de Gruyter Studies in Mathematics 35 Editors: Carsten Carstensen · Nicola Fusco Niels Jacob · Karl-Hermann Neeb
de Gruyter Studies in Mathematics 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
Riemannian Geometry, 2nd rev. ed., Wilhelm P. A. Klingenberg Semimartingales, Michel Me´tivier Holomorphic Functions of Several Variables, Ludger Kaup and Burchard Kaup Spaces of Measures, Corneliu Constantinescu Knots, 2nd rev. and ext. ed., Gerhard Burde and Heiner Zieschang Ergodic Theorems, Ulrich Krengel Mathematical Theory of Statistics, Helmut Strasser Transformation Groups, Tammo tom Dieck Gibbs Measures and Phase Transitions, Hans-Otto Georgii Analyticity in Infinite Dimensional Spaces, Michel Herve´ Elementary Geometry in Hyperbolic Space, Werner Fenchel Transcendental Numbers, Andrei B. Shidlovskii Ordinary Differential Equations, Herbert Amann Dirichlet Forms and Analysis on Wiener Space, Nicolas Bouleau and Francis Hirsch Nevanlinna Theory and Complex Differential Equations, Ilpo Laine Rational Iteration, Norbert Steinmetz Korovkin-type Approximation Theory and its Applications, Francesco Altomare and Michele Campiti Quantum Invariants of Knots and 3-Manifolds, Vladimir G. Turaev Dirichlet Forms and Symmetric Markov Processes, Masatoshi Fukushima, Yoichi Oshima and Masayoshi Takeda Harmonic Analysis of Probability Measures on Hypergroups, Walter R. Bloom and Herbert Heyer Potential Theory on Infinite-Dimensional Abelian Groups, Alexander Bendikov Methods of Noncommutative Analysis, Vladimir E. Nazaikinskii, Victor E. Shatalov and Boris Yu. Sternin Probability Theory, Heinz Bauer Variational Methods for Potential Operator Equations, Jan Chabrowski The Structure of Compact Groups, 2nd rev. and aug. ed., Karl H. Hofmann and Sidney A. Morris Measure and Integration Theory, Heinz Bauer Stochastic Finance, 2nd rev. and ext. ed., Hans Föllmer and Alexander Schied Painleve´ Differential Equations in the Complex Plane, Valerii I. Gromak, Ilpo Laine and Shun Shimomura Discontinuous Groups of Isometries in the Hyperbolic Plane, Werner Fenchel and Jakob Nielsen The Reidemeister Torsion of 3-Manifolds, Liviu I. Nicolaescu Elliptic Curves, Susanne Schmitt and Horst G. Zimmer Circle-valued Morse Theory, Andrei V. Pajitnov Computer Arithmetic and Validity, Ulrich Kulisch Feynman-Kac-Type Theorems and Gibbs Measures on Path Space, Jo´zsef Lörinczi, Fumio Hiroshima and Volker Betz Integral Representation Theory, Jaroslas Lukesˇ, Jan Maly´, Ivan Netuka and Jirˇ´ı Spurny´ Introduction to Harmonic Analysis and Generalized Gelfand Pairs, Gerrit van Dijk Bernstein Functions, Rene´ Schilling, Renming Song and Zoran Vondracˇek
Jaroslav Lukesˇ Jan Maly´ Ivan Netuka Jirˇ´ı Spurny´
Integral Representation Theory Applications to Convexity, Banach Spaces and Potential Theory
≥
Walter de Gruyter Berlin · New York
Authors Jaroslav Lukesˇ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail:
[email protected]
Jan Maly´ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic and Department of Mathematics Faculty of Science J. E. Purkyneˇ University ˇ eske´ mla´dezˇe 8 C ´ stı´ nad Labem, Czech Republic 40096 U E-Mail:
[email protected]
Ivan Netuka Mathematical Institute Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail:
[email protected]
Jirˇ´ı Spurny´ Department of Mathematical Analysis Faculty of Mathematics and Physics Charles University Sokolovska´ 83 18675 Prague 8, Czech Republic E-Mail:
[email protected]
Series Editors Carsten Carstensen Department of Mathematics Humboldt University of Berlin Unter den Linden 6 10099 Berlin, Germany E-Mail:
[email protected]
Niels Jacob Department of Mathematics Swansea University Singleton Park Swansea SA2 8PP, Wales, United Kingdom E-Mail:
[email protected]
Nicola Fusco Dipartimento di Matematica Universita` di Napoli Frederico II Via Cintia 80126 Napoli, Italy E-Mail:
[email protected]
Karl-Hermann Neeb Department of Mathematics Technische Universität Darmstadt Schloßgartenstraße 7 64289 Darmstadt, Germany E-Mail:
[email protected]
Mathematics Subject Classification 2000: 46-02, 31-02, 52-02, 46A55, 52A07, 46B99, 31B05, 31A05, 31A25, 31B20, 31C05, 31D05, 35K05, 35K20, 28A05, 54H05 Keywords: Convex sets, Choquet theory, Banach spaces, descriptive set theory, measure theory, potential theory, Dirichlet problem
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ISBN 978-3-11-020320-2 쑔 Copyright 2010 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Printed in Germany. Cover design: Martin Zech, Bremen. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen.
Introduction
In many branches of mathematics, one encounters the question of how to reconstruct a convex set from information on its vertices. This idea successfully emerged as the Krein–Milman theorem for compact convex subsets of locally convex spaces since any such set has plenty of extreme points. For any point of a compact convex set, a reformulation of the Krein–Milman theorem provides a representing measure that is concentrated in some sense on the set of extreme points. The goal of our book is to present a more general approach to integral representation theory based upon a notion of a function space and apply the obtained results to the theory of convex sets, Banach spaces and potential theory. We point out that this approach is far from being new, but we hope that our exposition may be profitable both for students interested in the basics of integral representation theory as well as for more advanced readers. The former group could be attracted by a self-contained presentation of the Choquet theory, the latter by a substantial amount of results of fairly recent origin or appearing in a book form for the first time. We also try to incorporate more techniques from descriptive set theory into subject, which further supports our belief that the book will be worth reading even for those well acquainted with the monographs by E. M. Alfsen [5], R. R. Phelps [374], Z. Semadeni [414], L. Asimow and A. J. Ellis [24] or V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. Let us continue by looking briefly at the contents of the book. After a prologue on the Korovkin theorem, we present basic facts on the extremal structure of finitedimensional compact convex sets. Then we move on to infinite-dimensional spaces and prove the Krein–Milman theorem and several of its consequences. The second part of Chapter 2 studies the concept of measure convex and measure extremal sets. Chapter 3 is devoted to cornerstones of the Choquet theory of functions spaces such as the Choquet order and its properties and integral representation theorems due to G. Choquet and E. Bishop and K. de Leeuw. Even though the results are standard, the key limiting process is established by means of the Simons lemma, which allows us to present later on several of its applications. The chapter is finished by a discussion on deeper properties of the Choquet ordering. The next chapter studies basic properties of affine functions on compact convex sets and characterizations of functions satisfying the barycentric formula. A link between the theory of function spaces and compact convex sets starts to emerge at the end of the chapter. Chapter 5 is crucial for the subsequent application of descriptive set theory; it describes a hierarchy of Borel sets and functions in topological spaces and proves their
vi
Introduction
basic properties. The most important fact is that many descriptive properties are stable with respect to perfect mappings, which allows us to transfer abstract Borel affine functions to the setting of compact convex sets. Simplicial function spaces are studied in Chapter 6. We discuss several classes of simplicial function spaces, namely the Bauer and Markov simplicial function spaces and spaces with boundary of type Fσ . Among other results, the abstract Dirichlet problem for continuous and non-continuous functions is considered. Choquet simplices are presented at the end of the chapter. Next we generalize the basic concepts for function cones since they are indispensable in potential theory. We focus in particular on ordered compact convex sets. Analogues of faces in a non-convex setting, so-called Choquet sets, are investigated in Chapter 8. The main result is a characterization of simplicial spaces by means of Choquet sets. Suitably chosen families of closed extremal sets generate interesting boundary topologies on the set of extreme points. Chapter 9 studies these topologies and functions continuous with respect to them. It turns out that maximal measures induce measures on sets of extreme points that are regular with respect to boundary topologies. The last section is devoted to a study of a facial topology and facially continuous functions. Chapter 10 collects several deeper results on function spaces and compact convex sets. Among others, study of Shilov and James boundaries, Lazar’s improvement of the Banach–Stone theorem, results on automatic boundedness of affine and convex functions, embedding of `1 in Banach spaces, metrizability of compact convex sets and their open images and some topological properties of the set of extreme points. The Lazar selection theorem and its consequences occupy the first part of Chapter 11. The second part is devoted to a presentation of Debs’ proof of Talagrand’s theorem on measurable selectors. Chapter 12 is concerned with two methods of constructing new function spaces: products and inverse limits. We show that both operations preserve simpliciality and describe resulting boundaries. The inverse limits lead to an interesting description of metrizable simplices as inverse limits of finite-dimensional simplices. The general results are illustrated by a construction of the Poulsen simplex and a couple of compact convex sets due to Talagrand. In Chapter 13, general results from the Choquet theory are applied to potential theory and several of its basic notions are investigated from this perspective. Important function cones and spaces appearing in potential theory are studied in detail, in particular, in connection to various solution methods for the Dirichlet problem. The functional analysis approach makes it possible to provide an interesting interpretation, for instance, of balayage and regular points in terms of representing measures and the Choquet boundary of suitable spaces and cones. The exposition covers potential the-
Introduction
vii
ory for the Laplace equation and the heat equation as well as a more general setting (harmonic spaces, fine potential theory etc.). The final Chapter 14 presents several applications of the integral representation theorems, such as for doubly stochastic matrices, the Riesz–Herglotz theorem, the Lyapunov theorem on the range of a vector measure, the Stone–Weierstrass theorem, positive-definite functions and invariant and ergodic measures. Each chapter concludes with a series of exercises with sketches of proofs and with concluding notes and comments where we try to give precise references and due credits for the results presented in the main body of the text, and discuss additional material which is related to the topics of the chapter in question, but was not included with complete proofs. Open problems are also mentioned. Since the presented material originates in an amalgamation of functional analysis, measure theory, topology, descriptive set theory and potential theory, we collect the needed notions and facts in the Appendix, sometimes even with proofs. We selected the following books for each subject as the key references: W. Rudin [403] and M. Fabian, P. Habala, P. H´ajek, V. Montesinos Santaluc´ıa, J. Pelant and V. Zizler [173] for functional analysis, D. H. Fremlin [182], [181] and [183] for measure theory, R. Engelking [169] and K. Kuratowski [285] for topology, A. S. Kechris [262] and C. A. Rogers and J. E. Jayne [394] for descriptive set theory, D. H. Armitage and S. J. Gardiner [21] for classical potential theory and J. Bliedtner and W. Hansen [66] for abstract potential theory. Next we point out what is omitted from the book. First, we focus on integral representation theorems for compact sets, and thus the readers interested in theory of sets with the Radon–Nikodym property are referred to R. D. Bourgin [82], and those interested in Choquet theory in sets of measures are referred to G. Winkler [473]. Second, although we consider several geometric aspects of simplicial spaces, they are not at the center of our attention. They are thoroughly investigated in H. E. Lacey [290] and P. Harmand, D. Werner and W. Werner [216]. Further, we do not pursue applications of integral representation theory in C ∗ -algebras and thus we refer the interested reader to E. M. Alfsen and F. W. Schultz [10] and [9], M. Rørdam [395] and M. Rørdam and E. Størmer [396], B. Blackadar [59] and H. Lin [303] and the references therein. And last but not least, our applications to potential theory do not require the full strength of abstract potential theory and thus we restrict ourselves to a less general framework than the one presented in J. Bliedtner and W. Hansen [66]. Except on a few explicitly stated occasions, we consider only real vector spaces and apart from Chapter 9 we deal only with Hausdorff topologies and Radon measures. We use the standard notation and terminology: •
N, Q, Z, R, C denote the usual sets of numbers,
•
Re z and Im z denote the real and imaginary part of a complex number z, respectively,
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•
cA is the characteristic function of a set A (sometimes we write 1 for the characteristic function of a space),
•
A4B is the symmetric difference of sets A and B,
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Ac is the complement of a set A,
•
•
• •
f ∧ g, f ∨ g denote the infimum and supremum of functions f, g, respectively, (usually they are considered pointwise), f + , f − , |f | denote the positive and negative parts, and absolute value of a function f , respectively, f |A is the restriction of a function f to a set A, if F is a system of functions, F b and F + are the families of all bounded and positive elements from F, respectively,
•
ω0 and ω1 are the first infinite and first uncountable ordinals, respectively,
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A, Int A, ∂A are the closure, interior and boundary of a set A in a topological space, respectively,
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dist(A, B) denotes the distance of sets in a metric space,
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diam A is the diameter of a set A in a metric space,
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U (x, r), B(x, r) and S(x, r) are the open ball, closed ball and sphere centered at x with radius r > 0, respectively,
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co A and span A are the convex and linear hull of a set A in a vector space, respectively, co A is the closed convex hull of a set A in a topological vector space,
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ker T denotes the kernel of an operator between linear spaces,
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BE , UE and SE are the closed unit ball, open unit ball and sphere of a normed linear space E, respectively,
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E/F is the quotient space of a locally convex space with respect to a closed subspace F ,
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E ⊕ F is the sum of locally convex spaces E and F ,
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E ∗ is the dual space of a topological linear space E,
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(x, y) stands for the scalar product of vectors x, y in a Hilbert space,
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c0 is the space of sequences converging to 0,
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C(X) is the space of real-valued continuous functions on a topological space X,
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C b (X) is the space of bounded continuous functions on a topological space X,
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`p and Lp (µ), p ∈ [1, ∞], are the usual Lebesgue spaces (see Section A.3),
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C n (U ), C n , C ∞ (U ), C ∞ stand for the space of n-times continuously differentiable functions on U or infinitely differentiable functions on U , respectively,
Introduction • • •
ix
R −A f (y) dy is the integral mean value of f over a set A, R d S(x,r) f (y) dS(y) is the surface integral of f over the sphere S(x, r) ⊂ R , ∇f is the gradient of f .
A function f is positive if f ≥ 0, it is strictly positive if f > 0. Similarly we use increasing, strictly increasing and so on. If µ is a measure, we often write µ(f ) for R the integral f dµ. In preparation of the present book, we have received many valuable suggestions from many colleagues. In particular, we would like to express our thanks to P. H´ajek, P. Holick´y, M. Johanis, O. Kalenda, P. Kaplick´y, M. Kraus, E. Murtinov´a, P. Simon, J. Tiˇser, L. Zaj´ıcˇ ek and M. Zelen´y for stimulating and fruitful discussions, and to E. Crooks for linguistic assistance. We are also indebted to the publishers for their care and cooperation. The preparation of the manuscript was supported by the grant 201/07/0388 of the Grant Agency of the Czech Republic and partly by the grant MSM21620839 of the Czech Ministry of Education. Finally, our thanks go to Jana, Jarka, Hana and Hanka for encouragement and patience during the preparation of this book. Prague, Summer 2009
Jaroslav Lukeˇs Jan Mal´y Ivan Netuka Jiˇr´ı Spurn´y
Contents
Introduction
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1
Prologue 1.1 The Korovkin theorem . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . .
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Compact convex sets 2.1 Geometry of convex sets . . . . . . 2.1.A Finite-dimensional case . . 2.1.B The Krein–Milman theorem 2.1.C Exposed points . . . . . . . 2.2 Interlude: On the space M(K) . . . 2.3 Structures in convex sets . . . . . . 2.3.A Extremal sets and faces . . . 2.3.B Measure convex sets . . . . 2.3.C Measure extremal sets . . . 2.4 Exercises . . . . . . . . . . . . . . 2.5 Notes and comments . . . . . . . .
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Choquet theory of function spaces 3.1 Function spaces . . . . . . . . . . . 3.2 More about Korovkin theorems . . . 3.3 On the H-barycenter mapping . . . 3.4 The Choquet representation theorem 3.5 In-between theorems . . . . . . . . 3.6 Maximal measures . . . . . . . . . 3.7 Boundaries and the Simons lemma . 3.8 The Bishop–de Leeuw theorem . . . 3.9 Minimum principles . . . . . . . . 3.10 Orderings and dilations . . . . . . . 3.11 Exercises . . . . . . . . . . . . . . 3.12 Notes and comments . . . . . . . .
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Affine functions on compact convex sets 107 4.1 Affine functions and the barycentric formula . . . . . . . . . . . . . . 107 4.2 Barycentric theorem and strongly affine functions . . . . . . . . . . . 113
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4.3 4.4 4.5 4.6 5
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State space and representation of affine functions Affine Baire-one functions on dual unit balls . . Exercises . . . . . . . . . . . . . . . . . . . . . Notes and comments . . . . . . . . . . . . . . .
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Perfect classes of functions and representation of affine functions 5.1 Generation of sets and functions . . . . . . . . . . . . . . . . 5.2 Baire and Borel sets . . . . . . . . . . . . . . . . . . . . . . . 5.3 Baire and Borel mappings . . . . . . . . . . . . . . . . . . . 5.4 Perfect classes of functions . . . . . . . . . . . . . . . . . . . 5.5 Affinely perfect classes of functions . . . . . . . . . . . . . . 5.6 Representation of H-affine functions . . . . . . . . . . . . . . 5.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Notes and comments . . . . . . . . . . . . . . . . . . . . . .
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135 136 142 146 149 150 154 159 166
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168 169 176 178 180 182 185 185 188 190 192 196 198
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Simplicial function spaces 6.1 Basic properties of simplicial spaces . . . . . . . . . . . . . . . 6.2 Characterizations of simplicial spaces . . . . . . . . . . . . . . 6.3 Simplicial spaces as L1 -preduals . . . . . . . . . . . . . . . . . 6.4 The weak Dirichlet problem and Ac (H)-exposed points . . . . . 6.5 The Dirichlet problem for a single function . . . . . . . . . . . 6.6 Special classes of simplicial spaces . . . . . . . . . . . . . . . 6.6.A Bauer simplicial spaces . . . . . . . . . . . . . . . . . . 6.6.B Markov simplicial spaces . . . . . . . . . . . . . . . . . 6.6.C Simplicial spaces with Lindel¨of boundaries . . . . . . . 6.6.D Simplicial spaces with boundaries of type Fσ . . . . . . 6.7 The Daugavet property of simplicial spaces . . . . . . . . . . . 6.8 Choquet simplices . . . . . . . . . . . . . . . . . . . . . . . . 6.8.A Simplicial function spaces and the classical definition of Choquet simplices . . . . . . . . . . . . . . . . . . . . 6.8.B Prime function spaces and prime compact convex sets . 6.8.C Characterization of Bauer simplices by faces . . . . . . 6.8.D Fakhoury’s theorem . . . . . . . . . . . . . . . . . . . 6.9 Restriction of function spaces . . . . . . . . . . . . . . . . . . 6.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . Choquet theory of function cones 7.1 Function cones . . . . . . . 7.2 Maximal measures . . . . . 7.3 Representation theorem . . . 7.4 Simplicial cones . . . . . .
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7.5 7.6 7.7 8
Ordered compact convex sets and simplicial measures . . . . . . . . . 232 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 243
Choquet-like sets 8.1 Split and parallel faces . . . . . . . . . . . . . . 8.2 H-extremal and H-convex sets . . . . . . . . . . 8.3 Choquet sets, M -sets and P -sets . . . . . . . . . 8.4 H-exposed sets . . . . . . . . . . . . . . . . . . 8.5 Weak topology on boundary measures . . . . . . 8.6 Characterizations of simpliciality by Choquet sets 8.7 Exercises . . . . . . . . . . . . . . . . . . . . . 8.8 Notes and comments . . . . . . . . . . . . . . .
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Topologies on boundaries 9.1 Topologies generated by extremal sets . . . . . . 9.2 Induced measures on Choquet boundaries . . . . 9.3 Functions continuous in σext and σmax topologies 9.4 Strongly universally measurable functions . . . . 9.5 Facial topology generated by M -sets . . . . . . . 9.6 Exercises . . . . . . . . . . . . . . . . . . . . . 9.7 Notes and comments . . . . . . . . . . . . . . .
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10 Deeper results on function spaces and compact convex sets 10.1 Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.A Shilov boundary . . . . . . . . . . . . . . . . . . . . 10.1.B Boundaries in Banach spaces . . . . . . . . . . . . . . 10.2 Isometries of spaces of affine continuous functions . . . . . . 10.3 Baire measurability and boundedness of affine functions . . . 10.3.A The Cantor set and its properties . . . . . . . . . . . . 10.3.B Automatic boundedness of affine and convex functions 10.4 Embedding of `1 . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Metrizability of compact convex sets . . . . . . . . . . . . . . 10.6 Continuous affine images . . . . . . . . . . . . . . . . . . . . 10.7 Several topological results on Choquet boundaries . . . . . . . 10.7.A The Choquet boundary as a Baire space . . . . . . . . 10.7.B Polish spaces as Choquet boundaries . . . . . . . . . . 10.7.C K-countably determined boundaries . . . . . . . . . . 10.8 Convex Baire-one functions . . . . . . . . . . . . . . . . . . 10.9 Function spaces with continuous envelopes . . . . . . . . . . 10.9.A Stable compact convex sets . . . . . . . . . . . . . . . 10.9.B CE-function spaces . . . . . . . . . . . . . . . . . . .
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10.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 10.11 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 384 11 Continuous and measurable selectors 11.1 The Lazar selection theorem . . . . . . . . . . 11.2 Applications of the Lazar selection theorem . . 11.3 The weak Dirichlet problem for Baire functions 11.4 Pointwise approximation of maximal measures 11.5 Measurable selectors . . . . . . . . . . . . . . 11.5.A Multivalued mappings . . . . . . . . . 11.5.B Selection theorem . . . . . . . . . . . . 11.5.C Applications of the selection theorem . 11.6 Exercises . . . . . . . . . . . . . . . . . . . . 11.7 Notes and comments . . . . . . . . . . . . . .
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389 389 394 398 400 402 402 406 409 412 416
12 Constructions of function spaces 12.1 Products of function spaces . . . . . . . . . . . . . . . . . . . . . . . 12.1.A Definitions and basic properties . . . . . . . . . . . . . . . . 12.1.B Maximal measures and extremal sets . . . . . . . . . . . . . 12.1.C Partitions of unity and approximation in products of function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1.D Products of simplicial spaces . . . . . . . . . . . . . . . . . . 12.2 Inverse limits of function spaces . . . . . . . . . . . . . . . . . . . . 12.2.A Admissible mappings . . . . . . . . . . . . . . . . . . . . . . 12.2.B Construction of inverse limits . . . . . . . . . . . . . . . . . 12.2.C Inverse limits of simplicial function spaces . . . . . . . . . . 12.2.D Structure of simplices . . . . . . . . . . . . . . . . . . . . . 12.3 Several examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.A The Poulsen simplex . . . . . . . . . . . . . . . . . . . . . . 12.3.B A big simplicial space . . . . . . . . . . . . . . . . . . . . . 12.3.C Functions of affine classes and Talagrand’s example . . . . . 12.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . .
428 436 440 440 442 445 447 455 455 465 470 477 486
13 Function spaces in potential theory and the Dirichlet problem 13.1 Balayage and the Dirichlet problem . . . . . . . . . . . . . . 13.1.A Essential solution of the generalized Dirichlet problem 13.2 Boundary behavior of solutions . . . . . . . . . . . . . . . . 13.2.A Regular points for the Laplace equation . . . . . . . . 13.2.B Regular points for the heat equation . . . . . . . . . . 13.3 Function spaces and cones in potential theory . . . . . . . . . 13.3.A Function spaces and cones: Laplace equation . . . . .
489 491 494 496 497 503 504 505
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419 420 420 424
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13.4
13.5
13.6 13.7
13.3.B Function spaces and cones in parabolic potential theory and harmonic spaces . . . . . . . . . . . . . . . . . . . . . . . . 510 13.3.C Continuity properties of H(U )-concave functions . . . . . . . 513 13.3.D Separation by functions from H(U ) . . . . . . . . . . . . . . 515 Dirichlet problem: solution methods . . . . . . . . . . . . . . . . . . 517 13.4.A PWB solution of the Dirichlet problem . . . . . . . . . . . . 518 13.4.B Cornea’s approach to the Dirichlet problem . . . . . . . . . . 523 13.4.C The Wiener solution . . . . . . . . . . . . . . . . . . . . . . 530 13.4.D Fine Wiener solution . . . . . . . . . . . . . . . . . . . . . . 532 13.4.E PDE solutions in Sobolev spaces . . . . . . . . . . . . . . . . 534 Generalized Dirichlet problem and uniqueness questions . . . . . . . 537 13.5.A Lattice approach . . . . . . . . . . . . . . . . . . . . . . . . 538 13.5.B Uniqueness for the Laplace equation . . . . . . . . . . . . . . 540 13.5.C Keldysh theorems in parabolic and axiomatic potential theories 542 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 Notes and comments . . . . . . . . . . . . . . . . . . . . . . . . . . 555
14 Applications 14.1 Representation of convex functions . . . . . . 14.2 Representation of concave functions . . . . . . 14.3 Doubly stochastic matrices . . . . . . . . . . . 14.4 The Riesz–Herglotz theorem . . . . . . . . . . 14.5 Typically real holomorphic functions . . . . . . 14.6 Holomorphic functions with positive real part . 14.7 Completely monotonic functions . . . . . . . . 14.8 Positive definite functions on discrete groups . 14.9 Range of vector measures . . . . . . . . . . . . 14.10 The Stone–Weierstrass approximation theorem 14.11 Invariant and ergodic measures . . . . . . . . . 14.12 Exercises . . . . . . . . . . . . . . . . . . . . 14.13 Notes and comments . . . . . . . . . . . . . .
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563 564 567 572 573 575 580 586 589 593 595 597 603 605
A Appendix A.1 Functional analysis . . . . . . . . . . . . . . . . . . . . . A.1.A Locally convex spaces . . . . . . . . . . . . . . . A.1.B Banach spaces . . . . . . . . . . . . . . . . . . . A.1.C Ordered Banach spaces and lattices . . . . . . . . A.2 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . ˇ A.2.A Compact spaces and Cech–Stone compactification A.2.B Baire and Borel sets . . . . . . . . . . . . . . . . A.2.C Semicontinuous functions . . . . . . . . . . . . . A.2.D Baire spaces and sets with the Baire property . . .
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608 608 608 609 610 615 616 619 621 623
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A.3 Measure theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3.A Measure spaces . . . . . . . . . . . . . . . . . . . . . . . . A.3.B Radon measures on locally compact σ-compact spaces . . . A.3.C Images, products and inverse limits of Radon measures . . . A.3.D Kernels and disintegration of measures . . . . . . . . . . . A.4 Descriptive set theory . . . . . . . . . . . . . . . . . . . . . . . . . A.5 Resolvable sets and Baire-one functions . . . . . . . . . . . . . . . A.6 The Laplace equation . . . . . . . . . . . . . . . . . . . . . . . . . A.6.A Weak solutions of the Laplace equation . . . . . . . . . . . A.7 The heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8 Axiomatic potential theory . . . . . . . . . . . . . . . . . . . . . . A.8.A Bauer’s axiomatic theory . . . . . . . . . . . . . . . . . . . A.8.B Hyperharmonic and superharmonic functions . . . . . . . . A.8.C Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8.D Superharmonic functions and Green potentials . . . . . . . A.8.E Superharmonic functions and potentials for the heat equation A.8.F Balayage . . . . . . . . . . . . . . . . . . . . . . . . . . . A.8.G Thinness, base and fine topology . . . . . . . . . . . . . . . A.8.H Polar and semipolar sets . . . . . . . . . . . . . . . . . . .
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624 624 626 632 636 637 640 645 647 649 652 653 654 656 657 660 661 663 665
Bibliography
669
List of symbols
695
Index
703
Chapter 1
Prologue
1.1
The Korovkin theorem
We start with the famous Weierstrass approximation theorem. Theorem 1.1 (The Weierstrass approximation theorem). The space of all polynomial functions on the interval [0, 1] is uniformly dense in the space C([0, 1]). There are several different proofs of this result and several methods for how to associate to a given continuous function f ∈ C([0, 1]) a sequence of polynomials {Pn } that converges uniformly to f on [0, 1]. For example, given f ∈ C([0, 1]) and n ∈ N, we define the corresponding Bernstein polynomial Bn f by n X n j j Bn f : x 7→ f x (1 − x)n−j , x ∈ [0, 1]. j n j=0
The task is to show that the sequence {Bn f } converges uniformly to f on [0, 1]. This can be easily verified in the case when f (x) = 1, x or x2 , since
n−1 2 1 x + x. n n Surprisingly, this is all that we need to compute, since these three tests are enough to guarantee the uniform convergence of Bn f to f for all f in C([0, 1]). Indeed, one of the current proofs of the classical Weierstrass approximation theorem is based on the Korovkin theorem about linear operators. The Weierstrass theorem is an easy consequence since the mappings Bn 1 = 1,
Bn x = x
and
Bn : f 7→ Bn f,
Bn x2 =
f ∈ C([0, 1]),
are positive linear operators on C([0, 1]). Theorem 1.2 (Korovkin). Let pj , j = 0, 1, 2, denote the monomial function pj : x 7→ xj and let {Tn } be a sequence of positive linear operators on the space C([0, 1]). Assume that Tn pj → pj uniformly on [0, 1] as n → ∞ for j = 0, 1, 2. Then Tn f → f uniformly on [0, 1] as n → ∞ for all f ∈ C([0, 1]).
2
1 Prologue
Proof. Pick f ∈ C([0, 1]) and ε > 0. By the uniform continuity of f , there exists δ ∈ (0, 1) such that |f (s) − f (t)| ≤ ε for any s, t ∈ [0, 1], |s − t| ≤ δ. Now fix t ∈ [0, 1] and set p∗ (x) := f (t) − ε − (x − t)2
2kf k , δ2
x ∈ [0, 1],
and
2kf k , x ∈ [0, 1]. δ2 Dealing separately with the cases |x − t| ≤ δ and |x − t| > δ, we get p∗ (x) := f (t) + ε + (x − t)2
|f (x) − f (t)| ≤ ε + 2(x − t)2
kf k δ2
for each x ∈ [0, 1]. Hence, p∗ (x) ≤ f (x) ≤ p∗ (x),
x ∈ [0, 1],
and, therefore, for any n ∈ N, Tn p∗ ≤ Tn f ≤ Tn p∗ . We find N ∈ N such that for any n ≥ N
Tn pj − pj < εδ 2 ,
j = 0, 1, 2.
Since 2kf k 4kf kt 2kf k p∗ (x) = f (t) + ε + t2 2 − x + 2 x2 , 2 δ δ δ
x ∈ [0, 1],
for n ≥ N we have the estimate kTn p∗ − p∗ k ≤ Cε
with C := 9kf k + ε.
In particular, Tn f (t) ≤ Tn p∗ (t) ≤ p∗ (t) + Cε = f (t) + Cε + ε,
n ≥ N,
and similarly Tn f (t) ≥ Tn p∗ (t) ≥ p∗ (t) − Cε = f (t) − Cε − ε, Hence kTn f − f k ≤ ε(C + 1), and the proof is complete.
n ≥ N,
n ≥ N.
1.2 Notes and comments
3
In stating the Korovkin theorem, which sometimes bears the name the first Korovkin theorem, it is possible to go further, replacing the interval [0, 1] by a suitable space, and the set of three functions {p0 , p1 , p2 } by a more general family of functions. In the sequel, we will take a deeper look at this issue. Definition 1.3 (Korovkin closure). Let K be a (metrizable) compact space and P a family of continuous functions on K (sometimes called test functions). We say that a sequence {Tn } of positive operators on C(K) is P-admissible if kTn ϕ − ϕk → 0 for any ϕ ∈ P, and define the Korovkin closure of P as Kor(P) := {f ∈ C(K) : kTn f − f k → 0 for any P-admissible sequence {Tn }} . Let H be the linear span of P. It is simple to check that H ⊂ Kor(H) = Kor(P). Two questions immediately arise: (a) How can Kor(P) be characterized ? (b) Under what conditions does the equality Kor(P) = C(K) hold ? In what follows, we will give answers to these questions and will also study analogous problems. To these ends, the framework of abstract linearity and convexity will turn out to be useful and efficient.
1.2
Notes and comments
The Korovkin theorem was proved independently by H. Bohman in [74] for a kind of special positive operators, and by P. P. Korovkin in [277] for integral-type operators. Korovkin extended his theory in [278] and we followed his proof from this monograph. The Korovkin theorem 1.2 sometimes bears the name of the Bohman– Korovkin theorem. Excellent sources for the Korovkin material are the monograph of F. Altomare and M. Campiti [13], and Chauvenet’s prize paper of H. Bauer [42].
Chapter 2
Compact convex sets
We begin our exposition with classical results on convex sets in finite-dimensional spaces. After showing Carath´eodory’s theorem 2.6, we define extreme points and prove Minkowski’s theorem 2.11 stating that any compact convex set in Rd is the convex hull of its extreme points. An amalgamation of these two results contained in Theorem 2.12 is a starting point leading to generalizations in infinite-dimensional spaces. So the next section is devoted to the study of the Krein–Milman theorem and related results. In particular we are interested in its reformulation known as the Integral representation theorem. The basic idea of representing points of a compact convex set as barycenters of probability measures is a central topic of the whole book. Thus after the proof of the Krein–Milman theorem 2.22 and Bauer’s minimum principle 2.24 we define the barycenter of a probability measure on a compact convex set and show its existence and uniqueness (see Theorem 2.29). Then the Integral representation theorem 2.31 and properties of the barycentric mapping are proved. We finish this part by some classical facts: Bauer’s characterization 2.40 of extreme points of a compact convex set, Choquet’s observation on extreme points of a compact convex set contained in Proposition 2.41 and the Milman theorem 2.43. The aim of Subsection 2.1.C is to show that a metrizable compact convex set has abundance of exposed points, namely, that a metrizable compact convex set is the closed convex hull of its exposed points and that exposed points are dense in the set of extreme points. Section 2.2 prepares the ground for examples concerning extremal sets and faces of compact convex sets presented in Section 2.3. We prove several facts on probability measures on compact spaces and show how they lead to a construction of affine functions on compact convex sets that do not satisfy the barycentric formula (see Proposition 2.63). Subsection 2.3.A investigates more closely extremal sets and faces of compact convex sets. The main result contained in Proposition 2.69 shows that a closed extremal set is a union of closed faces. We generalize the concept of convexity and extremality in Subsections 2.3.B and 2.3.C by introducing measure convex and measure extremal sets. The main tool is Theorem 2.75 due to D. H. Fremlin and J. D. Pryce that characterizes measure convex sets. Then we show that convex sets of low Borel complexity are also measure convex, but that there are examples of Fσ or Gδ faces that are not measure convex. Analogous results are proved in the next section for extremal and measure extremal sets.
2.1 Geometry of convex sets
2.1 2.1.A
5
Geometry of convex sets Finite-dimensional case
Throughout this subsection, let W be a real vector space. Definition 2.1 (Convex sets in vector spaces). A set C ⊂ W is convex if λx + (1 − λ)y ∈ C whenever x, y ∈ C and λ ∈ (0, 1). Let A be an arbitrary subset of W . The convex hull of A, denoted by co A, is the intersection of all convex sets of W that contain A. Since W is a convex set and the intersection of any family of convex sets is convex, the set co A is the smallest convex set containing A. It is easy to check that co A =
n nX
λj xj : n ∈ N, x1 , . . . , xn ∈ A and λ1 , . . . , λn ≥ 0,
j=1
n X
o λj = 1 .
j=1
Definition 2.2 (Affine independence and n-simplices). Recall that vectors e0 , . . . , en of W are said to be affinely independent if e1 −e0 , . . . , en −e0 are linearly independent. In other words, if whenever λ0 e0 + · · · + λn en = 0
and λ0 + · · · + λn = 0,
then λ0 = · · · = λn = 0. In this case, the convex hull co {e0 , . . . , en } is termed an n-simplex with vertices e0 , . . . , en . In Rd , there exist at most d + 1 affinely independent points. Definition 2.3 (Affine hulls and subspaces, hyperplanes). For a set A ⊂ W , the affine hull of A, denoted by aff A, is the set of all affine combinations of points of A. (A linear combination α1 x1 + · · · + αn xn , where α1 + · · · + αn = 1, is called an affine combination of points x1 , . . . , xn .) A set A ⊂ W is said to be an affine subspace of W if aff A = A. Affine subspaces are just the translations (of type) x + F , where F is a linear subspace of W and x ∈ W . By definition, the dimension (or, the codimension) of x + F is the dimension (or, the codimension, respectively) of F . A set H is a hyperplane if there exists a nonzero linear functional f on W and α ∈ R such that H = {w ∈ W : f (w) = α}. Since a subspace F of W is a maximal proper subspace of W if and only if there exists a nonzero linear functional f on W such that F = ker f , we see that H is a hyperplane if and only if there exist a maximal proper subspace F of W and w ∈ W such that H = w + F . In other words, hyperplanes are exactly affine subspaces of codimension 1.
6
2 Compact convex sets
Let C be a subset of W and H := {w ∈ W : f (w) = α} be a hyperplane. We say that H is a support hyperplane of C if C ∩ H 6= ∅ and either C ⊂ {w ∈ W : f (w) ≤ α}
or C ⊂ {w ∈ W : f (w) ≥ α}.
Any point c ∈ C ∩ H is called an H-support point of C. We also say that H supports C at c. In the sequel, we need the following assertion. Proposition 2.4. Let C be a closed convex subset of Rd with a nonempty interior and c ∈ ∂C. Then there exists a hyperplane H such that H supports C at c. Proof. See, for example, A. Barvinok [31], Corollary 2.8. Remark 2.5. In what follows, we direct our attention to geometry of compact convex sets in the Euclidean d-dimensional space Rd . All results of this subsection remain valid in any finite-dimensional topological vector space, since any such space is isomorphic to a suitable space Rd . Theorem 2.6 (Carath´eodory). Let A be an arbitrary subset of Rd . Then each point of co A is a convex combination of at most d + 1 points of A which are affinely independent. Proof. Assume that x ∈ co A, x = λ1 x1 + · · · + λn xn where x1 , . . . , xn ∈ A, λj > 0 for all j = 1, . . . , n (which we may suppose) and λ1 + · · · + λn = 1. If the vectors x1 , . . . , xn are affinely independent, then n ≤ d + 1 and we are done. Otherwise, n > d + 1 and there is (α1 , . . . , αn ) 6= (0, . . . , 0) such that α1 x1 + · · · + αn xn = 0 and α1 + · · · + αn = 0. Let k ∈ {1, . . . , n} be such that α α j k ≤ λj λk
for all j = 1, . . . , n.
Setting ηj := λj −
λk αj , αk
j = 1, . . . , n,
we have x=
X j6=k
η j xj ,
X
ηj = 1
and ηj ≥ 0 for j = 1, . . . , n.
j6=k
Thus, x is a convex combination of n − 1 points. If these points are affinely independent, the proof is finished. If not, the above argument can be repeated, and after finitely many steps x can be represented as a convex combination of affinely independent points of A.
2.1 Geometry of convex sets
7
Corollary 2.7. The convex hull co A of any set A ⊂ Rd is the union of all n-simplices (n ≤ d) with vertices in A. Proof. Obviously, any n-simplex with vertices in A, where n ≤ d, is a subset of co A. The reverse inclusion immediately follows from Carath´eodory’s theorem 2.6. Corollary 2.8. The convex hull of any compact subset of Rd is compact. Proof. Let K be a compact subset of Rd and d n o X d+1 D := λ ∈ R : λ = (λ0 , . . . , λd ), λj = 1 and λj ≥ 0 for j = 0, . . . , d . j=0
The mapping F : D × K × · · · × K → K defined as F : (λ, x0 , . . . , xd ) 7→
d X
λ j xj
j=0
is continuous. By Carath´eodory’s theorem 2.6, co K = F (D × K × · · · × K). Hence, co K, as a continuous image of the compact set D × K × · · · × K, is compact. Definition 2.9 (Extreme points). A point z of a set C ⊂ W is called an extreme point of C if z is not an internal point of any segment having its endpoints in C. In other words, z is an extreme point of C if x, y ∈ C, λ ∈ (0, 1) and z = λx + (1 − λ)y, implies x = y. It is easy to check that z is an extreme point of a convex set C if and only if the set C \ {z} is convex, and this is the case if and only if z is not a midpoint of any nondegenerate segment having its endpoints in C. We denote by ext C the set of all extreme points of C. Lemma 2.10. Let S be an n-simplex in Rd with vertices e0 , . . . , en . Then ext S = {e0 , . . . , en } . Proof. Let x ∈ ext S. Since S = co {e0 , . . . , en } and the set S \ {x} is convex, x = ek for some k ∈ {0, 1, . . . , n}. Conversely, select ek and assume that 1 1 ek = s + t 2 2 where s, t ∈ S = co {e0 , . . . , en }. Write s=
n X j=0
αj ej ,
t=
n X j=0
βj ej ,
8
2 Compact convex sets
with αj , βj ≥ 0,
n X j=0
Then ek =
n X 1 j=0
or
X1 j6=k
2
2
αj =
n X
βj = 1.
j=0
(αj + βj )ej
(αj + βj )(ej − ek ) = 0.
Consequently, αj + β j = 0
for j ∈ {0, . . . , n} \ {k} ,
thus s = t = ek . The following assertion shows the prominent role of extreme points in finite-dimensional compact convex sets. Infinite-dimensional situation is more complicated, see Example 2.15 and the Krein–Milman theorem 2.22. Theorem 2.11 (Minkowski). Each point of a compact convex set C ⊂ Rd is a convex combination of extreme points of C. Proof. We proceed by induction on the dimension d. For the dimension d = 0, the set C reduces to a one-point set and the assertion holds. So assume that d > 0 and that the assertion is valid for compact convex sets in spaces of dimension smaller than d. We may also assume that the interior of C is nonempty, for otherwise C is a subset of an affine subspace of a smaller dimension (cf. Exercise 2.107(c)) and the assertion follows by the induction assumption. We distinguish two cases. If x is a boundary point of C, then by Proposition 2.4 there exists a support hyperplane L of C at x. Then the compact convex set F := C ∩ L lies in the affine subspace L of dimension smaller than d. By the induction assumption, x ∈ co ext F . Since obviously ext F ⊂ ext C, the induction step is finished. Now suppose that x ∈ Int C. There exists a segment [a, b] ⊂ C such that x ∈ (a, b) and a, b ∈ ∂C. Since a, b ∈ co ext C by the previous argument, we see that x can be expressed as a convex combination of extreme points of C. Theorem 2.12 (Minkowski–Carath´eodory). Each point of a compact convex set C ⊂ Rd is a convex combination of (at most d + 1) affinely independent extreme points of C. Proof. A consequence of Theorems 2.6 and 2.11. Indeed, given a point c ∈ C, by the Minkowski theorem 2.11, there exists a set A ⊂ ext C such that c ∈ co A. Now, it suffices to apply Carath´eodory’s theorem 2.6.
2.1 Geometry of convex sets
9
Corollary 2.13. Let x be a point of a compact convex set C ⊂ Rd . Then there exists an n-simplex S, n ≤ d, such that x ∈ S ⊂ C and ext S ⊂ ext C. Proof. The assertion is a rewording of the previous Minkowski–Carath´eodory theorem. Remark 2.14. In [1] E. M. Alfsen constructed a non-simplicial compact polyhedron in `1 , showing that the conclusion of the previous Corollary 2.13 in infinite-dimensional spaces fails. At the same time, he posed a question that “it would be of some interest to find sufficient conditions for a compact convex set X to admit a decomposition” as in Corollary 2.13: Given x ∈ X, there would exists a set S ⊂ X such that x ∈ S, ext S ⊂ ext C, and a unique representing measure for x carried by ext S. We present in Exercise 6.93 an example illustrating this phenomenon.
2.1.B
The Krein–Milman theorem
Recall that a point z of a subset C of a vector space W is an extreme point of C if z is not an internal point of any nondegenerate segment having endpoints in C and that ext C denotes the set of all extreme points of C. In the Euclidean space Rd , the set ext C is of fundamental importance. The Minkowski theorem 2.11 says that each point of a compact convex set C ⊂ Rd is a convex combination of extreme points of C. Thus, C = co(ext C). The aim of this subsection is to examine an analogous result and its relatives in the framework of infinite-dimensional spaces. Note, that in infinite-dimensional spaces, a compact convex set need not be a convex hull of its extreme points, as Example 2.15 shows. 2 Example 2.15. Let {en }∞ n=1 be the orthonormal basis in ` formed by the standard unit vectors en , and let
1 1 B := {0, e1 , e2 , e3 , . . . } and C := co B. 2 3 Since B is clearly compact, it is easy to see that C is a compact convex set (see [173], Exercise 1.56). By the Milman theorem 2.43, ext C ⊂ B. (In fact, it is easy to verify that ext C = B.) Defining xn := (1 − 2−n )−1
n X k=1
1 2−k ek , k
n ∈ N,
we have xn ∈ co ext C and xn → x :=
∞ X k=1
1 2−k ek ∈ C. k
Since every element of co ext C has only a finite number of nonzero coordinates, x∈ / co ext C.
10
2 Compact convex sets
Definition 2.16 (Extremal sets and faces). A generalization of the notion of extreme points leads to an important concept: A nonempty subset F of a set C ⊂ W is an extremal subset of C if x, y ∈ F provided that x, y ∈ C and λx + (1 − λ)y ∈ F for some λ ∈ (0, 1). It is needless to say that one-point extremal sets are exactly extreme points of C, and that C itself is an extremal set. Convex extremal sets are called faces. Definition 2.17 (Affine, concave and convex functions). Let C be a convex subset of a vector space W . A real-valued function s on C is said to be concave if s λx + (1 − λ)y ≥ λs(x) + (1 − λ)s(y) for each x, y ∈ C and λ ∈ [0, 1]. A real-valued function f on C is convex if −f is concave and f is called affine if both f and −f are concave. Obviously, the restriction of any linear functional on W to C is an affine function. Lemma 2.18. If H is an extremal subset of F and F is an extremal subset of D, then H is an extremal subset of D. Proof. Obvious. Lemma 2.19. If X is a nonempty compact convex subset of a locally convex space E, s is a concave lower semicontinuous function on X and L := {x ∈ X : s(x) = min s(X)}, then L is a compact extremal subset of X. If K is a nonempty compact subset of E, f ∈ E ∗ and H := {x ∈ K : f (x) = min f (K)}, then H is a compact extremal subset of K. Proof. It is clear that L is compact and nonempty. Choose x, y ∈ X, λ ∈ (0, 1). If a := λx + (1 − λ)y and a ∈ L, then s(a) ≥ λs(x) + (1 − λ)s(y) ≥ λs(a) + (1 − λ)s(a) = s(a). From this it easily follows that x, y ∈ L. The proof of the second assertion is similar. Proposition 2.20. Let K be a nonempty compact subset of a locally convex space E and F a compact extremal subset of K. Then F ∩ ext K 6= ∅. In particular, ext K 6= ∅.
2.1 Geometry of convex sets
11
Proof. Consider the family F of all closed extremalTsubsets of F ordered by the reverse inclusion. If R is a chain in F, then Y := {R : R ∈ R} is nonempty in view of the compactness of K. Since it is easy to check that Y is an extremal subset of K, Y is an upper bound for R. Zorn’s lemma now provides a maximal element of this family, call it D. An appeal to the Hahn–Banach theorem reveals that D is a one-point set. Indeed, assuming that D were to contain distinct points x and y, the Hahn–Banach theorem would provide f ∈ E ∗ such that f (x) < f (y). By Lemma 2.19, the set {z ∈ D : f (z) = min f (D)} would be a proper closed extremal subset of D and, in view of Lemma 2.18, also a closed extremal subset of F . This contradicts the maximality of D. Since, as mentioned above, one-point extremal sets are exactly extreme points of F , we can find a point x ∈ ext F . Again, by Lemma 2.18, x ∈ ext K and we are done. Theorem 2.21. Let K be a compact subset of a locally convex space E. Then K ⊂ co ext K. Proof. Assume that there exists a point x ∈ K \co ext K. Using the geometric version of the Hahn–Banach theorem, there exists f ∈ E ∗ such that f (t) > f (x) for any t ∈ co ext K. If H := {z ∈ K : f (z) = min f (K)}, then H is a (nonempty) extremal subset of K by Lemma 2.19. Hence, by Proposition 2.20, H ∩ ext K 6= ∅. Since H ∩ co ext K = ∅, this is impossible. Therefore, K ⊂ co ext K. Theorem 2.22 (Krein–Milman). Let X be a nonempty compact convex subset of a locally convex space E. Then X = co ext X. Proof. It is pretty clear that co ext X ⊂ X. The reverse inclusion follows from Theorem 2.21. Remarks 2.23. (a) The Krein–Milman theorem also holds in locally convex spaces over complex numbers. For a proof see, for example, W. Rudin [403], Theorem 3.21. (b) If K is a compact subset of a locally convex space, then co K = co ext K. This is an immediate consequence of Theorem 2.21. Corollary 2.24 (Bauer’s concave minimum principle). Let s be a lower semicontinuous concave function on a nonempty compact convex subset X of a locally convex space. Then there exists z ∈ ext X such that s(z) = min s(X).
12
2 Compact convex sets
Proof. Denote D := {x ∈ X : s(x) = min s(X)} . Then D is a nonempty compact subset of X and, by Lemma 2.19, an extremal subset of X. By Proposition 2.20, D ∩ ext X 6= ∅, and thus the proof is complete. Remarks 2.25. (a) The Krein–Milman theorem is an easy consequence of Bauer’s concave minimum principle. Indeed, if X is a nonempty compact convex subset of a locally convex space E, the constant function 1 on X attains its minimum on ext X. Thus the set ext X is nonempty. If x ∈ X \ co ext X, then, by the geometric version of the Hahn–Banach theorem, there exists f ∈ E ∗ such that f (x) < 1
and f ≥ 1
on
co ext X.
This contradicts Bauer’s minimum principle since f |X is a continuous affine function. (b) In Section 3.9 we prove a generalization of Bauer’s concave minimum principle by a different method. Integral representation. Now we would like to show how to use the Krein–Milman theorem for establishing integral-type representation theorems. Let x be a point of a compact convex set X in a locally convex space. Our aim is to find a Radon measure µ on X so that Z f (x) = f dµ X
for any continuous affine function f on X. Of course, the Dirac measure εx at x is one such measure. However, we try to find other “representing” measures, preferably concentrated on a very small part of X. More precisely, we are looking for a measure µ such that (a) the support spt µ of µ is contained in the closure ext X of the set of all extreme points of X, or even (b) µ is carried by the set ext X. Definition 2.26 (Barycenter of a measure). Let X be a compact convex set in a locally convex space E. Denote by Ac (X) the set of all continuous affine functions on X. A point x ∈ X is said to be the barycenter of a probability Radon measure µ ∈ M1 (X) if the following barycentric formula Z f (x) = f dµ X
holds for any f ∈ Since the functionals from E ∗ separate the points of E, and since the restrictions to X of such functionals are elements of Ac (X), we see that the Ac (X).
2.1 Geometry of convex sets
13
barycenter r(µ) of µ (which exists by Theorem 2.29), is uniquely determined. Note that Z r(µ) = t dµ(t), X
where the integral is to be understood as the Pettis integral. In the case when r(µ) = x, we also say that the measure µ represents the point x. In other words, the equality x = r(µ) means that the Integral representation theorem holds for the point x. We denote by Mx (Ac (X)), or for short Mx (X), the set of all measures representing the point x. Obviously, as noted above, the Dirac measure εx always represents the point x. In what follows, we answer the following questions: (a) Does any Radon measure have a barycenter ? (b) Is any point of X a barycenter of a Radon measure carried by ext X ? Proposition 2.27. The space M1 (K) consisting of all probability measures on a compact space K is a compact convex subset of M(K) and ext M1 (K) = {εx : x ∈ K} . The mapping ε : x 7→ εx , x ∈ K, is a homeomorphism of K onto ext M1 (K). Proof. It is easy to verify that M1 (K) is a convex subset of M(K). By Theorem A.85(a), it is compact. If x ∈ K and εx = αµ + (1 − α)ν
where µ, ν ∈ M1 (K) and α ∈ (0, 1),
then 1 = αµ({x}) + (1 − α)ν({x}). This implies that µ({x}) = ν({x}) = 1, and therefore µ = ν = εx . If µ ∈ M1 (K) is not a Dirac measure, then there exists a compact set F ⊂ K such that the measures ν := µ|F and λ := µ|K\F are nontrivial and distinct from µ. Since µ = ν(F )
λ ν + λ(K \ F ) , ν(F ) λ(K \ F )
we see that µ is not an extreme point of M1 (K). The mapping x 7→ εx , x ∈ K, is an injective continuous mapping from K onto {εx : x ∈ K} and hence K and ext M1 (K) are homeomorphic. Corollary 2.28. The set of all convex combinations of Dirac measures is dense in M1 (K).
14
2 Compact convex sets
Proof. As an easy consequence of the Krein–Milman theorem 2.22 and the characterization of extreme points given by Proposition 2.27, we have M1 (K) = co ext M1 (K) = co {εx : x ∈ K} , which finishes the proof. Theorem 2.29. Let X 6= ∅ be a compact convex subset of a locally convex space E. Then each Radon measure from M1 (X) has a (unique) barycenter in X. Proof. With the uniqueness part already out of the way, we nowPconcentrate on an λj εxj , where existence proof. If a measure µPin question is molecular, µ = nj=1 P xj ∈ X, λj ≥ 0, j = 1, . . . , n, nj=1 λj = 1, then obviously r(µ) := nj=1 λj xj ∈ X is the barycenter of µ. Given a measure µ ∈ M1 (X), there exists a net {µγ } of molecular measures on X such that µγ → µ. Since X is a compact set, there exists a subnet {r(µα )} of {r(µγ )} converging to an element z ∈ X. Pick f ∈ Ac (X). Then Z Z f dµ, f dµα = f (z) = lim f (r(µα )) = lim α
α
X
X
and therefore z is a barycenter of µ. Definition 2.30 (Barycenter mapping). Let X be a compact convex subset of a locally convex space. The mapping r : µ 7→ r(µ), assigning to each measure µ ∈ M1 (X) its barycenter, is called the barycenter mapping. In Proposition 2.38 we show that the barycenter mapping is a continuous and affine mapping from M1 (X) into X. This mapping is surjective since r(εx ) = x for any x ∈ X. Theorem 2.31 (Integral representation theorem). Let X be a compact convex subset of a locally convex space E and let x ∈ X. Then there exists a measure µ ∈ M1 (X) such that r(µ) = x and spt µ ⊂ ext K. Proof. From the Krein–Milman theorem 2.22, we can see the following fact: if f ∈ Ac (X) and f = 0 on ext X, then f = 0 on X. We denote by B the subspace of C(ext X) consisting of all restrictions of functions in Ac (X) to ext X. Then, for every h ∈ B, there exists, by the above mentioned fact, a unique function b h ∈ Ac (X) which coincides with h on ext X. We fix x ∈ X and set ϕ : h 7→ b h(x),
h ∈ B.
Evidently ϕ ∈ B ∗ and kϕkB = 1. The functional ϕ can be extended by the Hahn– Banach theorem from B to a functional Φ ∈ (C(ext X))∗ with the same norm. Since,
2.1 Geometry of convex sets
15
in addition, Φ(1) = ϕ(1) = 1, Φ is a positive functional. Indeed, if f ∈ C(ext X), f ≥ 0, a = 12 sup f (ext X), then ka − f k ≤ a and so a − Φ(f ) = Φ(a) − Φ(f ) = Φ(a − f ) ≤ ka − f k ≤ a. This yields Φ(f ) ≥ 0. By the Riesz representation theorem, there exists a probability R measure µ on ext X such that Φ(f ) = X f dµ for every function f ∈ C(ext X). The measure µ can be regarded as a measure on X carried by the set ext X. Since obviously Z X
g dµ = Φ(g) = ϕ(g|ext X ) = gb(x) = g(x)
for every g ∈ Ac (X), we see that the barycenter of the measure µ is exactly the point x. Remarks 2.32. (a) Theorem 2.31 can be proved by an alternative manner. In fact, we can follow the proof of Proposition 2.39 step by step. (b) In concrete applications, we are often able to characterize the set ext X. However, the character of the elements of the set ext X \ ext X is generally rather obscure. Consequently, the information concerning the support of the measure from the theorem on integral representation is problematic, unless the set ext X is closed. Moreover, there is another problem. Let us imagine that the set of extreme points of a compact convex set X is dense in this set, that is, ext X = X. Then, naturally, the Krein–Milman theorem says nothing, and equally useless is the theorem on integral representation. Indeed, it suffices to take the Dirac measure εx at the point x for the measure representing the point x . This situation can actually occur. As an example, we can take the closed unit ball B in an arbitrary infinite-dimensional Hilbert space, which we of course consider to be equipped with the weak topology. The extreme points of B are then the points of the unit sphere, and its (weak) closure is equal to the whole ball B. A more sophisticated example with ext X = X is the Poulsen simplex in the Hilbert space `2 (see Subsection 12.3.A). However, much more is known. Namely, if we consider the so-called Hausdorff metric on the set F of all nonempty compact convex subsets ofa given Banach space X of infinite dimension, the space F is complete and the set C ∈ F : ext C 6= C is merely meager in F. This assertion was proved by V. L. Klee in [271]. Thus, in a certain sense, for the majority of compact convex sets we have ext C = C. Hence the problem of whether it is possible to find a measure µ which is carried just on the set of extreme points in the theorem on integral representation is crucial. This problem was solved successfully by G. Choquet in the fifties of the 20th century and laid the foundations of the Choquet theory. We will devote the next chapters to it, in the more general setting of function spaces. The Choquet theory has provided many insights for abstract analysis, infinite-dimensional geometry, descriptive set theory,
16
2 Compact convex sets
potential theory and other fields of mathematics. It has remained fruitful ever since and has found new applications again and again in deriving new results. (c) In Sections 14.5 and 14.6, we exceptionally consider locally convex spaces over the field of complex numbers. Note that the Integral representation theorem 2.31 extends trivially to the complex case. Indeed, consider a compact convex subset X of a complex locally convex space E. Of course, E can be regarded as a locally convex space over the field of real numbers. Then, given x ∈ X, there exists, by Theorem 2.31, a measure µ ∈ M1 (X) carried by ext X such that Z f dµ (2.1) f (x) = X
for every real continuous functional f on E. Given a complex continuous functional F on E, we can apply (2.1) to Re F and Im F to conclude that Z F dµ. F (x) = X
The Integral representation theorem will be applied several times in the sequel. For this purpose, the following easy consequence of the previous theorem will be useful. Proposition 2.33 (Krein–Milman theorem with transfer). Let E be a locally convex space of real-valued functions on a set M such that, for every x ∈ M , the evaluation functional Fx : f 7→ f (x) is continuous on E. Let K ⊂ E be a compact convex set. Let Q be a compact space and Φ : y 7→ ϕy , y ∈ Q, be an injective continuous mapping of Q onto ext K. Then there exists a probability measure µ on Q such that, for every f ∈ K, Z f (x) =
ϕy (x) dµ(y),
x ∈ M.
Q
Proof. Since ext K is a continuous image of a compact set, it is closed. Let f ∈ K. By Theorem 2.31, there exists a probability measure µ˜ carried by ext K such that Z F (f ) = F dµ˜ (2.2) ext K
for each continuous linear functional F on E. Let us define µ = Φ−1 ] µ. ˜ By Proposition A.92, Z Z Z F dµ˜ = (F ◦ Φ) ◦ Φ−1 dµ˜ = F ◦ Φ dΦ−1 ] µ˜ ext K
ext K
Z F ◦ ϕy dµ(y).
= Q
Q
(2.3)
2.1 Geometry of convex sets
17
Applying (2.2) and (2.3) to the evaluation functional Fx , we obtain Z Z ϕy (x) dµ(y), x ∈ M. Fx dµ˜ = f (x) = Q
ext K
Lemma 2.34. Let X be a compact convex subset of a locally convex space E. Then the space (E ∗ + R)|X is dense in Ac (X). Proof. Let X be a nonempty compact convex set. Fix a function h ∈ Ac (X) and ε > 0. Denote J1 := {(x, t) ∈ X × R : t = h(x)} and J2 := {(x, t) ∈ X × R : t = h(x) + ε} . Then J1 , J2 is a pair of disjoint nonempty compact convex subsets of E × R. By the Hahn–Banach theorem, there exist a functional F ∈ (E × R)∗ and λ ∈ R such that sup F (J1 ) < λ < inf F (J2 ). There are ϕ ∈ E ∗ and β ∈ R such that F (t, r) = ϕ(t) + βr for any t ∈ E and any r ∈ R. From the separation of J1 and J2 it easily follows that β 6= 0. Put ψ(t) := β1 (λ − ϕ(t)) for t ∈ E. Since ϕ(t) + βh(t) < λ < ϕ(t) + βh(t) + βε,
t ∈ X,
we easily get kh − ψk < ε. Remark 2.35. In general, there might exist a continuous affine function on a compact convex set X that is not of the form (E ∗ + R)|X ; see Exercise 2.111. Proposition 2.36. A Radon measure µ ∈ M1 (X) represents x ∈ X if and only if Z ϕ(x) = ϕ dµ for any ϕ ∈ E ∗ . X
R Proof. Recall that, by definition, µ represents x ∈ X if h(x) = X h dµ for any h ∈ Ac (X). Hence the assertion is an easy consequence of Lemma 2.34 and the Lebesgue dominated convergence theorem. Definition 2.37 (The barycenter revisited). Proposition 2.36 enables us to extend slightly the definition of a barycenter to the case of nonconvex sets and points not belonging to this set. Let K be a compact subset of a locally convex space E. We say that x ∈ E is the barycenter of a Radon measure µ ∈ M1 (K), or that µ represents x, in a symbol x = r(µ), if Z ϕ(x) =
ϕ dµ for any K
ϕ ∈ E∗.
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2 Compact convex sets
Proposition 2.38. If X is a compact convex subset of a locally convex space, the barycenter mapping r : M1 (X) → X is affine and continuous. Hint. Obviously, r is affine. If {µγ } is a net in M1 (X), µγ → µ and f ∈ E ∗ , then f (r(µγ )) = µγ (f ) → µ(f ) = f (r(µ)). Consequently, the net {r(µγ )} is weakly converging in X to r(µ). Since X is a compact set, r(µγ ) → r(µ) by Proposition A.28. Proposition 2.39. If K is a compact subset of a locally convex space E and x ∈ E, then the following statements are equivalent: (i) x ∈ co K, (ii) there exists a Radon measure µ ∈ M1 (K) such that r(µ) = x. Proof. Let µ ∈ M1 (K) satisfy r(µ) = x. Assuming that x ∈ / co K, by the Hahn– Banach theorem there exist ϕ ∈ E ∗ and λ ∈ R such that ϕ(x) < λ ≤ ϕ(t)Rfor any t ∈ co K. Then, obviously, no Radon measure µ ∈ M1 (K) with ϕ(x) = K ϕ dµ exists. Conversely, assume that x ∈ co K. There exists a net {xα } of points in co K such that xα → x. We can write xα =
nα X
λαj xαj ,
where nα ∈ N, xαj ∈ K, λαj ≥ 0,
nα X
λαj = 1.
j=1
j=1
We define for each α µα :=
nα X
λαj εxαj .
j=1
Then µα ∈ M1 (K) and r(µα ) = xα . Since the set M1 (K) is compact, we may assume that {µα } converges to µ ∈ M1 (K). For any ϕ ∈ E ∗ , ϕ(r(µα )) = µα (ϕ) → µ(ϕ) and
ϕ(r(µα )) = ϕ(xα ) → ϕ(x).
Hence ϕ(x) = µ(ϕ) and r(µ) = x. Theorem 2.40 (Bauer’s characterization of ext X). Let X be a compact convex subset of a locally convex space and x ∈ X. Then x ∈ ext X if and only if the Dirac measure εx is the only measure from M1 (X) with a barycenter x. Proof. If x = 12 (a+b) where a, b ∈ X, a 6= b, then the measure 12 (εa +εb ) ∈ M1 (X), which differs from εx , has the barycenter x. Assume now that x ∈ ext X and µ ∈ M1 (X) with r(µ) = x are given. It must be shown that µ = εx . To see this, it suffices to show that spt µ = {x}. Assume that
2.1 Geometry of convex sets
19
z ∈ spt µ \ {x}. Let U ⊂ X be a closed convex neighborhood of z in X such that x∈ / U . The set U is compact and convex, µ(U ) < 1 in view of Proposition 2.39, and obviously µ(U ) > 0. Set µ1 :=
1 µ|U µ(U )
and
µ2 :=
1 µ| . µ(X \ U ) X\U
Then µ1 , µ2 are in M1 (X). Since µ1 (U ) = 1, we get r(µ1 ) ∈ U (again from Proposition 2.39). Hence r(µ1 ) 6= x. We have µ = µ(U )µ1 + (1 − µ(U ))µ2 and we see that x = µ(U )r(µ1 ) + (1 − µ(U ))r(µ2 ) is a convex combination of r(µ1 ) and r(µ2 ). This contradicts the assumption that x is an extreme point of X and yields the required conclusion. Proposition 2.41 (Choquet). Let X be a compact convex subset of a locally convex space E and x ∈ ext X. Then the family {y ∈ X : f (y) < λ},
f ∈ E ∗ and λ ∈ R with f (x) < λ,
form a base of neighborhoods of x in X. Proof. Let U be an open neighborhood of x. Then X \ U is a compact set so that x∈ / co(X \ U ). Indeed, if this were not true, then Proposition 2.39 would provide a probability measure µ carried by X \ U with r(µ) = x. But this would contradict our assumption that x is an extreme point because the only representing measure for x is the Dirac measure εx (see Theorem 2.40). Now the Hahn–Banach separation theorem yields the existence of a continuous linear functional f and λ ∈ R such that f (x) < λ and f > λ on co(X \ U ). Thus x ∈ {y ∈ X : f (y) < λ} ⊂ U and the proof is finished. Proposition 2.42. Let K1 , . . . , Kn be compact convex subsets of a topological vector space. Then co(K1 ∪ · · · ∪ Kn ) is compact. Proof. We follow the same lines as in the proof of Corollary 2.8. Since co(K1 ∪ · · · ∪ Kn ) = F (D × K1 × · · · × Kn ) where n n o X D := λ ∈ Rn : λ = (λ1 , . . . , λn ), λj = 1 and λj ≥ 0 for j = 1, . . . , n j=1
20
2 Compact convex sets
and F : (λ, x1 , . . . , xn ) 7→
n X
(λ, x1 , . . . , xn ) ∈ D × K1 × · · · × Kn
λ j xj ,
j=1
is continuous, it follows that co(K1 ∪ · · · ∪ Kn ) is compact. Theorem 2.43 (Milman). Let B be a subset of a locally convex space E such that the set X := co B is compact. Then ext X ⊂ B. Proof. Assume that x ∈ ext X \ B. By Proposition 2.41, there exist f ∈ E ∗ and λ ∈ R such that f (x) > λ ≥ inf f (B). Since f ∈ E ∗ , inf f (B) = inf f (co B). Therefore we get x ∈ / co B = X, which is a contradiction. Proposition 2.44. Let X be a compact convex subset of a locally convex space E and ∅ 6= B ⊂ X. Then the following assertions are equivalent: (i) co B = X, (ii) inf f (B) = min f (X) for each f ∈ E ∗ , (iii) ext X ⊂ B. Proof. Assertions (i) and (iii) are equivalent by the Krein–Milman and Milman theorems 2.22 and 2.43. The implication (i) =⇒ (ii) is obvious. Conversely, the proof of the implication (ii) =⇒ (i) follows the same lines as the end of the proof of the Krein–Milman theorem: if co B 6= X, the Hahn–Banach separation theorem yields the assertion. Proposition 2.45. Any metrizable compact convex subset X of a locally convex is affinely homeomorphic to a compact convex subset of the Hilbert space `2 . Proof. Since X is a metrizable compact set, the space C(X) is separable, so it is the space Ac (X) of all continuous affine functions on X. Let {fn : n ∈ N} be a dense subset of the closed unit ball of Ac (X). If T : x 7→
n1 n
o fn (x) ,
x ∈ X,
then T is a continuous affine injective mapping of X into `2 . Thus T is an affine homeomorphism of X onto a compact convex subset T (X) of `2 .
2.1 Geometry of convex sets
2.1.C
21
Exposed points
Definition 2.46 (Exposed points). If X is a compact convex set, a point x ∈ X is exposed if there exists a function f ∈ Ac (X) such that f (x) > f (y) for each y ∈ X \ {x}. We call such a function f an exposing function for x and the set of all exposed points of X is denoted as exp X. We point out the following simple, but important fact. Proposition 2.47. For any compact convex set, exp X ⊂ ext X. Proof. The proof follows by a straightforward verification. Definition 2.48 (Farthest points). Let D be a subset of a normed linear space E. A point z ∈ D is called a farthest point of D if there exists x ∈ E such that kx − zk = sup{kx − tk : t ∈ D}. The set of all farthest points of D is denoted by far D. Lemma 2.49. If X is a nonempty compact convex subset of a Hilbert space H, then far X ⊂ exp X and it is a nonempty set. Proof. The inclusion far X ⊂ exp X follows from the fact that any point x in the closed unit ball BH of norm 1 is an exposed point of BH (we recall that BH is a compact convex set in the weak topology of H). Indeed, f (y) := (y, x), y ∈ H, is an exposing function for x. If y ∈ H is chosen arbitrarily, a simple compactness argument yields the existence of a point z ∈ X such that ky − zk = sup{ky − xk : x ∈ X}. Hence far X is nonempty. Proposition 2.50. Let X be a nonempty compact convex subset of a Hilbert space H. Then X = co far X = co exp X. Proof. Since far X ⊂ exp X, it suffices to show that X = co far X. Assume that x ∈ X \co far X. By the Hahn–Banach theorem and the Fr´echet–Riesz representation theorem, there exist h ∈ H and λ ∈ R such that (x, h) < λ ≤ (y, h) for each y ∈ co far X. Denote s := sup{ktk : t ∈ X}. Let α > 0 be such that 2α λ − (h, x) > s2 − kxk2 . By compactness, there exists z ∈ X such that kαh − zk = sup{kαh − tk : t ∈ X}. Of course, z ∈ far X. Since kαh − zk2 = α2 khk2 − 2α(h, z) + kzk2 ≤ α2 khk2 − 2αλ + s2 < α2 khk2 − 2α(h, x) + kxk2 = kαh − xk2 , we get kαh − zk < kαh − xk, which is a contradiction.
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2 Compact convex sets
Theorem 2.51. Let X be a metrizable compact convex subset of a locally convex space. Then X = co exp X. Proof. By Proposition 2.45, X is affinely homeomorphic to a compact convex subset of `2 . Obviously, exposed points are preserved by an affine homeomorphism and thus the assertion follows from Proposition 2.50. Corollary 2.52. Let X be a metrizable compact convex subset of a locally convex space. Then ext X ⊂ exp X. Proof. The assertion is an immediate consequence of Theorem 2.51 and Proposition 2.44.
2.2
Interlude: On the space M(K)
Let K be a compact space and M(K) the space of all signed Radon measures on K. We emphasize that on the space M(K) we always consider the weak∗ -topology given by the duality of C(K) and M(K). In this chapter we collect some properties of this space and its subspaces. Example 2.53 (Image of a measure from M1 (K)). If K is a compact space, it follows from Proposition 2.27 that the mapping ε : x 7→ εx , x ∈ K, is a homeomorphism of K onto ext M1 (K). Let Λ := ε] λ be the image of a probability measure λ with spt λ ⊂ K under ε. The following proposition will be useful in many examples. Proposition 2.54. The measure Λ from Example 2.53 is carried by the (closed) set ε(spt λ) and its barycenter equals λ. Proof. Since Λ(ε(spt λ)) = λ(ε−1 (ε(spt λ))) = λ(spt λ) = 1, we see that Λ is carried ∗ by ε(spt λ). Pick ϕ ∈ M(K) . By duality theory, there exists f ∈ C(K) such that ϕ(µ) = µ(f ) for any µ ∈ M(K). Then (cf. Proposition A.92), Z Λ(ϕ) = (ε] λ)(ϕ) = λ(ϕ ◦ ε) = K
= λ(f ) = ϕ(λ), and r(Λ) = λ.
Z ϕ(εx ) dλ(x) =
f (x) dλ(x) K
2.2 Interlude: On the space M(K)
23
Proposition 2.55. Let F be a closed subset of a compact space K and b > 0. Then the function ϕb : µ 7→ µ({x ∈ F : µ({x}) ≥ b}),
µ ∈ M1 (K),
is upper semicontinuous on M1 (K). Proof. Let c > 0 and µ ∈ G := {ν ∈ M1 (K) : ϕb (ν) < c} be given. We will show that G contains a neighborhood W of µ. The set L := {x ∈ F : µ({x}) ≥ b} is finite. Let U be an open subset of K such that L ⊂ U and µ(U ) < c. For every x ∈ F \ U we find an open neighborhood Vx of x so that µ(V x ) < b. Using compactness, we select finitely many points x1 , . . . , xn such that F \ U ⊂ V x1 ∪ · · · ∪ V xn . Since the function ν 7→ ν(H) is upper semicontinuous on M1 (K) for every closed set H ⊂ K (see Theorem A.85(b)), the set W := {ν ∈ M1 (K) : ν(U ) < c, ν(V xi ) < b, i = 1, . . . , n} is open and contains µ. It remains to show that ϕb (ν) < c for every ν ∈ W . Given a measure ν ∈ W , let Lν := {x ∈ F : ν({x}) ≥ b}. It follows from the choice of W that Lν ⊂ U . Thus ϕb (ν) = ν(Lν ) ≤ ν(U ) < c, and ϕb is upper semicontinuous. Proposition 2.56. Let F be a closed subset of a compact space K. Then the function X ψ : µ 7→ µd (F ) := µ({x}), µ ∈ M1 (K), x∈F
is a limit of an increasing sequence of positive upper semicontinuous functions on M1 (K). Proof. For n ∈ N, let ψn be the function ϕb from Proposition 2.55 for b = 1/n. Then it is easy to check that ψn % ψ on M1 (K). As ψn are upper semicontinuous and positive functions, the proof is finished. Definition 2.57 (Fσ and Gδ faces). A face which is simultaneously an Fσ set is called an Fσ face. Analogously, a Gδ face is a face which is a Gδ set.
24
2 Compact convex sets
Proposition 2.58. Let F be a closed subset of a compact space K. Then the set G := µ ∈ M1 (K) : µ|F is continuous is a Gδ face of M1 (K) such that G ∩ ext M1 (K) = ∅. Proof. Let ψ : µ 7→ µd (F ), µ ∈ M1 (K), where µd is the discrete part of µ. According to Proposition 2.56, there is a sequence {ψn } of positive upper semicontinuous functions such that ψn % ψ on M1 (K). Then ∞ \ ∞ \ 1 G = µ ∈ M1 (K) : ψ(µ) = 0 = µ ∈ M1 (K) : ψn (µ) < . k n=1 k=1
It follows that G is a Gδ set which is obviously convex and extremal. Since ext M1 (K) = {εx : x ∈ K} (cf. Proposition 2.27) and Dirac measures are discrete, we have G ∩ ext M1 (K) = ∅. Proposition 2.59. Let H :=
∞ [
µ ∈ M1 ([0, 1]) : spt µ ⊂
1
n, 1
.
n=2
Then H is an Fσ face of M1 ([0, 1]) which is not of type Gδ . Proof. Obviously, H is a convex set and extremal. Moreover, H is of type Fσ . Since both sets H and M1 ([0, 1]) \ H are dense in M1 ([0, 1]), H is not a Gδ set (cf. Theorem A.58). Proposition 2.60. Let K be a compact space and ω ∈ M1 (K). Define ψ : µ 7→ µs (K),
µ ∈ M1 (K),
where µs is the singular part of µ with respect to the measure ω. Then ψ is a limit of a decreasing sequence of lower semicontinuous functions on M1 (K). Proof. For n ∈ N, set 1 ψn (µ) := sup µ(G) : G ⊂ K open and ω(G) < , n
µ ∈ M1 (K).
Obviously, {ψn } is a decreasing sequence of lower semicontinuous functions. Recall that, by Theorem A.85(b), the function µ 7→ µ(G),
µ ∈ M1 (K),
is lower semicontinuous on M1 (K) for any open set G ⊂ K.
2.2 Interlude: On the space M(K)
25
Pick n ∈ N and µ ∈ M1 (K). There exists a Borel set B ⊂ K such that µs (B) = µs (K) = ψ(µ) and ω(B) = 0. Let G ⊂ K be an open set containing B for which ω(B) < n1 . Then ψ(µ) = µs (B) ≤ µs (G) ≤ µ(G) ≤ ψn (µ). Hence, ψ ≤ ψn for any n ∈ N. It remains to show that limn→∞ ψn = ψ. To this end, pick µ ∈ M1 (K) and c > ψ(µ). Since µac ω (recall that µac denotes the absolutely continuous part of µ with respect to ω, see Proposition A.65), there exists n ∈ N so that µac (B) < c − ψ(µ) whenever B is a Borel set with ω(B) < ω(G) < n1 , then
1 n.
Now, if G ⊂ K is an open set satisfying
µ(G) = µs (G) + µac (G) ≤ µs (K) + c − µs (K) = c. Thus, ψn (µ) ≤ c, and therefore ψn → ψ. Proposition 2.61. If λ denotes Lebesgue measure on [0, 1] and L := µ ∈ M1 ([0, 1]) : µ ⊥ λ , then L is a Gδ face of M1 ([0, 1]) which is not of type Fσ . Proof. It is easy to check that L is convex, extremal and dense in M1 ([0, 1]). Therefore all that needs to be proved is that L is a Gδ set. Let {ψn } be a sequence of functions as in Proposition 2.60 for K := [0, 1] and ω := λ. The assertion then follows from the following equalities ∞ \ L = µ ∈ M1 ([0, 1]) : µs ([0, 1]) = 1 = µ ∈ M1 ([0, 1]) : ψn (µ) = 1 n=1
=
∞ \ ∞ \ n=1 k=1
1 µ ∈ M ([0, 1]) : ψn (µ) > 1 − k 1
,
because the functions ψn are lower semicontinuous. Since L is dense, it is not an Fσ set. Remark 2.62. In Exercise 2.118 we indicate another reasoning of the fact that L is a Gδ set.
26
2 Compact convex sets
Proposition 2.63 (Choquet’s examples). Let ϕ : µ 7→ µd ([0, 1]),
µ ∈ M1 ([0, 1]),
ψ : µ 7→ µs ([0, 1]),
µ ∈ M1 ([0, 1]).
and Then ϕ and ψ are bounded affine functions on M1 ([0, 1]) of the second Baire class and there exists a probability measure Λ on M1 ([0, 1]) such that Λ(ϕ) 6= ϕ(r(Λ)) and
Λ(ψ) 6= ψ(r(Λ)).
Proof. Obviously, ϕ and ψ are bounded and affine. By Propositions 2.60, 2.56 and A.53, both functions ϕ and ψ are of the second Baire class on M1 ([0, 1]) because M1 ([0, 1]) is metrizable by Theorem A.85. Let Λ be the image of Lebesgue measure λ on [0, 1] under ε. According to Proposition 2.54, Λ is carried by the (closed) set ε([0, 1]) and its barycenter is equal to λ. It remains to show that the “barycentric formula” Z ϕ dΛ Λ(ϕ) = M1 (K)
does not hold. Indeed, Z
Z
Λ(ϕ) =
ϕ dΛ = M1 (K)
ϕ dΛ ε([0,1])
Z 1 dΛ = 1 6= 0 = ϕ(λ).
= ε([0,1])
The same argument can be used in the case of the function ψ.
2.3
Structures in convex sets
Throughout this section, X will be a compact convex subset of a locally convex space E.
2.3.A
Extremal sets and faces
Recall that a nonempty set F ⊂ X is called extremal if x, y ∈ F whenever x, y ∈ X, λ ∈ (0, 1) and
λx + (1 − λ)y ∈ F.
One-point extremal sets are just extreme points of X. It is simply checked that a set F is extremal if and only if [ {(λF − (λ − 1)X) ∩ X : λ ≥ 1} ⊂ F.
2.3 Structures in convex sets
27
Proposition 2.64. Let F be a subset of X. Then (a) F ∩ ext X ⊂ ext F , (b) if F is extremal, then ext F = F ∩ ext X. Proof. The assertion is a straightforward consequence of definitions. Recall that convex extremal sets are called faces. Closed extremal sets occasionally bear the name absorbent sets. Definition 2.65 (Generated faces). It is easy to see that any intersection of faces of X is again a face. Hence, given a set A ⊂ X, there exists the smallest face of X containing A. It equals the intersection of all faces containing A and is denoted by face A. Given x ∈ X, we will write simply face x instead of face {x} where no confusion can arise. Proposition 2.66. If F is a convex subset of X, then [ face F = {(λF − (λ − 1)X) ∩ X : λ ≥ 1} . Moreover, if F is closed, face F is an Fσ set. Proof. Let Fλ := (λF + (λ − 1)X) ∩ X,
λ ≥ 1.
First we notice that, given λ > 1, y ∈ Fλ if and only if there exists x ∈ F such that y + (λ − 1)−1 (x − y) ∈ X. S Hence Fλ ⊂ Fλ0 if 1S≤ λ ≤ λ0 . It follows that λ≥1 Fλ is a convex set. Since it is easy to observe that λ≥1 Fλ is extremal, we have [ face F ⊂ Fλ . λ≥1
On the other hand, Sit is immediate to verify from the definition the converse inclusion. Hence face F = λ≥1 Fλ . If F is closed, a routine verification yields that each Fλ is closed as well. Hence face F =
∞ [
Fn
n=1
is an Fσ set. Corollary 2.67. If x ∈ X, then face x = {y ∈ X : there exists z ∈ X and λ ∈ [0, 1) such that x = λz + (1 − λ)y} and face x is an Fσ set.
28
2 Compact convex sets
Proof. Use Proposition 2.66. Proposition 2.68. Let F be a subset of X. Then the following assertions are equivalent: (i) F is extremal, (ii) the characteristic function cF of F is convex, (iii) F is a union of faces. Proof. The equivalence of (i) and (ii) is clear from the definition. Since it is easy to check that any union of extremal sets is extremal, we have (iii) =⇒ (i). If (i) holds, then [ F = face x, x∈F
since face x ⊂ F for any x ∈ F . Proposition 2.69. Let F be a subset of X. The following assertions are equivalent: (i) F is a closed extremal set, (ii) the characteristic function cF of F is upper semicontinuous and convex, (iii) there exists a positive, lower semicontinuous and concave function f on X such that F = {x ∈ X : f (x) = 0}, (iv) F is closed and spt µ ⊂ F whenever µ ∈ M1 (X) and r(µ) ∈ F , (v) F is closed and a union of faces, (vi) F is closed and a union of closed faces. Proof. The equivalence of (i), (ii) and (v) follows immediately; see Proposition 2.68. Assertions (ii) and (iii) are obviously equivalent. To see that (i) =⇒ (iv), suppose that F is a closed extremal set and µ ∈ M1 (X) such that r(µ) ∈ F . If spt µ is not a subset of F , then there exists a closed convex set C ⊂ X \ F such that µ(C) > 0. We have µ(C) < 1, since otherwise r(µ) = r(µ|C ) ∈ / F . Set µ1 :=
1 µ|C µ(C)
and
µ2 :=
1 µ| . 1 − µ(C) X\C
Now, µ = µ(C)µ1 + (1 − µ(C))µ2 , hence r(µ) = µ(C)r(µ1 ) + (1 − µ(C))r(µ2 ) ∈ F. Since F is extremal, r(µ1 ), r(µ2 ) ∈ F , which is a contradiction.
2.3 Structures in convex sets
29
Now let F be a closed set and let x ∈ F , x = λy + (1 − λ)z where y, z ∈ X, λ ∈ (0, 1). If µ := λεy + (1 − λ)εz , then r(µ) = x and spt µ = {y, z}. Hence y, z ∈ F , which shows that F is extremal and proves that (iv) =⇒ (i). Since (vi) obviously implies (v), all that remains to be proved is that (i) =⇒ (vi). So let F be a closed extremal subset of X and x ∈ X. The set F := {C ⊂ F : x ∈ C, C convex} ordered by inclusion satisfies the assumptions of Zorn’s lemma. Indeed, if R ⊂ F is a chain, then the set [ {R : R ∈ R} belongs to F and it is an upper bound of R. Hence there exists a maximal element C ∈ F. Since x ∈ C ⊂ C ⊂ F and C ∈ F, the maximality of C implies that C = C. It remains to show that C is a face. Since C is convex, we must verify only that C is extremal. So, we are given a, b ∈ X and λ ∈ (0, 1) such that z := λa + (1 − λ)b ∈ C, and we wish to show that e be the convex hull of C and {a, b}. The proof a, b ∈ C. In order to prove this, let C e ⊂ F . Then, due to the maximality that a, b ∈ C will be achieved by showing that C e of C, C = C. e Then Choose e c ∈ C. e c = λ1 (λ2 a + (1 − λ2 )b) + (1 − λ1 )c, where c ∈ C and λ1 , λ2 ∈ [0, 1]. It is clearly sufficient to assume that λ1 ∈ (0, 1). (If λ1 = 0, e c = c. If λ1 = 1, e c belongs to the segment joining a and b, and thus e c ∈ F by the extremality of F .) Moreover, we may assume that λ2 ≥ λ (otherwise, 1 − λ2 ≥ λ). With λ1 and λ2 chosen in this manner, we set α :=
λ1 λ2 λ1 λ2 + λ(1 − λ1 )
and
β :=
λ . λ1 λ2 + λ(1 − λ1 )
Since β > 0, βe c + (1 − β)b = αz + (1 − α)c ∈ C ⊂ F and since F is extremal, we get e c ∈ F. Remarks 2.70. (a) Let F be an extremal set and x ∈ F . If µ ∈ M1 (X) and x = r(µ), then spt µ ⊂ F . The proof of this assertion can be obtained as a slight modification of the proof of the implication (i) =⇒ (iv) in Proposition 2.69. (b) In general, the closure of a face need not be a face. An example can be found in E. M. Alfsen [1], Theorem 1. In Exercise 4.52 we present an example of a compact convex set X and a point x ∈ X such that face x is not a face.
30
2 Compact convex sets
Corollary 2.71. Let F be a closed convex subset of X. Then F is a face if and only if for every measure µ ∈ M1 (X) with barycenter r(µ) in F , we have spt µ ⊂ F . Proof. A closed convex set is a face if and only if it is extremal. Hence the assertion follows immediately from Proposition 2.69, (i) ⇐⇒ (iv). Proposition 2.72. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . (a) If H ⊂ Y is extremal, then ϕ−1 (H) is an extremal set of X. (b) If H ⊂ Y is a face, then ϕ−1 (H) is a face of X. (c) ϕ(ext X) ⊃ ext Y . Proof. Let H ⊂ Y be an extremal set and αx1 + (1 − α)x2 ∈ ϕ−1 (H), where x1 , x2 ∈ X, α ∈ [0, 1]. Then αϕ(x1 ) + (1 − α)ϕ(x2 ) ∈ H, which yields ϕ(x1 ), ϕ(x2 ) ∈ H. Hence x1 , x2 ∈ ϕ−1 (H), concluding the proof of (a). Since (b) is a straightforward consequence of (a), we proceed to the proof of (c). Given a point y ∈ ext Y , the set ϕ−1 (y) is a closed extremal set by (a). By Proposition 2.20, it intersects ext X. Hence y ∈ ϕ(ext X), and we are done.
2.3.B
Measure convex sets
In the sequel, stronger versions of convexity and extremality are investigated. Constructions of counterexamples use properties of sets of probability measures studied in Section 2.2. As above, throughout this subsection, X will be a compact convex subset of a locally convex space E. Definition 2.73 (Measure convex sets). A universally measurable set F ⊂ X is measure convex if the barycenter r(µ) belongs to F for any measure µ ∈ M1 (X) carried by F . We will show that, for a universally measurable set F ⊂ X, the relations between “F measure convex” (labelled as MC) and “F convex” (labelled as C) are as follows: MC ⇒ C MC : C MC ⇐ C MC ⇐ C MC ⇐ C MC ⇐ C MC : C MC : C
F is closed or open F is resolvable dim E < ∞ F is a resolvable face F is Fσ face F is Gδ face
(2.74) (2.81) (2.74, 2.76) (2.80) (2.77) (2.91) (2.82, 2.83) (2.84)
2.3 Structures in convex sets
31
Proposition 2.74. Every measure convex universally measurable subset of X is convex, and every closed convex subset of X is measure convex. Proof. Let F ⊂ X be measure convex, x, y ∈ F , λ ∈ [0, 1] and z = λx + (1 − λ)y. Since the measure µ := λεx + (1 − λ)εy is carried by F and r(µ) = z, we get z ∈ F . Therefore, F is convex. If F is a closed convex set and if a measure µ ∈ M1 (X) has its support contained in F , then the implication (ii) =⇒ (i) of Proposition 2.39 tells us that r(µ) ∈ F . Accordingly, F is measure convex. Theorem 2.75. Let A be a universally measurable subset of X. Then A is measure convex if and only if co K ⊂ A for any compact set K ⊂ A. Proof. Let A be a measure convex subset of X and K ⊂ A a compact set. If x ∈ co K, then there exists a measure µ ∈ M1 (K) such that x = r(µ) (see Proposition 2.39). By the assumption, x ∈ A. For the proof of the converse implication, let µ ∈ M1 (X) be a probability measure with µ(A) = 1. If µ(K) = 1 for some compact set K ⊂ A, then by Proposition 2.39 and by the assumption, r(µ) ∈ co K ⊂ A. So assume that µ(K) < 1 for each compact set K ⊂ A. In this case, there exists an increasing sequence {Kn }∞ n=0 of compact sets in A satisfying K0 = ∅,
αn := µ(Kn+1 \ Kn ) > 0, n ≥ 0,
and
µ(Kn ) → 1.
P ∞ Since ∞ n=1 αn = 1 − α0 < 1, there is a sequence {βn }n=1 of real numbers in (0, 1] such that ∞ X αn = 1. βn → 0 and βn n=1
Put L := K1 ∪
∞ [
(βn Kn+1 + (1 − βn )K1 ) .
n=1
Then L ⊂ A. Since each Kn is compact and βn → 0, L is compact as well. Pick f ∈ E ∗ and set sn := sup f (Kn+1 ), n ≥ 0. Then s := sup f (L) = max s0 , sup βn sn + (1 − βn )s0 n∈N
= s0 + sup (sn − s0 )βn . n∈N
32
2 Compact convex sets
Hence sn ≤ s0 + βn−1 (s − s0 ) for each n ∈ N. Since f ≤ sn on Kn+1 \ Kn , we get Z f dµ =
f (r(µ)) = X
≤
∞ X
∞ Z X
f dµ
Kn+1 \Kn
n=0
αn s n ≤ α0 s 0 +
n=0
∞ X
αn (s0 + βn−1 (s − s0 ))
n=1
= α0 s0 + (s − s0 )
∞ X n=1
∞
X αn + s0 αn βn n=1
= s0 + s − s0 = s. Thus f (r(µ)) ≤ sup f (L). As f is arbitrary, Proposition 2.44 gives r(µ) ∈ co L. Since our assumption ensures that co L ⊂ A, the proof is finished. Proposition 2.76. Any open convex subset of X is measure convex. Proof. Let G ⊂ X be an open convex set. By Theorem 2.75, it is enough to show that co K ⊂ G whenever K ⊂ G is a compact set. So fix such K. For every x ∈ K there exists a compact convex neighborhood Vx such that x ∈ Vx ⊂ G. By compactness, the set K can be covered by finitely many compact convex sets Vx1 , . . . , Vxn . Then, by Proposition 2.42, co K ⊂ co (Vx1 ∪ · · · ∪ Vxn ) = co (Vx1 ∪ · · · ∪ Vxn ) ⊂ G, which is the required inclusion. Proposition 2.77. Let X be a subset of a finite-dimensional space. Then any universally measurable convex set A ⊂ X is measure convex. Proof. We again use Theorem 2.75. If K ⊂ A is a compact set, then co K ⊂ A is compact by Theorem 2.8 (see also Remark 2.5). Lemma 2.78. Let λ be a probability measure on X. If T := {µ ∈ M+ (X) : µ ≤ λ, µ 6= 0} and S := {r( then the closure of S equals co spt λ.
µ ) : µ ∈ T }, kµk
2.3 Structures in convex sets
33
Proof. It is easy to see that S = {r(µ) : µ ∈ M1 (X), there exists c ∈ R so that µ ≤ cλ}, from which it follows that S is convex. Set L := co spt λ. To show that S ⊂ L, let µ be a nontrivial measure on X with µ ) ∈ L because L is a closed convex set. µ ≤ λ. Then µ is carried by L and thus r( kµk Thus S ⊂ L and consequently S ⊂ L. Conversely, assuming that λ(S) < 1, we can find a compact set K ⊂ X \ S such that λ(K) > 0. For every x ∈ K we choose its closed convex neighborhood Vx not intersecting S. Using a compactness argument we select finitely many points x1 , . . . , xn of K so that Vx1 ∪· · ·∪Vxn covers K. As λ(K) > 0, there is i ∈ {1, . . . , n} so that λ(Vxi ) > 0. We set V := Vxi and µ := λ|V . Then µ is nontrivial and µ ≤ λ. µ µ belongs to S. On the other hand, r( kµk ) ∈ V because Hence the barycenter of kµk V is a closed convex set. This contradiction shows that λ(S) = 1. Thus spt λ ⊂ S which gives L ⊂ S. We recall that resolvable sets are defined and their basic properties presented in Section A.5. (In particular we note that any resolvable set in a compact space is universally measurable by Proposition A.118.) Lemma 2.79. Let F ⊂ X be a resolvable convex set and let λ ∈ M1 (X) be carried by F . Then there exists a nonempty set G ⊂ F ∩ co spt λ which is open in co spt λ. Proof. Let L := co spt λ. In order to find the required set G we note that L = F ∩ L because the latter set is a closed convex set containing the support of λ. In particular, F ∩ L is a dense resolvable set in L. Due to Proposition A.117(c), F ∩ L has a nonempty interior (relative to L). Hence, the interior of F ∩ L is the sought set G. Proposition 2.80. Any resolvable convex subset of X is measure convex. Proof. Let F be a resolvable convex subset of X and let λ be a probability measure on X carried by F . We set λ0 := λ and let L0 := co spt λ0 . Let S0 , T 0 and G0 be sets obtained from Lemma 2.78 and Lemma 2.79 when applied to the measure λ0 . Since S0 is dense in L0 and G0 is nonempty and open in L0 , there is a measure µ0 ∈ T 0 with µ0 ) ∈ G0 ⊂ F. r( kµ0 k We set λ1 := λ0 − µ0 and construct by transfinite induction a sequence {λα } of positive measures on X such that, for every ordinal number α ≥ 1, (i) λα+1 ≤ λα , (ii) either λα = 0 or kλα+1 k < kλα k,
34
2 Compact convex sets
(iii) if λα − λα+1 6= 0, then r(
λα − λα+1 ) ∈ F. kλα − λα+1 k
Suppose that the construction has been completed up to an ordinal α. If λα = 0, we set λα+1 := 0. If λα is nontrivial, we apply Lemma 2.78 and Lemma 2.79 to the measure kλλαα k (which is carried by F ) and get relevant sets Lα , T α , Sα and Gα with the properties described there. In particular, we have Gα ⊂ F ∩ L0 . As in the first step of the proof we choose a nontrivial measure ν ∈ T α such that ν ∈ Gα . r kνk By setting λα+1 := λα − ν we finish the inductive step for an isolated ordinal number. Let α be a limit ordinal number. Assume that λβ has been defined for every β < α. Since {λβ }β<α is a decreasing sequence of positive measures, by the Riesz representation theorem, the mapping λα : g 7→ inf λβ (g),
g ∈ C(X), g ≥ 0,
β<α
defines the measure λα . This step finishes the inductive construction. Let γ be the first ordinal number for which λγ = 0. Since {kλα k : α < γ} is a strictly decreasing transfinite sequence, the ordinal number γ is countable. We enumerate {λα − λα+1 }1≤α<γ into a (possibly finite) sequence {µn }, and obtain that X λ = µ0 + µn n≥1
and kλk = kµ0 k +
X
kµn k.
n≥1
If the sequence {µn } is finite, the equality λ = kµ0 k ·
X µ0 µn + kµn k · kµ0 k kµn k n≥1
yields that λ is a finite convex combination of probability measures having their barycenters in F . Thus, in this case, r(λ) ∈ F . Now, assume that the sequence {µn } is infinite. For every k ∈ N we set X c0 := kµ0 k, ck := kµn k n≥k
and c0 µ0 ck ωk := · + · c0 + ck c0 c0 + ck
P
n≥k
ck
µn
.
