ∈ N (e)} = {Ng : N ∈ N (e)}. There is a base of symmetric neighbourhoods of e. If σ is a topology on a group G which satisfies (a), and the mapping g → g−1 : (G, τ ) → (G, τ ) is continuous at e,

then (G, σ ) is a topological group. Let us give some examples. First, the general linear group GLn of invertible linear mappings of Rn into itself. To anticipate Chapter 7, if S ∈ GLn , let S = sup{d∞ (S(x), 0) : d∞(x, 0) ≤ 1}. Suppose that S0 , T0 ∈ GLn , that  > 0 and that S−S0 <  and T − T0 <  then ST − S0 T0  = (S − S0 )T0 + S0 (T − T0 ) + (S − S0 )(T − T0 ) ≤ (S − S0 )T0  + g0 (T − T0 ) + (S − S0 )(T − T0 ) ≤ (S0  + T0  + ), from which it follows that multiplication is continuous at (S0 , T0 ). If S ∈ L(E)  j and S − I < 1 then S ∈ GL(E) and S−1 = I + ∞ j=1 (I − S) (the Neumann series). Then

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I − S−1  ≤

∞  j=1

I − Sj =

I − S , (1 − I − S)

from which it follows that the mapping S → S−1 is continuous at I. It therefore follows that GL(E) is a topological group. Since a subgroup of a topological group, with the subspace topology, is also a topological group, the special linear group SLn = {T ∈ GLn : det T = 1} is also a topological group, as is the orthogonal group On = {T ∈ GLn : T  T = I}. Similarly the unitary group Un = {T ∈ GL(Cn ) : T ∗ T = I} is a topological group. A group is abelian if gh = hg for all g, h ∈ G. The group operation of an abelian group is often written as +, as in the group (R, +), which is a locally compact topological group under its usual topology. On the other hand, T = {z ∈ C : |z| = 1} is an abelian group under multiplication, and is a compact topological group when it is given the subspace topology. The Bernoulli sequence space (N) is also a compact abelian group, when addition  is defined co-ordinatewise (mod 2); (N) = ∞ n=1 (Z2 )n . Suppose that (X, U) is a uniform space. The set Homeo(X) of uniform homeomorphisms of X onto itself is a group under composition. If U ∈ U , let H(U) = {(f , g) ∈ Homeo(X)2 : (f (x), g(x)) ∈ U for all x ∈ X}. Then it is easy to check that {H(U) : U ∈ U} is the base for a uniformity H(U ) on Homeo(X). Theorem 5.3.3 The group Homeo(X) of homeomorphisms of a uniform space (X, U) with the topology τH(U ) is a topological group. Proof Suppose that g0 , h0 ∈ Homeo(X) and that U is a symmetric entourage in U . In order to show that multiplication is continuous, we must show that there exist V and W in U such that if (g, g0 ) ∈ H(V) and (h, h0 ) ∈ H(W) then (gh, g0 h0 ) ∈ H(U). There exists V ∈ U such that V.V ⊆ U. Since g0 is uniformly continuous, there exists W ∈ U such that if (x, x ) ∈ W then (g(x), g(x )) ∈ V. Now suppose that (g, g0 ) ∈ H(V) and that (h, h0 ) ∈ H(W). If x ∈ X then (gh(x), gh0 (x)) ∈ V and (g0 h(x), g0 h0 (x)) ∈ V, so that (gh(x), g0 h0 (x)) ∈ U. Thus (gh, g0 h0 ) ∈ H(U). If (g, e) ∈ H(U) and x ∈ X then (g(x), x) ∈ U. Applying this to g−1 (x), (x, g−1 (x)) ∈ U, and so (e, g−1 ) ∈ H(U). Thus inversion is continuous at e, and so Homeo(X), τH(U ) is a topological group, by Exercise 5.3.2(iii). Recall that if (X, d) is a metric space then we give the space C(X, X) the metric d∞ : d∞ (f , g) = supx∈X d(f (x), g(x)). Homeo(X) is a subset of C(X, X).

5.3 Topological Groups

63

Corollary 5.3.4 Suppose that (X, d) is a compact metric space. Then (Homeo(X), d∞ ) is a topological group. If k ∈ Homeo(X), the mapping g → gk is an isometry of (Homeo(X), d∞ ). Proof Let U be the metric uniformity defined by d. Then the collection of sets {(g, h) ∈ Homeo(X) × Homeo(X) : d∞ (g, h) < α}α>0 forms a basis for H(U), and so (Homeo(X), d∞ ) is a topological group. Further, if g, h, k ∈ Homeo(X) then d(gk(x), hk(x)) ≤ d∞ (g, h), so that d∞ (gk, hk) ≤ d∞ (g, h). Similarly, d(g(x), h(x)) = d(gkk−1 (x), hkk−1 (x)) ≤ d∞ (gk, hk), so that d∞ (g, h) ≤ d∞ (gk, hk). Thus d∞ (g, h) = d∞ (gk, hk). Proposition 5.3.5 Suppose that T : X → X is an isometry of a compact metric space (X, d) into itself. Then T is surjective. Proof Suppose not, so that there exists x ∈ X \ T(X). Since T(X) is compact, it is closed, and so α = d(x, T(X)) > 0. If m < n then d(T m (x), T n (x)) = d(x, T n−m (x)) > α, and so the sequence (T n (x))∞ n=1 has no convergent subsequence, contradictiong the sequential compactness of (X, d). Thus the set Iso(X) of isometries of (X, d) is a group, under composition. Theorem 5.3.6 The group Iso(X) of isometries of a compact metric space (X, d) is a compact subgroup of (Homeo(X), d∞ ). If R ∈ Iso(X), the mappings T → RT and T → TR and the mapping T → T −1 are isometries of (Iso(X), d∞ ). Proof Iso(X) is a closed subset of the complete metric space (C(X, X), d∞ ), and is an equicontinuous set of functions, and so it is compact, by the Arzel`a– Ascoli theorem. If x ∈ X then d(S−1 (x), T −1 (x) = d(x, ST −1 (x) = d(TT −1 (x), ST −1 (x) ≤ d∞ (T, S), so that d∞ (S−1 , T −1 ) ≤ d∞ (S, T). Replacing S by S−1 and T by T −1 , it follows that d∞ (S, T) ≤ d∞ (S−1 , T −1 ). Thus d∞ (S−1 , T −1 ) = d∞ (S, T), and the mapping T → T −1 is an isometry. Finally, d∞ (RS, RT) = d∞ (S−1 R−1 , T −1 R−1 ) = d∞ (S−1 , T −1 ) = d(S, T).

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5.4 The Uniformities of a Topological Group Suppose that (G, τ ) is a topological group. If U is a symmetric neighbourhood of the identity e, let l(U) = {(g, h) : g−1 h ∈ U} and r(U) = {(g, h) : hg−1 ∈ U}. There exists a symmetric neighbourhood V of e such that V 2 ⊆ U. Then l(V).l(V) ⊆ l(U) and r(V).r(V) ⊆ r(U), and so the sets {l(U) : U a symmetric neighbourhood of e} form a base for a uniformity L(G) on G, the left uniformity of G. Similarly, the sets {r(U) : U a symmetric neighbourhood of e} form a base for a uniformity R(G) on G, the right uniformity of G. Further, τ = τL(G) = τR(G) . If k ∈ G and (g, h) ∈ l(U), then (kg)−1 (kh) = g−1 h ∈ U, so that (kg, kh) ∈ l(U): the mapping g → kg is a uniform homeomorphism of (G, L(G)) onto itself. The mapping g → g−1 is a uniform homeomorphism of (G, L(G)) onto (G, R(G)). Proposition 5.4.1 The left uniformity L(G) and the right uniformity R(G) of a topological group are Hausdorff uniformities if and only if {e} is closed in G. Proof If L(G) or R(G) is a Hausdorff uniformity, then G is completely regular, and so it is a Hausdorff topological space. Conversely, suppose that {e} is closed. Since left multiplication is a homeomorphism, each point of G is closed. Thus if g = h there exists a symmetric neighbourhood U of e such that g−1 h ∈ U. Hence (g, h) ∈ l(U). Consequently the left uniformity is Hausdorff. Similarly for the right uniformity. Corollary 5.4.2 A Hausdorff topological group is completely regular. Suppose that G is a topological group. Let N = {e}. Then it is easy to see that N is a normal subgroup of G, and that the quotient group G/N, with the quotient topology, is a Hausdorff topological group. If G is a compact topological group, then, since there is only one uniformity on G which defines the topology, the left and right uniformities are the same. This is not the case for locally compact groups, as the next example shows. This is one of the reasons why the representation theory of locally compact groups is in general much more complicated than the representation theory of compact groups.

5.4 The Uniformities of a Topological Group

Let

 G=

a b 0 1

65





∈ GL2 : a > 0, b ∈ R .

G is a closed subgroup of GL2 , and so it is a locally compact topological group. Exercise 5.4.3 Show that G acts on the line {(x, 1) : x ∈ R}:     a 0 1 b acts by dilation, and acts by translation. 0 1 0 1 Let

 V=

a 0

b 1



 ∈ G : 2/3 < a < 3/2 and |b| < 2 .

V is a symmetric neighbourhood of e. We shall show that if U is any symmetric neighbourhood of e then l(U) ⊆ r(V). There exists  > 0 such that if |a − 1| <  then   a 0 ∈ U. 0 1 Choose a such that |1 − a| < , and choose b = 2/(1 − a). Now 

a−1 0

b 1

−1 

so that



1 b 0 1



 =

a−1 0

b 1

−ab 1

a 0

     1 b a 0 . = , 0 1 0 1

   1 b , ∈ l(U). 0 1

On the other hand, 

so that

1 b 0 1

  −1 a . 0



b 1

a−1 0

−1

b 1



 −ab = 1     a (1 − a)b a 2 = = , 0 1 0 1 1 b 0 1

  1 , 0

b 1

  a . 0

 ∈ r(V).

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5.5 Group Actions Suppose that G is a group and that X is a set. Let bij(X) denote the set of bijective mappings of X onto itself; bij(X) is a group, under composition of mappings. An action of G on X is a homomorphism of G into bij(X). Thus a(gh)(x) = a(g)(a(h)(x)), for g, h ∈ G and x ∈ X. The action is transitive if {a(g)(x) : g ∈ G} = X, for all x ∈ X. For example, GLn acts transitively on Rn \ {0}, and On and SOn act transitively on the unit sphere Sn−1 = {x : x = 1}. A group G acts transitively on itself, on the left and on the right. If g, h ∈ G, let λg (h) = gh and let ρg (h) = hg−1 ; λ is the left action of G on itself, and ρ is the right action. An action a of a topological group G on a topological space X is continuous if the mapping (g, x) → a(g)(x) from G × X to X is jointly continuous. Each of the examples of group actions that we have given is a continuous action. The next result is technically very convenient. Proposition 5.5.1 If a is an action of a topological group G on a topological space X then the action is continuous if and only if the mapping x → a(g)(x) : X → X is continuous for each g ∈ G, and the action is continuous at (e, x) for each x ∈ X. Proof The condition is certainly necessary. Suppose that it is continuous and that W is a neighbourhood of a(g)(x). Then there is a neighbourhood U × V of (e, a(g)(x)) such that if (h, y) ∈ U × V then a(h)(y) ∈ W. But then Ug × a(g−1 )V is a neighbourhood of (g, x), and if (h, y) ∈ Ug × a(g−1 )V then a(h)(y) ∈ W. Suppose that a is a continuous action of a compact Hausdorff group G on a compact Hausdorff space X. Then G acts linearly on C(X); if g ∈ G and f ∈ C(X) then we set πg (f )(x) = f (a(g)x). Then πg (f ) ∈ C(X) and πg is an isometry of (C(X)). For example, if we consider the left action of G on itself, then we get the left regular representation l of G, given by lg (f )(h) = f (gh); similarly for the right regular representation r, given by rg (f )(x) = f (xg−1 ). Is the action continuous? We need a lemma. Lemma 5.5.2 Suppose that a is a continuous action of a compact Hausdorff group G on a compact Hausdorff space X. If C is a closed subset of X, U an open subset of X and C ⊆ U then there exists a neighbourhood V of the identity e of G such that a(g)(C) ⊆ U for g ∈ V. Proof For each x ∈ C there exist open sets Vx in G and Wx in X with e ∈ Vx , x ∈ Wx ⊆ U such that a(g)(y) ∈ U for g ∈ Vx , y ∈ Wx . Since C is compact, there is a finite subset F of C such that C ⊆ ∪x∈F Wx . Take V = ∩x∈F Vx .

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Proposition 5.5.3 Suppose that a is a continuous action of a compact Hausdorff group G on a compact Hausdorff space X. Then the action π of G on C(X) is continuous. Proof By Proposition 5.5.1, it is enough to show that the action is jointly continuous at (e, f ), where f ∈ C(X). Suppose that  > 0. For each x there is an open neighbourhood Ux of x such that |f (x) − f (y)| < /3 for y ∈ Ux . Since X is normal, there exists a closed neighbourhood Cx of x contained in Ux , and by Lemma 5.5.2 there is a neighbourhood Vx of e such that a(g)(Cx ) ⊆ Ux for g ∈ Vx . Since X is compact, there is a finite set Z in X such that X = ∪z∈Z Cz . Let V = ∩z∈Z Vz . Suppose that g ∈ V and d∞ (f , h) < /3. If x ∈ X then x ∈ Cz for some z ∈ F, and so a(g)(x) ∈ Uz . Thus |f (x) − πg (h)(x)| = |f (x) − h(a(g)x)| ≤ |f (x) − f (a(gx)| + /3 ≤ |f (x) − f (z)| + |f (z) − f (a(g)x)| + /3 ≤ . Thus d∞ (f , πg (h)) ≤ .

5.6 Metrizable Topological Groups Theorem 5.6.1 Suppose that (G, τ ) is a first countable Hausdorff topological group. Then there exists a left-invariant metric dl on G (that is, dl (gh, gk) = dl (h, k) for all g, h, k ∈ G), which defines the left uniformity of G. Similarly, there exists a right-invariant metric dr on G which defines the right uniformity of G. Proof There exists a basic sequence (Un ) of left-invariant entourages in L(G). The construction of Theorem 5.1.6 now provides a left-invariant metric which defines L(G). Similarly for the right-invariant case. If (G, τ ) is a compact Hausdorff topological group, then there is only one unifomity on G which defines the topology, and so the left and right uniformities are the same. There are other topological groups with this property. Theorem 5.6.2 Suppose that (G, τ ) is a first countable Hausdorff topological group for which the left and right uniformities are the same. Then there exists a metric d on G which is both left- and right-invariant, and which satisfies d(g, h) = d(g−1 , h−1 ) for all g, h ∈ G. Proof Let U be the left uniformity. The mapping g → g−1 is a uniform homeomorphism of (G, U ) onto itself. There therefore exists a basic sequence

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(Un ) of left-invariant relations in L(G) for which (g−1 , h−1 ) ∈ Un whenever (g, h) ∈ Un . If d is the corresponding left-invariant metric, then d(g, h) = d(g−1 , h−1 ) for all g, h ∈ G. In particular, if g, h, k ∈ G then d(hg, kg) = d((hg)−1 , (kg)−1 ) = d(g−1 h−1 , g−1 k−1 ) = d(h−1 , k−1 ) = d(h, k), so that d is also right-invariant. The topology of a topological group is determined by the neighbourhoods of the identity; we consider this in the metrizable case. A function ν : G → R+ on a group G is called a group-norm if (a) ν(g) ≥ 0 if and only if g = e; (b) ν(g) = ν(g−1 ) for all g ∈ G; (c) ν(gh) ≤ ν(g) + ν(h) for all g, h ∈ G. This is clearly similar to, but should not be confused with, a norm on a vector space. Theorem 5.6.3 (i) Suppose that ν is a group-norm on a group G. Let dL (g, h) = ν(g−1 h) and let dR (g, h) = ν(gh−1 ), for g, h ∈ G. Then dL is a left-invariant metric on G, dR is a right-invariant metric on G, and dL and dR define the same topology τ on G. (G, τ ) is a topological group, and the collection of sets {{g : ν(g) < α} : α > 0} forms a base of τ -neighbourhoods of e. (ii) Suppose that d is a left-invariant or right-invariant metric on a group G. Let ν(g) = d(e, g). Then ν is a group-norm on G. Proof (i) Suppose that g, h, k ∈ G. Then dL (g) = 0 if and only if g = h, dL (g, k) = ν(g−1 k) = ν((g−1 h)(h−1 k)) ≤ ν(g−1 h) + ν(h−1 k) = dL (g, h) + dL (h, k), and dL (h, g) = ν(h−1 g) = ν(g−1 h) = dL (g, h), so that dL is a metric on G. Further, dL (kh, kg) = ν(g−1 k−1 kh) = ν(g−1 h) = dL (g, h), so that dL is left-invariant. Similarly for dR . If α > 0 then {g : dL (g, e) < α} = {g : ν(g) < α} = {g : dR (g, e) < α}, so that dL and dR define the same topology τ on G and (G, τ ) is a topological group. (ii) Suppose that d is left-invariant. (a) is trivially satisfied. Since d is leftinvariant,

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ν(g−1 ) = d(e, g−1 ) = d(g, gg−1 ) = d(g, e) = d(e, g) = ν(g), so that (b) holds. Further, ν(gh) = d(e, gh) ≤ d(e, g) + d(g, gh) = ν(g) + d(e, h) = ν(g) + ν(h), so that (c) holds. Similarly if d is right-invariant. Exercise 5.6.4 Show that the function t → | log t| is a group-norm on ((0, ∞), ×). Is the corresponding left- and right-uniformity on ((0, ∞), ×) the same as the uniformity defined by the usual metric on (0, ∞)? We can also consider the problem of finding a group-norm on GLn , when n > 1. GLn contains subgroups of translations and dilations, and the previous exercise shows that corresponding group-norms are very different. As a consequence, any explicit group-norm on GLn will be rather artificial. Here is one. Proposition 5.6.5 If S ∈ GLn , let 1

1

ν(S) = min(max(S − I 2 , S−1 − I 2 ), 1). Then ν is a group-norm on GLn . Proof Certainly ν(S) = 0 if and only if S = I, and ν(S−1 ) = ν(S). Suppose that S, T ∈ GL(E). We must show that ν(ST) ≤ ν(S) + ν(T). This is certainly   −1  < 2 the case  −1ifν(S) = 1 or ν(T) = 1. Otherwise, S < 2, T < 2, S and T  < 2. Without loss of generality, we can suppose that S − I ≤ T − I. Then ST − I ≤ ST − S + S − I ≤ 2 TI  + SI  . √ √ √ If 0 < a ≤ b then 2a + b ≤ a + b, and so 1

1

1

ST − I 2 ≤ S − I 2 + TI  2 . The same argument applies when we consider pairs (S, T −1 ), (S−1 , T) and (S−1 , T −1 ). We can also consider group-norms on the groups Homeo(X, d) and, more particularly, on Homeo+ ([0, 1], d). Proposition 5.6.6 Suppose that (X, d) is a compact metric space. Then the function ν(g) = g − e∞ = sup d(g(x), x) x∈X

is a group-norm on Homeo(X, d).

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Proof Suppose that g, h ∈ Homeo(X, d). Then clearly ν(g) = 0 if and only if g = e. Next, ν(g) = sup d(g(x), x) = sup d(g(g−1 (x)), g−1 (x)) = d(e, g−1 (x)) = ν(g−1 ), x∈X

x∈X

and finally ν(gh) = sup d(gh(x), x) ≤ sup d(gh(x), h(x)) + sup d(h(x), x) = ν(g) + ν(h). x∈X

x∈X

x∈X

Thus the right-invariant metric on Homeo(X) is dR (g, h) = ν(gh−1 ) = sup d(gh−1 (x), x) x∈X

= sup d(g(x), h(x)) = g − h∞ x∈X

  and the left-invariant metric is dL (g, h) = g−1 − h−1 ∞ . Let us consider the important case where X = [0, 1] with the usual metric. In this case, Homeo[0, 1] has two pathwise-connected components: Homeo+ [0, 1] = {g : g strictly increasing, g(0) = 0, g(1) = 1} and Homeo− [0, 1] = {g : g strictly decreasing, g(0) = 1, g(1) = 0}. Homeo+ [0, 1] is a closed pathwise-connected subgroup of Homeo[0, 1]. If n ∈ N, let  2t/n for 0 ≤ t ≤ 12 , hn (t) = (2 − 2/n)t − (1 − 2/n) for 12 ≤ t ≤ 1. + Then (hn )∞ n=1 is a dR -Cauchy sequence in Homeo [0, 1] which converges in + (C[0, 1], d∞ ) to a function not in Homeo [0, 1]; thus (Homeo+ [0, 1], dR ) is not complete. Further dL (hn , h2n ) = 14 , and so the left and right uniformities are different. On the other hand, let J = {(r, s) : r, s rational, 0 ≤ r < s ≤ 1}. Then

1 1 , f (r) < f (s), |f (1) − 1| < n n for n ∈ N, (r, s) ∈ J},

Homeo+ [0, 1] = {f ∈ C[0, 1] : |f (0)|
0 there exists a dissection D in D such that ω(D) (f ) < . Note that the intervals that we consider are closed on the left and open on the right. Proof Suppose that f is a c`adl`ag function and that  > 0. Let T = {t ∈ [0, 1] : there exists a suitable dissection of [0, t]}, and let T = sup T . Then T > 0, since f is continuous on the right at 0. Since f has finite limits on the left, T ∈ T . Suppose that T < 1. Since f is continuous on the right at T, there exists T < T0 ≤ 1 such that |f (x) − f (y)| <  for x, y ∈ [T, T0 ), and so T0 ∈ T , giving a contradiction. Thus T = 1. Conversely, if f is not continuous on the right at t ∈ [0, 1), there exists  > 0 such that if t < s ≤ 1, then there exist x, y ∈ [t, s) such that |f (x) − f (y)| ≥ , so that f ∈ D[0, 1]. A similar argument applies if f does not have a finite limit on the left at some x ∈ (0, 1]. Thus the condition is sufficient. Corollary 6.1.2 If f ∈ D[0, 1] then f is bounded. If f ∈ D[0, 1] and x ∈ (0, 1], let jf (x) = f (x) − f− (x). jf (x) is the jump of f at x. Thus f (x) = |jf (x)|, and f is continuous at x if and only if jf (x) = 0. Exercise 6.1.3 If f ∈ D[0, 1] and  > 0 show that {x ∈ (0, 1] : |jf (x)| > } is finite, deduce that f has only countably many points of discontinuity.

6.2 The Space (D[0, 1], d∞ ) The space D[0, 1] is a linear subspace of the space (B[0, 1], d∞ ) of bounded functions on [0, 1], and contains C[0, 1] as a linear subspace. Theorem 6.2.1 (D[0, 1], d∞ ) is complete. Proof It is enough to show that D[0, 1] is closed in (B[0, 1], d∞ ). Suppose that f ∈ D[0, 1], that t ∈ [0, 1] and that  > 0. There exists g ∈ D[0, 1] with d∞ (f , g) < /3, and there exists 0 < δ < 1 − t such that |g(s) − g(t)| < /3 for t ≤ s < t + δ. If t ≤ s < t + δ, then | f (s) − f (t)| ≤ |f (s) − g(s)| + |g(s) − g(t)| + |g(t) − f (t)| < , so that f is continuous on the right at t. The existence of finite limits on the left is as easy to prove.

6.3 The Skorohod Topology

73

A step function g on [0, 1] is a function of the form g=

k 

gj I[tj−1 ,tj ) + g{1} I{1} ,

j=1

where D = (0 = t0 < . . . < tk = 1) ∈ D. Exercise 6.2.2 The step functions are dense in (D[0, 1], d∞ ). Proof This follows easily from Theorem 6.1.1. Thus D[0, 1] can be thought of as the completion of the space of step functions, with the uniform norm. The uniform norm is however too strong to be useful. If 0 < s < t ≤ 1 then d∞ (I[0,s) , I[0,t) ) = 1. Thus (D[0, 1], d∞ ) is not separable, and does not reflect the geometric properties of [0, 1].

6.3 The Skorohod Topology In this section, we introduce a topology τS , the Skorohod topology, on D[0, 1]. This is weaker than the topology of uniform convergence, but agrees with the topology of uniform convergence on C[0, 1]. The topology is defined by constructing a metric dS , the Skorohod metric on D[0, 1]; τS is the corresponding metric topology. In terms of stochastic processes, the idea behind the construction is to allow small perturbations of the time variable; this is appropriate, since time cannot be measured with complete accuracy. We consider the group Homeo+ [0, 1] of increasing homeomorphisms of [0, 1], and its group-norm νS (h) = sup{|h(t) − t| : t ∈ [0, 1]}. If a > 0, let Va = {h ∈ Homeo+ [0, 1] : νS (h) ≤ a}. Let Ua = {(f , g) ∈ D[0, 1] × D[0, 1] : inf{d∞ (f , g ◦ h) : h ∈ Va } ≤ a}, and let dS (f , g) = inf{a : (f , g) ∈ Ua }. Theorem 6.3.1 dS is a metric on D[0, 1]. Proof Suppose that f , g, k ∈ D[0, 1]. Certainly dS (f , g) = 0 if and only if f = g. If a > dS (f , g) there exists h ∈ Va such that d∞ (f , g ◦ h) < a. Then h−1 ∈ Va , and d∞ (g, f ◦ h−1 ) = d∞ (f , g ◦ h) < a, so that dS (g, f ) ≤ dS (f , g). Exchanging f and g, it follows that dS (f , g) = dS (g, f ).

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Suppose that a > dS (g, k), so that there exists h ∈ Va such that d∞ (g, k ◦ h ) < a . Then νS (h h) ≤ νS (h ) + νS (h) < a + a , and d∞ (f , k ◦ (h h)) ≤ d∞ (f , g ◦ h) + d∞ (g ◦ h, k ◦ (h h)) = d∞ (f , g ◦ h) + d∞ (g, k ◦ h ) < a + a , so that dS (f , k) ≤ dS (f , g) + dS (g, k). Clearly dS (f , g) ≤ d∞ (f , g), so that the identity mapping from (D[0, 1], d∞ ) to (D[0, 1], dS ) is a 1-Lipschitz mapping. In particular, the step functions are dense in (D[0, 1], τS ). Theorem 6.3.2 Suppose that (fn )∞ n=1 is a sequence in D[0, 1] which converges in the Skorohod topology to f , and that f ∈ C[0, 1]. Then fn converges uniformly to f as n → ∞. Proof Suppose that  > 0. Since f is uniformly continuous on [0, 1], there exists 0 < δ < /2 such that if h ∈ Vδ then d∞ (f , f ◦ h) < /2. There exists N ∈ N such that if n ≥ N then dS (fn , f ) < δ. If n ≥ N, there exists hn ∈ Vδ such that fn − f ◦ hn ∞ < δ. Thus d∞ (fn , f ) ≤ d∞ (fn , f ◦ hn ) + d∞ (f ◦ fn , f ) ≤ δ + /2 < . Corollary 6.3.3 The restriction of τS to C[0, 1] is the topology of uniform convergence. Proposition 6.3.4 (D[0, 1], τS ) is separable.  Proof Suppose that g = kj=1 gj I[tj−1 ,tj ) + g{1} I{1} is a step function, and that  > 0. There exist rational 0 = s0 < . . . < sk = 1 such that tj < sj < tj+1 and sj − tj < /2 for 1 ≤ j ≤ k − 1. There is a piecewise linear h ∈ Homeo+ [0, 1] such that h(tj ) = sj for 0 ≤ j ≤ k. Then νS (h) < /2, so that dS (g, g◦h) < /2. There exist rational f1 , . . . fk and fr such that |fj − gj | < /2 for 1 ≤ j ≤ k and |f{1} − g{1} | < /2.  Let f = kj=1 fj I[sj−1 ,sj ) + f{1} I{1} . Then dS (f , g ◦ h) ≤ d∞ (f , g ◦ h) < /2, and so dS (f , g) < . Since the step functions are dense in (D[0, 1], dS ), and since there are countably many step functions taking rational values, and with rational points of dissection, (D[0, 1], τS ) is separable. Is (D[0, 1], τS ) a Polish space? Unfortunately, (D[0, 1], dS ) is not complete. Exercise 6.3.5 Let fn = I[ 1 , 1 + 1 ) . Show that dS (fm , fn ) = |1/2m − 1/2n|, so 2 2 2n ∞ that (fn )∞ n=1 is a dS -Cauchy sequence. Show that (fn )n=1 is not dS -convergent to any elememt of D[0, 1]. We need to introduce another metric.

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75

6.4 The Metric dB In this section, we introduce another metric, dB , on D[0, 1]. This is a complete metric equivalent to dS , so that the Skorohod topology τS is the corresponding metric topology: thus (D[0, 1], τS ) is a Polish space. The metric dS considers elements of Homeo+ [0, 1] which are uniformly close to the identity. But such homeomorphisms may behave badly locally. We need to consider homeomorphisms with good local behaviour.  For example,   we could consider differentiable homeomorphisms h for which h − 1∞ is small. But these homeomorphisms do not have good limiting properties, and so we consider a rather larger subgroup of Homeo+ [0, 1]; homeomorphisms for which the slope (h(t) − h(s))/(t − s) of chords is uniformly bounded away from 0. For technical reasons, it is convenient to consider   h(t) − h(s) = log(h(t) − h(s)) − log(t − s). l(h)(s, t) = log t−s We shall need the following elementary inequalities, which we shall use without comment. Exercise 6.4.1 Show that if 0 < α
0, let Mα = {h ∈ GB : νB (h) ≤ α}. Proposition 6.4.3 The inclusion (GB , ρB ) → (Homeo+ [0, 1], ρS ) is uniformly continuous; if 0 < α < 12 and h ∈ Mα then h ∈ V2α . Proof Set s = 0. If 0 < t ≤ 1 then | log h(t) − log t| ≤ α, so that e−α t ≤ h(t) ≤ eα t and |h(t) − t| ≤ (eα − 1)t ≤ eα − 1. Thus νS (h) ≤ eα − 1 < 2α. Theorem 6.4.4 The topological group (GB , ρB ) is complete. Proof It follows from Theorem 2.4.5 that it is sufficient to show that Mα is uniformly closed in C[0, 1], for α ≤ 12 . Suppose that h ∈ Mα , and that (hn )∞ n=1 is a sequence in Mα which converges uniformly to h. If 0 ≤ s < t ≤ 1, then log(hn (t)−hn (s)) → log(h(t)−h(s)) as n → ∞, so that sup{| log(h(t)−h(s))− log(t − s)| : 0 ≤ s < t ≤ 1} ≤ α. Consequently |(h(t) − h(s)) − (t − s)| ≤ (eα − 1)(t − s), and so h(t) − h(s) ≥ (2 − eα )(t − s) > 0; h is strictly increasing and continuous on [0, 1]. Since h(0) = 0 and h(1) = 1, it follows that h is a homeomorphism of [0, 1] onto itself. Finally νB (h) ≤ α, so that h ∈ Mα . On the other hand, (GB , ρB ) is not separable. Let tn = 1 − 1/2n , let sn = (tn + tn+1 )/2 and let bn = sn + 1/2n+2 for n ∈ N. If A ∈ P(N) let fA (tn ) = tn and let  bn if n ∈ A fA (sn ) = sn if n ∈ A. Let fA be defined linearly between these values, and let fA (1) = 1. Then fA ∈ GB , and ρB (fA , fC ) = log 2 if A = C. We now define a new metric on D[0, 1] which is equivalent to dS . Let Wa = {(f , g) ∈ D[0, 1] × D[0, 1] : inf{f − g ◦ h∞ : h ∈ Ma } ≤ a}, and let dB (f , g) = inf{a : (f , g) ∈ Wa }.

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77

Theorem 6.4.5 dB is a metric on D[0, 1]. Proof The proof is exactly similar to the proof of Theorem 6.3.1. Theorem 6.4.6 The metric space (D[0, 1], dB ) is complete. Proof Suppose that (fn )∞ n=1 is a Cauchy sequence in D[0, 1]. By extracting a subsequence if necessary, we may suppose that dB (fn−1 , fn ) < 1/2n , for n > 1. Thus for each n > 1 there exists hn ∈ M1/2n such that fn−1 − fn ◦ hn ∞ < 1/2n . If m < n, let hm,n = hm+1 ◦· · ·◦hn . Then hm,n ∈ M1/2m , and if m < n < p then ρB (hm,n , hm,p ) = νB (hn,p ) ≤ 1/2n . Thus the sequence (hm,n )∞ n=m+1 is a ρB -Cauchy sequence, which converges to an element hm,∞ . If m < n, let fm,n = fm ◦ hm,n . Then   dB (fm,n , fm ) ≤ 1/2m and fm,n − fn ∞ ≤ 1/2m . Similarly, let fm,∞ = fm ◦ hm,∞ . Again, dB (fm,n , fm ) ≤ 1/2m and   dB (fm,∞ , fm ) ≤ 1/2m and fm,∞ − f∞ ∞ ≤ 1/2m . Now fm,∞ = fm ◦ hm,n ◦ hn,∞ , so that fm,∞ − fn,∞ = (fm ◦ hm,n − fn ) ◦ hn,∞ . Consequently d∞ (fm,∞ , fn,∞ ) = d(fm ◦ hm,n , fn ) ≤ 1/2m . Thus the sequence (fm,∞ )∞ m=1 is a uniform Cauchy sequence, which by Theorem 6.2.1 converges uniformly to an element f∞ of D[0, 1]. Further, dB (fm,∞ , f∞ ) ≤ 1/2m , so that dB (fm , f∞ ) ≤ dB (fm , fm,∞ ) + dB (fm,∞ , f∞ ) ≤ 2/2m . Thus fm → f∞ as m → ∞. Theorem 6.4.7 The metrics dS and dB on D[0, 1] are equivalent. Proof First we show that the identity mapping (D[0, 1], dB ) → (D[0, 1], dS ) is uniformly continuous. Suppose that 0 < α < 12 . If dB (f , g) < α there exists h ∈ Mα such that d∞ (f , g ◦ h) < α. By Proposition 6.4.3, νs (h) < 2α, and so dS (f , g) < 2α. The identity mapping (D[0, 1], dS ) → (D[0, 1], dB ) cannot be uniformly continuous, since (D[0, 1], dB ) is complete, and (D[0, 1], ds ) is not. We show that it is continuous. Suppose that f ∈ D[0, 1] and that  > 0.

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By Theorem 6.1.1, there exists a dissection D = (0 = t0 < · · · < tk = 1) such that ω(D) < /2. Let δ = min1≤j≤k (tj − tj−1 ), and let η = δ/4. Suppose that dS (f , g) < η, so that there exists h ∈ Homeo+ [0, 1] with νS (h) < η for which f − g ◦ h∞ < η. Now let k(tj ) = h(tj ) for 1 ≤ j ≤ k, and let k be linear between these values. Thus h−1 k([tj−1 , tj )) = [tj−1 , tj ), so that |f (t) − g ◦ k(t)| ≤ |f (t) − f ◦ h−1 k(t)| + |f ◦ h−1 k(t) − g ◦ k(t)| ≤ ω(D) (f ) + dS (f , g) ≤ /2 + η < , and so f − g)∞ < . Further, if 1 ≤ j ≤ k then |(k(tj ) − k(tj−1 )) − (tj − tj−1 )| = |(h(tj ) − h(tj−1 )) − (tj − tj−1 )| ≤ η ≤ (tj − tj−1 ) so that, since k is piecewise linear, |(k(t) − k(s)) − (t − s)| ≤ (t − s)/4 for 0 < s < t < 1, and so |l(k)(s, t)| < /2. Thus k ∈ M , and so dB (f , g) ≤ . Thus the identity mapping (D[0, 1], dS ) → (D[0, 1], dB ) is continuous at f . Corollary 6.4.8 (D[0, 1], τS ) is a Polish space.

7 Banach Spaces

7.1 Normed Spaces and Banach Spaces Many of the metric spaces that we shall consider are real or complex vector spaces, or subsets of such spaces. Let us denote the underlying field R or C by K. Suppose that E is a real or complex vector space. It is natural to consider metrics d on E which are (i) translation invariant (d(x + a, y + a) = d(x, y) for all x, y, a ∈ E); (ii) scaling homogeneous (d(λx, λy) = |λ|d(x, y) for all x, y ∈ E) and λ ∈ K. Note that (iii) d is inversion invariant; d(−x, −y) = d(x, y) for all x, y ∈ E. Suppose that d is a metric on E with these properties. If x ∈ E, let x = d(x, 0). Then d(x, y) = d(x − y, 0) = x − y, so that . determines d. The function . then has the following properties: (a) (b) (c) (d)

x + y ≤ x + y, for x, y ∈ E (subadditivity); λx = |λ| x, for λ ∈ K and x ∈ E (positive homogeneity); x = −x for x ∈ E (symmetry); and x = 0 if and only if x = 0.

(a) follows, since x + y = d(x + y, 0) ≤ d(x + y, y) + d(y, 0) = d(x, 0) + d(y, 0) = x + y . (b) is a consequence of (ii) and (iii), and (c) follows, since d(x, 0) = 0 if and only if x = y. A function . on E which satisfies (a), (b), (c) and (d) is called a norm, and (E, .) is called a normed space. A function which satisfies (a), (b) and (c) is called a seminorm. 79

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If (E, .) is a normed space, then the function d(x, y) = x − y is a metric on E which satisfies (i), (ii) and (iii), the metric defined by the norm .. The mappings (x, y) → x+y from E ×E → E and (λ, x) → λx from R×E to E are jointly continuous, and the mapping x → −x is an isometry of E onto itself. A subset A of a normed space (E, .) is bounded in norm if sup{a : a ∈ A} < ∞. Since sup{a : a ∈ A} ≤ diam(A) ≤ 2 sup{a : a ∈ A}, A is bounded in norm if and only if it is bounded in the metric d. The open unit ball U(E) is the set U(E) = N1 (0) = {x : x < 1} and the closed unit ball B(E) is the set B(E) = M1 (E) = {x : x ≤ 1}. A set A is bounded if there exists λ > 0 such that λA ⊆ B(E). A Banach space is a normed space which is complete under the metric defined by the norm. Exercise 7.1.1 Suppose that (E, .) is a normed space. Use the proof of Proposition 2.5.3 to show that Ca(E) is a vector space, that if p(x) = φ(x, 0) then p is a seminorm on Ca(E) which determines φ, and that x ∼ y if and only if x−y ∈ N = {z : zn → 0 as n → ∞}. Conclude that N is a linear subspace of Ca(E) and q is the linear quotient mapping of Ca(E) onto Eˆ = Ca(E)/N; Eˆ is ˆ v) = u − v∧ a vector space. Further, there is a norm .∧ on Eˆ such that d(u, ˆ .∧ ) is a Banach space. Finally, j is a linear isometry of (E, .) onto and (E, a dense subspace of (Eˆ .∧ ): the completion of (E, .) is a Banach space. If a normed space is topologically complete, it must be a Banach space. Theorem 7.1.2 (Klee) Suppose that (E, .) is a topologically complete normed space. Then (E, .) is a Banach space. ˆ . ˆ ) be the completion of (E, .). By Alexandroff’s theorem Proof Let (E, (Theorem 2.6.2), E is the intersection of a sequence (On )∞ n=1 of open dense ˆ Suppose that E = E, ˆ so that there exists xˆ ∈ Eˆ \ E. Then subsets of E. ˆ ˆ∞ (On + x) n=1 is a sequence of open dense subsets of E, and ∞ (∩∞ ˆ = E ∩ (E + x) ˆ = ∅, n=1 On ) ∩ (∩n=1 (On + x))

contradicting Baire’s category theorem. Let us give a few examples of Banach spaces that we shall need. (i) If S is a non-empty set and (E, .) is a normed space, then the space B(S, E) of bounded functions taking values in E is a vector space with the operations of co-ordinate addition and scalar multiplication, and the function f ∞ = sup{f (s) : s ∈ S} is a norm on it, which defines the metric d∞ . It is a Banach space if and only if (E, .) is complete (Proposition 2.3.1). In particular the spaces B(S) and l∞ are Banach spaces.

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(ii) If (X, τ ) is a Hausdorff topological space and (E, .) is a normed space, then the space Cb (X, E) of bounded continuous functions on X taking values in E is a closed linear subspace of B(X, E); we give Cb (X, E) the norm .∞ . Then (Cb (X, E), .∞ ) is a Banach space if and only if E is. In particular (Cb (X), .∞ ) is a closed linear subspace of (B(X), .∞ ), and is a Banach space. In particular, if (K, τ ) is a compact Hausdorff space then C(K) = Cb (K) is Banach space. Note that if f ∈ C(K) then there exists x ∈ K such that |f (x)| = f . Exercise 7.1.3 Let c = {x ∈ K N : xn tends to a limit as n → ∞}. If x ∈ c, N let x∞ = sup∞ n=1 |xn |. Show that c is a linear subspace of K , that .∞ is a norm on c and that (c, .∞ ) is a Banach space. Let c0 = {x ∈ c : xn → 0 as n → ∞}. Show that c0 is a closed hyperplane in c. (iii) Suppose that (X, τ ) is a compact topological space and that (E, .) is a normed space. If f is a continuous function on X taking values in E, then f (X) is a norm bounded subset of E, and so C(X, E) = Cb (X, E). Again, we give C(X, E) the norm .∞ .  1 Exercise 7.1.4 Let l1 = {x ∈ K N : x1 = ∞ i=1 |xi | < ∞}. Show that l is a N 1 1 linear subspace of K , that .1 is a norm on l and that (l , .1 ) is a Banach space. n n Exercise 7.1.5 If x ∈ K n , let x1 = i=1 |xi |. Show that (K , .1 ) is a locally compact Banach space. Exercise 7.1.6 Show that any two norms on a finite-dimensional space E are Lipschitz equivalent, and make E a complete locally compact space. Exercise 7.1.7 Suppose that F is a closed linear subspace of a normed space (E, .), and that q : E → E/F is the quotient mapping. Let q(x)q = inf{x + f  : f ∈ F}. Show that .q is a norm on E/F and that q : (E, .) → (E/F, .q ) is continuous. Show that if (E, .) is complete, so is (E/F, .q ). Show that if (F, .) and (E/F, .q ) are complete, then so is (E, .). Exercise 7.1.8 Show that a finite-dimensional subspace of a normed space is closed. Show that if F is a closed linear subspace of a normed space (E, .) and D is a finite-dimensional subspace of E then F + D is closed. Exercise 7.1.9 Suppose that (E, .) is a locally compact normed space with closed unit ball BE . Show that there is a finite set F such that BE ⊆ F + 12 BE . Let G = span(F). Show inductively that BE ⊆ G + BE /2n , and deduce that E = G, so that E is finite-dimensional. Exercise 7.1.10 If (X, d) is a metric space, show that the space Ub (X, E) of bounded uniformly continuous functions on X taking values in a normed space

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(E, .) is a closed linear subspace of Cb (X, E). Show that (Ub (X, E), .∞ ) is a Banach space if and only if E is. Exercise 7.1.11 Let C(1) ([0, 1]) = {f ∈ C([0, 1] : f is continuously differentiable}. Show that C(1) ([0, 1]) is a meagre subset of (C([0, 1]), .∞ ). Show that f (1) = supx∈[0,1] (|f (x)| + |f  (x)|) is a norm on C(1) ([0, 1]) under which C(1) ([0, 1]) is a Banach space.

7.2 The Space BL(X) of Bounded Lipschitz Functions Spaces of Lipschitz functions provide further examples of Banach spaces. Suppose that (X, d) is a metric space. The space L(X) of Lipschitz functions on X is a linear subspace of C(X) and the space BL(X) of bounded Lipschitz functions on X is a linear subspace of Cb (X). If f ∈ BL(X), we set f BL = pL (f ) + f ∞ . Theorem 7.2.1 Suppose that (X, d) is a metric space. pL is a seminorm on L(X). .BL is a norm on BL(X), and its unit ball BBL is closed in (Cb (X), .∞ ). (BL(X), .BL ) is a Banach space. Further, BL(X) is a Banach algebra under pointwise multiplication: if f , g ∈ BL(X), then fg ∈ BL(X), and fgBL ≤ f BL . gBL . Proof It is clear that pL is a seminorm and that .BL is a norm on BL(X). Suppose that (fn )∞ n=1 is a sequence in BBL which converges uniformly to f ∈ Cb (X). Then pL (f ) ≤ 1, by Corollary 2.8.3, and so f ∈ BL(X). Since fn ∞ → f ∞ as n → ∞, given  > 0 there exist n0 such that fn ∞ > f ∞ − , for n ≥ n0 . Hence pL (fn ) < 1 − f ∞ +  for n ≥ n0 . By Corollary 2.8.3, pL (f ) ≤ 1 − f ∞ + , and so f BL ≤ 1 + . Since  is arbitrary, f ∈ BBL . Thus BBL is closed in (Cb (X), .∞ ). Completeness follows from Theorem 2.4.5. If f , g ∈ BL(X) and x, y ∈ X, then |f (x)g(x) − f (y)g(y)| ≤ |f (x)(g(x) − g(y))| + |(f (x) − f (y))g(y)| ≤ (f ∞ . gL + f L . g∞ )d(x, y), so that fg ∈ BL(X) and fgL ≤ f ∞ . gL + f L . g∞ . Thus fgBL ≤ f ∞ . g∞ + f ∞ . gL + f L . g∞ ≤ f BL . gBL .

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Proposition 7.2.2 Suppose that (X, d) is compact. Then the unit ball BBL (X) is a compact subset of (Cb (X), .∞ ). Proof For BBL is a bounded equicontinuous subset of Cb (X), and so is a totally bounded subset of (Cb (X), .∞ ), by the Arzel`a–Ascoli theorem. It is also complete, and so it is compact. Theorem 7.2.3 Let (X, d) be a compact metric space. There exists a countable subset G of BL(X), with supg∈G pL (g) ≤ 1, which is dense in C(X). Proof For each n ∈ N, there exists a finite 1/n-net An in X. Let Gn be the countable set of rational-valued functions on An with pL (g) ≤ 1, and let G = ∪∞ n=1 Gn . By the McShane–Whitney extension theorem, for each g ∈ Gn there exists a Lipschitz function hg on X with pL (hg ) = pL (g) ≤ 1 which extends g. We show that the set H = {hg : g ∈ Gn , n ∈ N} is .∞ -dense in C(X). Suppose then that f ∈ C(X) and that  > 0. Then f is uniformly continuous, and so there exists n > 3/ such that if d(x, y) < 1/n then |f (x) − f (y)| < /3. There exists g ∈ Gn such that max{|g(a) − f (a)| : a ∈ An } < /3. Suppose that x ∈ X. There exists a ∈ An with d(x, a) < 1/n. Then |f (x) − hg (x)| ≤ |f (x) − f (a)| + |f (a) − hg (a)| + |hg (a) − hg (x)| < /3 + /3 + 1/n < .

On the other hand, if (X, d) is an infinite compact metric space, then (BL(X), .BL ) is not separable. There exists a sequence (xn )∞ n=1 of distinct points which converges to x∞ , say. By picking a subsequence, we can suppose that d(xn , x∞ ) is a decreasing sequence and that d(xn , xm ) > 12 d(xn , x∞ ) for m > n. Let Y = {xn : n ∈ N}. Suppose that A ⊆ N. Let fA (x2n ) = d(xn , x∞ ) if n ∈ A, let fA (x2n ) = −d(xn , x∞ ) if n ∈ A, and let fA (x2n+1 ) = 0. Then fA ∈ BL(Y), and by the McShane–Whitney extension theorem (Theorem 2.8.6), we can extend each fA to a Lipschitz function gA on X. Then gA − gB BL ≥ fA − fB BL ≥ 1 if A = B, and so (BL(X), .BL ) is not separable. Exercise 7.2.4 Suppose that (X, d) is an infinite metric space. Show that BL(X) is a meagre subspace of Cb (X).

7.3 Introduction to Convexity If (E, .) is a normed space, then the function . is a convex function on E, and the closed unit ball B(E) and the open unit ball U(E) are convex sets.

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These simple facts have a profound effect on analysis. Here we begin the study of convexity; we shall return to it several times. A subset C of a real or complex vector space is convex if (1 − λ)x + λy ∈ C whenever x, y ∈ C and 0 ≤ λ ≤ 1. The intersection of a set of convex sets is convex. If S is any subset of a vector space E, then (S) = ∩{C : C convex, S ⊆ C} is a convex set, the smallest convex set containing S: (S) is the convex cover of S. Exercise 7.3.1 Show that if S is a non-empty subset of a vector space E then ⎫ ⎧ n ⎬ ⎨  αj = 1 . (S) = α1 s1 + · · · + αn sn : n ∈ N, sj ∈ S, αj ≥ 0, ⎭ ⎩ j=1

A convex set C is absolutely convex if αC ⊆ C for all |α| ≤ 1. If (E, .) is a normed space, then the closed unit ball B(E) and the open unit ball U(E) are absolutely convex sets. If C is a convex set and α ∈ K then αC is convex. If A is any subset of E, and α, β > 0, then (α + β)A ⊆ αA + βA. Exercise 7.3.2 Show that if C is a convex set and α, β > 0 then (α + β)C = αC + βC. We now turn to convex functions. A function f on a vector space E taking values in (−∞, ∞] is a convex function if f ((1 − λ)x + λy) ≤ (1 − λ)f (x) + λf (y) whenever x, y ∈ C and 0 < λ < 1. Why do we allow the value ∞? Proposition 7.3.3 Suppose that F is a set of convex functions on a convex subset C of a real vector space E. Then g = sup{ f : f ∈ F} is a convex function. Proof Suppose that x, y ∈ C and that 0 < λ < 1. If t < g((1 − λ)x + λy), there exists f ∈ F with t < f ((1 − λ)x + λy). Then t < f ((1 − λ)x + λy) ≤ (1 − λ)f (x) + λf (y) ≤ (1 − λ)g(x) + λg(y). Since this holds for all t < g((1 − λ)x + λy), g((1 − λ)x + λy) ≤ (1 − λ)g(x) + λg(y).

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Since the supremum of finite-valued functions can be infinite, this explains why the value ∞ is allowed. Recall that if f is a function taking values in [−∞, ∞], its effective domain f is the set where f is finite, and that the function f is proper if its effective domain is non-empty. Exercise 7.3.4 Show that the effective domain of a proper convex function is convex. If f is a convex function on a convex subset C of a vector space E – that is, f ((1 − λ)x + λy) ≤ (1 − λ)f (x) + λf (y) whenever x, y ∈ C and 0 < λ < 1 - it can be extended to a convex function on E by setting f (x) = ∞ for x ∈ E \ C. For example, if C is a convex subset of E, we define the function 0C by setting 0C (x) = 0 for x ∈ C, and 0C (x) = ∞ for x ∈ C. A function on E taking values in [−∞, ∞) is concave if −f is convex, and a real-valued function on E is affine if it is both convex and concave: that is, f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) for x, y ∈ E and 0 ≤ λ ≤ 1. f is affine if and only if f = φ + c, where c = f (0) and φ is a linear functional on E. Exercise 7.3.5 Suppose that f is a convex function on a convex subset C of a real vector space E. Show that the sets {x ∈ C : f (x) ≤ λ} and {x ∈ C : f (x) < λ} are convex, for each λ ∈ (−∞, ∞]. A function p from a vector space E to (−∞, ∞] which satisfies (a) p(x + y) ≤ p(x) + p(y), for x, y ∈ E (subadditivity); (b) p(λx) = λp(x), for λ > 0 and x ∈ E (positive homogeneity); (c) p(0) = 0, is called an extended sublinear functional. If it is real-valued, it is called a sublinear functional. A seminorm is an example of a sublinear functional. Extended sublinear functionals are proper convex functions. If p is an extended sublinear functional, then p(x) + p(−x) ≥ p(0) = 0. If p is a seminorm, this implies that p(x) ≥ 0 for all x ∈ E. This need not be the case when p is a sublinear functional; for example, a function p on R is a sublinear functional if and only if p(x) = αx for x ≤ 0 and p(x) = βx for x ≥ 0, where α ≤ β. Suppose that C is a convex subset of a real vector space E, and that 0 ∈ C. If x ∈ E, then IC (x) = {λ ≥ 0 : λx ∈ C} is an interval in [0, ∞). If IC (x) = {0}, set pC (x) = ∞, and if IC (x) = [0, ∞) set pC (x) = 0. Otherwise let pC (x) = inf{α : α > 0, x ∈ αC}. Suppose that x, y ∈ E. If pC (x) = ∞ or pC (y) = ∞, then trivially pC (x + y) ≤ pC (x) + pC (y). Otherwise, if α > pC (x) and β > pC (y), then x ∈ αC and y ∈ βC, so that x + y ∈ αC + βC = (α + β)C. Thus pC (x + y) ≤ α + β,

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and so pC (x + y) ≤ pC (x) + pC (y). Clearly, pC (λx) = λpC (x) for λ > 0, and so pC is an extended sublinear functional on E; it is the gauge, or Minkowski functional of C. Exercise 7.3.6 Suppose that C is a convex subset of a vector space E, and that 0 ∈ C. Show that {x : pC (x) < 1} ⊆ C ⊆ {x : pC (x) ≤ 1}. We shall consider convex functions further in Chapters 10 and 12.

7.4 Convex Sets in a Normed Space We now consider convex sets in a normed space. Proposition 7.4.1 Suppose that A is a convex subset of a normed space (E, .). Then A and Aint are convex sets. Proof Suppose that x, y ∈ A and that  > 0. Let z = (1 − λ)x + λy. There exist a, b ∈ A such that x − a <  and y − b < . Then c = (1 − λ)a + λc ∈ A and z − c = (1 − λ)(x − a) + λ(y − b) ≤ (1 − λ) x − a + λ y − b < (1 − λ) + λ = , so that z ∈ A. Suppose that a, b ∈ Aint , and that 0 ≤ λ ≤ 1. Let c = (1 − λ)a + λb. There exists δ > 0 such that if x − a < δ then x ∈ A and if y − b < δ then y ∈ A. Suppose that z − c < δ. Then a + (z − c) ∈ A and b + (z − c) ∈ A, so that z = (z − c) + (1 − λ)a + λb = (1 − λ)(a + (z − c)) + λ(b + (z − c)) ∈ A; thus c ∈ Aint . If S is any subset of a normed space (E, .), then (S) = ∩{C : C convex and closed, S ⊆ C} is a closed convex set, the smallest closed convex set containing S: (S) is the closed convex cover of S. Exercise 7.4.2 Show that, if S is any subset of a normed space (E, .), then (S) = (S). Proposition 7.4.3 If S is a totally bounded subset of a normed space (E, .), then (S) is totally bounded.

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Proof S is norm bounded: let M = sup{x : x ∈ S}. Suppose that  > 0. There exist a finite /2-net F in S. Let η = /2(M + 1)|F|. There exists a  finite η-net B in [0, 1]. Let G be the finite set { y∈F βy y : βy ∈ B}. We show n that (S) ⊆ ∪g∈G N (g). Suppose that u = j=1 αj xj ∈ (S),  αj ≥ 0  with  and nj=1 αj = 1. For each j there exists yj ∈ F such that xj − yj  < /2. n Let v = v ∈ (S), and u − v < /2. Gathering terms j=1 αj yj . Then   together, we can write v = y∈F γy y, where γy = {αj : yj = y}; thus γy ≥ 0  and y∈F γy = 1. For each y there exists βy ∈ B with |βy − γy | ≤ η. Then  u − v ≤ y∈F |βy − γy |. y < /2. Thus (S) ⊆ ∪g∈G N (g), and so (S) is totally bounded. Since the closure of a totally bounded set is totally bounded, (S) is totally bounded. Exercise 7.4.4 Show that if S is a compact subset of a Banach space (E, .) then (S) is compact.Give an example of a compact subset S in a normed space (E, .) for which (S) is not compact. Proposition 7.4.5 Suppose that C is a closed convex subset of a normed space (E, .), and that 0 ∈ C. Then C = {x ∈ E : pC (X) ≤ 1} (where pC is the gauge of C). Proof By Exercise 7.3.6, C ⊆ {x ∈ E : pC (X) ≤ 1}. On the other hand, if x ∈ C, then x = 0 and there exists  > 0 such that (x + B(E)) ∩ C is empty. Thus x ∈ (1 − / x)−1 C, so that pC (x) > 1. Proposition 7.4.6 Suppose that p is a sublinear functional which is bounded above in a neighbourhood of 0. Then p is a Lipschitz function on E. Proof There exist  > 0 and M > 0 such that if h ≤ , then p(h) ≤ M.   Suppose that x, y ∈  that x = y. Let x = x/ x − y and let y =  E, and    y/ x − y. Then x − y = , so that p(x ) − p(y ) ≤ p(x − y ) ≤ M, and p(y ) − p(x ) ≤ p(y − x ) ≤ M, so that |p(x ) − p(y )| ≤ M. By positive homogeneity, |p(x) − p(y)| =

x − y M x − y . |p(x ) − p(y )| ≤  

Thus p is an (M/)-Lipschitz function on E. Proposition 7.4.7 Suppose that C is an open convex subset of a normed space (E, .) and that 0 ∈ C. Then pC is real-valued and continuous on E, C = {x ∈ E : pC (x) < 1} and C = {x ∈ E : pC (x) ≤ 1}.

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Proof There exists  > 0 such that B(E) ⊆ C. If x ∈ E, then x/(x + 1) ∈ C, and so pC (x) ≤ (x + 1)/, and pC is finite-valued. Further, pC (x) ≤ 1/ for x ∈ B(E), and so pC is continuous, by Proposition 7.4.6. {x ∈ E : pC (x) < 1} ⊆ C, by Exercise 7.3.6. If x ∈ C \ {0}, there exists  > 0 such that x + B(E) ⊆ C. In particular, (1 + / x)x ∈ C, so that pC (x) < 1. Consequently, C ⊆ {x ∈ E : pC (x) < 1}. Since pC is continuous, {x ∈ E : pC (x) ≤ 1} is closed. Since C ⊆ {x ∈ E : pC (x) ≤ 1}, by Exercise 7.3.6, C ⊆ {x ∈ E : pC (x) ≤ 1}. On the other hand, if pC (x) ≤ 1 then pC ((1 − 1/n)x) < 1, so that (1 − 1/n)x ∈ C. Since (1 − 1/n)x → x as n → ∞, x ∈ C. Thus {x ∈ E : pC (x) ≤ 1} ⊆ C. Exercise 7.4.8 If A is a convex subset of a normed space (E, .) and Aint = ∅ then A = Aint . A subset B of a normed space (E, .) is a convex body if it is convex, norm bounded and has a non-empty interior. If b0 ∈ Bint and B0 = B − b0 , then pB0 is a non-negative sublinear functional, and by positive homogeneity there exist positive constants m and M such that m x ≤ pB0 (x) ≤ M x , for x ∈ E. Exercise 7.4.9 A convex body in a normed space (E, .) is symmetric if it is absolutely convex. Show that if B is a symmetric convex body then pB is a norm, uniformly equivalent to ..

7.5 Linear Operators We now return to the study of normed spaces. It is natural to consider continuous linear mappings between normed spaces. We prove some standard results. Theorem 7.5.1 Suppose that (E1 , .1 ) and (E2 , .2 ) are normed spaces and that T is a linear mapping from E1 to E2 . The following are equivalent: (i) (ii) (iii) (iv) (v) (vi)

K = sup{Tx2 : x1 ≤ 1} < ∞; there exists C ∈ R such that Tx2 ≤ C x1 , for all x in E1 ; T is Lipschitz; T is uniformly continuous on E1 ; T is continuous on E1 ; T is continuous at 0.

Proof (i) implies (ii): (ii) is trivially satisfied if x = 0. Otherwise, let x1 = x/ x1 . Then T(x)2 = T(x1 x1 )2 = x1 T(x1 )2 = x1 T(x1 )2 ≤ K x1 .

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89

(ii) implies (iii): T(x1 ) − T(x2 )2 = T(x1 − x2 )2 ≤ C x1 − x2 1 . Obviously (iii) implies (iv), (iv) implies (v) and (v) implies (vi). (vi) implies (i): there exists δ > 0 such that if x1 ≤ δ then T(x)2 ≤ 1. If x1 ≤ 1 then δx1 ≤ δ, so that T(x)2 = δ −1 T(δx)2 = δ −1 T(δx)2 ≤ δ −1 . Two norms .1 and .2 on a vector space E are equivalent if they define the same topology on E. Corollary 7.5.2 Suppose that .1 and .2 are two norms on a vector space E. They are equivalent if and only if they are uniformly equivalent, and if and only if there exist positive c and C such that c x1 ≤ x2 ≤ C x1 for each x ∈ E. Corollary 7.5.3 Suppose that .1 and .2 are two equivalent norms on a vector space E. Then (E, .1 ) is a Banach space if and only if (E, .2 ) is a Banach space. We denote the set of continuous linear mappings from E1 to E2 by L(E1 , E2 ). We write L(E) for L(E, E). A linear mapping from E1 to E2 is also called a bounded linear mapping, or a linear operator; a continuous linear mapping from E to itself is called an operator on E. We have the following extension theorem. Theorem 7.5.4 Suppose that F is a dense linear subspace of a normed space (E, .E ), and that T is a continuous linear mapping from F to a Banach space (G, .G ). Then there is a unique continuous linear mapping T˜ from E to G ˜ ˜ which extends T: T(y) = T(y) for y ∈ F. If T is an isometry then so is T. Proof By Theorem 7.5.1, T is uniformly continuous, and so by Theorem 2.4.4 ˜ which is an isometry if T is. We must there is a unique continuous extension T, show that T˜ is linear. Suppose that x, y ∈ E and that α, β are scalars. There ∞ exist sequences (xn )∞ n=1 and (yn )n=1 in F such that xn → x and yn → y as n → ∞. Then αxn + βyn → αx + βy as n → ∞, and so ˜ ˜ T(αx + βy) = lim T(αx n + βyn ) = lim T(αxn + βyn ) n→∞

n→∞

= lim (αT(xn ) + βT(yn )) = α lim T(xn ) + β lim T(yn ) n→∞

n→∞

n→∞

˜ ˜ ˜ n ) + β lim T(y ˜ n ) = α T(x) + β T(y). = α lim T(x n→∞

n→∞

90

Banach Spaces

Theorem 7.5.5 (i) L(E1 , E2 ) is a linear subspace of the vector space of all linear mappings from E1 to E2 . (ii) If T ∈ L(E1 , E2 ), set T = sup{T(x)2 : x1 ≤ 1}. Then T is a norm on L(E1 , E2 ), the operator norm. (iii) If T ∈ L(E1 , E2 ), and x ∈ E1 then T(x)2 ≤ T . x1 . Proof (i) We use condition (i) of Theorem 7.5.1. Suppose that S, T ∈ L(E1 , E2 ) and that α is a scalar. Then sup{(αT)(x)2 : x1 ≤ 1} = |α| sup{T(x)2 : x1 ≤ 1}, so that αT ∈ L(E1 , E2 ) and sup{(S + T)(x)2 : x1 ≤ 1} ≤ ≤ sup{S(x)2 : x1 ≤ 1} + sup{T(x)2 : x1 ≤ 1}, so that S + T ∈ L(E1 , E2 ). (ii) If T = 0, then T(x) = 0 for x with x ≤ 1, and so T(x) = 0 for all x; thus T = 0. αT = |α| T and S + T ≤ S + T, by the equation and inequality that we have established to prove (i). (iii) This is true if x = 0. Otherwise, let y = x/ x1 . Then y1 = 1, so that T(x)2 = T(x1 y) = x1 T(y)2 ≤ T x1 .

Theorem 7.5.6 If (E1 , .1 ) is a normed space and (E2 , .2 ) is a Banach space then L(E1 , E2 ) is a Banach space under the operator norm. Proof Let (Tn ) be a Cauchy sequence in L(E1 , E2 ). First we identify what the limit must be. Since, for each x ∈ E1 , Tn (x) − Tm (x)2 ≤ Tn − Tm  x1 , (Tn (x)) is a Cauchy sequence in E2 , which converges, by the completeness of E2 , to T(x), say. Secondly, we show that T is a linear mapping from E1 to E2 . This follows, since T(αx + βy) − αT(x) − βT(y) = lim (Tn (αx + βy) − αTn (x) − βTn (y)) = 0, n→∞

for all x, y ∈ E1 and all scalars α, β. Thirdly we show that T is continuous. There exists N such that Tn − Tm  ≤ 1, for m, n ≥ N. Then (T − TN )(x) = lim (Tn − TN )(x) ≤ x1 , n→∞

for each x ∈ E1 , so that T − TN ∈ L(E1 , E2 ). Since L(E1 , E2 ) is a vector space, T = (T − TN ) + TN ∈ L(E1 , E2 ). Finally we show that Tn → T. Given  > 0 there exists M such that Tn − Tm  ≤ , for m, n ≥ M. Then if m ≥ M, and x ∈ E1 ,

7.6 Five Fundamental Theorems

91

(T − Tm )(x) = lim (Tn − Tm )(x) ≤  x1 , n→∞

so that T − Tm  ≤ . A linear functional on a vector space E is a linear mapping from E into K. The vector space of continuous linear functionals on a normed space (E, .) is called the dual space E ; it is given the dual norm φ = {sup |φ(x)| : x ≤ 1}. This is simply the operator norm from (E, .) into the scalars. Corollary 7.5.7 The dual space (E , . ) of a normed space (E, .) is a Banach space. Exercise 7.5.8 Show that the dual space of (c0 , .∞ ) is naturally linearly isometrically isomorphic to (l1 , .1 ), and that the dual space of (l1 , .1 ) is naturally linearly isometrically isomorphic to (l∞ , .∞ ). What about the dual of (c, .∞ )?  Exercise 7.5.9 Let cs = {x ∈ l∞ : ∞ i=1 xi is convergent}. If x ∈ cs let xcs = n supn (| i=1 xi |). Show that .cs is a norm on cs, and that (cs, .cs ) is a Banach space. To what space is it naturally linearly isometrically isomorphic? Find natural representations of its dual and its bidual. We can extend these results to multilinear mappings. It is clearly enough to consider bilinear mappings. Exercise 7.5.10 Suppose that (E, .E ), (F, .F ) and (G, .G ) are normed spaces and that B is a bilinear mapping from E × F into G. Show that B is continuous if and only if there exists M ≥ 0 such that B(x, y)G ≤ M xE yF for all (x, y) ∈ E × F. Exercise 7.5.11 Suppose that (E, .E ), (F, .F ) and (G, .G ) are normed spaces. If T ∈ L(E, L(F, G)), and x ∈ E, y ∈ F, let j(T)(x, y) = (T(x))(y). Show that j(T) is a continuous bilinear mapping from E × F into G; show that j is a bijective linear mapping of L(E, L(F, G)) onto the vector space B(E, F; G) of continuous bilinear mappings from E × F into G. If b ∈ B(E, F; G), let b = sup{b(x, y)G : xE ≤ 1, yF ≤ 1}. Show that this is a norm on B(E, F; G), and with this norm the mapping j is an isometry. Thus if G is a Banach space, then so is B(E, F; G).

7.6 Five Fundamental Theorems We now prove five fundamental theorems, each of which depends on Baire’s category theorem.

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Theorem 7.6.1 (The Principle of Uniform Boundedness) Suppose that (E, .E ) is a Banach space, that (F, .F ) is a normed space and that T ⊆ L(E, F). Then T is norm bounded in L(E, F) if and only if {T(x) : T ∈ T } is norm bounded in F, for each x ∈ E. Proof If T is norm bounded in L(E, F) then certainly {T(x) : T ∈ T } is norm bounded in F, for each x ∈ E. If x ∈ E and T ∈ T , let pT (x) = T(x)F . Then pT is a continuous seminorm on (E, .E ). If {T(x) : T ∈ T } is norm bounded in F, for each x ∈ E, then P = sup{pT : T ∈ T } is a lower semi-continuous seminorm on (E, .E ), and An = {x : P(x) ≤ n} is closed, for each n. Since E = ∪∞ n=1 An , it follows from Baire’s 1 is not empty, for some n. If x ∈ Aint category theorem that Aint n n , then 0 = 2 x + 1 int 2 (−x) ∈ An . Thus there exists  > 0 such that B(E) ⊆ An . Hence, if x ≤ 1 then T(x) ≤ n/, for each T ∈ T ; that is, T ≤ n/, for each T ∈ T . We shall prove a non-linear version of this theorem in Theorem 10.2.4. Theorem 7.6.2 (The open mapping theorem) Suppose that (E, .E ) and (F, .F ) are two Banach spaces, and that T ∈ L(E, F) is surjective. Then T is an open mapping: if U is open in E then T(U) is open in F. Proof Since translation and dilation are homeomorphisms, it is sufficient to show that there exists r > 0 such that if y ∈ F and yF < r then there exists x ∈ B(E) such that T(x) = y. Let Fn = T(nB(E)). Then F = ∪∞ n=1 Fn . By Baire’s category theorem, there int exists n ∈ N such that Fn = ∅. By homogeneity, F1int = ∅. Thus there exists y ∈ F1 and r > 0 such that y+2rB(F) ⊆ F1 . By symmetry, −y+2rB(F) ⊆ F1 , and by convexity, 2rB(F) ⊆ 12 ((y + rB(F)) + (−y + rB(F))) ⊆ F1 . Suppose now that y ∈ F and that y < r. There exists x1 ∈ 12 B(E) such that y − T(x1 ) ≤ r/2. Arguing inductively, there exists a sequence (xn )∞ n=1 in E with xn ≤ 1/2n such that ⎛  ⎞   n    n+1 ⎝y −  ⎠ − T(x T(x ) ) , for n ∈ N. j n+1  < r/2    j=1  But then ∞ j=1 xj converges in E to an element x of B(E), and ⎛ ⎞ n n   T(x) = lim T ⎝ xj ⎠ = lim T(xj ) = y. n→∞

j=1

n→∞

j=1

7.6 Five Fundamental Theorems

93

Theorem 7.6.3 (The isomorphism theorem) Suppose that (E, .E ) and (F, .F ) are two Banach spaces, and that T ∈ L(E, F) is bijective. Then T is a homeomorphism: T −1 is continuous. Proof This is an immediate consequence of the open mapping theorem. Suppose that (E, .) and (F, .F ) are Banach spaces, and that T is a linear mapping from E to F. Then the graph T = {(x, T(x)) : x ∈ E} is a linear subspace of E × F. If we give E × F the product norm (x, y) = xE + yF , and if T is continuous, then T is closed in E × F. The closed graph theorem says that the converse is true. Theorem 7.6.4 (The closed graph theorem) Suppose that (E, .) and (F, .F ) are Banach spaces, and that T is a linear mapping from E to F. If T is closed in E × F, then T is continuous. Proof We give T the norm inherited from E × F. Since T is a closed linear subspace of E × F, it is a Banach space. If (x, T(x)) ∈ T , let π1 (x, T(x)) = x, and π2 (x, T(x)) = T(x). Then π1 is a norm-decreasing bijection of T onto E and π2 is a norm-decreasing injective mapping of T into F. By the isomorphism theorem, it follows that π1−1 is continuous, and so therefore is T = π2 ◦ π1−1 . A sequence (en )∞ (E, .) is a Schauder basis if each n=1 in a Banach space  x ∈ E can be written uniquely as x = ∞ n=1 xn en , where the co-efficients xn are scalars, and convergence is in the norm topology. Let φn (x) = xn : then φn is a  linear functional on E. Similarly, let Pn (x) = nj=1 xj ej : then Pn is a projection of E onto an n-dimensional subspace En of E, and Pm Pn = Pn Pm = Pm , for m ≤ n. Theorem 7.6.5 Suppose that (en )∞ n=1 is a Schauder basis for a Banach space (E, .). Let ! x! = supn∈N Pn (x), for x ∈ E. Then ! . ! is a norm on E, (E, ! . ! ) is a Banach space, and ! . ! is equivalent to the norm .. The co-ordinate functions φn and the projections Pn are continuous on (E, .); sup φn  < ∞ and sup Pn  < ∞. Proof If x ∈ E, then Pn (x) → x as n → ∞, and so supn∈N Pn (x) < ∞. It follows from this that ! . ! is a norm on E, and x ≤ ! x !, for x ∈ E. We shall show that ! . ! is a complete norm on E. It then follows from the isomorphism theorem that ! . ! and . are equivalent norms. The remaining results then follow from this. Suppose then that (x(k) )∞ k=1 is a ! . !-Cauchy sequence in E. For each n ∈ N, is a Cauchy sequence in Pn (E), which converges to an element (Pn (x(k) ))∞ k=1

94

Banach Spaces

x[n] =

n

[n] j=1 xj ej

of Pn (E). If m ≤ n, then xj[m] = xj[n] for 1 ≤ j ≤ m, since

Pm Pn = Pn Pm = Pm . Thus there exist xj ∈ E such that xj[n] = xj for 1 ≤ j ≤ n.  (k) → x We shall show that ∞ j=1 xj ej converges to an element x of E, and that x in ! . !-norm as k → ∞. (k) l) Suppose  > 0. There  exists K such that ! x − x ! < /3 for k, l ≥ K,  that (k) (l) so that Pn (x ) − Pn (x ) < /3 for all n, for k, l ≥ N. Letting l → ∞,     (∗) Pn (x(k) ) − x[n]  ≤ /3, for k ≥ K, and for all n.   There exists N such that Pn (x(K) ) − Pm (x(K) ) < /3 for m, n ≥ N. Consequently, if N ≤ m < n then  [n]  x − x[m]              ≤ x[n] − Pn (x(K) ) + Pn (x(K) ) − Pm (x(K) ) + Pm (x(K) ) − x[m]  < , so that x[n] converges in (E, .) as n → ∞ to an element x of E, and x[n] = Pn (x). It therefore follows from (*) that ! x(k) − x! ≤ , for k ≥ K. If (en )∞ n=1 is a Schauder basis for a Banach space (E, .), . is a monotone basis if Pn ≤ 1 for each n, and is a bimonotone basis if it is monotone, and if Pn − Pm  ≤ 1 for m, n ∈ N. Corollary 7.6.6 There exists an equivalent bimonotone norm |||.||| on E. Proof Let |||x||| = sup{max(Pn (x) − Pm (x), Pn (x)) : m, n ∈ N}, for x ∈ E. Then |||.||| is a bimonotone norm and ! x! ≤ |||x||| ≤ 2! x!, so that |||.||| is equivalent to .. A Banach space with a Schauder basis is separable. The converse is not true, but is hard to prove. The separable Banach spaces that one meets in practice have Schauder bases, though it can sometimes be difficult to give an explicit construction. Exercise 7.6.7 Let (ω, τ ) denote the linear space K N with the product topology, and let φ be the subspace of sequences with only finitely many nonzero terms. A Banach sequence space (E, .) is a linear subspace E of ω which contains φ, with a complete norm . for which the inclusion mapping (E, .) → (ω, τ ) is continuous. Suppose that (E, .) and (F .) are two Banach sequence spaces and that (aij ) is a doubly-infinite matrix for which

7.7 The Petal Theorem and Daneˇs’s Drop Theorem

95

 ∞ ( ∞ j=1 aij xj )i=1 ∈ F for each x ∈ E. Use the closed graph theorem to show that (aij ) defines a continuous linear mapping from E into F. Exercise 7.6.8 Suppose that . is a norm on C([0, 1]) with the property that if fn  → 0 then fn (t) → 0 as n → ∞, for each t ∈ [0, 1]. Show that if . is a complete norm then . is equivalent to the uniform norm .∞ , and give an example for which . is not complete, and is therefore not equivalent to .∞ .

7.7 The Petal Theorem and Daneˇs’s Drop Theorem We now give two applications of Ekeland’s variational theorem. If A is a nonempty bounded closed subset of a Banach space, and z ∈ A, then d(z, A) = infa∈A d(z, a) is not necessarily achieved. Both theorems show that there is a (weaker) alternative. We need two definitions. Suppose that a and b are distinct points of a Banach space (E, .) and that γ > 0 and 0 < r < a − b. Then the petal Pγ (a, b) is the set {x ∈ E : γ x − a + x − b ≤ a − b} and the drop Dr (a, b) is the set {x ∈ E : x ∈ (Mr (a) ∪ b) = {x = (1 − t)c + tb : c − a ≤ r}. Exercise 7.7.1 Draw some pictures to illustrate the shape of a petal and a drop (raindrop). Theorem 7.7.2 (The petal theorem) Suppose that Y is a non-empty closed subset of a Banach space (E, .), that y0 ∈ Y and that z ∈ E \ Y. Suppose that 0 < r < d(z, Y) and that γ > 0. Then there exists y˜ in Y, with y˜ − y0  < ˜ b) = {y}. ˜ (y0 − z − r)/γ such that y˜ ∈ Pγ (y0 , z) and Y ∩ Pγ (y, Proof Consider the function f (y) = y − z on Y, put α = γ and  = f (z0 )−r in Ekeland’s variational principle (Theorem 4.4.1). Then there exists y˜ ∈ Y satisfying y˜ − y0  < (y0 − z − r)/γ , (ii) and (iii). But (ii) implies that ˜ z) = {y}. ˜ y˜ ∈ Pγ (y0 , z), and (iii) implies that Y ∩ Pγ (y, Theorem 7.7.3 (Daneˇs’s drop theorem) Suppose that Y is a non-empty closed subset of a Banach space (E, .) and that z ∈ E \ Y. Suppose that 0 < r < d(z, Y) < ρ. Then there exists a point y˜ of Y such that y˜ − z ≤ ρ, and if ˜ ∩ Y then y = y. ˜ y ∈ (Mr (z), y) Proof We may assume that z = 0. Let Yρ = Y ∩ Mρ (0), let R = d(0, Y) and let α = (R − r)/2(ρ + r). Applying Ekeland’s variational principle to the

96

Banach Spaces

function ., the set Yρ and the parameter α, there exists y˜ ∈ Yρ such that if y ∈ Yρ then y ˜ < y + α y − y ˜ . Suppose now that y ∈ Yρ belongs to the drop Dr (0, y) ˜ = (Mr (0), y) ˜ and that y = y. ˜ There exist w ∈ Mr (0) and 0 ≤ t ≤ 1 such that y = (1 − t)y˜ + tw, so that y˜ − y = t(y˜ − w). Note that in fact 0 < t < 1 and that y ≤ (1 − t) y ˜ + t w ≤ (1 − t)ρ + tr. Thus t(ρ − r) ≤ y ˜ − y. Now y ˜ − y < α y − y ˜ = tα y˜ − w   y ˜ + w ˜ − y) .(t(ρ − r)) ≤ 12 (y ≤ 2(ρ + r) giving a contradiction. Corollary 7.7.4 Suppose that f is a proper lower semi-continuous function on a Banach space (E, .), that z ∈ E and that s < f (z). Suppose that 0 < r < inf{max(d(z, x), |s − t|) : x ∈ f , t ≥ f (x)} < ρ. Then there exists a point y˜ of f and t˜ ∈ R such that t˜ ≥ f (y), ˜ with y˜ − z < ρ and |t˜−s| < ρ such that if y ∈ f and t ≥ f (y) and (y, t) ∈ ((Mr (z), |t−s| ≤ r), (y, ˜ z˜)) then y = y˜ and t = t˜. If f is continuous on E, then t˜ = f (y). ˜ Proof The first statement follows by considering the episum Af when E × R is given the norm x, u = max(x , |u|). If f is continuous, and f (y) ˜ < t˜ then the line segment [(y, ˜ f (t˜)), (z, s)] contains more than one point of the episum, giving a contradiction.

8 Hilbert Spaces

In this chapter, we consider Hilbert spaces. These are Banach spaces with a great deal of symmetry, and important geometrical properties.