2.3 Structures in convex sets
35
Then {ωk } is a sequence of probability measures tending to c−1 0 µ0 . Moreover, µ0 + P n≥k µn is obviously an element of T 0 , and thus the barycenter r(ωk ) of ωk is contained in L0 . As r( µc00 ) ∈ G0 , which is a relatively open subset of L0 , we can find a sufficiently large k ∈ N such that r(ωk ) ∈ G0 ⊂ F . Then k−1
λ = c0
X µn µn µ0 X + kµn k + kµn k c0 kµn k kµn k n=1
=
k−1 X n=1
kµn k
n≥k
µn + (c0 + ck ) · ωk , kµn k
and the last formula shows that λ is a finite convex combination of measures which have their barycenters in F . Since F is convex, the barycenter of λ belongs to F as well. There are convex Borel subsets of a compact convex set which are not measure convex. We present some of them. Proposition 2.81. Let X := M1 ([0, 1]) and B := µ ∈ M1 ([0, 1]) : µ is discrete . Then B is a convex Borel set containing ext X which is not measure convex. Proof. It is clear that B is convex. Further, by Proposition 2.27, ext X = ext X = {εx : x ∈ [0, 1]} ⊂ B, and B is a Borel set by Proposition 2.56, since B = µ ∈ M1 ([0, 1]) : µd ([0, 1]) = 1 . Assume that B is measure convex. Pick ω ∈ X and find a measure Ω ∈ M1 (X) with barycenter ω carried by ext X (see Theorem 2.31). Since B is supposed to be measure convex, ω ∈ B. Hence B = X, which is a contradiction. Proposition 2.82. Let X :=
1 {xn } ∈ ` : 0 ≤ xn ≤ 2 for any n ∈ N n 1
and B := {xn } ∈ X : the set {n ∈ N : xn 6= 0} is finite . Then X is a compact convex set and B is an Fσ face of X which is not measure convex.
36
2 Compact convex sets
Proof. Since
∞ [ B= {xn } ∈ X : xn = 0 for all n ≥ j , j=1
B is an increasing countable union of closed faces. Therefore, B is an Fσ face. If an := n12 en , n ∈ N, where en is the standard unit vector in `1 , then an ∈ B for each n. Setting ∞ X 1 εa , µ := 2n n n=1 P 1 µ is carried by B whereas r(µ) = ∞ / B. n=1 2n an ∈ Proposition 2.83. If H :=
∞ [
µ ∈ M1 ([0, 1]) : spt µ ⊂ [ n1 , 1] ,
n=2
then H is an Fσ face of M1 ([0, 1]) which is not measure convex. Proof. By Proposition 2.59, H is an Fσ face of M1 ([0, 1]). Define the measure ω on [0, 1] as ∞ X 1 ε1. ω := 2n n n=1 / H. This shows If Ω := ε] ω, then Ω ∈ M1 M1 ([0, 1]) , Ω(H) = 1 and r(Ω) = ω ∈ that H is not measure convex. Proposition 2.84. There exists a Gδ face which is not measure convex. Proof. Let λ be Lebesgue measure on [0, 1] and L := µ ∈ M1 ([0, 1]) : µ ⊥ λ . By Proposition 2.61, L is a Gδ face. If Λ := ε] λ (cf. Example 2.53), then Λ ∈ M1 M1 ([0, 1]) , Λ(L) = 1 and r(Λ) = λ ∈ / L. Hence, L is not measure convex.
2.3.C
Measure extremal sets
Definition 2.85 (Measure extremal sets). Recall (see condition (iv) of Proposition 2.69) that a closed set F is extremal if and only if spt µ ⊂ F for every measure µ ∈ M1 (X) having its barycenter r(µ) in F . Hence, the following definition seems to be quite natural. A universally measurable set F ⊂ X is measure extremal if every measure µ ∈ M1 (X) with barycenter r(µ) in F is carried by F .
2.3 Structures in convex sets
37
We will show that, for a universally measurable set F ⊂ X, the relations between “F measure extremal” (labelled as ME) and “F extremal” (labelled as E) are as follows: ME ⇒ E ME ⇐ E ME ⇐ E ME ⇐ E ME : E ME : E
F is closed or open F is resolvable dim E < ∞ F is Fσ face F is Gδ face
(2.86) (2.86, 2.88) (2.92) (2.89) (2.94, 2.96) (2.93, 2.95)
Proposition 2.86. Every measure extremal universally measurable subset of X is extremal, and every closed extremal subset of X is measure extremal. Proof. Let F ⊂ X be measure extremal. Suppose that x, y ∈ X, λ ∈ (0, 1) and z := λx + (1 − λ)y ∈ F . Then the barycenter of the measure µ := λεx + (1 − λ)εy is z, hence in F . Since F is measure extremal, µ is carried by F . Therefore, x, y ∈ F . As has been already mentioned, any closed extremal set is measure extremal. Proposition 2.87. Let F be a universally measurable extremal subset of X. Then F is measure extremal if and only if X \ F is measure convex. Proof. Assume that F is measure extremal and µ ∈ M1 (X) is carried by X \ F . According to the hypothesis, r(µ) ∈ X \ F , which gives that X \ F is measure convex. Conversely, let X \ F be measure convex. Pick µ ∈ M1 (X) with r(µ) ∈ F . Note that µ(F ) > 0 since otherwise r(µ) would be contained both in F and in X \ F . Assume that µ(X \ F ) > 0 and set µ1 :=
1 µ|F µ(F )
and
µ2 :=
1 µ| . µ(X \ F ) X\F
Then r(µ2 ) ∈ X \ F and r(µ) = µ(F )r(µ1 ) + µ(X \ F )r(µ2 ). This is a contradiction since F is assumed to be extremal. Hence µ(X \ F ) = 0 and F is measure extremal. Since X \F is convex if F is extremal, Propositions 2.77, 2.74 and 2.76 yield using Proposition 2.87 the following two corollaries. Corollary 2.88. Every open or closed extremal subset of X is measure extremal.
38
2 Compact convex sets
Corollary 2.89. If A is a universally measurable extremal subset of a compact convex set in a finite-dimensional space, then A is measure extremal. Lemma 2.90. If X is a compact convex subset of a locally convex space E and F a proper extremal subset of X, then it has empty interior in X. Proof. Assume that z ∈ Int F and let x be any point of X \ F . By the continuity of the vector operations in E, there is α ∈ (0, 1) so that y := αx + (1 − α)z ∈ Int F . Since F is extremal, x ∈ F , which is a contradiction. Proposition 2.91. A resolvable face is closed and, consequently, it is measure convex. Proof. Let F be a nonempty resolvable face such that F \ F 6= ∅. Notice that F is a convex compact set. By Lemma 2.90, F has empty interior in F . Thus F \ F is dense in F . Since F and F \ F are disjoint nonempty dense subsets of the compact space F , Proposition A.117(e) and Theorem A.58 yield a contradiction. Proposition 2.92. Any resolvable extremal set is measure extremal. Proof. This follows from Proposition 2.80 and Proposition 2.87. Proposition 2.93. There exists a Gδ face which is not measure extremal. Proof. Let G := µ ∈ M1 ([0, 1]) : µ = µc . By Proposition 2.58, G is a Gδ face of M1 ([0, 1]) such that G ∩ ext M1 ([0, 1]) = ∅. Let λ denote Lebesgue measure on [0, 1]. If Λ := ε] λ (cf. Example 2.53), then by Proposition 2.54, r(Λ) = λ ∈ G whereas Λ(G) = 0 since Λ is carried by ε([0, 1]). Whence, G is not measure extremal. Proposition 2.94. There exists an Fσ face which is not measure extremal. Proof. Let again λ denote Lebesgue measure on [0, 1] and Λ = ε] λ (see Example 2.53). If F := face λ is the face generated by λ, then by Corollary 2.67, F is an Fσ face. Assume that µ ∈ F ∩ ε([0, 1]). Hence, µ = εx for some x ∈ [0, 1] and by Corollary 2.67, there exist ν ∈ M1 ([0, 1]) and α ∈ [0, 1) so that λ = αν + (1 − α)εx . Then 0 = λ({x}) = αν({x}) + (1 − α), which implies that α = 1, a contradiction. Therefore, F ∩ ε([0, 1]) = ∅. We see that Λ(F ) = 0 while r(Λ) = λ ∈ F . Therefore F is not measure extremal.
2.3 Structures in convex sets
39
Proposition 2.95. There exists a Gδ face which is neither measure convex nor measure extremal. Proof. We combine Examples 2.84 and 2.93. Let λ be Lebesgue measure on [0, 1] and C the Cantor ternary set. Set G := G1 ∩ G2 where G1 := µ ∈ M1 ([0, 1]) : µ ⊥ λ and G2 := µ ∈ M1 ([0, 1]) : µ|C is continuous . It follows from Propositions 2.58 and 2.61 that G is a Gδ set in M1 ([0, 1]). Further, as an intersection of faces, G is a face. Let Λ denote again the image ε] λ of Lebesgue measure λ on [0, 1] (see Example 2.53). Then r(Λ) = λ by Proposition 2.54 and the barycenter r(Λ) does not belong to G, although Λ(G) = λ(ε−1 (G)) = λ([0, 1] \ C) = 1. Thus G is not measure convex. Let Ω := ε] ν, where ν is a continuous probability measure carried by C. Then Ω is carried by ε(C), and consequently Ω(G) = 0. On the other hand, r(Ω) = ν ∈ G, and consequently G is not measure extremal. Proposition 2.96. There exists an Fσ face which is neither measure convex nor measure extremal. Proof. Here we combine examples constructed in Propositions 2.83 and 2.94. Set F := F1 ∩ F2 , where F1 :=
∞ [ µ ∈ M1 ([0, 1]) : spt µ ⊂ [ n1 , 1]
and
F2 := face λ
n=2
(here, λ is again Lebesgue measure on [0, 1] and face λ denotes the face generated by λ). According to the aforementioned examples, F is an Fσ face in M1 ([0, 1]). Let ω := 2λ|[ 1 ,1]
and
2
Ω := ε] ω.
Then spt Ω = ε [ 21 , 1] , thus spt Ω ∩ F = ∅. Hence Ω(F ) = 0, but r(Ω) = ω is contained in F . Hence F is not measure extremal. For the proof of the second statement, we define for n ∈ N λn :=
n λ| 1 n − 1 [ n ,1]
and
Ω :=
∞ X 1 ελ . 2n n n=1
40
2 Compact convex sets
Since λn ∈ F for every n ∈ N, Ω(F ) = 1. On the other hand, r(Ω) =
∞ X 1 λn 2n n=1
is not contained in F . Thus F is not measure convex and the proof is finished.
2.4
Exercises
Exercise 2.97. Let {Cα }α∈A be a collection of nonempty convex subsets of a vector space W . Prove that ∞ n n n [ [ X X co Cα = x∈W: x= λ j x αj , λj = 1, λj ≥ 0, xαj ∈ Cαj α∈A
n=1
j=1
j=1
o for j = 1, . . . , n and αj 6= αk for j 6= k . Exercise 2.98. Prove that x is an extreme point of a convex set X if and only if x x1 = x2 = · · · = xm whenever x = λ1 x1 + λ2 x2 + · · · + λm xm with m ∈ N, P= m j=1 λj = 1, xj ∈ X and λj > 0 for each j = 1, 2, . . . , m. Exercise 2.99. Let X 6= ∅ be a compact convex subset of Rd . Prove directly (do not use the Minkowski theorem 2.11) that ext X 6= ∅. Hint. Find x, y ∈ X such that |x − y| = diam X (here |x − y| denotes the Euclidean distance between points x and y). It easily follows that x, y ∈ ext X. Indeed, assume that 1 x = (x1 + x2 ) where x1 , x2 ∈ X, x1 6= x2 . 2 Since the vectors x1 − y and x2 − y are linearly independent, we get 1 1 diam X = |x − y| = (x1 − y) + (x2 − y) 2 2 1 < |x1 − y| + |x2 − y| ≤ diam X. 2
Another hint. Prove that any point z ∈ X having the property that |z| ≥ |x| for any x ∈ X is an extreme point of X. Exercise 2.100 (Radon). Assume that a set M ⊂ Rd contains at least d + 2 points. Then M = M1 ∪ M2 where M1 ∩ M2 = ∅ and co M1 ∩ co M2 6= ∅.
2.4 Exercises
41
Hint. Suppose that x1 , . . . , xn ∈ M , n ≥ d+2. Then there exists a nontrivial solution (α1 , . . . , αn ) to the system of d + 1 equations n X
αj = 0
n X
and
j=1
αj xj = 0.
j=1
Set I + := {j : αj ≥ 0}
and
I − := {j : αj < 0}
M1 := xj : j ∈ I +
and
M2 := xj : j ∈ I − .
and Then M1 ∩ M2 = ∅. Since λ :=
X
αj > 0,
j∈I +
we get co M1 3
X αj X αj xj = − xj ∈ co M2 . λ λ + −
j∈I
j∈I
Exercise 2.101 (Helly). Assume that K is a family of at least d + 1 convex sets in Rd such that either K is finite or the sets of K are in addition closed and one of them is compact. If each d + 1 sets of K have nonempty intersection, then \ {K : K ∈ K} 6= ∅. Hint. For a finite family K = {K1 , . . . , Kn }, suppose first that n = d + 2. By our assumption, there exist xi ∈
\
Kj ,
1 ≤ i ≤ n.
j∈{1,...,n}\{i}
T If there exist indices i 6= k with xi = xk , then xi ∈ nj=1 Ki . Otherwise we use Exercise 2.100 to find disjoint sets M1 , M2 such that M1 ∪ M2 = {x1 , . . . , xn } and co M1 ∩ co M2 contains a point y. Let i ∈ {1, . . . , n} be arbitrary. If xi ∈ M1 , then Ki ⊃ M2 . Hence y ∈Tco M2 ⊂ Ki . Analogously we get that y ∈ Ki in the case when xi ∈ M2 . Hence y ∈ ni=1 Ki . Assume now that the assertion has been proved for each family K in Rd consisting of n − 1 sets, where n ≥ d + 2. Let K = {K1 , . . . , Kn }. By the first part, each family K0 ⊂ K of at most d + 2 elements has nonempty intersection. Thus the family {K1 , . . . , Kn−2 , Kn−1 ∩ Kn }
42
2 Compact convex sets
satisfies that any subfamily with d + 1 elements has nonempty intersection. Hence the inductive assumption yields that the latter, and, consequently, the former family has nonempty intersection. Assume now that the family K consisting of closed convex sets is infinite and a set Z ∈ K is compact. By the first part, every finite subfamily of K has nonempty intersection. Hence K, and consequently {Z ∩K : K ∈ K}, has the finite intersection property. By compactness, \ \ {Z ∩ K : K ∈ K} = {K : K ∈ K} is nonempty. Exercise 2.102. (a) Let G be an open subset of Rd . Prove that co G is an open set. (b) If F is a closed subset of Rd , the convex hull co F need not be closed. Hint. Consider, for example, the following sets {(x, y) ∈ R2 : x > 0, xy = 1 or xy = −1} or {(x, y) ∈ R2 : x = 0} ∪ (1, 0). Exercise 2.103. (a) Let C be a compact convex subset of R2 . Prove that the set ext C is closed. (b) The set ext C need not be closed if C is a compact convex subset of Rd for d ≥ 3. Hint. Let {xn } be a sequence of points in ext C converging to x. Assuming that x is not an extreme point of C, let x = 21 (a + b) for some points a, b ∈ C, a 6= b. By passing to a subsequence, we may assume that all points xn are contained in the same open halfplane determined by the line passing through a and b. Then the interior of the triangle co{x1 , a, b} contains xn for a suitable n ∈ N, a contradiction with xn ∈ ext C. For (b), consider the convex hull in R3 of the set (x, y, z) : (x − 1)2 + y 2 = 1, z = 0 ∪ {(0, 0, 1) ∪ (0, 0, −1)} .
Exercise 2.104. Let C be a closed convex subset of Rd containing a line. Prove that ext C = ∅. If C contains no line, then ext C 6= ∅. Hint. If C contains a line L and x ∈ C, then C contains also a line passing through x parallel to L. If C contains no line then use induction on the dimension d. Find a boundary point of C and follow the reasoning of the proof of the Minkowski theorem 2.11.
2.4 Exercises
43
Exercise 2.105 (Closed convex hulls). Let C be a subset of a locally convex space E. Prove that the closed convex hull co C defined as \ co C := {F : F is a convex closed subset of E, F ⊃ C} is the closure of the convex hull co C, that is, co C = co C. Exercise 2.106 (Exposed points in Rd ). (a) Construct a compact convex set K ⊂ R2 for which ext K \ exp K 6= ∅. (b) Construct an example of a nonempty closed convex subset C of Rd such that the set exp C is not closed and co exp C 6= C. Hint. For the proof of (a) consider K to be the convex hull of the union of two circles {(x, y) ∈ R2 : (x + 1)2 + y 2 = 1} ∪ {(x, y) ∈ R2 : (x − 1)2 + y 2 = 1} then (−1, 1) ∈ ext K \ exp K. To show (b) consider again the example from (a). Exercise 2.107. (a) Let S be a d-simplex in Rd determined by affinely independent points e0 , . . . , ed . Prove that S has nonempty interior. (b) Using (a) prove that there are no d + 1 affinely independent points in any convex subset of Rd with empty interior. (c) Let C be a convex subset of Rd with empty interior. Prove that there is an affine subspace A of Rd containing C such that dim A < d. Hint. For (a) show that
e0 + · · · + ed ∈ Int S. d+1
To verify (c), let e0 , . . . , en be affinely independent points in C where n < d + 1 is the maximum number of affinely independent points in C. Consider now the affine hull of e0 , . . . , en . Exercise 2.108. Let X be a nonempty compact convex subset of a locally convex space E and x ∈ X. Prove that Mx (X) is a nonempty convex compact subset of M1 (X). Hint. A straightforward verification. Exercise 2.109. Let F be a closed subset of a compact convex set X such that ext X ⊂ F . Assume that for any x ∈ X there exists a unique measure µ ∈ M1 (F ) such that r(µ) = x. Prove that ext X = F .
44
2 Compact convex sets
Hint. Let x ∈ F \ ext X. Then x = y+z 2 , y 6= z, for some y, z ∈ X. The Integral representation theorem 2.31 yields measures µy , µz ∈ M1 (ext X) such that r(µy ) = y and r(µz ) = z. If µ = 12 (µy + µz ), then µ ∈ M1 (ext X),
r(µ) = x
and
µ 6= εx .
This contradicts the assumption, hence ext X = F . Exercise 2.110 (Proof of the Milman theorem 2.43). Verify the following indication of an alternative proof of Theorem 2.43. Hint. Let x ∈ ext X. We note that co B = co B. By Proposition 2.39 applied to K = B, there exists a measure µ ∈ M1 (B) representing the point x. Since x ∈ X, Bauer’s characterization in 2.40 asserts that µ = εx . The measure µ is carried by the (closed) set B, which yields that x ∈ B. Exercise 2.111. Find an example of a compact convex set X in a locally convex space E such that (E ∗ + R)|X 6= Ac (X). P n Hint. Let E := `2 , X := {x ∈ E : 0 ≤ xn ≤ 4−n , n ∈ N} and f (x) := ∞ n=1 2 xn , x ∈ X. Exercise 2.112. Prove that in any infinite-dimensional Banach space E, there exists a compact convex set X such that (E ∗ + R)|X 6= Ac (X). Hint. Find inductively points xn ∈ SE and functionals ϕn ∈ E ∗ , n ∈ N, such that ( 1, n = m, ϕn (xm ) = n, m ∈ N. 0, n 6= m, P −n on X, n Set X := co{4−n xn : n ∈ N} and f := ∞ n=1 2 ϕn . Since 0 ≤ ϕn ≤ 4 n ∈ N, f is well defined. If f = ϕ + c on X for some ϕ ∈ E ∗ and c ∈ R, then c = 0, because f (0) = 0. Further, since f (4−n xn ) = 2−n , we get ϕ(xn ) = 2n for each n ∈ N. But this is impossible as ϕ is bounded on BE . Exercise 2.113. Find a compact convex set X ⊂ R2 such that far X 6= exp X. Hint. Let X = {(x, y) ∈ R2 : |x|3 ≤ y ≤ 1, −1 ≤ x ≤ 1}. Then the point (0, 0) is obviously exposed. However, (0, 0) ∈ / far X as an easy geometrical argument shows. Exercise 2.114. Find an example of a compact convex subset X of a locally convex space E and a point z ∈ X that is exposed but there is no f ∈ E ∗ exposing z.
2.4 Exercises
45
First hint. Let H be a Hilbert space, M be a dense proper subspace of H ∗ and let Φ : H ∗ → H be the mapping assigning to each ϕ ∈ H ∗ a point y ∈ H such that ϕ(h) = (h, y) for any h ∈ H. Let X := BH be equipped with the σ(H, M )-topology. Since BH is w-compact and σ(H, M ) is weaker and Hausdorff, w = σ(H, M ) on X. Choose x ∈ SH \ Φ(M ) and denote ψ := Φ−1 (x). Show that ψ is not σ(H, M )continuous. Further show that ψ|X ∈ Ac ((X, σ(H, M )) and that ψ exposes x. On the other hand, no functional from (H, σ(H, M ))∗ = M exposes x. (This easily follows from the fact that, given x, y ∈ SH , (x, y) = 1 if and only if y = x.) Second hint. Let {en } be a sequence of standard unit vector in c0 . For 1 ≤ i ≤ j denote ui,j := −ei + 2ej . Let further X := cok·k C
where
C := 2−i−j ui,j : 1 ≤ i ≤ j .
Define the function h on X as h : x 7→
∞ X
xn ,
x = {xn } ∈ X.
n=1
It is easy to show that h is an affine continuous function on X. Obviously, C ∪ {0} is a compact subset of c0 and C ⊂ {{xn } ∈ c0 : |xn | ≤ 2−n }. It easily follows that X is compact. Let x ∈P X \ {0}. Since ext X ⊂ C ∪ {0}, by the Krein–Milman theorem 2.22 we have x = i≤j αi,j 2−i−j ui,j , where αi,j ≥ 0 and their sum is smaller or equal to 1. There exists a pair i ≤ j such that αi,j > 0. Hence h(x) =
X
αi,j 2−i−j h(ui,j ) =
i≤j
X
αi,j 2−i−j > 0.
i≤j
It follows that 0 is an exposed point of X. On the other hand, 0 is not exposed by any functional from (c0 )∗ = `1 . Indeed, assume that 0 6= f = {fn } ∈ (c0 )∗ = `1 . There exists i such that fi 6= 0. If fi < 0, then f (ui,i ) = f (ei ) = fi < 0. If fi > 0, then f (ui,j ) = f (−ei + 2ej ) = −fi + 2fj . Since limj→∞ fj = 0, we have limj→∞ f (ui,j ) = −fi . Hence, there exists j ≥ i such that f (ui,j ) < 0. If x := 2−i−j ui,j , then x ∈ X, x 6= 0, and f (x) < 0 = f (0). Exercise 2.115. Keeping the notation of the first hint in Exercise 2.114, consider locally convex spaces E1 := (H, w) and E2 := (H, σ(H, M )). Prove that the compact convex sets (BH , σ(H, M )) and (BH , w) are affinely homeomorphic (by the identity mapping), and that the point x is exposed by a functional from E1∗ whereas it is not exposed by a functional from E2∗ . (Compare with the proof of Theorem 2.51.) Hint. Follow the reasoning of Exercise 2.114.
46
2 Compact convex sets
Exercise 2.116. Let X be a nonempty compact convex set and K ⊂ X \ ext X be compact. Then co K ∩ ext X = ∅. Hint. For any x ∈ co K \ K there exists µ ∈ Mx (K) (see Proposition 2.39). Obviously, µ 6= εx , and thus x ∈ / ext X by Theorem 2.40. Exercise 2.117. Let F be a closed face of a compact convex set X and let U ⊂ X be an open set containing F . Then F ∩ co(X \ U ) = ∅. Hint. Proceed as in Exercise 2.116. Exercise 2.118. Let ν be a Radon probability measure on a compact space K. Let L := µ ∈ M1 (K) : µ ⊥ ν . Prove that L is a Gδ set. Hint. For each n ∈ N and each open subset G of K, let L(n, G) := µ ∈ M1 (K) : µ(G) > 1 − 2−n and Ln :=
[
L(n, G) : G ⊂ K open, ν(G) < 2−n .
Since each set L(n, G) is open in M1 (K) by Theorem A.85(b), the set Gδ set. The proof will be complete once we show that L=
∞ \
T∞
n=1 Ln
is a
Ln .
n=1
Pick an arbitrary index n ∈ N and µ ∈ L. As µ ⊥ ν, there is a Borel set A ⊂ K such that ν(A) = 0 and µ(A) = 1. Due to the regularity of ν it follows that there exists an open set G ⊃ A such that ν(G) < 2−n . Since µ(G) ≥ µ(A) = 1 > 1 − 2−n , we get µ ∈ Ln . T Conversely, assume that µ ∈ ∞ n=1 Ln . There is a sequence {Gn } of open subsets of K such that ν(Gn ) < 2−n and µ(Gn ) > 1 − 2−n T S∞ for each n ∈ N. Set G := ∞ m=1 n=m Gn . Then ν(G) = 0. Since 1 ≥ µ(G) = lim µ( m→∞
∞ [
n=m
Gn ) ≥ lim sup µ(Gm ) ≥ lim sup(1 − 2−m ) = 1, m→∞
m→∞
it follows that µ ⊥ ν. Hence, µ ∈ L, as required. Exercise 2.119. If F is a face of X and G is a face of F , then G is a face of X.
2.4 Exercises
47
Hint. A straightforward verification. Exercise 2.120. Prove that a set A ⊂ X is convex if X \ A is extremal. Hint. A straightforward verification. Exercise 2.121. Find an Fσ face that is not a countable union of closed faces. Hint. Use the set F from Proposition 2.94. If F were a countable union of closed faces, F would be measure extremal by Proposition 2.86. But this is not the case. Exercise 2.122. If F is a closed face of X, then there exists a set A ⊂ ext X such that F = co A. In particular, if F is a nonempty closed face of X, then F ∩ ext X 6= ∅ (cf. Theorem 2.20). Hint. Set A := F ∩ ext X. Then A = ext F by Proposition 2.64(b) and it suffices to apply the Krein–Milman theorem 2.22. Exercise 2.123. Let X be a convex subset of a locally convex space E. Prove that the boundary ∂X is an extremal subset of X. Hint. If Int X = ∅, the assertion is obvious. Hence, assume that Int X 6= ∅ and that ∂X is not extremal. In this case, there exist z ∈ ∂X, x ∈ X and y ∈ Int X so that z = λx + (1 − λ)y for some λ ∈ (0, 1). Find a neighborhood V of 0 such that y + V ⊂ Int X. For every ε > 0, there exists zε such that zε ∈ (z + εV ) \ X. Let wε :=
zε − λx . 1−λ
Show that we can choose ε small enough such that wε ∈ y + V . Since zε = λx + (1 − λ)wε , this implies that zε ∈ X, which is a contradiction. Exercise 2.124. Let X be a compact convex set and x ∈ X. Prove that \ D := {A ⊂ X : A is a closed extremal set containing x} is convex. Hint. The set D is obviously closed and extremal. Then use characterization (vi) of Proposition 2.69. Exercise 2.125. Find a continuous affine surjection of a compact convex set X onto a compact convex set Y such that ext Y 6= ϕ(ext X). Hint. Consider a triangle given as the convex hull of points (0, 0), (1, 0), ( 12 , 1) and its projection onto the unit segment [0, 1].
48
2 Compact convex sets
Exercise 2.126. Prove that the set G := µ ∈ M1 ([0, 1]) : µ = µc of all continuous Radon measures on [0, 1] is measure convex (cf. also Proposition 2.93). Hint. Let Ω ∈ M1 M1 ([0, 1]) , Ω(G) = 1 with r(Ω) = ω be given. For a bounded Borel function f on [0, 1], define an affine function If on M1 ([0, 1]) by the formula If (µ) := µ(f ),
µ ∈ M1 ([0, 1]).
Obviously, If is continuous whenever f is. Thus Ω(If ) = If (r(Ω)) = ω(f ) for any f ∈ C[0, 1].
(2.4)
Now, use the Lebesgue dominated convergence theorem to show that equality (2.4) holds for any bounded Borel function f on [0, 1]. Pick x ∈ [0, 1]. Then Z Ic{x} (µ) dΩ(µ) = 0. ω({x}) = Ic{x} (ω) = G
Thus ω({x}) = 0 for any x ∈ [0, 1], and therefore ω ∈ G. Exercise 2.127. Let K be a compact space, X := M1 (K), ω ∈ M1 (K), n µ o A1 := : µ ∈ M+ (K), µ ≤ ω and µ(K) > 0 µ(K) and A2 :=
n fω o : 0 ≤ f ≤ 1 Borel, ω(f ) > 0 . ω(f )
Prove that (a) face ω = A1 = A2 , (b) face ω = M1 ([0, 1]) if K = [0, 1] and ω is Lebesgue measure on [0, 1]. Hint. (a) Let µ ∈ face ω be given. Then there exist α ∈ (0, 1] and ν ∈ M1 (K) such that ω = αµ + (1 − α)ν. Then 0 < αµ ≤ ω. If µ ≤ ω, by the Radon–Nikodym theorem there exists a Borel measurable function f such that µ = f ω. Then 0 ≤ f ≤ 1 ω-almost everywhere and ω(f ) = µ(K) > 0. Hence µ fω = µ(K) ω(f ) and µ ∈ A2 .
2.5 Notes and comments
49
fω Finally, let µ = ω(f ) for some Borel function f with 0 ≤ f ≤ 1 and ω(f ) > 0. We may assume that ω(f ) < 1. Then
ω = ω(f )
(1 − f )ω fω + ω(1 − f ) , ω(f ) ω(1 − f )
and µ ∈ face ω. (b) Assume that λ is Lebesgue measure on [0, 1]. Given a point x ∈ [0, 1], it is easy to find a sequence of measures in face λ that converges to εx . Hence face λ is a convex set containing all extreme points of M1 ([0, 1]), and thus face λ = M1 ([0, 1]).
2.5
Notes and comments
The material of the Subsection 2.1.A on convexity in finite-dimensional spaces is standard and can be found in many textbooks; see, for example, Barvinok’s monograph [31]. Besides Minkowski’s result, at the beginning of 20th century, three important theorems on convex sets in Euclidean spaces associated with the names of Carath´eodory, Helly and Radon appeared. H. Minkowski proved Theorem 2.11 in the period 1901-03. The result appeared for the first time in a chapter on convex bodies (where the origin of the notion of extreme points can be traced) included in his collected works published in 1911; see J. J. Saccoman [405]. An exhaustive survey on Helly’s theorem 2.101 and its proofs, variants and applications is presented in the paper by L. Dantzer, B. Gr¨unbaum and V. Klee [130]. The Krein–Milman theorem 2.22 was proved by M. Krein and D. Milman in [283] for the case of w∗ -compact convex sets in the dual to a Banach space using the transfinite induction. The proof of a more general version contained in Theorem 2.22, based on an application of Zorn’s lemma, goes back to J. L. Kelley [266]. A similar proof was given by E. Artin (a letter from Artin to his former student M. Zorn published in the Picayune Sentinel of Indiana University in 1950; cf. [23]), A. Hotta [240] and by K. Yosida and M. Fukamiya [477]. The Krein–Milman theorem is one of the fundamental theorems of functional analysis and has rich applications. For instance, recall its use in de Branges’ proof of the Stone–Weierstrass theorem, Lindenstrauss’ proof of the Lyapunov theorem on the range of a vector measure, and in the proof of the Banach–Stone theorem on isometrically isomorphic spaces of continuous functions. See Chapter 14 for more applications. Bauer’s concave minimum principle 2.24 (even in a more general form) appeared in H. Bauer [36]. The Integral representation theorem 2.31 is a reformulation of the Krein–Milman theorem. It has wide applications in several areas of analysis. We gave two different proofs of this theorem (besides Theorem 2.31 it is its more general form in Proposition 2.39). Still another proof of Theorem 2.29, avoiding the notion of a net, is presented in Phelp’s book [374], Proposition 1.1.
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2 Compact convex sets
Bauer’s characterization of ext X in Theorem 2.40 presented in [37] is used later on as the definition of the Choquet boundary of function spaces in 3.4. Proposition 2.41 can be found as Proposition 25.13 in G. Choquet [108] and it is used several times in our text. For example, Milman’s converse 2.43 of the Krein–Milman theorem, which goes to V. P. Milman [346], is an easy consequence of it. Proposition 2.45 enables to solve certain problems concerning metrizable compact convex sets reducing them to subsets of Hilbert spaces. It was shown by O. -H. Keller [265] that any infinite-dimensional compact convex subset of a Hilbert space is N homeomorphic to the Hilbert cube [0, 1] (which is affinely homeomorphic to the set 1 2 {xn } ∈ ` : |xn | ≤ n , n ∈ N ). Moreover, V. L. Klee observed that any compact (convex) subset of a Banach space is affinely homeomorphic to a subset of a Hilbert space. An infinite-dimensional compact convex subset C of a topological vector space is said to be a Keller set if C is affinely homeomorphic to a compact convex subset of a Hilbert space `2 . Hence, Proposition 2.45 shows that any infinite-dimensional metrizable compact convex set is a Keller set. Moreover, it has been shown by C. Bessaga and T. Dobrowolski in [56] that any locally compact convex subset C of a topological vector space with a countable family of continuous affine functions on C separating points of C can be affinely embedded into `2 . The concept of exposed points in the case of Euclidean spaces was introduced by S. Straszewicz [441] in 1935. In this case, Corollary 2.52 and example in Exercise 2.103(b) are due to him. The proof of Proposition 2.50 is taken from V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. In the paper [53] by S. K. Berberian, the Krein–Milman theorem in Hilbert spaces is derived. This result can be deduced from more general statements concerning the exposed points in normed linear spaces; see, for example, the paper of V. L. Klee [272]. In fact, from the proof of Klee’s result it follows that in any smooth and strictly convex normed linear space any compact convex set is the closed convex hull of its set of so-called bare points (cf. a review of the paper [53] in Mathematical Reviews). See also the paper by M. V. Balashov [30]. We also refer the reader to the paper [151] by M. Edelstein and J. E. Lewis on exposed and farthest points. In [18], R. F. Arens and J. L. Kelley described extreme points of the unit ball of the ∗ space C(K) as Dirac measures εx and their antipodes −εx (cf. Proposition 2.27). For an alternative proof of Corollary 2.28 see, for example, Theorem 30.4 and Corollary 30.5 in Bauer’s monograph [45]. In Proposition 2.56 we present a simplified version of the proof of [5], Proposition I.2.8. Examples described in Proposition 2.63 are due to G. Choquet [106]; see also Alfsen’s monograph [5], Example I.2.10. Any union of faces was labelled in Goullet de Rugy in [200] as a σ-face. Closed extremal sets were studied by E. M. Alfsen in [1] under the name “stable sets”. The equivalence of (i) and (v) in Proposition 2.69 was proved by E. M. Alfsen in [1]; the idea of the proof (i) =⇒ (vi) is from D. P. Milman [346] (§ 4, Theorem 7).
2.5 Notes and comments
51
In part, the material Subsections 2.3.B and 2.3.C concerning a more detailed study of measure convex and measure extremal sets is taken from the paper [146] by P. Dost´al, J. Lukeˇs and J. Spurn´y. Theorem 2.75 and its proof are taken from D. H. Fremlin and J. D. Pryce [185]. Alfsen’s example 2.81 is taken from [5], p. 130. The proof of Proposition 2.92 can also be proved by means of a result of J. Saint Raymond 10.75 without recourse to Proposition 2.80. Counterexamples contained in Propositions 2.93 and 2.94 are just suitable modifications of the example by G. Choquet in [106]. The examples of Propositions 2.95 and 2.96 partially use a construction of H. v. Weizs¨acker (see [463]). We also refer the reader to Lecture Notes [473] by G. Winkler where a thorough investigation of measure convex sets is given. The result of Exercise 2.103(a) is due to G. B. Price [377]. A characterization of continuous affine functions on X (cf. Lemma 2.34 and Exercise 2.112) belonging to E ∗ + R|X even for noncompact convex sets X is given in a paper [281] by M. Kraus. Examples of Exercises 2.114 and 2.115 are due to M. Kraus and O. Kurka.
Chapter 3
Choquet theory of function spaces
This chapter lays the groundwork for the rest of the book by presenting the foundations of the Choquet theory of function spaces. The central concept of a function space is defined and its basic properties investigated in Section 3.1. We generalize the framework of spaces of affine continuous functions on compact convex sets by taking a subspace H of the space C(K) of continuous functions on a compact space K such that H contains constants and separates points of K. Then we introduce H-representing measures, H-affine and H-convex functions, and so on. A suitable substitute for the set of extreme points turns out to be the Choquet boundary and we show in Proposition 3.15 its nonemptiness and prove a minimum principle in Theorem 3.16. A crucial notion for obtaining integral representation theorems is the Choquet ordering introduced in Definition 3.19. This ordering somehow indicates how close to the Choquet boundary a measure is situated. A key tool for handling function spaces is Lemma 3.21 which serves as a substitute for the Hahn–Banach theorem. Several of its applications are shown afterwards, along with Bauer’s characterization 3.24 of the Choquet boundary. These abstract results are then applied to a reexamination of Korovkin’s theorems; Theorems 3.32, 3.34 and 3.36 are proved by means of the Choquet theory. After generalizing the concept of the barycenter mapping in Section 3.3, we turn our attention to the Choquet representation theorem 3.45 for function spaces on metrizable compact spaces. Our approach is to use the existence of a “strictly convex” function. Next, Section 3.5 indicates how the Key lemma 3.21 enables us to prove analogues of classical results on approximation of semicontinuous convex functions on compact convex sets. More precisely, we show that semicontinuous H-convex functions can be approximated by continuous H-convex functions (see Propositions 3.48 and 3.54). An important corollary is the fact that the Choquet ordering of measures can be extended to semicontinuous H-convex functions (see Proposition 3.56). Measures maximal with respect to the Choquet ordering are investigated in Section 3.6. First we prove Mokobodzki’s characterization in Theorem 3.58. Then we show that the set of maximal measures is rich enough (see Theorem 3.65) and we finish the section with Theorem 3.70, which describes the space of boundary measures. In order to prove the most important properties of maximal measures, namely, that they are carried by any Baire set containing the Choquet boundary, we need some kind of Fatou’s lemma (see Lemma 3.77). This task is accomplished in Section 3.7 by presenting the important Simons inequality 3.74 and a couple of its applications.
3.1 Function spaces
53
Then we can prove the integral representation theorem for nonmetrizable spaces in Theorem 3.81. The existence of representing measures “carried” by the Choquet boundary opens the way for the proof of several variants of the minimum principle, as shown in Section 3.9. The last section is devoted to a characterization of the fact that a pair of measures µ, ν is related in the Choquet ordering. Theorem 3.92 shows that the important notion of a dilation plays the key role here.
3.1
Function spaces
Definition 3.1 (Function spaces). By a function space H on a compact topological space K we mean a (not necessarily closed) linear subspace of C(K) containing the constant functions and separating points of K. We introduce some main examples of function spaces. Examples 3.2. (a) Continuous functions. The space C(K) of all continuous functions on a compact space K represents a simple example of a function space. Clearly, the space C(K) separates points of K. (b) Quadratic polynomials. The space P2 ([0, 1]) of all quadratic polynomials on the interval [0, 1] is a further example of a function space. We considered this example in Chapter 1. (c) Convex case – affine functions. Let X be a compact convex subset of a locally convex space E. The linear space Ac (X) consisting of all continuous affine functions on X is a function space. (d) Harmonic case – harmonic functions. Let U be a bounded open subset of the Euclidean space Rd . The function space H(U ) consisting of all continuous functions on U which are harmonic on U is another example of a function space. More generally, we can consider a relatively compact open subset U of a Bauer harmonic space (cf. Section A.8) and the function space H(U ), the linear subspace of C(U ) of functions which are harmonic on U . We tacitly assume that constant functions are harmonic and H(U ) separates points of U . (e) If K is a compact subset of Rd , we define [ H0 (K) := {H(U )|K : U is relatively compact open set, K ⊂ U }. Generally, the function space H0 (K) is not a closed subspace of C(K) (see Exercise 13.155). (f) Further examples can be found in 3.47, 3.82, 3.83, 3.103, 3.106, 3.111, 3.119, 6.67, 6.76, 6.77, 6.94, 7.65, 8.11(a), 8.29, 8.80, 9.11, 9.56, 10.97, 10.98, 10.99 and 12.3.B.
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3 Choquet theory of function spaces
(g) Let K be a compact space and T be a Markov operator on K (cf. Subsection 6.6.B). Let HT := {h ∈ C(K) : T (h) = h} . If HT separates points of K, then HT is a function space. Convention.
In what follows, H denotes a function space on a compact space K.
Definition 3.3 (H-representing measures). Recall that M1 (K) denotes the set of all probability Radon measures on K. We denote by Mx (H) the set of all H-representing measures for x ∈ K, that is, Z 1 f dµ for any f ∈ H}. Mx (H) := {µ ∈ M (K) : f (x) = K
Of course, the Dirac measure εx at the point x always belongs to Mx (H). Definition 3.4 (Choquet boundary). Define the Choquet boundary ChH (K) of a function space H as the set of those points x ∈ K for which the Dirac measure εx is the only H-representing measure for x; that is, ChH (K) := {x ∈ K : Mx (H) = {εx }} . Examples 3.5. We describe the Choquet boundary of our main examples from 3.2. We postpone the proofs to later sections. (a) Continuous functions. In the case when H = C(K) (K is a compact space), the equality ChH (K) = K follows immediately from the definition. (b) Quadratic polynomials. Let H = P2 ([0, 1]) be the function space of all quadratic polynomials on the interval [0, 1]. Then ChH ([0, 1]) = [0, 1], as easily follows by Proposition 3.7. (c) Convex case – affine functions. If H is the linear space Ac (X) of all continuous affine functions on a compact convex set X, then ChAc (X) (X) = ext X, by Bauer’s characterization of ext X in Theorem 2.40. (d) Harmonic case – harmonic functions. If H(U ) consists of all continuous functions on U ⊂ Rd which are harmonic on U , then ChH(U ) U = ∂ reg U (cf. Theorem 13.35). In the general situation of an abstract Bauer harmonic space, the situation is more delicate, cf. Theorem 13.41. (e) The Choquet boundary of H0 (K) (see Example 3.2(e)) consists of stable points of K (see Definition 13.4 and Theorem 13.35).
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55
(f) Let H := {f ∈ C([−1, 1]) : 2f (0) = f (−1) + f (1)}. Then ChH ([−1, 1]) = [−1, 1] \ {0}. Further examples of Choquet boundaries can be found in the examples of function spaces mentioned in Examples 3.2(f). Definition 3.6 (H-exposing functions and H-exposed points). Let x ∈ K. A function h ∈ H such that 0 = h(x) < h(t) for any t ∈ K, t 6= x, is said to be an Hexposing function for x. A point x ∈ K is called an H-exposed point if there exists an H-exposing function for x. The set of all H-exposed points of K will be denoted by expH (K). Proposition 3.7. Any H-exposed point belongs to the Choquet boundary of H. Proof. Let x ∈ K and let h ∈ H for which 0 = h(x) < h(t) for any t ∈ K \ {x}. If µ ∈ Mx (H), then 0 = h(x) = µ(h). Hence spt µ ⊂ {x}, and therefore µ = εx . We see that x ∈ ChH (K). Definition 3.8 (H-affine, H-convex and H-concave functions). We define the family A(H) of all H-affine functions as the family of all universally measurable functions f : K → [−∞, ∞] such that µ(f ) exists for every µ ∈ Mx (H), x ∈ K, and the following barycentric formula holds: Z f (x) = f dµ for each x ∈ K, µ ∈ Mx (H). K
Further, let Ac (H) be the family of all continuous H-affine functions on K. Similarly, we say that a universally measurable function f : K → [−∞, ∞] is H-convex if µ(f ) exists for every µ ∈ Mx (H), x ∈ K, and f (x) ≤ µ(f ). A function f is H-concave if −f is H-convex. We denote by Kc (H), Kusc (H), and Klsc (H) the family of all continuous, upper semicontinuous, and lower semicontinuous H-convex functions, respectively. We write S c (H), S usc (H) and S lsc (H) for the analogous families of H-concave functions. Remark 3.9. As we will see later in Chapter 4, in the convex case of Example 3.2(c), a continuous function is Ac (X)-concave if and only if it is concave in the usual sense. Definition 3.10 (Cone W(H)). We denote by W(H) the smallest min-stable cone generated by H, that is, W(H) consists of all functions of the form h1 ∧ · · · ∧ hn , h1 , . . . , hn ∈ H, n ∈ N. The following proposition collects several easy facts. Proposition 3.11. (a) Ac (H) is a closed function space on K containing H.
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3 Choquet theory of function spaces
(b) The families W(H), S c (H), S usc (H) and S lsc (H) form min-stable convex cones. (c) The space W(H) − W(H) is dense in C(K). Proof. Since (a) is obvious, we proceed to the proof of (b). Let F be any of the considered families. Obviously, F is a convex cone and the required topological property is stable with respect to taking finite minima (for F = W(H) we use identity f1 ∧ · · · ∧ fn + g1 ∧ · · · ∧ gk =
n ^ k ^
(fi + gj ).)
i=1 j=1
If k1 , k2 ∈ F and µ ∈ Mx (H), then µ(k1 ∧ k2 ) exists and µ(k1 ∧ k2 ) ≤ µ(k1 ) ∧ µ(k2 ) ≤ (k1 ∧ k2 )(x). For the proof of (c), we just use the lattice version of the Stone–Weierstrass theorem from Proposition A.31. Definition 3.12 (H-extremal sets). Let H be a function space on a compact space K. A universally measurable set F ⊂ K is called H-extremal if any measure representing a point in F is carried by F . If X is a compact convex set in a locally convex space and H = Ac (X), then a closed set F ⊂ X is Ac (X)-extremal if and only if F is extremal (see Proposition 2.69). Lemma 3.13. Let F be a closed H-extremal set and f ∈ S lsc (H). Then the set H := {x ∈ F : f (x) = min f (F )} is H-extremal. Proof. Obviously, H is closed. To show that H is H-extremal, pick x ∈ H and µ ∈ Mx (H). Since F is H-extremal, spt µ ⊂ F . Then from the inequalities Z min f (F ) = f (x) ≥
f (t) dµ(t) ≥ min f (F ), F
it follows that spt µ ⊂ H. Proposition 3.14. The family F of all closed H-extremal sets is stable under finite unions and arbitrary intersection.
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57
Proof. Obviously, F is stable under finite unions. T If {Fa }a∈A are closed H-extremal sets, then their intersection F := a∈A Fa is a closed set as well. Pick x ∈ F and µ ∈ Mx (H) and choose a compact set H ⊂K \F =
[
(K \ Fa )
a∈A
arbitrarily. There are sets Fa1 , . . . , Fan so that H ⊂ (K \ Fa1 ) ∪ · · · ∪ (K \ Fan ). Since µ(K \ Faj ) = 0 for each j = 1, . . . , n, we get µ(H) = 0. From the inner regularity of µ we obtain µ(K \ F ) = 0. The proof of Proposition 3.15 follows along the lines to that of the Krein–Milman theorem 2.22. Proposition 3.15. The Choquet boundary ChH (K) intersects any nonempty closed H-extremal set. In particular, the Choquet boundary ChH (K) is nonempty if K 6= ∅. Proof. Let F be a nonempty closed H-extremal set. We partially order the family S of all nonempty closed H-extremal subsets of F by the reverse inclusion. If Z is a T chain in S, then {C : C ∈ Z} is nonempty since it is the intersection of a downdirected family of nonempty compact sets. Moreover, it is H-extremal according to the previous Proposition 3.14. Zorn’s lemma now provides a maximal element H ∈ S. Assume that H contains two distinct points x and y. Since H is a function space, there exists a function h ∈ H such that h(x) 6= h(y). According to Lemma 3.13, the set {z ∈ H : h(z) = min h(H)} is a closed H-extremal set strictly contained in H, which contradicts the fact that H is maximal. Hence, H = {x} for some x ∈ F . Since one-point H-extremal sets are in ChH (K), it follows that x ∈ ChH (K). Theorem 3.16 (Minimum principle for S lsc (H)). Let f ∈ S lsc (H) be a lower semicontinuous H-concave function, f ≥ 0 on ChH (K). Then f ≥ 0 on K. Proof. Let K 6= ∅ and f ∈ S lsc (H) satisfy f ≥ 0 on ChH (K). Then F := {x ∈ K : f (x) = min f (K)} is a closed H-extremal set by Lemma 3.13. By Proposition 3.15, F ∩ ChH (K) 6= ∅, which implies that f ≥ 0 on K.
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3 Choquet theory of function spaces
Definition 3.17 (Envelopes and sublinear functionals). Let µ be a Radon measure on K and F be a subset of K containing ChH (K). Then for every extended real-valued function f on F , we define Qµ,F (f ) := inf{µ(k) : k ∈ S c (H), k ≥ f on F }. If x ∈ K and µ = εx , we define F ∗
f (x) := Qεx ,F (f ) and
f∗ (x) := −(F(−f )∗ (x)).
F
If F = K, we write for short Qµ (f ), f ∗ and f∗ . Lemma 3.18. Let µ ∈ M+ (K), F ⊃ ChH (K), and f be an upper bounded function on F . Then (a) Ff ∗ is an upper semicontinuous H-concave function, (b) Qµ,F (f ) = µ(Ff ∗ ), (c) if f is bounded, then −kf k ≤ Ff∗ ≤ Ff ∗ ≤ kf k, (d) if f is upper semicontinuous on F and f (x), g := lim sup f (y),
x ∈ F, x ∈ F \ F,
y→x,y∈F
then g is upper semicontinuous on F and Qµ,F (f ) = Qµ,F (g). (e) g 7→ Qµ,F (g) is a sublinear functional on `∞ (F ). Proof. The proof of (a) is obvious. To prove (b), let µ ∈ M+ (K) and an upper bounded function f on F be given. By Lemma 3.11(b), the family {k ∈ S c (H) : k ≥ f on F } is down-directed. Hence by Theorem A.84, Qµ,F (f ) = µ(Ff ∗ ). We proceed to (c). Let −k1 , k2 ∈ S c (H) be such that k1 ≤ f ≤ k2 on F . By Theorem 3.16, k1 ≤ k2 . Hence Ff∗ ≤ Ff ∗ on K. Since H contains constant functions, −kf k ≤ Ff∗ ≤ Ff ∗ ≤ kf k. Let g be defined as in (d). By a routine verification, g is upper semicontinuous on F . If k ∈ S c (H), then k ≥ f on F if and only if k ≥ g on F . This proves (d). Since the proof of (e) is straightforward, the proof is finished.
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Definition 3.19 (Choquet ordering). Let H be a function space on a compact space K. The convex cone Kc (H) of all continuous H-convex functions on K determines the Choquet ordering on M+ (K): µ ≺H ν
if
µ(f ) ≤ ν(f ) for each f ∈ Kc (H).
(Where no confusion can occur, µ ≺H ν will be abbreviated by µ ≺ ν.) Since, by Proposition A.31, the set Kc (H) − Kc (H) is uniformly dense in C(K), µ = ν whenever µ ≺ ν and ν ≺ µ. Proposition 3.20. Let µ ∈ M+ (K). (a) If ν satisfies µ ≺ ν, then µ − ν ∈ H⊥ and kµk = kνk. (b) If x is a point of K and εx ≺ µ, then µ ∈ Mx (H). Proof. Let h ∈ H. If µ ≺ ν, then µ(h) ≤ ν(h) and µ(−h) ≤ ν(−h). Hence µ(h) = ν(h) and kµk = µ(1) = ν(1) = kνk. This proves (a). Since (b) follows from (a), the proof is finished. Lemma 3.21 (Key lemma). Let µ ∈ M+ (K), F be a closed set in K containing ChH (K) and f be an upper semicontinuous function on K. Then there exists a measure ν ∈ M+ (F ) such that µ ≺ ν and Qµ,F (f ) = ν(f ). In particular, for any x ∈ K there exists a measure ν ∈ Mx (H) ∩ M+ (F ) such that Ff ∗ (x) = ν(f ). Proof. Given µ ∈ M+ (K), we assume first that f ∈ C(K). From Lemma 3.18(e) we know that the mapping p : g 7→ Qµ,F (g) is a sublinear functional on C(F ). We define a functional ϕ : span{f } → R as ϕ(tf ) = tQµ,F (f ),
t ∈ R.
Since 0 = Qµ,F (0) = Qµ,F (f − f ) ≤ Qµ,F (f ) + Qµ,F (−f ), we get ϕ(−f ) = −Qµ,F (f ) ≤ Qµ,F (−f ). Hence ϕ ≤ p on span{f }. The Hahn–Banach theorem provides a linear functional ν on C(F ) such that ν(f ) = Qµ,F (f ) and ν ≤ p on C(F ). Since ν(g) ≤ p(g) ≤ 0 whenever g ∈ C(F ) and g ≤ 0, we see that ν is, according to the Riesz representation theorem A.75, a Radon measure on F . Let k ∈ S c (H). Then ν(k) ≤ p(k) = Qµ,F (k) ≤ µ(k). Hence µ ≺ ν and the proof is finished in the case when f is continuous on F .
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Now let f be an upper semicontinuous function on F . Denote by G the downdirected set {g ∈ C(F ) : g ≥ f on F }. By the first part of the proof, for any g ∈ G there exists a measure νg ∈ M+ (F ) such that µ ≺ νg and νg (g) = Qµ,F (g). Given ψ ∈ G, let Mψ = {νg : g ∈ G, g ≤ ψ}. Since Mψ ⊂ {λ ∈ M+ (F ) : kλk = kµk} and the family {Mψ : ψ ∈ G} has the T finite intersection property, a compactness argument yields the existence of ν ∈ {M ψ : ψ ∈ G}. A moment’s reflection shows that µ ≺ ν. We observe that inf λ(ψ) : λ ∈ Mψ = inf λ(ψ) : λ ∈ M ψ ≤ ν(ψ) for each ψ ∈ G. Hence, using Theorem A.84 for verification of the equalities, Qµ,F (f ) ≤ inf Qµ,F (g) : g ∈ G = inf {νg (g) : g ∈ G} ≤ inf {inf {νg (ψ) : g ∈ G, g ≤ ψ} : ψ ∈ G} ≤ inf {ν(ψ) : ψ ∈ G} = ν(f ) ≤ inf {ν(k) : k ∈ S c (H), k ≥ f on F } ≤ inf {µ(k) : k ∈ S c (H), k ≥ f on F } = Qµ,F (f ), which yields the required equality. The second part of the assertion follows by taking µ = εx and using Proposition 3.20. Lemma 3.22. If f ∈ C(F ), F ⊃ ChH (K) is a closed set and x ∈ K, then F f∗ (x), Ff ∗ (x) = {µ(f ) : µ ∈ Mx (H) ∩ M+ (F )}. Proof. If µ ∈ Mx (H) ∩ M+ (F ) and k ∈ S c (H) satisfies k ≥ f on F , then µ(f ) ≤ µ(k) ≤ k(x). Thus µ(f ) ≤ Ff ∗ (x). Analogously, we get the reverse inequality. Conversely, if α ∈ [Ff∗ (x), Ff ∗ (x)] is given, we proceed as in the proof of Lemma 3.21. We set p(g) = Fg ∗ (x), g ∈ C(F ), and notice that the functional ϕ : span{f } → R defined as ϕ(tf ) = tα, t ∈ R, satisfies ϕ ≤ p on span{f }. Now we can finish the proof as in Lemma 3.21. Corollary 3.23. The following assertions hold:
3.1 Function spaces
61
(a) Ac (H) = {f ∈ C(K) : f ∗ = f∗ }. (b) For an upper bounded upper semicontinuous function f on ChH (K), we have ChH (K) f ∗ = f on Ch (K). H Proof. For the proof of (a), let f ∈ C(K) satisfy f∗ = f ∗ . Then f ∈ S usc (H) ∩ Klsc (H) = Ac (H). Conversely, if f ∈ Ac (H), then f∗ = f ∗ by definition. For the proof of (b), let f be an upper bounded upper semicontinuous function on ChH (K). Let g be the upper semicontinuous extension of f to ChH (K) defined as in Lemma 3.18(d). By Lemma 3.22, for each x ∈ ChH (K) there exists µ ∈ Mx (H) carried by ChH (K) such that µ(g) = ChH (K)g ∗ (x). Since x ∈ ChH (K), µ = εx and f (x) = g(x) = ChH (K)g ∗ (x) ≥ ChH (K)f ∗ (x). This concludes the proof. Theorem 3.24 (Bauer’s characterization of the Choquet boundary). Let x be a point of K. Then the following assertions are equivalent: (i) x ∈ ChH (K), (ii) f ∗ (x) = f (x) for each f ∈ −W(H), (iii) f ∗ (x) = f (x) for each f ∈ Kc (H), (iv) f ∗ (x) = f (x) for each f ∈ C(K), (v) f ∗ (x) = f (x) for each upper semicontinuous function f on K. Proof. For the proof of (i) =⇒ (v), let x ∈ ChH (K) be given and f be an upper semicontinuous function on K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) such that f ∗ (x) = µ(f ). Then µ = εx , and thus f ∗ (x) = µ(f ) = f (x). Since (v) =⇒ (iv) =⇒ (iii) =⇒ (ii) are obvious, we close the chain of implications by proving (ii) =⇒ (i). To this end, let a point x ∈ K satisfy f ∗ (x) = f (x) for each f ∈ −W(H). Then for each µ ∈ Mx (H) and f ∈ −W(H), f ∗ (x) = f (x) ≤ µ(f ) ≤ f ∗ (x). By Lemma 3.11(c), µ = εx and we are done. Proposition 3.25. Let f be an upper bounded function on K and F ⊃ ChH (K) be a closed set.
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3 Choquet theory of function spaces
(a) Then F ∗
f = inf {g ∈ H : g ≥ f on F } = inf {g ∈ Ac (H) : g ≥ f on F } = inf {g ∈ W(H) : g ≥ f on F } = inf {g ∈ S usc (H) : g ≥ f on F } .
(b) If f is upper semicontinuous, then F ∗ f = inf k ∈ S lsc (H) : k ≥ f on F . Proof. For the proof of (a), we first use Lemma 3.21 to conclude that inf {g ∈ S usc (H) : g ≥ f on F } = inf {g ∈ S c (H) : g ≥ f on F } .
(3.1)
Indeed, given g ∈ S usc (H) with g ≥ f on F , x ∈ K and ε > 0, by Lemma 3.21 there exists a measure µ ∈ Mx (H) ∩ M+ (F ) such that Fg ∗ (x) = µ(g) ≤ g(x). Thus there exists a function g 0 ∈ S c (H) with g 0 ≥ g on F such that g 0 (x) ≤ g(x) + ε. Hence (3.1) follows. Further, we fix a point x ∈ K and define p : C(F ) → R as p(g) = inf {g(x) : g ∈ H, g ≥ f on F } . Then p is a sublinear functional on C(F ) and we may follow the beginning of the proof of Lemma 3.21 to conclude that for any function g ∈ C(F ), there exists a measure µ in Mx (H) ∩ M+ (F ) such that µ(g) = p(g). Thus for every k ∈ S c (H), p(k) = µ(k) ≤ k(x), so that k(x) = inf{h(x) : h ∈ H, h ≥ f on F }. Hence inf{h ∈ H : h ≥ f on F } = inf{k ∈ S c (H) : k ≥ f on F }. Obviously, inf{k ∈ S c (H) : k ≥ f on F } ≤ inf{k ∈ W(H) : k ≥ f on F } ≤ inf{k ∈ H : k ≥ f on F }. Thus F ∗
f ≤ inf{k ∈ S usc (H) : k ≥ f on F } ≤ inf{h ∈ Ac (H) : h ≥ f on F } ≤ inf{h ∈ H : h ≥ f on F } = inf{k ∈ W(H) : k ≥ f on F } = inf{k ∈ S c (H) : k ≥ f on F } = Ff ∗ .
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63
This concludes the proof of (a). Let f be upper semicontinuous and x ∈ K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) ∩ M+ (F ) such that Ff ∗ (x) = µ(f ). Then, for k ∈ S lsc (H) with k ≥ f on F , we have F ∗ f (x) = µ(f ) ≤ µ(k) ≤ k(x). Hence F ∗
f = inf k ∈ S lsc (H) : k ≥ f on F ,
which concludes the proof. Corollary 3.26. If x ∈ K and µ ∈ Mx (H), then εx ≺ µ. Proof. Given a function k ∈ S c (H) and ε > 0, let h ∈ H be such that k ≤ h and h(x) ≤ k(x) + ε (use Proposition 3.25(a)). Then εx (k) = k(x) ≥ h(x) − ε = µ(h) − ε ≥ µ(k) − ε. Hence εx ≺ µ, as needed. Theorem 3.27 (Bauer’s theorem). Let H be a function space on K. Then H = Ac (H) if and only if there exists a min-stable closed set W ⊂ C(K) such that H = W ∩ (−W). Proof. Since the family S c (H) is min-stable and closed in C(K) and Ac (H) = S c (H) ∩ (−S c (H)), necessity immediately follows. As for the sufficiency, we show that S c (H) ⊂ W. Then H = W ∩ (−W) ⊃ S c (H) ∩ (−S c (H)) = Ac (H) ⊃ H. Take any s ∈ S c (H). Then, with the aid of Lemma 3.25(a), s = s∗ = inf {ϕ ∈ H : ϕ ≥ s} ≥ inf {ϕ ∈ W : ϕ ≥ s} ≥ s. Hence, given x ∈ K and ε > 0, there exists wx ∈ W such that s(x) ≤ wx (x) < s(x) + ε. There exists an open neighborhood Ux of x such that wx < s + ε on Ux . By the compactness of K, there are x1 , . . . , xn ∈ K such that K = Ux1 ∪ · · · ∪ Uxn . Set w := wx1 ∧ · · · ∧ wxn . Then w ∈ W and w ≥ s on K. Given y ∈ K, there exists j ∈ {1, . . . , n} so that y ∈ Uxj . Then s(y) ≤ w(y) ≤ wxj (y) < s(y) + ε. Therefore ks − wk ≤ ε. We see that s ∈ W = W.
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3.2
More about Korovkin theorems
Definition 3.28 (Korovkin closures). Let P be a family of continuous functions on a compact space K. A sequence {Ln } of positive linear operators on C(K) is said to be P-admissible if kLn ϕ − ϕk → 0 for any ϕ ∈ P. Recall that the Korovkin closure Kor(P) of P is the set of those functions f ∈ C(K) such that kLn f −f k → 0 for any P-admissible sequence {Ln } of operators on C(K). Proposition 3.29. If H is a function space on a metrizable compact space K, then Kor(H) = Ac (H). Proof. Pick f ∈ Ac (H) and ε > 0. By Proposition 3.25(a) and Corollary 3.23, for any x ∈ K, there exists a function hx ∈ H and a neighborhood Ux of x such that f ≤ hx
and hx (t) − f (t) < ε for any t ∈ Ux .
Select x1 , . . . , xk ∈ K so that Ux1 ∪ · · · ∪ Uxk = K and denote hj := hxj for j = 1, . . . , k. If h := h1 ∧ · · · ∧ hk , then f ≤ h and h − f < ε on K. To show that f ∈ Kor(H), choose an H-admissible sequence {Ln } of positive linear operators on C(K). There exists n0 ∈ N such that kLn hj − hj k < ε for all n ≥ n0
and j = 1, . . . , k.
Now fix n ≥ n0 . Using the monotonicity of Ln , we have Ln f ≤ Ln hj ≤ hj + ε,
j = 1, . . . , k,
and therefore Ln f ≤ h1 ∧ · · · ∧ hk + ε = h + ε. Given x ∈ K, we see that Ln f (x) − f (x) ≤ (Ln f (x) − h(x)) + (h(x) − f (x)) ≤ 2ε. Analogously, using lower envelopes we can prove the existence of n1 ∈ N such that f (x) − Ln f (x) ≤ 2ε for all n ≥ n1 . It follows that kLn f − f k → 0, so f ∈ Kor(H). Now let f be in Kor(H). Given x ∈ K and µ ∈ Mx (H), our aim is to prove that µ(f ) = f (x). Let ρ be a metric on K compatible with the topology of K. For n ∈ N let Bn := {t ∈ K : ρ(x, t) < 1/n}
3.2 More about Korovkin theorems
65
and define gn (t) := 1 − nρ(x, t) ∨ 0,
t ∈ K.
Then gn ∈ C(K) and 0 ≤ gn ≤ 1,
gn (x) = 1
and gn = 0 on K \ Bn .
If Ln : ϕ 7→ gn µ(ϕ) + (1 − gn )ϕ,
ϕ ∈ C(K),
then Ln : C(K) → C(K) is a positive linear operator. Let h ∈ H and t ∈ K. Since |Ln h(t) − h(t)| = |gn (t)µ(h) + (1 − gn (t))h(t) − h(t)| = |gn (t)µ(h) − gn (t)h(t)| = gn (t)|h(x) − h(t)|, it follows that kLn h − hk ≤ sup {|h(x) − h(t)| : t ∈ Bn } . Since the function h is continuous at x, we get kLn h − hk → 0. We see that the sequence {Ln } is H-admissible, and therefore kLn f − f k → 0. In particular, Ln f (x) → f (x). But Ln f (x) = µ(f ), and the proof is complete. Remark 3.30. Notice that in the proof of the inclusion Ac (H) ⊂ Kor(H) we did not use linearity of Ln , while the metrizability condition was used only in the course of the proof that Kor(H) ⊂ Ac (H). Now, we are in a position to answer our query from Chapter 1, where we asked under which conditions the equality Kor(H) = C(K) holds. Proposition 3.31. Let H be a function space on a metrizable compact space K. Then Kor(H) = C(K) if and only if ChH (K) = K. Proof. The preceding Proposition 3.29 shows that Kor(H) = Ac (H), and therefore we only need to examine when Ac (H) = C(K). According to Bauer’s characterization 3.24 of the Choquet boundary, we know that x ∈ ChH (K) if and only if f∗ (x) = f ∗ (x) for any f ∈ C(K). Hence, ChH (K) = K if and only if f∗ = f ∗ on K for any f ∈ C(K); that is, if and only if Ac (H) = C(K), by Corollary 3.23. (First Korovkin theorem revisited). Let H be the linear span of funcTheorem 3.322 tions 1, id, id on [a, b]. Then Kor(H) = C([a, b]). Proof. Let t ∈ [a, b]. Since x 7→ (t − x)2 , x ∈ [a, b], is an H-exposing function for t, we may apply Proposition 3.7 to see that ChH ([a, b]) = [a, b]. Proposition 3.31 then shows that Kor(H) = C([a, b]).
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Theorem 3.33. Let S := {x ∈ Rd : |x| = 1} and
ϕi (x) := xi , x ∈ S,
i = 1, . . . , d.
If P := {1, ϕ1 , . . . , ϕd }, then Kor(P) = C(S). Proof. For the function space H = Lin{1, ϕ1 , . . . , ϕd } we have ChH (S) = S. Indeed, given w ∈ S, it is straightforward to check that the function x 7→ 1 − x · w,
x ∈ S,
is a P-exposing function for w. Corollary 3.34 (Second Korovkin theorem). Let H be the linear span of {1, cos, sin} on [0, 2π]. Then the Korovkin closure Kor(H) equals the Banach space of all continuous functions f on [0, 2π] with f (0) = f (2π) (equipped with the sup-norm). Proof. The circle S := {x ∈ R2 : |x| = 1} can be topologically identified with the interval [0, 2π] after having identified the points 0 and 2π via the mapping ω : t 7→ (cos t, sin t), t ∈ [0, 2π]. Then the functions cos, sin correspond to the functions ϕ1 , ϕ2 from Theorem 3.33. Remark 3.35. There is also a direct proof of this corollary following the lines of the proof of Theorem 1.2, considering the function x 7→ sin2
t−x 1 = (1 − cos x cos t − sin x sin t), 2 2
x ∈ [0, 2π],
instead of the function x 7→ (x − t)2 . Theorem 3.36 (Third Korovkin theorem). Let H be a linear span of {1, ϕ} on [a, b], where ϕ is a continuous function on [a, b]. Then Kor(H) 6= C([a, b]). Proof. The assertion is obvious if the function ϕ does not separate points of [a, b]. Otherwise, there exists x ∈ (a, b) such that ϕ(x) = 21 ϕ(a) + ϕ(b) . Then the measure 12 (εa + εb ) represents x, and therefore x does not belong to ChH ([a, b]). In view of Proposition 3.31, Kor(H) 6= C([a, b]).
3.3
On the H-barycenter mapping
Proposition 3.37. Let H be a function space on a compact space K. Denote [ M(H) := Mx (H). x∈K
Then M(H) is a compact subset of M1 (K).
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67
Proof. It is easy to show that M(H) is a closed subset of M1 (K). Remark 3.38. Note that in the “convex case”, M(Ac (X)) = M1 (X) (see Theorem 2.29). If H = C([0, 1]), then M(H) = {εx : x ∈ [0, 1]} 6= M1 ([0, 1]). See also Exercise 3.106. Let us observe that if µ ∈ Mx (H) ∩ My (H) for some x, y ∈ K, then h(x) = µ(h) = h(y) for any h ∈ H, and hence x = y since H separates points of K. Definition 3.39 (H-barycenter mapping). We define the H-barycenter mapping r from M(H) onto K by r : µ ∈ Mx (H) 7→ x,
µ ∈ M(H).
(By the previous observation, the mapping r is defined correctly.) Proposition 3.40. The H-barycenter mapping r : M(H) → K is continuous. Proof. The mapping µ 7→ h(r(µ)),
µ ∈ M(H),
is continuous for any h ∈ H. Hence r is continuous as a mapping from M(H) into K, where on K we consider the topology generated by the family H. But this one coincides with the topology of K.
3.4
The Choquet representation theorem
In this section we will give a proof of the Choquet representation theorem in the metrizable case. First we establish a construction of strictly H-convex functions. Definition 3.41 (Strictly H-convex functions). A continuous H-convex function f on a compact space K is called strictly H-convex if Z f (x) <
f dµ
for any x ∈ K, µ ∈ Mx (H), µ 6= εx .
K
Theorem 3.42. Let H be a function space on a metrizable compact space K. Then K admits a continuous strictly H-convex function.
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3 Choquet theory of function spaces
Proof. Since the Banach space C(K) is separable, there exists a countable set {hn : n ∈ N} that is dense in {h ∈ H : 0 ≤ h ≤ 1}. We claim that the function h2n is H-convex for any n ∈ N. To this end, pick x ∈ K and µ ∈ Mx (H). Then Z 2 Z Z h2n dµ = µ(h2n ). (3.2) 1 dµ hn dµ ≤ h2n (x) = K
K
K
Further, given x ∈ K and µ ∈ Mx (H), µ 6= εx , there exists n ∈ N such that h2n (x) < µ(h2n ). If we establish this, then the function f :=
∞ X 1 2 h 2n n n=1
is clearly continuous and strictly H-convex on K. So suppose, for the sake of contradiction, that h2n (x) = µ(h2n ) for any n ∈ N and some x ∈ K and µ ∈ Mx (H), µ 6= εx . Then for any n ∈ N, the sign of equality holds in (3.2). Hence Z Z Z 2 2 hn − hn (x) dµ = hn dµ − 2hn (x) hn dµ + h2n (x) = 0, K
K
K
so that hn = hn (x) µ-almost everywhere. Therefore ∞ \ µ K\ {y ∈ K : hn (y) = hn (x)} = 0. n=1
Since µ 6= εx , there exists y ∈ K, y 6= x, such that hn (y) = hn (x) for any n ∈ N. It follows that h(y) = h(x) for each h ∈ H, 0 ≤ h ≤ 1. This contradicts the assumption that the set H, and hence also the set {h ∈ H : 0 ≤ h ≤ 1}, separates points of K. Proposition 3.43. Let H be a function space on a compact space K and let there exist a continuous strictly H-convex function h on K. Then ChH (K) = {x ∈ K : h(x) = h∗ (x)} and the Choquet boundary ChH (K) is a Gδ subset of K. Proof. By Bauer’s characterization of the Choquet boundary in 3.24, \ ChH (K) = {x ∈ K : f (x) = f ∗ (x)} ⊂ {x ∈ K : h(x) = h∗ (x)} . f ∈C(K)
Let x ∈ K \ ChH (K). There exists a measure µ ∈ Mx (H), µ 6= εx . Then, using the strict H-convexity of h and Corollary 3.22, we have h(x) < µ(h) ≤ h∗ (x).
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3.4 The Choquet representation theorem
Thus, we have ∗
ChH (K) = {x ∈ K : h(x) = h (x)} =
∞ \ n=1
1 x ∈ K : h (x) − h(x) < n ∗
.
Since h∗ is upper semicontinuous on K, the set x ∈ K : h∗ (x) − h(x) < n1 is open for each n ∈ N. Thus ChH (K), being the intersection of these sets, is a Gδ set. Remarks 3.44. (a) In the case of the “convex” function space Ac (X), there is an alternative proof of Theorem 3.42, cf. Exercise 3.107 and Remark 3.9. (b) If a compact convex set X admits a continuous strictly convex function, then X is metrizable (see Exercise 4.40). This assertion is no longer true for general function spaces; see Example 3.47. Theorem 3.45 (Choquet representation theorem). Assume that H is a function space on a compact space K and h is a continuous strictly H-convex function on K (for example, let K be metrizable). Then for each x ∈ K there exists a measure µ ∈ Mx (H) such that µ(K \ ChH (K)) = 0. Proof. Let x ∈ K. Corollary 3.22 provides a representing Radon measure µ ∈ Mx (H) such that µ(h) = h∗ (x). If ϕ ∈ H, ϕ ≥ h, then h∗ (x) = µ(h) ≤ µ(h∗ ) ≤ µ(ϕ) = ϕ(x). Taking the infimum over all possible choices of ϕ, we see that h∗ (x) = µ(h) ≤ µ(h∗ ) = h∗ (x). We conclude that µ(h) = µ(h∗ ). Since h∗ ≥ h, we get µ {t ∈ K : h∗ (t) > h(t)} = 0. In view of Proposition 3.43, ChH (K) = {t ∈ K : h(t) = h∗ (t)} , and the proof is complete. Remark 3.46. By Theorem 3.42, any metrizable compact set admits a continuous strictly H-convex function. Therefore, Proposition 3.43 and Theorem 3.45 hold under the assumption that the compact space K is metrizable. On the other hand, there are function spaces on nonmetrizable compact spaces which admit a continuous strictly H-convex function. As a trivial example consider a nonmetrizable compact space K and H = C(K). In this example, any continuous function on K is strictly H-convex. We present the following slightly less trivial example.
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3 Choquet theory of function spaces
Example 3.47. There exists a function space on a nonmetrizable compact space K that admits a strictly H-convex continuous function. Proof. Let K := D ∪ {d∞ } be the one-point compactification of an uncountable discrete topological space D and x and y be distinct points of D. Then K is not metrizable (since K is not separable). If h(x) + h(y) H := h ∈ C(K) : h(d∞ ) = , 2 then c{x,y} is a continuous strictly H-convex function on K.
3.5
In-between theorems
Throughout this section, H will be a function space on a compact space K. There are plenty of insertion problems occurring in various fields of analysis. They can be formulated as follows: Given families of functions S, T and F, find f ∈ F such that s ≤ f ≤ t (or, s < f < t) whenever s ∈ S, t ∈ T and s ≤ t (or, s < t). Perhaps the most famous result in this direction goes back to H. Hahn, who proved that for every pair of real-valued functions s, t on a metric space, where s is upper semicontinuous, t is lower semicontinuous and s ≤ t, there exists a continuous function f such that s ≤ f ≤ t. In this section, we will examine in-between theorems and approximation theorems for the case of H-convex and H-concave functions. Proposition 3.48. Let f be an upper semicontinuous H-concave function on K. If K := {g ∈ S c (H) : g ≥ f on K} and L := {g ∈ S c (H) : g > f on K} , then K and L are down-directed families and f = inf K = inf L. Proof. Since the cone S c (H) is min-stable (cf. Proposition 3.11), the sets K and L are down-directed. It suffices to show that f = inf L. To this end, choose x ∈ K and numbers f (x) < c < d. By Proposition 3.25(a), f ∗ = f and there exists h ∈ H such that h≥f
and
h(x) < c.
Now, if g := h+d−c, then g ∈ H, g(x) < d and g > f . Since d is chosen arbitrarily, f = inf L. Corollary 3.49. If g is a lower semicontinuous function on K, f ∈ S usc (H), f < g on K, then there exists a function k ∈ S c (H) such that f < k < g on K.
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71
Proof. By Proposition 3.25(a), f ∗ = f . With the aid of Proposition 3.48, for any x ∈ K we select a function gx ∈ S c (H) such that gx ≥ f on K
and gx (x) < g(x).
Adding a small constant to gx , we may suppose that gx > f on K
and gx (x) < g(x).
Now appeal to the upper semicontinuity of f and the lower semicontinuity of g to find an open neighborhood Ux of x such that f (y) < gx (y) < g(y) for any y ∈ Ux . By the compactness of K, there exist x1 , . . . , xn ∈ K such that K = Ux1 ∪ · · · ∪ Uxn . Then the function k := gx1 ∧ · · · ∧ gxn belongs to S c (H) and it is a routine argument to show that f < k < g on K. Indeed, given y ∈ K, there exists j ∈ {1, . . . , n} such that y ∈ Uxj . Then f (y) < k(y) ≤ gxj (y) < g(y).
Remark 3.50. There exist a function space H on a metrizable compact space K, a lower semicontinuous function g on K and a function f ∈ S usc (H) such that f ≤ g on K and there is no function from S c (H) between f and g. See Exercise 3.111. Proposition 3.51. If g is an upper semicontinuous function on K, f ∈ S lsc (H), g < f on K, then there exists a function k ∈ S c (H) such that g < k < f on K. Proof. Fix x ∈ K. By Lemma 3.21, there exists a measure µ ∈ Mx (H) such that g ∗ (x) = µ(g). Then g ∗ (x) = µ(g) < µ(f ) ≤ f (x). Therefore, by Proposition 3.25(a) there exists hx ∈ H such that hx ≥ g on K
and
hx (x) < f (x).
Adding a small constant to hx , we may assume that hx > g everywhere on K and we still have hx (x) < f (x). We infer from the lower semicontinuity of f − hx and a compactness argument that there exist points x1 , . . . , xn ∈ K such that k := hx1 ∧ · · · ∧ hxn < f . The function k has all the required properties. Remark 3.52. A hint to an alternative (slightly different) proof of Proposition 3.51 can be found in Exercise 3.110.
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3 Choquet theory of function spaces
Proposition 3.53. If g is an upper semicontinuous function on K, f ∈ S lsc (H), g ≤ f on K, then there exists a function k ∈ S c (H) such that g ≤ k ≤ f on K. Proof. The first order of business is to construct, for each ε > 0, a pair of functions uε and vε such that uε is upper semicontinuous, vε ∈ S lsc (H), g ≤ uε ≤ vε ≤ f
on K
and vε − uε < ε on K. This can be done quite easily. Namely, by Proposition 3.51, there exists a function k ∈ S c (H) such that g < k < f + ε. It is easy to check that the functions uε := (k − ε) ∨ g
and
vε := f ∧ k
do the job. Now, using the above construction we can construct inductively two sequences {un } and {vn } so that, for each n ∈ N, un is upper semicontinuous, vn ∈ S lsc (H), g ≤ un ≤ un+1 ≤ vn+1 ≤ vn ≤ f and
1 . 2n Then the sequence {vn } converges uniformly to an H-concave continuous function k. The inequalities 1 1 g ≤ vn ≤ un + n ≤ f + n 2 2 ensure that g ≤ k ≤ f . vn − un <
Proposition 3.54. Let f be a lower semicontinuous H-concave function on K. If S := {g ∈ S c (H) : g < f on K} and T := {g ∈ S c (H) : g ≤ f on K} , then S and T are up-directed families and f = sup S = sup T. Proof. It suffices to establish that f = sup S. Since f = sup {g ∈ C(K) : g < f } , (cf. Proposition A.50(b)), we conclude that f = sup S using Proposition 3.51. Now we are given g1 , g2 ∈ S, and we wish to find g ∈ S such that g > g1 ∨ g2 . Since (g1 ∨ g2 )∗ < f (cf. the proof of Proposition 3.51), a new application of
3.6 Maximal measures
73
Proposition 3.51 asserts the existence of g ∈ S c (H) such that g1 ∨ g2 ≤ (g1 ∨ g2 )∗ < g < f. All that is left to be shown is that T is up-directed. To finish the proof, consider g1 , g2 ∈ T. By Proposition 3.53, there exists g ∈ S c (H) such that g1 ∨g2 ≤ g ≤ f . Proposition 3.55. Let f ∈ S c (H) and ε > 0. Then there exists h ∈ W(H) such that f ≤ h ≤ f + ε. Proof. By Proposition 3.25, f = inf {h ∈ W(H) : h ≥ f } , where the family W(H) is down-directed. Since f is supposed to be continuous, we may use Dini’s theorem to obtain the desired function. Proposition 3.56. Let H be a function space on K. Then the following assertions are equivalent for µ, ν ∈ M+ (K): (i) µ ≺ ν, (ii) µ(f ) ≤ ν(f ) for any f ∈ −W(H), (iii) µ(f ) ≤ ν(f ) for any f ∈ Kusc (H), (iv) µ(f ) ≤ ν(f ) for any f ∈ Klsc (H). Proof. The implications (i) =⇒ (iii) and (i) =⇒ (iv) follow from Proposition 3.54, Corollary 3.49 and Theorem A.84. Obviously, (i) =⇒ (ii), (iii) =⇒ (i) and (iv) =⇒ (i). By Proposition 3.55, (ii) =⇒ (i).
3.6
Maximal measures
Throughout this section, H will be a function space on a compact space K. Definition 3.57 (H-maximal measures). If H is a function space on a compact space K, we recall that the convex cone Kc (H) of all continuous H-convex functions on K determines the Choquet ordering ≺. Maximal elements of M+ (K) with respect to the Choquet ordering are called maximal measures (or, more precisely, H-maximal measures). The set of all H-maximal measures on K is denoted by Mmax (H). We denote by Mbnd (H) the set of all signed Radon measures on K such that the variation |µ| of µ is a maximal measure. Elements of Mbnd (H) are called Hboundary measures or, simply, boundary measures. Theorem 3.58 (Mokobodzki’s maximality test). The following assertions on a measure µ ∈ M+ (K) are equivalent: (i) µ is maximal,
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3 Choquet theory of function spaces
(ii) µ(f ) = µ(f ∗ ) for any f ∈ −W(H), (iii) µ(f ) = µ(f ∗ ) for any f ∈ Kc (H), (iv) µ(f ) = µ(f ∗ ) for any f ∈ C(K), (v) µ(f ) = µ(Ff ∗ ) for any upper semicontinuous function f on K and any set F ⊃ ChH (K). Proof. Let µ ∈ M+ (K) be maximal, F ⊃ ChH (K) and f be an upper semicontinuous function on K. By Lemma 3.18(d), Qµ,F (f ) = Qµ,F (f ). By Lemma 3.21, there exists a measure ν ∈ M+ (F ) such that µ ≺ ν and ν(f ) = Qµ,F (f ). Since µ is maximal, we have µ = ν, and thus by Lemma 3.18(b), µ(f ) = ν(f ) = Qµ,F (f ) = Qµ,F (f ) = µ(Ff ∗ ). Hence (i) =⇒ (v). It is obvious that (v) =⇒ (iv) =⇒ (iii) =⇒ (ii). To see that (ii) =⇒ (i), assume that a measure µ ∈ M+ (K) satisfies µ(f ) = µ(f ∗ ) for each f ∈ −W(H). Let ν ∈ M+ (K), µ ≺ ν and fix f ∈ −W(H). Then µ(f ) ≤ ν(f ) and, using Lemma 3.18(b) and Theorem A.84, we get µ(f ) = µ(f ∗ ) = µ(inf {k ∈ S c (H) : k ≥ f }) = inf {µ(k) : k ∈ S c (H), k ≥ f } ≥ inf {ν(k) : k ∈ S c (H), k ≥ f } = ν(f ∗ ) ≥ ν(f ). Hence µ(f ) = ν(f ). Since the space W(H) − W(H) is uniformly dense in C(K), we conclude that µ = ν. We see that µ is maximal. Corollary 3.59. A measure µ ∈ M+ (K) is maximal if and only if µ({x ∈ K : f (x) < f ∗ (x)}) = 0
for any f ∈ −W(H).
(3.3)
Proof. The assertion follows immediately from Mokobodzki’s maximality test 3.58. Corollary 3.60. If µ ∈ M+ (K) and µ(F ) = 0 for any closed F ⊂ K \ ChH (K), then µ is a maximal. Proof. Let µ be as in the hypothesis. For any f ∈ −W(H), B := {x ∈ K : f ∗ (x) = f (x)} is a Borel set containing ChH (K). By the assumption and regularity of µ, µ(K \ B) = 0 and µ is maximal. Corollary 3.61. Let h be a continuous strictly H-convex function on K and µ ∈ M+ (K). Then the following assertions are equivalent: (i) µ is maximal,
3.6 Maximal measures
75
(ii) µ(h) = µ(h∗ ) , (iii) µ(K \ ChH (K)) = 0. Proof. Recall that, by Proposition 3.43, ChH (K) is a Gδ set. The implication (i) =⇒ (ii) follows from Mokobodzki’s maximality test 3.58. Assume that µ(h) = µ(h∗ ). Then µ({x ∈ K : h∗ (x) − h(x) > 0}) = 0. Since by Proposition 3.43 K \ ChH (K) = {x ∈ K : h∗ (x) − h(x) > 0} , we get (ii) =⇒ (iii). Finally, assume (iii) and choose v ∈ Kc (H). Since A := {x ∈ K : v ∗ (x) > v(x)} ⊂ K \ ChH (K), we have
Z
Z v dµ +
µ(v) = A
v dµ = µ(v ∗ ),
B
where B := {x ∈ K : v ∗ (x) = v(x)}. Thus (iii) =⇒ (i) by Mokobodzki’s maximality test 3.58. Corollary 3.62. Let K be a metrizable compact space and µ ∈ M+ (K). Then µ is maximal if and only if µ is carried by the Choquet boundary ChH (K). Proof. The assertion is a direct consequence of Theorem 3.42 and Corollary 3.61. Remark 3.63. We know from Corollary 3.62 that in the metrizable case any maximal measure is carried by the Choquet boundary, which is a Borel set. This assertion is no longer valid for nonmetrizable situations. Section 3.8 focuses on the Choquet theory in the nonmetrizable setting. Nevertheless, in this case any maximal measure is carried at least by the closure of the Choquet boundary; see the next Proposition 3.64. On the other hand, a measure carried by ChH (K) need not be maximal. Indeed, consider a function space H on a metrizable compact space K with a nonclosed Choquet boundary (cf. Exercise 2.103(b)). For a point x ∈ ChH (K) \ ChH (K), the Dirac measure at x is clearly carried by ChH (K) and it fails to be maximal in view of Corollary 3.62. A more striking example of this phenomenon is the Poulsen simplex (see Subsection 12.3.A), where the set of extreme points is dense in the whole set. Proposition 3.64. Any maximal measure is carried by ChH (K). Proof. Let µ be a maximal measure on K and C a compact subset of K \ ChH (K). Urysohn’s lemma ensures the existence of a function f ∈ C(K), 0 ≤ f ≤ 1 on K, such that f = 0 on ChH (K) and f = 1 on C.
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3 Choquet theory of function spaces
Since f∗ = sup {h ∈ H : h ≤ f }, we have h ≤ 0 on ChH (K) for any h ∈ H with h ≤ f . By Theorem 3.16, h ≤ 0 on K, and therefore f∗ ≤ 0 on K. Because obviously f∗ ≥ 0, we have f∗ = 0 on K. Mokobodzki’s maximality test 3.58 then yields µ(C) ≤ µ(f ) = µ(f∗ ) = 0. By regularity of µ we get µ(K \ ChH (K)) = 0. Theorem 3.65. For every measure µ ∈ M+ (K) there exists a maximal measure λ such that µ ≺ λ. Proof. Set M := {ν ∈ M+ (K) : µ ≺ ν}. Since for any ν ∈ M we have kνk = kµk, it follows that M is contained in the compact set {η ∈ M+ (K) : kηk = kµk}. The assertion will follow from Zorn’s lemma once it is shown that every chain in M has an upper bound. Let R be such a chain. Then R is a net in a compact set (directed by ≺) and hence there exists a subnet J of R which converges to some element ν0 ∈ M+ (K). Since M is closed, ν0 ∈ M . It remains to show that ν ≺ ν0 for any ν ∈ R. To this end, fix ν ∈ R, f ∈ Kc (H) and ε > 0. There is η ∈ J such that ν ≺ η and |ν0 (f ) − η(f )| < ε. Then ν0 (f ) ≥ η(f )−ε ≥ ν(f )−ε. Therefore ν ≺ ν0 since ε is an arbitrary positive number. We conclude by Zorn’s lemma: the ordered set M has a maximal element. Proposition 3.66. If µ ∈ M+ (K) is maximal, then µ({x}) = 0 for each x ∈ K \ ChH (K). Proof. Let µ ∈ M+ (K) be maximal and x ∈ K \ ChH (K). By Theorem 3.24, there exists a function f ∈ −W(H) such that f ∗ (x) > f (x). Since µ({y ∈ K : f ∗ (y) > f (y)}) = 0 (see Corollary 3.59), µ({x}) = 0. Proposition 3.67. If µ, ν ∈ M+ (K), then µ ≺H ν if and only if µ ≺Ac (H) ν. Hence, a measure µ ∈ M+ (K) is H-maximal if and only if it is Ac (H)-maximal. Proof. Since Mx (H) = Mx (Ac (H)) (see Exercise 3.95), a function f on K is H-convex if and only if it is Ac (H)-convex. Now the first assertion is a direct consequence of this fact, and the second assertion follows immediately from the first one. Theorem 3.68. The following assertions on a measure µ ∈ M(K) are equivalent: (i) µ is a boundary measure, (ii) µ(f ) = µ(f ∗ ) for any f ∈ C(K), (iii) µ(f ) = µ(f ∗ ) for any upper semicontinuous function f on K.
3.6 Maximal measures
77
Proof. By Theorem 3.58, (i) =⇒ (ii). Assume (ii), choose an upper semicontinuous function f on K and ε > 0. Using Theorem 3.48, we can find a function g ∈ S c (H) such that g > f ∗ and µ+ (g) < µ+ (f ∗ ) + ε
and
µ− (g) < µ− (f ∗ ) + ε.
Since f is upper semicontinuous, by Proposition A.50 and Theorem A.84 there exists h ∈ C(K) such that h > f and µ+ (h) < µ+ (f ) + ε
and
µ− (h) < µ− (f ) + ε.
Set ϕ := g ∧ h. Then ϕ ∈ C(K), ϕ > f , f ∗ ≤ ϕ∗ ≤ g and µ+ (ϕ) < µ+ (f ) + ε
and
µ− (ϕ) < µ− (f ) + ε.
µ+ (ϕ∗ ) ≤ µ+ (g) < µ+ (f ∗ ) + ε
and
µ− (ϕ∗ ) ≤ µ− (g) < µ− (f ∗ ) + ε,
It follows,
and therefore,
−2ε = µ(ϕ∗ ) − µ(ϕ) − 2ε ≤ µ(f ∗ ) − µ(f ) ≤ µ(ϕ∗ ) − µ(ϕ) + 2ε = 2ε.
We see that µ(f ) = µ(f ∗ ); hence (ii) =⇒ (iii). It remains to show that (iii) =⇒ (i). We have to verify that |µ| = µ+ + µ− is carried by the set Af := {x ∈ K : f ∗ (x) = f (x)} for any f ∈ C(K). To this end, let f ∈ C(K) be given. Let again µ = µ+ − µ− be the decomposition of µ into its positive and negative part, µ+ be carried by a set P and µ− be carried by a set N , where P ∩ N = ∅. It suffices to prove that µ+ (L) = 0 whenever L ⊂ P \ Af is a compact set. So, fix such a set L and set ( f on K \ L, g := ∗ f on L. ∗ ∗ Then g is upper semicontinuous on µ(g) = R K∗ and g =+ f . By the assumption, ∗ ∗ µ(g ) = µ(f ) = µ(f ), hence L (f − f ) dµ = 0. Since f ∗ − f > 0 on L, µ+ (L) = 0. The proof is complete.
We recall that the space M(K), as a dual space to the ordered Banach space C(K), is an ordered Banach space. Moreover, it is a lattice. Theorem 3.70 below shows that Mbnd (H) is a lattice with the structure inherited from M(K). Proposition 3.69. If µ ∈ Mmax (H) and ν ∈ M(K) is absolutely continuous with respect to µ, then ν ∈ Mbnd (H).
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3 Choquet theory of function spaces
Proof. It is enough to consider the case when ν is positive. If g is a continuous function, let A := {x ∈ K : g ∗ (x) = g(x)}. Since µ is a boundary R measure, |µ|(K \ A) = 0 (see Theorem 3.58). Let h ∈ L1 (µ) be such that ν(B) = B h dµ for any Borel set B ⊂ K. Then Z ν(K \ A) = h dµ = 0. K\A
Hence ν(g) = ν(g ∗ ) for any g ∈ C(K), and thus ν is maximal by Theorem 3.58. Theorem 3.70. The set Mbnd (H) is a norm closed subspace of M(K). Moreover, it is a lattice considered with the order inherited from M(K). Proof. We first show that Mbnd (H) is a norm closed subspace of M(K). To this end, let µ1 , µ2 ∈ Mbnd (H) and g ∈ C(K) be given. Since |µ1 | + |µ2 | is carried by the set {x ∈ K : g ∗ (x) = g(x)}, it is easy to see that |µ1 + µ2 | is carried by the set {x ∈ K : g ∗ (x) = g(x)}. Thus µ1 + µ2 is a boundary measure by Theorem 3.58. Analogously we show that cµ ∈ Mbnd (H) for any c ∈ R and µ ∈ Mbnd (H). If {µn } is a sequence of boundary measures norm converging to µ ∈ M(K), we use again Theorem 3.58 to show that µ is a boundary measure. To verify the lattice property we use Proposition A.15. Obviously, any measure µ ∈ Mbnd (H) can be written as a difference of positive maximal measures, and hence Mbnd (H) = (Mbnd (H))+ − (Mbnd (H))+ = Mmax (H) − Mmax (H). Let µ1 , µ2 ∈ Mmax (H) be given. Since M(K) is a lattice, there exists a measure ν ∈ M+ (K) such that ν = µ1 ∧ µ2 (with respect to the ordering of M(K)). Since ν ≤ µ1 + µ2 , ν is absolutely continuous with respect to the maximal measure µ1 + µ2 . By Proposition 3.69, ν ∈ Mmax (H) as well. Hence ν is a supremum of {µ1 , µ2 } in Mbnd (H) and Mbnd (H) is a lattice.
3.7
Boundaries and the Simons lemma
Definition 3.71 (σ-convex hull). Let A be a bounded subset of a Banach space E. We denote by ∞ ∞ nX o X coσ (A) := cn xn : cn ≥ 0, cn = 1, xn ∈ A n=1
n=1
the σ-convex hull of the set A. We notice the following elementary fact. Lemma 3.72. Let A be a bounded subset of a Banach space E. Then coσ (coσ (A)) = coσ (A).
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3.7 Boundaries and the Simons lemma
Proof. It follows by a straightforward verification. Definition 3.73 (Boundary). Let K be a set and let F be a subset of `∞ (K). We say that a nonempty set B in K is a boundary for F if every function from F attains its maximum at a point of B. Lemma 3.74 (Simons inequality). Let {fn } be a bounded sequence of functions on a set K. Let B ⊂ K be a boundary for coσ ({fn : n ∈ N}). Then sup lim sup fn (b) ≥ inf sup f (K) : f ∈ coσ ({fn : n ∈ N}) .
b∈B n→∞
Proof. Let {fn } be a bounded sequence on K. We start theP proof by fixing ε > 0 and a sequence {ρi } of strictly positive numbers such that ∞ i = 1. We set i=1 ρP P∞ εk σk := i=k ρi and find strictly positive numbers εk , k ∈ N, such that ∞ k=1 σk+1 < ε. Let Ck := coσ ({fn : n ≥ k}), k ∈ N. For a bounded function f on K, we write p(f ) := sup f (B) and q(f ) := sup f (K). We inductively find functions gi ∈ Ci , i ∈ N, such that p(ρ1 g1 ) ≤ infh∈C1 p(ρ1 h) + ε1 , Pk−1 Pk • p( i=1 ρi gi ) ≤ infh∈Ck p( i=1 ρi gi + ρk h) + εk , k ≥ 2. P We set g := ∞ i=1 ρi gi . P Claim. For every n ∈ N, p(g) ≥ p( nk=1 ρk gk ) + σn+1 (p(g) − ε). •
Proof of the claim. Let n ∈ N be given. Throughout the proof we interpret 0. We pick a number k ∈ {1, . . . , n}. Since ∞
k−1
i=1
i=1
P0
i=1
as
X 1 X ρi gi − ρi gi ∈ Ck , σk we get p
k X
ρi g i ≤ p
i=1
k−1 X
ρi gi + ρk
i=1
∞ k−1 X 1 X ( ρi gi − ρi gi ) σk i=1
! + εk
i=1
k−1 ρk X ρk p(g) + 1 − p ρi gi + εk . ≤ σk σk i=1
Hence p(g) 1 −
k k−1 X σk+1 X σk+1 ≥p ρi gi − p ρi g i − ε k . σk σk i=1
i=1
(3.4)
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3 Choquet theory of function spaces
We divide (3.4) by σk+1 and sum these inequalities for k = 1, . . . , n. Using p(0) = 0, we have n n X X 1 1 εk p(g) −1 ≥ p ρk gk − . σn+1 σn+1 σk+1 k=1
k=1
By the choice of the numbers εk , we get p(g) ≥ p
n X
ρk gk + σn+1 p(g) − ε .
k=1
This finishes the proof of the claim. We apply the assumption for the function g, and find b ∈ B such that g(b) = p(g) = q(g). Then, for n ∈ N, the claim yields ∞ X
ρi gi (b) ≥
i=1
Since
ρi gi (b) + σn+1 (q(g) − ε).
i=1
P∞ 1 σn+1 ( i=n+1 ρi gi ) q(g) − ε ≤
n X
∈ Cn+1 , we get 1
σn+1
∞ X
ρi gi (b)
i=n+1
≤ sup{gi (b) : i ≥ n + 1} ≤ sup{fi (b) : i ≥ n + 1}, since, for every i ≥ n + 1, gi ∈ Ci , and hence gi (b) ≤ sup{fi (b) : i ≥ n + 1}. By letting n tend to infinity, we get q(g) − ε ≤ lim sup fn (b) ≤ p(lim sup fn ). n→∞
n→∞
This concludes the proof. Corollary 3.75 (Simons lemma). Let {fn } and B ⊂ K be as in Lemma 3.74. Then sup lim sup fn (b) = sup lim sup fn (x). b∈B n→∞
x∈K n→∞
Proof. Let {fn } be as in Lemma 3.74. We again adopt the notation p(f ) = sup f (B) and q(f ) = sup f (K) for each f ∈ `∞ (K). Assume that p(lim supn→∞ fn ) < c < d < q(lim supn→∞ fn ). Let x ∈ K satisfy lim supn→∞ fn (x) > d. Let n1 < n2 < . . . be natural numbers such that fnk (x) > d, k ∈ N. Clearly, B is a boundary for coσ {fnk : k ∈ N}. By Lemma 3.74, there exists f ∈ coσ {fnk : k ∈ N} such that q(f ) ≤ p(lim sup fnk ) + d − c ≤ p(lim sup fn ) + d − c < d. k→∞
n→∞
Obviously, q(f ) ≥ f (x) > d. This contradiction finishes the proof.
3.8 The Bishop–de Leeuw theorem
81
Corollary 3.76. Let {fn } be a bounded sequence of functions in `∞ (K), g ∈ `∞ (K), A := {fn : n ∈ N} ∪ {g} and let B ⊂ K be a boundary for span A. (a) If {fn (b)} is convergent for each b ∈ B, then {fn (x)} is convergent for each x ∈ K. (b) If fn → g on B, then fn → g on K. Proof. (a) Let {fn } and B ⊂ K be as in the statement of the corollary. Note that B is a boundary for coσ (span A). We again write p(f ) := sup f (B) for f ∈ `∞ (K). Let {fn (b)} be convergent for each b ∈ B, and assume that there exists x ∈ K such that {fn (x)} is not Cauchy. By choosing a subsequence if necessary, we may assume that there exists a strictly increasing sequence {nk } of natural numbers and c > 0, such that c < fn2k (x) − fn2k−1 (x), k ∈ N. Then the functions gk := fn2k − fn2k−1 , k ∈ N, satisfy p(lim supk→∞ gk ) = 0, yet 0 < c ≤ lim supk→∞ gk (x). This contradicts Corollary 3.75 and proves (a). (b) Let fn → g on B. Then, for each x ∈ K, Corollary 3.75 gives lim sup(fn (x) − g(x)) ≤ p(lim sup(fn − g)) = 0. n→∞
n→∞
If we apply Corollary 3.75 to g − fn , we get lim inf(fn (x) − g(x)) ≥ 0. n→∞
This concludes the proof.
3.8
The Bishop–de Leeuw theorem
Throughout this section, H will be a function space on a compact space K. Lemma 3.77. Let {fn } be an upper bounded sequence of lower semicontinuous Hconvex functions on K such that lim supn→∞ fn ≤ 0 on ChH (K). Then we have lim supn→∞ fn ≤ 0 on K. Proof. Let {fn } be as in the statement of the lemma. Without loss of generality we may assume that {fn } is a bounded sequence, sup{kfn k : n ∈ N} ≤ C (otherwise we would consider the functions −1∨fn ). Fix x ∈ K and ε > 0. Using Proposition 3.48, we find functions kn ∈ Kc (H) such that kn ≤ fn and kn (x) ≥ fn (x) − ε. Without loss of generality we may assume that sup{kkn k : n ∈ N} ≤ C. By Theorem 3.16, each function in coσ ({kn : n ∈ N}), as an H-convex continuous function, attains its maximum on ChH (K). Since lim sup kn ≤ lim sup fn ≤ 0 n→∞
n→∞
on ChH (K),
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3 Choquet theory of function spaces
the Simons lemma 3.75 yields lim sup fn (x) ≤ ε + lim sup kn (x) ≤ ε. n→∞
n→∞
Since ε > 0 is arbitrary, the proof is finished. Lemma 3.78. Let µ be a maximal measure on K and F ⊂ K \ ChH (K) be a compact Gδ set. Then µ(F ) = 0. Proof. Given a maximal measure µ ∈ M+ (K) and a compact Gδ set F ⊂ K \ ChH (K), we find a sequence {fn } of continuous functions such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cF . Then {(fn )∗ } is a bounded sequence of H-convex lower semicontinuous functions such that lim sup(fn )∗ ≤ 0 on ChH (K). By Lemma 3.77, lim supn→∞ (fn )∗ ≤ 0 on K. Thus Theorem 3.58(v) together with Fatou’s lemma yields 0 ≤ µ(F ) = lim µ(fn ) = lim µ((fn )∗ ) n→∞
n→∞
≤ lim sup µ((fn )∗ ) ≤ µ(lim sup(fn )∗ ) ≤ 0. n→∞
n→∞
This concludes the proof. Theorem 3.79. Let µ ∈ M+ (K) be a maximal measure on K. Then (a) µ(B) = 0 for every Baire set B disjoint from ChH (K), (b) µ∗ (K \ L) = 0 for every Lindel¨of set L ⊃ ChH (K) (here µ∗ denotes the inner measure induced by µ, see Definition A.62), (c) µ(K \ A) = 0 for every K-analytic set A ⊃ ChH (K). Proof. Let µ ∈ M+ (K) be a maximal measure and B ⊂ K \ ChH (K) be a Baire set. Then µ(B) = 0 by Lemma A.89 and Lemma 3.78. This shows (a). Assume now that the set L ⊃ ChH (K) is Lindel¨of and F ⊂ K \ L is a compact set. For each x ∈ L we find an open Fσ set U (x) such that x ∈ Ux ⊂ K \ F . Using the Lindel¨ select countably many points xn ∈ L, n ∈ N, such that S of property we S∞ U L⊂ ∞ . Since n=1 xn n=1 Uxn is a Baire set containing ChH (K), µ(F ) = 0, by (a). Hence µ∗ (K \ L) = 0. For the proof of (c), let A ⊃ ChH (K) be a K-analytic set. By Theorem A.111(d), (g) we get that A is Lindel¨of and µ-measurable. Hence µ(K \ A) = 0, by (b). Proposition 3.80. Assume that ChH (K) is a Lindel¨of set and µ ∈ M+ (K). Then the following assertions are equivalent: (i) µ∗ K \ ChH (K) = 0, (ii) µ is maximal.
3.8 The Bishop–de Leeuw theorem
83
Proof. Since (ii) =⇒ (i) is proved in Theorem 3.79, it remains to show the reverse implication. Let µ ∈ M+ (K) be a measure satisfying µ∗ (K \ ChH (K)) = 0 and let f ∈ Kc (H) be fixed. By Theorem 3.24, {x ∈ K : f (x) < f ∗ (x)} is disjoint from ChH (K). Since µ({x ∈ K : f (x) < f ∗ (x)}) = 0, by assumption, Corollary 3.59 concludes the proof. Theorem 3.81 (Bishop–de Leeuw). For any point x ∈ K, there exists a measure µ ∈ Mx (H) such that µ(A) = 1 for every K-analytic set A ⊃ ChH (K). Proof. Given x ∈ K, Theorem 3.65 provides a maximal measure µ ∈ Mx (H). By Theorem 3.79(c), the proof is complete. Example 3.82. There exist a function space H on a compact space K and a maximal measure on K carried by a compact set disjoint from ChH (K). Proof. Let K := [0, 1]×{−1, 0, 1} be equipped with the topology defined as follows: points of [0, 1] × {−1, 1} are open (hence [0, 1] × {−1, 1} is discrete), and the base of neighborhoods of (x, 0) ∈ [0, 1] × {0} are of type (U × {−1, 0, 1}) \ F , where U ⊂ [0, 1] is a Euclidean neighborhood of x, and F is finite. If f is a continuous function on this compact space K, we notice that, for each δ > 0, the set {x ∈ [0, 1] : |f (x, 0) − f (x, 1)| + |f (x, 0) − f (x, −1)| > δ}
(3.5)
is finite. We define a function space H as 1 H := {f ∈ C(K) : f (x, 0) = (f (x, −1) + f (x, 1)), x ∈ [0, 1]}. 2 Fix x ∈ [0, 1]. Then the function c(x,1) − c(x,−1) shows that {(x, −1), (x, 1)} is a subset of ChH (K). Further, the function (t, j) 7→ |t − x|, (t, j) ∈ K, yields that the only µ ∈ M(x,0) (H) with µ({(x, 0)}) = 0 equals 21 (ε(x,−1) + ε(x,1) ). Thus ChH (K) = K \ ([0, 1] × {0}) and for each (x, 0) ∈ K, there exists a unique measure µ ∈ M(x,0) (H) with µ({(x, 0)}) = 0. Claim. If f ∈ Kc (H) and δ > 0, then {x ∈ [0, 1] : f ∗ (x, 0) − f (x, 0) > δ} is finite. Proof of the claim. Let f ∈ Kc (H) and δ > 0 be given. If x ∈ [0, 1], by Lemma 3.21, there exists a measure µ ∈ M(x,0) (H) such that f ∗ (x, 0) = µ(f ). Since f is Hconvex, µ(f ) ≤ ν(f ) ≤ f ∗ (x, 0) for any ν ∈ Mx (H) with µ ≺ ν. Hence we may assume that µ is maximal. By Proposition 3.66, µ({(x, 0)}) = 0. As shown above, µ = 21 (ε(x,−1) + ε(x,1) ). Thus 1 f ∗ (x, 0) = (f (x, −1) + f (x, 1)), 2 By (3.5), the proof of the claim is finished.
x ∈ [0, 1].
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3 Choquet theory of function spaces
Now let µ ∈ M+ ([0, 1] × {0}) be a continuous measure. By the claim, µ({x ∈ [0, 1] : f ∗ (x, 0) − f (x, 0) > 0}) = 0. Thus by Corollary 3.59, µ is maximal. On the other hand, µ is carried by the compact set [0, 1] × {0}, which is disjoint from ChH (K). Example 3.83. There exists a function space H on a compact space K and a measure on K carried by any Baire set containing ChH (K) which is not maximal. Proof. Let H be the function space from Example 3.47. Then ChH (K) = K \ {d∞ } and εd∞ is not a maximal measure. On the other hand, if A ⊃ ChH (K) is a Baire set, then A = K. Indeed, any Baire set in K is Lindel¨of (see Theorem A.111(i) and (d)) and hence K \ {d∞ } is not a Baire set. Otherwise, as an open set it would be an Fσ set, which it obviously is not. Hence εd∞ (A) = 1 for each Baire set A containing ChH (K).
3.9
Minimum principles
In the preceding sections, we already met several minimum principles. Recall Bauer’s convex minimum principle in 2.24 or the minimum principle for lower semicontinuous H-concave functions in Theorem 3.16. Definition 3.84 (Minimum principles). Let H be a function space on a compact space K. We say that a function f on K satisfies the minimum principle if f ≥ 0 on K provided f ≥ 0 on ChH (K), and f satisfies the strict minimum principle if f > 0 on K whenever f > 0 on ChH (K). Theorem 3.85 (Minimum principle for S usc (H)). Let f ∈ S usc (H) be an upper semicontinuous H-concave function. Then f satisfies the minimum principle. Proof. Since f is upper semicontinuous, f ≥ 0 on ChH (K). Pick x ∈ K. By the Choquet–Bishop–de Leeuw theorem 3.81, there exists a maximal measure µ in Mx (H) carried by ChH (K). Hence Z Z f dµ ≥ 0. f (x) ≥ f dµ = K
ChH (K)
Theorem 3.86 (Minimum principle for Baire H-concave functions). Let f be a Baire H-concave function. Then f satisfies the minimum principle.
3.9 Minimum principles
85
Proof. Assume that f ≥ 0 on ChH (K) and f (x) < 0 for some x ∈ K. The set F := {y ∈ K : f (y) ≤ f (x)} is a nonempty Baire subset of K, disjoint from ChH (K). Let µ ∈ Mx (H) be a maximal measure. By the Choquet–Bishop–de Leeuw theorem 3.81, µ(F ) = 0. Hence Z Z f (x) dµ(y) = f (x), f dµ > f (x) ≥ µ(f ) = K\F
K\F
which is a contradiction. Proposition 3.87 (Strict minimum principles). Let f be a lower bounded H-concave function on K. If f is lower or upper semicontinuous, or if f is a Baire function on K, then f satisfies the strict minimum principle. Proof. Let f ∈ S lsc (H), f > 0 on ChH (K). By Theorem 3.16, f ≥ 0 on K. If Z := {x ∈ K : f (x) = 0}, then Z is a closed H-extremal set. Since Z ∩ ChH (K) = ∅, it follows from Proposition 3.15 that Z = ∅. Assume now that f ∈ S(H) is a Baire function. The set Z := {x ∈ K : f (x) = 0} is a Baire set disjoint from ChH (K). If Z contains a point x, choose a maximal measure µ ∈ Mx (H). Then Z 0 = f (x) ≥ µ(f ) = f dµ > 0, K\Z
a contradiction. Finally, let f ∈ S usc (H) be upper semicontinuous, and let f (x) = 0 for some x ∈ K. By Proposition 3.48, there exists a sequence {fn } in S c (H) such that fn ≥ fn+1 ≥ f for each n ∈ N and fn (x) → 0. If h := inf {fn : n ∈ N}, then h is a Baire H-concave function such that h > 0 on ChH (K). By the strict minimum principle for H-concave Baire functions which we have just proved, h > 0 on K. However, h(x) = 0, a contradiction that finishes the proof. Proposition 3.88 (Minimum principle for semicontinuous functions). Assume that H is a function space on a compact space K and f , g are bounded functions on K such that f , −g are H-convex, f ≤ g on ChH (K) and f, g are semicontinuous. Then f ≤ g on K. Proof. Assume, for example, that both functions are upper semicontinuous. We fix x ∈ K and ε > 0 and find a continuous H-concave function g 0 on K such that g ≤ g 0 and g 0 (x) ≤ g(x) + ε (see Proposition 3.48). Then g 0 − f ≥ 0 on ChH (K) and thus g 0 − f ≥ 0 on K by Theorem 3.16. Hence f (x) ≤ g 0 (x) ≤ g(x) + ε. Since ε is arbitrary, the proof is complete. The remaining cases can be dealt with by a similar argument.
86
3.10
3 Choquet theory of function spaces
Orderings and dilations
Proposition 3.89. Let F1 , F2 be closed sets in K such that F2 ⊃ ChH (K) and µ ∈ M1 (F1 ), ν ∈ M1 (F2 ). Let M := {(εx , λ) ∈ M1 (F1 ) × M1 (F2 ) : εx ≺ λ}. Then M is closed and the following assertions are equivalent: (i) µ ≺ ν, (ii) there exists Λ ∈ M1 (M ) such that (µ, ν) is the barycenter of Λ. Proof. For the proof (i) =⇒ (ii), we show that (µ, ν) ∈ co M. Assume that this is not the case. Then we use the Hahn–Banach theorem to find continuous functions f1 ∈ C(F1 ), f2 ∈ C(F2 ) and c ∈ R such that µ(f1 ) + ν(f2 ) > c > εz (f1 ) + λ(f2 ),
(εz , λ) ∈ M.
The assumption µ ≺ ν yields ν(F2f2∗ ) ≤ µ(F2f2∗ ). We fix x ∈ F1 . By Lemma 3.21, sup{εx (f1 ) + λ(f2 ) : λ ∈ M1 (F2 ), εx ≺ λ} = f1 (x) + sup{λ(f2 ) : λ ∈ Mx (H) ∩ M1 (F2 )} = f1 (x) + F2f2∗ (x). Hence
µ(f1 + F2f2∗ ) ≥ µ(f1 ) + ν(F2f2∗ ) ≥ µ(f1 ) + ν(f2 ) > c ≥ sup{εz (f1 ) + λ(f2 ) : (εz , λ) ∈ M } ≥ f1 (x) + F2f2∗ (x).
Thus µ(f1 + F2f2∗ ) > c ≥ sup (f1 (x) + Ff2∗ (x)). x∈F1
Since µ is a probability measure, this is impossible. Hence (µ, ν) ∈ co M , which yields the desired measure Λ ∈ M1 (M ) in view of Proposition 2.39. For the proof of the converse implication, let Λ ∈ M1 (M ) represent (µ, ν). Then for each f ∈ Kc (H), Z Z µ(f ) = εx (f ) dΛ(εx , λ) ≤ λ(f ) dΛ(εx , λ) = ν(f ). M
This concludes the proof.
M
3.10 Orderings and dilations
87
Proposition 3.90. Let P := M1 (K) × M1 (K) and Λ ∈ M1 (P ) satisfy r(Λ) = (µ1 , µ2 ). Then, for any bounded universally measurable functions f1 , f2 on K, the function (λ1 , λ2 ) 7→ λ1 (f1 ) + λ2 (f2 ), (λ1 , λ2 ) ∈ P, is universally measurable on P , and Z µ1 (f1 ) + µ2 (f2 ) = (λ1 (f1 ) + λ2 (f2 )) dΛ(λ1 , λ2 ).
(3.6)
P
Proof. Let us first notice that (3.6) is clear for f1 , f2 ∈ C(K). To extend this formula to universally measurable functions, we will first consider characteristic functions. So, let A be the family of all universally measurable sets A ⊂ K such that (3.6) holds for pairs of functions (cA , 0) and (0, cA ). Step 1. We show that the family A contains all Borel subsets of K. Clearly, K ∈ A. If G ⊂ K is open, Proposition A.50 yields cG = sup{ϕ ∈ C(K) : 0 ≤ ϕ ≤ cG }. By Theorem A.84, ω(G) = sup{ω(ϕ) : ϕ ∈ C(K), 0 ≤ ϕ ≤ cG },
ω ∈ M1 (K).
Thus the functions (λ1 , λ2 ) 7→ λ1 (G),
(λ1 , λ2 ) 7→ λ2 (G),
(λ1 , λ2 ) ∈ P,
are suprema of up-directed families of continuous functions. Hence we may use Theorem A.84 again for the measure Λ to obtain validity of (3.6) for the pairs (cG , 0) and (0, cG ). Hence A contains all open subsets of K. Obviously K ∈ A and K \A ∈ A whenever A ∈ A. Since A is closed with respect to the formation of unions of two pairwise disjoint elements of A, and the Lebesgue monotone convergence theorem implies that the countable union of an increasing sequence of sets from A also belongs to A, we see that A is a Dynkin system. Since A contains all open subsets of K, the Dynkin lemma A.68 yields that A contains any Borel subset of K. Step 2. To show that any universally measurable subset of K belongs to A, let A be such a set. We write A = H ∪ N,
H is an Fσ set,
µ1 (N ) = µ2 (N ) = 0,
H ∩ N = ∅.
Since we know that H ∈ A, to finish the argument we need to prove that the functions (λ1 , λ2 ) 7→ λ1 (N ), equal zero Λ-almost everywhere.
(λ1 , λ2 ) 7→ λ2 (N ),
(λ1 , λ2 ) ∈ P,
88
3 Choquet theory of function spaces
We find a decreasing sequence {Gn } of open subsets of K such that N ⊂ Gn and (µ1 + µ2 )(Gn ) → 0. For a rational number q ∈ (0, 1) and n ∈ N, let Nq := {(λ1 , λ2 ) ∈ P : λ1 (N ) ≥ q} ∪ {(λ1 , λ2 ) ∈ P : λ2 (N ) ≥ q} and
Nqn :={(λ1 , λ2 ) ∈ P : λ1 (Gn ) ≥ q} ∪ {(λ1 , λ2 ) ∈ P : λ2 (Gn ) ≥ q}.
Since the sets Gn belong to A, for any n ∈ N and q ∈ (0, 1) ∩ Q, we get Z Z 1 n (λ1 (Gn ) + λ2 (Gn )) dΛ(λ1 , λ2 ) 1 dΛ(λ1 , λ2 ) ≤ Λ(Nq ) = q P Nqn 1 = (µ1 + µ2 )(Gn ). q Since Nq ⊂ Nqn , n ∈ N, the choice of the sets Gn yields that Nq is Λ-measurable and Λ(Nq ) = 0. Thus Λ ({(λ1 , λ2 ) ∈ P : λ1 (N ) + λ2 (N ) > 0}) [ = Λ(Nq ) = 0, q∈Q∩(0,1)
which is the desired conclusion. Step 3. Since A contains all universally measurable subsets of K, (3.6) holds for any pair of bounded universally measurable functions on K, by a standard approximation argument. This finishes the proof. Definition 3.91 (µ-dilations and dilations). If µ ∈ M+ (K) is given, a mapping T = {Tx }x∈K : K → M1 (K) is called a µ-dilation, if • x 7→ T (f ), x ∈ K, is µ-measurable for any bounded universally measurable x function f on K, for each h ∈ H, µ({x ∈ K : h(x) 6= Tx (h)}) = 0. If f is universally measurable, we define a function T f by •
T f (x) = Tx (f ). By the conditions above, the formula Z (T µ)(f ) = T f dµ, K
defines a measure T µ ∈ M+ (K).
f ∈ C(K),
3.10 Orderings and dilations
89
A mapping T = {Tx }x∈K : K → M1 (K) is termed a dilation, if • x 7→ T (f ), x ∈ K, is Borel for any f ∈ C(K), x Tx ∈ Mx (H), x ∈ K. Hence T is a particular example of a kernel as defined in Subsection A.3.D. As above, we use the notation T f : x 7→ Tx (f ) if f is universally measurable. We can extend the mapping T to a mapping from M(K) to M(K) by the formula Z T f dν, f ∈ C(K), ν ∈ M(K). (T ν)(f ) = •
K
Before stating our next S theorem, we will recall and introduce some notation. Remember that M(H) := x∈K Mx (H) (cf. Proposition 3.37) and r : M(H) → K is the H-barycenter mapping (cf. Definition 3.39). Let P := M1 (K) × M1 (K). We will also consider the barycenter mapping rP : M(P ) → P , here the subscript P serves only for easier distinguishing which barycenter mapping we are just operating with. We denote by πi : K × K → K the i-the projection, i = 1, 2. Then they induce mappings (πi )] : M1 (K × K) → M1 (K), i = 1, 2. Analogously, let pi : P → M1 (K) be the i-th projection, i = 1, 2. As above, we obtain mappings (pi )] : M1 (P ) → M1 (M1 (K)),
i = 1, 2.
If L is a compact set and F ⊂ L is closed, we write µ ∈ M1 (F ) ⊂ M1 (L) if µ ∈ M1 (L) is carried by F and we want to emphasize that we consider µ both as an element of M1 (F ) and M1 (L). Theorem 3.92. Let µ, ν be probability measures on K and M := {(εx , λ) ∈ P : εx ≺ λ, x ∈ K}. Then the following assertions are equivalent: (i) µ ≺ ν, (ii) there exists Ω ∈ M1 (M ) ⊂ M1 (P ) such that (µ, ν) = rP (Ω), (iii) there exists λ ∈ M1 (K × K) such that • •
(π1 )] λ = µ, (π2 )] λ = ν, R R A×K h ◦ π1 dλ = A×K h ◦ π2 dλ, whenever A ⊂ K is universally measurable and h ∈ H,
(iv) there exists a µ-dilation such that T µ = ν, (v) there exists Λ ∈ M1 (M(H)) such that
90
3 Choquet theory of function spaces •
•
R ν(f ) = M(H) λ(f ) dΛ(λ), for any bounded universally measurable function f on K, r] Λ = µ,
P µi ∈ M+ (K), there exist measures νi ∈ M+ (K), (vi) whenever µ = ni=1 µi withP i = 1, . . . , n, such that ν = ni=1 νi and µi − νi ∈ H⊥ . If, moreover, K is metrizable, then (i) is equivalent with the following assertion: (iv’) there exists a dilation T such that T µ = ν. Proof. We already know from Proposition 3.89 that (i) ⇐⇒ (ii). We will prove (i) =⇒ (iii) =⇒ (iv) =⇒ (i), (ii) =⇒ (v) =⇒ (i), (ii) =⇒ (vi) =⇒ (i) and (ii) =⇒ (iv’) =⇒ (i) in the case when K is metrizable. For the proof of (i) =⇒ (iii), assume that µ ≺ ν. For any functions f, g : K → R, we denote by f ⊗ g : K × K → R, f ⊗ g : (x, y) 7→ f (x)g(y),
(x, y) ∈ K × K.
If f is a function on K × K and x ∈ K, we write fx for the function y 7→ f (x, y), y ∈ K. Analogously define f y , y ∈ K. We define a subspace V ⊂ C(K × K) by V := {f ⊗ 1 + 1 ⊗ g : f, g ∈ C(K)} and a linear functional ρ : V → R by ρ : f ⊗ 1 + 1 ⊗ g 7→ µ(f ) + ν(g), We define σ : C(K × K) → R by Z ∗ σ(f ) := (fx )∗ (x) dµ(x),
f, g ∈ C(K).
f ∈ C(K × K).
K
R∗ ( is the upper integral, see Definition A.69). Then σ is a sublinear functional on C(K × K) satisfying σ(f ) ≤ 0 for f ≤ 0. (3.7) Further, ρ ≤ σ on V . Indeed, if f, g ∈ C(K) are given, we have (f ⊗ 1 + 1 ⊗ g)∗x (x) = f (x) + g ∗ (x),
x ∈ K,
and µ(g ∗ ) ≥ ν(g ∗ ) (see Proposition 3.56 and Lemma 3.18). Thus Z ∗ σ(f ⊗1 + 1 ⊗ g) = ((f ⊗ 1 + 1 ⊗ g)x )∗ (x) dµ(x) K Z = (f (x) + g ∗ (x)) dµ(x) ≥ µ(f ) + ν(g ∗ ) K
≥ µ(f ) + ν(g) = ρ(f ⊗ 1 + 1 ⊗ g).
3.10 Orderings and dilations
91
By the Hahn–Banach theorem, there exists a functional λ ∈ (C(K × K))∗ such that λ = ρ on V and λ ≤ σ on C(K × K). By (3.7), λ is a Radon measure on K × K. If k ∈ S c (H) and g ∈ C(K) positive are given, then Z ∗ ((g ⊗ k)x )∗ (x) dµ(x) λ(g ⊗ k) ≤ σ(g ⊗ k) = K Z Z g(x)k(x) dµ(x) = λ(gk ⊗ 1). g(x)k ∗ (x) dµ(x) = = K
K
This gives λ(gh ⊗ 1) = λ(g ⊗ h),
g ∈ C(K), h ∈ H.
(3.8)
(If g is not positive, let c ≥ 0 be such that g + c is positive, and apply (3.8) to (g + c) − c.) We claim that λ ∈ M+ (K ×K) possesses all the required properties. First, λ(1) = µ(1) = 1, and hence λ ∈ M1 (K × K). Further, for f ∈ C(K), we have (π1 )] λ(f ) = λ(f ◦ π1 ) = λ(f ⊗ 1) = ρ(f ⊗ 1) = µ(f ), and (π2 )] λ(f ) = λ(f ◦ π2 ) = λ(1 ⊗ f ) = ρ(1 ⊗ f ) = ν(f ). Finally, (3.8) and standard measure-theoretic techniques yield Z Z g(x)h(x) dλ(x, y) = g(x)h(y) dλ(x, y) K×K
K×K
for h ∈ H and a bounded universally measurable function g on K. By setting g = cA , where A ⊂ K is universally measurable, we get Z Z h(x) dλ(x, y) = h(y) dλ(x, y), h ∈ H. A×K
A×K
For the proof of (iii) =⇒ (iv), let λ ∈ M1 (K × K) be as in (iii). By Theorem A.106, there exists a family {Tx }x∈K of Radon probability measures on K such that Z Z Z f dλ = f (x, y) dTx (y) dµ(x) (3.9) K×K
K
K
for every λ-integrable function f on K × K. We claim that the mapping T : x 7→ Tx , x ∈ K, is a µ-dilation such that T µ = ν. To prove this, let f be a bounded universally measurable function on K. Then the function fe := 1 ⊗ f , is λ-measurable, and so (3.9) together with the properties of λ give Z Z Z µ(T f ) = f (y) dTx (y) dµ(x) = fe(x, y) dλ(x, y) K
K
= ((π2 )] λ)(f ) = ν(f ). Hence T f is µ-measurable and T µ = ν.
K×K
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3 Choquet theory of function spaces
To finish the proof of the implication, let h ∈ H be given. For any A ⊂ K Borel, we define e h := cA ⊗ h. Then e h is λ-measurable and using (iii) we obtain Z Z Z Z e e T h dµ = h(x, y) dTx (y) dµ(x) = h(x, y) dλ(x, y) A
K
K
K×K
Z
Z
h(x) dλ(x, y)
h(y) dλ(x, y) =
=
A×K
A×K
Z
Z h(x) d((π1 )] λ) =
=
h(x) dµ(x). A
A
This shows that T h = h for µ-almost all x ∈ K. We proceed with the proof of (iv) =⇒ (i). Let T = {Tx }x∈K : K → M1 (K) be a µ-dilation such that T µ = ν. Let f ∈ −W(H) be given; that is, f = h1 ∨ · · · ∨ hn , h1 , . . . , hn ∈ H. We set A0i := {x ∈ K : f (x) = hi (x)}, A1 := A01 , Ai := A0i \ (A1 ∪ · · · ∪ Ai−1 ),
i = 1, . . . , n, i = 2, . . . , n.
(3.10)
Let N ⊂ K be a set of µ-measure zero such that T hi (x) = hi (x) for each x ∈ K \ N and i = 1, . . . , n. Then Z n Z X µ(f ) = f dµ = hi dµ K\N
=
n Z X Ai \N
i=1
i=1
Ai \N
T hi dµ ≤
n Z X i=1
T f dµ
Ai \N
Z ≤
T f dµ = T µ(f ) = ν(f ). K
Hence µ ≺ ν by Proposition 3.56. For the proof of (ii) =⇒ (v), let Ω ∈ M1 (M ) ⊂ M1 (P ) satisfy rP (Ω) = (µ, ν). We claim that the measure Λ := (p2 )] Ω satisfies our requirements. First, we observe that p2 (M ) ⊂ M(H), so that Λ ∈ M1 (M(H)) ⊂ M1 (M1 (K)). Let f be a bounded universally measurable function on K. Then Proposition 3.90 yields Z Z ν(f ) = λ(f ) dΩ(εx , λ) = λ(f ) dΛ(λ) M(H)
M
and
Z
Z
(r] Λ)(f ) =
f (r(λ)) dΛ(λ) = M(H)
f (r(λ)) dΩ(εx , λ) M
Z =
εx (f ) dΩ(εx , λ) = µ(f ). M
Hence r] Λ = µ.
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93
To prove (v) =⇒ (i), let Λ ∈ M1 (M(H)) be as in (v). For each f ∈ Kc (H), we have Z Z µ(f ) = r] Λ(f ) = f (r(λ)) dΛ(λ) ≤ λ(f ) dΛ(λ) = ν(f ). M(H)
M(H)
Hence µ ≺ ν. 1 1 For the proof of (ii) Pn=⇒ (vi), let Ω ∈ M (M ) ⊂ M (P ) satisfy rP (Ω) = (µ, ν). Suppose that µ = i=1 µi , where Radon measures µi are nontrivial. According to the Radon–Nikodym theorem, there exist positive Borel measurable functions fi , i = 1,P. . . , n, such that µi (ϕ) = µ(fi ϕ) for each bounded Borel function ϕ on K. ThenP ni=1 fi = 1 µ-almost everywhere. Without loss of generality we may assume that ni=1 fi = 1 on K. For each i ∈ {1, . . . , n}, we define the function gi : M → R by gi : (εx , λ) 7→ εx (fi ),
(εx , λ) ∈ M,
and the measure Ωi ∈ M1 (P ) by Ωi :=
gi Ω . Ω(gi )
Note that, by Proposition 3.90, Z Ω(gi ) = gi dΩ(εx , λ) = µ(fi ) = µi (K) > 0. M
Let (λ1i , λ2i ) ∈ P be the barycenter of Ωi , i = 1, . . . , n, and let ϕ be a bounded Borel function on K. Then Z Z 1 1 λi (ϕ) = εx (ϕ) dΩi (εx , λ) = εx (ϕ)gi (εx , λ) dΩ(εx , λ) Ω(gi ) M M Z 1 1 1 = εx (fi ϕ) dΩ(εx , λ) = µ(fi ϕ) = µi (ϕ). Ω(gi ) M Ω(gi ) Ω(gi ) Thus µi = Ω(gi )λ1i ,
i = 1, . . . , n.
νi := Ω(gi )λ2i ,
i = 1, . . . , n.
We set By Proposition 3.89, λ1i ≺ λ2i , and hence µi − νi ∈ H⊥ ,
i = 1, . . . , n.
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3 Choquet theory of function spaces
Finally, for any ϕ ∈ C(K) we get (using (3.6) from Proposition 3.90) n X
n X
νi (ϕ) =
i=1
Z
i=1 n Z X
=
λ(ϕ) dΩi (εx , λ) =
Ω(gi ) M
i=1
gi (εx , λ) λ(ϕ) dΩ(εx , λ)
M
Z εx (fi )λ(ϕ) dΩ(εx , λ) =
M
i=1
n Z X
εx
n X fi λ(ϕ) dΩ(εx , λ)
M
i=1
Z λ(ϕ) dΩ(εx , λ) = ν(ϕ).
= M
P Hence ν = ni=1 νi , as required. To show that (vi) =⇒ (i), let µ, ν satisfy (vi). If f = h1 ∨ · · · ∨ hn with h1 , . . . , hn ∈ H, let the sets Ai be defined as in (3.10). We set µi := µ|Ai ,
i = 1, . . . , n,
and use the assumptionP to find Radon measures νi , i = 1, . . . , n, such that µi − νi ∈ ⊥ H , i = 1, . . . , n, and ni=1 νi = ν. Then µ(f ) =
n Z X i=1
Ai
hi dµ =
n X
µi (hi ) =
i=1
n X
νi (hi ) ≤
i=1
n X
νi (f ) = ν(f ).
i=1
Thus µ ≺ ν, by Proposition 3.56. We proceed by showing (ii) =⇒ (iv’). In this case, we use the Disintegration theorem A.107; this is the point where we use metrizability. Assume that Ω ∈ M1 (M ) and rP (Ω) = (µ, ν). We define Ψ : M → K by Ψ(εx , λ) := x,
(εx , λ) ∈ M.
Then Ψ is a continuous surjection of M onto K. We pick f ∈ C(K) and set ˜ λ) := λ(f ˜ ), ϕ(λ,
˜ λ) ∈ P. (λ,
Then ϕ ∈ Ac (P ) and we have Z Z Z ˜ ) dΩ(λ, ˜ λ) λ(f f (Ψ(εx , λ)) dΩ(εx , λ) = εx (f ) dΩ(εx , λ) = Ψ] Ω(f ) = M M M Z ˜ λ) dΩ(λ, ˜ λ) = ϕ(µ, ν) = µ(f ). = ϕ(λ, M
This shows that µ = Ψ] Ω. According to the Disintegration theorem A.107, there exists a mapping S = {Sx }x∈K : K → M1 (M ) ⊂ M1 (P ) such that •
the mapping x 7→ Sx (ψ), x ∈ K, is Borel for each ψ ∈ C(M ),
3.11 Exercises • •
95
spt Sx ⊂ Ψ−1 (x), x ∈ K, R Ω(ψ) = K Sx (ψ) dµ(x), ψ ∈ C(M ).
We define the required mapping T = {Tx }x∈K : K → M1 (K) as T : x 7→ p2 (rP (Sx )),
x ∈ K.
Since T is a composition of S with continuous mappings, T f is Borel for each f ∈ C(K). Further, given f ∈ C(K), we set ˜ λ) := λ(f ), ψ(λ,
˜ λ) ∈ P. (λ,
(3.11)
Then ψ ∈ Ac (P ) and Z Sx (ψ) =
˜ λ) dSx (λ, ˜ λ) = ψ(rP (Sx )) ψ(λ,
M
x ∈ K.
= p2 (rP (Sx ))(f ) = Tx (f ), Hence Z T µ(f ) =
Z Tx (f ) dµ(x) =
Sx (ψ) dµ(x) = Ω(ψ) = ψ(rP (Ω))
K
K
= ψ(µ, ν) = ν(f ). This means T µ = ν. For each x ∈ K we have spt Sx ⊂ Ψ−1 (x) = {εx } × Mx (H). Since {εx } × Mx (H) is closed and convex, it follows rP (Sx ) ∈ {εx } × Mx (H) and
Tx = p2 (rP (Sx )) ∈ Mx (H).
To show that (iv’) =⇒ (i), let T = {Tx }x∈K be a dilation such that T µ = ν. Then, for f ∈ Kc (H), Z Z ν(f ) = (T µ)(f ) = Tx f dµ(x) ≥ f (x) dµ(x) K
K
yields µ ≺ ν. This concludes the proof.
3.11
Exercises
Exercise 3.93 (Bauer functionals). For a measure µ ∈ M+ (K) and an upper bounded real-valued function f on K denote B µ (f ) := inf {µ(h) : h ∈ H, h ≥ f } .
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3 Choquet theory of function spaces
The mapping f 7→ B µ (f ) is called a Bauer functional on `∞ (K). Of course, µ(f ) ≤ B µ (f ) and B εx (f ) = f ∗ (x) for x ∈ K. Given µ ∈ M+ (K), we denote n o Mµ (H) := ν ∈ M+ (K) : ν − µ ∈ H⊥ . Prove that Mµ (H) is a convex compact subset of M+ (K) and B µ (f ) = sup {ν(f ) : ν ∈ Mµ (H)} for any upper semicontinuous function f on K. The supremum is attained. Hint. The proof is almost the same as the proof of Lemma 3.21. Exercise 3.94. Given a measure µ ∈ M+ (K), we defined in Definition 3.17 the functional Qµ := inf{µ(k) : k ∈ S c (H), k ≥ f on K}. By Lemma 3.21, Qµ (f ) = sup ν(f ) : ν ∈ M+ (K), µ ≺ ν , where the supremum is attained. Prove that, in general, B µ 6= Qµ . Hint. Let H be the space of all affine continuous functions on the closed interval [−1, 1]. Then S c (H) equals the set of all continuous concave functions on [−1, 1]. Show that for 1 f (x) := −|x| + 1 and µ := (ε−1 + ε1 ) 2 we get Qµ (f ) = 0 and B µ (f ) = 1.
Exercise 3.95. Let H be a function space on K and x ∈ K. Prove that Mx (H) = Mx (Ac (H)) and ChH (K) = ChAc (H) (K). Exercise 3.96. Prove that, for any upper bounded function f on K, the inequalities “≥” in the families of functions in Proposition 3.25(a) may be replaced by “>”. Hint. If F is a family of functions and f = inf{h ∈ F : h ≥ f }, then f = inf{h + ε : h ∈ F, ε > 0}. Now use Proposition 3.25.
3.11 Exercises
97
Exercise 3.97. If f = g on ChH (K), then ChH (K)f ∗ = ChH (K)g ∗ . Hint. Obviously, h ≥ f on ChH (K) if and only h ≥ g on ChH (K). Exercise 3.98. Prove that kFf ∗ − Fg ∗ k ≤ kf − gk`∞ (F ) for any set F ⊂ K containing ChH (K) and f, g ∈ `∞ (F ). Hint. Pick x ∈ F . Then Ff ∗ ≤ kf k since kf k ∈ H and f ≤ kf k. Hence F ∗
f (x) = F((f − g) + g)∗ (x) ≤ F(f − g)∗ (x) + Fg ∗ (x).
Thus F ∗
f (x) − Fg ∗ (x) ≤ F(f − g)∗ (x) ≤ kf − gk,
and the assertion easily follows. Exercise 3.99. Let {fα }α∈A be a down-directed net of upper semicontinuous functions on K, fα & f . Prove that fα∗ & f ∗ . Hint. It is clear that {fα∗ }α∈A is a down-directed net of functions satisfying f ∗ ≤ fα∗ for every α ∈ A. Since by Exercise 3.96 f ∗ = inf {h ∈ H : h > f } , it is enough to verify that for any h > f , h ∈ H, there exists α ∈ A such that fα ≤ h. Since fα & f , we can use the upper semicontinuity of the functions {fα } and the compactness of K to find α1 , . . . , αn ∈ A such that fα1 ∧ · · · ∧ fαn < h. Then fα∗ < h for every α ∈ A satisfying α ≥ αi , i = 1, . . . , n. Exercise 3.100. Let {xα } be a net of points converging to x ∈ ChH (K). Let µα ∈ Mxα (H). Then µα → εx . Hint. If we suppose the contrary, then there exist a measure µ 6= εx and a subnet {µβ } so that µβ → µ. It is straightforward to verify that µ is an H-representing measure for x. Since µ is not the Dirac measure at x, we have arrived at a contradiction with the assumption that x ∈ ChH (K). Exercise 3.101. Prove that a closed set F ⊂ K is H-extremal if and only if cK\F ∈ S lsc (H). Exercise 3.102. Let X be a compact convex subset of a locally convex space and B b (X) be the family of all bounded Borel functions on X.
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3 Choquet theory of function spaces
(a) If f is upper semicontinuous on X, then there exists a bounded concave upper semicontinuous function g on X such that f = g on ext X. (b) Let f be a bounded concave upper semicontinuous functions on X. Prove that the set {x ∈ ext X : f (x) > f∗ (x)} is meager (in the relative topology of ext X). (c) Let g be a Borel function on X with 0 ≤ g ≤ 1 and f an upper semicontinuous function on X such that {x ∈ ext X : f (x) − g(x) 6= 0} is meager in ext X. Find an upper semicontinuous concave function h with 0 ≤ h ≤ 1 such that {x ∈ ext X : g(x) − h(x) 6= 0} is meager in ext X. (d) Let g ∈ B b (X). Prove that there exists a bounded concave upper semicontinuous function f on X such that the set {x ∈ ext X : g(x) 6= f (x)} is meager. Hint. For the proof of (a) use Theorem 3.24. To verify (b), for any n ∈ N set Mn := {x ∈ ext X : f (x) − f∗ (x) ≥ 1/n}. Each set Mn is closed, so it suffices to show that Int Mn = ∅. Suppose a contrary. By Proposition 2.41, there exists a ∈ Ac (X), a ≤ 1/n, such that ∅ 6= {x ∈ ext X : a(x) > 0} ⊂ Mn . Show that the compact convex set F := {x ∈ X : f (x) − f∗ (x) − a(x) ≥ 0} contains ext X. Hence, by the Krein–Milman theorem 2.22, F = X and f ≥ f∗ + a on X. Then f∗ ≥ f∗ + a and it follows that a ≤ 0 on X, which is an obvious contradiction. To verify (c), let f and g be as in (c). Then h = (0 ∨ (f ∧ 1))∗ is an upper semicontinuous concave function with 0 ≤ h ≤ 1 that satisfies h = 0 ∨ (f ∧ 1) on ext X. Hence {x ∈ ext X : g(x) − h(x) 6= 0} is meager in ext X. Finally show (d). Let V be the family of all g ∈ B b (X) for which there exists a bounded concave upper semicontinuous function f on X such that the set {x ∈ ext X : g(x) 6= f (x)} is meager in ext X. Obviously, V is closed under addition and multiplication by positive scalars. It follows from (b) that −g ∈ V whenever g ∈ V. If {gn } is a bounded decreasing sequence in V and fn , n ∈ N, are bounded upper semicontinuous functions witnessing it, by (c) we can assume that all fn are bounded by the same constants. Then the function infn∈N fn shows that limn→∞ gn is contained in V. Hence V is closed with respect to taking monotone limits of bounded sequences. By (a), V contains bounded upper semicontinuous functions and thus V contains all bounded Borel functions by Proposition A.46. Exercise 3.103. Let H be the linear span of 1, id, id3 on [−1, 1]. Prove that Kor(H) 6= C([−1, 1]). Find a continuous function on [−1, 1] which does not belong to Kor(H). Hint. Show that ChH ([−1, 1]) = [−1, − 21 ] ∪ [ 12 , 1]. Exercise 3.104. Let H be the linear span of {1, id, ϕ} on an interval [a, b] , where ϕ is a continuous function on [a, b]. Prove that Kor(H) = C([a, b]) if and only if ϕ is either strictly convex or strictly concave on [a, b].
3.11 Exercises
99
Hint. If ϕ is strictly convex and a ∈ [0, 1], then the function ϕ − h exposes a for each affine function h supporting ϕ at a. For the converse, set G = {[a, b, c] ∈ [0, 1]3 : a < c < b}, Φ(a, b, c) = (b − c)ϕ(a) + (c − a)ϕ(b) − (b − a)ϕ(c). The set G is convex, thus connected, and ϕ is continuous on G. If ϕ is neither strictly convex, nor strictly concave, then Φ is neither strictly positive, nor strictly negative. It follows that there exist a, b, c ∈ [0, 1] such that a < c < b and Φ(a, b, c) = 0. If we b−c , then λ ∈ (0, 1), c = λa + (1 − λ)b and ϕ(c) = λϕ(a) + (1 − λ)ϕ(b). denote λ = b−a Then h(c) = λh(a)+(1−λ)h(b) for each h ∈ H and thus the measure λεa +(1−λ)εb represents c. Exercise 3.105 (Fej´er’s theorem). Let f be a continuous 2π-periodic function on R and let sk (f ) be the k-th partial sum of the Fourier series of f on R. Prove the following Fej´er’s theorem: If Tn f :=
s0 (f ) + s1 (f ) + · · · + sn−1 (f ) , n
then the sequence Tn f converges uniformly to f on R. Hint. Since the operators Tn are obviously linear and positive in view of 1 Tn f (x) = 2πn
Z
2π
f (x + t)
0
sin( nt 2 ) sin( 2t )
2 dt,
x ∈ [0, 2π],
it suffices to use the second Korovkin theorem 3.34 to show that the sequence {Tn } is H-admissible if H equals the linear span of {1, cos, sin}. Exercise 3.106. If Z H := f ∈ C([0, 1]) : f (0) =
1
f
0
and λ is Lebesgue measure on [0, 1], prove that M(H) = co {ε0 , λ} ∪ {εx : x ∈ (0, 1]} 6= M1 ([0, 1]). Exercise 3.107. Let X be a compact convex metrizable subset of a locally convex space. Then there exists a continuous strictly convex function on X. Hint. According to Proposition 2.45, we may assume that X is a norm compact convex subset of `2 . Then x 7→ kxk2 , x ∈ X, is the required strictly convex function.
100
3 Choquet theory of function spaces
Exercise 3.108 (Rosenthal’s proof). In the following, supply the details of Rosenthal’s proof of the Choquet representation theorem in the setting of metrizable compact convex sets. More precisely, prove that given a metrizable compact convex subset X of a locally convex space E and x ∈ X, then there exists a probability measure µ ∈ Mx (X) such that µ is carried by extreme points of X. Hint. Exercise 3.107 provides a strictly convex function Q on X. For n ∈ N, define 1 1 . Fn := x ∈ X : there are y, z ∈ X, 2x = y + z, (Q(y) + Q(z)) ≥ Q(x) + 2 n Obviously, each Fn is closed. First, prove the following assertion. Claim. Let ρ ∈ M1 (Fn ). Then there exists a measure σ ∈ M1 (X) such that Z Z Z Z 1 Q dρ + . Q dσ ≥ f dρ, f ∈ E ∗ , and f dσ = n Fn X X Fn Proof of the claim. First, the claim holds in the case when ρ = εx for some x ∈ Fn . It suffices to put σ = 21 (εy + εz ), where y, z are as in the definition of the set Fn . The assertion then easily follows for any molecular measure. In general, use approximation by molecular measures (cf. Proposition 2.27). Proof of the Choquet theorem. Using the compactness of M1 (X) (cf. again Proposition 2.27) show that there exists a representing measure µ ∈ Mx (X) such that Z nZ o Q dµ = sup Q dν : ν ∈ Mx (X) =: m. X
K
Then assume that µ(X \ ext X) > 0. Since X \ ext X =
∞ [
Fn ,
n=1
find n ∈ N so that r := µ(Fn ) > 0. By the claim, there exists a measure σ for ρ := r1 µ|Fn . Denote κ := rσ + µ|X\Fn and show that κ ∈ Mx (X). Then deduce that Z Z r r m≥ Q dκ ≥ Q dµ + = m + , n n X X which is an obvious contradiction. Exercise 3.109. Let H be a function space on a compact space K, g an upper semicontinuous function on K, f ∈ S lsc (H), g ≤ f on K and ε > 0. Find a function k ∈ S c (H) such that g ≤ k ≤ f + ε on K.
3.11 Exercises
101
Hint. Look at the set F := {f − k + ε : k ∈ S c (H), k ≥ g} . Then F is a nonempty convex set of lower semicontinuous functions on K. The proof will be completed by showing that F contains a positive element. Let us see why this is so. Take any nonzero positive Radon measure µ on K. By Corollary 3.25, g ∗ ≤ f on K. Appeal to Theorem A.84 to find a function k ∈ S c (H) such that k ≥ g and µ(k) < µ(g ∗ ) + εµ(K). Then µ(k) < µ(g ∗ ) + εµ(K) ≤ µ(f ) + εµ(K) = µ(f + ε). Hence µ(f − k + ε) > 0, and Lemma A.88 concludes the proof. Exercise 3.110. If g is an upper semicontinuous function on K, f ∈ S lsc (H), and g < f on K, then there is a function k ∈ S c (H) such that g < k < f on K. Hint. By compactness, f − g > ε for some ε > 0. Now use Exercise 3.109. Exercise 3.111. Prove that if f < g is replaced by f ≤ g in the hypothesis of Corollary 3.49, then a function k ∈ S c (H) satisfying f ≤ k ≤ g on K need not exist. 1 . Let K be a (metrizable) compact space on R Hint. For n ≥ 2, denote rn = 2n consisting of points 0, 1, rn , 1 − rn , 1 + rn , n ≥ 2,
and let H be the space of all continuous functions f on K satisfying f (1 − rn ) =
1 1 f (rn ) + (1 − )f (1 + rn ), n n
n ≥ 2.
Since the function f (x) = x belongs to H, the vector space H separates points of K. Thus H is a function space on K. Since, for any n ≥ 2, the function x = rn , 1, 1 f (x) := − n−1 , x = 1 + rn , 0 elsewhere, exposes the points rn and 1 + rn , these points belong to the Choquet boundary ChH (K). Also the points 0 and 1 are exposed, as the functions f (x) := x and x ∈ [0, 21 ] ∩ K, 1, f (x) := 3n−1 , x = 1 − rn , n ≥ 2, 2n2 x − 1, x ∈ [1, 2] ∩ K, are exposing 0 and 1, respectively.
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3 Choquet theory of function spaces
Any measure from M1−rn (H) is carried by the set {rn , 1 − rn , 1 + rn }. Hence the function ( 1, x = 0, f := 0 elsewhere is H-concave. It is clear that f is also upper semicontinuous. Let g be a lower semicontinuous function defined as ( 1, on K ∩ [0, 12 ], g := 0 elsewhere. Then it is impossible to insert a continuous H-concave function between f and g. Indeed, if h were such a function, then h(rn ) > 0 for some rn , by the continuity of h at the point 0. But then 0 = h(1 − rn ) ≥
1 1 1 h(rn ) + (1 − )h(1 + rn ) ≥ h(rn ) > 0. n n n
This obvious contradiction proves the result. Exercise 3.112. Prove that Mmax (H) is a norm closed cone in M(K). Hint. Use Mokobodzki’s maximality test 3.58. Exercise 3.113. Prove that the set of all probability maximal measures is a norm closed face of M1 (K). Hint. Use Mokobodzki’s maximality test 3.58. Exercise 3.114. Let {fn } be a bounded sequence of continuous functions on [0, 1] R1 such that 0 ≤ fn ≤ 1, n ∈ N, and fn → 0. Prove that 0 fn → 0 without recourse to Lebesgue integration theory. R1 Hint. Assume that there exists δ > 0 such that 0 fn ≥ 2δ for any n ∈ N. By Lemma 3.74, there exist λn ≥ 0, n ∈ N, such that ∞ X
λn = 1
∞ X
and
n=1
Let k ∈ N be such that Z δ≥
λn fn ≤ δ on [0, 1].
n=1
Pk
n=1 λn
∞ 1X
> 12 . Then Z
λn fn ≥
0 n=1
This contradiction finishes the proof.
k 1X
0 n=1
λn fn ≥ 2δ
k X n=1
λn > δ.
3.11 Exercises
103
Exercise 3.115. Let {fn } be a bounded sequence of continuous functions on a compact space K that P pointwise converges toP0. Then, for every ε > 0, there exist λi > 0, i = 1, . . . , n, ni=1 λi = 1, such that k ni=1 λi fi k < ε. Prove this without recourse to Mazur’s theorem A.2. Hint. Use the Simons lemma 3.74. Exercise 3.116. Let {xn } be a sequence in a Banach space E that converges weakly P to x ∈ E. P Then for every ε > 0 there exist λi > 0, i = 1, . . . , n, ni=1 λi = 1, such that k ni=1 λi xi − xk < ε. Prove this without the recourse to Mazur’s lemma (cf. W. Rudin [403], Theorem 3.13). Hint. By the uniform boundedness principle (see Theorem 2.5 in W. Rudin [403]), the sequence {xn } is bounded. If we view elements of E as continuous functions on the dual unit ball BE ∗ endowed with the w∗ -topology, then the assertion follows from Exercise 3.115. Exercise 3.117 (Bishop–de Leeuw preordering). Let H be a function space on a compact space K. Given µ, ν ∈ M+ (K), we define the Bishop–de Leeuw preordering l by declaring that µ l ν if µ(h2 ) ≤ ν(h2 ) and
µ(h) = ν(h)
for any h ∈ H. A measure µ ∈ M+ (K) is called l-maximal if given ν ∈ M+ (K), µ l ν, then µ(h2 ) = ν(h2 ) for any h ∈ H. Prove the following assertions hold. (a) The relation l is reflexive and transitive. (b) If µ ≺ ν, then µ l ν. (c) For any measure µ ∈ M+ (K) there exists a l-maximal measure λ ∈ M+ (K) such that µ l λ. (d) Any l-maximal measure is ≺-maximal. Hint. The verification of reflexivity and transitivity in assertion (a) is straightforward. For the proof of (b), we just note that h2 ∈ Kc (H) for any h ∈ H, and assertion (c) can be proved by an analogous argument to that in Theorem 3.65. For the proof of (d), let µ be a l-maximal measure. According to Theorem 3.58, it is enough to show that µ(f ) = µ(f ∗ ) for any f ∈ −W(H). Let ε > 0 and f ∈ −W(H) be given; that is, f = h1 ∨ · · · ∨ hn ,
h1 , . . . , hn ∈ H.
Let H := {x ∈ K : f ∗ (x) − f (x) ≥ ε}
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3 Choquet theory of function spaces
and, for each m ∈ N and i = 1, . . . , n, 1 + h2i (x) . Lm,i := x ∈ K : there exists ν ∈ Mx (H) with ν(h2i ) ≥ m Then each Lm,i is a closed set and H⊂
∞ [ n [
Lm,i .
m=1 i=1
Sn S Indeed, let x ∈ H be given. Suppose that x is not contained in ∞ m=1 i=1 Lm,i . By the Key lemma 3.21, there exists a measure λx ∈ Mx (H) such that λx (f ) = f ∗ (x). Since x ∈ K \ Lm,i , we get, for each i = 1, . . . , n, Z 2 Z 2 2 2 λx (hi ) ≤ hi (x) = hi (t) dλx (t) ≤ |hi (t)| dλx (t) ≤ λx (h2i ). K
K
Hence hi = hi (x) λx -almost everywhere. Thus the set A :=
n \
{y ∈ K : hi (y) = hi (x)}
i=1
has λx -measure 1. We define the sets Ai , i = 1, . . . , n, as in (3.10). Then Z n Z X ∗ f (x) = λx (f ) = f (t) dλx (t) ≤ hi (t) dλx (t) A
i=1
A∩Ai
Z ≤
f (x) dλx (t) = f (x). A
S Sn This contradicts our assumption that f ∗ (x) − f (x) ≥ ε. Thus x ∈ ∞ m=1 i=1 Lm,i . To conclude the proof, it is enough to show that µ(Lm,i ) = 0 for every m ∈ N and i = 1, . . . , n. Suppose that this is not the case for some m ∈ N and some i = 1, . . . , n. For the measure σ := µ|Lm,i find a net of molecular measures {σα }α∈A carried by Lm,i , converging to σ and satisfying σα (1) = σ(1). According to the definition of Lm,i , there exist Radon measures {κα }α∈A such that σα l κα
and
κα (h2i ) ≥
σ(K) + σα (h2i ). m
Without loss of generality we may assume that κα → κ. Then σ l κ and κ(h2i ) ≥ σ(K) 2 m + σ(hi ). Hence we get µ l (µ − σ) + κ and µ 6= (µ − σ) + κ, which is a contradiction with the l-maximality of µ. Thus µ(H) = 0. Since ε > 0 is arbitrary, we get ! ∞ [ 1 ∗ ∗ = 0. µ ({x ∈ K : f (x) − f (x) > 0}) = µ x ∈ K : f (x) − f (x) ≥ n n=1
Hence µ(f ) =
µ(f ∗ )
and µ is ≺-maximal. This concludes the proof.
3.12 Notes and comments
105
Exercise 3.118. Prove that the preordering l need not be antisymmetric. Hint. Let H be the space of all affine functions on K := [0, 1]. Choose a nonzero measure µ ∈ M([0, 1]) satisfying µ(h) = 0 for any h ∈ {1, id, id2 } and decompose µ into its positive and negative part µ+ and µ− . Then µ+ l µ− and µ− l µ+ but µ+ 6= µ− . Thus l need not be antisymmetric. Exercise 3.119. A ≺-maximal measure µ need not be l-maximal. Hint. Let X ⊂ R2 be defined by 1 1 1 1 X := (0, 0), (0, 1), (1, 0), (1, 1), ( , −δ), ( , 1 + δ), (1 + δ, ), (−δ, ) , 2 2 2 2 where 0 < δ < 81 . Set K := co X = co X and let H consist of all affine functions on K. Then ext K = ChH (K) = X. Set µ :=
1 ε(0,0) + ε(1,0) + ε(1,1) + ε(0,1) 4
and
1 ε( 1 ,−δ) + ε(1+δ, 1 ) + ε( 1 ,1+δ) + ε(−δ, 1 ) . 2 2 2 2 4 Then r(µ) = r(ν), ν is ≺-maximal, νlµ, but there exists h ∈ H with ν(h2 ) < µ(h2 ). Obviously, ν is ≺-maximal because ν is carried by extreme points of K. The verification of ν l µ can be done by a straightforward computation. Finally, if we define h(x, y) := x + y, (x, y) ∈ K, ν :=
then ν(h2 ) <
3.12
3 2
= µ(h2 ).
Notes and comments
Section 3.2 on Korovkin theorems follows the paper [42] by H. Bauer. Theorem 3.24 comes from H. Bauer’s paper [38]. The “convex case” appeared in M. Herv´e [223]. As stated in [5], p. 43, the assertion is already implicitly contained in Kadison’s memoir [256]. Further, Bauer’s theorem 3.27 is in [38] (Korollar 1, p. 106). As concerns the metrizability of compact spaces (cf. Remark 3.44(b)), we do not know whether the following conjecture is true. Problem 3.120. Let K be a compact space. Are the following assertions equivalent? (i) K is metrizable, (ii) for any function space H on K there exists a continuous strictly H-convex function on K, (iii) for any function space H on K, the Choquet boundary ChH (K) is a Gδ set.
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3 Choquet theory of function spaces
In [109], G. Choquet discloses the following three sources of motivation for his investigation on integral representation in compact convex sets: study of certain capacities (alternating set functions of infinite order), R.S. Martin’s representation of positive harmonic functions in a Euclidean domain and R. Godement’s result on positive operators on Hilbert spaces. G. Choquet completed his proof on integral representation Theorem 3.45 (the metrizable case) in July 1956. The result, examples and an application to a new proof of the ergodic theorem were presented in G. Choquet [111]. New proofs appeared soon afterwards (M. Herv´e [223], L. H. Loomis [311], F. F. Bonsall [75]). Theorem 3.42 was proved for the convex case by M. Herv´e in [223]. The characterization of maximal measures contained in Theorem 3.58 is due to G. Mokobodzki [347]. The Simons inequality in 3.74 and the Simons lemma in 3.75 go back to S. Simons’ paper [417]. Proof of Lemma 3.74 follows the lines of a proof given by M. Ruiz Gal´an and S. Simons [404]. We use the Simons lemma 3.75 for the proof of Lemma 3.77. Another proof of this lemma can be found in E. M. Alfsen [5], Proposition 1.4.10. (see also Exercise 4.45). For generalizations and applications of the Simons inequality, as well as for its simplified proofs, we refer the reader to E. Oja [367], K. Kiviso and E. Oja [270], G. Godefroy [199] and R. Deville and C. Finet [143]. For a related lemma by V. Pt´ak we refer the reader to [380]. The approach proposed by E. Bishop and K. de Leeuw in [58] turned out to be particularly important. It relies on introducing partial ordering comparing measures by means of squares of continuous affine functions. Maximal measures, which in the metrizable case are carried by the set of extreme points, then play a role of desired representing measures. Using a different ordering on measures induced by all continuous convex functions, G. Choquet obtained a similar representation theorem for nonmetrizable case in [105]. Example 3.82 is due to E. Bishop and K. de Leeuw [58]. Example 3.83 is a variant of an example due to G. Mokobodzki (see [374], Chapter 10, Example). The characterization given in (vi) of Theorem 3.92 was taken as the definition of the strong ordering of measures by L. H. Loomis in [311]. The equivalence of his strong ordering with the Choquet ordering was later established in P. Cartier, J. M. G. Fell and P. -A. Meyer [97]. Theorem 3.92 follows their result and Theorem 1.3.6 in G. Winkler [473] dealing with noncompact case. Conditions of Theorem 3.92 and the proof of their equivalence follows R. D. Bourgin [82], Chapter 6, G. A. Edgar [152], [153], J. L. Doob [144], J. Hoffmann-Jørgensen [227], H. v. Weizs¨acker [462] and the paper H. v. Weizs¨acker and G. Winkler [464]. Exercise 3.102 is taken from J. D. Maitland Wright [476]. For details of Rosenthal’s proof of the Choquet representation theorem indicated in Exercise 3.108, see the paper H. P. Rosenthal [397]. Exercises 3.114 and 3.115 use techniques from S. Simons [418], Exercise 3.117 is taken from E. Bishop and K. de Leeuw [58] and Exercise 3.119 is taken from the paper [311] by L. H. Loomis.
Chapter 4
Affine functions on compact convex sets
The preceding chapter deals with abstract function spaces on compact spaces. Here we focus on a particular example of a function space, namely, on the space Ac (X) of continuous affine functions on a compact convex set X, and show how classical concepts from convex analysis fit into the general framework of function spaces. We first establish an easy, but important, Lemma 4.3 on approximation of probability measures on X by molecular measures with the same barycenter. Next we introduce a notion of strongly affine functions, that is, the functions satisfying the barycentric formula. Corollary 4.8 then identifies semicontinuous convex and affine functions on X with semicontinuous H-convex and H-affine functions. As a consequence, we obtain classical results on approximation of semicontinuous convex functions (see Propositions 4.9, 4.10 and 4.12). Next Section 4.2 characterizes strongly affine functions. The main result contained in Theorem 4.19 shows that strongly affine functions can be characterized by means of the Haydon condition and a Lusin type property. A list of conditions characterizing strongly affine functions is further enriched by Theorem 4.34. As a corollary of Theorem 4.19 we get Choquet’s barycentric theorem 4.22 and Mokobodzki’s approximation theorem 4.24. Section 4.3 introduces the state space S(H) of a function space H. This provides an efficient link between the theory of function spaces and the theory of compact convex sets. In a series of propositions we describe basic facts that will enable us later on to transfer properties of function spaces to the framework of compact convex sets and vice versa. First glimpses of this kind of reasoning can be observed in a reformulation of Goldstine’s lemma 4.33 and in Theorem 4.35. As an application of Mokobodzki’s approximation theorem, we prove in Section 4.4 that affine Baire-one functions on the dual unit ball of a Banach space E are pointwise limits of a sequence of elements from E. This also shows that w∗ -sequential closure of E in its bidual is norm closed.
4.1
Affine functions and the barycentric formula
In this section we will be working in a compact convex subset X of a locally convex space E.
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4 Affine functions on compact convex sets
Definition 4.1 (Affine, concave and convex functions). A real-valued function f is said to be affine on X if f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y) for each x, y ∈ X and for each λ ∈ (0, 1). We denote by A(X) the set of all affine functions on X, by Ab (X) the space of all bounded affine functions on X, and by Ac (X) the set of all continuous functions from A(X). It is obvious that Ac (X) is a function space. We also recall that the Choquet boundary of Ac (X) coincides with the set ext X of all extreme points of X. More generally, a function f is called concave if f is upper or lower finite and f (λx + (1 − λ)y) ≥ λf (x) + (1 − λ)f (y) whenever
x, y ∈ X, λ ∈ (0, 1).
The collection of all concave functions on X will be denoted by S(X). Further, let Sc (X), Susc (X) and Slsc (X) denote the families of all continuous, upper semicontinuous and lower semicontinuous concave functions on X, respectively. A function f is convex if −f is concave. We write K(X) for the family of all convex functions on X. The notation Kc (X), Kusc (X) and Klsc (X) should be clear. Recall that we abbreviated Mx (Ac (X)) to Mx (X) and notice that µ ∈ Mx (X) means exactly that x is a barycenter of µ; for short, r(µ) = x. Analogously we write Mmax (X) and Mbnd (X) instead of Mmax (Ac (X)) and Mbnd (Ac (X)). Proposition 4.2. A continuous function f on X is affine if and only if f is Ac (X)affine. For short, Ac (X) = Ac (Ac (X)). Proof. Obviously, Ac (X) ⊂ Ac (Ac (X)). If f is a continuous Ac (X)-affine function and x, y ∈ X, λ ∈ (0, 1), then λεx + (1 − λ)εy ∈ Mλx+(1−λ)y (X), so that f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y). By Corollary 2.28, Radon measures on compact spaces can be approximated by molecular measures. Proposition 4.3 says that in the convex case a Radon measure µ representing a point x can be approximated by molecular measures having the same barycenter x. Proposition 4.3. Let x ∈ X and µ ∈ Mx (X). Then there exists a net {µα } of molecular measures in Mx (X) such that µα → µ.
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109
Proof. Let a neighborhood of µ be determined by ε > 0 and f1 , . . . , fn ∈ C(X), that is, it consists of all Radon measures ν for which |µ(fi ) − ν(fi )| ≤ ε,
i = 1, . . . , n.
By the continuity of the functions fi , there exists a closed convex neighborhood W of 0 in E such that |fi (y) − fi (y 0 )| ≤ ε, (4.1) whenever i = 1, . . . , n, y, y 0 ∈ X and y − y 0 ∈ W . By compactness, there exist points x01 , . . . , x0m ∈ X such that X⊂
m [
(x0j + W ).
j=1
Denote W0 := ∅, Wj :=
x0j
+ W for j = 1, . . . , m and set
Aj := X ∩ (Wj \ (W1 ∪ · · · ∪ Wj−1 )),
j = 1, . . . , m.
Let J be the set of indices for which µ(Aj ) 6= 0. Let λj := µ(Aj ) for j = 1, . . . , m and j ∈ J. µj := λ−1 j µ|Aj , Clearly, P µj are probability Radon measures, spt µj ⊂ Wj ∩ X for j ∈ J and µ = j∈J λj µj . Let xj be the barycenter of the measure µj , and observe that xj ∈ X ∩ Wj , since µj is carried by the compact convex set X ∩ Wj for every j ∈ J. If j ∈ J and z ∈ Wj ∩ X, then z − xj = (z − x0j ) + (x0j − xj ), and so by (4.1) |fi (z) − fi (xj )| ≤ 2ε,
i = 1, . . . , n.
Since µj is a probability measure carried by X ∩ Wj , we obtain |µj (fi ) − fi (xj )| ≤ ε,
i = 1, . . . , n, j ∈ J.
(4.2)
P If ν := j∈J λj εxj , then ν is a molecular probability measure with the same barycenter as µ. By (4.2), X Z X fi dµ − λj fi (xj ) ≤ λj |µj (fi ) − fi (xj )| |µ(fi ) − ν(fi )| = j∈J
≤ 2ε, This completes the proof.
Aj
i = 1, . . . , n.
j∈J
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4 Affine functions on compact convex sets
Definition 4.4 (Barycentric formula). A function f : X → R satisfies the barycentric formula if f is universally measurable and, given x ∈ X and µ ∈ Mx (X), µ(f ) exists and f (x) = µ(f ). Sometimes we label functions satisfying the barycentric formula as strongly affine functions. We denote by Abar (X) the space of all functions satisfying the barycentric formula. Lemma 4.5 yields that any strongly affine function is bounded. Thus, a universally measurable function f satisfies the barycentric formula if and only if f is Ac (X)affine. Lemma 4.5. Any strongly affine function on X is bounded. Proof. Let f be a strongly affine function on X. Assume that f is not bounded from above. Then we can find points xn ∈ X such that f (xn ) ≥ 22n forP any n ∈ N. Then P∞ 1 the measure µ := n=1 21n εxn belongs to Mr(µ) (X), and µ(f ) = ∞ n=1 2n f (xn ) = ∞. This contradicts the requirement that µ(f ) = f (r(µ)). Definition 4.6 (Superbarycentric formula). We say that a universally measurable function f on X satisfies the superbarycentric formula if µ(f ) exists for every µ ∈ Mx (X), x ∈ X, and f (x) ≥ µ(f ). As above, a function satisfies the superbarycentric formula if and only if it is Ac (X)-concave. A universally measurable function f on X satisfies the subbarycentric formula if −f satisfies the superbarycentric formula. Proposition 4.7. Let f be a semicontinuous concave function on X. Then f satisfies the superbarycentric formula. Proof. Let x ∈ X and µ ∈ Mx (X). Assume first that f is lower semicontinuous and pick c < µ(f ). Let g ∈ C(X) be such that g ≤ f and c < µ(g). By Proposition 4.3, there exists a molecular measure µ0 ∈ Mx (X) such that µ0 (g) > c. Then c < µ0 (g) ≤ µ0 (f ) ≤ f (x). Since c is arbitrary, µ(f ) ≤ f (x). Assume now that f is a real-valued upper semicontinuous function. Again choose ε > 0. There exists a function h ∈ Ac (X) such that h ≥ f and h(x) < f (x) + ε. Indeed, the closed convex set F := {(t, y) ∈ X × R : y ≤ f (t)} can be separated by the Hahn–Banach theorem from the point (x, f (x) + ε). Then µ(f ) ≤ µ(h) = h(x) < f (x) + ε, and the conclusion follows. An upper semicontinuous function f may attain the infinite value −∞, and then the reasoning above is not correct; namely, the set F need not be well defined. Thus we must use a more subtle argument. We will prove the following claim.
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111
Claim. Let x ∈ X and c > f (x) be given. Then there exists n ∈ N such that c > sup {t ∈ R : (x, t) ∈ co (Fn ∪ Hn )} , where Fn := {(y, t) ∈ X × R : −n ≤ t ≤ f (y)}
and
Hn := X × {−n}.
Proof of the claim. Note that both sets Fn and Hn are compact convex sets in E × R, and hence co (Fn ∪ Hn ) = λu + (1 − λ)v : u ∈ Fn , v ∈ Hn , λ ∈ [0, 1] . Suppose that the claim does not hold. Then, for any n ∈ N, there exists tn ∈ R such that c ≤ tn and (x, tn ) ∈ co (Fn ∪ Hn ) . Hence there exist yn , zn ∈ X, λn ∈ [0, 1], sn ∈ R, satisfying (x, tn ) = (λn yn + (1 − λn )zn , λn sn + (1 − λn )(−n))
and
(yn , sn ) ∈ Fn .
Using compactness we find subnets {yni }, {zni } and {λni } of {yn }, {zn } and {λn }, respectively, and points y, z ∈ X and λ ∈ [0, 1] such that yni → y, zni → z and λni → λ. Then x = λy + (1 − λ)z and, for each index i, f (x) < c ≤ tni = λni sni + (1 − λni )(−ni ) ≤ λni f (yni ) + (1 − λni )(−ni ). (4.3) If λ = 1, we get from (4.3) f (x) < c ≤ f (x). It does not matter whether f (x) is finite or infinite; in both cases we obtain a contradiction. If λ < 1, we get c ≤ −∞, which again is a contradiction with the choice of c. The proof of the claim is thus completed. Now, the proof of the proposition can be finished as in the case of a real-valued upper semicontinuous functions. Let x ∈ X, µ ∈ Mx (X) and c > f (x) be given. Find an integer n ∈ N with the property guaranteed by the claim. Then the point (x, c) can be separated from the set co (Fn ∪ Hn ); in other words, there exists a continuous affine function h on X such that f ≤ h and h(x) < c. Then µ(f ) ≤ µ(h) = h(x) < c, and the proof is finished. Proposition 4.7 asserts that a continuous (or semicontinuous) concave function f on X is Ac (X)-concave, since f satisfies the superbarycentric formula if and only if f ∈ S(Ac (X)). Similarly, as in Proposition 4.2, we can show that any Ac (X)concave function is concave. Moreover we get that any semicontinuous affine function f on X is Ac (X)-affine, because both f and −f are Ac (X)-concave. We bring together these observations in the following corollary.
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4 Affine functions on compact convex sets
Corollary 4.8. Let X be a compact convex set. Then Slsc (X) = S lsc (Ac (X)),
Susc (X) = S usc (Ac (X)),
Ausc (X) = Ausc (Ac (X)),
Alsc (X) = Alsc (Ac (X)),
and
Sc (X) = S c (Ac (X)).
Corollary 4.8 allows us to apply the general approximation results of Propositions 3.48 and 3.54 to get the following assertions. Proposition 4.9. Let f be an upper semicontinuous concave function on X. If K := {g ∈ Sc (X) : g ≥ f on X}
and
L := {g ∈ Sc (X) : g > f on X} ,
then K and L are down-directed families and f = inf K = inf L. Proposition 4.10. Let f be a lower semicontinuous concave function on K. If S := {g ∈ Sc (X) : g < f on X}
and
T := {g ∈ Sc (X) : g ≤ f on X} ,
then S and T are up-directed families and f = sup S = sup T. Lemma 4.11. Let f , −g be upper semicontinuous functions on X with f < g. Then there exists a function h ∈ (E ∗ +R)|X such that f < h < g if and only if µ(f ) < ν(g) for any µ, ν ∈ M1 (X) satisfying r(µ) = r(ν). Proof. The necessity of the condition is obvious. Concerning the sufficiency, let f, −g be upper semicontinuous functions on X such that µ(f ) < ν(g) for any µ, ν ∈ M1 (X) satisfying r(µ) = r(ν). We claim that f ∗ < g∗ . Indeed, pick x ∈ X. By Lemma 3.21, there exist measures µ, ν ∈ Mx (X) such that f ∗ (x) = µ(f )
and
g∗ (x) = ν(f ).
By the assumption, f ∗ (x) < g∗ (x). By Lemma 3.18(a), f ∗ , −g∗ are upper semicontinuous Ac (X)-concave functions. Using Corollary 3.49, there exist continuous Ac (X)-concave functions f1 , −g1 such that f ∗ < f1 < g1 < g∗ . Let m, M ∈ R be numbers satisfying m ≤ f1 < g1 ≤ M on X. Consider the following compact convex sets in E × R: J1 := {(x, t) ∈ E × R : m ≤ t ≤ f1 (x)} and J2 := {(x, t) ∈ E × R : g1 (x) ≤ t ≤ M }.
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113
Since J1 , J2 is a pair of disjoint sets, there exist a functional F ∈ (E × R)∗ and λ ∈ R such that sup F (J1 ) < λ < inf F (J2 ). As in the proof of Lemma 2.34, there exist ϕ ∈ E ∗ and β ∈ R so that F (x, t) = ϕ(x) + βt for any x ∈ E and t ∈ R. Then h : x 7→
1 (λ − ϕ(x)), β
x ∈ X,
fulfils f < h < g, as needed. Proposition 4.12. Let f be a lower semicontinuous affine function on X and A := {g ∈ Ac (X) : g < f on X} . Then A is up-directed and f = sup A. Proof. Let h be a continuous function on X, h < f , and let µ, ν ∈ M1 (X) with r(µ) = r(ν) be given. Then µ(h) < µ(f ) = f (r(µ)) = f (r(ν)) = ν(f ), because f is Ac (X)-affine, by Corollary 4.8. An appeal to Lemma 4.11 provides a continuous affine function g with h < g < f . Since f = sup{h ∈ C(X) : h < f }, we get that f = sup A. It remains to prove that A is up-directed. If g1 , g2 ∈ A are given, according to the previous considerations, there exists g ∈ A such that g1 ∨ g2 < g < f . This observation concludes the proof.
4.2
Barycentric theorem and strongly affine functions
If f : X → R is a function and V ⊂ X, we recall that oscV f (x) := inf{diam f (U ∩ V ) : U is a neighborhood of x},
x ∈ V,
is the oscillation of f on V at x (see Definition A.119). Definition 4.13 (The Haydon condition). Let f be a bounded affine function on a compact convex set X. We say that f satisfies the Haydon condition if for any Radon measure µ on X and for any pair of real numbers a < b, there exists a compact convex set L such that µ(L) > 0 and L is contained either in {x ∈ X : f (x) < b} or in {x ∈ X : f (x) > a}. Later on, we will see that the Haydon condition is strongly related to the following notion.
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4 Affine functions on compact convex sets
Definition 4.14 (Convex inner measure). For a Radon measure µ on X, let the convex inner measure µ for A ⊂ X be defined as µ(A) := sup {µ(L) : L ⊂ A, L is compact convex} . We note that the convex inner measure µ need not be a measure at all. Lemma 4.15. Let µ ∈ M1 (X) and {An } be a decreasing sequence of convex subsets of X. Then ∞ \ µ An = lim µ(An ). n=1
n→∞
Proof. Since µ(A) ≤ µ(B) whenever A ⊂ B, we obtain that limn→∞ µ(An ) exists and ∞ \ µ( An ) ≤ lim µ(An ). n=1
n→∞
For the converse, fix ε > 0. By induction, we will construct a decreasing sequence of compact convex sets Kn ⊂ An such that µ(Kn ) > µ(An ) − ε,
n ∈ N.
First find a compact convex set K1 ⊂ A1 such that µ(K1 ) > µ(A1 ) − 2ε . If the sets K1 , . . . , Kn−1 satisfying the required condition have been constructed, select a real number δ such that 0 < δ < ε and µ(Kn−1 ) > µ(An−1 ) − ε + δ. There exists a compact convex set L ⊂ An such that µ(L) > µ(An ) − δ. If we set Kn := L ∩ Kn−1 , we obtain the desired compact convex set. Indeed, since µ(L \ Kn−1 ) = µ(L ∪ Kn−1 ) − µ(Kn−1 ) ≤ µ(co(L ∪ Kn−1 )) − µ(Kn−1 ) ≤ µ(An−1 ) − µ(Kn−1 ) < ε − δ, we get µ(Kn ) = µ(Kn−1 ∩ L) = µ(L) − µ(L \ Kn−1 ) > µ(An ) − δ − (ε − δ) = µ(An ) − ε. Since
T∞
n=1 Kn
µ(
is a compact convex set contained in
∞ \
n=1
An ) ≥ µ(
∞ \
n=1
T∞
n=1 An ,
we obtain
Kn ) = lim µ(Kn ) ≥ lim µ(An ) − ε. n→∞
Since ε > 0 is arbitrary, the proof is complete.
n→∞
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115
Lemma 4.16. Assume that a bounded affine function f on X satisfies the Haydon condition. Then for any µ ∈ M1 (X) and any b ∈ R, µ ({x ∈ X : f (x) ≥ b}) + µ ({x ∈ X : f (x) < b}) = 1.
(4.4)
Proof. Let µ be a Radon probability measure on X. We will prove that for any pair of real numbers a < b, the inequality µ(A) + µ(B) ≥ 1 holds, where A := {x ∈ X : f (x) > a}, B := {x ∈ X : f (x) < b}. If this is not true, we find compact convex sets Ln , Kn , n ∈ N, such that Ln ⊂ A, Kn ⊂ B, µ(Ln ) → µ(A) and µ(Kn ) → µ(B). Due to the assumption, λ := µ|X\S∞ n=1 (Ln ∪Kn ) is a nonzero Radon measure. Hence we may apply the Haydon condition in order to obtain a compact convex set L such that λ(L) > 0 and such that L is contained either in A or in B. Suppose that L ⊂ A. Since A is convex, co(L ∪ Ln ) is contained in A for any n ∈ N. Thus we obtain µ(A) ≥ µ (co(L ∪ Ln )) ≥ µ(L ∪ Ln ) = µ(Ln ) + µ(L \ Ln ) ∞ [ ≥ µ(Ln ) + µ L \ (Ln ∪ Kn ) = µ(Ln ) + λ(L). n=1
Letting n → ∞, we get a contradiction. In order to obtain the desired equality, it is enough to realize that {x ∈ X : f (x) ≥ b} =
∞ \ n=1
1 x ∈ X : f (x) > b − n
.
Thus using the monotonicity of a convex inner measure (Lemma 4.15), we obtain µ ({x ∈ X : f (x) ≥ b}) + µ ({x ∈ X : f (x) < b}) ≥ 1. Since the converse inequality is obvious, the proof is finished. Definition 4.17 (Saturated sets). Let f be a real-valued function on a compact convex set X. A subset F of X is saturated with respect to f if for any Radon measure µ on X and any ε > 0, there exists a sequence {Kn } of compact convex sets in X such that ∞ [ diam f (Kn ) ≤ ε for each n ∈ N and µ F \ Kn = 0. n=1
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4 Affine functions on compact convex sets
Lemma 4.18. Let f be a function on a compact convex set X such that X is saturated with respect to f . Then f is universally measurable. Proof. Let µ ∈ M1 (X) be given. We fix a sequence {εk } of strictly positive numbers converging to 0. For each k ∈ N, we find a sequence {Knk }∞ of compact convex n=1 S k k sets in X such that diam f (Kn ) ≤ εk for each n ∈ N and µ(X \ ∞ n=1 µ(Kn )) = 0. Sn−1 k k k k k We set An := Kn \ i=1 Ki , n ∈ N, and pick numbers an ∈ f (An ) if the set Akn is nonempty. We define ( akn , if x ∈ Akn , n ∈ N, fk (x) := S k 0, if x ∈ X \ ∞ n=1 An . Then the functions fk are µ-measurable and fk → f uniformly µ-almost everywhere on X. Hence f is universally measurable. Theorem 4.19. Let f be a bounded affine function on X. Then the following conditions are equivalent: (i) f satisfies the barycentric formula, (ii) for any µ ∈ M+ (X) and any ε > 0, there exists a compact convex set L ⊂ X such that µ(L) > µ(X) − ε and f |L is continuous, (iii) f satisfies the Haydon condition, (iv) X is saturated with respect to f. Proof. (i) =⇒ (ii): Suppose that f satisfies the barycentric formula and µ is a Radon measure on X. Clearly we may assume that µ is a probability measure. For given ε > 0, we use Lusin’s theorem A.76 to find a compact set F ⊂ X such that f |F is continuous and µ(F ) ≥ 1 − ε. If we define L := coF, we obtain the required compact convex set. Indeed, we want to prove that f |L is continuous. Let x be a point in L and {xα }α∈A be a net of points of L converging to x. By the Milman theorem 2.43, ext L is a subset of F. The Integral representation theorem 2.31 ensures the existence of a net of probability measures {µα }α∈A carried by ext L with barycenters xα . Without loss of generality, we may assume that {µα }α∈A converges to a probability measure ν. Since the net {xα }α∈A converges to x, the measure ν represents x. Since f is continuous on ext L and µα and ν are carried by ext L, we get µα (f ) → ν(f ). Now we apply the assumption to obtain f (xα ) = µα (f ) → ν(f ) = f (x), which completes the proof. For the proof of the implication (ii) =⇒ (iii), let a nonzero measure µ ∈ M+ (X) and a < b be given. Using (ii), there exists a compact convex set L such that f |L is
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117
continuous and µ(L) > 0. We select a point x ∈ spt µ|L . Without loss of generality we may assume that f (x) < b. By the continuity of f |L we can find a closed convex neighborhood V of x such that f < b on V ∩L. Then V ∩L satisfies our requirements. For the proof of the implication (iii) =⇒ (iv) we will use Lemma 4.16. Let a Radon probability measure µ on X and ε > 0 be given. Fix δ > 0. Since f is bounded, we may suppose that 0 ≤ f < 1. Choose N ∈ N such that N1 < ε and define P n Fn := {x ∈ X : n−1 4.16, N n=1 µ(Fn ) = 1. N ≤ f < N }, n = 1, . . . , N . By Lemma P N Hence there exist compact convex sets Kn ⊂ Fn such that n=1 µ(Kn ) > 1 − δ. Then 1 < ε. diam f (Kn ) ≤ diam f (Fn ) ≤ N If we put together all compact convex sets obtained by choosing δl = 1l , l ∈ N, we get the required sequence. It remains to prove (iv) =⇒ (i). Let a Radon probability measure µ on X with the barycenter x and ε > 0 be given. By condition (iv), there exists a sequence {Kn } of closed convex subsets of X such that diam f (Kn ) < ε
and
∞ [ Kn = 0. µ X\ n=1
Find N ∈ N such that µ X \ (K1 ∪ · · · ∪ KN ) < ε
(4.5)
and define En := Kn \
n−1 [
Ki ,
n = 1, . . . , N,
E0 := X \
N [
Ki ,
i=1
i=1
λn := µ(En ),
n = 0, . . . , N.
We define Radon probability measures µn , n = 0, . . . , N , by ( 1 µ|En if λn > 0, µn := λn εx if λn = 0. Let xn be the barycenter of µn , n = 0, . . . , N . Then xn ∈ co En ⊂ Kn if n = 1, . . . , N satisfies λn > 0. The oscillation properties of f give |µ0 (f ) − f (x0 )| ≤ 2kf k
and
|µn (f ) − f (xn )| < ε,
n = 1, . . . , N. (4.6)
Obviously N X n=0
λn = 1,
N X n=0
λn xn = x
and
N X n=0
λn µn = µ.
(4.7)
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4 Affine functions on compact convex sets
Since f is affine, by (4.5), (4.6) and (4.7) we have N N N X X X |f (x) − µ(f )| = f λ n xn − λn µn (f ) = λn (f (xn ) − µn (f )) n=0
n=0
≤ λ0 |f (x0 ) − µ0 (f )| +
n=0 N X
λn |f (xn ) − µn (f )| ≤ ε(2kf k + 1).
n=1
Letting ε → 0, we conclude the proof. Lemma 4.20. Let f be a convex real-valued function on a compact convex set X that is lower bounded on some nonempty open subset of X. Then f is lower bounded on X. In particular, if f has a point of continuity, then f is lower bounded on X. Proof. Given f as in the lemma, let U be a nonempty open subset of X such that f ≥ C on U for some C ∈ R. Assume that f is not lower bounded on X; that is, that there exist points xn ∈ X, n ∈ N, such that f (xn ) → −∞. Let x be a cluster point of the sequence {xn } and let y ∈ U be chosen arbitrarily. Then there exists α ∈ (0, 1) such that αx + (1 − α)y ∈ U . Since x is a cluster point of {xn }, for infinitely many indices n ∈ N we have αxn + (1 − α)y ∈ U . Then for these indices we get C ≤ f (αxn + (1 − α)y) ≤ αf (xn ) + (1 − α)f (y), which yields a contradiction. Theorem 4.21. Let f be an affine function on a compact convex set X such that f is fragmented. Then f satisfies the barycentric formula. Proof. First, we note that f is bounded by Lemma 4.20 and Theorem A.121. We will show that X is saturated with respect to f . To this end, let µ ∈ M+ (X) and ε > 0 be given. Let U := {U ⊂ X : U is open and there are compact convex sets Kn ⊂ X such that µ U \
∞ [
Kn = 0 and diam f (Kn ) < ε}.
(4.8)
n=1
S It is not difficult to verify that V := {U : U ∈ U} ∈ U. Indeed, V is obviously open. Due to the inner regularity of µ, there exists a sequence {Hk } of compact sets such that µ(Hk ) % µ(V ). By a compactness argument, we can cover each Hk by a finite family U1 , . . . , Unk of sets contained in U. For every Ui , k ∈ N and 1 ≤ i ≤ nk , we find a countable family of compact convex sets guaranteed by (4.8). Putting together all these families, we obtain a countable family L of compact convex sets which covers µ-almost all of V and diam f (K) < ε for each K ∈ L.
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119
Let K be the family of all closed convex subsets of X whose complement in X is contained in U. Let Z be the intersection of K. By the argument above, Z is the smallest element of K. Set Y := {x ∈ Z : oscZ f (x) ≥ ε}. Then Y is a closed convex subset of Z. If x ∈ Z \ Y , then there exists an open convex neighborhood U of x such that U ∩ Y = ∅ and diam f (U ∩ Z) < ε. Since U \ Z ∈ U and U ∩ Z contains U ∩ Z, we observe that U ∈ U. Appealing again to the properties of U, it follows that Y is a closed convex subset of Z whose complement in X is contained in U. By the minimality of Z, we have Y = Z. There is no point of continuity of f |Z , and, by our assumption and Theorem A.121, it follows that Z = ∅. Hence X ∈ U and X is saturated with respect to f , which concludes the proof by Theorem 4.19(iv). Corollary 4.22 (Choquet’s barycentric theorem). Let f be a Baire-one affine function on a compact convex set X. Then f is strongly affine. Proof. By Theorem A.124 and Theorem A.121, any Baire-one function on X is fragmented. Thus the assertion follows from Theorem 4.21. Corollary 4.23 (Affine B1 -minimum principle). Let f be a Baire-one affine function on a compact convex set X. If f ≤ 0 on ext X, then f ≤ 0 on X. Moreover, sup {|f (x)| : x ∈ X} = sup {|f (x)| : x ∈ ext X} . Proof. Choquet’s barycentric theorem 4.22 asserts that any affine Baire-one function is Ac (X)-affine. Thus the affine B1 -minimum principle is a particular case of Theorem 3.86. The second assertion is then an immediate consequence. Another simple but important consequence of Choquet’s barycentric theorem is the following approximation theorem due to G. Mokobodzki (cf. M. Rogalski [392], Appendix). Notice that taking Mokobodzki’s theorem for granted, Choquet’s barycentric theorem is its immediate consequence. Theorem 4.24 (Mokobodzki’s approximation theorem). Let X be a compact convex set and f be a Baire-one affine function on X. Then there exists a bounded sequence of continuous affine functions on X pointwise converging to f on X. Proof. Let f be a Baire-one affine function on X. By Theorem A.121, Lemma 4.20 and Theorem A.124, there are bounded sequences {fn } and {gn } such that both fn and −gn are upper semicontinuous functions, fn % f
and
gn & f.
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4 Affine functions on compact convex sets
Fix n ∈ N. By Choquet’s barycentric theorem 4.22, µ fn −
1 1 < ν gn + n n
for every µ, ν ∈ M1 (X) with r(µ) = r(ν). Lemma 4.11 yields the existence of a function hn ∈ Ac (X) such that fn −
1 1 < hn < gn + . n n
Then {hn } is a bounded sequence of continuous affine functions such that hn → f on X.
4.3
State space and representation of affine functions
In this section we introduce a notion of the state space and present its basic properties. It represents a natural and efficient link between function spaces and convex analysis. Definition 4.25 (State space). Let H be a function space on a compact space K and S(H), the state space of H, be defined as S(H) := {ϕ ∈ H∗ : ϕ ≥ 0, ϕ(1) = 1} . Clearly, S(H) is a w∗ -compact convex subset of the dual space H∗ . We denote by π the quotient mapping from M(K) onto H∗ ; that is, π(µ) = µ|H , µ ∈ M(K). As a simple consequence of the Hahn–Banach theorem, we obtain that S(H) = π(M1 (K)).
(4.9)
Further, we define the evaluation mapping φ : K → S(H) as φ : x 7→ φx ,
x ∈ K,
where φx : h 7→ h(x),
h ∈ H.
Let Φ : H → Ac (S(H)) be the mapping defined for h ∈ H as Φ(h) : s 7→ s(h),
s ∈ S(H).