8.1 Inner-product Spaces Suppose that E is a real or complex vector space. An inner product on E is a scalar-valued function (x, y) → x, y on E × E which satisfies the following: (i) (bilinearity) α1 x1 + α2 x2 , y = α1 x1 , y + α2 x2 , y , x, β1 y1 + β2 y2  = β1 x, y1  + β2 x, y2  , for all x, x1 , x2 , y, y1 , y2 in E and all real α1 , α2 , β1 , β2 ; (ii) (skew-symmetry) y, x = x, y for all x, y in E; (iii) (positive definiteness) x, x > 0 for all non-zero x in E. A function which satisfies (i) and (ii) is called a skew-symmetric bilinear form. A vector space E equipped with an inner product is called an inner-product space. Note that if (E, ., .) is a complex inner-product space then ., .R , its real part, is an inner product on ER , the underlying real space, and x, x = x, xR . Here is a most important example of an inner-product space. Let l2 denote ∞ 2 the set of all sequences (an )∞ n=1 |an | is finite. Then l2 is a n=1 for which 97

98

Hilbert Spaces

vector space (with the algebraic operations defined pointwise), such that if  a, b ∈ l2 then ∞ n=1 an bn converges absolutely, and that the function (a, b) →  a, b = ∞ n=1 an bn is an inner product on l2 . Similarly, if E = Cd , we define the usual inner product by setting z, w = d i=1 zi wi for z = (zi ), w = (wi ). As another example, the space C[a, b] of continuous real-valued functions on the closed interval [a, b] is an inner-product space when the inner product is defined by  b  f , g = f (x)g(x) dx. a 1

If x is a vector in E, we set x = x, x 2 . We shall show that . is a norm on E, the inner-product norm on E. Note that, in the complex case, x = x, xR . Thus, when we consider metric properties of E we can frequently suppose that we are dealing with a real inner-product space. Certainly x = 0 if and only if x = 0, and λx = |λ| x. We now establish some basic properties of inner product spaces. Proposition 8.1.1 (The Cauchy–Schwarz inequality) If x and y are vectors in an inner-product space E then | x, y | ≤ x . y , with equality if and only if x and y are linearly dependent. Proof This depends upon the quadratic nature of the inner product. The inequality is trivially true if x, y = 0. If x = 0, then x = 0 and x, y = 0, so that the inequality is true, and the same holds if y = 0. Otherwise, if λ is a scalar, then 0 ≤ x + λy2 = x + λy, x + λy = x, x + λ x, y + λ y, x + |λ|2 y, y . Put λ=−

x x, y x . , so that |λ| = . y | x, y | y

It follows that

  x | x, y |2 x x2 2 2 y = 2 x − | x, y |. . + , 0 ≤ x − 2 y | x, y | y y2 2

so that | x, y | ≤ x . y. If x = 0 or y = 0, then equality holds, and x and y are linearly dependent.

8.1 Inner-product Spaces

99

Otherwise, if equality holds, then x + λy = 0, so that x + λy = 0, and x and y are linearly dependent. Conversely, if x and y are linearly dependent, then x = αy for some scalar α, and so | x, y | = |α| y2 = x . y . Corollary 8.1.2 x + y ≤ x + y, with equality if and only if y = 0 or x = αy, with α real and non-negative. Proof We have x + y2 = x2 + x, y + y, x + y2 ≤ x2 + 2 x . y + y2 = (x + y)2 . Equality holds if and only if x, y + y, x = 2 x2 . y2 , which is equivalent to the condition stated. Thus . is a norm on E. Note also that the inner product is determined by the norm: we have the polarization formulae. Exercise 8.1.3 Show that x, y = 12 (x + y2 − x2 − y2 ) = 14 (x + y2 − x − y2 ) (in the real case), and

 2   ik x + ik y − (x2 + y2 ) k=1  2π  2 1   = eit x + eit y dt − (x2 + y2 ) 2π 0

x, y =

1 4

4

(in the complex case). Exercise 8.1.4 Suppose that T is a linear mapping of a complex inner-product space into itself. Establish a polarization formula for T(x), y, and deduce that if T(x), x = 0 for all x, then T = 0. Give a two-dimensional example to show that the same is not true for real inner-product spaces. We also have the following. Exercise 8.1.5 (The parallelogram law) By expanding the first two terms, show that if x and y are vectors in an inner-product space E, then x + y2 + x − y2 = 2x2 + 2y2 .

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Hilbert Spaces

Proposition 8.1.6 If y ∈ E, an inner-product space, then the mapping   ly defined by ly (x) = x, y is a continuous linear functional on E, and ly  = y. l is a linear (in the real case) or antilinear (in the complex case) isometry of E into its dual space E .      Proof Since | x, y | ≤ x. y,  ly ∈ E , and ly ≤ y. If y = 0, then   y/ y , y = y and so ly = y. Trivially, l0  = 0 = 0. The antilinearity of l in the complex case follows from the sesquilinearity of the inner product. Many of the geometric and metric properties of inner-product spaces can be expressed in terms of orthogonality. Vectors x and y in an inner-product space E are said to be orthogonal if x, y = 0; if so, we write x⊥y. In the complex case, x and y are orthogonal in the underlying real space if and only if x, y is pure imaginary; the notions of orthogonality are different. Proposition 8.1.7 (Pythagoras’ theorem) If x and y are vectors in a real innerproduct space E then x + y2 = x2 + y2 if and only if x⊥y. If x and y are vectors in a complex inner-product space E then x + y2 = x2 + y2 if and only if x, y is pure imaginary. Proof For x + y2 = x2 + y2 + x, y + y, x. If A is a subset of an inner-product space E, we set A⊥ = {x ∈ E : a, x = 0 for all a ∈ A}. We write x⊥ for {x}⊥ . A⊥ is the annihilator of A. Proposition 8.1.8 If A is a subset of an inner-product space E, then A⊥ is a closed linear subspace of E. Proof If a ∈ A then a⊥ = {x : la (x) = 0} is closed, since la is continuous, and is clearly a linear subspace of E. Then A⊥ = ∩a∈A a⊥ is closed, and a linear subspace of E. Exercise 8.1.9 Suppose that A and B are subsets of an inner-product space E. Show the following. (i) (ii) (iii) (iv) (v)

A⊥ = {x ∈ E : x, a = 0 for all a ∈ A}. If A ⊆ B then B⊥ ⊆ A⊥ . A ⊆ A⊥⊥ . A⊥ = A⊥⊥⊥ . A ∩ A⊥ ⊆ {0}.

Suppose that x is a unit vector in E, and that z ∈ E. Let λ = z, x and let y = z − λx. Then y, x = z, x − z, x x, x = 0. Thus z = λx + y, where

8.2 Hilbert Space; Nearest Points

101

λx ∈ span(x) and y ∈ x⊥ . If z = μx + w, with w ∈ x⊥ , then z, x = μ, so that μ = λ and w = y; the decomposition is unique. Here is an application. Proposition 8.1.10 Suppose that x, y and z are elements of an inner-product space E and that z − x = y − x + z − y. Then there exists 0 ≤ λ ≤ 1 such that y = (1 − λ)x + λz. Proof By translation, we can suppose that x = 0. Let y = λz + w, where w⊥z. Then z = λz + w + (1 − λ)z − w ≥ λz + (1 − λ)z = (|λ| + |1 − λ|) z , with equality throughout if and only if w = 0 and 0 ≤ λ ≤ 1. A corresponding result does not hold for general Banach spaces; in (R2 , .∞ ), take x = (0, 0), y = (1, 1) and z = (2, 0).

8.2 Hilbert Space; Nearest Points An inner-product space which is complete under the inner-product norm is called a Hilbert space. A finite-dimensional real Hilbert space is called a Euclidean space, and a finite-dimensional complex Hilbert space is called a Hermitian space. The following result is very important; the corresponding result for Banach spaces is not in general true. Theorem 8.2.1 Suppose that C is a non-empty closed convex subset of a real Hilbert space H and that x ∈ H. Then there exists a unique point c(x) in C such that x − c(x) = d(x, C)

(= inf{x − y : y ∈ C}).

(c(x) is the nearest point to x in C.) x, c(x) = sup{x, c : c ∈ C} so that C ⊆ H− = {z : x, z ≤ c(x)}. The mapping x → c(x) is a retraction of H onto C. Proof Let d = d(x, C). For each n, there exists cn ∈ C with x − cn  < d + 1/n. By the parallelogram law,

102

Hilbert Spaces 4 x − (cm + cn )/22 + cm − cn 2 = 2 x − cm 2 + 2 x − cn 2 .

Since, by convexity, 12 (cm + cn ) ∈ C, 4d2 + cm − cn 2 ≤ 4 x − (cm + cn )/22 + cm − cn 2 = 2 x − cm 2 + 2 x − cn 2 ≤ 2(d + 1/m)2 + 2(d + 1/n)2 , so that 4d 4d 2 2 + 2+ + 2 →0 m n m n as m, n → ∞. Thus (cm ) is a Cauchy sequence in C. Since H is complete, (cm ) converges, to c(x), say, and c(x) ∈ C, since C is closed. Then d ≤ x − c(x) = lim x − cn  ≤ d, so that x − c(x) = d. If c and c are two nearest points,       2x − c − c 2 + c − c 2 = 2 x − c2 + 2 x − c 2 = 4d2 and (cm − cn )2 ≤

 2     2x − c − c 2 = 4 x − 1 (c + c ) ≥ 4d2 ,   2 2  since 12 (c + c ) ∈ C, so that c − c  ≤ 0, and c = c . By translation, and then scaling, we can suppose that x = 1 and that c(x) = 0. Suppose that c ∈ C. We can write c = αx + w, where α ∈ R and w⊥x. Suppose, if possible, that α > 0. Then if β > 0, x − βc2 = 1 − 2αβ + β 2 c2 < 1 for small values of β. But βc ∈ C, for 0 < β ≤ 1, and so we have a contradiction; thus α ≤ 0, and C s contained in the half-space H− = {z : x, z ≤ 0 = c(x)}. We must show that the mapping x → c(x) is continuous. Suppose that 0 <  < 1. Let δ = /5 and suppose that y − x < δ. Let a(y) = y − c(y). Then a(y) ≤ y − c(x) = y ≤ 1 + δ. Let λ = a(y), x. Then λ = y, x − c(y), x ≥ 1 + y − x, x ≥ 1 − δ, so that a(y) − λx ≤ a(y) − λ ≤ 2δ. Now a(y) − x = (a(y) − λx) − (1 − λ)x and (a(y) − λx)⊥x, so that, using Pythagoras’ theorem, a(y) − x2 = a(y) − λx2 + (1 − λ)2 ≤ 5δ 2 ≤  2 /4.

8.2 Hilbert Space; Nearest Points

103

Thus c(y) = a(y) − y ≤ a(y) − x + x − y ≤ /2 + /2 = , and so c is continuous. Suppose that V is a vector space. A projection P is a linear mapping: V → V such that P2 = P. Proposition 8.2.2 If P is a projection, and U = P(V), W = P−1 (0), then V = U ⊕ W (U + W = V and U ∩ W = {0}) and U = {x ∈ V : P(x) = x}, W = {x ∈ V : P(x) = 0}. Proof If x = P(y) ∈ U then P(x) = P2 (y) = P(y) = x, while if P(x) = x then x ∈ U; thus U = {x ∈ V : P(x) = x}. If x ∈ V, then x = P(x) + (x − P(x)): P(x) ∈ U and x − P(x) ∈ W. If x ∈ U ∩ W, P(x) = x, since x ∈ U, and P(x) = 0, since x ∈ W. Thus U ∩ W = {0}, and V = U ⊕ W. P is the projection onto U along W. I − P is then the projection onto W along U. Theorem 8.2.3 Suppose that M is a closed linear subspace of a Hilbert space H. If x ∈ H, let PM (x) be the unique nearest point to x in M. (i) x − PM (x) ∈ M ⊥ , and PM (x) is the only point in M with this property. (ii) PM is linear and continuous, with PM  = 1 (unless M = {0}, when PM = 0). (iii) PM is a projection onto M, along M ⊥ . (iv) I − PM = PM ⊥ , so that H = M ⊕ M ⊥ . Thus M = M ⊥⊥ . (v) If x, y ∈ H, then PM (x), y = PM (x), PM (y) = x, PM (y). Proof (i) Let z = x − PM (x). Suppose that y ∈ M and that z, y = 0. Then y = 0. Let w = y/ y. Then w ∈ M, w = 1 and z, w = α = 0. Then z − αw, αw = |α|2 − |α|2 = 0. By Pythagoras’ theorem, z2 = z − αw2 + αw2 = z − αw2 + |α|2 > z − αw2 . But z − αw = x − (PM (x) + αw), and PM (x) + αw ∈ M, so that we get a contradiction. Thus z ∈ M ⊥ . Suppose that u ∈ M, and that x − u ∈ M ⊥ . Then PM (x) − u ∈ M and PM (x) − u = (x − u) − (x − PM (x)) ∈ M ⊥ , so that PM (x) − u2 = PM (x) − u, PM (x) − u = 0, and PM (x) = u. (ii) and (iii) If x, y ∈ H and α, β are scalars, (αx + βy) − (αPM (x) + βPM (y)) = α(x − PM (x)) + β(y − PM (y)) ∈ M ⊥ , so that αPM (x) + βPM (y) = PM (αx + βy), by (i). Thus PM is linear. Since (x − PM (x)) ⊥ PM (x), x2 = PM (x)2 + x − PM (x)2 , so that PM  ≤ 1

104

Hilbert Spaces

and PM is continuous. Since PM (x) = x if and only if x ∈ M, PM  = 1 (unless M = {0}), and PM is a projection onto M. By (i), PM (x) = 0 if and only if x ∈ M ⊥ , so that PM is the projection onto M along M ⊥ . (iv) If x ∈ H, x − (I − PM )(x) = PM (x) ∈ M ⊆ M ⊥⊥ , so that (I − PM )(x) = PM ⊥ (x), by (i). (v) If x, y ∈ H, then PM (x) ∈ M and y − PM (y) ∈ M ⊥ , so that PM (x), y − PM (y) = 0; thus PM (x), y = PM (x), PM (y). Similarly, x, PM (y) = PM (x), PM (y). Theorem 8.2.4 Suppose that H is a Hilbert space. (i) C ⊆ H is a closed linear subspace if and only if C = C⊥⊥ . (ii) If A is a non-empty subset of H then span(A) = A⊥⊥ . Proof (i) The condition is sufficient, by Proposition 8.1.8. If C is a closed linear subspace of H, then H = C ⊕ C⊥ . If x ∈ C⊥⊥ , we can write x = y + z, with y ∈ C and z ∈ C⊥ . But then z, z = x, z − y, z = 0, since x ∈ C⊥⊥ and z ∈ C⊥ , so that z = 0 and x = y ∈ C. Thus C⊥⊥ ⊆ C. Since the reverse inclusion always holds, the condition is also necessary. (ii) Let M = span(A). Then A⊥⊥ ⊆ M ⊥⊥ = M, by (i). But A⊥⊥ is a closed linear subspace of H containing A, and so A⊥⊥ ⊇ M.

8.3 Orthonormal Sequences; Gram–Schmidt Orthonormalization A finite or infinite sequence (xn ) in an inner-product space is an orthogonal sequence if xm , xn  = 0 if m = n; it is an orthonormal sequence if, in addition, xm , xm  = 1 for all m. Exercise 8.3.1 Let CC ([0, 1]) be given the inner product  1 f , g = f (t)g(t) dt, 0

and let γ (t) = sequence.

e2πit .

Show that the sequence (γ n )∞ n=−∞ is an orthonormal

Exercise 8.3.2 Let CR ([a, b]) be given the inner product  b f , g = f (x)g(x)w(x) dx, a

where w is a positive continuous weight function. Let (Qn )∞ n=1 be defined inductively by Q0 = 1,

Q1 = x − c1 ,

Qn = (x − cn )Qn−1 − dn Qn−2

8.3 Orthonormal Sequences; Gram–Schmidt Orthonormalization 105

where cn = xQn−1 , Qn−1  / Qn−1 2

and

dn = xQn−1 , Qn−2  / Qn−2 2 .

Show that (Qn )∞ n=0 is an orthogonal sequence of monic polynomials, with each Qn of degree n. Exercise 8.3.3 In Exercise 8.3.2, show that Qn  = inf {Q : Q a monic polynomial of degree n}. Exercise 8.3.4 In Exercise 8.3.2, show that Qn has n distinct roots in [a, b].  (Consider Qn , P, where P = ki=1 (x − ri ).) Exercise 8.3.5 If [a, b] = [−1, 1] and w = 1 in Exercise 8.3.2 the resulting polynomials are called the Legendre polynomials and are usually denoted by (Xn )∞ n=0 . Calculate Xn for 0 ≤ n ≤ 4. Proposition 8.3.6 Suppose that (e1 , . . . en ) is a finite orthonormal sequence in an inner product space H. Let An = {e1 , . . . , en } and let Mn = span(An ). If  x ∈ H then PMn (x) = ni=1 x, ei  ei . !  Proof x − ni=1 x, ei  ei , e j = 0 for 1 ≤ j ≤ n, so that x−

n 

⊥ x, ei  ei ∈ A⊥ n = Mn .

i=1

Also

n

i=1 x, ei  ei

∈ Mn , so that the result follows from Theorem 8.2.3.

Theorem 8.3.7 (Gram–Schmidt orthonormalization) Let (xn )∞ n=1 be a linearly independent sequence in an inner product space V and let Vn = span(x1 , . . . , xn ). Then there exists an orthonormal sequence (en )∞ n=1 in V such that (i) span(e1 , . . . , en ) = Vn , for n = 1, 2, . . . , and (ii) en ⊥ Vn−1 , for n = 2, 3, . . . . The sequence (en )∞ n=1 is fairly unique: if (fn ) is another orthonormal sequence which satisfies (i) and (ii) then fn = λn en for each n, where |λn | = 1. Proof Let yn = xn − PVn−1 (xn ). yn = 0, since xn ∈ Vn−1 . Let en = yn / yn . Then en is a unit vector in Vn orthogonal to Vn−1 , and so (en )∞ n=1 is an orthonormal sequence. In particular, (e1 , . . . , en ) is a linearly independent sequence in the n-dimensional space Vn , and so span(e1 , . . . , en ) = Vn . If (fn ) is another orthonormal sequence satisfying (i) and (ii) then, since fn ∈ Vn , we can write fn = α1 e1 + · · · + αn en , and, since fn ⊥ Vn−1 , αi = 0 for i < n; thus fn = αn en and |αn | = fn  = 1.

106

Hilbert Spaces

Theorem 8.3.8 (The Riesz–Fischer theorem) Suppose that (en )∞ n=1 is an  c e converges in norm orthonormal sequence in a Hilbert space H. Then ∞ i i=1 2 i  ∞ ∞ ∞ 2 2   if and only if i=1 |ci | < ∞. If so, then = i=1 |ci | . i=1 ci ei n Proof Let sn = i=1 ci ei . Suppose that sn converges in norm, to c say. Then n 

|ci |2 = sn 2 → c2 as n → ∞

i=1

 2 so that ∞ i=1 |ci | < ∞.  2 Suppose conversely that ∞ i=1 |ci | < ∞. Then if m > n, sm − sn  =    m 2 1/2   m , so that (sn ) is a Cauchy sequence. By i=n+1 ci ei = ( i=n+1 |ci | ) completeness, it converges. Theorem 8.3.9 Suppose that (en )∞ n=1 is an orthonormal sequence in a Hilbert space H. Let A = span{en : n ∈ N} and M = A. If x ∈ H then PM (x) = ∞ i=1 x, ei  ei . n Proof M = A⊥⊥ and M ⊥ = A⊥ . Let sn = i=1 x, ei  ei . Then sn , x = n 2 = s , s , so that s ⊥(x − s ). Thus s 2 = x2 − n n n n n i=1 | x, ei  | x − sn 2 ≤ x2 , and we can apply the Riesz–Fischer theorem. Let s = ∞ i=1 x, ei  ei . Then s ∈ M, and sn → s. As x − sn , em  = 0 for n ≥ m, x − s, em  = 0, and so x − s ∈ A⊥ = M ⊥ . The result now follows from Theorem 8.2.3. ∞ 2 2 Corollary 8.3.10 n=1 | x, en  | ≤ x (Bessel’s inequality). ∞ 2 2 2 Proof n=1 | x, en  | = PM (x) ≤ x . When do we have equality in Bessel’s inequality? Corollary 8.3.11 The following are equivalent: (i) x ∈ M;  x, e  e (Parseval’s equation); (ii) x = ∞ i=1 ∞ i i 2 x = i=1 | x, ei  |2 . (iii) Proof (i) and (ii) are equivalent, and imply (iii), by the Riesz–Fischer theorem. If (iii) holds then 2  k k       2 x x,  = − | x, ei  |2 → 0, e e x −  i i   i=1

so that (i) holds.

i=1

8.4 Orthonormal Bases

107

8.4 Orthonormal Bases An orthonormal sequence (en )∞ n=1 in a Hilbert space H is an orthonormal basis for H if H is the closed linear span of (en )∞ n=1 . We consider the case where H is infinite-dimensional; the finite-dimensional case is simply a matter of linear algebra. Proposition 8.4.1 An orthonormal sequence (en )∞ in a Hilbert space H is  n=1 x, an orthonormal basis for H if and only if x = ∞ en  en for each x ∈ V. n=1  x,  e , y. If so, and if x, y ∈ V, then x, y = ∞ e i i i=1 Proof (en )∞ n=1 is an orthonormal basis for H if and only if M = span{en :n ∈ N}, and this happens if and only if x = PM (x) for each x ∈ H. Apply Theorem 8.3.9. If the condition is satisfied, and if x, y ∈ H, then ∞ ∞ ∞    x, y = ly (x) = ly x, ei  ei = x, ei  ly (ei ) = x, ei  ei , y . i=1

i=1

i=1

Note that we use the continuity of ly in an essential way. Theorem 8.4.2 A separable Hilbert space H has an orthonormal basis. Proof There exists a dense sequence (yk ) in H. Let / span(y1 , . . . , yk−1 )} = {k1 < k2 < . . . }, K = {k : yk ∈ and let xn = ykn . Then (xn ) is a linearly independent sequence; applying Gram–Schmidt orthonormalization, there exists an orthonormal sequence (en )∞ n=1 satisfying the conditions of Theorem 8.3.7. Then span{en : n ∈ N} = span{xn : n ∈ N} = span{yn : n ∈ N} = H, so that (en )∞ n=1 is an orthonormal basis for H. Corollary 8.4.3 If H is a separable Hilbert space, there is a linear isometry of H onto l2 . Proof Let (fn ) be an orthonormal basis for H. Let J(x) = (x, fn )∞ n=1 . Applying Corollary (iii), we see that J is an isometry into l2 . It is surjective, by the Riesz–Fischer theorem. Unfortunately, at this stage we do not have examples of Hilbert spaces other than l2 . When we have defined Lebesgue measure in Part II, it will follow that 2 (γ n )∞ n=−∞ is an orthonormal basis for LC ([0, 1]), and the normalized Legendre ∞ 2 ([−1, 1]). polynomials (Xn / Xn )n=0 form an orthonormal basis for LR

108

Hilbert Spaces

8.5 The Fr´echet–Riesz Representation Theorem; Adjoints Theorem 8.5.1 (The Fr´echet–Riesz representation theorem) Suppose that H is a Hilbert space. If y ∈ H, let ly (x) = x, y. Then l is an antilinear (in the complex case) or linear (in the real case) isometry of H onto the dual space H  . Proof We have seen (Proposition 8.1.6) that l is a linear isometry of H into H  . It remains to see that it is surjective. Suppose that φ is a non-zero element of H  . Let N be its null-space. Then N is a proper closed linear subspace of H, and so H = N ⊕ N ⊥ . Since φ is one-one on N ⊥ , dim(N ⊥ ) = 1. Thus if y is a unit vector in N ⊥ , any element x of H can be written uniquely as x = n + αy, with n ∈ N, and then φ(x) = αφ(y). But if z = φ(y)y, then lz (x) = n, z + α y, z = αφ(y) = φ(x), so that φ = lz . Theorem 8.5.2 Suppose that H1 and H2 are Hilbert spaces and that T ∈ L(H1 , H2 ). There exists a unique T ∗ ∈ L(H2 , H1 ) such that ! T(x), y = x, T ∗ (y) for all x ∈ H1 , y ∈ H2 . In the complex case, the mapping T → T ∗ is an antilinear isometry of L(H1 , H2 ) onto L(H2 , H1 ), and in the real case it is a linear isometry. T ∗∗ = T. Further, T2 = T ∗ T = TT ∗ . Proof Suppose that y ∈ H2 . Let ψy (x) = T(x), y, for x ∈ H1 . ψy is a linear functional on H1 , and |ψy (x)| ≤ T(x) y ≤ T y x ,   so that ψy ∈ H1 , and ψy  ≤ T y. By the Fr´echet–Riesz representation theorem, there exists a unique of H1 , T ∗ (y) say, such that ψy (x) =   element ∗ ∗ x, T (y) and T (y) = ψy . It is easily verified that the mapping T ∗ : H2 → H1 is linear, and so  ∗ "  # T  = sup ψy  : y ≤ 1 ≤ T . Since

! x, (αS + βT)∗ y = (αS + βT)x, y = α S(x), y + β T(x), y ! ! ! ¯ ∗ (y) = x, (αS ¯ ∗ )(y) , ¯ ∗ + βT = x, αS ¯ ∗ (y) + x, βT

the mapping T → T ∗ is antilinear. Also ! ! ! y, T ∗∗ (x) = T ∗ (y), x = x, T ∗ (y) = T(x), y = y, T(x) , so that by uniqueness, T = T ∗∗ . Thus T ≤ T ∗ , and so T = T ∗ .

8.5 The Fr´echet–Riesz Representation Theorem; Adjoints

109

Finally, T ∗ T ≤ T ∗  . T = T2 and T2 = sup{T(x)2 : x ≤ 1} = sup{T(x), T(x) : x ≤ 1} ! = sup{| x, T ∗ T(x) | : x ≤ 1}   ! = sup{| T ∗ T(x), x | : x ≤ 1} ≤ T ∗ T  , so that T ∗ T = T2 . Applying this to T ∗ , we see that TT ∗  = T2 , as well. T ∗ is called the adjoint of T. Exercise 8.5.3 Suppose that H is a Hilbert space, that S, T are linear mappings from H to itself which satisfy S(x), y = x, T(y) for all x, y ∈ H. Show that S and T are continuous, and that T = S∗ . Exercise 8.5.4 Give an example where E is an inner-product space and S, T are discontinuous linear mappings from E to itself which satisfy S(x), y = x, T(y) for all x, y ∈ E. Exercise 8.5.5 Suppose that H1 , H2 and H3 are Hilbert spaces, that S ∈ L(H1 , H2 ) and T ∈ L(H2 , H3 ). Show that (TS)∗ = S∗ T ∗ . Suppose that H is a Hilbert space and that T ∈ L(H). T is normal if TT ∗ = T ∗ T. Exercise 8.5.6 Suppose that H is a complex Hilbert space and that T ∈ L(H). Show the following. (i) T is normal if and only if T(x) = T ∗ (x) for each x ∈ H. (ii) If T is normal then T = T ∗ , and T n  = Tn for each n ∈ N. (iii) If T is normal, then T(x) = 0 if and only if x ∈ T(H)⊥ , if and only if T ∗ (x) = 0, and if and only if x ∈ T ∗ (H)⊥ . Suppose that H is a Hilbert space and that T ∈ L(H). T is self-adjoint if T = T ∗ . A self-adjoint operator is clearly normal. In the complex case, a selfadjoint operator is also called a Hermitian operator, and in the real case is called a symmetric operator. Exercise 8.5.7 Suppose that T ∈ L(H), where H is a complex Hilbert space. By considering T1 = 12 (T + T ∗ ) and T2 = − 12 i(T − T ∗ ), show that T can be written as T1 + iT2 , where T1 and T2 are Hermitian. Show that such a representation is unique. Exercise 8.5.8 Suppose that T ∈ L(H), where H is a real Hilbert space. T is skew-symmetric if T(x), y = − x, T(y) for all x, y ∈ H. If T ∈ L(H), show

110

Hilbert Spaces

that T can be written uniquely as T = T1 + T2 , where T1 is symmetric, and T2 is skew-symmetric. Exercise 8.5.9 Suppose that H is a complex Hilbert space and that T ∈ L(H). (i) Show that TT ∗ and T ∗ T are Hermitian. (ii) Show that if T is Hermitian then T(x), x ∈ R for each x ∈ H. A Hermitian operator is positive if T(x), x ≥ 0 for all x ∈ H. Exercise 8.5.10 Show that the set of all Hermitian operators on a complex Hilbert space H is a closed real linear subspace of L(H). Suppose that H is a Hilbert space. Recall that an element P of L(H) is a projection if P2 = P. By long tradition, a projection which is self-adjoint is called an orthogonal projection. Theorem 8.5.11 If P is an orthogonal projection, then M = P(H) is closed, P = PM and H = P(H) ⊕ (I − P)(H). Conversely if N is a closed linear subspace of H, then PN is an orthogonal projection. Proof M = {x : (I − P)(x) = 0} is closed (this is generally true for continuous projections on Banach spaces). If x ∈ H and y ∈ M, x − P(x), y = x, y − P(x), y = x, y − x, P(y) = x, y − x, y = 0. The results now follow from Exercise 8.1.9. Exercise 8.5.12 Suppose that P is a projection on a Hilbert space. Show that the following are equivalent. (i) P is an orthogonal projection. (ii) P is normal. (iii) P(x)2 = P(x), x for all x ∈ H. An element U ∈ L(H), where H is a complex Hilbert space is unitary if it is an invertible isometry of H. The unitary elements of L(H) form a group under composition, which is closed in L(H). Exercise 8.5.13 Suppose that U ∈ L(H), where H is a complex Hilbert space. Show that U is unitary if and only if it is invertible and U(x), U(y) = x, y for all x, y ∈ H, and if and only if it is invertible and U ∗ = U −1 . Exercise 8.5.14 Suppose that T ∈ L(H), where H is a complex Hilbert space. By considering power series in T, show how to define eiT ∈ L(H). Show that if T is Hermitian then eiT is unitary.

8.5 The Fr´echet–Riesz Representation Theorem; Adjoints

111

An element O ∈ L(H), where H is a real Hilbert space is called an orthogonal isometry if it is an invertible isometry of H. Most of the properties of unitary mappings carry over to orthogonal isometries. The orthogonal isometries of L(H) form a group under composition, which is closed in L(H). Theorem 8.5.15 (von Neumann’s theorem) Let T be an isometric linear mapping of a Hilbert space H into itself. Let F = {y : T(y) = y} and let S = {y − T(y) : y ∈ H}. Then ¯ (i) F = S⊥ and H = F ⊕ S. Let An (x) = 1n (x + T(x) + · · · + T n−1 (x)). Then (ii) (iii) (iv) (v)

An (x) = x for x ∈ F; An (x) → 0 for x ∈ S; An (x) → 0 for x ∈ S; An (x) → PF (x) for x ∈ H (where PF is the orthogonal projection onto F).

Proof (i) and (ii) are easy. For (iii) if y − T(y) ∈ S, An (y) =

1 (y − T n+1 (y)) → 0 as n → ∞. n

(iv) Note that An (x) ≤ x for x ∈ H. If z ∈ S and  > 0, there exists y ∈ S with z − y < . Then 1 An (z) ≤ An (z − y) + An (y) ≤ ( + (An (z) + ), n from which the result follows. (v) If x ∈ H, by (i) we can write x = y + z, where y ∈ F and z ∈ S. Then An (x) → y ∈ F, and y⊥(x − y), so that y = PF (x).

9 The Hahn–Banach Theorem

9.1 The Hahn–Banach Extension Theorem In the remaining chapters, we consider convex sets and convex functions. The first and most important theorems here are the Hahn–Banach extension theorem, and its geometric equivalent, the separation theorem. We begin by proving a one-dimensional extension theorem, which is at the heart of the matter. We then use the axiom of choice to prove a general extension theorem. (This, with Tychonoff’s compactness theorem, is one of the first times that analysts meet the axiom of choice.) When we consider separable Banach spaces, it is however possible to use a version of the Hahn–Banach theorem which does not use the axiom of choice, and we also give an account of this. Theorem 9.1.1 Suppose that p is a convex function on a real vector space E, that F0 is a linear subspace of E, that w ∈ E\F0 and that F1 = span(F0 ∪{w}). Suppose that f is a linear functional on F0 satisfying f (x) ≤ p(x) for all x ∈ F0 . Then there exists a linear functional g on F1 such that g(x) = f (x) for all x ∈ E0 (g extends f ) and such that g(y) ≤ p(y) for all y ∈ F1 (control is maintained). Proof If y ∈ F1 , we can write y = x + θ w uniquely, with x ∈ F0 . Let α ∈ R and let g(y) = f (x) + θ α. Then g is a linear functional on F1 which extends f , and we need to choose α to maintain control. We require that f (x) + θ α ≤ p(x + θ w) for all x ∈ En and all real θ . The inequality is certainly satisfied if θ = 0. For θ > 0 we require that α≤

p(x + θ w) − f (x) , θ

for all x ∈ F0 , and for θ < 0, writing φ = −θ, we require 112

9.1 The Hahn–Banach Extension Theorem

α≥

113

f (y) − p(y − φw) , φ

for all y ∈ En . If θ and φ are positive, then p(x + θ w) p(y − φw) + θ φ   φ θ θ +φ p(x + θ w) + p(y − φw) = θφ θ +φ θ +φ   θ +φ θ φ ≥ p (x + θw) + (y − φw) θφ θ +φ θ +φ     θ +φ θ +φ φx + θ y φx + θ y = p ≥ f θφ θ +φ θφ θ +φ f (x) f (y) = + . θ φ Thus



 f (y) − p(y − φw) : φ > 0, y ∈ F1 φ   p(x + θw) − f (x) ≤ inf : θ > 0, x ∈ F1 , θ

sup

and we can indeed find an α which satisfies both requirements. We now use the axiom of choice, in the form of Zorn’s lemma, to prove the general Hahn–Banach extension theorem. Theorem 9.1.2 (The Hahn–Banach extension theorem) Suppose that p is a convex function on a real vector space E and that F is a linear subspace of E. Suppose that f is a linear functional on F satisfying f (x) ≤ p(x) for all x ∈ F. Then there exists a linear functional g on E such that g(x) = f (x) for all x ∈ E (g extends f ) and such that g(y) ≤ p(y) for all y ∈ E (control is maintained). Proof Let P be the set of all pairs (G, h), where G is a linear subspace of E containing F, and h is a linear functional on G which extends f ; h(x) = f (x) for x ∈ F. Partially order P by setting (G, h) ≺ (G , h ) if G ⊂ G and h extends h; h (x) = h(x) for x ∈ G. Then it is immediate that if C is a chain in P then (G , h ) is an upper bound for C, where G = ∪{G : (G, h) ∈ C} and h (x) is the common value of h(x) for those (G, h) for which x ∈ G. Thus we can apply Zorn’s lemma; there exists a maximal (Gmax , hmax ) in P. We claim that Gmax = E, which completes the proof. If not, there exists w ∈ E \Gmax . But then, by Theorem 9.1.1, we can extend hmax to span(Gmax ∪{w}), retaining control, contradicting the maximality of (Gmax , hmax ).

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Here is a classical application of the Hahn–Banach theorem.  ∞ , let p(x) = lim sup(( n x )/n). Exercise 9.1.3 (Banach limits) If x ∈ lR i=1 i ∞ . Show that there is a linear Show that p is a sublinear functional on lR ∞ (a Banach limit) such that lim inf(x ) ≤ LIM(x) ≤ functional LIM on lR n lim sup(xn ), LIM(x) = LIM(L(x)), where L(x)n = xn+1 , and LIM(x) = limn→∞ (xn ) if x ∈ c. Do we need the axiom of choice? When E is a separable normed space and p is continuous, we can avoid it. We need an easy preliminary result. Proposition 9.1.4 Suppose that E0 is a linear subspace of a separable normed space (E, .). Then either there exists a finite sequence (x1 , . . . , xn ) in E such that, setting Ej = span(E0 , x1 , . . . , xj ), the sequence (Ej )nj=0 is a strictly increasing sequence of linear subspaces of E, with En = E, or there exists an infinite sequence (xn )∞ n=1 in E such that, setting Ej = span(E0 , x1 , . . . , xj ), is a strictly increasing sequence of linear subspaces of E, the sequence (Ej )∞ j=0 E dense in E. with ∪∞ n=1 n Proof Let (en )∞ n=1 be a dense sequence in E. Let n1 = inf{n : en ∈ E0 }, and inductively let nj+1 = inf{n : en ∈ Ej = span(E0 , x1 , . . . , xj )}, let xj+1 = enj , and let Ej+1 = span(Ej , enj ) = span(E0 , x1 , . . . , xj+1 ). If the process terminates, we are in the first case. Otherwise {ej : j ∈ N} ⊆ ∪∞ n=1 En , .). E is dense in (E, so that ∪∞ n n=1 Theorem 9.1.5 Suppose that p is a convex function on a separable real normed space (E, .) with p(0) = 0 and that p is continuous at 0. Suppose that E0 is a linear subspace of E and that f is a linear functional on E0 satisfying f (x) ≤ p(x) for all x ∈ E0 . Then there exists a continuous linear functional g on E such that g(x) = f (x) for all x ∈ E0 (g extends f ) and such that g(y) ≤ p(y) for all y ∈ E (control is maintained). Proof Let (En ) and (xn ) be sequences satisfying the conclusions of Proposition 9.1.4. Using Theorem 9.1.1, we can inductively define a linear function h on ∪∞ n=1 En which extends f and is controlled by p. Let q(x) = max(p(x), p(−x)); q is continuous at 0. Then |h(x + y) − h(x)| = |h(y)| ≤ q(y), so that h is continuous; it therefore extends to a continuous linear functional g on E. g certainly extends f , and, since g and p are both continuous, g(x) ≤ p(x) for all x ∈ E. In what follows, if a result is true in general, but can be proved in the separable case without using the axiom of choice, we shall include [separable] in its statement. Since a sublinear functional is convex, the next theorem follows immediately.

9.1 The Hahn–Banach Extension Theorem

115

Theorem 9.1.6 (The separable Hahn–Banach extension theorem) Suppose that p is a continuous sublinear functional on a [separable] real normed space (E, .), that F is a linear subspace of E and that f is a linear functional on E0 satisfying f (x) ≤ p(x) for all x ∈ F. Then there exists a continuous linear functional g on E such that g(x) = f (x) for all x ∈ F (g extends f ) and such that g(y) ≤ p(y) for all x ∈ F (control is maintained). Corollary 9.1.7 Suppose that p is a continuous sublinear functional on a [separable] real normed space (E, .), and that x ∈ E. There exists a linear functional g on E with g(x) = p(x), and with g(y) ≤ p(y) for all y ∈ E. Thus p(x) = sup{g(x) : g linear, and g(y) ≤ p(y) for all y ∈ E}, and the supremum is achieved. Proof Take E0 = span(x), and let f (λx) = λp(x). Then if λ ≥ 0 we have f (λx) = p(λx), while if λ < 0 then f (x) = λp(x) = −p(|λ|x) ≤ p(λx), since 0 = p(0) ≤ p(λx) + p(−λx). Recall that if (E, .) is a normed space, then the dual space E of continuous linear functionals on E is a Banach space under the norm f  = sup{|f (x)| : x ∈ E, x ≤ 1}. Theorem 9.1.8 Suppose that f is a continuous linear functional on a linear subspace F of a [separable] real normed space (E, .). Then there exists a continuous linear functional g on E which extends f , with g = f  . Proof Let p(y) = f  y, and apply Theorems 9.1.2 and 9.1.6. Then |g(y)| ≤ p(y) = f  y, so that g ≤ f  . Of course, g ≥ f  . Corollary 9.1.9 Suppose that x is a non-zero element of (E, .). There exists a continuous linear functional g on E such that g(x) = x and g = 1. Thus x = sup{|g(x)| : g = 1}, and the supremum is attained. Proof Let F = span(x), and if λx ∈ F let f (λx) = λ x. Then f ∈ F  and f  = 1. Let g be an extension with the same norm. We can also consider the dual E of the dual E : E is the bidual of E. There is a natural linear map j : E → E , given by j(x)(f ) = f (x). Corollary 9.1.10 The mapping j : E → E is an isometry of E into E . For j(x) = sup{|j(x)(f )| : f  ≤ 1} = sup{|f (x)| : f  ≤ 1} = x .

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The mapping j need not be surjective. If it is, we say that (E, .) is reflexive. Hilbert space is an example of a reflexive space. The bidual of the separable space (c0 , .∞ ) of null sequences is isomorphic to the non-separable space (l∞ , .∞ ), and so (c0 , .∞ ) is not reflexive.

9.2 The Separation Theorem What does the Hahn–Banach theorem mean geometrically? We need some definitions. A subset A of a real vector space E is absorbent if whenever x ∈ E there exists λ > 0 such that αx ∈ E for 0 ≤ α < λ. A subset A of a real vector space E is radially open if A − a is absorbent for each a ∈ A. The collection of radially open subsets of E is a topology on E, the radially open topology. It follows that if A is convex and absorbent then the gauge pA is a real-valued non-negative sublinear functional on E and that {x : pA (x) < 1} ⊆ A ⊆ {x : pA (x) ≤ 1}. Proposition 9.2.1 Suppose that A is a radially open convex subset of a real vector space E and that 0 ∈ A. Then pA is continuous in the radially open topology, and A = {x : pA (x) < 1}. Proof If x ∈ E and  > 0 then x + A is radially open. If y ∈ x + A then pA (x + y) − pA (x) ≤ pA (y) < ; pA is continuous. If x ∈ A, then there exists η > 0 such that x + ηx ∈ A. Thus (1 + η)x ∈ A, so that pA (x) < 1. Theorem 9.2.2 (The separation theorem I) Suppose that A and B are nonempty disjoint convex subsets of a real vector space (E, .) and that A is radially open. Then there exists a linear functional g on E and a scalar λ such that g(a) < λ for a ∈ A and g(b) ≥ λ for b ∈ B. Proof Let C = A − B = {a − b : a ∈ A, b ∈ B}. C is convex, and 0 ∈ C. Since C = ∪b∈B (A − b), C is radially open. Pick c0 = a0 − b0 ∈ C, and let D = C − c0 . Then D is convex and radially open, 0 ∈ D and −c0 ∈ D. Let pD be the gauge of D. Then pD (−c0 ) ≥ 1. Let W = span(c0 ), and set f (αc0 ) = −α, for αc0 ∈ W. Then f is a linear functional on W, and f (αc0 ) ≤ pD (αc0 ). By the Hahn–Banach theorem, there exists a linear functional g on B extending f and such that g(x) ≤ pD (x) for all x ∈ V. Now if a ∈ A and b ∈ B then a − b − c0 ∈ D, and so g(a − b − c0 ) = g(a) − g(b) + 1 < 1, by Proposition 9.2.1; thus g(a) < g(b). Let λ = inf{g(b) : b ∈ B}; λ is finite, and g(a) ≤ λ for a ∈ A. Finally, if a ∈ A

9.2 The Separation Theorem

117

then a − γ c0 ∈ A for some γ > 0, so that g(a − γ c0 ) = g(a) + γ ≤ λ. Thus g(a) < λ. Since an open subset of a normed space (E, .) is radially open, we obtain the more familiar separation theorem for normed spaces. Corollary 9.2.3 (The separation theorem for normed spaces) Suppose that A and B are non-empty disjoint convex subsets of a [separable] real normed space (E, .) and that A is open. Then there exists a continuous linear functional g on E and a scalar λ such that g(a) < λ for a ∈ A and g(b) ≥ λ for b ∈ B. Proof For then pD is continuous, and so therefore is g. It is also of interest to prove the separation theorem directly. We start with a slightly weaker result. Theorem 9.2.4 Suppose that A is a non-empty radially open convex set in a real vector space E and that F is a linear subspace of E disjoint from A. Then there exists a hyperplane G of E containing F and disjoint from A. Proof We use the axiom of choice by a simple application of Zorn’s lemma. Partially order the linear subspaces of E which contain F and are disjoint from A by direct inclusion. By Zorn’s lemma, there exists a maximal element G. We must show that G is a hyperplane. Suppose not. Let q : E → E/G be the quotient mapping. Then q(A) is radially open and convex. Let a0 be any element of A. If G is not a hyperplane in E, there exists an element e0 of E/G linearly independent of q(a0 ). Let us consider the two-dimensional space P = span(q(a0 ), e0 ). Then q(A) ∩ P is a non-empty radially open subset of P. If θ ∈ [0, 2π], let c(θ ) = q(a0 ) cos θ +e0 sin θ and let J = {θ ∈ [0, 2π ) : there exists r > 0 such that rc(θ ) ∈ q(A)}. Then J is a non-empty open subset. Let K = {( j + π ) mod(2π ) : j ∈ J}. Then K is also open. But J ∩ K = ∅, since if rc(θ ) ∈ q(A) then −sc(θ ) ∈ q(A) for any s > 0, since q(A) is convex and 0 ∈ q(A). But [0, 2π ) is connected, and so there exists m ∈ J ∪ K. Thus span(c(m) ∩ q(A)) is empty. Consequently q−1 span(c(m)) is a linear subspace of E which contains G strictly, which contains F and is disjoint from A. Thus the theorem depends upon the connectedness of [0, 2π ). Corollary 9.2.5 (The separation theorem II) Suppose that A and B are disjoint non-empty convex subsets of a real vector space E, and that A is radially open. Then there exists a linear functional φ on E and a constant λ such that

118

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φ(a) > λ for a ∈ A and φ(b) ≤ λ for b ∈ B. If (E, .) is a normed space and A is open, then φ is continuous. Proof Let a0 ∈ A, b0 ∈ B, and let c0 = a0 − b0 . Let C = A − B; then 0 ∈ C, so that there exists a hyperplane G such that G ∩ C = ∅. Since c0 ∈ C, c0 ∈ G; consequently E = G ⊕ span(c0 ), and if x ∈ E, we can write x uniquely as g + αc0 . Let φ(x) = α; then φ is a linear functional on E with null-space G and φ(c0 ) = 1. If c ∈ C, then φ(c) > 0; for otherwise φ(c − φ(c)c0 ) = 0. Thus if a ∈ A and b ∈ B then φ(a) > φ(b); in particular, φ is bounded below on A. Let λ = infa∈A φ(a), so that supb∈B φ(b) ≤ λ. Since A is radially open, φ(A) is an open interval, and so φ(a) > λ for a ∈ A. We can deduce the analytic Hahn–Banach theorem from the geometric separation theorem. Proof of Theorem 9.1.2. The set Sp = {(x, λ) : λ > p(x)} is radially open in V × R. It is disjoint from the linear subspace G( f ) ⊆ F × R ⊆ V × R. By the separation theorem, there is a hyperplane H in E × R which contains G(f ) and is disjoint from Sp . If (h, k1 ) and (h, k2 ) are two elements of H, so is (h, (1 − t)k1 + tk2 ) for all t ∈ R, which is only possible if k1 = k2 . Thus there exists a linear functional g on E such that H = G(g). Then g extends f and g(x) ≤ p(x) for all x ∈ E. This corollary applies when p is a sublinear function, or semi-norm or norm.

9.3 Weak Topologies Corollary 9.1.9 shows that a [separable] real normed space E and its dual E form a dual pair; a dual pair (E, F) consists of two vector spaces E and F, over the same field K, together with a bilinear mapping E × F → K, written as (e, f ) → e, f , with the properties that if e = 0 then there exists f ∈ F with e, f  = 0 and that if f = 0 then there exists e ∈ E with e, f  = 0. If (E, F) is a dual pair we can consider E as a vector space of functions on F – that is, E is a linear subspace of K F, the vector space of all scalar-valued functions on F – and F as a vector space of functions on E. When K is R or C we can give K F the product topology – that is, the topology of pointwise convergence on the points of F. The subspace topology on E is then called the weak topology σ (E, F). Similarly, we have the weak topology σ (F, E) on F. Thus a basic neighbourhood of 0 is a set of the form {e ∈ E : | e, fi  | < 1 fi ∈ F, 1 ≤ i ≤ n}; note that this is a convex set. A neighbourhood of a point e of E is a set of the form e + N, where N is a neighbourhood of 0.

9.4 Polarity

119

Exercise 9.3.1 Show that (E, σ (E, F)) is a topological vector space; the mapping (e1 , e2 ) : (E, σ (E, F)) × (E, σ (E, F)) → (E, σ (E, F)) is jointly continuous, and so is the mapping (λ, e) → λe : K ×(E, σ (E, F)) → (E, σ (E, F)). Proposition 9.3.2 If (E, F) is a real or complex dual pair, then F can be identified with the space of σ (E, F)-continuous linear functionals on E. Proof Since σ (E, F) is the topology of pointwise convergence, the evaluation functional e → e, f  is continuous, for each f ∈ F. Suppose that φ is a continuous linear functional on (E, σ (E, F)). Since φ is continuous at 0, there exist f1 , . . . , fr in F and  > 0 such that if | e, fi  | <  for 1 ≤ i ≤ r then |φ(e)| < 1. By homogeneity, if | e, fi  | = 0 for 1 ≤ i ≤ r then |φ(e)| = 0. Let T : E → Rr be defined by T(e) = (e, f1  , . . . , e, fr ). If T(e1 ) = T(e2 ) then φ(e1 ) = φ(e2 ), and so there exists a linear functional θ on T(E) such that φ(e) = θ (T(e)) for all e ∈ E. By linear algebra, we can extend θ to a linear functional ψ on Rr . There exist α1 , . . . , αr such that ψ(x1 , . . . , xr ) = α1 x1 + · · · + αr xr . If e ∈ E, then φ(e) = θ (T(e)) = ψ(T(e)) = α1 e, f1  + · · · + αr e, fr  = e, α1 f1 + · · · + αr fr  . Thus we can identify φ with α1 f1 + · · · + αr fr . Exercise 9.3.3 Show that if C is convex, then so is its σ (E, F)-closure. If A ⊆ E, let (A) denote the smallest convex set containing A; (A) is the convex cover of A. Show that (A) is the smallest σ (E, F)-closed convex set containing A. Exercise 9.3.4 Suppose that (E1 , F1 ) and (E2 , F2 ) are real or complex dual pairs and that T : E1 → E2 is linear. Let T  ( f2 )(e1 ) = T(e1 ), f2  for e1 ∈ E1 and f2 ∈ F2 . Then T is weakly continuous (that is, continuous from (E1 , σ (E1 , F1 )) to (E2 , σ (E2 , F2 ))) if and only if T  (F2 ) ⊆ F1 .

9.4 Polarity When we have duality, it is natural to consider polarity. Here we restrict attention to the real case. Suppose that (E, F ) is a real dual pair and that A ⊆ E. We define Ao = { f ∈ F : a, f  ≤ 1 for all a ∈ A} = ∩a∈ A {a}o . Similarly if B ⊆ F we define Bo ⊆ E.

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Proposition 9.4.1 Suppose that (E, F) is a real dual pair, and that A ⊆ E. Then Ao is σ (F, E)-closed, convex and contains 0. Proof For each of the sets { f ∈ F : a, f  ≤ 1} is σ (F, E)-closed, convex and contains 0. Exercise 9.4.2 Suppose that (E, F) is a real dual pair, and that A, A1 , A2 ⊆ E. Show the following. (i) If A1 ⊆ A2 then Ao1 ⊇ Ao2 . (ii) A ⊆ Aoo and Ao = Aooo . (iii) If λ is real and positive then (λA)o = (1/λ)Ao . Theorem 9.4.3 (The theorem of bipolars) Suppose that (E, F) is a real dual pair and that A ⊆ E. Then Aoo is the smallest σ (E, F)-closed convex set containing A and 0: that is, Aoo = (A ∪ {0}). Proof Certainly Aoo ⊇ (A ∪ {0}). Suppose that x ∈ (A ∪ {0}). Then there exists non-zero f1 , . . . , fn such that if M = {y : | x − y, fi  | < 1 for 1 ≤ i ≤ n}, then M is disjoint from (A∪{0}). Let T(e) = (e, fi )ni=1 . T is a linear mapping from E into Rn . Let U = {y ∈ Rn : |yi − T(x)i | < 1 for 1 ≤ i ≤ n}. Then U is a convex open subset of Rn disjoint from the convex set B = T((A ∪ {0}), and so by the separation theorem (applied to Rn , with its usual topology) there exists a linear functional φ on Rn and λ ∈ R such that φ(y) > λ for y ∈ U and φ(z) ≤ λ for z ∈ B. Since x ∈ B, λ ≥ 0. Let φ(ei ) = φi and let  ψ = ni=1 φi fi ∈ F. Then x, ψ > λ and y, ψ) ≤ λ for y ∈ (A ∪ {0}). Choose λ < μ < ψ(x), and let θ = ψ/μ. Then θ ∈ Ao . Since x, θ  > 1, x ∈ Aoo . The following corollary is a convenient consequence of the theorem of bipolars. Corollary 9.4.4 If A is a non-empty σ (E, F)-closed convex subset of E and x ∈ A then there exists f ∈ F such that x, f  > sup{a, f  : a ∈ A}. Proof Choose a0 ∈ A, and apply the theorem of bipolars to A − a0 .

9.5 Weak and Weak* Topologies for Normed Spaces When (E, .) is a normed space, we can consider the weak topology σ (E, E ) on E. There are two weak topologies on E , the topology σ (E , E), which is called the weak*-topology and the weak topology σ (E , E ). They must not be

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121

confused. How are the weak and weak* topologies related to the corresponding norm topologies? Proposition 9.5.1 The weak topology σ (E, E ) on a real normed space (E, .E ) is weaker than the norm topology on E. The following are equivalent. (i) The weak topology is the same as the norm topology on E. (ii) The restriction of the weak topology to the unit ball B(E) is the same as the restriction of the norm topology on E. (iii) E is finite-dimensional. Proof Since the norm topology and the weak topology are translation invariant, it is sufficent to consider neighbourhoods of 0. Suppose that U = {x : | x, φi | < 1, 1 ≤ i ≤ n} is a fundamental weak neighbourhood of 0. Let M = max1≤i≤n φi . If x < 1/M then x ∈ U, and so the weak topology is weaker than the norm topology. (i) certainly implies (ii). If (ii) holds, there exist φ1 , . . . , φn in E such that if x, φi  | < 1 then x < 1. Suppose that x, φi  = 0 for 1 ≤ i ≤ n. If α ∈ R then αx, φi  = 0 for 1 ≤ i ≤ n, so that |α|. x < 1. Since this holds for all α ∈ R, x = 0. Thus the linear mapping x → (x, φi )ni=1 : E → Rn is injective, and E is finite-dimensional; (ii) implies (iii). Suppose that (E, .) is finite-dimensional, and that (e1 , . . . , en ) is a basis for E, with dual basis (φ1 , . . . , φn ). Let U = {x : | x, φi  | < ei  /n for ≤ i ≤ n}.  If x ∈ U then x ≤ ni=1 | x, φi  | ei  < 1, and so the weak topology and the norm topology are the same on E; (iii) implies (i). Things can be different on the unit sphere. A real normed space (E, .) is uniformly convex if whenever  > 0 there exists  δ > 0 such that whenever 1  x = y = 1 and x − y ≥  then  2 (x + y) ≤ (1 − δ). Proposition 9.5.2 Suppose that (E, .) is a uniformly convex Banach space. If  > 0 then there exists δ > 0 such that if x = 1 and if x∗ is a continuous linear functional on E with x∗  = x, x∗  = 1 then ! # " y : y, x∗ > 1 − δ} ⊆ {y : y = 1, y − x ≤  , so that the weak topology and the norm topology coincide on the unit sphere S(E) = {x ∈ E : x = 1}. Proof Suppose that  > 0. Let δ be the quantity assured% by the definition of $ 1 ∗ uniform convexity. If y, x  ≥ 1 − δ, then 2 (x + y), x∗ ≥ 1 − δ/2. Hence x − y ≤ .

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Theorem 9.5.3 A Hilbert space H is uniformly convex. Proof Suppose that x = y = 1. Then (x + y)/22 + x − y2 = 1, by the parallelogram law, and this gives the result. We shall meet more examples of uniformly convex spaces later; in Section 15.5 we shall define the Lp spaces, for 1 < p < ∞, and show that they are uniformly convex. If (E, F) is a dual pair and B ⊆ E, we say that B is σ (E, F)-bounded or weakly bounded if {b, f  : b ∈ B} is bounded, for each f ∈ F. A set which is bounded for the weak* topology is said to be weak* bounded. Proposition 9.5.4 A subset B of the dual E of a Banach space (E, .) is weak* bounded if and only if it is norm bounded. Proof If B is norm bounded, then {b(x) : b ∈ B} is bounded for each x ∈ (E, .), since the mapping φ → φ(x) is a continuous linear functional on E ; thus B is weak* bounded. Conversely if B is weak* bounded, it is a pointwise bounded set of continuous linear functionals on (E, .), and so it is norm bounded, by the principle of uniform boundedness (Theorem 7.6.1). Corollary 9.5.5 A subset A of a [separable] normed space (E, .) is weakly bounded if and only if it is norm bounded. Proof As before, A is weakly bounded if it is norm bounded. If A is weakly bounded then j(A) is weak* bounded in E , where j is the canonical isometric embedding of (E, .) into E . Thus j(A) is norm bounded in E , and so A is norm bounded. Exercise 9.5.6 Suppose that (E1 , .1 ) and (E2 , .2 ) are [separable] real Banach spaces and that T is a linear map from (E1 , .1 ) to (E2 , .2 ). Then T is norm continuous if and only if it is weakly continuous. If (E, F) is a dual pair of vector spaces, a sequence (xn )∞ n=1 is a σ (E, F) Cauchy sequence if whenever N is a σ (E, F) neighbourhood of 0 there exists n0 such that xm − xn ∈ N for m, n ≥ n0 . That is to say, (xn )∞ n=1 is a Cauchy sequence in the uniformity defined by σ (E, F). A σ (E, F)-convergent sequence is clearly a σ (E, F) Cauchy sequence. When appropriate, we use the terms ‘weakly Cauchy’ and ‘weak* Cauchy’. Although the norm topology and the weak topology on the Banach space l1 , and on its unit ball B(l1 ), are different, the two topologies have the same Cauchy sequences and convergent sequences. Theorem 9.5.7 A sequence (y(n) )∞ n=1 in l1 is weakly Cauchy if and only if it is norm convergent.

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123

Proof A norm convergent sequence is certainly weakly Cauchy. Suppose, if possible, that (x(n) )∞ n=1 is weakly Cauchy, but not norm convergent. By choosing a subsequence if necessary, we can suppose that there exists  > 0 such that if y(n) = x2n − x2n+1 then (y(n) )∞ n=1 converges weakly to 0, and  (n)  y  ≥  for all n ∈ N. Note that y(n) → 0 as n → ∞ for all j ∈ N. j 1 We use a ‘sliding hump’ argument. Let j0 = 0 and let n1 = 1. There exists j1  (n1 ) | < /5. We now show that there are strictly increasing such that ∞ j=j1 +1 |y ∞ sequences (ni )i=1 and (ji )∞ i=1 in N such that (n )

(i) |yj i | < /5ji−1 for i > 1 and 1 ≤ j ≤ ji−1 , and ∞ (ni ) (ii) j=ji +1 |yj | < /5. Suppose that we have found nk and jk satisfying (i) and (ii), for 1 ≤ k ≤ i. By co-ordinatewise convergence, we can find ni+1 satisfying (i), and we can then find ji+1 satisfying (ii). For each i there exist φj for ji−1 < j ≤ j1 , with |φj | = 1  ji  ji (n ) (n ) such that j=j φj (yj i ) = j=j |yj i |. Then φ = (φj )∞ j=1 ∈ l∞ , and i−1 +1 i−1 +1 φ∞ = 1. Now ji 

(n ) |yj i |

ji−1 ∞      (ni )  (n ) (n ) ≥ y  − |yj i | − |yj i | ≥ 3/5, 1

j=ji−1 +1

j=ji +1

j=1

and so |φ(y(ni ) )| ≥

ji  j=ji−1 +1

(ni )

φj (yj

)−

ji−1 

(ni )

φj (yj

j=1

)−

∞ 

(ni )

φj (xj

)

j=ji +1

≥ 3/5 − /5 − /5 = /5. Thus y(ni ) does not converge weakly to 0, giving a contradiction. Exercise 9.5.8 Deduce that the unit ball B(l1 ), with the weak topology σ (l1 , l∞ ), is not metrizable. Proposition 9.5.9 A subset L of l1 is weakly compact (that is, σ (l1 , l∞ )compact) if and only if it is norm compact. Proof A norm compact set is certainly weakly compact. Suppose that L is weakly compact. The metrizable topology of co-ordinatewise convergence is weaker than the weak topology on l1 , and so is the same as the weak topology on L. Consequently, L is a compact metrizable space in the weak topology, and so is weakly sequentially compact. Thus if (xn )∞ n=1 is a sequence in L then there exists a weakly convergent subsequence (xnk )∞ k=1 . But this subsequence converges in norm, and so L is norm compact.

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Since the weak topology on a normed space is weaker than the norm topology, a weakly closed set is norm closed. For convex sets, the converse holds. Theorem 9.5.10 A norm closed convex subset B of a real [separable] normed space (E, .) is weakly closed. Proof Since translation is a homeomorphism in either topology, we can suppose that 0 ∈ B. Suppose that x ∈ B. There exists  > 0 such that U = {y : y − x < } is disjoint from B. U is open and convex, and so, by the separation theorem, there exist a continuous linear functional g on (E, .) and a real number λ such that g(y) < λ for y ∈ U and g(b) ≥ λ for b ∈ B. Since 0 ∈ B, λ ≤ 0. Choose g(x) < α < λ, and let h = g/α. Then h(b) ≤ 1 for b ∈ B, so that h ∈ Bo , while h(x) > 1, so that x ∈ Boo . Thus B = Boo , and B is weakly closed.

9.6 Banach’s Theorem and the Banach–Alaoglu Theorem Proposition 9.6.1 Suppose that G is a subset of a normed space (E, .) and that span(G) is dense in E. Then the weak* topology on B(E ) is the same as the topology of pointwise convergence on the elements of G. Proof The topology of pointwise convergence on the elements of G is weaker than the weak* topology, and is the same as the topology of pointwise convergence on the elements of span(G). Suppose that φ ∈ B(E ) and that N = {ψ ∈ B(E ) : |ψ(yi ) − φ(yi )| < 1, 1 ≤ i ≤ j} is a basic σ (E , E) neighbourhood of φ in B(E ). For each 1 ≤ i ≤ j there exists ci ∈ span(G) with ci − yi  < 1/3. Let M = {ψ ∈ B(E ) : |ψ(ci ) − φ(ci )| < 1/3, 1 ≤ i ≤ j}. M is a neighbourhood of φ in B(E ). If ψ ∈ M and 1 ≤ i ≤ j then |ψ(yi ) − φ(yi )| ≤ |ψ(yi − ci )| + |ψ(ci ) − φ(ci )| + |φ(ci − yi )| < 1/3 + 1/3 + 1/3 = 1, so that M ⊆ N. Thus the two topologies are the same. Exercise 9.6.2 If (E, .) is separable then B(E ) is metrizable in the weak* topology.

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Theorem 9.6.3 (Banach’s theorem) The unit ball B(E ) of the dual of a separable real normed space (E, .) is compact and metrizable in the weak* topology. Proof Let G be a countable dense subset of E. If φ ∈ B(E ) and g ∈ G let j(φ)g = φ(g). Then j is a homeomorphism of B(E ) into K =   g∈G ([− g , g]). K is compact, and so it is enough to show that j(B(E ))  is closed in K. An easy approximation argument shows that u ∈ j(B(E )) if (and only if) αug + βuh = uαg+βh for all α, β ∈ Q and g, h ∈ G. Thus if u ∈ K \ j(B ), there exist α, β ∈ Q and g, h ∈ G such that |(αug + βuh ) − uαg+βh | = γ > 0. Let  = γ /(|α| + |β| + 1) and let V = {v ∈ K : |vg − ug | < , |vh − uh | < , |vαg+βg − uαg+βg | < }. Then V is a neighbourhood of u in K which is disjoint from j(B(E )), and so j(B(E )) is closed in K. If we accept the axiom of choice, so that we can use Tychonoff’s theorem, we can improve Banach’s theorem. Theorem 9.6.4 (The Banach–Alaoglu theorem) If (E, .) is a normed space, and we accept the axiom of choice, then B(E ) is compact with the weak* topology σ (E , E).  Proof For then g∈E ([− g , g]) is compact, and the argument of Theorem 9.6.3 goes through. Theorem 9.6.5 A [separable] Banach space (E, .) is reflexive if and only if its unit ball B(E) is weakly compact. Proof The condition is necessary, by the Banach–Alaoglu theorem [or Banach’s theorem]. Suppose that B(E) is weakly compact. Let j : E → E be the canonical embedding. Then j(B(E)) is σ (E , E )-compact, and so σ (E , E )-closed. By the theorem of bipolars, this means that j(B(E)) = j(B(E))oo = B(E ), the unit ball of E , and so j is surjective.

9.7 The Complex Hahn–Banach Theorem The Hahn–Banach theorem is essentially a real theorem, but it can be usefully applied in the complex case. Usually this is done by considering the underlying real space ER . The complex linear functionals on a complex vector space E are

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closely related to the real linear functionals on the real space ER . If ψ is a complex linear functional on E then ψR = (ψ) is a real linear functional on ER . Proposition 9.7.1 Suppose that E is a complex vector space, and that ER is the underlying real space. If φ is a linear functional on ER and x ∈ E let j(φ)(x) = φ(x) − iφ(ix). Then j(φ) is a complex linear functional on E. If ψ is a complex linear functional on E, then j(ψR ) = ψ. Proof j(φ)(y1 ) + j(φ)(y2 ) = j(φ)(y1 + y2 ) and j(φ)(ix) = φ(ix) − iφ(−x) = i(φ(x) − i(φ((x)) = ij(φ)(x), so that j(φ) is a complex linear functional on E. If ψ(x) = a+ib, then ψ(ix) = −b+ia, so that ψR (x) = a and ψR (ix) = −b, so that j(ψR )(x) = a − i(−b) = a + ib = ψ(x).

Theorem 9.7.2 (The complex separation theorem) Suppose that (E, .) is a [separable] complex normed space, that A and B are disjoint convex sets, and that A is open. Then there exist a continuous linear functional f and a real number λ such that  a, f  > λ for a ∈ A and supb∈B  b, f  ≤ λ. Proof For we can separate A and B by a real linear functional k, and then set f (x) = k(x) − ik(ix). The analytic Hahn–Banach theorem takes the following form. Theorem 9.7.3 Suppose that p is a continuous semi-norm on a [separable] complex normed space (E, .), and that f is a linear functional on a subspace F of E for which | f (x)| ≤ p(x) for x ∈ F. Then there exists a linear functional g on E such that f (x) = g(x) for x ∈ F (g is an extension of f ) and |g(y)| ≤ p(y) for y ∈ E (control is maintained). Proof Let ER and FR be the underlying real spaces, and let fR be the real part of f . Then fR is a linear functional on FR and fR (x) ≤ p(x) for x ∈ FR . By Theorem 9.1.6, fR can be extended to a real linear functional k such that k(y) ≤ p(y) for y ∈ E. By Proposition 9.7.1, g = j(k) is a complex linear functional on E. If x ∈ F and f (x) = reiθ , then f (x) = eiθ f (e−iθ x) = eiθ k(e−iθ x) = eiθ g(e−iθ x) = g(x),

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127

so that g extends f . If y ∈ E and g(y) = reiθ then |g(y)| = r = g(e−iθ y) = k(e−iθ y) ≤ p(e−iθ y) = p(y), so that control is maintained. If (E, F) is a complex dual pair and A ⊆ E, we set Ao = { f ∈ F :  a, f  ≤ 1 for all a ∈ A}. Then all the results corresponding to those of Sections 9.4–9.6 go through.

10 Convex Functions

Exercise 10.0.1 Suppose that f is a real-valued convex function on an open interval (a, b) of R. Show the following. (i) f is continuous. (ii) If x ∈ (a, b) then the right derivative f+ (x) = limt 0 ( f (x + t) − f (x))/t and the left derivative f− (x) = limt 0 ( f (x) − f (x − t))/t exist and are finite. (iii) The functions f+ and f− are increasing functions on (a, b). f+ is right continuous and f− is left continuous. (iv) If a < x < y < b then f− (x) ≤ f+ (x) ≤ f− (y) ≤ f+ (y). (v) The set J = {x ∈ (a, b) : f+ (x) = f− (x)} is countable. If x ∈ J then f+ and f− have a jump discontinuity at x, of size f+ (x) − f− (x). If x ∈ J then f+ and f− are continuous at x, and f is differentiable at x, with derivative f+ (x) (= f− (x)). Our aim is to see how far these results extend to convex functions on a vector space, and in particular to convex functions on a separable Banach space. We begin by considering a dual pair of real vector spaces.

10.1 Convex Envelopes Suppose that (E, F) is a real dual pair. We denote the vector space of σ (E, F)continuous affine functions on E (functions of the form a(x) = x, φ + c, where φ ∈ F and c ∈ R) by A(E). Suppose that f is a proper function on E taking values in (−∞, ∞]. We set L( f ) = {a ∈ A(E), a ≤ f }. If L( f ) = ∅, we set f = sup{a : a ∈ L( f )}. (If L( f ) = ∅, we set f = −∞.) The function f , which is convex, is the lower convex envelope of f . Note that if L( f ) is not empty and a ∈ L( f ) then g = f − a ≥ 0 and f = g + a, so that in many circumstances, we can consider the case where f ≥ 0. 128

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Exercise 10.1.1 Suppose that (E, F) is a dual pair, and that f and g are functions on E taking values in (−∞, ∞]. Show the following. (i) f is convex and σ (E, F) lower semi-continuous. (ii) f + g ≥ f + g, with equality if f or g is affine, and rf = rf for r ≥ 0. (iii) If f is bounded below then f ≥ infx∈E f (x). Theorem 10.1.2 Suppose that (E, F) is a dual pair, and that f and g are functions on E taking values in (−∞, ∞]. (i) f ≤ f , and f = f if and only if f is convex and σ (E, F) lower semicontinuous. (ii) If (E, .) is a normed space and F = E then f is lower semi-continuous in the norm topology. f = f if and only if f is convex and lower semicontinuous in the norm topology. (iii) If f and g are bounded then |f − g| ≤ f − g∞ . Proof (i) If f = f then f is convex and lower semi-continuous. Suppose that f is convex and lower semi-continuous. Then the epigraph Af = {( f (x), r) : f (x) ≤ r} is a σ (E, F)-closed convex subset of E × R. Suppose that s < f (x). By Corollary 9.4.4, there exists (φ, t) ∈ F × R+ such that φ(x) + ts < λ = inf{φ(y) + tr : (y, r) ∈ Af }. Let ψ = φ/t and let μ = λ/t, so that ψ(x) + s < μ = inf{ψ(y) + r : (y, r) ∈ S}. Let a(y) = ψ(x) − ψ(y) + s :then a ∈ A(K), and a(x) = s. Further, a(y) ≤ ψ(x) − (ψ(y) + f (y)) + f (y) + s ≤ ψ(x) − μ + f (y) + s ≤ f (y), and so f (x) ≥ s. Since this holds for all s < f (x), f (x) ≥ f (x). (ii) Since f is the supremum of norm-continuous affine functions, it is lower semi-continuous in the norm topology. If f is convex and lower semicontinuous in the norm topology then Af is a convex norm closed subset of E × R, and so it is weakly closed. Thus f is σ (E, E )-lower semi-continuous, and so f = f by (ii). (iii) f = ( f − g) + g ≥ f − g + g, so that g − f ≤ −(f − g) = inf (g(x) − f (x)) ≤  f − g∞ , x∈E

and similarly f − g ≤  f − g∞ . Thus | f − g| ≤  f − g∞ . Corollary 10.1.3 −f ≥ (−f ). A proper σ (E, F)-lower semi-continuous convex function on E is called a regular convex function. This is unfortunate terminology, since the word ‘regular’ is much overused.

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The upper concave envelope f of a function in E taking values in [−∞, ∞) is defined similarly. We set U( f ) = {a ∈ A(E) : a ≥ f }. If U( f ) = ∅, we set f = inf{a : a ∈ U( f )}. (If L( f ) = ∅, we set f = ∞.) Then f is a σ (E, F)-upper semi-continuous convex function on E, which satisfies results corresponding to those of Theorem 10.1.2. Clearly, f = −(−f ). We shall consider the lower convex envelope further in Chapter 11.

10.2 Continuous Convex Functions We now consider convex functions on a normed space. When are they continuous? If f is a convex function on a normed space, we say that f is very regular if int f is non-empty (f is finite on a non-empty open set) and f is int continuous on f . Theorem 10.2.1 Suppose that f is a convex function on a normed space (E, .). The following are equivalent. (i) f is very regular. int int (ii) The graph f ∩ (int f × R) of f in f is closed in f × R. int (iii) int f is non-empty, and there is a non-empty open subset U of f on which f is bounded above. (iv) The strict epigraph Sfint is non-empty and Sfint = {(x, t) : x ∈ int f , t > f (x)}. int (v) Sfint is non-empty and Sf ∩ (int f × R) is open in f × R.

Proof If g is any continuous function, then its graph is closed and its strict epigraph is open, and so (i) implies (ii) and (v). int int Suppose that f ∩ (int f × R) is closed in f × R and that x ∈ f . int Suppose that  > 0. There exists a δ > 0 such that Nδ (x) ⊆ f , and such that (Nδ (x)×(, ∞))∩f = ∅. Thus f is bounded above on Nδ (x); (ii) implies (iii). Suppose that (iii) holds. By translation, and addition of a constant if necessary, we can suppose that 0 ∈ int f , and that f (y) < 0 for y ∈ U, so that Sf is a convex set containing the origin (0, 0) of E × R, and we can consider its gauge pSf . Since U × (0, ∞) is an open subset of Sf , Sfint is non-empty. Certainly, Sfint ⊆ {(x, t) : x ∈ int f , t > f (x)}. We need to prove the converse int inclusion. Suppose that x ∈ f and that t > f (x). There exists λ > 1 such that [0, λx) ⊆ int f . Let fx (s) = f (sx), for s ∈ [0, λ]; fx is a convex function on [0, λ], and fx (s) ≤ (1 − s)f (0) + sf (x) < st, for s ∈ [0, 1]. Since fx is a

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continuous function on [0, λ], there exists 1 < μ ≤ λ such that fx (s) < st for s ∈ [0, μ]; that is, (sx, st) ∈ Sf for 0 ≤ s ≤ μ. Thus pSf (x, t) < 1, and so int (x, t) ∈ Sfint . Thus {(x, t) : x ∈ int f , t > f (x)} ⊆ Sf , and (iv) holds. Clearly, (iv) implies (v). Suppose that (v) holds, and that x ∈ int f . We can suppose that x = 0 and that f (x) = 0. There exists 0 < δ < 1 such that Uδ = {(y, μ) : y < δ, |μ − 1| < δ} ⊆ Sf . Suppose that 0 <  < 1. Let η = δ. If z < η, let y = z/, so that y < δ and f (y) < 1. Hence f (z) ≤ f (y) < . Similarly f (−z) < , so that f (z) ≥ −f (−z) > −; f is continuous at x, and (i) holds. Corollary 10.2.2 Suppose that f is a convex function on a normed space int (E, .). Then int f is non-empty and f is continuous on f if and only if f is upper semi-continuous at a point of f . is non-empty, and f is continuous at a point of int then f is If int f f . continuous on int f Proposition 10.2.3 Suppose that f is a very regular convex function on a int normed space (E, .). Then f is locally Lipschitz on int f ; if x ∈ f , there exists δ > 0 such that f is 2/δ-Lipschitz on x + δU(E). Proof There exists δ > 0 such that x + 2δU(E) ⊆ int f and |f (y) − f (x)| ≤ 1  are distinct elements of x + δU(E). If on x + 2δU(E). Suppose that y and y      y − y ≥ δ then f (y ) − f (y) ≤ 2 ≤ (2/) y − y. Otherwise, let z =    y + δ/ y − y (y − y). Then z − y = δ, and z − x < 2δ. By convexity, f (y ) − f (y) f (z) − f (y) |f (z) − f (x)| + |f (y) − f (x)| 2 ≤ ≤ ≤ .  y − y z − y z − y δ     Similarly, f (y) − f (y ) ≤ (2R/δ) y − y. When (E, .) is a Banach space, things work well. The following result is related to the principle of uniform boundedness (Theorem 7.6.1). Theorem 10.2.4 Suppose that f is a convex function on a Banach space (E, .) which is lower semi-continuous on a non-empty open convex subset U of int f . Then f is very regular. Proof Suppose that x ∈ U. As before, we can suppose that x = 0 and that f (x) = 0. There exists δ > 0 such that V = {y : y < δ} ⊆ U. If y ∈ V let g(y) = max( f (y), f (−y)). Then g is a non-negative lower semi-continuous real-valued convex function on V, g(0) = 0 and f (y) ≤ g(y). If n ∈ N, let Vn = {y ∈ V : g(y) ≤ n}. Since g is lower semi-continuous, Vn is a closed subset

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of V. Since g is real-valued, V = ∪∞ n=1 Vn . But V is topologically complete, and so, by Baire’s category theorem, there exists n ∈ N such that Vnint is not empty. But Vnint is convex, and if z ∈ Vnint then −z ∈ Vnint , and so 0 ∈ Vnint . Thus there exists 0 < η < δ such that ηU(E) ⊆ Vn ; that is, if y < η then 0 ≤ g(y) ≤ n. Suppose that  > 0. Since g(λy) ≤ λg(y) for 0 ≤ λ ≤ 1, if y < η/n then 0 ≤ g(y) < ; g is continuous at 0. Since f (x) ≤ g(x) for x ∈ V, f is bounded above on an open subset W of V, and so f is continuous on int f , by Theorem 10.2.1. Corollary 10.2.5 Suppose that f is a convex function on a Banach space (E, .) and that U is a non-empty open subset of f . Then f is very regular if and only if the epigraph Af ∩ (U × R) of f in U is closed in U × R. Corollary 10.2.6 Suppose that F is a set of convex functions on a Banach space (E, .) which are real-valued and continuous on an open convex subset U of E, and that g(x) = supf ∈F f (x) is finite, for each x ∈ U. Then g is continuous on U. Proof For g is lower semi-continuous.

11 Subdifferentials and the Legendre Transform

11.1 Differentials and Subdifferentials We need several definitions, generalizing what happens in one dimension. Suppose that f is a real-valued function defined on a radially open subset U of a real vector space E, that x ∈ U and that h is a non-zero element of E. Then f has a directional derivative d+ fx (h) at x if f (x + th) − f (x) 0. → d+ fx (h) as t t f is Gˆateaux differentiable at x, with Gˆateaux derivative Dfx , if there is a linear functional Dfx on E for which f (x + th) − f (x) 0 for each h ∈ E \ {0}. → Dfx (h), as t t Clearly, if f is Gˆateaux differentiable at x then it has directional derivatives in every direction, but the converse is not true. Suppose now that f is a real-valued proper convex function on a vector space E, and that x ∈ f . The subdifferential ∂∨ f (x) is the set of linear functionals on E for which f (x + h) − f (x) − φ(h) ≥ 0 for all h ∈ E. So far, we have not considered the case where there is a topology on E. Suppose that (E, .) is a [separable] normed space. Since the epigraph Af of a convex function f is closed in the norm topology of E × R if and only if it is weakly closed, a regular convex function is lower semi-continuous in the norm topology. Exercise 11.1.1 Suppose that f is a regular convex function on a normed space (E, .) and that x ∈ int f . Show that f has directional derivatives in all directions at x. 133

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Proposition 11.1.2 If f is a regular convex function on a normed space (E, .)  and x ∈ int f , then ∂∨ f (x) is a weak*-closed convex subset of E . Proof Let g(h) = f (x + h) − f (x). Then g(0) = 0, and g is lower semicontinuous, so there exists a neighbourhood N of 0, with N ⊆ int f such that g(h) > −1 for h ∈ N. If φ is a discontinuous linear functional on E, then φ(N) = R, so that φ ∈ ∂∨ f (x). Thus ∂∨ f (x) ⊆ E . It is then clearly weak* closed and convex. Proposition 11.1.3 Suppose that f is a regular convex function on a [separaateaux differentiable at x ble] normed space (E, .) and that x ∈ int f . f is Gˆ if and only if ∂∨ f (x) is a singleton. Proof If f is Gˆateaux differentiable at x, then g(h) = f (x+h)−f (x)−Dfx (h) is a int non-negative Gˆateaux differentiable function on int g = f −x, with g(0) = 0 and Dg0 = 0. Clearly ∂∨ g(0) = {0}, which shows that ∂∨ f (x) is a singleton. Conversely, suppose that ∂∨ f (x) = {φ}. Let g(h) = f (x + h) − f (x) − φ(h); g is a non-negative convex function, and g(0) = 0. Suppose that the directional derivative d+ g(h0 ) = 0 for some h0 ; then d+ g(h0 ) > 0. Define a linear functional on span(h0 ) by setting ψ(λh0 ) = λd+ g(h0 ). Then ψ(λh0 ) ≤ g(λh0 ); extend by the Hahn–Banach theorem to a linear functional ψ dominated by g. But then ∂∨ g(0) is not a singleton, and neither is ∂∨ f (x).