Proposition 4.26. The evaluation mapping φ is a homeomorphism of K into S(H) and the following assertions hold. (a) S(H) = co(φ(K)), (b) Φ(H) is dense in Ac (S(H)),
4.3 State space and representation of affine functions
121
(c) r(φ] µ) = π(µ) for any µ ∈ M1 (K); in particular, r(φ] µ) = φ(x) for µ ∈ Mx (H), (d) φ(ChH (K)) = ext S(H), (e) Φ is a positive linear mapping and kΦ(h)k = khk for each h ∈ H, (f) Φ is surjective if and only if H is closed, (g) if Φ is surjective, then the inverse mapping is realized by Φ−1 (F ) = F ◦ φ,
F ∈ Ac (S(H)).
(4.10)
Proof. Since φ is a continuous injective mapping of a compact Hausdorff space K into S(H), it is a homeomorphism between K and φ(K). We prove that S(H) = co(φ(K)). To this end, suppose that there exists a point s ∈ S(H) \ co(φ(K)). By the Hahn–Banach theorem, there exists H ∈ (H ∗ , w∗ )∗ such that H(s) < H(t) for any t ∈ co(φ(K)). Further, there exists h ∈ H such that H(ψ) = ψ(h) for every ψ ∈ H∗ . Thus s(h) < φ(x)(h) = h(x),
x ∈ K.
(4.11)
The Hahn–Banach theorem provides a Radon measure µ on K such that kµk = ksk = 1 and s(g) = µ(g) for each g ∈ H. Since µ(1) = s(1) = 1, the measure µ belongs to M1 (K). Thus s(h) = µ(h) ∈ co h(K). But this contradicts (4.11). Thus S(H) = co(φ(K)). Lemma 2.34 yields (b). Indeed, Lemma 2.34 ensures that ((H ∗ , w∗ )∗ + R)|S(H) is dense in Ac (S(H)). Hence the space Φ(H) + R is dense in Ac (S(H)). Since H contains constant functions, Φ(H) + R = Φ(H). For the proof of (c), let µ ∈ M1 (K) be given. By (b), it is enough to verify that Φ(h)(r(φ] µ)) = Φ(h)(π(µ)) for any h ∈ H. To this end, fix h ∈ H. Then Φ(h)(r(φ] µ)) = φ] µ(Φ(h)) = µ(Φ(h) ◦ φ) = µ(h) = π(µ)(h). We proceed with the proof of (d). Pick s ∈ ext S(H). Since S(H) = co(φ(K)), the Milman theorem 2.43 implies that ext S(H) ⊂ φ(K) = φ(K). Hence s = φ(x) for some point x ∈ K. If µ ∈ M1 (K) is an H-representing measure for the point x, then φ] µ represents φ(x). By Bauer’s characterization of extreme points 2.40, φ] µ = εs . Hence µ = εx and x is in the Choquet boundary ChH (K).
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4 Affine functions on compact convex sets
Conversely, let x ∈ ChH (K) be given and φ(x) = 12 (s1 + s2 ), s1 , s2 ∈ S(H). Find Radon probability measures µi , i = 1, 2, on K with π(µi ) = si , i = 1, 2. Then µ = 12 (µ1 + µ2 ) represents x. Hence µ1 = µ2 = εx , which implies s1 = s2 = s. This completes the proof of (d). Obviously, Φ is positive and linear. Further, the inequalities kΦ(h)k = sup |Φ(h)(s)| = sup |s(h)| ≥ sup |φx (h)| s∈S
s∈S
x∈K
= sup |h(x)| = khk ≥ sup |s(h)| = kΦ(h)k x∈K
s∈S
give (e). If Φ is onto, the function space H is closed because Ac (S(H)) is a complete space and Φ is an isometric mapping. Conversely, suppose that H is a closed space. Since Φ(H) is a dense subspace of Ac (S(H)), H is complete and Φ is an isometry, the space Φ(H) is closed in Ac (S(H)). Hence Ac (S(H)) ⊂ (Φ(H)) = Φ(H). It is straightforward to check the validity of the inversion formula (4.10), so the proof of the proposition is finished. Lemma 4.27. Let F be a continuous convex function on S(H) and φ : K → S(H) be the evaluation mapping. Then F ◦ φ is an H-convex continuous function on K. Proof. Let F ∈ Sc (S(H)), x ∈ K and µ ∈ Mx (H) be given. Then µ(F ◦ φ) = φ] µ(F ) ≥ F (r(φ] µ)) = F (φ(x)), because r(φ] µ) = φ(x) by Proposition 4.26(c). Proposition 4.28. Let H be a function space on a compact space K and S(H) its state space. Then the following assertions hold. (a) If µ ∈ Mφ(x) (Ac (S(H))) with spt µ ⊂ φ(K), then (φ−1 )] µ ∈ Mx (H). (b) π(M1 (F )) = co φ(F ) for any closed set F ⊂ K. (c) If µ, ν ∈ M+ (K), then µ ≺ ν if and only if φ] µ ≺ φ] ν. (d) A measure Λ ∈ M+ (S(H)) is Ac (S(H))-maximal if and only if Λ = φ] λ for some H-maximal measure λ ∈ M+ (K). (e) A measure Λ ∈ M(S(H)) is a boundary measure if and only if Λ = φ] λ for some H-boundary measure λ ∈ M(K).
4.3 State space and representation of affine functions
123
Proof. Assertion (a) is a direct consequence of the definitions and the density of Φ(H) in Ac (S(H)). For the proof of (b), if s ∈ co φ(F ), then there exists a measure µ ∈ M1 (φ(F )) which represents s (see Proposition 2.39). By assertion (a), π((φ−1 )] µ) = r(φ] ((φ−1 )] µ)) = r(µ) = s, and s ∈ π(M1 (F )). Conversely, let µ ∈ M1 (F ) be such that the point s := π(µ) does not belong to co φ(F ). Then we may separate s and co φ(F ) by a continuous affine function on S(H). Due to the density of Φ(H) in Ac (S(H)), we may suppose that there exists a function h ∈ H such that s(h) > sup{t(h) : t ∈ co φ(F )}. Then we obtain µ(h) = s(h) >
sup t∈co φ(F )
t(h) ≥ sup t(h) = sup h(F ) ≥ µ(h), t∈φ(F )
which is a contradiction. Hence π(M1 (F )) = co φ(F ). To prove (c), suppose that µ, ν ∈ M+ (K) with µ ≺ ν and F ∈ Kc (S(H)) are given. By Lemma 4.27, F ◦ φ ∈ Kc (H). Thus φ] µ(F ) = µ(F ◦ φ) ≤ ν(F ◦ φ) = φ] µ(F ), and hence φ] µ ≺ φ] ν. Conversely, assume that φ] µ ≺ φ] ν. By Proposition 3.56, it is enough to show that µ(f ) ≤ ν(f ) for any f = h1 ∨ · · · ∨ hn , where h1 , . . . , hn ∈ H. By the assumption and the definition of the image of a measure, we get Z Z h1 ∨ · · · ∨ hn dµ = Φ(h1 ) ∨ · · · ∨ Φ(hn ) d(φ] µ) K
φ(K)
Z ≤
Z Φ(h1 ) ∨ · · · ∨ Φ(hn ) d(φ] ν) =
φ(K)
h1 ∨ · · · ∨ hn dν, K
which is the desired conclusion. In order to check assertion (d), pick an H-maximal measure µ in M+ (K). Let Λ ∈ M+ (S(H)) be an Ac (S(H))-maximal measure with φ] µ ≺ Λ. Then Λ is carried by φ(K), and thus Λ = φ] λ for some λ ∈ M+ (K). By assertion (c), µ ≺ λ. Since µ is H-maximal, µ = λ and φ] µ is Ac (S(H))-maximal. Conversely, let φ] µ be an Ac (S(H))-maximal measure and λ ∈ M+ (K) satisfying µ ≺ λ. Then φ] µ ≺ φ] λ, which gives that φ] µ = φ] λ, and consequently µ = λ. Since assertion (e) is a direct consequence of (d), the proof is finished.
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4 Affine functions on compact convex sets
Proposition 4.29. Let H be a closed function space on a compact space K and X := S(H). Then the following assertions hold: (a) X is contained in a closed hyperplane not containing 0, S (b) if (H∗ )+ denotes the set of all positive elements of H∗ , then (H∗ )+ = c≥0 cX, (H∗ )+ ∩ −(H∗ )+ = {0} and H∗ = (H∗ )+ − (H∗ )+ , (c) for any F ∈ Ac (X) there exists a unique Fb ∈ (H∗ , w∗ )∗ such that F = Fb on X. Proof. The first assertion (a) is obvious since X ⊂ {ϕ ∈ H∗ : ϕ(1) = 1}. S For the proof of (b) we first notice that clearly c≥0 cX ⊂ (H∗ )+ . On the other hand, if ϕSis a nonzero positive functional on H and c = kϕk, then ϕ = c(c−1 ϕ), and thus ϕ ∈ c≥0 cX. Obviously, (H∗ )+ ∩ −(H∗ )+ = {0}. If ϕ ∈ H∗ is given, extend ϕ to a signed measure µ ∈ M(K) with µ = ϕ on H. If we decompose µ into its positive and negative part µ+ and µ− and set ϕ+ := µ+ |H and ϕ− := µ− |H , we obtain the desired decomposition of ϕ. Thus H∗ = (H∗ )+ − (H∗ )+ . For the third part of the proof, pick F ∈ Ac (X). Since H is closed, F = Φ(f ) for some f ∈ H by Proposition 4.26(f). Setting Fb(ϕ) := ϕ(f ),
ϕ ∈ H∗ ,
we obviously obtain an extension of F . We have to show that this extension is unique. But this easily follows from assertion (b). This concludes the proof. Notation 4.30. We recall that τX is the topology of pointwise convergence on a set X; see Notation A.27. Proposition 4.31. Let X be a compact convex set in a locally convex space, E := Ac (X) and φ : X → S(Ac (X)) be the evaluation mapping. Then the following assertions hold: (a) The mapping φ is an affine homeomorphism of X onto the set S(Ac (X)) and S(Ac (X)) ⊂ SE ∗ . (b) For every η ∈ E ∗ there exist numbers a1 , a2 ≥ 0 and points x1 , x2 ∈ X such that η = a1 φ(x1 ) − a2 φ(x2 ) and kηk = a1 + a2 . Moreover, a1 , a2 are uniquely determined. (c) φ(X) ∪ −φ(X) ⊂ SE ∗ and BE ∗ = co(φ(X) ∪ −φ(X)). (d) For the set (E ∗ )+ of all positive elements of E ∗ we have (E ∗ )+ =
S
c≥0 cφ(X).
(e) For every affine continuous function f on X there exists a unique function fb ∈ (E ∗ , w∗ )∗ such that f = fb ◦ φ on X. (f) The mapping id : (Ac (X), τX ) → (E, w) is a homeomorphism.
125
4.3 State space and representation of affine functions
Proof. We already know from Proposition 4.26 that φ is a homeomorphism. A routine argument shows that φ is affine. To check its surjectiveness, note that for any s ∈ S(Ac (X)) there exists µ ∈ M1 (X) such that π(µ) = s. Then x := r(µ) satisfies φ(x) = s. Obviously, S(Ac (X)) ⊂ SE ∗ , which concludes the proof of (a). For the proof of (b), let η ∈ E ∗ be given. Let µ ∈ M(X) be an extension of η with kηk = kµk. We decompose µ into its positive and negative part as µ = a1 µ1 − a2 µ2 , where a1 , a2 ≥ 0 and µ1 , µ2 ∈ M1 (X). Let x1 := r(µ1 ) and x2 := r(µ2 ). Then η(h) = a1 h(x1 ) − a2 h(x2 ) for any h ∈ Ac (X), and thus η = a1 φ(x1 ) − a2 φ(x2 ) and
kηk = kµk = a1 + a2 .
(4.12)
If η = b1 φ(y1 ) − b2 φ(y2 ) is another representation of η with kηk = b1 + b2 , we apply η to the constant function 1 to get a1 − a2 = b1 − b2 . This, along with (4.12), yields a1 = b1 and a2 = b2 . For the proof of (c), we note that φ(X) ∪ −φ(X) ⊂ SE ∗ . Further, SE ∗ ⊂ co(φ(X) ∪ −φ(X)) by (b). Since 0 ∈ co(φ(X) ∪ −φ(X)), BE ∗ ⊂ co(φ(X) ∪ −φ(X)). Since the reverse inclusion is obvious, the proof of (c) is complete. Since (d) and (e) follow from Proposition 4.29(b) and (c), respectively, we proceed to the proof of (f). The topology τX on Ac (X) is generated by functionals h 7→ h(x), h ∈ Ac (X), where x ∈ X. On the other hand, by (b), the weak topology on Ac (X) is generated by functionals of the form h 7→ a1 h(x1 ) − a2 h(x2 ), h ∈ Ac (X), where a1 , a2 ≥ 0 and x1 , x2 ∈ X. It follows that these topologies coincide on Ac (X). This finishes the proof of (f). Proposition 4.32. Let X be a compact convex set in a locally convex space and E := Ac (X). (a) For every η ∈ E ∗∗ there exists a unique fη ∈ Ab (X) such that kηk = kfη k and η(a1 φ(x1 ) − a2 φ(x2 )) = a1 fη (x1 ) − a2 fη (x2 ),
a1 , a2 ≥ 0, x1 , x2 ∈ X.
(b) The mapping η 7→ fη is an isometric isomorphism of E ∗∗ onto Ab (X) which is a (w∗ –τX )-homeomorphism. Moreover, it preserves order (that is, η ≥ 0 if and only if fη ≥ 0). Proof. We start with (a). Given a functional η ∈ E ∗∗ , we define a function fη : X → R as fη : x 7→ η(φ(x)), x ∈ X. Obviously, fη is affine and kfη k ≤ kηk. To prove the converse inequality, Proposition 4.31(c) gives kηk = sup{η(a1 φ(x1 ) − a2 φ(x2 )) : a1 , a2 ≥ 0, a1 + a2 ≤ 1, x1 , x2 ∈ X} = sup{a1 fη (x1 ) − a2 fη (x2 ) : a1 , a2 ≥ 0, a1 + a2 ≤ 1, x1 , x2 ∈ X} ≤ sup{a1 kfη k + a2 kfη k : a1 , a2 ≥ 0, a1 + a2 ≤ 1} ≤ kfη k.
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4 Affine functions on compact convex sets
Uniqueness of fη immediately follows from Proposition 4.31(d). For the proof of (b), we start by noticing that (a) yields that the mapping T : η 7→ fη is an isometric mapping of (E ∗∗ , k · k) into (Ab (X), k · k). If f ∈ Ab (X) is given, the function η(a1 φ(x1 ) − a2 φ(x2 )) := a1 f (x1 ) − a2 f (x2 ),
a1 , a2 ≥ 0, x1 , x2 ∈ X,
is in E ∗∗ (see Exercise 4.48) and f = fη . Hence T is surjective. As in Proposition 4.31(f), for the proof that T is a (w∗ –τX )-homeomorphism, we look at the defining functionals. By Proposition 4.31(b), the w∗ -topology on E ∗∗ is generated by functionals η 7→ η(a1 φ(x1 ) − a2 φ(x2 )), η ∈ E ∗∗ , where a1 , a2 ≥ 0 and x1 , x2 ∈ X. The topology τX on Ab (X) is generated by functionals of the form f 7→ f (x), f ∈ Ab (X), where x ∈ X. Hence it follows from (a) that T is a (w∗ –τX )homeomorphism. Finally, the fact that the function fη is positive if and only if η is positive, follows from the definition of the mapping T . This concludes the proof. Theorem 4.33 (Goldstine’s lemma). If X is a compact convex set, then BAc (X) is τX -dense in BAb (X) . Proof. The assertion is just a reformulation of Goldstine’s lemma (see Theorem A.4). Indeed, if E stands for the Banach space Ac (X), then the unit ball BE is w∗ -dense in BE ∗∗ . Let ε : E → E ∗∗ be the canonical embedding and T : E ∗∗ → Ab (X) be the mapping from Proposition 4.32(b). If I : Ac (X) → Ab (X) denotes the identity mapping, then I = T ◦ ε. From this the assertion follows. Theorem 4.34. For a real-valued function f on a compact convex set X, the following assertions are equivalent: (i) f is strongly affine, (ii) f is bounded and for any measure µ ∈ M1 (X) there exists a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, and fn → f µ-almost everywhere. Proof. To prove (i) =⇒ (ii), let f be a strongly affine function. Since f is bounded by Lemma 4.5, we may assume without loss of generality that kf k = 1. Given a measure µ ∈ M1 (X), Theorem 4.19 implies the existence of compact convex sets Kn ⊂ X, n ∈ N, such that µ(Kn ) → 1 and f |Kn is continuous. For a fixed natural number n ∈ N, consider the restriction operator R : Ab (X) → Ab (Kn ) defined as Rf := f |Kn ,
f ∈ Ab (X).
By Goldstine’s lemma 4.33, BAc (X) is τX -dense in BAb (X) . Since the operator R is (τX –τKn )-continuous, R(BAc (X) ) is τKn -dense in R(BAb (X) ). Hence we have Rf ∈
4.4 Affine Baire-one functions on dual unit balls
127
τK
R(BAc (X) ) n . Since Rf is continuous on Kn and τKn coincides with the weak topology on Ac (Kn ) (see Proposition 4.31(f)), Mazur’s theorem (see Theorem A.2 or Exercise 3.116) yields k·kAc (Kn )
Rf ∈ co R(BE )
k·kAc (Kn )
= R(BE )
.
It follows that there exists a function g ∈ BAc (X) that is uniformly close to f on Kn as much as we wish. Applying this argument to each Kn , we construct the required sequence {fn } in BAc (X) . To show (ii) =⇒ (i), assume that f satisfies (ii). First we notice that the assumption ensures that given finitely many measures µ1 , . . . , µk in M+ (X), we can find a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, and fn → f µi -almost everywhere for each i = 1, . . . , k (we just apply the assumption to a multiple of µ1 + · · · + µk ). Let µ ∈ M1 (X) with x = r(µ) be given. We find a sequence {fn } of continuous affine functions on X such that kfn k ≤ kf k, n ∈ N, fn → f µ-almost everywhere and fn (x) → f (x) (use the preceding argument for the measures µ and εx ). Then µ(f ) = lim µ(fn ) = lim fn (x) = f (x), n→∞
n→∞
and f is strongly affine. This concludes the proof. Theorem 4.35. Let H be a closed function space on a metrizable compact space K. The the set of H-exposed points is dense in ChH (K). Proof. Let X := S(H). Since K is metrizable, H is separable, and thus X is metrizable as well (see Theorem A.5). Let φ : K → X and Φ : H → Ac (X) be as in Definition 4.25. Since Φ is a positive and surjective isometry (see Proposition 4.26(e) and (f)), it is easy to verify that φ(expH (K)) = exp X. Since φ(ChH (K)) = ext X by Proposition 4.26(d), the conclusion follows from Corollary 2.52.
4.4
Affine Baire-one functions on dual unit balls
In what follows, we present an application of Mokobodzki’s approximation result 4.24. For this reason we need a characterization of affine functions on unit balls in Banach spaces. If E is a Banach space, we recall that E is canonically embedded in E ∗∗ . The first proposition is a particular version of the Banach–Dieudonn´e theorem (see Theorem A.7). Proposition 4.36. Let E be a Banach space, BE ∗ be the closed unit ball of E ∗ equipped with the w∗ -topology, and let f be a function on BE ∗ . Then f is a continuous affine function on BE ∗ if and only if there exist x ∈ E and λ ∈ R such that f = x + λ on BE ∗ .
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4 Affine functions on compact convex sets
Proof. Assume that f ∈ Ac (BE ∗ ). Since (E ∗ , w∗ )∗ = E, by Lemma 2.34 there exist xn ∈ E and λn ∈ R, n ∈ N, such that xn + λ n → f
uniformly on BE ∗ .
Since f (0) = lim (xn (0) + λn ) = lim λn , n→∞
n→∞
we see that there exists limn→∞ λn and it equals f (0). Given ε > 0, there are n, k ∈ N such that |xn (ϕ) − xk (ϕ)| < ε for all ϕ ∈ BE ∗ . Hence ||xn − xk || ≤ ε for those n, k, and therefore there exists x := limn→∞ xn . It follows that f = x + f (0) on BE ∗ . Obviously, the function (x + λ)|BE∗ , where x ∈ E and λ ∈ R, is a continuous affine function on BE ∗ . Theorem 4.37. Let f be a function on BE ∗ , f (0) = 0. The following assertions are equivalent: (i) f is an affine Baire-one function on BE ∗ , (ii) there exists a sequence {xn } in E such that xn → f on BE ∗ , (iii) there exists s∗∗ in the w∗ -sequential closure of E such that f = s∗∗ on BE ∗ . Proof. To show (i) =⇒ (ii), let f be an affine Baire-one function on BE ∗ with f (0) = 0. By Mokobodzki’s approximation theorem 4.24, there exists a sequence {fn } of continuous affine functions on BE ∗ such that fn → f . By Proposition 4.36, each fn is of the form xn + λn on BE ∗ for suitable xn ∈ E and λn ∈ R. Since f (0) = 0, we may assume that λn = 0 for each n ∈ N and (ii) follows. We proceed with (ii) =⇒ (iii). If xn ∈ E and xn → f on BE ∗ , then f is an affine Baire-one function on BE ∗ . It is easy to see that f is bounded (cf. Theorem A.121 and Lemma 4.20). By Exercise 4.48, there exists s∗∗ ∈ E ∗∗ such that f = s∗∗ on BE ∗ . Hence s∗∗ (ϕ) = f (ϕ) = lim xn (ϕ) n→∞
s∗∗ (ϕ)
for every ϕ ∈ BE ∗ . It follows that = limn→∞ xn (ϕ) even for each ϕ ∈ E ∗ , ∗∗ ∗ and we see that s belongs to the w -sequential closure of E. The implication (iii) =⇒ (i) is obvious. Corollary 4.38. Let E be a Banach space. Then the w∗ -sequential closure of E is norm closed in E ∗∗ . Proof. By Theorem 4.37, the w∗ -sequential closure of E equals the set M := s∗∗ ∈ E ∗∗ : s∗∗ |BE∗ is a Baire-one function , and the assertion follows since M is norm closed (see Proposition A.126).
4.5 Exercises
4.5
129
Exercises
Exercise 4.39. Let f be a continuous function on a compact convex set X and x ∈ X. Then f ∗ (x) = sup{µ(f ) : µ ∈ Mx (X), µ is molecular}. Hint. Use Lemma 3.21 and Proposition 4.3. Exercise 4.40. If a compact convex set X in a locally convex space E admits a strictly convex continuous function, then X is metrizable. Hint. In order to show that X is metrizable, it suffices to find a countable family B ⊂ C(K) such that B separates points on X (see Lemma 10.45). So, suppose that f is a continuous strictly convex function on X. By Corollary 4.8 and Proposition 3.55, there exist kn ∈ N, n ∈ N, and functions fn = hn1 ∨ · · · ∨ hnkn , hni ∈ Ac (X), i = 1, . . . , kn , such that fn < f < fn + n1 , n ∈ N. We want to show that the family B := {hni : i = 1, . . . , kn , n ∈ N} separates points on X. Assume that there are distinct x, y ∈ X such that a(x) = a(y) for each a ∈ B. Let z := 12 (x + y). For ε > 0, find a1 , a2 ∈ B such that f (x) < a1 (x) + ε and f (y) < a2 (y) + ε. Then 1 1 f (z) ≤ (f (x) + f (y)) ≤ (a1 (x) + a2 (y)) + ε 2 2 1 ≤ (f (z) + f (z)) + ε. 2 1 Hence f (z) = 2 (f (x) + f (y)) and f is not strictly convex. Exercise 4.41. Prove that there exists a strictly convex function f on a compact convex set X that is not Ac (X)-strictly convex. Hint. Consider the compact convex set M1 ([0, 1]) and a strictly convex continuous function f on M1 ([0, 1]), 0 ≤ f ≤ 21 . If g : µ 7→ µd ([0, 1]),
µ ∈ M1 ([0, 1]),
where µd denotes the discrete part of µ, then g is a Borel (in fact, by Proposition 2.63, of the second Baire class) affine function on M1 ([0, 1]). Show that f − g is a strictly convex function on M1 ([0, 1]) which is not strictly Ac (M1 ([0, 1]))-convex. By Proposition 2.27, the mapping ε : x 7→ εx is a homeomorphism of [0, 1] onto ext M1 ([0, 1]). Let Λ := ε] λ be the image of Lebesgue measure λ by ε. By Proposition 2.54, λ is a barycenter of Λ. Moreover, 1 Λ(f − g) = Λ(f ) − Λ(g) ≤ − 1 < 0 ≤ f (λ) = f (λ) − g(λ) 2 = (f − g)(λ).
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4 Affine functions on compact convex sets
Exercise 4.42. Let f, −g be upper semicontinuous concave functions on a compact convex set X, with f < g. Then there exists a function h ∈ Ac (X) such that f < h < g. Hint. By Corollary 4.8, µ(f ) ≤ f (r(µ)) < g(r(µ)) ≤ ν(g) for any µ, ν ∈ M1 (X) with r(µ) = r(ν). Thus we may apply Lemma 4.11. Exercise 4.43. Prove that f ∈ Susc (X) if and only if f = inf {w ∈ W(Ac (X)) : w ≥ f } . Hint. Combine Corollary 4.8, Proposition 3.25 and Corollary 3.23. Exercise 4.44. If f is a real-valued convex function on compact convex set X, then its lower semicontinuous regularization defined as fb(x) = f (x) ∧ lim infy→x f (y), x ∈ X, is convex as well (see Subsection A.2.C). Hint. Obvious from the definition. Exercise 4.45. Let X be a compact convex set. Prove Lemma 3.77 for this particular case without the Simons lemma 3.75. Hint. Let {fn } be an upper bounded sequence of lower semicontinuous convex functions on X satisfying lim supn→∞ fn ≤ 0 on ext X. Without loss of generality we may assume that {fn } is a bounded sequence. Given x ∈ X and ε > 0, we find continuous affine functions hn , n ∈ N, such that hn ≤ fn and hn (x) > fn (x) − ε (use Corollary 3.23(b) and Proposition 3.25(a)). We define a continuous affine mapping ϕ : X → RN by ϕ(x) = {hn (x)}n∈N , x ∈ X, and let Y := ϕ(X). Then Y is a metrizable compact convex set and ϕ(ext X) ⊃ ext Y (see Proposition 2.72(c)). Find a maximal measure µ ∈ Mϕ(x) (Y ) and let h0n ∈ Ac (Y ) be such that h0n ◦ ϕ = hn , n ∈ N. Then lim supn→∞ h0n ≤ 0 on ext Y . By Corollary 3.61 and Fatou’s lemma, lim sup hn (x) = lim sup h0n (ϕ(x)) = lim sup µ(h0n ) n→∞
n→∞
n→∞
≤ µ(lim sup fn ) ≤ 0. n→∞
Hence lim sup fn (x) ≤ ε + lim sup hn (x) ≤ ε. n→∞
This concludes the proof.
n→∞
4.5 Exercises
131
Exercise 4.46 (Mokobodzki). Let f be a real-valued lower bounded concave function on a compact convex set X. Then f is upper bounded on X. Hint. Without loss of generality we can suppose that f ≥ 0 on X. We suppose that f is not upper bounded and we will deduce a contradiction. There are xn ∈ K, n ∈ N, so that f (xn ) > 2n . Let ∞ X 1 x := r εxj j 2 j=1
and sn :=
∞ X
2−j ,
yn := r
∞ X j=n+1
j=n+1
Then x=
1 εxj , j sn 2
n ∈ N.
n X 1 x j + s n yn . 2j
(4.13)
j=1
Indeed, for any h ∈ Ac (X) we have h(x) =
n ∞ X X 1 1 ε (h) h(xj ) + sn h(yn ). = x 2j j 2j j=1
j=1
As functions from Ac (X) separate points of X, we have equality (4.13). Since x is a convex combination of points x1 , . . . , xn , yn , we get f (x) ≥
n X 1 f (xj ) + sn f (yn ) ≥ n. 2j j=1
Hence f (x) = ∞, which is obviously a contradiction. Exercise 4.47. Prove that there exists a function f on a compact convex set X satisfying the barycentric formula such that f |F has no point of continuity for some nonempty closed F ⊂ X. Hint. Take X := M1 ([0, 1]) and f (µ) = µ(g), µ ∈ X, where g is a bounded Borel function on [0, 1] with no point of continuity on [0, 1]. To prove the barycentric formula, start with a continuous function and then use transfinite induction along with the Lebesgue dominated convergence theorem. Exercise 4.48. Let X be a convex subset of a vector space E and f : X → R be an affine function on X. Assume that •
either 0 ∈ X and f (0) = 0,
•
or, there exists ϕ : E → R linear such that X ⊂ {x ∈ E : ϕ(x) = 1}.
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4 Affine functions on compact convex sets
Then f has a unique linear extension on span X. Hint. If 0 ∈ X and f (0) = 0, then span X = {a1 x1 − a2 x2 : a1 , a2 ≥ 0, x1 , x2 ∈ X} and the required extension is defined by fb(a1 x1 − a2 x2 ) := a1 f (x1 ) − a2 f (x2 ) a1 , a2 ≥ 0, x1 , x2 ∈ X. Since f is affine, fb is a well-defined unique linear extension of f . If the second assumption is satisfied, then X ∩ −X = ∅ and f has a unique affine extension on co(X ∪ −X) given by fb(ax − (1 − a)y) := af (x) − (1 − a)f (y),
a ∈ [0, 1], x, y ∈ X.
Now the extension to span X can be defined as above. Exercise 4.49. Let X be a compact convex set and A ⊂ Ac (X) be a vector space containing the constant functions and separating points of X. Then A is dense in Ac (X). Hint. By the Hahn–Banach theorem it is enough to show that ϕ = 0 whenever ϕ ∈ (Ac (X))∗ satisfies ϕ = 0 on A. By Proposition 4.31(b) there exist numbers a1 , a2 ≥ 0 and x1 , x2 ∈ X so that ϕ(f ) = a1 f (x1 ) − a2 f (x2 ), f ∈ Ac (X). By the assumption, 0 = ϕ(1) = a1 − a2 . Hence f (x1 ) = f (x2 ) for each f ∈ A. Since A separates points of X, x1 = x2 and ϕ = 0. Exercise 4.50. Let X be a compact convex set and A ⊂ Ac (X) be convex. Prove that τ f ∈ Ac (X) is in the norm closure of A if and only if f ∈ A X . Hint. This follows from Theorem A.2 and Proposition 4.31(f). Exercise 4.51. Let H be a function space on a compact space K and let {1, f1 , . . . , fd } be a basis of H. Let ψ : K → Rd be defined as ψ(x) = (f1 (x), . . . , fd (x)),
x ∈ K.
Prove that S(H) is affinely homeomorphic to co ψ(K). Hint. Let ϕ : S(H) → Rd be defined by ϕ(s) := (s(f1 ), . . . , s(fd )),
s ∈ S(H).
Then ϕ is a homeomorphism of S(H) into Rd such that ϕ ◦ φ = ψ on K. By Proposition 4.26(a), ϕ(S(H)) = ϕ(co φ(K)) = co(ϕ ◦ φ(K)) = co ψ(K). Since co ψ(K) = co ψ(K) by Corollary 2.8, S(H) is affinely homeomorphic to co ψ(K).
4.6 Notes and comments
133
Exercise 4.52. Find a compact convex set X and a point x ∈ X such that face x is not a face. Hint. Let K := {−1} ∪ [0, 1] ∪ {2} and 1 1 H := {f ∈ C(K) : f ( ) = (f (−1) + f (2))}. 2 2 Let X := S(H) and π : M1 (K) → X be the quotient mapping from Proposition 4.26. Let s := π(λ), where λ is Lebesgue measure on [0, 1]. Consider the face λ in M1 ([0, 1]) ⊂ M1 (K). Then π(face λ) = face s.
(4.14)
Indeed, the inclusion face s ⊂ π(face λ) follows from the fact that π −1 (face s) is a face containing λ (see Proposition 2.72). For the converse inclusion, let t ∈ face s be given. Then there exist α ∈ (0, 1] and t0 ∈ X such that s = αt + (1 − α)t0 . Let µ, µ0 ∈ M1 (K) satisfy π(µ) = t and π(µ0 ) = t0 . We set A := {−1, 12 , 2} and find a sequence {fn } of functions from H such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cA . Then the equality λ(fn ) = αµ(fn ) + (1 − α)µ0 (fn ) yields 0 = αµ(A). Hence µ ∈ M1 ([0, 1]). We claim that αµ ≤ λ. To verify this, let a positive function f ∈ C([0, 1]) be given. We find a positive function g ∈ H such that f = g on [0, 1]. Then λ(f ) = s(g) ≥ αt(g) = αµ(f ). Hence µ ∈ face λ by Exercise 2.127(a) and t ∈ π(face λ). This proves (4.14). By (4.14), Proposition A.40 and Exercise 2.127(b), face s = π(face λ) = π(face λ) = π(M1 ([0, 1])). Hence φ 1 ∈ face s, φ 1 = 21 (φ−1 + φ2 ) and φ−1 , φ2 are not contained in face s. Thus 2
2
face s is not a face.
4.6
Notes and comments
Basic properties of affine functions on compact convex sets are well known and can be found for example in E. M. Alfsen [5] and L. Asimow and A. J. Ellis [24]. The proof of Lemma 4.3 is taken from E. M. Alfsen [5], Proposition I.2.3, Lemma 4.5 is a variant of U. Krause [282], Satz 2.1.
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4 Affine functions on compact convex sets
Section 4.2 follows mainly the paper R. Haydon [220]. Choquet’s barycentric theorem 4.22 is proved by G. Choquet in [106] (see also E. M. Alfsen [5], Theorem I.2.6, or R. R. Phelps [374]). A slightly simplified proof can be found in J. Lukeˇs, J. Mal´y, I. Netuka, M. Smrˇcka and J. Spurn´y [319]. G. Choquet himself also showed that the barycentric formula is no longer valid for affine functions of the second Baire class (see Proposition 2.63). Theorem 4.24 due to G. Mokobodzki can be found in M. Rogalski [392]. Results on the state space contained in Section 4.3 are standard; they are variants of results in E. M. Alfsen [5], Chapter 2, §1 and L. Asimow and A. J. Ellis [24], Chapter 2. Theorem 4.34 is taken from S. Teleman [449]. Section 4.4 is a well-known application of G. Mokobodzki’s theorem 4.24; it can be found for example in S. A. Argyros, G. Godefroy and H. P. Rosenthal [19] and E. Odell and H. P. Rosenthal [366]. Exercise 4.40 is proved in M. Herv´e [223]. Exercise 4.45 is the classical way of proving Lemma 3.77, see E. M. Alfsen [5], Proposition I.4.10. In A. Goullet de Rugy [200], Exercise 4.46 is ascribed to G. Mokobodzki.
Chapter 5
Perfect classes of functions and representation of affine functions
The aim of this chapter is to introduce a hierarchy of Borel sets in topological spaces, show its relation to inductive generation of Baire and Borel functions, and apply these facts to get a link between topological properties of function spaces and compact convex sets. The first part is devoted to an abstract generation of classes of sets and functions from a given family of sets. It turns out that a suitable starting point for this procedure is an algebra of sets, because then the standard properties of Borel sets hold (see Proposition 5.4) as well as the characterization of Baire classes of functions via level sets (see Theorem 5.9). The next section employs an abstract procedure for creating classes of Baire and Borel sets in topological spaces. The main result is Theorem 5.16, showing that Borel classes are “quotient” with respect to perfect mappings. The key ingredient of the proof is elementary Lemma 5.15. Abstract results obtained so far are applied in Section 5.3 to get a characterization of Baire and Borel mappings via their measurability (see Corollary 5.22). As another consequence we get Theorem 5.23 characterizing Baire-one functions on compact spaces. The nice behavior of our Borel classes with respect to perfect mappings is pointed out by Theorem 5.26, which is a result widely used throughout the book. The next section begins to describe the interplay between topological results and properties of functions on compact convex sets. To emphasize crucial properties of affine functions, we introduce in Definition 5.28 a notion of “affinely perfect functions”. Theorem 5.31 and Corollary 5.32 show to what extent the behavior of an affine function is determined by its properties on the closure of extreme points and Theorem 5.33 provides examples of affinely perfect classes. The last section is crucial for the transfer of properties of function spaces to the context of compact convex sets. A precise description of this transfer is contained in Theorem 5.40 and Corollary 5.41. First we recall several definitions. Definition 5.1 (Sublattices and algebras of sets). If X is a set and F is a family of its subsets, then F is a sublattice if ∅, X ∈ F and F is closed with respect to finite unions and finite intersections. The family F is an algebra if F is a sublattice closed with respect to complements. If A is a family of sets in a set X, we write Aσ (or, Aδ ) for all countable unions (or, intersections, respectively) of sets from A. If f : X → Y
136
5 Perfect classes of functions and representation of affine functions
is a mapping from X to a topological space Y , we say that f is F-measurable, if f −1 (U ) ∈ F for each open U ⊂ Y .
5.1
Generation of sets and functions
Definition 5.2 (Abstract Borel classes of sets). If F is a family of sets in a set X, we define Borel classes generated by F as follows: Let Σ1 (F) := F, Π1 (F) := {X \ F : F ∈ F}, and for each countable ordinal α ∈ (1, ω1 ), let [ Σα (F) := Πβ (F) σ
β<α
and Πα (F) :=
[
Σβ (F) .
β<α
δ
Elements of the family Σα (F) are termed the sets of abstract additive class α and elements of the family Πα (F) are called the sets of abstract multiplicative class α. The sets in ∆α (F) = Σα (F) ∩ Πα (F) are the sets of abstract ambiguous class α. Remark 5.3. The definition and notation of abstract Borel classes is chosen in such a way that it coincides with the definition of Borel sets in metrizable spaces; see Definition A.113, Proposition 5.14 or Section 11.B of [262]. Proposition 5.4. Let F be an algebra of sets in a set X. Then the following assertions hold: (a) Σα (F) ∪ Πα (F) ⊂ ∆β (F), 0 < α < β < ω1 , (b) a set A ⊂ X is in Σα (F) if and only if X \ A ∈ Πα (F), α ∈ (0, ω1 ), (c) the family Σα (F) is stable with respect to countable unions and finite intersections, Πα (F) is stable with respect to finite unions and countable intersections and ∆α (F) is an algebra, α ∈ (1, ω1 ), S S (d) α<ω1 Σα (F) = α<ω1 Πα (F) is the σ-algebra generated by F, (e) Σα (F) = (∆α (F))σ , α ∈ (1, ω1 ), (f) for each α ∈ (0, ω1 ), Σα (F) has the generalized reduction property; that is, if An ∈ Σα (F), n ∈ N, then there existS pairwise disjoint sets A0n ∈ Σα (F), n ∈ N, S∞ 0 0 such that An ⊂ An and n=1 An = ∞ n=1 An , (g) if α ∈ (0, ω1 ) and A ∈ ∆α+1 (F), then there exists a sequence {An } of sets such that An ∈ ∆α (F), n ∈ N, and cA = limn→∞ cAn , (h) if α ∈ (1, ω1 ) is a limit ordinal S and A ∈ ∆α+1 (F), then there exists a sequence {An } of sets such that An ∈ β<α ∆β (F), n ∈ N, and cA = limn→∞ cAn .
5.1 Generation of sets and functions
137
Proof. Assertions (a), (b) and (c) follow by a straightforward transfinite induction, (d) follows from (a) and (c), and (e) follows from (a). For the proof of (f), let An ∈ Σα (F), n ∈ N, be given. If α = 1, we use the fact that F S is an algebra to get the required sets A0n , n ∈ N. Otherwise, using (e), we write An = ∞ k=1 Ank , where Ank ∈ ∆α (F). We enumerate the family {Ank : n, k ∈ N} 0 as {Bj } and define B1 := B10 ,
0 Bj := Bj0 \ (B10 ∪ · · · ∪ Bj−1 ),
j ≥ 2.
Then the sets Bj are contained in ∆α (F). We set [ A01 := {Bj : Bj ⊂ A1 } and [ A0n := {Bj : Bj ⊂ An , Bj * A01 ∪ · · · ∪ A0n−1 },
n ≥ 2.
Then A0n ⊂ An and A0n ∈SΣα (F), n ∈ N. Moreover, {A0n : n ∈ N} is a disjoint family whose union equals ∞ n=1 An . ToSverify (g), let A ∈ ∆ (F) for some α ∈ (0, ω1 ) be given. By definition, α+1 T∞ B = C , where Bn ∈ Πα (F), Cn ∈ Σα (F), n ∈ N. We fix n ∈ N A= ∞ n=1 n n=1 n and use (f) for the pair {X \ Bn , Cn } to find a set An ∈ ∆α (F) such that Bn ⊂ An ⊂ Cn . Then limn→∞ cAn = cA , as required. ForSthe proof ofT(h), let α be a countable limit ordinal and A ∈ ∆α+1 (F). We write ∞ A= ∞ n=1 Cn , where Bn ∈ Πα (F), Cn ∈ Σα (F), n ∈ N. By (a), there n=1 Bn = exist ordinals αnk < α and sets Bnk , Cnk ∈ ∆αnk (F), k ∈ N, such that Bn =
∞ \
Bnk
and
Cn =
∞ [
Cnk ,
n, k ∈ N.
k=1
k=1
Without loss of generality we may assume that the sequences {Bn }, {Cnk }∞ k=1 are are decreasing. Then the sets increasing and the sequences {Cn }, {Bnk }∞ k=1 An =
n [ p=1
(Bpn ∩
p \
Cjn ),
n ∈ N,
j=1
are contained in ∆αn (F) for some αn < α and limn→∞ cAn = cA . Indeed, let x ∈ A be given. First we show that x is contained in An for all n ∈ N sufficiently large. To this end, let w ∈ N be chosen in such a way that x ∈ Bw . Then x ∈ Bwp for all p ∈ N. Let r ∈ N be such that x ∈ Cml
for each m = 1, . . . , w and l > r.
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5 Perfect classes of functions and representation of affine functions
Then
w \
x ∈ Bwn ∩
Cjn ,
if n > r.
j=1
Hence for all n > r + w we get n [
x∈
(Bpn ∩
p=1
p \
Cjn ).
j=1
Conversely, let x ∈ / A be given. We find q ∈ N such that x ∈ / Cq . Then x ∈ / S∞ C . Let s ∈ N be such that x ∈ / B for any m = 1, . . . , q and k > s. Let qn mk n=1 T n > s be arbitrary. For any m ∈ N, x ∈ / Bmn if m = 1, . . . , q, or x ∈ / m j=1 Cjn if m > q. In both cases, x∈ /
n [
(Bpn ∩
p=1
p \
Cjn ) = An .
j=1
This concludes the proof. Definition 5.5 (Abstract Baire classes of mappings). Let Φ be a family of mappings from a set X to a topological space Y . We define abstract Baire classes generated by Φ inductively as follows: Let Φ0 := Φ and for each countable S ordinal α ∈ (0, ω1 ), let Φα be the family of all pointwise limits of sequences from β<α Φβ . Remark 5.6. Later on (see Theorem 5.9 and Definition 5.18), it will be sometimes convenient to denote the starting family of the inductive definition by Φ1 . More precisely, we start from a family S denoted by Φ1 and then Φα consists of all pointwise limits of sequences from 1≤β<α Φβ , α ∈ (1, ω1 ). The purpose of this convention is that we want to start the generation of mappings between topological spaces from “Baire-one” mappings. Lemma 5.7. Let F be an algebra of sets in a set X, α ∈ (0, ω1 ) and f : X → Y be a Σα+1 (F)-measurable mapping from X to a precompact metric space (Y, ρ). Then, for every ε > 0, there exist a partition {A1 , . . . , An } of X consisting of sets in ∆α+1 (F) and points y1 , . . . , yn ∈ Y such that the function g(x) = yi ,
x ∈ Ai , i = 1, . . . , n.
is Σα+1 (F)-measurable and satisfies ρ(f (x), g(x)) < ε, x ∈ X. Proof. Let f : X → Y be a Σα+1 (F)-measurable mapping and ε > S 0. Since Y is precompact, we may find a finite set {y1 , . . . , yn } ⊂ Y , so that Y ⊂ ni=1 U (yi , ε). Then {f −1 (U (yi , ε)) : i = 1, . . . , n} is a cover of X consisting of sets in Σα+1 (F).
5.1 Generation of sets and functions
139
Using Lemma 5.4(f), we can find a partition {A1 , . . . , An } of X consisting of sets in ∆α+1 (F) so that Ai ⊂ f −1 (U (yi , ε)), i = 1, . . . , n. Then the function x ∈ Ai , i = 1, . . . , n,
g(x) = yi ,
is Σα+1 (F)-measurable and satisfies ρ(f (x), g(x)) < ε, x ∈ X. The proof of Theorem 5.9 below closely follows the proof of Theorem 24.3 in [262]; however, we include the following lemma. Lemma 5.8. Let F be an algebra of sets in a set X and let 0 < ε < ε0 . Let f, g : X → Y be mappings from a set X to a separable metric space (Y, ρ) such that ρ(f (x), g(x)) < ε, x ∈ X. Let fn , gn : X → Y , n ∈ N, be Σγn (F)-measurable mappings such that fn → f , gn → g and γn ∈ (1, ω1 ). Then there exist Σγn (F)-measurable mappings hn : X → Y , n ∈ N, so that hn → g and ρ(fn (x), hn (x)) < ε0 for each x ∈ X and n ∈ N. Proof. Given the objects as in the lemma, for n ∈ N we set A1n := {x ∈ X : ρ(fn (x), gn (x)) < ε0 },
A2n := {x ∈ X : ρ(fn (x), gn (x)) > ε}.
By the separability of the space Y we get that the sets A1n , A2n are in Σγn (F). Indeed, we select countably many open sets Uk , Vk , k ∈ N, in Y such that ∞ [
(Uk × Vk ) = {(y1 , y2 ) ∈ Y × Y : ρ(y1 , y2 ) < ε0 }.
k=1
Hence A1n =
∞ [
(fn−1 (Uk ) ∩ gn−1 (Vk ))
k=1
is contained in Σγn (F). We similarly argue for the set A2n . Thus we may use Proposition 5.4(f) to find disjoint sets Bn1 , Bn2 ∈ Σγn (F) such that n ∈ N. Bn1 ⊂ A1n , Bn2 ⊂ A2n , and Bn1 ∪ Bn2 = X, By setting ( gn hn := fn
on Bn1 , on Bn2 ,
n ∈ N,
we get Σγn (F)-measurable mappings satisfying our requirements. Theorem 5.9. Let F be an algebra of sets in a set X and let Y be a separable metrizable space. Let Φ1 stand for the family of all Σ2 (F)-measurable mappings from X to Y and, for α ∈ (1, ω1 ), let Φα be defined from Φ1 as in Definition 5.5. Then, for each α ∈ (0, ω1 ) and f : X → Y , the following assertions are equivalent:
140
5 Perfect classes of functions and representation of affine functions
(i) f ∈ Φα , (ii) f is Σα+1 (F)-measurable. Proof. We prove (i) =⇒ (ii) by transfinite induction. If α = 1, the assertion holds by the definition of Φ1 . We assume its validity for all β < α, where α ∈ (1, ω1 ). If f ∈ Φα , then f = limn→∞ fn , where fn ∈ ΦS αn for some αn < α, n ∈ N. Given an S ∞ U = open set U ⊂ Y , we write U = ∞ m m=1 Um , where the sets Um are open. m=1 Then ∞ [ ∞ \ ∞ [ f −1 (U ) = fk−1 (Um ) m=1 n=1 k=n
and the sets fk−1 (Um ) ∈ Παk +1 (F) by the inductive assumption. Thus f −1 (U ) ∈ Σα+1 (F). For the proof of (ii) =⇒ (i) we use transfinite induction again. The case α = 1 follows from the definition. Assume that α ∈ (1, ω1 ) and the assertion holds for all β < α. Let f : X → Y be a Σα+1 (F)-measurable function. We assume first that f : X → Y has only finitely many values; that is, there exist pairwise disjoint sets Ai ⊂ X and points yi ∈ Y , i = 1, . . . , n, such that Ai ∈ ∆α+1 (F) and f = yi on Ai , i = 1, . . . , n. Assume first that α is isolated, that is, α = α0 + 1 for some α0 ∈ [1, ω1 ). According to Proposition 5.4(g), there exist sequences {Aki }∞ k=1 of sets in ∆α (F), i = 1, . . . , n, such that cAk → cAi . We may assume that {Ak1 , . . . , Akn } is a disjoint family for each i k ∈ N. (Otherwise we would take B1k = Ak1 and Bjk = Akj \ (Ak1 ∪ · · · ∪ Akj−1 ), j = 2, . . . , n.) Then fk (x) = yi , x ∈ Aki , i = 1, . . . , n, k ∈ N, are Σα (F)-measurable mappings that converge pointwise to f . By the inductive assumption, the mappings fk are contained in Φα0 , and so f ∈ Φα . Assume that α is limit. In this case we use Proposition 5.4(h) and find sequences {Aki }∞ k=1 of sets in ∆αk (F), i = 1, . . . , n, where αk < α for every k ∈ N, such that cAk → cAi , i = 1, . . . , n. The rest of the argument is analogous to that above. i Assume now that f : X → Y is an arbitrary Σα+1 (F)-measurable mapping. Since Y is separable and metrizable, we can fix on Y a compatible metric ρ such that (Y, ρ) is precompact. Hence we may find finite sets Y k := {y1k , . . . , ynk k } ⊂ Y , k ∈ N, so S k U (yik , 2−k ). that Y k ⊂ Y k+1 and Y ⊂ ni=1 −1 We fix k ∈ N. Then {f (U (yik , 2−k )) : i = 1, . . . , nk } is a cover of X consisting of sets in Σα+1 (F). Using Proposition 5.4(f) we find a partition {Aki : i = 1, . . . , nk } of X consisting of sets in ∆α+1 (F) so that Aki ⊂ f −1 (U (yik , 2−k )), i = 1, . . . , nk . Then the function f k (x) = yi , x ∈ Aki , i = 1, . . . , nk ,
5.1 Generation of sets and functions
141
is Σα+1 (F)-measurable. According to the reasoning above, we may find mappings fnk ∈ Φαnk so that fnk → f k as n → ∞ and αnk < α. By implication (i) =⇒ (ii), fnk ∈ Σαnk +1 (F),
n, k ∈ N.
(5.1)
If x ∈ X, from ρ(f k (x), f (x)) < 2−k it follows that ρ(f k (x), f k+1 (x)) < 2−k+1 . Now we employ Lemma 5.8 to replace {fnk } by {hkn } in such a way that • h1 = f 1 , n ∈ N, n n •
−k+2 , x ∈ X, n, k ∈ N, ρ(hkn (x), hk+1 n (x)) < 2
•
hkn is Σγnk +1 (F)-measurable for some γnk < α, n, k ∈ N,
hkn → f k , k ∈ N. To do this, we proceed inductively. First we set •
h1n := fn1 ,
n ∈ N,
and having hjn : X → Y defined for all n ∈ N and j = 1, . . . , k − 1, we use Lemma 5.8 to get mappings hkn : X → Y , n ∈ N, so that k −(k−1)+2 ρ(hk−1 , n (x), hn (x)) < 2
x ∈ X,
n ∈ N,
and hkn is Σγnk +1 (F)-measurable, where γnk = max{γn(k−1) , αnk }, n ∈ N. Having done this, we set hk (x) := hkk (x),
x ∈ X,
k ∈ N.
Then the mappings hk are Σγkk +1 (F)-measurable, where γkk < α, and hk → f . By the inductive assumption, hk ∈ Φγkk , k ∈ N. Hence f ∈ Φα , as required. Theorem 5.10. Let F be an algebra of sets in a set X and let Φ1 stand for the family of all Σ2 (F)-measurable functions from X to R and, for α ∈ (1, ω1 ), let Φα be the abstract Baire class defined from Φ1 . Let α ∈ (0, ω1 ). Then Φα is a vector space containing constant functions that is closed with respect to uniform convergence, finite minima, multiplications and inversions (if defined). Proof. Let α ∈ (0, ω1 ) be given. By Theorem 5.9, it is enough to show that the family M of all Σα+1 (F)-measurable functions on X possesses all the required properties. Obviously, M contains constant functions. Further, if f, g are Σα+1 (F)-measurable functions on X, the mapping ψ : X → R2 defined as ψ(x) = (f (x), g(x)),
x ∈ X,
is Σα+1 (F)-measurable. (This follows from the fact that any open set in R2 is a countable union of open rectangles.) If ϕ : R2 → R is any of the following mappings (a, b) 7→ ab,
(a, b) 7→ a + b,
(a, b) 7→ a ∨ b,
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5 Perfect classes of functions and representation of affine functions
then ϕ is continuous. Hence ϕ ◦ ψ is Σα+1 (F)-measurable, and thus M is stable with respect to addition, multiplication and finite minima. Using the mappings t 7→ ct,
1 t 7→ , t
t ∈ R,
t ∈ R \ {0},
we get that cf ∈ M provided f ∈ M and c ∈ R, and f1 ∈ M provided f ∈ M and f does not attain 0 on X. Let {fn } be a sequence of functions in M that converges uniformly to f . Without loss of generality we may assume that supx∈X |f (x) − fn (x)| < n1 , n ∈ N. Then, for any c ∈ R, we get {x ∈ X : f (x) > c} =
∞ [ ∞ [ k=1 n=k
1 {x ∈ X : fn (x) > c + }. k
Thus f is Σα+1 (F)-measurable and the proof is complete.
5.2
Baire and Borel sets
Notation 5.11. If X is a topological space, we write G(X) for the sublattice of all open subsets of X. We recall the following notions from Subsection A.2.B. A subset A of X is called a zero set if A = f −1 ({0}) for a continuous real-valued function f on X. It is clear that such a function f can be chosen with values in [0, 1]. We recall that a subset A of a topological space X is Fσ if A can be written as a countable union of closed sets. The complement of an Fσ set is called a Gδ set. Lemma 5.12. Let F be a sublattice of sets in a set X. Then the algebra generated by F consists of all finite disjoint unions of differences of sets from F. Proof. Let B denote the family of all finite disjoint unions of differences of sets from F and let A S be the algebra generated by F. Obviously, B ⊂ A. Let B := ni=1 (Fi \Hi ), where the sets Fi , Hi are contained in F and {Fi \Hi : i = 1, . . . , n} is a disjoint family. Since F is a sublattice, we may assume that Hi ⊂ Fi for each i = 1, . . . , n. Then n \ [ \ [ Hi \ X \ B = (Hi ∪ (X \ Fi )) = Fi , i=1
I⊂{1,...,n}
i∈I
i∈{1,...,n}\I
where the union on the right-hand side is disjoint. Indeed, let I, J ⊂ {1, . . . , n} be distinct sets. Assume that j ∈ I \ J. If \ [ \ [ x ∈ Hi \ Fi ∩ Hi \ Fi , i∈I
i∈{1,...,n}\I
i∈J
i∈{1,...,n}\J
5.2 Baire and Borel sets
143
then x ∈ Hj ⊂ Fj
and
[
x∈ /
Fi .
i∈{1,...,n}\J
Hence x ∈ / Fj , a contradiction. Thus X \ B ∈ B and B is stable with respect to complements. Obviously, B is stable with respect to finite unions of disjoint sets. Let {Fi \ Hi : i ∈ I} and {Gj \ Vj : j ∈ J} be disjoint finite families and Fi , Hi , Gj , Vj ∈ F. Then [
(Fi \ Hi ) ∩
i∈I
[
(Gj \ Vj ) =
j∈J
[
(Fi \ Hi ) ∩ (Gj \ Vj )
(i,j)∈I×J
=
[
(Fi ∩ Gj ) \ (Hi ∪ Vj ),
(i,j)∈I×J
where the last set is a finite disjoint union of differences of sets from F. Thus intersection of any pair of sets from B is again in B, and, consequently, B is stable with respect to finite intersections. Thus B is an algebra and A ⊂ B. Definition 5.13 (Baire and Borel sets). We consider the following families of subsets of X. (a) The algebra Bas(X) generated by zero sets. By Lemma 5.12, Bas(X) =
n [
(Fi \ Hi ) : Fi , Hi is a zero set in X, n ∈ N .
i=1
(b) The algebra Bos(X) generated by closed subsets of X. As above, Bos(X) =
n [
(Fi \ Hi ) : Fi , Hi closed in X, n ∈ N .
i=1
(c) If we start the Borel hierarchy as defined in Section 5.1 from the sublattice G(X) of all open subsets of X, for metrizable spaces we get the standard Borel classes Σ0α (X) and Π0α (X) as defined in Definition A.113. We denote as Σ0α (G(X)) and Π0α (G(X)) the families obtained by this procedure. We show below its relation to the families defined in (a), (b). We just mention that a set A belongs to Σ02 (G(X)) if and only if A = F ∪ G, where F is Fσ and G is open. For both families (a) and (b) above we consider the abstract Borel classes defined in Section 5.1 and call them the sets of additive, or multiplicative, Borel class α and the sets of additive, or multiplicative, Baire class α for α < ω1 , respectively.
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5 Perfect classes of functions and representation of affine functions
Lemma 5.14. Let X be a topological space. Then the following assertions hold. (a) Σα (Bas(X)) ⊂ Σα (Bos(X)), α ∈ (0, ω1 ). S (b) The family α<ω1 Σα (Bas(X)) is the σ-algebra of all Baire sets in X and the S family α<ω1 Σα (Bos(X)) is the σ-algebra of all Borel sets in X. (c) If A is a subset of a normal space, then A ∈ ∆2 (Bas(X)) if and only if A is both Fσ and Gδ . (d) If X is metrizable, then (d1) Σα (Bas(X)) = Σα (Bos(X)), α ∈ (0, ω1 ), (d2) Σα (Bos(X)) = Σ0α (G(X)) = Σ0α (X), α ∈ (1, ω1 ). Proof. Assertion (a) follows from the fact that Bas(X) ⊂ Bos(X) and (b) is obvious. is both Fσ and Gδ in any For the proof of (c), notice that any set in ∆S2 (Bas(X)) T ∞ F = topological space. If X is normal and A = ∞ n n=1 Gn , where Fn ⊂ X n=1 closed, Gn ⊂ X open, n ∈ N, then there exist continuous functions fn : X → [0, 1] so that ( 1 on Fn , fn = n ∈ N. 0 on X \ Gn , Then fn → cA , and hence A=
∞ \ ∞ n ∞ ∞ [ 1o \ [ n 1o x ∈ X : fn (x) ≥ = x ∈ X : fn (x) > 2 2
k=1 n=k
k=1 n=k
is in ∆2 (Bas(X)). Assertions (d1) and (d2) follow from the fact that any closed subset of a metrizable space is a zero set. This concludes the proof. Lemma 5.15. Let X, Y be compact topological spaces and Φ : Y → 2X be a mapping such that Φ(y) is nonempty compact for every y ∈ Y and Φ−1 (F ) = {y ∈ Y : Φ(y) ∩ F 6= ∅} is closed in Y for each closed set F in X. Let {Fn } be a sequence of closed subsets of X. Then there exists a mapping φ : Y → X such that (a) φ(y) ∈ Φ(y) for every y ∈ Y , (b) φ−1 (Fn ) ∩ φ−1 (X \ Fn ) = ∅ for each n ∈ N, and (c) φ−1 (Fn ) ∈ Bos(Y ) for each n ∈ N. Proof. Set F0 := X. We will use induction to construct a sequence {Φn }n≥0 of mappings from Y into X such that, for every n ≥ 0,
5.2 Baire and Borel sets
145
(d) Φn (y) is a nonempty compact set for each y ∈ Y , (e) Φn+1 (y) ⊂ Φn (y) ⊂ Φ0 (y) for each y ∈ Y , (f) (Φn )−1 (F ) ∈ Bos(Y ) for every closed set F ⊂ X, (g) (Φn )−1 (Fn ) ∩ (Φn )−1 (X \ Fn ) = ∅. By the assumptions on Φ, we fulfill all the conditions needed for Φ0 by setting Φ0 := Φ. Suppose that Φk satisfying the required conditions have been constructed for all k ≤ n. Set A := (Φn )−1 (Fn+1 ). Condition (f) ensures that A ∈ Bos(Y ). Let Φn+1 be defined as ( Φn (y) ∩ Fn+1 , y ∈ A, Φn+1 (y) := Φn (y), y ∈ Y \ A. Conditions (d), (e) and (g) are obviously satisfied. For a closed subset F of X we have (Φn+1 )−1 (F ) = {y ∈ Y : Φn (y) ∩ Fn+1 ∩ F 6= ∅} ∩ A ∪ {y ∈ Y : Φn (y) ∩ F 6= ∅} ∩ (Y \ A) , which yields that condition (f) holds. Thus the construction is complete. Set ∞ \ b φ(y) := Φn (y), y ∈ Y. n=0
b Since {Φn (y)} is a decreasing sequence of nonempty compact subsets of X, φ(y) is a nonempty compact set contained in Φ(y) for every y ∈ Y . Let φ : Y → X be some b selection for φ. Since φ−1 (Fn ) ⊂ (Φn )−1 (Fn )
and
φ−1 (X \ Fn ) ⊂ (Φn )−1 (X \ Fn ),
condition (b) is satisfied and φ−1 (Fn ) = (Φn )−1 (Fn ). Thus (c) follows from condition (f). Theorem 5.16. Let ϕ : X → Y be a continuous mapping of a compact space X onto a compact space Y , B ⊂ Y and α ∈ (0, ω1 ). Then ϕ−1 (B) ∈ Σα (Bos(X)) if and only if B ∈ Σα (Bos(Y )). Proof. If B ∈ Σα (Bos(Y )), then ϕ−1 (B) ∈ Σα (Bos(X)) by the continuity of ϕ. Conversely, let A := ϕ−1 (B) be contained in Σα (Bos(X)). It follows by a straightforward transfinite induction argument that it is possible to find a countable family F in Bos(X) such that A ∈ Σα (F). By Lemma 5.12, any set in F can be written as a finite union of differences of closed sets. We enumerate all these closed sets as a single sequence {Fn } and use Lemma 5.15 to get a mapping φ : Y → X such that φ(y) ∈ ϕ−1 (y), y ∈ Y , and φ−1 (Fn ) ∈ Bos(Y ) for each n ∈ N. (We recall that ϕ(F )
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5 Perfect classes of functions and representation of affine functions
is a closed set in Y for each closed F ⊂ X and hence the assumptions of Lemma 5.15 are satisfied.) Since B = φ−1 (A), B ∈ Σα ({φ−1 (F ) : F ∈ F}) ⊂ Σα (Bos(Y )). This concludes the proof. Theorem 5.17. Let A be a Baire subset of a compact space X and α ∈ (0, ω1 ). Then A ∈ Σα (Bas(X)) if and only if A ∈ Σα (Bos(X)). Proof. Assume that A is a Baire subset of a compact space X and α ∈ (0, ω1 ). Obviously, we need to prove that A ∈ Σα (Bas(X)) provided A ∈ Σα (Bos(X)). Since A is a Baire set, we can find a countable family {fn : n ∈ N} of continuous functions on X with values in [0, 1] such that A belongs to the σ-algebra generated by {fn−1 ((0, 1]) : n ∈ N} (see Proposition A.48). We define a mapping ϕ : X → [0, 1]N as ϕ(x) = {fn (x)}n∈N , x ∈ X. Then ϕ is a continuous mapping of X onto a compact metrizable space Y := ϕ(X). Further, ϕ−1 (ϕ(A)) = A. By Theorem 5.16, ϕ(A) ∈ Σα (Bos(Y )). Since Y is a metrizable space, Σα (Bos(Y )) = Σα (Bas(Y )). Hence A = ϕ−1 (ϕ(A)) ∈ Σα (Bas(X)). This concludes the proof.
5.3
Baire and Borel mappings
Definition 5.18 (Baire and Borel measurable mappings and functions). Let us consider the following classes of mappings from X to a topological space Y : (a) Let Baf1 (X, Y ) be the family of all Σ2 (Bas(X))-measurable mappings from X to Y and for α ∈ (1, ω1 ), let Bafα (X, S Y ) := (Baf1 (X, Y ))α (see Section 5.1 and Remark 5.6). We call the elements of α<ω1 Bafα (X, Y ) Baire measurable mappings. (b) Let Bof1 (X, Y ) be the family of all Σ2 (Bos(X))-measurable mappings from X to Y and for α S ∈ (1, ω1 ), as above we set Bofα (X, Y ) := (Bof1 (X, Y ))α . We call the elements of α<ω1 Bofα (X, Y ) Borel measurable mappings. S Obviously, α<ω1 Bafα (X, Y ) is the family of all Baire measurable mappings and S Bof (X, Y ) is the family of all Borel measurable mappings as defined in Defα α<ω1 inition A.44. In the case when Y = R we speak about Baire and Borel measurable functions. The following theorem justifies the term “measurability” in the definition above. Theorem 5.19. Let f be a mapping from a topological space X to a separable metrizable space Y and α ∈ (0, ω1 ). Then the following assertions hold:
5.3 Baire and Borel mappings
147
(a) f ∈ Bafα (X, Y ) if and only if f is Σα+1 (Bas(X))-measurable. (b) f ∈ Bofα (X, Y ) if and only if f is Σα+1 (Bos(X))-measurable. Proof. This follows from Theorem 5.9. Definition 5.20 (Baire mappings). Let α ∈ (0, ω1 ). A mapping f : X → Y between topological spaces X and Y is said to be of Baire class α if f ∈ (C(X, Y ))α , where C(X, Y ) denotes the set of all continuous mappings from X to Y . We write B α (X, Y ) for the family of all mappings of Baire class α. If Y = R, we simply write B α (X) and speak about functions of Baire class α. Theorem 5.21. Let X be a compact space, Y be a separable metrizable space and f : X → Y be a function such that f −1 (U ) is a Baire subset of X for every open U ⊂ Y . Then the following assertions hold: (a) There exists β ∈ (0, ω1 ) such that f is Σβ+1 (Bas(X))-measurable. (b) For α ∈ (0, ω1 ), f ∈ Bafα (X, Y ) if and only if f ∈ Bofα (X, Y ). Proof. Let f : X → Y be a function as in the theorem and let {Un : n ∈ N} be a base of open sets in Y . If αn ∈ (0, ω1 ) are chosen in such a way that f −1 (Un ) ∈ Σαn (Bas(X)), n ∈ N, then f is Σα+1 (Bas(X))-measurable, where α = sup{αn : n ∈ N}. By Theorem 5.17, the function f is Σα+1 (Bas(X))-measurable if and only if f is Σα+1 (Bos(X)))-measurable. Thus Theorem 5.19 concludes the proof. Corollary 5.22 (Lebesgue–Hausdorff). Let f : X → R be a function on a compact space X such that f −1 (U ) is a Baire subset of X for every open U ⊂ R. Let α ∈ (0, ω1 ). Then there exists β ∈ (0, ω1 ) such that f is Σβ+1 (Bas(X))-measurable and the following assertions are equivalent: (i) f ∈ Bafα (X, R), (ii) f ∈ Bofα (X, R), (iii) f ∈ B α (X). Proof. By Theorem 5.21, f is Σβ+1 (Bas(X))-measurable for some β ∈ (0, ω1 ) and assertions (i)–(ii) are equivalent. If α = 1 and f : X → R is a Σ2 (Bas(X))measurable function, Theorem A.124 yields that f ∈ B 1 (X). Thus f ∈ Bafα (X, Y ) if and only if f ∈ B α (X, Y ) for any α ∈ [1, ω1 ), by Theorems 5.19 and 5.9. Theorem 5.23. Let f : X → R be a function on a compact space X such that f −1 (U ) is a Baire subset of X for every open U ⊂ R. Then the following assertions are equivalent: (i) f ∈ Baf1 (X, R), (ii) f has the point of continuity property,
148
5 Perfect classes of functions and representation of affine functions
(iii) f is fragmented, (iv) the set {x ∈ F : f |F discontinuous at x} is meager in F for each F ⊂ X, (v) f ∈ B 1 (X). Proof. It follows from Theorem A.121 that (i) =⇒ (ii) ⇐⇒ (iii) ⇐⇒ (iv). By Theorem A.124, (i) ⇐⇒ (v). To finish the proof we show that (iii) =⇒ (v). Since f −1 (U ) is a Baire set in X for each U ⊂ R open, Corollary 5.22 gives that f ∈ B α (X) for some α < ω1 . It is easy to see that there exists a family Φ := {fn : n ∈ N} of continuous functions on X such that f ∈ Φα (see Definition 5.5 and cf. Proposition A.48). If ϕ : X → RN is defined as ϕ(x) = {fn (x)}n∈N ,
x ∈ X,
then Y := ϕ(X) is a metrizable compact space. We can define a function g : Y → R by g(y) := f (x),
x ∈ ϕ−1 (y), y ∈ Y,
and get a function satisfying f = g ◦ ϕ. We aim to prove that g is fragmented. Let H ⊂ Y be a given closed set and ε > 0. Using Proposition A.38 we find a minimal closed F ⊂ X such that ϕ(F ) = H. Let U ⊂ F be a relatively open set such that diam f (U ) < ε. Then V := H \ ϕ(F \ U ) is a nonempty relatively open set in H such that diam g(V ) < ε. Hence g is fragmented. By Theorem A.127, g is a Baire-one function on Y , and thus f is a Baire-one function on X. This finishes the proof. Theorem 5.24. Let f : X → R be a Baire function on a compact space X. Then the following assertions are equivalent: (i) f is a limit of an increasing sequence of continuous functions, (ii) f is lower semicontinuous. Proof. For the proof of the nontrivial implication (ii) =⇒ (i), notice that A := f −1 ((c, ∞)) is an open Baire set in X for each c ∈ R. Since any Baire subset of a compact space is Lindel¨of, A is a cozero set. By Proposition A.53, f is a limit of an increasing sequence of continuous functions.
5.4 Perfect classes of functions
5.4
149
Perfect classes of functions
The aim of this section is to capture topological properties of functions on compact spaces that are invariant with respect to compositions with perfect mappings. If C denotes a class of functions and X is a topological space, we write C(X) for the set of all functions on X that belong to C. Definition 5.25 (Perfect classes of functions). We say that a class C of functions is perfect if, given X, Y compact spaces and ϕ : X → Y continuous, then f ◦ϕ ∈ C(X) if and only if f ∈ C(Y ). Theorem 5.26. Let X, Y be compact spaces and ϕ : X → Y be a continuous surjection. Let g : Y → R and denote f := g ◦ ϕ. Then (a) f is the limit of an increasing sequence of continuous functions on X if and only if g is the limit of an increasing sequence of continuous functions on Y , (b) f is lower semicontinuous if and only if g is lower semicontinuous, (c) f ∈ B α (X) if and only if g ∈ B α (Y ), α ∈ (0, ω1 ), (d) f ∈ Bofα (X) if and only if g ∈ Bofα (Y ), α ∈ (0, ω1 ), (e) f is universally measurable if and only if g is universally measurable. Proof. The “if” parts of the statements of the theorem follow from the continuity of the mapping ϕ. Assertions (a) and (b) follow from the characterization via level sets (see Definition A.49 and Proposition A.53) and from the fact that ϕ maps closed sets in X to closed sets in Y . For the proof of (c), we need to show that g ∈ B α (Y ) provided f ∈ B α (X). Combining Corollary 5.22 with Theorem 5.19 yields that we need to show that g −1 (U ) ∈ Σα+1 (Bas(Y )) for any U ⊂ R open. By Theorem 5.16, g −1 (U ) = ϕ(f −1 (U )) ∈ Σα+1 (Bos(Y )). Since both the sets g −1 (U ) = ϕ(f −1 (U )) and
Y \ g −1 (U ) = ϕ(f −1 (R \ U ))
are continuous images of Baire sets in a compact space, they are K-analytic (see Theorem A.111(a)) and hence g −1 (U ) is a Baire set by the separation principle (see Theorem A.112). By Theorem 5.17, g −1 (U ) ∈ Σα+1 (Bas(Y )). The proof of (d) follows the same line of argument as the proof of (c); in fact, it is enough to use Theorem 5.16 together with Theorem 5.19. It remains to prove the last assertion. Similarly to the previous paragraph, we need to prove that A ⊂ Y is universally measurable if ϕ−1 (A) is universally measurable in X. Let ν be a probability measure on Y . Since the mapping ϕ] : M1 (X) →
150
5 Perfect classes of functions and representation of affine functions
M1 (Y ) is surjective, we can find a measure µ ∈ M1 (X) such that ν = ϕ] µ (see Theorem A.94). Since B := ϕ−1 (A) is µ-measurable, we can write B as a disjoint union of an Fσ set H and a µ-null set N . If we enlarge H by ϕ−1 (ϕ(H)), we may suppose that N = ϕ−1 (ϕ(N )) and H = ϕ−1 (ϕ(H)). (5.2) We find an Fσ set F ⊂ X disjoint from N such that µ(F ) = 1. By virtue of (5.2), ϕ(N ) ∩ ϕ(F ) = ∅. Hence ν(ϕ(F )) = ϕ] µ(ϕ(F )) = µ(ϕ−1 (ϕ(F ))) ≥ µ(F ) = 1. Thus ν(ϕ(N )) = 0 and A is a disjoint union of an Fσ set ϕ(H) and ν-null set ϕ(N ). Hence A is ν-measurable and the proof is finished. Corollary 5.27. The following classes of functions on topological spaces are perfect: (a) increasing limits of continuous functions, (b) lower semicontinuous and upper semicontinuous functions, (c) Baire functions of class α and Baire functions, (d) Borel measurable functions of class α and Borel functions, (e) universally measurable functions.
5.5
Affinely perfect classes of functions
Definition 5.28 (Affinely perfect classes). If C is a class of functions on compact convex sets, we say that C is affinely perfect if the following assertions hold for any compact convex set X: (AP1) for any affine continuous mapping ϕ : X → Y of X onto a compact convex set Y , a function f on Y satisfies f ◦ ϕ ∈ C(X) if and only if f ∈ C(Y ), (AP2) Ac (X) ⊂ C(X) ⊂ Abar (X) and C(X) is a subspace of Abar (X), (AP3) for any f ∈ C(X), the function fb: M1 (ext X) → R defined as fb(µ) = µ(f ),
µ ∈ M1 (ext X),
belongs to C(M1 (ext X)), (AP4) C(X) satisfies the minimum principle; that is, if f is a function from C(X) such that f ≥ 0 on ext X, then f ≥ 0 on X. Proposition 5.29. Let ϕ : X → Y be an affine continuous mapping of a compact convex set X onto a compact convex set Y . Let g be a function on Y . Then g ∈ Abar (Y ) if and only if g ◦ ϕ ∈ Abar (X).
5.5 Affinely perfect classes of functions
151
Proof. Let µ ∈ M1 (X). Proposition A.92 yields that for every bounded universally measurable function h on Y we have (ϕ] µ)(h) = µ(h ◦ ϕ).
(5.3)
We observe that r(ϕ] µ) = ϕ(r(µ)). Indeed, for any h ∈ Ac (Y ) we get h(r(ϕ] µ)) = (ϕ] µ)(h) = µ(h ◦ ϕ) = (h ◦ ϕ)(r(µ)) = h(ϕ(r(µ))). Assume now that g ∈ Abar (Y ). Then, for any µ ∈ M1 (X), we obtain µ(g ◦ ϕ) = (ϕ] µ)(g) = g(r(ϕ] µ)) = (g ◦ ϕ)(r(µ)). Thus g ◦ ϕ ∈ Abar (X). Conversely, let g ◦ ϕ be in Abar (X). By Theorem 5.26(e), the function g is universally measurable on Y . It remains to prove the validity of the barycentric formula for g. To this end, let ν be a probability measure on Y . We find a probability measure µ ∈ M1 (X) with ϕ] µ = ν. By the assumption, the function f := g ◦ ϕ satisfies the barycentric formula on X. Then (5.3) gives ν(g) = (ϕ] µ)(g) = µ(g ◦ ϕ) = µ(f ) = f (r(µ)) = g(ϕ(r(µ))) = g(r(ν)), which concludes the proof. Proposition 5.30. Let f be a bounded universally measurable function on a compact space K. Then the function If : M1 (K) → R defined by If (µ) := µ(f ),
µ ∈ M1 (K),
is universally measurable on M1 (K) and satisfies the barycentric formula. Moreover, if f belongs to any class listed in Corollary 5.27, then If belongs to the same class. Proof. Given a bounded universally measurable function on K, the function fe(λ1 , λ2 ) = λ1 (f ),
(λ1 , λ2 ) ∈ M1 (K) × M1 (K),
is universally measurable on M1 (K) × M1 (K) by Proposition 3.90. If Λ is a e ∈ probability measure on M1 (K) with the barycenter µ, we define a measure Λ 1 1 1 e e M (M (K) × M (K)) as Λ = Λ ⊗ Λ. Then r(Λ) = (µ, µ) and, by Proposition 3.90, Z e 1 , λ2 ) If (µ) = µ(f ) = fe(µ, µ) = fe(λ1 , λ2 ) dΛ(λ M1 (K)×M1 (K)
Z = M1 (K)
λ1 (f ) dΛ(λ1 ) = Λ(If ).
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5 Perfect classes of functions and representation of affine functions
Hence If satisfies the barycentric formula on M1 (K). Concerning the second part of the statement, we notice that the assertion for classes (a) and (c) of Corollary 5.27 follows from the Lebesgue dominated convergence theorem, (b) follows from Theorem A.85(b) and (e) was proved above. To prove (d), we first show that the function If is Σ2 (Bos(M1 (K)))-measurable whenever f is Σ2 (Bos(K))-measurable. Let F, H be closed subsets of K. Then IcF \H = I(cF ) − I(cF ∩H ) is Σ2 (Bos(M1 (K)))-measurable by Theorem A.85(b) and Theorem 5.10. S If A ∈ Bos(K), by Lemma 5.12 we can write A = ni=1 (Fi \ Hi ), where Hi ⊂ Fi , the sets Hi and Fi are closed, and the family {Fi \ Hi : i = 1, . . . , n} is disjoint. Then IcA =
n X
IcFi \Hi
i=1
is Σ2 (Bos(M1 (K)))-measurable. S If A ∈ ∆2 (Bos(K)), then A = ∞ n=1 An , where A1 ⊂ A2 ⊂ · · · and An ∈ Bos(K), n ∈ N. Then, for c ∈ R, the set ∞ [
{µ ∈ M1 (K) : IcA (µ) > c} =
{µ ∈ M1 (K) : µ(An ) > c}
n=1
belongs to Σ2 (Bos(M1 (K))). Since IcK\A = 1 − IcA and K \ A ∈ Σ2 (Bos(K)), we get that the set {µ ∈ M1 (K) : IcA (µ) < c} = {µ ∈ M1 (K) : IcK\A (µ) > 1 − c} is in Σ2 (Bos(M1 (K))) by the argument above. Hence IcA is Σ2 (Bos(M1 (K)))measurable. Again, by Theorem 5.10, we get that If is Σ2 (Bos(M1 (K)))-measurable for every f=
n X
ai cAi ,
(5.4)
i=1
where n ∈ N, a1 , . . . , an ∈ R, and A1 , . . . , An ∈ ∆2 (Bos(K)) are pairwise disjoint. If f is an arbitrary bounded Σ2 (Bos(K))-measurable function, using Lemma 5.7 we find functions {fn } of type (5.4) that converge uniformly to f . Then {Ifn } converges uniformly to If and thus If is Σ2 (Bos(M1 (K)))-measurable by Theorem 5.10. Thus we have proved the desired assertion for functions in Bof1 (X, R). The proof for the remaining classes now follows from Theorem 5.19 using transfinite induction. This concludes the proof. Theorem 5.31. Let H be a function space on a compact space K and f be a bounded function on K such that f |ChH K is universally measurable on ChH K.
5.5 Affinely perfect classes of functions
153
(a) If for any x ∈ K and µ ∈ M1 (ChH K) ∩ Mx (H), µ(f ) = f (x), then f is H-affine on K. (b) If moreover, for any boundary measure µ ∈ H⊥ , µ(f ) = 0, then ν(f ) = 0 for any ν ∈ H⊥ . (c) If, in addition to (a), the function f belongs to a class listed in Corollary 5.27, if f |ChH K belongs to this class. Proof. Let r : M1 (ChH K) → K be the barycentric mapping (see Section 3.3). Then r is a surjective continuous mapping. The function F : µ 7→ µ(f |ChH K ),
µ ∈ M1 (ChH K),
satisfies the barycentric formula on M1 (ChH K) due to Proposition 5.30. Pick x ∈ K and µ ∈ M1 (ChH K) ∩ Mx (H). By the assumption we have F (µ) = µ(f |ChH K ) = µ(f ) = f (x). Thus F = f ◦ r. By virtue of Theorem 5.26(e), f is universally measurable on K. To show that f is H-affine, choose x ∈ K and µ ∈ Mx (H). Let ν ∈ M+ (K) be an H-maximal measure such that µ ≺ ν (see Theorem 3.65). Then ν ∈ M1 (ChH K) (see Theorem 3.81). In Proposition 3.89, we set F1 := K and F2 := ChH K and M := {(εx , λ) ∈ M1 (K) × M1 (ChH K) : εx ≺ λ}, and find Λ ∈ M1 (M ) such that (µ, ν) ∈ M1 (K) × M1 (K) is the barycenter of Λ. By Proposition 3.90 and the hypotheses, Z Z µ(f ) = εx (f ) dΛ(εx , λ) = λ(f ) dΛ(εx , λ) = ν(f ) = f (x). M
M
Hence f is H-affine, concluding the proof of (a). For the proof of (b), let ν = ν1 − ν2 with ν1 , ν2 ∈ M1 (K) be a measure in H⊥ . We find H-maximal measures µ1 , µ2 such that νi ≺ µi , i = 1, 2. By the argument above, µi (f ) = νi (f ), i = 1, 2. The assumption yields ν1 (f ) = µ1 (f ) = µ2 (f ) = ν2 (f ), finishing the proof of (b). The last assertion (c) follows from Theorem 5.26, Proposition 5.30 and the fact that F = f ◦ r. Corollary 5.32. Let X be a compact convex set in a locally convex space and f be a bounded function on X such that, for any µ ∈ M1 (ext X), f is µ-measurable and µ(f ) = f (r(µ)). Then f satisfies the barycentric formula on X. Moreover, such a function f belongs to a class listed in Corollary 5.27 if f |ext X belongs to this class.
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5 Perfect classes of functions and representation of affine functions
Proof. If f satisfies the assumptions, then f is universally measurable on ext X = ChAc (X) (X) and we can apply Theorem 5.31. Theorem 5.33. The classes of strongly affine Baire-α functions and the class of strongly affine Baire functions are affinely perfect. Proof. The property (AP1) follows from Proposition 5.29, (AP2) is obvious, (AP3) follows from Proposition 5.30 and (AP4) from Theorem 3.86. Remarks 5.34. (a) We recall that a function f on a topological space X has the Baire property if f is measurable with respect to the σ-algebra of the sets with the Baire property. A function f on a topological space X has the restricted Baire property if, for any closed F ⊂ X, f |F has the Baire property. A set A ⊂ X is perfectly meager if A ∩ P is meager in P for every perfect P ⊂ X. To see that neither class of functions is perfect, we recall an example of W. Sierpinski who constructed in Th´eor`eme of [416] a perfectly meager set A ⊂ [0, 1] and a continuous function ϕ : [0, 1] → [0, 1] such that ϕ−1 (A) does not have the Baire property. Hence the characteristic function cA has the restricted Baire property and cA ◦ ϕ does not have even the Baire property. (b) Later on we will present several examples of classes of functions that are not affinely perfect, see Exercise 9.59, Subsections 12.3.B and 12.3.C.
5.6
Representation of H-affine functions
Definition 5.35 (Classes of H-affine functions). If f is a bounded universally measurable function on a compact space K, we recall that it can be considered as an element of (C(K))∗∗ via the identification µ 7→ µ(f ), µ ∈ M(K). If H is a function space on K and H⊥ ⊂ M(K) is its annihilator, U(K) ∩ H⊥⊥ is the space of all bounded universally measurable functions f such that µ(f ) = 0 for each µ ∈ H⊥ . For any countable ordinal α and a family of functions H ⊂ C(K), we define the functions of H-affine class α similarly as in Definition 5.5 but we consider only bounded sequences. More precisely, we set H0,b := H and having Hβ,b defined for all β < α for some α ∈ (0, ω1 ), we letSHα,b to be the space of all pointwise limits of bounded sequences of functions from β<α Hβ,b . If we take H to be the space C(K) of all continuous functions on K, (C(K))α,b is the usual space B bα (K) of all bounded Baire-α functions on K (see Section 5.3). Lemma 5.36. For a compact convex set X and α ∈ [0, ω1 ), (Ac (X))α,b = (Ac (X))α . Proof. Given a compact convex set X, denote E := Ac (X). By Proposition 4.32, any bounded affine function on X can be viewed as an element of E ∗∗ and the τX topology coincides with the w∗ -topology. Hence any pointwise convergent sequence
5.6 Representation of H-affine functions
155
{fn } of bounded affine functions on X is bounded by the uniform boundedness principle (see [403], Theorem 2.5). This concludes the proof. Definition 5.37 (Classes of affine functions). If X is a compact convex set and α ∈ [0, ω1 ), we write Aα (X) for the space (Ac (X))α and call it the functions of affine class α. If X is a symmetric compact convex set, we say that f is odd on X if f (−x) = −f (x), x ∈ X. We write Aodd (X) for the space of all odd affine functions on X. We note that an affine function f on a symmetric compact convex set is odd if and only if f (0) = 0. Definition 5.38 (Embedding of abstract H-affine functions). If H is a function space on a compact space K, let X stand for the dual unit ball BH∗ endowed with the w∗ topology and let π : BM(K) → X be the restriction mapping. Let φ : K → X be the evaluation mapping; that is, φ(x)(h) = h(x), h ∈ H, x ∈ K. We define I : U(K) ∩ H⊥⊥ → A(X) as If (s) = µ(f ),
µ ∈ π −1 (s).
We remark that φ(x) = π(εx ), x ∈ K, and φ is a homeomorphism of K into BH∗ . Further, I is an extension of the canonical embedding of H into H∗∗ . In order to obtain properties of I needed for our purposes, we have to establish several lemmas. Lemma 5.39. Let F1 , F2 be closed convex subsets of a compact convex set X such that co(F1 ∪ F2 ) = X. Let f be an affine function on X. Then the following assertions hold. (a) The function f is universally measurable (or of any class listed in Corollary 5.27, respectively) on X if and only if f is universally measurable (or of any class listed in Corollary 5.27, respectively) on F1 ∪ F2 . (b) The function f belongs to Abar (X) if and only if f |F1 ∈ Abar (F1 ) and f |F2 ∈ Abar (F2 ). Proof. We set Y := F1 × F2 × [0, 1] and define a continuous surjection ϕ : Y → X by ϕ(x1 , x2 , λ) = λx1 + (1 − λ)x2 . Let fb : Y → R be defined as fb(x1 , x2 , λ) = λf (x1 ) + (1 − λ)f (x2 ),
(x1 , x2 , λ) ∈ Y.
Then fb inherits topological properties of f |F1 ∪F2 and fb = f ◦ ϕ. Thus Corollary 5.27 yields (a).
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5 Perfect classes of functions and representation of affine functions
For the proof of the nontrivial part of (b), let f be an affine function on X such that f |Fi ∈ Abar (Fi ), i = 1, 2. By (a), f is universally measurable on X. Let x ∈ X and µ ∈ M1 (ext X) with r(µ) = x be given. Since X ⊂ co(F1 ∪ F2 ), the Milman theorem 2.43 yields ext X ⊂ F1 ∪ F2 . We set a1 := µ(F1 ) and
a2 := µ(F2 \ F1 ).
Obviously, we may assume that a1 , a2 > 0. We define µ1 := a−1 1 µ|F1
and µ2 := a−1 2 µ|F2 \F1 .
Then r(µi ) ∈ Fi , i = 1, 2, a1 + a2 = 1
and a1 r(µ1 ) + a2 r(µ2 ) = x.
By the hypotheses, µ(f ) = a1 µ1 (f ) + a2 µ2 (f ) = a1 f (r(µ1 )) + a2 f (r(µ2 )) = f (x). Thus f satisfies the barycentric formula with respect to any measure carried by ext X. By Corollary 5.32, f ∈ Abar (X). This concludes the proof. Theorem 5.40. Let H be a function space on a compact space K and X = BH∗ . Then the mapping I has the following properties: (a) I is an isometric isomorphism of U(K) ∩ H⊥⊥ onto Abar (X) ∩ Aodd (X), (b) I −1 (F ) = F ◦ φ, F ∈ Abar (X) ∩ Aodd (X), (c) I(U(K) ∩ H⊥⊥ ) = U(X) ∩ Abar (X) ∩ Aodd (X), (d) I(C(K) ∩ H⊥⊥ ) = C(X) ∩ Abar (X) ∩ Aodd (X), whenever C is any class of functions listed in Corollary 5.27, (e) if {fn } ⊂ U(K) is a bounded sequence pointwise converging to f , then Ifn → If , (f) I(Hα,b ) = Aodd (X) ∩ Aα (X), α ∈ [0, ω1 ). Proof. Obviously, I is linear and I(U(K)∩H⊥⊥ ) ⊂ Aodd (X). Further, the equalities kIf k = sup |If (s)| = s∈X
sup
|If (π(µ))| =
µ∈BM(K)
= sup |f (x)| = kf k x∈K
imply that I is an isometry. Let Ib : U b (K) → U b (BM(K) ) be defined as b (µ) = µ(f ), If
sup µ∈BM(K)
µ ∈ BM(K) .
First we check that Ib maps U b (K) into U b (BM(K) ).
|µ(f )|
5.6 Representation of H-affine functions
157
b | 1 If f ∈ U b (K), then Proposition 5.30 yields that If M (K) is universally measurable and satisfies the barycentric formula. Let C be any class of functions listed in 1 b | 1 Corollary 5.27. By Proposition 5.30, If M (K) ∈ C(M (K)) if f ∈ C(K). b | By symmetry, If satisfies analogous properties. Since 1 −M (K)
BM(K) = co(M1 (K) ∪ −M1 (K)), b ∈ Abar (BM(K) ) and If b ∈ C(BM(K) ) if f ∈ C(K). Lemma 5.39 yields that If Since b (µ) = If (π(µ)), µ ∈ BM(K) , If (5.5) Corollary 5.27 implies inclusions “⊂” in (a), (c) and (d). For the proof of (b), let F ∈ Abar (X) ∩ Aodd (X) be given and let f := F ◦ φ. Then f inherits all topological properties from F because φ is a homeomorphism. We pick a measure µ ∈ H⊥ . Without loss of generality we may assume that µ = µ1 − µ2 , µ1 , µ2 ∈ M1 (K). Then r(φ] µi ) = π(µi ), i = 1, 2 (see Proposition 4.26(c)). Then µi (f ) = µi (F ◦ φ) = φ] µi (F ) = F (r(φ] µi )) = F (π(µi )),
i = 1, 2.
(5.6)
Since π(µ1 ) = π(µ2 ), µ(f ) = 0 and f ∈ H⊥⊥ . Further, given s ∈ X, we select a measure µ ∈ π −1 (s). Let µ = aµ1 − (1 − a)µ2 , a ∈ [0, 1], µ1 , µ2 ∈ M1 (K). Since F is affine, by (5.6) we obtain If (s) = µ(f ) = a1 µ1 (f ) − a2 µ2 (f ) = a1 F (π(µ1 )) − a2 F (π(µ2 )) = F (π(µ)) = F (s). Thus I −1 (F ) = F ◦ φ and we get the inclusions “⊃” in (a), (c) and (d). Using the Lebesgue dominated convergence theorem, we get (e). By transfinite induction and (e), I(Hα,b ) = Aodd (X) ∩ Aα (X) for α ∈ [0, ω1 ), which finishes the proof. Corollary 5.41. Let H be a function space on a compact space K and X = S(H). Given f ∈ U(K) ∩ H⊥⊥ , let If denote the function s 7→ µ(f ),
µ ∈ π −1 (s),
s ∈ X.
Then I is an isometric isomorphism between U(K) ∩ H⊥⊥ and Abar (X) such that I = Φ on H and I preserves natural order of functions. Its inverse is given by I −1 F = F ◦ φ,
F ∈ Abar (X).
(5.7)
Further, for any class of functions C listed in Corollary 5.27, If ∈ C(X) if and only if f ∈ C(K).
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5 Perfect classes of functions and representation of affine functions
Proposition 5.42. Let H be a function space on a compact space K. Then the following assertions hold: (a) U(K) ∩ H⊥⊥ ∩ B 1 (K) = H1,b , (b) If f ∈ U(K) ∩ H⊥⊥ is lower semicontinuous, then f = sup{h ∈ H : h < f } and this family is up-directed. Proof. Let X be the state space of H, Φ : H → Ac (X) be the isometry into and I be as in Corollary 5.41. For the proof of (a) we notice that If ∈ B 1 (X) ∩ Abar (X). By Theorem 4.24, If ∈ A1 (X); that is, there exists a bounded sequence {hn } ⊂ Ac (X) pointwise converging to If . Since Φ(H) is dense in Ac (X), we may assume that hn ∈ Φ(H), n ∈ N. Then Φ−1 hn → f , and f ∈ H1 . Similarly we verify (b). Indeed, If is lower semicontinuous and affine provided f ∈ H⊥⊥ is lower semicontinuous. By Proposition 4.12, If = sup{h ∈ Ac (X) : h < If }, where this family is up-directed. Thus the set {h ∈ H : h < f } is up-directed as well and its supremum equals f . Lemma 5.43. Let H be a function space on a compact space K. If f, −g are upper semicontinuous bounded functions on K, then the following assertions are equivalent: (i) there exists h ∈ H with f < h < g, (ii) µ(f ) < ν(g) for any pair of measures µ, ν ∈ M1 (K) with µ − ν ∈ H⊥ . Proof. The implication (i) =⇒ (ii) is obvious. To show the converse implication, let f, −g be upper semicontinuous functions on K satisfying (ii). Then f < g. Define functions F and G on the state space X := S(H) as follows: ( f ◦ φ−1 on φ(K), F := min f (K) on S(H) \ φ(K), and
( g ◦ φ−1 on φ(K), G := max g(K) on S(H) \ φ(K).
Then F , −G are upper semicontinuous functions on X with F < G. Let µ1 , µ2 ∈ M1 (X) with r(µ1 ) = r(µ2 ) be given. Let νi ∈ M+ (X) be maximal such that µi |X\φ(K) ≺ νi ,
i = 1, 2.
Then λi := µi |φ(K) + νi ,
i = 1, 2,
5.7 Exercises
159
are measures in M1 (φ(K)) and satisfy r(λ1 ) = r(λ2 ). By Proposition 4.26(c), (φ−1 )] λ1 − (φ−1 )] λ2 ∈ H⊥ . By the assumption, µ1 (F ) ≤ λ1 (F ) = λ1 (f ◦ φ−1 ) = ((φ−1 )] λ1 )(f ) < ((φ−1 )] λ2 )(g) = λ2 (g ◦ φ−1 ) = λ2 (G) ≤ µ2 (G). Lemma 4.11 provides a function H ∈ ((H∗ , w∗ )∗ + R) |X = Φ(H)|X such that F < H < G. Then h := H ◦ φ ∈ H and f < h < g.
5.7
Exercises
Exercise 5.44. For every α ∈ (1, ω1 ) there exists a Baire subset A of a normal space Xα such that A is of additive Borel class 1 and A is not of additive Baire class α + 1. Hint. If denotes the Euclidean topology of the real line R, let A ⊂ R be a set in Πα (Bas(R, )) \ Σα (Bas(R, )) (see [262], Theorem 22.4). Let τ be the topology on R such that its open sets are of the form U ∪ V , where U is -open and V is any subset of R \ A. It is easy to see that (R, τ ) is a Hausdorff regular space. It is even a normal space. Indeed, let U be an open cover of (R, τ ). Since (A, τ ) = (A, ), we can find a countable open family U 1 ⊂ U covering A. Then V = U 1 ∪ {{x} : x ∈ R \ A} is a σ-discrete refinement of U, and thus (R, τ ) is paracompact (see Theorem 5.1.12 in [169]). Since any paracompact space is normal (see Theorem 5.1.5 in [169]), the claim follows. Further, since τ is finer that , A is a Baire subset of (R, τ ). The set A is also τ -closed. We claim that A ∈ / Σα (Bas(R, τ )). To see this, we notice that for any τ -open V ⊂ R there exists an -open set W ⊂ R such that V ∩ A = W ∩ A and W ⊂ V . Thus it follows that if A were in Σβ (Bas(R, τ )) for some β < α, then A would be in Σβ (Bas(R, )), which is not the case. Exercise 5.45. There exists a normal space X such that for each α ∈ (1, ω1 ) there exists a Baire set A ⊂ X such that A ∈ Σ1 (Bos(X)) \ Σα+1 (Bas(X)). Hint. For every α ∈ (1, ω1 ), we construct Xα as in Exercise 5.44. Then the topological sum of these spaces is the space with the required properties.
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5 Perfect classes of functions and representation of affine functions
Exercise 5.46. Let X be a normal space and F1 , . . . , Fn be its closed subsets. Then βX βX βX F1 ∩ · · · ∩ Fn = F1 ∩ · · · ∩ Fn . βX
Hint. First we show the following claim: If F1 , F2 ⊂ X are closed and F1 ∩ βX F2 6= ∅, then F1 ∩ F2 6= ∅. To show this assume that F1 ∩ F2 = ∅. By Tietze’s theorem A.26, there exists a bounded continuous function f on X such that f = 0 on F1 and f = 1 on F2 . Let fb ∈ C(βX) be the continuous extension of f . Then F1
βX
βX ⊂ {x ∈ βX : fb(x) = 0} and F2 ⊂ {x ∈ βX : fb(x) = 1},
βX
βX
and thus F1 ∩ F2 = ∅, a contradiction. This proves the claim. βX βX Now let F1 , F2 ⊂ X be closed with x ∈ F1 ∩ F2 . Let U 3 x be an open set. Then x ∈ F1 ∩ (X ∩ U
βX
βX
)
∩ F2 ∩ (X ∩ U
βX
βX
)
.
By the claim, F1 ∩ F2 ∩ (X ∩ U
βX
) 6= ∅.
βX
Since U is arbitrary, x ∈ F1 ∩ F2 . βX By induction we prove the assertion for finitely many sets in X. If x ∈ F1 ∩ βX · · · ∩ Fn+1 , where F1 , . . . , Fn+1 are closed in X, then x ∈ F1 ∩ · · · ∩ Fn
βX
∩ Fn+1
and thus x ∈ F1 ∩ · · · ∩ Fn+1
βX
βX
,
.
Exercise 5.47. Let (NN , ) denote the space NN endowed with the usual product topology and let X = NN ∪ {p} equipped with the following topology τ : the points of NN are discrete and a base of neighborhoods of the point p is formed by the sets {p} ∪ (NN \ F ),
F ⊂ NN is closed and discrete in (NN , ).
Prove the following assertions: (a) X is a regular space. (b) If ϕ : NN → 2X is defined as ϕ(σ) = {σ, p}, σ ∈ NN , then ϕ is a usco mapping with ϕ(NN ) = X (see Section A.4). (c) X is a normal space. T S∞ (d) X is a Kσδ subset of βX (that is, X = ∞ n=1 k=1 Knk for some compact sets Knk in βX), and thus it is a K-analytic subset of βX.
5.7 Exercises
161
(e) An infinite set K ⊂ X is compact if and only if K = F ∪{p}, where F ⊂ (NN , ) is relatively compact. (f) X is not σ-compact in βX. (g) X ∈ Bos(βX) and X is not a Baire subset of βX. Hint. Assertions (a) and (b) are easy to verify. Since X is the image of a Lindel¨of space under a usco mapping, X is Lindel¨of (see Proposition A.108). Since any regular Lindel¨of space is normal (see Theorem 3.8.2 in [169]), (c) follows. To verify (d), let s ∈ N
Us := {σ ∈ NN : σ||s| = s}, and
Is = ϕ(Us ) = {σ ∈ NN : σ||s| = s} ∪ {p},
s ∈ N
If s, t ∈ N
βX
βX
∩ It
We claim that
= Is ∩ It ∞ [ \
X=
βX
Is
βX
= {p}.
(5.8)
.
(5.9)
n=1 |s|=n
Indeed, let x ∈ βX \ X be contained in the right-hand side of (5.9). For each n ∈ N βX we select sn ∈ Nn such that x ∈ Isn . By (5.8), there exists σ ∈ NN such that n σ|n = s . βX βX Let U 3 x and V ⊃ ϕ(σ) be open sets in βX such that U ∩V = ∅. By the upper semicontinuity of ϕ there exists n ∈ N such that ϕ(Uσ|n ) ⊂ V ∩ X. Then βX
ϕ(Uσ|n )
= Iσ|n βX
βX
⊂V
βX
.
But this contradicts the fact that x ∈ Isn . Hence (5.9) holds and X is Kσδ set in βX. By Theorem A.111(a), X is K-analytic. For the proof of (e), let F ⊂ (NN , ) be relatively compact. If U ⊂ X is an open set containing p, then X \ U intersects F in a finite set. Hence F ∪ {p} is compact in X. Conversely, let K ⊂ X be an infinite compact set. Obviously, p ∈ K and K \ {p} contains no closed discrete set in (NN , ) (otherwise K would not be τ -compact). Hence K \ {p} is relatively compact in (NN , ). Since (NN , ) is not σ-compact, (f) follows from (e). Finally, X is a union of a discrete set and a singleton, and thus X ∈ Bos(βX). Assume that S X is a Baire subset of βX. By Theorem 5.17, X ∈ Σ1 (Bas(βX)), and thus X = ∞ n=1 Kn for some compact sets Kn ⊂ βX, n ∈ N. But this is impossible by (f).
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5 Perfect classes of functions and representation of affine functions
Exercise 5.48. Let X be a completely regular space and α ∈ (1, ω1 ). Prove that the following assertions are equivalent: (i) X ∈ Σα (Bos(cX)) for any compactification cX of X, (ii) X ∈ Σα (Bos(cX)) for some compactification cX of X, ˇ (iii) X ∈ Σα (Bos(βX)) for the Cech–Stone compactification βX of X. Hint. Obviously, (i) =⇒ (ii). If X ∈ Σα (Bos(cX)) for some compactification cX of X, let ϕ : βX → cX be the mapping with ϕ−1 (cX \ X) = βX \ X (use Theorem A.37). By the continuity of ϕ, X ∈ Σα (Bos(βX)). Thus (ii) =⇒ (iii). If X ∈ Σα (Bos(βX)) and cX is a compactification of X, then there exists a mapping ϕ : βX → cX such that ϕ−1 (cX \ X) = βX \ X (use Theorem A.37). By Theorem 5.16, X ∈ Σα (Bos(cX)). Hence (iii) =⇒ (i). Exercise 5.49. Let X be a completely regular space and α ∈ (0, ω1 ). Prove that the following assertions are equivalent: (i) X ∈ Σα (Bas(cX)) for any compactification cX of X, (ii) X ∈ Σα (Bas(cX)) for some compactification cX of X, ˇ (iii) X ∈ Σα (Bas(βX)) for the Cech–Stone compactification βX of X. Hint. Follow the proof of Exercise 5.48 along with Theorem 5.26(c). Exercise 5.50. We recall that a compact space K is scattered if every subset of K has an isolated point. Verify the following assertions: (a) A compact space is scattered if and only if every closed subset of K has an isolated point. (b) A continuous image of a scattered compact space is scattered. (c) If K is not scattered, then there exists a perfect subset P ⊂ K and a continuous surjection ϕ : P → {0, 1}N . (d) If K is a metrizable compact scattered space, then it is countable. Hint. The verification of (a) is easy. If ϕ : K → L is a continuous surjection of a compact scattered space K onto a compact space L, let H ⊂ L be a closed set. By Proposition A.38, there exists a minimal closed set F ⊂ K such that ϕ(F ) = H. Let x ∈ F be an isolated point. Then H \ ϕ(F \ {x}) is a nonempty open set. Thus ϕ(x) is an isolated point of H. This proves (b). To verify (c), let a closed set F ⊂ K have no isolated points. Inductively we construct relatively open sets Us ⊂ F , s ∈ {0, 1}
5.7 Exercises •
Us∧ 0 ∪ Us∧ 1 ⊂ Us , s ∈ {0, 1}N ,
•
Us∧ 0 ∩ Us∧ 1 = ∅, s ∈ {0, 1}N .
163
This is possible in view of the fact that F has no isolated point. Let ∞ [ ∞ \ [ \ H := Us = Uσ|n . n=1 |s|=n
By the compactness of F , {0, 1}N by the formula
T∞
n=1 Uσ|n
ϕ(x) := σ
σ∈{0,1}N n=1
6= ∅ for each σ ∈ {0, 1}N . We define ϕ : H →
for x ∈
∞ \
Uσ|n ,
σ ∈ {0, 1}N .
n=1
Then ϕ is a continuous surjection of H onto {0, 1}N . Let P ⊂ H be a minimal closed set with ϕ(P ) = {0, 1}N (see Proposition A.38). Then P is perfect. Indeed, if x were an isolated point of P , {0, 1}N \ ϕ(P \ {x}) would be an isolated point of {0, 1}N , a contradiction. Assertion (d) is a classical fact. If K is a metrizable uncountable compact space, let U be the union of all open countable sets in K. Since K is metrizable, we can choose countably many open sets whose union is U . Thus K \ U is an uncountable closed set whose any relatively open subset is uncountable. In particular, K \ U has no isolated points. Hence K is not scattered. Exercise 5.51. Prove that a compact space K is scattered if and only if every f ∈ C(K) has countable range. Hint. This follows from (b) and (d) of Exercise 5.50. Exercise 5.52. Any scattered compact space K has a base consisting of clopen sets. Hint. Let x, y ∈ K be distinct points. We find a continuous function f : K → [0, 1] such that f (x) = 0 and f (y) = 1. By Exercise 5.51, f (K) is a countable set. Hence there exists r ∈ (0, 1) \ f (K). Then {z ∈ K : f (z) < r} = {z ∈ K : f (z) ≤ r} and {z ∈ K : f (z) > r} = {z ∈ K : f (z) ≥ r} are clopen sets separating x and y. From this the assertion follows.
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5 Perfect classes of functions and representation of affine functions
Exercise 5.53. Let F be a closed subset of a compact space K. If B ⊂ F is a Baire set in F , then there exists a Baire set A ⊂ K such that A ∩ F = B. Hint. If B ⊂ F is a zero set, by Tietze’s theorem there exists a zero set A ⊂ K with A ∩ F = B. If F denote the family of all Baire sets in F that are traces of Baire sets in K, it is easy to verify that F is a σ-algebra. Hence F contains all Baire sets in F. Exercise 5.54. Prove that for a compact space K the following assertions are equivalent: (i) K is not scattered, (ii) for any α < ω1 there exists a Baire set A ∈ / Σα (Bas(K)). Hint. Assume that K is scattered and A ⊂ K is a Baire set. Using Proposition A.48 we find a countable family {fn : n ∈ N} of continuous functions such that A is contained in the σ-algebra generated by {fn−1 ((0, 1]) : n ∈ N}. Then ϕ : K → RN defined as ϕ(x) := {fn (x)}n∈N , x ∈ K, is a continuous surjection of K on a metrizable compact space ϕ(K). By Exercise 5.50(b), (d), ϕ(K) is a countable compact space. Since A = ϕ−1 (ϕ(A)) and ϕ(A) ∈ Σ02 (ϕ(K)) = Σ2 (Bas(ϕ(K))), A ∈ Σ2 (Bas(K)). Hence any Baire set in K is contained in Σ2 (Bas(K)). Assume now that K is not scattered. By Exercise 5.50(c), we can find a closed set P ⊂ K and a continuous surjection ϕ : P → {0, 1}N . By Theorem 22.4 of [262], for any α < ω1 we can find a Borel set B ⊂ {0, 1}N that is not contained in Σ0α ({0, 1}N ). By Theorem 5.16, ϕ−1 (B) ∈ / Σα (Bas(P )). If A ⊂ K is a Baire subset −1 satisfying A ∩ P = ϕ (B) (see Exercise 5.53), then A ∈ / Σα (Bas(K)). Exercise 5.55. If K is a compact space that is not scattered, then for any α < ω1 there exists a Borel set A ∈ / Σα (Bos(K)). Hint. If K is not scattered and α < ω1 , Exercise 5.54 provides a Baire set A ⊂ K not contained in Σα (Bas(K)). By Theorem 5.17, A is not contained in Σα (Bos(K)). Exercise 5.56. Let f : X → R be a function on a normal space X. Prove that the following assertions are equivalent: (i) f is Baire-one, (ii) there exist functions fn : X → R, n ∈ N, such that fn → f and osc fn (x) ≤ n1 , x ∈ K, n ∈ N.
5.7 Exercises
165
Hint. If f : X → R, then u(x) := f (x)∨lim supy→x f (y), x ∈ X, is upper semicontinuous on X, and l(y) := f (x) ∧ lim infy→x f (y), x ∈ X, is lower semicontinuous on X. If osc f ≤ δ, u − δ < l + δ. Since X is normal, there exists a continuous function g with u − δ < g < l + δ. Now it is easy to construct continuous functions gn pointwise converging to f . Exercise 5.57. Prove the following variant of the Kuratowski–Ryll-Nardzewski selection theorem: If X, Y are metrizable separable spaces, Φ : Y → 2X is an usco mapping (see Section A.4), then there exists φ ∈ Bof1 (Y, X) such that φ(y) ∈ Φ(y) for y ∈ Y . Hint. If {Un : n ∈ N} is a countable base of open sets in X, we consider a sequence {Fn } of closed sets in X, where Fn := Un , n ∈ N. We construct inductively multivalued mappings Φn : Y → 2X , n ≥ 0, as follows: Let Φ0 (y) := Φ(y), If Φn has been defined for n ≥ 0, let ( Φn (y) ∩ Fn+1 , Φn+1 (y) := Φn (y),
y ∈ Y.
y ∈ Φ−1 (Fn+1 ), y∈ / Φ−1 (Fn+1 ).
As in the proof of Lemma 5.15 we get a decreasing sequence {Φn }∞ n=0 of mappings with nonempty compact values such that (Φn )−1 (F ) ∈ Bos(Y ), Let φ(y) :=
∞ \
F ⊂ X closed, n ≥ 0.
Φn (y),
y ∈ Y.
n=0
Then φ(y) is a singleton. Indeed, if x1 , x2 ∈ φ(y), let n ≥ 0 be such that x1 ∈ Fn+1 and x2 ∈ / Fn+1 . Since x1 ∈ φ(y) ⊂ Φn (y), Φn (y) ∩ Fn+1 6= ∅. Hence Φn+1 (y) ⊂ Fn+1 , and thus x2 ∈ / Φn+1 (y). Since φ(y) ⊂ Φn+1 (y), we get a contradiction. Further, φ(y) ⊂ Φ(y) for each y ∈ Y . This follows from the first step of the construction. Finally, since φ−1 (Fn+1 ) = (Φn )−1 (Fn+1 ) ∈ Bos(Y ), and any open set in X is a countable union of elements from {Fn : n ∈ N}, φ is Σ2 (Bos(Y ))-measurable mapping. In other words, φ ∈ Bof1 (X, Y ).
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5 Perfect classes of functions and representation of affine functions
Exercise 5.58. Let E be a Banach space and X := BE ∗ with the w∗ -topology. We set E0 := E ⊂ E ∗∗ and inductively define [ Eα := {x∗∗ |X : x∗∗ is a w∗ -limit of a sequence from Eβ }, α ∈ (0, ω1 ). β<α
Prove that Aα (X) = Eα + R. Hint. The case α = 0 is settled by Proposition 4.36. For higher ordinals use transfinite induction.
5.8
Notes and comments
The theory of Borel sets is a deeply studied topic, in particular, in separable metrizable spaces. Since it is not the main theme of our book, we refer the reader to the following books and papers and to the references therein. The following sources R. Baire [29], K. Kuratowski [285], F. Hausdorff [217], Z. Frol´ık [187], S. M. Srivastava [434] and A. S. Kechris [262] provide detailed surveys of results in separable metrizable spaces and A. H. Stone [439], R. W. Hansell [207], [206], P. Holick´y and J. Pelant [235] and P. Holick´y [231] of results in nonseparable metrizable spaces. The question of how to develop a satisfactory theory of Borel classes in general topological spaces has been studied less extensively (see for example D. Mauldin [335], Z. Frol´ık [186], P. Holick´y [229], [230], R. W. Hansell [209], [210], M. Raja [381], [382] and P. Holick´y and J. Spurn´y [236]). The results of Section 5.1 are rather standard and our exposition follows the book by A. S. Kechris [262] and the paper J. Spurn´y [421]. The idea of generating Borel classes from algebras of sets is due to P. Holick´y. The key Lemma 5.15 is from P. Holick´y and J. Spurn´y [236], where a more general version for resolvable sets is proved. Theorem 5.16 can be also found in [236], its metrizable predecessor is in J. Saint Raymond [406]. Theorem 5.17 is proved in C. A. Rogers and J. E. Jayne [394], Theorem 5.9.13. It is a natural question whether the results of Theorem 5.17 can be proved for more general spaces than just compact ones. A natural candidate is the class of K-analytic spaces. In particular, we do not know an answer to the following question. Problem 5.59. Let A be a subset of a K-analytic space X such that A is a Baire subset of X and A ∈ Σ1 (Bos(X)). Does A have to be in Σ1 (Bas(X))? Theorem 5.19 and Corollary 5.22 are classical, see for example K. Kuratowski [285] and A. S. Kechris [262]. Theorem 5.23 is based upon results of G. Koumoullis [280]. The results of Section 5.4 are immediate consequences of the previous ones, hence they can be found in aforementioned P. Holick´y and J. Spurn´y [236], J. Saint Raymond [406] and C. A. Rogers and J. E. Jayne [394].
5.8 Notes and comments
167
Theorem 5.31 is a generalization of J. Saint Raymond [406], Corollaire 8, to nonmetrizable compact spaces. Section 5.6 follows the papers by J. Spurn´y [425] and [420]. Affine classes are very strongly related to the so-called Baire classes of Banach spaces; we refer the reader to S. A. Argyros, G. Godefroy and H. P. Rosenthal [19] and the references therein. Exercise 5.44 uses the examples from E. A. Michael [343] and S. Willard [471], Exercise 5.47 is taken from M. Talagrand [444] (see also C. A. Rogers and J. E. Jayne [394], Section 5.2). Exercises 5.48 and 5.49 can be found in P. Holick´y and J. Spurn´y [236]. There are many descriptive properties that are “absolute” in the sense that a comˇ pletely regular space X has a property (P) in its Cech–Stone compactification if and ˇ only if X has (P) in every Y containing X. Cech completeness serves as a classical ˇ example, that is, X is a Gδ subset of its Cech–Stone compactification if and only if X is a Gδ subset of every completely regular space Y containing X (see [169], Section 3.9). For more involved examples of this phenomenon, see J. E. Jayne [246], Theorem 8, P. Holick´y [229], Theorem 3, [230], Theorem 2, R. W. Hansell [209], Theorem 5.3 and Theorem 6.14(d), or M. M. Choban [103], Sections 5 and 6. The same can be proved, for example, for a K-analytic space (see C. A. Rogers and J. E. Jayne [394], Theorem 5.8.8); for a space obtainable as a result of the Souslin operaˇ ˇ tion applied to Borel sets in its Cech–Stone compactification (so-called Cech analytic spaces, see D. H. Fremlin [184] and P. Holick´y and J. Spurn´y [236], Theorem 5); for a space obtainable as a result of the Souslin operation applied to resolvable sets in its ˇ Cech–Stone compactification (such spaces are called scattered-K-analytic spaces in R. W. Hansell [209], Theorem 5.3 and Theorem 6.14(d), and [236], Theorem 5, almost K-descriptive in P. Holick´y [229], Theorem 3, and cover-analytic in I. Namioka ˇ and R. Pol [353], §2); for a space of some resolvable class in its Cech–Stone compactification (see P. Holick´y and J. Spurn´y [236], Theorem 5). Exercises 5.50, 5.51 and 5.52 are rather standard; we refer the reader, for example, to [173], pp. 398–400. Exercise 5.54 can be found in P. R. Meyer [340], Exercise 5.55 is taken from J. Spurn´y [421] and Exercise 5.57 is a particular version of a selection theorem by K. Kuratowski and C. Ryll-Nardzewski [287].
Chapter 6
Simplicial function spaces
This chapter is devoted to a presentation of results on simplicial function spaces and Choquet simplices. We recall that, given a function space H on a compact space K and x ∈ K, there exists an H-maximal measure H-representing x (see Section 3.6). If this measure is uniquely determined for every x ∈ K, H is called a simplicial function space. We explicitly wish to emphasize that this condition and the condition that the state space S(H) is a Choquet simplex are not equivalent (see Remark 6.2). The first useful characterization of simplicial function spaces is Theorem 6.5, followed by the Edwards in-between theorem 6.6. The mapping assigning to each x a unique H-maximal H-representing measure turns out to be a dilation as defined in Definition 3.91 (see Theorem 6.8). The rest of the section presents basic facts on this mapping. Next we turn our attention to the characterization of simplicial function spaces by means of lattice properties of the ordered Banach space Ac (H). This is contained in Theorem 6.16, Theorem 6.18 and Theorem 6.20. The important Theorem 6.25 classifies simplicial spaces as a particular example of L1 -preduals. Section 6.4 deals with the solvability of the so-called weak Dirichlet problem in simplicial spaces and shows that any point of the Choquet boundary of such a space is Ac (H)-exposed. The question of an affine extension of a function defined on the set of extreme points is studied in Section 6.5. Theorem 6.31 provides a necessary and sufficient condition for a function to be extendable to a function of affine class α. A particular case is then a result of E. M. Alfsen on continuous affine extensions (see Theorem 6.32). Theorem 6.35 shows how extension results can be formulated in an abstract way using the concept of affinely perfect classes. Special classes of simplicial function spaces are studied in Section 6.6. The first example is the class of Bauer simplicial spaces; they are characterized in Theorem 6.37 and the subclass of Markov simplicial spaces is described in Theorem 6.42. A more general class of spaces are spaces with Lindel¨of Choquet boundaries. Some of their features are described in Theorem 6.45 and Theorem 6.46. Simplicial spaces on metrizable compact spaces with boundaries of type Fσ are investigated in Subsection 6.6.D, where Theorem 6.49 provides an analogue of Theorem 6.37 for these spaces. The Daugavet property of simplicial spaces is presented in Theorem 6.53. The convex counterpart of simplicial function spaces, namely Choquet simplices, is studied in Section 6.8. Theorem 6.54 provides a basic link between simplices and simplicial function spaces. Prime simplices are characterized in Theorem 6.60, Bauer
6.1 Basic properties of simplicial spaces
169
simplices by means of their facial structure are described in Theorem 6.69. Fakhoury’s characterization of simplices is contained in Theorem 6.70. We conclude the chapter with a simple result in Section 6.9 on restrictions of function spaces. As usually, H will stand for a function space on a compact space K.
6.1
Basic properties of simplicial spaces
Definition 6.1 (Simplicial function spaces and simplices). A function space H on a compact space K is called simplicial if for each x ∈ K there exists a unique maximal measure δx ∈ Mx (H). In the convex case when H = Ac (X) we say simply that X is a Choquet simplex or briefly a simplex. (Section 6.8 below is devoted to a study of Choquet simplices.) Remark 6.2. We point out that a function space H might be simplicial and yet the state space S(H) need not be a simplex (see Exercise 6.79). In a paper [44] by H. Bauer, the spaces H whose state space is a simplex are called simplicial, whereas spaces satisfying Definition 6.1 are called weakly simplicial. By Theorem 6.54, H is simplicial if and only if S(Ac (H)) is a simplex. Lemma 6.3. Let H be a function space on K. (a) If H is simplicial, x ∈ K and µ ∈ Mx (H), then µ ≺ δx . In particular, εx ≺ δx . (b) The function space H is simplicial if and only if Ac (H) is simplicial. Proof. (a) Let µ ∈ Mx (H). By Theorem 3.65, there exists a maximal measure λ such that µ ≺ λ. Using Proposition 3.20 we get λ ∈ Mx (H). As H is simplicial, λ = δx . Hence µ ≺ δx . For the proof of (b) notice that H-maximal measures coincide with Ac (H)-maximal measures (see Proposition 3.67). This concludes the proof. Lemma 6.4. If H is simplicial, then f ∗ (x) = δx (f ) for any x ∈ K and f ∈ Kusc (H). Proof. Let f ∈ Kusc (H) and x ∈ K be given. Key Lemma 3.21 provides a measure µ ∈ Mx (H) so that µ(f ) = f ∗ (x). By Lemma 6.3(a), µ ≺ δx . Hence f ∗ (x) = µ(f ) ≤ δx (f ) ≤ f ∗ (x), which concludes the proof. Theorem 6.5. The following assertions are equivalent: (i) H is simplicial, (ii) f ∗ is H-affine for any f ∈ −W(H), (iii) f ∗ is H-affine for any f ∈ Kc (H), (iv) f ∗ is H-affine for any f ∈ Kusc (H).
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6 Simplicial function spaces
Proof. For the proof of (i) =⇒ (iv), let f ∈ Kusc (H) be given. We fix x ∈ K and µ ∈ Mx (H). By Lemma 6.3(a), µ ≺ δx . Using Lemma 3.18(b), Lemma 6.4, and Lemma 3.18(a) we get µ(f ∗ ) = inf{µ(k) : k ∈ S c (H), k ≥ f } ≥ inf{δx (k) : k ∈ S c (H), k ≥ f } = δx (f ∗ ) ≥ δx (f ) = f ∗ (x) ≥ µ(f ∗ ). Hence f ∗ is H-affine. Obviously, (iv) =⇒ (iii) =⇒ (ii). To close the chain of implications, assume that (ii) holds. Let x ∈ K and µ, ν ∈ Mx (H) be maximal measures. Then, for each f ∈ −W(H), Theorem 3.58 yields µ(f ) = µ(f ∗ ) = f ∗ (x) = ν(f ∗ ) = ν(f ). By Lemma 3.11(c), µ = ν, as required. This concludes the proof. Theorem 6.6 (Edwards in-between theorem). If H is a function space on a compact space K, then the following assertions are equivalent: (i) H is simplicial, (ii) for any pair s, t, where s ∈ Kc (H), t ∈ S c (H), s ≤ t on K, there exists h ∈ Ac (H) with s ≤ h ≤ t on K, (iii) for any pair s, t, where s ∈ Kusc (H), t ∈ S lsc (H), s ≤ t on K, there exists h ∈ Ac (H) with s ≤ h ≤ t on K. Proof. (i) =⇒ (iii): Assume that H is simplicial, s ∈ Kusc (H), t ∈ S lsc (H) and s ≤ t on K. With the aid of Proposition 3.53, find s0 ∈ Kc (H), t0 ∈ S c (H) in such a way that s ≤ s0 ≤ t0 ≤ t on K. We see that there is no loss of generality in assuming that s, −t ∈ Kc (H). Given ε > 0, denote F := {f − g + ε : s ≤ f ≤ g ≤ t, f, −g ∈ Kc (H)} . In order to find a strictly positive element of F, we employ Lemma A.88. To this end, choose a nonzero measure µ ∈ M+ (K). Using Theorem A.84 we get a function k ∈ S c (H) such that s∗ ≤ k
and µ(k − s∗ ) < εµ(K).
Since the cone S c (H) is min-stable, by Proposition 3.53 there exists a function g ∈ S c (H) such that s∗ ≤ g ≤ k ∧ t. By Theorem 6.5, s∗ is an H-affine function. Now, by an analogous reasoning, there exists f ∈ Kc (H) such that s∗ ≤ f ≤ g ≤ t. Then µ(g − f ) ≤ µ(g − s∗ ) < εµ(K), and therefore µ(f − g + ε) = εµ(K) + µ(f − g) > 0.
6.1 Basic properties of simplicial spaces
171
With this in hand, it is now possible to construct sequences {fn } in Kc (H) and {gn } in S c (H) so that s ≤ fn ≤ fn+1 ≤ gn+1 ≤ gn ≤ t
and
0 < gn − fn <
1 2n
for each n ∈ N. Both sequences converge uniformly on K to an H-affine function h satisfying s ≤ h ≤ t. Since obviously (iii) =⇒ (ii), all that remains to be proved is that (ii) =⇒ (i). Pick x ∈ K and consider maximal measures µ, ν ∈ Mx (H). If f ∈ Kc (H), then by assumption (ii), it follows that the set {h ∈ Ac (H) : h ≥ f } is down-directed. Hence, using Proposition 3.25(a) and Theorem A.84, µ(f ∗ ) = µ(inf {h ∈ Ac (H) : h ≥ f }) = inf {µ(h) : h ∈ Ac (H), h ≥ f } = inf {h(x) : h ∈ Ac (H), h ≥ f } = f ∗ (x). Mokobodzki’s maximality test 3.58 now yields µ(f ) = µ(f ∗ ) = f ∗ (x) = ν(f ∗ ) = ν(f ). Since the space Kc (H)−Kc (H) is uniformly dense in C(K), it follows that µ = ν. Definition 6.7 (Abstract Dirichlet problem). Let H be a simplicial function space on a compact space K. For any bounded universally measurable function f on K, we define Z T f : x 7→ f dδx , x ∈ K. K
If f is defined only on a universally measurable subset A of K, we consider f to be defined by 0 on K \ A, and define T f as above. The function T f is an (abstract) solution of the Dirichlet problem for the function f . We show in Theorem 6.8 that the mapping x 7→ δx is a kernel as defined in Subsection A.3.D; furthermore, it is a dilation as defined in Section 3.10. Theorem 6.8. Let H be simplicial. Then the following assertions hold. (a) If f ∈ C(K) and ε > 0, then there exist functions u, v ∈ Kc (H) so that kT (u − v) − T f k < ε. Hence, there exist sequences {fn }, {gn } of bounded upper semicontinuous functions on K so that the sequence {fn − gn } converges uniformly to T f . In particular, if K is metrizable, T f is a Baire-one function. (b) T f is a Borel function for any bounded Baire function f on K. (c) T f ∈ (Ac (H))⊥⊥ , hence T f is H-affine, for any bounded Baire function f on K.
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6 Simplicial function spaces
Proof. For the proof of (a), let f ∈ C(K) be given. By Proposition 3.11(c), there exist sequences {fn }, {gn } of H-convex continuous functions such that fn − gn → f uniformly. Then T (fn −gn ) → T f uniformly, and so Lemma 6.4 concludes the proof of (a) (note that semicontinuous functions on metrizable spaces are Baire-one and Baire-one functions are stable with respect to uniform convergence, see Section A.5). To verify (b), let B consist of all bounded Baire functions f on K, for which T f is Borel. It is easy to check that B is vector space that is stable with respect to taking pointwise limits of bounded sequences. Since C(K) ⊂ B by (a), B contains all bounded Baire functions by Proposition A.47. For the proof of (c), we fix f ∈ Kc (H). Let µ, ν ∈ M+ (K) satisfy µ − ν ∈ (Ac (H))⊥ . By Theorem 6.6, the family inf{h ∈ Ac (H) : h ≥ f } is down-directed. Using Proposition 3.25(a) and Theorem A.84, µ(f ∗ ) = inf{µ(h) : h ∈ Ac (H), h ≥ f } = inf{ν(h) : h ∈ Ac (H), h ≥ f } = ν(f ∗ ). Hence f ∗ = T f (see Lemma 6.4) is contained in (Ac (H))⊥⊥ . Let f ∈ C(K) and ε > 0 be given. We use (a) to get functions u, v ∈ Kc (H) so that kT (u − v) − T f k < ε. It follows from the argument above that T f ∈ (Ac (H))⊥⊥ . Now we finish the proof as above to conclude that T f ∈ (Ac (H))⊥⊥ for any bounded Baire function on K. Proposition 6.9. Let H be a function space on a compact space K. Then the following conditions are equivalent: (i) H is simplicial, (ii) if µ, ν ∈ Mmax (H) with µ − ν ∈ (Ac (H))⊥ , then µ = ν, (iii) (Ac (H))⊥ ∩ Mbnd (H) = {0}. Proof. (i) =⇒ (ii): Let H be simplicial and µ and ν maximal measures on K, µ − ν ∈ (Ac (H))⊥ . Using Theorem 6.8(c) along with Lemma 6.4 and Theorem 3.58, we get µ(f ) = µ(f ∗ ) = µ(T f ) = ν(T f ) = ν(f ∗ ) = ν(f ),
f ∈ Kc (H).
By Proposition 3.11(c), µ = ν. Obviously, (ii) =⇒ (iii). Assuming (iii), let x ∈ K and µ, ν ∈ Mx (H) be maximal. Since µ − ν ∈ (Ac (H))⊥ , µ = ν by the assumption. Hence (iii) =⇒ (i).
Remark 6.10. Example 3.2(b) shows that a closed function space H can be simplicial even though H⊥ ∩ Mbnd (H) 6= {0}.
6.1 Basic properties of simplicial spaces
173
Theorem 6.11. Let H be simplicial and T : K → M1 (K) be the dilation defined as T : x 7→ δx , x ∈ K, and T µ defined as in Definition 3.91. Then the following assertions hold: (a) For any µ ∈ M+ (K), (a1) T µ ∈ M+ (K), (a2) kT µk = kµk, (a3) T µ is H-maximal, (a4) µ ≺ T µ. (b) T µ is a boundary measure for any µ ∈ M(K). (c) T εx = δx , x ∈ K. Proof. (a): If µ ∈ M+ (K) and f ≥ 0 is continuous on K, then Z T µ(f ) = µ(T f ) = δx (f )dµ(x) ≥ 0 K
yields that T µ is a positive measure on K. Further, kµk = µ(1) = µ(T 1) = T µ(1) = kT µk gives (a2). If f is an H-convex continuous function, then the equalities Z Z T µ(f ) = δx (f )dµ(x) = δx (f ∗ )dµ(x) = T µ(f ∗ ) K
K
imply that T µ is H-maximal (see Theorem 3.58). Finally, for such a function f ∈ Kc (H) we get Z Z µ(f ) = f (x)dµ(x) ≤ δx (f )dµ(x) = T µ(f ). K
K
Hence µ ≺ T µ. This concludes the proof of (a). Since (b) follows from (a3), we proceed to the proof of (c). We fix x ∈ K and notice that T εx (f ) = εx (T f ) = T f (x) = δx (f ), f ∈ C(K). Hence (c) follows. Corollary 6.12. Let H be a simplicial function space and f be a bounded universally measurable function such that δx (f ) = f (x) for each x ∈ K. Then f ∈ (Ac (H))⊥⊥ , and thus f ∈ A(H). In particular, any bounded function in A(H) is contained in (Ac (H))⊥⊥ .
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6 Simplicial function spaces
Proof. Let f be a universally measurable bounded function satisfying the assumption and let µ ∈ (Ac (H))⊥ . Obviously, T µ ∈ (Ac (H))⊥ , and since T µ is a boundary measure by Theorem 6.11(b), T µ = 0 by Proposition 6.9. Hence Z Z f (x) dµ(x) = µ(f ). δx (f ) dµ(x) = 0 = T µ(f ) = K
K
Thus f ∈ (Ac (H))⊥⊥ . We use Example 3.82 as a starting point for a general construction leading to various examples of simplicial function spaces. In particular, we get that the space constructed in Example 3.82 is a simplicial function space. Definition 6.13 (Stacey function spaces). Let L be a compact space, B ⊂ L, {Lb : b ∈ B} be a family of σ-compact locally compact spaces and µb ∈ M1 (Lb ), b ∈ B, be probability Radon measures such that no µb is a Dirac measure. We define a set K as the discrete union of L and the spaces Lb , b ∈ B, and endow it with the following topology: A point x ∈ Lb , b ∈ B, has a neighborhood of sets consisting of all sets U ⊂ Lb open in Lb , a point x ∈ L has a neighborhood of sets of the form U∪
[
Lb ∪
b b∈(B∩U )\B
[
(Lb \ Fb ),
b b∈B∩U
b ⊂ B is finite, U ⊂ L is an open neighborhood of x in L and Fb ⊂ Lb , where B b are compact in Lb . b ∈ B, The space H := {f ∈ C(K) : f (b) = µb (f ), b ∈ B} a Stacey function space. Notice that K is a compact space and H is a well-defined function space. Lemma 6.14. Let H be a Stacey function space on a compact space K. Then H is simplicial, H = Ac (H), ChH (K) = K \ B and δb = µb for every b ∈ B. Proof. For a point x in some Lb and a relatively compact open set U ⊂ Lb containing x, let fb : Lb → [0, 1] be a continuous function such that fb (x) = 0 and fb = 1 on Lb \ U . Then the function ( fb fU := 1
on Lb , on K \ Lb ,
belongs to H, and thus shows that the support of any measure H-representing x is contained in U . Hence Mx (H) = {εx } and K \ L ⊂ ChH (K).
6.1 Basic properties of simplicial spaces
175
We fix x ∈ L \ B. For any open neighborhood U ⊂ L of x, we find a function g ∈ C(L) such that f (x) = 0 and f = 1 on L \ U . Then the function ( g(t), f (t) := g(b),
t ∈ L, t ∈ Lb ,
(6.1)
belongs to H,Sand shows that the support of any measure H-representing x is contained in U ∪ b∈B∩U Lb . Hence L \ B ⊂ ChH (K). Obviously, B ⊂ K \ ChH (K) and hence ChH (K) = K \ B. Let b ∈ B be arbitrary. Similar functions as in (6.1) show that any measure µ ∈ Mb (H) is carried by {b} ∪ Lb . Let µ ∈ Mb (H) satisfy µ({b}) = 0. Let g1 : Lb → [0, 1] be a continuous function with compact support and ε > 0. We find a continuous function g2 : {b} ∪ Lb → [0, 1] such that |g2 (b) − µb (g1 )| < ε and and g1 = g2 on the support of g1 . Then the function ( g2 (x), f (x) := g2 (b),
x ∈ {b} ∪ Lb , x ∈ K \ ({b} ∪ Lb ),
belongs to H. Thus µb (f ) = f (b) = µ(f ). Since g1 and ε > 0 are arbitrary, µb = µ. Hence, Mb (H) = co({εb , µb }).
(6.2)
By (6.2), H is simplicial, H = Ac (H) and δb = µb , b ∈ B. This concludes the proof. Example 6.15. There exists a simplicial function space on a compact space K and a bounded Borel function f on K such that T f is not universally measurable. Proof. Let L := [0, 1], B ⊂ L be a Lebesgue nonmeasurable set, Lb := {0, 1} and µb := 21 (ε0 + ε1 ), b ∈ B. Let H be the Stacey function space on the compact space K constructed in Definition 6.13. By Lemma 6.14, H is simplicial and ChH (K) = K \ B. Let f := cL . Then f is Borel, but T f |L = cL\B . Hence T f is not universally measurable.
176
6.2
6 Simplicial function spaces
Characterizations of simplicial spaces
For this section, recall that if f, g are functions from a function space H, then f ∨ g denotes the pointwise supremum of f and g and that H endowed with the pointwise ordering is an ordered vector space (Example A.19(a)). Theorem 6.16 (Riesz interpolation property). For a function space H on a compact space K, the following assertions are equivalent: (i) H is simplicial, (ii) for any pair f ≤ g, f, −g ∈ Kc (H), there is h ∈ Ac (H) such that f ≤ h ≤ g, (iii) for any pair f < g, f, −g ∈ Kc (H), there is h ∈ Ac (H) such that f < h < g, (iv) for any pair f ≤ g, −f, g ∈ W(Ac (H)), there is h ∈ Ac (H) such that f ≤ h ≤ g, (v) for any pair f < g, −f, g ∈ W(Ac (H)), there is h ∈ Ac (H) such that f < h < g, (vi) for any quadruple, f1 , f2 , g1 , g2 ∈ Ac (H), f1 ∨f2 ≤ g1 ∧g2 , there is h ∈ Ac (H) such that f1 ∨ f2 ≤ h ≤ g1 ∧ g2 , (vii) for any quadruple, f1 , f2 , g1 , g2 ∈ Ac (H), f1 ∨f2 < g1 ∧g2 , there is h ∈ Ac (H) such that f1 ∨ f2 < h < g1 ∧ g2 . Proof. The equivalence (i) ⇐⇒ (ii) is just the Edwards in-between theorem (see Theorem 6.6). The implications (ii) =⇒ (iv) =⇒ (vi) and (iii) =⇒ (v) =⇒ (vii) are obvious. For the proof of (vi) =⇒ (vii), we select f1 , f2 , g1 , g2 ∈ Ac (H) so that f1 ∨ f2 < g1 ∧g2 . By the compactness of K, there exists ε > 0 such that f1 ∨f2 +ε ≤ g1 ∧g2 −ε. Then assumption (vi) yields the existence of a function h ∈ Ac (H) satisfying f1 ∨ f2 < f1 ∨ f2 + ε ≤ h ≤ g1 ∧ g2 − ε < g1 ∧ g2 . To prove (vii) =⇒ (v), pick f1 , . . . , fm ∈ Ac (H) and g1 , . . . , gn ∈ Ac (H) such that f1 ∨ · · · ∨ fm < g1 ∧ · · · ∧ gn . We use assumption (vii) and find a function h1 ∈ Ac (H) with f1 ∨ f2 < h1 < g1 ∧ g2 . Inductively choose functions h2 , . . . , hm−1 ∈ Ac (H) satisfying fl+1 ∨ hl−1 < hl < g1 ∧ g2 ,
2 ≤ l ≤ m − 1.
Then the function b h1 := hm−1 fulfills f1 ∨ · · · ∨ fm < b h1 < g1 ∧ g2 . Thus f1 ∨ · · · ∨ b fm < h1 ∧ g3 . Use the previous construction to find a function b h2 ∈ Ac (H) with b b f1 ∨ · · · ∨ fm < h2 < h1 ∧ g3 . After (n − 1)-th step of this construction we arrive at the desired function h := b hn−1 ∈ Ac (H) satisfying f1 ∨ · · · ∨ fm < h < g1 ∧ · · · ∧ gn .
6.2 Characterizations of simplicial spaces
177
For the proof of (v) =⇒ (iii), let f, −g ∈ Kc (H) with f < g be given. Find ε > 0 so that f + ε < g − ε. By Proposition 3.55, there exist functions fb, −b g ∈ W(Ac (H)) with f − ε ≤ fb ≤ f and g ≤ gb ≤ g + ε. An appeal to (v) provides a function h ∈ Ac (H) satisfying f ≤ fb + ε < h < gb − ε ≤ g, which is the desired function. To finish the proof we verify condition (iii) of Theorem 6.5 provided that (iii) holds. Let f ∈ Kc (H) be given. By Exercise 3.96, f ∗ = inf{h ∈ Ac (H) : h > f }. Since the latter family is down-directed by our assumption, we can employ Theorem A.84 and deduce that f ∗ is H-affine. Hence (iii) =⇒ (i), which concludes the proof. Remark 6.17. The conditions from Theorem 6.16(vi) and (vii) express that the space Ac (H) has the Riesz interpolation property and the weak Riesz interpolation property, respectively. Theorem 6.18 (Riesz decomposition property). For a function space H on a compact space K, the following assertions are equivalent: (i) H is a simplicial function space, (ii) for any positive functions f , g1 , g2 ∈ Ac (H) with f ≤ g1 +g2 there exist positive functions h1 , h2 ∈ Ac (H) such that h1 + h2 = f and h1 ≤ g1 , h2 ≤ g2 . Proof. By Proposition A.11 we know that condition (ii) is equivalent to condition (vi) of Theorem 6.16. Remark 6.19. If the function space Ac (H) satisfies the requirements of Theorem 6.18(ii), we say that Ac (H) has the Riesz decomposition property. Theorem 6.20 (Finite binary intersection property). Let H be a function space on a compact space K. Then the following assertions are equivalent: (i) H is simplicial, (ii) if n ∈ N and {Bi }ni=1 is a family ofTclosed balls in Ac (H) satisfying Bi ∩Bj 6= ∅ whenever i, j ∈ {1, . . . , n}, then ni=1 Bi 6= ∅.
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6 Simplicial function spaces
Proof. Assume that H is a simplicial function space and {B(hi , ri )}ni=1 is a finite family of closed balls in Ac (H) with centers hi and radii ri , i = 1, . . . , n, such that B(hi , ri ) ∩ B(hj , rj ) 6= ∅, i, j = 1, . . . , n. Then the functions fi = hi − ri and gi = hi + ri , i = 1, . . . , n, satisfy f1 ∨ · · · ∨ fn ≤ g1 ∧ · · · ∧ gn . By Theorem 6.16, there exists a function h ∈ Ac (H) so that f1 ∨ · · · ∨ fn ≤ h ≤ g1 ∧ · · · ∧ gn . T Then h ∈ ni=1 B(hi , ri ), concluding the proof of (i) =⇒ (ii). For the proof of (ii) =⇒ (i), let f, g1 , g2 ∈ Ac (H) be positive functions such that f ≤ g1 + g2 . Without loss of generality we may assume that g1 + g2 ≤ 1 on K. Then the closed balls B1 := B(1, 1), B2 := B(g1 − 1, 1), B3 := B(f − 1, 1), B4 := B(1 + f − g2 , 1), satisfy Bi ∩ Bj 6= ∅, i, j ∈ {1, . . . , 4}. Indeed, it is straightforward to verify that 0 ∈ B1 ∩ B2 ∩ B3 ,
f ∈ B1 ∩ B3 ∩ B4
and
g1 ∈ B2 ∩ B4 .
T By the assumption, there exists f1 ∈ 4i=1 Bi . We set f2 := f − f1 . Then f1 ≥ 0 (because f1 ∈ B1 ), f1 ≤ g1 (because f1 ∈ B2 ), f2 ≥ 0 (because f1 ∈ B3 , and so f1 ≤ f ), and f2 ≤ g2 (because f1 ∈ B4 , and so f1 ≥ f − g2 ). By Theorem 6.18, H is simplicial. Remark 6.21. If the function space Ac (H) satisfies the requirements of Theorem 6.20(ii), we say that Ac (H) has the finite binary intersection property. Corollary 6.22. Let Hi be a function space on a compact space Ki , i = 1, 2, and let the spaces Ac (H1 ) and Ac (H2 ) be isometrically isomorphic. Then H1 is simplicial if and only if H2 is simplicial. Proof. The proof follows from Theorem 6.20, because the finite binary intersection property of a Banach space is preserved by an isometric isomorphism.
6.3
Simplicial spaces as L1 -preduals
Lemma 6.23. Let L ∈ H∗ be a nonzero functional on H. Then there exists a boundary measure µ so that kµk = kLk and µ(h) = L(h) for any h ∈ H. Moreover, if L is positive, the measure µ can be taken to be positive.
6.3 Simplicial spaces as L1 -preduals
179
Proof. Let ν ∈ M(K) be a measure corresponding to a Hahn–Banach extension of L to C(K), kνk = kLk 6= 0. If ν = ν + − ν − is the decomposition of ν into its positive and negative part, let µ1 , µ2 be maximal measures such that ν + ≺ µ1 and ν − ≺ µ2 . If we set µ := µ1 − µ2 , then µ is a boundary measure and ν − µ ∈ H⊥ . If µ = µ+ − µ− is the decomposition of µ into the positive and negative part, we get kµk = µ+ (K) + µ− (K) ≤ µ1 (K) + µ2 (K) = ν + (K) + ν − (K) = kνk. On the other hand, kνk = kLk = sup |L(h)| h∈BH
= sup |ν(h)| = sup |µ(h)| ≤ kµk. h∈BH
h∈BH
Hence kµk = kνk. If L is positive, then kLk = L(1). If µ ∈ Mbnd (H) is the measure obtained by the argument above, we have kµk ≥ µ(1) = L(1) = kLk = kµk. It follows that µ is a positive measure. Proposition 6.24. Let H be a simplicial function space on a compact space K. Then there exists an order preserving isometry R : Mbnd (H) → (Ac (H))∗ . In particular, (Ac (H))∗ is a lattice. Proof. Let R : Mbnd (H) → (Ac (H))∗ be the restriction mapping. By Proposition 6.9, R is injective. Using Lemma 6.23 applied to Ac (H) and Proposition 3.67, we infer that R is a surjective isometry. It is easy to check that Rµ is a positive functional on Ac (H) if and only if µ ∈ Mbnd (H) is a positive measure (use Lemma 6.23). Hence R preserves order, which concludes the proof. Theorem 6.25. Let H be a simplicial function space on a compact space K. Then (Ac (H))∗ is order isometric to a space L1 (X, Σ, σ) for a suitable measure space (X, Σ, σ). Proof. By Proposition 6.24 it is enough to show that Mbnd (H) is order isometric to L1 (X, Σ, σ) for a suitable measure space (X, Σ, σ). We use throughout the proof the identification of Radon measures on K with finite measures defined on Borel sets that are inner regular with respect to compact sets (see Proposition A.73). Using Zorn’s lemma, we find a set {µi : i ∈ I} of pairwise singular probability measures in Mmax (H) that is maximal with respect to inclusion. For each i ∈ I we consider the measure space (K, Σi , σi ), where Σi denotes the σ-algebra of Borel sets
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6 Simplicial function spaces
in K and σi is the restriction of µi to Σi . Let (X, Σ, σ) be the disjoint union of the spaces (K, Σi , σi ), i ∈ I. (More precisely, let X be the disjoint union of spaces Ki , i ∈ I, where each Ki equals K. Further, P Σ consists of all sets A ⊂ X satisfying A ∩ Ki ∈ Σi for each i ∈ I and let σ(A) = i∈I σi (A ∩ Ki ) for any A ∈ Σ.) To construct an order isometry T : Mbnd (H) → L1 (X, Σ, σ), let ν ∈ Mbnd (H) be arbitrary. For any i ∈ I, let νi denote the absolutely continuous part of the restriction of ν to the σ-algebra of Borel sets with respect to σi and let hi ∈ L1 (K, Σi , σi ) be a Borel function that is the Radon–Nikodym derivative of νi with respect to σi . We define T ν ∈ L1 (X, Σ, σ) as (T ν)|Ki := hi ,
i ∈ I.
Since the measures νi , i ∈ I, are pairwise singular, Z X XZ |hi | dµi = |νi |(K) ≤ |ν|(K). |T ν| dσ = X
i∈I
K
(6.3)
i∈I
Thus the function T ν is indeed in L1 (X, Σ, σ) and kT νk ≤ kνk. Conversely, we need to show P that kνk ≤ kT νk. By (6.3) it is enough to show that P ν = i∈I νi . Let λ := ν − i∈I νi and λ+ , λ− be the positive and negative part of λ, respectively. Assume that λ+ 6= 0. Since λ ⊥ µi for any i ∈ I, (λ+ (K))−1 λ+ is a probability measure contradicting Pthe maximality of {µi : i ∈ I}. Analogously we can show that λ− = 0. Thus ν = i∈I νi and T is an isometry. P To check that T is surjective, let h ∈ L1 (X, Σ, σ) be given. Then ν := i∈I hσi is in M(K) and T ν = h. If ν ∈ Mbnd (H) is positive, each νi is positive as well, and thus T ν is a positive function on X. This finishes the proof.
6.4
The weak Dirichlet problem and Ac (H)-exposed points
Definition 6.26 (The weak Dirichlet problem). We say that the weak Dirichlet problem for a function space H on a compact space K is solvable if for every compact set D ⊂ ChH (K) and every f ∈ C(D) there is a “solution” h ∈ Ac (H) such that h=f
on
D
and khk = kf k.
The solution h of the weak Dirichlet problem for f , if it exists, is by no means uniquely determined. Theorem 6.27. If a function space H on a compact space K is simplicial, then the weak Dirichlet problem for H is solvable.
6.4 The weak Dirichlet problem and Ac (H)-exposed points
181
If, moreover, K is metrizable, then H is simplicial if and only if the weak Dirichlet problem for H is solvable. Proof. The proof of the first part of the theorem is an immediate consequence of the Edwards in-between theorem 6.6. Indeed, assume that H is simplicial, D ⊂ ChH (K) compact and f ∈ C(D). Set ( ( min f (D), x ∈ K \ D, max f (D), x ∈ K \ D, s(x) := t(x) := f (x), x ∈ D, f (x), x ∈ D. Then s ∈ Kusc (H), t ∈ S lsc (H), s ≤ t on K, so by the Edwards in-between theorem 6.6 there exists h ∈ Ac (H) such that s ≤ h ≤ t on K and the proof is complete. Assume now that H is a function space on a metrizable compact space K and x ∈ K. If µ and ν are maximal measures from Mx (H), our aim is to show that µ = ν. Select f ∈ C(K). Now if ε > 0, then Corollary 3.61 and the regularity of measures µ and ν yield the existence of a compact set D ⊂ ChH (K) with (µ + ν)(K \ D) < ε. Since the weak Dirichlet problem for H is solvable, there exists h ∈ Ac (H) such that h=f
on D
and khk = kf k.
Then |µ(f ) − ν(f )| ≤ |µ(f ) − h(x)| + |h(x) − ν(g)| = |µ(f ) − µ(h)| + |ν(h) − ν(f )| Z Z Z ≤ |f − h| dµ + |h − f | dν = |f − h| d(µ + ν) K K K Z = |f − h| d(µ + ν) ≤ 2kf kε. K\D
As ε > 0 is arbitrary, we have µ(f ) = ν(f ), as needed. Theorem 6.28. Let H be a simplicial function space on a metrizable compact set K. Then any point of the Choquet boundary ChH (K) is Ac (H)-exposed. Proof. Let x ∈ ChH (K). For y ∈ ChH (K)\{x}, let hy ∈ Ac (H) denote the solution of the weak Dirichlet problem for the set {x, y} such that 0 ≤ hy ≤ 1,
hy (x) = 0,
hy (y) = 1.
In view of Theorem 6.27, such a solution exists. Then [ ChH (K) \ {x} ⊂ {t ∈ K : hy (t) > 0} . y∈ChH (K)\{x}
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6 Simplicial function spaces
Since ChH (K) \ {x} is Lindel¨of, there are yn ∈ ChH (K), n ∈ N, such that ChH (K) \ {x} ⊂
∞ [
{t ∈ K : hyn (t) > 0} .
n=1
Set h :=
∞ X 1 hy . 2n n n=1
Ac (H),
h ≥ 0 on K and h(x) = 0. It remains to show that h > 0 on Then h ∈ K \ {x}. Let z ∈ K satisfy h(z) = 0. Since h > 0 on ChH (K) \ {x} and the maximal measure δz at z is carried by ChH (K), it follows that spt δz ⊂ {x}. Thus z = x. Remark 6.29. There is a more general theorem even in a nonmetrizable case: If H is a simplicial function space on a compact space K, then any Gδ point of the Choquet boundary ChH (K) is Ac (H)-exposed, cf. Exercise 8.75. In general, there is no hope to have an “exposing function” from H. Consult examples of Exercises 6.76 and 6.77.
6.5
The Dirichlet problem for a single function
If K is a topological space, we recall that B bα (K) is the space of all bounded functions of Baire class α on K (see Definition 5.20). Definition 6.30 (Envelopes defined by non-continuous functions). Let H be a function space on a compact space K and α ∈ [0, ω1 ). For a bounded function f defined at least on the set ChH (K), we set ChH (K) ∗,α
f
:= inf{h ∈ A(H) ∩ B bα (K) : h ≥ f on ChH (K)} and
f∗,α := −ChH (K)(−f )∗,α .
ChH (K)
We remark that ChH (K)f∗,α ≤ ChH (K)f ∗,α due to the Minimum principle 3.86. Theorem 6.31. Let α ∈ [0, ω1 ), f be a bounded function on ChH (K) and let g := ChH (K)f ∗,α . Then the following assertions are equivalent: (i) The function f satisfies (i1)
ChH (K)f ∗,α
= ChH (K)f∗,α on ChH (K),
(i2) g ∈ Bbα (ChH (K)), (i3) if µ1 , µ2 ∈ Mx (H) are maximal measures for x ∈ K \ ChH (K), then µ1 (g) = µ2 (g).
6.5 The Dirichlet problem for a single function
183
(ii) There exists a unique function h ∈ A(H)∩B bα (K) such that h = f on ChH (K). (iii) There exists a function h ∈ A(H) ∩ B bα (K) such that h = f on ChH (K). Proof. Let a bounded function f on ChH (K) be given. To start the proof we notice that (ii) =⇒ (iii) is obvious. For the proof of (iii) =⇒ (i), let h ∈ B bα (K) ∩ A(H) extend f . Let b h ∈ B bα (K) ∩ A(H) be any function such that f ≤ b h on ChH (K). By Ch (K) ∗,α b H the Minimum principle 3.86, h ≤ h on K. Hence h ≤ f . Analogously we obtain ChH (K)f∗,α ≤ h. On the other hand, h ≤ ChH (K)f∗,α ≤ ChH (K)f ∗,α ≤ h by the definition. Hence h = ChH (K)f ∗,α = ChH (K)f∗,α on ChH (K) and property (i1) follows. Since (i2) and (i3) are obvious, the proof of (iii) =⇒ (i) is finished. To close the chain of implications, we have to verify that (i) =⇒ (ii). It follows from the Minimum principle 3.86 that the extension is unique, provided it exists. Let f and g be as in (i). We denote Z := ChH (K). Step 1: For any z ∈ Z and µ ∈ Mz (H) ∩ M1 (Z), we have µ(g) = g(z). Indeed, let z ∈ Z and µ ∈ Mz (H) ∩ M1 (Z) be given. Then µ(g) = µ inf{h ∈ A(H) ∩ B bα (K) : h ≥ f on ChH (K)} ≤ inf{µ(h) : h ∈ A(H) ∩ B bα (K)), h ≥ f on ChH (K)} = inf{h(z) : h ∈ A(H) ∩ B bα (K), h ≥ f on ChH (K)} = ChH (K)f ∗,α (z) = g(z). Similarly we deduce that µ(g) ≥ g(z). This finishes the proof of Step 1. Step 2: If µ, ν ∈ M1 (Z) satisfy µ ≺ ν, then µ(g) = ν(g). Let µ ≺ ν be probability measures on K carried by Z. We denote M := {(εx , λ) ∈ M1 (Z) × M1 (Z) : εx ≺ λ}. By Proposition 3.89, there exists Λ ∈ M1 (M ) such that Z µ(f1 ) + ν(f2 ) = (εx (f1 ) + λ(f2 )) dΛ(εx , λ),
f1 , f2 ∈ C(Z).
(6.4)
M
By Proposition 3.90, formula (6.4) holds for any bounded universally measurable functions on Z. Hence we can apply (6.4) along with Step 1 to pairs (g, 0) and (0, g) consecutively to get Z Z µ(g) = εx (g) dΛ(εx , λ) = λ(g) dΛ(εx , λ) = ν(g). M
This concludes the proof of Step 2.
M
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6 Simplicial function spaces
Step 3: If µ1 , µ2 ∈ Mx (H) ∩ M1 (Z) for some x ∈ K, then µ1 (g) = µ2 (g). For i = 1, 2, let νi be a maximal measure such that µi ≺ νi . Combining Step 2 and Step 1 with assumption (i3), we get µ1 (g) = ν1 (g) = ν2 (g) = µ2 (g). This finishes the proof of Step 3. Now we can define a desired extension h : K → R as h(x) := µ(g),
µ ∈ M1 (Z) ∩ Mx (H), x ∈ K.
It follows from Step 3 that h is well defined. Obviously, h = f on ChH (K). Finally, Theorem 5.31 yields that h ∈ B bα (K) ∩ A(H). This finishes the proof. Theorem 6.32. Let f be a bounded function on ChH (K) and let g := ChH (K)f ∗ . Then the following assertions are equivalent: (i) The function f satisfies (i1)
ChH (K)f ∗
= ChH (K)f∗ on ChH (K),
(i3) µ(g) = 0 for any boundary measure µ ∈ H⊥ . (ii) There exists a unique function h ∈ H such that h = f on ChH (K). (iii) There exists a function h ∈ H such that h = f on ChH (K). Proof. The proof follows from Theorem 6.31 once we realize that (i1) implies the continuity of g on ChH (K). Hence we can find a function h ∈ Ac (H) extending g. To finish the proof we have to verify that h ∈ H. Since µ(f ) = 0 for any boundary µ ∈ H⊥ , Theorem 5.31 gives h ∈ H⊥⊥ . Hence h ∈ H by the Hahn–Banach theorem. This concludes the proof. We recall that H1,b stands for the space of all functions of H-affine class 1 (see Definition 5.35). Theorem 6.33. Let f be a bounded function on ChH (K) and let g := Then the following assertions are equivalent:
ChH (K)f ∗,1 .
(i) The function f satisfies (i1)
ChH (K)f ∗,1
= ChH (K)f∗,1 on ChH (K),
(i2) g ∈ Bb1 (ChH (K)), (i3) µ(g) = 0 for any boundary measure µ ∈ H⊥ . (ii) There exists a unique function h ∈ H1,b such that h = f on ChH (K). (iii) There exists a function h ∈ H1,b such that h = f on ChH (K).
6.6 Special classes of simplicial spaces
185
Proof. The proof again follows from Theorem 6.31. The only fact we need to verify is that the obtained extension h is contained in H1,b . We already know that h ∈ B b1 (K) ∩ A(H). Using assumption (i3) we get from Theorem 5.31(b) that h ∈ H⊥⊥ . It follows from Proposition 5.42(a) that h ∈ H1,b . This concludes the proof. Remark 6.34. Inspection of the previous proofs shows that the whole procedure can be done for any affinely perfect class C of functions on compact convex sets (see Section 5.5). We define envelopes similarly to above, namely, for a bounded function f defined at least on the set ext X, we set ext X ∗,C
f
:= inf{h ∈ C(X) : h ≥ f on ext X}
and
f∗,C := −ext X(−f )∗,C .
ext X
Thus we obtain the following result. Theorem 6.35. Let X be a compact convex set, C be an affinely perfect class of functions, f be a bounded function on ext X and let g := ext Xf ∗,C . Then the following assertions are equivalent: (i) The function f satisfies (i1)
ext Xf ∗,C
= ext Xf∗,C on ext X,
(i2) the function µ 7→ µ(g), µ ∈ M1 (ext X), is contained in C(M1 (ext X)), (i3) µ(g) = 0 for any boundary measure µ ∈ A(X)⊥ . (ii) There exists a unique function h ∈ C(X) such that h = f on ext X. (iii) There exists a function h ∈ C(X) such that h = f on ext X.
6.6 6.6.A
Special classes of simplicial spaces Bauer simplicial spaces
Definition 6.36 (Bauer simplicial spaces and simplices). A simplicial function space H on K with a closed Choquet boundary ChH (K) is called a Bauer simplicial space. A Bauer simplex X is a Choquet simplex whose set of extreme points is closed. There are plenty of conditions saying that a compact convex set X is a Bauer simplex. In the following proposition we collect some of them for the case of general function spaces. Theorem 6.37. Let H be a function space on a compact space K. Then the following conditions are equivalent: (i) H is a Bauer simplicial space, (ii) for any x ∈ K there exists a unique representing measure µ ∈ Mx (H) such that µ is carried by ChH (K),
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6 Simplicial function spaces
(iii) H is a simplicial space and the mapping x 7→ δx , x ∈ K, from K to M1 (K) is continuous, (iv) given a function f ∈ Kc (H), the function f ∗ is continuous and H-affine on K, (v) for any function f ∈ C(ChH (K)) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (vi) for any f ∈ C(ChH (K)) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (vii) for any bounded continuous function f on ChH (K) there exists a function h ∈ Ac (H) such that f = h on ChH (K), (viii) the space Ac (H) is a lattice in its natural ordering. Proof. For the proof that (i) =⇒ (ii), suppose that x ∈ K and µ ∈ Mx (H) with spt µ ⊂ ChH (K) are given. If f ∈ C(K), then by Bauer’s characterization of the Choquet boundary 3.24, {x ∈ K : f (x) < f ∗ (x)} ⊂ K \ ChH (K) = K \ ChH (K). Hence µ {x ∈ K : f (x) < f ∗ (x)} ≤ µ K \ ChH (K)) = 0. By Corollary 3.59, µ is maximal, and therefore µ = δx . Suppose that (ii) holds. Since any maximal measure is carried by ChH (K) by Proposition 3.64, (ii) implies that H is a simplicial space. Let {xα }α∈A be a net in K, xα → x. We show that δx is the only cluster point of {δxα }α∈A . We lose no generality by supposing that δxα → ν. As spt δxα ⊂ ChH (K), we have spt ν ⊂ ChH (K). Since ν ∈ Mx (H) by Proposition 3.40, it follows from (ii) that ν = δx , which proves that (ii) =⇒ (iii). Let f ∈ Kc (H). Theorem 6.5 along with Lemma 6.4 assures that f ∗ is an H-affine function and that f ∗ (x) = δx (f ) for any x ∈ K. Since the mapping x 7→ δx is continuous, f ∗ is continuous. This shows that (iii) =⇒ (iv). For the proof (iv) =⇒ (v), suppose that f is a continuous function on ChH (K). By Tietze’s theorem extend f continuously to the whole of K. We know from Theorem 6.5 that H is simplicial. Hence for each ε > 0, Theorem 6.8(a) provides functions f1 , f2 ∈ Kc (H) so that kT f − (T f1 − T f2 )k ≤ ε (for the definition of T see Definition 6.7). Since T (Kc (H)) ⊂ C(K) by our assumption, T f , as a uniform limit of continuous functions, is continuous as well. By Theorem 6.8(c), T f is H-affine. Since f = T f on ChH K and both functions are continuous, f = T f on ChH K. Suppose that (v) holds. It is enough to show that ChH (K) = ChH (K). Assume, in order to obtain a contradiction, that there is x ∈ ChH (K) \ ChH (K). Using Bauer’s
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187
characterization of the Choquet boundary 3.24, we find a function f ∈ Kc (H) for which f (x) < f ∗ (x). By the assumption, there is h ∈ Ac (H) such that h = f on ChH (K). Since h − f ≥ 0 on ChH (K), the minimum principle for S c (H) (see Theorem 3.85) yields h−f ≥ 0 on K. Since h ≥ f on K, it follows from Proposition 3.25 that h = h∗ ≥ f ∗ . Hence h(x) = f (x) < f ∗ (x) ≤ h(x). This obvious contradiction yields the required conclusion, and therefore (v) =⇒ (vi). Obviously, (vi) =⇒ (vii). Suppose that (vii) holds and select h1 , h2 ∈ Ac (H). By (vii), there is h ∈ Ac (H) such that h = h1 ∨ h2 on ChH (K). If k ∈ Ac (H), k ≥ h1 ∨ h2 on K, then k ≥ h on ChH (K). Using the Minimum principle 3.85 again, we get k ≥ h on K. We see that every pair of elements of Ac (H) has a least upper bound, in other words, Ac (H) is a lattice in its natural ordering, hence (viii) holds. The proof will be completed once it is shown that (viii) =⇒ (i). To this end, let h1 , . . . , hn ∈ Ac (H) and let h ∈ Ac (H) be their least upper bound in Ac (H). By Proposition 3.25(a), h = (h1 ∨ · · · ∨ hn )∗ . We see that the function (h1 ∨ · · · ∨ hn )∗ is H-affine, and thus Theorem 6.5 yields that H is simplicial. It remains to prove that the Choquet boundary ChH (K) is closed. Assume that x ∈ ChH (K) \ ChH (K). With the aid of Bauer’s characterization of ChH (K) (see Theorem 3.24) we can find functions h1 , . . . , hn ∈ Ac (H) such that f := h1 ∨· · ·∨hn satisfies f (x) < f ∗ (x). As above, f ∗ ∈ Ac (H) is the least upper bound of h1 , . . . , hn in Ac (H) and f ∗ = f on ChH (K). The continuity of f and f ∗ yields the equality f (x) = f ∗ (x), which is an obvious contradiction. Proposition 6.38. The set M1 (K) of all probability Radon measures on a compact space K is a Bauer simplex. Proof. We use condition (vi) of Theorem 6.37 as a characterization of Bauer simplices. So assume that F ∈ C(ext M1 (K)) = C({εx : x ∈ K}) is given. We set H := µ(F ◦ ϕ),
µ ∈ M1 (K),
where ϕ : x 7→ εx is the homeomorphic mapping of K onto ext M1 (K) (cf. Proposition 2.27). Then H ∈ Ac (M1 (K)) by Proposition 5.30, and also H = F on ext M1 (K). This concludes the proof. Proposition 6.39. Let X be a Bauer simplex. Then there exists a compact space K such that X is affinely homeomorphic to M1 (K). Proof. Choose K as ext X and consider the mapping x 7→ δx , x ∈ K, where δx is the unique maximal measure representing x.
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6 Simplicial function spaces
6.6.B
Markov simplicial spaces
Recall that a function space H on a compact space K is simplicial if and only if (Ac (H))⊥ ∩ Mbnd (H) = {0} (see Proposition 6.9), and that simplicial function spaces H having a closed Choquet boundary ChH (K) were labeled as Bauer simplicial spaces. The equivalence (i) ⇐⇒ (v) of Theorem 6.37 says that H is a Bauer simplicial space if and only if Ac (H)|ChH (K) = C(ChH (K)). In this subsection we will examine function spaces for which H|ChH (K) = C(ChH (K)). Before this we recall the notion of a Markov operator. Definition 6.40 (Markov operators and projections). Let K be a compact space. A Markov operator T on K is a bounded linear operator on C(K) with T (1) = 1 = kT k. It is easy to check that any Markov operator on K is positive (see the proof of Theorem 2.31). If, moreover, T 2 = T , the operator T is called a Markov projection on K. If T is a Markov operator on K and HT := {f ∈ C(K) : T f = f } , then HT is a function space on K provided it separates points of K. It is obvious that T f ∈ HT whenever T is a Markov projection and f ∈ C(K). Let T be a Markov operator on K. Given x ∈ K, the functional Φ : f 7→ T f (x),
f ∈ C(K),
is a positive linear functional on C(K). By the Riesz representation theorem A.72 there exists a unique Radon measure µTx on K such that Z T f (x) = f dµTx for any f ∈ C(K). K
Obviously, µTx ∈ Mx (HT ) for every x ∈ K. Lemma 6.41. Let T be a Markov projection on K such that the space HT separates points of K. Then ChHT (K) = x ∈ K : µTx = εx . In particular, the Choquet boundary ChHT (K) is closed.
6.6 Special classes of simplicial spaces
189
Proof. Pick x ∈ K. If x ∈ ChHT (K), then Mx (HT ) = {εx }, hence µTx = εx . Assume now that x ∈ K \ ChHT (K). Then there exists s ∈ Kc (HT ) such that s(x) < s∗ (x) (see Theorem 3.24). Then T s ∈ HT and T s ≥ s. It follows that s(x) < s∗ (x) = inf {h(x) : h ∈ HT , h ≥ s} ≤ T s(x) = µTx (s), and therefore εx 6= µTx . Since ChHT (K) =
\
{x ∈ K : T f (x) = f (x)} ,
f ∈C(K)
the set ChHT (K) is closed. Theorem 6.42. Let H be a function space on K. The following assertions are equivalent: (i) there exists a Markov projection on K such that H = HT , (ii) H is a Bauer simplicial space and H = Ac (H), (iii) H|ChH (K) = C(ChH (K)). Proof. If (i) holds and f ∈ C(ChH (K)) = C(ChHT (K)) = C(ChHT (K)) is given, there exists g ∈ C(K) such that g = f on ChH (K). Then T g ∈ HT = H and T g = g = f on ChH (K), so (i) =⇒ (iii). Suppose now that (iii) holds. According to (i) ⇐⇒ (v) of Theorem 6.37, H is a Bauer simplicial space. It remains to show that H = Ac (H). To this end, let f ∈ Ac (H). There exists a function h ∈ H so that h = f on ChH (K). Since h − f ∈ Ac (H), the Minimum principle 3.85 yields the equality h = f on K. Hence, f ∈ H. This shows that (iii) =⇒ (ii). Finally, suppose that H is as in (ii). Given x ∈ K, let δx be the unique maximal measure in Mx (H). By Theorem 6.37, the mapping x 7→ δx is continuous, hence the function T f defined as T f (x) = δx (f ), x ∈ K, is continuous for each f ∈ C(K). Since T f is an H-affine function (cf. Theorem 6.8(c)), T is a Markov projection on K. To show that H = HT , let h ∈ H. Then T h(x) = δx (h) = h(x) for each x ∈ K. Thus, H ⊂ HT . Conversely, pick g ∈ HT . Since the function T g is in Ac (H), we get as above g = T g ∈ Ac (H) = H which shows that HT ⊂ H. This proves that (ii) =⇒ (i) and finishes the proof. Definition 6.43 (Markov simplicial spaces). Function spaces for which one of the equivalent conditions of Theorem 6.42 holds are labeled as Markov simplicial spaces. Corollary 6.44. If H is a Markov simplicial space on K, then the function space H is closed and H⊥ ∩ Mbnd (H) = {0}. Proof. The assertion easily follows from the facts that H = Ac (H) and that Ac (H) is closed (see Proposition 3.11(a)).
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6 Simplicial function spaces
6.6.C
Simplicial spaces with Lindel¨of boundaries
We obtained that the solution T f of the Dirichlet problem for a continuous function f on K is a pointwise limit of a sequence of H-affine continuous functions on K provided K is metrizable (see Definition 6.7 and Theorem 6.8(a)). We are able to extend this result to a general simplicial function space H, if we suppose that ChH (K) is a Lindel¨of topological space. The precise statement is provided by Theorem 6.45. Theorem 6.45. Let H be a simplicial function space on a compact space K such that ChH (K) is Lindel¨of. (a) Let a bounded function f ∈ Kusc (H) be a pointwise limit of a decreasing sequence of continuous functions. Then f ∗ is a pointwise limit of a decreasing sequence of continuous H-affine functions. (b) Any function f ∈ C b (ChH (K)) can be extended to a Baire-one H-affine function on K. Moreover, this extension is unique. Proof. Let f ∈ Kusc (H) be as in the statement of the theorem. Due to the Edwards inbetween theorem (Theorem 6.6), the set {h ∈ Ac (H) : h ≥ f } is down-directed and its pointwise infimum is f ∗ (see Corollary 3.25(a)). For x ∈ ChH (K), f ∗ (x) = f (x) and so, by Lemma A.54, there is a decreasing bounded sequence {fn } of continuous H-affine functions converging pointwise to f on ChH (K). The set B := {x ∈ K : fn (x) → f (x)} is a Baire set containing ChH (K), and therefore carries any maximal measure on K by Theorem 3.79(a). Hence, for all y ∈ K, Z Z ∗ f (y) = f (x) dδy (x) = lim fn (x) dδy (x) = lim fn (y). n→∞ B
B
n→∞
Thus the sequence {fn } decreases to f ∗ on K and the first part of the proof is finished. Let f be a bounded continuous function on ChH (K). We extend f to an upper semicontinuous function g on X := ChH (K) by the formula
g(x) :=
lim sup
f (y),
x ∈ X \ ChH (K),
y→x,y∈ChH (K)
f (x),
x ∈ ChH (K).
By Lemma 3.18(d), X ∗
g (x) = g(x) = f (x),
x ∈ ChH (K).
By Lemma 3.21 and Theorem 3.24, f (x) = inf{h(x) : h ∈ S c (H), h ≥ g on X},
x ∈ ChH (K).
6.6 Special classes of simplicial spaces
191
Since the latter set is down-directed, it follows from Lemma A.54 that there exist a decreasing sequence of continuous H-concave functions {gn } and an increasing sequence of continuous H-convex functions {hn } such that hn % f
and
gn & f
on ChH (K).
We have hn ≤ gn and so, by Theorem 6.6, for each n ∈ N, there is a continuous H-affine function fn with hn ≤ fn ≤ gn . Then {fn } is a bounded sequence from Ac (H) pointwise converging on ChH (K) to f , and thus, by Corollary 3.76(a), it converges pointwise on K. Hence the function h := lim fn n→∞
is the required extension of f to a Baire-one H-affine function on K. The uniqueness of the extension follows from the Minimum principle 3.86. Theorem 6.46. The following assertions are equivalent: (i) H is simplicial and T (C(K)) ⊂ (Ac (H))1,b , (ii) H is simplicial and T (C(K)) ⊂ B 1 (K), (iii) H is simplicial and T f is a Baire function for each f ∈ C(K), (iv) f ∗ is an H-affine Baire function for any continuous H-convex function f on K, S (v) F := α<ω1 (Ac (H))α,b is a lattice in the natural ordering. Proof. We start the proof by noticing that the implications (i) =⇒ (ii) =⇒ (iii) are obvious. To check (iii) =⇒ (iv), let f be a continuous H-convex function on K. By Theorem 6.5, f ∗ is H-affine. By Lemma 6.4, f ∗ = T f , hence by the hypothesis, T f is a Baire function. To verify (iv) =⇒ (i), notice that H is simplicial by Theorem 6.5. Since Kc (H) − Kc (H) is uniformly dense in C(K), T (Kc (H)) ⊂ B 1 (K) and the space of Baire-one functions is closed with respect to the uniform convergence (see Proposition A.126), we have T (C(K)) ⊂ B 1 (K). By Theorem 6.8(c), T (C(K)) ⊂ (Ac (H))⊥⊥ . By Proposition 5.42(a), T (C(K)) ⊂ (Ac (H))1,b . We proceed with the proof by showing (iii) =⇒ (v). We notice that a straightforward transfinite induction gives that T f is in F for any bounded Baire function f on K. Given functions f1 , f2 ∈ F, the function T (f1 ∨ f2 ) ≥ f1 ∨ f2 by the Minimum principle 3.86 and T (f1 ∨ f2 ) = f1 ∨ f2 on ChH (K). Again by the Minimum principle 3.86, T (f1 ∨ f2 ) is the supremum of f1 ∨ f2 in the natural ordering. We conclude the proof by showing (v) =⇒ (iii). Let f ∈ −W(H) be given, that is, f = h1 ∨ · · · ∨ hn , h1 , . . . , hn ∈ H. By the assumption, there exists a function h ∈ F such that h is the supremum of h1 , . . . , hn in the natural ordering. If g ∈ H is a function with g ≥ f , then h ≤ g. Hence h ≤ f ∗ .
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6 Simplicial function spaces
On the other hand, given x ∈ K, we find µ ∈ Mx (H) such that µ(f ) = f ∗ (x) (see Lemma 3.21). Then f ∗ (x) = µ(f ) ≤ µ(h) = h(x). Hence f ∗ = h is H-affine and H is simplicial by virtue of Theorem 6.5. Thus we have proved that f ∗ = T f is a Baire function for each f ∈ −W(H). Again we use the density of W(H)−W(H) to conclude that T f is a Baire function for each f ∈ C(K). This concludes the proof. Remark 6.47. A function space satisfying condition (i) of Theorem 6.46 are sometimes termed analytic simplicial spaces or Lion simplicial spaces. By Theorem 6.45, any simplicial function space with a Lindel¨of Choquet boundary is Lion. Example 6.48 below shows that there exists a Lion simplicial space such that its Choquet boundary is not Lindel¨of. Example 6.48. There exists a Lion simplicial space such that ChH (K) is not Lindel¨of. Proof. Let L := [0, 1], B := L, Lb := {0, 1} and µb := 21 (ε0 + ε1 ) for b ∈ B, and let H be the Stacey simplicial function space on the compact space K from Definition 6.13. Then ChH (K) is not Lindel¨of, because it is an uncountable discrete space. On the other hand, let f ∈ Kc (H) be given. For any δ > 0, the set {x ∈ L : f ∗ (x) − f (x) > δ} is finite. Hence it follows that the sets {x ∈ L : f ∗ (x) − f (x) > c}
and
{x ∈ L : f ∗ (x) − f (x) ≥ c}
are finite for each c ∈ R. Since the points of K are Gδ , the function f ∗ − f is Baire-one by Theorem A.124. Hence f ∗ is Baire-one as well. By Theorem 6.8(a), T (C(K)) ⊂ B 1 (K), and thus H is Lion by Theorem 6.46.
6.6.D
Simplicial spaces with boundaries of type Fσ
We know that (in a simplicial case) the Dirichlet problem on the Choquet boundary is solvable for every continuous function on it exactly when this boundary is closed. More precisely, Theorem 6.37 yields that H is a simplicial space with closed Choquet boundary ChH (K) (Bauer simplicial space) if and only if given a bounded continuous function f on ChH (K) there is a continuous H-affine function h such that h = f on ChH (K). We now consider the question, under what conditions every bounded Baire-one function on ChH (K) can be extended to a bounded Baire-one H-affine function on K. To this end, we need an analogue of envelopes defined similarly as in Section 6.5. For a function f on K we define f ∗,1 := inf{h ∈ A(H) ∩ B b1 (K) : h ≥ f } and
f∗,1 := −(−f )∗,1 .
6.6 Special classes of simplicial spaces
193
Theorem 6.49. Let H be a function space on a metrizable compact space K. Consider the following assertions: (i) H is simplicial and ChH (K) is an Fσ set, (ii) for any bounded Baire-one function f on ChH (K) there exists a bounded Haffine Baire-one function h such that f = h on ChH (K), (iii) for any bounded Baire-one function f on K there exists a bounded H-affine Baire-one function h such that f = h on ChH (K), (iv) H is simplicial and T f ∈ B 1 (K) for any f ∈ B b1 (K), (v) f ∗,1 is an H-affine Baire-one function for every bounded H-convex function f ∈ B 1 (K), (vi) A(H) ∩ B b1 (K) is a lattice in the natural ordering. Then assertions (i)–(v) are equivalent and (v) =⇒ (vi). Proof. For the proof of the implication (i) =⇒ (ii), suppose S∞ that H is simplicial and ChH (K) is an Fσ set. Thus we can write ChH (K) = n=1 Fn where {Fn } is an increasing sequence of compact sets. Let f be a bounded Baire-one function on ChH (K) and {fn } be a sequence of continuous functions on ChH (K) converging pointwise to f . We may assume that kf k, kfn k are all bounded by a positive number M . By Theorem 6.27, there exists a sequence {hn } of H-affine continuous functions on K such that hn = fn on Fn and khn k = kfn k. The proof will be completed by showing that the sequence {hn } converges pointwise to the function h := T f . Then T f is Baire-one and H-affine by the Lebesgue dominated convergence theorem. Notice that the definition is meaningful since maximal measures are carried by ChH (K) due to Theorem 3.81. R For a fixed point x ∈ K and ε > 0 we find n0 ∈ N such that ChH (K) |f −fn | dδx < ε and δx (Fn ) > 1 − ε for all n ≥ n0 . Then, for n ≥ n0 , we have Z |T f (x) − hn (x)| = (f − hn ) dδx K Z Z ≤ |f − fn | dδx + Fn0
2M dδx
K\Fn0
≤ ε + ε2M, which proves the required statement and concludes the proof of (i) =⇒ (ii). Since the implication (ii) =⇒ (iii) is obvious, we proceed to the proof of (iii) =⇒ (iv). Let H be a function space on a compact space K satisfying condition (iii). First we verify that H is simplicial. Indeed, for a given continuous H-convex function f on K we find an H-affine Baire-one function h with h = f on ChH (K).
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6 Simplicial function spaces
By Theorem 3.86, h ≥ f on K. For a given x ∈ K, Lemma 3.21 yields the existence of a measure µ ∈ Mx (H) such that µ(f ) = f ∗ (x). Then f ∗ (x) = µ(f ) ≤ µ(h) = h(x). On the other hand, let g ∈ H satisfy g ≥ f . Then g ≥ h on ChH (K), and thus g ≥ h on K, which again follows from Theorem 3.86. Hence h(x) ≤ inf{g(x) : g ∈ H, g ≥ f } = f ∗ (x). Thus f ∗ = h is H-affine for every H-convex continuous function f and H is a simplicial function space by Theorem 6.5. It remains to check that T f is a Baire-one function on K for every f ∈ B b1 (K). If f is a bounded Baire-one function on K, let h be an H-affine Baire one function on K with f = h on ChH (K). Then h = T f on ChH (K) and the application of Theorem 3.86 yields that h = T f on K. Thus T f is a Baire-one function as required. The next step will be a proof of the implication (iv) =⇒ (v). Let f be an H-convex Baire-one function on K. By condition (iv), T f is an H-affine Baire-one function. Moreover, T f = f on ChH (K) and T f ≥ f on K by the Minimum principle 3.86. Thus T f ≥ f ∗,1 . On the other hand, given an H-affine Baire-one function h with h ≥ f , the minimum principle laid down in Theorem 3.86 gives h ≥ T f . Thus T f ≤ f ∗,1 and f ∗,1 = T f is an H-affine Baire-one function. In order to prove (v) =⇒ (vi), let f and g be H-affine Baire-one functions. Since h = (f ∨ g)∗,1 is an H-affine Baire-one function, h is obviously the least upper bound for the pair f and g in A(H) ∩ B b1 (K). Thus A(H) ∩ B b1 (K) is a lattice in its natural ordering. It remains to prove the implication (v) =⇒ (i). First we check that H is simplicial. Let f ∈ Kc (H) be given. Then f ∗,1 ≤ f ∗ by the definition. On the other hand, given x ∈ K, let µ ∈ Mx (H) satisfy µ(f ) = f ∗ (x). Then f ∗ (x) = µ(f ) ≤ inf{µ(h) : h ∈ A(H) ∩ B b1 (K), h ≥ f } = f ∗,1 (x). Hence f ∗ = f ∗,1 is H-affine and H is simplicial by Theorem 6.5. Assume that ChH (K) is not an Fσ set in K. We fix on K a compatible metric ρ. Since ChH (K) is a Gδ subset of a compact metrizable space, Theorem A.115 yields the existence of a homeomorphic copy C ⊂ K of the Cantor set {0, 1}N such that C ∩ ChH (K) = C \ ChH (K) = C
and
C \ ChH (K) is countable.
We enumerate C \ ChH (K) as {qn : n ∈ N} and, using regularity of Radon measures, we find compact sets Kn ⊂ ChH (K), n ∈ N, such that δqn (Kn ) > 1 − n1 , n ∈ N (see Proposition 3.80). Then there is no Fσ set F ⊂ K satisfying C\
∞ [ n=1
Kn ⊂ F ⊂ K \ {qn : n ∈ N}.
6.6 Special classes of simplicial spaces
195
(Indeed, otherwise the set C ∩ ChH (K) would be an Fσ set, which is absurd.) S Another use of Theorem A.115 provides a homeomorphic copy D ⊂ C \ ∞ n=1 Kn of the Cantor set {0, 1}N such that D ∩ {qn : n ∈ N} = D ∩ ChH (K) = D. Then cD is a Baire-one function on K and T cD (x) = 1, T cD (x) ≤
1 , n
x ∈ D \ ChH (K), x = qn , n ∈ N.
Hence T cD has no point of continuity on D, and thus it is not a Baire-one function (see Theorem A.127). This concludes the proof. Remark 6.50. We remark that only the proof of (v) =⇒ (i) in the preceding theorem used the assumption of metrizability of K. Example 6.51. There exists a simplicial space H on a metrizable space K such that B b1 (K) ∩ A(H) is a lattice in the natural ordering but ChH (K) is not an Fσ set. Proof. We take a suitable Stacey simplicial function space from Definition 6.13. Namely, we take L := [0, 1] and B := Q ∩ [0, 1] and K as in Definition 6.13. We enumerate B as {qn : n ∈ N} and take Lqn := {0, 1} with µqn := n1 ε0 + (1 − n1 )ε1 , n ∈ N. By Lemma 6.14, H is simplicial and ChH (K) = K \ B. Hence ChH (K) is not an Fσ set. Further we know from Lemma 6.14 that δqn = µqn . For brevity we write qn+ for the point 1 in Lqn and qn− for the point 0 in Lqn . In order to check that A(H)∩B b1 (K) is a lattice in the natural ordering, it is enough to prove that T (f ∨g) is a Baire-one function for every pair f and g of H-affine Baireone functions. Let f and g be such functions with values in [0, 1] and set h := f ∨ g. We claim that 2 for every n ∈ N. |h(qn+ ) − h(qn )| ≤ (6.5) n Indeed, for a fixed n ∈ N we have 1 1 f (qn− ) − (1 − )f (qn+ )| n n 2 1 ≤ |f (qn+ ) − f (qn− )| ≤ . n n
|f (qn+ ) − f (qn )| = |f (qn+ ) −
By the same argument, |g(qn+ ) − g(qn )| ≤ n2 . We need to check this inequality for the function h. The only nontrivial case is when h(qn+ ) = f (qn+ ) and h(qn ) = g(qn ) (or
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6 Simplicial function spaces
vice versa). Then h(qn+ ) = f (qn+ ) = f (qn+ ) − f (qn ) + f (qn ) ≤ f (qn+ ) − f (qn ) + g(qn ) ≤ and
2 2 + g(qn ) = + h(qn ), n n
h(qn ) = g(qn ) = g(qn ) − g(qn+ ) + g(qn+ ) ≤ g(qn ) − g(qn+ ) + f (qn+ )
2 2 + f (qn+ ) = + h(qn+ ). n n Combining these inequalities together we get (6.5). Applying inequality (6.5) we obtain ≤
1 1 |T h(qn ) − h(qn )| = | h(qn− ) + (1 − )h(qn+ ) − h(qn )| n n 1 = | (h(qn− ) − h(qn+ )) + h(qn+ ) − h(qn )| n 2 4 2 ≤ + = . n n n Hence the set {x ∈ K : |T h(x) − h(x)| ≥ ε} = {qn ∈ [0, 1] : |T h(qn ) − h(qn )| ≥ ε, n ∈ N} is finite for every ε > 0. By Lemma A.128, T h is a Baire-one function and the space A(H) ∩ B b1 (K) is a lattice in the natural ordering.
6.7
The Daugavet property of simplicial spaces
In 1963, I. K. Daugavet in [131] showed that any compact operator S on the Banach space C([0, 1]) satisfies the identity kI + Sk = 1 + kSk.
(6.6)
(Here, I stands for the identity operator.) Afterwards, Banach spaces and operators satisfying the equality (6.6) were studied. A Banach space E satisfying (6.6) for any compact operator S on E is said to satisfy the Daugavet property, or, for short, the Daugavet space. Among other properties, it was shown that a Banach space satisfies the equality (6.6) for any compact operator S if and only if the equality (6.6) is satisfied for the class of all rank-one operators, and that the space C(K) has the Daugavet property if and only if the compact space K has no isolated points (see, for example, V. M. Kadets, R. V. Shvidkoy, G. G. Sirotkin and D. Werner [255]).
6.7 The Daugavet property of simplicial spaces
197
Definition 6.52 (The Daugavet property). A Banach space E has the Daugavet property if kI + F k = 1 + kF k for each rank-one operator F on E. Theorem 6.53. Let H be a simplicial function space on a compact space K such that ChH (K) has no isolated points. Then Ac (H) has the Daugavet property. Proof. Let a rank-one operator F : Ac (H) → Ac (H) be given. There exists a functional ϕ ∈ Ac (H)∗ and a function h0 ∈ Ac (H) such that F (h) = ϕ(h)h0 ,
h ∈ Ac (H).
Using Lemma 6.23 we find a boundary measure µ ∈ M(K) such that kµk = kϕk and
µ = ϕ on Ac (H).
Fix ε > 0. By Theorem 3.85, there exists a point y0 ∈ ChH (K) (= ChAc (H) (K) by Exercise 3.95) such that h0 (y0 ) = kh0 k. Let W be an open neighborhood of y0 such that h0 > kh0 k − ε on W . Since ChH (K) has no isolated points, there exists a point x0 ∈ W ∩ ChH (K) and an open neighborhood U of x0 such that |µ|(U ) < ε. Let f be a continuous function on K such that kf k = 1 and µ(f ) > kµk − ε. According to Tietze’s theorem, there exists a continuous function g on K such that kgk = 1,
g=f
on K \ U
and g(x0 ) = 1.
Then µ(g) > kµk − 2ε. Since µ is a boundary measure, µ(g) = µ(g ∗ ). For the positive part µ+ of µ we find by Theorem A.84 a continuous H-concave function k + such that g ≤ k+ ≤ 1
and µ+ (k + ) − µ+ (g ∗ ) < ε.
Similarly we choose a continuous H-concave function k − such that g ≤ k− ≤ 1
and µ− (k − ) − µ− (g ∗ ) < ε.
Then k := k + ∧ k − is a continuous H-concave function satisfying |µ(k) − µ(g ∗ )| < 2ε, kkk ≤ 1 and k(x0 ) = g ∗ (x0 ) = g(x0 ) = 1. Another use of Theorem A.84 along with Theorem 6.6 provides a function h ∈ Ac (H) such that −1 ≤ h ≤ k,
|µ(h) − µ(k)| < ε and
|h(x0 ) − k(x0 )| < ε
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6 Simplicial function spaces
(recall that k∗ (x0 ) = k(x0 ) by Theorem 3.24). Combining all the estimates together we get (I + F )(h)(x0 ) = h(x0 ) + µ(h)h0 (x0 ) ≥ k(x0 ) − ε + (µ(k) − ε)(kh0 k − ε) ≥ 1 − ε + (µ(g ∗ ) − 3ε)(kh0 k − ε) = 1 − ε + (µ(g) − 3ε)(kh0 k − ε) ≥ 1 − ε + (kµk − 5ε)(kh0 k − ε) = 1 + kµkkh0 k − ε(kµk − 5kh0 k − 1 − 5ε). Since ε > 0 is arbitrary, this shows that kI + F k = 1 + kµkkh0 k = 1 + kF k, finishing the proof.
6.8 6.8.A
Choquet simplices Simplicial function spaces and the classical definition of Choquet simplices
We recall that a compact convex set X is a simplex if the function space Ac (X) is a simplicial function space on X. Theorem 6.54 shows how the simpliciality of a function space is related to properties of the state space of Ac (H). Theorem 6.54. For a function space H, the following assertions are equivalent. (i) H is simplicial. (ii) The ordered Banach space Ac (H)∗ is a vector lattice. (iii) The compact convex set S(Ac (H)) is a simplex. Proof. We start the proof by noticing that (i) =⇒ (ii) due to Theorem 6.18 and Theorem A.24. The next step is showing that (ii) =⇒ (iii). Let X := S(Ac (H)). We want to prove that Ac (X) is simplicial, which means, by virtue of Theorem 6.5 and Corollary 4.8, to check that f ∗ is affine for any continuous convex function f on X. Since f ∗ is always concave, we need to check its convexity. To this end, let s := α1 s1 + α2 s2 be a nontrivial convex combination of points s1 , s2 ∈ X. We fix ε > 0 P and use Lemma 3.21 along with Proposition 4.3 to find a convex combination s = nj=1 βj tj , t1 , . . . , tn ∈ X, β1 , . . . , βn strictly positive, such that n X βj f (tj ) > f ∗ (s) − ε. j=1
6.8 Choquet simplices
199
Using Proposition A.11, we find positive functionals u0ij ∈ (Ac (H)∗ , i = 1, 2, j = 1, . . . , n, so that αi si =
n X
u0ij ,
i = 1, 2,
and
βj tj = u01j + u02j ,
j = 1, . . . , n.
j=1
Let γij ≥ 0 and uij ∈ X satisfy u0ij = γij uij , i = 1, 2, j = 1, . . . , n. Then we get the following convex combinations si =
n X γij
αi
j=1
tj =
uij ,
i = 1, 2,
γ1j γ2j u1j + u2j , βj βj
and
j = 1, . . . , n.