11.2 The Legendre Transform Suppose that (E, F) is a real dual pair, and that f is a proper function on E taking values in (−∞, ∞], and with the property that there exists a0 ∈ A(E) such that f ≥ a0 . Then the Legendre transform f ∗ is the function on F defined as f ∗ (φ) = sup(x, φ − f (x)). x∈E

f ∗ (φ)

= supx∈f (x, φ − f (x)), and if x0 ∈ f then f ∗ (φ) ≥ x0 , φ − Clearly f (x0 ), so that f ∗ takes values in (−∞, ∞]. Suppose that a0 (x) = x, φ0  + c0 . Then f ∗ (φ0 ) ≤ −c0 , so that f ∗ is a proper function. Since it is the supremum of a set of σ (F, E)-continuous affine functions, it is a regular convex function on F, when F is given the topology σ (F, E). Exercise 11.2.1 Suppose that (E, F) is a real dual pair. Establish the following. (i) The Legendre transform is order reversing; if f ≤ g, then f ∗ ≥ g∗ . (ii) 0 ∈ f ∗ if and only if f is bounded below.

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If g(x) = f (x) + c then g∗ (φ) = f ∗ (φ) − c. If g(x) = f (x + c) then g∗ (φ) = f ∗ (φ) − c, φ. If g(x) = αf (x), where α > 0, then g∗ (φ) = αf ∗ (φ/α). If g(x) = f (βx), where β > 0, then g∗ (φ) = f ∗ (φ)/β. If g(x) = f (−x), then g∗ (φ) = f ∗ (−φ); in particular, if f is an even function, then so is f ∗ . (viii) If (x, φ) ∈ E × F, then f ∗ (φ) + f (x) ≥ x, φ. (iii) (iv) (v) (vi) (vii)

In the same way, if g is a proper function on F taking values in (−∞, ∞], and with the property that there exists a0 ∈ A(F) such that g ≥ a0 , we define the Legendre transform of g as g∗ (x) = sup(x, φ − g(φ)). φ∈F

We can therefore construct the function f ∗∗ on E. Theorem 11.2.2 Suppose that (E, F) is a real dual pair, and that f is a proper function on E taking values in (−∞, ∞], and with the property that there exists a0 ∈ A(E) such that f ≥ a0 . Then f ∗∗ = f , the lower convex envelope of f . Proof If a ∈ A(E) and a(x) = x, φ + c for x ∈ E, then a ≤ f if and only if −c ≥ x, φ − f (x) for all x ∈ E; that is, if and only if −c ≥ f ∗ (φ). Thus f (x) = sup{x, φ + c : −f ∗ (φ) ≥ c} = sup{x, φ − f ∗ (φ) : φ ∈ F} = f ∗∗ (x). Exercise 11.2.3 Show that f is a regular convex function on E if and only if f = f ∗∗ . This clearly corresponds to the theorem of bipolars, for convex sets. Theorem 11.2.4 Suppose that (E, F) is a real dual pair, that f is a regular convex function on E, that x ∈ f and that φ ∈ f ∗ . Then the following are equivalent: (i) φ ∈ ∂∨ f (x); (ii) x ∈ ∂∨ f ∗ (φ); (iii) f (x) + f ∗ (x) = x, φ. Proof φ ∈ ∂∨ f (x) if and only if f (y) − f (x) ≥ y − x, φ for all y ∈ E if and only if y, φ − f (y) ≤ x, φ − f (x) for all y ∈ E if and only if f ∗ (φ) + f (x) ≤ x, φ .

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Since f ∗ (y) + f (x) ≥ x, φ, (i) and (iii) are equivalent, and similarly (ii) and (iii) are equivalent. Exercise 11.2.5 Show that ∂∨ f ∗ is the transpose of ∂∨ f : (φ, x) ∈ ∂∨ f ∗ if and only if (x, φ) ∈ ∂∨ f . It is convenient to consider concave functions in a similar way. If f is a proper function on E taking values in [−∞, ∞), and with the property that there exists a0 ∈ A(E) such that f ≤ a0 . we define the concave Legendre transform f † of f to be f † (φ) = inf (x, φ − f (x)). x∈E

Proposition 11.2.6 Suppose that f is a proper function on E taking values in [−∞, ∞), and with the property that there exists a0 ∈ A(E) such that f ≤ a0 , Let S( f )(x) = −f (−x). Then f † (φ) = −(S( f ))∗ (φ). Proof For f † (φ) = − sup(− x, φ + f (x)) = − sup(−x, φ − (−f )(x)) x∈E

x∈E

= − sup(x, φ − (−f )(−x)) = −(S( f ))∗ (φ). x∈E

Thus f † is a proper concave function, and (S( f ))† = −(S2 ( f ))∗ = −f ∗ . Arguing as in Theorem 11.2.2, if a ∈ A(E) and a(x) = x, φ + c for x ∈ E, then a ≥ f if and only if −c ≤ x, φ − f (x) for all x ∈ E; that is, if and only if −c ≤ f † (φ). Thus f (x) = inf{x, φ + c : −f † (φ) ≤ c} = inf{x, φ − f † (φ) : φ ∈ F} = f †† (x), and other properties of f † can be derived from corresponding properties of the Legendre transform. For example, we have the following. Theorem 11.2.7 Suppose that (E, F) is a real dual pair. If f is a regular concave function on E, if x ∈ f and if φ ∈ f † , then the following are equivalent: (i) φ ∈ ∂ ∧ f (x); (ii) x ∈ ∂ ∧ f † (φ); (iii) f (x) + f † (x) = x, φ.

11.3 Some Examples of Legendre Transforms

137

11.3 Some Examples of Legendre Transforms Let us give some examples. Suppose first that f = 0C , where C is a σ (E, F)closed convex subset of E; f (x) = 0 if x ∈ C, and f (x) = ∞ otherwise. Then f ∗ (φ) = supx∈C x, φ. This is a regular convex function on F, since f ∗ (0) = 0. If 0 ∈ C of E then f ∗ (φ) = sup{x, φ : x ∈ C} = pC◦ (φ, ) where pC◦ is the gauge of C◦ . In particular, if E is a normed space, F = E and C = B(E) then f ∗ (φ) = φ . Similarly if C = B(E ) then f ∗ (x) = x. Exercise 11.3.1 Show directly that if (E, .) is a normed space, and n1 (x) = x then n∗1 = 0B(E ) . Suppose that (E, F) is a real dual pair and that f is a regular convex function on E. How do we calculate the Legendre transform of f ? The function x, φ − f (x) is a regular concave function. If it attains its supremum at x, then Theorem 11.2.4 implies that f ∗ (φ) ∈ ∂∨ f (x), and that f ∗ (φ) = x, φ − f (x). We use this to calculate the Legendre transform of convex functions on R, and concave Legendre transforms of concave functions. Here are some examples. (a) Suppose that 1 < p < ∞ and that q = p/(p − 1), so that 1/p + 1/q = 1 and (p − 1)(q − 1) = 1. Let f (x) = |x|p /p. If s > 0 then sx − |x|p /p attains its p−1 supremum at x0 , where s = x0 . Thus x0 = sq−1 and f ∗ (s) = s.sq−1 − s(q−1)p /p = sq /q. If s < 0 then f ∗ (s) = |s|q /q. (b) Suppose that 0 < p < 1 and that 1/p + 1/q = 1 (so that q < 0). Let  p x /p for x > 0, f (x) = −∞ for x ≤ 0. Then f is a regular concave function on R. If t ≤ 0 then f † (t) = −∞. If t > 0 then tx − f (x) attains its infimum when x = tq−1 and then f † (t) = tq /q = −tq /|q|. Thus  −∞ for t ≤ 0, f † (t) = q q t /q = −t /|q| for t > 0. (c) Suppose that r > 0 and that 1/s − 1/r = 1 (so that 0 < s < 1). Let  −r x /r for x > 0, f (x) = ∞ for x < 0.

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Then it follows from the previous example that 

∗

f (t) =

∞ −|t|s /s

for t ≥ 0, for t < 0.

Exercise 11.3.2 Verify the details of the next three examples. (d) Let f (x) = ex−1 . Then ⎧ ⎨ t log t f ∗ (t) = 0 ⎩ ∞

for t > 0, for t = 0, for t < 0.

⎧ ⎨ x log x f (x) = 0 ⎩ ∞

for x > 0, for x = 0, for x < 0.

(e) Let

Then f ∗ (t) = et−1 . ( f) Let f (x) = (|x| + 1) log(|x| + 1). Then ∗



f (t) =

0 e|t|−1 − |t|

for |t| ≤ 1, for |t| > 1.

We can apply results concerning convex functions on R to certain convex functions on a normed space. Proposition 11.3.3 Suppose that f is a non-negative convex function on [0, ∞) for which f (t) = 0 if and only if t = 0, that (E, .) is a normed space and that F(x) = f (x) for x ∈ E. Then F ∗ (φ) = f ∗ (φ ) for φ ∈ E . Proof Since sup{φ(x) : x = α} = α φ , F ∗ (φ) = sup(sup{φ(x) − f (α) : x = α}) α>0

= sup(α φ − f (α)) = f ∗ (φ ). α>0

If (E, .) is a normed space, and 1 ≤ p < ∞, we set np (x) = xp /p for x ∈ E and np (φ) = (φ )p /p for φ ∈ E ; np is a continuous convex function on E and np is a continuous convex function on E . Corollary 11.3.4 If p > 1 then n∗p = nq , where 1/p + 1/q = 1.

11.4 The Episum

139

11.4 The Episum What about the Legendre transform of a sum? Suppose that f and g are proper functions on E, each of which is bounded below. The episum or inf-convolution f ! g of f and g is defined as f ! g(x) = inf{f (y) + g(z) : y + z = x} = inf {f (y) + g(x − y)}. y∈E

Note that f ! g = g ! f and that f !g = f + g . Proposition 11.4.1 If (E, F) is a real dual pair and f and g are regular convex functions on E, each of which is bounded below, then f ! g is convex. Proof Suppose that x0 , x1 ∈ f !g , that 0 ≤ λ ≤ 1 and that  > 0. There exists y0 , y1 ∈ f such that f (y0 ) + g(x0 − y0 ) < f ! g(x0 ) + /2 and f (y1 ) + g(x1 − y1 ) < f ! g(x1 ) + /2. Let xλ = (1 − λ)x0 + λx1 and yλ = (1 − λ)y0 + λy1 . Then f (yλ ) + g(xλ − yλ ) ≤ ((1 − λ)f (y0 ) + λf (y1 )) + ((1 − λ)g(x0 − y0 ) + λg(x1 − y1 )) ≤ (1 − λ)f ! g(x0 ) + λf ! g(x1 ) + . Since  is arbitrary, the result follows. On the other hand, f ! g need not be lower semi-continuous. If f is a regular convex function on E which is bounded below, and if C is closed convex subset of E, then f ! 0C (x) = inf{f (y) : y ∈ x − C} for x ∈ f + C. In particular, if C and D are closed convex subsets of E, then 0C ! 0D = 0C+D . But C + D need not be closed, as the following example shows; it shows that the sum of two closed linear subspaces of a Hilbert space need not be closed. If x ∈ l2 , let T(x) = (xn /n)∞ n=1 . Then T is a bounded linear operator on l2 , whose image T(l2 ) is a dense subspace of l2 which is not closed in l2 . Let H = l2 ⊕l2 , let F = {(x, y) ∈ H : y = 0} and let G = T = {(x, T(x)) : x ∈ l2 }. Then F and G are closed linear subspaces of H, and F + G = l2 ⊕ T(l2 ) is a proper dense subspace of H. Why is the episum important? Proposition 11.4.2 Suppose that (E, F) is a real dual pair and that f and g are proper functions on E, each of which is bounded below. Then ( f ! g)∗ = f ∗ + g∗ .

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Proof If φ ∈ F then

 ( f ! g) (φ) = sup x, φ − inf ( f (y) + g(x − y)) 

∗

x∈E

y∈E

= sup (x, φ − f (y) − g(x − y)) x,y∈E

= sup ((y, φ − f (y)) + (x − y, φ − g(x − y))) x,y∈E

= f ∗ (φ) + g∗ (φ).

Corollary 11.4.3 Suppose that f and g are regular convex functions on E, and that 0 ∈ f ∩ g . Then f + g is a regular convex function on E and ( f + g)∗ = ( f ∗ ! g∗ )∗∗ . Proof The function f + g is certainly a regular convex function on E. The condition implies that if φ ∈ F then f ∗ (φ) ≥ 0, φ − f (0) = −f (0) and similarly g∗ (φ) ≥ −g(0), so that f ∗ ! g∗ is defined. Then ( f ∗ ! g∗ )∗ = f ∗∗ + g∗∗ = f + g, and ( f + g)∗ = ( f ∗∗ + g∗∗ )∗ = ( f ∗ ! g∗ )∗∗ . In fact, the episum is most useful when E is a normed space and g is a very regular convex function, as we shall see in the next section (Proposition 11.5.5).

11.5 The Subdifferential of a Very Regular Convex Function Theorem 11.5.1 Suppose that f is a very regular convex function on a [separable] normed space (E, .) and that x ∈ int f . Then the subdifferential ∂∨ fx at x is a non-empty convex weak*-compact subset of E . If x ∈ int f and h ∈ E then d+ fx (h) = sup{φ(h) : φ ∈ ∂∨ fx }, and the supremum is attained. Proof By Proposition 10.2.3, the function f is locally Lipschitz, and so there exist δ > 0 and M > 0 such that |f (x + h) − f (x)| ≤ M h for h < δ. Consequently, d+ fx (h) = lim( f (x + th) − f (x))/t ≤ M h for h = 0, t

0

and so d+ fx is a sublinear functional on E, and is continuous, by Proposition 7.4.6. If h ∈ E, then by the Hahn–Banach theorem there exists a linear functional φh on E such that φh (h) = d+ fx (h) and φh (k) ≤ d+ fx (k) for k ∈ E. Since d+ fx

11.5 The Subdifferential of a Very Regular Convex Function

141

is continuous, so is φh , and so φh ∈ ∂∨ fx ; thus ∂∨ fx is a non-empty convex set. Also, d+ fx (h) = sup{φ(h) : φ ∈ ∂∨ fx }, and the supremum is attained. If φ ∈ ∂∨ fx and k ∈ E, then φ(k) ≤ d+ fx (k) ≤ M k so that sup{φ : φ ∈ ∂∨ fx } ≤ M: ∂∨ fx is bounded in E . Since it is weak* closed, it is therefore weak* compact, by Banach’s theorem. Corollary 11.5.2 The function f is determined, up to a constant, on int f by ∂∨ f . int Proof Suppose that x0 ∈ int f . If x1 is another element of f , let h =  x1 −x0  ∂∨ f + (t) dt. (x1 − x0 )/ x1 − x0 . Then f (x1 ) = f (x0 ) + 0

Suppose that A is a closed convex subset of a normed space (E, .) with non-empty interior Aint . A is strictly convex if whenever x, y are distinct points of A and 0 < λ < 1 then (1 − λ)x + λy ∈ Aint . Thus A is strictly convex if and only if the boundary ∂A contains no non-trivial line segments. A Banach space (E, .) is strictly convex if its unit ball B(E) is strictly convex; it is easy to see that a uniformly convex Banach space is strictly convex. A very regular convex function f on a Banach space (E, .) is said to be strictly convex if its epigraph Af is strictly convex; that is, if x, y are distinct points of f and 0 < λ < 1 then f ((1 − λ)x + λy) < (1 − λ)f (x) + λf (y). Proposition 11.5.3 Suppose that f is a strictly convex very regular function on a normed space (E, .), and that x, y are distinct points of f . Then ∂∨ f (x) and ∂∨ f (y) are disjoint. Proof Suppose if possible that φ ∈ ∂∨ f (x) ∩ ∂∨ f (y). If 0 < λ < 1 then f ((1 − λ)x + λy) ≥ f (x) + φ(λ(y − x)) and f ((1 − λ)x + λy) ≥ f (y) + φ((1 − λ)(x − y)), so that f ((1 − λ)x + λy) ≥ (1 − λ)f (x) + λf (y), giving a contradiction. Proposition 11.5.4 Suppose that f is a very regular convex function on a normed space (E, .). Then the function (x, h) → d+ fx (h) is upper semicontinuous on int f × E. + Proof Suppose that (x, h) ∈ int f × E, and that μ > d fx (h). There exists t > 0 such that ( f (x + th) − f (x))/t < μ. Since f is continuous there exists a neighbourhood N of (x, h) in int f × E such that

( f (y + tk) − f (y))/t < μ for (y, k) ∈ N. Then d+ fy (k) ≤ ( f (y + tk) − f (y))/t < μ for (y, k) ∈ N.

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In fact, we can improve the result of Theorem 11.5.1. The episum of a regular convex function and a very regular function is well-behaved. Proposition 11.5.5 If f is a regular convex function on a normed space (E, .) and g is a very regular function on E, each of which is bounded below, then f ! g is very regular. Thus ( f + g)∗ = f ∗ ! g∗ . Proof For int f !g = ∅, and f ! g is the infimum of continuous functions, and is therefore upper semi-continuous on int f !g . But it is convex, and so it is int continuous on f !g , by Corollary 10.2.2. Thus f ! g is very regular, and so ( f ∗ ! g∗ ) = ( f ∗ ! g∗ )∗∗ . Theorem 11.5.6 If f and g are regular convex functions on a [separable] Banach space (E, .) and ∂∨ f = ∂∨ g then there is a constant c such that f = g + c. Proof Recall that n2 (x) = 12 x2 , for x ∈ E. Then n2 is a very regular convex function on E, and n∗2 (φ) = n2 (φ), for φ ∈ E , by Corollary 11.3.4. Let f1 = f + n2 and g1 = g + n2 . Then ∂∨ f1 = ∂∨ f + ∂∨ n2 = ∂∨ g + ∂∨ n2 = ∂∨ g1 . Thus ∂∨ f1∗ = ∂∨ g∗ , by Theorem 11.2.4. But f1∗ = f ∗ ! n∗2 is continuous, and so is g∗1 . Thus it follows from Corollary 11.5.2 that there is a constant c such that f1∗ = g∗1 − c. But then f1 = f1∗∗ = g∗∗ 1 + c = g1 + c, and so f = g + c. A very regular convex function f on a separable Banach space is Gˆateaux differentiable at many points of int f . Theorem 11.5.7 (Mazur) Suppose that f is a very regular convex function on a real separable Banach space (E, .). Let G = {x ∈ φfint : f is Gˆateaux differentiable at x}. Then G is a dense Gδ subset of int f . Proof Let (xn ) be a dense subset of E. Let Gn,m = {x ∈ int f : φ(xn ) − ψ(xn ) < 1/m for φ, ψ ∈ ∂∨ fx }, and let Bn,m = int f \ Gn,m . Then G ⊆ ∩m,n Gm,n . On the other hand, if x ∈ ∩m,n Gm,n , then φ(xn ) = ψ(xn ) for φ, ψ ∈ ∂∨ fx , and so φ = ψ; thus x ∈ G. int We show that each Bn,m is closed in int f and that Gn,m is dense in f . It then follows from Baire’s category theorem that G = ∩m,n Gm,n is a dense Gδ subset of int f .

11.6 Smoothness

143

Suppose that yk ∈ Bn,m and that yk → y ∈ int f as k → ∞ . There exists a neighbourhood N of y such that f is a Lipschitz function on N, and so there exist k0 and M > 0 such that ∂∨ fyk ⊆ MB(E ) for k ≥ k0 . For each k, there exist φk , ψk in ∂∨ fyk such that φk (xn ) − ψk (yn ) ≥ 1/m. Since MB(E ) is weak* compact and metrizable, by extracting a subsequence if necessary we can suppose that there exist φ, ψ such that φk → φ and ψk → ψ in the weak* topology. Suppose that z ∈ E. Since the sequence (φk )∞ k=1 is norm bounded, φk (z − yk ) − φk (z − y) − φk (y − yk ) → 0 as k → ∞, and so φ(z − y) = lim φk (z − y) = lim φk (z − yk ) k→∞

k→∞

≤ lim ( f (z) − f (yk )) = f (z) − f (y), k→∞

so that φ ∈ ∂∨ fy ; similarly ψ ∈ ∂∨ fy . Further, φ(xn ) − ψ(xn ) = lim (φk (xn ) − ψk (xn )) ≥ 1/n, k→∞

so that y ∈ Bn,m . Thus Bn,m is closed in int f . int Next we show that Gn,m is dense in int f . Suppose that x ∈ f and that  > of 0. Let fn (t) = f (x + txn ). Then fn is a convex function   t in a neighbourhood of 0, and so there exists x = x + t xn with x − x <  such that fn is differentiable at t . Suppose now that φ, ψ ∈ ∂∨ fx . Then φ(sxn ) = sfn (t ) = ψ(sxn ) for s ∈ R, and so φ(xn ) = ψ(xn ); thus x ∈ Gn,m .

11.6 Smoothness Besides considering Gˆateaux differentiability, we can also consider Fr´echet differentiability. A real-valued function is Fr´echet differentiable at a point x of echet derivative Df (x) if int f , with Fr´ f (x + h) = f (x) + Df (x)(h) + r(h), where h ∈ f , Df (x) ∈ E and r(h) / h → 0 as h → 0. If E is finite dimensional and f is convex (or concave), then f is Fr´echet differentiable at x if and only if it is Gˆateaux differentiable at x. In the case where E is a Hilbert space H, we denote the Fr´echet derivative by ∇fx , and call it the gradient or grad of f at x.

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Subdifferentials and the Legendre Transform

A very regular convex function f is Gˆateaux smooth if it is Gˆateaux echet smooth if it is Fr´echet differentiable at each point of int f , and is Fr´ int differentiable at each point of f . Theorem 11.6.1 If f is a Gˆateaux smooth very proper convex function on a normed space (E, .), then the function (x, h) → Dfx (h) is continuous on int f × E. Proof Since Dfx (h) = d+ fx (h), the function is upper semi-continuous, by Proposition 11.5.4. But Dfx (h) = −Dfx+ (−h), and so it is also lower semicontinuous. Corollary 11.6.2 If f is a Gˆateaux smooth very proper convex function on a normed space (E, .), then Df is a continuous function from int f , with the norm topology, to E , with the weak* topology. Theorem 11.6.3 If f is a Gˆateaux smooth very proper convex function on a normed space (E, .), then f is Fr´echet smooth if and only if Df is a  continuous function from int f , with the norm topology, to E , with the dual norm topology. Proof Suppose first that Df is norm-to-norm continuous. Suppose that x ∈ φfint and that  > 0. There exists δ > 0 such that if x − y < δ then y ∈ int f and   Dfy − Dfx  < . For such y, Dfx (y − x) ≤ f (y) − f (x) and Dfy (x − y) ≤ f (x) − f (y), so that 0 ≤ f (y) − f (x) − Dfx (y − x) ≤ (Dfy − Dfx )(y − x)   ≤ Dfy − Dfx  . y − x ≤  y − x , and so f is Fr´echet smooth. Conversely suppose that f is Fr´echet smooth, and suppose if possible that Df is not continuous at x ∈ int f . Thus there exists 0 < α < 1      such that limr 0 (supy∈Nr (x) Dfy − Dfx ) > α. Since f is locally Lipschitz, Df is locally bounded, and so  there  exist δ > 0 and L ≥ 1 such that Dfy  ≤ L. There exists 0 < η < δ such and sup Nδ (x) ⊆ int y∈Nδ (x) f that |f (y) − f (x) − Dfx (y − x)| < (α/4) y − x for y ∈ Nη (x). There exists 0 < θ < η such that |f (y) − f (x)| < αη/4 for y ∈ Nθ (x), and there exists z ∈ Nθ (x) such that z − x < αδ/4L and Dfz − Dfx  > α. Consequently, there exists h ∈ E with h = 1 such that Dfz (h) − Dfx (h) > α. Let k = ηh. Then Dfz (x + k − z) ≤ f (x + k) − f (z). Thus

11.6 Smoothness

145

αη ≤ Dfz (k) − Dfx (k) ≤ ( f (x + k) − f (x) − Dfx (k)) + Dfz (x − z) + ( f (x) − f (z)) ≤ αη/4 + αη/4 + αη/4, giving the necessary contradiction. Exercise 11.6.4 Show that if f is a Gˆateaux smooth very proper convex function on Rd , then f is Fr´echet differentiable on int f , and Df is continuous int on f . As a simple and important example of a very regular function, we consider the norm of a [separable] normed space. Theorem 11.6.5 Suppose that (E, .) is a [separable] normed space. Recall that n1 (x) = x. Then (∂∨ n1 )0 = B(E ). If x = 0 then φ ∈ (∂∨ n1 )x if and only if φ(x) = x and φ = 1. Proof (∂∨ n1 )0 = B(E ), since (∂∨ n1 )0 = {φ ∈ E : φ(h) ≤ h for h ∈ E )} = {φ ∈ E : |φ(h)| ≤ h for h ∈ E} = B(E ). Suppose that x = 0, so that (∂∨ n1 )x = {φ ∈ E : φ(h) ≤ x + h − x for h ∈ E}. If φ ∈ (∂∨ n1 )x then, since x + h − x ≤ h, it follows that φ ≤ 1. On the other hand if 0 < α < 1 then αφ(x) = φ(αx) ≤ x + αx − x = α x and − αφ(x) = φ(−αx) ≤ x − αx − x = −α x , so that φ(x) = x and φ ≥ 1. Thus φ = 1. Conversely, suppose that φ ∈ E , that φ = 1 and that φ(x) = x. If h ∈ E then φ(h) = φ(x + h) − φ(x) = φ(x + h) − x ≤ x + h − x , so that φ ∈ (∂∨ n1 )x . Note that if x = 0 then (∂∨ n1 )x = (∂∨ n1 )x/x . Exercise 11.6.6 Suppose that (E, .) is a [separable] normed space. Determine ∂∨ np for p > 1. A Banach space (E, .) is said to be smooth if the function n1 is Gˆateaux smooth on E \ {0}.

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Subdifferentials and the Legendre Transform

Corollary 11.6.7 A [separable] Banach space is smooth if and only if for each x ∈ E with x = 1 there exists a unique φ ∈ E for which φ(x) = φ = 1. The following result is an immediate consequence of Mazur’s theorem (Theorem 11.5.7). Exercise 11.6.8 Use Mazur’s theorem to show that if (E, .) is a separable Banach space, then the set of smooth points of S(E) is a dense Gδ subset of S(E). The norm of a Hilbert space H is smooth. For if x = y = 1 and x, y = 1 then x − y2 = x2 − 2 x, y + y2 = 0, so that ∂∨ n(x) = {x} and Dfx = x. We shall give further examples in Chapter 15. Smoothness and strict convexity of Banach spaces are related in the following way. Theorem 11.6.9 Suppose that (E, .) is a [separable] reflexive Banach space. (i) If (E , . ) is strictly convex then (E, .) is smooth. (ii) If (E , . ) is smooth then (E, .) is strictly convex. Proof (i) Suppose that (E, .) is not smooth. Then there exists x with x = 1, and distinct φ, ψ ∈ E such that φ(x) = φ = ψ(x) = ψ = 1. If 0 < λ < 1 then ((1 − λ)φ + λψ)(x) = 1, so that (1 − λ)φ + λψ = 1 and [φ, ψ] is a proper line segment in S(E ). Thus E is not strictly convex. (ii) Suppose that (E, .) is not strictly convex, so that there exists a proper line segment [x, y] in S(E). By the Hahn–Banach theorem, there exists φ ∈ E with φ(x) = φ(y) = φ = 1. Thus (E , . ) is not smooth. Even if (E, .) is strictly convex, the function n1 is not strictly convex, since (1 − λ)x + λ(2x) = (1 − λ) x + λ 2x = 1 + λ. What about the function np , for p > 1? ∞ 2 ateaux smooth and strictly If x ∈ l2 , let f (x) = n=1 xn /n. Then f is Gˆ  ∞ 2 , so that  ∗ = {x : 2 ∗ nx convex. Then f ∗ (y) = ∞ f n n=1 n=1 nxn < ∞}, and f is not very regular. We now consider the Legendre transform of certain strictly convex very regular functions. Theorem 11.6.10 Suppose that f is a non-negative finite-valued strictly convex very regular function on a [separable] reflexive Banach space (E, .) for which • f (x) = 0 if and only if x = 0; • f is bounded on the bounded sets of E; and • f (x)/ x → ∞ as x → ∞.

11.6 Smoothness

147

Then f ∗ is a non-negative finite-valued Gˆateaux smooth function on E with f ∗ (0) = 0. Df ∗ is a surjection of E onto E. Proof Suppose that φ ∈ E . Then f ∗ (φ) ≥ φ(0) − f (0) = 0, and if φ(0) = 0 then f ∗ (φ) = 0. There exists R > 0 such that f (x) ≥ φ . x for x ≥ R, so that φ(x) − f (x) ≤ φ . x − f (x) ≤ 0, for x ≥ R. Thus f ∗ (φ) = supx≤R φ(x) − f (x) ≤ R φ, and so f ∗ (φ) is finite, for all φ ∈ E . Let S = supx≤R f (x), and let A = {(x, t) : x ≤ R, f (x) ≤ t ≤ S}. Then A is a bounded norm-closed convex subset of E × R, and so, since E × R is reflexive, it is weakly compact. Thus the linear functional (φ, −1) attains its supremum on A at a point (x0 , t0 ) of A, and, since f (x0 ) ≤ t0 , it follows that t0 = f (x0 ). Thus φ ∈ ∂∨ fx0 . Since f is strictly convex, x0 is unique, and so f ∗ is Gˆateaux differentiable at φ. If x ∈ E then ∂∨ fx = ∅; if φ ∈ ∂∨ fx , then x = Df ∗ (φ), by Theorem 11.2.4, and so Df ∗ is surjective. Corollary 11.6.11 Suppose that f is also Gˆateaux smooth. Then Df is a bijection of E onto E , with inverse Df ∗ , and if E is finite-dimensional, then ∇f is a homeomorphism of E onto E , with inverse ∇f ∗ . As an example, suppose that (E, .) is a smooth and strictly convex reflexive Banach space, and that ψ is a strictly increasing, strictly convex differentiable function on [0, ∞) for which ψ(0) = ψ  (0) = 0 and ψ(t)/t → ∞ as t → ∞ (for example, ψ(t) = tp /p, where p > 1, or ψ(t) = t log(t + 1)). Let us check that the function f (x) = ψ ◦ n1 satisfies the conditions of the theorem. Clearly f (x)/ x → ∞ as x → ∞. It is straightforward to verify that d+ fx (h) = ψ  ((Dn1 )x (h)) if x = 0, and that Df0 (h) = 0, so that  f is    Gˆateaux smooth. Suppose that x = x and that 0 < λ < 1. If x = x  then   f ((1 − λ)x + λx ) = ψ((1 − λ)x + λx    ≤ ψ((1 − λ) x + λ x )   ≤ (1 − λ)ψ(x) + λψ(x ) = (1 − λ)f (x) + λf (x ),       while if x = x  = r then (1 − λ)x + λx  < (1 − λ) x + λ x  = r, so that f ((1 − λ)x + λx ) < ψ(r) = (1 − λ)f (x) + λf (x ). Thus f is strictly convex.

148

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11.7 The Fenchel–Rockafeller Duality Theorem Theorem 11.7.1 (The Fenchel–Rockafeller duality theorem) Suppose that (E, .) is a [separable] normed space, that f is a regular convex function on E, that g is a very regular convex function on E, that f ∩ int g = ∅ and that infx∈E ( f (x) + g(x)) = c > −∞. Then f ∗ (−φ) + g∗ (φ) ≥ −c for φ ∈ E , and there exists φ0 ∈ E such that f ∗ (−φ0 ) + g∗ (φ0 ) = −c. Proof Setting x = y, f ∗ (−φ) + g∗ (φ) = sup (φ(x − y) − f (x) − g(y)) x,y∈E

≥ − inf ( f (x) + g(x)) = −c. x∈E

Once again, we use the separation theorem. Let h(x) = f (x) − c, and let A(h) = {(x, t) : h(x) + t ≤ 0}. The convex sets A(h) and S(g) are disjoint, since if h(x) + t ≤ 0 and t > g(x) then f (x) + g(x) < c. Since S(g) is open in E × R, by the separation theorem there exists (ψ0 , s) ∈ E × R and λ ∈ R such that ψ0 (x) + st > λ for (x, t) ∈ S(g), and ψ0 (y) − sr ≤ λ for (y, r) ∈ A(h). By considering x0 ∈ f ∩ int g , it follows that s = 0, and, since if (x, t) ∈ Sf and t > t then (x, t ) ∈ Sf , it follows that s > 0. Let φ0 = ψ0 /s and let μ = λ/s. If x ∈ f and y ∈ int g then φ0 (x) + t > μ if t > f (x) and φ0 (y) − r ≤ μ if r ≥ g(y) − c, so that ((−φ0 )(x) − f (x)) + (φ0 (y) − g(y)) ≤ −c; that is, f ∗ (−φ0 ) + g∗ (φ0 ) ≤ −c. Corollary 11.7.2 If x0 ∈ f ∩ int g , then ∂∨ ( f + g)(x0 ) = ∂∨ f (x0 ) + ∂∨ g(x0 ). Proof In general, ∂∨ f (x0 ) + ∂∨ g(x0 ) ⊆ ∂∨ ( f + g)(x0 ). Suppose that φ ∈ ∂∨ ( f + g)(x0 ). By replacing f and g by f (x + x0 ) − f (x0 ) − φ(x) and g(x + x0 ) − g(x0 ), we may suppose that x0 = 0, that φ = 0 and that f (0) = g(0) = 0. Thus infx∈E f (x) + g(x) = 0 ≥ 0. By the theorem, there exists φ0 ∈ E such that f ∗ (−φ0 ) = g∗ (φ0 ) = 0; that is,

11.8 The Bishop–Phelps Theorem

149

sup(−φ0 (x) − f (x)) + sup(φ0 (y) − g(y)) = 0. x∈E

y∈E

Setting y = 0, we see that f (x) − f (0) = f (x) ≥ −φ0 (x) for all x ∈ E, and, setting x = 0, we see that g(y) − g(0) = g(y) ≥ φ0 (y) for all y ∈ E; that is, −φ0 ∈ ∂∨ f (0) and φ0 ∈ ∂∨ g(0). Thus φ = 0 ∈ ∂∨ f (x0 ) + ∂∨ g(x0 ). int Corollary 11.7.3 If x0 ∈ int ateaux differentiable at f ∩ g , then f + g is Gˆ x0 if and only if f and g are Gˆateaux differentiable at x0 .

11.8 The Bishop–Phelps Theorem Suppose that f is a regular convex function on a [separable] normed space (E, .), that x0 ∈ f and that  > 0. The -subdifferential ∂ ( f )(x0 ) of f at x is defined as ∂ ( f )(x0 ) = {φ ∈ E : φ(h) ≤ f (x0 + h) − f (x0 ) +  for all h ∈ E}. Exercise 11.8.1 Use the separation theorem to show that ∂ ( f )(x0 ) is not empty, for all x ∈ f ; use the definition of f ∗ to show that φ ∈ ∂ ( f )(x0 ) if and only if f (x0 ) + f ∗ (φ) ≤ . Theorem 11.8.2 Suppose that f is a regular convex function on a [separable] Banach space (E, .). Suppose that x0 ∈ f , that  > 0, that α > 0 and that ˜ such that φ0 ∈ ∂ ( f )(x0 ). Then there exist x˜ ∈ f and φ˜ ∈ ∂∨ f (x)     x0 − x ˜ ≤ /α and φ0 − φ˜  ≤ α. Proof First, we simplify the problem. By considering f (x + x0 ) − f (x0 ) − φ0 (x) + , we can suppose that x0 = 0, that f (x) ≥ 0 and that f (0) = . Let k = infx∈E f (x), and let h = f − k. Then h is a non-negative regular convex function on E, infx∈E h(x) = 0, and h(0) ≤ . It is then sufficient   to show that   ˜ ˜ ≤ /α and φ ∈ ∂∨ h(x) ˜ with φ˜  ≤ α. there exists x˜ ∈ h with x We apply Ekeland’s variational principle (Theorem 4.4.1). There exists x˜ ∈ E with x ˜ ≤ /α such that h(x) ˜ < h(x) + α x − x ˜ for all x ∈ E \ {x}; ˜ x˜ ∈ h . Let n(x) = α x − x. ˜ Then n is a very regular convex function on ˜ But ∂∨ (h + n)(x) ˜ = ∂∨ h(x) ˜ + ∂∨ n(x), ˜ by Corollary E, and 0 ∈ ∂∨ (h + n)(x). ˜ ∈ E such that −φ˜ ∈ {φ ∈ E : φ ≤ α}. Thus 11.7.2, and so there exists φ     ˜ and φ˜  ≤ α. φ˜ ∈ ∂∨ h(x) Corollary 11.8.3 The set {x ∈ f : ∂∨ f (x) = ∅} is dense in f .

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Suppose that A is a non-empty subset of a normed space (E, .). If φ ∈ E , we set pA (φ) = sup{φ(a) : a ∈ A}

b(A) = {φ ∈ E : pA (φ) < ∞}.

A non-zero element φ of b(A) is a support functional of A if there exists a0 ∈ A such that φ(a0 ) = pA (φ), and an element a of A is a support point of A if there exists φ ∈ b(A) such that φ(a) = pA (φ). We denote the set of support points of A by s(A) and the set of support functionals of A by s (A). Theorem 11.8.4 (The Bishop–Phelps theorem) Suppose that C is a non-empty proper closed convex subset of a [separable] Banach space (E, .). (i) s(C) is dense in the boundary ∂C of C. (ii) s (C) is norm dense in b(C). Proof (i) We apply Theorem 11.8.2 to the regular convex function 0C . Suppose that x0 ∈ ∂C and that 0 <  < 1. There exists x1 ∈ E\C such that x0 − x1  < /2. By the separation theorem, there exists φ0 ∈ E , with φ0  = 1, such that φ0 (x1 ) > pC (φ). If x ∈ C then φ0 (x − x0 ) = φ0 (x) − φ0 (x0 ) < φ0 (x1 ) − φ0 (x0 ) = φ0 (x1 − x0 ) ≤ /2 = /2 + 0c (x) − 0c (x0 ), so that φ0 ∈ ∂/2 0C (x0 ). By Theorem 11.8.2, there exist x˜ ∈ C and φ˜ ∈     ˜ such that x0 − x ˜ ≤  and φ0 − φ˜  ≤ . Thus φ˜ = 0. Since ∂∨ 0C (x) ∂∨ 0C (x) = {0} for x ∈ Cint , x˜ ∈ ∂C. (ii) Suppose that φ0 ∈ b(C). If φ0 = 0, and ψ is any element of s (C), then ψ/n → 0 as n → ∞, so that φ0 ∈ s (C). Otherwise, since λφ ∈ b(C) if φ ∈ b(C) and λ > 0, we can suppose that φ0  = 1. Suppose that 0 <  < 1. There exists x0 ∈ C such that φ0 (x0 ) > pC (φ0 ) − . Thus if x ∈ C then φ0 (x − x0 ) <  = 0C (x) − 0C (x0 ) + , and so φ0 ∈ ∂ 0C (x0 ). Take α = . Then by Theorem 11.8.2 there  exist x˜ ∈    C = φ0C and φ˜ ∈ ∂∨ 0C (x) ˜ such that x˜ − x0  ≤  and φ˜ − φ0  ≤ . Thus φ˜ = 0, and if x ∈ C then ˜ ˜ ˜ x), φ(x) ≤ φ(x) + (0C (x) − 0C (x)) ˜ = φ( ˜ so that φ˜ is a support functional of C. The following corollary is also often referred to as the Bishop–Phelps theorem.

11.9 Monotone and Cyclically Monotone Sets

151

Corollary 11.8.5 The set {φ ∈ E : φ(x) = φ for some x ∈ B(E)} is norm dense in E . Proof Take C to be the unit ball B(E) of E. A closed convex subset C of a separable Banach space is determined by its support functionals. Proposition 11.8.6 Suppose that C is a proper non-empty closed convex subset of a separable Banach space (E, .). Let C˜ = ∩{x ∈ E : φ(x) ≤ pC (x) for p ∈ s (C)}. ˜ Then C = C. ˜ Suppose that y0 ∈ C. Let d = d(y0 , C). By the Proof Certainly C ⊆ C. separation theorem, there exists 0 ∈ E with φ = 1, and λ ∈ R such that φ0 (x) ≤ λ if x ∈ C and φ0 (y) > λ if y − y0  < d. Thus if h < d then φ0 (y0 ) − φ0 (h) = φ0 (y0 − h) > pC (φ0 ) and so d = sup{φ0 (h) : h < d} ≤ φ0 (y0 ) − pC (y0 ). Suppose that 0 <  < d/4(d + 1). Choose x0 ∈ C with x0 − y0  < d + /2. Note that φ0 (y0 − x0 ) < d + /2. If x ∈ C, then φ(x − x0 ) = φ0 (x − y0 ) + φ0 (y0 − x0 ) < −d + (d + /2) = /2, so that φ0 ∈ ∂/2 0C (x0 ). By Theorem 11.8.2, there exist x˜ ∈ C and φ˜ ∈     ˜ with x˜ − x0  ≤  and φ˜ − φ0  ≤ . In particular φ˜ = 0, so that ∂∨ 0C (x) x˜ ∈ ∂C. Further, if x ∈ C then ˜ x˜ − y0 ) = (φ˜ − φ0 )(x˜ − y0 ) + φ0 (x˜ − y0 ) ˜ − y0 ) ≤ φ( φ(x ≤  x˜ − y0  + pC (φ) − φ0 (y0 ) ≤ ( + d + /2) − d < −d/2. ˜ ˜ 0 ) > pC (φ), ˜ and so y0 ∈ C. Thus φ(y Exercise 11.8.7 Suppose that f is a regular convex function on a [separable] Banach space (E, .). By considering the epigraph Af of f in E × R, show that if x ∈ f , then f (x) = sup{f (y) + φ(x − y) : φ ∈ ∂∨ f (y) for some y ∈ f }.

11.9 Monotone and Cyclically Monotone Sets Suppose that (E, .) is a normed space. A subset A of E × E is said to be monotone if φ1 (x1 ) + φ2 (x2 ) ≥ φ1 (x2 ) + φ2 (x1 ) for all (x1 , φ1 ), (x2 , φ2 ) in

152

Subdifferentials and the Legendre Transform

  A, and is said to be cyclically monotone if nj=1 φj (xj ) ≥ nj=1 φj (xj+1 ), for all (x1 , φ1 ), . . . , (xn , φn ) in A, where we set xn+1 = x1 . Clearly a cyclically monotone set is monotone, and a set A is cyclically monotone if and only if n n j=1 φj (xj ) ≥ j=1 φj (xσ (j) ) for all (x1 , φ1 ), . . . , (xn , φn ) in A and for every permutation σ of {1, . . . , n}. These notions are frequently defined in terms of operators. A mapping T from a normed space (E, .) into the subsets of its dual E is said to be a monotone operator (cyclically monotone operator) if φ1 (x1 ) + φ2 (x2 ) ≥  φ (x ) + φ2 (x1 ) for all x1 , x2 ∈ E and φ1 ∈ T(x1 ), φ2 ∈ T(x2 ) ( nj=1 φj (xj ) ≥ 1 n 2 j=1 φj (xj+1 ), for all x1 , . . . , xn ∈ E and φi ∈ T(xi ) for 2 ≤ i ≤ n and φn+1 = φ1 ). Exercise 11.9.1 Show that a mapping T from a normed space (E, .) into the subsets of its dual E is a monotone operator (cyclically monotone operator) if and only if its graph G(T) = {(x, φ) : x ∈ E, φ ∈ T(x)} is a monotone set (cyclically monotone set). Exercise 11.9.2 Suppose that f is a real-valued function on R. Show that the graph of f is monotone if and only if f is non-decreasing. Exercise 11.9.3 Suppose that T ∈ L(H), where H is a Hilbert space. Show that the graph of T is monotone if and only if T(x), x ≥ 0 for all x ∈ E. Suppose that f is a regular convex function on a Banach space (E, .), with subderivative ∂∨ f . Recall that G(∂∨ f ) = {(x, φ) : x ∈ f , φ ∈ ∂∨ fx }. Proposition 11.9.4 Suppose that f is a regular convex function on a Banach space (E, .). Then G(∂∨ f ) is cyclically monotone. Proof Suppose that (x1 , φ1 ), . . . , (xn , φn ) ∈ G(∂∨ f ), and that xn+1 = x1 . Then n  j=1

φj (xj+1 ) −

n 

φj (xj ) =

j=1

≤

n 

φj (xj+1 − xj )

j=1 n 

( f (xj+1 ) − f (xj )) = 0.

j=1

Exercise 11.9.5 Suppose that T ∈ L(H) is positive and self-adjoint (where H is a real Hilbert space). Show that the graph of T is cyclically monotone.

11.9 Monotone and Cyclically Monotone Sets

153

Exercise 11.9.6 Suppose that Rθ is a rotation of R2 by θ , where 0 < θ ≤ π/2. Show that the graph of Rθ is monotone, but not cyclically monotone. We can use a cyclically monotone set to define a regular convex function. Theorem 11.9.7 Suppose that (E, .) is a normed space and that A is a nonempty cyclically monotone subset of E × E . Then there exists a regular convex function f on E such that A ⊆ G(∂∨ f ). Proof Choose an element (x0 , φ0 ) of A. If x ∈ E, set ⎧ ⎫ n−1 ⎨ ⎬ f (x) = sup φj (xj+1 − xj ) + φn (x − xn ) : ((xj , φj ))nj=1 ∈ An , n ∈ N . ⎩ ⎭ j=0

Taking x = x0 , n = 1 and (x1 , φ1 ) = (x0 , φ0 ), we see that f (x0 ) ≥ 0. But f (x0 ) ≤ 0, by cyclic monotonicity, and so f (x0 ) = 0. Since f is the supremum of a set of continuous affine functions on E it is a lower semi-continuous convex function on E; f is a regular convex function. Suppose that (x, φ) ∈ A; we must show that f (x) < ∞. Suppose that α < f (x) and that y ∈ E. There exist (x1 , φ1 ), . . . , (xn , φn ) in A such that φn (x − xn ) + φn−1 (xn − xn−1 ) + · · · + φ0 (x1 − x0 ) > α. Set (xn+1 , φn+1 ) = (x, φ). Then f (y) ≥ φn+1 (y − xn+1 ) + φn (xn+1 − xn ) + · · · + φ0 (x1 − x0 ) = φ(y − x) + φn (x − xn ) + · · · + φ0 (x1 − x0 ) > φ(y − x) + α. In particular, 0 = f (x0 ) ≥ φ(x0 − x) + α, so that f (x) ≤ φ(x − x0 ) < ∞, and x ∈ f . Consequently, if y ∈ E then f (y) − f (x) ≥ φ(y − x). Thus (x, φ) ∈ G(∂∨ f ). If (E, .) is a normed space, the collection of monotone subsets of E × E is ordered by inclusion. Theorem 11.9.8 If f is a regular convex function on a normed space (E, .), then G(∂∨ f ) is a maximal monotone set. Proof Suppose that (x0 , φ0 ) ∈ E × E and that φ(x) + φ0 (x0 ) ≥ φ(x0 ) + φ0 (x) for all (x, φ) ∈ G(∂∨ f ). We must show that (x0 , φ0 ) ∈ G(∂∨ f ). First we simplify the problem. By replacing f (x) by f (x + x0 ) − φ0 (x), we can suppose that x0 = 0 and that φ0 = 0. Thus we must show that 0 ∈ ∂∨ f (0), or equivalently that f (0) + f ∗ (0) = 0.

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Subdifferentials and the Legendre Transform

We apply the Fenchel–Rockafeller theorem (Theorem 11.7.1), taking g = n2 . Thus there exists φ1 ∈ E such that inf{f (x) + n2 (x) : x ∈ E} = −f ∗ (φ1 ) − n2 (φ1 ). First we show that φ1 = 0. Suppose not, and suppose that 0 <  < φ1  . Then there exists x1 ∈ E such that f (x1 ) + n2 (x1 ) + f ∗ (φ1 ) + n2 (φ1 ) ≤  2 /2. Since f (x1 ) + f ∗ (φ1 ) ≥ φ1 (x1 ) it follows that 0 ≤ 12 (x1  − φ1  )2 ≤ n2 (x1 ) + φ1 (x1 ) + n2 (φ1 ) <  2 /2. Thus | x1  − φ1  | <  and f (x1 ) + f ∗ (φ1 ) − φ1 (x1 ) <  2 /2. By Theorem 11.8.2, there exists (x2 , φ2 ) ∈ G(∂∨ f ) with x2 − x1  <  and φ1 − φ2  < . In particular | x2  − φ1  | < 2, so that x2  ≤ 3 φ1  . By hypothesis φ2 (x2 ) ≥ 0 and φ2 (x2 ) ≤ φ1 (x1 ) + x2  . φ1 − φ2  + x1 − x2  . φ1  ≤ φ1 (x1 ) + 4 φ1  , so that  2 1 2 (φ1  )

= n2 (φ1 ) ≤  2 /2 − φ1 (x1 ) ≤ 4 φ1  .

Consequently, φ1  ≤ 8. Since  can be chosen to be arbitrarily small, φ1 = 0. Thus inf{ f (x) + f ∗ (0) + n2 (x) : x ∈ E} = 0. If  > 0 there exists x1 such that f (x1 )+ f ∗ (0)+n2 (x) <  2 /2. But then x1  <  and 0 ≤ f (x1 )+ f ∗ (0) <  2 /2. Since f is lower semi-continuous, it follows that f (0) = 0. Corollary 11.9.9 If f is a regular convex function on a normed space (E, .), (x0 , φ0 ) ∈ G(∂∨ f ) and x ∈ E then f (x) − f (x0 ) = ⎫ ⎧ n−1 ⎬ ⎨ φj (xj+1 − xj ) + φn (x − xn ) : ((xj , φj )) ∈ G(∂∨ f )n , n ∈ N . sup ⎭ ⎩ j=0

Proof By Theorem 11.9.7, the expression in the formula defines a regular convex function g, and G(∂∨ f ) ⊆ G(∂∨ g). Since G(∂∨ f ) is maximal, G(∂∨ f ) = G(∂∨ g). Since f (x0 ) = g(x0 ), it follows from Theorem 11.5.6 that f = g.

12 Compact Convex Polish Spaces

12.1 Compact Polish Subsets of a Dual Pair Suppose that (E, F) is a dual pair, and that K is a σ (E, F)-compact metrizable subset of E. We shall investigate the geometric and topological properties of K. Theorem 12.1.1 Suppose that (E, F) is a dual pair, and that K is a σ (E, F)compact subset of E. The following are equivalent. (i) K is metrizable. (ii) K is separable. (iii) There is an affine map T : E → RN , continuous for the weak topologiy σ (E, F) and the product topology on RN such that T : K → T(K) is an affine homeomorphism of K onto a compact subset of the Hilbert cube. (iv) There is a linear map R : E → RN , continuous for the weak topology σ (E, F) and the product topology on RN , such that R : K → S(K) is a homeomorphism of K onto a closed subset of the compact subset L = {x : |xn | ≤ 1/n for n ∈ N} of l2 . If (E, .) is a normed space and F is a closed linear subspace of (E , . ) then R can be chosen to be a compact linear operator from (E, .) into (l2 .2 ), which is also continuous for the weak topology σ (E, F) on E and the norm topology on l2 . Proof Clearly (i) implies (ii). Suppose that (ii) holds, and that (xn )∞ n=1 is a σ (E, F)-dense sequence in K. Then there exists an array {(φm,n ) : 1 ≤ m < ∞, 1 ≤ n < ∞} & ! ! !& in F such that xm , φm,n = xn , φm,n for xm = xn . Let Mm,n = supx∈K & x, φm,n &; Mm,n < ∞, since K is compact and φm,n is σ (E, F)-continuous. Let j : N → N × N be an enumeration of N × N, and let 155

156

Compact Convex Polish Spaces ! x, φj(n) + 1; T(x)n = 2Mj(n) + 1 2

then T is an affine map of E into RN . T(K) is contained in the Hilbert cube, and the restriction of T to K is injective, so that ! it is a homeomorphism. Thus (ii) implies (iii). Similarly, if S(x)n = x, φj(n) /n(Mj(n) + 1), then S satisfies the conditions of (iv). Clearly, (iii) and (iv) each imply (i). (∞,∞) The final statement follows by taking (φm,n )(m,n)=(1,1) in the unit ball of F ! and R(x)n = x, φj(n) /n, for then if y ∈ l2 it follows that T(x)n yn ∈ l1 . Here are three circumstances in which we can apply the theorem. 1. K is a bounded weak*-closed subset of the dual of a separable Banach space (G, .). Then K is weak* separable, and is weak* compact, by Banach’s theorem. We take (E, F) as (G , G). 2. K is a weakly compact subset of a separable normed space (E, .). By Banach’s theorem, B(E ) is compact and metrizable in the weak* topology; therefore it is separable in the weak* topology. Let G be a countable weak*dense subset of B(E ). Then G◦◦ = B(E ) and so G separates the points of K. Thus the topology of pointwise convergence on G is Hausdorff, and so the weak topology on K is the same as the topology of pointwise convergence on G, which is metrizable. 3. K is a compact subset of a normed space (E, .). Then K is σ (E, E )compact and metrizable. The great advantage of this theorem is that we can apply simple Euclidean geometrical ideas to S(K), and then transfer them back to K. Here is a simple example of this, which shows how we can avoid using the separation theorem. Proposition 12.1.2 Suppose that (E, .) is a normed space and that K1 and K2 are disjoint separable weakly compact convex subsets of E. Then there exist φ ∈ E and α ∈ R such that supk∈K1 φ(k) < α and infk∈K2 φ(k) > α. Proof Let R : E → l2 be a compact linear operator which satisfies the requirements of Theorem 12.1.1, and let L1 = R(K1 ), L2 = R(K2 ). Let d = d(L1 , L2 ) = inf{l1 − l2 2 : (l1 , l2 ) ∈ L1 × L2 }; by compactness, there exists (m1 , m2 ) ∈ L1 × L2 with d = m1 − m2 2 . Let y = m2 −m1 and let α = (m1 + m2 )/2, y. Then m1 , y < α and m2 , y > α. Suppose, if possible, that there exists l ∈ L1 such that l, y > m1 , y. Then m − l1 = βy + w, where β > 0 and w, y = 0. Let mλ = (1 − λ)m1 + λl; by convexity, mλ ∈ L1 . Then l2 − mλ = (1 − λβ)y − λw, so that

12.2 Extreme Points

157

l2 − mλ 22 = (1 − λβ)2 d2 + λ2 w22 < d2 for small positive λ, giving a contradiction. Thus m1 , y = supl∈L1 l, y. Similarly, m2 , y = infl∈L2 l, y. Then φ = R (y) satisfies the requirements of the proposition.

12.2 Extreme Points We need some definitions. A variety V in a vector space E is a subset of E of the form x + F, where x ∈ E and F is a linear subspace of E; a variety V is the translate of a linear subspace. If A is a non-empty subset of E, a support variety H of A is a variety in E for which H ∩ A = ∅ and for which if x ∈ H ∩ A, y, z ∈ A and x ∈ (y, z) = {(1 − λ)y + λz : 0 < λ < 1} then y, z ∈ H. If so, then H ∩ A is called a face of A. If H ∩ A = {x}, then x is an extreme point of A. A support variety is proper if it is a proper subset of E. Exercise 12.2.1 If A is a radially open subset of a vector space E, then there is no proper support variety of A. On the other hand, we have the following. Proposition 12.2.2 Suppose that A is a convex neighbourhood of 0 in a [separable] normed space (E, .) and that H is a support variety of A. Then there is a closed support hyperplane V of A which contains H. Proof Suppose that H = x + F, where x ∈ A ∩ H. Define a linear functional φ on span(H) by setting φ(λ(x + f )) = λ. If λ > 1 then x + f ∈ (0, λ(x + f )), so that λ(x + f ) ∈ A. Therefore φ(y) < 1 for y ∈ A ∩ span(H), and so φ(y) ≤ pA (y) for y ∈ span(H), where pA is the gauge of A. By the Hahn– Banach theorem, φ can be extended to a linear functional ψ on E satisfying ψ(z) ≤ pA (z) for z ∈ E. Since A is a neighbourhood of 0, pA and ψ are continuous. Then V = {z ∈ E : ψ(z) = 1} is a closed support hyperplane V of A which contains H. Proposition 12.2.3 Suppose that A is a subset of a real vector space E. If L is a face of A and M is a face of L, then M is a face of K. Proof Suppose that x ∈ M, and that x = (1 − λ)y + λz with y, z ∈ A, and 0 < λ < 1. Then y, z ∈ L, since M ⊆ L and since L is a face of A. Consequently y, z ∈ M, since M is a face of L.

158

Compact Convex Polish Spaces

Proposition 12.2.4 Suppose that (E, F) is a real dual pair and that K is a [separable] σ (E, F)-compact convex subset of E and that L is a closed face of K. Then L contains an extreme point of K. Proof By the preceding proposition, it is enough to show that L has an extreme point. First suppose that K is separable. By Theorem 12.1.1, L is affinely homeomorphic to a compact subset of a Hilbert space H, and so we can suppose that L is a compact subset of a Hilbert space H. Let d be the diameter of L. If L = {l} is a singleton, then l is an extreme point of L. Otherwise, d > 0. The function (x, y) → x − y is continuous on L × L, and so there exist x, y ∈ L such that x − y = d. We shall show that x and y are extreme points of K. Let J = y+(y−x)⊥ . If z = y+w ∈ J, then z − x2 = d2 +w2 , so that J ∩ L = {y}, and y is an extreme point of L; similarly for x. When K is not separable, we need the axiom of choice. Order the closed faces of L by inverse inclusion; F ≤ G if and only if G ⊂ F. If C is a chain of closed faces, then ∩F∈C F is non-empty and closed, since L is compact, and it is easily verified that it is a face of L, which is an upper bound for C. Thus we can apply Zorn’s lemma; there is a maximal face G. We claim that G = {g} is a singleton, so that g is an extreme point. If not, suppose that a and b are two distinct points of G. There exists φ ∈ F such that a, φ > b, φ. Let α = supc∈G c, φ. Then G ∩ {x : φ(x) = c} is a closed face of L properly contained in G, contradicting the maximality of G. Theorem 12.2.5 (The Krein–Mil’man theorem) Suppose that (E, F) is a dual pair of real vector spaces and that K is a non-empty [separable] σ (E, F)compact subset of E. K ⊆ (Ex(K)), the closed convex cover of the set Ex(K) of extreme points of K. Proof Suppose not. Then there exists x ∈ K \ (Ex(K)). It follows from Corollary 12.1.2, or the separation theorem, that there exists f ∈ F with x, f  > sup{y, f  : y ∈ (Ex(K))}. Let M = sup{k, f  : k ∈ K}. Then J = {y ∈ E : y, f  = M} is a closed support hyperplane of K disjoint from (Ex(K)), and J ∩ K is a face of K; by the preceding proposition, it contains an extreme point of K, giving a contradiction. Corollary 12.2.6 Suppose that (E, .) is a [separable] normed space, and that x ∈ E. Then there exists φ ∈ Ex(B(E )) for which φ(x) = x. Proof The result is trivially true if x = 0. Otherwise, by Corollary 9.1.9 there exists ψ ∈ B(E ) for which ψ(x) = x. Then H = {θ ∈ E : θ (x) = x} is a weak*-closed support hyperplane of B(E ). If φ is an extreme point of H ∩ B(E ) then φ is an extreme point of B(E ), and φ(x) = x.

12.2 Extreme Points

159

Exercise 12.2.7 Suppose that (E, F) is a dual pair of real vector spaces and that K is a non-empty [separable] convex σ (E, F)-compact subset of E. Show that K = (Ex(K)). Exercise 12.2.8 Suppose that (E, .) is a [separable] normed space. Show that B(E ) = (Ex(B(E ))) (the closure being taken in the weak* topology). We have a partial converse to the Krein–Mil’man theorem. We need a lemma. Lemma 12.2.9 If K1 , . . . , Kn are compact convex sets, then (∪nj=1 Kj ) is compact. Proof Let =

⎧ ⎨ ⎩

δ = (δ1 , . . . , δn ) ∈ Rn : δj ≥ 0 for 1 ≤ j ≤ n,

n  j=1

⎫ ⎬

δj = 1 . ⎭

is compact. Let T : (K1 × · · · × Kn × ) → E be defined as T(k1 , . . . , kn , δ) =

n 

δj kj .

j=1

Since K1 × · · · × Kn × is compact and T is continuous, the image, which is (∪nj=1 Kj ), is compact. Theorem 12.2.10 (Mil’man’s theorem) Suppose that (E, F) is a dual pair of real vector spaces and that K is a non-empty metrizable σ (E, F)-compact convex subset of E. Suppose that D is a closed subset of K, and that K = (D). Then Ex(K) ⊆ D. Proof Suppose that x is an extreme point of K. We show that x ∈ D = D. Suppose that N is a closed convex σ (E, F) neighbourhood of 0. The sets {N + d : d ∈ D} cover D, and so there exists a finite  subset D0 of D such  that D ⊆ ∪d∈D0 ((N + d) ∩ K). By the lemma, K ⊆  ∪d∈D0 ((N + d) ∩ K) .   Thus we can write x = d∈D0 λd yd , where λd ≥ 0 and d∈D0 λd = 1. There exists d ∈ D0 for which λd > 0. Then x = λd yd + (1 − λd )z, with z ∈ K. Since x is an extreme point, x = yd = n + d, so that x − n ∈ D. Thus x ∈ D. Here is a finite-dimensional example, which illustrates the importance of extreme points. The set Sn of doubly stochastic matrices is the set of n × n matrices m with entries in [0, 1] for which n  i=1

mij =

n  j=1

mij = 1 for all i, j.

160

Compact Convex Polish Spaces

Theorem 12.2.11 (Birkhoff’s theorem) The set Pn of permutation matrices is the set of extreme points of the set Sn of doubly stochatic matrices. Proof It is easy to see that the permutation matrices are extreme points of Sn . Suppose that m ∈ Sn \ Pn . Let V = {(i, j) : 0 < mij < 1}. Choose (i0 , j0 ) ∈ V. Then the i0 -th row must contain another element (i0 , j1 ) of V, and the j1 -th column must contain another element (i1 , j1 ) of V. Continuing in this way, there exists a sequence (ik , jk )∞ k=0 in V with (ik , jk+1 ) ∈ V. Since {1, . . . , n} × {1, . . . , n} is finite, there is a least k such that there exists l < k such that either (il , jl ) = (ik , jk ) or (il , jl+1 ) = (ik , jk+1 ). From this it follows that there exists a sequence (ik , jk )lk=0 of distinct elements in V such that, setting jl+1 = j0 , (ik , jk+1 ) ∈ V for 0 ≤ k ≤ l; thus we have a circuit with no two terms in common. Let k = min{ci ,j , ci ,j , 1 − ci ,j , 1 − ci ,j } and let  = k k k k+1 k k k k+1 min0≤k≤l k . Then  > 0. We now define a matrix d by setting ⎧ ⎪ if (i, j) = (ik , jk ) for some 0 ≤ k ≤ l, ⎨  dij = − if (i, j) = (ik , jk+1 ) for some 0 ≤ k ≤ l, ⎪ ⎩ 0 otherwise. Then m + d and m − d are doubly stochastic matrices, and so m is not an extreme point of Sn . Exercise 12.2.12 Identify the extreme points of the unit balls of the following Banach spaces: c0 ;

l1 ;

l∞ ;

C[0, 1]; Hilbert space;

and draw appropriate conclusions.

12.3 Dentability Can we slice small bits off a convex set? Theorem 12.3.1 Suppose that (E, F) is a dual pair, that K is a non-empty metrizable convex σ (E, F)-compact subset of E, and that p is a seminorm on E with the property that B = {x : p(x) ≤ 1} is σ (E, F)-closed. If  > 0, there exists a closed convex proper subset C of K such that if x, y ∈ A = K \ C then p(x − y) ≤ . Proof First we show that p is bounded on K. The closed sequence of sets (nB)∞ n=1 cover K. By Baire’s category theorem there exists n such that nB ∩ K has a non-empty interior; there exist x ∈ K and a σ (E, F)-neighbourhood U of 0 such that (x+U)∩K ⊆ nB∩K. Since K−x is compact, there exists R ≥ 1 such

12.3 Dentability

161

that (K − x) ⊆ RU. Thus if y = x + h ∈ K then x + h/R ∈ (x + U) ∩ K ⊆ nB, so that p(y) ≤ p(x) + p(h) = p(x) + Rp(h/R) ≤ p(x) + R(p(x + h/R) + p(x)) ≤ (R + 1)p(x) + nR. Let D = Ex(K) and let d = sup{p(x − y) : x, y ∈ K}. If d ≤  we can take C to be a singleton set. Otherwise, D is σ (E, F)-separable; let H be a countable dense subset of D. The weakly closed sets {(h + (/4)B) ∩ D : h ∈ H} cover D. By Baire’s category theorem, one of them has a non-empty interior W. Then D \ W is a proper non-empty σ (E, F)-closed subset of D. Let K1 = (D \ W) and let K2 = (W). If k, l ∈ K2 then p(k − l) ≤ /2. By Theorem 12.2.10, K1 = K and K2 = K, and by Lemma 12.2.9, K = (K1 ∪ K2 ). Suppose that 0 < r < min(/4d, 1). Let Cr = {(1 − λ)k1 + λk2 : k1 ∈ K1 , k2 ∈ K2 , λ ∈ [0, 1 − r]}. As in Lemma 12.2.9, Cr is a convex weakly compact subset of K. If x = (1 − λ)k1 + λk2 is an extreme point of K in Cr , then x = k1 ∈ K1 , so that Cr = K. If y ∈ K \ Cr , then y = (1 − λ)k1 + λk2 , with λ > 1 − r, so that y − k2 = (1 − λ)(k1 − k2 ), and p(y − k2 ) ≤ rd. Thus if y, y ∈ K \ Cr then p(y − y ) ≤ /2 + 2rd < . If L is a σ (E, F)-bounded subset of a dual pair (E, F), a slice of L is a set of the form {y ∈ L : y, f  ≥ supx∈L x, f  − α}, where f ∈ F and α > 0. Corollary 12.3.2 If  > 0, there exists a slice S of K such that p(x − y) ≤  for x, y ∈ S. Proof By Corollary 9.4.4, there exists f ∈ E such that if r = sup{f (x) : x ∈ K} and s = sup{f (x) : x ∈ C}, then r > s. Choose 0 < α < r − s. A bounded subset L of a normed space is dentable if whenever  > 0 there exists a slice of L with diameter less than . Corollary 12.3.3 A non-empty metrizable weakly compact convex subset of a Banach space is dentable. In particular, the unit ball of the dual of a separable Banach space is dentable, and the unit ball of a separable reflexive Banach space is dentable.

13 Some Fixed Point Theorems

13.1 The Contraction Mapping Theorem This is the most familiar fixed point theorem, included for completeness’ sake. If f is a mapping of a set X into itself, then an element x of X is a fixed point of f if f (x) = x. A mapping f : (X, d) → (X, d) of a metric space into itself is a contraction mapping of (X, d) if there exists 0 ≤ K < 1 such that d( f (x), f (y)) ≤ Kd(x, y) for all x, y ∈ X; that is, f is a Lipschitz mapping with constant strictly less than 1. The fact that the constant K is strictly less than 1 is of fundamental importance. Theorem 13.1.1 (The contraction mapping theorem) If f is a contraction mapping of a non-empty complete metric space (X, d) then f has a unique fixed point x∞ . Proof Let K be the Lipschitz constant of f . Let x0 be any point of X. Define the n sequence (xn )∞ n=0 recursively by setting xn+1 = f (xn ). Thus xn = f (x0 ), and d(xn , xn+1 ) ≤ Kd(xn−1 , xn ) ≤ K 2 d(xn−2 , xn−1 ) ≤ · · · ≤ K n d(x0 , x1 ). We show that (xn )∞ n=0 is a Cauchy sequence. Suppose that  > 0. There exists n0 ∈ N such that K n < (1−K)/(d(x0 , x1 )+1) for n ≥ n0 . If n > m ≥ n0 then d(xm , xn ) ≤ d(xm , xm+1 ) + d(xm+1 , xm+2 ) + · · · + d(xn−1 , xn ) ≤ K m d(x0 , x1 ) + K m+1 d(x0 , x1 ) + · · · + K n−1 d(x0 , x1 ) ≤ K m d(x0 , x1 )/(1 − K) < . Since (X, d) is complete, there exists x∞ ∈ X such that xn → x∞ as n → ∞. Since f is continuous, xn+1 = f (xn ) → f (x∞ ) as n → ∞, and so x∞ = f (x∞ ); x∞ is a fixed point of f . If y is any fixed point of f then 162

13.1 The Contraction Mapping Theorem

163

d(y, x∞ ) = d( f (y), f (x∞ )) ≤ Kd(y, x∞ ); hence d(y, x∞ ) = 0, and y = x∞ ; x∞ is the unique fixed point of f . Corollary 13.1.2 Suppose that g is a mapping from X to X which commutes with f : f ◦ g = g ◦ f . Then x∞ is a fixed point of g. Proof f (g(x∞ )) = g( f (x∞ )) = g(x∞ ); g(x∞ ) is a fixed point of f , and so g(x∞ ) = x∞ . We can strengthen the contraction mapping in the following way. Corollary 13.1.3 Suppose that h : (X, d) → (Xd ) is a mapping of a complete metric space into itself, and suppose that hk is a contraction mapping for some k ∈ N. Then h has a unique fixed point. Proof Let f = hk . Then f has a unique fixed point x∞ . As f ◦ h = h ◦ f = f k+1 , x∞ is a fixed point of h. If y is a fixed point of h then f (y) = hk (y) = y, so that y = x∞ ; x∞ is the unique fixed point of f . Three points are worth making about this proof. First, we start with any point x0 of X, and obtain a sequence which converges to the unique fixed point x∞ ; further, d(x0 , x∞ ) ≤ d(x0 , x1 )/(1 − K). Secondly, d(xn+1 , x∞ ) = d( f (xn ), f (x∞ )) ≤ Kd(xn , x∞ ), so that d(xn , x∞ ) ≤ K n d(x0 , x∞ ); the convergence is exponentially fast. Thirdly, the condition that d( f (x), f (y)) < d(x, y) for x = y is not sufficient for f to have a fixed point. The function f (x) = x + e−x : [0, ∞) → [0, ∞) does not have a fixed point, but satisfies the condition; if 0 ≤ y < x < ∞ then, by the mean-value theorem, f (x)−f (y) = (1−e−c )(x−y) for some x < c < y, so that | f (x)−f (y)| < |x−y|. The most familiar application of this theorem gives a proof of the existence and uniqueness of solutions of certain ordinary differential equations. Theorem 13.1.4 Suppose that M ≥ 0 and that L ≥ 0. Suppose that H is a continuous real-valued function on the triangle T = {(x, y) ∈ R2 : 0 ≤ x ≤ b, |y| ≤ Mx}, that |H(x, y)| ≤ M and that |H(x, y) − H(x, y )| ≤ L|y − y |, for (x, y) ∈ T and (x, y ) ∈ T. Then there exists a unique continuously differentiable function f on [0, b] such that (x, f (x)) ∈ T for x ∈ [0, b] and df (x) = H(x, f (x)) for all x ∈ [0, b], and f (0) = 0. dx Proof If f is any solution, then  x  x H(t, f (t))dt| ≤ |H(t, f (t))|dt ≤ Mx, | f (x)| = | f (x) − f (0)| = | 0

0

164

Some Fixed Point Theorems

for 0 ≤ x ≤ b, so that the graph of f is contained in T. The second condition is a Lipschitz condition, which is needed to enable us to use the contraction mapping theorem. First let us observe that the fundamental theorem of calculus shows that solving this differential equation is equivalent to solving an integral equation. If f is a solution, then,  x  x f  (t) dt = H(t, f (t)) dt for all x ∈ [0, b]. f (x) = f (x) − f (0) = 0

0

Conversely, if f is a continuous function which  x satisfies this integral equation, then f (0) = 0 and the function J( f )(x) = 0 H(t, f (t)) dt is differentiable, with continuous derivative H(x, f (x)). Thus f  (x) = H(x, f (x)) for x ∈ [0, b]. Let X = {g ∈ C[0, b] : |g(x)| ≤ Mx for x ∈ [0, b]}. X is a closed subset of the Banach space (C[0, b], .∞ ), and so is a complete metric space under the metric defined by the norm. We define a mapping J : X → C[0, b] by setting  x H(t, g(t)) dt, for x ∈ [0, b]. J(g)(x) = 0

Then J(g) is a continuous function on [0, b] and  x M dt = Mx, |J(g)(x)| ≤ 0

so that J(g) ∈ X. We now show by induction that, for each n ∈ Z+ , Ln xn g − h∞ , for g, h ∈ X and 0 ≤ x ≤ b. n! The result is certainly true for n = 0. Suppose that it is true for n. Then  x |H(t, J n (g)(t)) − H(t, J n (h)(t))| dt |J n+1 (g)(x) − J n+1 (h)(x)| ≤ 0  x L|J n (g)(t) − J n (h)(t)| dt ≤ |J n (g)(x) − J n (h)(x)| ≤

0



x

Ln+1 g − h∞ tn dt n! 0 Ln+1 xn+1 g − h∞ . = (n + 1)! ≤

Thus

 n  J (g) − J n (h)

∞

≤

Ln bn g − h∞ . n!

Now Ln bn /n! → 0 as n → ∞, and so there exists k ∈ N such that Lk bk /k! < 1. Thus J k is a contraction mapping of X. We apply Corollary 13.1.3. J has a unique fixed point f , and f is the unique solution of the integral equation.

13.2 Fixed Point Theorems of Caristi and Clarke

165

Suppose that (X, d) and (Y, ρ) are metric spaces and that f : X × Y → Y is continuous. Can we solve the equation y = f (x, y) for each x ∈ X? In other words, is there a function φ : X → Y such that φ(x) = f (x, φ(x)) for each x ∈ X? If so, is it unique? Is it continuous? Our next application of the contraction mapping theorem gives sufficient conditions for these questions to have a positive answer. It can be thought of as a contraction mapping theorem with a continuous parameter. Theorem 13.1.5 (The Lipschitz implicit function theorem) Suppose that (X, d) is a metric space, that (Y, ρ) is a complete metric space and that f : X × Y → Y is continuous. If there exists 0 < K < 1 such that ρ( f (x, y), f (x, y )) ≤ Kρ(y, y ) for all x ∈ X and y, y ∈ Y then there exists a unique mapping φ : X → Y such that φ(x) = f (x, φ(x)) for each x ∈ X. Further, φ is continuous. Proof The proof of existence and uniqueness follows easily from the contraction mapping theorem. If x ∈ X then the mapping fx : Y → Y defined by fx (y) = f (x, y) is a contraction mapping, which has a unique fixed point φ(x). Then f (x, φ(x)) = fx (φ(x)) = φ(x). It remains to show that φ is continuous. Suppose that x ∈ X and that  > 0. There exists δ > 0 such that if d(x, z) < δ then ρ(φ(x), f (z, φ(x))) = ρ( f (x, φ(x)), f (z, φ(x))) < (1 − K). For such z, let z0 = φ(x) and let zn = fz (zn−1 ) for n ∈ N. Then zn → φ(z), and ρ(φ(x), φ(z)) ≤ ρ(φ(x), z1 ) + ρ(z1 , φ(z)) ≤ ρ(φ(x), z1 ) + Kρ(φ(x), φ(z)), so that ρ(φ(z), φ(x)) ≤

ρ(φ(x), z1 ) ρ(φ(x), f (z, φ(x))) = < . 1−K 1−K

Exercise 13.1.6 Suppose that (x, d) is a compact metric space and that f is a mapping from X to X which satisfies d( f (x), f (y)) < d(x, y) whenever x, y ∈ X and x = y. Show that f has a unique fixed point. Exercise 13.1.7 Suppose that f and g are contractions of a complete metric space (X, d). Show that there are x, y ∈ X such that f (x) = y and g(y) = x.

13.2 Fixed Point Theorems of Caristi and Clarke We prove two fixed point theorems which depend upon Ekeland’s variational principle.

166

Some Fixed Point Theorems

Theorem 13.2.1 (Caristi’s fixed point theorem) Suppose that φ is a realvalued lower semi-continuous function on a complete metric space (X, d) which is bounded below and that f is a mapping from X into itself which satisfies d(x, f (x)) ≤ φ(x) − φ( f (x)) for all x ∈ X. Then f has a fixed point. Proof By adding a constant, we can suppose that infx∈X φ(x) = 0. By Ekeland’s variational principle, there exists a point x˜ such that φ(x) ≥ φ(x) ˜ − 12 d(x, x) ˜ for all x ∈ X. Thus d(x, ˜ f (x)) ˜ ≤ φ(x) ˜ − φ( f (x)) ˜ ≤ 12 d(x, ˜ f (x)), ˜ so that d(x, ˜ f (x)) ˜ = 0, and x˜ is a fixed point. Note that no continuity conditions are placed on f . Exercise 13.2.2 Suppose that (X, d) is a complete metric space, that ψ is a lower semi-continuous function on X × X and that f is a continuous mapping from X to X which satisfies ψ( f (x), f (y)) ≤ ψ(x, y) − d(x, y) for all (x, y) ∈ X × X. Show that f has a unique fixed point. Clarke’s theorem is an extension of the contraction mapping theorem (Theorem 13.1.1). Theorem 13.2.3 (Clarke’s fixed point theorem) Suppose that g is a continuous mapping of a complete metric space (X, d) into itself, with the property that there exists 0 ≤ k < 1 such that if x ∈ X then there exists 0 < t ≤ 1 and xt in X such that d(x, xt ) = td(x, g(x)) and d(xt , g(x)) = (1 − t)d(x, g(x)), and d(g(x), g(xt )) ≤ kd(x, xt ). Then g has a fixed point. Proof Let α = (1 − k)/2. Let f (x) = d(x, g(x)). Then f is continuous, and so by Ekeland’s variational principle (Theorem 4.4.1) there exists x˜ ∈ X such that d(x, ˜ g(x)) ˜ ≤ d(x, g(x)) + αd(x, x) ˜ for all x ∈ X. Setting x = x˜t˜ we see that d(x, ˜ g(x)) ˜ ≤ d(x˜t˜, g(x˜t˜)) + αd(x˜t˜, x) ˜ ˜ ≤ d(x, ˜ g(x)) ˜ + d(g(x), ˜ g(x˜t˜)) + αd(x˜t˜, x) ˜ ≤ (1 − t˜)d(x, ˜ g(x)) ˜ + k(d(x, ˜ x˜t˜) + αd(x˜t˜, x)

13.3 Simplices

167

so that t˜d(x, ˜ g(x)) ˜ ≤ (k + α)t˜d(x, ˜ g(x)) ˜ which ensures that d(x, ˜ g(x)) ˜ = 0, and x˜ is a fixed point of g. Exercise 13.2.4 Let g be a continuous mapping of a closed subset X of a Banach space (E, .) into itself for which [x, g(x)] ⊆ X for each x and for which there exists 0 < k < 1 such that if x ∈ X there exists y ∈ (x, g(x)] such that d(g(x), g(y)) ≤ kd(x, y). Show that g has a fixed point.

13.3 Simplices Our aim is to prove Brouwer’s fixed point theorem; for this we need to consider simplices an triangulations. Suppose that V = {v0 , . . . , vn } is a finite subset of a real normed vector space (E, .), with the property that {vi − v0 : 1 ≤ i ≤ n} is a linearly independent set. Then the convex cover = (V) = (V) is an n-simplex in E. The elements of V are the vertices of (V). If x ∈ , then x  can be written uniquely as x = ni=0 xi vi , where xi ≥ 0 for 1 ≤ i ≤ n and n i=0 xi = 1. The numbers xi are the barycentric co-ordinates of x. The set s(x) = {i : xi = 0} is the support of x. As an example, let the basic unit vectors of Rn+1 be denoted by e0 , e1 , . . . , en , and let E = {e0 , e1 , . . . , en }. Then (E) is the fundamental n-simplex. Suppose that F is a non-empty subset of {0, . . . , n}. Then F = {x ∈ : s(x) ⊂ F} is a face of ; it is a (|F| − 1)-simplex. Thus the 0-faces of are the singletons {vi }, the 1-faces are line segments [vi , vj ], the 2- faces are triangles, the 3-faces are tetrahedra, and so on. An (n − 1)-face is called a facet. The interior int is the set {x ∈ : s(x) = {0, . . . , n}} and the boundary ∂ is the union of the n + 1 facets of . For example the barycentre b = b(V) = n int i=0 vi /(n + 1) is an element of . A triangulation T(V) of (V) is a finite set of n-simplices whose union is , with the property that if 1 and 2 are two distinct elements of T(V) int then int 1 ∩ 2 = ∅. An important example is provided by the barycentric triangulation. This is defined inductively. If n = 1 then the barycentre b = 12 (v0 + v1 ), and the barycentric triangulation is [v0 , b], [v1 , b]. If the barycentric triangulation has been defined for (n − 1)-simplices, then the barycentric triangulation of an n-simplex (V) is defined as follows. Each facet F of (V) is an (n − 1)-simplex, and therefore there is a barycentric triangulation of T(F). If  ∈ F then (  ∪ {b(V)) is an n-simplex. The collection

168

Some Fixed Point Theorems {(  ∪ {b(V)) :  ∈ T(F), F a facet of }

is then a triangulation of , the barycentric triangulation Tb (V) of . The barycentric triangulation Tb (V) has (n + 1)! elements. Since the norm is a convex function,   # diam( (V)) = max v − v  : v, v ∈ V . It is an important fact that the diameters of the barycentric triangulation are smaller than the diameter of V. Proposition 13.3.1 If  = (V  ) ∈ Tb (V), then n diam(  ) ≤ diam( ). n+1 Proof Again, the proof is by induction. The result is true if n = 1. Suppose that it is true for n − 1. Let b be the barycentre of  , and let v be the vertex in V which does not belong to  . Then b=

v nb + , n+1 n+1

so that if v is a vertex of  then, using the inductive hypothesis,           1 n b − v  ≤ v − v  + v − b  n+1 n+1   n n−1 1 + . diam( ) ≤ n+1 n+1 n   n diam( ). = n+1 On the other hand, if v1 , v2 ∈ V  then     n n−1 v1 − v2  ≤ diam( ) ≤ diam( ). n n+1 Thus iterating the construction of barycentic triangulations, we can obtain a triangulation of for which each simplex of the triangulation has small diameter.