We use the convexity of f , concavity of f ∗ and preceding equalities to get ∗
f (s) − ε <
n X
βj f (
j=1
≤
n X
γ1j γ2j u1j + u2j ) βj βj
γ1j f (u1j ) + γ2j f (u2j )
j=1
= α1
n X γ1j j=1
≤ α1
α1
n X γ1j j=1
α1
f (u1j ) + α2
n X γ2j j=1
f ∗ (u1j ) + α2
α2
f (u2j )
n X γ2j j=1
α2
f ∗ (u2j )
≤ α1 f ∗ (s1 ) + α2 f ∗ (s2 ). Since ε is arbitrary, f ∗ (s) = α1 f ∗ (s1 ) + α2 f ∗ (s2 ), and f ∗ is affine. Now we verify (iii) =⇒ (i). Let φ : K → X be the embedding from Definition 4.25. Given a point x ∈ K, let µ, ν ∈ Mx (Ac (H)) be Ac (H)-maximal measures. Then φ] µ and φ] ν are maximal measures on X (see Proposition 4.28(d)) that represent the point φ(x) (use Proposition 4.26(c)). Hence φ] µ = φ] ν, and µ = ν. Thus Ac (H) is simplicial. By Lemma 6.3(b), H is simplicial. Remark 6.55. We remark that the original definition of Choquet simplices is different from ours. We recall that X is affinely homeomorphic to the state space S(Ac (X)), which is contained in the dual space E = (Ac (X))∗ . This identification is called a regular embedding of X (see p. 82 in E. M. Alfsen [5]). A regularly embedded compact convex set X is called a simplex in Chapter II, §3 of [5] if E is a vector lattice with the ordering given by the cone E + . Thus from Theorem 6.54 we get the following classical definition of Choquet simplices.
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6 Simplicial function spaces
Theorem 6.56. Let X be a compact convex set that is regularly embedded in the locally convex space E = (Ac (X))∗ . Then X is a Choquet simplex if and only if E is a vector lattice.
6.8.B
Prime function spaces and prime compact convex sets
Definition 6.57 (Antilattices). We recall that a closed function space H with its natural order is an example of an ordered Banach space (see Example A.19(a)). It is called an antilattice if whenever f, g ∈ H and the infimum f f g exists in H then either f ≤ g or g ≥ f . Definition 6.58 (Exposed and relatively exposed sets). A set F in a compact convex set X is called relatively exposed if for any x ∈ X \ F there exists f ∈ Ac (X) so that f ≥ 0 on X, f = 0 on F and f (x) > 0 (cf. Definition 8.47). A set F ⊂ X is said to be exposed if there is a positive function f ∈ Ac (X) such that f (x) = 0 if and only if x ∈ F . We remark that any exposed set is relatively exposed and any relatively exposed set is a closed face (cf. Proposition 8.48). Definition 6.59 (Prime compact convex sets). A compact convex set X is said to be prime if whenever F and G are closed relatively exposed faces of X such that co(F ∪ G) = X, then either F = X or G = X. Theorem 6.60. A compact convex set X is prime if and only if the function space Ac (X) is an antilattice. Proof. Assume that Ac (X) is an antilattice and X = co(F ∪ G) where F and G are proper closed relatively exposed faces. We find nonzero positive continuous affine functions f and g so that f = 0 on F and g = 0 on G. Due to Minimum principle 3.85, the infimum f f g exists and equals 0. Since Ac (X) is an antilattice, either f ≤ g or g ≤ f . We have arrived at the conclusion that either f = 0 or g = 0 which is a contradiction with our choice of f and g. Conversely, suppose that X is a prime compact convex set and f, g ∈ Ac (X) are such that f f g exists in Ac (X). It follows from the definition of the lower envelope that f f g = (f f g)∗ . By Theorem 3.24, f f g = f ∧ g on ext X. If we set F := {x ∈ X : (f f g)(x) = f (x)}
and
G := {x ∈ X : (f f g)(x) = g(x)},
we obtain two exposed faces satisfying ext X ⊂ F ∪ G. Thus X = co(F ∪ G). Since X is supposed to be prime, either X = F or X = G. The former equality yields f ≤ g, the latter one implies g ≤ f . Thus we have verified that Ac (X) is an antilattice and the proof is finished.
6.8 Choquet simplices
201
Lemma 6.61. Let X be a simplex and F ⊂ X be a closed Gδ face of X. Then F is exposed. Proof. Let F be a closed Gδ face of X. For any x ∈ X \F we find a continuous affine function fx so that sup fx (F ) ≤ 0 < fx (x). Since X \ FSis a Lindel¨of space, we can ∞ select countably many points {xn }∞ n=1 so that X \ F ⊂ n=1 {y ∈ X : fxn (y) > 0}. Set ∞ X 1 hn . hn := fxn ∨ 0, n ∈ N, and h := 2n khn k n=1
Then h is a continuous convex function on X such that h = 0 on F and h > 0 on X \ F . If we define g := cX\F , we obtain a lower semicontinuous concave function with h ≤ g. Using Theorem 6.6 we obtain an affine continuous function f ∈ Ac (X) so that h ≤ f ≤ g. Obviously f exposes F , which concludes the proof. The previous Lemma 6.61 shows that the following definition of prime simplices is a particular case of the general definition of prime compact convex sets. Definition 6.62 (Prime simplices). A simplex X is said to be prime if given closed Gδ faces F and G in X with X = co(F ∪ G), then either F = X or G = X. Definition 6.63 (Prime simplicial spaces). A simplicial function space H is prime if Ac (H) is an antilattice. Proposition 6.64. Let H be a simplicial space. The following assertions are equivalent: (i) H is prime, (ii) the simplex X := S(Ac (H)) is prime. Proof. Let Φ : Ac (H) → Ac (X) be the surjective isometric isomorphism from Definition 4.25. By Corollary 5.41, Φ preserves order of functions and hence for any f, g ∈ Ac (H), f f g exists if and only if Φ(f ) f Φ(g) exists. Thus the conclusion follows from Theorem 6.60. Lemma 6.65. Let H be a simplicial space. If there exists a point x ∈ ChH (K) such that ChH (K) ⊂ spt δx , then H is prime. Proof. Suppose that for given f, g ∈ Ac (H) there exists f f g ∈ Ac (H). It follows from the definition of the lower envelope that f fg = (f ∧g)∗ . Thus f fg = f ∧g on ChH (K), by Theorem 3.24. By continuity, this equality also holds for every point in ChH (K). In particular, (f fg)(x) = (f ∧g)(x). Let us assume that (f ∧g)(x) = f (x). From the equalities Z Z (f ∧ g) dδx = (f f g) dδx = (f f g)(x) ChH (K)
ChH (K)
Z = (f ∧ g)(x) = f (x) =
f dδx ChH (K)
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6 Simplicial function spaces
we get that f = f ∧ g almost everywhere with respect to δx . Since ChH (K) ⊂ spt δx , the equality f = f ∧ g holds on ChH (K). In other words, f ≤ g on ChH (K). The Minimum principle 3.85 yields the desired conclusion that f ≤ g on K. Lemma 6.66. Let X be a compact convex set such that ext X = X. Then X is prime. Proof. Suppose that for functions f, g ∈ Ac (X) there exists f f g in Ac (X). As in the previous proof we have f f g = (f ∧ g)∗ . Since (f ∧ g)∗ = f ∧ g on X and ext X is dense in X, we obtain that f ∧ g = f f g on X. We claim that this is possible only if either f ≤ g or g ≤ f . Indeed, suppose that this is not the case, that is, we can find points x, y ∈ X so that f (x) < g(x) and f (y) > g(y). Then it is obvious that f ∧ g = f f g cannot be affine on the segment joining x and y. This contradiction finishes the proof. Example 6.67. There exists a simplicial function space H on a compact space K such that ChH (K) = K and H is not prime. Proof. Let K := [0, 1] ∪ {2} ∪ {3} and H := {f ∈ C(K) : f (0) = 21 (f (2) + f (3))}. Then H = Ac (H), H is simplicial and ChH (K) = K \ {0}. Nevertheless, H is not an antilattice. Indeed, if ( x(1 − x) x ∈ [0, 1], f (x) := 0 otherwise, and ( g(x) :=
1 4x
0
x ∈ [0, 1], otherwise,
then f, g ∈ Ac (H) and f ∧ g ∈ Ac (H) as well and neither f ≤ g nor g ≤ f . Thus f f g exists although ChH (K) is dense in K.
6.8.C
Characterization of Bauer simplices by faces
In this subsection we present another characterization of Bauer simplices by means of faces. We recall that a compact convex set X is a Bauer simplex if X is a simplex whose set of extreme points is closed (see Definition 6.36), and that δx denotes the unique maximal measure in Mx (X). Lemma 6.68. Let X be a simplex and H ⊂ ext X. Then co H is a face of X. Proof. Since Pnco H is convex, it is enough to show that co H is extremal. So assume that x := j=1 λj hj = αy + (1 − α)z, where hj ∈ H, λj ≥ 0, j = 1, . . . , n,
6.8 Choquet simplices
203
P = 1, α ∈ (0, 1) and y, z ∈ X. Then nj=1 λj εhj ∈ Mx (X) is maximal by Corollary 3.59 and αδy + (1 − α)δz ∈ Mx (X). Hence Pn
j=1 λj
n X
λj εhj = αδy + (1 − α)δz ,
j=1
which yields y, z ∈ co H. Theorem 6.69. Let X be a simplex. The following assertions are equivalent: (i) X is a Bauer simplex, (ii) F is a face for any face F ⊂ X. Proof. For the proof of (i) =⇒ (ii), let X be a Bauer simplex and F ⊂ X be a face. Since F is convex, we only have to show that F is extremal, that is, our aim is to prove that spt δx ⊂ F for every x ∈ F (see Exercise 6.82). First, suppose that x ∈ F and that δx is not carried by F . Let V be a closed convex neighborhood not intersecting F such that δx (V ) > 0. If δx (V ) = 1, then x ∈ V and thus V ∩ F 6= ∅, which is impossible. Setting c := δx (V ) ∈ (0, 1), we have δx = cδx |V + (1 − c)δx |X\V , hence x = cr(δx |V ) + (1 − c)r(δx |X\V ). Since F is a face, r(δx |V ) ∈ F and thus r(δx |V ) ∈ V ∩ F , which is a contradiction. Assuming that x ∈ F , there exists a net {xα } in F such that xα → x. By (iii) of Theorem 6.37, δxα → δx . Since spt δxα ⊂ F , we get spt δx ⊂ F . To show that (ii) =⇒ (i), assume that there exists x ∈ ext X \ ext X. There exist y, z ∈ X, y 6= x, such that 2x = y + z. The Hahn–Banach theorem asserts the existence of f ∈ E ∗ and c ∈ R such that f (y) < c < f (x). Denote H := ext X ∩ {t ∈ X : f (t) ≥ c}. According to Lemma 6.68, co H is a face. On the other hand, co H is not a face. Indeed, co H is not extremal since x ∈ H ⊂ co H and y∈ / co H.
6.8.D
Fakhoury’s theorem
Theorem 6.70. Let X be a compact convex set. Then the following conditions are equivalent: (i) X is a simplex, (ii) there exists an affine mapping T : X → M1 (X) such that T (x) ∈ Mx (X).
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6 Simplicial function spaces
Proof. If X is a simplex, it is enough to check that the mapping x 7→ δx is affine. Let x and y be points of X and α ∈ [0, 1]. It suffices to verify that δαx+(1−α)y (f ) = αδx (f ) + (1 − α)δy (f ) for any f ∈ Kc (X). By Corollary 4.8 and Lemma 6.4, the previous equality can be rewritten as f ∗ (αx + (1 − α)y) = αf ∗ (x) + (1 − α)f ∗ (y), which holds because the function f ∗ is affine by Theorem 6.5. For the proof of the reverse implication, let T be an affine mapping provided by condition (ii). We write Tx for the measure T (x). We need to prove that f ∗ is affine for any f ∈ Kc (X). Since T is assumed to be affine, it is enough to verify that Tx (f ) = f ∗ (x) for any f ∈ Kc (X) and x ∈ X. Let f and x be as above and ε > 0. By Lemma 3.22 and Proposition 4.3, there ∗ exists − ε. Then µ = Pnthat µ(f ) ≥ fP(x) Pn a molecular measure µ ∈ Mx (X) such n i=1 αi xi = x. Since i=1 αi = 1 and i=1 αi εxi where xi ∈ X, αi > 0 such that f is convex and T is affine, Proposition 3.20(b) gives ∗
f (x) − ε ≤ µ(f ) =
n X
αi εxi (f ) ≤
i=1
n X
αi Txi (f )
i=1
= TPni=1 αi xi (f ) = Tx (f ) ≤ f ∗ (x). As ε is arbitrary, Tx (f ) = f ∗ (x). This concludes the proof.
6.9
Restriction of function spaces
Definition 6.71 (Restricted function space). Let H be a function space on a compact space K and L ⊂ K be a closed subset. Then G := {h|L : h ∈ H} is the restricted function space. Proposition 6.72. The following assertions hold: (a) If µ, ν ∈ M+ (L), then µ ≺H ν if and only if µ ≺G ν. (b) Mx (G) = Mx (H) ∩ M1 (L) for any x ∈ L. (c) ChH (K) ∩ L ⊂ ChG (L). (d) If L ⊃ ChH (K), then G-maximal measures coincide with H-maximal measures and ChG (L) = ChH (K). Proof. For the proof of (a), we notice that for µ, ν ∈ M+ (L), µ ≺H ν if and only µ(f ) ≤ ν(f ) for each f ∈ −W(H) (see Proposition 3.56), and this is the case if and only if µ(f ) ≤ ν(f ) for any f ∈ −W(G). Further, (b) and (c) are obvious.
6.10 Exercises
205
Assume that L ⊃ ChH (K). Let µ ∈ M+ (K) be an H-maximal measure. Then µ is carried by L (see Proposition 3.64). Let ν ∈ M+ (L) be a G-maximal measure with µ ≺G ν. Then µ ≺H ν, and thus µ = ν is G-maximal. Conversely, let µ ∈ M+ (L) be a G-maximal measure. If ν ∈ M+ (K) is Hmaximal measure with µ ≺H ν, then ν ∈ M+ (L). For any function f ∈ −W(H), we have µ(f |L ) = µ(f ) ≤ ν(f ) = ν(f |L ). Thus µ(g) ≤ ν(g) for any function g ∈ −W(G). By Proposition 3.56, µ ≺G ν. Hence µ = ν is H-maximal. Since x ∈ ChH (K) if and only if εx is H-maximal, the second part of (d) follows from the previous argument.
6.10
Exercises
Exercise 6.73. Let H be a simplicial space and x ∈ K. Prove that δx ∈ ext(Mx (H)). Hint. By Exercise 3.113, H-maximal probability measures form a face F in M1 (K). Since Mx (H) ∩ F = {δx }, the assertion follows. Exercise 6.74. Let X be a simplex in Rd . Prove that X has at most d + 1 extreme points and they are affinely independent. Hint. If {x0 , . . . , xd+1 } ⊂ ext X are distinct points, Pd+1dependent. P they are affinely a = 0 and Hence there exist numbers a0 , . . . , ad+1 such that d+1 i i=0 ai xi = 0 i=0 (see Definition 2.2). We assume that a0 , . . . , an ≥ 0 and an+1 , . . . , ad+1 < 0. Then 0 < c :=
n X
ai =
i=0
d+1 X
−ai ,
i=n+1
and x :=
n X i=0
c−1 ai xi =
d+1 X
−c−1 ai xi .
i=n+1
Hence x has two different maximal representing measures, a contradiction. Analogously we get that the extreme points of X are affinely independent. Exercise 6.75. Let X be a compact convex set in Rd . Prove that X is a simplex if and only if X is an n-simplex for some n ≤ d (see Definition 2.2). Hint. If X is an n-simplex with vertices {x0 , . . . , xn }, then ext X = {x0 , . . . , xn } by Lemma Pn 2.10 and any point x ∈ X is expressed by a unique convex combination x = i=0 λi xi (the points {x0 , . . . , xn } are affinely independent).
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6 Simplicial function spaces
Conversely, let X be a simplex. By Exercise 6.74, ext X = {x0 , . . . , xn } for some n ≤ d and the set of extreme points consists of affinely independent points. Hence X is an n-simplex by Definition 2.2. Exercise 6.76. Let H := {f ∈ C([0, 1]) : there is af > 0 such that f is constant on [0, af ]} . Since the function space H is dense in C([0, 1]), we have ChH ([0, 1]) = ChH ([0, 1]) = ChC([0,1]) ([0, 1]) = [0, 1]. It follows that H is simplicial and {0} ∈ ChH ([0, 1]). Nevertheless, {0} is not Hexposed. Exercise 6.77. Let X be the convex hull of two sets (x, y) ∈ R2 : (x + 1)2 + y 2 = 1 and (x, y) ∈ R2 : (x − 1)2 + y 2 = 1 and let K be the boundary of X. If H consists of restrictions of all affine functions on R2 to K, then H is a closed function space on K. Prove that ChH (K) = {(x, y) ∈ K : x ≤ −1 or x ≥ 1} and that a function f ∈ C(K) belongs to Ac (H) if and only if f is affine on each segment contained in K. Further show that H is simplicial and that the point (1, 1) is Ac (H)-exposed but it is not H-exposed. Hint. In order to show that the point (1, 1) is Ac (H)-exposed, consider, for example, the function (x, y) 7→ | − x − y + 2| + |1 − y|, (x, y) ∈ K.
Exercise 6.78. Let H be a closed function space on a compact space K possessing the weak Riesz interpolation property. Then Ac (H) = H. Hint. If f ∈ Ac (H) and ε > 0, for any x ∈ K we use Corollary 3.23(a) and Proposition 3.25(a) to find a function hx ∈ H such that f < hx and hx (x) < f (x) + ε. A simple compactness argument yields the existence of finitely many functions h1 , . . . , hn ∈ H such that f < h1 ∧ · · · ∧ hn < f + ε. Analogously, we find functions g1 , . . . , gm ∈ H with f > g1 ∨ · · · ∨ gm > f − ε.
6.10 Exercises
207
By the weak Riesz interpolation property, there exists a function h ∈ H such that g1 ∨ · · · ∨ gm < h < f1 ∧ · · · ∧ fn . Hence kf − hk < 2ε, which gives f ∈ H = H. Exercise 6.79. Find an example of a simplicial function space whose state space is not a simplex. Hint. Consider the function space H := P 2 ([0, 1]) from Example 3.2(b). Then ChH ([0, 1]) = [0, 1] (see Example 3.5(b)), and thus H is simplicial. On the other hand, if ψ : [0, 1] → R2 is defined by ψ(t) = (t, t2 ),
t ∈ [0, 1],
S(H) is affinely homeomorphic to co ψ([0, 1]) (use Exercise 4.51). Hence S(H) is not a simplex. Exercise 6.80. Let X be a compact convex set, µ, ν ∈ M+ (X), µ ≺ ν. Prove that spt µ ⊂ co spt ν. Hint. If z ∈ spt µ \ co spt ν, find a continuous affine function f on X such that f (z) = −1 and inf f co spt ν ≥ 0. If k := f ∧ 0, then k is continuous concave on X, k = 0 on spt ν and ν(k) ≤ µ(k) < 0. Exercise 6.81. Let F be a closed extremal subset of a compact convex set X, µ, ν ∈ M+ (X), µ ≺ ν. If spt µ ⊂ F , then spt ν ⊂ F . Hint. The characteristic function cX\F is concave (cf. Proposition 2.68) and lower semicontinuous. By Proposition 3.56, ν(cX\F ) ≤ µ(cX\F ). Exercise 6.82. Let F be a closed convex subset of a compact convex set X. Prove that F is extremal if and only if spt ν ⊂ F for any x ∈ F and any maximal measure ν ∈ Mx (X). Hint. If F is a closed convex extremal set, it carries any maximal measure with barycenter in F by Proposition 2.69. Conversely, let x ∈ F and µ ∈ Mx (X). Let ν ∈ M1 (X) be maximal with µ ≺ ν. Then ν ∈ Mx (X). By Exercise 6.80, spt µ ⊂ co spt ν ⊂ F . Hence spt µ ⊂ F and F is extremal. Exercise 6.83. Let F be a closed face in a compact convex set X. Prove that Ac (F )maximal measures coincide with Ac (X)-maximal measures carried by F .
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6 Simplicial function spaces
Hint. Let µ ∈ M1 (F ) be Ac (X)-maximal and ν ∈ M1 (F ) satisfy µ ≺Ac (F ) ν. Then µ ≺Ac (X) ν and thus µ = ν. If µ ∈ M1 (F ) is Ac (F )-maximal and ν ∈ M1 (X) with µ ≺Ac (X) ν, then ν is carried by F due to Exercise 6.81. If f is a positive convex function on F , the function g := f on F and 0 on X \ F is upper semicontinuous and convex on X. By Proposition 3.56, µ(f ) = µ(g) ≤ ν(g) = ν(f ). Hence µ ≺Ac (F ) ν, and thus µ = ν. It follows that µ is Ac (X)-maximal. Exercise 6.84. Let X be a simplex and H a closed extremal subset of X. Then co H is a closed face of X (cf. Proposition 8.31 and Lemma 6.68). Hint. By Exercise 6.82, we only have to show that spt δx ⊂ co H for any x ∈ co H. Fix x ∈ co H. Using Proposition 2.39 find a measure µ ∈ Mx (X) carried by H. Then µ ≺ δx . In view of Exercise 6.81, we get spt δx ⊂ H. Exercise 6.85. Let X be a simplex and F ⊂ X be a closed face. Prove that F is a simplex. Hint. This follows from Exercise 6.83. Exercise S 6.86. Let X be a simplex and {Fα } a family of closed faces of X. Then co Fα is a face of X. Hint. Similar to the proof of Lemma 6.68. Exercise 6.87. Let X be a compact convex subset of a locally convex space. (a) If X is metrizable and F is a face of X, then F = co(F ∩ ext X). In particular, F ∩ ext X 6= ∅. (b) Prove that there exists a simplex X and x ∈ X such that face x ∩ ext X = ∅. Hint. For the proof of (a), let x ∈ F . By Theorem 3.45, there exists µ ∈ Mx (X) carried by ext X. Since, by Remark 2.70(a), µ is carried by F , we see that µ is carried by F ∩ ext X. Hence, x ∈ co(F ∩ ext X). It follows that F ⊂ co(F ∩ ext X), and therefore F ⊂ co(F ∩ ext X) ⊂ F . To verify (b), we use the existence of a simplex X and a maximal measure µ ∈ M1 (X) such that µ is carried by a compact set K disjoint from ext X. (The construction of X is based on the state space of the function space H of Example 3.82. In fact, the state space X of S(H) is a simplex by Lemma 6.14 and Theorem 6.54.) By Exercise 2.116, co K ∩ ext X = ∅. Let x be the barycenter of µ. Then x ∈ co K (Proposition 2.39). We have face x ∩ ext X = ∅. Indeed, choose y ∈ face x. By Proposition 2.66, there exist z ∈ X and λ ∈ [0, 1) such that x = λz + (1 − λ)y. Let νy and νz be maximal measures representing points y and z, respectively. Then λνz + (1 − λ)νy is a maximal measure representing x. Since X is a simplex, µ = λνz + (1 − λ)νy , and therefore spt νy ⊂ co K. Hence y ∈ co K.
6.10 Exercises
209
Exercise 6.88. Verify the details of the following example to provide an example of a face in a simplex whose closure is not a face. Hint. Let P := {xn } ∈ c0 : xn ≥ 0 for all n ∈ N and Q := {n−1 en : n ∈ N}
and X := co {−e1 } ∪ Q ,
where en are the standard unit vectors in c0 . Show that X is a simplex and ext X = {−e1 } ∪ Q. Let F := co Q. Then F is a face of X by Lemma 6.68. Since P is a closed convex subset of c0 and F ⊂ P , we have F ⊂ P . On the other hand, F is not a face since 0 ∈ F and 0 = 12 (−e1 ) + 12 e1 . Exercise 6.89. Prove that there exist metrizable Bauer simplices X, Y and affine continuous surjections ϕ : X → Y and ψ : Y → X such that X, Y are not affinely homeomorphic. Hint. Let K := [0, 1]. There exist continuous surjections f : K → K 2 and g : K 2 → K. Let X := M1 (K) and Y := M1 (K 2 ). Then f] is an affine surjection of X onto Y and g] is an affine surjection of Y onto X. The sets X and Y are metrizable Bauer simplices (Proposition 6.38). If we assume that X and Y are affinely homeomorphic, the sets ext X and ext Y are homeomorphic. Then, by Proposition 2.27, K and K 2 are homeomorphic, a contradiction. Exercise 6.90. Let X be a simplex and Y := {s ∈ (Ab (X))∗ : s ≥ 0, s(1) = 1} be endowed with the w∗ -topology Let φ : X → Y be the evaluation embedding, that is, φ(x)(f ) = f (x), f ∈ Ab (X), x ∈ X. Prove the following assertions. (a) The ordered Banach space Ab (X) is a lattice. (b) The set φ(X) is dense in Y . (c) The set Y is a Bauer simplex. Hint. For the proof of (a) use Theorem A.24. More precisely, since X is a simplex, Ac (X) has the Riesz decomposition property and (Ac (X))∗ = ((Ac (X))∗ )+ − ((Ac (X))∗ )+ , (Ac (X))∗ is a lattice by Theorem A.24. By Proposition A.17 and Theorem A.24, (Ac (X))∗∗ is a lattice (we remark that (Ac (X))∗∗ can be identified with Ab (X) by Proposition 4.32, and thus satisfies the condition (Ac (X))∗∗ = ((Ac (X))∗∗ )+ − ((Ac (X))∗∗ )+ .) Hence Ab (X) is a lattice.
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To show (b), notice that Ab (X) can be identified with Ac (Y ) via the canonical embedding of Ab (X) into (Ab (X))∗∗ . Thus assuming the existence of s ∈ Y \ φ(X), we can find a function f ∈ Ab (X) such that s(f ) > sup f (X). Without loss of generality we may suppose that f is positive. Then s(f ) ≤ kf k = sup f (X), a contradiction. The last assertion (c) follows now from (a) and Theorem 6.37. Indeed, Ac (Y ), as well as Ab (X), is a lattice, and thus (viii) of Theorem 6.37 applies. Exercise 6.91. Let X be a compact convex set, A ⊂ Ac (X) be a subspace containing constant functions and F ⊂ A be a τX -dense in A. Let S := {s ∈ A∗ : s ≥ 0, ksk = 1} and ϕ : X → RF be defined as ϕ(x) = {f (x)}f ∈F ,
x ∈ X.
Let Y := ϕ(X). Prove the following assertions: (a) The set Y is a compact convex set. (b) If g ∈ Ac (Y ), then g ◦ ϕ ∈ A. (c) For any f ∈ A there exists g ∈ Ac (Y ) such that f = g ◦ ϕ. (d) If T : Ac (Y ) → A is defined as g 7→ g ◦ ϕ, then T is a positive isometric isomorphism of Ac (Y ) onto A and T 1 = 1. (e) The sets Y and S are affinely homeomorphic. (f) If A has the weak Riesz interpolation property, then so does Ac (Y ) and Y is a simplex. (g) If A is a lattice, then so is Ac (Y ) and Y is a Bauer simplex. Hint. Obviously, Y is a compact convex set. Let g ∈ Ac (Y ) be given and assume that g ◦ ϕ ∈ / A. By the Hahn–Banach theorem and Proposition 4.31(b) there exist x1 , x2 ∈ X and a1 , a2 ∈ R positive such that a1 f (x1 ) − a2 f (x2 ) = 0 for all f ∈ A and a1 g(ϕ(x1 )) 6= a2 g(ϕ(x2 )). Since 1 ∈ A, a1 = a2 . By the definition of ϕ, ϕ(x1 ) = ϕ(x2 ), which is a contradiction. This proves (b). Let f ∈ A be given. Since F is τX -dense in A, f is constant on every ϕ−1 (y), y ∈ Y . Hence there exists a function g : Y → R such that f = g ◦ ϕ. By Theorem 5.33, g ∈ Ac (Y ) and (c) follows. If T is defined as in (d), then T is obviously a positive isometric isomorphism of c A (Y ) into A. By (c), T is surjective.
6.10 Exercises
211
For the proof of (e), let T ∗ : (A)∗ → (Ac (Y ))∗ be the dual operator of T from (d). Then T ∗ is an affine homeomorphism of S onto S(Ac (Y )), and thus S is affinely homeomorphic to Y by Proposition 4.31(a). If A has the weak Riesz interpolation property, then so does A. Since T from (d) preserves order, Ac (Y ) possesses the weak Riesz interpolation property as well. Consequently, Y is a simplex by Theorem 6.16. Analogously we get that Ac (Y ) is a lattice provided A is. In this case, Y is a Bauer simplex by Theorem 6.37. Exercise 6.92 (The ESP property). A function space H on K has the equal support property, for short ESP, if for any x ∈ K and any two maximal measures µ1 , µ2 ∈ Mx (H) representing the point x we have spt µ1 = spt µ2 . It is clear that any simplicial function space has the ESP. There are examples of function spaces having the ESP without being simplicial. Construct a compact convex set X with ext X closed such that X is not a simplex and X has the ESP. Hint. Let µ be any nonzero signed Radon measure on [0, 1] such that µ([0, 1]) = 0, kµ+ k = kµ− k = 1
and
spt µ+ = spt µ− = [0, 1].
Consider the set H := {f ∈ C([0, 1]) : µ(f ) = 0} which is, obviously, a linear subspace of C([0, 1]). Then H⊥ = {µ ∈ M([0, 1]) : ν = αµ for some α ∈ R}. The space H separates points of [0, 1]: Given x, y ∈ [0, 1], assume that h(x) = h(y) for any h ∈ H. Then εx −εy ∈ H⊥ and, consequently, εx −εy = αµ for some α ∈ R. Since spt µ+ = spt µ− = [0, 1], we get x = y. We see that H is a (closed) function space. Moreover, ChH ([0, 1]) = [0, 1]. To show this, select any x ∈ [0, 1] and ν ∈ Mx (H). Then ν − εx ∈ H⊥ , so that ν − εx = αµ for some α ∈ R. Hence ν = εx , and therefore x ∈ ChH ([0, 1]). Let X be the state space of H. Since s := π(µ+ ) = π(µ− ) ∈ X and r(φ] µ+ ) = π(µ+ ) = π(µ− ) = r(φ] µ− ) by Proposition 4.28(a), we see that the point s has two different representing measures φ] µ+ and φ] µ− carried by ChH ([0, 1]) (hence maximal by Corollary 3.61). Therefore, X is not a simplex. On the other hand, X has the ESP property: Suppose that Λ1 and Λ2 are maximal probability measures on X representing a point x ∈ X. By Proposition 4.28(d), there exist H-maximal measures λ1 , λ2 ∈ M1 ([0, 1]) such that Λ1 = φ] λ1 and Λ2 = φ] λ2 . Then λ1 −λ2 ∈ H⊥ . Hence, there exists α ∈ R such that λ1 −λ2 = αµ = αµ+ −αµ− .
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6 Simplicial function spaces
We may assume that α > 0. Since λ1 ≥ αµ+ , we have α ≤ 1. Then either λ1 = µ+ and λ2 = µ− (in the case α = 1), or there exists η ∈ M1 ([0, 1]) such that λ1 = αµ+ + (1 − α)η. In this case, [0, 1] = spt µ+ ⊂ spt λ1 ⊂ [0, 1]. Analogously, we get spt λ2 = [0, 1]. In either case, spt λ1 = spt λ2 , and therefore spt Λ1 = spt Λ2 . Notice that Λ1 is maximal and spt Λ1 = spt φ] µ+ = φ([0, 1]) = φ ChH ([0, 1]) = ext X. Exercise 6.93. Let X be a compact convex set as given by Exercise 6.92. Recall that X is a metrizable set with ext X closed such that X is not a simplex and such that X has the ESP property. Prove that there exists x ∈ X such that, if F is a compact convex set, x ∈ F ⊂ X and ext F ⊂ ext X, then F is not a simplex. (Compare with Remark 2.14.) Hint. Keep the notation of Exercise 6.92. Let x := π(µ+ ) and let F ⊂ X be a compact convex set containing x such that ext F ⊂ ext X. There exists Λ ∈ Mx (F ) carried by ext F . Since Λ is carried by (a closed set) ext X, it is maximal. Hence, spt Λ = ext X (see the hint of Exercise 6.92). It follows that ext F ⊃ ext X, and therefore F = X. Example 6.94. Let K = [0, 1] ∪ {2} and Z 1 Z H := f ∈ C(K) : f (2) = 2tf (t) dt = 0
1
2(1 − t)f (t) dt .
0
Prove that (a) ChH (K) = [0, 1], Ac (H) = H and H is not simplicial, (b) H has the ESP, (c) the state space S(H) has not the ESP. Hint. Assertion (a) is easy to verify and (b) follows by a similar argument as to that in Exercise 6.92. The only difference is that H⊥ = span{ε2 − 2tλ, ε2 − 2(1 − t)λ} (here λ denotes Lebesgue measure on [0, 1]). To verify (c), let s := π(c[ 1 ,1] 2(2t − 1)λ) 2
and µ1 := φ] (c[ 1 ,1] 2(2t − 1)λ), 2
µ2 := φ] (c[0, 1 ] (1 − 2t)λ + c[ 1 ,1] (2t − 1)λ). 2
2
Then µi is a maximal measure on S(H) and represents s, i = 1, 2. Hence S(H) has not ESP.
6.11 Notes and comments
213
Exercise 6.95. If T is a rank-one operator on a Banach space E, then kI + T k = 1 + kT k if and only if kI + αT k = 1 + αkT k for each α ≥ 0. Hint. If kI + T k = 1 + kT k, then 1 + αkT k = (1 + kT k) − (1 − α)kT k = kI + T k − (1 − α)kT k ≤ kI + T − (1 − α)T k = kI + αT k ≤ 1 + αkT k. The reverse implication is obvious.
6.11
Notes and comments
The basic facts on simplicial spaces can be found in many sources; we mention for example E. M. Alfsen [5], Chapter 1, §5, G. Choquet [108], Chapter 6, H. Bauer [37] and [38], N. Boboc and A. Cornea [73], J. Bliedtner and W. Hansen [66] or M. Rogalski [393], [391]. Theorem 6.5 can be found in papers by G. Choquet and P. -A. Meyer [106], [105], [113], [339]. Theorem 6.6 is due to D. A. Edwards [155] (see also [157]) and Theorem 6.8 follows the same argument as E. M. Alfsen [5], Proposition II.3.14. Example 6.15 follows the classical example from E. Bishop and K. de Leeuw [58]. This example was generalized in P. J. Stacey [436]. The question of constructing function spaces with given Choquet boundaries was studied by E. Briem [93] and A. Clausing and G. M¨agerl [120]. We also mention that simplicial spaces are called weakly simplicial in H. Bauer [44], whereas spaces H with S(Ac (H)) being a simplex are called simplicial there (see Remark 6.2). Characterizations of simplicial spaces in Section 6.2 can be found in J. Lindenstrauss [304] and Z. Semadeni [412], see also E. M. Alfsen [5], Corollary II.3.11. Section 6.3 follows J. Lindenstrauss [304] and Z. Semadeni [412]. Theorem 6.27 can be found in G. Choquet [108], Theorem 28.6; see also E. M. Alfsen [5], Theorem II.3.12. Proposition II.3.17 of [5] provides an example of a nonmetrizable compact convex set X, that is not a simplex, and yet every continuous function defined on a compact subset of ext X can be extended to an affine continuous function with preservation of norm. Theorem 6.28 can be found, for example, in E. M. Alfsen [5] as Corollary II.5.20; we follow the reasoning of J. Bliedtner and W. Hansen [62]. The results of Section 6.5 for continuous functions (see Theorem 6.32) were proved by E. M. Alfsen [4]. The proof of this theorem can be also found in [5], Theorem II.4.5 and its several predecessors in E. M. Alfsen [2], E. G. Effros [163], Theorem 2.4 and A. J. Lazar [292], Theorem 2.2. We also refer the reader to J. B. Bednar [48], N. Boboc and Gh. Bucur [67] and J. N. McDonald [337] for results on continuous affine extensions in the setting of function spaces. The generalization for Baire functions is due to J. Spurn´y [426] and [423].
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6 Simplicial function spaces
Subsection 6.6.A follows the paper by H. Bauer [38], see also E. M. Alfsen [5], Theorem II.4.1. The definition of a Markov simplex appeared in a paper [419] by R. Sine. Theorem 6.45 is taken from F. Jellett [250] (see also H. Fakhoury [175]), Theorem 6.46 from M. Rogalski [392] and A. Goullet de Rugy, C. Schol-Cancelier and B. Taylor-MacGibbon [201]. We mention the following problem on the converse implication in Theorem 6.45. Problem 6.96. Let X be a compact convex set such that any f ∈ C b (ext X) has an affine Baire-one extension. Is then ext X a Lindel¨of space? Subsection 6.6.D follows the papers by J. Spurn´y [424] and [427]. The easy implication of Theorem 6.49 can be found in M. Rogalski [392] and F. Jellett [250]. The case of nonmetrizable spaces was solved in the paper J. Spurn´y and O. Kalenda [432] by proving the following theorem. For a compact convex set X, the following assertions are equivalent: • X is a simplex and ext X is a Lindel¨ of resolvable set, any bounded Baire-one function on ext X has an affine Baire-one extension. The proof uses a result on extension of Baire-one functions from the set of extreme points proved in O. Kalenda and J. Spurn´y [259]. We mention here a problem posed in [259]. •
Problem 6.97. Let X be a compact convex set with ext X Lindel¨of. Is then ext X hereditarily Baire? It follows from these results and Theorem 6.37 that the question of extending continuous and Baire-one functions from the set of extreme points is determined by a topological quality of the set of extreme points. The following result from J. Spurn´y [431] shows that this is no longer true for functions of higher Baire classes. There exist metrizable simplices X1 , X2 and a homeomorphism ϕ : ext X1 → ext X2 such that • ϕ(ext X ) = ext X , 1 2 •
there exists a bounded Baire-two function on ext X1 that cannot be extended to a Baire-two affine function on X1 ,
if α ∈ [2, ω1 ), any bounded Baire-α function on ext X2 can be extended to a function of affine class α on X2 . For the Daugavet property and its relatives, it is possible to consult a survey paper by D. Werner [466]; in particular, see Example (e) on p. 79. Prime simplices were introduced in E. G. Effros and J. L. Kazdan [165] and we follow in Subsection 6.8.B their paper. In [1], E. M. Alfsen proved that in Bauer simplices, closures of faces are faces. Prime compact convex sets were also investigated in C. H. Chu [117]. The characterization of Bauer simplices in Theorem 6.69 •
6.11 Notes and comments
215
was given by E. Størmer in [440] (see also A. K. Roy [399] and [400]). Note that E. StørmerSin [440] proved that a compact convex set X is a Bauer simplex if and only if co( α Fα ) is a closed split face of X whenever {Fα } is any family of closed split faces of X (Størmer’s axiom). Theorem 6.70 is from H. Fakhoury [174] The example of Exercise 6.77 is taken from Bauer’s paper [44]. It represents a slight modification of an earlier Bauer’s example from [38] and is attributed to S. Papadopoulou. Exercise 6.88 is due to M. Kraus. Exercise 6.90 is taken from C. H. Chu ˚ Lima [215], Example 3.3(b). and B. Cohen [119] and P. Harmond and A. The ESP property in Exercise 6.92 was introduced by I. Feinberg in his thesis (cf. J. N. McDonald [336]). In the same paper, J. N. McDonald investigated the ESP property and constructed an example of the simplex which fails the ESP property (Exercise 6.92 follows this example; see also J. N. McDonald [338]). Exercises 6.80 and 6.81 can also be found in [336]. S. Alpay proved in [12] that if X is a compact convex set and x ∈ X, then all maximal measures on X with barycenter x have the same support if and only if x ∈ ext(co spt µ) whenever µ is a maximal measure on X with barycenter x. The ESP property is also studied in M. W. Grossman [204]. Note that J. K¨ohn [274] localized the notion of a simplex in the convex setting and gave several statements that a point has a unique maximal representing measure. We also refer the reader to C. Cho-Ho [118] for similar results. Examples and a study of points of simpliciality in the context of function spaces is presented in the paper [27] by M. Baˇca´ k.
Chapter 7
Choquet theory of function cones
The notion of a function cone generalizes the concept of a function space. Even though we focus mainly on function spaces, the Choquet theory of function cones is an indispensable tool for us, since later on we investigate typical function cones arising in potential theory. Another motivation is the selection theorem from Section 11.5 and a description of boundary measures contained in Section 8.5. Since the basic results of the Choquet theory of function cones very often use the same techniques as their analogues from the theory of function spaces, we present them in Sections 7.1– 7.6 in a brief but, we hope, comprehensible way. We point out that Theorem 7.27 provides a measure on the Choquet boundary induced by a maximal measure. This concept will be strengthened considerably in Chapter 9. Other results not encountered before are Theorem 7.38 and Theorem 7.40, which clear up the relation between simplicial function cones and function spaces generated by them. An important concept of ordered compact convex sets is studied in Section 7.5. Main facts are summarized in Theorem 7.54 and an application of Theorem 7.27 is presented in Theorem 7.55. We show in Theorem 7.58 how an ordered compact convex set can be recovered from the set of maximal elements by a method imitating the Krein–Milman theorem. The Douglas characterization of simplicial measures is given in Theorem 7.60.
7.1
Function cones
Definition 7.1 (Function cones). Let K be a compact space and S a convex cone of continuous functions on K. We call S a function cone if S contains the constant functions and separates points of K. A function cone S is called min-stable if W(S) = {s1 ∧ · · · ∧ sn : s1 , . . . , sn ∈ S, n ∈ N} = S. In this section, S is a function cone on a compact space K 6= ∅. Examples 7.2. (a) Any function space can serve as an example of a function cone. (b) Let H be a function space on a compact space K. Then W(H) is a function cone on K. Also the family S c (H) of all continuous H-concave functions is an example of a function cone.
7.1 Function cones
217
(c) If S is a function cone on a compact space K, then W(S) is a function cone. If F is a closed subset of K, then S|F is a function cone as well. (d) Let U be a bounded open subset of Rd . Then the family S(U ) := s ∈ C(U ) : s is superharmonic on U is a function cone. Definition 7.3 (S-representing measures, ChS (K) and expS (K)). For a point x ∈ K we denote by Mx (S) the set of all S-representing measures for x; more precisely, Mx (S) := {µ ∈ M+ (K) : s(x) ≥ µ(s) for all s ∈ S}. We remark that Mx (S) ⊂ M1 (K) because S contains constant functions. We define the Choquet boundary ChS (K) of S as ChS (K) := {x ∈ K : Mx (S) = {εx }} and the set of S-exposed points as expS (K) := {x ∈ K : there exists s ∈ S with s(x) < s(y) for all y ∈ K \ {x}}. Definition 7.4 (S-convex, concave and affine functions). We define the system A(S) of all S-affine functions to be the family of all universally measurable functions f : K → [−∞, ∞] such that µ(f ) exists for every µ ∈ Mx (S), x ∈ K, and Z f (x) = f dµ for each x ∈ K and µ ∈ Mx (S). K
Ac (S)
Further, let be the family of all continuous S-affine functions on K. Similarly, we say that a universally measurable function f : K → [−∞, ∞] is Sconvex if µ(f ) exists for every µ ∈ Mx (S), x ∈ K, and f (x) ≤ µ(f ). A function f is S-concave if −f is S-convex. We denote by Kc (S), Kusc (S) and Klsc (S) the family of all continuous, upper semicontinuous, and lower semicontinuous S-convex functions, respectively. We write S c (S), S usc (S) and S lsc (S) for the analogous families of S-concave functions. Definition 7.5 (Choquet ordering, S-maximal and S-boundary measures). Let S be a function cone on a compact space K. We define an ordering ≺S on M+ (K) by setting µ ≺S ν if µ(s) ≤ ν(s) for each s ∈ Kc (S). Sometimes we write ≺ instead of ≺S . A measure µ ∈ M+ (K) is S-maximal (briefly maximal), if it is maximal with respect to the partial order ≺S . A measure µ ∈ M(K) is S-boundary (briefly boundary), if |µ| is S-maximal. The set of all S-maximal measures on K is denoted by Mmax (S), the set of all S-boundary measures is denoted by Mbnd (S).
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7 Choquet theory of function cones
Proposition 7.6. The following assertions hold: (a) expS (K) ⊂ ChS (K), (b) S ⊂ W(S) ⊂ S c (S), (c) W(S), S c (S), S usc (S), S lsc (S) are min-stable convex cones and, consequently, W(S) and S c (S) are function cones, (d) Mx (S) = Mx (W(S)) = Mx (S c (S)) for any x ∈ K, (e) S c (S) = S c (W(S)) = S c (S c (S)), (f) ChS (K) = ChW(S) (K) = ChS c (S) (K), (g) for µ, ν ∈ M+ (K), µ ≺S ν if and only if µ ≺W(S) ν and this is the case if and only if µ ≺S c (S) ν, (h) the space W(S) − W(S) is dense in C(K). Proof. Assertions (a) and (b) are obvious, the first part of (c) follows by the same argument as in Proposition 3.11(b) and the second part from (b). Assertion (d) follows from the definition, (e), (f) are consequences of (d) and (g) follows from (e). Finally, (h) follows from the lattice version of the Stone–Weierstrass theorem contained in Proposition A.31. Definition 7.7 (Envelopes and sublinear functionals). Given f : K → [−∞, ∞] and a measure µ ∈ M+ (K), we define Qµ (f ) := inf{µ(s) : s ∈ S c (S), s ≥ f }. If µ = εx for x ∈ K, we write f ∗ (x) instead of Qεx (f ) and define f∗ (x) := −(−f )∗ (x). Lemma 7.8. Let µ ∈ M+ (K) and f be an upper bounded function on K. Then (a) f ∗ is an upper semicontinuous S-concave function, (b) Qµ (f ) = µ(f ∗ ), (c) if f is bounded, then −kf k ≤ f∗ ≤ f ∗ ≤ kf k, (d) g 7→ Qµ (g) is a sublinear functional on `∞ (K). Proof. Assertion (a) is obvious, (b) follows from Theorem A.84, (c) holds because S contains constant functions and (d) follows by a straightforward verification. The following observation shows that if we consider a function space H as a function cone, our new objects defined for H so far in Section 7.1 coincide with those defined for H in Section 3.1. Proposition 7.9. Let H be a function space on a compact space K. If we consider H as a function cone S, then
7.1 Function cones •
219
Mx (H) = Mx (S), S c (S) = S c (H), ≺H coincides with ≺S , expS (K) = expH (K),
•
S-maximal measures coincide with H-maximal measures,
•
ChS (K) = ChH (K).
Hint. The proof follows easily from the definitions. Lemma 7.10 (Key lemma). Let µ ∈ M+ (K) and f be an upper semicontinuous function on K. Then there exists a measure ν ∈ M+ (K) such that µ ≺ ν and Qµ (f ) = ν(f ). In particular, for any x ∈ K there exists a measure ν ∈ Mx (S) such that f ∗ (x) = ν(f ). Proof. Given µ ∈ M+ (K), we assume first that f ∈ C(K). We define a functional ϕ : span{f } → R as ϕ(tf ) = tQµ (f ), t ∈ R, and a sublinear functional p : C(K) → R as p : g 7→ Qµ (g),
g ∈ C(K).
Since ϕ ≤ p on span{f }, the Hahn–Banach theorem provides a linear functional ν on C(K) such that ν(f ) = Qµ (f ) and ν ≤ p on C(K). Since ν(g) ≤ p(g) ≤ 0 whenever g ∈ C(K) is negative, ν ∈ M+ (K). Let s ∈ S c (S). Then ν(s) ≤ p(s) = Qµ (s) ≤ µ(s). Hence µ ≺ ν and the proof is finished in the case of a continuous function. Now let f be an upper semicontinuous function. We define a down-directed family G := {g ∈ C(K) : g ≥ f } and use the first part of the proof to find a measure νg ∈ M+ (K) such that µ ≺ νg and νg (g) = Qµ (g), g ∈ G. Given h ∈ G, let Mh := {νg : g ∈ G, g ≤ h}. Since Mh ⊂ {λ ∈ M+ (K) : kλk = kµk} and the family {Mh : h ∈ G} has T the finite intersection property, a compactness argument yields the existence of ν ∈ {M h : h ∈ G}. Then µ ≺ ν and inf λ(h) : λ ∈ Mh = inf λ(h) : λ ∈ M h ≤ ν(h) for each h ∈ G.
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Hence
Qµ (f ) ≤ inf {Qµ (g) : g ∈ G} = inf {νg (g) : g ∈ G} ≤ inf {inf {νg (h) : g ∈ G, g ≤ h} : h ∈ G} ≤ inf {ν(h) : h ∈ G} = ν(f ) ≤ inf {ν(k) : k ∈ S c (S), k ≥ f } ≤ inf {µ(k) : k ∈ S c (S), k ≥ f } = Qµ (f ),
which yields the required equality. The second part of the assertion follows by taking µ = εx and observing that ν ∈ Mx (S) provided εx ≺ ν. Proposition 7.11. Let f be an upper bounded function on K. Then (a) f ∗ = f if and only if f ∈ S usc (S), (b) we have
f ∗ = inf {g ∈ S : g ≥ f } = inf {g ∈ W(S) : g ≥ f } = inf {g ∈ S c (S) : g ≥ f } = inf {g ∈ S usc (S) : g ≥ f } ,
(c) if f is upper semicontinuous, then f ∗ = inf{k ∈ S lsc (S) : k ≥ g}. Proof. Let f be an upper bounded function on K. If f ∈ S usc (S) and x ∈ K, let µ ∈ Mx (S) be chosen such that f ∗ (x) = µ(f ). Then f ∗ (x) = µ(f ) ≤ f (x) ≤ f ∗ (x). Since the converse implication follows from Lemma 7.8(a), assertion (a) is proved. For the proof of (b) we first show that inf{s(x) : s ∈ S, s ≥ g} = max{µ(g) : µ ∈ Mx (S)},
x ∈ K, g ∈ C(K). (7.1)
The inequality “≥” being obvious, we proceed with the proof of “≤”. Let x ∈ K and g ∈ C(K) be fixed. The formula p : h 7→ inf{s(x) : s ∈ S, s ≥ h},
h ∈ C(K),
defines a sublinear functional on C(K). Analogously as in the proof of Lemma 7.10 we find a measure µ ∈ M+ (K) such that p(g) = µ(g) and µ ≤ p on C(K). Then µ(s) ≤ p(s) ≤ s(x), and thus µ ∈ Mx (S). Hence (7.1) follows.
s ∈ S,
7.1 Function cones
221
Let s ∈ S c (S) be arbitrary. By (7.1), p(s) = s(x),
x ∈ K.
Hence inf{g ∈ S : g ≥ f } = inf{g ∈ S c (S) : g ≥ f }.
(7.2)
inf{g ∈ S c (S) : g ≥ f } = inf{g ∈ S usc (S) : g ≥ f }
(7.3)
Further, by (a). Combining (7.2) and (7.3) we get f ∗ = inf{g ∈ S c (S) : g ≥ f } = inf{g ∈ S usc (S) : g ≥ f } ≤ inf{g ∈ W(S) : g ≥ f } ≤ inf{g ∈ S : g ≥ f } = inf{g ∈ S c (S) : g ≥ f }. To verify (c), let f be upper semicontinuous and s ∈ S lsc (S). Let x ∈ K and µ ∈ Mx (S) be such that f ∗ (x) = µ(f ). Then f ∗ (x) = µ(f ) ≤ µ(s) ≤ s(x), which yields “≤” in (c). The reverse inequality being obvious, the proof is finished. Proposition 7.12. Let µ ∈ M+ (K). (a) If ν ∈ M+ (K) satisfies µ ≺ ν, then kµk = kνk. (b) If x is a point of K, then εx ≺ µ if and only if µ ∈ Mx (S). Proof. Assertion (a) follows from inequalities µ(1) ≤ ν(1) and µ(−1) ≤ ν(−1). To show (b), εx ≺ µ clearly implies µ ∈ Mx (S). Conversely, if µ ∈ Mx (S), s ∈ S c (S) and ε > 0, let t ∈ S be such that s ≤ t and t(x) < s(x) + ε (use Proposition 7.11(a), (b)). Then µ(s) ≤ µ(t) ≤ t(x) ≤ s(x) + ε. Hence εx ≺ µ, and the proof is complete. Proposition 7.13. (a) If f ∈ S usc (S) and g is lower semicontinuous with f < g, then there exists s ∈ W(S) such that f < s < g. (b) If f ∈ S usc (S), then f = inf{g ∈ S c (S) : g > f } and this family is downdirected.
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(c) If f ∈ S lsc (S) and g is upper semicontinuous with g < f , then there exists s ∈ W(S) such that g < s < f . (d) If f ∈ S lsc (S), then f = sup{g ∈ S c (S) : g < f } and this family is up-directed. (e) W(S) is dense in S c (S). Proof. To show (a), let f < g as in (a) be given. For any x ∈ K we use Proposition 7.11(a),(b) to find a function sx ∈ S such that f < s and f (x) < sx (x) < g(x). By the semicontinuity of f and g there exists an open set Ux 3 x such that f < sx < g on Ux . By the compactness of K, we can select finitely many points x1 , . . . , xn such that f < sx1 ∧ · · · ∧ sxn < g. Hence s := sx1 ∧ · · · ∧ sxn is the required function. Since (b) follows immediately from Proposition 7.11(a),(b) and Proposition 7.6(c), we proceed to the proof of (c). Let f ∈ S lsc (S) and upper semicontinuous g with g < f be given. For a fixed x ∈ K, let µ ∈ Mx (S) be such that g ∗ (x) = µ(g) (use Lemma 7.10). Then g ∗ (x) = µ(g) < µ(f ) ≤ f (x). Hence g ∗ < f , and we may use (a) to find a function s ∈ W(S) with g ∗ < s < f . Since f = sup{g ∈ C(K) : g < f } (see Proposition A.50(ii)), assertion (d) follows from (c). Finally, let f ∈ S c (S) be given. Combining (a) with (b) we see that f = inf{s ∈ W(S) : s > f } and the latter family is down-directed. By Dini’s theorem, for any ε > 0 there exists s ∈ W(S) such that f < s < f + ε. This concludes the proof.
7.2
Maximal measures
In this section, S is a function cone on a compact space K 6= ∅. Definition 7.14 (S-extremal sets). A universally measurable set F ⊂ K is called Sextremal if any measure representing a point in F is carried by F . Theorem 7.15. (a) The family of all closed S-extremal sets is stable with respect to finite unions and arbitrary intersections. (b) A closed set F ⊂ K is S-extremal if and only if cK\F ∈ S lsc (S). (c) If f ∈ S lsc (S), then F := {x ∈ K : f (x) = min f (K)} is S-extremal. (d) A point x ∈ K is in ChS (K) if and only if the set {x} is S-extremal. (e) Any nonempty closed S-extremal set intersects ChS (K); in particular, ChS (K) is nonempty. (f) If f ∈ S lsc (S) is positive on ChS (K), then f ≥ 0 on K.
7.2 Maximal measures
223
Proof. The proof of (a) is analogous to the proof of Proposition 3.14 and (b), (c) and (d) follows by a straightforward verification (cf. Lemma 3.13). To show (e) we use Zorn’s lemma again (cf. Proposition 3.15 and Theorem 2.22). Given a closed Sextremal set F , we consider the family Z of all nonempty closed S-extremal sets contained in F endowed with the partial ordering given by the reverse inclusion. Zorn’s lemma provides a maximal element H ∈ Z. If x1 , x2 ∈ H are distinct points, let s ∈ S be such that s(x1 ) 6= s(x2 ). Then it is easy to see that {z ∈ H : s(z) = min s(H)} is a closed S-extremal set strictly smaller than H, a contradiction with the maximality of H. Hence H is a singleton, say H = {x}, and thus x ∈ ChS (K) by (d). This proves (e). Finally, let f ∈ S lsc (S) be positive on ChS (K). Then F := {x ∈ K : f (x) = min f (K)} is a closed S-extremal set, and thus F ∩ ChS (K) is nonempty by (e). Hence f ≥ 0 on K, which concludes the proof. Proposition 7.16. For measures µ, ν ∈ M+ (K), the following assertions are equivalent: (i) µ ≺ ν, (ii) ν(s) ≤ µ(s) for all s ∈ W(S), (iii) ν(s) ≤ µ(s) for all s ∈ S lsc (S), (iv) ν(s) ≤ µ(s) for all s ∈ S usc (S). Proof. Obviously, (i) =⇒ (ii), (iii) =⇒ (i), (iv) =⇒ (i). Proposition 7.13(b),(d) yields (i) =⇒ (iii) and (i) =⇒ (iv). Finally, Proposition 7.13(e) yields (ii) =⇒ (i). This concludes the proof. Corollary 7.17. Let S be a min-stable function cone on a compact space K and µ, ν ∈ M+ (K). Then µ ≺ ν if and only if ν(s) ≤ µ(s) for each s ∈ S. Proof. Since S = W(S) by the assumption, the proof follows from Proposition 7.16. Theorem 7.18. For every measure µ ∈ M+ (K) there exists a maximal measure λ such that µ ≺ λ. Proof. It suffices to follow step by step the proof of Theorem 3.65. Theorem 7.19. For a measure µ ∈ M+ (K), the following assertions are equivalent: (i) µ is maximal, (ii) µ(f ) = µ(f ∗ ) for any f ∈ −W(S), (iii) µ(f ) = µ(f ∗ ) for any f ∈ Kc (S),
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7 Choquet theory of function cones
(iv) µ(f ) = µ(f ∗ ) for any f ∈ Kusc (S), (v) µ(f ) = µ(f ∗ ) for any upper semicontinuous function f on K. Proof. Obviously, (v) =⇒ (iv) =⇒ (iii) =⇒ (ii). To verify (i) =⇒ (v), let f be an upper semicontinuous function. Using Lemma 7.10 we find ν ∈ M+ (K) such that µ ≺ ν and Qµ (f ) = ν(f ). Since µ is maximal, µ = ν. Hence µ(f ) = ν(f ) = Qµ (f ) = µ(f ∗ ) by Lemma 7.8(b). For the proof of (ii) =⇒ (i), let ν ∈ M+ (K) satisfy µ ≺ ν. For any f ∈ −W(S) we get, using Theorem A.84, that µ(f ) = µ(f ∗ ) = inf{µ(s) : s ∈ S c (S), s ≥ f } ≥ inf{ν(s) : s ∈ S c (S), s ≥ f } = ν(f ∗ ) ≥ ν(f ). Since µ(f ) ≤ ν(f ), µ(f ) = ν(f ). Thus µ = ν on W(S) − W(S), which yields µ = ν by Proposition 7.6(h). Lemma 7.20. If x ∈ K, then εx is maximal if and only if x ∈ ChS (K). Proof. The assertion follows from Proposition 7.12(b). Theorem 7.21. For a point x ∈ K, the following assertions are equivalent: (i) x ∈ ChS (K), (ii) f (x) = f ∗ (x) for any f ∈ −W(S), (iii) f (x) = f ∗ (x) for any f ∈ Kc (S), (iv) f (x) = f ∗ (x) for any f ∈ Kusc (S), (v) f (x) = f ∗ (x) for any upper semicontinuous function f on K. Proof. It suffices to combine Lemma 7.20 and Theorem 7.19.
7.3
Representation theorem
In this section, S is a function cone on a compact space K 6= ∅. Lemma 7.22. Let {fn } be an upper bounded sequence of lower semicontinuous Sconvex functions on K such that lim supn→∞ fn ≤ 0 on ChS (K). Then we have lim supn→∞ fn ≤ 0 on K.
7.3 Representation theorem
225
Proof. Let {fn } be as in the lemma. By taking functions −1 ∨ fn if necessary we may assume that {fn } is a sequence bounded by a constant C. For a fixed x ∈ K and ε > 0 we use Proposition 7.11(b) to find functions kn ∈ Kc (S) such that kn ≤ fn and kn (x) ≥ fn (x) − ε. Without loss of generality we may assume that sup{kkn k : n ∈ N} ≤ C. By Theorem 7.15(c)(e), each function in coσ ({kn : n ∈ N}), as an S-convex continuous function, attains its maximum on ChS (K). Since lim sup kn ≤ lim sup fn ≤ 0 n→∞
on ChS (K),
n→∞
the Simons lemma 3.75 yields lim sup fn (x) ≤ ε + lim sup kn (x) ≤ ε. n→∞
n→∞
Since ε > 0 is arbitrary, the proof is finished. Lemma 7.23. Let µ be a maximal measure on K and F ⊂ K \ ChS (K) be a compact Gδ set. Then µ(F ) = 0. Proof. Given a maximal measure µ ∈ M+ (K) and a compact Gδ set F ⊂ K \ ChS (K), we find a sequence {fn } of continuous functions such that 0 ≤ fn ≤ 1, n ∈ N, and fn → cF . Then {(fn )∗ } is a bounded sequence of S-convex lower semicontinuous functions such that lim supn→∞ (fn )∗ ≤ 0 on ChS (K). By Lemma 7.22, lim supn→∞ (fn )∗ ≤ 0 on K. Thus Theorem 7.19 and Fatou’s lemma yield 0 ≤ µ(F ) = lim µ(fn ) = lim µ((fn )∗ ) n→∞
n→∞
≤ lim sup µ((fn )∗ ) ≤ µ(lim sup(fn )∗ ) ≤ 0. n→∞
n→∞
This concludes the proof. Theorem 7.24. Let µ ∈ M+ (K) be a maximal measure on K. Then (a) µ(B) = 0 for every Baire set B disjoint from ChS (K), (b) µ∗ (K \ L) = 0 for every Lindel¨of set L ⊃ ChS (K) (here µ∗ denotes the inner measure induced by µ, see Definition A.62), (c) µ(K \ A) = 0 for every K-analytic set A ⊃ ChS (K). Proof. The proof follows from Lemma 7.23 as Theorem 3.79 follows from Lemma 3.78. Corollary 7.25. Let S be a function cone on a compact space K. Then for every x ∈ K there exists µ ∈ Mx (S) such that µ(B) = 1 for any K-analytic set containing ChS (K). Proof. Combine Theorem 7.18 with Theorem 7.24(c).
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Theorem 7.26. Let S be a function cone on a metrizable compact space K. Then the following assertions hold: (a) ChS (K) is of type Gδ . (b) A measure µ ∈ M+ (K) is maximal if and only if µ(K \ ChS (K)) = 0. (c) For any x ∈ K there exists µ ∈ Mx (S) such that µ(K \ ChS (K)) = 0. Proof. We select a dense set {kn : n ∈ N} in Kc (S). To show (a), we verify that ChS (K) =
∞ \
{x ∈ K : kn∗ (x) = kn (x)}.
(7.4)
n=1
Indeed, the inclusion “⊂” follows from Theorem 7.21. Conversely, if x is contained in the right-hand side of (7.4), f ∗ (x) = f (x) for any f ∈ Kc (S) due to the density of {kn : n ∈ N} and Lemma 7.8(c). Thus Theorem 7.21 again implies x ∈ ChS (K). Now (7.4) yields assertion (a). To verify (b), we first notice that any measure µ ∈ M+ (K) satisfying µ(K \ ChS (K)) = 0 is maximal due to Theorem 7.19 and Theorem 7.21. Conversely, any maximal measure µ ∈ M+ (K) satisfies µ(f ) = µ(f ∗ ) for all f ∈ Kc (S), and thus ChS (K), being the intersection of countably many sets {x ∈ K : kn∗ (x) = kn (x)}, n ∈ N, is of full measure µ. The last assertion (c) follows from Theorem 7.18 and (b). Theorem 7.27. Let S be a function cone on a compact space K. Let Σ := {A ∩ ChS (K) : A ⊂ K is a Baire set} and µ ∈ M+ (K) be maximal. Then there exists a measure µ0 on the σ-algebra Σ such that (a) f |ChS (K) is µ0 -measurable for any Baire function f on K, (b) Rf |ChS (K) is µ0 -integrable for any µ-integrable Baire function f on K and µ(f ) = 0 ChS (K) f (t) dµ (t). Proof. Given a maximal measure µ ∈ M+ (K) and A0 ∈ Σ, we define µ0 (A0 ) := µ(A),
A ⊂ K is a Baire set with A ∩ ChS (K) = A0 .
It follows from Theorem 7.24 that the value µ0 (A0 ) does not depend on the choice of A. It is easy to verify that µ0 is a measure on the σ-algebra Σ. If f is a Baire function on K, let f 0 := f |ChS (K) . Given an open set U ⊂ R, {x ∈ ChS (K) : f (x) ∈ U } = ChS (K) ∩ {x ∈ K : f (x) ∈ U } shows that (f 0 )−1 (U ) ∈ Σ, and thus f 0 is µ0 -measurable. If, moreover, f is µ-integrable, we define Ft := {x ∈ K : f (x) ≥ t} for any t ∈ R. Then µ0 (Ft ∩ ChS (K)) = µ(Ft ),
7.4 Simplicial cones
and we get from Fubini’s theorem that Z Z µ(f ) = f (x) dµ(x) = µ(Ft ) dt K R Z Z 0 µ (Ft ∩ ChS (K)) dt = = R
227
f (x) dµ0 (x).
ChS (K)
This concludes the proof.
7.4
Simplicial cones
In this section, S is a function cone on a compact space K 6= ∅. Definition 7.28 (Simplicial function cones). A function cone S on a compact space K is called simplicial if for each x ∈ K there exists a unique maximal measure δx ∈ Mx (S). Lemma 7.29. Let S be a function cone on a compact space K. (a) If S is simplicial, x ∈ K and µ ∈ Mx (S), then µ ≺ δx . In particular, εx ≺ δx . (b) The function cone S is simplicial if and only if W(S) is simplicial and this is the case if and only if S c (S) is simplicial. (c) If S is simplicial, then f ∗ (x) = δx (f ) for any x ∈ K and f ∈ Kusc (S). Proof. To show (a), let µ ∈ Mx (S). By Theorem 7.18, there exists a maximal measure λ such that µ ≺ λ. Using Proposition 7.12(b) we get λ ∈ Mx (S). Since S is simplicial, λ = δx . Hence µ ≺ δx . For the proof of (b), we note that the notions of S-maximal, W(S)-maximal and S c (S)-maximal measures coincide (see Proposition 7.6(g)) and that representing measures are the same for all these cones (see Proposition 7.6(d)). If S is simplicial, f ∈ Kusc (S) and x ∈ K are given, we use Lemma 7.10 to find a measure µ ∈ Mx (S) so that µ(f ) = f ∗ (x). By (a), µ ≺ δx . Using Proposition 7.16 we get f ∗ (x) = µ(f ) ≤ δx (f ) ≤ inf{δx (s) : s ∈ S c (S), s ≥ f } ≤ inf{s(x) : s ∈ S c (S), s ≥ f } = f ∗ (x), which concludes the proof. Theorem 7.30. The following assertions are equivalent: (i) S is simplicial, (ii) f ∗ is S-affine for any f ∈ −W(S), (iii) f ∗ is S-affine for any f ∈ Kc (S), (iv) f ∗ is S-affine for any f ∈ Kusc (S).
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7 Choquet theory of function cones
Proof. Obviously, (iv) =⇒ (iii) =⇒ (ii). For the proof of (i) =⇒ (iv), let f ∈ Kusc (S) be given. Let x ∈ K and µ ∈ Mx (S) be given. By Lemma 7.29(a), µ ≺ δx . Using Lemma 7.29(c) we get µ(f ∗ ) = inf{µ(k) : k ∈ S c (S), k ≥ f } ≥ inf{δx (k) : k ∈ S c (S), k ≥ f } = δx (f ∗ ) ≥ δx (f ) = f ∗ (x). By Proposition 7.8(a), µ(f ∗ ) ≤ f ∗ (x). Hence f ∗ is S-affine. Let us assume that (ii) holds, let x ∈ K and µ, ν ∈ Mx (S) be maximal measures. Then, for each f ∈ −W(S), Theorem 7.19 yields µ(f ) = µ(f ∗ ) = f ∗ (x) = ν(f ∗ ) = ν(f ). By Proposition 7.6(h), µ = ν as required. This concludes the proof. Theorem 7.31 (In-between theorem). The following assertions are equivalent: (i) S is simplicial, (ii) for any s, −t ∈ Kc (S) with s < t there exists h ∈ Ac (S) such that s ≤ h ≤ t, (iii) for any s, −t ∈ Kusc (S) with s ≤ t there exists h ∈ Ac (S) such that s ≤ h ≤ t, (iv) for any s, −t ∈ Kusc (S) with s < t there exists h ∈ Ac (S) such that s < h < t. Proof. Clearly, (iii) =⇒ (ii). To show (i) =⇒ (iv), assume that S is simplicial and s, t are as in (iv). Given ε > 0, denote F := {f − g + ε : s < f < g < t, f, −g ∈ Kc (S)} . We want to use Lemma A.88 in order to find a strictly positive function in F. To this end, choose a nonzero measure µ ∈ M+ (K). By Theorem A.84 and Lemma 7.8(b) there exists a function k ∈ S c (S) such that s∗ < k
and µ(k − s∗ ) < εµ(K).
Using Proposition 7.13(a) we find a function g ∈ S c (S) such that s∗ < g < k ∧ t. By Theorem 7.30, s∗ is an S-affine function. Proposition 7.13(c) yields the existence of a function f ∈ Kc (S) such that s∗ < f < g. Then µ(g − f ) ≤ µ(g − s∗ ) < µ(k − s∗ ) < εµ(K),
7.4 Simplicial cones
229
and thus µ(f − g + ε) = εµ(K) + µ(f − g) > 0. This proves the existence of a strictly positive element in F and thus we are able to find a pair of functions f, −g ∈ Kc (S) such that s < f < g < t and g − f < ε. Now we construct inductively functions fn , −gn ∈ Kc (S), n ∈ N, such that • s < f < f n n+1 < gn+1 < gn < t, gn − fn < 2−n . Both sequences {fn } and {gn } converge uniformly to an S-affine continuous function h satisfying s < h < t. To verify (iv) =⇒ (iii), let s, −t ∈ Kc (S) satisfy s ≤ t. We inductively construct a sequence {hn }∞ n=0 of S-affine functions such that • s − 1 < h < t + 1, 0 •
(hn−1 − 2−n ) ∨ (s − 2−n ) < hn < (hn−1 + 2−n ) ∧ (t + 2−n ), n ∈ N. To start the construction, we use (iv) to find h0 ∈ Ac (S) with s − 1 < h0 < t + 1. Assume that we have constructed functions h0 , . . . , hn−1 . From •
s − 2−(n−1) < hn−1 < t + 2−(n−1) we get (hn−1 − 2−n ) ∨ (s − 2−n ) < (hn−1 + 2−n ) ∧ (t + 2−n ). By Lemma 7.29(c), ((hn−1 − 2−n ) ∨ (s − 2−n ))∗ < ((hn−1 + 2−n ) ∧ (t + 2−n ))∗ . Moreover, both functions are S-affine by Theorem 7.30. Hence we can use (iv) again to find a function hn ∈ Ac (S) satisfying (hn−1 − 2−n ) ∨ (s − 2−n ) < hn < (hn−1 + 2−n ) ∧ (t + 2−n ). This completes the construction. It is obvious that {hn } is a Cauchy sequence converging to h ∈ Ac (S) that satisfies s ≤ h ≤ t. To finish the proof we have to verify (ii) =⇒ (i). Pick x ∈ K and consider a maximal measure µ ∈ Mx (S). If f ∈ Kc (S), it follows from (ii) that f ∗ = inf{h ∈ Ac (S) : h ≥ f } and the latter family is down-directed. Hence µ(f ∗ ) = inf{µ(h) : h ∈ Ac (S), h ≥ f } = inf{h(x) : h ∈ Ac (S), h ≥ f } = f ∗ (x). Thus f ∗ is S-affine, which concludes the proof by Theorem 7.30.
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7 Choquet theory of function cones
Corollary 7.32. If S is a simplicial function cone and f ∈ Kc (S), then f ∗ = inf{h ∈ Ac (S) : h ≥ f } and this family is down-directed. Hence f ∗ ∈ (Ac (S))⊥⊥ . Proof. The first part follows from Theorem 7.31(iii) and the second part is an immediate consequence of the first one. Definition 7.33 (Kernel T ). Let S be a simplicial function cone on a compact space K. For any bounded universally measurable function f on K, we define a function T f as Z T f : x 7→
f dδx ,
x ∈ K.
K
If f is defined only on a universally measurable subset A of K, we consider f to be defined by 0 on K \ A, and define T f as above. We show in Theorem 7.34 below that the mapping T : x 7→ δx is a kernel as defined in Subsection A.3.D. Theorem 7.34. Let S be simplicial. Then the following assertions hold: (a) If f ∈ C(K) and ε > 0, then there exist functions u, v ∈ Kc (S) so that kT (u − v) − T f k < ε. Hence, there exist sequences {fn }, {gn } of bounded upper semicontinuous functions on K such that the sequence {fn − gn } converges uniformly to T f . In particular, if K is metrizable, T f is a Baire-one function. (b) T f is a Borel function for any bounded Baire function f on K. (c) T f ∈ (Ac (S))⊥⊥ , and hence T f is Ac (S)-affine, for any bounded Baire function f on K. Proof. The proof is identical with the proof of Theorem 6.8; we only have to use Lemma 7.29(c) and Corollary 7.32. Proposition 7.35. The following conditions are equivalent: (i) S is simplicial, (ii) (Ac (S))⊥ ∩ Mbnd (S) = {0}. Proof. The implication (i) =⇒ (ii) follows from Corollary 7.32 and (ii) =⇒ (i) is obvious. Theorem 7.36. Let S be a simplicial cone and T : K → M1 (K) be the kernel from Definition 7.33. Then the following assertions hold: (a) For any µ ∈ M+ (K),
7.4 Simplicial cones
231
(a1) T µ ∈ M+ (K), (a2) kT µk = kµk, (a3) T µ is S-maximal, (a4) µ ≺ T µ. (b) T µ is a boundary measure for any µ ∈ M(K). (c) T εx = δx , x ∈ K. Proof. The proof is identical to the proof of Theorem 6.11. Corollary 7.37. Let S be simplicial and f be a bounded universally measurable function on K such that δx (f ) = f (x) for each x ∈ K. Then f ∈ (Ac (S))⊥⊥ . In particular, any bounded function in A(S) belongs to (Ac (S))⊥⊥ . Proof. Follow the proof of Corollary 6.12. Theorem 7.38. Suppose that H := Ac (S) separates points of K. (a) Then H is a closed function space, Mx (S) ⊂ Mx (H) for x ∈ K, S c (H) ⊂ S c (S), Mbnd (H) ⊂ Mbnd (S), ChH (K) ⊂ ChS (K) and Ac (H) = H. (b) If S is moreover simplicial, then H is simplicial, Mbnd (S) = Mbnd (H) and ChS (K) = ChH (K). Proof. We start the proof of (a) by noticing that H is indeed a closed function space. Inclusion Mx (S) ⊂ Mx (H) follows by the definition, inclusions S c (H) ⊂ S c (S), Mbnd (H) ⊂ Mbnd (S) and ChH (K) ⊂ ChS (K) are its consequences and Ac (H) = H follows from Theorem 3.27. Now assume that S is simplicial. Let s, −t ∈ Kc (H) with s ≤ t be given. Since c K (H) ⊂ Kc (S), Theorem 7.31 yields the existence of a function h ∈ H such that s ≤ h ≤ t. By Theorem 6.6, H is simplicial. Let µ ∈ M+ (K) be an S-maximal measure and let ν ∈ M+ (K) satisfy µ ≺H ν. Then µ − ν ∈ H⊥ . For any f ∈ Kc (H) ⊂ Kc (S) we get from Theorem 7.19 and Corollary 7.32 that µ(f ) = inf{µ(h) : h ∈ H, h ≥ f } = inf{ν(h) : h ∈ H, h ≥ f } = ν(f ). Hence µ = ν and µ is H-maximal. Thus Mbnd (S) ⊂ Mbnd (H), which gives Mbnd (S) = Mbnd (H) by (a). Since the equality ChS (K) = ChH (K) is then an immediate consequence, the proof is complete. Lemma 7.39. If f ∈ Kusc (S) and g ∈ S usc (S) with f ≤ g on ChS (K), then f ≤ g on K.
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Proof. Given x ∈ K and ε > 0, Proposition 7.13(b) provides a function g 0 ∈ S c (S) such that g < g 0 and g 0 (x) ≤ g(x) + ε. Then g 0 − f ∈ S lsc (S) and g 0 − f ≥ 0 on ChS (K). By Theorem 7.15(f), g 0 − f ≥ 0 on K. In particular, 0 ≤ g 0 (x) − f (x) ≤ g(x) − f (x) + ε. Since ε is arbitrary, the assertion follows. Theorem 7.40. Suppose that Ac (S) separates points of K. If Ac (S) is simplicial and ChAc (S) (K) = ChS (K), then S is simplicial. Proof. Let H stand for the space Ac (S). The aim of the proof is to show that any Smaximal measure is H-maximal. To this end, let f ∈ Kc (H) ⊂ Kc (S) be given. We temporarily write f ∗,H and f ∗,S for the respective envelopes. We claim that f ∗,H = f ∗,S on K. Indeed, f ∗,H = f on ChH (K) by Theorem 3.24 and f ∗,S = f on ChS (K) by Theorem 7.21. Since H is simplicial, f ∗,H is H-affine (see Theorem 6.5) and thus also S-affine. Since f ∗,S ∈ S usc (S) and f ∗,H = f ∗,S on ChS (K), Lemma 7.39 yields f ∗,H ≤ f ∗,S on K. Since f ∗,H ≥ f ∗,S by the definitions and Theorem 7.38(a), the claim is proved. Let µ ∈ Mmax (S) and f ∈ Kc (H) be given. Then µ(f ) = µ(f ∗,S ) by Theorem 7.19, and thus µ(f ) = µ(f ∗,H ) by the previous step. Theorem 3.58 gives that µ is H-maximal. Since Mx (S) ⊂ Mx (H) for all x ∈ K, the simpliciality of S follows.
7.5
Ordered compact convex sets and simplicial measures
Definition 7.41 (Ordered compact convex set). Let E be a locally convex space provided with a partial ordering ≤ given by a closed convex cone E + ⊂ E that is proper (that is, E + ∩ −E + = {0}). A compact convex set X ⊂ E with the ordering inherited from E is called an ordered compact convex set. A real-valued function l on X is called isotone if l(x) ≤ l(y) whenever x ≤ y, x, y ∈ X. We denote by Lc (X) the convex cone of all continuous isotone affine functions on X. Proposition 7.42. Let X be an ordered compact convex set and x, y ∈ X. Then x ≤ y if and only if l(x) ≤ l(y) for each l ∈ Lc (X). In particular, Lc (X) separates points of X. Proof. The first part is nothing else than Lemma A.20. If x, y ∈ X are distinct points, we may assume that x y. By the first part, f (y) < f (x) for a suitable f ∈ Lc (X). This concludes the proof.
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Example 7.43. Let H be a function space on a compact space K. Consider the locally convex space M(K) ordered by the cone {µ ∈ M(K) : µ(f ) ≥ 0 for all f ∈ Kc (H)}. Then the ordering induced by this cone on M+ (K) is nothing else than the Choquet ordering ≺H . The sets M1 (K), Mx (H) where x ∈ K, Mµ (H) := {ν ∈ M+ (K) : µ − ν ∈ H⊥ } and {ν ∈ M+ (K) : µ ≺ ν ∈ H⊥ } where µ ∈ M+ (K) are ordered compact convex sets. In these examples, the cone of affine isotone continuous functions coincides with the set of all H-convex continuous functions on K, where we view C(K) as a subspace of (M(K))∗ . Notation 7.44 (Maximal elements). If X is an ordered compact convex set, the set of all points of X, which are maximal in the given ordering, is denoted by Xmax . Proposition 7.45. Let X be an ordered compact convex set. Then the following assertions hold: (a) For every point x ∈ X, there exists a maximal element z ∈ X so that x ≤ z. (b) The set Xmax is an extremal subset of X and ext Xmax = Xmax ∩ ext X. (c) If Xmax is convex, it is a face of X. Proof. If x ∈ X is given, let F := {y ∈ X : x ≤ y}. This is a partially ordered set and we need to show that any chain in F has an upper bound. But this easily follows from the compactness of X. Indeed, if R ⊂ F is a chain, find a subnet J ⊂ R and a point y0 ∈ X such that J converges to y0 . Fix y ∈ R. Let l ∈ Lc (X) and ε > 0. Find y1 ∈ J so that y ≤ y1 and |l(y1 ) − l(y0 )| ≤ ε. Then l(y0 ) ≥ l(y1 ) − ε ≥ l(y) − ε. Since ε is arbitrary, l(y) ≤ l(y0 ) and thus y ≤ y0 due to Proposition 7.42. Since we have verified that any chain in F has an upper bound, Zorn’s lemma concludes the proof of (a). Concerning (b), let z ∈ Xmax and x, y ∈ X with z = 21 (x + y) be given. Our aim is to show that x, y ∈ Xmax . If not, let y0 be a point of X distinct from y and satisfying y ≤ y0 . Then z ≤ 21 (x + y0 ) and z 6= 21 (x + y0 ), which contradicts maximality of z. Thus Xmax is an extremal subset of X. The second part of (b) follows from Proposition 2.64. Finally, (c) follows from (b). Definition 7.46 (Hereditary upwards sets). A subset F of an ordered compact convex set X is said to be hereditary upwards if x ∈ F , y ∈ X and x ≤ y implies that y ∈ F . We remark that X itself is hereditary upwards.
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Proposition 7.47. If F is a nonempty closed face which is hereditary upwards, then F ∩ ext Xmax 6= ∅. In particular, ext Xmax is nonempty if X 6= ∅. Proof. Let the collection F := {H ⊂ F : H is a nonempty hereditary upwards closed face } be partially ordered by the reverse T inclusion. Then F is nonempty since F ∈ F. If R ⊂ F is a chain, then M := R is also a nonempty hereditary upwards closed face contained in F . Thus F has a maximal element H by Zorn’s lemma. If we suppose that H contains two distinct points x and y, we can find a continuous isotone affine function l in X such that l(x) 6= l(y), say l(x) < l(y). Then N := {z ∈ H : l(z) = max l(H)} is a hereditary upwards closed face properly contained in H, which contradicts maximality of H. Thus H is a singleton, say H = {z}. Since H is a face, z is an extreme point of X. Moreover, z is maximal because H is hereditary upwards. Thus z ∈ ext Xmax ∩ F as required. Proposition 7.48. Let ϕ : X → Y be a continuous affine surjection of an ordered compact convex set X onto an ordered compact convex set Y such that ϕ(x1 ) ≤ ϕ(x2 ) whenever x1 ≤ x2 are elements of X. Then ϕ(Xmax ) ⊃ Ymax . Proof. Given y ∈ Ymax , we select x ∈ ϕ−1 (y). Using Proposition 7.45(a) we find z ∈ Xmax with x ≤ z. Since ϕ−1 (y) is hereditary upwards due to the assumption on ϕ, z ∈ ϕ−1 (y), and the proof is complete. Proposition 7.49. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Then ϕ] (Mmax (X)) ⊃ Mmax (Y ). Proof. If ϕ is as in the hypothesis, ϕ] : M1 (X) → M1 (Y ) satisfies the assumptions of Proposition 7.48 (here M1 (X) and M1 (Y ) are ordered compact convex sets when considered with the Choquet ordering). Hence any Ac (Y )-maximal measure ν ∈ M1 (Y ) is the image of some measure µ ∈ Mmax (X) ∩ M1 (X), and the assertion follows. In the sequel we apply results of previous sections on function cones. It will be more convenient to work with the function cone −Lc (X) instead of Lc (X). Proposition 7.50. Let X be an ordered compact convex set. Then −Lc (X) is a closed function cone. Proof. A straightforward verification using Proposition 7.42.
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Definition 7.51 (Monotone envelopes). If X is an ordered compact convex set, we define a monotone envelope of a function f on X by fe(x) := inf{l(x) : l ≥ f, −l ∈ Lc (X)},
x ∈ X.
Example 7.52. If X = M1 (K) as in Example 7.43 with ordering determined by some function space H on K, then the family of continuous isotone affine functions on X consists of functions of the form µ 7→ µ(f ), where f is an H-convex continuous function on K. If F ∈ Ac (X), then F (µ) = µ(f ), µ ∈ M1 (K), for some continuous function f on K. Then Fe(µ) = inf{L(µ) : L ≥ F, −L ∈ Lc (X)} = inf{µ(l) : l ≥ f, l ∈ S c (H)} = µ(f ∗ ). Lemma 7.53. Let X be an ordered compact convex set and f be a function on X. Then fe(x) = Qεx (f ), x ∈ X, where Qεx (f ) is the functional from Definition 7.7 if the role of S is played by the cone −Lc (X). Proof. The proof follows from Proposition 7.11(b). Theorem 7.54. Let X be an ordered compact convex set and −Lc (X) be considered as a function cone on X. Then the following assertions hold: (a) If µ, ν ∈ M1 (X) satisfy µ ≺−Lc (X) ν, then r(µ) ≤ r(ν). (b) The function −fe is isotone for any f ∈ C(X). (c) A point x is in Xmax if and only if fe(x) = f (x) for any f ∈ Ac (X). (d) Ch−Lc (X) (X) = ext Xmax . (e) For every x ∈ X there exists a −Lc (X)-maximal measure µ ∈ Mx (−Lc (X)). Proof. Given µ, ν as in (a), let l ∈ Lc (X) be arbitrary. Then l(r(µ)) = µ(l) ≤ ν(l) = l(r(ν)). By Proposition 7.42, r(µ) ≤ r(ν). Assertion (b) is easy to verify because the supremum of a family of isotone functions is again an isotone function. To verify (c), let x ∈ Xmax be given. For any f ∈ Ac (X), there exists a measure µ ∈ Mx (−Lc (X)) such that fe(x) = µ(f ) (see Lemma 7.10). Then εx ≺−Lc (X) µ (see Proposition 7.12(b)) and thus x ≤ r(µ) by (a). Hence x = r(µ) due to the maximality of x. This yields fe(x) = µ(f ) = f (r(µ)) = f (x).
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Conversely, let x ∈ X \ Xmax be given. It means that there exists y ∈ X such that x < y. Using Proposition 7.42, we find f ∈ Lc (X) such that f (x) < f (y). Then f (x) < f (y) ≤ fe(y) ≤ fe(x) (the last inequality holds due to (b)). This proves (c). To show (d), assume first that x ∈ ext Xmax . Given an arbitrary function f ∈ C(X) and ε > 0, let h ∈ Ac (X) be such that f ≤ h and h(x) < f (x)+ε (use Theorem 3.24 and Proposition 3.25 for the function space Ac (X)). By (c), e h(x) = h(x), and thus c there exists a function l ∈ −L (X) such that h ≤ l and l(x) < h(x) + ε. Thus f ≤ l and l(x) ≤ f (x) + 2ε, which gives fe(x) = f (x). By Theorem 7.21, x ∈ Ch−Lc (X) (X). Conversely, assume that x ∈ Ch−Lc (X) (X). If y ∈ X satisfies x ≤ y, then εx ≺−Lc (X) εy . Since x ∈ Ch−Lc (X) (X), x = y and thus x ∈ Xmax . Let µ ∈ Mx (X) be arbitrary. Then εx ≺Ac (X) µ by Corollary 3.26. Since W(−Lc (X)) ⊂ W(Ac (X)), εx ≺−Lc (X) µ (use Proposition 3.56 and Proposition 7.16). Hence µ = εx and x ∈ ext X, by Theorem 2.40. Thus x ∈ Xmax ∩ ext X = ext Xmax by Proposition 7.45(b). Since (e) is a consequence of Theorem 7.18, the proof is complete. Theorem 7.55. Let X be an ordered compact convex set and −Lc (X) be considered as a function cone on X. If Σ := {A ∩ ext Xmax : A ⊂ X is a Baire set}, then for every x ∈ Xmax there exists a measure µ0 on the σ-algebra Σ such that Z f (x) = f (t) dµ0 (t), f ∈ Ac (X). ext Xmax
Proof. Given the objects as in the hypothesis, let µ ∈ Mx (−Lc (X)) be a −Lc (X)maximal measure (use Theorem 7.18) and let µ0 be the induced measure on Σ given by Theorem 7.27. Since εx ≺−Lc (X) µ, x ≤ r(µ) by Theorem 7.54(a). Since x is a maximal element of X, x = r(µ). Hence Theorem 7.27 gives Z Z f (t) dµ0 (t) = f (t) dµ0 (t) ext Xmax
Ch−Lc (X)
= µ(f ) = f (r(µ)) = f (x) for any f ∈ Ac (X).
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237
Definition 7.56 (Weak topology generated by monotone envelopes). Let X be an ordered compact convex set in a locally convex space E. Assume that, for every h ∈ E ∗ , g the monotone envelope h| X is the restriction of some linear functional ϕh : E → R. ∗ Let D := span(E ∪ {ϕh : h ∈ E ∗ }). We denote by τ the locally convex topology on E generated by D. Example 7.57. If X is any of the ordered compact convex sets from Example 7.43, then D is the linear span of functions of the form µ 7→ µ(f ), µ ∈ M(K),
f ∈ C(K) ∪ {g ∗ : g ∈ C(K)}.
Theorem 7.58. Let X 6= ∅ be an ordered compact convex set such that the topology τ from Definition 7.56 is well defined. Then Xmax is convex and Xmax = coτ ext Xmax . Proof. It follows from Theorem 7.54(c) that \ \ g Xmax = {x ∈ X : h| {x ∈ X : ϕh (x) = h(x)}. X (x) = h(x)} = h∈E ∗
h∈E ∗
Thus Xmax is convex and it is a closed set in the topology τ . Hence we have Xmax ⊃ coτ ext Xmax . We know from Proposition 7.47 that ext Xmax is a nonempty set. Assume that z ∈ Xmax \ coτ ext Xmax . By the Hahn–Banach separation argument, there exists a function f ∈ D such that f (z) > sup f (ext Xmax ). P Then f = h0 + ni=1 ci ϕhi , where h0 , . . . , hn ∈ E ∗ , c1 , . . . , cn ∈ R and ϕhi are linear functionals on E such that ϕhi = hg i |X on X. Thus f = h0 +
n X
ci hg i |X
i=1
on X. By Theorem 7.54(c), f = h0 +
Pn
i=1 ci hi
h := h0 +
n X
on Xmax . Thus
ci hi
i=1
is a function from E ∗ such that h(z) > sup h(ext Xmax ). ^ Since z ∈ Xmax , −h| X (z) = −h(z) by Theorem 7.54(c). Hence there exists an isotone function l ∈ Lc (X) so that l≤h
and
l(z) > sup h(ext Xmax ).
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7 Choquet theory of function cones
Then F := {x ∈ X : l(x) = max l(X)} is a nonempty closed face which is hereditary upwards. By Proposition 7.47, F intersects ext Xmax , say at a point y. Then l(z) > sup h(ext Xmax ) ≥ sup l(ext Xmax ) ≥ l(y) = max l(X) ≥ l(z). This contradiction finishes the proof. Definition 7.59 (Simplicial measures). Let H be a function space on a compact space K. For µ ∈ M+ (K), let Mµ (H) := {ν ∈ M+ (K) : ν − µ ∈ H⊥ }. A measure ν ∈ Mµ (H) is called simplicial if ν is an extreme point of Mµ (H). We remark that Mµ (H) is an ordered compact convex set by Example 7.43. If x ∈ K, then Mεx (H) is nothing else than Mx (H). Theorem 7.60 (Douglas). Let H be a function space on a compact space K and µ ∈ M+ (K). A measure ν ∈ Mµ (H) is simplicial if and only if the space H is dense in L1 (ν). Proof. Notice that H is a subspace of L1 (ν) and that H is dense in L1 (ν) if and only if F ∈ (L1 (ν))∗ : F (h) = 0 for all h ∈ H = {0} . If H is not dense in L1 (ν), then there exists f ∈ L∞ (ν) such that kf kL∞ (ν) = 1 and Z for all h ∈ H.
hf dν = 0 K
Define ν1 and ν2 by setting ν1 := (1 − f )ν
and
ν2 := (1 + f )ν.
Then ν1 , ν2 ∈ M+ (K) and Z Z Z ν1 (h) = (1 − f )h dν = h dν = (1 + f )h dν = ν2 (h) K
K
K
for any h ∈ H. Hence ν1 , ν2 ∈ Mµ (H). Since ν=
ν1 + ν2 2
the measure ν is not simplicial.
and
ν1 6= ν 6= ν2 ,
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239
For the converse, assume that H is dense in L1 (ν) and that ν ∈ Mµ (H) satisfies 2 where ν1 , ν2 ∈ Mµ (H). Since 0 ≤ ν1 ≤ 2ν and 0 ≤ ν2 ≤ 2ν, the Radon– ν = ν1 +ν 2 Nikodym theorem yields positive functions f1 , f2 ∈ L∞ (ν) such that f1 + f2 = 2 ν-almost everywhere and νj = fj ν for j = 1, 2. For any h ∈ H we have Z Z f1 h dν = ν1 (h) = µ(h) = ν2 (h) = f2 h dν. K
K
Since H is dense in L1 (ν), the same equalities hold for any h ∈ L1 (ν). We see that f1 = f2 = 1 ν-almost everywhere. Therefore ν = ν1 = ν2 . Theorem 7.61. Let H be a function space on a compact space K, µ ∈ M+ (K) and X := Mµ (H) be the ordered compact convex set from Definition 7.59. Let τ be the locally convex topology on M(K) induced by the family D from Example 7.57. Then (a) ν ∈ Xmax if and only if ν ∈ X is H-maximal, (b) Xmax is a face of X, (c) ext Xmax = Xmax ∩ ext X, (d) Xmax = coτ ext Xmax , (e) for any H-maximal measure ν ∈ X there exists a measure Ω0 defined on the σ-algebra Σ := {A ∩ ext Xmax : A ⊂ X is a Baire set} such that
Z
λ(f ) dΩ0 (λ) = ν(f ),
f ∈ C(K).
ext Xmax
Proof. If ν ∈ Xmax and λ ∈ M+ (K) with ν ≺H λ, then λ ∈ X and thus ν = λ. Hence ν is H-maximal. Obviously, any H-maximal measure ν ∈ X is in Xmax . This proves (a). Assertions (b), (c) and (d) follow from Theorem 7.58 and Proposition 7.45; (e) is a consequence of Theorem 7.55. Theorem 7.62. Let H be a function space on a compact space K. Then for any x ∈ K and f ∈ Kc (H) there exists a simplicial measure µ ∈ Mx (H) such that µ(f ) = f ∗ (x). Proof. Let X denote the ordered compact convex set Mx (H). Then F := {µ ∈ X : µ(f ) = f ∗ (x)} is a closed hereditary upwards face of X. Moreover, it is nonempty by Lemma 3.21. Hence F ∩ ext Xmax 6= ∅ by Proposition 7.47, which finishes the proof.
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7.6
7 Choquet theory of function cones
Exercises
Exercise 7.63. Let X be a compact convex set and S := Sc (X) (see Definition 4.1). Prove that + • ≺ c A (X) coincides with ≺S on M (X), • S-maximal measures coincide with Ac (X)-maximal measures, • Ch (X) = ext X, M (S) = M (X) for x ∈ X. x x S Hint. Since W(S) = S and S is closed, S c (S) = S by Proposition 7.13(e). Hence S c (S) = S = S c (Ac (X)). Thus ≺S coincides with ≺Ac (X) . This implies Mx (X) = Mx (S) for x ∈ X and coincidence of S-maximal and Ac (X)-maximal measures. Exercise 7.64. Consider the function cone S := {f ∈ C([0, 1]) : f is increasing}. Prove that • S is closed and min-stable, M (S) = {µ ∈ M1 ([0, 1]) : spt µ ⊂ [0, x]} for x x ∈ [0, 1], ChS ([0, 1]) = {0}, • Ac (S) is not a function space. Hint. A straightforward verification. Exercise 7.65. Find an example of a function cone S on a compact space K such that Ac (S) is a function space and ChAc (S) (K) 6= ChS (K). Hint. For n ∈ N, consider the following objects in R2 : In := [−2, −1] × {n−1 }, xn := (0, n−1 ), I := [−2, −1] × {0},
Jn := [1, 1 + n−1 ] × {n−1 }, yn := ((2n)−1 , n−1 ), x := (0, 0),
z := (1, 0).
Let µn be one-dimensional Lebesgue measure on In and νn be one-dimensional Lebesgue measure on Jn normalized in such a way that it is a probability measure. Let ∞ [ K := I ∪ {x} ∪ {z} ∪ (In ∪ Jn ∪ {xn } ∪ {yn }) n=1
and
S := {f ∈ C(K) : f (xn ) ≥ (1 − n−1 )f (yn ) + n−1 µn (f ), f (xn ) ≥ (1 − n−1 )f (yn ) + n−1 νn (f ), n ∈ N}.
Then S is a closed min-stable function cone on a compact space K and S c (S) = S. Further, the function 0 at z, f = n−1 on Jn , n ∈ N, 1 otherwise, shows that z ∈ ChS (K).
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241
Further, Ac (S) = {f ∈ C(K) : f (xn ) = (1 − n−1 )f (yn ) + n−1 µn (f ), f (xn ) = (1 − n−1 )f (yn ) + n−1 νn (f ), n ∈ N}. is a function space and z ∈ / ChAc (S) (K). Indeed, for any function f ∈ Ac (S) and n ∈ N we have (1 − n−1 )f (yn ) + n−1 µn (f ) = (1 − n−1 )f (yn ) + n−1 νn (f ). Hence µn (f ) = νn (f ),
n ∈ N,
which in turn yields Z f (t) dt = f (z). I
This concludes the proof. Exercise 7.66. Find an example of an ordered compact convex set X such that Xmax is not convex. Hint. Consider E := R2 ordered by the cone E + = {(x, y) ∈ E : x ≥ 0, y ≥ 0}, and let X := {(x, y) ∈ E : x2 + y 2 ≤ 1}. Then Xmax = {(x, y) ∈ X : x ≥ 0, y ≥ 0, x2 + y 2 = 1} is not convex. Exercise 7.67. If X is an ordered compact convex set, x ∈ Xmax and µ ∈ Mx (X), then µ(fe) = µ(f ) for any f ∈ Ac (X). Hint. Given f ∈ Ac (X), −fe is upper semicontinuous and concave. Using Corollary 4.8 and Theorem 7.54(c), we get f (x) = µ(f ) ≤ µ(fe) ≤ fe(x) = f (x).
Exercise 7.68. If X is a metrizable ordered compact convex set, then Xmax is a measure extremal subset of X. Hint. Let {fn : n ∈ N} be a dense countable subset in BAc (X) . From Lemma 7.8(c) and Theorem 7.54(c) we deduce that Xmax =
∞ \
{x ∈ X : fe(x) = f (x)}.
n=1
Thus Xmax is a Gδ set in X. If x ∈ Xmax and µ ∈ Mx (X), Exercise 7.67 yields µ(fe) = µ(f ). Thus we get µ(Xmax ) = 1 and Xmax is measure extremal.
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7 Choquet theory of function cones
Exercise 7.69. Let X be a metrizable compact convex set and f be a bounded universally measurable function on X such that, for any maximal µ ∈ M1 (X), µ(f ) = f (r(µ)). Prove that f is strongly affine. Hint. Given a measure µ ∈ Mx (X), let ν ∈ M1 (X) be maximal with µ ≺ ν. Then ν ∈ Mx (X) as well. We consider Y := M1 (X) as an ordered compact convex set and let πi : Y × Y → Y be the projections, i = 1, 2. Let M := {(εx , λ) ∈ Y × Y : εx ≺ λ}. By Theorem 3.92(ii), there exists Ω ∈ M1 (M ) such that Ω represents the point (µ, ν) ∈ Y × Y . Then (π2 )] Ω represents ν ∈ Ymax , and thus Exercise 7.68 gives that ((π2 )] Ω)(Ymax ) = 1. By assumption, ν(f ) = f (x). We apply Proposition 3.90 and the assumption to the pairs (f, 0) and (0, f ) to get Z Z µ(f ) = λ1 (f ) dΩ(λ1 , λ2 ) = λ1 (f ) dΩ(λ1 , λ2 ) M ∩(Y ×Ymax )
M
Z =
λ1 (f ) dΩ(λ1 , λ2 ) {(εx ,λ)∈Y ×Y :εx ≺λ,λ∈Ymax }
Z =
λ2 (f ) dΩ(λ1 , λ2 ) {(εx ,λ)∈Y ×Y :εx ≺λ,λ∈Ymax }
Z =
λ2 (f ) dΩ(λ1 , λ2 ) M
= ν(f ) = f (x). This concludes the proof. Exercise 7.70. Find an example of a function space H on a compact space K and an ordered compact convex set X ⊂ M(K) such that Xmax is a convex set that is strictly smaller then co ext Xmax . Hint. Let H be a function space on a compact space K such that there exists x ∈ ChH (K) \ ChH (K). If X := M1 (K), then Xmax = M1 (K) ∩ Mmax (H),
ext Xmax = {εy : y ∈ ChH (K)}.
Hence εx ∈ co ext Xmax \ Xmax . Exercise 7.71. Find an example of a function space H on a compact space K and an ordered compact convex set X ⊂ M(K) such that cok·k ext Xmax is strictly smaller then Xmax .
7.7 Notes and comments
243
Hint. Let K := [0, 1], H := C(K) and X := M1 (K). Then Xmax = M1 (K),
ext Xmax = {εx : x ∈ K}
and cok·k ext Xmax consists of discrete probability measures on K. Hence Xmax 6= cok·k ext Xmax .
7.7
Notes and comments
In the literature, there are various treatises of Choquet theory of function cones in a diverse level of generality. Let us mention, for example, cones of lower semicontinuous functions, cones on locally compact spaces, weakly complete cones, well-capped cones, or cones of potentials. The reader can find material concerning Choquet’s theory of convex cones in several books such as E. M. Alfsen [5], R. Becker [47], J. Bliedtner and W. Hansen [66], N. Boboc and Gh. Bucur [70], G. Choquet [108], D. Sibony [415], and in papers written by many authors. We mention here J. Bliedtner and W. Hansen [62], N. Boboc and Gh. Bucur [68], N. Boboc and A. Cornea [73], D. A. Edwards [157], G. Mokobodzki [347], G. Mokobodzki and D. Sibony [348]. We incorporated many ideas from these sources in Sections 7.1 – 7.4. Section 7.5 follows E. M. Alfsen [5] (see also E. M. Alfsen and C. F. Skau [11]). The characterization of simplicial measures in Theorem 7.60 is in R. G. Douglas [147] (see also N. Boboc and Gh. Bucur [69]). Recall that the Minkowski–Carath´eodory theorem 2.12 states that every point of a compact convex subset of Rd is a convex combination of its at most d + 1 affinely independent extreme points. For a proof based on properties of simplicial measures, see [5], Corollary I.6.13. Problem 7.72. We do not know whether the assertion of Exercise 7.69 holds without the metrizability assumption.
Chapter 8
Choquet-like sets
The aim of this chapter is to transfer the concept of faces of compact convex sets to the framework of general function spaces. The most important examples are P -sets and M -sets, which are analogues of parallel and split faces. We start by recalling in Theorems 8.5 and 8.7 measure theoretic characterizations of split and parallel faces that guide us to the general definition. In order to be able to handle “faces” in function spaces as in the framework of compact convex sets, we need several analogues of classical tools from convex analysis. This task is accomplished in Section 8.2, where we define H-convex and H-extremal sets and prove results imitating the Krein–Milman theorem, the Milman theorem and the Hahn–Banach separation theorem (see Corollary 8.19, Corollary 8.20 and Proposition 8.23). In Section 8.3, we present analogues of faces in function spaces termed Choquet sets. Particular examples of P -sets and M -sets are defined there and their basic characterizations are proved (see Theorem 8.39 and Theorem 8.44). An important result due to C. J. K. Batty on characterization of H-maximal measures is given in Theorem 8.32. H-exposed sets are investigated in Section 8.4. The next section provides preparatory results on boundary measures that are necessary in Section 8.6. The central result of this chapter is Theorem 8.60 and its consequences contained in Theorem 8.62 and Corollary 8.63. They provide characterization of simplicial function spaces by means of Choquet sets. In the context of compact convex sets we get a characterization of simplices by means of split and parallel faces. Throughout this chapter, we frequently use the notation and results from Section 4.3. Hence H is a function space on a compact space K, φ is the evaluation mapping from K to the state space S(H) and X denotes a compact convex subset of a locally convex space.
8.1
Split and parallel faces
In the developing of Choquet’s theory we would like to find a counterpart of the notions of split and parallel faces studied in the convex setting. In Theorems 8.5 and 8.7 we present without proofs known characterizations of these faces fitted to our purposes. Definition 8.1 (Complementary sets). Let F ⊂ X. The union of all faces of X disjoint from F is called the complementary set of F and it is denoted by F 0 .
8.1 Split and parallel faces
245
A complementary set F 0 is always an extremal set, and it is a face if and only if it is convex. Moreover, x ∈ F 0 if and only if F ∩ face x = ∅. Proposition 8.2. If F is a closed face of a compact convex set X, then F = {x ∈ X : (cF )∗ (x) = 1}
and F 0 = {x ∈ X : (cF )∗ (x) = 0} .
Every x ∈ X can be written as a convex combination x = λy + (1 − λ)z,
(8.1)
where y ∈ F , z ∈ F 0 and the “barycentric coefficient” λ = (cF )∗ (x) ∈ [0, 1]. Proof. See [5], Proposition II.6.5. Remarks 8.3. (a) Since (cF )∗ is an upper semicontinuous function, the complementary set F 0 of a closed face F is a Gδ set. (b) Notice that {x ∈ X : cF (x) = c∗F (x)} ⊂ F ∪ F 0 . In general, the decomposition x = λy + (1 − λ)z in (8.1) is not unique. The question of uniqueness of the decomposition leads to notions of parallel and split faces and is investigated next. Definition 8.4 (Split faces). A face F of X is said to be a split face if its complementary set F 0 is convex and if any point in x ∈ X \ (F ∪ F 0 ) can be uniquely represented as a convex combination x = λy + (1 − λ)z,
y ∈ F, z ∈ F 0
and
λ ∈ (0, 1).
The following theorem provides a measure theoretic characterization of split faces. Theorem 8.5. The following statements about a closed face F of X are equivalent: (i) F is a split face, (ii) µ|F ∈ (Ac (X))⊥ for any boundary measure µ ∈ (Ac (X))⊥ . Proof. See E. M. Alfsen [5], Theorem II.6.12. Definition 8.6 (Parallel faces). A face F is called parallel if its complementary set F 0 is convex, X = co(F ∪ F 0 ), and if for every point x ∈ X \ (F ∪ F 0 ) the barycentric coefficient λ in the convex combination x = λy + (1 − λ)z, is uniquely determined.
where
y ∈ F, z ∈ F 0 , λ ∈ (0, 1),
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8 Choquet-like sets
As in the case of split faces we have the following measure theoretic characterization of parallel faces. Theorem 8.7. The following statements about a closed face F of X are equivalent: (i) F is a parallel face, (ii) µ(F ) = 0 for any boundary measure µ ∈ (Ac (X))⊥ . Proof. See B. Hirsberg [226], Theorem 2.12 or L. Asimow and A. J. Ellis [24], Theorem 2.10.10.
8.2
H-extremal and H-convex sets
Definition 8.8 (H-extremal sets). We recall that a universally measurable subset F of K is H-extremal if, given x ∈ F and µ ∈ Mx (H), then µ is carried by F (see Definition 3.12). Example 8.9. Any universally measurable subset of the Choquet boundary ChH (K) is H-extremal. This follows from the fact that Mx (H) = {εx } for any x ∈ ChH (K). Lemma 8.10. A universally measurable set B ⊂ K is H-extremal if and only if its characteristic function cB is H-convex, and B is an H-extremal closed set if and only if cB is upper semicontinuous and H-convex. Moreover, if H is a measure extremal subset of the state space S(H), then φ−1 (H ∩ φ(K)) is H-extremal in K. Proof. The former assertion follows immediately from the definitions. If H ⊂ S(H) is measure extremal and µ is a representing measure for x ∈ F := φ−1 (H ∩ φ(K)), then φ] µ represents φ(x) ∈ H. Since H is measure extremal, φ] µ is carried by H. Therefore, µ is carried by F . Examples 8.11. (a) If F is an H-extremal subset of K, then φ(F ) need not be an extremal subset of S(H). Consider the following example: Let λ1 and λ2 denote the restrictions of Lebesgue measure on [0, 1] and [3, 4], respectively. If K := [0, 1] ∪ {2} ∪ [3, 4] and
H :=
1 f ∈ C(K) : f (2) = (λ1 (f ) + λ2 (f )) , 2
then K is an H-extremal set. On the other hand, sj = π(λj ), j = 1, 2, are elements of S(H), 1 φ(2) = (s1 + s2 ) ∈ φ(K), 2 while s1 and s2 do not belong to φ(K) since φ(K) = {π(εx ) : x ∈ K}.
8.2 H-extremal and H-convex sets
247
(b) Example 8.29 illustrates the fact that there is an H-extremal set H in K such that co φ(H) is not an extremal subset of S(H). If F is a closed face of a compact convex set X, then the complementary set F 0 of F equals to {x ∈ X : (cF )∗ (x) = 0} (see Proposition 8.2). Definition 8.12 (Complementary sets). If H is a function space on a compact space K, we associate to each subset F of K its complementary set F 0 by F 0 := {x ∈ K : (cF )∗ (x) = 0} . Since (cF )∗ is an upper semicontinuous function, the complementary set F 0 is a Gδ set. Notice that {x ∈ K : cF (x) = (cF )∗ (x)} ⊂ F ∪ F 0 . Lemma 8.13. Let F be a closed subset of K. Then any maximal measure on K is carried by F ∪ F 0 . Proof. According to Corollary 3.58, µ(cF ) = µ (cF )∗ . Since K \ (F ∪ F 0 ) ⊂ {x ∈ K : cF (x) < (cF )∗ (x)} , it follows that µ is carried by F ∪ F 0 . Definition 8.14 (H-convex sets). We say that a universally measurable set B ⊂ K is H-convex if x ∈ B whenever x ∈ K and µ ∈ Mx (H) satisfies µ(K \ B) = 0. Definition 8.15 (closed H-convex hull). Let F be a subset of K. The closed H-convex hull of F is the set \ coH F := {C : C ⊃ F, C is closed and H-convex} . Obviously, coH F is a closed H-convex set. Conversely, if F is a closed H-convex set, then coH F = F . Also, it is easy to verify that coH F = coH F . Lemma 8.16. Let F ⊂ K. Then we have: (a) The set F is H-convex or H-extremal if and only if it is Ac (H)-convex or Ac (H)extremal, respectively. (b) coH F = coA
c (H)
F.
Proof. The assertions follow directly from the definitions. Proposition 8.17. Let {Fi : i ∈ I} be a family of H-convex sets. If sally measurable, it is H-convex as well. Proof. The proof follows by a straightforward verification.
T
i∈I
Fi is univer-
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8 Choquet-like sets
Proposition 8.18. Let F be a subset of K. Then coH F = {x ∈ X : |h(x)| ≤ sup |h|(F ) for all h ∈ H} = x ∈ X : there is µ ∈ Mx (H) such that spt µ ⊂ F . Proof. Denote by Fe := {x ∈ X : |h(x)| ≤ sup |h|(F ) for all h ∈ H} and Fb := x ∈ X : there is µ ∈ Mx (H) such that spt µ ⊂ F . The set Fe is closed and contains F . Moreover, Fe is H-convex. Indeed, given x ∈ K and µ ∈ Mx (H) such that spt µ ⊂ Fe, we have Z Z |h(x)| = |µ(h)| = h dµ ≤ |h| dµ ≤ sup {|h(t)| : t ∈ F } Fe
Fe
for every h ∈ H. Hence x ∈ Fe. It follows that coH F ⊂ Fe. Assume that x ∈ Fe. Define Tx : f 7→ hf (x) for
f ∈ H|F ,
where hf is any extension of f to a function from H. The functional Tx is well defined, because if h1 and h2 are functions from H such that h1 |F = h2 |F = f , then h1 − h2 = 0 on F . Since x ∈ Fe, we get h1 (x) = h2 (x). Moreover, Tx is positive on H|F and kTx k = 1. By the Hahn–Banach theorem, there exists a Radon measure µ ∈ M1 (K) such that spt µ ⊂ F and µ(f ) = Tx (f ) for any f ∈ H. Since µ(h) = Tx (h) = h(x) for any h ∈ H, we have µ ∈ Mx (H), and therefore x ∈ Fb. We have shown that Fe ⊂ Fb. The next step is to show that Fb ⊂ coH F . To this end, assume that C ⊃ F is a closed H-convex set. Pick x ∈ Fb. There exists a measure µ ∈ Mx (H) such that spt µ ⊂ F ⊂ C. Hence from the H-convexity of C, it follows that x ∈ C. We can conclude that Fb ⊂ coH F . Since coH F = coH F (see Definition 8.15), from foregoing we get coH F ⊂ Fe ⊂ Fb ⊂ coH F = coH F, which established the assertion and finishes the proof. Corollary 8.19 (Krein–Milman type theorem). For any function space H on K, K = coH (ChH (K)). Proof. Given x ∈ K, let µx be a maximal measure in Mx (H). Since maximal measures are carried by ChH (K) (see Proposition 3.64), x ∈ coH (ChH (K)) by Proposition 8.18. Hence K = coH (ChH (K)) = coH (ChH (K)).
8.2 H-extremal and H-convex sets
249
Corollary 8.20 (Milman type theorem). Let F be a subset of K such that K = coH F . Then ChH (K) ⊂ F . Proof. Take x ∈ ChH (K). Since x ∈ coH F , by Proposition 8.18 there exists µ ∈ Mx (H) such that spt µ ⊂ F . Since x ∈ ChH (K), µ = εx , which implies that x ∈ F. Remarks 8.21. (a) Let F be a subset of a compact convex set X and H = Ac (X). Then coH F is nothing else than the closed convex hull of F . (b) It is not true that, for a Borel subset F of a compact convex set X and H = Ac (X), coH F = {x ∈ X : there is µ ∈ Mx (H) such that µ(X \ F ) = 0} . Consider the compact convex set X := M1 ([0, 1]) and F := {εx : x ∈ S} where S is a countable dense subset of [0, 1]. Then co F = X while for any measure Λ ∈ M1 (X) representing Lebesgue measure λ ∈ X we have Λ(F ) = 0. Proposition 8.22. A closed set F ⊂ K is H-convex if and only if F = φ−1 (co φ(F ) ∩ φ(K)). Proof. Let F be a closed H-convex set and let x be in K such that φ(x) ∈ co φ(F ). By Proposition 4.28(b), there exists a measure µ ∈ M1 (φ(F )) representing the point φ(x). Then (φ−1 )] µ ∈ Mx (H) and spt µ ⊂ F . Thus x ∈ coH F = F . Conversely, let F = φ−1 (co F ∩ φ(K)) and let µ be a measure representing a point x ∈ K such that spt µ ⊂ F . Then spt φ] µ ⊂ φ(F ), and hence φ(x) ∈ co φ(F ). Due to the assumption, φ(x) ∈ φ(F ). Proposition 8.23 (Hahn–Banach type theorem). If F is a nonempty subset of K and x∈ / coH F , then there exists h ∈ H such that h(x) > max h(F ). Proof. This readily follows from Proposition 8.18. Proposition 8.24. If µ and ν are positive measures on K, µ ≺ ν, then spt µ ⊂ coH spt ν. Proof. Let µ ≺ ν and x ∈ spt µ \ coH spt ν. By Proposition 8.23, there exists a function h ∈ H such that h(x) > max h(coH spt ν). By adding a suitable constant, we may assume that max h(coH spt ν) = 0. Then h ∨ 0 is a continuous H-convex function and µ(h ∨ 0) > 0 ≥ ν(h ∨ 0) which contradicts the assumption µ ≺ ν. Lemma 8.25. Let F be a closed set in K. Then F ∩ ChH (K) = (coH F ) ∩ ChH (K).
250
8 Choquet-like sets
Proof. The inclusion “⊂” being obvious, we need to verify the reverse one. Let x ∈ (coH F ) ∩ ChH (K) be given. By Proposition 8.18 there exists a measure µ ∈ Mx (H) ∩ M1 (F ). Since x ∈ ChH (K), µ = εx and x ∈ F , which completes the proof. Lemma 8.26. Let µ be a boundary measure and F a closed subset of K. Then |µ|(coH F \ F ) = 0. In particular, if F is a closed subset of ChH (K), then F = (coH F ) ∩ ChH (K). Proof. Let L be a compact subset of coH F \ F . Given ε > 0, there exists a compact set M ⊂ K \ coH F such that |µ|(M ) > |µ|(K \ coH F ) − ε. Let g ∈ C(K) be such that 0 ≤ g ≤ 1 on K, g=1
on L and
g=0
on F ∪ M.
Then g∗ = 0 on coH F (see Proposition 8.18) and (obviously) on M . Since µ is a boundary measure, using Corollary 3.58 we get Z Z Z g∗ d|µ| + g∗ d|µ| + g∗ d|µ| |µ|(L) ≤ |µ|(g) = coH F
K\(M ∪coH F )
M
H
≤ |µ|(K \ (M ∪ co F )) < ε. As ε > 0 is arbitrary, |µ|(L) = 0. Hence, |µ|(coH F \ F ) = 0. The second assertion follows immediately from Lemma 8.25.
8.3
Choquet sets, M -sets and P -sets
Definition 8.27 (Choquet sets). A universally measurable subset F of K is a Choquet set if F is both H-extremal and H-convex. Hence, a closed set F is a Choquet set if and only if the following two conditions are satisfied: (a) given x ∈ F and µ ∈ Mx (H), then spt µ ⊂ F , (b) if x ∈ K \ F and µ ∈ Mx (H), then spt µ is not contained in F . Example 8.28 (Convex case). Let X be a compact convex subset of a locally convex space. A closed set F ⊂ X is a Choquet set if and only if F is a closed face of X. This assertion follows by the fact that a closed set F in X is convex or extremal if and only if F is Ac (X)-convex or Ac (X)-extremal, respectively.
8.3 Choquet sets, M -sets and P -sets
251
Example 8.29 (State space). Let H be a closed face of S(H). Then the set F := φ−1 (H ∩ φ(K)) is a closed Choquet set. Indeed, F is H-convex due to Proposition 8.22 and H-extremal by Lemma 8.10. The converse implication is false in general. Let K := [0, 1] ∪ [3, 4]. Denote by λ1 and λ2 restrictions of Lebesgue measure on [0, 1] and [3, 4], respectively. Let H := {f ∈ C(K) : λ1 (f ) = λ2 (f )}. Then F := [0, 1] is a Choquet set but the set H := co φ(F ) is not Ac (S(H))-extremal in S(H). Indeed, the state s := π(λ1 ) belongs to H, the measure φ] λ2 represents s and (φ] λ2 )(H) = 0. Proposition 8.30. (a) If F is a closed H-extremal set, then F ⊂ coH (F ∩ ChH (K)). (b) For any closed Choquet set F ⊂ K we have F = coH (F ∩ ChH (K)). (c) The family of closed Choquet sets is stable with respect to arbitrary intersections. Proof. For the proof of (a), let F be a closed H-extremal set and let x ∈ F be given. First, we show that h(x) ≤ sup h(F ∩ ChH (K)) (8.2) holds for any strictly positive h ∈ H. Let h ∈ H be a strictly positive function. Then f := hcF is an upper semicontinuous H-convex function, and hence it attains its maximum at some point y ∈ ChH (K) (see Lemma 3.13 and Proposition 3.15). Since h is strictly positive, y ∈ F ∩ ChH (K), and (8.2) follows. If h ∈ H is arbitrary, we apply the previous argument to h + c for a suitable c ∈ R to obtain (8.2) for any h ∈ H. Given h ∈ H, (8.2) applied to h and −h yields |h(x)| ≤ sup |h|(F ∩ ChH (K)). By Proposition 8.18, x ∈ coH (F ∩ ChH (K)). To verify (b), we first notice that obviously F ⊃ coH (F ∩ ChH (K)). Since the reverse inclusion follows by (a), the proof of (b) is complete. By combining Propositions 3.14 and 8.17 we get (c). Proposition 8.31. If H is a simplicial function space, then coH F is a Choquet set for any compact F ⊂ ChH (K). Proof. If F is a compact subset of ChH (K), coH F is obviously an H-convex set. To show its H-extremality, by Proposition 8.24 it is enough to show δx is carried by coH F for any x ∈ coH F . For such a point x we find ν ∈ Mx (H) ∩ M1 (F ) (see Proposition 8.18). Since ν is carried by ChH (K), it is H-maximal (use Corollary 3.59), and thus ν = δx . Hence δx ∈ M1 (coH F ) and the proof is finished. S Theorem 8.32. For any x ∈ K, let Fx (H) := ν∈Mx (H) spt ν. For a measure µ ∈ M+ (K) the following assertions are equivalent: (i) µ is maximal,
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8 Choquet-like sets
(ii) there exists a set S ⊂ C(K) separating points of K such that each function in S is constant on Fx (H) for µ-almost all x ∈ K, (iii) any continuous function on K is constant on Fx (H) for µ-almost all x ∈ K. Proof. We start the proof by showing (i) =⇒ (ii). Let µ be a maximal measure and let h ∈ H be arbitrary. Then µ((h2 )∗ ) = µ(h2 ) by Theorem 3.58, and thus (h2 )∗ = h2 on a set A ⊂ K with µ(K \ A) = 0. Let x ∈ A be arbitrary and ν ∈ Mx (H). Then, by the H¨older inequality, ν(h2 ) ≤ (h2 )∗ (x) = h2 (x) = (ν(h))2 ≤ ν(h2 ). Thus
Z
2
Z
(h − h(x)) dν =
2
h dν − 2h(x) K
K
Z
h dν + h2 (x) = 0.
K
Hence h = h(x) on spt ν, and h is constant on Fx (H). Thus we can take H to be the required family S of continuous functions. For the proof of (ii) =⇒ (iii), let S ⊂ C(K) be as in (ii). Then the family F := {f ∈ C(K) : f is constant on Fx (H) for µ-almost all x} is a closed linear space stable with respect to pointwise multiplication and maxima. Since S ⊂ F, F = C(K) by the Stone–Weierstrass theorem. Assume now that (iii) holds. Given a function f ∈ C(K), let x ∈ K be a point such that f is constant on Fx (H). We find a measure ν ∈ Mx (H) such that ν(f ) = f ∗ (x) (see Lemma 3.21). Since x ∈ Fx (H) and spt ν ⊂ Fx (H), f = f (x) on Fx (H). Thus f ∗ (x) = ν(f ) = f (x). Thus f (x) = f ∗ (x) for µ-almost all x ∈ K, and thus µ(f ) = µ(f ∗ ). By Theorem 3.58, µ is maximal. Definition 8.33 (Conditions (M) and (P)). We say that a closed set F ⊂ K satisfies condition (M) if (M)
µ|F ∈ H⊥
for any µ ∈ H⊥ ∩ Mbnd (H),
µ(F ) = 0
for any
and condition (P) if (P)
µ ∈ H⊥ ∩ Mbnd (H).
Proposition 8.34. Let F ⊂ K be a closed set. Then F satisfies condition (M) or (P) if and only if the set coH F satisfies (M) or (P), respectively. Proof. If µ ∈ H⊥ ∩ Mbnd (H) is given, Lemma 8.26 asserts that µ|F = µ|coH F , from which the assertion readily follows.
8.3 Choquet sets, M -sets and P -sets
253
Example 8.35 (State space). If a closed set F satisfies (M), then φ(F ) satisfies (M) in the state space as well. Indeed, given µ ∈ Ac (S(H))⊥ ∩ Mbnd (S(H)), we have (φ−1 )] µ ∈ H⊥ ∩ Mbnd (H) which yields (φ−1 )] µ|F ∈ H⊥ . Then µ|φ(F ) ∈ Ac (S(H))⊥ . Similarly, we see that φ(F ) satisfies (P) provided F satisfies (P). Definition 8.36 (P-sets). We say that a set F is a P -set if F is a closed Choquet set satisfying condition (P). Example 8.37 (Convex case). If F is a closed face in a compact convex set X, then F is a P -set if and only if F is a parallel face. This follows from a measure theoretic characterization of parallel faces stated in Theorem 8.7. Proposition 8.38 (State space). The following assertions hold: (a) If H is a closed parallel face of S(H), then the set F := φ−1 (H ∩ φ(K)) is a P -set. (b) If F is a P -set, then co φ(F ) is a parallel face of S(H). Proof. (a) We already know from Example 8.29 that F is a Choquet set. Let µ ∈ H⊥ ∩ Mbnd (H) be given. Then φ] µ ∈ Ac (S(H))⊥ ∩ Mbnd (S(H)). Thus we get µ(F ) = (φ] µ)(H) = 0, which verifies condition (P) for F . (b) By Example 8.35, the set φ(F ) satisfies condition (P). By Proposition 8.34, co φ(F ) satisfies (P) as well. We have to check that co φ(F ) is extremal. Let a maximal measure Λ ∈ M1 (S(H)) and s ∈ co φ(F ) with r(Λ) = s be given. By Proposition 4.28(d), Λ = φ(λ) for some H-maximal measure λ on K. By Proposition 4.28(b), there exists ν ∈ M1 (F ) such that π(ν) = s. Let µ be an H-maximal measure with ν ≺ µ. Since F is a Choquet set, cF is an upper semicontinuous Hconvex function and thus (see Proposition 3.56) 1 = ν(F ) = ν(cF ) ≤ µ(cF ) = µ(F ). Thus µ is carried by F likewise. Then π(µ) = π(ν) = s = π(λ), in other words, µ − λ ∈ H⊥ . Hence condition (P) implies 1 = µ(F ) = λ(F ), and thus Λ is carried by φ(F ), and therefore also by co φ(F ). Let now Λ ∈ M1 (S(H)) with the barycenter in co φ(F ) be given arbitrarily. We find an Ac (S(H))-maximal measure Ω with Λ ≺ Ω. Then Ω is carried by co φ(F ) by the considerations above. By Proposition 8.24, spt Λ ⊂ co spt Ω ⊂ co φ(F ), which gives the desired conclusion that co φ(F ) is extremal and consequently a face. Hence, it is a parallel face by Example 8.37.
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8 Choquet-like sets
Theorem 8.39. Let F be a closed H-convex subset of K. Then the following assertions are equivalent: (i) F is a P -set, (ii) µ(c∗F ) = ν(c∗F ) for any probability measures µ and ν with µ − ν ∈ H⊥ , (iii) the set {h ∈ H : h > cF } is down-directed, (iv) the function c∗F belongs to H⊥⊥ , (v) there exists a parallel face H ∈ S(H) such that F = φ−1 (H ∩ φ(K)). Proof. The equivalence (i) ⇐⇒ (v) follows from Proposition 8.38 and Proposition 8.22. Assume now that F is a P -set and choose µ, ν ∈ M1 (K) such that µ − ν ∈ H⊥ . Lemma 3.21 provides a measure µ0 ∈ M1 (K) so that µ ≺ µ0 and Qµ (cF ) = µ0 (cF ). Let µm be a maximal measure satisfying µ0 ≺ µm . Since F is a closed Choquet set, cF is upper semicontinuous and H-convex, and thus µ0 (cF ) ≤ µm (cF ). Let ν m be a maximal measure with ν ≺ ν m . Then µm − ν m ∈ H⊥ . Then, using Lemma 3.18, Proposition 3.56, Theorem 3.58, and condition (P) we get µ(c∗F ) = Qµ (cF ) = µ0 (cF ) ≤ µm (cF ) = µm (F ) = ν m (F ) = ν m (cF ) = ν m (c∗F ) ≤ ν(c∗F ). By interchanging µ with ν we get that µ(c∗F ) = ν(c∗F ). Hence, (i) =⇒ (ii). Suppose next that (ii) holds and that h1 , h2 ∈ H, h1 > cF , h2 > cF . Then ∗ cF < h1 ∧ h2 . Let µ, ν ∈ M1 (K) with µ − ν ∈ H⊥ be given. Since µ(c∗F ) = ν(c∗F ) < ν(h1 ∧ h2 ), the assumptions of Lemma 5.43 are satisfied. Hence there exists h ∈ H such that cF ≤ c∗F < h < h1 ∧ h2 . Statement (ii) therefore implies (iii). Further, suppose that (iii) holds. Pick µ ∈ H⊥ , µ = µ1 − µ2 where µ1 and µ2 are positive. Using Exercise 3.96 and Theorem A.84 we get µ1 (c∗F ) = µ1 (inf {h ∈ H : h ≥ cF }) = µ1 (inf {h ∈ H : h > cF }) = inf {µ1 (h) : h ∈ H, h > cF } = inf {µ2 (h) : h ∈ H, h > cF } = · · · = µ2 (c∗F ). Thus µ(c∗F ) = 0 as needed. Finally, let (iv) hold. For a boundary measure µ ∈ H⊥ we get from Theorem 3.58 µ(cF ) = µ(c∗F ) = 0. Thus, µ(F ) = 0, which proves (iv) =⇒ (i).
8.3 Choquet sets, M -sets and P -sets
255
Definition 8.40 (M-sets). A set F ⊂ K is said to be an M -set if F is a closed Choquet set satisfying condition (M). Example 8.41 (Convex case). Let X be a compact convex subset of a locally convex space. Then it follows from Theorem 8.5 that a closed face F ⊂ X is an M -set if and only if F is a split face of X. Proposition 8.42 (State space). The following assertions hold: (a) If H is a closed split face of S(H), then the set F := φ−1 (H ∩φ(K)) is an M -set. (b) If F is an M -set, then co φ(F ) is a closed split face of S(H). Proof. To show (a) we first notice that F is a Choquet set by Example 8.29. If µ ∈ Mbnd (H) ∩ H⊥ , then φ] µ ∈ Mbnd (S(H)) ∩ (Ac (S(H)))⊥ . Thus (φ] µ)|H belongs to (Ac (S(H)))⊥ , which yields µ|F ∈ H⊥ . To verify (b), we use Proposition 8.38(b) to get that co φ(F ) is a (parallel) face. By Example 8.35, φ(F ) possesses condition (M), and thus co φ(F ) has condition (M) by Proposition 8.34. Hence co φ(F ) is a split face by Theorem 8.5. Lemma 8.43. Let F ⊂ K be an M -set, f an upper semicontinuous H-convex function on K, g ∈ H and f0 , g0 ∈ H|F . If f
and
f |F ≤ f0 < g0 ≤ g|F ,
then there is h ∈ H such that f
and
f0 < h|F < g0 .
Proof. We define the following functions ( f0 on F, fe := and f on K \ F,
( g0 ge := g
on F, on K \ F.
The functions fe and −e g are upper semicontinuous and H-convex. Let µ and ν be probability measures on K, µ − ν ∈ H⊥ . There are maximal measures µm and ν m such that µ ≺ µm and ν ≺ ν m . Then µm − ν m ∈ H⊥ . Let G be the complementary set of F . Since, by Lemma 8.13, any maximal measure on K is carried by F ∪ G, we have µm = µm |F + µm |G
and
ν m = ν m |F + ν m |G .
By condition (M), µm |F − ν m |F ∈ H⊥ .
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8 Choquet-like sets
It follows that µm |G − ν m |G ∈ H⊥ . Then µ(fe) ≤ µm (fe) = µm |F (f0 ) + µm |G (f ) = ν m |F (f0 ) + ν m |G (f ) < ν m |F (g0 ) + ν m |G (g) = ν m (e g ) ≤ ν(e g ). Using Lemma 5.43 we can find a function h ∈ H such that fe < h < ge. This function possesses all the required properties. Theorem 8.44. Let F be a closed H-convex subset of K. Then the following assertions are equivalent: (i) F is an M -set, (ii) µ((f cF )∗ ) = ν((f cF )∗ ) for any measures µ, ν with µ−ν ∈ H⊥ and any f ∈ H satisfying f |F ≥ 0, (iii) the set {h ∈ H : h > f cF } is down-directed for any f ∈ H satisfying f |F ≥ 0, (iv) for any f ∈ H satisfying f |F ≥ 0, we have (f cF )∗ ∈ H⊥⊥ , (v) there exists a closed split face H ∈ S(H) such that F = φ−1 (H ∩ φ(K)). Proof. As in the proof of Theorem 8.39 we notice that (i) ⇐⇒ (v) by Proposition 8.42 and Proposition 8.22. For the proof of (i) =⇒ (ii), let f ∈ H with f |F ≥ 0 be given and let µ, ν ∈ M1 (K) satisfy µ − ν ∈ H⊥ . We proceed similarly as in Theorem 8.39. Lemma 3.21 provides a measure µ0 ∈ M1 (K) so that Qµ (f cF ) = µ0 (f cF ) and µ ≺ µ0 . Let µm be a maximal measure satisfying µ0 ≺ µm . Since F is a closed Choquet set, f cF is upper semicontinuous and H-convex, and thus µ0 (f cF ) ≤ µm (f cF ). Let ν m be a maximal measure with ν ≺ ν m . Then, using Lemma 3.18, Proposition 3.56 and Theorem 3.58, we get µ((f cF )∗ ) = Qµ (f cF ) = µ0 (f cF ) ≤ µm (f cF ) = ν m (f cF ) = ν m ((f cF )∗ ) ≤ ν((f cF )∗ ). By interchanging µ with ν we get that µ((f cF )∗ ) = ν((f cF )∗ ). Hence, (i) =⇒ (ii). Assume now that f ∈ H satisfy f |F ≥ 0 and h1 , h2 ∈ H be such that f cF < h1 ∧ h2 . Again we verify that µ((f cF )∗ ) = ν((f cF )∗ ) < ν(h1 ∧ h2 ), whenever µ − ν ∈ H⊥ . An application of Lemma 5.43 yields the existence of a function h ∈ H with f cF < h < h1 ∧ h2 . The proof of (iii) =⇒ (iv) follows again from Theorem A.84 and (iv) =⇒ (i) is straightforward. Thus the proof is finished.
8.4 H-exposed sets
257
Definition 8.45 (Archimedean sets). A closed set F is called Archimedean if for each g ∈ H with g ≥ 0 on F there exists a function h ∈ H+ such that h = g on F and h ≥ g on K. Proposition 8.46. Let H be a closed function space on a compact space K. Then any M -set is Archimedean. Proof. Let F ⊂ K be an M -set, g ∈ H, g ≥ 0 on F . By Lemma 8.43, there is h1 ∈ H such that g ∨ 0 < h1 on K
g < h1 < g + 2−1 on F.
and
Again, using Lemma 8.43, there exists h2 ∈ H such that (g ∨ 0 ∨ (h1 − 2−1 )) < h2 < h1
on K
and g < h2 < g + 2−2 on F.
This can be done by setting in Lemma 8.43 f := (g ∨ 0 ∨ (h1 − 2−1 )),
g := h1 ,
f0 := g|F
and g0 := (g + ε1 )|F ,
where ε1 > 0 is so small that ε1 < 2−1 and g + ε1 < h1 on F . By induction, we find a sequence {hn } from H+ such that, for n ≥ 2, (g ∨ 0 ∨ (hn−1 − 2−n+1 )) < hn < hn−1
on K
and
g < hn < g + 2−n
on F.
Since khn − hn−1 k < 2−n , there is a positive function h ∈ H such that hn → h uniformly on K. Obviously, h = g on F and h ≥ g on K.
8.4
H-exposed sets
We recall that a point x ∈ K is said to be an H-exposed point if there exists a function f ∈ H which attains a strict minimum (or a strict maximum) at x. It is easy to see that each H-exposed point belongs to the Choquet boundary (see Proposition 3.7). Definition 8.47 (H-exposed sets). A set F ⊂ K is called H-exposed if there is a function h ∈ H such that h=0
on
F
and
h>0
on
K \ F.
A set F is relatively H-exposed if for each x ∈ K \ F there is a function hx ∈ H+ such that hx = 0 on F and hx (x) > 0. Proposition 8.48. A set F is relatively H-exposed if and only if F is the intersection of H-exposed sets. Furthermore, any relatively H-exposed set F is a closed Choquet set.
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8 Choquet-like sets
Proof. Assume that F is relatively H-exposed. Then \ F = {t ∈ K : hx (t) = 0} , x∈K\F
where, for any x ∈ K \ F , hx is a function from H+ such that hx (x) > 0 and hx = 0 on F . Conversely, any intersection of H-exposed sets is relatively H-exposed. Since any H-exposed set is clearly closed, a relatively H-exposed set F (being an intersection of H-exposed sets) is itself closed. Let µ ∈ M1 (F ) be a measure representing x ∈ K. If x ∈ / F , then there exists a function hx ∈ H+ with hx (x) > 0 and hx = 0 on F . Then 0 = µ(hx ) = hx (x) > 0, which is a contradiction. Thus F is H-convex. Let x ∈ F and µ ∈ Mx (H) be given. If µ(K \F ) > 0, we find a point y ∈ spt µ\F and hy ∈ H+ such that hy = 0 on F and hy (y) > 0. Then the inequalities Z Z hy (t) dµ(t) > 0 0 = hy (x) = hy (t) dµ(t) ≥ spt µ
spt µ∩{z∈K:hy (z)>0}
lead to a contradiction. Thus F is H-extremal. Proposition 8.49. Any H-convex Archimedean set is relatively H-exposed. Proof. If F ⊂ K is an H-convex Archimedean set, pick x ∈ K \ F . The function f := cF − 1 is upper semicontinuous. By Lemma 3.21, there exists a Radon measure ν ∈ Mx (H) so that f ∗ (x) = ν(f ). Since F is H-convex and x ∈ / F , we see that ν is not carried by F . Hence ν(f ) < 0, and therefore f ∗ (x) < 0. The definition of f ∗ yields a function g ∈ H such that g ≥ 0 on F and g(x) < 0. Since F is supposed to be Archimedean, there is h ∈ H+ such that h = g on F and h ≥ g on K. If u := h − g, then u ∈ H is positive, u = 0 on F and u(x) > 0. Corollary 8.50. Let H be a closed function space on a compact space K. Then any M -set is relatively H-exposed. Proof. This follows immediately from Propositions 8.46 and 8.49. Summary. Let H be a closed function space. If we summarize results on various Choquet like sets, we get the following sequence of implications for a closed Hconvex set F : F is an M -set =⇒ F is Archimedean =⇒ F is relatively H-exposed =⇒ F is a Choquet set .
8.5 Weak topology on boundary measures
8.5
259
Weak topology on boundary measures
Definition 8.51 (Ordered set of measures). Given a function space H on a compact space K, let M := {(µ, ν) ∈ M+ (K) × M+ (K) : kµk ≤ 1, kνk ≤ 1, µ − ν ∈ H⊥ }. We define a partial ordering ≺ on M: (µ1 , ν1 ) ≺ (µ2 , ν2 ) if µ1 ≺H µ2 and ν1 ≺H ν2 . Then M is an ordered compact convex set (see Definition 7.41). (The cone E + ⊂ M(K) × M(K) required by Definition 7.41 consists of all pairs (µ, ν) ∈ M(K) × M(K) satisfying µ(f ) ≥ 0 and ν(f ) ≥ 0 whenever f ∈ Kc (H).) It can be easily observed that (µ, ν) is maximal in M if and only if both µ and ν are H-maximal measures. If Mmax denotes the set of all ≺-maximal elements of M, then Mmax is a convex and extremal subset of M, that is, Mmax is a face of M. By Proposition 7.47, the set ext Mmax is nonempty. Lemma 8.52. If (µ, ν) ∈ ext Mmax , then either µ = ν or kµ − νk = 2. Proof. Suppose that (µ, ν) ∈ ext Mmax is a pair of nontrivial measures. First we realize that µ, ν ∈ M1 (K). Indeed, since (µ, ν) ∈ M, then µ(K) = ν(K) ≤ 1. Suppose that 0 < µ(K) = ν(K) < 1. Then µ ν (µ, ν) = µ(K) , + (1 − µ(K))(0, 0), µ(K) µ(K) which contradicts the extremality of (µ, ν). Assume that µ 6= ν and kµ − νk < 2. Let µ − ν = η + − η − be the decomposition of µ − ν into the positive and negative part. Then kη + k = kη − k and 2 > kµ − νk = kη + k + kη − k = 2kη + k. Set λ := ν − η − = µ − η + . Since 1 = kη + k + kλk and kη + k < 1, λ is a nontrivial measure. It follows from the equality + η η− λ λ + (µ, ν) = kη k , + kλk , , kη + k kη + k kλk kλk that the pair (µ, ν) cannot be an extreme point of Mmax . Notation 8.53. We denote ⊥ H⊥ 2 := H ∩ {µ ∈ M(K) : kµk ≤ 2} ⊥ and notice that H⊥ 2 ∩ Mbnd (H) 6= {0} if and only if H ∩ Mbnd (H) 6= {0}.
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8 Choquet-like sets
Lemma 8.54. If H⊥ 2 ∩ Mbnd (H) 6= {0}, then ext H⊥ ∩ M (H) = {µ − ν : (µ, ν) ∈ ext Mmax , kµ − νk = 2}. bnd 2 + − Proof. Let η = η + − η − ∈ ext(H⊥ 2 ∩ Mbnd (H)) be given (η and η is the positive and negative part of η, respectively). Then kη + k = kη − k and
2 = kηk = kη + k + kη − k. It easily follows that (η + , η − ) ∈ ext Mmax . Indeed, if (η + , η − ) = α(µ1 , µ2 ) + (1 − α)(ν1 , ν2 ) for some α ∈ (0, 1) and (µ1 , µ2 ), (ν1 , ν2 ) ∈ Mmax , then η = α(µ1 − µ2 ) + (1 − α)(ν1 − ν2 ). Thus µ1 − µ2 = ν1 − ν2 = ν. Since µ1 , ν1 ≤ ν + , µ2 , ν2 ≤ ν − and all these measures are probability, µ1 = ν + = ν1 and µ2 = ν − = ν2 . Thus (η + , η − ) ∈ ext Mmax . Conversely, let (µ, ν) ∈ ext Mmax with kµ − νk = 2 be given. Then η := µ − ν ∈ H⊥ . Since η + ≤ µ, η − ≤ ν and all these measures are probability, we have η + = µ, η − = ν. If η = αη1 + (1 − α)η2 for some α ∈ (0, 1) and η1 , η2 ∈ H⊥ 2 ∩ Mbnd (H), then 2 = kη1 k = kη2 k = kηk and kη1+ k = kη1− k = kη2+ k = kη2− k = 1. Since
µ − ν = η = α(η1+ − η1− ) + (1 − α)(η2+ − η2− ) = αη1+ + (1 − α)η2+ − (αη1− + (1 − α)η2− ),
we get µ ≤ αη1+ + (1 − α)η2+
and η ≤ αη1− + (1 − α)η2− .
Since all these measures are probability, (µ, ν) = α(η1+ , η1− ) + (1 − α)(η2+ , η2− ). Since (µ, ν) ∈ ext Mmax , we get η1+ = η2+ and η1− = η2− . Thus η1 = η2 , and hence µ − ν ∈ ext(H⊥ 2 ∩ Mbnd (H)). This finishes the proof. ⊥ Lemma 8.55. If H⊥ 2 ∩ Mbnd (H) 6= {0}, then ext(H2 ∩ Mbnd (H)) 6= ∅.
8.5 Weak topology on boundary measures
261
Proof. By Lemma 8.52 and Lemma 8.54 it suffices to show that the set A := {(µ, ν) ∈ ext Mmax : µ 6= ν} is nonempty. Let D be the linear span of functionals on M(K) × M(K) of the form (µ, ν) 7→ µ(f ) + ν(g),
f, g ∈ C(K),
and (µ, ν) 7→ µ(f ∗ ) + ν(g ∗ ),
f, g ∈ C(K).
Consider the locally convex topology σ on M(K) × M(K) generated by functionals from D. It follows from Theorem 7.58 that Mmax = coσ ext Mmax . Now the proof of the lemma follows easily. If the set A were empty, then every pair (µ, ν) ∈ Mmax would satisfy µ = ν. But our hypothesis assures that there is + a nonzero measure µ ∈ H⊥ 2 ∩ Mbnd (H). Then its positive and negative parts µ , − + − + − µ satisfy (µ , µ ) ∈ Mmax and µ 6= µ , a contradiction. This concludes the proof. Definition 8.56 (Weak topology generated by upper envelopes). Let D be the linear span of the family C(K)∪{f ∗ : f ∈ C(K)} in the space of all bounded Borel functions on K. We denote by τ the (locally convex) topology on M(K) generated by the functionals µ 7→ µ(f ), f ∈ D. Theorem 8.57. Let H be a function space on K. Then τ ⊥ co ext H⊥ ∩ M (H) = H ∩ M (H) . bnd bnd 2 2 Proof. By Theorem 3.68 ⊥ H⊥ 2 ∩ Mbnd (H) = H ∩
\
{µ ∈ M(K) : kµk ≤ 2, µ(f ∗ − f ) = 0} ,
f ∈C(K)
and thus the set H⊥ 2 ∩ Mbnd (H) is τ -closed. ⊥ Suppose that H⊥ 2 ∩ Mbnd (H) 6= {0}. Then ext(H2 ∩ Mbnd (H)) 6= ∅ by Lemma 8.55. Assume that there exists τ ⊥ η ∈ H⊥ 2 ∩ Mbnd (H) \ co (ext(H2 ∩ Mbnd (H)).
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8 Choquet-like sets
Then there exists f ∈ D so that n o η(f ) > s := sup µ(f ) : µ ∈ ext(H⊥ 2 ∩ Mbnd (H)) = sup {(µ − ν)(f ) : (µ, ν) ∈ ext Mmax , µ 6= ν} , where the latter equality follows by Lemma 8.54 and Lemma 8.52. Since (µ, ν) ∈ ext Mmax if and only if (ν, µ) ∈ ext Mmax , it follows that s ≥ 0. By the definition of τ there are continuous functions f0 , . . . , fn on K such that f = f0 +
n X
fi∗ .
i=1
Pn
Set g := f0 + i=1 fi and let η + and η − be the positive and the negative part of η. Since η is a boundary measure, we obtain s < η(f ) = η(g) = η + (g) + η − (−g) = η + (g∗ ) + η − ((−g)∗ ). An appeal to Lemma 3.18(b) provides a continuous H-convex functions k1 and k2 such that k1 ≤ g, k2 ≤ −g and η + (k1 ) + η − (k2 ) > s. (8.3) If M denotes the set from Definition 8.51, let F := {(µ, ν) ∈ M : µ(k1 ) + ν(k2 ) = sup{λ1 (k1 ) + λ2 (k2 ) : (λ1 , λ2 ) ∈ M}} . Then F is a nonempty closed face in M which is hereditary upwards, that is, if (µ, ν) ∈ F , (b µ, νb) ∈ M and (µ, ν) ≺ (b µ, νb), then (b µ, νb) ∈ F . By Proposition 7.47, there exists (µ, ν) ∈ ext Mmax ∩ F . If µ = ν, then 0 ≤ s < µ(k1 ) + µ(k2 ) ≤ µ(g) + µ(−g) = 0, which is impossible. Thus µ 6= ν and µ − ν ∈ ext(H⊥ 2 ∩ Mbnd (H)) by Lemma 8.54. By (8.3) we get µ(k1 ) ≤ µ(g) and ν(k2 ) ≤ ν(−g). Then s ≥ µ(f ) − ν(f ) = µ(g) − ν(g) ≥ µ(k1 ) + ν(k2 ) ≥ η + (k1 ) + η − (k2 ) > s yields a contradiction, and the proof is complete.
8.6
Characterizations of simpliciality by Choquet sets
Lemma 8.58. Let µ be a boundary measure on K and F be a P -set. Then Z |µ(F )| ≤ sup h dµ : h ∈ H and khk ≤ 2 . K
8.6 Characterizations of simpliciality by Choquet sets
263
Proof. Let F be a P -set. Then c∗F = inf{h ∈ H : cF < h < 2}. Indeed, c∗F = inf{h ∈ H : cF ≤ h} = inf{h ∈ H : cF < h}. The latter set is down-directed by Theorem 8.39. Thus for any h ∈ H with cF < h we can find g ∈ H with cF < g < h ∧ 2. Pick ε > 0. An appeal to Theorem A.84 provides a function h ∈ H so that cF < h < 2 and |µ|(c∗F ) − |µ|(h) < ε. Then |µ(F )| = |µ(cF )| = |µ(c∗F )| ≤ |µ(c∗F − h)| + |µ(h)| ≤ |µ|(c∗F − h) + |µ(h)| ≤ |µ(h)| + ε Z ≤ sup h dµ : h ∈ H and khk ≤ 2 + ε. K
As ε is arbitrary, we get the assertion. Lemma 8.59. Let H be a closed function space on a compact space K. Assume that any H-exposed set satisfies condition (P). If µ ∈ Mbnd (H) satisfies µ(F ) = 0 for any H-exposed set F , then µ ∈ H⊥ . Proof. For a contradiction, we suppose µ(h) = 6 0 for some h ∈ H and define a continuous functional on H as Z L(h) := h dµ, h ∈ H. K
Since L 6= 0, we may assume that kLk = 1. By the Bishop–Phelps theorem A.8, there are sequences {Ln } in H∗ and {hn } in H so that kLn − Lk → 0, kLn k = Ln (hn ) and khn k = 1 for each n ∈ N. Set Fn+ := {x ∈ K : hn (x) = 1}
and
Fn− := {x ∈ K : hn (x) = −1}.
Let {µn } be a sequence given by Lemma 6.23, that is, for any n ∈ N we have kµn k = kLn k and µn (h) = Ln (h) for any h ∈ H. Fix now n ∈ N. Since khn k = 1 and kµn k = kLn k = Ln (hn ) = µn (hn ), we obtain + − spt µ+ and spt µ− n ⊂ Fn n ⊂ Fn . As each set Fn+ is H-exposed, it satisfies condition (P), and so by Lemma 8.58 we have Z + + |µ(Fn ) − µn (Fn )| ≤ sup h d(µ − µn ) {h∈H:khk≤2}
=
sup {h∈H:khk≤2}
K
|(L − Ln )(h)| = 2 kL − Ln k.
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8 Choquet-like sets
Thus |µ(Fn+ ) − µn (Fn+ )| → 0. Since Fn+ is an H-exposed set, µ(Fn+ ) = 0 due to the + − + assumption. Recall that spt µ+ n ⊂ Fn and spt µn ∩ Fn = ∅. Hence, + + + + − + + kµ+ n k = µn (Fn ) = µn (Fn ) − µn (Fn ) = µn (Fn )
= µ(Fn+ ) + µn (Fn+ ) − µ(Fn+ ) ≤ |µn (Fn+ ) − µ(Fn+ )|, from which it follows that kµ+ n k → 0. Similarly we obtain kµ− n k → 0. Therefore − 1 = kLk = lim kLn k = lim kµn k = lim kµ+ n k + kµn k = 0, n→∞
n→∞
n→∞
which is a contradiction. Thus µ ∈ H⊥ as required. Theorem 8.60. Let H be a closed function space on a compact space K. Then the following assertions are equivalent: (i) H⊥ ∩ Mbnd (H) = {0}, (ii) any closed Choquet subset of K is an M -set, (iii) any closed Choquet subset of K is a P -set, (iv) any H-exposed subset of K is a P -set. Proof. The implications (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) are obvious. It remains to prove that (iv) =⇒ (i). By Theorem 8.57 it is enough to show that there are no nonzero measures in ⊥ ext(H⊥ 2 ∩ Mbnd (H)). Suppose that µ is a nonzero element of ext(H2 ∩ Mbnd (H)). Let µ = µ+ − µ− be the decomposition of µ into the positive and negative part. Since spt µ+ cannot be a singleton, we can find a Borel subset B of K so that µ+ (K) 6= µ+ (B) 6= 0. Set λ1 := µ+ |B −
µ+ (B) − µ µ− (K)
and
λ2 := µ − λ1 .
Then kµk = kλ1 k + kλ2 k and both λ1 , λ2 are nontrivial boundary measures. If λ1 ∈ H⊥ , then both λ1 and λ2 are contained in H⊥ 2 ∩ Mbnd (H) and µ is a nontrivial convex combination of suitable multiples of λ1 and λ2 . Since this contradicts the extremality of µ, we have λ1 ∈ / H⊥ . Set Z L(h) := h dλ1 , h ∈ H. K
Since L is a nonzero functional on H, an appeal to the Bishop–Phelps theorem provides a sequence {Ln } of nonzero functionals in H∗ and functions hn ∈ H such that kLn − Lk → 0, kLk = kLn k = Ln (hn ) and khn k = 1.
8.6 Characterizations of simpliciality by Choquet sets
265
Put Fn+ := {x ∈ K : hn (x) = 1}
and
Fn− := {x ∈ K : hn (x) = −1}.
Let {µn } be a sequence of boundary measures corresponding to Ln as in Lemma 6.23, − in particular kµn k = kLn k for each n ∈ N. Let µ+ n and µn be the positive and negative part of µn . As in the proof of Lemma 8.59 we get + spt µ+ n ⊂ Fn
and
− spt µ− n ⊂ Fn .
Claim 1. There exists n0 ∈ N so that spt µ * Fn+ for any n ≥ n0 . Proof of Claim 1. Suppose the contrary. Hence there exists an increasing sequence {nk } in N so that spt µ ⊂ Fn+k . As in the proof of Lemma 8.59 we obtain that − − − − |µ− nk (Fnk ) − λ1 (Fnk )| = |µnk (Fnk ) − λ1 (Fnk )| → 0.
Since we assume that spt µ ⊂ Fn+k , we have spt λ1 ⊂ Fn+k as well. Thus we get lim kµ− nk k = 0.
k→∞
With this fact in hand, from the equalities − 0 = L(1) = lim Lnk (1) = lim µ+ nk (1) − µnk (1) k→∞
= lim
k→∞
µ+ nk (1)
k→∞
= lim kµnk k = lim kLnk k = kLk k→∞
k→∞
we obtain an obvious contradiction. Thus the proof of Claim 1 is finished. Similarly we deduce that spt µ * Fn− for all but finitely many n ∈ N. Claim 2. For all but finitely many n ∈ N, we have µ|Fn+ ∪Fn− = 0. Proof of Claim 2. Let n0 be an integer provided by Claim 1, n ≥ n0 and let F be an H-exposed set. Since F ∩ Fn+ is also H-exposed, by using the assumption we get µ|Fn+ (F ) = µ(F ∩ Fn+ ) = 0. By Lemma 8.59, we have µ|Fn+ ∈ H⊥ . Suppose that µ|Fn+ 6= 0. By Claim 1, µ is a nontrivial convex combination of suitable multiples of measures µ|Fn+ and µ|K\Fn+ , which contradicts the extremality of µ. Similarly we obtain that µ|Fn− = 0. Hence µ|Fn+ ∪Fn− = 0 and the Claim 2 is proved.
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8 Choquet-like sets
Thus it follows that λ1 |Fn+ ∪Fn− = 0. According to Lemma 8.58, we get |µn (Fn+ )
−
λ1 (Fn+ )|
n Z o ≤ sup h d(µn − λ1 ) : h ∈ H, khk ≤ 2 K
≤ 2kL − Ln k. As in the proof of Lemma 8.59 we obtain + + + − + + µ+ n (Fn ) = µn (Fn ) − µn (Fn ) = µn (Fn )
≤ λ1 (Fn+ ) + |µn (Fn+ ) − λ1 (Fn+ )|. Since λ1 (Fn+ ) = 0, we get + + lim kµ+ n k = lim µn (Fn ) = 0.
n→∞
n→∞
Similarly it can be deduced that lim kµ− n k = 0. Putting this together we have − kLk = lim kLn k = lim kµn k = lim kµ+ n k + kµn k = 0, n→∞
n→∞
n→∞
which contradicts the hypothesis L 6= 0 and finishes the proof. Remark 8.61. Before proceeding, we have a need of a short note. Given a function space H on K, the space Ac (H) of all continuous H-affine functions on K can be considerable larger than H. Although a lot of objects determined by H and by Ac (H) coincide (for example, representing measures, boundary measures, Choquet sets), this coincidence is no longer valid for M -sets, P -sets and exposed sets. In order to obtain next Theorem 8.62 in a full generality, we must specify whether we think over M sets and P -sets with respect to H or with respect to Ac (H). In order to avoid an inaccuracy, we point out that conditions (iii)–(v) of the next theorem are laid down for the function space Ac (H). More precisely, for example, condition (iv) means that any closed Choquet set F ⊂ K satisfies µ(F ) = 0 for any measure µ ∈ (Ac (H))⊥ ∩ Mbnd (H). With these preliminaries out of the way, we may state the main result of this chapter. Theorem 8.62. Let H be a function space on a compact space K. Then the following assertions are equivalent: (i) H is simplicial, (ii) (Ac (H))⊥ ∩ Mbnd (H) = {0} , (iii) any closed Choquet subset of K is an M -set with respect to Ac (H), (iv) any closed Choquet subset of K is a P -set with respect to Ac (H), (v) any Ac (H)-exposed subset of K is a P -set with respect to Ac (H).
8.6 Characterizations of simpliciality by Choquet sets
267
Proof. We already know from Theorem 8.60 that assertions (ii)–(v) are equivalent (note that the space Ac (H) is closed). Since (i) ⇐⇒ (ii) by Proposition 6.9, the proof is finished. Corollary 8.63. For a compact convex set X the following assertions are equivalent: (i) X is a simplex, (ii) any closed face of X is a split face, (iii) any closed face of X is a parallel face, (iv) any exposed subset of X is a parallel face. Proof. This follows from Theorem 8.62, if we use the characterization of split and parallel faces from Example 8.41 and Example 8.37. Corollary 8.64. Let H be a simplicial function space. Then closed Choquet sets coincide with relatively Ac (H)-exposed sets. Proof. If F is a closed Choquet set, then by Theorem 8.62, F is an M -set with respect to Ac (H). Corollary 8.50 asserts that F is relatively Ac (H)-exposed. The converse implication follows by Proposition 8.48 and the observation that a set F is a Choquet set with respect to H if and only if F is a Choquet set with respect to Ac (H) (in virtue of the equality Mx (H) = Mx (Ac (H)) for any x ∈ K). Corollary 8.65. Let H be a Markov simplicial function space on a compact space K. Then the class of relatively H-exposed sets coincides with the class of closed Choquet sets. Proof. Since we have H = Ac (H) by Theorem 6.42, the assertion follows from Corollary 8.64. There are simplicial function spaces where supports of maximal measures are big enough. In this case we are able to give a better characterization of both closed Choquet sets and M -sets. Proposition 8.66. Let H be a simplicial function space so that spt δx = ChH (K) for any x ∈ K \ ChH (K). Then a closed set F ⊂ K is Choquet if and only if either F = K or F is a proper subset of ChH (K). Proof. Assume first that F is a Choquet set which is not a proper subset of ChH (K). We show that ChH (K) ⊂ F . If z ∈ F \ ChH (K), then spt δz = ChH (K) by the assumption. Since F is an H-extremal set, we have spt δz ⊂ F . Hence ChH (K) ⊂ F . By Corollary 8.19, K = coH (ChH (K)) ⊂ coH F = F. Conversely, assume that F is a closed proper subset of ChH (K). By Example 8.9, F is H-extremal. We now wish to show that F is H-convex. To this end, fix z ∈ K
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8 Choquet-like sets
and µ ∈ Mz (H) satisfying spt µ ⊂ F . Our aim is to show that z ∈ F . Since spt µ ⊂ F ⊂ ChH (K), we see that µ is a maximal measure. Because H is a simplicial space, µ = δz . Since F is a proper subset of ChH (K), z ∈ ChH (K) according to the assumption. Hence µ = εz , and therefore z ∈ F . Proposition 8.67. Let H be a simplicial space such that H = Ac (H) and spt δx = ChH (K) for any x ∈ K \ ChH (K). Then a closed set F ⊂ K is an M -set if and only if either F is a proper subset of ChH (K) or equals K. Proof. If F is an M -set, then either F equals K or F is a proper subset of ChH (K) by Proposition 8.66. Conversely, if a closed set F is a proper subset of ChH (K), then F is a Choquet set by Proposition 8.66. By Theorem 8.62, F is an M -set with respect to Ac (H) = H. This finishes the proof.
8.7
Exercises
Exercise 8.68. Let F ⊂ K be a closed H-extremal set and µ a Radon measure carried by F . If ν is a Radon measure, µ ≺ ν, then ν is carried by F as well. Hint. We may assume that µ is a probability measure on K. Since cF ∈ Kusc (H), an appeal to Proposition 3.56 reveals that 1 ≥ ν(F ) = ν(cF ) ≥ µ(cF ) = 1. Hence ν is carried by F . Exercise 8.69. Let F ⊂ K and let µ be a Radon measure carried by the complementary set F 0 . If ν is a Radon measure such that µ ≺ ν, then ν is carried by F 0 . Hint. Since the function c∗F is upper semicontinuous and H-concave, we have by Proposition 3.56 0 ≤ ν(c∗F ) ≤ µ(c∗F ) = 0. It follows that ν(c∗F ) = 0, which yields the conclusion that ν is carried by F 0 . ∗ Exercise 8.70. If F ⊂ K, then c∗F = cco HF .
Hint. Obviously, c∗F ≤ c∗coH F . If h ∈ H, h ≥ cF , then h ≥ ccoH F according to ∗ Proposition 8.18. Hence, c∗F ≥ cco HF . Exercise 8.71. Let F be a P -set. Then c∗F ∈ H if and only if the complementary set F 0 is closed.
8.7 Exercises
269
Hint. Clearly, if c∗F ∈ H, then the complementary set F 0 is closed since c∗F is continuous and F 0 = {x ∈ K : c∗F (x) = 0}. Conversely, suppose that F 0 is closed. Choose ε > 0. By (i) =⇒ (iii) of Theorem 8.39 and Dini’s theorem, there exists a function h ∈ H so that h > cF and h<1+ε
on F,
h < ε on F 0 .
Now, for any x ∈ K, find a maximal measure µx ∈ Mx (H). Then µx is carried by F ∪ F 0 by Lemma 8.13 and we have 0 < h(x) − c∗F (x) ≤ µx (h) − µx (cF ) Z (h(t) − cF (t)) dµx (t) < ε. = F ∪F 0
Therefore, kh − c∗F k < ε, proving the reverse implication. Exercise 8.72. Find an example of a compact set F ⊂ ext X such that co F is not a face. Hint. Consider the unit square in R2 . Exercise 8.73. Let H be a closed function space. If F is a P -set for which its complementary set F 0 is closed, then F is Archimedean. Hint. Let g ∈ H, g ≥ 0 on F . Consider the function u := 1 − c∗F . Then u ∈ H by Exercise 8.71, 0 ≤ u ≤ 1, and u=1
on F 0
and
u=0
on F.
There is α > 0 so that αu ≥ − min g(K) on F 0 . Set h := g + αu. Then h ∈ H, h ≥ g on K and h = g on F . Further, u ≥ 0 on F ∪ F 0 . Given x ∈ K, let µ ∈ Mx (H) be maximal. By Lemma 8.13, Z Z u(x) = u dµ = u dµ ≥ 0. K
F ∪F 0
Hence u ∈ H+ , which proves that the set F is Archimedean. Exercise 8.74. Clearly, any H-exposed set F is a Gδ subset of K. Prove the following converse assertions. (a) Let H be a closed function space. Then any relatively H-exposed Gδ set F is H-exposed. (b) Relatively H-exposed sets are H-exposed provided the compact space K is metrizable.
270
8 Choquet-like sets
Hint. Since F is relatively H-exposed, for any x ∈ K \ F we can find a function hx ∈ H+ so that hx = 0 on F and hx (x) > 0. Since the complement K \ F is a Lindel¨of space, we may find a countable set {xn : n ∈ N} ⊂ K \ F such that K \F ⊂
∞ [
{z ∈ K : hxn (z) > 0}.
n=1
Then the function h :=
∞ X n=1
1 hx 2n khxn k n
exposes the set F . The second part follows from the fact that closed subsets of metrizable spaces are Gδ . Exercise 8.75. Let H be a simplicial function space on a compact space K. Prove that any Gδ point x ∈ ChH (K) is Ac (H)-exposed (cf. also Exercise 8.74). Hint. You can imitate the proof of Lemma 6.61 for the case of function spaces. A different reasoning could be as follows: By Proposition 8.31, {x} is a (closed) Choquet set. By Theorem 8.62, {x} is an M -set with respect to Ac (H). By Corollary 8.50, {x} is relatively Ac (H)-relatively exposed. Finally, use Exercise 8.74. Exercise 8.76. Let X be a compact convex set, E := Ac (X) and φ : X → S(Ac (X)) be the evaluation mapping. Prove that φ(X) is a parallel face of BE ∗ . Hint. It is easy to show that φ(X) and −φ(X) are closed faces. By Proposition 4.31(c), BE ∗ = co(φ(X) ∪ −φ(X)). Hence, φ(X) and −φ(X) are complementary faces. They are parallel by Proposition 4.31(b). Exercise 8.77 (Weak peak points). If X is a compact convex set, x ∈ X is called a weak peak point, if for any ε > 0 and any neighborhood U of x there exists h ∈ Ac (X) such that khk ≤ 1, |h(x)| > 1 − ε and |h(y)| < ε for each y ∈ ext X \ U . (a) Prove that any weak peak point of X is an extreme point. (b) Prove that any weak peak point of X is a split face. (c) If ext X is a closed set and {x} is a parallel face, x is a weak peak point. Hint. To show (a), let x be a weak peak point and µ ∈ Mx (X) be a maximal measure. If U 3 x is an arbitrary closed neighborhood of x and ε > 0, let h be a function guaranteed by the assumption. Since µ is carried by ext X and |h| ≤ ε on ext X \ U , we get Z Z 1 − ε < |µ(h)| ≤
|h| dµ + U
|h| dµ ≤ µ(U ) + ε. ext X\U
Thus µ is carried by {x}, in other words, x is an extreme point.
8.7 Exercises
271
To verify (b), we use Theorem 8.5. Let µ ∈ (Ac (X))⊥ be a boundary measure of norm 1 and assume that µ = cεx + ν, where c 6= 0 and |ν|({x}) = 0. We select ε ∈ (0, 1) such that |c|(1 − ε) − 2ε > 0 and a closed neighborhood U of x with |ν|(U ) < ε. Let h ∈ Ac (X) be chosen for U and ε. Then |h| ≤ ε on ext X \ U , and hence Z Z Z |h| d|ν| + |h| d|ν| = |h| d|ν| ≤ ε + ε. X
U
ext X\U
Thus 0 = |µ(h)| ≥ |cεx |(h) − |ν|(|h|) ≥ |c|(1 − ε) − 2ε > 0, a contradiction. For the proof of (c), let {x} be a parallel face, ext X closed and U 3 x along with ε > 0 be given. We choose η > 0 with (1 + η)−1 > 1 − ε. By Theorem 8.39, (c{x} )∗ is an affine function, the family {h ∈ Ac (X) : h > c{x} } is down-directed and its infimum equals (c{x} )∗ . Thus also the family {h ∈ Ac (X) : 1 + η > h > c{x} } is down-directed and its infimum equals (c{x} )∗ . Since (c{x} )∗ = 0 on ext X \ {x} and ext X is a closed set, Dini’s theorem yields the existence of a function h ∈ Ac (X) such that c{x} < h < 1 + η and h < η on ext X \ U . Then (1 + η)−1 h is the sought function witnessing that x is a weak peak point. Exercise 8.78. Find an example of an extreme point of a simplex that is not a weak peak point. Hint. Consider an extreme point x of the Poulsen simplex S (see Definition 12.56). Let x0 ∈ S be arbitrary and y = 13 x + 23 x0 . If U is a closed convex neighborhood of x not containing y, then there is no h ∈ Ac (S) such that khk ≤ 1, h(x) > 1 − 71 and |h| < 17 on ext S \ U . Indeed, assume that h is such a function. Since x0 ∈ / U and ext S is dense in S, h(x0 ) ≤ 71 . Then 4 1 2 h(y) = h(x) + h(x0 ) ≥ . 3 3 21 Let {xn } be a sequence in ext S converging to y. Since all but finitely many points xn are in X \ U , h(y) ≤ 71 . Thus we get a contradiction. Exercise 8.79. Let X ⊂ Rd be a compact convex set such that {x} is a split face for each x ∈ ext X. Prove that X is a simplex.
272
8 Choquet-like sets
Hint. First we show To Punique. Pnthat convex combinations of extreme points of X are n this end, let x = i=1 ai xi , where ai arePstrictly positive numbers with i=1 ai = 1 and xi are extreme points of X. If a := ni=2 ai , x = a1 x1 + a
n X
a−1 ai xi .
i=2
P Since {x1 } is a split face and ni=2 a−1 ai xi is contained in the complementary face {x1 }0 , there is only one possibility how to decompose x as a convex combination of x1 and an element from {x1 }0 . Proceeding inductively we get that for any x ∈ X there is only one possibility how to write it as a convex combination of extreme points. Further we show that X has at most d + 1 extreme points. Assume that this is not the case and {x0 , . . . , xd+1 } are distinct extreme points of X. Then they are not affinely thus there exist nonzero real numbers a0 , . . . , ad+1 with Pd+1 independent, and Pd+1 0. We assume that a0 , . . . , aj are positive and i=0 ai = 0 such that i=0 ai xi = P aj+1 , . . . , ad+1 are negative. If a := ji=0 ai , then j X
a−1 ai xi =
i=0
d+1 X
a−1 ai xi
i=j+1
are different convex combinations of extreme points expressing the same point. Hence we have arrived to a contradiction with the first part of the proof. Combining both parts together we get that ext X is finite and any x ∈ X can be written uniquely as a convex combination of ext X. Thus X is a simplex. Exercise 8.80. Find an example of a compact convex set X such that each extreme point of X is a weak peak point and X is not a simplex. Hint. Consider the compact space K := [0, 1] ∪ [2, 3] and the function space Z n H := f ∈ C(K) : 0
1
Z f (t) dt =
3
o f (t) dt ,
2
and let X := S(H). We denote by λ1 and λ2 Lebesgue measure on [0, 1] and [2, 3], respectively. The set X is not a simplex since the point π(λ1 ) has two different measures carried by ext X = φ(ChH (K)), namely φ] λ1 and φ] λ2 (here we use the notation and results from Section 4.3). On the other hand, if x ∈ K, ε > 0 and an open set U 3 x are given, it is not difficult to construct a function f ∈ H such that f (x) = 0, 0 ≤ f ≤ 1 and f > 1 − ε on K \ U . Hence any extreme point of X is a weak peak point.
8.8 Notes and comments
8.8
273
Notes and comments
The presentation in this chapter closely follows the paper [322] by J. Lukeˇs, T. Mocek, M. Smrˇcka and J. Spurn´y; however, the method of the proof of Theorem 8.62 is analogous to the techniques used in the convex setting. The notion of a split face was introduced by E. M. Alfsen and T. B. Andersen in [6] and independently by F. Perdrizet in [370] and [371]. The measure theoretic characterization of split faces in Theorem 8.5 can be found in E. M. Alfsen and T. B. Andersen [7] (see also Theorem II.6.12 in E. M. Alfsen [5] and Theorem 2.10.9 in L. Asimow and A. J. Ellis [24]). The idea of parallel faces appears in B. Hirsberg’s paper [226]. Conditions (M) and (P) appeared under the labels (A.1) and (A.2); see, for example, a paper by E. M. Alfsen and B. Hirsberg [8]. Theorem 8.32 is contained in C. J. K. Batty [32], the characterization of P -sets is in [226], Theorem 2.12 or in Theorem 2.10.10 of [24]. Lemma 8.43 is an easier variant of Theorem II.6.15 in [5] and Theorem 2.10.5 in [24] (see also E. M. Alfsen and T. B. Andersen [6] and E.G. Effros [162]). Theorems 8.44 and Proposition 8.46 are variants of Theorem II.6.18 and Corollary II.6.16 in [5] for general function spaces. E. Størmer introduced the concept of an Archimedean face in [440] in the convex setting. In [3], E. M. Alfsen gave a characterization of Archimedean faces which suggested our definition in the framework of function spaces. Proposition 8.49 can be found as Proposition II.5.17 in [5]. We do not know the answer to the following question. Problem 8.81. Let H be a closed function space on a compact space K and F ⊂ K be a P -set. Is F Archimedean? Section 8.5 is based upon E. Briem’s paper [92]. The central Section 8.6 combines some techniques of A. J. Ellis and A. K. Roy [167] (see also Theorem 3.2.5 in [24]), E. Briem [94] and E. G. Effros [163], [164]. We also mention another characterization of simplices given by C. J. K Batty [33] which reads as follows: A compact convex set X is a simplex if and only if there exists C > 0 such that, whenever F ⊂ X is a closed extremal set, each positive f ∈ Ac (coF ) has a positive extension g ∈ Ac (X) with kgk ≤ Ckf k. Corollary 8.64 is a slightly more general version of Theorem 2.5 in R. E. Atalla [25], where it is proved that for Markov simplicial function spaces (see Definition 6.43) the class of relatively H-exposed sets coincides with the class of closed Choquet sets. Exercises 8.77 and 8.78 are taken from C. H. Chu and B. Cohen [119].
Chapter 9
Topologies on boundaries
In this chapter we further explore families of H-extremal and Choquet sets and show that they generate compact topologies on Choquet boundaries. The fact that these topologies are typically non Hausdorff turns out to be a minor inconvenience. The first section introduces the topologies σext and σmax generated by closed H-extremal and H-maximally extremal sets. Theorem 9.10 shows that these topologies are Hausdorff if and only if the Choquet boundary is closed. Theorem 9.12 provides a key tool for introducing measures on Choquet boundaries induced by maximal measures. This procedure is clarified by Theorem 9.19 which shows that the induced measures are regular in some sense. As a consequence we get that a bounded H-affine Baire function satisfies the barycentric calculus with respect to the induced measures (see Corollary 9.20). Next we characterize functions continuous with respect to the topologies σext (see Theorem 9.25) and σmax (see Theorem 9.29). Section 9.4 is devoted to a study of H-strongly universally measurable functions. The main result, Theorem 9.36, proves that upper semicontinuous affine functions on compact convex sets are measurable with respect to the completion of the induced measure from Theorem 9.19 and that they satisfy the barycentric calculus with respect to these measures. As a consequence we obtain that H-strongly universally measurable functions satisfy the barycentric calculus with respect to these measures (see Theorem 9.38). The last section describes properties of the so-called facial topology generated by M -sets. It turns out in Theorem 9.48 that facially continuous functions on Choquet boundaries can be identified with the center of H (see Definition 9.46). In the case of prime simplicial function spaces, there are no nonconstant facially continuous functions (see Corollary 9.50).
9.1
Topologies generated by extremal sets
In the sequel, H is a function S space on a compact space K 6= ∅. We recall from Section 3.3 that M(H) = x∈K Mx (H) and r : M(H) → K is the barycentric mapping. Definition 9.1. Let M be a subset of M(H). We say that a universally measurable set F ⊂ K is M-extremal, if µ(K \ F ) = 0 whenever µ ∈ M and r(µ) ∈ F .
9.1 Topologies generated by extremal sets
275
Examples 9.2. (a) If M = {εx : x ∈ K}, then any universally measurable subset is M-extremal. (b) If M = M(H), then M-extremal sets are exactly H-extremal sets (see Definition 3.12). (c) If M = M(H) ∩ Mmax (H), we get the family of H-maximally extremal sets. Proposition 9.3. If M ⊂ M(H), then the family of all closed M-extremal sets is stable with respect to arbitrary intersections and finite unions. Proof. The proof follows by an easy verification from the definitions. Proposition 9.4. The following assertions hold: (a) If F is a closed H-convex set, then F is H-extremal if and only if F is Hmaximally extremal, and this is the case if and only if F is a closed Choquet set. (b) Any closed H-extremal or H-maximally extremal set intersects ChH (K). Proof. We start the proof by noticing that a closed H-convex set F is H-extremal if and only if F is a closed Choquet set, by definition. To finish the proof we have to check that F is H-extremal provided it is H-maximally extremal. Let x ∈ F and µ ∈ Mx (H) be given. We find an H-maximal measure ν with µ ≺ ν and use the assumption to infer that ν ∈ M1 (F ). Since F is H-convex, Proposition 8.24 yields spt µ ⊂ coH spt ν ⊂ F. This concludes the proof of (a). For the proof of (b), let F be a closed set in K. If F is an H-extremal set, the assertion is proved in Proposition 3.15. If F is an H-maximally extremal set, we consider the family F := {H ⊂ F : H is nonempty closed H-maximally extremal}. By compactness, there exists a minimal element (with respect to inclusion) H ∈ F. If H is not a singleton, we find a function h ∈ H that is nonconstant on H. Then H 0 := {x ∈ H : h(x) = max h(H)} is easily seen to be an H-maximally extremal set. But this contradicts the minimality of H. Hence H = {x} is a singleton, and this is the case if and only if x ∈ ChH (K). This concludes the proof. Definition 9.5 (Topologies generated by M-extremal sets). If M is a subset of the set M(H), we define the following family of sets in ChH (K) as {F ∩ ChH (K) : F is closed and M-extremal}.
(9.1)
It follows from Proposition 9.3 that the family (9.1) is the collection of closed sets for a topology on ChH (K). We denote this topology as τM .
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9 Topologies on boundaries
Examples 9.6. (a) If M = {εx : x ∈ K}, then the topology τM is the original topology on ChH (K). (b) If M = M(H), we denote the resulting topology as σext . (c) If M = M(H) ∩ Mmax (H), we write σmax for the resulting topology. Proposition 9.7. Both σext and σmax are compact topologies on ChH (K) such that one-point subsets of ChH (K) are closed. Proof. Let M stand for M(H) or M(H) ∩ Mmax (H) and let σ stand for σext or σmax . Since any point of ChH (K) is M-extremal, it is a σ-closed set. To check the compactness of (ChH (K), σ), let {Hi : i ∈ I} be a family of σ-closed sets in ChH (K) having the finite intersection property. Let Fi , i ∈ I, be closed Mextremal sets with Hi = Fi ∩ ChHT(K), i ∈ I. Then {Fi } has the finite intersection property as well, and hence F := i∈I Fi is a nonempty closed M-extremal set (see Proposition 9.3). Thus \ \ Hi = Fi ∩ ChH (K) = F ∩ ChH (K) i∈I
i∈I
is nonempty by Proposition 9.4. Hence ChH (K) is compact in σ as required. Proposition 9.8. Let L be a closed H-maximally extremal set and G := {h|L : h ∈ H} be the restricted function space. Then the following assertions hold: (a) if µ ∈ M+ (L), then µ is G-maximal if and only if it is H-maximal, (b) ChG (L) = ChH (K) ∩ L, (c) if H is simplicial, then G is simplicial. Proof. For the proof of (a), let µ ∈ M+ (L) be an H-maximal measure. If ν ∈ M+ (L) with µ ≺G ν, then µ ≺H ν by Proposition 6.72. Hence µ = ν and µ is G-maximal. Conversely, let µ be a G-maximal measure on L. We pick an arbitrary f ∈ Kc (H) and denote by (f |L )∗,G the upper envelope of f |L with respect to G. For any x ∈ L we find a measure ν1 ∈ Mx (H) such that f ∗ (x) = ν1 (f ) (see Lemma 3.21). If ν2 is an H-maximal measure with ν1 ≺H ν2 , then f ∗ (x) = ν1 (f ) ≤ ν2 (f ) ≤ f ∗ (x). Since L is H-maximally extremal, ν2 is carried by L. Hence f ∗ (x) = ν2 (f ) ≤ (f |L )∗,G (x). Obviously, (f |L )∗,G (x) ≤ f ∗ (x) for x ∈ L, and thus µ(f ) = µ((f |L )∗,G ) = µ(f ∗ ). Hence µ is H-maximal by Theorem 3.58.
9.1 Topologies generated by extremal sets
277
To verify (b), let x ∈ ChG (L). Then εx is G-maximal, and thus H-maximal by (a). It follows that x ∈ ChH (K). Since the reverse inclusion follows from Proposition 6.72(c), assertion (b) is proved. Since (c) follows immediately from (a), the proof is finished. Proposition 9.9. A set H ⊂ ChH (K) is σmax -closed if and only if H is relatively closed in ChH (K) and H is H-maximally extremal. Proof. Let H be a subset of ChH (K). If H is relatively closed and H is H-maximally extremal, then H ∩ ChH (K) = H and H is σmax -closed by definition. Assume that H is σmax -closed and let F ⊂ K be an H-maximally extremal set such that H = F ∩ ChH (K). Let x ∈ H and an H-maximal measure µ ∈ Mx (H) be arbitrary. We consider the restricted function space G := {h|F : h ∈ H}. Since µ ∈ M1 (F ), µ is G-maximal by Proposition 9.8(a). By Proposition 3.64 and Proposition 9.8(b), µ is carried by ChG F = ChH (K) ∩ F = H. Hence H is H-maximally extremal, which finishes the proof. Theorem 9.10. The following assertions are equivalent: (i) ChH (K) is closed, (ii) σext is Hausdorff, (iii) σmax is Hausdorff, (iv) σext coincides with the original topology, (v) σmax coincides with the original topology. Proof. Let τ stand for the original topology on ChH (K). Obviously we have σext ⊂ σmax ⊂ τ.
(9.2)
We prove (i) =⇒ (iv) =⇒ (ii) =⇒ (iii) =⇒ (i) and (iv) =⇒ (v) =⇒ (iii). If ChH (K) is closed, any τ -closed subset of ChH (K) is H-extremal and hence σext -closed. Hence τ = σext , and thus (i) =⇒ (iv). Obviously, (iv) =⇒ (ii) and (ii) =⇒ (iii) because of (9.2). To verify (iii) =⇒ (i), assume that there exists x ∈ ChH (K) \ ChH (K). Let F be the smallest closed H-maximally extremal set containing x (such a set exists by virtue of Proposition 9.3). Then H := F ∩ ChH (K) is σmax -closed and nonempty by Proposition 9.4(b). Moreover, H contains two different points, say x1 , x2 . We claim that they cannot be separated by disjoint σext -open sets. Indeed, assume that U1 , U2 are such sets. Let Fi , i = 1, 2, be closed H-maximally extremal sets such that ChH (K) \ Ui = Fi , i = 1, 2. Then ChH (K) ⊂ F1 ∪ F2 . / F. We suppose that x ∈ F1 . By the minimality of F , F ⊂ F1 and thus x1 ∈
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9 Topologies on boundaries
Analogously, x ∈ F2 implies x2 ∈ / F . In both cases we get a contradiction. Hence σmax is not Hausdorff and (iii) =⇒ (i). If σext = τ , then σext = σmax by (9.2). Hence (iv) =⇒ (v). Obviously, (v) =⇒ (iii), which finishes the proof. Example 9.11. There exists a compact convex set X such that σext is strictly weaker than σmax . Proof. Let x0 , x2 , x2 be distinct points in [0, 1]. We identify (x0 , 0) and (x2 , 0) in [0, 1]×[0, 21 ] and let K := p([0, 1]×[0, 12 ]), where p is the quotient mapping. We write ω0 for the point p(x0 , 0) and identify K \ {ω0 } with [0, 1] × [0, 21 ] \ {(x0 , 0), (x2 , 0)}. Let H := {f ∈ C(K) : f (ω0 ) = tf (x1 , t) + (1 − t)f (x2 , t), t ∈ (0, 12 ], Z f (x1 , 0) + f (ω0 ) = 4 f (x, t) dx dt}. K
Then H is a closed function space on K whose Choquet boundary equals K \ {ω0 }. Let X stand for the state space of H and let φ : K → X be the homeomorphic embedding from Definition 4.25. Let H := {t(x1 , t) + (1 − t)(x2 , t) : t ∈ (0, 12 ]}
and F := φ({ω0 } ∪ H).
Since any H-maximal measure representing ω0 is carried by H, using Proposition 9.9 we get that F is an Ac (X)-maximally extremal set and hence φ(H) is a σmax -closed subset of X. We show that φ(H) is not a σext -closed set in ext X. To this end, let C be a closed extremal subset of X containing φ(ω0 ). Since C is a union of closed faces by Proposition 2.69, there exists a closed face D ⊂ C containing φ(ω0 ). Since φ(ω0 ) = tφ(x1 , t) + (1 − t)φ(x2 , t),
t ∈ (0, 21 ],
C contains both φ(x1 , t) and φ(x2 , t) for each t ∈ (0, 12 ]. Since D is a face, we get 1 (φ(x1 , t) + φ(x2 , t)) ∈ D, 2
t ∈ (0, 21 ],
as well. Thus 12 (φ(x1 , 0) + φ(ω0 )) ∈ D because D is closed. If µ = p] (2λ), where λ is Lebesgue measure on [0, 1] × [0, 21 ], then 21 (φ(x1 , 0) + φ(ω0 )) is the barycenter of φ] µ. Thus D, and consequently C, contains the support of φ] µ. Hence C ⊃ φ(K) and F is not σext -closed.
9.2
Induced measures on Choquet boundaries
Theorem 9.12. Let H be a function space on a compact space K such that H = Ac (H). Let {µn } be a sequence of H-maximal measures, {An } be a sequence of
9.2 Induced measures on Choquet boundaries
279
Baire subsets of K and F ⊂ C(K) be a norm separable set. Then there exist a metrizable compact space K 0 , a function space H0 on K 0 , a family F 0 ⊂ C(K 0 ), and a continuous surjection ϕ : K → K 0 such that (a) h ◦ ϕ ∈ H for every h ∈ H0 , (b) ϕ] µn is H0 -maximal, n ∈ N, (c) ϕ(An ) is a Baire subset of K 0 and ϕ−1 (ϕ(An )) = An , n ∈ N, (d) for every f ∈ F there exists f 0 ∈ F 0 so that f = f 0 ◦ ϕ. Moreover, if H is simplicial, H0 can be chosen to be simplicial as well and then Ac (H0 ) = H0 . Proof. Given the objects as in the hypotheses, we find a dense countable subset D of F. Next we select a countable family F 1 ⊂ C(K) such that each An is contained in the σ-algebra generated by {f −1 (U ) : U ⊂ R open, f ∈ F 1 } (see Proposition A.48). Without loss of generality we may assume that 1 ∈ F 1 . For each function f ∈ D ∪F 1 we find a countable family Hf ⊂ H such that f ∈ W(Hf ) − W(Hf ). Let [
F 2 :=
Hf .
f ∈D∪F 1
Now we inductively construct countable families F k ⊂ H, k ≥ 3, as follows. Assume that F k has been constructed for some k ≥ 2. Using Theorem 3.58, Proposition 3.25 and Theorem A.84 we construct a family F k+1 in such a way that (e) F k ⊂ F k+1 , Pm (f) i=1 qi fi ∈ F k+1 whenever m ∈ N, q1 , . . . , qm ∈ Q and f1 , . . . , fm ∈ F k , (g) for each function f ∈ −W(F k ), any measure µn and ε > 0, there exists a function g ∈ W(F k+1 ) such that f ≤ g and µn (g − f ) < ε. Finally, we define F ∞ :=
∞ [
Fk
k=2
and ϕ : K → RN as x 7→ {f (x)}f ∈F ∞ . Then K 0 := ϕ(K) is a metrizable compact space and for any function f ∈ F ∞ there exists a function gf ∈ C(K 0 ) such that f = gf ◦ ϕ. Let H0 := {gf : f ∈ F ∞ }. It follows from (f), that F ∞ is a linear space, hence H0 is a linear space as well. Obviously, it contains constant functions and separates points of K 0 , so it is a function space. Since F ∞ ⊂ H, h ◦ ϕ ∈ H for every h ∈ H0 , and (a) therefore holds.
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9 Topologies on boundaries
In the next step we show that ϕ] µn is H0 -maximal for every n ∈ N. Let n ∈ N be fixed and g ∈ −W(H0 ) be arbitrary. For a fixed ε > 0, we can find a function f ∈ −W(F ∞ ) such that kf − g ◦ ϕk < ε. By (g), there exists f 0 ∈ W(F ∞ ) such that f ≤ f 0 and µn (f 0 − f ) < ε. Let g 0 ∈ W(H0 ) satisfy g 0 ◦ ϕ = f 0 . Then g ≤ g 0 and (ϕ] µn )(g) = µn (g ◦ ϕ) ≥ µn (f ) − ε ≥ µn (f 0 ) − 2ε = µn (g 0 ◦ ϕ) − 2ε = (ϕ] µn )(g 0 ) − 2ε ≥ (ϕ] µn )(g ∗ ) − 2ε. Hence ϕ] µn is H0 -maximal. To verify (c) and (d), we notice that, by the construction, any function from D ∪ F 1 is contained in the closure of the family {g ◦ ϕ : g ∈ W(H0 ) − W(H0 )}. Hence any function from D ∪ F 1 can be expressed as g ◦ ϕ for a suitable function g ∈ C(K 0 ). From this observation the existence of the required family F 0 ⊂ C(K 0 ) as well as (c) and (d) follow. Moreover, if H is simplicial, we adjust the construction of the families F k , k ≥ 3, in the following way. We moreover assure in the inductive step that the family F k+1 possesses the following property: (h) for each f1 , −f2 ∈ −W(F k ) with f1 ≤ f2 there exists f ∈ F k+1 with f1 ≤ f ≤ f2 . Then the resulting family F ∞ , and consequently H0 , possesses the weak Riesz interpolation property, which yields that H0 is simplicial and that Ac (H0 ) = H0 (see Theorem 6.16 and Exercise 6.78). This concludes the proof. Lemma 9.13. Let (K, H), (K 0 , H0 ) be function spaces and let ϕ : K → K 0 be a continuous surjection satisfying h0 ◦ ϕ ∈ H for any h0 ∈ H0 . Then ϕ−1 (F 0 ) is Hextremal for any H0 -extremal set F 0 ⊂ K 0 . In particular, ChH0 (K 0 ) ⊂ ϕ(ChH (K)). Proof. Set F := ϕ−1 (F 0 ), where F 0 ⊂ K 0 is a universally measurable H0 -extremal set. Then F is universally measurable as well. Let x ∈ F and µ ∈ Mx (H) be given. Then ϕ] µ ∈ Mϕ(x) (H0 ), and hence (ϕ] µ)(K 0 \ F 0 ) = 0. Thus µ(K \ F ) = 0. Proposition 9.14. Let µ ∈ M+ (K) be a maximal measure. Then (a) for any Baire set B ⊂ K, µ(B) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal}, (b) for any G ⊂ K of type Gδ , µ(G) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal}.
9.2 Induced measures on Choquet boundaries
281
Proof. Given a maximal measure µ ∈ M+ (K), a Baire set B ⊂ K and ε > 0, let (K 0 , H0 ) and ϕ : K → K 0 be constructed as in Theorem 9.12. Then ϕ] µ, as an H0 -maximal measure on a metrizable compact space K 0 , satisfies (ϕ] µ)(ϕ(B)) = (ϕ] µ)(ϕ(B) ∩ ChH0 (K 0 )). Using the regularity of ϕ] µ, we find a compact set F 0 ⊂ ϕ(B) ∩ ChH0 (K 0 ) such that (ϕ] µ)(ϕ(B)) ≤ (ϕ] µ)(F 0 ) + ε. By Example 8.9, F 0 is H0 -extremal. Then F := ϕ−1 (F 0 ) is H-extremal (see Lemma 9.13), of type Gδ and µ(F ) = (ϕ] µ)(F 0 ) ≥ (ϕ] µ)(ϕ(B)) − ε = µ(B) − ε. This finishes the proof of (a). T Let G ⊂ K be a Gδ set, say G = ∞ n=1 Gn , where the sets Gn are open in K. If F is a closed subset of an open set U ⊂ K, we can find a closed Gδ set F 0 with F ⊂ F 0 ⊂ U . Hence it follows from (a) that µ(Gn ) = sup{µ(F ) : F ⊂ B is Gδ , closed and H-extremal},
n ∈ N.