13.4 Sperner’s Lemma Suppose that A is a subset of an n-simplex (V). A mapping l from A into {0, . . . , n} is a Sperner mapping if l(a) ∈ s(a), the support of a, for each a ∈ A. Thus if a is in a face F then l(a) is a vertex of F.

13.4 Sperner’s Lemma

169

Suppose that T(V) is a triangulation of (V). We denote the set of vertices of the n-simplices in T(V) by W(T(V)). Suppose that l is a Sperner mapping on W(T(V)), and that  ∈ T(V). Then  is completely labelled if l maps the vertices of  onto {0, . . . , n}. Theorem 13.4.1 (Sperner’s Lemma) Suppose that T(V) is a triangulation of an n-simplex (V), and that l is a Sperner mapping on W(T(V)). Then the number N(T(V)) of completely labelled n-simplices in T(V) is odd, and so there exists at least one completely labelled n-simplex in T(V). Proof The proof is by induction on n. First suppose that n = 1. Then W(T(V)) = {v0 = q0 , . . . , qj = v1 }, where the vertices are listed in order on [v0 , v1 ]. Let dj = l(qj )−l(qj−1 ), let A = { j : dj = 1} and let B = {j : dj = −1}. Then the number of completely numbered 1-simplices in T(V) is |A| + |B|. But 1=

j  i=1

di =

 i∈A

1+

 (−1) = |A| − |B|, i∈B

so that |A| + |B| = 1 + 2|B|, and |A| + |B| is odd. Suppose now that the result holds for n − 1, that T(V) is a triangulation of an n-simplex (V), and that l is a Sperner mapping on W(T(V)). Suppose that  =  (V  ) ∈ T(V), and that F is a facet of  . We set φ(F) = 1 if {l(v) : v a vertex of F} = {0, . . . , n − 1},  and set φ(F) = 0 otherwise. We set ψ(  ) = {φ(F) : F a facet of  }.  We count S = {ψ(  ) :  ∈ T(V)} in two different ways. Let R(  ) = {l(v) : v ∈ V  }. If  is completely labelled, then F(  ) = 1. If R(  ) = {1, . . . , n − 1} then just one value is taken twice, so that ψ(  ) = 2, and otherwise ψ(  ) = 0. Adding, S − N(T(V)) is even. On the other hand, let F = {F : F a facet of  , for some  ∈ T(V)}. Let F1 = {F ∈ F : F ⊆ ∂ },

F2 = {F ∈ F : F ∩ int = ∅}.

If F ∈ F1 , then F is a facet of exactly one  in T(V), and φ(F) = 1 if and only if it is a completely numbered n − 1-simplex in the triangulation of the facet {v0 , . . . , vn−1 }. By the inductive hypothesis, there is an odd number of these. If F ∈ F2 , then F is a facet of exactly two elements of T(V), and so φ(F) is counted twice. Consequently, S is odd. Since S − N(T(V)) is even it follows that N(T(V)) is odd.

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Some Fixed Point Theorems

13.5 Brouwer’s Fixed Point Theorem Theorem 13.5.1 (Brouwer’s fixed point theorem) Suppose that K is a compact convex subset of Rn and that f : K → K is continuous. Then f has a fixed point; there exists x ∈ K such that f (x) = x. Proof The result is a topological one. Since K is homeomorphic to a simplex, we may assume that K = (E), the fundamental n-simplex. We give Rn+1 the  metric defined by the norm x = ni=0 |xi |. Suppose that the result is false. Let g(x) = (g0 (x), . . . , gn (x)) = x − f (x).  Then ni=0 gi (x) = 0, for all x ∈ (E). Since g is continuous and K is compact,  = infx∈K g(x) > 0, and there exists δ > 0 such that if x − y < δ then g(x) − g(y) < /2(n + 1). We now define a Sperner mapping on the set (E). Suppose that x ∈   (E). Since ni=0 gi (x) = 0 and ni=0 |gi (x)| ≥ , it follows that m(x) = maxni=0 gi (x) ≥ /2(n + 1). We set l(x) = inf{i : gi (x) = m(x), so that gl(x) (x) = m(x). Then xl(x) = gl(x) (x) + fl(x) (x) ≥ m(x) > 0, and so l is a Sperner mapping. There is a triangulation T(E) such that all the simplices have diameter less than δ. It follows from Sperner’s lemma that one of the n-simplices in T(E),  (V  ), say, is completely labelled, and we can label the vertices so that l(vi ) = i. Then g0 (v0 ) ≥ /2(n + 1), and if 1 ≤ i ≤ n then gi (v0 ) ≥ gi (vi ) − |gi (vi ) − gi (v0 )| ≥ /2(n + 1) − /2(n + 1) = 0. Thus

n

 i=0 gi (v0 )

> 0, giving a contradiction.

Here is another application of Sperner’s lemma. Theorem 13.5.2 Suppose that K is a compact convex subset of Rm , that f : K → K is continuous and that f (x) = x for x ∈ ∂K. Then f is surjective: f (K) = K. Proof Once again, we can suppose that K = (E). It is clearly sufficient to show that if y ∈ (E)int then there exists x ∈ (E) with f (x) = y. If x ∈ (E) let l(x) = inf{i : fi (x)/yi = max{fj (x)/yj : 0 ≤ j ≤ n}}. Then l is a Sperner mapping. Note that f (x)l(x) ≥ yl(x) .

13.6 Schauder’s Fixed Point Theorem

171

By Sperner’s lemma, there exists a decreasing sequence ( k (Vk ))∞ k=1 of completely labelled n-simplices, with diam( k (Vk )) → 0. Then ∩∞ k=1 k (Vk ) is a singleton {x}. We show that f (x) = y. (k) (k) (k) We label Vk = {v0 , . . . vn } in such a way that l(vi ) = i for 0 ≤ i ≤ n. If 0 ≤ i ≤ n then (k)

(k)

(k)

fi (x) = fi (vi ) + ( fi (x) − fi (vi )) ≥ yi − | fi (x) − fi (vi )|. But f (v(k) ) → fi (x) as k → ∞, and so fi (x) ≥ yi , for 0 ≤ i ≤ n. But n n i i i=0 fi (x) = 1 = i=0 yi , and so fi (x) = yi for 0 ≤ i ≤ n. Corollary 13.5.3 There is no retract of K onto ∂K. In fact, we can do better. Suppose that K is a compact convex subset of Rm . Let F = span{x − y : x, y ∈ K}, and let ∂rel K be the boundary of K in F; ∂rel K is the relative boundary of K. Corollary 13.5.4 Suppose that K is a compact convex subset of Rm , that f : K → K is continuous and that f (x) = x for x ∈ ∂rel K. Then f is surjective: f (K) = K. Thus, unless K is a singleton, there is no retract of K onto ∂rel (K). We can use Theorem 13.5.2 to give another proof of Brouwer’s fixed point theorem. This time, we can take K to be the unit ball B(Rn ) of Rn with its Euclidean norm. Suppose that f : K → K does not have a fixed point. If x ∈ K, the ray {f (x) + t(x − f (x)) : t ≥ 1} meets ∂K. Let tx = inf{t ≥ 0 : x + t(x − f (x)) ∈ ∂K}, and let r(x) = x + tx (x − f (x)). Then r is a retract of K onto ∂K. It remains to show that r is continuous. It is easy to see that the function x → tx is continuous when x ∈ K int ; let us consider the case where x ∈ ∂K. Suppose that  > 0. There exists 0 < δ <  such that |x − f (x)| > 2δ, and there exists 0 < η < δ such that if |x − y| < η then |r(y) − x| = |r(y) − r(x)| < δ < . Thus r is a continuous retract of K onto ∂K, giving a contradiction.

13.6 Schauder’s Fixed Point Theorem We now extend the Brouwer fixed point theorem to infinite-dimensional spaces. Theorem 13.6.1 (Schauder’s fixed point theorem) Suppose that (E, F) is a dual pair, and that K is a σ (E, F)-compact convex metrizable subset of E. If f : K → K is continuous, then it has a fixed point; there exists x ∈ K with f (x) = x.

172

Some Fixed Point Theorems

Proof By Theorem 12.1.1, we can suppose that K is a compact subset of L, where L = {x : x ∈ l2 , |xn | ≤ 1/n for n ∈ N}. Let (en )∞ n=1 be the standard orthonormal basis for l2 , and let En = span(e1 , . . . , en ). Let jn : En → l2 be the inclusion mapping, and let Pn be the orthogonal projection of l2 onto En . Then Ln = Pn (L) ⊆ L. Let fn = Pn ◦ f ◦ jn . Then fn is a continuous mapping from Ln to itself, and so, by Brouwer’s fixed point theorem, it has a fixed point xn . Note that

xn − f (xn )2 = Pn w( f (xn )) − f (xn )2 ≤

∞ 

| f (xn )|2j ≤ 1/n.

j=n+1

Since M is compact, there exists a convergent subsequence (xnj )∞ j=1 , x ) → f (x), so that − f (x) = convergent to x say. But then f (x nj     limj→∞ xnj − g(xnj ) = 0, and f (x) = x. Exercise 13.6.2 Give an example of a continuous mapping f from the unit sphere S(H) of a Hilbert space H into itself which has no fixed point. Exercise 13.6.3 Let T be a continuous mapping of a closed convex subset A of a Banach space (E, .) into itself for which T(A) is compact. Show that T has a fixed point. Theorem 13.6.4 Suppose that (E, .) is a Banach space, that F is a linear subspace of E for which (E, F) is a dual pair and the unit ball B(E) is σ (E, F)compact and separable. If f : B(E) → B(E) is continuous for the topology σ (E, F) and f (x) = x for x ∈ S(E) = {x : x = 1} then f maps B(E) onto B(E). Proof By Theorem 12.1.1, there exists a linear map R : E → l2 such that K = R(B(E)) ⊆ L = {x : |xn | ≤ 1/n for n ∈ N}, and such that R is a homeomorphism of (B(E), σ (E, F)) onto the compact set (K, .2 ). Let M = R(S(E)), and let g = R ◦ f ◦ R−1 : K → K. Let Pn : l2 → l2 be the projection l2 → l2 described in Theorem 7.6.5, let Kn = Pn (K), let Mn = Pn (M) and let gn = Pn ◦ g : K → Kn . Suppose that x ∈ K. Let xn = Pn (x). Note that ∂rel Kn ⊆ Mn . Thus gn maps Kn continuously into Kn , and gn (y) = y for y ∈ ∂rel Kn . It therefore follows from Brouwer’s theorem that there exists wn ∈ K such that if zn = Pn (wn ), then gn (zn ) = xn . Now the sequence (wn )∞ n=1 is contained in the compact set K, and so there is a subsequence (wnj )∞ which j=1 converges in norm to an element w of K; then zn → w as well. If n < nj then Pn (xnj ) = xn , so that gn (znj ) = xn . Thus gn (w) = xn , and so g(w) = x; g maps K onto K, and so f maps B(E) onto B(E).

13.7 Fixed Point Theorems of Markov and Kakutani

173

Corollary 13.6.5 There is no retract of (B(E), σ (E, F)) onto the unit sphere S(E) = {x ∈ E : x = 1}. This theorem applies to weak* duals of separable Banach spaces, and so to separable reflexive Banach spaces.

13.7 Fixed Point Theorems of Markov and Kakutani Schauder’s fixed point theorem shows that there is at least one fixed point, but gives no further information about the set of fixed points. The situation is different when the mapping is affine; here we do not need to suppose that K is metrizable. Theorem 13.7.1 (The Markov–Kakutani fixed point theorem) Suppose that (E, F) is a separated dual pair, that K is a non-empty σ (E, F)-compact convex subset of E and that T is a weakly continuous affine mapping of E into itself for which T(K) ⊆ K. Then the set of fixed points of T in K is a non-empty convex compact subset of K. Proof If n ∈ N and T ∈ T , let An (T) = (1/n)(I + T + T 2 + · · · + T n−1 ). Then An (T)(K) is a σ (E, F)-compact subset of K. If T1 , . . . , Tk ∈ T , (n1 , . . . , nk ) ∈ Zk , and σ is a permutation of {1, . . . , n} with σ (1) = j then Anj Tj (K) ⊇ Anσ (1) (Tσ (1) ) . . . Anσ (k) (Tσ (k) )(K) = An1 (T1 ) . . . Ank (Tk )(K), so that ∩kj=1 Anj (Tj )(K) ⊇ An1 (T1 ) . . . Ank (Tk )(K) = ∅. Using the finite intersection property, C = ∩{An (T)(K) : T ∈ T , n ∈ N} is a non-empty closed convex subset of K. We show that if x ∈ C then T(x) = x for all T ∈ T . Suppose not, and that T(x0 ) = x0 , for x0 ∈ C. There exists f ∈ F such that T(x0 ) − x0 , f  = 0. f is bounded on K; let M = sup{| x − y, f  | : x, y ∈ K}. Since x0 ∈ C, if n ∈ N, there exists yn ∈ K such that An (T)(yn ) = x0 . But then x0 − T(x0 ) = (1/n)(yn − T n (yn )), so that | T(x) − x, f  | =

! M 1 | yn − T n (yn ), f | ≤ . n n

Since this holds for all n ∈ N, we have a contradiction.

174

Some Fixed Point Theorems

This is a start; but the non-commutative case remains. We need another fixed point theorem. First, we obtain some easy standard results about weak* topologies. Exercise 13.7.2 Suppose that (E, .) is a normed space, and that A is the collection of totally bounded subsets of E. For φ ∈ E let Btb (φ) = {φ + A◦ : A ∈ A}. There is a topology τtb (E ) on E , such that Btb is a base of neighbourhoods of φ, for each φ ∈ E . The topology ttb is finer than the weak* topology σ (E , E). On bounded subsets of E they are the same. Proposition 13.7.3 Suppose that (E, .) is a normed space and that B is a norm-bounded subset of E . Then the restrictions of τtb and σ (E , E) to B are the same. Proof By scaling, we can assume that B is contained in the unit ball of E . Suppose that φ ∈ B and that φ + A◦ is a τtb neighbourhood of φ. We must show that there exists a finite set A0 in E such that if ψ ∈ B and maxx∈A0 | x, φ − ψ | < 12 then ψ ∈ φ + A◦ . There exists a finite 1/4-net A0 in A. Let N = {ψ ∈ B : | a0 , φ − ψ | < 1/2 for a0 ∈ A0 }. N is a weak* neighbourhood of φ in B. If a ∈ A, there exists a0 ∈ A0 such that a − a0  ≤ 1/4. Thus if ψ ∈ N | a, φ − ψ | ≤ | a0 , φ − ψ | + | a − a0 , φ | + | a − a0 , ψ | ≤ 1, and so N ⊂ φ + A◦ . Theorem 13.7.4 (Kakutani’s fixed point theorem) Suppose that (E, .) is a separable Banach space, that K is a non-empty bounded closed convex subset of the dual E , and that S is a bounded semigroup in L(E) with the property that {T(x) : T ∈ S} is totally bounded, for each x ∈ E, and such that T  (K) ⊆ K for each T ∈ S. Then there exists φ ∈ K such that T  (φ) = φ, for each T ∈ S. Proof We give a proof which avoids the use of the axiom of choice. We can suppose that I ∈ S. Let (xn )∞ n=1 be a dense sequence in B(E). Then it is easy to see the set A = {T(xn )/2n : n ∈ N, T ∈ S} is a totally bounded subset of E. Let pA (φ) = sup{|φ(a)| : a ∈ A}. Since (xn )∞ n=1 is a dense sequence in B(E), pA is a norm on E ; applying Proposition 13.7.3, it defines the weak* topology on K. Let K be the set of non-empty weak*-closed (and so weak*-compact) convex S invariant subsets of K. We order K by inverse inclusion. If C is a chain in K, then ∩L∈C L is an upper bound for C, and so we can apply the Br´ezis–Browder

13.8 The Ryll–Nardzewski Fixed Point Theorem

175

lemma (Theorem 4.3.1) to the function d, where d(L) = diam(L); there exists L ∈ K such that if M ∈ K and M ⊆ L then diam(M) = diam(L). We now show that L is a singleton set, which establishes the result. Suppose not, so that δ = diam(L) > 0. Let D = {φ : pA (φ) ≤ δ}. Thus L − L ⊆ D, and if 0 < α < 1 then L − L ⊆ αD. Note that if T ∈ S then T  (D) ⊆ D: D is S-invariant. The sets {D/2 + φ : φ ∈ L} cover L, and so there is a finite subcover {D/2 + φj : 1 ≤ j ≤ k}. Let φ0 = (1/k)(φ1 + · · · + φk ). Then p0 ∈ L. If φ ∈ K0 , there exists 1 ≤ j ≤ k such that φ ∈ D/2 + φj . Thus φ0 − φ =

1 1 ((φ1 − φ) + · · · + (φk − φ)) ∈ (1 − )D, k 2k

and so φ0 ∈ (1 − 1/2k)D + φ. Now let   1 L1 = L ∩ ∩{(1 − )D + φ : φ ∈ K0 } . 2k Then L1 is a compact convex subset of L0 , and is non-empty, since φ0 ∈ L1 . Since D is S-invariant, so is L1 . If φ, ψ ∈ L1 then φ ∈ (1−1/2k)D+ψ, so that L1 −L1 ⊆ (1−1/2k)D, and diamL1 < δ, giving the required contradiction. We shall use this in Section 16.12 to show that if (G, ×) is a compact metrizable topological group, there exist a left Haar measure and right Haar measure on G.

13.8 The Ryll–Nardzewski Fixed Point Theorem We prove another fixed point theorem. First, we need a definition. Suppose that E is a Banach space, that S is a semigroup in L(E) and that K is a convex S-invariant subset of E. Then S is non-contracting on K if for each k1 , k2 ∈ K, with k1 = k2 , there exists δ > 0 such that T(k1 ) − T(k2 ) ≥ δ, for each T ∈ S. Theorem 13.8.1 (The Ryll-Nardzewski fixed point theorem) Suppose that E is a separable Banach space, that S is a semigroup in L(E), that K is a non-empty weakly compact convex S-invariant subset of E and that S is non-contracting on K. Then S has a fixed point in K. Proof First observe that it is enough to prove the result when S is finitely generated. If S1 is a finitely generated sub-semigroup, then {k ∈ K : T(k) = k for T ∈ S1 } is a non-empty weakly closed convex subset of K, and the

176

Some Fixed Point Theorems

collection of all such sets has the finite intersection property. Any point in the intersection is then a fixed point for S. Suppose then that S is generated by T1 , . . . , Tk . Let A = (T1 + · · · + Tk )/k. Then K is A-invariant, and so it has a fixed point x0 , by the Markov fixed point theorem. We claim that Ti (x0 ) = x0 for 1 ≤ i ≤ k. Suppose not. By relabelling, we can suppose that Ti (x0 ) = x0 for 1 ≤ i ≤ j and that Ti (x0 ) = x0 for j < i ≤ k. Let B be the average (T1 + · · · + Tj )/j. Then kx0 = kA(x0 ) = jB(x0 ) + (k − j)x0 , so that B(x0 ) = x0 . Let S2 be the semigroup generated by T1 , . . . , Tj . There exists δ > 0 such that TTi (x0 ) − T(x0 ) > δ for all T ∈ S1 and for 1 ≤ i ≤ j. Let L = {T(x0 ) : T ∈ S2 }: then L is a separable weakly compact convex subset of K, and x0 = B(x0 ) ∈ L. By Corollary 12.3.3 there exists a slice M of L of diameter less than δ, and there exists T ∈ S1 such that T(x0 ) ∈ M. But T(x0 ) = TB(x0 ) = (TT1 (x0 ) + · · · + TTj (x0 ))/j, so that there exists 1 ≤ i ≤ j such that TTi (x0 ) ∈ M. Thus T(x0 ) − TTi (x0 ) < δ, giving a contradiction. The Ryll-Nardzewski theorem can be established in other settings. Exercise 13.8.2 Suppose that S is a semigroup in L(E), that {T(x) : T ∈ S} is totally bounded, for each x ∈ E, and that S  = {T  : T ∈ S} and that K is a nonempty weak*-compact convex S  -invariant subset of E , with the following non-contraction property: for each k1 , k2 ∈ K, with k1 = k2 there exists a totally bounded subset A of E such that pA (T  (k1 ) − T  (k2 )) ≥ 1, for each T ∈ S. Let A be the totally bounded subset of E described in Theorem 13.7.4. Show that there exists a weak*-closed slice M in K for which PA (x − y) ≤ 1 for x, y ∈ M. Arguing as in Theorem 13.8.1, deduce that S  has a fixed point in K. This again will give a proof of the existence of Haar measure on a compact Hausdorff topological group.

PART TWO Measures on Polish Spaces

14 Abstract Measure Theory

In this chapter, we give a brief account of the theory of measures on abstract sets. It contains definitions and statements of results, and the notation that is used. Abstract measure theory proceeds by many simple steps; many of these are set as exercises, although some comments are made. Proofs can be found in [G III], [Bi II],[H] or [B].

14.1 Measurable Sets and Functions First, we describe the notation that we shall use. If f : X → Y is a mapping, if c ∈ Y and if A ⊆ Y, we write ( f = c) for the set {x ∈ X : f (x) = c} and ( f ∈ A) for the set {x ∈ X : f (x) ∈ A}, and use similar notation for other such sets. A non-empty collection  of subsets of a non-empty set X is called a σ -ring if ∞ (i) if (Aj )∞ j=1 is a sequence of sets in  then ∪j=1 Aj ∈ , and (ii) if A, B ∈  then A \ B ∈ .

If (iii) X ∈ , then  is a σ -field. Exercise 14.1.1 Suppose that  is a non-empty collection of subsets of a nonempty set X. The following are equivalent. (i)  is a σ -field. (ii) If A ∈  then X \A ∈ , and if (An )∞ n=1 is a sequence of disjoint elements ∞ of  then ∪n=1 An ∈ . (iii) If A ∈  then X \ A ∈ , and if (An )∞ n=1 is an increasing sequence in  A ∈ . then ∪∞ n n=1 179

180

Abstract Measure Theory

Exercise 14.1.2 Let s be a collection of σ -fields on a set X. Show that ∩{ :  ∈ s} = {A : A ∈ , for all  ∈ s} is a σ -field. Is the union of two σ -fields necessarily a σ -field? A measurable space is a pair (X, ), where X is a set and  is a σ -field of subsets of X. Proposition 14.1.3 Suppose that F is a set of subsets of a set S. Then there is a smallest σ -field σ (F) of subsets of S which contains F. Proof Let s be the collection of those σ -fields of subsets of S which contain F. It is non-empty, since the collection P(S) of all subsets is in s. Let σ (F) = ∩{ :  ∈ s} = {A : A ∈ , for all  ∈ s}. Then σ (F) is a σ -field which belongs to s. It is clearly the smallest element of s. The σ -field σ (F) is called the σ -field generated by F. An important feature of this proposition is that its proof is indirect, and gives no indication of the structure of sets in σ (F). This fact gives a particular flavour to much of measure theory. A mapping f : (X1 , 1 ) → (X2 , 2 ) from a measurable space (X1 , 1 ) to a measurable space (X2 , 2 ) is said to be measurable if f −1 (A) ∈ 1 for each A ∈ 2 . If so, then f = {f −1 (A) : A ∈ 2 } is a sub-σ -field of 1 . If 2 = σ (F), then f is measurable if and only if f −1 (F) ∈ 1 for each F ∈ F. The σ -field B generated by the collection of open subsets of a topological space (X, τ ) is called the Borel σ -field of (X, τ ). In the preceding definition, if X1 is a topological space, and 1 is its Borel σ -field, then f is said to be Borel measurable. When f is real-valued, then we always assume that 2 is the Borel σ -field; thus f is measurable if and only if f −1 (c, ∞) ∈ 1 for each c ∈ Q; that is, if and only if ( f > c) ∈ 1 for each c ∈ Q. If f is a continuous map from a topological space (X, τ ) into a topological space (Y, σ ), then f is a Borel measurable map of X into Y, with its Borel σ -field, but a continuous map does not necessarily map Borel sets to Borel sets. Lebesgue mistakenly claimed that if A is a Borel set in the plane R2 , then p1 (A) is a Borel subset of the line R (where p1 is the projection onto the first co-ordinate). Suslin invented the notion of analytic set (which we shall not consider) to show that this is not so. Borel sets are complicated! The composition of measurable mappings is measurable.

14.1 Measurable Sets and Functions

181

Exercise 14.1.4 Suppose that (X, ) is a measurable space, that Y is a set and that f : X → Y is a mapping. Show the following. (i) {A ⊆ Y : f −1 (A) ∈ } is a σ -field. (ii) If F is a collection of subsets of Y, and f −1 (A) ∈  for A ∈ F, then f −1 (A) ∈  for A ∈ σ (F); the mapping f : (X, ) → (Y, σ (F)) is measurable. Exercise 14.1.5 Let An = {ω ∈ (N) : ωk = 1 for k > n}, and let B = (N)\ ∞ −n ∪∞ n=1 2 ωn n=1 An . Show that B is a Polish subspace of (N), that if j(ω) = then j is a bijection of B onto [0, 1], and that C ⊂ B is a Borel set if and only if j(C) is. Is j−1 continuous? Exercise 14.1.6 Show that if  is a σ -field in X and A ⊆ X, then the indicator function IA of A is -measurable if and only if A ∈ . Exercise 14.1.7 Suppose that f and g are real-valued measurable functions on (X, ) and that α ∈ R. Show that each of the functions αf , f + , f − , |f |, f + g, fg, f ∨ g, and f ∧ g is -measurable. The sets ( f < g), ( f ≤ g) and ( f = g) are in . For example, ( f ∨ g > c) = ( f > c) ∪ (g > c), and ( f + g > c) = ∪r∈Q (( f > r) ∩ (g > c − r)). Thus the set L0 = L0 (X, ) of real-valued measurable functions on X is a real vector space, when addition and scalar multiplication are defined pointwise. We also consider extended real-valued functions, taking values in R = {−∞} ∪ R ∪ {∞}. We say that such a function f is measurable if ( f ∈ A) ∈  for each Borel set A in R and both ( f = −∞) and ( f = ∞) are in . Exercise 14.1.8 Suppose that ( fn )∞ n=1 is a sequence of extended real-valued -measurable functions on X. Then each of the extended real-valued functions supn∈N fn , infn∈N fn , lim supn∈N fn , and lim infn∈N fn is -measurable. The set C∗ = {x : fn (x) converges in R as n → ∞} is in . Let f (x) = limn→∞ fn (x) if x ∈ C∗ , and let f (x) = 0 otherwise. Then f is -measurable. If each fn is real-valued, then the set C = {x : fn (x) converges in R as n → ∞} is in . Let f (x) = limn→∞ fn (x) if x ∈ C, and let f (x) = 0 otherwise. Then f is -measurable. For example, ∞ ∞ (lim sup fn > c) = ∪∞ k=1 ∩n=1 ∪m=n ( fm > c + 1/k). n→∞

182

Abstract Measure Theory

14.2 Measure Spaces A finite measure space is a triple (X, , μ), where (X, ) is a measurable space and μ is a countably additive, or σ -additive mapping of  into R+ : if (An ) is ∞ a sequence of disjoint elements of  then μ(∪∞ n=1 μ(An ). (Note n=1 An ) = that, since all the summands are non-negative, the sum does not depend upon the order of summation.) The function μ is called a measure. A probability space is a finite measure space (, , P) for which P() = 1; P is called a probability measure, and a measurable function on a probability space is called a random variable. Exercise 14.2.1 Suppose that (X, , μ) is a finite measure space. Suppose that A, B ∈ , and that (An )∞ n=1 is a sequence in . (i) μ(A ∪ B) + μ(A ∩ B) = μ(A) + μ(B). (ii) If B ⊆ A then μ(B) ≤ μ(A). (iii) (Upwards continuity) If (An )∞ n=1 is an increasing sequence then μ(∪∞ n=1 An ) = sup μ(An ). n∈N

(iv) (Downwards continuity) If

(An )∞ n=1

is a decreasing sequence then

μ(∩∞ n=1 An ) = inf μ(An ). (v) (vi)

supn∈N μ(An ) ≤ μ(∪∞ n=1 An ) ∞ A ) ≤ inf μ(∩∞ n=1 μ(An ). n=1 n

≤

∞

n∈N

n=1 μ(An ).

For example, if (An )∞ n=1 is an increasing sequence, let D1 = A1 and Dn = An \ An−1 for n > 1. Then ∞ μ(∪∞ n=1 An ) = μ(∪n=1 Dn ) =

∞  n=1

μ(Dn ) = lim

n→∞

n 

μ(Dj ) = lim μ(An ). n→∞

j=1

Exercise 14.2.2 Show that if f is a measurable real-valued function on X, then μ(|f | > n) → 0 as n → ∞. Recall that if (An )∞ n=1 is a sequence of subsets of a set X then



 ∞ ∞ ∞ lim sup An = ∩∞ n=1 ∪j=n Aj and lim inf An = ∪n=1 ∩j=n Aj . n→∞

n→∞

Theorem 14.2.3 (The first Borel–Cantelli lemma) If μ(lim supn→∞ An ) = 0. Proof For μ(lim supn→∞ An ) ≤ μ(∪∞ n=m An ) ≤ n → ∞.

∞

n=1 μ(An )

∞

m=n μ(Am )

< ∞ then → 0 as

14.2 Measure Spaces

183

A σ -finite measure space is a measurable space (X, ), together with a sequence (Ik )∞ k=1 of disjoint elements of  whose union is X, and a function μ on  (a σ -finite measure), taking values in [0, ∞], with the properties that (i) μ is countably additive; if (An )∞ n=1 is a sequence of disjoint elements of   μ(A ), and then μ(∪n∈N An ) = ∞ n n=1 (ii) μ(Ik ) < ∞ for k ∈ N. Thus A ∈  if and only if A ∩ Ik ∈  for each k ∈ N; if so, then μ(A) = ∞ k=1 μ(A ∩ Ik ). In future we shall use the term ‘measure’ to mean either a finite measure or a σ -finite measure, adding ‘finite’ when necessary. Proposition 14.2.4 Suppose that φ is a measurable mapping from a finite measure space (X, , μ) into a measurable space (Y, T). If A ∈ T, let φ∗ μ(A) = μ(φ −1 (A)). Then φ∗ μ is a finite measure on T. Suppose that φ is a measurable mapping from a σ -finite measure space (X, , μ) into a measurable space (Y, T). Suppose also that there exists an −1 increasing sequence (An )∞ n=1 in T with union Y such that μ(φ (An )) < ∞ −1 for each n ∈ N. If A ∈ T, let φ∗ μ(A) = μ(φ (A)). Then φ∗ μ is a σ -finite measure on T. Exercise 14.2.5 Prove Proposition 14.2.4. The measure φ∗ μ is called the image measure, or push-forward measure. If φ is real-valued, then φ∗ μ is called the distribution of φ. In the case where μ is a probability measure, it is also called the law of φ. Suppose that (X, , μ) is a measure space. An element N of  is a null set if μ(N) = 0. The collection of null sets is a σ -ring contained in . A real-valued measurable function f is a simple function if it only takes finitely many values, each on a measurable set of finite measure; f = n i=1 λi IAi , where λi ∈ R and Ai is a measurable set of finite measure, for each i. Theorem 14.2.6 Suppose that f is a real-valued measurable function on a σ -finite measurable space (X, ). There exists a sequence (gn )∞ n=1 of simple functions which converges pointwise to f . If f is non-negative, then the sequence (gn )∞ n=1 can be taken to be a pointwise increasing sequence of nonnegative simple functions. Proof Let (An )∞ n=1 be a disjoint sequence of sets of finite measure which cover n X, let Cn = ∪j=1 Aj and let Bn,j = (j/2n < f ≤ (j + 1)/2n ) for |j| ≤ 2n . Then j=2n −1 set gn = j=−2n (j/2n )ICn ∩Bn,j .

184

Abstract Measure Theory

14.3 Convergence of Measurable Functions If a property holds on a measure space, except possibly on a null set, we say that it holds almost everywhere (probabilists use the term almost surely). Thus 0 0 if ( fn )∞ n=1 is a sequence in L (X, , μ) and if f ∈ L (X, , μ) then fn → f almost surely, as n → ∞ if there exists a null set N such that fn (x) → f (x) for all x ∈ X \ N. Suppose that (X, , μ) is a finite measure space and that ( fn )∞ n=1 is a 0 sequence in L (X, , μ). We say that fn → f almost uniformly if for each  > 0 there exists A ∈  with μ(A) <  such that fn → f uniformly on X \ A as n → ∞. This is bad terminology, but is well established. Theorem 14.3.1 (Egorov’s theorem) Suppose that (X, , μ) is a finite measure 0 0 space, that ( fn )∞ n=1 is a sequence in L (X, , μ) and that f ∈ L (X, , μ). Then fn → f almost everywhere as n → ∞ if and only if fn → f almost uniformly as n → ∞. Proof Suppose that fn → f almost everywhere as n → ∞ and that  > 0. Let Aj,k = ∩i≥j (|fi − f | ≤ 1/k) for j, k ∈ N. Then (Aj,k )∞ j=1 is an increasing A ) = 0, so that for each k ∈ N there exists jk such sequence and μ(X \ ∪∞ n,j j=1 ∞ k that μ(X \ Ajk ,k ) < /2 . Then if A = ∪k=1 Ajk ,k , μ(X \ A) <  and fn → f uniformly on A as n → ∞. If fn → f almost uniformly then for each k ∈ N there exists a set Bk with μ(X \ Bk ) < 1/k such that fn → f uniformly on Bk as n → ∞. Let B = ∪∞ k=1 Bk . Then μ(X \ B) = 0 and fn → f pointwise on B as n → ∞. Suppose that (X, , μ) is a finite measure space, that ( fn )∞ n=1 is a sequence in L0 and that f ∈ L0 . We say that fn → f in measure if, for each c > 0, μ((|fn − f | > c) → 0 as n → ∞. In the case where (X, , P) is a probability space, ‘convergence in measure’ is called convergence in probability. Theorem 14.3.2 Suppose that (X, , μ) is a finite measure space, that ( fn )∞ n=1 is a sequence in L0 and that f ∈ L0 . If fn → f almost everywhere as n → ∞, then fn → f in measure. If fn → f in measure as n → ∞, there exists a subsequence ( fnk )∞ k=1 which converges almost everywhere to f as k → ∞. Proof If fn → f almost everywhere then fn → f in measure, by Egorov’s theorem. Conversely, suppose that fn → f in measure. For each k ∈ N there exists nk such that μ(|fn − f | > 1/k) < 1/2k for n ≥ nk , and we can assume that (nk )∞ k=1 is an increasing sequence. Let Bk = (|fnk − f | > 1/k), so that ∞ k=1 μ(Bk ) < ∞. By the first Borel–Cantelli lemma, μ(lim supk→∞ Bk ) = 0. But if x ∈ lim supk→∞ Bk then fn (x) → f (x) as n → ∞.

14.3 Convergence of Measurable Functions

185

If f ∈ L0 (X, , μ), f is called a null function if f = 0 almost everywhere. The set N 0 (X, , μ) of null functions is a linear subspace of L0 (X, , μ); the quotient is denoted by L0 (X, , μ). As is customary, we identify a measurable function f with its equivalence class [f ]. Although the elements of L0 (X, , μ) are equivalence classes of functions, we shall consider them and treat them as functions, identifying functions that are equal almost everywhere. For 0 example, if ( fn )∞ n=0 is a sequence in L (X, , μ), we say that fn → f0 almost surely if whenever gn is an element of L0 (X, , μ) which is in the equivalence class fn , then gn (x) → g0 (x) for almost all x. It is easy to see that this holds for one sequence of representatives if and only if it holds for any sequence of representatives. The next result lies at the heart of measure theory. If μ is a finite measure on (X, σ ) then convergence in measure in L0 (X, , μ) can be characterized by a pseudometric, and the corresponding metric on L0 (X, , μ) is complete. Theorem 14.3.3 Suppose that (X, , μ) is a finite measure space. If f , g ∈ L0 (X, , μ), let ρ0 ( f , g) = inf{ > 0 : μ(|f − g| > ) ≤ }. Then ρ0 is a pseudometric and ρ( f , g) = 0 if and only if f = g almost 0 everywhere. If ( fn )∞ n=1 is a sequence in L (X, , μ), then ρ( fn , f ) → 0 if and only if fn → f in measure. If d0 is the corresponding metric on L0 , then (L0 (X, , μ), d0 ) is complete. The simple measurable functions are dense in (L0 (X, , μ), d0 ). Proof Suppose that f , g, h ∈ L0 (X, , μ). Certainly ρ0 ( f , g) = ρ0 (g, f ). Suppose that 1 > ρ0 ( f , g) and that 2 > ρ(g, h). Then there exist ρ( f , g) ≤ η1 ≤ 1 and ρ(g, h) ≤ η2 ≤ 2 such that if A1 = {x : |f (x) − g(x)| > η1 then μ(A1 ) ≤ η1 , and similarly for A2 . Then {x : |f (x) − h(x)| > 1 + 2 } ⊆ A1 ∪ A2 so that μ({x : |f (x) − h(x)| > 1 + 2 }) ≤ η1 + η2 ≤ 1 + 2 and so ρ0 ( f , h) ≤ ρ0 ( f , g) + ρ0 (g, h). Thus ρ0 is a pseudometric. Clearly, ρ( fn , f ) → 0 if and only if fn → f in measure, and ρ( f , g) = 0 if and only if f = g almost everywhere. Suppose now that ( fn )∞ n=1 is a d0 -Cauchy sequence. Then there exists a such that d0 ( fnk , fnk+1 ) ≤ 2−k , for each k. Let Ck = {x : subsequence ( fnk )∞ k=1 ∞ −k |fnk (x) − fnk+1 (x)| > 2 , and let C = lim sup∞ k=1 μ(Ck ) < ∞, k=1 Ck . Since it follows from the first Borel–Cantelli lemma that μ(C) = 0. Suppose that (X, , μ) is a finite measure space. If x ∈ C then there exists k0 such that |fnk (x) − fnk+1 (x)| ≤ 2−k for k ≥ k0 . Thus |fnk (x) − fnl (x)| ≤ 2.2−k for l > k > k0 , and ( fnk (x))∞ k=1 is a real Cauchy sequence, which converges to f (x), say.

186

Abstract Measure Theory

Set f (x) = 0 for x ∈ C. Then f is measurable, and fnk → f almost everywhere. Thus fnk → f almost uniformly, by Egorov’s theorem, and so d0 ( fnk , f ) → 0. Consequently d( fn , f ) → 0 as n → ∞, and (L0 (X, , μ), d0 ) is complete. Suppose that f ∈ L0 (X, , μ). Let Aj,n = (j/2n ≤ f < (j + 1)/2n ), and let fn be the simple function fn =

n n−1 2

(j/2n )IAj,n .

j=−2n n

Then |f − fn | > 1/n only if |f | > n, so that d0 ( f , fn ) → 0 as n → ∞, since μ(|f | > n) → 0 as n → ∞. As an immediate result of Theorem 14.3.2, we have the following. Theorem 14.3.4 A subset A of a finite measure space (X, , μ) is d0 -compact if and only if every sequence in A has a subsequence which converges almost everywhere to an element of A. Proposition 14.3.5 Suppose that (X, , μ) is a finite measure space, that ∞ 0 0 ( f n )∞ n=1 and (gn )n=1 are sequences in L (X, , μ) and that f , g ∈ L (X, , μ). If fn → f and gn → g almost everywhere, as n → ∞, then fn +gn → f +g and fn gn → fg almost everywhere, as n → ∞. If fn → f and gn → g in measure, as n → ∞, then fn + gn → f + g, fn2 → f 2 and fn gn → fg in measure, as n → ∞. Proof Addition is easy, and left to the reader. Given  > 0, there exists C > 1 such that μ(|f | > C) < /2, and there exists n0 such that μ(|fn −f |) > /(2C+ ) < /2 for n ≥ n0 . Since |fn2 − f 2 | = |fn − f |.|fn + f |, μ(|fn2 − f 2 | ≥ ) <  if n ≥ n0 ; fn2 → f 2 in measure, as n → ∞. It then follows from polarization that fn gn → fg in measure, as n → ∞. We can extend these ideas to σ -finite measure spaces. Suppose that (X, , μ) is a σ -finite measure space, with a sequence (Ik )∞ k=1 of disjoint elements of  of finite positive measure, whose union is X. Suppose that ( fn )∞ n=1 is a sequence in L0 (X, , μ), and that f ∈ L0 (X, , μ). Let πk ( f ) = f|Ik . If ( fn )∞ n=1 is a sequence in L0 (X), then fn is said to converge to f locally in measure if and only if πk ( fn ) → πk ( f ) in measure, as n → ∞, for each k ∈ N. Local convergence in measure can be defined by a complete product metric, such as, for example, d(0,loc) ( f , g) =

∞  k=1

2−k inf{ > 0 : μ((|f − g| > ) ∩ Ik ) < }.

14.4 Integration

187

14.4 Integration Suppose that f is a non-negative extended-real-valued measurable function defined on a finite or σ -finite measure space (X, , μ). The tail distribution function λf is defined as λf (t) = μ(f > t) = f∗ (μ)((t, ∞]), for t ∈ [0, ∞).  If λf (t) = ∞ for some t > 0 we define X f dμ = ∞. Otherwise, the function λf is a decreasing right-continuous real-valued function on [0, ∞). We define   ∞ f dμ = λf (t) dt. X

0

Here, the integral on the right is an improper Riemann integral, taking values in [0, ∞]. n Exercise 14.4.1 Suppose that f = i=1 vi IAi is a non-negative  simple measurable function on a measure space (X, , μ). Show that X f dμ = n i=1 vi μ(Ai ), and that the integral does not depend on the representation of f . Show that if f and g are non-negative simple functions on  measurable   a measure space (X, , μ) then X ( f + g) dm = X f dμ + X g dμ.   If A ∈ , we set A f dμ = X fIA , dμ, where IA is the indicator function of A.   If 0 ≤ f ≤ g almost everywhere then λf ≤λg , and so X f dμ ≤ X g dμ. In particular, if f = g almost  everywhere then X f dμ = X g dμ, and if f = 0 almost everywhere then X f dμ = 0. We now come to the fundamental theorem of integration theory. First we need an elementary result about Riemann integrals. Exercise 14.4.2 Suppose that (λn )∞ n=1 is an increasing sequence of decreasing real-valued functions on an interval [a, b] which converges pointwise to the real-valued  b function λ.b By considering upper and lower Riemann sums, show that a λn (t)dt → a λ(t)dt. (Here the integrals are Riemann integrals.) Theorem 14.4.3 (The monotone convergence theorem) Suppose that ( fn )∞ n=1 is an increasing sequence of non-negative measurable functions on a finite or σ -finite measure space (X, , μ) which converges pointwise almost everywhere to a function f . Then   fn dμ → f dμ as n → ∞. X

X

Proof After approximating at 0 and ∞, this follows from the exercise.

188

Abstract Measure Theory

Corollary 14.4.4 (Fatou’s lemma) Suppose that ( fn )∞ n=1 is a sequence of nonnegative measurable functions on a finite or σ -finite measure space (X, , μ). Then   fn dμ. lim inf fn dμ ≤ lim inf n→∞

X

n→∞

In particular, if fn → f almost everywhere then

 X

f dμ ≤ lim infn→∞



fn dμ.

Proof Let gn = infj≥n fj . Then gn ≤ fn and (gn )∞ n=1 increases pointwise to lim inf fn . Thus    lim inf fn dμ = lim gn dμ ≤ lim inf fn dμ. n→∞ X

X

n→∞ X

14.5 Integrable Functions Suppose that f is a real-valued, but not necessarily non-negative, measurable function f on a finite or σ -finite measure space (X, , μ). Then we set      • X f dμ = X f +dμ − X f − dμ if X f + dμ < ∞ and X f − dμ < ∞; • X f dμ = ∞ if Xf + dμ = ∞ and Xf − dμ < ∞; ∞ and X f − dμ ∞; • X f dμ = −∞ if X f + dμ  <  = + − • X f dμ is not defined if X f dμ = ∞ and X f dμ = ∞.  +  − The function f is said to be integrable if f dμ < ∞ and X X f dμ < ∞;  this is clearly the case if and  only if X |f | dμ < ∞. If (, , P) is a probability space, we write E( f ) for  f dP; then E( f ) is the expectation of f . Proposition 14.5.1 If f and g are integrable, if h is a bounded measurable function and if α ∈ R then hf , αf and f + g are integrable, and      αf dμ = α f dμ and ( f + g) dμ = f dμ + g dμ. X

X

If f ∈ N 0 then f is integrable and



X X

X

X

f dμ = 0.

Proof Scalar multiplication is easy. The addition formula holds for nonnegative simple functions, then by the monotone convergence theorem for nonnegative integrable functions, and finally for arbitrary functions by considering positive and negative parts. Exercise 14.5.2 Suppose that f is a real-valued measurable function f on a  finite or σ -finite measure space (X, , μ). Show that X |f | dμ = 0 if and only if f is a null function.

14.5 Integrable Functions

189

We denote the set of integrable functions on (X, , μ) by L1 (X, , μ). The 1 f → proposition shows that L (X, , μ) is a vector space and the mapping 1 0 X f dμ is a linear functional on it. We denote the quotient space L /N by 1 1 [f ]. If f and g L = L (X, , μ). Again, weidentify f with  its equivalence class 1 (X, , μ) then f dμ = g dμ, and so if f ∈ L are equivalent functions then X X   1 X f dμ is a well-defined linear functional on L . The function f 1 = X |f | dμ is then a norm on L1 (X, , μ). Theorem 14.5.3 Suppose that (X, , μ) is a σ -finite measure space. Then (L1 (X, , μ), .1 ) is complete and the simple measurable functions are dense in L1 . Proof L1 (X, , μ) ⊆ L0 (X, , μ), and if (yn )∞ n=1 is a .1 -Cauchy sequence in (L1 (X, , μ), .1 ), then it is a Cauchy sequence in the complete metric space (L0 (X, , μ), d0,loc ). Suppose that (yn )∞ n=1 is a sequence in Mη (y0 ) and that yn − y1 = X |yn − y| dμ → 0 as n → ∞. By extracting a subsequence if necessary, we can suppose that yn → y almost everywhere. But then, by Fatou’s lemma,   |y − y0 | dμ ≤ lim inf |yn − y0 | dμ ≤ η, n→∞ X

X

so that Mη (y0 ) is d0 -closed. The result therefore follows from Theorem 2.4.5. The proof that the simple measurable functions are dense in L1 is just the same as that of Theorem 14.3.3. Although this result can be proved directly just as easily, the proof underlines the importance of the completeness of (L0 (X, , μ), d0,loc ). Theorem 14.5.4 (The dominated convergence theorem) Suppose that ( fn )∞ n=1 is a sequence of measurable functions which converges pointwise almost everywhere to f , and that g is an integrable function such that |fn | ≤ |g|, for each n. Then   fn dμ → Further,

X

 X

f dμ as n → ∞. X

|fn − f | dμ → 0 as n → ∞.

Proof The functions |g| − fn and |g| − f are non-negative and integrable. Appling Fatou’s lemma,   (|g| − f ) dμ ≤ lim inf (|g| − fn ) dμ, n→∞ X

X

so that



 f dμ ≥ lim sup X

n→∞ X

 fn dμ ≥ lim inf

n→∞ X

 fn dμ ≥

f dμ, X

190

Abstract Measure Theory

so that all the terms are equal, and the first result follows. Since |fn − f | ≤ 2|g| and |fn − f | → 0 as n → ∞, so does the second. Corollary 14.5.5 (The bounded convergence theorem) Suppose that ( fn )∞ n=1 is a uniformly bounded sequence of measurable functions on a finite measure space (X, , μ) which converges pointwise almost everywhere to f . Then    fn dμ → f dμ and |fn − f | dμ → 0 as n → ∞. X

X

X

Exercise 14.5.6 Suppose that (X, , μ) is a measure space, and that (An )∞ n=1  is a sequence in  with μ(An ) > 0 for each n and ∞ n ) < ∞. Let n=1 μ(A fn = IAn /μ(An ) − 1, and let an = ( f1 + · · · + fn )/n. Show that X an = 0 for each n, but that an → −1 almost everywhere. Exercise 14.5.7 Suppose that (X, , μ) is a measure space, that (Y, T) is a measurable space, and that φ is a measurable map from X to Y. Suppose 1 1 that  f ∈ L (Y, T, φ∗ μ). Show that f ◦ φ ∈ L (μ), and that X ( f ◦ φ) dμ = Y f d(φ∗ μ).

15 Further Measure Theory

In this chapter, we consider further properties of measures on abstract sets. This material may be less familiar, and so fuller details are given than in the previous chapter.

15.1 Riesz Spaces Many of the Banach spaces that we consider have a natural partial order, which interacts with the topological, metric and geometric properties of the space. For example, a partial order on a C(X) space is given by setting f ≤ g if and only if f (x) ≤ g(x) for all x ∈ X, and a partial order on L0 is given by setting f ≤ g if and only if f (x) ≤ g(x) for almost all x. A partially ordered vector space (E, ≤) is a real vector space E and a partial order ≤ on E which satisfies (i) if x ≤ y then x + z ≤ y + z and (ii) if x ≤ y then ax ≤ ay for all x, y, z ∈ E and all a ≥ 0. Thus x ≤ y if and only if y − x ≥ 0. The set {x ∈ E : x ≥ 0} is the positive cone in E; it is a convex subset of E. An element of E+ is called a positive element. A partially ordered vector space (E, ≤) is Archimedean if whenever x, y ∈ E and nx ≤ y for all n ∈ N then x ≤ 0. We shall assume, without saying so, that all the partially ordered spaces that we shall consider are Archimedean. If x, y, z are elements of a partially ordered set, then z is the least upper bound of x and y if x ≤ z and y ≤ z and if whenever x ≤ z and y ≤ z then z ≤ z . If a least upper bound exists, it is denoted by x ∨ y. A greatest lower bound is defined in the same way; when it exists, it is denoted by x ∧ y. A lattice is 191

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a partially ordered set in which each pair of elements has a least upper bound and a greatest lower bound. A Riesz space is a partially ordered vector space which is a lattice. C(X), L0 (X, , μ) and L1 (X, , μ) are Riesz spaces under their natural orderings. Exercise 15.1.1 Suppose that x, y, z ∈ E, where (E, ≤) is a Riesz space. Show that (x + z) ∨ (y + z) = (x ∨ y) + z. Exercise 15.1.2 Suppose that (E, ≤) is a partially ordered vector space. Show that E is a Riesz space if and only if x and 0 have a least upper bound, for each x ∈ E. If (E, ≤) is a Riesz space then x∨0 is denoted by x+ , and (−x)∨0 is denoted by x− . Proposition 15.1.3 Suppose that (E, ≤) is a Riesz space and that x ∈ E. Then x = x+ − x− and |x| = x+ + x− . Proof Since x+ − x ≥ 0 and x+ − x ≥ −x, x+ − x ≥ x− , and so x+ − x− ≥ x. Similarly, x− + x ≥ x and x− + x ≥ 0, so that x− + x ≥ x+ , and x+ − x− ≤ x. Consequently, x = x+ − x− . Further, x ∨ (−x) = (2x) ∨ 0 − x = 2x+ − x and x ∨ (−x) = 0 ∨ (−2x) + x = − 2x − x; averaging, |x| = x+ + x− . Consequently, E = E+ − E+ . Proposition 15.1.4 Suppose that (E, ≤) is a Riesz space and that f is a realvalued function on E+ which satisfies f (x + y) = f (x) + f (y) and f (ax) = af (x) for x, y ∈ E+ and a ≥ 0. Then there exists a unique linear functional φ on E which extends f ; that is φ(x) = f (x) for x ∈ E+ . Proof If x ∈ E let φ(x) = f (x+ ) − f (x− ). Then φ(−x) = −φ(x), so that φ(ax) = aφ(x) for a ∈ R. Since (x + y)+ + x− + y− = (x + y)− + x+ + y+ , f ((x + y)+ ) + f (x− ) + f (y− ) = f ((x + y)− ) + f (x+ ) + f (y+ ) so that φ(x + y) = φ(x) + φ(y); φ is a linear functional on E which extends f , and φ is unique, since E = E+ − E+ . The following decomposition result has important consequences. Proposition 15.1.5 Suppose that (E, ≤) is a Riesz space, and that x1 , x2 , y ∈ E+ and that 0 ≤ y ≤ x1 + x2 . Then there exist y1 , y2 in E+ such that 0 ≤ y1 ≤ x1 , 0 ≤ y2 ≤ x2 and y = y1 + y2 .

15.1 Riesz Spaces

193

Proof Let y1 = y ∧ x1 , so that 0 ≤ y1 ≤ x1 and let y2 = y − y1 . Then y2 ≥ 0 and x2 − y2 = (x2 − y) + (y ∧ x1 ) = x2 ∧ (x1 + x2 − y) ≥ 0.

Suppose that (E, ≤) is a partially ordered vector space. If x ≤ y, the order interval [x, y] is the set {z : x ≤ z ≤ y}. A subset of E is order bounded if it is contained in an order interval. A linear functional φ on E is positive if φ(x) ≥ 0 whenever x ≥ 0, and is order bounded if it is bounded on each order interval. If φ is positive and [x, y] is an order interval, then φ(x) ≤ φ(z) ≤ φ(y) for z ∈ [x, y], so that a positive linear functional is order bounded. We denote the vector space of order bounded linear functionals by E ˜, and define a partial order on it by setting φ ≤ ψ if ψ − φ is positive. Theorem 15.1.6 If (E, ≤ 0) is a Riesz space, then (E ˜, ≤) is a Riesz space. If φ ∈ E ˜ and x ≥ 0 then φ + (x) = sup{φ(z) : z ∈ [0, x]}. Proof If φ ∈ E ˜ and x ∈ E+ , let f (x) = sup{φ(z) : z ∈ [0, x]}. Since φ is order bounded, f (x) is finite. Clearly f (ax) = af (x) for a ≥ 0. Suppose that x1 , x2 ∈ E+ and that 0 ≤ z ≤ x1 + x2 . By Proposition 15.1.5 there exist z1 ∈ [0, x1 ] and z2 ∈ [0, x2 ] with z1 + z2 = z. Thus f (x1 + x2 ) = sup{φ(z1 + z2 ) : z1 ∈ [0, x1 ], z2 ∈ [0, x2 ]} = f (x1 ) + f (x2 ). By Proposition 15.1.4, there exists a linear function ψ which extends f , and ψ is positive. Suppose that θ ∈ E ˜ and that θ ≥ φ. If x ∈ E+ then θ (z) ≥ φ(z) for z ∈ [0, x], so that θ ≥ ψ. Consequently, ψ = φ + , and (E ˜, ≤) is a Riesz space. Exercise 15.1.7 Suppose that x is a positive element of a Riesz space (E, ≤). Show that Bx = {y ∈ E : |y| ≤ x} is a convex set. A positive element e of a Riesz space (E, ≤) is an order unit if whenever y ∈ E then there exists λ > 0 such that λy ∈ Bx ; in other words, Be is absorbent, and the gauge of Be is a norm .e on E. Exercise 15.1.8 Suppose that e is an order unit in a Riesz space (E, ≤). Show that E ˜ = (E, .e ) . The function 1 is an order unit in the Riesz space Cb (X), where (X, τ ) is a topological space; the corresponding norm is simply .∞ .

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15.2 Signed Measures So far we have been concerned with measures which take non-negative values. We now drop this requirement. A signed measure σ on a measurable space (X, ) is a real-valued function on  which is σ -additive; if (An )∞ n=1 is a sequence of disjoint elements of , then σ (∪∞ n=1 An )

=

∞ 

σ (An ).

n=1

An important feature of this definition is that infinite values are not allowed. A finite measure is a signed measure; in this setting, we call such a measure a positive measure. Proposition 15.2.1 Suppose that σ is a signed measure on a measurable space (X, ). (i) σ (∅) = 0. ∞ (ii) If (An )∞ n=1 σ (An ) n=1 is a sequence of disjoint elements of , then converges absolutely. (iii) If (An )∞ n=1 is an increasing sequence in  with union A then σ (A) = limn→∞ σ (An ). (iv) If (Bn )∞ n=1 is a decreasing sequence in  with intersection B then σ (B) = limn→∞ σ (Bn ). (v) σ () = {σ (A) : A ∈ } is a bounded subset of R.  Proof (i) Take An = ∅ for n ∈ N. Then ∞ n=1 σ (An ) converges, so that σ (∅) = 0.  (ii) If τ is any permutation of N then ∞ n=1 σ (Aτ (n) ) converges, so that the sum is absolutely convergent. (iii) Let C1 = A1 and let Cn = An \ An−1 for n > 1. Then A is the disjoint union of the sequence (Cn )∞ n=1 , so that σ (A) =

∞ 

σ (Cn ) = lim

n=1

n→∞

n  j=1

σ (Cj ) = lim σ (An ). n→∞

(iv) Since (X \ Bn )∞ n=1 increases to X \ B, σ (B) = σ (X) − σ (X \ B) = σ (X) − lim σ (X \ Bn ) n→∞

= lim (σ (X) − σ (X \ Bn )) = lim σ (Bn ). n→∞

(v) We need a lemma.

n→∞

15.2 Signed Measures

195

Lemma 15.2.2 Let H = {H ∈  : {σ (C) : C ∈ , C ⊆ H} is unbounded}. If H ∈ H, then there exist H  ∈ H and C ∈  such that H is the disjoint union H  ∪ C and |σ (C)| ≥ 1. Proof There exists D ∈  such that D ⊆ H and |σ (D)| ≥ |σ (H)| + 1. Then |σ (H \ D)| ≥ 1. If D ∈ H, take H  = D and C = H \ D. Otherwise, H \ D must be in H; take H  = H \ D and C = D. Suppose that X ∈ H. Let A0 = X. Applying the lemma repeatedly, there exists a decreasing sequence (An )∞ n=1 in H, such that if Cn = An−1 \ An then is a sequence of disjoint elements of , and so |σ (Cn )| ≥ 1. But (Cn )∞ n=1 ∞ n=1 σ (Cn ) converges. This gives a contradiction. Thus X ∈ H, and so σ () is bounded. The set M(X, ) of signed measures on a measure space (X, ) contains the finite measures, and is a linear subspace of the Riesz space of all realvalued functions on . Thus if π and ν are positive measures then π − ν is a signed measure. We can decompose a signed measure σ as the difference of two positive measures, in a canonical way. Theorem 15.2.3 If σ is a signed measure on a measurable space (X, ), then there exist disjoint P and N in , with X = P ∪ N, such that σ (A) ≥ 0 for A ⊆ P and σ (A) ≤ 0 for A ⊆ N. Let σ + (A) = σ (A ∩ P), σ − (A) = −σ (A ∩ N). Then σ + and σ − are positive measures on , and σ = σ + − σ − . Further, the decomposition is essentially unique; if X = P ∪ N  , where P and N  are disjoint elements of  for which π  (A) = σ (A ∩ P ) ≥ 0, ν  (A) = −σ (A ∩ N) ≥ 0 for A ∈ , then π = σ + and ν = σ − . Finally, σ + = σ ∨ 0 and σ − = σ ∧ 0. Proof Say that A is strictly non-negative if σ (B) ≥ 0 for all B ⊆ A. First we show that if A ∈  then there exists strictly non-negative C ⊆ A with σ (C) ≥ σ (A). Suppose not. If σ (A) ≤ 0 then we can take C = ∅. Suppose that σ (A) > 0. Let l0 = inf{σ (B) : B ⊆ A}: −∞ < l0 < 0. Choose B1 ⊆ A such that σ (B1 ) < l0 /2, and let A1 = A \ B1 . Then σ (A1 ) > σ (A), and if l1 = inf{σ (B) : B ⊆ A1 } then l0 /2 < l1 < 0.

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Repeating the process, we obtain a decreasing sequence (An ) such that σ (An ) is increasing, and ln = inf{σ (B) : B ⊆ An } → 0. Then σ (∩n (An )) ≥ σ (A) and ∩n (An ) is strictly non-negative. It follows that M = sup{σ (A) : A ∈ } = sup{σ (A) : A strictly non-negative}. There exist strictly non-negative Pn such that σ (Pn ) > M − 1/n. Then P = ∪n Pn is strictly non-negative, and σ (P) = M. It follows that if A ∩ P = ∅ then σ (A) ≤ 0, so that we can take N = X \ P. It is then immediate that σ + and σ − are positive measures on (X, ), and that σ = σ + − σ − . Finally, suppose that P , N  , π and ν satisfy the conditions of the theorem, and that A ∈ . If B ⊆ A∩P then σ (B) ≥ 0, so that σ + (A∩P ) = σ (A∩P ) = π(A). Similarly, if B ⊆ A ∩ N  then σ (B) ≤ 0, so that σ + (A ∩ N  ) = 0. Consequently, σ + (A) = σ (A ∩ P ) = π(A ∩ P ) = π(A), so that π = σ + . Similarly, ν = σ − . Finally, it is clear that σ + = σ ∨ 0 and σ − = −(σ ∧ 0). Thus M(X, ) is a Riesz space, and the notation is consistent; σ + = σ ∨ 0, σ − = (−σ ) ∨ 0 and |σ | = σ ∨ (−σ ). The decomposition σ = σ + − σ − of this theorem is called the Jordan decomposition of σ . Proposition 15.2.4 If σ ∈ M(X, ) and A ∈  then |σ (A)| ≤ |σ |(A). Proof |σ (A)| = |σ (A ∩ P) + σ (A ∩ N)| ≤ |σ (A ∩ P)| + |σ (A ∩ N)| = |σ |(A ∩ P) + |σ |(A ∩ N) = |σ |(A). Exercise 15.2.5 Show that if σ ∈ M(X, ) and (An )∞ n=1 is a sequence of ∞ disjoint elements of  with union A then n=1 |σ (An )| ≤ |σ |(A).

15.3 M(X), L1 and L∞ The space M(X, ) can be given a norm .TV under which it is a Banach space. Theorem 15.3.1 If (X, ) is a measurable space and σ ∈ M(X, ), let σ TV = |σ |(X). Then .TV is a norm on the vector space M(X, ) of signed measures on (X, ) under which M(X, ) is complete.

15.3 M(X), L1 and L∞

197

Proof Let σ = σ + − σ − be the Jordan decomposition of σ . If λ ≥ 0 then λσ = λσ + − λσ − is the Jordan decomposition of λσ , so that λσ  = λ σ . If λ < 0 then λσ = |λ|σ − − |λ|σ + is the Jordan decomposition of λσ , so that λσ TV = |λ|σ − (X) + |λ|σ + (X) = |λ| σ TV . If σ1 , σ2 are signed measures then there exists a positive measure ν such that (σ1 + σ2 )+ = σ1+ + σ2+ − ν and (σ1 + σ2 )− = σ1− + σ2− − ν so that σ1 + σ2 TV = σ1 TV + σ2 TV − 2ν(X) ≤ σ1 TV + σ2 TV . Thus .TV is a norm on M(X, ). ∞ is a Cauchy sequence in M(X, ). If  > 0 there exists Suppose that   (σk )k=1 K such that σj − σk TV ≤ /2 for j, k ≥ K. If A ∈  then, by Proposition 15.2.4,   |σj (A) − σk (A)| ≤ |σj − σk |(A) ≤ σj − σk TV , so that (σk (A))∞ k=1 is a Cauchy sequence in R, which converges to σ (A), say. Note that   |σ (A) − σk (A)| = lim |σj (A) − σk (A)| ≤ sup σj − σk TV ≤ /2 j→∞

j≥K

for k ≥ K. We shall show that σ is a signed measure and that σk → σ in norm as k → ∞. Suppose that (An )∞ n=1 is a sequence of disjoint  of  with union A,  elements and that  > 0. There exists K ∈ N such that σj − σk TV < /2 for j, k ≥ K. By Exercise 15.2.5, ∞ 

  |σj (An ) − σK (An )| ≤ |σj − σK |(A) ≤ σj − σK TV < /2,

n=1

 for j ≥ K. Letting j → ∞, it follows that ∞ n=1 |σ (An ) − σK (An )| ≤ /2, and similarly |σ (A) − σK (A)| ≤ /2. Thus & & ∞ & &∞ & &  & & & & & & σ (An ) − σ (A)& = & (σ (An ) − σK (An )) − (σ (A) − σK (A))& & & & & & n=1 n=1 ∞  (|σ (An ) − σK (An )|) + |σ (A) − σK (A)| ≤ . ≤ n=1

Since  is arbitrary, it follows that σ is σ -additive. Finally, if k ≥ K and X = Pk ∪ Nk is the Jordan decomposition for σ − σk then

198

Further Measure Theory σ − σk TV = (σ (Pk ) − σk (Pk )) − (σ (Nk ) − σk (Nk )) < ;

thus σk → σ as k → ∞. The norm .TV is called the total variation norm. Exercise 15.3.2 Suppose that μ ∈ M(X, ). Show that μTV = supA∈ {μ(A) − μ(X \ A)}. If μ is a finite or σ -finite measure, then elements of L1 (X, ,μ) can be considered as signed measures on (X, ). First observe that f 1 = X | f |dμ is a norm on L1 (X, , μ) which defines the metric d1 , so that (L1 (X, , μ), .1 ) is a Banach space. 1 Exercise 15.3.3 Suppose that (fn )∞ n=1 is a sequence in L (X, , μ) which 1 converges almost everywhere to f ∈ L , and that fn 1 → f 1 as n → ∞. Show that fn → f in (L1 , .1 ). 1 Theorem 15.3.4 Suppose that f ∈ L (X, , μ). If A ∈ , let f .dμ(A) = A f dμ. Then f .dμ is a signed measure, and the mapping f → f .dμ is an isometric linear mapping of L1 (X, , μ) onto a closed subspace of the space M(X, ) of signed measures on (X, ).

Proof Suppose that (Bn )∞ n=1 is an increasing sequence in , with union B. Then | fIBn | ≤ |f | for each n ∈ N, and fIBn → fIB pointwise as n → ∞. By the theorem of dominated convergence,  f .dμ(Bn ) =

 f dμ →

Bn

f dμ = f .dμ(B), B

so that f .dμ is a signed measure. Clearly f .dμ = f + .dμ − f − .dμ, so that     f .dμTV = f + .dμTV + f − .dμTV    = f + dμ + f − dμ = |f | dμ = f 1 . X

X

X

Thus the mapping is an isometry. Since (L1 (X, , μ), .1 ) is complete, the image is closed. Suppose that (X, , μ) is a finite or σ -finite measure space. A function f in L0 (X, , μ) is essentially bounded if there exists M such that |f | < M almost everywhere; that is, there exists M such that λ|f | (M) = μ(|f | > M) = 0. We define L∞ = L∞ (X, , μ) to be {f ∈ L0 : f is essentially bounded}. L∞ is a linear subspace of L0 .

15.4 The Radon–Nikodym Theorem

199

Theorem 15.3.5 The function p(f ) = ess sup(f ) is a seminorm on L∞ (X, , μ), and {f : p(f ) = 0} = N ∞ (X, , μ) = {f ∈ L∞ : f = 0 almost everywhere}. Let .∞ be the corresponding norm on the quotient space L∞ (X, , μ). Then (L∞ (X, , μ), .∞ ) is a Riesz space and a Banach space, and the inclusion (L∞ (X, , μ), .∞ ) → (L0 (X, , μ), ρ0 ) is continuous. If (X, , μ) is a finite measure space, the subspace S of simple measurable functions is dense in (L∞ (X, , μ), .∞ ). Proof It is easily seen that .∞ is a norm on L∞ , that L∞ is a Riesz space and that (L∞ , .∞ ) is complete. If fn − f ∞ → 0 as n → ∞ then fn (x) → f (x) almost everywhere, and so ρ0 (fn , f ) → 0 as n → ∞. Suppose that f ∈ L∞ (X, , μ) and that f ∞ < m ∈ N. Let An,k =  {x : k/n < f (x) ≤ (k + 1)/n} for |k| ≤ mn, and let fn = mn k=−mn (k/n)IAn,k . Then fn ∈ S and f − fn ∞ ≤ 1/n. Norm convergence in L∞ (X, , μ) is called uniform convergence almost everywhere. Unless it is finite-dimensional, ((L∞ , , μ), .∞ ) is not separable. For if is a disjoint sequence of sets in  of positive measure, then the set (An )∞  n=1 ω { ∞ n=1 n IAn : ω ∈ (N)} is an uncountable set for which the distance between any two distinct points is 2.

15.4 The Radon–Nikodym Theorem Suppose that (X, , μ) is a measure space and that ν is a finite measure on . When does there exist f ∈ L1 (X, , μ) such that ν = f .dμ? This is answered by the Radon–Nikodym theorem, which is a fundamental theorem of measure theory. But first we must introduce the Hilbert space L2 (X, , μ). This is defined as L2 = L2 (X, , μ) = {f ∈ L0 : f 2 ∈ L1 (X, , μ)}. Theorem 15.4.1 Suppose that (X, σ , μ) is a measure space. If f , g ∈ L2 (X, , μ) then fg ∈ L1 (X, , μ), and the function (f , g) → X fg dμ is an inner product on L2 (X, , μ), under which L2 (X, , μ) is a Hilbert space, whose unit ball is closed in L0 (X, , μ). 1 2 2 1 Proof Since , μ), and it then follows that  |fg| ≤ 2 (f + g ), fg ∈ L (X, 2 (f , g) → X fg dμ is an inner product on L (X, , μ). If fn 2 ≤ 1 for n ∈ N

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and fn converges to f locally in measure, then there exists a subsequence (fnk )∞ k=1 which converges almost everywhere to f . Then   f 22 = fn2k dμ ≤ 1, f 2 dμ ≤ lim inf k→∞

X

by Fatou’s lemma, so that the unit ball is closed in L0 (X, , μ). It therefore follows from Theorem 2.4.5 that (L2 (X, , μ), .2 ) is complete.  Corollary 15.4.2 If f , g ∈ L2 (X, , μ), let lg (f ) = X fg dμ. Then g → lg is an isometric isomorphism of L2 onto its dual (L2 ) . Proof This is simply a special case of the Fr´echet–Riesz theorem. Theorem 15.4.3 (The Lebesgue decomposition theorem) Suppose that (X, , μ) is a finite or σ -finite measure space, and that ν is a finite measure on 1 . Then there exists  a non-negative f ∈ L (μ) and a set B ∈  with μ(B) = 0 such that ν(A) = A f dμ + ν(A ∩ B) for each A ∈ . If we define νB (A) = ν(A ∩ B) for A ∈ , then νB is a measure. The measures μ and νB are mutually singular; we decompose X as B ∪ (X \ B), where μ(B) = 0 and νB (X \ B) = 0; μ and νB live on disjoint sets. 2 Proof Let π(A)  = μ(A)+ν(A); π is a measure on . Suppose that g ∈ LR (π ). Let L(g) = g dν. Then, by the Cauchy–Schwarz inequality,  |L(g)| ≤ (ν(X))1/2 ( |g|2 dν)1/2 ≤ (ν(X))1/2 gL2 (π ) , X

2 (π ). By Corollary 15.4.2, there so that L is a continuous linear functional on LR 2 g, h, for each g ∈ L2 (π ); that exists  h ∈ LR (π  ) such that L(g) =  an element is, X g dν = X gh dμ + X gh dν, so that   g(1 − h) dν = gh dμ. (∗) X

X

Taking g as an indicator function IA , we see that    ν(A) = L(IA ) = h dπ = h dμ + h dν A

A

A

for each A ∈ . Now let N = (h < 0), Gn = (0 ≤ h ≤ 1 − 1/n), G = (0 ≤ h < 1) and B = (h ≥ 1). Then   h dμ + h dν ≤ 0, so that μ(N) = ν(N) = 0, ν(N) = N N and ν(B) = h dμ + h dν ≥ ν(B) + μ(B), so that μ(B) = 0. B

B

15.4 The Radon–Nikodym Theorem

201

Let f (x) = h(x)/(1 − h(x)) for x ∈ G, and let f (x) = 0 otherwise. Note that if x ∈ Gn then 0 ≤ f (x) ≤ 1/(1 − h(x)) ≤ n. If A ∈ , then, using (∗),    1−h ν(A ∩ Gn ) = fIA∩Gn dμ = f dμ. IA∩Gn dν = X 1−h X A∩Gn Applying the monotone convergence theorem, we see that ν(A ∩ G) = A∩G f dμ = A f dμ. Thus  ν(A) = ν(A ∩ G) + ν(A ∩ B) + ν(A ∩ N) = f dμ + ν(A ∩ B). Taking A = X, we see that

A

 X

f dμ < ∞, so that f ∈ L1 (μ).

This beautiful proof is due to von Neumann. Exercise 15.4.4 Suppose that μ and (νn )∞ n=1 are finite measures on a mea∞ ν = ν and that each νn has Lebesgue surable space (X, ), that n n=1 decomposition νn = fn .dμ+πn with respect to μ. Suppose that ν has Lebesgue  decomposition ν = f .dμ + π. Show that f = ∞ n=1 fn μ-almost everywhere, ∞ and that π = n=1 πn . We can now identify the image of the inclusion mapping L1 (X, , μ) → M(X, ). Suppose that (X, , μ) is a measure space and that ψ is a real-valued function on . We say that ψ is absolutely continuous with respect to μ if, given  > 0, there exists δ > 0 such that if μ(A) < δ then |ψ(A)| < . Corollary 15.4.5 (The Radon–Nikodym theorem) Suppose that (X, , μ) is a finite measure space and that ν is a finite measure on . Then the following are equivalent. (i) ν is absolutely continuous with respect to μ. (ii) If A ∈  and μ(A) = 0 then ν(A) = 0.  (iii) There exists a non-negative f ∈ L1 (μ) such that ν(A) = A f dμ for each A ∈ . Proof If ν is absolutely continuous with respect to μ, then (ii) holds. If (ii) holds, and B is the subset of the theorem, then ν(B) = 0, and so (iii) holds. Finally, suppose that (iii) holds, and that  > 0.  Let Bn = (f > n). Then by the dominated convergence theorem, ν(Bn ) = Bn f dμ → 0 as n → ∞, and so there exists n such that ν(Bn ) < /2. Let δ = /2n. Then if μ(A) < δ,  ν(A) = ν(A ∩ Bn ) + f dμ < /2 + nδ = , A∩(0≤f ≤n)

so that (i) holds. There is also a ‘signed’ version of this corollary.

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Theorem 15.4.6 Suppose that (X, , μ) is a finite measure space and that ψ is a bounded absolutely continuous real-valued function on  which is additive: if A, B are disjoint sets in  then  ψ(A ∪ B) = ψ(A) + ψ(B). Then there exists f ∈ L1 (μ) such that ψ(A) = A f dμ, for each A ∈ . Proof If A ∈ , let ψ + (A) = sup{ψ(B) : B ⊆ A}. ψ + is a bounded additive non-negative function on . We shall show that ψ + is countably additive. ∅, and Suppose that A is the disjoint union of (Ai ). Let Rj = ∪i>j Ai . Then Rj so μ(Rj ) → 0 as j → ∞. By absolute continuity, sup{|ψ(B)| : B ⊆ Rj } → 0 as j → ∞, and so ψ + (Rj ) → 0 as j → ∞. This implies that ψ + is countably additive. Thus ψ + is a measure on , which is absolutely continuous with respect to μ, and so it is represented by some f + ∈ L1 (μ). But now ψ + − ψ is additive, non-negative and absolutely continuous with respect to μ, and so is represented by a function f − . Let f = f + − f − . Then f ∈ L1 (μ) and    ψ(A) = ψ + (A) − (ψ + (A) − ψ(A)) = f + dμ − f − dμ = f dμ. A

A

A

We can use this result to represent the dual of L1 . Theorem 15.4.7 Suppose that (X, , μ) is a finite measure space. If f ∈ L∞ (μ) and g ∈ L1 (μ), let jf (g) = X fg dμ. Then j is an isometric isomorphism of (L∞ (μ), .∞ ) onto the dual space ((L1 ) , .1 ). Proof The map j is clearly a norm-decreasing linear map of (L∞ (μ), .∞ ) into the dual ((L1 ) , .1 ). If f ∈ L∞ (μ) and  > 0, let A = {x : ||f (x)| > f ∞ − , and g =

sgn(f ) .IA . μ(A ) 

 Then g1 = 1 and fg dμ ≥ f ∞ − , so that j is an isometry. Finally, if φ ∈ (L1 ) and A ∈ , let ν(A) = φ(IA ). Then ν is a signed measure on  which is absolutely continuous with respect to μ, and so is represented by some f ∈ L1 (μ). But if A ⊆ {x : |f | > φ1 }, then |ν(A)| ≤ μ(A), from which it follows that μ(A) = 0. Thus f ∈ L∞ and j(f ) = φ. All these results extend easily to σ -finite measure spaces. In particular, we have the following result. Proposition 15.4.8 Suppose that μ and ν are σ -finite measures on a measurable space (X, ), and that 0 ≤ ν(A) ≤ μ(A) for each  A ∈ . Then there exists f ∈ L∞ (μ) such that 0 ≤ f ≤ 1 and ν(A) = A f dμ for each A ∈ .

15.5 Orlicz Spaces and Lp Spaces

203

∞ Proof There exists an increasing sequence (Bn )∞ n=1 in  such that X = ∪n=1 Bn and μ(Bn ) < ∞ for each n ∈ N. Let n = {A ∈  : A ⊆ Bn }, and let μn and νn be the restrictions of μ and ν to n . By the Radon–Nikodym theorem,  for each n there exists fn ∈ L1 (μn ) such that μn (A) = A fn dμn for each A ∈ n . Then 0 ≤ fn≤ 1, since 0 ≤ νn ≤ μn . If m > n and A ∈ n then νn (A) = A fn dμn = A fm dμm , from which it follows that fm|Bn = fn . There therefore exists f ∈ L∞ (μ) such that f|Bn = fn . Thus if A ∈  then   ν(A) = lim ν(A ∩ Bn ) = lim f dμ = f dμ. n→∞

n→∞ A∩B n

A

Exercise 15.4.9 Show how the results of this section can be extended to complex-valued functions and measures.

15.5 Orlicz Spaces and Lp Spaces We now use the Legendre transform to define a large class of function spaces. An N-function  on R is an even (that is, (s) = (−s)) convex nonnegative real-valued function for which (i) (s) = 0 if and only if s = 0, (ii)  is differentiable at 0, with  (0) = 0, and (iii) (s)/|s| → ∞ as |s| → ∞. Exercise 15.5.1 Suppose that  is an N-function. Let p(s) = + (s) = limu s ((u)  s − (s))/(u − s) be the right-hand derivative of , so that (s) = 0 p(u) du. Let q(t) = infv>t p(v) be the right-continuous inverse of p. Then  t p and q are continuous increasing functions on R; show that if (t) = 0 q(v) dv, then  is an N-function. Show that  is the Legendre transform of . Deduce that uv ≤ (u) + (v), with equality if and only if p(u) = v, (or equivalently q(v) = u).  is called the complementary N-function of . The inequality uv ≤ (u)+ (v), with equality if and only if p(u) = v (or equivalently q(v) = u), is known as Young’s inequality. Suppose now that (X, , μ) is a measure space. We define B(L (X)) to be the set  B(L (X)) = {f ∈ L0 (X) :

(f (x)) ≤ 1}. X

Proposition 15.5.2 B(L (X)) is a convex subset of L0 (X), which is closed in (L0 (X), d0 ).

204

Further Measure Theory

Proof Since  is a convex function, B(L (X)) is a convex subset of L0 (X). ∞ Suppose that (fn )∞ n=1 is a d0 -Cauchy sequence in B(L ). Then (fn )n=1 has a subsequence (fnk )∞ k=1 which converges almost everywhere to a function f ∈ (X). Then (f ) converges almost surely to (f ), and so, by Fatou’s lemma, L 0 n k  (f ) dμ ≤ 1; f ∈ B(Lφ )), d0 (fnk , f ) → 0 as k → ∞, and so d0 (fn , f ) → 0 X as n → ∞. We define the Orlicz space L (X) to be the span of B(L (X)), and define . to be the gauge of B(L (X)).  Thus B(L (X)) is the unit ball of (L (X), . ), and f  = inf{α ≥ 0 : X (f (x)/α ≤ 1)}. L (X) is a Riesz space, and . is known as the Luxemburg norm on Lφ (X). Proposition 15.5.3 Suppose that  is an N-function.  (i) If f  ≤ 1 then X (f ) dμ ≤ f  . It follows from Proposition 15.5.2 that (L , . ) is complete.  (ii) If 1 ≤ X (f ) dμ = k ≤ ∞ then f  ≤ X (f ) dμ. Proof (i) By convexity,      1 f 1 1=  (f ) dμ = (f ) dμ. ≥ f  f  X X X f  (ii) By convexity again,   X

   f 1 (f ) dμ = 1, dμ ≤ k k X

so that f  ≤ k.  Exercise 15.5.4 (i) Show that f  = 1 if and only if X (f ) dμ = 1.  (ii) Show that if fn  → 0 as n → ∞ then X (fn ) dμ → 0 as n → ∞. Proposition 15.5.5 If μ is a finite measure then L ⊆ L1 , and the inclusion mapping is continuous. Proof There exists t0 such that (t)/t ≥ 1 for t ≥ t0 . Suppose that f ∈ L and that f  = 1. Then    |f | dμ ≤ |f | dμ + |f | dμ X (|f | n, while fgn 1 > n. Then gn → g ∈ L (X), and fg1 = ∞, giving a contradiction. The quantity f  = supg∈B(L (X)) fg1 is a norm on L˘  (X). It is called the Orlicz norm on L˘  (X). Theorem 15.5.7 L˘  (X) = Lφ (X), and if f ∈ L (X) then f  ≤ f  ≤ 2 f  . Proof It follows from Young’s inequality that if f ∈ L (X) and g ∈ B(L (X)) and f = 0 then    |fg| f dμ ≤ ( ) dμ + g dμ ≤ 2, f φ X f φ X X so that f  ≤ 2 f  , and L (X) ⊆ L˘  (X). On the other hand, it follows from the convexity of  that if h ≥ 1 then   h h = h  dμ ≤ (h) dμ, h X X  so that in general h ≤ 1 + X (h) dμ. Further, if g ≤ 1 then   |hg| |hg| dμ = h dμ ≤ h . X X h Suppose now that f ∈ L˘  (X), and f  ≤ 1. Let g(x) = p(f (x)). Then     (f ) dμ + (g) dμ = |fg| dμ ≤ g ≤ 1 + (g) dμ, X

so that f  .

X

 X

X

X

(f ) dμ ≤ 1 and f  ≤ 1, Thus L˘  (X) ⊂ Lφ (X), and f  ≤

Suppose that  is an N-function and that α > 0. We set α (x) = (αx). Then α is also an N-function, and {α : α > 0} is an increasing family of N-functions. The function conjugate to α is 1/α , and {α : α > 0} is also an increasing family of N functions. An N-function  is said to satisfy the 2 condition if there exists L > 0 such that 2 ≤ L for all t > 0. If  satisfies the 2 condition and 2k−1 ≤ α ≤ 2k , k then α ≤ 2 ≤ Lk  ≤ 2αL.

206

Further Measure Theory

Exercise 15.5.8 Show that if  satisfies the 2 condition with constant L then the conjugate function  also satisfies the 2 condition, with the same constant L. Theorem 15.5.9 Suppose that the N-function  satisfies the 2 condition.  (i) If f ∈ L0 , then f ∈ L if and only if X (f ) dμ < ∞. (ii) If (fn )∞ n=1 is a sequence in L then X (fn ) dμ → 0 as n → ∞ if and only if fn  → 0 as n → ∞.  Proof  (i) Certainly, if X (f ) dμ < ∞ then f ∈ L , and if f ∈ B(L ) then X (f ) dμ ≤ 1 < ∞. If f ∈ L and f  = α > 1, let αf = g. Then    (f ) dμ = (g/α) dμ ≤ 2Lα g dμ = 2Lα < ∞. X

X

X

 (ii) By Exercise 15.5.4, if fn  → 0 as n → ∞, then X (fn ) dμ → 0. Suppose that 0 <  < 1, so that  α = 1/ > 1, and that X (fn ) dμ → 0. Thus there exists N such that X (fn ) dμ < /2L for n ≥ N. If n ≥ N then   (αfn ) dμ ≤ 2Lα X

(fn ) dμ ≤ 1, X

so that fn  ≤ 1/α = . Exercise 15.5.10 Suppose that the N-function  satisfies the 2 condition. Show that the simple measurable functions are dense in L . Theorem 15.5.11 Suppose that (X, , μ) is a finite or σ -finite measure space, and that  is an N-function, with conjugate function , which satisfies the 2  condition. If f ∈ L and g ∈ L let j(g)(f ) = X fg dμ. Then the mapping j is an isometry of (L , . ) onto the dual space ((L ) , (. )). Proof It follows from the definitions that j is an isometric injection. We need to show that it is surjective. By a standard approximation argument, we can suppose that (X, , μ) is a finite measure space. Suppose that θ in (L ) , and  ˜ θ  (IA ). If μ(A that θ n ) → 0 as n → ∞,   = 1. If A ∈ , let θ (A) =  ˜ then An  dμ → 0 as n → ∞, and so θ (An ) = IAn  → 0 as n → ∞. It therefore follows that θ˜ is a signed measure on , and that it is absolutely continuous with respect to μ. By the Radon–Nikodym  theorem, there exists g 1 in L (X, , μ) such that μ˜ = g dμ. But then θ (f ) = X fg dμ for each f ∈ Lφ , and so g = j(θ ) ∈ L . Corollary 15.5.12 If  satisfies the 2 condition then L is reflexive. We now consider a more familiar special case.

15.5 Orlicz Spaces and Lp Spaces

207

If 1 < p < ∞ then the function (t) = |t|p is an N-function which satisfies the 2 condition, with conjugate function (t) = |t|q , where 1/p + 1/q = 1. We denote Orlicz function by Lp (X, , μ). The function  thep corresponding 1/p f p = X |f | dμ is positive homogeneous, and is therefore the gauge of B(Lp ). Thus f  = f p . Exercise 15.5.13 Suppose that (X, , μ) is a finite or σ -finite measure space, that 1 < p < ∞ and that 1/p + 1/q = 1. Show the following.  (i) Lp is a linear subspace of L0 , and the function .p = ( X |f |p dμ)1/p is a norm on Lp under which Lp is a reflexive Banach space, with dual Lq . 2 . (ii) Show that  the space (L , 2 ) is a Hilbert space, with inner product f , g = X fg dμ. (iii) Suppose that f ∈ Lp and g ∈ Lq . Then fg ∈ L1 , and & &  & & & fg dμ& ≤ |fg| dμ ≤ f p gq . & & Equality holds throughout if and only if either f p gp = 0, or g = λsgn(f ).|f |p−1 almost everywhere, where λ = 0. The Lp spaces are uniformly convex, for 1 < p < ∞. Proposition 15.5.14 Suppose that (X, , μ) is a measure space and that 1 < p < ∞. Then (Lp (X, , μ), .p ) is uniformly convex. Proof Let δ() = 1 − 2(1 − /2)p /(1 + (1 − )p ), for 0 <  < 2. Then (1 + (1 − )p )(1 − /2)p−1 − 2(1 − /2)p (1 − )p−1 (1 + (1 − )p )2 (1 − /2)p−1 = (1 + (1 − )p − 2(1 − /2)(1 − )p−1 ) > 0 (1 + (1 − )p )2

δ  () =

so that δ is a strictly increasing non-negative function. It follows by homogeneity that if |x| ≥ |y| and |x − y| ≥ |x| then 

x+y 2

p ≤ (1 − δ())(|x|p + |y|p ).

Now let M = {x ∈ X : |f (x) − g(x)|p ≥  p (|f (x)|p + |g(x)|p )/4}, let fM = f .IM , let GM = g.IM and let N = X \ M. Then   p p |f (x) − g(x)|p dμ(x) ≤ |f (x)|p + |g(x)|p dμ(x) = 4 N 2 N

208

Further Measure Theory

 so that M |f (x) − g(x)|p dμ(x) ≥  p /2. Thus fM − gM  ≥ /21/p , and max(fM p , gM p ) ≥ /2(1/p + 1). But then  |f (x) + g(x)|p p p p 1 (f 1− + g)/2 = dx 2 (|f (x)| + |g(x)| ) − 2p X |f (x) + g(x)|p p p 1 ≥ dx 2 (|f (x)| + |g(x)| ) − 2p M ≥ δ(/41/p )( 12 |f (x)|p + |g(x)|p ) dx M

≥ δ(

p ) , 41/p 2p+2 

which gives the result. Corollary 15.5.15 (Lp (X, , μ), .p ) is smooth. Proof This follows from Theorem 11.6.9. Corollary 15.5.16 The weak topology and norm topology coincide on the unit sphere of Lp . On the other hand, the space (L1 (X, , μ), .1 ) is not uniformly convex (consider f = IA /μ(A) and g = IB /μ(B), where A and B are two disjoint measurable subsets of positive measure; then f 1 = g1 = 1 and f − g1 = f + g1 = 2). In particular, l1 is not uniformly convex. Exercise 15.5.17 If f ∈ S(L1 (X, , μ)), then the norm is smooth at f if and only if μ(f = 0) = 0. We can relate Orlicz spaces to Lp spaces. Proposition 15.5.18 Suppose that  is an N-function which satisfies the 2 condition, with (2t) ≤ L(t) = 2α L (t) for t > 0. Then (t) ≤ L(1)tα for t > 1 and (t) ≥ ((1)/L)tα for 0 < t < 1. Thus if (X, , μ) is a finite measure space then Lφ (X, , μ) ⊆ Lα (X, , μ), and the inclusion is continuous. Similarly, if N is given counting measure then lα ⊆ l , and the inclusion is continuous. Proof First observe that (2n ) ≤ Ln = 2αn . Suppose that t > 1 and that 2n < t ≤ 2n+1 . Then (t) ≤ (2n+1 ) ≤ L(2n ) ≤ L(2αn ) ≤ Ltα . Thus if f ∈ Lα (X, , μ) then   (f (x)) dμ(x) ≤ μ(X) + X

(|f |>1)

|f |α (x) dμ(x),

15.5 Orlicz Spaces and Lp Spaces

209

and so f ∈ L (X, , μ). Since the inclusion mapping has a closed graph, the inclusion is continuous. The case where N is given counting measure is proved similarly. We can also define Lp spaces for 0 < p < 1; we again set  Lp (X, , μ) = { f ∈ L0 : dp ( f ) = | f |p dμ < ∞}. X

In this case, the function t →

tp

is a concave function on [0, ∞).

Exercise 15.5.19 If f , g ∈ Lp (X, , μ), where 0 < p < 1, show that the function dp (f , g) = X |f − g|p dμ is a complete translation-invariant metric on Lp . Theorem 15.5.20 Suppose that (X, , μ) is a finite measure space and 0 < p < r < ∞. Then Lr ⊆ Lp , and the inclusion mapping is uniformly continuous. If p ≥ 1 and μ is a probability measure, then the inclusion mapping is normdecreasing. Proof Suppose that f ∈ Lr . Let t = r/(r − p), so that p/r + 1/t = 1. Applying H¨older’s inequality, with exponents t and r/p , to IX and | f |p , we find that  p/r  p 1/t r | f | dμ ≤ (μ(X)) . | f | dμ , X

X

from which the theorem follows. Theorem 15.5.21 Suppose that (X, , μ) is a finite measure space, and that  is an N-function which satisfies the 2 condition. Then L∞ (X, , μ) ⊆ L (X, , μ), and the inclusion is continuous.  Proof For if  f ∞ → 0 as n → ∞, then X ( fn ) → 0.

16 Borel Measures

16.1 Borel Measures, Regularity and Tightness Recall if (X, τ ) is a topological space, then the Borel σ -field B of X is the σ -field generated by the open sets of X. If (X, τ ) and (Y, σ ) are topological spaces and f : X → Y is continuous, then it follows from Exercise 14.1.4 that if B is a Borel set in Y then f −1 (B) is a Borel set in X. Proposition 16.1.1 Suppose that  is σ -field of subsets of a metrizable space (X, d). The following are equivalent. (i)  = B, the Borel σ -field of X. (ii)  = 1 , the smallest σ -field for which every continuous real-valued function on X is measurable. (iii)  = 2 , the smallest σ -field for which every lower semi-continuous realvalued function on X is measurable. Proof Clearly 1 ⊆ 2 . If f is lower semi-continuous, then ( f > c) is open, for each c ∈ R, and so f is Borel measurable; hence 2 ⊆ B. Finally, if A is closed, then A = {x ∈ X : d(x, A) = 0}, so that A ∈ 1 , and B ⊆ 1 . This proposition depends on the fact that a closed subset of a metrizable space is a Gδ set. This is not necessarily the case for more general topological spaces. Here it is necessary to consider the Baire σ -field – the σ -field generated by the closed Gδ sets – and the theory is consequently more complicated. In applications, the space is usually metrizable, and indeed is usually a Polish space, where the results are stronger, and easier to prove, and so we shall restrict our attention to such spaces. For example, if μ is a Borel measure on a separable metrizable space (X, τ ), we can define its support supp(μ). If μ is a finite Borel measure on a topological space (X, τ ), a closed subset C of X is the support of μ, if μ(X \ C) = 0 and C is the smallest closed subset of X with this property. 210

16.1 Borel Measures, Regularity and Tightness

211

Proposition 16.1.2 Suppose that μ is a Borel measure on a separable metrizable space (X, τ ). Then μ has a support. Proof (X, τ ) is second countable; let (Un )∞ n=1 be a base of open sets, and let K = {n ∈ N : μ(Un ) = 0}. Let U = ∪n∈K Un . Then μ(U) = 0, and U is the largest open subset of X with this property. Thus X \ U is the support of μ. Exercise 16.1.3 Suppose that μ is a Borel measure on a metrizable space (X, τ ) and that x ∈ X. Show that x ∈ supp(μ) if and only if μ(N) > 0 for each open neighbourhood of x. A Finite Borel measure on a metrizable space has good regularity properties. Theorem 16.1.4 Suppose that μ is a finite Borel measure on a metrizable space (X, τ ). Then μ is closed-regular; that is, if A ∈ B then μ(A) = sup{μ(B) : B closed, B ⊆ A} = inf{μ(C) : C open, C ⊇ A}. Proof Let d be a metric on X which defines the topology, and let T be the collection of Borel sets for which the result holds. Suppose first that A is closed, and that Uj = {x ∈ X : d(x, A) < 1/j}. Then (Uj )∞ j=1 is a decreasing sequence of open sets with intersection A. Then μ(A) = limj→∞ μ(Uj ) = infj∈N μ(Uj ), and so A ∈ T. It is therefore enough to show that T is a σ -field. Since A ∈ T if and only if X \ A ∈ T, it is enough to show that if (An )∞ n=1 is a sequence in T then A = A ∈ T. Suppose that  > 0. Then for each n there exist Fn ⊆ An ⊆ Un ∪∞ n=1 n n+1 (Fn closed, Un open) with μ(An \ Fn ) < /2 and μ(Un \ An ) < /2n . Then  U = ∪n Un is open, and U \ A ⊂ ∪n (Un \ An ), so that μ(U \ A) ≤ n μ(Un \ An ) < . Let Bn = ∪ni=1 Ai . Then Bn $ A, and so there exists N such that N μ(A \ BN ) < /2. Then GN = ∪N i=1 Fj is closed, and BN \ GN ⊆ ∪i=1 (Ai \ Fi ), so that N  μ(BN \ GN ) ≤ μ(Ai \ Fi ) < /2. i=1

Thus μ(A \ GN ) < . Exercise 16.1.5 Suppose that U is an open subset of X and that F is a closed subset of X. Then  μ(U) = sup{ f dμ : f ∈ C(X), 0 ≤ f ≤ IU } X μ(F) = inf{ f dμ : f ∈ C(X), f ≥ IF }. X

212

Borel Measures

For the next few exercises, F is the linear span of the indicator functions of closed sets, and G is the linear span of the indicator functions of open sets. Exercise 16.1.6 Show that if μ is a finite Borel measure on a metrizable space (X, τ ) then Cb (X), F and G are each dense in (L1 (μ), .1 ). If (X, τ ) is separable, then (L1 (μ), .1 ) is separable. Exercise 16.1.7 Show that if μ is a finite Borel measure on a metrizable space (X, τ ) then Cb (X), F and G are each dense in (L0 (μ), d0 ). If (X, τ ) is separable, then (L0 (μ), d0 ) is separable. Exercise 16.1.8 Show that if μ is a finite Borel measure on a metrizable space (X, τ ) and  is an N-function which satisfies the 2 condition then Cb (X), F and G are each dense in (L (μ), . ). If (X, τ ) is separable, then (L (μ), . ) is separable. In particular, this result holds for the Lp spaces, for 1 < p < ∞. Exercise 16.1.9 Show that if μ is a finite Borel measure on a metrizable space (X, τ ) then Cb (X) is closed in (L∞ , .∞ ), and is dense if and only if it is finite-dimensional. We consider an even stronger property. A mapping f from the Borel sets of a metrizable space (X, τ ) to [0, ∞] is tight if f (K) < ∞ for each compact K in X and f (A) = sup{f (K) : K compact , K ⊆ A}, for each A ∈ B(X). Tightness is very powerful, as the next result shows. We consider nonnegative functions on the Borel subsets of a metric space (X, τ ) which can take infinite values. As before, a mapping f : B(X) → [0, ∞] is additive if f (A ∪ B) = f (A) + f (B) whenever A and B are disjoint, and is σ -additive if ∞ ∞ f (∪∞ n=1 f (An ) for each sequence (An )n=1 of disjoint Borel sets. n=1 An ) = Proposition 16.1.10 Suppose that f is a non-negative additive tight function on the Borel sets of a metrizable space (X, τ ). Then f is σ -additive, and so it is a tight Borel measure on X. Proof Suppose that (An )∞ n=1 is a sequence of disjoint Borel sets whose union is A. First we consider the case where f (A) < ∞. Let Bn = ∪nj=1 Aj . Since   Bn ⊆ A, nj=1 f (Aj ) ≤ f (A), and so ∞ j=1 f (Aj ) ≤ f (A).  f (A ) Suppose, if possible that ∞ j = s < f (A). Let  = ( f (A) − s)/2 and j=1 let Cn = A \ Bn . Thus f (Cn ) ≥ 2, for n ∈ N. By combining blocks of terms, we can suppose that f (Bn ) > s − /2n+1 , for n ∈ N. For each n ∈ N there exists a subset Kn of Cn with f (Kn ) > f (Cn ) − /2n+1 . Let Ln = ∩nj=1 Kj .

16.1 Borel Measures, Regularity and Tightness

213

We show by induction that f (Ln ) ≥ (1 + 1/2n ) (so that f (Cn \ Ln ) ≤ (1 − 1/2n )). The result is true when n = 1; suppose that it is true for n. Now f (Cn \ Cn+1 ) = f (Bn+1 \ Bn ) = f (Bn+1 ) − f (Bn ) ≤ s − (s − /2n+1 ) = /2n+1 ; since Cn \ Kn+1 ⊆ (Cn \ Cn+1 ) ∪ (Cn+1 \ Kn+1 ), it follows that f (Cn \ Kn+1 ) ≤ f (Cn \ Cn+1 ) + f (Cn+1 \ Kn+1 ) < /2n . Since Cn = (Cn \ Kn+1 ) ∪ (Cn \ Ln ) ∪ (Ln ∩ Kn+1 ) 2 ≤ f (Cn ) ≤ f (Cn \ Kn+1 ) + f (Cn \ Ln ) + f (Ln ∩ Kn+1 ) ≤ /2n+1 + (1 − 1/2n ) + f (Ln+1 ), so that f (Ln+1 ) ≥ (1 + 1/2n+1 ). This establishes the induction. ∞ But ∩∞ n=1 Ln ⊆ ∩n=1 Cn = ∅; since the sets Ln are compact, it follows that there exists N ∈ N for which LN = ∅, so that f (Ln ) = 0, giving a contradiction. Finally, suppose that f (A) = ∞. If M < ∞, there exists a compact subset K of A with f (K) > M. Then ∞ 

f (An ) ≥

n=1

so that

∞

n=1 f (An )

∞ 

f (An ∩ K) = f (K) > M,

n=1

= ∞.

Proposition 16.1.11 A finite Borel measure μ on a metric space (X, d) is tight if sup{μ(K) : K compact, K ⊆ X} = μ(X). Proof The condition is certainly necessary. Suppose that it is satisfied. Suppose that A is a Borel set, and that  > 0. There exists a closed set B ⊆ A such that μ(B) ≥ μ(A) − /2 and there exists a compact K such that μ(K) > μ(X)−/2. Then B∩K is a compact subset of A, and μ(B∩K) ≥ μ(A)−. Exercise 16.1.12 Suppose that μ is a tight Borel measure on a metrizable space (X, τ ). Show that there exists a σ -compact, and therefore separable Borel subset X0 of X, for which μ(X0 ) = μ(X). If μ is a tight measure on a metrizable space, we can prove a result whch goes beyond the monotone convergence theorem. We need a definition. A subset A of a partially ordered set is directed upwards if whenever a1 , a2 ∈ A there exists b ∈ A such that b ≥ a1 and b ≥ a2 .

214

Borel Measures

Theorem 16.1.13 Suppose that μ is a tight Borel measure on a metrizable space (X, τ ), and that A is a set of lower semi-continuous function on X which  is directed  upwards. Let s(x) = supa∈A a(x) and let J = X s dμ. Then J = supa∈A X a dμ. Proof First observe that s is lower semi-continuous, and so Borel measurable,  so that J exists. Let I = supa∈A X a dμ. Clearly I ≤ J; we must show that equality holds. An easy induction shows that there exists an increasing  in A such that a dμ sequence (an )∞ n=1 X n  → I as n → ∞. Let h = supn an . By the monotone convergence theorem, X h dμ = I. Suppose now that I < J, so that, in particular, I < ∞. Then there exists a Borel set B and u < v such that μ(B) > 0 and h(y) < u < v < s(y) for y ∈ B. If C is a Borel set, let μB (C)−μ(B∩C); μB is a tight Borel measure. Let y ∈ supp(μB ). There exists a ∈ A such that a (y) > v, and there exists an open neighbourhood N of y such that a (x) > v for x ∈ N. By induction, we can find an increasing sequence      (an )∞ n=1 in A such that an ≥ a and an ≥ an for all n. Let h = supn an . Then    I= h dμ = h dμ + h dμ X B∩N X\(B∩N)   ≥ h + (v − u) dμ + h dμ B∩N X\(B∩N)  = h dμ + (v − u)μ(B ∩ N) = I + (v − u)μ(B ∩ N), X

giving a contradiction, since μ(B ∩ N) > 0.

16.2 Radon Measures Suppose that f is a function on the Borel subsets of a metric space X taking values in [0, ∞]. f is locally finite if for each x ∈ X there exists a neighbourhood N of x with f (N) < ∞. Proposition 16.2.1 If f : B(X) → [0, ∞] is a locally finite additive function on a metrizable space (X, τ ) and K is a compact subset of X then f (K) < ∞. Proof For each x ∈ K there exists a neighbourhood Nx of x with f (Nx ) < ∞. The sets {Nx : x ∈ K} cover K, and so there is a finite subcover. Additivity then ensures that f (K) < ∞. A Radon measure μ on a metrizable space (X, τ ) is a tight additive function from B(X) to [0, ∞] which is locally finite. By Proposition 16.1.10, a Radon

16.3 Borel Measures on Polish Spaces

215

measure μ is σ -additive, and, by the preceding property, μ(K) < ∞ if K is compact. Proposition 16.2.2 If μ is a Radon measure on a separable metrizable space, then μ is σ -finite; there exists a countable set W of open sets for which X = ∪W∈W W and f (W) < ∞ for all W ∈ W. Proof Let d be a metric on X which defines the topology τ . For each n ∈ N, let Un = {x ∈ X : there exists rx > 1/n such that f (Nrx (x)) < ∞}. If x ∈ Un and d(x, y) < rx − 1/n, let sy = rx − d(x, y). Then sy > 1/n and Nsy (y) ⊆ Nrx (x), so that y ∈ Un : Un is open. Let Cn be a countable dense subset of Un , and let Wn (c) = N1/n (c), for c ∈ Cn . Then f (Wn (c)) < ∞ for each c ∈ Cn . If x ∈ Un then there exists c ∈ Cn with d(x, c) < 1/n, so that Un ⊆ ∪c∈Cn Wn (c). Let W = {Wn (c) : n ∈ N, c ∈ Cn }. Then X = ∪W∈W W. Suppose that X and Y are metric spaces, that f : X → Y is continuous and that μ is a Radon measure on X. Then the push-forward measure f∗ (μ) need not be a Radon measure on Y; let μ be counting measure on N, and let f : N → [0, ∞] be the inclusion mapping. Then f∗ (μ) is not locally finite at ∞.

16.3 Borel Measures on Polish Spaces A Borel measure on a compact metric space is tight. More generally, if (X, d) is a σ -compact metric space, then every Borel measure on X is tight. We can say more. Theorem 16.3.1 (Ulam’s theorem) A finite Borel measure on a Polish space (X, τ ) is tight. Proof We give two proofs of this important theorem. By Theorem 3.1.1 there is a homeomorphism j of X onto a Gδ subset Y of the Hilbert cube H. Let j∗ (μ) be the push-forward measure on H. Since H is compact, j (μ) is tight. Thus j∗ (μ)(Y) = sup{j∗ (μ)(K) : K compact, K ⊆ Y}; since j is a homeomorphism, μ(X) = sup{μ(K) : K compact, K ⊆ X}. Here is the direct proof, given by Ulam. Let d be a complete metric on X which defines the topology τ . Let (cj )∞ j=1 be a dense sequence in X, and let

216

Borel Measures j

Mj,n = {x ∈ X : d(x, cj ) ≤ 1/n}. Let Aj,n = ∪i=1 Mi,n . Suppose that  > 0. If n ∈ N, each Aj,n is closed, and Aj,n $ X as j → ∞; thus there exists Jn such n is a finite that if Cn = AJn ,n then μ(Cn ) > (1 − /2n )μ(X). Further, (Mj,n )Jj=1 n 1/n-net which covers the closed set Cn . Let Dn = ∩j=1 Cj . Then (Dn )∞ n=1 is a decreasing sequence of closed sets, and μ(Dn ) > (1 − (1 − 1/2n ))μ(X) > (1 − )μ(X), for each n ∈ N. Let D = ∩∞ n=1 Dn . Then μ(D) ≥ (1 − )μ(X) and D is closed and totally bounded, and is therefore compact. Corollary 16.3.2 If μ is a σ -finite Borel measure on a Polish space (X, τ ), then there exists an increasing sequence (Kn )∞ n=1 of compact sets such that K ) = 0, and μ is tight. μ(X \ ∪∞ n n=1 Proof Let (An ) be an increasing sequence of sets of finite measure such that μ(X \ ∪∞ n=1 An ) = 0. Consider the finite measures IAn .dμ; there exists an ∞ increasing sequence (Kn )∞ n=1 of compact sets such that μ(X \ ∪n=1 Kn ) = 0. If A is a Borel set, then μ(A) = limn→∞ μ(A ∩ Kn ). Since μ(A ∩ Kn ) = sup{μ(K) : K compact K ⊆ A ∩ Kn }, the result follows. If (X, τ ) is a Polish space, then Radon measures can be defined locally. Theorem 16.3.3 Suppose that (X, τ ) is a Polish space, and that (Ui )∞ i=1 is a ∞ sequence of open sets in X which covers X; ∪i=1 Ui = X. Suppose that for each i, μi is a finite measure on the Borel sets of Ui and that these measures are compatible; if A is a Borel set of Ui ∩ Uj then μi (A) = μj (A). Then there exists a unique Radon measure π on X for which π(A) = μi (A) for each Borel set A in of Ui , for each i ∈ N. j

Proof Let Vj = ∪i=1 Ui . The compatiblity condition ensures that we can define a finite positive Borel measure νj on Vj such that νi (A) = μi (A) if 1 ≤ i ≤ j and A ⊆ Ui . Further, if A is a Borel subset of Vj and j ≤ k then νj (A) = νk (A). If A is a Borel subset of X, (νj (A∩Vj ))∞ j=1 is an increasing sequence; let π(A) = limn→∞ νj (A ∩ Vj ); then it is easily verified that π is tight, locally finite and additive.

16.4 Lusin’s Theorem Borel measurable functions on a Polish space are well-behaved. Theorem 16.4.1 (Lusin’s theorem) Suppose that X and Y are Polish spaces, that μ is a finite Borel measure on X, that f : X → Y is Borel measurable and that  > 0. Then there exists a compact subset K of X, with μ(X \ K) < , such that the restriction of f to K is continuous.

16.4 Lusin’s Theorem

217

Proof Let d be a metric on Y which defines the topology of Y, and let (yn )∞ n=1 be a dense sequence in Y. Suppose that j ∈ N. Let An,j = {x ∈ X : d( f (x), yn ) < 1/j}, Bn,j = An,j \ (∪n−1 m=1 Am,j ), Cn.j = ∪nm=1 Am,j = ∪nm=1 Bm,j . Then (Cn,j )∞ n=1 is an increasing sequence of Borel subsets of X whose union is X, and so there exists Nj such that μ(X \ CNj ) < /2j+1 . For each 1 ≤ n ≤ Nj there exists a compact subset Kn,j of Bn.j such that μ(Bn.j \ Kn,j ) < /2j+1 Nj . N

j Kn.j . Then μ(X \ Kj ) < /2j . If x ∈ Kn,j , let fj (x) = yn . Since Let Kj = ∪n=1 the sets Kn,j are disjoint and closed, fj is a continuous function on Kj . Further, d( fj (x), f (x)) < 1/j for x ∈ Kj . Now let K = ∩∞ j=1 Kj . Then K is a compact subset of X and μ(X \ K) ≤ ∞ j=1 μ(X \ Kj ) < . The restriction of each fj to K is continuous, and fj → f uniformly on K as j → ∞, and so f is continuous on K.

In these circumstances, we can improve on Egorov’s theorem. Corollary 16.4.2 Suppose that ( fi )∞ i=1 is a sequence of Borel measurable functions from X → Y, that fi → f almost everywhere as i → ∞ and that  > 0. Then there exists a compact subset K of X with μ(X \ K) <  such that each fi is continuous on K and such that fi → f uniformly on K as i → ∞. Proof By Egorov’s theorem, there exists a Borel subset E of X with μ(X \E) < /2, such that fi → f uniformly on E, and there exists a compact subset L of E with μ(X \ L) < /2. For each i there exists a compact subset Ki of L with μ(L \ Ki ) < /2i+1 such that fi is continuous on Ki . Then K = ∩∞ i=1 Ki has the required properties. Corollary 16.4.3 Suppose that f is a non-negative real-valued Borel measurable function on X. Then   f dμ = sup{ f dμ : K compact, K ⊆ X, f|K continuous}. X

K

Proof There exists an increasing sequence (Kj )∞ j=1 of compact sets of X, with μ(X \ Kj ) → 0 as j → ∞, such that f|Kj is continuous. Let fj (x) = f (x) for x ∈ Kj and fj (x) = 0 otherwise. Then    f dμ = lim fj dμ = lim f dμ, X

j→∞ X

j→∞ Kj

by monotone convergence. The result follows from this.

218

Borel Measures

Exercise 16.4.4 Suppose that A is a closed subset of [0, 1], and that μ is a Borel measure on [0, 1]. Using the fact that ∂C is countable if C is closed, prove Lusin’s theorem for IA . Use this, and tightness, to prove Lusin’s theorem for a Borel function on [0, 1].

16.5 Measures on the Bernoulli Sequence Space (N) So far, we have not constructed any measures. Here we show how to construct Borel measures on (N). This is rather simpler than constructing Borel measures on R, because the geometry and topology is technically simpler. We can then consider measures on other spaces, such as R, as push-forward measures of measures on (N). We begin by considering the set Cyl() of cylinder sets in . Recall that the cylinder set C,m of rank m is the set C,m = {ω ∈ (N) : ωj = j for 1 ≤ j ≤ m}. Recall also that each cylinder set is open and closed, and that Cyl() is a countable base for the topology of . Proposition 16.5.1 If C,m and Cη,n are cylinder sets, and m ≤ n, then either Cη,n ⊆ C,m or Cη,n ∩ C,n = ∅. Proof If ηj = j for 1 ≤ j ≤ m, then Cη,n ⊆ C,m . Otherwise, Cη,n ∩ C,m = ∅. Corollary 16.5.2 If U is an open subset of , then U is union of a disjoint sequence of cylinder sets. Corollary 16.5.3 If U is an open and closed subset of , and C is a set of disjoint cylinder sets whose union is U, then C is finite. Proof For C is an open cover of the compact set U. A cylinder set C of rank n is the disjoint union of two cylinder sets C(0) = {x ∈ C : xn+1 = 0} and C(1) = {x ∈ C : xn+1 = 1} of rank n + 1. We denote by A() the set of real-valued functions α on Cyl() which satisfy α(C) = α(C(0) ) + α(C(1) ) for each C ∈ Cyl(). Elements of A() are called dyadic martingales. We also set α(∅) = 0, and set A+ () = {α ∈ A() : α ≥ 0}.

16.5 Measures on the Bernoulli Sequence Space (N)

219

Proposition 16.5.4 Suppose that C is a set of disjoint cylinder sets whose union  is the cylinder set D. If α ∈ A+ () then α(D) = C∈C α(C). Consequently, 0 ≤ α(D) ≤ α(). Proof |C| is finite. The proof is by induction on |C|. The result is true if |C| = 1 or 2. Suppose that it holds when |C| = j, and that |C| = j + 1. There exists (0) (1) C = C,n ∈ C with n maximal. Then C,n−1 = C,n−1 ∪C,n−1 = C ∪C ; then     C ⊆ D, so that C ∈ C. Replace C and C by C,n−1 to obtain a new set C  .   Since α(C,n−1 ) = α(C ) + α(C ), C∈C α(C) = C∈C  α(C) = α(D). If μ ∈ M + () = M + (, B), let j(μ)(C) = μ(C). Then j maps M + () into A+ (). Since B = σ (Cyl()), j is injective. More importantly, it is also surjective. Theorem 16.5.5 The mapping j : M + () → A+ () is bijective. Proof The proof is similar to, but simpler than, the proof of the existence of Lebesgue measure, and Lebesgue–Stieltjes measures. Suppose that α ∈ A+ (). If U is an open subset of , U = ∪i Ci , where (Ci ) is a finite or infinite sequence of disjoint cylinder sets. We set lα (U) =  j α(Cj ). We must show that this is well-defined. Suppose that (Dj ) is another finite or infinite sequence of disjoint cylinder sets whose union is U. Then U = ∪i,j (Ci ∩Dj ) = ∪k Ek , where (Ek ) is a finite or infinite sequence of disjoint cylinder sets. For each i, Ci = ∪k∈Fi Ek , where Fi is a finite set of indices. But  then lα (Ci ) = k∈Fi lα (Ek ), so that     lα (Cj ) = lα (Ek ) = lα (Ek ). 

i



i

k∈Fi

k

Similarly j lα (Dj ) = k lα (Ek ), and so l(U) is well-defined. Here are the basic properties of the function lα . Lemma 16.5.6 Suppose that U, (Un )∞ n=1 and V are open subsets of , and U . that U = ∪∞ n=1 n  (i) If Un ∩ Um = ∅ for m = n then lα (U) = ∞ n=1 lα (Un ). (ii) If V ⊆ U then lα (V) ≤ lα (U). (iii) If (Un )∞ n=1 is an increasing sequence, then lα (U) = limn→∞ lα (Un ). (iv) lα (U ∪ V) + lα (U ∩ V) = lα (U) + lα (V).  (v) lα (U) ≤ ∞ n=1 lα (Un ). Proof Let U = ∪i Ci , where (Ci ) is a sequence of disjoint cylinder sets, and let V = ∪j Dj , where (Dj ) is a sequence of disjoint cylinder sets. (i) For each n, Un = ∪∞ i=1 Cn,i , where (Cn,i ) is a sequence of disjoint cylinder sets. Then U = ∪n,i Cn,i so that

220

Borel Measures

lα (U) =

∞   n=1

 lα (Cn,i ) =

i

∞ 

lα (Un ).

n=1

(ii) V = ∪i,j (Dj ∩ Ci ), so that    α(Dj ∩ Ci ) ≤ α(Ci ) = lα (U). lα (V) = i

j

i

(iii) Since Un ⊆ U for each n, limn→∞ lα (Un ) ≤ lα (U). For each i, {Un } is an open cover of Ci , and so there exists ni such that Ci ⊆ Uni . Let j mj = maxi≤j ni . Then ∪1≤j≤i Cj ⊆ Umj , so that i=1 α(Ci ) ≤ lα (Umj ). Thus lα (U) ≤ limn→∞ lα (Un ). (iv) By (iii), it is enough to prove the result when U = ∪m i=1 Ci and V = ∪nj=1 Dj are unions of finitely many disjoint cylinder sets. We prove this by induction on m + n. The result is true when m + n = 2. Suppose that it is true when m + n = k, and that m + n = k + 1. Without loss of generality, we can suppose that m ≥ 2. Let U  = ∪m−1 i=1 Ci . We consider three possibilities. First, suppose that Cm ∩ V = ∅. Then U ∪ V = (U  ∪ V) ∪ Cm and U ∩ V = U  ∩ V, so that lα (U ∪ V) + lα (U ∩ V) = lα (U  ) + lα (V) + lα (Cm ) = lα (U) + lα (V). Secondly, suppose that Cm ⊆ Dj for some j. Then U ∪ V = U  ∪ V and U ∩ V = (U  ∩ V) ∪ Cm , so that lα (U ∪ V) + lα (U ∩ V) = lα (U  ) + lα (V) + lα (Cm ) = lα (U) + lα (V). Thirdly, suppose that Cm ∩ V = ∅, but that Cm ⊆ Dj , for any j. Let V  = V \ Cm . Then U ∪ V = U ∪ V  and U ∩ V = (U ∩ V  ) ∪ (V ∩ Cm ), so that lα (U ∪ V) + lα (U ∩ V) = lα (U) + lα (V  ) + lα (V ∩ Cm ) = lα (U) + lα (V). (v) Let Wn = ∪ni=1 Ui . Then, using (iii) and (iv), lα (U) = lim lα (Wn ) ≤ lim n→∞

n→∞

n  i=1

lα (Ui ) =

∞  n=1

lα (Un ).

16.5 Measures on the Bernoulli Sequence Space (N)

221

Suppose now that A is a subset of . We define the α-outer measure μ∗α (A) to be μ∗α (A) = inf{lα (U) : U open, A ⊆ U}. The function μ∗α has the following properties. Theorem 16.5.7 Suppose that A, {An : n ∈ N} and B are subsets of , that A = ∪n∈N An and that U is an open subset of . (i) (ii) (iii) (iv)

μ∗α (U) = lα (U). If A ⊆ B then μ∗α (A) ≤ μ∗α (B).  ∗ μ∗α (A) ≤ ∞ n=1 μα (An ). ∗ ∗ μα (A ∪ B) + μα (A ∩ B) ≤ μ∗α (A) + μ∗α (B).

Proof (i) and (ii) follow immediately from Lemma 16.5.6 (ii). (iii) Suppose that  > 0. For each n ∈ N there exists an open Un with An ⊆ Un for which lα (Un ) < μ∗α (An ) + /2n . Then μ∗α A)

≤

lα (∪∞ n=1 Un )

≤

∞  n=1

lα (Un ) ≤

∞ 

μ∗α (An ) − ,

n=1

by Lemma 16.5.6 (v). Since  is arbitrary, the result follows. (iv) Suppose that  > 0. There exist open sets U and V with A ⊆ U, B ⊆ V and lα (U) ≤ μ∗α (A) + /2, lα (V) ≤ μ∗α (B) + /2. Then μ∗α (A) + μ∗α (B) ≥ lα (U) + lα (V) −  = lα (U ∪ V) + lα (U ∩ V) −  ≥ μ∗α (A ∪ B) + μ∗α (A ∩ B) − , by Lemma 16.5.6 (iv). Since  is arbitrary, the result follows. In particular, if A ⊆  then μ∗α (A) + μ∗α ( \ A) ≥ μ∗α () = lα (). We say that A is α-measurable if μ∗α (A) + μ∗α ( \ A) = μ∗α () = lα (), and denote the set of α-measurable sets by α . If A ∈ , we set μα (A) = μ∗α (A). If C is a cylinder set then C ∈ α . If A, B ∈ , let A =  \ A and B =  \ B. Then A , B ∈ α and μ∗α (A ∪ B) + μ∗α (A ∩ B) ≥ 2lα () − μ∗α (A ∩ B ) − μ∗α (A ∪ B ) ≥ 2lα () − μα (A ) − μα (B ) = μα (A) + μα (B) ≥ μ∗α (A ∪ B) + μ∗α (A ∩ B), and so there is equality throughout. Thus μ∗α (A ∪ B) + μ∗α ( \ (A ∪ B)) = lα () and μ∗α (A ∩ B) + μ∗α ( \ (A ∩ B)) = lα (),

222

Borel Measures

and so A ∪ B and A ∩ B are in α . Consequently A \ B ∈ α , and μα (A) = μα (A\B)+μα (A∩B). Suppose that (An )∞ n=1 is a sequence of disjoint elements ∞  of α , and that A = ∪n=1 An . Let An =  \ An . Then μ∗α (A) ≤

∞  n=1

μα (An ) =

∞  (lα () − μα (A )) n=1

≤ lα () − μ∗α (A) ≤ μ∗α (A),

 and so there is equality throughout; A ∈ α , and μα (A) = ∞ n=1 μα (An ). Consequently, α is a σ -field, and μα is a finite measure on it. Since the cylinder sets are in α , α contains the Borel σ -field B() of . If we also denote the restriction of μα to B() by μα , then j(μα ) = α; the mapping j is a bijection of M + () onto A(). Note that (, α , μα ) is a complete measure space; it is the completion of (, B(), μα ). We can now push forward measures on (N) to construct measures on compact metrizable spaces. Here are some examples. Exercise 16.5.8 Let α(C,m = 2−m ) for each . Verify that α ∈ A(). Let γ be the corresponding Borel measure. (N) is a compact Abelian topological group, under co-ordinatewise addition mod 2. Show that γ (B + ) = γ (B) for every Borel set B and  ∈ (N). ∞ n Exercise 16.5.9 If  ∈ (N) let f () = n=1 n /2 . Show that f maps (N) continuously onto [0, 1]. Let λ = f∗ (γ ). Show that λ is a Borel measure (Lebesgue measure) on [0, 1] for which λ([a, b]) = b − a for each interval [a, b]. We can extend Lebesgue measure to the whole real line; if A is a Borel set  in R, we set λ(A) to be ∞ n=−∞ λ((A − n) ∩ [0, 1]).  n Exercise 16.5.10 If  ∈ (N) let g() = 2 ∞ n=1 n /3 . Show that g maps (N) continuously onto the Cantor set. Let μ = g∗ (γ ). Show that λ and μ are mutually singular.

16.6 The Riesz Representation Theorem Suppose that (X, τ ) is a Polish space. If μ = μ+ − μ− is a signed Borel we set j(μ)( f ) = X f dμ+ − X f dμ− . We measure b (X),  f ∈ C   on X and write X f dμ for X f dμ+ − X f dμ− .

16.6 The Riesz Representation Theorem

223

Proposition 16.6.1 Suppose that (X, τ ) is a Polish space. The mapping j : (M(X, B), .TV ) → (Cb (X) , . ) is a linear isometry of (M(X, B), .TV ) into (C(X) , . ), and j(M(X, B) is weak* dense in Cb (X) . Proof It is clear that j(αμ) = αμ for α ≥ 0 and that j(−μ) = −μ. If μ1 , μ2 ∈ M(X), then, since + + − + − μ+ 1 + μ2 = (μ1 + μ2 ) + ν and μ1 − μ2 = (μ1 + μ2 ) − ν,

for some ν ∈ M + (X), it follows from the  definition that j(μ1 + μ2 ) = j(μ1 ) + j(μ2 ); j is linear. Since | j(μ( f ))| ≤ X | f | dμ ≤  f ∞ . μTV , j is normdecreasing. If μ ∈ M(X), there exist disjoint Borel sets P and N, such that μ+ (P) = μ+ (X) and μ− (P) = μ− (X). If  > 0, there exist a closed subset K of P and a closed subset L of N such that μ+ (K) > μ+ (P) − /4 and μ− (L) > μ− (N) − /4. There exists f ∈ C(X) with  f ∞ ≤ 1 for which f (x) = 1 for x ∈ K and f (x) = −1 for x ∈ L. Then j(μ)( f )) ≥ μTV − ;  μTV . Thus j is an isometry. since  is arbitrary j(μ)  ≥  Since  f ∞ = sup{ X f dδx : x ∈ X} = sup{ X f dμ : μ ∈ M1 (X, B)}, it follows from the theorem of bipolars that j(M1 (X, B)) is weak* dense in B(E ). Consequently, j(M(X, B)) is weak* dense in Cb (X) . When (X, τ ) is compact, the mapping j is surjective; every continuous linear function is represented by a signed Borel measure on X. Theorem 16.6.2 (The Riesz representation theorem for compact metrizable spaces) Suppose that (X, τ ) is a compact metrizable topological space. The mapping j : (M(X, B), .TV ) → (C(X) , . ) is a linear isometry of (M(X, B), .TV ) onto (C(X) , . ). Proof First we consider the case where X = . Suppose that φ ∈ C() . If C is a cylinder set, let α(C) = φ(IC ). Then α ∈ A() and |α(C)| ≤ φ . Let α + (C) = sup{φ( f ) : 0 ≤ f ≤ IC }. Then 0 ≤ α + (C) ≤ φ . Since C(0) and C(1) are disjoint, it follows that α(C) = α(C(0) ) + α(C(1) ); thus α + ∈ A+ (). By Theorem 16.5.5, there exists a positive Borel measure μ+ on  such that if C is a cylinder set then α + (C) = μ+ (C). The function α − = α + − α is also in A+ (), and so there exists a positive Borel measure μ− on  such that if C is a cylinder set then α − (C) = μ− (C). Then μ = μ+ − μ− is a signed Borel measure on , and j(μ) = φ. Now we consider the general case. By Theorem 3.3.11, there exists a continuous surjection ψ of  onto X. If f ∈ C(X), let Tψ ( f ) = f ◦ ψ. Then Tψ is an isometric isomorphism of C(X) into C(), and so, by the Hahn–Banach

224

Borel Measures

 is a surjection of C() onto C(X) . Thus if φ ∈ C(X) , there theorem, T exists θ ∈ C() with Tψ (θ ) = φ. There exists a signed Borel measure μ = μ+ − μ− on  which represents θ. Let ν = ψμ+ and π = ψμ− be the push-forward Borel measures on X. If f ∈ C(X) then   φ( f ) = Tψ (θ )( f ) = θ ( f ◦ φ) = ( f ◦ φ) dμ+ − ( f ◦ φ) dμ−     = f dν − f dπ = j(ν − π )( f ), X

X

so that φ = j(U − π ). Let ρ = ν − π. Of course π need not equal ρ + , nor need π equal ρ − . Corollary 16.6.3 (M(X), .TV ) is a Banach space. The weak* topology σ ∗ (M(X, B), C(X)) is denoted by w, and called the w topology; it is also, misleadingly, called the weak topology on M(X, B). As an immediate consequence of Banach’s theorem, we have the following corollary. Corollary 16.6.4 M1 (X, B) is w-compact and metrizable, and is therefore wseparable. Exercise 16.6.5 Suppose that φ : (K, τ ) → (L, σ ) is a continuous mapping from a compact metrizable space (K, τ ) to a compact metrizable space (L, σ ). If f ∈ C(L), let Tφ ( f )(x) = f (φ(x)), for x ∈ K. Show that Tφ ( f ) ∈ C(K), and Tφ ( f )∞ ≤  f ∞ . If μ ∈ M(K), show that Tφ (μ) = φ∗ (μ), the pushforward measure of μ. If φ is surjective, show that Tφ is an isometry, and Tφ maps M(K) onto M(L). In fact, it is convenient to have a more explicit description of separability. Proposition 16.6.6 Suppose that (X, τ ) is a compact metrizable space. Let C be a countable dense subset of X. Then the countable set  n  n   λi δci : n ∈ N, λi ∈ Q, |λi | ≤ 1, ci ∈ C AC = i=1

i=1

is w-dense in M1 (X). Proof Suppose that μ ∈ M(X), that F is a finite subset of C(X) and that  > 0. Let d be a metric on X which defines the topology of X, and let M = max{ f ∞ : f ∈ F}. Since each f is uniformly continuous on X, there exists δ > 0 such that if d(x, y) < δ then | f (x) − f (y)| < /2, for f ∈ F. Since X is totally bounded, there exists a finite partition D of X into non-empty Borel sets, each of diameter at most δ/2. For each D ∈ D there exists cD ∈ C such  that d(cD , y) < δ for y ∈ D. Now let ν = D∈D μ(D)δcD . If f ∈ F then

16.7 The Locally Compact Riesz Representation Theorem

225

& & & &   & & & & & f dμ − f dν & = & & f dμ − μ(D)f (c ) D & & & & X

X

D

D∈D

& &  & & & ( f − fD ) dμ&& =& D∈D D

≤ (/2)



|μ(D)| ≤ /2.

D∈D

For each D ∈ D there exists λD ∈ Q such that |λD − μ(D)| < /2(M + 1)   and such that D∈D |λD | ≤ 1. Let π = D∈D λD δcD . If f ∈ F then & & & &  & & & & & & f dμ − f dπ & = & (μ(D) − λ )f (c ) D D & ≤ /2, & & & X

so that |

 X

X

f dμ −

 X

D∈D

f dπ| < .

Exercise 16.6.7 Suppose that (X, τ ) is a compact metrizable space. Bearing in mind that (C(X), ≤) is a Riesz space with order unit 1, establish the following. (i) If φ is a positive linear functional on C(X) then  there exists a unique positive Borel measure μ on  such that φ( f ) = X f dμ, for all f ∈ C(X).  there exists a Borel probability measure μ on C(X) such that (ii) If φ ∈ C(X)  φ( f ) = X f dμ, for all f ∈ C(X) if and only if φ(1) = 1 = φ . The set P(X) = {μ ∈ M1 (X), μ(X) = 1} is a w-closed convex subset of M1 (X) and is therefore w-compact. Exercise 16.6.8 Suppose that (X, τ ) is a compact metrizable space. Let C be a countable dense subset of X. Show that the countable set  n  n   AC = λi δci : n ∈ N, λi ∈ Q, λi ≥ 0, λi = 1, ci ∈ C i=1

i=1

is w-dense in P(X).

16.7 The Locally Compact Riesz Representation Theorem We now extend the Riesz representation theorem in a straightforward way to a locally compact Polish space (X, τ ) which is not compact. By Theorem 3.4.1, (X, τ ) is σ -compact, and there exists an increasing sequence (Kn )∞ n=1 int for each n ∈ N, and such that of compact subsets of X such that Kn ⊆ Kn+1 X = ∪∞ n=1 Kn . Recall that if f is a continuous real-valued function on a topological space (X, τ ), the support of f is the closure of the set {x ∈ X : f (x) = 0}. If s( f ) is the

226

Borel Measures

support of f , then X \ s( f ) is the largest open subset of X on which f is equal to 0. If (X, τ ) is a locally compact Polish space, we set Cc (X) to be the set of all continuous real-valued functions on X with compact support. Cc (X) is a Riesz space, under the usual pointwise ordering. If μ is a Radon measure on X, and  f ∈ Cc (X), let j(μ)( f ) = X f dμ. Then j(μ) is a positive linear functional on Cc (X). The Riesz representation theorem for locally compact Polish spaces says that the converse is true. Theorem 16.7.1 (The Riesz representation theorem for locally compact Polish spaces) Suppose that (X, τ ) is a locally compact Polish space which is not compact, and that φ is a positive linear functional on Cc (X).  Then there exists a unique Radon measure μ on X such that φ( f ) = X f dμ, for all f ∈ Cc (X). Proof There exists an increasing sequence (Kn )∞ n=1 of compact subsets of X int for each n ∈ N, and such that X = ∪∞ K . Let us such that Kn ⊆ Kn+1 n=1 n set En = {f ∈ Cc (X) : supp( f ) ⊆ Kn }, so that (En )∞ n=1 is an increasing sequence of subspaces of Cc (X) whose union is Cc (X). If f ∈ En+1 and x ∈ Kn let Tn ( f )(x) = f (x). Then Tn is positive linear mapping of En+1 onto C(Kn ). Using Tietze’s extension theorem, it follows that there exists a sequence (gn.r )∞ r=1 in En+1 which decreases pointwise to IKn . If h ∈ C(Kn+1 ), let ψn (h) = limr→∞ φ(gn,r h); ψn is a positive linear functional on C(Kn+1 ), and if Tn (h1 ) = Tn (h2 ) then ψ(h1 ) = ψ(h2 ). Thus if f = Tn (h) ∈ C(Kn ) and we set θn ( f ) = ψn (h) then θn is a properly defined positive linear functional on C(Kn ), which is represented by a finite Borel measure μn , by the Riesz representation theorem. If A is a Borel set in X, we set μ(A) = limn→∞ μn (A ∩ Kn ). It then follows from Theorem  16.3.3 that μ is a Radon measure on X; further, if f ∈ Cc (X) then φ( f ) = X f dμ.

16.8 The Stone–Weierstrass Theorem We now use the Riesz representation theorem and the Krein–Mil’man theorem to prove the Stone–Weierstrass theorem, at least for compact metrizable spaces. Theorem 16.8.1 (The Stone–Weierstrass theorem) Suppose that K is a compact metrizable space and that A is a linear subspace of C(K) which is an algebra under pointwise multiplication, which contains the constant functions and which separates points; if x1 = x2 , there exists g ∈ A such that g(x1 ) = g(x2 ). Then A is dense in (C(K), .∞ ).

16.8 The Stone–Weierstrass Theorem

227

Proof Suppose that A = C(K). It then follows from the Hahn–Banach theorem that A◦ = {φ ∈ C(K) : φ(a) = 0 for a ∈ A} is a non-zero weak*-closed linear subspace of C(K) , and B(A◦ ) = {φ ∈ A◦ ; φ ≤ 1} is a non-zero weak*compact convex subset of C(K) . By the Krein–Mil’man theorem, there exists an extreme point φ of B(A◦ ), and this is represented by a non-zero signed measure μ, by the Riesz representation theorem. Let μ = μ+ − μ− . Since μ(X) = 0 and |μ|(X) = 1, μ+ (X) = μ− (X) = 12 . Since μ(K) = φ(1) = 0, supp(|μ|) is not a singleton. Let x1 and x2 be two distinct points of S = supp(|μ|). By hypothesis there exists g ∈ A with g(x1 ) < g(x2 ). By adding  a constant, we can assume that g ≥ 0, and by scaling we can suppose that K g d|μ| = 1. Since A is an algebra, K g dμ = 0; so   1 + − K g dμ = K g dμ = 2 . Thus sup{g(x) : x ∈ supp(|μ|)} = α ≥ 1. There exist x0 ∈ supp(|μ|) such that g(x0 ) = α. We consider two cases. If α = 1 then    |1 − g| d|μ| = 1 − g d|μ| = |μ|(K) − g d|μ| = 0, K

K

K

so that g = 1 μ almost everywhere on supp(|μ|). But g is continuous, and so g = 1 on supp(|μ|). But g(x1 ) = g(x2 ), giving a contradiction. If α > 1, let μ2 = g.dμ. There is an open neighbourhood U of x0 such that g(x) > (1 + α)/2 for x ∈ U. Then there exists a Borel subset B of U such that μ(B) = μ2 (B), so that μ = μ2 . Let λ = 1/α, and let   1 − λg μ1 = dμ so that μ = (1 − λ)μ1 + λμ2 . 1−λ Then μ1 and μ2 are in B(A◦ ). Since μ2 = μ, again we have a contradiction. Thus A is dense in C(K). Corollary 16.8.2 Suppose that K1 and K2 are compact metrizable spaces. Then the set   n fj ⊗ gj : n ∈ N, fj ∈ C(K1 ), gj ∈ C(K2 ) j=1

is dense in C(K1 × K2 ). Corollary 16.8.3 The space BL(K) is dense in C(K). Proof For BL(K) is an algebra which contains the constants and separates points. Here is the usual way to prove the general Stone–Weierstrass theorem.

228

Borel Measures

Exercise 16.8.4 Define a sequence of polynomials (pn )∞ n=0 by setting p0 = 0 and pn+1 (x) = pn (x) + 12 (x2 − (pn (x))2 ). Show that if |x| ≤ 1 then 0 ≤ pn (x) ≤ pn+1 (x) ≤ |x|, and that pn (x) tends uniformly to |x|. Exercise 16.8.5 Suppose that A is a subalgebra of C(X), where (X, τ ) is a compact Hausdorff space. Show that A is a lattice. Exercise 16.8.6 Suppose that L is a linear subspace of C(X), where (X, τ ) is a compact Hausdorff space, and that L is a lattice, that L contains the constants and that L separates points. Show that L is dense in C(X).

16.9 Product Measures We now use the Riesz representation theorem to develop the theory of product measures on Polish spaces. This is rather simpler than the abstract theory. Theorem 16.9.1 Suppose that X1 and X2 are two compact metrizable spaces, that μ1 ∈ P(K1 ) and μ2 ∈ P(X2 ). Then there exists a unique μ1 ⊗ μ2 in P(X1 × X2 ) such that (μ1 ⊗ μ2 )(B1 × B2 ) = μ1 (B1 ).μ2 (B2 ) for B1 , B2 Borel sets in X1 , X2 . Proof X1 and X2 are homeomorphic to Gδ -subsets of compact metrizable spaces K1 and K2 . Push the measures forward to K1 and K2 ; it is enough to prove the result for K1 and K2 . Let A be the algebra   n fj ⊗ gj : n ∈ N, fj ∈ C(K1 ), gj ∈ C(K2 ) . j=1

If h =

n

j=1 fj

⊗ gj , let φ(h) =



Now (

n   j=1

 fj dμ1 . K1

gj dμ2 . K2

∈ C(K2 ), so that    φ(h) = h(x, y) dμ1 (x) d μ2 , (y)

K1 fj dμ1 )gj

K2

K1

and φ(h) does not depend upon the representation of h, and is properly defined, and unique. Further, φ is positive, so that φ is a continuous linear functional

16.9 Product Measures

229

on A. Since A is dense in C(K1 × K2 ) (Corollary 16.8.2), φ extends uniquely to a positive linear functional on C(K1 × K2 ), which, by the Riesz representation theorem, is represented by an element μ1 ⊗μ2 of P(K1 ×K2 ). Since φ is unique, so is μ1 ⊗μ2 . Approximation arguments then show first that (μ1 ⊗μ2 )(C1 ×C2 ) for C1 , C2 compact sets, and then (μ1 ⊗μ2 )(B1 ×B2 ) for B1 , B2 Borel sets. We can extend this result to finite products, and to σ -finite measures. For example, we denote by λd the product λ ⊗ · · · ⊗ λ the product of d copies of Lebesgue measure λ; λd is Lebesgue measure on Rd . How can we evaluate K1 ×K2 f d(μ1 × μ2 )? We can do this by repeated integration. Theorem 16.9.2 (Tonelli’s theorem) Suppose that X1 and X2 are two Polish spaces, that μ1 ∈ P(X1 ) and μ2 ∈ P(X2 ). If f is a non-negative Borel measurable function on X1 × X2 then X1 f (x, y) dμ1 (x) is a Borel measurable function on X2 , and   f d(μ1 ⊗ μ2 ) = (f (x, y) dμ1 (x)) dμ2 (y). X1 ⊗X2

X2

Proof Once again, it is enough to prove the result for compact metrizable spaces K1 and K2 .First, suppose that f is continuous. Since P(K1 ) ⊆ C(K1 ) , the function y → K1 f (x, y) dμ1 (x) is continuous. Since   f d(μ1 ⊗ μ2 ) = (f (x, y) dμ1 (x)) dμ2 (y) K1 ⊗K2

K2

for f ∈ A, and A is dense in C(K1 × K2 ), the equation holds for continuous f . If f is lower semi-continuous, there exists an increasing sequence ( fn )∞ n=1 in C(K1 × K2 ) which converges pointwise to f . Then K1 fn (x, y) dμ1 (x) converges pointwise to K1 f (x, y) dμ1 (x), so that K1 f (x, y) dμ1 (x) is lower semi-continuous. Further,   f d(μ1 ⊗ μ2 ) = lim fn d(μ1 ⊗ μ2 ) n→∞ K ×K K1 ×K2   1 2 = lim fn (x, y) dμ1 (x) dμ2 (y) n→∞ K K    2  1 = fn (x, y) dμ1 (x) dμ2 (y) lim K2 n→∞ K1    = f (x, y) dμ1 (x) dμ2 (y). K2

K1

In particular, the result holds for indicator functions of open sets. Now let S be the collection of Borel sets E whose indicator functions satisfy the theorem.

230

Borel Measures

Then S is a σ -field which contains the open sets, and so is equal to the Borel σ -field B. Finally, a standard approximation argument shows that it holds for non-negative Borel measurable functions. This theorem can also be extended to σ -finite measures, by summing over a disjoint sequence of sets of finite measure, in the usual way. Exercise 16.9.3 Suppose that f is a positive integrable Borel measurable function on (X, B, μ), where μ is a σ -finite measure on a Polish space X. Let  S = {(x, t) : x ∈ X, 0 ≤ t ≤ f (x)}. Show that X f dμ = μ(S). (The integral is the ‘area under the curve’.) By considering positive and negative parts, we obtain the following. Theorem 16.9.4 (Fubini’s theorem) Suppose that μ1 and μ2 are Borel probability measures on Polish spaces X1 and X2 , and that f is a Borel measurable function on X1 × X2 . Then the following are equivalent: (i) f ∈ L1 (μ1 ⊗ μ2 ); (ii) the function fy(x) = f (x, y) is in L1 (μ1 ) for μ2 -almost all y, and the function y → X1 fy dμ1 is μ2 -measurable, and in L1 (μ2 ). If so, then 

 X1 ×X2





f d(μ1 ⊗ μ2 ) =

fy (x) dμ1 (x) X2

dμ2 (y).

X1

 It may however happen that X1 fy dμ1 is infinite, or undefined, on a set of μ2 -measure 0. What about infinite products? Suppose now that for each n ∈ N, πn is a Borel  probability measure on a Polish space (Xn , τn ). Let (X, τ ) = ∞ n=1 (Xn , τn ). Can we define an infinite product measure on (X, τ )? Again, we use the Riesz representation theorem to show that we can. Theorem 16.9.5 Suppose that, for each n ∈ N, (Xn , τn ) is a Polish space and  πn ∈ P(Xn ). Let X = ∞ n=1 Xn , with the product topology, and let pn (x) = (x1 , . . . , xn ), for each n ∈ N and x ∈ X. Then there exists a unique π ∈ P(X) such that pn ∗ (π ) = π1 ⊗ · · · ⊗ πn . Proof For each n, there is a homeomorphism in of Xn onto a dense Gδ subset  of a compact metrizable space (X˜ n , τ˜n ); then ∞ n=1 in (Xn ) is a Gδ subset of ∞ X˜ = n=1 X˜ n . By considering the push-forward measures (in )∗ πn we can therefore suppose that each Xn is compact. If x ∈ X, let pn (x) = (x1 , . . . , xn ). Let Fn = {μ ∈ P(X) : pn∗ (μ) = π1 ⊗ · · · ⊗ πn }.

16.10 Disintegration of Measures

231

Then it follows from Exercise 16.6.5 that Fn is a non-empty w-compact subset ∞ of P(X), and the sequence ( fn )∞ n=1 is decreasing, so that F = ∩n=1 Fn is not empty. The set of functions {f ◦ pn : f ∈ C(X1 × · · · × X), n ∈ N} is dense in C(X), from which it follows that F is a singleton {π }. Finally, π is unique, since the sets {p−1 n (A) : A a Borel set in X1 × · · · × Xn , n ∈ N} generate the Borel sets in X.

16.10 Disintegration of Measures An important feature of the theorems of Tonelli and Fubini is that integrals with repect to product measures can be evaluated by repeated integration. We shall show in Theorem 16.10.2 that this extends to more general measures on products. For this, we need the idea of the disintegration of a measure. Suppose that X and Y are Polish spaces, that μ ∈ P(X) and that T is a Borel measurable mapping from X into Y. Let ν = T∗ (μ) be the push-forward measure: ν(A) = μ(T −1 (A)). A T-disintegration of μ is a family {λy : y ∈ T(X)} of Borel probability measures on X such that (i) if y ∈ T(X) then λy (T −1 {y}) = 1 for ν-almost all y, and (ii) if f ∈ L1 (X, μ) then (a) f ∈ L1 (λy ) for almost  all y ∈ T(X), (b) the function y →   X f dλ   y is ν-measurable, and (c) X f dμ = Y X f dλy dν. Theorem 16.10.1 Suppose that X and Y are Polish spaces, that μ ∈ P(X) and that T is a Borel measurable mapping from X into Y. Then a T-disintegration of μ exists, and is essentially unique; if {λy : y ∈ T(X)} and {λy : y ∈ T(X)} are two T-disintegrations, then λy = λy for almost all y ∈ T(X). Proof Again, let ν = T∗ (μ). First we consider the case where X and Y are compact metric spaces and T is a continuous surjection of X onto Y. We consider the product space X × Y, and denote the projection of X × Y onto Y by p. If f and g are Borel measurable functions on X and Y respectively, we set ( f ⊗ g)(x, y) = f (x)g(y); then f ⊗ g is a Borel measurable function on X × Y, which is continuous if f and g are. If x ∈ X, let γ (x) = (x, T(x)). γ is a homeomorphism of X onto T , the graph of T. Let π = γ∗ (μ): π ∈ P(X × Y), and the support of π is contained in T . Further, ν = p∗ (π ).

232

Borel Measures

Suppose now that f ∈ L1 (X, μ). Then γ∗ ( f .dμ) = ( f ⊗ 1).dπ . Let νf = T∗ ( f .dμ); then T∗ ( f .dμ)  = p∗ (( f ⊗ 1).dπ ). If  > 0, there exists δ > 0 such that if μ(E) < δ then E | f | dμ < . Thus if ν( f ) < δ then & &  & & & f dμ&& ≤ | f | dμ < . |νf ( f )| = & T −1 ( f )

T −1 ( f )

Thus νf is absolutely continuous with respect to ν. By the Radon–Nikodym theorem, there exists lf ∈ L1 (ν) such that νf = lf . dν. We now need to restrict attention to a suitable countable dense subset of C(X). Let C be a countable dense subset of C(X) which is a vector space over Q, and which contains 1. Then there exists a Borel subset YC of Y, with ν(Y \ YC ) = 0, such that if y ∈ YC , f , g ∈ C and α ∈ Q then lαf (y) = αlf (y), l1 (y) = 1,

lf +g (y) = lf (y) + lg (y), lf (y) ≥ 0 if f ≥ 0.

It follows that if y ∈ YC and f , g ∈ C then − f − g∞ .1 ≤ f (y) − g(y) ≤ f − g∞ .1, and so l. (y) extends by continuity to a positive linear functional my on C(X), with my (1) = 1. By the Riesz representation theorem, there exists λy ∈ P(X) such that my ( f ) = X f dλy , for  f ∈ C(X).  If f ∈ C and y ∈ YC then X f dλy = lf (y), so that X f dλy is measurable, and       f dμ = dνf = lf dν = fdλy (x) dν(y). X

Y

Y

Y

X

This extends by bounded convergence to all f in C(X). By Exercise 16.1.5, it extends to indicator functions of open sets and closed sets, and by regularity to indicator functions of Borel sets. It then extends by monotone convergence if f ∈ L1 (μ), there may be a subset of YC of to functions in L1 (μ) (though  ν-measure 0 on which X f dλy is infinite, or undefined). If f ∈ C(X) and g ∈ C(Y) then      f ⊗ g dπ = g dνf = ( f ⊗ g)(x, y) dλy dν(y). X×Y

Y

Y

X

Again, this extends by linearity and continuity to all h ∈ C(X × Y), to give     h dπ = h(x, y) dλy dν(y). X×Y

Y

X

16.10 Disintegration of Measures

233

By Exercise 16.1.5, this extends to indicator functions of closed subsets of X × Y. Thus     I dπ = I (x, y) dλy (x) dν(y), 1= X×Y

so that



Y

X

I (x, y) dλy (x) = λy (T −1 {y}) = 1 X

for ν almost all y. Next, we establish uniqueness. Suppose that {λy : y ∈ T(X)} and {λy : y ∈ T(X)} are two T-disintegrations. If f ∈ C and A is a Borel subset in Y then     f dλy dν(y) = f (x)IA (T(x)) dμ(x) A X X     f dλy dν(y), = 



A

X

dλy ,

for all f ∈ C, for almost all y, and so λy = λy for so that X f dλy = X f almost all y. We now turn to the general case, where X and Y are Polish spaces, and T is Borel measurable. There exist a disjoint sequence (Kn )∞ n=1 of compact subsets K ) = 1, such that the restriction of T to Kn is continuous, of X with μ(∪∞ n n=1 for each n. We can disintegrate μ on each Kn , and add the disintegrations together, to obtain the result. Again, uniqueness follows from the uniqueness on each Kn . The fact that we consider the measure π on the product X × Y suggests that we should consider more general measures on products. Theorem 16.10.2 Suppose that X, Y are Polish spaces, and that π ∈ P(X ×Y). Let ν = p∗ (π ), where p is the projection of X × Y onto Y. (ν is the Y-marginal distribution of π .) Then there exists a Borel subset YC of Y, with ν(Y \YC ) = 0, and a family {μy : y ∈ YC } in P(X) such that if f ∈ L1 (π ) then there is a Borel subset Yf of YC with ν(Y \ Yf ) = 0 such that (i) f (., y) ∈ L1 (μy ) for y ∈ Yf , is a measurable function on Yf , and (ii) X f (x, y) dμy (x)   (iii) X×Y f dπ = Yf X f (x, y) dμy (x) dν(y). Proof In the theorem, we replace X by X × Y, μ by π and T by p. Since the support of λy is contained in X × {y}, there exists μy ∈ P(X) such that μy (A) = λy (A × {y}) for each Borel set A in X. The family {μy : y ∈ YC } clearly satisfies the corollary.

234

Borel Measures

Let us reconsider the formula in Theorem 16.10.1. Let us set BT = T −1 B(T(X)). Then BT is a sub σ -field of B. Since ν = T∗ (μ), let us set μz = λT(z) , for z ∈ X. Note that if T(z) = T(z ) then μz = μz . We then see that if f ∈ L1 (X, μ) then (a) f ∈ L1 (μz ) for almost  all z ∈ X, (b) the function z →   X f dμz is B   T -measurable, and (c) X f dμ = X X f (x) dμz (x) dμ(z).

 The important feature is (b). The function E( f |BT )(z) = X f dμz is the conditional expectation of f with respect to BT and the function z → μz is the regular conditional probability of μ, with respect to BT . As an example, let X = C([0, S]), and let μ ∈ P(X). Thus μ is a probability measure on the continuous paths on [0, S]. Suppose that 0 < R < S. Let Y = C([0, R]), and let T be the restriction mapping. Then if y ∈ Y, λy is the conditional probability of a path g ∈ C([0, S]) for which g(x) = f (x) for 0 ≤ x ≤ R.

16.11 The Gluing Lemma We shall need a fundamental result about measures on products. Suppose that X and Y are Polish spaces, that μ ∈ P(X) and ν ∈ P(Y). If (x, y) ∈ X × Y, let p1 (x, y) = x and p2 (x, y) = y. We set μ,ν = {π ∈ P(X × Y) : (p1 )∗ π = μ, (p2 )∗ π = ν}. μ,ν is the set of Borel probability measures on X × Y with marginals μ and ν. Theorem 16.11.1 (The gluing lemma) Suppose that X, Y, Z are Polish spaces, that μ ∈ P(X), ν ∈ P(Y), π ∈ P(Z) and that α ∈ μ,ν and β ∈ ν,π . Then there exists γ ∈ P(X × Y × Z) with marginals α on X × Y and β on Y × Z. ˜ Y˜ Proof We can embed X, Y and Z as Gδ subsets of compact metric spaces X, ˜ push μ, ν and π forward to elements μ, ˜ P(Y), ˜ P(Z), ˜ and Z, ˜ ν˜ and π˜ of P(X), and  . Thus we can suppose that X, Y and α, β to elements α, ˜ β˜ of μ,˜ ˜ ν ν˜ ,π˜ and Z are compact metric spaces. Let V = {f + g : f ∈ C(X × Y), g ∈ C(Y × Z)}; V is a linear subspace of C(X × Y × Z). If f + g ∈ V, we set φ( f + g) = X×Y f dα + Y×Z g dβ.

16.11 The Gluing Lemma

235

First we show that φ is well-defined. If f + g = f  + g then f − f  = g − g, so that f − f  and g − g are functions of y only. Hence        ( f − f ) dα = ( f − f ) dν = (g − g) dν = (g − g) dβ, X×Y

Y

Y

Y×Z

and φ( f + g) = φ(f  + g ). Thus φ is a linear functional on V. Secondly, we show that if f + g ≥ 0 then φ( f + g) ≥ 0. Let h(y) = infx∈X f (x, y); h is an upper semi-continuous function on Y. Let fˆ = f − h; then fˆ ≥ 0. Let gˆ = h + g. Then g(x, ˆ y) = inf (f (x, y) + g(y, z)) ≥ 0. x∈X

fˆ and gˆ are semi-continuous, and therefore integrable, and   φ( f + g) = fˆ dα + gˆ dβ ≥ 0. X×Y

Y×Z

Thirdly, we observe that φ(1) = 1, so that φ = 1. We now apply the Hahn–Banach theorem. There exists a linear functional ψ on C(X × Y × Z), with ψ = φ, which extends φ. Since ψ(1) = 1, ψ is a positive linear functional on C(X × Y × Z); by the Riesz  representation theorem, there exists γ ∈ P(X × Y × Z) such that ψ(k) = (X×Y×Z) k dγ for k ∈ C(X × Y × Z). If f ∈ C(X × Y) then   ψ( f ) = f dα = f dγ , X×Y

X×Y×Z

from which it follows by the usual arguments that γ (A × Z) = α(A), for A a Borel set in X × Y. Similarly, γ (X × B) = β(B), for B a Borel set in Y × Z. Exercise 16.11.2 Suppose that a(x, y) and b(y, z) are non-negative functions in L1 (Q, B, λ3 ), where Q is the  1 unit cube {(x, y, z)1 : 0 ≤ x, y, z ≤ 1} and λ3 is Lebesgue measure, and that 0 a(x, y) dλ(x) = 0 b(y, z) dλ(z) = h(y) for all 1 y ∈ [0, 1]. Find a function g in L1 (λ3 ) such that 0 g(x, y, z) dλ(z) = a(x, y) 1 and 0 g(x, y, z) dλ(x) = b(y, z) for all x, y, z. The gluing lemma can also be proved, using the disintegration of measures. (1)

Exercise 16.11.3 In the setting of the gluing lemma, let {λy : y ∈ Y} be the (2) p1 -disintegration ofα and {λy : y ∈ Y} be the p2 -disintegration of β. Let (1) γ (A × B × C) = B λ (A)y .λ(2) (C) dν(y), where A, B, C are Borel sets in X, Y, Z. Show that γ extents to a Borel measure on X × Y × Z which meets the requirements of the gluing lemma.

236

Borel Measures

16.12 Haar Measure on Compact Metrizable Groups We end this chapter by establishing the existence of Haar measure on compact and locally compact metrizable topological groups. This is a Borel measure which is left or right translation invariant. We begin with the compact metrizable case, such as the group On of orthogonal transformations of Rn , which is, for good reasons, much the simpler. Theorem 16.12.1 Suppose that G is a compact metrizable group. Then there exist unique Borel probability measures μl and μr on G such that μl (gA) = μl (A) and μr (Ag) = μr (A) for each g ∈ G and each Borel set A. Further, μl = μr . Proof We give two proofs. The first uses Kakutani’s fixed-point theorem. Let lg be the left regular representation of G in C(G), so that lg ( f )(h) = f (gh), for f , g ∈ G. Then the mapping (g, f ) → lg ( f )(h) : G × C(G) → C(G) is jointly continuous (Proposition 5.5.1), and so {lg ( f ); g ∈ G} is a compact  subset of    C(G), for each f ∈ C(G). If μ ∈ P(X) ⊆ C(X) then lg (μ) ≤ 1 and lg (μ)(1) = μ(1) = 1, so that lg (μ) ∈ P(X). Thus the conditions of Kakutani’s fixed point theorem are satisfied, and so there exists μl ∈ P(X) for which  lg (μl ) = μl , for each g ∈ G. That is, G f dμ = G lg ( f ) dμ for each g ∈ G and f ∈ C(G). From this it follows that μl (gA) = μl (A) for each g ∈ G and each Borel set A. The existence of a right-invariant Borel probabilty measure is proved in an exactly similar way. For the second proof of existence, we use Hall’s marriage theorem, and consider minimal -nets; recall that these are -nets with as few elements as possible. Let d be a metric on G for which d(gx, gy) = d(x, y) = d(xg, yg) for all x, y, g ∈ G. G is d-totally bounded. For each k ∈ N, let nk be a minimal 1/k net in G. If f ∈ C(G), let φk ( f ) = (1/|nk |) g∈nk f (g). Then φk is a positive linear functional in C(G) , φk  ≤ 1 and φk (1) = 1. If f ∈ C(G) then f is uniformly continuous, so that if  > 0 then there exists K ∈ N such that if d(g, h) < 2/K then | f (g) − f (h)| < /2. Suppose that k, l ≥ K and that h ∈ nl . Then nk h = {gh : g ∈ nk } is also a minimal 1/k-net. By Proposition 3.2.5 there is a bijection ψh : nk → nk h such that d(g, ψ(g)) < 2/k, for each g ∈ nk . Thus & & & &  & & 1 &φk ( f ) − f (gh)&& < /2. & |n | k g∈n & & k Averaging over nl ,

& & & &  & & 1 &φk ( f ) − & < /2. f (gh) & & |nk |.|nl | & & g∈nk ,h∈nl

16.12 Haar Measure on Compact Metrizable Groups

Similarly,

237

& & & &  & & 1 &φl ( f ) − f (gh)&& < /2, & |nk |.|nl | & & g∈nk ,h∈nl

and so |φk ( f ) − φl ( f )| < . Thus φk ( f ) converges, as k → ∞, to φ( f ), say. Then φ is a positive linear functional on C(K), φ ≤ 1 and φ(1) = 1, so that φ is represented by a Borel probability measure μ. If g ∈ G, then gnk and nk g−1 are minimal 1/k-nets, and so, if k ≥ K, then |φk (lg ( f )) − φk ( f )| < /2 and |φk (rg ( f )) − φk ( f )| < /2. Thus φ(lg ( f )) = φ(rg ( f )) = φ( f ), and φ, and therefore μ, are left and right invariant. It remains to show that there is a unique left-invariant Borel probability measure and a unique right-invariant Borel probability measure, and that they are equal. Suppose that μl is a left-invariant Borel probability measure and that Borel probability measure. If f ∈ C(G) and g ∈ G, then μ   r is a right-invariant f dμ = f (gh) dμ l l (h), a constant, so that G G     f dμl = f (gh) dμl (h) dμr (g). G

Similarly

G

G

 

 f dμr = G

 f (gh) dμr (g) dμl (h).

G

G

The two repeated integrals are equal, and so μl = μr ; uniqueness follows from this. Suppose now that a is a continuous action of a compact metrizable group G on a compact metrizable space X. An orbit of a is a set of the form a(G)(x) = {a(g)(x) : g ∈ G}, where x is an element of X. Orbits are compact, and the orbits form a partition of X. The action is transitive if X is an orbit; that is, if whenever x, y ∈ X there exists g ∈ G such that a(g)(x) = y. In this case X is called a homogeneous space for G, and G is called a group of symmetries of X. Theorem 16.12.2 Suppose that a is a continuous action of a compact metrizable group G on a compact metrizable space X. Then there exists a Borel probability measure μ on X, which is invariant under a; if A is a Borel subset of X and g ∈ G then μ(a(g)(A)) = μ(A). If X is homogeneous, they are unique. Proof Let x ∈ X. The mapping φ : g → a(g)(x) is a continuous mapping of G onto an orbit of the action. Let μ = φ∗ (ν), where ν is Haar measure on G. If A is a Borel subset of X and g ∈ G, then

238

Borel Measures φ −1 (a(g)(A)) = {h ∈ G : a(h)(x) ∈ a(g)(A)} = {h ∈ G : a(g−1 h)(x) ∈ (A)} = {h ∈ G : g−1 h ∈ φ −1 (A)} = gφ −1 (A),

so that μ(a(g)(A)) = μ(A). The support of μ is contained in the orbit a(G)(x), and so the measure μ is unique only if the action a is transitive. In fact the support of μ = a(G)(x), since if y is in the support of μ then, by invariance, so is a(g)(y), for each g ∈ G. Suppose now that the action istransitive, so that the support of μ is X. This implies that if g ∈ L∞ (μ) and X fg dμ = 0 for all f ∈ C(X), then g = 0. Suppose now that π is a Borel probability measure which is invariant under a. Let σ = 12 (π + μ). Then σ is invariant under a, the support of σ is X, and by the Radon–Nikodym theorem there exists k ∈ L∞ , with 0 ≤ k ≤ 2 such that π = k dσ . If f ∈ C(X) and g ∈ G then    fk dσ = f dπ = a(g)f dπ X X  X a(g)( f ).k dσ = f .a(g−1 )(k) dσ . = X

X

Since this holds for all f ∈ C(X), k = a(g−1)(k) for all g ∈ G. Thus k is constant, so that π = kσ . Since π(X) = σ (X) = 1, k = 1 and π = μ. Exercise 16.12.3 Suppose that (X, d) is a compact metric space, and that IX is the group of isometries of X onto itself and that it acts transitively on X. (X is then called an isometrically homogeneous space.) Show that there is a unique Borel probability measure μ on X such that μ(i(A)) = μ(A) for each Borel set A and each isometry i of X onto itself. Exercise 16.12.4 Let B(Rn ) be the unit ball in Rn , and let On be the group of rotations and reflections of B(Rn ). Show that On is the group of isometries of B(Rn ). What are the orbits of On ? Show that there is an On -invariant probability measure on B(Rn ) whose support is B(Rn ). Is such a measure unique?

16.13 Haar Measure on Locally Compact Polish Topological Groups Suppose now that (G, τ ) is a locally compact Polish topological group, which is not compact. We shall show that there exist a left-invariant Radon measure ψl and a right-invariant Radon measure ψl on G, each unique up to scaling,

16.13 Haar Measure on Locally Compact Polish Topological Groups 239

but that they may be essentially different. (Of course, if G is an Abelian group, then they are the same.) As we shall see, the proof is more complicated than in the compact case. First let us give a two-dimensional example where the left- and rightinvariant measures are essentially different. Let Hr be the right half plane {(a, b) ∈ R2 : a > 0}. If (a, b) ∈ Hr , let   a b φ((a, b)) = . 0 1 Then φ is a homeomorphism of Hr onto a subgroup of GL2 (R), and if we define (a, b)(c, d) = φ −1 (φ((a, b)), φ((c, d))) = (ac, ad + b), then Hr becomes a locally compact Polish topological group. Let λ be Lebesgue measure on Hr . Since (a, b).(x, y) = (ax, ay + b), the measure λ(x, y)/xy is a left-invariant measure on Hr . Since (x, y).(a, b) = (ax, bx + y), the measure λ(x, y)/x is a right-invariant measure on Hr . Theorem 16.13.1 Suppose that (G, τ ) is a locally compact Polish topological group, which is not compact. There exist a non-zero left-invariant Radon measure ψl and a right-invariant Radon measure ψr on G, each unique up to scaling. Proof By Theorem 5.6.3 there exists a group-norm ν on G which defines the topology τ . Let Nj = {g ∈ G : ν(g) ≤ 1/2j } for j ∈ Z+ . Since G is locally compact, we may suppose, by scaling d if necessary, that each Nj is compact. Let bj = (2 − 2j ν)+ . Then bj ≥ INj and if j > 0 then supp(bj ) ⊆ Nj−1 . Let Cc (G) be the Riesz space of continuous functions on G of compact support. Our aim is to define a translation-invariant positive linear functional φ on Cc (G). Suppose that f ∈ Cc+ (G) and that j ∈ Z+ . Recall that if f ∈ Cc (G) and g ∈ G then lg ( f (h)) = f (g−1 h). If f ∈ Cc+ (G), there exists a finite set Fj in G such that supp( f ) ⊆ ∪g∈Fj g−1 Nj , and so if we set    ψj ( f ) = inf tg : F ⊆ G, F finite, tg ≥ 0, tg lg (bj ) ≥ f , g∈F

g∈F

then ψj ( f ) < ∞. Then ψj has the following easy properties, which we shall frequently use without comment. Exercise 16.13.2 (i) ψj (lg ( f )) = ψj ( f ); (ii) ψj ( f1 + f2 ) ≤ ψj ( f1 ) + ψj ( f2 ); (iii) ψj (αf ) = αψj ( f ) for α ≥ 0;

240

Borel Measures

(iv) if f1 ≤ f2 then ψj ( f1 ) ≤ ψj ( f2 ); (v) ψj ( f ) ≥ 12  f ∞ ; (vi) if j ≤ k then ψk ( f ) ≤ ψk (bj ).ψj ( f ). Proof Here is the proof of (vi). Suppose that f ≤  bj ≤ h∈F th lh (bk ). Then    tg lg sh lh (bk ) f ≤ g∈F



= so that ψk ( f ) ≤





g∈F tg lg (bj ),

and that

h∈F 

tg sh lgh (bk ),

g∈F h∈F  g∈F h∈F  tg sh .

Taking infima, (vi) follows.

The next lemma provides the key to the proof of the theorem. Lemma 16.13.3 Suppose that f1 , f2 ∈ Cc+ (G), that j ∈ N+ and that  > 0. Then there exists j0 ∈ Z+ such that ψj ( f1 ) + ψj ( f2 ) ≤ ψj ( f1 + f2 ) + ψj (b0 ), for j ≥ j0 . Proof There exists a ∈ Cc+ (G) such that a(g) = f1 + f2 ∞ for g ∈ supp( f1 + f2 ). Choose 0 < δ < 1 such that 5δψ0 (a) < . Let q = f1 + f2 + δa; then q ≤ 2a and ψ0 (q) ≤ (1 + δ)ψ0 (a) ≤ 2ψ0 (a). Let p1 = f1 /q, p2 = f2 /q, so that p1 + p2 ≤ 1. The functions p1 and p2 are uniformly continuous, and so there exists j0 such that if ν(g−1 h) ≤ 2/2j0 then |p1 (g) − p1 (h)| < δ and  |p2 (g) − p2 (h)| < δ . Suppose now that j ≥ j0 and that q ≤ g∈F tg lg (bj ), where F is a finite subset of G. If h ∈ G then lg (bj )(h) = 0 unless ν(g−1 h) ≤ 2/2j , in which case |p1 (g) − p1 (h)| < δ and |p2 (g) − p2 (h)| < δ. Thus   f1 (h) = p1 (h)q(h) ≤ tg p1 (h)lg (bj )(h) ≤ tg p1 (g + δ)lg (bj )(h) g∈F



g∈F

so that ψj ( f1 ) ≤ (p1 (g) + δ) g∈F tg . Similarly ψj ( f2 ) ≤ p2 (g + δ)  and so ψj ( f1 ) + ψj ( f2 ) ≤ (1 + 2δ) g∈F tg . Taking the infimum,



g∈F tg ,

ψj ( f1 ) + ψj ( f2 ) ≤ (1 + 2δ)ψj (q) ≤ (1 + 2δ)(ψj ( f1 + f2 ) + δψj (a)) = (ψj ( f1 + f2 ) + δ(2ψj ( f1 + f2 ) + 3ψj (a)) ≤ (ψj ( f1 + f2 ) + 5δψj (a). But ψj (a) ≤ ψ0 (a).ψj (b0 ), and so ψj ( f1 ) +ψj ( f2 ) ≤ ψj ( f1 + f2 ) + ψj (b0 ). We now set φj = ψj /ψj (b0 ). By Exercise 16.13.2 (vi), φj ( f ) ≤ φ0 ( f ). The group G is σ -compact; let (Kn )∞ n=1 be a fundamental sequence of compact subsets, let Cn (G) = {f ∈ Cc (G) : supp(f ) ⊆ Kn } and let

16.13 Haar Measure on Locally Compact Polish Topological Groups 241 + Cn+ (G) = {f ∈ Cn (G) : f ≥ 0}. For each n let cn be a function in Cn+1 (G) with cn (g) = 1 for g ∈ Kn and 0 ≤ cn ≤ 1. Each Cn (G) is a separable Banach space under the uniform norm, and so there exists a countable dense subset Dn of Cn+ , closed under addition. Let D = ∪n∈N Dn . Since φj ( f ) ≤ φ0 ( f ) for f ∈ Cc+ (G), a diagonal argument shows that there exists an increasing sequence (jk )∞ k=1 such that φjk ( f ) converges, to φ( f ), say, for each f ∈ D. Note that if f1 , f2 ∈ D then φ( f1 + f2 ) = φ( f1 ) + φ( f2 ). We now show that φjk ( f ) converges, to φ( f ), say, for each f ∈ Cc+ (G). Suppose that f ∈ Cn+ and that  > 0. Let η = /5(1 + φ0 (cn )). There exists d ∈ Dn such that f − d∞ < η. Let u = ( f ∨ d) − f and v = ( f ∨ d) − d, so that f ∨ d = f + u = d + v, u∞ < η and v∞ < η. Thus φj (u) ≤ ηφ0 (cn ) and φj (v) ≤ ηφ0 (cn ). By Lemma 16.13.3, there exists k0 such that

|φjk ( f + u) − φjk (f ) − φjk (u)| < η and |φjk (d + v) − φjk (d) − φjk (v)| < η for k ≥ k0 , and so |φjk (f ) − φjk (d)| < 2η(1 + φ0 (cn )) for k ≥ k0 . There exists k1 ≥ k0 such that |φjk (d) − φjl (d)| < η for k, l ≥ k1 , and so |φjk (f ) − φjl (f )| < 5η(1 + φ0 (cn )) =  for k, l ≥ k1 : φjk ( f ) converges, to φ( f ) say, as k → ∞. It follows from the construction that φ( f1 + f2 ) = φ( f1 ) + φ( f2 ), φ(αf ) = αφ( f ) and φ(lg ( f )) = φ( f ) for f1 , f2 , f ∈ Cc+ (G), α ≥ 0 and g ∈ G; by Proposition 15.1.4, φ extends to a positive linear functional, again denoted by φ, on Cc (G), and φ(lg ( f )) = φ( f ) for f ∈ Cc (G) and g ∈ G. Since φj (b0 ) = 1 for all j, φ(b0 ) = 1 and φ is nonzero. It now follows from the Riesz representation theorem for locally compact spaces (Theorem 16.7.1) that there is a unique Radon measure μl on G which represents φ, and the left invariance of φ implies that μl is left invariant. A right-invariant Radon measure μr is defined in exactly the same way. It remains to show that μl is essentially unique; we use the Radon–Nikodym theorem for this. Suppose that ν is a left-invariant Radon measure on G. Let π = μl + ν, so that 0 ≤ ν ≤ π. ByProposition 15.4.8, there exists h ∈ L∞ (π ) with 0 ≤ h ≤ 1 for which ν(A) = A h dπ for every Borel set A. Suppose that g ∈ G and that A ∈ B(G). Then    h(x) dπ(x) = ν(A) = ν(g−1 A) = h(x) dπ(x) = h(g−1 x) dπ(x). g−1 A

A

h(g−1 x)

A

Since this holds for all A ∈ B(G), h(x) = for π -almost all x. Suppose that A and B are Borel sets of finite π measure. Then, by Fubini’s theorem,    0= |h(x) − h(g−1 x)| dπ(x) dπ(g) B A    |h(x) − h(g−1 x)| dπ(g) dπ(x), = A

B

242

Borel Measures

 so that B |h(x) − h(g−1 x)| dπ(g) = 0 for π -almost all x. For such x, h(x) = h(g−1 x) for π -almost all g, so that h is a constant in L∞ (π ). If A ∈ B(G) then ν = h(μl + ν), so that 0 < h < 1 and ν = (h/(1 − h))μl . Note also that it follows from the essential uniqueness of μl that φj ( f ) → f as j → ∞ for each f ∈ Cc (G). A continuous action a of a locally compact metrizable group G on a locally compact metrizable space X is a Radon action if it is transitive and if for each x in X there exists a neighbourhood N(x) such that μl (a−1 N(x)) < ∞ and μr (a−1 N(x)) < ∞. Exercise 16.13.4 Show that if a is a Radon action of a locally compact metrizable group G on a locally compact metrizable space X then the pushforward measures a∗ (μl ) and a∗ (μr ) are Radon measures. Exercise 16.13.5 Suppose that a is a Radon action of a locally compact metrizable group G on a locally compact metrizable space, and that H is a closed subgroup of G for which the restriction of a also acts transitively. Let νl and νr be the Haar measures on H. Show that, up to a scaling factor, νl = μl and νr = μr . Exercise 16.13.6 Use the preceding exercise to show that Lebesgue measure on Rd is rotation invariant.

17 Measures on Euclidean Space

Euclidean space is one of the most important examples of a Polish space. In this chapter, we consider measures on Euclidean space, and their interaction with the geometry of the space.

17.1 Borel Measures on R and Rd We begin by considering Borel measures on R. Suppose that μ is a σ -finite measure on R. Let ⎧ for t > 0, ⎨ μ((0, t]) Jμ (t) = 0 for t = 0, ⎩ −μ((t, 0]) for t < 0. t > 0 then The function Jμ is an increasing function on R. If tn μ((0, tn ]) → μ((0, t]) as n → ∞, so that Jμ is continuous on the right at t. Similar arguments apply when t = 0 and when t < 0. The set of points of discontinuity of Jμ is countable, and the points of continuity are dense in R. t is a point of discontinuity of Jμ if and only if t is an atom of μ; that is, if and only if μ({t}) = J(t) − J(t− ) > 0, where J(t− ) = lims$t J(s). We now establish a converse result. Theorem 17.1.1 Suppose that G is an increasing right continuous real-valued function on R and that G(0) = 0. Then there exists a unique σ -finite Borel measure on R such that G = Jμ . Proof Suppose that n ∈ N and that f ∈ C([−n, n]). Suppose that D is a dissection (−n = a0 < a1 < · · · < ak = n) of [−n, n], with mesh size δ(D) = max1≤j≤k (aj − aj−1 ). Let 243

244

Measures on Euclidean Space

sD ( f ) = f (a0 )(G(a0 ) − G((a0 )− ) +

k 

f (aj )(G(aj ) − G(aj−1 )).

j=1

n Exercise 17.1.2 Show that sD ( f ) tends to a limit φn ( f ) = −n f dg, the Riemann–Stieltjes integral of f with respect to G, as the mesh size δ(D) tends to 0 and that φn is a positive linear functional on C([−n, n]), and |φn ( f )| ≤ f ∞ (G(n) − G((−n)− ). By the Riesz representation theorem, there exists a finite Borel measure μn on [−n, n] such that φn ( f ) = [−n.n] f dμn . If B is a Borel subset of [−n, n] and m > n, then μm (B) = μn (B), and so there exists a σ -finite Borel measure μ on R such that μ(B) = μn (B). Exercise 17.1.3 Shows that Jμ = G. The first and most important example is Lebesgue measure λ on R, which we have met earlier, and which is obtained by taking G(t) = t, so that λ((a, b)) = b − a for any open interval (a, b). (In fact, the term ‘Lebesgue measure’ is usually used for the completion of this measure, but this will not concern us.) Lebesgue measure is a Haar measure on R. If μ is a finite measure, it is natural to replace Jμ by the cumulative distribution function Fμ , defined as Fμ (t) = μ((−∞, t]) = Gμ (t) + μ((−∞, 0]). Thus Fμ is a right continuous function on R, and Fμ (t) → 0 as t → −∞ and Fμ (t) → μ(R) as t → ∞. In particular, if μ ∈ P(R) then Fμ is an increasing function and Fμ (t) → 1 as t → ∞. Note that if f is a non-negative measurable function on a probability space (X, , μ) then the push-forward measure f∗ (μ) is a Borel probability measure on R, and the tail distribution function λf satisfies λf (t) = μ( f > t) = 1 − Ff (μ) (t). We can without difficulty extend these ideas to Borel measures on Rd . Let us restrict our attention to Borel probability measures. Suppose that μ ∈ P(Rd ). If  (a1 , . . . , ad ) ∈ Rd , let Jμ (a1 , . . . , ad ) = μ( dj=1 (−∞, aj ]). If C ⊆ {1, . . . , d} let C(j) = aj if j ∈ C and let C(j) = bj otherwise. Then Jμ satisfies the following conditions. (i) (ii) (iii) (iv)

If inf aj → ∞, then Jμ (a1 , . . . , ad ) → 1. If inf aj → −∞, then Jμ (a1 , . . . , ad ) → 0. a˜ for 1 ≤ j ≤ d then Jμ (a1 , . . . , ad ) Jμ (a˜1 , . . . , a˜d ). If aj  j 0 ≤ C⊆{1,...,d} (−1)|C| Jμ (C(1), . . . , C(d)) ≤ 1.

17.2 Functions of Bounded Variation

245

Exercise 17.1.4 Verify this. A function which satisfies (i)–(iv) is called a Stieltjes (probability) function. Conversely, if G is a Stieltjes probability function, the preceding arguments show that there is a unique Borel probability measure on Rd such that  μ( dj=1 (−∞, aj ]) = G(a1 , . . . , ad ). Recall that the product measure λd = λ ⊗ · · · ⊗ λ on Rd is also called Lebesgue measure. Exercise 17.1.5 Let (e1 , . . . , ed ) be the standard basis for Rd . Since the orthogonal group On is generated by rotations which fix e3 , . . . , ed , the reflection which fixes e2 , . . . , ed and isometries which permute e1 , . . . , ed , give another proof that λd is invariant under the orthogonal group; if J ∈ On and B is a Borel set, then λd (J(B)) = λd (B). Now suppose that E is a d-dimensional Euclidean space, and that ( f1 , . . . , fd )   is an orthonormal basis for E. If x = dj=1 xj ej ∈ Rd , let T(x) = dj=1 xj fj , and let μd = T∗ (λd ). Then μd is a σ -finite measure on E, and it follows from the invariance of λd under the orthogonal group that μd does not depend upon the choice of orthogonal basis. μd is also called Lebesgue measure, and is frequently also denoted by λd .

17.2 Functions of Bounded Variation We now extend the results of the preceding section to signed measures on R. This involves functions of bounded variation. Let I be an interval in R. We denote the set of all finite strictly increasing sequences T = (t0 < t1 < · · · < tk ) in I by T (I). Suppose that f is a realvalued function on R. If T = (t0 < t1 < · · · < tk ) ∈ T (I), we set v+ T (f)

k  = ( f (tj ) − f (tj−1 ))+ , v+ ( f , I) = sup v+ T ( f ), j=1

v− T (f) =

T∈T (I)

k  ( f (tj ) − f (tj−1 ))− , v− ( f , I) = sup v− T ( f ), j=1

vT ( f ) =

k  j=1

T∈T (I)

|f (tj ) − f (tj−1 )|, v( f , I) = sup vT ( f ). T∈T (I)

− + − Clearly vT ( f ) = v+ T ( f ) + vT ( f ) and f (tk ) − f (t0 ) = vT ( f ) − vT ( f ). + − Consequently v( f , I) = v ( f , I) + v ( f , I) and if I = [a, b] then f (b) − f (a) = v+ ( f , I) − v− ( f , I).

246

Measures on Euclidean Space

The quantity v+ ( f , I) is the positive variation of f on I, v− ( f , I) is the negative variation of f on I, and v( f , I) is the total variation of f on I. A real-valued function f on I is of bounded variation if v( f , I) is finite. Exercise 17.2.1 Suppose that f is a function of bounded variation on R. Let − + − v+ f (t) = v ( f , (−∞, t]), vf (t) = v ( f , (−∞, t]) and vf (t) = v( f , (−∞, t]). Show that each is a bounded increasing non-negative function on R, vf = − v+ f − vf , and that each tends to 0 as t → −∞. Show that the function f is bounded, and tends to limits f (−∞) and f (∞) − as t → −∞ and as t → ∞. Show further that f (t) = f (−∞) + v+ f (t) − vf (t). Theorem 17.2.2 Suppose that f is a function of bounded variation on R which − is right-continuous at t0 . Then v+ f ,vf and vf are right-continuous at t0 . − 1 1 Proof Since v+ f = 2 (vf + f ) and vf = 2 (vf − f ), it is enough to show that vf is right-continuous. Suppose that  > 0. There exists δ > 0 such that |f (s) − f (t0 )| < /2 for t < s < t + δ. Choose t0 < r < t0 + δ. There exists T = (t0 < t1 < · · · < tk = r) ∈ T ([t, r]) for which vT ( f ) > v( f , [t0 , r]) − /2. Then

vf (t1 ) − vf (t0 ) = v( f , [t0 , r]) − v( f , [t1 , r]) ≤ (vT ( f ) + /2) −

k 

|f (tj ) − f (tj−1 )|

j=2

= |f (t1 ) − f (t0 )| + /2 < . Since vf is an increasing function, this shows that f is right-continuous. We denote by bv0 (R) the vector space of right-continuous functions of bounded variation on R which tend to 0 as t → −∞. Exercise 17.2.3 Establish the following. (i) The function v(., R) is a norm on bv0 (R); we denote v( f , R) by f bv . (ii) If f is an increasing functionin bv  0 (R),  f bv = f (+∞) = f ∞ .  then  +  − (iii) If f ∈ bv0 (R) then f bv = vf  + vf  . bv

bv

Recall that if σ ∈ (M(R, B)) then Fσ (t) = σ ((−∞, t]). Theorem 17.2.4 The mapping F : σ → Fσ is a linear isometry of (M(R, B), .TV ) onto (bv0 (R), .bv ), with inverse mapping μ : f → μf , where μf = μv+ − μv− . Further, if f ∈ bv0 then (μf )+ = μv+ and (μf )− = μv− . f

f

f

f

Proof If σ ∈ M(R), then Fσ = Fσ + −Fσ − so that Fσ ∈ bv0 (R). If Fσ = 0 then Fσ + = Fσ − . It therefore follows from Theorem 17.1.1 that σ + = σ − = 0, so that F is injective.

17.3 Spherical Derivatives

247

− If f ∈ bv0 (R), then f = v+ − μv− ∈ M(R). If σ ∈ f − vf . Then μf = μv+ f f f → μf is the inverse M(R) then σ = μFσ , so that F is bijective; the mapping   of the mapping F. If μ is a positive measure, then Fμ bv = μ(R) = μTV . Thus if σ ∈ M(R), then

Fσ bv = Fσ + − Fσ − bv ≤ Fσ + bv + Fσ − bv       = σ +  + σ −  = σ +  , −TV

TV

TV

so that F is norm decreasing. Similarly, if f ∈ bv0 (R), then              μf  =  + − μ − + − ≤ + μ μ μ v v v v TV f f f f TV TV TV      +  − = vf  + vf  = f bv , bv

so that

bv

F −1

is also norm decreasing. Thus F is an isometry. Further,              −   + μf  = f bv =  + − + = + ,   v v μ μ v v f f TV bv

bv

f

TV

f

TV

so that (μf )+ = μv+ and (μf )− = μv− , by Exercise 15.3.2. f

f

17.3 Spherical Derivatives We now consider finite Borel measures on Rd and functions in L1 (Rd , Bd , λd ). As usual, let Nr (x) denote the open Euclidean ball {y : |y − x| < r} of radius r with centre x. Then λd (Nr (x)) = rd d , where d is the Lebesgue measure of the unit ball in Rd . Exercise 17.3.1 Calculate d . If μ is a finite Borel measure on Rd , we set Ar (μ)(x) =

μ(Nr (x)) μ(Nr (x)) . = λd (Nr (x)) rd d

Proposition 17.3.2 The function Ar (μ) is lower semi-continuous. Proof Suppose that x ∈ Rd , and that  > 0. Let rn increase to r as n → ∞. By upwards continuity, there exists n ∈ N such that μ(Nrn (x)) > μr (x) − . If y − x < r − rn then Nrn (x) ⊆ Nr (y), so that Ar (y) > Ar (x) − . We say that μ has a spherical derivative Ds μ(x) at x if, given  > 0, there exists r0 > 0 such that if 0 < r < r0 and x ∈ Nr (y) then |Ar (μ)(y)−Ds μ(x)| < . It is important that in this definition we consider spheres to which x belongs, and not just spheres centred at x.

248

Measures on Euclidean Space

Similarly, if f ∈ L1 (Rd , B, λd ), we set   1 N (x) f dλd f dλd . = d Ar ( f )(x) = r λ(Nr (x)) r d Nr (x) Ar ( f )(x) is the average value of f over the ball Nr (x). Again, we say that f has a spherical derivative Ds f (x) at x if, given  > 0, there exists r0 > 0 such that if 0 < r < r0 and x ∈ Nr (y) then |Ar ( f )(y) − Ds f (x)| < . Thus the spherical derivative of the function f is the same as the spherical derivative of the measure f .dλd . First we consider a function f in L1 (Rd , B, λd ). We set mu ( f )(x) = sup (sup{Ar (|f |)(y) : y ∈ Nr (x)}) . r>0

An extended real-valued function φ on L1 (Rd , B, λd ) is said to be of weak type (1, 1) with constant k if, whenever α > 0 and Eα = (φ( f ) > α), then  k λd (Eα ) ≤ |f | dλd . α Eα Theorem 17.3.3 If f ∈ L1 (Rd , B, λd ) then mu ( f ) is a lower semi-continuous extended sublinear functional on L1 (Rd , B, λd ), which is of weak type (1, 1) with constant 3d . Proof Since mu is the supremum of lower semi-continuous functions, it is lower semi-continuous, and it is clearly an extended sublinear functional on L1 (Rd , B, λd ). The key result for the second statement is the following covering lemma. Lemma 17.3.4 (Wiener’s lemma) Suppose that G is a finite set of open balls in Rd . Then there is a finite subcollection F of disjoint balls such that   ( (  1 λd (U) = λd U ≥ d λd U . 3 U∈F

U∈F

U∈G

Proof We use a greedy algorithm. If U = Nr (x) is an open ball, let U ∗ = N3r (x) be the ball with the same centre as U, but with three times the radius. Let U1 be a ball of maximal radius in G. Let U2 be a ball of maximal radius in G, disjoint from U1 . Continue, choosing Uj of maximal radius, disjoint from U1 , . . . , Uj−1 , until the process stops, with the choice of Uk . Let F = {U1 , . . . , Uk }. Suppose that U ∈ G. There is a least j such that U∩ Uj = ∅. Then the radius of U is no greater than the radius of Uj (otherwise we would have chosen U to ) ) be Uj ) and so U ⊆ Uj∗ . Thus U∈G U ⊆ U∈F U ∗ and

17.4 The Lebesgue Differentiation Theorem  λd

( U∈G

U

 ≤ λd

( U∈F

U

∗

≤

 U∈F

λd (U ∗ ) = 3d



249

λ(U).

U∈F

We now return to the proof of Theorem 17.3.3. Suppose that f ∈ L1 (Rd ), that α > 0 and that Eα = (mu ( f )(x) > α). Let K be a compact subset of Eα . For each x ∈ K, there exist yx ∈ Rd and rx > 0 such that x ∈ Nrx (yx ) and Arx (|f |)(yx ) > α. It follows from the definition of mu that Nrx (yx ) ⊆ Eα . The sets Nrx (yx ) cover K, and so there is a finite subcover G. By the lemma, there is a subcollection F of disjoint balls such that  (  1 λd (K) λd (U) ≥ d λd U ≥ . 3 3d U∈F U∈G  ) But if U ∈ F, αλd (U) ≤ U |f | dλd , so that since U∈F U ⊆ Eα ,    1  1 λd (U) ≤ |f | dλd ≤ |f | dλd . α α Eα U∈F U∈F U  Thus λd (K) ≤ 3d ( Eα |f | dλ)/α, and  3d |f | dλd . λd (Eα ) = sup{λd (K) : K compact, K ⊆ Eα } ≤ α Eα Exercise 17.3.5 Let m( f ) = max(mu ( f ), |f |). Show that m is a sublinear mapping of weak type (1, 1), with constant 3d + 1. Exercise 17.3.6 Suppose that V is a bounded open subset of Rd . A Vitali covering V is a collection of open balls with the property that if F is a finite subset of V and CF = ∪{B : B ∈ F}, then V \ CF = ∪{B ∈ V : B ∩ CF = ∅}. Use Wiener’s lemma to show that there exists a disjoint sequence (Bn )∞ n=1 in V  λ (B ). such that λd (V) = ∞ d n n=1 Exercise 17.3.7 Suppose that V is a bounded open subset of Rd . Show that ∞ there exists a disjoint sequence (Bn )∞ n=1 λd (Bn ). n=1 in V such that λd (V) = Deduce that λd is rotation invariant.

17.4 The Lebesgue Differentiation Theorem Theorem 17.4.1 (The Lebesgue differentiation theorem) Suppose that g ∈ L1 (Rd , B, λd ). Then g(x) is the spherical derivative of g at x, for λd -almost every x ∈ Rd .

250

Measures on Euclidean Space

Proof The result holds for f ∈ Cc (Rd ), the space of continuous functions on Rd of compact support, and Cc (Rd ) is norm dense in L1 (Rd , B, λd ). We use the first Borel–Cantelli lemma. For each n there exists fn ∈ Cc (Rd ) with g − fn 1 ≤ 1/2n . Let m( f ) = max(mu ( f ), |f |), and let Bn = {x ∈ Rd : m(g − fn )(x) > 1/n}. Bn is open, and, using Corollary 17.3.5, λd (Bn ) ≤ (3d +1)n/2n . Let B = lim sup(Bn ). Then λd (B) = 0, by the first Borel–Cantelli lemma. If x ∈ B, there exists n0 such that x ∈ Bn for n ≥ n0 , so that |Ar (g)(x) − Ar ( fn )(x)| ≤ m(g − fn )(x) ≤ 1/n, for r ≥ 0, and |g(x) − fn (x)| < 1/n. Thus if n ≥ n0 , then |Ar (g)(x) − g(x)| ≤ |Ar (g)(x) − Ar ( fn )(x)| + |Ar ( fn )(x) − fn (x)| + |fn (x) − g(x)| ≤ 2/n + |Ar ( fn )(x) − fn (x)| ≤ 3/n for small enough r, and so Ar (g)(x) → g(x) as r → 0. Corollary 17.4.2 (The Lebesgue density theorem) If E is a measurable subset of Rd then 1

=

λd (Nr (x) ∩ E) → 1 as r → 0 for almost all x ∈ E λd (Nr (x))

1

=

λd (Nr (x) ∩ E) → 0 as r → 0 for almost all x ∈ / E. λd (Nr (x))

λd (Nr (x) ∩ E) rd d and λd (Nr (x) ∩ E) rd d

Proof Apply the theorem to the indicator functions IE∩Nk (0) , for k ∈ N.

17.5 Differentiating Singular Measures Next we consider a finite measure μ for which μ and λd are mutually singular. Theorem 17.5.1 Suppose that μ is a finite Borel measure on Rd for which μ and λd are mutually singular. Then μ has spherical derivative 0 at λd -almost every point of Rd . Proof There exists a Borel λd -null set A for which μ(Rd \ A) = 0. Since μ is tight, there exists an increasing sequence (Kn )∞ n=1 of compact subsets of A with μ(Kn ) > μ(A) − 1/4n for n ∈ N. Let Un be the open set Rd \ Kn ; then μ(Un ) < 1/4n . Let Hn = {(x, r) : x ∈ Un , 0 < r < min(1/2n , d(x, Kn )), Ar (μ)(x) > 1/2n }

17.6 Differentiating Functions in bv0

251

and let Vn = ∪(x,r)∈Hn Nr (x). Suppose that L is a compact subset of Vn . There exists a finite subset G of Hn such that L ⊆ ∪(x,r)∈G Nr (x). By Wiener’s lemma, there exists a subset F of G such that the sets {Nr (x) : (x, r) ∈ F} are disjoint and λd (∪(x,r)∈F Nr (x)) ≥ (1/3d )λd (∪(x,r)∈G Nr (x)). Then λd (L) ≤ λd (∪(x,r)∈G Nr (x)) ≤ 3d λd (∪(x,r)∈F Nr (x))   = 3d λd (Nr (x)) ≤ 3d .2n μ(Nr (x)) (x,r)∈F

(x,r)∈F

= 3 .2 μ(∪(x,r)∈G Nr (x)) ≤ 3 .2n μ(Vn ) ≤ 3d .2−n . d

n

d

Since μ is tight, μ(Vn ) ≤ 3d .2−n . Let B = lim supn→∞ Vn . It follows from the first Borel–Cantelli lemma that λd (B) = 0. Consequently λd (A ∪ B) = 0. If x ∈ A ∪ B then there exists N such that x ∈ Vn , for n ≥ N. If n ≥ N, and 0 < r < 12 min(1/2d , d(x, Kn )) then A2r (μ)(x) < 1/2n . If d(x, y) < r then Nr (y) ⊆ N2r (x), so that Ar (μ)(y) ≤ 2d /2n . Thus μ has spherical derivative 0 at x. Combining Theorems 17.4.1 and 17.5.1, we obtain the following. Theorem 17.5.2 Suppose that μ is a finite Borel measure on Rd . Then μ has d a spherical derivative Ds μ at λd -almost  every point of R . The function Ds μ is λd -integrable. Set ν(A) = μ(A) − A Ds μ dλd . Then ν is a Borel measure on Rd , ν and λd are mutually singular, and μ = Ds μ.dλd + ν is the Lebesgue decomposition of μ. Proof Let μ = f .dλd + ν be the Lebesgue decomposition of μ. Then μ has spherical derivative f .dλd -almost everywhere.

17.6 Differentiating Functions in bv0 Once again, let us return to the one-dimensional case. A real-valued function f on R is absolutely continuous if whenever  > 0 there exists δ > 0 such that if (Ij )kj=1 = ((aj , bj ))kj=1 is a sequence of disjoint intervals of total length k k k j=1 l(Ij ) = j=1 (bj − aj ) less than δ then j=1 |f (bj ) − f (aj )| < . An absolutely continuous function is clearly uniformly continuous. Theorem 17.6.1 A positive signed Borel measure ν on R is absolutely continuous with respect to Lebesgue measure λ if and only if its cumulative distribution function Fν is an absolutely continuous function on R.

252

Measures on Euclidean Space

Proof It is clearly enough to consider the case where ν is a finite positive measure. Suppose first that ν is absolutely continuous with respect to λ. Given  > 0, there exists δ > 0 such that if A ∈ B and λ(A) < δ then ν(A) < . If (Ij )kj=1 = ((aj , bj ))kj=1 is a sequence of disjoint intervals of total length k k k j=1 l(Ij ) = j=1 (bj − aj ) then λ(∪j=1 (aj , bj ]) < δ, so that ν(∪kj=1 (aj , bj ]) =

k 

|Fν (bj ) − Fν (aj )| < ,

j=1

and Fν is an absolutely continuous function. Suppose conversely that Fν is an absolutely continuous function. We use the Radon–Nikodym theorem (Corollary 15.4.5). Suppose that A is a Borel set for which λ(A) = 0, and that  > 0. There exists δ > 0 for which the absolute continuity condition is satisfied. There then exists an open set U containing A, ∞ with λ(U) < δ. Suppose that U = ∪∞ j=1 Ij = ∪j=1 (aj , bj ) is a disjoint union of an infinite sequence of open intervals. Then ν(A) ≤ ν(U) = lim ν(∪kj=1 (aj , bj )) k→∞

≤ lim ν(∪kj=1 (aj , bj ]) = lim k→∞

k→∞

k 

(Fν (bj ) − Fν (aj )) ≤ .

j=1

(The case where U is a finite union is even easier.) Since  is arbitrary, ν(A) = 0, and so ν is absolutely continuous with respect to λ. We now apply the results of the previous section to functions in bv0 . Theorem 17.6.2 Suppose that F ∈ bv0 . Then F is differentiable almost  everywhere. If f is the derivative of F, then f is integrable, and (−∞,t] fdλ ≤ F(t) for almost all t ∈ R. Equality holds for all t ∈ R if and only if R f dλ = limt→+∞ F(t), and if and only if F is an absolutely continuous function on R. Proof It is clearly enough to consider the case where F is increasing. Then F is continuous except on a countable set J, and the discontinuities are all jump discontinuities. F is the cumulative distribution function of a finite Borel measure μ, and μ has a spherical derivative Ds μ except on a null-set N, which clearly includes J. Thus if x ∈ N then lim

h,k

0

F(x + h) − F(x − k) μ([x − k, x + h]) = lim = Ds μ(x). h,k 0 h+k h+k

17.6 Differentiating Functions in bv0

253

Since f is continuous at x, F(x + h) − F(x − k) F(x + h) − F(x) → as k h+k h F(x) − F(x − k) F(x + h) − F(x − k) → as h and h+k k

0, 0.

Thus F is differentiable at x, with derivative f = Ds μ. By Theorem 17.5.2, f is integrable, and μ = f .dλ + ν, where ν and λ are mutually singular. If t ∈ J, then 



F(t) = μ((−∞, t]) =

f dλ + ν((−∞, t]) ≥ (−∞,t]

f dλ. (−∞,t]

Equality holds for all t if and only if ν = 0. This happens if and only if F is absolutely continuous, and if and only if  μ(R) = lim F(t) = t→+∞

f dλ. R

Let us consider the structure of a finite Borel measure μ on R. By the Lebesgue decomposition theorem, μ = f .dλ + ν, where f ∈ L1 (R, B,  λ), and ν and λ are mutually singular. The cumulative distribution function (−∞,t] f dλ of f .dλ is absolutely continuous, so that Fν has the same set J of discontinuities as Fμ . If x ∈ J, let j(x) be the size of the jump at x. If A is a Borel set, let α(A) =  {j(x) : x ∈ A ∩ J}. Then α is an atomic Borel measure; α({x}) = j(x) > 0 if x ∈ J, and α(R \ J) = 0. Further, α and λ are mutually singular. Now let π = ν−α. Then π is a finite measure, π and λ are mutually singular, as are π and α. The cumulative distribution function Fπ has no jumps, and is therefore a continuous function. Since π and λ are mutually singular, Fπ is differentiable almost everywhere, and its derivative is 0 almost everywhere. A Borel measure on R such as π, which has a continuous cumulative distribution function, but for which π and λ are mutually singular, is called a continuous singular measure. Summing up, if μ is a finite Borel measure on R, then μ can be written as the sum of an absolutely continuous measure f .dλ, an atomic measure α, and a continuous singular measure π. It is easy to see that this decomposition is uniquely determined.

254

Measures on Euclidean Space

17.7 Rademacher’s Theorem Suppose that f is an L-Lipschitz function on a bounded closed interval [a, b] in R. Extend f to a function on R by setting f (t) = f (a) for t < a and f (t) = f (b) for t > b. Then f ∈ bv0 and f is absolutely continuous on [a, b]. Thus f is differentiable almost everywhere on [a, b], f  ∈ L1 ([a, b], B, λ) and f (t) =   f (a) + [a,b] f dλ) for t ∈ [a, b]. In fact it follows from the Lipschitz condition that f  ∈ L∞ ([a, b]). Rademacher’s theorem extends this to locally Lipschitz functions on Rd . Theorem 17.7.1 (Rademacher’s theorem) A real-valued locally Lipschitz function f on an open subset U of Rd is differentiable λd -almost everywhere. Proof By localization and extension we can suppose that U = Rd and that f is a Lipschitz function on Rd . We need two lemmas. Suppose that h is a unit vector in Rd . If it exists, the directional derivative is f (x + th) − f (x) . 0 t

∂h+ f = lim t

Lemma 17.7.2 Suppose that h is a unit vector in Rd . Let Ah = {x ∈ Rd : ∂h+ f exists}. Then Ah is a Borel set, and λd (Rd \ Ah ) = 0. Proof If x ∈ Rd , and s, t > 0, let & & & f (x + sh) − f (x) f (x + th) − f (x) & &, k(x; s, t) = && − & s t and let hn (x) = sup{k(x, s, t) : 0 < s, t < 1/n}. Then hn is a lower semicontinuous function on Rd , and 0 ≤ hn (x) ≤ 2L, where L is the Lipschitz constant of f . Thus if  > 0 then the set Bn () = {x ∈ Rd : hn (x) ≤ } ∞ is closed, so that C = ∪∞ n=1 Bn () is an Fσ set, and Ah = ∩m=1 C(1/m) is a Borel set. But it follows from the Lebesgue density theorem that if lx = {x + sh : s ∈ R} is a line in Rd parallel to h then λ(lx \ Ah ) = 0, and then it follows from Fubini’s theorem that λd (Rd \ Ah ) = 0. Consequently, the set G = {x ∈ Rd :

∂f (x) exists for 1 ≤ j ≤ d} ∂xj

is a Borel set, and λd (Rd \ G) = 0.

17.7 Rademacher’s Theorem

255

Lemma 17.7.3 Suppose that x ∈ Ah ∩ G. Then ∂ + fh (x) =

d 

hj

j=1

∂f (x). ∂xj

Proof Suppose that g is a continuously differentiable function of compact 0. Then support on Rd and that tn  1 ( f (x + tn h) − f (x))g(x) dλd (x) = tn Rd   1 ( f (x)(g(x − tn x) − g(x)) dλd (x) → f (x)∂h+ g(x)dλd (x) tn Rd Rd as n → ∞, so that, by bounded convergence,   + ∂h f .g dλd = f . ∂h+ g dλd . Rd

Similarly

 Rd

for 1 ≤ j ≤ d and so  Rd

Rd

∂f .g dλd = ∂xj

∂h+ f .g dλd = =



 Rd d 

f. Rd

f . ∂h+ g dλd  hj

j=1

 = Thus

∂h+ f

dλd =

d

∂f j=1 hj ∂xj

f. Rd

d  Rd

∂g dλd , ∂xj

j=1

hj

∂g dλd ∂xj

∂f . gdλd . ∂xj

dλd .

We now prove Rademacher’s theorem. Let H = (h(n) )∞ n=1 be a dense sequence in the unit sphere Sd−1 of Rd , and let J = G ∩ (∩∞ n=1 Ah(n) ). Then d d−1 λd (R \ J) = 0. If h ∈ S , let ⎞ ⎛ d  ∂f 1⎝ f (x + th) − f (x) − t hj (x)⎠ . k(x, h, t) = t ∂xj j=1

We show that if x ∈ J then k(x, h, t) → 0 uniformly on Sd−1 as t 0, which establishes the theorem. Suppose that  > 0. There exists N ∈ N such that {h(1) , . . . , h(N) } is an /2(d + 1)L-net in H. From Lemma 17.7.3, there exists 0 < δ < 1 such that

256

Measures on Euclidean Space

|k(x, h(n) , t)| < /2 for 0 < t < δ and 1 ≤ n ≤ N. Suppose that h ∈ Sd−1 . There exists 1 ≤ n ≤ N such that h − h(n)  < /2(d + 1)L. Consequently |k(x, h, t) − k(x, h(n) , t)| < /2 for 0 < t < δ, and so |k(x, h, t)| <  for 0 < t < δ. Corollary 17.7.4 Suppose that f is a very regular convex function on Rd . Then f is differentiable λd -almost everywhere on int f . Proof For f is a locally Lipschitz function on int f . This corollary does not need Lebesgue’s density theorem, since a convex function f on R is differentiable at all but countably many points of int f .

18 Convergence of Measures

Suppose that (X, τ ) is a Polish space. In this chapter, we consider topologies and metrics on • • • •

the space M(X) of signed Borel measures on X; the space M + (X) of finite positive Borel measures on X; the space M1 (X) of signed Borel measures μ on X with |μ|(X) ≤ 1; the space P(X) of Borel probability measures on X.

18.1 The Norm .TV Suppose that (X, τ ) is a Polish space. Recall (Proposition 16.6.1)that if f ∈  (X) and μ ∈ M(X) then we define j(μ)(f ) = f dμ to be X f dμ+ − C b X  − X f dμ , and that j is a linear isometry of the Banach space (M(X, B), .TV ), (where μTV = |μ|(X)), into (Cb (X) , .∞ ). M1 (X) is the unit ball of M(X), and P(X) = M1 (X) ∩ {μ : μ(X) = 1} ⊆ M + (X). The norm .TV is too strong for most purposes. If x ∈ X let δx be the Dirac measure, or point measureon X: δx (A) = 1 if x ∈ A, and δx (A) = 0 otherwise, so that if f ∈ Cb (X) then X f dδx = f (x). If x = y then δx − δy TV = 2, so that the norm takes no account of the topology of X. Proposition 18.1.1 Suppose that (X, τ ) is a Polish space. Then (M(X), .TV ) is separable if and only if X is countable. If so, then j is a surjective mapping of (M(X), .TV ) onto (l1 , .1 ) or (l1n , .1 ). Proof If X is uncountable, then δ(X) = {δx : x ∈ X} is not separable in the subspace topology, and so (M(X), .TV ) is not separable. Suppose that X = {xn : n ∈ N} (with xm = xn for m = n) is countably infinite. If μ ∈ M(X), let k(μ)n = μ({xn }). Then k is an isometry of (M(X), .TV ) into 257

258

Convergence of Measures

(l1 , .1 ). It is surjective, since if f ∈ l1 , f = k(μ), where μ(A) = Similarly if X is finite.



{fn : n ∈ A}.

18.2 The Weak Topology w We need a weaker topology than the .TV topology. We identify M(X) with j(M(X)), so that M(X) is a norm-closed linear subspace of Cb (X) ; we then give M(X) the weak* topology w: a base of neighbourhoods of μ if given by the sets & &    & & & & ν ∈ M(X) : & f dν − f dμ& <  for  > 0, f ∈ F(finite) ⊆ Cb (X) . X

X

(By scaling, we can if we like restrict  to the value 1.) This is the topology of pointwise convergence on the points of Cb (X). We write μn ⇒ μ if (and  only if) μn → μ in the w topology. Thus μn ⇒ μ if and only if X f dμn → X f dμ as n → ∞ for each f ∈ Cb (X). The subspace topologies on M1 (X), and more particularly on P(X), are also denoted by w. Confusingly, w is called the weak topology, rather than the weak* topology. Elements of P(X) are also called laws. Convergence in (P(X), w) is also called convergence in law. Similarly, if (fn )∞ n=1 is a sequence of measurable mappings from a probability space (, , P) into a Polish space (X, τ ), then fn converges to f in law if the push-forward measures (fn )∗ (P) converge in the w topology to f∗ (P). 0 Exercise 18.2.1 Suppose that (fn )∞ n=1 is a sequence in L (, , P), where (, , P) is a probability space. Show that if fn → f in probability, then fn → f in law. Show that the converse is not generally true. 0 Exercise 18.2.2 Suppose that (fn )∞ n=1 is a sequence in L (, , P), where (, , P) is a probability space. Show that if (fn )∗ (P) ⇒ δt , then fn → t in probability.

The following proposition suggests that w is a useful topology to consider. Proposition 18.2.3 Suppose that (X, τ ) is a Polish space. If x ∈ X and f ∈ Cb (X) let f , δ(x) = f (x). Then δ(X) ⊆ Cb (X) , and if Cb (X) is given the weak* topology then δ is a homeomorphism of (X, τ ) onto δ(X). Proof Certainly δ(x) ∈ Cb (X) . If x = y then there exists f ∈ Cb (X) with f (x) = f (y), so that δ is injective. If f ∈ Cb (X) then (f ◦ δ)(x) = f (x), so that δ is continuous. Suppose that x ∈ X and that U is an open neighbourhood of x. Since (X, τ ) is completely regular, there exists f ∈ Cb (X) such that f (x) = 0 and f (y) = 1 for y ∈ U. Then V = {φ ∈ Cb (X) : |φ(f )| < 1} is a

18.2 The Weak Topology w

259

weak* neighbourhood of δx , and if φ ∈ δ(X) ∩ V then δ −1 (φ) ∈ U, so that δ −1 : δ(X) → X is also continuous. Proposition 18.2.4 (i) M1 (X) is a w-closed subset of M(X). (ii) The set P(X) is a w-closed subset of M1 (X). (iii) The set δ(X) is a w-closed subset of P(X). Proof (i) If f ∈ Cb (X) then  f ∞ = sup{|δx (f )| : x ∈ X} = sup

 f dμ : μ ∈ M1 (X) ,

X

so that this follows from the theorem  of bipolars. (ii) For P(X) = {μ ∈ M1 (X) : x 1 dμ = 1}. (iii) Suppose that μ ∈ P(X) \ δ(X). Then there exist disjoint compact sets A and B such that μ(A) = α > 0 and μ(B) = β > 0. By Tietze’s extension theorem there exists f ∈ Cb (X) such that f (x) = 0 if x ∈ A, f (x) = 1 if x ∈ B and 0 ≤ f ≤ 1. By Tietze’s theorem again, there exist functions g1 and g2 in Cb (X) such that g1 = 0 on A, g2 = 0 on (f ≤

1 2 ),

g1 = 1 on (f ≥ 12 ),

0 ≤ g1 ≤ 1,

g2 = 1 on B,

0 ≤ g2 ≤ 1.

 Supposethat 0 <  < min(α, β)/2. Then  β ≤ X gi dμ ≤ 1 − α, for i = 1, 2. If | X g1 dμ − g1 (y)| <  then f (y) < 12 , and if | X g2 dμ − g2 (y)| <  then f (y) > 12 , so that 

& &   & & & & ν ∈ P(X) : & gi dν − gi dμ& <  for i = 1, 2 X

X

is a w-neighbourhood of μ disjoint from δ(X). Theorem 18.2.5 The following are equivalent: (a) P(X) is w-compact; (b) M1 (X) is w-compact; (c) X is compact. Proof If M1 (X) is w-compact, then P(X) is w-compact, since P(X) is w-closed in M1 (X). If P(X) is compact, then δ(X) is compact, and so X is compact. If X is compact, then the mapping j : M(X) → C(X) is surjective, by the Riesz representation theorem, and so M1 (X) is w-compact, by Banach’s theorem.

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Convergence of Measures

Exercise 18.2.6 Suppose that (X, τ ) is a compact metrizable space. Show that D(X) = {δx : x ∈ X} is the set of extreme points of the w-compact convex set P(X), and that x → δx is a homeomorphism of X onto D(X), so that D(X) is a closed subset of (P(X), w). Note that if X is a Polish space which is not finite, then (M(X), w) is not first countable, and so it is not metrizable. We can say more. Proposition 18.2.7 If (X, d) is a Polish metric space which is not compact then (M1 (X), w) is not metrizable. Proof Since (X, d) is not totally bounded, there exist  > 0 and a sequence 1 (xn )∞ X such that d(xn , xm ) ≥  for m = n. If α = (αn )∞ n=1 in n=1 ∈ l , let ∞ 1 1 ∞ T(α) = n=1 αn δn . Then T is a linear isomorphism of (B(l ), σ (l , l )) into (M1 (X), w). Since (B(l1 ), σ (l1 , l∞ )) is not metrizable, neither is (M1 (X), w).

18.3 The Portmanteau Theorem Suppose that (X, τ ) is a Polish space. We have defined the w topology on M(X) in terms of bounded continuous functions. We now characterize w-convergence in P(X) in terms of semi-continuous functions, and in terms of open and closed sets. Proposition 18.3.1 Suppose that (X, τ ) is a Polish space. If f is a bounded lower semi-continuous function on X then the real-valued function μ → X f dμ is lower semi-continuous on (P(X), w). Proof Let d be a metric on X defining the topology τ . By Theorem 4.2.9 there exists an increasing sequence (fn )∞ n=1 of Lipschitz functions on X which increases pointwise to f . Suppose that μ ∈ P(X) and that  the theorem   > 0. By f dμ > of bounded convergence, there exists n ∈ N such that X n X f dμ−/2.   If | X fn dν − X fn dμ| < /2 then     f dν > fn dν > fn dμ − /2 > f dμ − . X

X

X

X

Corollary 18.3.2 If A is an open subset of X then the real-valued function μ → μ(A) is lower semi-continuous on (P(X), w), and if B is a closed subset of X then the real-valued function μ → μ(A) is upper semi-continuous on (P(X), w). Proof For IA is lower semi-continuous, and μ(B) = 1 − μ(X \ B).

18.3 The Portmanteau Theorem

261

A Borel subset A of X is a continuity set for μ if μ(∂A) = 0. Corollary 18.3.3 If A is a continuity set for μ0 then the real-valued function μ → μ(A) is continuous at μ0 . Proof For μ(A) = μ(Aint ) = μ(A). Theorem 18.3.4 (The portmanteau theorem) Suppose that (X, τ ) is a Polish space, that (μn )∞ n=1 is a sequence in P(X) and that μ ∈ P(X). The following are equivalent. (i) μn ⇒ μ as n → ∞.  (ii) If f is a bounded lower semi-continuous function on X then X f dμ ≤  lim infn→∞ X f dμn .  (iii) If f is a bounded upper semi-continuous function on X then X f dμ ≥  lim supn→∞ X f dμn . (iv) If A is an open subset of X, then μ(A) ≤ lim infn→∞ μn (A). (v) If A is a closed subset of X, then μ(A) ≥ lim supn→∞ μn (A). (vi) If A is a continuity set A of μ then μn (A) → μ(A) as n → ∞. Proof (i) implies (ii), by Proposition 18.3.1 and (ii) and (iii) are equivalent. (ii) implies (iv), and (iv) and (v) are equivalent. (iv) and (v) together imply (vi). Suppose that (vi) holds, that f ∈ Cb (X) and that  > 0. Let F(t) = μ(f ≤ t). Then F is an increasing function on R, F(t) = 0 for t < − f ∞ and F(t) = 1 for t ≥ f ∞ . Suppose that  > 0. The set J of jump discontinuities of F is countable, and so therefore is the set Q.J = {qj : q ∈ Q, j ∈ J}. It follows that there exists 0 < η <  such that (k + 12 )η ∈ J, for all k ∈ Z. Let Ak = (f ≤ (k + 12 )η): Ak is a set of continuity for μ, and so is Bk = Ak \ Ak−1 . Thus if we set   



g= k + 12 ηIBk and g = k − 12 ηIBk , k∈Z

k∈Z

 (both of theses sums have only finitely many non-zero terms), then X g dμn →    g dμ and h dμ → h dμ as n → ∞. Since f −η ≤ h ≤ f ≤ g ≤ f +η, n X X X it follows that      f dμ − η ≤ h dμ ≤ f dμ lim sup n→∞ X X X     g dμ ≤ lim inf f dμ + η. ≤ X

n→∞ X

Since 0 < η <  and  is arbitrary, the result follows. Let us see how the results that we have established apply to Borel measures on the real line R.

262

Convergence of Measures

Theorem 18.3.5 (The Helly–Bray theorem) Suppose that (μn )∞ n=1 is a sequence in P(R) and that μ ∈ P(R). Then μn ⇒ μ if and only if Fmn (t) → Fμ (t) as n → ∞ for each point of continuity t of Fμ . Proof Since (−∞, t] is a continuity set for μ if and only if t is a point of continuity of Fμ , the condition is necessary, by the portmanteau theorem. Suppose that it is satisfied, that U is an open subset of R and that  > 0. Since μ is tight, there exists a finite set {[ai , bi ] : 1 ≤ i ≤ j} of disjoint closed j intervals in U such that i=1 μ([ai , bi ]) > μ(U) − . For each i there exist points ci , di of continuity of Fμ such that ci < ai , bi ≤ di and (ci , di ] ⊆ U. Let C = ∪ni=1 (ci , di ]. Then μn (C) → μ(C) as n → ∞. Thus lim inf μn (U) ≥ lim μn (C) = μ(C) > μ(U) − . n→∞

n→∞

Since this holds for all  > 0, lim infn→∞ μn (U) ≥ μ(U). Thus μn ⇒ μ, by the portmanteau theorem. Exercise 18.3.6 Suppose that (μn )∞ n=1 is a sequence in P(R) and that μ ∈ P(R) has a continuous distibution function F. Show that if μn ⇒ μ then Fn converges uniformly to F. Exercise 18.3.7 Let P be a probability measure on R which has an atom at each rational number. Let Pn (A) = P(A − 1/n), for each Borel set A. Show that Pn ⇒ P, but that Pn (q) → P(q), for each rational q. Exercise 18.3.8 Suppose that μ ∈ P(Rd ). If t ∈ Rd , let L(t) = {x ∈ Rd : xi ≤ ti for 1 ≤ i ≤ d} and let F(t) = μ(L(t)). By considering the atoms of μ, show that the points of discontinuity of L lie in a countable union of hyperplanes, and have Lebesgue measure 0. Prove a version of the Helly–Bray theorem for probabilities in Rd . Exercise 18.3.9 Suppose that for each n fn is a random variable on a probability space (, , Pn ), each taking values in a Polish space (X, τ ) and that f is a random variable on a probability space (, , P) taking values in X. Formulate a version of the portmanteau theorem for the convergence in law of (fn )∞ n=1 . Do the same for the Helly–Bray theorem. Recall (Theorem 3.1.1) that if (X, τ ) is a Polish space then there is a homeomorphism k of X onto a Gδ subset of the Hilbert cube H. Let X˜ = j(X), ˜ τ˜ ) is a compact metrizable space. If μ ∈ with the subspace topology τ˜ ; (X, ˜ thus μ(A) ˜ = μ(A ∩ X) P(X), we denote the push-forward measure k∗ (μ) by μ; ˜ We denote the weak topology on P(X) ˜ by w˜ and write for A a Borel set in X. ˜ if νn → ν in the topology w. ˜ νn ⇒ν

18.3 The Portmanteau Theorem

263

Theorem 18.3.10 Suppose that (X, τ ) is a Polish space, and that k is a ˜ homeomorphism of X onto a dense subset of a compact metrizable space X. The mapping k∗ : μ → μ˜ is a homeomorphism of (P(X), w) onto a dense ˜ w). subset of the compact metrizable space (P(X), ˜ Proof The mapping is certainly continuous, since the restriction of a function ˜ to X is in Cb (X). Suppose that μ˜ n ⇒ in C(X) ˜ μ. ˜ If A is a closed subset of X, ˜ Thus then A = A ∩ X, where A is the closure of A in X. μ(A) = μ(A) ˜ ≥ lim sup μ˜ n (A) = lim sup μn (A), n→∞

n→∞

and so μn → μ; the inverse mapping is also continuous. ˜ If ˜ set P(X). The set k(P(X)) is a convex subset of the convex w-compact ˜ it were a proper subset of P(X), it would follow from the Krein–Mil’man ˜ not in k(P(X)). But theorem that there would be an extreme point ν of P(X) ˜ and there exists a sequence (xn )∞ in X which then ν = δy , for some y ∈ X, n=1 ˜ y , giving a contradiction. converges to y. But then k∗ (δxn )⇒δ Corollary 18.3.11 (P(X), w) is a separable metrizable space, and there is a countable subset F of Cb (X) such that w is the topology of pointwise convegence on F. If C is a countable dense subset of X, then the countable set  n  n   AC = λi δci : n ∈ N, λi ∈ Q, λi ≥ 0, λi = 1, ci ∈ C i=1

i=1

is w-dense in P(X). ˜ w) Proof The first statement follows from the fact that (P(X), ˜ has these ˜ and so the second statement follows properties. The set k(C) is dense in X, from Exercise 16.6.8. This enables us to prove the empirical law of large numbers. Suppose that (gn )∞ n=1 is a sequence of identically distributed real-valued random variables on a probability space (, , P), each with distribution μ. For each n ∈ N  and ω ∈ , let μn (ω) = (1/n) ni=1 δgi (ω) . Then μn is a random measure, the nth empirical measure; it represents repeated sampling from a population which has distribution μ. Thus if A ∈  then μn (ω)(A) is the proportion of 1 ≤ i ≤ n} which lie in A, and if f ∈ L0 () then the elements {gi (ω) :  n i=1 f (gi (ω)). We would hope that almost surely the  f dμn (ω) = (1/n) empirical measures would give an approximation to the distribution μ; the empirical law of large numbers says that this is so.

264

Convergence of Measures

Theorem 18.3.12 (The empirical law of large numbers) Suppose that  is a Polish space, that P is a Borel probability measure on  and that (gn )∞ n=1 is a sequence of identically distributed random variables on , each with distribution μ. Let μn be the nth empirical distribution. Then μn (ω) ⇒ μ for almost all ω ∈ . Proof Let F be a countable set in Cb () which satisfies the conclusions of  Corollary 18.3.11. By the strong law of large numbers,  f dμn (ω) →  all ω, for each f ∈ F, and so, since F is countable,  f dμ for almost  f dμn (ω) →  f dμ for almost all ω, for all f ∈ F. Thus μn (ω) ⇒ μ almost surely. This theorem is usually applied in the following situation. Suppose that (X, τ ) is a Polish space, and that μ is an unknown Borel probability distribution μ on X. We sample repeatedly and independently from X, and hope that the resulting empirical distribution approximates to P. Thus, for each n ∈ N,  (n , Pn ) is a copy of (X, μ), (, P) is the countable product ( ∞ n=1 n , P ) and g is the outcome of the nth trial. ⊗∞ n n n=1

18.4 Uniform Tightness Throughout this section, we suppose that (X, τ ) is a Polish space. Recall that if μ ∈ P(X) then μ is tight; given  > 0 there exists a compact subset K of X such that μ(K) > 1 − . We extend this notion to subsets of P(X). A subset C of P(X) is uniformly tight if whenever  > 0 there exists a compact subset K of X such that μ(K) > 1 −  for all μ ∈ C. The proof of the next result is reminiscent of the proof of Ulam’s theorem. Proposition 18.4.1 Suppose that (X, d) is a complete Polish metric space and that C ⊆ P(X). Then C is uniformly tight if and only if for each  > 0 there exists a finite subset F of X such that μ(∪x∈F N (x)) > 1 − , for all μ ∈ C. Proof If C is uniformly tight, and  > 0 there exists a compact subset K of X such that μ(K) > 1 −  for all μ ∈ C. There exists a finite subset F of X such that K ⊆ ∪x∈F N (x). F clearly satisfies the requirements of the proposition. Conversely, suppose that the condition is satisfied, and that  > 0. For each j ∈ N there exists a finite subset Fj of X such that if Aj = ∪x∈Fj M/2j (x) then μ(Aj ) > 1 − /2j , for all μ ∈ C. Let B = ∩∞ j=1 Aj . Then B is closed and totally bounded, and is therefore compact. Further, μ(B) > 1 − .

18.4 Uniform Tightness

265

Here is a useful test for the distributions of a set of random variables to be uniformly tight. Proposition 18.4.2 Suppose that {(α , α , μα )}α∈A is a family of probability spaces and that for each α ∈ A fα is a random variable on α taking values in Rd . Then {L(fα )}α∈A is uniformly tight if and only if there is an increasing non-negative function φ on [0, ∞) for which φ(t) → ∞ as t → ∞ such that  { α φ(fα ) dμα }α∈A is bounded. Proof Suppose that the condition is satisfied, that  sup φ(fα ) dμα = M α∈A α

and that  > 0. There exists T such that φ(t) ≥ M/ for t ≥ T. If α ∈ A then  1 φ(fα ) dμα L(fα )(t > T) ≤ φ(T) fα >T   ≤ φ(fα ) dμα ≤ , M α so that {L(fα )}α∈A is uniformly tight. Conversely, suppose that {L(fα )}α∈A is uniformly tight. There therefore exists a strictly increasing sequence (tn )∞ n=1 in [0, ∞) such that L(fα ){x : x > tn } > 1 − 1/4n for each α ∈ A. Let φ be a non-negative increasing function on [0, ∞) for which φ(tn ) = 2n . If α ∈ A then  φ(fα ) dμα α

=

 fα ≤t1

≤ 2+

∞  n=1

φ(fα ) dμα +

∞   n=1

2n+1 4n

tn−1 0. Exercise 18.4.3 Suppose that (G, τ ) is a locally compact Polish topological group, and that ν is a group-norm on G which defines the topology, for which the sets Kn = {g : ν(g) ≤ n} are a fundamental sequence of compact sets. Formulate and prove a version of Proposition 18.4.2 in this setting.

266

Convergence of Measures

18.5 The β Metric Suppose that (X, τ ) is a Polish space. So far, we have made little use of a complete metric d on X which defines the topology τ . We now consider a complete Polish metric space (X, d); a complete separable metric space. Recall that BL(X) then denotes the space of bounded Lipschitz functions, with norm f BL = f ∞ + pL (f ). Proposition 18.5.1 Suppose that (X, d) is a Polish metric space. If μ and   ν are distinct elements of P(X), there exists f ∈ BL(X) such that X f dμ = X f dν. Proof Let π = μ − ν and let X = P ∪ N be a Lebesgue decomposition of X for π. There exist compact subsets K of P and L of N such that π(K) > π(P)− πTV /5 and π(L) < π(N) + π TV /5. Consequently, |π |(X \ (K ∪ L)) < 2 πTV /5 and |π|(K ∪ L) ≥ 3 πTV /5.  There exists f ∈ BL(X) such that f|K = 1, f|L = −1 and f ∞ = 1. Then X f dπ > π TV /5 > 0, so that f separates μ and ν.  Corollary 18.5.2 Let μβ = sup{| X f dμ| : f BL ≤ 1}. Then .β is a norm on M(X), and μβ ≤ μTV . We denote the metric that .β defines on P(X) by β. Suppose that (X, d) is a metric space. If A is a non-empty Borel set in X, let A = {x ∈ X, d(x, A) < }. Proposition 18.5.3 Suppose that (X, d) is a Polish metric space, and that 0 <  < 1. If β(μ, ν) ≤  2 /2 and A is a nonempty Borel set then ν(A ) ≥ μ(A)−. Proof Let f = (1 − d(x, A)/)+ . Then f BL ≤ 1 + 1/, so that & &  & & & f dμ − f dν & ≤  2 (1 + 1/)/2 < . & & X

Thus

X





ν(A ) ≥

f dν ≥ X

f dμ −  ≥ μ(A) − . X

Theorem 18.5.4 Suppose that (X, d) is a complete Polish metric space. If C is a non-empty β-totally bounded subset of P(X), then it is uniformly tight. Proof Suppose that  > 0. There exists a finite set D in C such that C ⊆ D 2 /8 . There exists a compact subset KD of X such that μ(KD ) > 1 − /2 for μ ∈ D. There exists a finite subset F of X such that KD ⊆ F/2 , so that (KD )/2 ⊆ F . If ν ∈ C, there exists μ ∈ D with β(ν, μ) ≤  2 /8, and so

18.5 The β Metric

267

ν(F ) ≥ ν((KD )/2 ) ≥ μ(KD ) − /2 ≥ 1 − . The result therefore follows from Proposition 18.4.1. Let wBL be the topology on M(X) of pointwise convergence on BL(X). Theorem 18.5.5 Suppose that (X, d) is a complete Polish metric space. The β-metric topology, the topology w and the topology wBL are the same on P(X). Proof First we show that the identity mapping (P(X), wBL ) → (P(X), β) is continuous. Suppose μ ∈ P(X), and that 0 <  < 1. By tightness, there exists a non-empty compact subset K of X such that μ(K) > 1 − /11. By the Arzel`a– Ascoli theorem, A = {f|K ; f BL ≤ 1} is totally bounded in C(K), and so there exist g1 , . . . , gk ∈ A such that A ⊆ ∪ki=1 B/11 (gi ). By the McShane– Whitney theorem, each gi can be extended without increasing the BL norm to fi ∈ BL(X). Also, let h(x) = (1−11d(x, K)/)+ . Let C = {h}∪{fi : 1 ≤ i ≤ k}. Let & &    & & & N = ν ∈ P(X) : & f dν − f dμ&& < /11 for f ∈ C ; X

X

N is a wBL -neighbourhood of μ in P(X). Suppose that ν ∈ N. Then   ν(K/11 ) ≥ h dν ≥ h dμ − /11 ≥ 1 − 2/11. X

X

If f BL ≤ 1 there exists fi such that |fi (x) − f (x)| ≤ /11 for x ∈ K. Using the Lipschitz condition, |fi (y) − f (y)| ≤ 3/11 for y ∈ K/11 . Now    |f − fi | dν ≤ |f − fi | dν + (|f | + |fi |) dν ≤ 3/11 + 4/11 X

and

K/11



X\K/11

 |f − fi | dμ =

X

 |f − fi | dμ +

K

(|f | + |fi |) dμ ≤ /11 + 2/11. X\K

Thus & & &  &    & & & & & f dν − f dμ& ≤ & fi dν − fi dμ& + | f − fi | dν + | f − fi | dμ & & & & X

X

X

X

X

X

≤ /11 + 7/11 + 3/11 = , and so β(ν, μ) ≤  for n ≥ N. Since β is a metric, to show that the identity (X, β) → (X, w) is continuous it is sufficient to show that it is sequentially continuous. We   must show that if β(μn , μ) → 0 as n → ∞ and f ∈ C(K) then X f dμn → X f dμ as n → ∞.

268

Convergence of Measures

The sequence {μn : n ∈ N} is β-totally bounded, and so is uniformly tight. Thus given  > 0 there exists a compact subset K of X such that μ(K) ≥ 1 − /7 and μn (K) ≥ 1−/7 for all n. Suppose that f ∈ Cb (X) and that f ∞ ≤ 1. By the corollary to the Stone–Weierstrass theorem (Corollary 16.8.3), there exists g ∈ BL(K) such that gC(K) ≤ 1 and f − gC(K) ≤ /7. We can extend g to h ∈ BL(X), with hC(X) ≤ 1. Then & &  & & & f dμn − f dμ& & & X X &  &   & & ≤ && h dμn − h dμ&& + |f − h| dμn + |f − h| dμ. X

Now 

X

X





| f − h| dμn ≤ X

X



| f − h| dμn + K

| f | dμn + C(K)

|h| dμn ≤ 3/7, C(K)

 and similarly X |f − h| dμ ≤ 3/7, and so & & & &   & & & & & f dμn − f dμ& ≤ & h dμn − h dμ& + 6/7 <  & & & & X

X

X

X

for large enough n. Finally, the identity mapping (P(X), w) → (P(X), wBL ) is trivially continuous. Corollary 18.5.6 If (X, d) is a compact metric space, then (P(X), β) is compact. Proof For (P(X), w) is compact, by Banach’s theorem. Theorem 18.5.7 If (X, d) is a compact metric space, then (M1 (X), β) is compact, and therefore complete, and the β-metric topology, the topology w and the topology wBL are the same on M1 (X). Proof Since (M1 (X), β) is the continuous image of (P(X), β)2 × [0, 1] under the mapping (μ, ν, α) to (1−α)μ−αν, (M1 (X), β) is compact. The β-topology and wBL therefore agree on M1 (X). But (M1 (X), w) is compact, by Banach’s theorem, and so w and wBL agree on M1 (X). Exercise 18.5.8 Show that if X = N, then M1 (X) = l1 , β(x) = x1 and w = σ (l1 , l∞ ). Consequently, the β topology and w may not agree on M1 (X). Theorem 18.5.9 If (X, d) is a complete Polish metric space then (P(X), β) is complete. Proof Suppose that (μn )∞ n=1 is a β-Cauchy sequence in P(X). Then {μn : n ∈ N} is totally bounded. There therefore exists an increasing sequence

18.6 The Prokhorov Metric

269

(Kj )∞ j=1 of compact subsets of X such that μn (Kj ) > 1 − 1/j, for j, n ∈ N. Fix j. Let βj be the β-norm on P(Xj ). Since every f ∈ BL(Kj ) can be extended to a bounded Lipschitz function g on X with gBL = f BL , it follows that the push-forward mapping from (M(Kj ), βj ) to (M(X), β) is an isometry. If A is a (j) (j) (j) Borel subset of Xj , let νn (A) = μn (A). Then βj (νm − νn ) ≤ β(μm − μn ) (j) (j) and so (νn )∞ n=1 is a βj -Cauchy sequence. By Theorem 18.5.7, νm therefore (j) (j) converges, to ν , say. Let μ be the push-forward measure on X. Note that if A is a Borel subset of X and j < k then 0 ≤ μ(j) ≤ μ(k) ≤ 1. Let μ(A) = limj→∞ μ(j) (A). Then μ is finitely additive, and if (Ar )∞ r=1 is an increasing sequence of Borel subsets of X, then μ(A) = sup μ(j) (A) = sup sup μ(j) (Ar ) = sup μ(Ar ), j∈N

j∈N r∈N

r∈N

so that μ ∈ M1 (X). Since μ(X) = sup ν (j) (Kj ) ≥ sup(1 − 1/j) = 1, j∈N

j∈N

μ ∈ P(X). Finally, if f ∈ Cb (X) and f ∞ ≤ 1 then & & & &&   & & & & & & f dμ − f dμn & ≤ & f dμ − f dμ & + 2/j n & & & & X X Kj Kj & &  & & & (j) (j) & f dνn & + 2/j ≤ 3/j = & f dν − & & Kj Kj for large enough n, so that μn → μ as n → ∞ in the w topology. Thus β(μn , μ) → 0 as n → ∞. Corollary 18.5.10 If (X, τ ) is a Polish space, then (P(X), w) is a Polish space. Proposition 18.5.11 If A ⊆ M1 (X) is uniformly tight, then A is β-totally bounded. Proof Suppose that  > 0. There exists a compact set K such that |μ|(X \K) < /2 for μ ∈ A, so that β(μ − IK .μ) ≤ /2 for μ ∈ A. But the set AK = {IK .μ : μ ∈ A} is totally bounded, and so there exists a finite set F such that AK ⊆ ∪μ∈F N/2 (IK .μ). Thus A ⊆ ∪μ∈F N (μ).

18.6 The Prokhorov Metric Suppose that (X d) is a complete Polish metric space. We now introduce another complete metric ρ on P(X) which is equivalent to the metric β.

270

Convergence of Measures

If μ, ν ∈ P(X), we set ρ(μ, ν) = inf{ > 0 : μ(A) < ν(A ) +  for all non-empty Borel sets A}. Since μ(A) = sup{μ(B) : B closed , B ⊆ A and N (A) = N (A), ρ(μ, ν) = inf{ > 0 : μ(B) < ν(B ) +  for all non-empty closed sets B}. Theorem 18.6.1 Suppose that (X d) is a complete Polish metric space. The function ρ is a metric on P(X), equivalent to β. Proof First we show that ρ is a metric. Suppose that ρ(μ, ν) > . Thus there exists A such that μ(A) > ν(A ) + . If y ∈ (X \ A ) , there exists z ∈ X \ A with d(y, z) < . Since z ∈ A , y ∈ A. Thus (X \ A ) ⊆ X \ A. Consequently ν(X \ A ) = 1 − ν(A ) > 1 − μ(A) +  ≥ μ((X \ A ) ) + , and so ρ(ν, μ) > . It follows from this that ρ(ν, μ) = ρ(μ, ν). Suppose that ρ(μ, ν) = 0. If A is closed, μ(A) ≤ ν(A1/n ) + 1/n, for all n ∈ N. But ν(A1/n ) + 1/n → ν(A) as n → ∞, and so μ(A) ≤ ν(A). Similarly ν(A) ≤ μ(A) and so μ(A) = ν(A). Since this holds for all closed A, it follows from regularity that μ = ν. Suppose that ρ(μ, ν) <  and ρ(ν, π ) < η. If A is a non-empty Borel set then (A )η ⊆ A+η , and so μ(A) < ν(A ) +  < π((A )η ) + η +  ≤ π(Aη+ ) + η + . Thus ρ(μ, π ) < η + , and so ρ(μ, π ) ≤ ρ(μ, ν) + ρ(ν, π ). Consequently, ρ is a metric on P(X). It follows from Proposition 18.5.3 that if β(μ, ν) <  2 /2 then ρ(μ, ν) < , and so the identity mapping id : (P(X), β) → (P(X), ρ) is uniformly continuous. On the other hand, suppose that ρ(μn , μ) → 0 as n → ∞ and that  > 0. Suppose that A is a continuity set for μ. There exists 0 < δ < /2 so that μ(Aδ ) < μ(A) + /2 and μ((X \ A)δ ) < μ(X \ A) + /2. There exists n0 such that ρ(μn , μ) < δ for n ≥ n0 . Thus if n ≥ n0 then μn (A) ≤ μ(Aδ ) + δ < μ(A) + δ + /2 < μ(A) + , and μn (X \ A) ≤ μ((X \ A)δ ) + δ < μ(X \ A) + δ + /2 < μ(X \ A) + . Consequently μn ⇒ μ, by the portmanteau theorem (Theorem 18.3.4). The metric ρ is called the Prokhorov metric.

18.7 The Fourier Transform and the Central Limit Theorem

271

Theorem 18.6.2 (Prokhorov’s theorem) Suppose that (X, d) is a Polish metric space, and that C ⊆ P(X). The following are equivalent. (i) (ii) (iii) (iv)

C is β-totally bounded. C is ρ-totally bounded. C is uniformly tight. C is w-compact.

Proof We have seen that (i), (iii) and (iv) are equivalent; (i) implies (ii) since the identity mapping (P(X), β) → (P(X), ρ) is uniformly continuous. We show that (ii) implies (iii). We use Theorem 18.4.1. Suppose that C is ρ-totally bounded, and that  > 0. There exists a finite subset D of C such that C ⊆ ∪μ∈D N/2 (μ). There exists a compact set K in X, such that μ(X \ K) < /2 for μ ∈ D, and there exists a finite subset F of K such that K ⊆ A = ∪x∈F N/2 (x). Let B = ∪x∈F N (x). Suppose now that ν ∈ C. There exists μ ∈ F such that ρ(ν, μ) < /2. Now (X \ B)/2 ∩ A = ∅, so that μ(X \ B) < /2. Consequently, ν(X \ B) ≤ μ(N/2 (X \ B)) + /2 < . Thus C is uniformly tight, by Theorem 18.4.1. Corollary 18.6.3 (P(X), ρ) is complete. Proof If (μn )∞ n=1 is a ρ-Cauchy sequence in P(X), then it is ρ-totally bounded, and so it is β-totally bounded, by Prokhorov’s theorem. Thus it has a β-convergent subsequence. Since the β and ρ topologies are the same, it has a ρ-convergent subsequence, so that the original sequence is ρ-convergent.

18.7 The Fourier Transform and the Central Limit Theorem We now illustrate the use of Prokhorov’s theorem (Theorem 18.6.2) by proving the central limit theorem. Suppose that f is a real-valued random variable on a probability space (, , μ). We define n independent copies f1 , . . . , fn of f on (n ,  n , ⊗n μ) by setting fj (ω) = f (ωj ), where ω = (ω1 , . . . , ωn ).  18.7.1 Suppose that f ∈ L2 (, , μ), that  f dμ = 0, that Proposition 2 that f1, . . . , fn are n independent  f dμ = 1 and  copies of f . Let cn = √ ( f1 + · · · + fn )/ n. Then n cn d(⊗n μ) = 0 and n c2n d(⊗n μ) = 1.

272

Convergence of Measures

Proof Certainly



n cn d(⊗

 n



1 n



1 n



j=1 n

=

j=1 n

=

j=1

= 0, and

   1 fj2 d(⊗n μ) + fi fj d(⊗n μ) n n n

1 n n

c2n d(⊗n μ) =

n μ)

i =j

    1 fj2 dμ + fi dμ. fj dμ n    i =j

n

  f 2 dμ =

f 2 dμ.

n

It follows from Proposition 18.4.2 that the sequence (L(cn ))∞ n=1 is uniformly tight, and so it has a limit point. The central limit theorem says that there is only one limit point γ , which is the law of a Gaussian random variable γ with mean 0 and variance 1. Consequently cn converges in distribution to γ . [A Gaussian random variable with mean 0 and variance 1 is a random √ γ −t 2 variable with distribution (1/ 2π )e /2 .dλ(t).] Theorem 18.7.2 (The central theorem) Suppose that f ∈ L2 (, , μ),  limit  √ 2 that  f dμ = 0 and that  f dμ = 1. Let cn = ( f1 + · · · + fn )/ n, where f1 , . . . , fn are n independent copies of f . Then cn converges in distribution to γ , where γ is a Gaussian random variable with mean 0 and variance 1. We shall prove a stronger version of this result later (Theorem 21.6.3). Proof There are several proofs of this fundamental theorem. The one that we give here uses Fourier transforms. The theory of Fourier transforms is enormous, but we shall only establish the results that we need. First, we introduce the Schwartz space S and the space S  of tempered distributions. A function g on R belongs to S if and only if it can be differentiated infinitely often, and the functions |x|j |dn g/dxn (x)| belong to C0 (R), for each j and n in Z+ . We give S the metrizable topology given by the seminorms pj,n , where & & n & g & pj,n (g) = sup |x| & n (x)&& . dx 

j &d

x∈R

 Then S is homeomorphic to a subspace of j,n C0 (R)j,n ; elementary arguments show that this is a closed subspace, and so S is a Polish space.

18.7 The Fourier Transform and the Central Limit Theorem

273

The space S  is the space of continuous linear functionals on S; elements of S  are called tempered distributions. We consider the dual pair (S, S  ). We can consider of S  ; if g ∈ S and ν ∈ M(R),  M(R) as a linear subspace 2 then g, ν = R g dν. Similarly, if k ∈ L (R, B, λ), we set g, k = R gk dλ. We now define the Fourier transform of a function g ∈ S; we set  e−2π ixt g(x) dλ(x), for t ∈ R. F(g)(t) = R

The Fourier transform is concerned with both translation and scaling. If j is a function on R, we set τh j(x) = j(x − h) for h ∈ R and σα j(x) = j(x/α) for α = 0. Theorem 18.7.3 Suppose that g, j ∈ S, h ∈ R and α = 0. (i) (ii) (iii) (iv)

= e−2π iht F(g)(t) and F(σα (g))(t) = αF(g)(αt). F(τ

h (g))(t)  dg F dx (t) = 2πitF(g)(t) and F(−2πixg(x))(t) = dFdt(g) (t). The  maps S continuously into S.  Fourier transform F F(g)(t)j(t) dλ(t) = R R g(x)F (j)(x) dλ(x).

Proof (i) and (ii) follow from elementary calculations, (iii) follows from them, and (iv) follows by changing the order of integration. Let us give two examples. Let f (x) = e−π x . Then f satisfies the differential equation df (x) + 2πxf (x) = 0, f (0) = 1. dx 2

It follows from Theorem 18.7.3 that F(f √ ) satisfies the same equation, and so 2 F(f ) = f . Similarly, if γ (x) = e−x /2 / 2π then F(γ )(t) = γ (2π t). Theorem 18.7.4 Suppose that g ∈ S. Then F 2 (g)(x) = g(−x), so that F is a homeomorphism of S onto itself, which has period 4. Proof Suppose that j ∈ S and that α > 0. Changing variables, and using Theorem 18.7.3,   g(α −1 x)F (j)(x) dλ(x) = α g(x)F (j)(x) dλ(x) R

R



(g)(x)j(α −1 x) dλ(x).

= R

Let α → 0. Then we obtain the general formula   g(0) F(j)(x) dλ(x) = j(0) F(g)(x) dλ(x). R

R

274

Convergence of Measures

 2 Now let j(x) = e−πx ; then g(0) = R F(g)(x) dλ(x), and so  g(−x) = (τ−x g)(0) = F(τ−x g)(t)dλ(t) R



F(g)(t)e−2π ixt dλ(t) = F 2 (g)(x).

= R

We also call the transposed map F  from S to itself the Fourier transform, and denote it by F; this is appropriate terminology, since if g ∈ S, then its Fourier transform as a tempered distribution is the same as its Fourier transfom as an element of S. Corollary 18.7.5 F is a bijective linear map of S  onto itself.  Note also that if ν ∈ M(R) then F(ν)(t) = R e−2πixt dν(x). Exercise 18.7.6 Use the theorem of bounded convergence, and the fact that |e2π ix | = 1, to show that if ν ∈ M(R) then F(ν) ∈ Cb (R). Show that the mapping ν → F(ν) is linear and injective, and that F(ν)∞ ≤ νTV . Suppose now that (, , μ) is a measure space and that f ∈ L0 (). We consider the distribution L(f ), and define the characteristic function of f to be fˆ = F(L(f )), the Fourier transform of its law. Thus  ˆf (t) = e−2π itf (ω) dμ(ω). 

Theorem 18.7.7 If f ∈ L2 (|, .μ) then fˆ is twice continuously differentiable, and  dfˆ f (ω)e−2π if (ω)t dμ, (t) = −2π i dt   2 ˆ d f (t) = −4π 2 f 2 (ω)e−2π if (ω)t dμ(ω). dt2  Proof Let a(y) =

e−iy − 1 + i for y ∈ (R\{0}). y

Then a ∈ Cb (R\{0}), a∞ ≤ 2 and a(y) → 0 as y → 0. Thus & & & & e−2π if (ω)s − 1 & & + 2πif (ω)& ≤ 4π |f (ω)|. & & & s

18.7 The Fourier Transform and the Central Limit Theorem

275

Multiplying by e−2πif (ω)t and integrating over , we find that & & & & fˆ(t + s) − fˆ(t)  & & −2π if (ω)t 2πif (ω)e dμ(ω)& ≤ 4π f 1 , + & & & s  and so, by the theorem of dominated convergence,  dfˆ f (ω)e−2π if (ω)t dμ. (t) = −2πi dt  Thus    −2πif (ω)s − 1 1 dfˆ dfˆ e (t + s) − (t) = (−2πif (ω))e−2π if (ω) dμ, s dt dt s  and the integrand is dominated pointwise by 4m2 π 2 f 2 (ω). Further applications of the theorem of dominated convergence gives the second equation, and show that the right-hand side of it is a continuous function of t. Corollary 18.7.8 dfˆ (0) = −2π i dt



d2 fˆ f dμ and 2 (0) = −4π 2 dt 

 f 2 dμ. 

This theorem can clearly be extended to functions in Ln (ω, , μ), where n ∈ N. We now prove Theorem 18.7.2. Applying Taylor’s theorem, fˆ(t) = 1 − 2π 2 2 t /2 + r(t), where r(t)/t2 → 0 as |t| → 0. Now  √ cˆn (t) = e−2π it(f1 +···+fn )/ n d(⊗n μ) n

=

n  * j=1

√ −2π itfj / n

e

 d(μ)

n

√ = ( fˆ(t/ n))n , so that if n ≥ t2 then

√ log cˆn (t) = n log(1 − 2π 2 t2 /2n + r(t/ n)) = −t2 /2 + sn (t),

where sn (t) → 0 as n → ∞. Consequently, if ν is a w-limit point of L(cn ) then Fν(t) = −2π 2 t2 /2, and so ν = L(γ ). Thus the sequence (L(cn )) has a unique limit point, and so cn → γ in distribution as n → ∞. [Terminology and notation vary; in particular, probabilists also call the Fourier transform of an element of P(R) its characteristic function. The choice of constants also varies; −2π is frequently replaced by 2π , by −1 or by 1.]

276

Convergence of Measures

18.8 Uniform Integrability We introduce the notion of uniform integrability in various settings. A function c on [0, ∞) is a growth function if it is an increasing continuous function for which c(t) = 0 if and only if t = 0, and which satisfies the 2 condition; there exists L > 0 such that c(2t) ≤ Lc(t) for all t > 0. Thus an N-function which satisfies the 2 condition is a growth function, as are the functions t → tp , for p > 0. If A ⊆ P([0, ∞)), A is c-uniformly integrable if  c dμ → 0 as R → ∞. sup μ∈A [R,∞)

If c(t) = tp , where 0 < p < ∞, we say that A is p-uniformly integrable, and if c(t) = t we say that A is uniformly integrable. Proposition 18.8.1 If A ⊆ P([0, ∞)) is c-uniformly integrable then supμ∈A [0,∞) c dμ < ∞, and A is uniformly tight.  Proof There exists R > 0 such that supμ∈A [R,∞) c dμ ≤ 1. If μ ∈ A, then 

 c dμ =

[0,∞)

 c dμ +

[0,R)

c dμ ≤ c(R) + 1. [R,∞)

 Suppose that  > 0. There exists R ≥ 1 such that supμ∈A [R,∞) c dμ < c(1). If μ ∈ A then  1 c(R) c dμ < . μ([R, ∞) ≤ μ([R, ∞)) ≤ c(1) c(1) [R,∞)

Next, suppose that (X, τ ) is a Polish space, that c is a non-negative continuous real-valued function on X, and that A ⊆ P(X). We say that A is c-uniformly integrable if the set {c∗ (μ) : μ ∈ A} of push-forward measures on [0, ∞) is uniformly integrable. That is,  c dμ → 0 as R → ∞. sup μ∈A (c≥R)

If so, then, as in Proposition 18.8.1, supμ∈A

 X

c dμ < ∞.

Theorem 18.8.2 Suppose that (X, τ ) is a Polish space, that c is a non-negative continuous real-valued function on X and that (μk )∞ k=0 is a sequence in P(X) for which μk ⇒ μ0 as k → ∞. Then {μk : k ∈ Z+ } is c-uniformly integrable if and only if supk∈Z+ X c dμk < ∞ and X c dμk → X c dμ0 as k → ∞.

18.8 Uniform Integrability

277

Proof Suppose first that {μk : k ∈ Z+ } is c-uniformly integrable. Certainly,  supk∈Z+ X c dμk < ∞. Suppose that  > 0. There exists R > 0 such that supk∈Z+ (c≥R) c dμk < . Then 





c dμk − X

c dμk <  for k ∈ Z+ ,

(c ∧ R) dμk ≤ X

(c≥R)

so that & & & &   & & & & & c dμk − c dμ0 & ≤ & (c ∧ R) dμk − (c ∧ R) dμ0 & + 2. & & & & X

X

X

X

    Since X (c ∧ R) dμk → X (c ∧ R) dμ0 as k → ∞, X c dμk → X c dμ0 as k → ∞.    Suppose conversely that supk∈Z+ X c dμk < ∞ and that X c dμk → X c dμ0 as k → ∞. Suppose that  > 0. Then there exists k0 such that & &  & & & c dμk − c dμ0 & < /3 for k ≥ k0 . & & X

X

By Tietze’s extension theorem, for each n ∈ N, there exists cn ∈ C(X) such that cn (x) = c(x) if c(x) ≤ n, cn (x) = 0 if c(x) ≥ n + 1 and 0 ≤ cn ≤ c. By monotone convergence, there exists n0 such that   0≤ c dμ0 − cn dμ0 < /3 for n ≥ n0 . X

X

Since μk ⇒ μ0 , there exists k1 ≥ k0 such that & &  & & & cn dμk − cn dμ0 & < /3 for k ≥ k0 . 0 0 & & X

X

Putting these inequalities together, if k ≥ k1 then   c dμk ≤ (c − cn0 ) dμk (c≥n+1) X & &   & & & ≤ (c − cn0 ) dμ0 + & c dμk − c dμ0 && X X X & &  & & + && cn0 dμk − cn0 dμ0 && < . X

X

By monotone convergence, there exists N ≥ n + 1 such that for 1 ≤ k ≤ k1 , and so the result follows.



(c≥N) c dμk

≤

278

Convergence of Measures

18.9 Uniform Integrability in Orlicz Spaces Suppose that c is a growth function, that μ ∈ P(X), where (X, τ ) is a Polish space, and that F is a set of real-valued random variables on X. We say that F is c-uniformly integrable if the set of distributions {| f |∗ (μ) : f ∈ F} is; that is, if  c(|f |) dμ → 0 as R → ∞. sup f ∈F

|f |≥R

If c = , an N-function which satisfies the 2 condition, and F is a uniformly integrable set of real-valued random variables on (X, μ), then F is a norm-bounded subset of (L (μ), . ). Exercise 18.9.1 Suppose that (X, τ ) is a Polish space, that μ ∈ P(X) and that c is a growth function. Suppose that F and G are -uniformly integrable sets of real-valued random variables on (X, μ), and that α ∈ R. Then F + G = {f + g : f ∈ F, g ∈ G} and αF are c-uniformly integrable. Theorem 18.9.2 Suppose that (X, τ ) is a Polish space, that μ ∈ P(X), that  is an N-function which satisfies the 2 condition, or that (t) = tp for some 0 < p ≤ 1. If (fk )∞ k=0 is a sequence of real-valued random variables on (X, μ), then the following are equivalent. (i) fk → f0 in probability as k → ∞ and (fk )∞ k=0 is -uniformly integrable. (ii) (fk )k∈Z+ is a norm-bounded sequence in L (X, μ), and fk − f0  → 0 as k → ∞. (iii) fk → f0 in probability as k → ∞, (fk )k∈Z+ is a norm-bounded sequence in L (X, μ), and fk  → f0  as k → ∞. Proof Suppose that (i) holds. Then (fk )k∈Z+ is a norm-bounded sequence in L (X, μ). To show that (ii) holds, it is sufficient to show that each subsequence has a sub-subsequence for which (ii) holds, and so by extracting a subsequence if neccessary, we may suppose that fk → f0 almost everywhere, as k → ∞. integrable Suppose that  > 0. The sequence (fk − f0 )k∈N is -uniformly  (Exercise 18.9.1), and so there exists M > 0 such that (|fk −f0 |≥M) (|fk − of bounded convergence,  f0 |) dμ < . Let M (t) = (|t|)∧M. By the theorem φ (f − f0 ) dμ → 0 as k → ∞. Since (|fk −f0 |≤M) (|fk − f0 )|p dμ ≤  X M k X φM (fk − f0 ) dμ, there exists k0 such that (|fk −f0 |≤M) (|fk − f0 |) dμ <  p /2 for k ≥ k0 . Combining this with the earlier inequality, it follows that fk − f0  → 0 as k → ∞. Condition (ii) certainly implies (iii). Suppose that (iii) holds; again we can suppose that fk → f0 almost everywhere as k → ∞. Suppose that  > 0. If

18.9 Uniform Integrability in Orlicz Spaces

279

n ∈ N, let n (t) = (t) for 0 ≤ t ≤ n and let n (t) = 0 for t > n. By the theorem of monotone convergence, there exists n0 such that X (|f0 | dμ ≤  X n (|f0 | dμ + /2 for n ≥ n0 . By the theorem of bounded convergence, there exists k0 such that X |n0 (fk ) − n0 (f0 )| dμ < /3 for k ≥ k0 . If k ≥ k0 then    (|fk |) dμ ≤ (|fk |) dμ − n0 (|fk |) dμ (|fk |≥n0 )

X

 ≤

X





(|f0 |) dμ + /3 − X

X

 n0 (|fk |) dμ − /3

< . Again, the first k0 terms are -uniformly integrable, and so (iii) implies (i). Exercise 18.9.3 Suppose that (X, τ ) is a Polish space, that μ ∈ P(X), that  is an N-function which satisfies the 2 condition, or that (t) = tp for some 0 < p ≤ 1, and that F ⊂ L . Show that F is -uniformly integrable if and only if F is . bounded and if  > 0 there exists δ > 0 such that if μ(A) < δ than A φ(|f |) dμ <  for f ∈ F. We can use Theorem 18.9.2 to characterize compactness in L and Lp . Exercise 18.9.4 Suppose that A is a bounded subset of L (or Lp ). The following are equivalent. (i) A is a compact subset of L (or Lp ). (ii) A is -uniformly integrable (or p-uniformly integrable) and compact in L0 . (iii) Whenever (fn )∞ n=1 is a sequence in A which converges in probability to f ∈ A then fn  → f  (or fn p → f p ) as n → ∞.

19 Introduction to Choquet Theory

Choquet theory is concerned with the properties of compact convex sets; either weakly compact convex sets of a space E, where (E, F) is a dual pair, or compact convex sets of a Banach space E. In general, the theory is complicated; we shall only consider the metrizable case, where the theory is much simpler.

19.1 Barycentres Theorem 19.1.1 Suppose that (E, F) is a dual pair of vector spaces, that K is a metrizable σ (E, F)-compact subset of E, and that L = (K) is also σ (E, F)compact and metrizable. Suppose that  ! μ ∈ P(K). Then there exists a unique βμ ∈ L such that K f dμ = βμ , f for each f ∈ F. The mapping μ → βμ is a continuous affine mapping of (P(K), w) onto L; that is, β(1−λ)μ+λν = (1 − λ)βμ + λβν for 0 ≤ λ ≤ 1 and μ, ν ∈ K.  Proof For each f ∈ F, let Hf = {e : e, f  = f dμ}. Suppose that f1 , . . . , fn ∈ E. If e ∈ E, let T(e) = (e, f1  , . . . , e, fn ); T is a continuous linear mapping of E into Rn , and so T(L) is a compact convex subset of Rn . Let    f1 dμ, . . . , fn dμ . p= K

K

We show that p ∈ T(L). If not, by the separation theorem in Rn , there exists α ∈ Rn such that n n   ! αj pj > sup αj x, fj . j=1

x∈L j=1

280

19.1 Barycentres

281

  Let f = nj=1 αj fj . Then K f dμ > sup{ f (x) : x ∈ K}, which is clearly not true. Consequently p ∈ T(L), and so L ∩ (∩nj=1 Hfi ) is a non-empty compact convex subset of L. By the finite intersection ! property, L0 = L ∩ (∩f ∈F Hf ) is non-empty. If l ∈ L0 then K f dμ = lμ , f for each f ∈ F. It remains to show that L0 is a singleton set. If l, m ∈ L0 and f ∈ F then  l, f  = K f dμ = m, f , so that since F separates the points of E, l = m.  !  Since βμ − βν , f = K f dμ − K f dν, it follows that the mapping μ → βμ from (P(K), w) to L is a continuous affine mapping. Thus β(K) is a compact convex subset of L containing K, and so β(P(K)) = L. The element βμ is called the barycentre of μ. Exercise 19.1.2 Suppose that L is a convex σ (E, F)-compact metrizable subset of E and that K = Ex(L) is closed. Then if x ∈ L there exists μ ∈ P(Ex(L)) with barycentre x. If (E, F) is a dual pair of vector spaces and L is a σ (E, F)-compact convex metrizable subset of E then Ex(L) need not be closed. We shall however show later (Theorem 19.3.4) that a corresponding result still holds. We can go in the opposite direction. Suppose that (E, F) is a dual pair of vector spaces, that K is a metrizable σ (E, F)-compact subset of E and that x ∈ K. We set Rep(x) = {μ ∈ P(K) : βμ = x}; Rep(x) is the set of probability measures on K which represent x. Of course, δx ∈ Rep(x). Theorem 19.1.3 Suppose that (E, F) is a dual pair of vector spaces, that K is a metrizable σ (E, F)-compact subset of E and that x ∈ K. Then Rep(x) = {δx } if and only if x is an extreme point of K. Proof Suppose that x is not an extreme point of K, and that x = (1 − α)y + αz, with y, z ∈ K and 0 < α < 1. Then (1 − α)δy + αδz ∈ Rep(x), so that Rep(x) = {δx }. Conversely, suppose that Rep(x) = {δx } and that μ ∈ Rep(x) \ {δx }. Let s(μ) be the support of μ. Then s(μ) = {x}, and so there exists y ∈ s(x) with y = x. Let U be a closed convex neighbourhood of y in E which does not contain x, and let V = K ∩ U. Then V is a compact convex subset of K, and r = μ(V) > 0. Suppose that μ(V) = 1; it would then follow that βμ ∈ V, which is not true. Thus r < 1. If A is a Borel subset of K, let μ1 (A) = μ(A ∩ V)/r and μ2 (A) = μ(A ∩ (K \ V))/(1 − r); then μ1 , μ2 ∈ P(K). Let y = βμ1 and z = βμ2 . Then y ∈ V, so that y = x, z ∈ K and x = ry + (1 − r)z, so that x is not an extreme point of K.

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19.2 The Lower Convex Envelope Revisited Suppose that (E, F) is a dual pair and that K is a convex σ (E, F)-compact metrizable subset of E. If f is a real-valued function on K, we extend f to a function on E by setting f (y) = +∞ for y ∈ E \ K. Recall that the lower convex envelope f is the greatest convex function less than or equal to f ; f is lower semi-continuous. Theorem 19.2.1 Suppose that (E, F) is a dual pair, that K is a convex σ (E, F)compact metrizable subset of E and that f ∈ C(K).  Supposethat μ ∈ P(K). Then there exists  such that βν = βμ , K g dμ ≤ K g dν for each  ν ∈ P(K) g ∈ C(K) and K f dν = K f dμ. Proof Let A(K) be the set of σ (E, F)-continuous affine functions on E,  restricted to to K. If g ∈ C(K), let p(g) = − K g dμ. Then p(αg) = αp(g) for α > 0 and p(g1 + g2 ) ≤ p(g1 ) + p(g2 ), since g1 + g2 ≤ g1 + g2 . Thus p is sublinear; also, p(a) = −a(βμ ) for a ∈ A(K). Let φ(αf ) = αp( f ) for α ∈ R: φ is a linear functional on span( f ), and φ(αf ) ≤ p(αf ). We apply the Hahn–Banach theorem. There exists a linear functional φ on C(K) such that φ( f ) = p( f ) and φ(g) ≤ p(g) for g ∈ C(K). Note that if a is affine, then φ(a) ≤ p(a) = −a(βμ ) and φ(−a) ≤ p(−a) = a(βμ ), so that φ(a) = −a(βμ ). In particular, φ(1) = −1. Further, if g ≥ 0 then p(−g) ≤ 0, so that φ(g) ≤ 0. Thus −φ is positive, and by the Riesz representation theorem there exists ν ∈ P(K) such  that −φ(g) = K g dν for all g ∈ C(K). In particular, K f dν = −φ( f ) = f dμ, and if a ∈ A(K) then  a(βμ ) = K a dν, so that βν = βμ . Exercise 19.2.2 Suppose that (E, F) is a dual pair, that K is a convex σ (E, F)compact metrizable subset of E and that f ∈ C(K).  Suppose that x0 ∈ K. = x , Show that there exists ν ∈ P(K) such that β ν 0 K g dν ≥ g(x0 ) for each  g ∈ C(K) and K f dν = f (x0 ). Proposition 19.2.3 Suppose that (E, F) is a dual pair, that K is a convex σ (E, F)-compact metrizable subset of E and that f ∈ C(K). Let f∗ (x) = inf{ K f dμ : μ ∈ Rep(x)}. Then f∗ = f .  Proof If x ∈ K, a ∈ A(K) and a ≤ f , and if μ ∈ Rep(x) then a(x) = K a dμ ≤ K f dμ, so that a(x) ≤ f∗ (x). Consequently, f∗ (x) ≥ f (x). On the other hand, f∗ ≤ f , by Exercise 19.2.2. Exercise 19.2.4 (Jensen’s inequality) If f is a convex lower semi-continuous  function on K, if x ∈ K and if μ ∈ Rep(x), then f (x) ≤ K f dμ.

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A real-valued function f on a convex set K is strictly convex if whenever x and y are distinct points of K and 0 < λ < 1 then f ((1 − λ)x + λy) < (1 − λ)f (x) + λf (y). Proposition 19.2.5 Suppose that (E, F) is a dual pair and that K is a convex σ (E, F)-compact metrizable subset of E. Then there exists a non-negative continuous strictly convex function f on K. Proof Let A1 (K) = {h ∈ A(K) : h∞ ≤ 1}. A1 (K) separates the points of K, and is separable in the uniform norm topology; let (hn )∞ n=1 be a dense sequence in A1 (K); then (hn ) separates the points of K. We set f (k) =

∞ 

h2n (k)/2n .

n=1

Then f ∈ C(K). If x and y are distinct points of K and 0 < λ < 1 then there exists n such that hn (x) = hn (y), so that h2n ((1 − λ)x + λy) < (1 − λ)h2n (x) + λh2n (y), while, for all m, h2m ((1 − λ)x + λy) ≤ (1 − λ)h2m (x) + λh2m (y), so that f ((1 − λ)x + λy) < (1 − λ)f (x) + λf (y); that is, f is strictly convex. Proposition 19.2.6 Suppose that (E, F) is a dual pair and that K is a convex σ (E, F)-compact metrizable subset of E and that x ∈ K. Then x is an extreme point of K if and only if f (x) = f (x) for each f ∈ C(K). Proof If x is an extreme point, then Rep(x) = {δx }, and so f (x) = f (x) for each f ∈ C(K), by Proposition 19.2.3. If x is not an extreme point of K, let f be a strictly convex function in C(K). There exist y, z ∈ K and 0 < α < 1 such that x = (1 − α)y + αz. Then f (x) ≤ (1 − α)f (y) + αf (z) ≤ (1 − α)f (y) + αf (z) < f (x).

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19.3 Choquet’s Theorem Suppose that (E, F) is a dual pair and that K is a σ (E, F)-compact subset of E. The set of extreme points of such a general compact set can be very unpleasant topologically, but this is not the case when K is metrizable. Proposition 19.3.1 Suppose that (E, F) is a dual pair and that K is a metrizable σ (E, F)-compact subset of E. Then Ex(K) is a Gδ subset of K. Proof Let d be a metric on K which defines the topology of K, and let An = {(x, y) ∈ K × K : d(x, y) ≥ 1/n} × [1/n, 1 − 1/n]. An is a compact subset of the set A = K × K × [0, 1]. If (x, y, λ) ∈ A, let f (x, y, λ) = (1 − λ)x + λy. Then f is a continuous function on A, and so Bn = f (An ) ∩ K is closed in K. But x is an extreme point of K if and only if x ∈ ∪n Bn . This enables us to show that a point of a convex compact set K can be represented by a measure which lives on the extreme points of K. Even when defined by simple conditions, the set of extreme points of a subset of (P(K), w) need not be closed. Suppose that J is a homeomorphism of a compact metric space K onto itself. A Borel probability measure μ is Jinvariant if J∗ μ = μ; that is, μ(J(A)) = μ(A) for each Borel set A. The set of J-invariant elements of P(K) is denoted by PJ (K). Exercise 19.3.2 Suppose that J is a homeomorphism of a compact metric space K onto itself. Show that PJ (K) is a w-closed convex subset of P(K). Exercise 19.3.3 Let K = T2 . If k = (eiθ , eiφ ) ∈ K, let J(k) = (eiθ , ei(θ+φ ). If eiθ ∈ T, let S(eiθ ) = (1, eiθ ). (i) Show that J is a homeomorphism of K onto itself. (ii) Show that if μ ∈ P(T) and the support of μ is contained in {1} × T then μ ∈ PJ (K).  (iii) Let μn = 1n nj=1 δ(e2π i/n ,e2π ij/n ) . Show that μn is an extreme point of PJ (K). (iv) Show that μn ⇒ S∗ (λ) (where λ is Haar measure on T). (v) Show that S∗ (λ) is not an extreme point of PJ (K). Theorem 19.3.4 (Choquet’s theorem) Suppose that (E, F) is a dual pair, that K is a metrizable σ (E, F)-compact convex subset of E and that x0 ∈ K. Then there exists ν ∈ P(K) with barycentre x0 such that μ(Ex(K)) = 1. Proof Let f be a strictly convex continuous non-negative function on K. By  Corollary 19.2.2, there exists ν ∈ P(K) such that βν = x0 , K g dν ≥ K g dμ for each g ∈ C(K) and K f dν = f (x0 ). We show that ν(Ex(K)) = 1. Since

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285

f (x0 ) = sup{a(x0 ) : a ∈ A(K), a ≤ f }    a dν : a ∈ A(K), a ≤ f = f dν, = sup K

K

 it follows that inf{ K ( f − a) dν : a ∈ A(K), a ≤  f } = 0. Thus for each n ∈ N there exists an ∈ A(K), with an ≤ f , such that K ( f − an ) dν ≤ 2−n . Now let   sn = nj=1 ( f − aj ), and let s = ∞ j=1 ( f − aj ). By monotone convergence, 

 0≤

s dν = lim K

n→∞ K

sn dν ≤ 1,

and so ν{k : s(k) = ∞} = 0. But if k is not an extreme point of K, then δ = f (k) − f (k) > 0, so that f (k) − an (k) ≥ δ for each n ∈ N, and s(k) = ∞. Thus K \ Ex(K) ⊆ {k : s(k) = ∞}, and ν(K \ Ex(K)) = 0.

19.4 Boundaries Although convexity is a real concept, it has an important role to play in complex analysis. (Recall that a set is convex in a complex vector space E if and only if it is a convex subset of the underlying real space ER .) In this section, we apply Choquet theory to subspaces of the complex Banach space (CC (K), .∞ ), where K is a compact metrizable space. Let us begin with a familiar example, which motivates the theory that we shall develop. Let A(D) denote the space of complex-valued functions which are continuous on the closed unit disc D, and analytic on the interior of D. Then it follows from the maximum modulus theorem that if f is a non-constant function in A(D) then the set where |f | attains its maximum is a subset of the boundary T = ∂D. Further, if 0 ≤ θ < 2π, let fθ (z) = 1 + e−iθ z. Then  fθ ∞ = 2, and |fθ (z)| = 2 if and only if z = eiθ . A closed linear subspace E of a space (CC (K), .∞ ) is a separating subspace of CC (K) if it contains the constant functions, and separates points; that is, if x1 and x2 are distinct points of K then there exists f ∈ E with f (x1 ) = f (x2 ). If E is a separating subspace of CC (K), we denote the inclusion mapping E → CC (K) by iE . If j is the isomorphism of MC (K, B) onto CC (K) given by the Riesz representation theorem, then iE ◦j is a surjection of MC (K, B) onto E ; we denote its restriction to π(K) by πE . If φ ∈ E , we denote πE−1 (φ) by Mφ . We denote πE ◦ δ by qE . Thus πE (μ)( f ) = K f dμ and qE (x)( f ) = f (x). Exercise 19.4.1 Show that qE is a homeomorphism of K onto qE (K).

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Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space. The state space S(E) of E is the set {φ ∈ E : φ(1) = 1 = φ }. Theorem 19.4.2 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space. Then S(E) = πE (P(K)). Proof If μ ∈ P(K) then πE (μ) ≤ μTV = 1. But πE (μ)(1) = 1, so that πE (μ) = 1 and πE (μ) ∈ S(E). Conversely, if φ ∈ S(E), it follows from the Hahn–Banach theorem that there exists μ ∈ MC (K) such that iE (μ) = φ and μTV = 1. But μ(K) = φ(1) = 1, and so μ ∈ P(K). Thus S(E) = πE (P(K)). If φ ∈ S(E), we define φ by setting (φ)( f ) = (φ( f )). Thus if μ ∈ Mφ then φ( f ) = K ( f ) dμ. We denote the real linear subspace {f : f ∈ E} of C(K) by E. Proposition 19.4.3 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space. Then S(E) = (qE (K)). Proof S(E) is a weak*-closed convex subset of the unit ball B1 (E ) of E , and so (qE (K)) ⊆ S(E). Suppose if possible that there exists φ ∈ S(E) \ (qE (K)). By the separation theorem, there exists h = f + ig ∈ CC (K) such that φ( f ) = (φ( f )) > sup{(ψ(h)) : ψ ∈ (qE (K))} = sup{(ψ(h)) : ψ ∈ qE (K)} = sup f (x). x∈K

Let α = f ∞ . Then φ( f + α1) > supx∈K f (x) + α = f + α1∞ , giving a contradiction. Theorem 19.4.4 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space, that f ∈ CC (K) and that φ ∈ S(E). Let m = sup{φ(g) : g ∈ E, g ≤ f } and M = inf{φ(h) : h ∈ E, h ≥ f }.  If m ≤ c ≤ M there exists μ ∈ Mφ such that K f dμ = c. Proof The result is trivially true if f ∈ E. Otherwise, let F = span(E, f ). If h = g + λf ∈ F let ψ(h) = φ(g) + λc. Thus ψ is a linear functional on F which extends φ. We show that ψ = 1. For this, it is enough to show that if h ≤ 1, then ψ(h) ≤ 1. If λ = 0 this is certainly true. Suppose next that λ > 0. Then f ≤ ((1−g)/λ), so that Reφ((1−g)/λ) ≥ M ≥ c. Thus ψ(h) = φ(g)+λc ≤ φ(1) = 1. Finally suppose that λ < 0. Then f ≥ ((1−g)/λ), so that φ((1 − g)/λ) ≤ m ≤ c. Thus ψ(h) = φ(g) + λc ≤ φ(1) = 1.

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287

We now apply the Hahn–Banach theorem; there exists μ ∈ M(K), with μTV = 1, which extends ψ. Since μ(K) = K 1 dμ = 1, μ ∈ P(K). Further,  πE (μ) = φ, so that μ ∈ Mφ , and K f dμ = ψ( f ) = c. Proposition 19.4.5 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space, and that φ ∈ S(E). Then φ is an extreme point of S(E) if and only if there exists x ∈ K such that Mφ = {δx }. Proof Suppose first that φ is an extreme point of S(E). It follows from the preceding proposition and Mil’man’s theorem (Theorem 12.2.10) that there exists x ∈ K such that φ = qE (x). Suppose that μ ∈ P(K) and that φ = πE (μ) that is, f (x) = φ( f ) = K f dμ for each f ∈ E. Let ν be the push-forward measure qE ∗ (μ). Then φ( f ) = S(E) ψ( f ) dν(ψ) for each f ∈ E, and so φ is the barycentre of ν. But φ is an extreme point of S(E), and so ν = δφ . Thus μ = πE (δx ) = qE (x). Conversely, suppose that φ is not an extreme point of S(E). Thus there exist distinct φ1 and φ2 in S(E) such that φ = 12 (φ1 + φ2 ). There exist μ1 and μ2 in P(K) such that φ1 = πE (μ1 ) and φ2 = πE (μ1 ). Let μ = 12 (μ1 + μ2 ). Then φ = πE (μ). But μ is not an extreme point of P(K), and so πE−1 (φ) = {δx }, for some x ∈ K. Exercise 19.4.6 Suppose that φ is an extreme point of S(E) and that Mφ = {x}. Use Theorem 19.4.4 to show that if f ∈ C(K) and a < f (x) then there exists g ∈ E with g(0) = a and (g) ≤ f . Suppose that E is separating subspace of CC (K), where K is a compact metrizable space. A subset B of K is a boundary for E if for each f ∈ E there exists x ∈ B for which |f (x)| = f ∞ . Exercise 19.4.7 Use the maximum modulus principle to show that T = ∂D is a boundary for A(D). Theorem 19.4.8 Suppose that E is separating subspace of CC (K), where K is a compact metrizable space. Let Ch(E) = {x ∈ K : qE (x) is an extreme point of S(E)}, then Ch(E) is a boundary for E. Proof Suppose that f ∈ E. There exists x ∈ K such that |f (x)| = f ∞ . Let H = {φ ∈ E : φ( f ) = qE (x)( f )}. Then H is a support hyperplane of S(E), and H ∩S(E) is a non-empty convex set. Let φ0 be an extreme point of H ∩S(E). By Proposition 12.2.4 it is an extreme point of S(E). There exists a unique x0 ∈ K such that φ0 = qE (x0 ). Then x0 ∈ Ch(E) and |f (x0 )| = f ∞ .

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The set Ch(E) is called the Choquet boundary of E. Exercise 19.4.9 By considering functions of the form 1 + eiθ , show that T is the Choquet boundary of A(D). Exercise 19.4.10 Let E = { f ∈ C([0, 1]) : f ( 12 ) = 12 ( f (0) + f (1))}. Show that the set [0, 1] \ { 12 } is a non-closed Choquet boundary for E. The closure Sh(E) = Ch(E) of the Choquet boundary is called the Shilov boundary of E. Theorem 19.4.11 Suppose that E is separating subspace of CC (K), where K is a compact metrizable space. Then the Shilov boundary Sh(E) of E is the smallest closed boundary of E; if B is a closed boundary of E then Sh(E) ⊆ B. Proof Suppose not, so that there exists x0 ∈ Ch(E)\B. Since qE (B) is σ (E , E)closed there exist f1 , . . . , fn in E such that U = {φ ∈ E : |φ( fj ) − fj (x0 )| < 1 for 1 ≤ j ≤ n} is a weak* neighbourhood of qE (x0 ) disjoint from qE (B). Let gj = fj − fj (x0 ), for 1 ≤ j ≤ n, and let ⎧ for 1 ≤ j ≤ n gj ⎪ ⎪ ⎨ −gj−n for n + 1 ≤ j ≤ 2n hj = ⎪ for 2n + 1 ≤ j ≤ 3n 'g ⎪ ⎩ j−2n −'gj−3n for 3n + 1 ≤ j ≤ 4n. Then V = {φ ∈ E : φ(hj ) < of qE (x0 ) contained in U. Let

1 2

for 1 ≤ j ≤ 4n} is a weak* neighbourhood

Lj = {φ ∈ S(E) : φ(hj ) ≥ 12 } for 1 ≤ j ≤ 4n. Then each Lj is weak* compact, and so is M = (∪4n j=1 Lj ). Now qE (x0 ) ∈ Lj for any j, and qE (x0 ) is an extreme point of S(E), so that qE (x0 ) ∈ M. By the separation theorem, there exists f ∈ E such that a = f (x0 ) = (qE (x0 )( f )) > b = sup{φ( f ) : φ ∈ M}. Let c = sup{|'φ( f )| : φ ∈ M}. If d > c2 /(a − b) then easy calculations show that |f (x0 ) + d| > sup{|φ( f ) + d)| : φ ∈ M}. But B ⊆ M, and so the function f + d1 does not attain its supremum on B, giving a contradiction.

19.5 Peak Points

289

19.5 Peak Points Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space. An element x of K is a peak point for E if there exists f ∈ E for which f (x) = 1 and  |f (y)| <  1 for y = x. If x is a peak point for E and μ ∈ MqE (x) then 1 = K f dμ ≤ K |f | dμ, so that μ(|f | < 1) = 0, and μ = δx . Thus x ∈ Ch(E). Theorem 19.5.1 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space, and that f is a smooth point of the unit sphere of E. Then there exists x ∈ K such that f (x) = 1 and such that |f (y)| < 1 for y = x, so that x is a peak point for E. Proof There exists a unique φ ∈ S(E) such that φ( f ) = 1. Thus φ is an extreme point of S(E), and so there exists x ∈ Ch(E) such that φ( f ) = f (x) = δx ( f ). Since φ is unique, |f (y)| < 1 for y = x, and so x is a peak point for E. We use this to show that there are plenty of peak points in the Choquet boundary of a separating space. Theorem 19.5.2 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space. The set Pe(E) of peak points for E is dense in Ch(E) in the weak* topology. Proof Suppose not. Since Ch(E) is the set of extreme points of the state space S(E), it follows from Mil’man’s theorem (Theorem 12.2.10) that (Pe(E)) is a proper subset of Ch(E). If φ ∈ Ch(E) \ (Pe(E)), there exists f ∈ E with f  = 1 such that φ( f ) > sup{ψ( f ) : ψ ∈ (Pe(E))} = sup{ψ( f ) : ψ ∈ Pe(E)}. By Exercise 11.6.8, the set of smooth points of the unit sphere of E is dense in the unit sphere, and so there exists a smooth point g such that φ(g) > sup{ψ(g) : ψ ∈ Pe(E)}. But there exists ψ ∈ Pe(E) for which ψ(g) = 1, and so φ(g) > 1, giving a contradiction. Here is a useful sufficient condition for x to be a peak point. Theorem 19.5.3 Suppose that E is a separating subspace of CC (K), where K is a compact metrizable space, and that x ∈ K. Then x is a peak point for E if the following condition holds. (*) There exist M ≥ 0 and 0 <  < 1 such that if U is any open neighbourhood of x there exists f ∈ E with f (x) = 1, f ∞ ≤ M + 1 and |f (y)| < 1 −  for y ∈ U.

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Proof The proof uses a classic ‘sliding hump’ argument. Choose s < 1 such that t = (M + )s − M > 0. Let (Un )∞ n=1 be a base of open neighbourhoods of x, and let Fn = X \ Un for n ∈ N. Let h0 = 1. Using condition (*), an inductive argument then shows that there exists a sequence (hn )∞ n=1 in E such that hn (1) = 1, hn ∞ ≤ M + 1 and if Bn = ∪nj=1 {y : |hj (y)| > 1 + sn+1 t then |hn+1 (y)| ≤ 1 −  for y ∈ Bn ∪ Fn . For if we have found h0 , . . . , hn then Bn ∪ Fn is a closed set disjoint from {x}, and so condition (*) enables us to find hn+1 .  n We now set h = (1−s) ∞ n=1 s hn . We show that h(x) = 1 and that |h(y)| < 1 for y = x. Certainly h(x) = 1. Suppose that y = x, so that y ∈ Fk , for some k. We consider two cases. First suppose that y ∈ ∪∞ n=1 Bn . If j ∈ N, then |hj (y)| ≤ n 1 + s t for all n, and so |hj (y)| ≤ 1. But |hk (y)| < 1 − , and so |h(y)| < 1. Secondly, suppose that y ∈ ∪∞ n=1 Bn . Then there exists a least n such that y ∈ Bn . Thus ⎧ n ⎨ 1 + s t for 1 ≤ j < n |hj (y)| ≤ M + 1 for j = n ⎩ 1− for j > n. Hence

⎛

⎡

|h(y)| ≤ (1 − s) ⎣(1 + sn t) ⎝

n−1 

⎛

⎞

sj ⎠ + sn (M + 1) + (1 − ) ⎝

j=1 n

⎞⎤ sj ⎠⎦

j=n+1

= (1 − s )(1 + s t) + (1 − s)s (M + 1) + s n

∞ 

n

n+1

(1 − )

= 1 + (1 − s )s t − s (s(M + ) − M) = 1 − s2n t < 1. n

n

n

Important examples of separating subspaces are provided by uniform algebras. If K is a compact metrizable topological space, a uniform algebra A on K is a separating subspace of CC (K) which is also an algebra under pointwise multiplication. An important feature of uniform algebras is that  n (since separating subspaces are closed) if f ∈ A, then ef = ∞ n=0 f /n! ∈ A; the following result illustrates this. Theorem 19.5.4 Suppose that A is a uniform algebra on a compact metrizable space K and that x ∈ Ch(A). Suppose that f ∈ CR (K), that f (x) > 0 for all x ∈ K and that 0 < a < f (x). Then there exists g ∈ A with g(x) = a and |g| ≤ f . Proof Let F = log f . By Exercise 19.4.6, there exists G ∈ A with G(0) = log a and G ≤ F. Let g = eG . Then g(x) = a and |g| ≤ f .

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291

When A is a uniform algebra, we can improve upon Theorem 19.5.2. Corollary 19.5.5 If x ∈ Ch(A) then x is a peak point for A. Proof Suppose that U is an open neighbourhood of x. By Urysohn’s lemma, there exists f ∈ C(X) with f (x) = 1, f (y) = 0 for y ∈ X \ U and 0 ≤ f ≤ 1. Apply the theorem to f + 1. There therefore exists g ∈ A with g(x) = 3/2, |g| ≤ 2 and |g(y)| ≤ 1 for y ∈ U. Then h = 2g/3 satisfies condition (*) of Theorem 19.5.3, with M =  = 1/3.

19.6 The Choquet Ordering Let us investigate Choquet’s theorem further. Suppose that (E, F) is a dual pair and that K is a metrizable σ (E, F)-compact convex subset of E. We denote the set of convex continuous functions on K by CC(K) and the set { f ∈ CC(K); f ∞ ≤ 1} by CC1 (K).  a partial  We define order, the Choquet ordering, on P(K) by setting μ  ν if K f dμ ≤ K f dν for every f ∈ CC(K). Intuitively, this suggests that ν lives closer to the boundary of K than μ does. Clearly, it is enough to verify the condition for functions in  (K). If f ∈ A(K) then f and −f are convex, and so if μ  ν then f dμ = CC 1 K  K f dν, so that βμ = βν . Note also that it follows from Jensen’s inequality that δβμ  μ. We consider measures which are maximal with respect to the Choquet ordering. Theorem 19.6.1 Suppose that (E, F) is a dual pair, that K is a metrizable σ (E, F)-compact convex subset of E and that μ ∈ P(K). Then there exists a maximal measure ν ∈ P(K) with μ  ν. Proof Since (C(K), .∞ ) is separable, so is CC1 (K). Let (cn )∞ n=1 be a dense sequence in CC(K). Let A0 = {π ∈ P(K) : μ  π }; A0 is a non-empty compact subset of (P(K), w). We define     A1 = π ∈ A0 : c1 dπ = sup c1 dρ : ρ ∈ A0 . K

K

Then A1 is a compact subset of A0 . We now repeat the process recursively; we set     An+1 = π ∈ An : cn+1 dπ = sup cn+1 dρ : ρ ∈ An . K

Then (An )∞ n=0

K

is a decreasing sequence of non-empty  compact subsets of P(K), and therefore there exists ν ∈ ∩n∈Z+ An . Then K cn dμ ≤ K cn dν for each

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Introduction to Choquet Theory

n ∈ N, and so, since (cn )∞ n=1  is a dense sequence in CC(K), μ  ν. Similarly, if ν  ν  , then K cn dν = K cn dν  for each n ∈ N, and so, since (cn )∞ n=1 is a  dense sequence in CC(K), ν = ν : ν is maximal. Theorem 19.6.2 Suppose that (E, F) is a dual pair, that K is a metrizable σ (E, F)-compact  convex subset of E. If μ is a maximal measure and f ∈ C(K) then f dμ = f dμ. Proof Suppose first that μ is a maximal measure. By Theorem 19.2.1 there   exists ν ∈ P(K) such that β = β , g dμ ≤ g dν for each g ∈ C(K) and ν μ K K   f dν = f dμ. K K   If h ∈ CC(K), then h = h, so that K h dμ ≤ K h dν, and μ  ν. Since μ is maximal, ν = μ. Thus    f dμ = f dν = f dμ. K

K

K

Note that in this theorem we do not need the axiom of choice. Exercise 19.6.3 By considering a strictly convex function, show that if μ is a maximal measure, then μ(Ex(K)) = 1. Thus Theorem 19.6.1 provides another proof of Choquet’s theorem. What about the converse? Suppose that f is a bounded function on K. Let D( f ) = {g ∈ C(K) : g convex, g ≤ f }. It follows from the definitions that f = sup{g : g ∈ D( f ), g ≤ f }. Theorem 19.6.4 Suppose that K is a metrizable σ -compact convex subset of E, on K and that f is a bounded function on K. Then   that μ is a finite measure f dμ = sup g dμ. g∈D( f ) K K Proof D( f ) is directed upwards, and so the result follows from Theorem 16.1.4. Theorem 19.6.5 Suppose that (E, F) is a dual  K is a metrizable  pair, that σ (E, F)-compact convex subset of E and that K f dμ = K f dμ for each f ∈ C(K). Then μ is a maximal measure. Proof Suppose that μ  ν. Then if f ∈ C(K),     f dν ≥ f dν = sup g dν : g ∈ D( f ) K K K   ≥ sup g dμ : g ∈ D( f ) (since μ  ν) K   = f dμ = f dμ, K

and so ν = μ.

K

19.7 Dilations

293

Exercise 19.6.6 Suppose that (E, F) is a dual pair, that K is a metrizable σ (E, F)-compact convex subset of E. Show that if μ ∈ P(K) and μ(Ex(K)) = 1, then μ is a maximal measure.

19.7 Dilations Suppose that (E, F) is a dual pair of vector spaces, that K is a metrizable σ (E, F)-compact convex subset of E. A mapping T from K to P(K) is a dilation of K if (i) βT(x) = x for all x ∈ K, and  (ii) for each f ∈ C(K), the function K f dT(x) is Borel measurable. If T is a dilation, we can extend T to a mapping from P(X) to P(X); if μ ∈ P(X) and f ∈ C(K), let    f dT(x) dλ(x). φ(λ)( f ) = K

K

Then φ(λ)(1) = 1, and φ(λ) is a positive linear functional on C(X), and so by the Riesz representation theorem there exists T(λ) ∈ P(K) for which     f dT(λ) = f dT(x) dλ(x). X

K

K

Note that T(δx ) = T(x), so that we have extended T. Proposition 19.7.1 Suppose that (E, F) is a dual pair of vector spaces, that K is a metrizable σ (E, F)-compact convex subset of E and that T is a dilation of K. If λ ∈ P(X), then λ  T(λ). Proof If x ∈ K then βT(x) = x and  so δx  T(δx ). Thus if f is a continuous convex function on K then f (x) ≤ K f dT(δx ), and so      f dλ ≤ f dT(δx ) dλ(x) = f dT(λ). X

K

K

K

Our main aim is to prove the converse. Let L = {(λ, μ) ∈ P(K) × P(K) : λ  μ}, M = {δx , μ) ∈ P(K) × P(K) : x ∈ K, δx  μ}. Then L is a convex subset of P(K) × P(K). Since      f dλ ≤ f dμ : f convex, continuous , L = ∩ (λ, μ) : K

K

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L is w × w-closed. Since M = L ∩ (δ(K) × P(K)), M is a w × w-closed subset of L. Lemma 19.7.2 L = (M). Proof By the separation theorem, it is enough to show that if ( f , g) ∈ C(K) × C(K) and  f (x) − g dμ ≤ t for all (δx , μ) ∈ M, K

then



 f dλ − K

g dμ ≤ t for all (λ, μ) ∈ L. K

If x ∈ X, then by Proposition 19.2.3,   g(x) = inf g dμ : βμ = x , 

K

 so  K f dλ − K g dλ ≤ t. But g is convex, and so  that f (x) − g(x) ≤ t and K g dλ ≤ K g dμ, so that K f dλ − K g dμ ≤ t. Corollary 19.7.3 If (λ, μ) ∈ L, there exists a probability measure θ on M such that (λ, μ) = βθ . Theorem 19.7.4 If λ, μ ∈ P(K) and λ  μ then there exists a dilation T such that T(λ) = μ. Proof Let θ be the measure of the preceding corollary. If (δx , μ) ∈ M, let φ((δx , μ)) = x and let ψ(δx , μ) = μ; φ is a continuous mapping of M onto K and ψ is a continuous mapping of M onto P(K). If f , g ∈ C(X), then       f dλ − g dμ = g dπ dθ (δx , π ). (∗) f (x) − K

K

M

K

Putting g = 0 we see that λ = φ∗ (θ ). We now apply the disintegration theorem (Theorem 16.10.1) to θ and φ; there exists a family {νx : x ∈ K} in P(M) such that if h ∈ C(M) then  (i) M h dνx is a measurable   function on K, (ii) M h dθ = K M h dνx dλ(x), and (iii) νx (φ −1 ({x})) = 1 for almost all x ∈ K. Let S(x) = ψ∗ (νx ); S(x) ∈ P(P(K)). Let T(x) = β(S(x)), the barycentre of S(x); thus T(x) ∈ P(K). If f ∈ C(K) then the mapping μ → K f dμ is a continuous affine mapping on P(K), so that        f dT(x) = f dπ dS(π ) = f (y) dπ dνx (δy , π ). K

P(K)

K

M

K

19.7 Dilations

295

 Thus K f dT(x) is measurable. If a ∈ A(K), then since λx is supported on {(δx , π ) : βπ = x},   a dT(x) = a(y) dνx (δy , μ). K

M



It therefore follows from (iii) that K a dT(x) = a(x). Consequently  x = βT(x) , , π ) = and T is a dilation. Suppose now that g ∈ C(K). Let h(δ y K g dπ ; then  h ∈ C(M) and it follows from (*) that K g dμ = M h dθ . But        g dT(λ) = g dT(x) dλ(x) = h dνx dλ(x) K K M K K  h dθ = g dμ, = M

K

so that T(λ) = μ. Theorem 19.7.5 Suppose that λ ∈ P(K), where K is a compact convex metrizable metric space. Suppose that T is a dilation such that T(λ) = μ is maximal. Then T(δx ) is maximal for λ-almost all x.   Proof We use the fact that ν is maximal if and only if K f dν = K f dν for every f ∈ C(K). Since C(K) is separable, there exists a dense sequence ( fn )∞ n=1 in C(K). For each n,     (f n − fn ) dT(δx ) dλ(x), 0 = (f n − fn ) dμ = K

so that



K

 K

f n dT(δx ) =

fn dT(δx ) K

for all n, for λ-almost all x. The mapping f → f is uniformly continuous (Theorem 10.1.2, (vi)) and so   f dT(δx ) = f dT(δx ) K

K

for all f ∈ C(K), for λ-almost all x. This gives the result.

PART THREE Introduction to Optimal Transportation

20 Optimal Transportation

20.1 The Monge Problem The study of optimal transportation was inaugurated by Gaspard Monge in 1781. He asked the question ‘What is the most economical way of transporting material (iron ore, say, or rubbish) from various sites to other sites (iron works, or landfill sites)?’ This is an optimization problem, with three ingredients: the original distribution of material, the distribution of the sites where the material needs to be taken and the cost of transportation from one place to another. Let us describe the problem in mathematical terms. Suppose that (X, τ ) and (Y, σ ) are two Polish spaces, and that μ and ν are Borel probability measures on X and Y respectively; μ represents the (normalized) distribution of the material that is to be moved, and ν the distribution of the location to which it has to be moved. Suppose also that c (the cost function) is a lower semi-continuous function on X × Y. In general, we shall suppose that c is non-negative, but we shall also consider the case where c can take negative values, provided that there exist a ∈ L1 (μ) and b ∈ L1 (ν) such that c(x, y) + a(x) + b(y) ≥ 0 for all x, y. If S is a Borel measurable mapping from X to Y such that S∗ (μ) = ν (S pushes forward the material in X to the right distribution in Y), then S is called a deterministic transport plan, or deterministic coupling. Monge’s problem is to find a deterministic transport plan T such that   c(x, T(x)) dμ ≤ c(x, S(x)) dμ X

X

for all deterministic transport plans S. (Of course, conditions are required to ensure that deterministic transport plans exist; for example if X = Y = {0, 1} and μ({0}) = 1/3, μ({1}) = 2/3, ν({0}) = ν({1}) = 1/2 then there are no deterministic transport plans.) 299

300

Optimal Transportation

Besides considering a deterministic transport plan S, we also consider its graph G(S) = {(x, S(x)) : x ∈ X} and the push forward measure π = G(S)∗ (μ) on X × Y; thus π(A × B) = μ(A ∩ S−1 B), and π has marginals μ and ν. π is called a deterministic coupling; it has the advantage that distinct deterministic transport plans determine distinct deterministic couplings. Proposition 20.1.1 Suppose that μ and ν are probability measures on Polish spaces (X, τ ) and (Y, σ ) respectively, and that S and S are distinct deterministic transport plans. Then the couplings π = G(S)∗ (μ) and π  = G(S )∗ (μ) are distinct. Proof For there exists a measurable subset A of X with μ(A) > 0 such that S(A) ∩ S (A) = ∅. Then G(S)∗ (μ)(A, S(A)) = μ(A) > 0 and G(S )∗ (μ) (A, S(A)) = 0.

20.2 The Kantorovich Problem Suppose that we consider the measures μ and ν on {0, 1} described in the previous section, and also consider the cost function c defined as follows: c(0, 0) = c(1, 1) = 0 (it costs nothing to stay put), c(0, 1) = 3 and c(1, 0) = 1. Then an obvious optimal procedure is to send a third of the material at 1 to 0, and to leave the rest fixed. This kind of procedure, which cannot be defined by a deterministic transport plan, was investigated in the early 1940s by the Russian mathematician and economist Leonid Kantorovich; his work led to the award of the Nobel prize for economics in 1975. Once again, we start with Borel probability measures μ and ν on Polish spaces (X, τ ) and (Y, σ ). An element of P(X × Y) is called a transport plan if it has marginal distributions μ and ν; the set of transport plans is denoted by μ,ν . μ,ν is non-empty, since μ ⊗ ν ∈ μ,ν . We also consider a lower semi-continuous cost function c. The Kantorovich problem is then ‘Is there an optimal transport plan? If so, how do we find it, and what are its properties?’ Let us show that, under reasonable conditions, the answer to the first question is ‘yes’. Theorem 20.2.1 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, and that μ and ν are Borel probability measures on X and Y respectively. Then μ,ν is a non-empty w-compact convex subset of P(X × Y). Proof μ,ν is clearly a convex subset of P(X × Y), and we have seen that it is non-empty.

20.2 The Kantorovich Problem

301

We show that μ,ν is also uniformly tight. For if  > 0, there exist compact subsets K of X and L of Y such that μ(X \ K) < /2 and ν(Y \ L) < /2. If π ∈ μ,ν , then π((X \ K) × Y) = μ(X \ K) < /2 and π(X × (Y \ L)) = ν(Y \ L) < /2. Since (X × Y) \ (K × L) ⊆ ((X \ K) × Y) ∪ (X × (Y \ L)), it follows that π((X × Y) \ (K × L)) < . Finally we show that μ,ν is a w-closed subset of P(X × Y). Suppose that (πn )∞ n=1 is a sequence in μ,ν , and that πn ⇒ π. If C is a closed subset of X, then π(C × Y) ≥ lim sup πn (C × Y) = μ(C). On the other hand, π((X \ C) × Y) = sup{π(K × Y) : K compact, K ⊆ X \ C} ≥ sup{μ(K) : K compact, K ⊆ X \ C} = μ(X \ C). Since π(C × Y) + π((X \ C) × Y) = 1 = μ(C) + μ(X \ C), it follows that π(C ×Y) = μ(C). By regularity, if A is a Borel subset of X, then π(A × Y) = μ(A). Similarly, if B is a Borel subset of Y then π(X × B) = ν(B). Thus π ∈ μ,ν . Corollary 20.2.2 If f is a w-lower semi-continuous function on μ,ν taking values in (−∞, ∞], f is bounded below and attains its bounds. Thus the Kantorovich problem has a solution. Similarly, we have the following: Proposition 20.2.3 Suppose that M is a uniformly tight subset of P(X) and that N is a uniformly tight subset of P(Y). Then ∪{μ,ν : μ ∈ M, ν ∈ N} is a uniformly tight subset of P(X × Y). Recall (Theorem 4.2.9) that if c is lower semi-continuous and if d is a metric on X × Y which defines the product topology on X × Y then there exists an approximating sequence: there is an increasing sequence (cn )∞ n=1 of bounded non-negative Lipschitz functions on X × Y which converges to c pointwise. Theorem 20.2.4 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that c is a proper lower semi-continuous cost function on X×Y and that μ and ν are Borel probability measures on X and Y respectively. If there exists π ∈ P(X × Y) for  which X×Y c dπ < ∞, then there exists an extreme point π0 of μ,ν such that    c dπ0 = inf c dπ : π ∈ μ,ν = Mc , say. X×Y

X×Y

302

Optimal Transportation

 Proof Let (cn )∞ n=1 be an approximating sequence, and let φn (π ) = X×Y cn dπ . Then (φn )∞ n=1 is an increasing sequence  of w-continuous functions on μ,ν , which converges pointwise to φ(π ) = X×Y c dπ, by the theorem of monotone convergence. Thus φ is a w-lower semi-continuous function on μ,ν , and so it attains its lower bound, by Corollary 20.2.2. The function φ is an affine function on μ,ν – that is, φ((1 − λ)μ1 + λμ2 ) = (1 − λ)φ(μ1 ) + λφ(μ2 ) for 0 ≤ λ ≤ 1. It follows that



π :



c dπ  = Mc



X×Y

is a face F of μ,ν , and we can take π0 to be an extreme point of F, by the Krein–Mil’man theorem. Let us consider the case where c can take negative values. Corollary 20.2.5 Suppose that c is a lower semi-continuous function on X × Y, and that there exist lower semi-continuous functions a ∈ L1 (μ) and b ∈ L1(ν) such that c = c + a + b ≥ 0. If there exists π ∈ P(X × Y) for which  then there exists an extreme point π0 of μ×ν such  X×Y c dπ < ∞, that X×Y c dπ0 = inf{ X×Y c dπ : π ∈ μ,ν }. Proof The function cis a cost function,and so there exists an extreme point π0 of μ×ν such that X×Y c dπ0 = inf{ X×Y c dπ : π ∈ μ,ν }. This has the required properties, since if π ∈ μ×ν then      c dπ0 = c dπ0 − a dμ + b dν X×Y X×Y X Y       ≤ c dπ − a dμ + b dν = c dπ . X×Y

X

Y

X×Y

A solution to the Kantorovich problem can often lead to a solution to the Monge problem, as the following elementary example shows. We suppose that X = Y = {1, . . . , n}, μ = ν is uniform probability measure on {1, . . . , n} and c is any function on X × Y. Probability measures on X × Y can be identified with n × n matrices (pij )  with non-negative entries for which i,j pij = 1. Then π ∈ μ,ν if and only if n n i=1 πij = 1/n for 1 ≤ j ≤ n and j=1 πij = 1/n for 1 ≤ i ≤ n; that is, nπ is a doubly stochastic matrix. The set Sn of doubly stochastic matrices is a closed convex subset of the space Mn of all n × n matrices. If c is a cost function on

20.3 The Kantorovich–Rubinstein Theorem

303

  X × Y, it is represented by a matrix (cij ), and X×Y c dπ = 1≤i,j≤n cij πij .  c m is a linear functional If m ∈ Mn then the mapping m → 1≤i,j≤n ij ij on Mn . Consequently there exists an extreme point π of Sn at which the linear functional attains its infimum on Sn . But it follows from Birkhoff’s theorem (Theorem 12.2.11) that the permutation matrices are the extreme points of the set of doubly stochastic matrices. Thus there is an optimal transport mapping given by a permutation matrix Iσ . This example shows another phenomenon that we shall consider later. If τ is another permutation of {1, . . . , n} then the corresponding cost of Iτ σ is greater than or equal to the optimal cost, and so n 

ci,σ (i) ≤

n 

i=1

ci,τ (σ (i)) .

i=1

20.3 The Kantorovich–Rubinstein Theorem We now give another proof of the existence of an optimal transport plan. The proof is quite different from that of Theorem 20.2.4; it uses the Hahn–Banach theorem and the Riesz representation theorem. Its most important features are that it considers approximation from below, and that it leads to the maximal Kantorovich potential. Theorem 20.3.1 (The Kantorovitch–Rubinstein theorem) Suppose that (X, d) and (Y, ρ) are Polish metric spaces, that μ and ν are Borel probability measures on X and Y respectively, that c is a cost function on X × Y and that Mc (μ, ν) = inf{ X×Y c dπ : π ∈ μ,ν } < ∞. Let d be a complete metric on X × Y which defines the product topology. Let L = { f (x) + g(y) : f ∈ BL(X), g ∈ BL(Y)} ⊆ C(X × Y) and let





mc = mc (μ, ν) = sup

f dμ + X

 g dν : f + g ∈ L, f + g ≤ c .

Y

Then Mc = mc , and there exists π ∈ μ,ν such that

 X×Y

c dπ = mc .

Proof If π ∈ μ,ν , f + g ∈ L and f + g ≤ c then     f dμ + g dν = f (x) + g(y) dπ(x, y) ≤ X

Y

X×Y

c dπ , X×Y

so that mc ≤ Mc . We need to find π for which the reverse inequality holds.

304

Optimal Transportation

First we consider the case where X and Y are  L is a linear subspace  compact. of C(X × Y). If f + g ∈ L let φ( f + g) = X f dμ + Y g dν. This is a well    defined linear functional   on L,since if f + g =f + g then f − f = g − g is a constant k, so that X f dμ = X f dμ − k and Y g dν = Y g dν + k. Further, φ(1) = 1, so that φ is non-zero. Now let U = {h ∈ C(X × Y) : h(x, y) < c(x, y) for all (x, y) ∈ X × Y}. U is a non-empty convex open subset of C(X × Y). U ∩ L is also non-empty. If f + g ∈ U ∩ L and π0 ∈ μ,ν then   ( f + g) dπ ≤ c dπ0 , φ( f + g) = X×Y

X×Y

so that φ is bounded above on U ∩ L by Mc . Let M = sup{φ(h) : h ∈ U ∩ L}, and let B = {l ∈ L : φ(l) ≥ M}. Then B is a non-empty convex set disjoint from U. By the Hahn–Banach theorem, there exists a non-zero continuous linear functional ψ on C(X×Y) such that if h ∈ U then ψ(h) < K = inf{ψ(b) : b ∈ B}. If h > 0 then −αh ∈ U for all sufficiently large α, and so ψ(h) ≥ 0. Thus ψ is a positive linear functional on C(X×Y). Since ψ = 0, ψ(1) > 0. Let θ = ψ/ψ(1). θ is a non-negative linear functional on C(X × Y), and θ (1) = 1; by the Riesz representation theorem θ is represented by a Borel probability measure π on X × Y. We shall show that π has the required properties. Note that if h ∈ U, θ (h) <  = inf{θ (b) : b ∈ B}. If l0 ∈ L and φ(l0 ) = 0, then φ(M.1+αl0 ) = M , so that M.1+αl0 ∈ B for all α, and so θ (M.1+αl0 ) = M + αθ (l0 ) ≥  for all α, and so θ (l0 ) = 0. If l ∈ L then l = φ(l)1 + l0 , this means that where φ(l0 ) = 0, and so θ (l) = φ(l): θ extends φ. In particular,   = M: if h ∈ U then X×Y h dπ < M. If f ∈ C(X) then X×Y f (x) dπ(x, y) = φ( f ) = X f dμ, and a similar result holds for g ∈ C(Y); thus π ∈ μ,ν . Further, mc = sup{φ(h) : h ∈ U ∩ L} = M. By Theorem 4.2.9, there exists an approximating sequence (hn )∞ n=1 in BL(X × Y) which increases pointwise to c. Then each hn is in U, and so    hn dπ ≤ sup{ h dπ : h ∈ U} ≤ M = mc . c dπ = lim n→∞

We now consider the case where X and Y are Polish spaces. We consider X as a dense Gδ subspace of a compact metrizable space X˜ and Y as a dense ˜ and consider the push forward Gδ subspace of a compact metrizable space Y, measures μ, ˜ ν˜ and π˜ 0 . By Theorem 4.2.8, there exists a lower semi-continuous ˜ Then extension c˜ of c on X˜ × Y.

20.4 c-concavity 

305

 ˜ Y˜ X×

c˜ dπ˜ 0 =

c dπ0 < ∞. X×Y

Thus there exists π˜ ∈ μ,˜ ˜ ν such that     ˜ g˜ ∈ C(Y), ˜ f˜ + g˜ ≤ c˜ . f˜ dμ˜ + g˜ dν˜ : f˜ ∈ C(X), c˜ dπ˜ = sup ˜ Y˜ X×

X

Y

Now ˜ = μ( π˜ ((X˜ \ X) × Y) ˜ X˜ \ X) = 0 and π˜ (X˜ × (Y˜ \ Y)) = ν˜ (Y˜ \ Y) = 0, so that π˜ is the push-forward measure of an element π of πμ,ν . Since    f dμ + g dν : f + g ∈ L, f + g ≤ c sup Y X   ˜ ˜ ˜ ˜ ˜ ≥ sup f dμ˜ + g˜ dν˜ : f ∈ C(X), g˜ ∈ C(Y), f + g˜ ≤ c˜ , X

Y

the measure π satisfies the conclusions of the theorem. Corollary 20.3.2 Let    m˜ c = sup f dμ + g dν : f ∈ L1 (μ), g ∈ L1 (ν), f + g ≤ c , X

Y

then Mc = m˜ c . The same holds if c takes negative values, and satisfies the conditions of Corollary 20.2.5. Proof Clearly mc ≤ m˜ c ≤ Mc . The argument of Corollary 20.2.5 deals with the case where c takes negative values.

20.4 c-concavity The Kantorivich–Rubinstein theorem raises many questions. What are the 1 fundamental properties of an optimal transport plan?   Can we find f ∈ L (μ) 1 and g ∈ L (ν) such that f + g ≤ c, with X f dμ + y g dν = mc ? When is there a deterministic transport plan? In this, and the next two sections, we show that we can use a cost function to introduce some geometric ideas, related to the Legendre transform, and to concavity. It is an unfortunate necessity that although the parallel with the Legendre transform is close, we need to consider infima rather than suprema. Further, we consider infima of sequences of lower semi-continuous

306

Optimal Transportation

functions, which, although Borel measurable, need be neither upper nor lower semi-continuous. Suppose that (X, d) and (Y, ρ) are complete Polish metric spaces, that μ and ν are Borel probability measures on X and Y respectively, and that c is a lower semi-continuous cost function on X × Y. Let (cn )∞ n=1 be an approximating sequence in BL(X × Y), increasing pointwise to c. If f is a proper function on X and y ∈ Y, we set f c (y) = inf{c(x, y) − f (x) : x ∈ X}. f c is the c-transform of f ; it takes values in [−∞, ∞). We shall suppose that f c is not identically −∞. If c is continuous, then f c is upper semicontinuous. If c is lower semi-continuous, then, if (cn )∞ n=1 is an approximating sequence, each f cn is upper semi-continuous, and f c = limn→∞ f cn , so that f c is a Borel measurable function on Y, taking values in [−∞, ∞]. Similarly, if g is a function on Y and x ∈ X, we define gc (x) = inf{c(x, y) − g(y) : y ∈ Y}. We say that f is c-concave if there exists a function g on Y such that f = gc . Proposition 20.4.1 Suppose that (X, τ ) and (Y, ρ) are complete Polish metric spaces and that c is a cost function on X × Y. If f is a proper function on X and g is a proper function on Y then f cc ≥ f , and gccc = gc . Proof f cc (x) = inf(c(x, y) − f c (y)) y

˜ y) − f (x))) ˜ = inf(c(x, y) − inf(c(x, x˜

y

˜ y) + f (x)) ˜ = inf sup(c(x, y) − c(x, y

x˜

≥ inf(c(x, y) − c(x, y) + f (x)) = f (x). y

Also gccc = (gc )cc ⊇ gc and gccc = (gcc )c ⊆ gc .

Thus f is c-concave if and only if f = f cc . If f is c-concave, we set ˜ y) − f (x) ˜ : x˜ ∈ X}}. ∂ c f = {(x, y) : c(x, y) − f (x) = inf{c(x, ∂ c f is the c-superdifferential of f ; it may be empty. In general, f (x) + f c (y) ≤ c(x, y), and ∂ c f = {(x, y) ∈ X × Y : f (x) + f c (y) = c(x, y)}.

20.4 c-concavity

307

As an example, which we shall consider further later, suppose that X = Y = H, a separable Hilbert space, and that μ˜ and ν˜ are Borel measures, on X and Y respectively, for which   x2 dμ(x) y2 dν˜ (y) = 2. ˜ = H

.2

H

.2

dμ˜ and ν = dν˜ . Then μ ∈ P(X) and ν ∈ P(Y). Let Let μ = c(x, y) = − x, y. Then c is a continuous cost function on X ×Y, since c(x, y)+ 1 1 1 2 2 2 2 x + 2 y = 2 x − y ≥ 0. If f is a function on X and y ∈ Y, then 1 2

1 2

f c (y) = inf (− x, y − f (x)) = − sup(x, y + f (x)) = −(−f )∗ . x∈X

x∈X

It therefore follows that f cc = f − , the upper concave envelope of f . Thus ∂ c f is the superdifferential of f . When the cost function is continuous, we can say more. Proposition 20.4.2 Suppose that (X, τ ) and (Y, σ ) are Polish spaces and that c is a continuous cost function. If f is c-concave then it is upper semi-continuous, and ∂ c f is a closed subset of X × Y. Proof Since f is the infimum of a set of continuous functions, it is upper semicontinuous. c Suppose that (xn , yn )∞ n=1 is a sequence in ∂ f converging to (x, y). By picking a subsequence if necessary, we can suppose that f (xn ) → l, where l ≤ f (x). If z ∈ X then cn (xn , yn ) − f (xn ) = f c (yn ) ≤ c(z, yn ) − f (z), so that f (x) ≥ l ≥ c(x, y) + f (z) − c(z, y). Since this holds for all z ∈ X, f (x) ≥ c(x, y) − f c (y), and (x, y) ∈ ∂ c f . One special but important case arises when X = Y and c is a lower semicontinuous metric on X (which need not necessarily define the topology τ ). If (X, c) is a metric space, and f is a proper function on X, we define the lower L-convex envelope fc,L of f to be sup{g : g is L-Lipschitz, g ≤ f }. Theorem 20.4.3 Suppose that X = Y and that c is a lower semi-continuous metric on X (which need not necessarily define the topology τ ). If f is a proper function on X then f c = (−f )c,1 , the lower 1-Lipschitz envelope of −f . Proof First, since (−f )c,1 (y) ≤ (−f )c,1 (x) + c(x, y) ≤ −f (x) + c(x, y), (−f )c,1 (x) ≤ f c . Secondly, putting x = y, f c ≤ −f . Thirdly, if x1 , x2 ∈ X, then c(x1 , y) − f (y) ≤ c(x2 , y) − f (y) + c(x1 , x2 ), from which it follows that

308

Optimal Transportation

f c (x1 ) ≤ f c (x2 ) + c(x1 , x2 ), so that f c is 1-Lipschitz with respect to c. Putting these three statements together, the result follows. Corollary 20.4.4 If f is c-concave, it is 1-Lipschitz with respect to c, and f c = −f . Proof For if f = gc then f = (−g)c,1 so that f is 1-Lipschitz and so therefore is −f . Thus fc = (−f )c,1 = −f .

20.5 c-cyclical Monotonicity We now consider subsets of X × Y on which a cost function behaves well. Let us consider the example at the end of Section 19.2 again. The final inequality suggests the following definition. Suppose that X and Y are sets and that c is a function on X × Y taking values in (−∞, ∞]. A subset  of X × Y is said to be c-cyclically monotone if whenever (xi , yi ) ∈  for 1 ≤ i ≤ n then n 

c(xi , yi ) ≤

i=1

n 

c(xi , yτ (i) ) for any τ ∈ n .

i=1

Thus in the example, the set  = {( j, σ ( j)) : 1 ≤ j ≤ n} is a c-cyclically monotone set. The next result explains the terminology that is used. Proposition 20.5.1 A set  is c-cyclically monotone if and only if whenever n ∈ N and (xi , yi ) ∈  for 1 ≤ i ≤ n then, setting xn+1 = x1 , n  i=1

c(xi , yi ) ≤

n 

c(xi+1 , yi ).

i=1

Proof For any permutation of {1, . . . , n} can be written as the product of disjoint cycles. We now see how c-concavity and c-cyclical monotonicity are related. We need another definition. Suppose that X and Y are Polish spaces and that c is a cost function on X × Y. A subset  of X × Y is said to be strongly c-monotone if there exists a c-concave function f on X such that f (x) + f c (y) = c(x, y) for all (x, y) ∈ . That is,  ⊆ ∂ c ( f ). Theorem 20.5.2 (R¨uschendorf’s theorem) Suppose that (X, τ ) and (Y, ρ) are Polish spaces, and that c is a cost function on X × Y. If  is a c-cyclically monotone subset of X × Y then γ is strongly c-monotone.

20.5 c-cyclical Monotonicity

309

Proof Since f (x) + f c (y) ≤ c(x, y) for (x, y) ∈ X × Y, we must find f for ˜ y) ˜ ∈ X × Y and that which f (x) + f c (y) ≥ c(x, y) for (x, y) ∈ . Suppose that (x, (x, y) ∈ . First, pick a base-point (x0 , y0 ) ∈ . If zn = ((x1 , y1 ), . . . , (xn , yn )) ∈  n , let xn+1 = x0 , and let ˜ = h(zn )(x)

n−1  (c(xi+1 , yi ) − c(xi , yi )) + c(x, ˜ yn ) − c(xn , yn ) i=0

n  = (c(xi+1 , yi ) − c(xi , yi )) + c(x, ˜ yn ) − c(x0 , yn ). i=0

In particular, h(zn )(x0 ) ≥ 0, and if z1 = ((x0 , y0 )), then h(z1 )(x0 ) = 0. Now suppose that (x, y) ∈ . Let (xn+1 , yn+1 ) = (x, y), and let z+ n = ((x1 , y1 ), . . . , (xn , yn ), (x, y)) ∈  n+1 . Then ˜ = h(z+ n )(x)

n  (c(xi+1 , yi ) − c(xi , yi )) + c(x, ˜ y) − c(x, y) i=0

=

n−1 

(c(xi+1 , yi ) − c(xi , yi )) + c(x, yn ) − c(xn , yn ) + c(x, ˜ y) − c(x, y)

i=0

˜ y) − c(x, y). = h(zn )(x) + c(x, Now let j(x) = inf{h(zn )(x) : n ∈ N, zn ∈  n } ∈ [−∞, ∞). Note that j(x0 ) = 0. Then j(x) ˜ ≤ inf{h(z+ ˜ : n ∈ N, zn ∈  n }, so that n )(x) j(x) ˜ ≤ j(x) + c(x, ˜ y) − c(x, y). Putting x˜ = x0 , we see that j(x) ≥ c(x, y) − c(x0 , y), so that if (x, y) ∈  then g(x) is real-valued. Then c(x, y) − j(x) ≤ inf{c(x, ˜ y) − j(x) ˜ : x˜ ∈ X} = jc (y), so that j(x) + jc (y) ≥ c(x, y). Now let f = jcc . Then f is c-concave, f c = jc , and c(x, y) ≥ f (x) + f c (y) ≥ j(x) + jc (x) ≥ c(x, y), so that f (x) + f c (y) = c(x, y). The function f is called a maximal Kantorovich potential for the set . If c is continuous, then the function j is upper semi-continuous, but if c is only lower semi-continuous it might not be Borel measurable (it is the infimum of an uncountable set of lower semi-continuous functions). On the other hand, since f is c-concave, it is Borel measurable.

310

Optimal Transportation

20.6 Optimal Transport Plans Revisited We now show how c-cyclic monotonocity and c-concavity relate to optimal transport plans. A transport plan π is said to be c-cyclically monotone if there is a c-cyclically monotone Borel subset  of X × Y with π(γ ) = 1, and is said to be strongly c-monotone if there is a strongly c-monotone Borel set  with π() = 1. Theorem 20.6.1 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that c is a cost function on X × Y, that μ and ν are Borel probability measures on X and Y respectively, and that π ∈ μ,ν . Then the following are equivalent. (i) π is optimal. (ii) π is c-cyclically monotone. (iii) π is strongly c-monotone. Proof Suppose that π is optimal. Let A = { f (x) + g(y) : f ∈ L1 (μ), g ∈ L1 (ν), f + g ≤ c}. ∞ By Corollary 20.3.2, there exists a sequence ( fk + gk )k=1 in A such that X×Y c(x, y) − ( fk (x) + gk (y)) dπ(x, y) → 0 as k → ∞. By extracting a subsequence if necessary, there exists a Borel measurable subset , with π() = 1, such that fk (x) + gk (y) → c(x, y) as k → ∞, for (x, y) ∈ . Suppose that (xi , yi ) ∈ , for 1 ≤ i ≤ n, and that τ ∈ n . Then if k ∈ N, n 

c(xi , yτ (i) ) ≥

i=1

n n n    ( fk (xi ) + gk (yi )) = fk (xi ) + gk (yτ (i) ) i=1

=

n  i=1

i=1

fk (xi ) +

n 

gk (yi ) →

i=1

n 

i=1

c(xi , yi ) as k → ∞.

i=1

Thus π is c-cyclically monotone. R¨uschendorfs theorem shows that a c-cyclically monotone measure is strongly c-monotone. Suppose that π is strongly c-monotone, and that  is a strongly c-monotone Borel subset of X × Y with π() = 1. Let f be a c-concave function such that  ⊆ ∂ c ( f ). Let An = {x : |f (x)| ≤ n} and let gn = IAn .f . Similarly, let Bn = {y : |f c (y)| ≤ n} and let hn = IBn .f c . Then gn ∈ L∞ (μ), hn ∈ L∞ (ν) and     gn dν + hn dν =  c dπ → c dπ X

as n → ∞. Thus

 X×Y

Y

 (An ×Bn )

c dπ = Mc , and π is optimal.

When c is continuous, we can say more.

X×Y

20.6 Optimal Transport Plans Revisited

311

Theorem 20.6.2 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that c is a continuous cost function on X ×Y, that μ and ν are Borel probability measures on X and Y respectively, and that π ∈ μ,ν . Then the following are equivalent. (i) π is optimal. (ii) supp(π ) is c-cyclically monotone. (iii) supp(π ) is strongly c-monotone. If so, if f is a c-concave function on X such that f + f c = c on supp(π ), then ( f , f c ) is continuous on supp(π ). Proof It follows from the previous theorem and R¨uschendorfs theorem that it is enough to show that if π is optimal, then supp(π ) is a c-cyclically monotone set. Suppose not. Then there exist distinct (x0 , y0 ), . . . , (xn , yn ) in supp(π ) such that, setting xn+1 = x0 , n 

c(xj , yj ) −

j=0

n 

c(xj+1 , yj ) = η > 0.

j=0

Since c is continuous, there exist closed neighbourhoods Wj = Uj × Vj for 0 ≤ j ≤ n such that if Wj = Uj+1 × Vj then |c(uj , vj ) − c(xj , yj )| < η/4(n + 1) for (uj , vj ) ∈ Wj and |c(uj+1 , vj ) − c(xj + 1j, yj )| < η/4(n + 1) for (uj+1 , vj ) ∈ Wj , for 0 ≤ j ≤ n. Since (xj , yj ) ∈ supp(π ), λj = π(Wj ) > 0. Let πj = (IWj /λj ) dπ . Then πj is a Borel probability measure on X × Y, with support in Wj . Let α =  (min0≤j≤n λj )/n + 1, and let σ = π − nj=0 απj . Then σ is a non-negative Borel measure on X × Y.  Let μj and νj be the marginals of πj , and let π  = σ + nj=0 αμj+1 ⊗ νj . Then π  is a Borel probability measure on X × Y, and π  ∈ μ,ν . Further, 

 c dπ −

X×Y

X×Y

c dπ = α

n   j=0

⎛ ≥ α⎝

Wj

n  j=0

contradicting the optimality of π.

c dμj − α

c(xj , yj ) −

n   j=0

n  j=0

Wj

c d(μj+1 ⊗ νj ) ⎞

c(xj+1 , yj ) − η/2⎠ ≥ αη/2,

312

Optimal Transportation

The function ( f , f c ) is upper semi-continuous on X × Y. But if (x, y) ∈ supp(π ) then ( f (x), f c (y)) = (c(x, y) − f c (y), c(x, y) − f (x)), so that ( f , f c ) is also lower semi-continuous on supp(π ). When the cost function c is continuous, there is a strongly c-monotone set which contains the support of every optimal measure. Theorem 20.6.3 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that c is a continuous cost function on X ×Y, that μ and ν are Borel probability measures  on X and Y respectively, and that there exists π ∈ μ,ν with X×Y c dπ < ∞. Then there exists a strongly c-monotone set  such that if π is an optimal measure then supp(π ) ⊆ . Proof Let  = ∪{supp(π ) : π optimal}. By R¨uschendorfs theorem, it is sufficient to show that  is c-cyclically monotone. Suppose that (x0 , y0 ), . . . , (xn , yn ) ∈ . There exist optimal measures π0 , . . . , πn such that (xj , yj ) ∈  supp(πj ), for 0 ≤ j ≤ n. Let π = ( nj=0 πj )/(n + 1). Then π is an optimal measure, and (xj , yj ) ∈ supp(π ), for 0 ≤ j ≤ n. Consequently, setting xn+1 =   x0 , nj=0 c(xj , yj ) ≤ nj=0 c(xj+1 , yj ), and  is c-cyclically monotone. Theorem 20.6.2 suggests another way of showing the existence of optimal transport plans, or constructing optimal transport plans, when the cost function c is continuous. Theorem 20.6.4 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that μ and ν are Borel probability measures on X and Y respectively, and that c is a continuous cost function on X × Y. Then there exists a c-cyclically monotone transport plan. Proof By the strong law of large numbers for empirical processes, there exist sequences  ∞ ∞  n n 1 1 δx i and νn = δy i μn = n n i=1

i=1

n=1

n=1

such that μn ⇒ μ and νn ⇒ ν. For each n there exists a c-cyclically monotone ∞ transport plan πn for the pair (μn , νn ). The sequences (μn )∞ n=1 and (νn )n=1 ∞ are uniformly tight, and so, by Proposition 20.2.3, is the sequence (πn )n=1 . It therefore follows from Prokhorov’s theorem that there is a subsequence (which we again denote by (πn )∞ n=1 ) which converges weakly to some π ∈ M(X × Y). If f ∈ Cb (X) then    f dμn = f (x) dπn (x, y) → f (x) dπ(x, y) as n → ∞. X

X×Y

X×Y

20.7 Approximation

313

    But X f dμn → X f dμ as n → ∞, so that X×Y f (x) dπ(x, y) = X f dμ. A similar property holds for Y, and so π ∈ μ,ν . Now fix k ∈ N. Let Fk be the set  k  (c(xi , yi ) − c(xi , yσ (i) )) ≤ 0, for σ ∈ k . ((x1 , y1 ), . . . , (xk , yk )) :



i=1

Since c is continuous, Fk is closed in (X × Y)n . Now (⊗ki=1 πn )(Fk ) = 1, and so (⊗ki=1 π )(Fk ) = 1. Thus, by the portmanteau theorem, supp(π )k ⊆ Fk . Since this holds for all k, π is a c-cyclically monotone transport plan.

20.7 Approximation In applications, it is often necessary (for example, when making calculations) to approximate the distributions μ and ν, and the cost function c. When the cost function is continuous, then we can use c-cyclic monotonicity to show that things work well. Theorem 20.7.1 Suppose that (X, τ ) and (Y, ρ) are Polish spaces, that μ ∈ P(X), ν ∈ P(Y) and that c is a continuous cost function. Suppose also that ∞ (μi )∞ i=1 is a sequence in P(X) such that μi ⇒ μ, that (νi )i=1 is a sequence in ∞ P(Y) such that νi ⇒ ν, that (ci )i=1 is a sequence of continuous cost functions such that sup{|ci (x, y) − c(x, y)| : (x, y) ∈ X × Y} → 0 as i → ∞  and that for each i there is an optimal transport plan πi with X×Y ci dπi ≤ m < ∞. Then there exists a subsequence (πik )∞ k=1 such that πik ⇒ π , an optimal transport plan for μ and ν. ∞ Proof The sequences (μi )∞ i=1 and (νi )i=1 are uniformly tight, from which it follows that the sequence (πi )∞ i=1 is also uniformly tight. By extracting a subsequence if necessary, we can therefore suppose that πi ⇒ π . ˜ = ⊗∞ Now let  = (X × Y)N , π˜ i = ⊗∞ j=1 πi and π j=1 π . By c-cyclic mono tonicity, for each i there exists Ci such that if (x, ˜ y) ˜ ∈ C˜ i then nj=1 ci (xj , yj ) ≤ n j=1 ci (xj , y(j+1)modn ), and πi (Ci ) = 1. Thus if

C(k) =

⎧ ⎨

(x, ˜ y) ˜ :

⎩

n  j=1

c(xj , yj ) ≤

n  j=1

⎫ ⎬

c(xj , y(j+1)modn ) + 1/k for n ∈ N , ⎭

314

Optimal Transportation

then Ci ⊆ C(k) for sufficiently large i. Since C(k) is closed, π(C(k) ) = 1. Thus (k) if C = ∩∞ k=1 C , π(C) = 1. But C is c-cyclically monotone, and so π is optimal. This theorem also gives a proof of the existence of optimal measures when the cost function c is continuous. Let μi be the ith empirical measure of μ and νi the ith empirical measure of ν, and let ci = c. Then there is certainly an optimal measure for each of the finite pairs of measures (μi , νi ), and the empirical law of large numbers ensures that μi ⇒ μ and νi ⇒ ν.

21 Wasserstein Metrics

21.1 The Wasserstein Metrics W p Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ). We consider Borel probability measures on X whose spread is controlled by the metric d. If 0 < p < ∞, let    Pp (X) = μ ∈ P(X) : Ap (X) = dp (x, y) dμ(x) dμ(y) < ∞ . X×X

Proposition 21.1.1 Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ) and that 0 < p < ∞. The following are equivalent. (i) μ  ∈ Pp (X). (ii) X dp (x, x0 ) dμ(x) < ∞ for some x0 ∈ X. (iii) X dp (x, x0 ) dμ(x) < ∞ for every x0 ∈ X. Proof If 0 < p < 1, then dp is a metric on X uniformly equivalent to d. Consequently, we only need to consider the case where 1 ≤ p < ∞. Certainly (i) implies (ii). Suppose that (ii) holds and that x1 ∈ X. Since the function |x|p is convex, if a, b ≥ 0 then p  p p a+b (a + b) = 2 ≤ 2p−1 (ap + bp ). 2 Thus





 d (x, x1 ) dμ(x) ≤ 2 p

d (x, x0 ) dμ(x) + d (x0 , x1 ) ,

p−1

X

p

p

X

and (iii) holds. If (iii) holds, then     dp (x, y) dμ(x)dμ(y) ≤ 2p−1 dp (x, x0 ) dμ(x) + dp (x0 , y) dμ(y) X×X

< ∞,

X

so that (i) holds. 315

X

316

Wasserstein Metrics

Theorem 21.1.2 Suppose that (X, d) is a lower semi-continuous metric on a Polish space (X, τ ) and that 1 < p < ∞. If μ, ν ∈ P(X), let   1/p p Wp (μ, ν) = inf d dπ : π ∈ μ×ν . X×X

Then the restriction of Wp to Pp (X) × Pp (X) is a metric on Pp (X). As we have seen, the infimum is attained. Proof Clearly Wp (μ, ν) = Wp (ν, μ). Let us show that Wp (μ, ν) = 0 if and only if μ = ν. If μ = ν, then there exists a compact set K such that μ(K) = ν(K) + 2 > ν(K). By upper continuity, there exists δ > 0 such that ν(Kδ ) < ν(K) + . Thus if π ∈ μ,ν then π(X × (X \ Kδ ) > , so that   p d dπ ≥ dp dπ ≥ δ p . X×X

K×(K\Kδ )

> 0. Conversely, if μ = ν then, setting π = GI∗ (μ), Thus Wp (μ, ν) ≥ where I(x) = x, X×X dp dπ = 0, so that Wp (μ, ν) = 0. Next we prove the triangle inequality. Suppose that μ, ν, π ∈ Pp (X). There exist α ∈ μ,ν and β ∈ ν,π such that  1/p  1/p Wp (μ, ν) = dp dα and Wp (ν, π ) = dp dβ . δ. 1/p

X×X

X×X

Let γ ∈ P(X × X × X) satisfy the conclusions of the gluing lemma (Theorem 16.11.1), and let γ1,3 be the marginal distribution on the first and third factors. Then, using H¨older’s inequality,  1/p p Wp (μ, π ) ≤ d(x, z) dγ1,3 (x, z) X×X  1/p = d(x, z)p dγ (x, y, z) X×X×X  1/p ≤ (d(x, y) + d(y, z))p dγ (x, y, z) X×X×X  1/p ≤ d(x, y)p dγ (x, y, z) X×X×X  1/p + d(y, z)p dγ (x, y, z) X×X×X  1/p  1/p = d(x, y)p dα(x, y) + d(y, z)p dβ(y, z) X×X

= Wp (μ, ν) + Wp (ν, π ).

X×X

21.2 The Wasserstein Metric W1

317

The metric Wp is called the Wasserstein p-metric. Proposition 21.1.3 Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ). If 1 ≤ p < q < ∞ and μ, ν ∈ Pq (X), then μ, ν ∈ Pp (X) and Wp (μ, ν) ≤ Wq (μ, ν). Proof It follows from H¨older’s inequality and Proposition 21.1.1 that μ, ν ∈ Pp (X). Let π be an optimal measure for dp in (μ, ν). By H¨older’s inequality again,  1/p  1/q p q Wp (μ, ν) ≤ d dπ ≤ d dπ = Wq (μ, ν). X×X

X×X

We can extend this to the case where 0 < p < 1. Suppose that (X, d) is a Polish metric space and that 0 < p < 1. Then dp is a strictly subadditive metric on X which is uniformly equivalent to d. If μ, ν ∈ p (X), we set   p d dπ : π ∈ μ,ν . Wp (μ, ν) = inf X×X

Then Wp is a metric on p (X). Let Lipp (X) be the space of functions on (p)

X which are Lipschitz for the metric dp and if f ∈ Lipp (X), let pL (f ) = sup{|f (x) − f (y)|/dp (x, y) : x = y}. Then Wp (μ, ν) = γp (μ, ν), where & &   & & (p) γp (μ, ν) = sup && f dμ − f dν && : f ∈ Lipp (X), pL (f ) ≤ 1 . X

X

The metric Wp is again called the Wasserstein p-metric.

21.2 The Wasserstein Metric W 1 Proposition 21.2.1 Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ). If μ ∈ P1 (X) then LipX ⊆ L1 (μ). Proof Suppose that f ∈ LipX and that x0 ∈ X. Then   |f | dμ ≤ |f (x) − f (x0 )| + |f (x0 )| dμ(x) X

X

 d(x, x0 ) dμ(x) + |f (x0 )| < ∞.

≤L X

318

Wasserstein Metrics

Theorem 21.2.2 Suppose that (X, τ ) is a Polish space, that c is a lower semicontinuous metric on X, that μ, ν are Borel probability measures on X and that  is a Borel subset of X × X. Then  is c-strictly monotone if and only if there exists a function f on X, 1-Lipschitz for the metric c, such that c(x, y) = f (x) − f (y) for (x, y) ∈ . Proof It follows from Corollary 2.8.5 that f c = (−f )c,1 , the lower 1-Lipschitz envelope of −f . Thus f is c-concave if and only if it is 1-Lipschitz for the metric c, and, if so then f c = −f . Thus the result follows from R¨uschendorf’s theorem (Theorem 20.5.2). If μ, ν ∈ P1 (X), let

& &   & & γ (μ, ν) = sup && f dμ − f dν && : f ∈ Lip(X), pL (f ) ≤ 1 . X

X

Corollary 21.2.3 If μ, ν ∈ P1 (X) then W1 (μ, ν) = γ (μ, ν). Then γ is a metric on P1 (X), and γ ≥ β|P1 (X) , so that the inclusion (P1 (X), γ ) → (P(X), w) is continuous. Corollary 21.2.4 Suppose that d is a bounded metric, bounded by M say, and that 1 < p < ∞. Then P(X) = P1 (X) = Pp (X), and the metrics β, W1 and Wp are uniformly equivalent metrics on P(X). Proof Since a Lipschitz function on X is bounded by M say, β and γ are uniformly equivalent. If π ∈ P(X × X) then 

 d dπ ≤ X×X

1/p dp dπ

1/p

 ≤ M 1−1/p

X×X

d dπ

,

X×X 1/p

from which it follows easily that W1 ≤ Wp ≤ M 1−1/p W1 , and W1 and Wp are uniformly equivalent.

21.3 W 1 Compactness Suppose that (X, τ ) is a Polish space, and that d is a lower semi-continuous metric on X. We now characterize the compact subsets of (P1 (X), W1 ). First, we consider convergent sequences. Theorem 21.3.1 Suppose that (X, τ ) is a Polish space, and that d is a lower semi-continuous metric on X. Suppose that (μk )∞ k=0 is a sequence in P1 (X), and that x0 ∈ X. Then the following are equivalent.

21.3 W1 Compactness

319

(i) W1 (μk , μ0 ) → 0 as k → ∞. ∞ (ii) μk ⇒ μ0 and (μ  k )k=0 is uniformly  integrable. (iii) μk ⇒ μ0 and X d(x, x0 ) dμk → X d(x, x0 ) dμ0 as k → ∞. Proof The equivalence of (ii) and (iii) follows from Theorem 18.8.2. Suppose that (i) holds. For each k ∈ N there exists an optimal πk ∈ μk ,μ0 . If x, y ∈ X, d(x, x0 ) ≤ d(y, x0 ) + d(x, y). Integrating with respect to πk , it follows that   d(x, x0 ) dμk (x) ≤ d(y, x0 ) dμ0 (y) + W1 (μk , μ0 ). X

Similarly,

X



 d(y, x0 ) dμ0 (y) ≤

X

d(x, x0 ) dμk (y) + W1 (μk , μ0 ), X

so that (iii) holds.  Suppose that (ii) holds and that  >+ 0. There exists R > 0 such that ˜ d(x,x0 )>R/2 d dμk (x) < /3, for k ∈ Z . Let d(x, y) = d(x, y) ∧ R and let ˜ W 1 be the corresponding metric on P(X). Then the W˜ 1 topology is the same as the w topology, so that W˜ 1 (μk , μ0 ) → 0 as k → ∞. If d(x, y) ≥ R then either d(x, x0 ) ≥ R/2 or d(y, x0 ) ≥ R/2. Thus d(x, y) ≤ d(x, y) ∧ R + d(x, y)I(d(x, x0 )≥R/2) + d(x, y)I(d(y, x0 )≥R/2) . ˜ Then Let πk be an optimal measure for the pair (μk , μ0 ), with cost d.  d dπk W1 (μk , μ0 ) ≤ X   d˜ dπk + ≤ d dμk (x) + d dμ0 X

(d(x, x0 )≥R/2)

(d(y, x0 )≥R/2)

≤ W˜ 1 (μk , μ0 ) + 2/3, <  for large enough k. Thus (ii) implies (i). Straightforward arguments now extend this to a characterization of compactness. Theorem 21.3.2 Suppose that (X, τ ) is a Polish space, and that d is a lower semi-continuous metric on X. Suppose that A ⊆ P1 (X). Then A is W1 compact if and only if it is w compact and uniformly integrable. Proof Suppose that A is W1 compact. Then it is w compact. Suppose if possible that it is not uniformly integrable. Then there is a sequence (μk )∞ k=1 in A ∞ with no uniformly integrable subsequence. But (μk )k=1 has a W1 convergent subsequence, which is uniformly integrable, by Theorem 21.3.1, giving a contradiction.

320

Wasserstein Metrics

If the conditions are satisfied, and (μk )∞ k=1 is a sequence in A, then it has a w convergent subsequence in A, and this converges in the metric W1 , by Theorem 21.3.1, so that A is W1 sequentially compact, and so is W1 compact.

21.4 W p Compactness Suppose that (X, τ ) is a Polish space, and that d is a lower semi-continuous metric on X. We now extend the results of the previous section to characterize the compact subsets of (Pp (X), Wp ) for 1 < p < ∞. Again, we first consider convergent sequences. Theorem 21.4.1 Suppose that (X, τ ) is a Polish space, and that d is a lower semi-continuous metric on X. Suppose that (μk )∞ k=0 is a sequence in Pp (X), where 1 < p < ∞, and that x0 ∈ X. Then the following are equivalent. (i) Wp (μk , μ0 ) → 0 as k → ∞. ∞ (ii) μk ⇒ μ0 and (μ  k )pk=0 is p-uniformly  integrable. (iii) μk ⇒ μ0 and X d (x, x0 ) dμk → X dp (x, x0 ) dμ0 as k → ∞. Proof Let φ(x) = dp (x, x0 ) and let φn = φn , for n ∈ N. Suppose that (i) holds. Then μk ⇒ μ0 , and so β(μk , μ0 ) → 0 as k → ∞. Thus   φ dμ0 = lim φn dμ0 n→∞ X X    = lim → ∞ lim φn dμk n k→∞ X  φ dμk . ≥ lim inf k→∞ X

To prove that (iii) holds, we must therefore show that   lim sup φ dμk ≤ φ dμ0 . k→∞ X

X

Suppose that  > 0. Let η = (1 + )1/p − 1 (so that (1 + η)p = 1 + ), and let C = (1 + 1/η)p . Suppose that a, b > 0. If β < ηa then (a + b)p ≤ ((1 + η)a)p = (1 + )ap , and if β > ηa then (a + b)p ≤ ((1 + 1/η)b)p = C bp , and so in any case (a + b)p ≤ 1 + ap + C bp . Thus if x, y ∈ X then φ(x) ≤ (d(y, x0 ) + d(x, y))p ≤ (1 + )φ(y) + C dp (x, y). Let πk be an optimal measure for μk and μ0 . Integrating with respect to μk , it follows that

21.4 Wp Compactness 

321

 φ dμk ≤ (1 + )

X

φ dμ0 + C Wp (μk , μ0 )p . X

Since Wp (μk , μ0 ) → 0 as k → ∞, we see that   φ dμk ≤ (1 + ) φ dμ0 , lim sup k→∞ X

X

and the result follows, since  is arbitrary. Now suppose that (ii) holds. Let d˜ = d ∧ 1, and let W˜ 1 and W˜ p be the corresponding metrics. It follows from Corollary 21.2.4 that W˜ p (μk , μ0 ) → 0 as k → ∞. Suppose that  > 0. There exists R > 1 such that  dp (x, x0 ) dμk < /6p+1 for k ∈ Z+ (d(x, x0 )≥R/2)

and there exists k0 such that (3R)p W˜ p (μk , μ0 ) < 2/3 for k ≥ k0 . If d(x, y) ≥ R and d(x, x0 ) ≥ d(y, x0 ), then d(x, x0 ) ≥ 2d(x, y), and a similar inequality holds if d(x, x0 ) < d(y, x0 ). Thus ˜ y) + 2d(x, x0 )I(d(x,x )≥R/2) + 2d(y, x0 )I(d(y, x )≥R/2) , d(x, y) ≤ Rd(x, 0 0 and so d(x, y)p ≤ 3p (Rp d˜p (x, y) + 2p dp (x, x0 )I(d(x, x0 )≥R/2 + 2p dp (y, x0 )I(d(y, x0 ) ≥ R/2 . Let πk be an optimal measure for μk and μ0 for the cost function d˜p . If k ≥ k0 , then, integrating with respect to π,  dp dπk Wp (μk , μ0 ) ≤ X×X  ≤ (3R)p W˜ p (μk , μ0 ) + 6p dp (x, x0 ) dμk (d(x, x0 )≥R/2)  + 6p dp (y, x0 ) dμ0 (d(y, x0 )≥R/2)

≤ 2/3 + /6 + /6 = . Thus (ii) implies (i). The next result follows from this, using the arguments used to prove Theorem 21.3.2. Theorem 21.4.2 Suppose that (X, τ ) is a Polish space, that d is a lower semicontinuous metric on X and that 1 < p < ∞. Suppose that A ⊆ Pp (X). Then A is Wp compact if and only if it is w compact and p-uniformly integrable.

322

Wasserstein Metrics

21.5 W p -Completeness Proposition 21.5.1 Suppose that d is a lower semi-continuous metric on a Polish  space (X, τ ), and that 0 < p < ∞. If ν ∈ Pp (X), μ ∈ P(X) and inf X×X dp dπ : π ∈ μ,ν } < ∞ then μ ∈ Pp (X). Proof Let π be an optimal measure in μ,ν . If x, y ∈ X, then dp (x, x0 ) ≤ 2p (dp (x, y) + dp (y, x0 ). Integrating with respect to π,   dp (x, x0 ) dμ(x) ≤ 2p inf π∈μ,ν

X

 dp dπ + X×X

dp (y, x0 ) dν(y) < ∞. X

Proposition 21.5.2 Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ), and that 0 < p < ∞. Suppose that (μn )∞ n=1 is a sequence in Pp (X), that ν ∈ Pp (X) and that μ ∈ P(X). Suppose that supn Wp (μn , ν) ≤ R < ∞ and that μn ⇒ μ. Then μ ∈ Pp (X), and Wp (μ, ν) ≤ R. Proof We use the Kantorovich–Rubinstein theorem. Let ρ be a complete metric on X which defines the topology, and let BL(X) be the space of bounded ρ-Lipschitz functions. Suppose that f (x) ∈ BL(X), g(y) ∈ BL(X) and f (x) + g(y) ≤ dp (x, y). Then      f dμ + g dν = lim f dμn + g dν ≤ sup Wpp (μn , ν) ≤ Rp . X



n→∞

X

Hence inf{ X×X that μ ∈ Pp (X).

dp dπ

X

n

X

: π ∈ μ,ν } ≤ R, and it follows from Proposition 21.5.1

Theorem 21.5.3 Suppose that d is a lower semi-continuous metric on a Polish space (X, τ ), and that 0 < p < ∞. (Pp (X), Wp ) is a complete metric space, for 0 < p < ∞. Proof Suppose that ν ∈ Pp (X) and that (μn )∞ n=1 is a sequence in Pp (X) such that Wp (μn , ν) ≤ r and such that μn ⇒ μ. By Proposition 21.5.2, μ ∈ Pp (X) and Wp (μ, ν) ≤ r. Thus Mr (ν) = {μ ∈ Pp (X) : Wp (μ, ν) ≤ r} is closed in (P(X), β), and the result follows from Theorem 2.4.5. Proposition 21.5.4 Suppose that (X, d) is a Polish metric space, that 0 < p < ∞ and that C is countable dense subset of X. Then the countable set  n  n   AC = λi δci : n ∈ N, λi ∈ Q, λi ≥ 0, λi = 1, ci ∈ C i=0

is Wp -dense in (Pp (X), Wp ).

i=1

21.6 The Mallows Distances

323

Proof Suppose that μ ∈ Pp (X) and that 0 <  < 1. Pick c0 ∈ C. There exists a compact subset K of X such that X d(x, c0 )p dμ < . There exists a finite partition D of K into non-empty Borel sets, each of diameter less than , and for each D ∈ D there exists cD ∈ C such that d(cD , x) <  for x ∈ D. Let  ν = μ(X \ K)δc0 + μ(D)δcD , D∈D

and let π = IX\K .dμ ⊗ δc0 +



ID .dμ ⊗ δcD .

D∈D

 Then π ∈ μ,ν , and X×X dp dπ <  p , so that Wp (μ, ν) < . We can approxi mate ν by a probability measure of the form λ0 δc0 + D∈D λD δcD , with rational coefficients λ0 and λD , which establishes the result. Thus the spaces (Pp (X), Wp ) are Polish spaces, and we can start analyzing functions and Borel measures on them.

21.6 The Mallows Distances We now consider the Wasserstein distances between the distributions of random variables. Suppose that μ1 and μ2 are Borel probability measures on Polish spaces (X1 , τ1 ) and (X2 , τ2 ) respectively. Suppose that (E, .) is a Banach space, that 1 ≤ p < ∞ and that f1 ∈ Lp (μ1 ; E) and f2 ∈ Lp (μ2 ; E). Let C(f1 , f2 ) be the collection of all pairs (f˜1 , f˜2 ) in Lp (μ; E), for some Borel probability measure μ on some Polish space (X, τ ), for which f1 and f˜1 have the same distribution and f2 and f˜2 have the same distribution. The Mallows p distance dp (f1 , f2 ) between f1 and f2 is then defined as dp (f1 , f2 ) = inf{f˜1 − f˜2 p : (f˜1 , f˜2 ) ∈ C(f1 , f2 )}. This gives a measure of the difference of the distributions of f1 and f2 . We can express the Mallows p-distance in terms of the Wasserstein metric Wp . Theorem 21.6.1 Suppose that f1 and f2 are as above. Let ν1 = f1 dμ1 and let ν2 = f2 dμ2 , and let cp the cost function cp (x, y) = x − yp on E × E. Then dp (f1 , f2 ) = Wp (ν1 , ν2 ). Let f˜1 (x, y) = x and let f˜2 (x, y) = y. If π ∈ ν1 ,ν2 , then f˜1 ∈ Lp (π ), and f˜1 has the same distribution ν1 as f1 ; similar properties are enjoyed by f˜2 . Further,   f˜1 − f˜2 p dπ = cp dπ , X×X

E×E

from which it follows that dp (f1 , f2 ) ≤ Wp (ν1 , ν2 ).

324

Wasserstein Metrics

On the other hand, suppose that (f˜1 , f˜2 ) ∈ C(f1 , f2 ), where f˜1 , f˜2 ∈ Lp (X, B, μ). Let f˜ = (f˜1 , f˜2 ), and let π = f˜∗ (μ). Then π ∈ ν1 ,ν2 , and    f˜1 − f˜2 p dμ = cp dπ , X

E×E

from which it follows that Wp (ν1 , ν2 ) ≤ dp (f1 , f2 ). Corollary 21.6.2 There exist ( f˜1 , f˜2 ) ∈ C( f1 , f2 ) for which dp (f1 , f2 ) =  f˜1 − f˜2 p . We have the following extension to the central limit theorem (Theorem 18.7.2). Theorem 21.6.3 (The central limit theorem, II) Suppose that μ is a probability on a Polish space (X, τ ), that f ∈ L2 (μ), that X f dμ = 0 and that measure √ 2 X f dμ = 1. Let cn = (f1 + · · · + fn )/ n, where f1 , . . . , fn are n independent copies of f . Then d2 (cn , γ ) → 0 as n → ∞, where γ is a Gaussian random variable γ with mean 0 and variance 1, and (cn )n∈N is 2-uniformly integrable. Proof Theorem 18.7.2 shows that cn → γ in distribution as n → ∞. Let νn be the distribution of cn and let νγ be the distribution of γ . Since cn 2 = γ 2 , it follows from Theorem 21.4.1 that d2 (cn , γ ) = W2 (νn , νγ ) → 0 as n → ∞ and that the sequence (νn )n∈N is 2-uniformly integrable – that is, the sequence (cn )n∈N is 2-uniformly integrable.

22 Some Examples

22.1 Strictly Subadditive Metric Cost Functions A metric d is strictly subadditive if d(x, z) < d(x, y) + d(y, z) whenever y is not equal to x or z. For example, if b is a strictly concave non-negative realvalued function on [0, ∞) for which b(0) = 0 (such as b(t) = tp for some 0 < p < 1), and (X, d) is a metric space, then the function b ◦ d is a strictly subadditive metric uniformly equivalent to d. We show that if a cost function is a lower semi-continuous strictly subadditive metric on a Polish space (X, τ ), and μ and ν are Borel probability measures on X, then an optimal measure π on X × X fixes mass common to μ and ν. We begin with an easy lemma about measures on X. Lemma 22.1.1 Suppose that μ and ν are Borel probability measures on a Polish space (X, τ ) and that π ∈ μ,ν . Let D = {(x, x) : x ∈ X} be the diagonal in X × X, let πD = ID dπ , so that if A is a Borel subset of X × X, then πD (A) = π(A ∩ D), and let β = p∗ (πD ), where p : X × X → X is the projection onto the first co-ordinate. Then β ≤ μ ∧ ν. Proof Since p|D is a homeomorphism of D onto X, if A is a Borel subset of X, then β(A) = πD (A × X) ≤ π(A × X) = μ(A). Similarly, β(A) ≤ ν(A), and so β ≤ μ ∧ ν. Theorem 22.1.2 Suppose that d is a strictly subadditive lower semicontinuous metric on a Polish space (X, τ ), that μ, ν are Borel probability measures on X and that π ∈ μ,ν has d-monotone support. Let πD = ID dπ (so that πd (A) = π(A ∩ D)), and let β = p∗ (πD ), where p : X × X → X is the projection onto the first co-ordinate. Then β = μ ∧ ν. Thus if μo = μ − β and νo = ν − β then μo and νo are mutually singular. 325

326

Some Examples

Proof We use the notation of the lemma. If A is a Borel set in X × X, let πo (A) = π(A \ D), so that π = πo + πD , and μ = μ0 + β, and similarly ν = νo + β. Let S = supp(π ) \ D, and let U = p(S). Since S is a σ -compact subset of X × X, U is a Borel subset of X, and μo (U) = μo (X). Similarly if V = q(S) (where q is the projection onto the second co-ordinate), then V = q(S) is a Borel subset of X, and νo (V) = νo (X). We show that U ∩ V = ∅, which establishes the theorem. If not, there exists z ∈ U ∩ V, and so there exist x, y ∈ S such that (x, z) ∈ S and (z, y) ∈ S. But then by d-monotonicity, d(x, z) + d(z, y) ≤ d(x, y) + d(z, z) = d(x, y). Since x = z and y = z, this contradicts the strict subadditivity of d. Thus πo (U × V) = π0 (X × X), and πo is an optimal measure for μo and νo . We only need solve the Kantorovich problem for mutually singular measures.

22.2 The Real Line We now consider the case where X = Y = R. First we consider the case where the cost function c is of the form c(x, y) = b(x − y), where b is a proper strictly convex non-negative function on R. Suppose that μ and ν are Borel probability measures on R, with distribution functions Fμ and Fν respectively. We suppose that inf{ R2 c dπ : π ∈ μ,ν } < ∞ so that there exists an optimal measure π . In this case, we show that π is unique, and that if μ is atom-free then it is a deterministic transport plan. Suppose that C is a c-cyclically monotone subset of R2 . First we show that if (x0 , y0 ) and (x1 , y1 ) are in C then (x1 − x0 )(y1 − y0 ) ≥ 0. Suppose not. Then without loss of generality we can suppose that x0 < x1 = x0 + s and y0 > y1 = y0 − t. Let α = x0 − y0 . Then [c(x0 , y0 ) + c(x1 , y1 )] − [c(x0 , y1 ) + c(x1 , y0 )] = [b(α + s + t) − b(α + s)] − [b(α + t) − b(α)] > 0, by strict convexity, giving a contradiction. If (x, y) ∈ R2 , let NW(x, y) = {(x , y ) : x < x, y > y} and let SE(x, y) = {(x , y ) : x > x, y < y}. It then follows that if (x, y) ∈ C then NW(x, y) ∩ C = SE(x, y) ∩ C = ∅; the open sets NW(x, y) and SE(x, y) are disjoint from C. For each x ∈ supp(μ) let yx = inf{y : Fν (y) > Fμ (x−)} and yx = inf{y : Fν (y) > Fμ (x)} and let C = {(x, y) : x ∈ supp(μ), yx ≤ y ≤ yx }. It then follows that if π is an optimal transport plan, supp(π ) ⊆ C. If μ is atomfree, then C is the graph of an increasing function T, and T is the unique

22.3 The Quadratic Cost Function

327

deterministic transport plan. If μ has an atom at x, again the transport plan is unique, but it is not deterministic; ν determines how the mass at x is distributed in the interval [yx , yx ]. Y

X

Note that supp(π ) depends on μ and ν, but not on the cost function. Suppose that b is a proper convex non-negative function on R, and that b(x) → ∞ as |x| → ∞. Let c be the corresponding cost function. Let h(t) = |t|+e−|t| −1. Let bn = b+h/n and let cn (x, y) = bn (x−y). Then bn is strictly convex, and so C is cn -cyclically monotone. But cn → c pointwise, and so C is c-cyclically monotone. Thus there exists an optimal transport plan π with supp(π ) ⊆ C. Note however that when c(x, y) = |x − y| then this solution is generally not unique. Next, suppose that d(x, y) = b(|x − y|), where b is a non-negative strictly concave function on [0, ∞) for which b(0) = 0 (for example, d(x, y) = |x−y|p , with 0 < p < 1), so that d is a strictly subadditive metric on R. Suppose also that μ and ν are probability measures on R for which sup(supp(μ)) ≤ inf(supp(ν)). Then a similar argument shows that there is a unique transport plan π. But here there is order reversal; for example, if μ is atom-free, there is a unique optimal deterministic transport plan T which is a decreasing function.

22.3 The Quadratic Cost Function We now consider the case where X = Y = H, a Hilbert space or Euclidean space, and where c(x, y) = 12 n2 (x−y), where n2 (x) = x2 . We consider Borel

328

Some Examples

 probability  measures μ and ν on H which satisfy K2 (μ) = X n2 dμ < ∞ and K2 (ν) = Y n2 dν < ∞. Thus if π ∈ μ,ν then   c dπ = 12 n2 (x − y) dπ(x, y) X×Y X×Y  n2 (x) + n2 (y) dπ ≤ K2 (μ) + K2 (ν) < ∞, ≤ X×Y

so that an optimal transport plan π exists. If π ∈ μ,ν then   1 1 c dπ = 2 n2 (x) − x, y + 2 n2 (y) dπ(x, y) X×Y X×Y  x, y dπ . = 12 (K2 (μ)) + K2 (ν) − X×Y

Let c(x, ˜ y) = − x, y. Then π is an optimal transport plan for the cost c if and only if it is an optimal transport plan for the cost c. ˜ Further, Mc = 1 (K (μ) + K (ν)) + M , where M is the optimal cost for the cost c. ˜ The cost 2 2 c˜ c˜ 2 c˜ is not non-negative, but we can apply Corollary 20.2.5. If f is a function on X and y ∈ Y, then f c˜ (y) = inf (− x, y − f (x)) = − sup(x, y + f (x)) = f † (y), x∈X

x∈X

f c˜

so that is an upper semi-continuous concave function on X, and f c˜c˜ is the upper semi-continuous concave envelope of f . Applying this to the original problem, we obtain the following result. Theorem 22.3.1 Suppose that X = Y = H, a Hilbert space or Euclidean space, and that c(x, y) = 12 n2 (x − y), where n2 (x) = x2 . Suppose  that μ and (μ) = ν are Borel probability measures on H which satisfy K 2 X n2 dμ < ∞  and K2 (ν) = Y n2 dν < ∞. Then there exists an optimal transport plan π . Further, there exist upper semi-continuous concave functions f and g on H, ≤ c(x, with g = f † , such that ( 12 x2 +f (x))+ 12 (y2 +g(y))   y), with equality on supp(π ), so that Mc = 12 (K2 (μ) + K2 (ν)) + X f dμ + Y g dν. Let us now consider the case where H is a d-dimensional Euclidean space, and μ is absolutely continuous with respect to Lebesgue measure λd . In particular, this implies that supp(μ) is not contained in a (d − 1)-dimensional affine subspace of E, and so C = (supp(μ)) has a non-empty interior. Since f and g are finite on supp(μ), this implies that they are finite on Cint . Further, μ(∂C) = 0. Now it follows from Rademacher’s theorem (Theorem 17.7.1) that f is differentiable μ-almost everywhere on Cint . Let D = {x ∈ f : f is differentiable at x}. If x ∈ D ∩ supp(μ), let T(x) = ∇f (x), so that

22.4 The Monge Problem on Rd

329

{T(x)} = ∂fx . Consequently π is the push-forward measure GT∗ (μ), and ν = T∗ (μ). Thus T solves the Monge problem. Let us show that the solution is unique. Suppose that π˜ is an optimal measure, and that f˜ and g˜ are the corresponding concave functions. Let T˜ be the corresponding transport mapping. Then   (f + g) dπ = ( f˜ + g) ˜ dπ˜ X×Y X×Y    f˜ dμ + g˜ dν = ( f˜ + g) ˜ dπ , = X

so that



Y

X×Y

( f˜(x) + g(y) ˜ + x, y) dπ(x, y) = 0;

X×Y

that is to say, 

( f˜(x) + g(∇f ˜ (x)) + x, ∇f (x)) dμ(x) = 0.

X

But the integrand is non-positive μ-almost everywhere, so that f˜(x) + g(∇f ˜ (x)) + x, ∇f (x) = 0 μ-almost everywhere. Thus ∇f ∈ ∂ f˜ μ-almost everywhere, and so it follows from Rademacher’s theorem that ∇f = ∇ f˜ μ-almost everywhere. Conse˜ quently, T = T.

22.4 The Monge Problem on Rd Suppose that c(x, y) = h(x − y) is a continuous cost function on Rd and that μ, ν ∈ P(Rd ). Can we find conditions on h, μ and ν which ensure that there is a solution to the Monge problem? If so, is the solution unique?  We shall always suppose that there exists π ∈ μ,ν with Rd ×Rd c dπ < ∞, and will also suppose that μ = g.dλd is absolutely continuous with respect to Lebesgue measure λd . The function h need be neither convex nor concave; we need to extend the notion of sub- and superdifferentiability to it. If x, y ∈∈ Rd , we say that (x, y) ∈ ∂ ∨ h, the subdifferential of h, if there exists a non-positive function s in a neighbourhood N of x such that if r(v) = h(x + v) − h(x) − v, y − s(x + v), for x + v ∈ N,

330

Some Examples

then |r(v)|/ v → 0 as v → 0. We set ∂ ∨ h(x) = {y : (x, y) ∈ ∂ ∨ h}. Similarly, we say that (x, y) ∈ ∂ ∧ h, the superdifferential of h, if there exists a non-negative function s in a neighbourhood N of x such that if r(v) = h(x + v) − h(x) − v, y − s(x + v), for x + v ∈ N, then |r(v)|/ v → 0 as v → 0. We set ∂ ∧ h(x) = {y : (x, y) ∈ ∂ ∧ h}. By Theorem 20.6.1, there exists a strongly c-monotone subset  of Rd × supp(ν) and a corresponding maximal Kantorovich potential f on Rd such that if π is an optimal solution to the Kantorovich problem then supp(π ) ⊆ ; f is c-concave, and  ⊆ ∂c ( f ), where ∂c (h) = {(x, y) : c(x, y) − f (x) = inf (c(x , y) − f (x ))}.  x ∈X

We shall see that properties of f and of h∗ are used to ensure that the Monge problem has a solution. In particular, we shall need f to be Fr´echet differentiable on a large subset of Rd , so that we can apply the following theorem. Theorem 22.4.1 Suppose that c, h,  and f are as above, that (x, y) ∈ ∂ c ( f ) and that f is Fr´echet differentiable at x. If h is convex, then ∇f (x) ∈ ∂h(x − y). If h is concave in a neighbourhood of x, then h is differentiable at x − y and ∇f (x) = ∇h(x − y). Proof Suppose that h is convex. If k ∈ Rd then f (x + k) − f (x) − ∇fx (k) = r(k), where r(k) = o(k). Since (x, y) ∈ ∂ c ( f ), h(x − y) − f (x) = f c (y) = inf (h(u − y) − f (u)) u∈Rd

≤ h(x + k − y) − f (x + k) = h(x + k − y) − f (x) − ∇fx (k) − r(k) so that h(x + k − y) − h(x − y) − ∇fx (k) ≥ r(k). Thus if h is convex then φ ∈ ∂h(x − y), and if h is concave in a neighbourhood of x, then h is differentiable at x − y and ∇f (x) = ∇h(x − y). Theorem 22.4.2 Suppose that c(x, y) = h(x − y) is a translation invariant continuous cost function on Rd , that μ, ν ∈ P(Rd ), that μ = g dλ is absolutely

22.5 Strictly Convex Translation Invariant Costs on Rd

331

continuous with respect to Lebesgue measure λd and that there exists π ∈ μ,ν  with Rd ×Rd c dπ < ∞. Let  ⊆ Rd ×supp(ν) and let f satisfy Theorem 20.6.1. Suppose that there exists a Borel measurable A in Rd , with μA) = 1, such that if x ∈ A then f is Fr´echet differentiable at x and ∂ c f (x) = ∅. Suppose also that there exists a continuous mapping J : Rd → Rd such that if (x, y) ∈ ∂h then J(y) = x. Then there exists a unique measurable s : Rd → Rd such that if π is any optimal measure in μ,ν then π = G(s)∗ (μ). Hence s∗ (μ) = ν and    c(x, s(x)) dμ(x) = inf c dπ : π ∈ μ,ν . X

Rd ×Rd

Then s is the unique solution to the Monge problem. Proof Since ∂f (x) = lim n( f (x + ei /n) − f (x)), n→∞ ∂xi ∇f is a Borel measurable mapping of A into Rd . Now let  x − J(∇f (x)) for x ∈ A s(x) = 0 otherwise. Since J is continuous, s is Borel measurable. By Theorem 22.4.1, if (x, y) ∈ ∂ c f (x) then ∇f (x) ∈ ∇h(x − y); that is, (x − y, ∇f (x)) ∈ ∂h. Hence J(∇f (x)) = x − y and y = s(x). We show that if π is an optimal measure then G(s)∗ (μ) = π (so that s∗ (μ) = ν). We must show that if B and C are Borel subsets of Rd , then π(B × C) = μ(B × s−1 (C)). Let S = {(x, y) : x ∈ A, (x, y) ∈ ∂ c f }; if (x, y) ∈ S then y = s(x), and so (B × C) ∩ S = ((B ∩ s−1 (C)) × Rd ) ∩ S. Then π(S) = 1, so that π(B × C) = π((B × C) ∩ S) = π((B ∩ s−1 (C)) × Rd ) = μ(B ∩ s−1 (C)) = G(s)∗ (B × C). It remains to show that s is unique. If s is another optimal transport mapping, then π  = G(s )∗ (μ) is an optimal transport plan, and so π  = G(s)∗ (μ). But then s = s , by Proposition 20.1.1.

22.5 Strictly Convex Translation Invariant Costs on Rd We consider a translation invariant cost c(x, y) = h(x−y) on Rd which satisfies three conditions.

332

Some Examples

(H1) h is a strictly convex real-valued non-negative function on Rd , and h(0) = 0. Then h is a locally Lipschitz function on Rd . If x ∈ Rd \ {0}, we set H(x) = {y ∈ Rd : h(y) ≤ h(x)}. Then H(x) is a closed strictly convex body in Rd , and H(x)int = {y ∈ Rd : h(y) < h(x)}. We need to put a rotundity condition on h. An element φ of Rd is tangent to H(x) at x if φ = 1 and x, φ = sup{y, φ : y ∈ H(x)}. If so, then, since h is strictly convex, y, φ < x, φ for y ∈ H(x) \ {x}. If φ is tangent to H(x) at x, let n(x) be the outward normal unit vector to H(x) at x. (H2) Suppose that r > 0 and 0 < θ < π . Then there exists R > 0 such that if x > R then the cone K = {y : y − x cos θ/2 ≤ x − y, n(x) ≤ r} with vertex at x and height r is contained in H(x). x + n (x) x

q

h(y ) = h (x) n(y ) = n (x) 0 n(y ) = n (x) – r

One specially important case occurs when h(x) = ψ(x), where ψ is a strictly convex non-negative function on [0, ∞) for which ψ(0) = 0. (For example, ψ(t) = tp for some p > 1, or ψ(t) = t log(t + 1).) In this case, H(x) is the ball {y ∈ Rd : y ≤ x}, and so H(2) is satisfied. We also need h to grow more than linearly. (H3) h(x)/ x → ∞ as x → ∞. Theorem 22.5.1 Suppose that c(x, y) = h(x − y), where h satisfies (H1)–(H3), and that f is a proper c-concave function on Rd . Let K = (f ). Then f is locally bounded on K int (so that K int ⊆ f ⊆ K). Proof Suppose that f is not locally bounded at x. We show that there exists a unit vector φ such that if y, φ < x, φ then f (y) = −∞. Thus f ⊆ {y : y, φ ≥ x, φ}, and so K ⊆ {y : y, φ ≥ x, φ} and K int ⊆ {y : y, φ > x, φ}. Hence x ∈ K int .

22.5 Strictly Convex Translation Invariant Costs on Rd

333

If f (x) > −∞ then, since f is upper semi-continuous, it is bounded above in a neighbourhood of x, and so it cannot be bounded below in any neighbourhood of x. There therefore exists a sequence (xn )∞ n=1 such that xn → x as n → ∞ and f (xn ) < −n for each n ∈ N. If f (x) = −∞, set xn = x for all n ∈ N. c There exists a sequence (yn )∞ n=1 such that c(xn , yn )−f (yn ) < −n for n ∈ N. Then f (x) ≤ c(x, yn ) − f c (yn ) = (c(x, yn ) − c(xn , yn )) + (c(xn , yn ) − f c (yn )) ≤ c(x, yn ) − c(xn , yn ) − n. Since c is continuous, it follows that yn  → ∞ as n → ∞. Let zn = xn − yn . Then zn  → ∞ as n → ∞. For each n there exists φn , tangent to H(zn ), such that the cone {y ∈ Rd : φn (y) ≥ 12 φn (zn ), φn (zn − y) ≥ β(h(zn )} is contained in H(zn ). Extracting a subsequence if necessary, we can suppose that φn → φ, say, as n → ∞. Suppose that w, φ < x, φ. Since xn → x and φn → φ as n → ∞, xn , φn  → x, y and w, φn  → w, φ as n → ∞. Thus w, φn  < xn , φn  for sufficiently large n. For such n, w − yn , φn  < xn − yn , φn  = zn , φn  . Now a straightforward geometric argument shows that w − yn ∈ H(zn ) for all sufficiently large n, as well. For such n, c(w, yn ) = h(w − yn ) < h(zn ) = c(xn , yn ), so that f (w) ≤ c(w, yn ) − f c (yn ) < c(xn , yn ) − f c (yn ) < −n. Consequently, f (w) = −∞. Theorem 22.5.2 Suppose that c(x, y) = h(x − y), where h satisfies (H1)–(H3), and that f is a c-concave function on Rd which is locally bounded on an open subset U of Rd . Then ∂ c f (x) is non-empty for all x ∈ U, and ∂ c f (U) is locally bounded in Rd × Rd . Proof Suppose that x ∈ U and that V = Nδ (x) is an open neighbourhood of x in U for which R = supy∈V |f (y)| < ∞. We need a preliminary lemma. Lemma 22.5.3 Suppose that (xn )∞ n=1 is a sequence in Nδ/2 (x) and that d × R which satisfies is a sequence in R (yn , θn )∞ n=1

334

Some Examples

(i) f (z) ≤ c(z, yn ) − θn for z ∈ Rd , and (ii) c(xn , yn ) − θn < R for all n ∈ N. Then (yn )∞ n=1 is a bounded sequence. Proof Suppose not. Since (xn )∞ n=1 is a bounded sequence, by extracting a subsequence if necessary we can assume that if vn = xn − yn then vn  > n, for n ∈ N. Let αn = δ/2 vn , so that αn → 0 as n → ∞. Let wn = xn − αn vn , so that wn ∈ Vand wn − yn = (1 − αn )vn . Then −R ≤ f (wn ) ≤ c(wn , yn ) − θn = h(wn − yn ) − θn = h((1 − αn )vn ) − θn while h(vn ) = h(xn − yn ) = c(xn , yn ) ≤ R + θn . Consequently, h(vn ) − h((1 − αn )vn ) ≤ 2R. We now use the convexity of h. Suppose that φn ∈ ∂h(1−αn )vn . Then / 0 δ vn α  , φn h(vn ) − h((1 − αn )vn ) ≥ n vn , φn = 2 vn  and h((1 − αn )vn ) = h((1 − αn )vn ) − h(0) ≤ (1 − αn )vn , φn  , so that

0 / h((1 − αn )vn ) 4R vn . ≤ , φn ≤ vn  (1 − αn ) vn  δ

But this contradicts (H3), since (1 − αn )vn  → ∞ as n → ∞. We now prove the theorem. First, let xn = x for all n ∈ N. Since f (x) < n, d c there exists a sequence (yn )∞ n=1 ∈ R such that c(xn , yn ) − f (yn ) → f (x) as c n → ∞ and c(x, yn ) − f (yn ) < R for all n. Since f is c-concave, f (z) ≤ c(z, yn ) − f c (yn ), and so the conditions of the lemma are satisfied. Thus the sequence (yn )∞ n=1 is bounded, and by extracting a subsequence if necessary, we can suppose that yn → y as n → ∞, for some y ∈ Rd . Then c(x, yn ) → c(x, y) as n → ∞ and so f c (yn ) → c(x, y) − f (x) as n → ∞. But f c is upper semicontinuous, and so f c (y) ≥ c(x, y) − f (x). Thus y ∈ ∂ c f (x), and so ∂ c f (x) is non-empty. Suppose now that (xn , yn ) is a sequence in ∂ c ( f (V)). Setting θn = f c (yn ), then, since f (xn ) = c(xn , yn ) − θn < R, the conditions of the lemma are satisc fied. Thus (yn )∞ n=1 is a bounded sequence, and so ∂ ( f (V)) is bounded. Corollary 22.5.4 f is locally Lipschitz on U. Proof If x ∈ U, there exists a neighbourhood V of X in U on which ∂ c f (V) is bounded, and so there exists S > 0 for which y ≤ S if (x, y) ∈ ∂ c f (U). Thus if x ∈ V and y ∈ ∂ c f (x) then

22.5 Strictly Convex Translation Invariant Costs on Rd

335

f (x) = c(x, y) − f c (y) = inf{c(x, z) − f c (z) : z ≤ S}. Now there exists L > 0 such that if z ≤ S then the function x → c(x, z) is L-Lipschitz on V. It therefore follows from Proposition 2.8.2 that f is L-Lipschitz on V. Theorem 22.5.5 Suppose that c(x, y) = h(x − y), where h satisfies (H1)–(H3), and that f : Rd → [−∞, ∞) is a proper function. If (x, y) ∈ ∂ c ( f ), then ∂f (x) ⊆ ∂h(x − y). Proof If f (x) = −∞ then ∂f (x) = ∅, and so the result is trivially true. Otherwise, suppose that φ ∈ ∂f (x), so that if k ∈ Rd then f (x + k) − f (x) − k, φ = s(k) + r(k), where s(k) ≥ 0 and r(k) = o(k). Since (x, y) ∈ ∂ c ( f ), h(x − y) − f (x) = f c (y) = inf (h(u − y) − f (u)) U∈Rd

≤ h(x + k − y) − f (x + k) = h(x + k − y) − f (x) − k, φ − s(k) − r(k) so that h(x + k − y) − h(x − y) − k, φ ≥ s(k) + r(k). Thus φ ∈ ∂h(x − y). We now show that when the cost function c on Rd satisfies the preceding conditions, and when μ = g.dλ ∈ P(Rd ) is absolutely continuous with respect to Lebesgue measure, then there exists a unique solution to the Monge problem. Theorem 22.5.6 Suppose that c(x, y) = h(x − y) is a cost function on Rd , where h satisfies (H1)–(H3), that μ, ν ∈ P(Rd ) and that μ = g.dλ ∈ P(Rd ) is absolutely continuous with  respect to Lebesgue measure. Suppose also that there exists π ∈ μ,ν with Rd ×Rd c dπ < ∞. Then there exists a unique measurable s : Rd → Rd such that s∗ (μ) = ν and such that Rd c(x, s(x)) dμ(x) = infπ∈μ,ν Rd ×Rd c dπ. Proof By Theorem 20.6.1, there exists a strongly c-monotone subset  in Rd × supp(ν) such that if π is an optimal measure, then supp(π ) ⊆ , and there exists a maximal Kantorovich potential f ; f is c-concave, and  ⊆ ∂ c ( f ). Let K = (f ). Then K ⊇ p1 (), where p1 is the projection of Rd × supp(ν) onto the first component, and so μ(K) = 1. Consequently λ(K) > 0, so that

336

Some Examples

λ(K int ) > 0 and λ(∂K) = 0. By Corollary 22.5.4, f is locally Lipschitz on K int , and so, by Rademacher’s theorem (Theorem 17.7.1), f is differentiable on a measurable subset C of K int of full λ measure: λ(A) = λ(K int ). Thus μ(C) = 1. Suppose that π is an optimal measure, and let B = p1 (supp(π )). Then B is a σ -compact subset of Rd , and so it is Borel measurable, and μ(B) = 1. If x ∈ B, then ∂ c f (x) = ∅. Let A = B ∩ C. Then A satisfies the conditions of Theorem 22.4.2. Further, h∗ is continuously differentiable on Rd . If (x, y) ∈ ∂h then (y, x) ∈ ∂h∗ . Thus if we set J = ∇h∗ the conditions of Theorem 22.4.2 are satisfied and the result follows.

22.6 Some Strictly Concave Translation–Invariant Costs on Rd We now consider a translation invariant cost function c on Rd of the form c(x, y) = h(x − y) = l(x − y), where l is a function on R which satisfies (L) l(t) = −∞ for t < 0, l is continuous on [0, ∞), l(0) = 0, l(t) is positive and strictly concave on (0, ∞), l(t)/t → ∞ as t 0, and l(t)/t → 0 as t → ∞. An important example is the case where l(t) = tp for t ≥ 0, where 0 < p < 1. The function l is differentiable at all but countably many points of (0, ∞), and ∂ ∧ l(t) = ∅ for all t ∈ (0, ∞). What about h? Theorem 22.6.1 Suppose that h(x) = l(x), where l satisfies (L), is a function on Rd . Then (x, y) ∈ ∂ ∧ h(x) if and only if x = 0 and y = sx/ x, where s ∈ ∂ ∧ l(x). Proof Clearly ∂ ∧ h(0) = ∅. Suppose that x = 0 and that (x , s) ∈ ∂ ∧ l. Since l is concave, s > 0 and l(x + t) ≤ l(x + st), for t ∈ R. Suppose that v < x and that 0 < t < 1. Then x + tv2 = x2 + 2 x, v t + v2 t2 , 1

so that since the function s → s 2 is concave on (0, ∞), x + tv ≤ x +

v2 2 x, v t . t+ x 2 x

It therefore follows, since l is concave, that l(x + v) ≤ s x, v / x, so that h is superdifferentiable at x and (x, sx/ x) ∈ ∂ ∧ h.

22.6 Some Strictly Concave Translation–Invariant Costs on Rd

337

Conversely, suppose that (x, y) ∈ ∂ ∧ h, so that h(x + v) ≤ h(x) + v, y, for small v. Let y = αx + v, where x, v = 0. Then if t is small and positive, x ≤ x − tv, so that h(x) ≤ h(x − tv) ≤ h(x) − t v2 + o(t), and so v = 0. Thus y = αx. Since h(x + tx) ≤ h(x) + αt x2 , it follows that y = sx/ x, with s ∈ ∂ ∧ l(x). Corollary 22.6.2 h is differentiable λ-almost everywhere. p2 (∂ ∧ h) = Rd \ {0} (where p2 (x, y) = y), and there exists a continuous mapping J : Rd \ {0} → Rd \ {0} such that if (x, y) ∈ ∂ ∧ h then J(y) = x. Proof h is differentiable except at 0 and on a countable union of spheres. The other results follow from the form of ∂ ∧ h, and the facts that p2 (∂ ∧ l) = (0, ∞) and that there is a continuous mapping j : (0, ∞) → (0, ∞) such that if (t, u) ∈ ∂ ∧ l then j(u) = t. Next, let us consider the concave Legendre transform l† of l. Proposition 22.6.3 Suppose that l is a non-negative strictly convex function on [0, ∞) which satisfies (L). Then l† (t) = ∞ for t < 0, and l† is a differentiable negative strictly increasing function on (0, ∞), for which l† (t) → −∞ as t 0 and l† (t) → 0 as t → ∞. Proof Certainly l† (t) = ∞ for t < 0. Since l is strictly concave, if t > 0 there is a unique xt > 0 at which tx − l(x) attains its infimum, and then f † (t) = txt − f (xt ) < 0. If 0 < t1 < t2 then xt1 > xt2 and f † (t1 ) < f † (t2 ). Further, 0, and xt → ∞ as t   f (xt ) † f (t) = xt t − → −∞ as t 0. xt Similarly xt → 0 as t → ∞, so that f (xt ) → 0 as t → ∞. Since −f (xt ) ≤ f † (xt ) < 0, f † (t) → 0 as t → ∞. Finally, since xt is unique, ∂ ∧ l† (t) = {xt }, so that l† is differentiable on (0, ∞), with derivative xt . We now consider the Monge problem. Since c is a strictly subadditive metric on Rd , it follows from the remarks in Section 22.1 that it is enough to consider the case where μ and ν are mutually singular. In fact we shall require a little more; we suppose that μ(supp(ν)) = 0. We need a definition. Suppose that f is a function from an open subset U of Rd into [−∞, ∞), and that x ∈ U. Then f is locally semi-concave at x if there exist δ > 0 and α > 0 such that Nδ (x) ⊆ U and the function y → f (y)−α y2 is concave on Nδ (x). Theorem 22.6.4 Suppose that c(x, y) = h(x − y) = l(x − y) is a cost function on Rd × V, where l satisfies (L), and where V is a closed subset

338

Some Examples

of Rd . If f is a c-concave function on Rd , then f is locally semi-concave on U = Rd \ V. Proof Suppose that x ∈ V and that N2δ (x) ⊆ U. Choose 0 < s < δ such that l is differentiable at s. Let α = l (s)/2s and let ls (t) = l(t) − αt2 for t ≥ s. Then ls is a strictly concave function on (0, ∞), and ls (s) = 0, so that ls is strictly decreasing on [s, ∞). Set ls (t) = ls (s) for 0 ≤ t < s, and let hs (x) = ls (x) for x ∈ Rd . If x, y ∈ Rd and 0 < θ < 1 then (1 − θ )x + θ y ≤ (1 − θ ) x + θ y , so that hs ((1 − θ )x + θ y) ≥ ls ((1 − θ ) x + θ y) ≥ (1 − θ )hs (x) + θ hs (y). Thus hs is a concave function on Rd . Suppose now that z ∈ Nδ (x), so that d(z, V) > s. Then f (z) − α z2 = inf (h(z − y) − f c (y)) − α z2 y∈V

= inf (h (z − y) + α z − y2 − f c (y)) − α z2 y∈V

= inf (h (z − y) − 2α z, y) + α y2 − f c (y). y∈V

But (h (z − y) − 2α z, y) + α y2 − f c (y) is a concave function on Nδ (x), for each y ∈ V, and so f (z) − α z2 is concave on Nδ (x). Corollary 22.6.5 f is differentiable λ-almost everywhere on U. Theorem 22.6.6 Suppose that c(x, y) = h(x − y) = l(x − y) is a cost function on Rd , where l satisfies (L), that μ, ν ∈ P(Rd ), that μ = g.dλ ∈ P(Rd ) is absolutely continuous with respect to Lebesgue measure  and that μ(supp(ν)) = 0. Suppose also that there exists π ∈ μ,ν with Rd ×Rd d c dπ < ∞. Then there exists a unique measurable s : Rd →   R such that s∗ (μ) = ν and such that Rd c(x, s(x) dμ(x)s(x)) = infπ∈μ,ν Rd ×Rd c dπ . Proof Let V = supp(ν) and let U = Rd \ V . Then there exists a subset A of U such that f is differentiable on A and λ(U \ A) = 0. Then μ(A) = 1. Let J be the function of Corollary 22.6.2. Then the conditions of Theorem 22.4.2 are satisfied, and so the result follows.

Further Reading

Topological and Metric Spaces All the results of Chapter 1 are proved in [G II]. An entertaining account of the anomalies of topology can be found in [SS]. The Br´ezis–Browder lemma was proved in [BB], and Ekeland’s variational principle and its uses is discussed in [E] and [P I].

Banach Spaces and Hilbert Space [Bo] contains an excellent account of basic linear analysis. The two volumes of [LT] contain more advanced material concerning the classical Banach spaces, and [W] considers applications of Banach space theory to other areas of analysis. [Y] gives an elementary account of Hilbert space theory.

Uniform Spaces Many algebraic objects, such as topological groups, have natural uniform structures. We follow the notation of [J].

C`adl`ag Functions The Skorohod topology is discussed in [Bi I].

Convexity Although it concentrates on finite-dimensional convex sets, [R] is the standard reference for convexity. Other details can be found in [P I], [Ss] and [S]. 339

340

Further Reading

Measure Theory Proofs of the results of Chapters 15 and 17 can be found in [G III]. The strong law of large numbers can be found in [B II]. [H] is a standard work on measure theory. Borel measures and the convergence of measures are dealt with in [D] and [Bi II]. There are many excellent books on Fourier transforms; I like [Duo].

Haar Measure The account given here is derived from the report [Pe] by Pederson.

Choquet Theory [P II] is a very good source for this, and so is [S].

Optimal Transportation The two fundamental references are the tomes [V I] and [V II]. Further information about the strictly convex and strictly concave costs considered in Sections 22.5 and 22.6 is given in [GMcC].

References [Bi I] [Bi II] [Bo] [BB]

[D] [Duo] [E] [GMcC] [G II] [G III] [H]

Patrick Billingsley, Convergence of Probability Measures, John Wiley, 1968. Patrick Billingsley, Probability and Measure, John Wiley, 1979. B´ela Bollob´as, Linear Analysis, Cambridge Mathematical Textbooks, 1990. H. Br´ezis and F.E. Browder, A General Principle on Ordered Sets in Nonlinear Functional Analysis, Advances in Mathematics 21 (1976), 355– 364. R.M. Dudley, Real Analysis and Probability, Cambridge University Press, 2005. Javier Duoandikoetxea, Fourier Analysis, AMS Graduate Studies in Mathematics 29, 2001. Ivar Ekeland, Nonconvex minimization problems, Bulletin of the American Mathematical Society (New Series) (1979), 443–474. Wilfrid Gangbo and Robert J. McCann, The Geometry of Optimal Transportation, Acta Mathematica 177 (1966), 113–161. D.J.H. Garling, A Course in Mathematical Analysis, Volume II, Cambridge University Press, 2013. D.J.H. Garling, A Course in Mathematical Analysis, Volume III, Cambridge University Press, 2014. Paul R. Halmos, Measure Theory, Van Nostrand Reinhold, 1969.

Further Reading

341

[J] I.M. James, Introduction to Uniform Spaces, L.M.S. Lecture Note Series 144 1990. [LT] Joram Lindenstrauss and Lior Tzafriri, Classical Banach Spaces, Volumes I and II, Springer-Verlag, 1977 and 1979. [Pe] Gert K. Pedersen, The Existence and Uniqueness of the Haar Integral on a Locally Compact Topological Group, Report, Preprint, University of Copenhagen, 2000. [P I] R.R. Phelps, Convex Functions, Monotone Operators and Differentiability, Springer Lecture Notes in Mathematics 1364, 1993. [P II] R.R. Phelps, Lecture Notes on Choquet’s Theorem, Springer Lecture Notes in Mathematics 1757, 2008. [R] R. Tyrrell Rockafellar, Convex Analysis, Princeton University Press, 1972. [S] Barry Simon, Convexity: An Analytic Viewpoint, Cambridge Tracts in Mathematics 187, 2011. [Ss] Stephen Simons, From Hahn–Banach to Monotonicity, Springer Lecture Notes in Mathematics 1693, 2008. [SS] Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology, Dover Publications Inc., 1995. [V I] C´edric Villani, Topics in Optimal Transportation, American Mathematical Society, 2003. [V II] C´edric Villani, Optimal Transport, Old and New, Springer-Verlag, 2009. [W] P. Wojtaszczyk, Banach Spaces for Analysts, Cambridge Studies in Advanced Mathematics, 1991. [Y] N.J. Young, An Introduction to Hilbert Space, Cambridge University Press, 1988.

Index

(Lp , .p ), 207 Gδ set, 22, 38, 210 L1 (X, , μ), 198 M(X, ), 196 N-function, 203 complementary, 203 TV, 257 T1 − T4 spaces, 12 W1 compact, 318 Wp compact, 320 Wp -complete, 322 2 condition, 205 α-measurable, 221 σ -additive, 182 σ -compact, 17, 34, 48 σ -field, 179 generated by F , 180 σ -ring, 179 c-concave, 318 p-adic metric, 21 β metric, 266 -net, 39 minimal, 40 -subdifferential, 149 σ -additive, 212 abelian, 62 absolutely continuous, 201, 206, 251 absolutely convex, 84, 88 absorbent, 116 accumulation point, 10, 11

action, 66 continuous, 66, 237 left, 66 right, 66 transitive, 66, 237 additive, 212 adjoint, 108, 109 affine, 85, 128 affine homeomorphism, 155 Alexandroff’s theorem, 32 almost everywhere, 184 almost surely, 184 almost uniformly, 184 analytic set, 180 annihilator, 100 antilinear, 100, 108 Archimedean, 191 Arzel`a–Ascoli theorem, 46, 63, 267 atom, 243 Baire σ -field, 210 Baire space, 33 Baire’s category theorem, 33, 34, 39, 80, 91, 132, 142, 160 Banach limits, 114 Banach sequence space, 94 Banach space, 79, 128 Banach’s theorem, 124, 141, 156, 259, 268 Banach–Alaoglu theorem, 125 barycentre, 167 barycentric co-ordinates, 167 base for a uniformity, 56 Bernoulli sequence space, 23, 38, 43, 44, 62, 218

342

Index

Bessel’s inequality, 106 bidual, 115 bilinear, 91, 118 bilinearity, 97 bimonotone basis, 94 bipolar, 120 Birkhoff’s theorem, 160 Bishop–Phelps theorem, 149 Borel σ -field, 180 Borel measurable, 180, 293 Borel measure, 210 boundary, 10, 150, 167, 285, 287 bounded, 19, 80 σ (E, F), 122 bounded convergence, 232, 255 bounded Lipschitz, 82 bounded variation, 245 Br´ezis–Browder Lemma, 53 c`adl`ag function, 71 Cantor set, 24 Cauchy sequence, 24, 39, 77, 106 σ (E, F), 122 Cauchy–Schwarz inequality, 98, 200 central limit theorem, 271, 324 Choquet boundary, 288 Choquet ordering, 291 Choquet’s theorem, 284, 292 closed, 10 closed -neighbourhood, 19 closed graph theorem, 93 closed unit ball, 80 closed-regular, 211 closure, 10 co-ordinate projection, 10, 11, 28 compact, 41, 48, 59, 125, 155 countably, 41 Hausdorff, 54 sequentially, 41 compactification, 17 one-point, 17 complete, 24, 32, 74, 75, 77, 80 completely labelled, 169 completely regular, 13, 59, 64, 258 completion, 29, 80 concave, 27, 85 conditional expectation, 234 conditional probability, 234 continued fraction, 38

continuity downwards, 182 upwards, 182 continuity set, 261, 270 continuous, 11, 88 sequentially, 11 uniformly, 57 continuous linear functional, 119 continuous on the right, 71 contraction mapping, 162 convergence, 11 convergence in law, 258 convex, 84, 86, 119 body, 88 cover, 84 function, 83 strictly, 141 convex envelope, 128 convex cover closed, 86 convex function, 128, 130, 151, 152 regular, 129 convexity, 83 countably additive, 182 countably compact, 16 countably inductive, 53 counting measure, 208 cross-section mapping, 11, 28 cumulative distribution function, 244, 251 cyclically monotone, 152 cyclically monotone operator, 152 cylinder set, 218 rank, 23 Daneˇs’s drop theorem, 95 dense Gδ , 142, 146 dentability, 160 diameter, 19, 44, 158, 161, 168, 176 Dieudonn´e, 70 differentiable almost everywhere, 252 differential, 133 differential equation, 164 dilation, 293 Dini’s theorem, 19 Dirac measure, 257 directional derivative, 133, 254 disintegration, 231 disintegration theorem, 294 dissection, 71, 243 distance, 22

343

344

distribution, 183 dominated convergence theorem, 201 doubly stochatic matrices, 160 drop, 95 dual pair, 118 dual space, 91, 115 dyadic martingale, 218 effective domain, 50, 85 Egorov’s theorem, 184, 217 Ekeland’s variational principle, 149, 166 entourages, 56 envelope convex, 128 epigraph, 50, 151 strict, 50 episum, 96, 139, 142 equicontinuous, 46, 63 essentially bounded, 198 Euclidean space, 38, 101, 243, 327, 328 extend, 112–114 extension, 89 extension theorem lower semi-continuous functions, 53 extreme point, 157, 283, 284, 287 face, 157, 167 facet, 167 Fatou’s lemma, 204 Fenchel–Rockafeller duality, 148 Fenchel–Rockafeller theorem, 154 filter, 15, 56 finite intersection property, 15, 60 first Borel–Cantelli lemma, 182, 250 first category, 34 first countable, 12, 19 fixed point, 162, 170 fixed point theorem Brouwer’s, 170 Caristi’s, 166 Clarke’s, 166 Kakutani’s, 174, 236 Markov–Kakutani, 173 Ryll–Nardzewski, 175 Schauder’s, 171 Fourier transform, 271, 272 Fr´echet differentiable, 143, 330

Index

Fr´echet–Riesz representation theorem, 108 Fr´echet–Riesz theorem, 200 frontier, 10 Fr´echet smooth, 144 Fubini’s theorem, 230, 241, 254 fundamental theorem of calculus, 164 Gˆateaux derivative, 133 Gˆateaux differentiable, 133 Gˆateaux smooth, 144 gauge, 86, 137, 204, 207 Gaussian random variable, 272 general principle of convergence, 24 gluing lemma, 234, 316 grad, 143 gradient, 143 Gram–Schmidt orthonormalization, 105 graph, 11, 93 greedy algorithm, 248 group compact Hausdorff, 66 orthogonal, 62 unitary, 62 group action, 66 group-norm, 68 growth function, 276 H¨older’s inequality, 209, 317 Haar measure, 40, 176, 236, 244 left, 175 locally compact, 238 right, 175 Hahn–Banach theorem, 140, 223, 227, 235, 282, 286, 287 complex, 125 Hall’s marriage theorem, 40 Hausdorff, 59 uniformity, 64 Hausdorff space, 12 Helly space, 16, 22, 46 Helly–Bray theorem, 262 Hermitian operator, 109 Hermitian space, 38, 101 Hilbert cube, 10, 23, 24, 38, 43, 262 Hilbert space, 101, 116, 122, 207, 327 homeomorphism, 11, 60, 61, 73 uniform, 57 homogeneous space, 237 hypercube, 10 hyperplane, 117

Index

implicit function theorem Lipschitz, 165 in measure, 184 in probability, 184 independent copies, 271 indicator function, 22 inf-convolution, 139 inner product, 97 space, 97 usual, 98 inner-product space, 97 integral equation, 164 interior, 10, 88, 167 inversion, 62 inversion invariant, 79 irrational, 38 isolated point, 10 isometrically homogeneous, 238 isometry, 19, 22, 29, 63, 89, 108, 247 isomorphism theorem, 93 Jensen’s inequality, 282 Jordan decomposition, 196, 197 jump, 72 Kantorovich problem, 326 Kantorovich–Rubinstein theorem, 322 Klee’s theorem, 80 Krein–Mil’man theorem, 158, 226, 263 law, 183, 258 Lebesgue decomposition, 251, 266 Lebesgue decomposition theorem, 200 Lebesgue density theorem, 250, 254 Lebesgue differentiation theorem, 249 Lebesgue measure, 219, 222, 229, 244 Lebesgue–Stieltjes, 219 left derivative, 128 Legendre polynomials, 105 Legendre transform, 134 concave, 136, 337 limit on the left, 71 limit point, 10, 11 line segment, 167 linear functional, 85, 91 linear mapping bounded, 89 continuous, 89 linear operator, 88 Lipschitz, 74, 82, 88, 164 Lipschitz constant, 162

345

Lipschitz function, 35, 87, 143 Lipschitz mapping, 162 local oscillation, 52 locally bounded, 144 locally compact, 16, 34, 47, 226 locally finite, 214 locally homogeneous, 34 locally in measure, 186 locally Lipschitz, 131, 140, 144, 334 locally semi-concave, 337 lower L-Lipschitz envelope, 36 lower convex envelope, 135, 282 lower semi-continuous, 50, 54, 153, 260, 282 lower semi-continuous envelope, 52 lower semi-continuous function, 96 Lusin’s theorem, 216 Luxemburg norm, 204 Mallows distance, 323 mapping graph, 11 marginal distribution, 233 maximal, 117 maximal Kantorovich potential, 330, 335 maximal monotone, 153 Mazur’s theorem, 142, 146 McShane–Whitney theorem, 37, 267 meagre, 34 mean-value theorem, 163 measurable, 180 function, 181 measurable space, 180 measure, 182 σ -finite, 183 atomic, 253 continuous singular, 253 image, 183 maximal, 291 positive, 194 signed, 194 measure space, 182 finite, 182 mesh size, 243 metric, 18, 59, 77 associated, 18 Cantor, 24 discrete, 19 equivalent, 21 Euclidean, 19 left-invariant, 67 operator, 21

346

metric (cont.) right-invariant, 67 Skorohod, 73 subspace, 19 uniform, 19 uniform product, 28 usual, 19 metrizable, 21, 60, 123, 124, 143 metrizable topological groups, 67 Mil’man’s theorem, 159, 287, 289 minimal -net, 236 Minkowski functional, 86 Monge problem, 329, 337 monotone, 151 monotone basis, 94 monotone operator, 152 multilinear, 91 multiplication, 61 nearest point, 101 neighbourhood, 10 base, 10 punctured, 10 Neumann series, 61 non-contracting, 175 non-contraction, 176 norm, 79 dual, 91 equivalent, 89 operator, 90 uniformly equivalent, 89 normal, 13, 60, 67, 109 normed space, 79, 86, 88, 130 nowhere dense, 34 null set, 183 one point compactification, 48 open, 19, 57 open -neighbourhood, 19 open -neighbourhood of A, 35 open r-neighbourhood, 22 open cover, 15 open mapping theorem, 92 open sets, 9 open unit ball, 80 operator, 89 orbit, 237 order unit, 193 ordinary differential equation, 163 Orlicz norm, 205 Orlicz space, 203, 278

Index

orthogonal, 100 orthogonal group, 245 orthogonal isometry, 111 orthogonal projection, 110 orthogonal sequence, 104 orthonormal basis, 107 orthonormal sequence, 104 oscillation, 19 local, 20 outer measure, 221 parallelogram law, 99, 101 Parseval’s equation, 106 partial order, 117 partially ordered sets, 53 partition of unity, 42 peak point, 289 period 4, 273 permutation matrices, 160 petal, 95 point measure, 257 pointwise convergence, 124 polarity, 119 polarization formulae, 99 Polish space, 38, 47, 71, 78 portmanteau theorem, 260, 262, 270 positive, 110 positive definite, 97 positive homogeneity, 79 precompact, 42 Principle of Uniform Boundedness, 92, 131 probability, 182 projection, 103 Prokhorov metric, 269 Prokhorov’s theorem, 271 proper, 50, 85 proper function, 50 pseudometric, 18, 30, 58 push-forward, 224, 231 push-forward measure, 183, 244, 262, 276 Pythagoras’ theorem, 100, 103 quadratic cost, 327 quadrilateral inequality, 18, 30 R¨uschendorf’s theorem, 318 Rademacher’s theorem, 254, 328 radially open, 116 Radon action, 242 Radon measure, 214, 226

Index

Radon–Nikodym theorem, 199, 201, 203, 206, 241, 252 random variable, 182 rational, 39 reflexive, 56, 116, 125, 206 regular, 24 regular conditional probability, 234 regular space, 13 regularity, 210 relation, 56 relative boundary, 171 represent, 281 representation left regular, 66, 236 right regular, 66 retract, 43, 171, 173 retraction, 43, 101 Riemann–Stieltjes integral, 244 Riesz representation theorem, 222, 226, 235, 244, 259, 282, 285, 293 locally compact, 225, 241 Riesz space, 226 Riesz–Fischer theorem, 106 right derivative, 128 scaling homogeneous, 79 Schauder basis, 93 Schwartz space, 272 second category, 34 second countable, 12, 21, 24, 48, 211 self-adjoint, 109 semigroup, 175 seminorm, 79 separable, 12, 21, 48, 74, 94, 107, 124, 128 separating subspace, 285 separation, 116 separation theorem, 116, 117, 148, 288, 294 complex, 126 sequentially compact, 16 Shilov boundary, 288 simple function, 183 simplex, 167 fundamental n-, 167 singleton, 161, 167 singular measure, 250 skew-symmetric, 109 skew-symmetric bilinear form, 97 skew-symmetry, 97 Skorohod function, 71

Skorohod metric, 73 Skorohod topology, 73 slice, 161, 176 sliding hump, 290 smooth, 145 smoothness, 143 space metric, 18 Sperner mapping, 169 Sperner’s lemma, 168 spherical derivative, 247, 250 standard orthonormal basis, 172 state space, 286 step function, 43, 73 stochastic process, 71 Stone–Weierstrass theorem, 226, 268 strictly concave, 336 strictly convex, 141, 283, 284, 331 strictly non-negative, 195 strictly subadditive, 27, 325 strongly c-monotone, 335 subadditivity, 79 subcover, 15 subdifferential, 133, 140 sublinear functional, 85 extended, 85 support, 210, 225 support functional, 150 support point, 150 supremum, 85 surjective, 63 symmetric, 56, 61, 64, 88 symmetric operator, 109 symmetry, 79, 237 tail distribution function, 244 tangent, 332 tempered distribution, 272, 273 tetrahedron, 167 theorem of bipolars, 120, 223, 259 theorem of bounded convergence, 278 theorem of monotone convergence, 279 Tietze’s extension theorem, 25, 277 tight, 212 tightness, 210 Tonelli’s theorem, 229 topological group, 61 topological space, 9 topological vector space, 119 topologically complete, 31, 34

347

348

topology, 9 completely regular, 22 base, 10, 21 coarser, 11 countable product, 23 discrete, 9 finer, 11 metric, 19 normal, 22 product, 10, 118 quotient, 10 right half-open, 10 Skorohod, 73 stronger, 11 subspace, 10 trivial, 9 usual, 9 weak, 118, 120 weak*, 120 weaker, 11 total variation norm, 198 totally bounded, 39, 46, 86 translation invariant, 79, 209 triangle, 167 triangle inequality, 18, 316 triangulation, 167 barycentric, 167 Tychonoff’s theorem, 15, 125 Ulam’s theorem, 215, 264 ultrametric, 21 uniform algebra, 290 uniform homeomorphism, 27, 39 uniform integrability, 276 uniform space, 56 uniformity, 56 left, 64 metric, 57 right, 64

Index

uniformly continuous, 27, 35, 39, 43, 45, 81, 88, 251 uniformly convex, 121, 207 uniformly equicontinuous, 46 uniformly equivalent, 27 uniformly integrable, 319 uniformly tight, 264, 265, 271 unit sphere, 173 unitary, 110 upper L-Lipschitz envelope, 36 upper semi-continuous, 20, 50, 141 upper semi-continuous envelope, 52 Urysohn’s lemma, 13, 291 usual topology, 48 variation negative, 246 positive, 246 total, 246 variety, 157 proper support, 157 support, 157 vertex, 167 very regular, 130 vicinities, 56 Vitali covering, 249 von Neumann, 201 von Neumann’s theorem, 111 Wasserstein metric, 315 weak topology w, 258 weak type (1, 1), 248 weak* compact, 140, 143 Wiener’s lemma, 248 Young’s inequality, 203 Zorn’s lemma, 54