For ε > 0 we find closed H-extremal sets Fn ⊂ Gn of type Gδ such that µ(Gn \Fn ) ≤ T F ε2−n . Then F := ∞ n=1 n is a closed H-extremal set of type Gδ with F ⊂ G, and µ(G \ F ) ≤ µ(
∞ [
(Gn \ Fn )) ≤ ε.
n=1
This concludes the proof. Notation 9.15. Let Σ stand for the σ-algebra generated by the family consisting of all Baire sets and closed H-extremal sets, and let Σ0 := {A ∩ ChH (K) : A ∈ Σ}. Proposition 9.16. For any A ∈ Σ and any maximal measure µ ∈ M+ (K), µ(A) = sup{µ(F ) : F ⊂ A closed H-extremal} = sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A}. Proof. We define µ1 , µ2 : Σ → [0, ∞) as µ1 (A) := sup{µ(F ) : F ⊂ A closed H-extremal}, µ2 (A) := sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A}, for any A ∈ Σ. Obviously, µ1 ≤ µ and µ1 ≤ µ2 on Σ.
(9.3)
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9 Topologies on boundaries
Let T := {A ∈ Σ : µ1 (A) = µ(A), µ1 (K \ A) = µ(K \ A)}. It follows from Proposition 9.14(a) that T contains all Baire sets. Further, any closed H-extremal set is in T , by Proposition 9.14(b). We show that T is an σ-algebra. Clearly, T is closed with respect to complements. S Let {An } be a sequence of sets in T and let A := ∞ n=1 An . For ε > 0 we find k ∈ N S such that µ(A \ kn=1 An ) < ε. For each n ∈ N we select closed H-extremal sets Fn , Fn0 ⊂ K such that Fn ⊂ An ,
Fn0 ⊂ K \ An
Then F :=
and µ(An \ Fn ) + µ(K \ (An ∪ Fn0 )) < ε2−n . k [
Fn
and
F 0 :=
n=1
∞ \
Fn0
n=1
are closed H-extremal sets. Further, F ⊂ A,
F0
⊂ K \ A and
µ(A \ F ) + µ(K \ (A ∪ F 0 )) < 2ε + ε. Hence A ∈ T , and T is a σ-algebra. It follows that Σ = T , which proves the first equality in (9.3). To prove the second equality, we have µ2 (A) ≥ µ1 (A) = µ(A), A ∈ Σ. (9.4) Let A ∈ Σ and F, F 0 be closed H-extremal set with F ∩ ChH (K) ⊂ A
and
F 0 ∩ ChH (K) ⊂ K \ A.
We claim that F ∩ F 0 = ∅. Indeed, assume that x ∈ F ∩ F 0 . Consider the function space G := {h|F ∩F 0 : h ∈ H}. Let λ be a G-maximal measure in Mx (G). By Propositions 3.64 and 9.8(b), λ is carried by ChG (F ∩ F 0 ) = ChH (K) ∩ F ∩ F 0 = ∅, which is impossible. Hence F ∩ F 0 = ∅. Given A ∈ Σ, let F be a closed H-extremal set with F ∩ ChH (K) ⊂ A. We fix ε > 0 and find a closed H-extremal set F 0 ⊂ K \ A such that µ(K \ A) − µ(F 0 ) < ε. By the argument above, F ∩ F 0 = ∅. Thus µ(F ) ≤ µ(K \ F 0 ) = µ(K) − µ(F 0 ) ≤ µ(K) − µ(K \ A) + ε = µ(A) + ε. Hence µ2 (A) ≤ µ(A) for each A ∈ Σ, which, in conjunction with (9.4), completes the proof.
283
9.2 Induced measures on Choquet boundaries
Corollary 9.17. If µ is maximal and A ∈ Σ disjoint from ChH (K), then µ(A) = 0. Proof. This follows from (9.3) and Proposition 9.4(b). Proposition 9.18. Let a measure µ ∈ M+ (K) satisfy µ(A) = sup{µ(F ) : F ⊂ A is Gδ closed and H-extremal},
A ⊂ K Baire.
Then µ is H-maximal. Proof. We verify condition (b) of Theorem 8.32; more precisely, we show that any function h ∈ H is constant on Fx (H) for µ-almost all x ∈ K (we recall that Fx (H) = S ν∈Mx (H) spt ν). Let h ∈ H be arbitrary and let [a, b] be a closed interval in R. BySthe assumption, there exist closed H-extremal sets Fn ⊂ h−1 ([a, b]) such that S ∞ ∞ µ( n=1 Fn ) = µ(h−1 ([a, b])). For a point x ∈ n=1 Fn and ν ∈ Mx (H), spt ν ⊂
∞ [
Fn ⊂ h−1 ([a, b]).
n=1
Hence, Fx (H) ⊂ h−1 ([a, b]) for µ-almost all x ∈ h−1 ([a, b]). Thus there exists a set K 0 ⊂ K satisfying • •
µ(K \ K 0 ) = 0, if x ∈ K 0 and [a, b] is a closed interval with rational endpoints such that x ∈ h−1 ([a, b]), then Fx (H) ⊂ h−1 ([a, b]).
It follows that, for any x ∈ K 0 , the function h is constant on Fx (H). Thus µ is maximal. Theorem 9.19. If µ ∈ M+ (K) is a maximal measure, then there exists a measure µ0 on (ChH (K), Σ0 ) such that µ0 (A ∩ ChH (K)) = µ(A),
A ∈ Σ.
(9.5)
Moreover, µ0 (A) = sup{µ0 (F ) : F ⊂ A is σext -closed},
A ∈ Σ0 ,
and µ0 (F ) = inf{µ0 (B ∩ ChH (K)) : B ∩ ChH (K) ⊃ F, B is Baire in K}, for any σext -closed set F ⊂ ChH (K).
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9 Topologies on boundaries
Proof. If µ is a maximal measure on K, it follows from Corollary 9.17 that µ0 defined by (9.5) is a well-defined measure on Σ0 . Moreover, for A ∈ Σ we have, by Proposition 9.16, µ0 (A ∩ ChH (K)) = µ(A) = sup{µ(F ) : F closed H-extremal, F ∩ ChH (K) ⊂ A ∩ ChH (K)} = sup{µ0 (F 0 ) : F 0 σext -closed, F 0 ⊂ A ∩ ChH (K)}. For a closed H-extremal set F ⊂ K, µ0 (F ∩ ChH (K)) = µ(F ) = inf{µ(B) : B Baire, F ⊂ B} ≥ inf{µ0 (B ∩ ChH (K)) : B Baire, F ∩ ChH (K) ⊂ B ∩ ChH (K)} ≥ µ0 (F ∩ ChH (K)). This proves the assertion. Corollary 9.20. Let µ ∈ Mx (H) be a maximal measure and f be a bounded Baire H-affine function. If µ0Ris the measure from Theorem 9.19, then f |ChH (K) is µ0 measurable and f (x) = ChH (K) f (t) dµ0 (t). Proof. Given µ ∈ Mx (H) and f as in the corollary, we assume that 0 ≤ f ≤ 1. For any t ∈ R, the set Ft := {y ∈ K : f (y) ≥ t} is a Baire set and µ0 (Ft ∩ ChH (K)) = µ(Ft ) by Theorem 9.19. Hence we get from Fubini’s theorem that Z f (x) =
Z f (y) dµ(y) =
K Z 1
1
µ(Ft ) dt 0
0
Z
µ (Ft ∩ ChH (K))) dt =
= 0
f (y) dµ0 (y).
ChH (K)
This concludes the proof.
9.3
Functions continuous in σext and σmax topologies
Notation 9.21. We write τext for the topology on K whose closed sets are closed Hextremal subsets of K and τmax for the topology on K whose closed sets are precisely closed H-maximally extremal sets. Then σext = τext |ChH (K) and σmax = τmax |ChH (K) .
9.3 Functions continuous in σext and σmax topologies
285
Lemma 9.22. Let f : ChH (K) → [0, 1] be a σext -upper semicontinuous function. Then there exist a closed H-extremal set K 0 ⊃ ChH (K) and a τext -upper semicontinuous function g : K 0 → [0, 1] that extends f . Proof. Let f : ChH (K) → [0, 1] be σext -upper semicontinuous. We fix n ∈ N and find closed H-extremal sets Fin , i = 0, . . . , 2n − 1, such that Fin ∩ ChH (K) = {x ∈ ChH (K) : f (x) ≥ i2−n }, Let Kn :=
S2n −1 i=0
i = 0, . . . , 2n − 1.
Fin and
gn (x) :=
sup i=0,...,2n −1
(i + 1)2−n cFin (x),
x ∈ Kn .
Then Kn ⊃ ChH (K) is a closed H-extremal set and gn , as a finite supremum of τext -upper semicontinuous functions, is τext -upper semicontinuous. By this procedure we get τext -upper semicontinuous functions gn defined on closed −n H-extremal T∞ sets Kn ⊃ ChH (K) such that f ≤ gn ≤ f + 2 on ChH (K). By setting 0 K := n=1 Kn we get a closed H-extremal set containing ChH (K). Then g(x) := inf gn (x), x ∈ K 0 , n∈N
is a τext -upper semicontinuous function extending f . Lemma 9.23. Let K 0 be a closed H-extremal set containing ChH (K) and g : K 0 → R a τext -upper semicontinuous function such that g = 0 on ChH (K). Then g ≤ 0 on K 0 and there exists a closed H-extremal set K 00 ⊂ K 0 containing ChH (K) such that g = 0 on K 00 . Proof. Assume that sup g(K 0 ) > 0. Since g is upper semicontinuous, g attains its maximum on K 0 . Then there exists a closed H-extremal set F ⊂ K such that F ∩ K 0 = {y ∈ K 0 : g(y) = max g(K 0 )}. Since K 0 is H-extremal, F ∩ K 0 is a nonempty closed H-extremal set disjoint from ChH (K), a contradiction with Proposition 3.15. For the proof of the second part, let Fn ⊂ K be closed H-extremal sets such that g −1 ([−n−1 , ∞)) = Fn ∩ K 0 ,
n ∈ N.
T 00 Then F := ∞ n=1 Fn is a closed H-extremal set containing ChH (K), and thus K := 0 K ∩ F is the required set. Proposition 9.24. Let K 0 ⊃ ChH (K) be a closed H-extremal set and g : K 0 → R be a τext -continuous function. Then
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9 Topologies on boundaries
(a) for every x ∈ K 0 and µ ∈ Mx (H), g = g(x) µ-almost everywhere on K 0 , (b) for every continuous function h : R → R, every x ∈ K 0 and µ ∈ Mx (H), we have Z Z h(g(y)) dµ0 (y), h(g(y)) dµ(y) = h(g(x)) = ChH (K)
K
where
µ0
is the induced measure from Theorem 9.19.
Proof. Let g : K 0 → R be a τext -continuous function on a closed H-extremal set K 0 ⊃ ChH (K). For any closed set H ⊂ R, g −1 (H) is a closed H-extremal set. Thus if x ∈ K 0 and µ ∈ Mx (H), then µ is carried by {y ∈ K 0 : g(y) = g(x)}. In other words, g = g(x) µ-almost everywhere. This shows (a). If h : R → R is a continuous function, x ∈ K 0 and µ ∈ Mx (H), then F := {y ∈ K 0 : g(y) = g(x)} is a closed H-extremal set of µ-measure 1. Thus µ0 (F ∩ ChH (K)) = µ(F ) and Z Z Z h(g(y)) dµ(y) = h(g(y)) dµ(y) = h(g(y)) dµ0 (y) K
F
Z =
F ∩ChH (K)
h(g(x)) dµ0 (y) = h(g(x)).
F ∩ChH (K)
This completes the proof. Theorem 9.25. Let f : ChH (K) → R. Then the following assertions are equivalent: (i) f is σext -continuous. (ii) there exist a closed H-extremal set K 0 ⊃ ChH (K) and a τext -continuous function g : K 0 → R extending f . Proof. Let f : ChH (K) → R be σext -continuous. By the compactness of ChH (K) in σext , f is bounded. Using Lemma 9.22, we find closed H-extremal sets K1 , K2 containing ChH (K), a τext -lower semicontinuous function g1 : K1 → R and a τext upper semicontinuous function g2 : K2 → R that extend f . Then K1 ∩ K2 is a closed H-extremal set containing ChH (K). By Lemma 9.23, there exists a closed H-extremal set K 0 ⊂ K1 ∩ K2 containing ChH (K) such that g1 = g2 on K 0 . Then g := g1 |K 0 is the sought extension. This proves (i) =⇒ (ii). Conversely, if g : K 0 → R is a τext -continuous function on a closed H-extremal set 0 K containing ChH (K), then f := g|ChH (K) is σext -continuous. Indeed, f −1 (H) = g −1 (H) ∩ ChH (K) is σext -closed for any closed H ⊂ R. Hence (ii) =⇒ (i). Lemma 9.26. Let f : ChH (K) → [0, 1] be a σmax -upper continuous function. Then there exists a τmax -upper semicontinuous function g : ChH (K) → [0, 1] extending f .
9.3 Functions continuous in σext and σmax topologies
287
Proof. The proof is similar to that of Lemma 9.22. Let f : ChH (K) → [0, 1] be σmax -upper semicontinuous function. We fix n ∈ N and define Fin := {x ∈ ChH (K) : f (x) ≥ i2−n },
i = 0, . . . , 2n − 1.
By Proposition 9.9, each Fin is an H-maximally extremal set. We define gn : ChH (K) → [0, 1] as gn := sup (i + 1)2−n cFin . i=0,...,2n −1
Then gn , as a supremum of finitely many τmax -upper semicontinuous functions, is τmax -upper semicontinuous as well. By this procedure we get a sequence {gn } of τmax -upper semicontinuous functions on ChH (K) such that f ≤ gn ≤ f + 2−n on ChH (K). Hence g(x) := inf gn (x), x ∈ ChH (K), n∈N
is a τmax -upper semicontinuous function extending f . Lemma 9.27. Let g : ChH (K) → [0, ∞) be a τmax -upper semicontinuous function such that g = 0 on ChH (K). Then g = 0 on ChH (K). Proof. Let g : ChH (K) → [0, ∞) be τmax -upper semicontinuous and assume that g(x) > δ > 0 at some point x ∈ ChH (K). Then {y ∈ ChH (K) : g(y) ≥ δ} is a nonempty closed H-maximally extremal set that does not intersect ChH (K), a contradiction with Proposition 9.4(b). Since g ≥ 0 by upper semicontinuity, the proof is finished. Proposition 9.28. Let g : ChH (K) → R be a τmax -continuous function. Then (a) for every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), g = g(x) µ-almost everywhere on ChH (K), (b) for every continuous function h : R → R, every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), we have Z h(g(x)) =
h(g(y)) dµ(y). K
Proof. Let g : ChH (K) → R be a τmax -continuous function and H be a closed set in R. Then g −1 (H) is a closed H-maximally extremal set. If x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), then µ is carried by {y ∈ ChH (K) : g(y) = g(x)}. Hence g = g(x) µ-almost everywhere. This shows (a).
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9 Topologies on boundaries
If h : R → R is a continuous function, x ∈ ChH (K) and µ ∈ Mx (H)∩Mmax (H), then F := {y ∈ ChH (K) : g(y) = g(x)} is a closed H-maximally extremal set of µ-measure 1. Thus Z Z h(g(y)) dµ(y) = h(g(y)) dµ(y) = h(g(x)). K
F
The proof is complete. Theorem 9.29. Let f : ChH (K) → R. Then the following assertions are equivalent: (i) f is σmax -continuous, (ii) there exists a continuous function g : ChH (K) → R extending f such that for every x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H), g = g(x) µ-almost everywhere. R Moreover, if g is the extension of f , then h(g(x)) = K h(g(y)) dµ(y) for each x ∈ ChH (K), µ ∈ Mx (H) ∩ Mmax (H) and h : R → R continuous. Proof. To show (i) =⇒ (ii), let f : ChH (K) → R be a σmax -continuous function. Since ChH (K) is σmax -compact, f is bounded. By Lemma 9.26, there exist bounded functions g1 , g2 : ChH (K) → R such that •
g1 = f = g2 on ChH (K),
•
g1 , −g2 are τmax -lower semicontinuous.
Since τmax is weaker than the original topology, g1 , −g2 are lower semicontinuous. Hence g1 ≤ g2 on ChH (K). By Lemma 9.27, g := g1 = g2 on ChH (K) is a τmax -continuous extension of f . By Proposition 9.28, g possesses the property required in (ii). Conversely, let g : ChH (K) → R be as in (ii). We show that F := g −1 ([a, b]) is H-maximally extremal for each closed interval [a, b] ⊂ R. For x ∈ F and µ ∈ Mx (H) ∩ Mmax (H), µ is carried by the set {y ∈ ChH (K) : g(y) = g(x)}. Hence µ is carried by F and F is H-maximally extremal. Thus g|ChH (K) is σmax -continuous. The last part of theorem follows from the fact that a function g satisfying condition (ii) is τmax -continuous, and hence Proposition 9.28 applies.
9.4
Strongly universally measurable functions
Definition 9.30. If H is a function space on a compact space K, a function f on K is called H-strongly universally measurable, if for each µ ∈ M1 (K) and ε > 0 there exist bounded lower semicontinuous functions g, −h in H⊥⊥ such that h ≤ f ≤ g and µ(g) − µ(h) < ε. If X is a compact convex set, we call Ac (X)-strongly universally measurable functions just strongly universally measurable.
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289
Proposition 9.31. Any H-strongly universally measurable function is in H⊥⊥ and any bounded semicontinuous function from H⊥⊥ is H-strongly universally measurable. Proof. If f is H-strongly universally measurable, then f is obviously universally measurable and bounded. Given µ, ν ∈ M1 (K) with µ − ν ∈ H⊥ and ε > 0, let g, h be as in Definition 9.30. Then the inequalities µ(f ) ≤ µ(g) ≤ µ(h) + ε = ν(h) + ε ≤ ν(f ) + ε
and
µ(f ) ≥ µ(h) ≥ µ(g) − ε = ν(g) − ε ≥ ν(f ) − ε show that f ∈ H⊥⊥ . Let f be a bounded lower semicontinuous function contained in H⊥⊥ . By Proposition 5.42(b), f is the supremum of an up-directed family of functions from H that are smaller than f . Hence for any measure µ ∈ M1 (K) and ε > 0, we can find h ≤ f in H such that µ(f − h) < ε. Thus f is H-strongly universally measurable. Proposition 9.32. Let X be a compact convex set. Then a function f on X is strongly universally measurable if and only if for each x ∈ X and ε > 0 there exist lower semicontinuous affine functions g, −h such that h ≤ f ≤ g and g(x) − h(x) < ε. In particular, f is strongly affine. Proof. Necessity of the condition is obvious. To show sufficiency, let f satisfy the condition and µ ∈ M1 (X) and ε > 0 be given. Let g, −h be lower semicontinuous affine functions such that h ≤ f ≤ g and g(r(µ)) − h(r(µ)) < ε. By Proposition 4.7 and Lemma 4.5, both g and h are strongly affine and hence bounded. Thus both functions are in Ac (X)⊥⊥ . Finally, µ(g) − µ(h) = g(r(µ)) − h(r(µ)) < ε, as required. From the considerations above it also follows that µ(f ) = f (r(µ)), and f is strongly affine. Proposition 9.33. Let H be a function space on a compact space K, X := S(H) and I be the mapping from Corollary 5.41. Then a function f on K is H-strongly universally measurable if and only if If is strongly universally measurable on X. Proof. If f is H-strongly universally measurable, s ∈ X and ε > 0, let µ ∈ M1 (K) be such that π(µ) = s (see Definition 4.25). Let g, h be functions given by Definition 9.30. Then Ig, −Ih are lower semicontinuous affine functions by Corollary 5.41, Ig ≤ If ≤ Ih and by Definition 5.38 Ih(s) − Ig(s) = µ(g) − µ(h) < ε. Hence If is strongly universally measurable by Proposition 9.32.
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9 Topologies on boundaries
Conversely, let If be strongly universally measurable. Given µ ∈ M1 (K) and ε > 0, let g, −h be lower semicontinuous affine functions on X such that f ≤ If ≤ g and g(π(µ)) − h(π(µ)) < ε. Then I −1 g, I −1 h are lower semicontinuous functions in H⊥⊥ by Corollary 5.41, I −1 g ≤ f ≤ I −1 h and µ(I −1 g) − µ(I −1 h) = g(π(µ)) − h(π(µ)) < ε. This concludes the proof. Notation 9.34 (Completion of induced measures). If µ is an H-maximal measure, let µ0 be the induced measure on (ChH (K), Σ0 ) from Theorem 9.19. We denote by (ChH (K), Σ00 , µ00 )) the completion of the measure space (ChH (K), Σ0 , µ0 ). Lemma 9.35. Let h1 , h2 be bounded affine functions on a compact convex set X with h1 ≤ h2 . Let Γ := {(x, t) ∈ X × R : h1 (x) ≤ t ≤ h2 (x)} and let π : X × R → X be the projection. (a) We have ext Γ = {(x, h1 (x)) : x ∈ ext X} ∪ {(x, h2 (x)) : x ∈ ext X}. (b) If h1 , −h2 are lower semicontinuous, then for each x ∈ X, µ ∈ Mx (X) and t ∈ [h1 (x), h2 (x)] there exists ν ∈ M1 (Γ) such that π] ν = µ and ν ∈ M(x,t) (Γ). (c) If h1 , −h2 are lower semicontinuous, then for each x ∈ X, a maximal measure µ ∈ Mx (X) and t ∈ [h1 (x), h2 (x)] there exists a maximal measure ν ∈ M1 (Γ) such that π] ν = µ and ν ∈ M(x,t) (Γ). Proof. Assertion (a) follows by a straightforward verification. To show (b), let µ ∈ Mx (X) be given. We assume first that h1 , h2 are continuous. Then the graph gr hi := {(x, hi (x)) : x ∈ X},
i = 1, 2,
is a compact convex set such that π : gr hi → X is an affine homeomorphism. Hence there exist measures νi ∈ M1 (gr hi ), i = 1, 2, such that π] νi = µ and r(νi ) = (x, hi (x)), i = 1, 2. Let α ∈ [0, 1] be such that t = αh1 (x) + (1 − α)h2 (x). Then ν := αν1 + (1 − α)ν2 ∈ M1 (Γ) satisfies π] ν = µ and (x, t) is the barycenter of ν. Assume now that h1 , −h2 are lower semicontinuous. By Proposition 4.12, h1 is a supremum of a bounded up-directed family F 1 of affine continuous functions. Similarly, h2 = inf F 2 for a bounded down-directed family F 2 ⊂ Ac (X). For each f1 ∈ F 1 , f2 ∈ F 2 , let Γf1 ,f2 := {(x, t) ∈ X × R : f1 (x) ≤ t ≤ f2 (x)}.
9.4 Strongly universally measurable functions
291
Then Γ ⊂ Γf1 ,f2 and thus we may use the first part to deduce that µ ∈ π] (M(x,t) (Γf1 ,f2 )). Since Γ=
\
{Γf1 ,f2 : f1 ∈ F 1 , f2 ∈ F 2 }
and the latter family is down-directed, \ M(x,t) (Γ) = {M(x,t) (Γf1 ,f2 ) : f1 ∈ F 1 , f2 ∈ F 2 } and the latter family is down-directed as well. By compactness and the continuity of π] , Proposition A.40 yields \ π] (M(x,t) (Γ)) = {π] (M(x,t) (Γf1 ,f2 )) : f1 ∈ F 1 , f2 ∈ F 2 }, and thus µ ∈ π] (M(x,t) (Γ)). To verify (c), let µ ∈ Mx (X) be a maximal measure. Using (b) we find λ ∈ M(x,t) (Γ) with π] λ = µ. Let ν ∈ M1 (Γ) be a maximal measure with λ ≺ ν. Then ν ∈ M(x,t) (Γ) and µ ≺ π] ν. Indeed, if k is a continuous convex function on X, k ◦ π is a continuous convex function on Γ. Hence µ(k) = π] λ(k) = λ(k ◦ π) ≤ ν(k ◦ π) = π] ν(k). Since µ is maximal, π] ν = µ. This finishes the proof. Theorem 9.36. Let f be a lower semicontinuous affine function on a compact convex set X. If µ ∈ M1 (X) is a maximal measure, let µ00 denote the measure from Notation 9.34. R Then f |ext X is µ00 -measurable and f (r(µ)) = ext X f (t) dµ00 (t). Proof. Since f is bounded (see Proposition 4.7 and Lemma 4.5), we may assume that f is a lower semicontinuous function on X with 0 ≤ f < 1. Let Γ := {(x, t) ∈ X × R : f (x) ≤ t ≤ 1}. Using Lemma 9.35(c), we find a maximal measure ν ∈ M1 (Γ) such that π] ν = µ
and r(ν) = (x, f (x)).
We claim that the set gr f is a Gδ face of Γ and gr 1 = {(x, 1) : x ∈ X} is a closed face of Γ. To verify this, we first notice that both sets are obviously faces of Γ and gr 1 is closed. Since gr f =
∞ n \ 1o (x, t) ∈ Γ : f (x) ≤ t < f (x) + n
n=1
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9 Topologies on boundaries
and the sets on the right-hand side are open in Γ by the lower semicontinuity of f , gr f is of type Gδ in Γ. Further, ν(gr f ) = 1. To see this, we consider a positive continuous affine function F on Γ defined as F (x, t) := t − f (x), (x, t) ∈ Γ. Then
Z 0 = F (x, f (x)) = F (r(ν)) =
F (y, s) dν(y, s). Γ
Hence F = 0 ν-almost everywhere, and thus ν(gr f ) = 1.
(9.6)
Let a ∈ R be arbitrary. We denote E := {y ∈ X : f (y) ≤ a},
E 0 := E ∩ ext X,
F := {y ∈ X : f (y) > a},
F 0 := F ∩ ext X.
Our aim is to prove that both E 0 and F 0 are µ00 -measurable. Let (µ0 )∗ and (µ0 )∗ denote the outer and inner measure on ext X induced by µ0 , respectively. We first notice that Proposition 9.14(b) and Theorem 9.19 applied to the open set F give µ(F ) = sup{µ(D) : D ⊂ F is closed, Baire and H-extremal} = sup{µ0 (D ∩ ext X) : D ⊂ F is closed, Baire and H-extremal}
(9.7)
≤ (µ0 )∗ (F 0 ). The most important step of the proof is the following claim. Claim. We have µ(E) ≤ (µ0 )∗ (E 0 ). Proof of the claim. Obviously we may assume that 0 ≤ a < 1. Since gr f is a Gδ set in Γ, Proposition 9.14 yields the existence of a closed extremal set D1 ⊂ Γ such that D1 ⊂ gr f
and
ν(D1 ) > 1 − ε.
Let G := {(x, t) ∈ Γ : t ≤ a},
G0 := G ∩ ext Γ.
Then G0 = G ∩ ext gr f and E = π(G),
E 0 = π(G0 ).
9.4 Strongly universally measurable functions
293
Since G is a closed Gδ set, Theorem 9.19 provides a closed extremal set D2 ⊂ Γ such that D2 ∩ ext Γ ⊂ G ∩ ext Γ = G0
and
ν(D2 ) = ν 0 (D2 ∩ ext Γ) > ν 0 (G0 ) − ε.
Then D := D1 ∩ D2 is a closed extremal set in Γ contained in gr f such that ν(D) > ν(G) − 2ε.
(9.8)
Since π : gr f → X is an affine bijection, π(D) is an extremal set in X. Moreover, it is closed by the continuity of π. From (9.8) we get µ(π(D)) = ν(π −1 (π(D))) = ν(π −1 (π(D)) ∩ gr f ) = ν(D) > ν 0 (G0 ) − 2ε. Since π(D) ∩ ext X ⊂ E 0 , we get (µ0 )∗ (E 0 ) ≥ µ0 (π(D) ∩ ext X) = µ(π(D)) > ν 0 (G0 ) − 2ε.
(9.9)
Since G is a Baire set, we get ν 0 (G0 ) = ν(G) ≤ ν(π −1 (π(G))) = ν(π −1 (π(G)) ∩ gr f ) = ν(G ∩ gr f ) ≤ ν(G). This implies µ(E) = ν(π −1 (E)) = ν(π −1 (π(G))) = ν 0 (G0 ).
(9.10)
From (9.9) and (9.10) we get (µ0 )∗ (E 0 ) ≥ ν 0 (G0 ) − 2ε = µ(E) − 2ε. Hence (µ0 )∗ (E 0 ) ≥ µ(E) as required. It follows from the claim that the set F 0 is µ00 -measurable. Indeed, the claim, along with (9.7), yields (µ0 )∗ (F 0 ) = 1 − (µ0 )∗ (E 0 ) ≤ 1 − µ(E) = µ(F ) ≤ (µ0 )∗ (F 0 ), which implies µ00 -measurability of F 0 . Moreover, µ00 (F 0 ) = µ(F ).
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9 Topologies on boundaries
Thus f is µ00 -measurable and to finish the proof we need to verify the barycentric formula. To this end, let Fa := {y ∈ X : f (y) > a},
Fa0 := Fa ∩ ext X,
a ∈ R.
By the first part of the proof, µ00 (Fa0 ) = µ(Fa ),
a ∈ R.
Since µ(f ) = f (x) by Proposition 4.7, Fubini’s theorem gives Z 1 Z µ(Fa ) da f (y) dµ(y) = f (x) = X 1
Z =
0
µ00 (Fa0 ) da =
Z
0
f (y) dµ00 (y).
ext X
This concludes the proof. Lemma 9.37. Let H be a function space on a compact space K, X := S(H) be its state space and φ : K → X be the evaluation mapping as in Definition 4.25. For a measure µ ∈ Mmax (H), let µ00 be the induced measure on ChH (K) as in Notation 9.34. Similarly, let (φ] µ)00 be the induced measure on ext X. We write Σ00 (ChH (K)) or Σ00 (ext X) for the respective σ-algebras. If φ0 : ChH (K) → ext X denotes the restriction of φ, then • φ0 : (Ch (K), Σ00 (Ch (K))) → (ext X, Σ00 (ext X)) is measurable; that is, H H (φ0 )−1 (A) ∈ Σ00 (ChH (K)) for any A ∈ Σ00 (ext X), • (φ µ)00 = (φ0 ) µ00 . ] ] Proof. Given the objects as in the statement of the lemma, let Σ(K) and Σ0 (ChH (K)) stand for the σ-algebras from Notation 9.15, where K indicates their relevance to the space K. Analogously, we write Σ(X) and Σ0 (ext X) for the respective σ-algebras on X. Now (ChH (K), Σ00 (ChH (K)), µ00 ) denotes the completion of the measure space (ChH (K), Σ0 (ChH (K), µ0 )), and (ext X, Σ00 (ext X), (φ] µ)00 ) denotes the completion of (ext X, Σ0 (ext X), (φ] µ)0 ). First we notice that (φ0 )−1 (A ∩ ext X) = φ−1 (A) ∩ ChH (K),
A ⊂ X.
(9.11)
Clearly, φ−1 (A) is a Baire set in K for any Baire set A ⊂ X and φ−1 (A) is a closed H-extremal set for any closed extremal set A ⊂ X (see Lemma 8.10). By (9.11), (φ0 )−1 (A0 ) ∈ Σ0 (ChH (K)),
A ∈ Σ0 (ext X).
From this it follows that φ0 : (ChH (K), Σ0 (ChH (K))) → (ext X, Σ0 (ext X)) is measurable.
9.4 Strongly universally measurable functions
295
Further, for any A ∈ Σ(X), we get from Theorem 9.19 and (9.11) that (φ] µ)00 (A ∩ ext X) = (φ] µ)(A) = µ(φ−1 (A)) = µ00 (φ−1 (A) ∩ ChH (K)) = µ00 ((φ0 )−1 (A ∩ ext X)) = ((φ0 )] µ00 )(A ∩ ext X). Hence ((φ0 )] µ00 )(A) = (φ] µ)00 (A),
A ∈ Σ0 (ext X).
(9.12)
It follows from (9.12) that, for any N ∈ Σ00 (ext X) with (φ] µ)00 (N ) = 0, (φ0 )−1 (N ) ∈ Σ00 (ChH (K)) and
µ00 ((φ0 )−1 (N )) = 0.
Hence ((φ0 )] µ00 )(A) = (φ] µ)00 (A),
A ∈ Σ00 (ext X),
which is nothing else than the desired result (φ0 )] µ00 = (φ] µ)00 . This concludes the proof. Theorem 9.38. Let H be a function space on a compact space K and f be an Hstrongly universally measurable function. If µ ∈ Mmax (H) and µ00 is the measure from Notation 9.34, the function f |ChH (K) is µ00 -measurable and Z µ(f ) = f (y) dµ00 (y). ChH (K)
In particular, if µ H-represents a point x ∈ K, then Z f (x) = f (y) dµ00 (y). ChH (K)
Proof. Let f be as in the hypotheses and µ ∈ M1 (K) be maximal. Let X := S(H) be the state space of H and let I be the mapping from Corollary 5.41. First we show that If |ext X is (φ] µ)00 -measurable (here (φ] µ)00 is the measure on ext X given by Notation 9.34). By Proposition 9.33, If is strongly universally measurable. Hence, there exist lower semicontinuous affine functions gn , −hn on X, n ∈ N, such that hn ≤ If ≤ gn
and (φ] µ)(gn − hn ) < n−1 .
We define an := h1 ∨ · · · ∨ hn ,
bn := g1 ∧ · · · ∧ gn ,
n ∈ N.
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9 Topologies on boundaries
Then sup an ≤ f ≤ inf bn , n∈N
n∈N
and by Theorem 9.36 we can write Z ( inf bn (y) − sup an (y)) d(φ] µ)00 (y) 0≤ ext X n∈N
Z = lim
n→∞ ext X
n∈N
(bn (y) − an (y)) d(φ] µ)00 (y)
≤ lim sup(φ] µ)(gn − hn ) n→∞
= 0. Hence If |ext X is (φ] µ)00 -measurable and hn (r(φ] µ)) = (φ] µ)00 (hn |ext X ) ≤ (φ] µ)00 (If |ext X ) ≤ (φ] µ)00 (gn |ext X ) = gn (r(φ] µ)),
n ∈ N,
gives (φ] µ)00 (If |ext X ) = If (r(φ] µ)) = µ(f ). Let φ0 : ChH (K) → ext X be the restriction of φ. By the first part of the proof, If is (φ] µ)00 -measurable. Lemma 9.37 yields (φ] µ)00 = (φ0 )] µ00 , and thus the function If |ext X is (φ0 )] µ00 -measurable. Hence Z µ(f ) = If (r(φ] µ)) = If (s) d(φ] µ)00 (s) ext X Z Z 0 00 If (s) d((φ )] µ )(s) = If (φ0 (y)) dµ00 (y) = ChH (K)
ext X
Z =
f (y) dµ00 (y),
ChH (K)
which is the desired conclusion. Since the particular assertion obviously follows from the general one, the proof is finished.
9.5
Facial topology generated by M -sets
This section studies properties of the topology generated by M -sets (see Definition 8.40).
9.5 Facial topology generated by M -sets
297
Proposition 9.39. The family of all M -sets is closed with respect to taking arbitrary intersections and H-convex hulls of finite unions. T Proof. Let F be a family of M -sets and F := F ∈F F . Given µ ∈ H⊥ ∩ Mbnd (H) and h ∈ H positive, let F 0 stand for the family of all intersections of finite families in T F. Then F = F ∈F 0 F and (µ|F )(h) = µ( inf 0 cF h) = inf 0 (µ|F )(h) F ∈F
F ∈F
by Theorem A.84. Hence it is enough to show that F possesses condition (M) when F is finite. By a simple induction, it follows that we need to verify that F1 ∩ F2 satisfies condition (M) whenever both F1 and F2 do. But this is obvious, because µ|F1 ∈ H⊥ ∩ Mbnd (H) and hence µ|F1 ∩F2 (h) = (µ|F1 )|F2 (h) = 0. Let F1 , F2 be M -sets and let F := coH (F1 ∪ F2 ). We first verify that F satisfies condition (M). To this end, let µ ∈ H⊥ ∩ Mbnd (H) and h ∈ H be given. Then µ|F1 ∪F2 (h) = µ|F1 (h) + µ|F2 (h) − µ|F1 ∩F2 (h) = 0, because F1 ∩ F2 is an M -set. Hence F = coH (F1 ∪ F2 ) satisfies condition (M) by Proposition 8.34. To finish the proof, we have to verify that F is H-extremal. To this end, let x ∈ F and µ ∈ Mx (H) be given. By Proposition 8.18, there exists a measure µ1 ∈ Mx (H) ∩ M1 (F1 ∪ F2 ). Let µ2 be an H-maximal measure with µ1 ≺ µ2 . Since F1 ∪ F2 is H-extremal (see Proposition 9.3), cF1 ∪F2 is H-convex and hence, by Proposition 3.56, 1 = µ1 (F1 ∪ F2 ) ≤ µ2 (F1 ∪ F2 ). Hence µ2 is carried by F1 ∪ F2 as well. Let ν be an H-maximal measure with µ ≺ ν. Then µ2 − ν ∈ H⊥ ∩ Mbnd (H), and thus (µ2 − ν)|F1 ∪F2 (1) = 0. In other words, 1 = µ2 (F1 ∪ F2 ) = ν(F1 ∪ F2 ). Hence ν is carried by F1 ∪ F2 . Finally, from Proposition 8.24 we get spt µ ⊂ coH spt ν ⊂ F. This finishes the proof.
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9 Topologies on boundaries
Definition 9.40 (Facial topology). It follows from Propositions 9.39 and 8.30(b) that the family {F ∩ ChH (K) : F is an M -set} defines the family of closed sets for a topology on ChH (K). We call it the facial topology and denote it σfac . In general, the topology σfac is not Hausdorff. Proposition 9.41. The topology σfac is compact. Proof. Let {Fi }i∈I be a family of M -sets such that {Fi ∩ ChH (K)}i∈I has the finite intersectionTproperty. Then {Fi }i∈I has the finite intersection property as well, and thus F := i∈I Fi is a nonempty M -set (see Proposition 9.39). Since F ∩ChH (K) 6= ∅ (see Proposition 9.4(b)), we get \ (Fi ∩ ChH (K)) = F ∩ ChH (K) 6= ∅. i∈I
This concludes the proof. Theorem 9.42. Let f : ChH (K) → R be a bounded σfac -upper semicontinuous function and a ∈ H be a positive function. Then there exists a unique upper semicontinuous function h ∈ H⊥⊥ such that h(x) = a(x)f (x),
x ∈ ChH (K).
Moreover, if we denote g := f · a|ChH (K) , then h = ChH (K)g ∗ . Proof. Assume first that f is as in the hypothesis and f (ChH (K)) ⊂ [0, 1]. We fix n ∈ N and find for i = 0, . . . , n an M -set Fi ⊂ K such that {x ∈ ChH (K) : f (x) ≥
i } = Fi ∩ ChH (K). n
By Theorem 8.44, fi := ( na · cFi )∗ ∈ H⊥⊥ , i = 0, . . . , n. Moreover, ( fi =
1 na
0
on Fi ∩ ChH (K), on ChH (K) \ Fi .
Thus the function hn :=
n X
fi
i=0
is upper semicontinuous, belongs to H⊥⊥ and satisfies g ≤ hn ≤ g +
1 n
on ChH (K).
9.5 Facial topology generated by M -sets
299
Thus we can construct a sequence {hn } of upper semicontinuous functions in H⊥⊥ that converges to g uniformly on ChH (K). By the minimum principle of Proposition 3.88, the sequence {hn } converges uniformly on K. Then its limit h is the required function. If f (ChH (K)) is contained in an interval [−m, m] for some m > 0, we can apply 1 (f + m) to get a function h0 satisfying h0 = the argument above to the function 2m 1 0 2m (f +m)·a|ChH (K) on ChH (K). Then the function h := 2mh −ma is the required one. Since the question of uniqueness is settled again by Proposition 3.88, we proceed to the last assertion. If u ∈ H satisfies u ≥ g on ChH (K), then u ≥ h on K by Proposition 3.88. Hence h ≤ ChH (K)g ∗ . On the other hand, ChH (K)g ∗ ≤ h by Proposition 3.25(a). This concludes the proof. Theorem 9.43. Let f : ChH (K) → R be a σfac -continuous function and a ∈ H. Then there exists a unique h ∈ H such that h(x) = a(x)f (x),
x ∈ ChH (K).
Proof. Given f and a as in the theorem, we first notice that f is bounded by Proposition 9.41. Let m > 0 satisfy a(K) ⊂ [−m, m]. By Theorem 9.42, there exist upper semicontinuous functions h1 , h2 ∈ H⊥⊥ such that h1 (x) = f (x)(a(x) + m),
and
h2 (x) = −f (x),
x ∈ ChH (K).
Then h := h1 +mh2 is upper semicontinuous, h ∈ H⊥⊥ and h = af on ChH (K). As in the proof of Theorem 9.42 we obtain that h = ChH (K)g ∗ , where g := f · a|ChH (K) . Analogously we get that ChH (K)g∗ is a lower semicontinuous function in H⊥⊥ . By the minimum principle of Proposition 3.88, we get that ChH (K) ∗
g = ChH (K)g∗
on K.
Thus h is the sought function. Since the uniqueness follows trivially from the minimum principle, the proof is finished. Corollary 9.44. Let f : ChH (K) → R be a σfac -continuous function. Then there exists a unique h ∈ H such that h = f on ChH (K). Proof. The assertion follows from Theorem 9.43 by setting a := 1. Theorem 9.45. The following assertions are equivalent: (i) the topology σfac coincides with the original topology on ChH (K), (ii) the topology σfac is Hausdorff, (iii) H is a Markov simplicial space.
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9 Topologies on boundaries
Proof. We start by proving (ii) =⇒ (iii). First we show that ChH (K) is a closed set. Assume that there exists x ∈ ChH (K) \ ChH (K). Let F be the smallest M -set containing x (see Proposition 9.39). Since x ∈ / ChH (K), F intersects ChH (K) in at least two distinct points, say x1 and x2 (see Proposition 3.15 and Proposition 8.30(b)). For i = 1, 2, let Ui be disjoint σfac -open sets containing xi , respectively. By the definition of σfac , there exist M -sets Fi such that ChH (K) \ Ui = Fi ∩ ChH (K),
i = 1, 2.
Since U1 ∩ U2 = ∅, then x ∈ ChH (K) ⊂ F1 ∪ F2 = F1 ∪ F2 . Assume, for example, that x ∈ F1 . Then F ⊂ F1 , and thus x1 ∈ / F , a contradiction. Analogously we get a contradiction in the case x ∈ F2 . Hence the points x1 and x2 cannot be separated, and σfac is not Hausdorff. Hence ChH (K) is closed. Now we know that σfac is a Hausdorff weaker topology then the original compact topology, and thus it coincides with it. Hence any continuous function on ChH (K) is σfac -continuous as well. Thus Corollary 9.44 yields that any continuous function on ChH (K) can be extended to a function from H. By Theorem 6.42, H is a Markov simplicial space. Assume now that H is a Markov simplicial space and H ⊂ ChH (K) is relatively closed in the original topology. Since ChH (K) is closed, it is a closed set in K. By Proposition 8.31 and Theorem 8.62, F := coH H is an M -set with respect to H = Ac (H). Further, H = F ∩ ChH (K) by Lemma 8.25, and thus H is σfac -closed as needed. Since the proof of the remaining implication (i) =⇒ (ii) is obvious, the proof is finished. Definition 9.46 (Center of H). We say that a function f ∈ H belongs to the center of H, if for every a ∈ H there exists h ∈ H such that h(x) = f (x)a(x),
x ∈ ChH (K).
We write Z(H) for the center of H. Proposition 9.47. If H is a closed function space, then the center Z(H) is a commutative Banach algebra and a vector lattice, where the multiplication and lattice operations are considered pointwise on ChH (K). Proof. First we notice that Z(H) is a linear subspace of H containing constant functions. Step 1. Next we show that Z(H) is closed in H. Let {fn } be a sequence of functions from Z(H) uniformly convergent to f ∈ H. For any a ∈ H and n ∈ N,
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301
there exists a function hn ∈ H such that hn = fn a on ChH (K). By the minimum principle of Theorem 3.16, {hn } is uniformly convergent on K, say to a function h ∈ H. Then h = f a on ChH (K). Step 2. Further we show that Z(H) is an algebra. Let f1 , f2 be functions in Z(H). By the definition of the center, there exists a function h ∈ H such that h = f1 f2
on ChH (K).
We claim that h ∈ Z(H) as well. Indeed, for a given a ∈ H we can find h1 ∈ H such that h1 = f2 a on ChH (K). Further we find h2 ∈ H with h2 = f1 h1 on ChH (K). Then h2 (x) = f1 (x)h1 (x) = f1 (x)f2 (x)a(x) = h(x)a(x),
x ∈ ChH (K).
Hence h ∈ Z(H). Step 3. If f ∈ Z(H), then f = (kf k + f ) − kf k and thus Z(H) = (Z(H))+ − (Z(H))+ . We show that for any f ∈ Z(H) there exists h ∈ Z(H) such that h = f ∨ 0 on ChH (K). To this end, let {pn } be a sequence of polynomials on R that converges to the function t 7→ t∨0, t ∈ R, uniformly on f (K). By the second step of the proof, we find functions hn ∈ Z(H), n ∈ N, satisfying hn = pn ◦ f on ChH (K). Again by the minimum principle, we deduce that {hn } converges uniformly on K to an element h ∈ Z(H). It easily follows that h = f ∨ 0 on ChH (K). Proposition A.15 yields that Z(H) is a vector lattice, which completes the proof. Theorem 9.48. The mapping R : Z(H) → C(ChH (K), σfac ) defined as the restriction to ChH (K) is an isometric isomorphism preserving multiplication and lattice operations. Proof. We first show that h|ChH (K) is σfac -continuous for each h ∈ Z(H). Since Z(H) is a linear space, to this end it is enough to show that the set H := {x ∈ ChH (K) : h(x) ≥ α} is σfac -closed for any h ∈ Z(H) with h(K) ⊂ [0, 1] and α ∈ [0, 1]. For any n ∈ N we use the algebraic structure of Z(H) to find a function hn ∈ Z(H) such that hn (x) = (((1 − α) + h(x)) ∧ 1)n ,
x ∈ ChH (K).
By the minimum principle, the sequence {hn } decreases to an upper semicontinuous function h ∈ H⊥⊥ . Moreover, ( 1, x ∈ H, h(x) = (9.13) 0, x ∈ ChH (K) \ H.
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We set F := {x ∈ K : h(x) = 1}. Obviously, F is a closed Choquet set and F ∩ ChH (K) = H. We need to verify that F is an M -set. To this end, let a positive function a ∈ H and µ ∈ H⊥ ∩ Mbnd (H) be given. We notice that (9.13) yields, using Proposition 3.25(a), that h = c∗F . (9.14) By the definition of the center, for any n ∈ N we can find a function fn ∈ H such that fn = hn a on ChH (K). As in the proof of Theorem 9.42 we infer that fn = ChH (K)(hn a)∗ . By Lemma 8.13, µ is carried by F ∪ F 0 , where F 0 = {x ∈ K : c∗F = 0}. By (9.14) and Theorem 3.58, 0 = lim µ(fn ) = lim µ(ChH (K)(hn a)∗ ) n→∞
n→∞
= lim µ(hn a) = µ(ha) n→∞ Z Z ∗ = µ(acF ) = a · 1 dµ + F
a · 0 dµ
F0
= µ(acF ). Hence µ|F ∈ H⊥ as needed, and F is an M -set. This shows that h|ChH (K) is σfac continuous. It is clear from the minimum principle that R is injective and by Theorem 9.43 we know that it is surjective. Since multiplication and lattice operation are pointwise on ChH (K), we get that R respects both multiplication and lattice operations. Finally, it is an isometry from the minimum principle again. This concludes the proof. Proposition 9.49. Let H be a simplicial function space and let σfac denote the topology on ChH (K) generated by M -sets with respect to Ac (H). Then the following assertions are equivalent: (i) H is prime, (ii) for any pair of nonempty σfac -open sets U1 , U2 ⊂ ChH (K), U1 ∩ U2 6= ∅. Proof. Assume that H is prime and U1 , U2 are nonempty disjoint σfac -open sets in ChH (K). By the definition of σfac we find M -sets F1 , F2 with respect to Ac (H) such that Ui = ChH (K) \ Fi , i = 1, 2. We pick xi ∈ Ui and use Corollary 8.50 to find positive functions hi ∈ Ac (H) such that hi (Fi ) = 0 and hi (xi ) = 1, i = 1, 2. Since h1 ∧ h2 = 0 on ChH (K), the function 0 is the greatest lower bound of h1 ∧ h2 and differs from both h1 and h2 . Hence Ac (H) is not an antilattice. For the proof of (ii) =⇒ (i), let h1 , h2 ∈ Ac (H) be such that h1 ∧ h2 admits the greatest lower bound h ∈ Ac (H) which differs from both h1 and h2 . It follows from Proposition 3.25 that h = (h1 ∧ h2 )∗ . Then Fi := {x ∈ K : h(x) = hi (x)},
i = 1, 2,
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303
are closed Choquet sets with ChH (K) ⊂ F1 ∪ F2 . Since H is simplicial, they are M -sets with respect to Ac (H) (see Theorem 8.62). Then Ui := ChH (K) \ Fi ,
i = 1, 2,
are σfac -open disjoint sets in ChH (K). Since neither F1 nor F2 covers ChH (K) (otherwise h = h1 or h = h2 ), the sets U1 , U2 are nonempty. This concludes the proof. Corollary 9.50. Let H be a prime simplicial space. Then the center of Ac (H) consists of constant functions. Proof. The assertion follows from Proposition 9.49 and Theorem 9.48.
9.6
Exercises
Exercise 9.51. Let X be a compact convex set, X 0 a minimal closed extremal set containing ext X and g : X 0 → R τext -continuous. Prove that g is constant on any segment and on any face contained in X 0 . Hint. Let x, y ∈ X 0 be distinct points such that the segment [x, y] ⊂ X 0 and g(x) < g(y). Then 1 F1 := {z ∈ X 0 : g(z) ≤ (g(x) + g(y))} 2 and 1 F2 := {z ∈ X 0 : g(z) ≥ (g(x) + g(y))} 2 are closed extremal subsets of X with X 0 ⊂ F1 ∪ F2 . On the other hand, if z = 1 / F1 ∪ F2 . This 2 (x + y), then it follows from the extremality of F1 and F2 that z ∈ contradiction shows that g is constant on the segment joining points x and y. If F ⊂ X 0 is a face, then g is constant on any segment contained in F . Hence g is constant on F . Exercise 9.52. Let H be a function space on a compact space K, K 0 a minimal closed H-extremal set containing ChH (K) and g : K 0 → R τext -continuous. Prove that g need not be constant on every closed Choquet set contained in K 0 . Hint. Consider H := C(K) on a compact space K (see Example 3.2(a)). Then any continuous function on K is τext -continuous and any closed set F ⊂ K is a closed Choquet set. Exercise 9.53. Let Σ0 be the σ-algebra on ChH (K) generated by the family {F ∩ ChH (K) : F is Baire or a closed H-maximally extremal set}.
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Prove that any maximal measure µ induces a measure µ0 on Σ0 such that µ0 (A) = sup{µ0 (F ) : F ⊂ A is σmax -closed},
A ∈ Σ0 ,
µ0 (F ) = inf{µ0 (B ∩ ChH (K)) : B ∩ ChH (K) ⊃ F, B is Baire in K}, F is a σmax -closed subset of ChH (K). Hint. Imitate the proof of Proposition 9.16 and Theorem 9.19. Exercise 9.54. Let Σ0 and µ0 be as in the previous Exercise 9.53. Let f : ChH (K) → R be σmax -continuous and g : ChH (K) → R be the extension provided by Theorem 9.29. Let h : R → R be continuous, x ∈ ChH (K) and µ ∈ Mx (H) ∩ Mmax (H). Prove that Z h(g(x)) = h(f (y)) dµ0 (y). ChH (K)
Hint. By Proposition 9.28, g = g(x) µ-almost everywhere and by the proof of Theorem 9.29, g is τmax -continuous. Hence f is µ0 -measurable and F := {y ∈ ChH (K) : g(y) = g(x)} is a τmax -closed set which carries µ. Thus µ0 (F ∩ ChH (K)) = µ(F ) = 1 and Z Z h(f (y)) dµ0 (y) = h(f (y)) dµ0 (y) F ∩ChH (K)
ChH (K)
Z =
h(g(x)) dµ0 (y)
F ∩ChH (K)
= h(g(x)).
Exercise 9.55. Prove that any H-strongly affine function satisfies the minimum principle. Hint. Use either Theorem 9.38 or the definition along with the Minimum principle 3.16. Exercise 9.56. Let φ0 : ChH (K) → ext S(H) be the mapping from Lemma 9.37. Prove that (a) φ0 : (ChH (K), σext ) → (ext S(H), σext ) is continuous, (b) φ0 : (ChH (K), σmax ) → (ext S(H), σmax ) is a homeomorphism, (c) φ0 : (ChH (K), σext ) → (ext S(H), σext ) need not be a homeomorphism.
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305
Hint. For the σext -continuity of φ0 use Lemma 8.10. To show that φ0 is a homeomorphism with respect to σmax -topologies, it is enough to show that F ⊂ K is H-maximally extremal if and only if φ(F ) is S(H)-maximally extremal. But this easily follows from the facts that µ ∈ M+ (K) is H-maximal if and only if φ] µ is Ac (S(H))-maximal (see Proposition 4.28(d)) and that Ac (S(H))-maximal measures on S(H) are carried by φ(K) (see Proposition 3.64). To find an example required by (c), consider the following compact space K ⊂ R2 . Let In := [−2, −1] × {n−1 }, Jn := [1, 2] × {n−1 }, n ∈ N, I0 := [−2, −1] × {0},
J0 := [1, 2] × {0},
I−1 := [−2, −1] × {−1}, and let µn , n ∈ N ∪ {0, −1} and νn , n ∈ N ∪ {0}, be copies of one-dimensional Lebesgue measure on the sets In and Jn , respectively. Let z := (1, 1) ∈ R2 . We define [ [ Jn In ∪ K := n∈N∪{0,−1}
and
n∈N∪{0}
n 1 1 H := f ∈ C(K) : f (z) = µn (f ) + (1 − )νn (f ), n ∈ N, n n o µ0 (f ) = µ−1 (f ) .
Then H is a function space on K and it can be easily seen that ChH (K) = K \ {z}. We claim that K \ I−1 is H-extremal. To this end, we consider functions hn defined as S on I−1 ∪ I0 ∪ ∞ 1 k=n Ik , S∞ −1 n ≥ 2. hn := −(k − 1) on k=n Jk , 0 otherwise, Then {hn } is a bounded sequence of functions from H converging pointwise to the function cI0 ∪I−1 . If µ ∈ Mz (H), then 0 = hn (z) = µ(hn ) → µ(I0 ∪ I−1 ). Hence µ is carried by K \ I−1 as needed. Thus F := ChH (K) ∩ (K \ I−1 ) is a σext -closed set. We show that φ0 (F ) is not σext -closed set in ext S(H). To this end, let φ : K → S(H) be the evaluation mapping and let H ⊂ S(H) be a closed Ac (S(H))-extremal set in S(H) such that φ0 (F ) ⊂ H ∩ ext S(H).
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Then φ(z) ∪ φ(I0 ) ⊂ H. Let π : M1 (K) → S(H) be the restriction mapping and denote sn := π(µn ), n ∈ N ∪ {0, −1}, tn := π(νn ), n ∈ N. Then
1 1 sn + (1 − )tn , n ∈ N, n n and thus sn , tn ∈ H, n ∈ N. Since H is closed, s0 ∈ H as well. We get that the measure φ] µ−1 ∈ Ms0 (S(H)), and hence its support is contained in H (see Proposition 2.86). Thus φ(I−1 ) ⊂ H ∩ ext(S(H)). φ(z) =
Hence φ0 (F ) is strictly smaller than ext(S(H)) ∩ H for any closed extremal set H in S(H) satisfying φ0 (F ) ⊂ ext(S(H)) ∩ H. Exercise 9.57. Let H be a function space on a compact space K, µ ∈ M+ (K) be a maximal measure and (ChH (K), Σ00 , µ00 ) be the measure space from Notation 9.34. Let f : ChH (K) → R be a µ00 -measurable function. Prove that for any ε > 0 there exists a σext -closed set F ⊂ ChH (K) such that µ00 (ChH (K) \ F ) < ε and f |F is σext -continuous. Hint. Let {Un : n ∈ N} be a countable base of open sets in R. Given ε > 0, using Theorem 9.19, we find σext -closed sets Fn ⊂ f −1 (Un ) and Hn ⊂ f −1 (R \ Un ),
n ∈ N,
such that µ00 (f −1 (Un ) \ Fn ) + µ00 (f −1 (R \ Un ) \ Hn ) < 2−n ε, Then F :=
∞ \
n ∈ N.
(Fn ∪ Hn )
n=1
is a σext -closed set such that µ00 (ChH (K) \ F ) < ε and f |F is σext -continuous. Exercise 9.58. Find a compact convex set X, a maximal measure µ ∈ M1 (X) and a strongly affine function f on X such that f |ext X is not µ00 -measurable. Hint. Let H be the function space from Example 3.82, X := S(H) be the state space of H and let f := c[0,1]×{1} − c[0,1]×{−1} . Let λ be Lebesgue measure on [0, 1] × {0} and λ00 be the induced measure on ChH (K) from Notation 9.34. Since H is simplicial, H = Ac (H) (see Lemma 6.14 and its proof) and f is Borel, f ∈ H⊥⊥ (see Corollary 6.12). Thus If is a strongly affine function on X (see Corollary 5.41). By Lemma 9.37, (φ] λ)00 = (φ0 )] λ00 . Further, If (φ(x)) = f (x),
x ∈ ChH (K),
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307
and hence (φ] λ)00 -measurability of If |ext X implies λ00 -measurability of f |ChH (K) . Thus it suffices to show that f |ChH (K) is not λ00 -measurable. Since Σ00 is the completion of Σ0 with respect to the measure λ0 , we know from Theorem 9.19 that λ00 (A) = sup{λ00 (F ) : F ⊂ A is σext -closed},
A ∈ Σ00 .
Thus to disprove that f |ChH (K) is µ00 -measurable it is enough to show that λ00 (F ) = 0 for any σext -closed set F ⊂ [0, 1] × {1} (and analogously for [0, 1] × {−1}). If F is such a set, then there exists a closed H-extremal set H ⊂ K such that H ∩ ChH (K) = F. If F is infinite, it is easy to realize that H intersects [0, 1] × {−1} as well. Hence F is finite and H = F . But then λ00 (F ) = λ(H) = 0. Exercise 9.59. Prove that the class of strongly universally measurable functions is not affinely perfect. More precisely, find compact convex sets X, Y , an affine continuous surjection π : X → Y and a function f : Y → R such that f ◦π is strongly universally measurable and f is not strongly universally measurable. Hint. Consider the function space H from Example 3.82 and set Y := S(H), X := M1 (K). Let π : M1 (K) → S(H) be the restriction mapping and let f and λ be as in Exercise 9.58. If I is the mapping from Corollary 5.41, the function If |ext Y is not (φ] λ)00 -measurable. The function g : X → R defined as g(µ) = µ(f ), µ ∈ X, is strongly universally measurable on X. Indeed, given ε > 0 and ν ∈ X, let h1 , −h2 be upper semicontinuous bounded functions on K such that h1 ≤ f ≤ h2 and ν(h2 ) − ν(h1 ) < ε. Then the functions b hi (µ) := µ(hi ), µ ∈ X, i = 1, 2, witness that g is strongly universally measurable. Finally, the formula If (π(µ)) = g(µ), µ ∈ X, shows that If is the desired function. Exercise 9.60 (Semiextremal sets). If X is a compact convex set X, a set F ⊂ X is semiextremal if X \ F is convex. Prove the following assertions. (a) Any extremal set is semiextremal. (b) Semiextremal sets are stable with respect to unions and intersections of downdirected families. (c) An intersection of an extremal and a semiextremal set is semiextremal. (d) A closed nonempty semiextremal set intersects ext X.
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9 Topologies on boundaries
Hint. Assertions (a), (b) and (c) are easy from the definitions. To prove (d), let F be a closed semiextremal set. We consider the family F := {F ∩ H : H closed extremal, F ∩ H 6= ∅} ordered by reverse inclusion. An application of Zorn’s lemma yields the existence of a minimal set in F. Hence we may assume that F has the property that any closed extremal set intersecting F contains F .
(9.15)
Take any point x ∈ F and a maximal measure µ ∈ Mx (X). Then µ(F ) > 0 (otherwise x ∈ X \ F by Proposition 2.76). Thus there exists a maximal measure ν carried by F . If ν = εy for some y ∈ X, y is an extreme point contained in F and the proof is finished. Now consider the case when ν is not a Dirac measure. Then we can find a pair of disjoint Baire sets A1 , A2 ⊂ K such that ν(Ai ) > 0, i = 1, 2. Using Proposition 9.14 we find closed extremal sets Fi ⊂ Ai such that ν(Fi ) > 0, i = 1, 2. In particular, the sets Fi intersect F . It follows from Proposition 2.69 that there exists a pair of closed faces Hi ⊂ X such that ∅ 6= Hi ∩ F ⊂ Fi ∩ F,
i = 1, 2.
But this contradicts (9.15).
9.7
Notes and comments
Our presentation of topologies on Choquet boundaries follows mainly papers of S. Teleman and C. J. K. Batty (see [449], [450], [453], [452], [451], [34] and [35]). We also refer the reader to the paper of M. Rogalski [393]. The general concept of M-extremal sets is explicitly defined in [35] in order to cover simultaneously notions of the Choquet and maximal topology. Proposition 9.9 and Example 9.11 can be found in [35], Theorem 9.10 in N. Boboc and Gh. Bucur [68]. The factorization theorem 9.12 appears as Theorem 1 in [451], properties of induced measures described in Section 9.2 are presented in [450], [35] and [34]. Results of Section 9.3 on functions continuous with respect to the topologies σext and σmax are taken from papers S. Teleman [455] and C. J. K. Batty [35]. The exposition in Section 9.4 follows the paper of S. Teleman [454]. The results of Section 9.5 are classical and can be found in Chapter II, §6 and 7 in E. M. Alfsen [5] and in L. Asimow and A. J. Ellis [24], Chapter 3, Section 1. We only transferred the techniques into the framework of function spaces. The motivation for the introduction of facial topology came from the theory of C ∗ -algebras. Since we do not pursue this topic in our treatise, we refer the reader to [5] and [24] for the references related to this area.
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309
Theorem 9.42 is attributed in [5] to T. Bai Andersen, an independently found proof is contained in a paper W. Wils [472] (see also T. B. Andersen and H. R. Atkinson [16] for a related result on σfac -continuous mappings with values in Banach algebras). Theorems 9.43, 9.45 and Corollary 9.44 are in E. M. Alfsen and T. Bai Andersen [6], Theorem 9.48 and Corollary 9.50 can be found as Corollary II.7.11 and Observation II.7.16 in [5] (see also Theorem 3 in E. Briem’s paper [91]). We refer the reader to papers [197], [196] and [195] of A. Gleit for topological properties of the topology σfac . Results on extending Borel affine functions on analytic split faces can be found in P. J. Stacey [435]. Exercise 9.51 is Th´eor`eme 2 in S. Teleman [455], Exercise 9.53 is a particular case of Theorem 3.2 in C. J. K. Batty [35] and Exercise 9.54 is Th´eor`eme 3 in [455]. Exercise 9.57 can be found as Theorem 7 in S. Teleman [450]. Assertion (d) of Exercise 9.60 is proved in J. D. Pryce [378] by geometric methods; we present it as a consequence of the results on induced measures on extreme points.
Chapter 10
Deeper results on function spaces and compact convex sets
The goal of this chapter is to present several important results on function spaces and compact convex sets. The first section is devoted to boundaries and to applications of this notion in Banach spaces. The first result is Theorem 10.2, proving the existence of the smallest closed boundary (so-called Shilov boundary) under a rather mild condition. Theorem 10.3 recalls the classical fact that the closure of the Choquet boundary is the Shilov boundary for a function space. Next we prove several results on James’ boundaries which is a concept particularly useful in the theory of Banach spaces. We present an approach based upon the notion of I-envelopes of sets in duals of Banach spaces. This leads us to Theorem 10.7 by V. Fonf and J. Lindenstrauss, which opens a way for the proof of Rode’s result on norm separability of dual spaces (see Theorem 10.8). As a consequence of the Simons lemma, we prove James’ theorem characterizing weakly compact sets in separable Banach spaces (see Theorem 10.9). Then we present Grothendieck’s result on angelicity of spaces of continuous functions on compact spaces when they are endowed with the pointwise topology. As its application we show Khurana’s proof of a theorem by J. Bourgain and M. Talagrand on angelicity of bounded sets in a Banach space with respect to the weak topology generated by extreme points of the dual unit ball (see Theorem 10.12). The next section contains the proof of Lazar’s generalization of the Banach–Stone theorem (see Theorems 10.17 and 10.18). Then we turn our attention to results on automatic boundedness and continuity of affine functions. First we need several important auxiliary facts on the Cantor set, such as the 0–1 laws of Theorem 10.24. The main tool needed later is Theorem 10.26 on countable additivity of finitely additive measures on the Cantor set. Then we present Christensen’s results on strongly linearly independent sequences in locally convex spaces and functionals on L∞ (µ) (see Theorems 10.30 and 10.32). Results on automatic boundedness of affine and convex functions on compact convex sets are proved in Theorems 10.31 and 10.37. Haydon’s characterization of Banach spaces not containing `1 by means of strongly affine functions is contained in Theorem 10.42. Metrizability of compact convex sets is investigated in Section 10.5. The first result in Theorem 10.51 shows that a compact convex set X is metrizable provided it has
10.1 Boundaries
311
a boundary with a countable network. Theorem 10.56 collects several topological conditions on extreme points that are equivalent to metrizability of X. Next we study continuous affine images of compact convex set. First we investigate under what conditions a continuous affine surjection induces a mapping preserving maximal measures (see Theorem 10.57 and 10.59), and then we apply the results to get a variant of the Stone–Weierstrass theorem for function spaces (see Theorem 10.60). The second part of this section presents results on openness of affine continuous surjections (see Theorem 10.66). As a corollary, we get that a continuous mapping ϕ of a compact space K onto a compact space L is open if and only if the induced mapping ϕ] : M1 (K) → M1 (L) is open (see Corollary 10.67). The aim of Section 10.7 is the proof of three topological results on Choquet boundaries. The first one asserts that any Choquet boundary is a Baire space (see Theorem 10.68), the second one constructs a metrizable simplex whose set of extreme points is a prescribed Polish space (see Theorem 10.70) and the last one shows that the set of extreme points is a Borel set of low complexity provided it is K-countably determined. Analogues of Theorems 4.21 and 4.24 for fragmented convex functions are proved in Theorems 10.75 and 10.77. Finally, we characterize stable compact convex sets (see Theorem 10.86) and function spaces for which the barycentric mapping is open (see Theorem 10.88).
10.1 10.1.A
Boundaries Shilov boundary
Definition 10.1 (Shilov sets). Let K be a nonempty compact space and F be a nonempty family of lower semicontinuous functions on K. A closed set S ⊂ K is called a Shilov set for F if for every u ∈ F there exists x ∈ S such that u(x) = inf u(K). If there exists a Shilov set, which is contained in any other Shilov set, it is called the Shilov boundary. Obviously, K is a Shilov set for F. Notice that if F contains the constants and u + c ∈ F whenever u ∈ F and c ∈ R, S is a Shilov set if and only if for every u ∈ F with u ≥ 0 on S we have u ≥ 0 on K. Consider the following simple example of the space K := {x1 , x2 , x3 } with the discrete topology. We identify functions on K with vectors in R3 and denote F := {(1, 0, 0), (0, 1, 0), (0, 0, 1)} . Then every two-point subset of K is a Shilov set for F. Consequently, there exists no smallest Shilov set for F.
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10 Deeper results on function spaces and compact convex sets
The existence of a smallest Shilov set is guaranteed under additional assumptions imposed on F. It is worth mentioning that this is the case when a minimum principle like in Theorem 3.16 holds. Then the Shilov boundary is simply the closure of the Choquet boundary. In this section, however, we provide a direct proof of the existence of a Shilov boundary. Theorem 10.2. Let K be a nonempty compact space, F be a family of lower semicontinuous functions and S ⊂ F be a convex cone of continuous functions containing the constant functions and separating points of K such that u + s ∈ F whenever u ∈ F and s ∈ S. Then there exists the Shilov boundary for F. Proof. A moment’s reflection shows that we may suppose that both F and S are minstable (otherwise we would consider a new family W(F) and a new convex cone W(S) formed by finite minima of elements of F and S, respectively). Claim. The system of Shilov sets is closed with respect to finite intersections. Proof of the claim. Let S and T be Shilov sets. Suppose that u ∈ F and u ≥ 0 on S ∩ T . Fix ε > 0 and define U := {x ∈ K : u(x) + ε > 0} . Then U is an open set containing S ∩ T . In order to show that S ∩ T is a Shilov set, it is sufficient to prove that v := u + ε ≥ 0 on K. Since L := S \ U and T are disjoint compact sets, there are open sets V and W such that L ⊂ V, T ⊂ W and V ∩ W = ∅. Obviously, D := {s − t : s, t ∈ S} is a vector lattice of continuous functions on K containing constant functions and separating points of K. By the Stone–Weierstrass theorem A.30, there exists g ∈ D such that g ≥ 1 on V and g ≤ −1 on W . Since g + = g ∨0 ∈ D, there exist s, t ∈ D such that s−t = g + . We may suppose that s ≥ 0 and v + s ≥ 0 on K, because S contains positive constants. Choose a ∈ (0, 1) such that (1 − a)s ≤ 1 on K. Then s − t ≥ 1 ≥ (1 − a)s
on V,
hence t ≤ as on V , and obviously s = t on W . Let B := {b ∈ [0, ∞) : v + bs ≥ 0 on K} .
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313
Fix b ∈ B. Then 0 ≤ v + bs = v + bt on W and since T ⊂ W is a Shilov set, we have 0 ≤ v + bt on K. In particular, 0 ≤ v + bt ≤ v + bas
on V.
Since v ≥ 0 on U and s ≥ 0 on K, we have 0 ≤ v + bas
on
U ∪ V ⊃ S.
By the assumption, S is a Shilov set, and thus we have 0 ≤ v + bas on K. We conclude that b ∈ B implies ba ∈ B. Since 1 ∈ B, we get a ∈ B and by iteration an ∈ B, hence 0 ≤ v + an s on K for every n ∈ N. Since a ∈ (0, 1), we get v ≥ 0 on K. This establishes the claim. Define now Σ := {S ⊂ K : S is a Shilov set for F} and denote S0 :=
\
{S : S ∈ Σ}.
We claim that S0 is a Shilov set. Choose u ∈ F, u ≥ 0 on S0 , fix ε > 0 and define, as above, the open set U := {x ∈ K : u(x) + ε > 0} . Since U ⊃ S0 and Σ is closed with respect to finite intersections, there exists T0 ∈ Σ with T0 ⊂ U . It follows that u + ε ≥ 0 on T0 . We conclude that u + ε ≥ 0 on K whenever ε > 0, thus u ≥ 0 on K. Consequently S0 is the smallest Shilov set for F, finishing the proof. Theorem 10.3. If H is a function space on a compact space K, then ChH K is the Shilov boundary for H. Proof. By Theorem 10.2, there exists the Shilov boundary S for H. By the Minimum principle 3.16, ChH (K) is a Shilov set for H. To show its minimality, assume that there exists x ∈ ChH (K) \ S. Then x ∈ / coH S (see Proposition 8.18), and thus there exists a function h ∈ H such that h(x) > max h(S) (see Proposition 8.23). But this contradicts the fact that S is a Shilov set.
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10 Deeper results on function spaces and compact convex sets
10.1.B
Boundaries in Banach spaces
Throughout this subsection, E denotes a Banach space. Definition 10.4 (I-envelope). If E is a Banach space and B ⊂ E ∗ is a set, we define its I-envelope as I-env(B) :=
∞ ∞ o \n [ [ ∗ Bn . cok·k cow (Bn ) : Bn ⊂ B, B = n=1
n=1
Lemma 10.5. If B ⊂ I-env(B) =
E∗
is a set, then
∞ ∞ o [ \n [ ∗ cok·k cow (Bn ) : B1 ⊂ B2 ⊂ · · · ⊂ B, B = Bn . n=1
n=1
Proof. Let x∗ ∈ / I-env(B), that is, there exist d > 0 and sets Bn ⊂ B, n ∈ N, such that ! ∞ [ w∗ ∗ co Bn = ∅. B(x , d) ∩ co n=1 ∗ By replacing the sets Bn by Bn ∩ kB SE , n, k ∗∈ N, if necessary, we may assume that the sets Bn are bounded. Since co( nk=1 cow (Bk )) is a w∗ -compact convex set for each n ∈ N, the set ! ! ∞ n [ [ w∗ w∗ w∗ co (B1 ∪ · · · ∪ Bn ) ⊂ co co Bk ⊂ co co Bn
n=1
k=1
does not intersect B(x∗ , d). Hence the sequence {B1 ∪ · · · ∪ Bn }∞ n=1 witnesses that x∗ ∈ /
\
cok·k
∞ [ ∗ Bn . cow (Bn ) : B1 ⊂ B2 ⊂ · · · ⊂ B, B =
∞ [
n=1
n=1
This concludes the proof. Proposition 10.6. For x∗ ∈ E ∗ and B ⊂ E ∗ , the following assertions are equivalent: (i) x∗ is not contained in I-env(B), (ii) there exists a sequence {xn } in BE such that inf x∗ (xn ) > sup lim sup b∗ (xn ).
n∈N
b∗ ∈B n→∞
Proof. For the proof of (ii) =⇒ (i), let x∗ and {xn } be as in (ii). We find c1 , c2 ∈ R satisfying inf x∗ (xn ) > c1 > c2 > sup lim sup b∗ (xn ). n∈N
b∗ ∈B n→∞
10.1 Boundaries
By setting Bn :=
∞ \
{b∗ ∈ E ∗ : b∗ (xk ) ≤ c2 },
315
n ∈ N,
k=n
S we get an increasing sequence of w∗ -closed convex sets satisfying B = ∞ n=1 Bn ∩B. S∞ S ∗ moreover, the norm distance of x∗ to this Hence n=1 cow Bn = ∞ B and, n n=1 union is greater than c1 − c2 . Indeed, assume the existence of b∗ in some Bn with kx∗ − b∗ k ≤ c1 − c2 . Then, for k ≥ n, we have x∗ (xk ) ≤ kx∗ − b∗ k + b∗ (xk ) ≤ c1 − c2 + c2 = c1 , a contradiction. S Hence x∗ is not contained in the norm closure of the convex hull of ∞ n=1 Bn , and ∗ therefore x ∈ / I-env(B). For the proof of the converse implication (i) =⇒ (ii), assume that x∗ ∈ / I-env(B). Without loss of generality we may assume that x∗ = 0. By the assumption and Lemma 10.5, there exist d > 0 and an increasing sequence {Bn } such that ! ∞ ∞ [ [ ∗ cow Bn = ∅. B= Bn and B(0, d) ∩ co n=1
n=1
By the Hahn–Banach separation theorem there exist elements xn ∈ E so that sup
b∗ (xn ) <
b∗ ∈B(0,d)
inf∗
b∗ ∈cow Bn
b∗ (xn ),
n ∈ N.
Without loss of generality we may assume that all points xn are of norm 1. Hence 0 = x∗ (xn ) < d =
sup
b∗ (xn ) <
b∗ ∈B(0,d)
b∗ (xn ),
inf
∗ b∗ ∈cow
Bn
n ∈ N.
Thus {xn } is the sought sequence. Theorem 10.7. Let B be a subset of a w∗ -compact convex set X ⊂ E ∗ such that each element x ∈ E attains its maximum on X at some point of B. Then X = I-env(B). Proof. Assume that there exists x∗ ∈ X \ I-env(B). By Proposition 10.6, there exist a sequence {xn } in BE and c1 , c2 ∈ R such that inf x∗ (xn ) > c1 > c2 > sup lim sup b∗ (xn ).
n∈N
b∗ ∈B n→∞
By the P Simons inequality 3.74, there exists a sequence {λn } of positive numbers such that ∞ n=1 λn = 1 and ∞ X sup y ∗ λn xn < c2 . y ∗ ∈X
n=1
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10 Deeper results on function spaces and compact convex sets
On the other hand, ∗
c1 < x
∞ X
λn xn ≤ sup y
∗
∞ X
y ∗ ∈X
n=1
λ n xn ,
n=1
a contradiction. Hence the assertion follows. Theorem 10.8. Let B be a norm separable subset of a w∗ -compact convex set X in E ∗ such that each element of E attains its maximum on X at some point of B. Then X is norm separable as well. ∗ : n ∈ N} ⊂ B be a norm Proof. Let B ⊂ X be as in the hypothesis. n S∞ Let {b dense subset and let ε > 0. Since B ⊂ n=1 B(b∗n , ε) and closed balls in E ∗ are w∗ -compact and convex, Theorem 10.7 yields
X ⊂ cok·k
∞ [
B(b∗n , ε) .
n=1
Hence the closed convex hull of {b∗n : n ∈ N} is norm dense in X, from which the theorem follows. Theorem 10.9 (James’ theorem for separable spaces). Let B ⊂ E be a closed convex subset of a separable Banach space E such that each element of E ∗ attains its supremum on B. Then B is weakly compact. Proof. If B is as in the hypothesis, we first notice that B is bounded since it is weakly bounded (see the uniform boundedness principle, Theorem 3.15 in [173]) and weakly w∗ closed by the Mazur theorem A.2. Thus we need to show that B ⊂ E ∗∗ is a subset w∗ of E. Indeed, if this is the case, B is a w∗ -compact subset of E considered as a w w∗ subset of E ∗∗ , and thus B = B = B is weakly compact. w∗ To this end, let ϕ ∈ B be given. By Theorem A.7, it is enough to show that ϕ is w∗ -continuous on BE ∗ . Since E is separable, it suffices to check the w∗ -sequential continuity of ϕ|BE∗ . Let {x∗n } be a sequence in BE ∗ w∗ -converging to 0. Assume that {ϕ(x∗n )} does not converge to 0. Without loss of generality we may assume that there exists c > 0 such that ϕ(x∗n ) > c for all n ∈ N. If we consider w∗ w∗ points of E ∗ as functions on B , the assumption ensures that B ⊂ B is a boundary for E ∗ . Hence B is a boundary for any x∗ ∈ coσ {x∗n : n ∈ N}. Since x∗n (b) → 0 for each b ∈ B, Corollary 3.74 gives 0 = sup lim sup x∗n (b) = sup lim sup ψ(x∗n ) ≥ lim sup ϕ(x∗n ) ≥ c, b∈B n→∞
ψ∈B
w∗
n→∞
a contradiction. Thus ϕ(x∗n ) → 0, which is the desired conclusion.
n→∞
10.1 Boundaries
317
Definition 10.10 (Relatively countably compact and angelic spaces). We recall that a subset A of a topological space X is relatively countably compact if any sequence {xn } of elements of A has a cluster point in X. A topological space X is angelic if each relatively countably compact set A ⊂ X is relatively compact and any point x ∈ A can be obtained as a limit of a sequence of points from A. Theorem 10.11. If X is a compact space, then (C(X), τX ) is angelic. Proof. Let A ⊂ C(X) be a relatively τX -countably compact set. Step 1: The set A is relatively τX -compact. It follows from the assumption that {f (x) : f ∈ A} is a bounded subset of R for every x ∈ X. By the Tychonoff theorem, we need to show that the τX -closure of A in RX is contained in C(X). To τ achieve this, assume that g ∈ A X is not continuous, say at a point x0 ∈ X. Then there exists ε > 0 such that for each open neighborhood U of x0 there exists y ∈ U so that |g(x0 )−g(y)| ≥ ε. We construct inductively points xn ∈ X, open neighborhoods Un of x0 and functions fn ∈ A, n ∈ N, such that, for each n ∈ N, (a) |g(xn ) − g(x0 )| ≥ ε, (b) |fn (xi ) − g(xi )| ≤ 2−n , i = 0, . . . , n, (c) xi ∈ Un , i ≥ n + 1, (d) Un+1 ⊂ Un and diam fn (Un ) ≤ 2−n . To start the construction, let x1 ∈ X be a point such that |g(x1 ) − g(x0 )| ≥ ε and let f1 ∈ A satisfy |f1 (xi ) − g(xi )| < 2−1 , i = 0, 1. We find a neighborhood U1 of x0 such that diam f1 (U1 ) ≤ 2−1 . This finishes the first step of the induction. Assume that the construction has been completed up to the n-th stage. We find xn+1 ∈ Un such that |g(xn+1 ) − g(x0 )| ≥ ε. Let fn+1 ∈ A be chosen in such a way that |fn+1 (xi ) − g(xi )| ≤ 2−(n+1) for i = 0, . . . , n + 1. Let Un+1 be a neighborhood of x0 such that Un+1 ⊂ Un and diam fn+1 (Un+1 ) ≤ 2−(n+1) . This concludes the construction. Let f be a τX -cluster point of {fn } and x be a cluster point of {xn }. For any i ∈ N, property (b) gives f (xi ) = g(xi ), and hence |f (xi ) − g(x0 )| ≥ ε by (a). By the continuity of f , |f (x) − g(x0 )| ≥ ε. It follows from property (c) that, for a fixed n ∈ N, x ∈ Un . By (d), |fn (x) − fn (x0 )| ≤ 2−n , and thus f (x) = f (x0 ). But by (b) again, f (x0 ) = g(x0 ). Hence we have arrived at a contradiction, because g(x0 ) = f (x0 ) = f (x) 6= g(x0 ). τ
This concludes the first step of the proof. Let now g ∈ A X be given. Step 2: There exists a sequence {fn } in A pointwise converging to g. To find a desired sequence, we construct countable sets Y ⊂ X and B ⊂ A such that
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10 Deeper results on function spaces and compact convex sets
(e) for any I ⊂ B ∪ {g} finite, ε > 0 and x ∈ X there exists y ∈ Y such that |f (y) − f (x)| ≤ ε for each f ∈ I, (f) for any I ⊂ Y finite and ε > 0 there exists f ∈ B such that |f (x) − g(x)| ≤ ε for every x ∈ I. The construction will be done inductively. First we notice that, for any finite set I ⊂ C(X), the set {(f (x))f ∈I : x ∈ X} is a separable subset of RI considered with the supremum norm. Hence there exists a countable set YI ⊂ X such that {(f (y))f ∈I : y ∈ Y } is dense in {(f (x))f ∈I : x ∈ X}. If I ⊂ X is finite, we can choose a sequence {fj(I) } from A pointwise converging to g on I. Let BI consists of the elements of the sequence {fj(I) }. We set inductively Yn :=
n−1 [n [ o YI : I ⊂ {g} ∪ Bi and
Bn :=
n−1 [n [ o BI : I ⊂ Yi ,
i=1
n ∈ N.
i=1
S∞
S∞
At the end of the construction we set Y := n=1 Yn and B := n=1 Bn . Then the requirements (e) and (f) are satisfied. By (f), there exists a sequence {fj } of functions from A pointwise converging to g on Y . We claim that fj (y) → g(y) for all y ∈ X. Assume that this is not the case. Then there exist y ∈ X and ε > 0 such that the set {j ∈ N : |fj (y) − g(y)| ≥ ε} is infinite. For I = {f1 , . . . , fm , g} we use (e) to find ym ∈ Y satisfying |u(ym ) − u(y)| < 2−m ,
u ∈ I.
(10.1)
Let x be a cluster point of {ym : m ∈ N} and h be a τX -cluster point of {fj : j ∈ N}. For each ym , fj (ym ) → g(ym ), and hence h(ym ) = g(ym ). By continuity, h(x) = g(x). Further, for each fi we have from (10.1) that fi (x) = fi (y), and thus h(x) = h(y). Also by (10.1), g(x) = g(y). Finally we notice that |g(y) − h(y)| ≥ ε. Putting all these information together, we get g(y) = g(x) = h(x) = h(y) 6= g(y), a contradiction. Hence fj → g as required, and the proof is finished. Theorem 10.12. Let X be a compact convex set. Then (BAc (X) , τext X ) is an angelic space. In particular, any norm bounded relatively τext X -countably compact set A is relatively τX -compact. Proof. Let A ⊂ BAc (X) be relatively τext X -countably compact. Since (C(X), τX ) is an angelic space, to conclude that A is relatively τX -compact it suffices to show that any sequence {fn } in A has a τX -cluster point. We select a τext X -cluster point
10.1 Boundaries
319
f0 of {fn } and show that f0 is also a τX -cluster point of {fn }. First we notice that kf0 k ≤ 1 by the minimum principle of Corollary 2.24. If we assume a contrary, after omitting finitely many elements from {fn } we may assume that there exist x1 , . . . , xk ∈ X and ε > 0 such that {fn : n ∈ N} ∩ {g ∈ C(X) : |f0 (xi ) − g(xi )| < ε, i = 1, . . . , k} = ∅.
(10.2)
We define ϕ : X → RN as ϕ(x) := {fn (x)}∞ n=0 , x ∈ X, and set Y := ϕ(X). Then Y is a metrizable compact convex set and there exist functions gn ∈ Ac (Y ) such that fn = gn ◦ ϕ, n ≥ 0 (namely, the coordinate functions). For i = 1, . . . , k, let yi := ϕ(xi ) and let νi ∈ Myi (Y ) be a maximal measure. By Theorem 3.79 there exists a compact set K ⊂ ext Y such that νi (K) > 1 − 4ε , i = 1, . . . , k. Pick y ∈ K. Then the set F := ϕ−1 (y) is a face of X (see Proposition 2.72), and thus ext F = F ∩ ext X by Proposition 2.64(b). Further, all functions fn are constant on F . It follows that if g is a continuous affine function on X contained in τext X τF , then g|F ∈ {fn |F : n ∈ N} . {fn : n ∈ N} Thus the set {fn |ϕ−1 (K) : n ∈ N} is relatively τϕ−1 (K) -countably compact. By Theorem 10.11, there exists a subsequence {fnk } of {fn } so that fnk → f0 on ϕ−1 (K). By the Lebesgue dominated convergence theorem there exists k0 ∈ N such that Z Z ε gn (t) dνi (t) − g0 (t) dνi (t) < , i = 1, . . . , k. k0 2 K K Then for i = 1, . . . , k, Z Z gnk0 (t) dνi (t) − g0 (t) dνi (t) |fnk0 (xi ) − f0 (xi )| = ext Y ext Y Z Z ≤ gnk0 (t) dνi (t) − g0 (t) dνi (t) K K Z + |gnk0 (t) − g0 (t)| dνi (t) ext Y \K
ε ε < + 2 = ε. 2 4 But this contradicts (10.2) and shows that A is relatively τX -compact. τ Thus A X is τX -compact, and hence also τext X -compact; in particular τext X -closed. Thus τ τ τ A ext X ⊂ A X ⊂ A ext X . τ
τ
It follows that the identity mapping id : (A ext X , τX ) → (A ext X , τext X ), as a continuous injective mapping, is a homeomorphism. Since the topology τX is angelic, τext X τ is angelic on A ext X , which concludes the proof.
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10 Deeper results on function spaces and compact convex sets
Corollary 10.13. Let E be a Banach space and A be a norm bounded relatively σ(E, ext BE ∗ )-countably compact set. Then A is relatively weakly compact. Proof. Let A be a norm bounded relatively σ(E, ext BE ∗ )-countably compact set in a Banach space E. Then A can be viewed as a relatively τext BE∗ -countably compact subset of the space Ac (BE ∗ ) (here BE ∗ is endowed with the w∗ -topology). Moreover, the weak topology coincide with the topology τBE∗ (see Proposition 4.31(f)). Hence the assertion follows from Theorem 10.12.
10.2
Isometries of spaces of affine continuous functions
Theorem 10.14. Let X, Y be compact convex sets and T : Ac (X) → Ac (Y ) be a positive isometry. Then X is affinely homeomorphic to Y . Proof. If T is as in the hypothesis, the dual operator T ∗ : (Ac (Y ))∗ → (Ac (X))∗ is a positive isometry as well and, moreover, it is a (w∗ –w∗ )-homeomorphism. Hence T ∗ is an affine homeomorphism of S(Ac (Y )) to S(Ac (X)). To see its surjectivity, let s ∈ S(Ac (X)) be arbitrary. Then t : g 7→ s(T −1 g), g ∈ Ac (Y ), is in S(Ac (Y )) and T ∗ t = s. Using the identification of X with S(Ac (X)) and Y with S(Ac (Y ) (see Proposition 4.31(a)) we finish the proof. Lemma 10.15. Let X1 , X2 be compact convex sets and, for i = 1, 2, let Fi be closed faces of Xi such that •
Fi is a split face of Xi ,
•
the complementary face Fi0 is closed.
Let ϕ : F1 ∪ F10 → F2 ∪ F20 be a continuous mapping that is affine on F1 and F10 and maps F1 to F2 and F10 to F20 . Then there exists a unique continuous affine mapping ψ : X1 → X2 extending ϕ. Moreover, if ϕ is injective or surjective, then ψ is injective or surjective, respectively. Proof. Given x ∈ X1 , let ψ(x) be defined as follows. Let x = αx1 + (1 − α)x2 be the unique decomposition of x with α ∈ [0, 1] and x1 ∈ F1 , x01 ∈ F10 (see Definition 8.4). We set ψ(x) := αϕ(x1 ) + (1 − α)ϕ(x2 ). It is a matter of routine verification that ψ : X1 → X2 is a well-defined affine mapping. To show its continuity, pick g ∈ Ac (X2 ). Then g ◦ ψ is an affine function that is continuous on F1 ∪ F10 . By Lemma 5.39, g ◦ ψ is continuous on X1 . Hence ψ is continuous with respect to the weak topology on X2 . Since X2 is compact, ψ is continuous.
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10.2 Isometries of spaces of affine continuous functions
Assume that ϕ is injective on F1 ∪ F10 and let ψ(x) = ψ(y) for some x, y ∈ X1 . Choose α, β ∈ [0, 1], x1 , y1 ∈ F1 and x2 , y2 ∈ F10 such that x = αx1 + (1 − α)x2
and
y = βy1 + (1 − β)y2 .
Then αϕ(x1 ) + (1 − α)ϕ(x2 ) = βϕ(y1 ) + (1 − β)ϕ(y2 ). Since F2 is a split face, this decomposition yields α = β and ϕ(xi ) = ϕ(yi ), i = 1, 2. Hence xi = yi , i = 1, 2, and x = y. If ϕ is surjective, then X2 = co(F2 ∪ F20 ) gives ψ(X1 ) = X2 . Lemma 10.16. Let X, Y be compact convex set and T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then the sets F1 := {y ∈ Y : T 1(y) = 1} and
F2 := {y ∈ Y : T 1(y) = −1}
are closed faces such that F1 is parallel and F2 is its complementary face. Proof. Since kT 1k = 1, both F1 , F2 are closed faces. We use the identifications from Proposition 4.31, namely, that span X = Ac (X)∗ and span Y = Ac (Y )∗ . Then T ∗ is an isometry, and thus T ∗ maps extreme points of BAc (Y )∗ onto ext BAc (X)∗ . Hence for any y ∈ ext Y there exists x ∈ ext X such that T ∗ y = x or T ∗ y = −x. Then either T 1(y) = T ∗ y(1) = x(1) = 1
or
T 1(y) = T ∗ y(1) = (−x)(1) = −1.
Hence Y = co ext Y = co(F1 ∪ F2 ). From this it follows that F10 = F2 (see Exercise 10.93). Let y = αy1 + (1 − α)y2 for yi ∈ Fi , i = 1, 2, and α ∈ [0, 1]. Then T 1(y) = T ∗ y(1) = αT ∗ (y1 )1 + (1 − α)T ∗ (y2 )1 = α − (1 − α) = 2α − 1. Hence the coefficient α is unique and F1 is a parallel face. Theorem 10.17. Let X, Y be compact convex sets with the following property: •
if F1 , F2 are closed faces such that F1 is parallel and F2 = F10 , then F1 is a split face.
Let T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then there exists an affine homeomorphism ϕ : Y → X and a function h ∈ Ac (Y ) such that T f (y) = h(y)f (ϕ(y)),
f ∈ Ac (X), y ∈ ext Y.
(10.3)
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10 Deeper results on function spaces and compact convex sets
Proof. If T : Ac (X) → Ac (Y ) is an isometry, we define G1 := {y ∈ Y : T 1(y) = 1},
G2 := {y ∈ Y : T 1(y) = −1}
and F1 := {x ∈ X : T −1 1(x) = 1},
F2 := {x ∈ X : T −1 1(x) = −1}.
By Lemma 10.16, G1 is a closed parallel face of Y such that the closed face G2 is its complementary face. Similarly, F1 is a closed parallel face of X such that F2 = F10 . By the assumption, both F1 and G1 are split faces. We use the identification of span X with Ac (X)∗ and span Y with Ac (Y )∗ (see Proposition 4.31). We claim that T ∗ (G1 ) = F1 and T ∗ (G2 ) = −F2 . Indeed, for any y ∈ G1 we see from T ∗ y(1) = T 1(y) = 1, that T ∗ y is a positive element of norm 1 in Ac (X)∗ and thus it belongs to X. Moreover, (T −1 1)(T ∗ y) = ((T −1 )∗ T ∗ y)(1) = 1. Thus T ∗ y ∈ F1 . If x ∈ F1 is given, the equalities ((T −1 )∗ x)(1) = T −1 1(x) = 1 yield that (T −1 )∗ x ∈ Y . Further, T 1((T −1 )∗ x) = ((T −1 )∗ x)(T 1) = (T ∗ (T −1 )∗ x)(1) = 1. Hence (T −1 )∗ x ∈ G1 and T ∗ (G1 ) = F1 . Similarly we obtain T ∗ (G2 ) = −F2 . Let φ : G1 ∪ G2 → F1 ∪ F2 be defined as ( T ∗ y, y ∈ G1 , φ(y) := ∗ −T y, y ∈ G2 . Then φ is continuous, injective, surjective and affine on both G1 and G2 . Using Lemma 10.15 we find an affine homeomorphism ϕ : Y → X extending φ. We set h := T 1 and show the validity of (10.3). Pick f ∈ Ac (X) and y ∈ ext Y . If y ∈ G1 , ϕ(y) = φ(y) ∈ F1 , and thus T f (y) = T ∗ y(f ) = f (φ(y)) = T 1(y)f (φ(y)). Similarly, for y ∈ G2 we get T f (y) = T ∗ y(f ) = −f (φ(y)) = T 1(y)f (φ(y)). Thus T f (y) = h(y)f (ϕ(y)) for all y ∈ ext Y , and the proof is complete.
10.3 Baire measurability and boundedness of affine functions
323
Theorem 10.18. Let X, Y be simplices and T : Ac (X) → Ac (Y ) be an isometric isomorphism. Then X is affinely homeomorphic to Y . Proof. The proof follows from Theorem 10.17 and Corollary 8.63. Corollary 10.19 (Banach–Stone). Let K, L be compact spaces and T : C(K) → C(L) be an isometric isomorphism. Then K is homeomorphic to L. Proof. If T is as in the hypothesis, it induces an isometry between Ac (M1 (K)) and Ac (M1 (L)). Since both sets M1 (K) and M1 (L) are Bauer simplices (see Proposition 6.38), Theorem 10.18 yields that they are affinely homeomorphic. Hence their sets of extreme points are homeomorphic and Proposition 2.27 finishes the proof.
10.3 10.3.A
Baire measurability and boundedness of affine functions The Cantor set and its properties
We recall the following notation from Section A.4. We write N
s ∈ {0, 1}
Further, {0, 1}N can be as a set identified with the family of all subsets of N as follows. A set A ⊂ N is mapped to its characteristic function cA ∈ {0, 1}N . Thus the group operation + on {0, 1}N is nothing else than the symmetric difference of sets. Further, the mapping σ 7→ N \ σ (= σ + 1),
σ ∈ {0, 1}N ,
is a homeomorphism of {0, 1}N onto itself. As usually, −σ is the inversion of an element σ ∈ {0, 1}N (of course, −σ = σ). compact group {0, 1}N possesses a unique Haar measure λ, namely λ = NThe ∞ n=1 λn is the Radon product measure (see Definition A.95), where each λn is a
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10 Deeper results on function spaces and compact convex sets
copy of the measure on {0, 1} assigning to both {0} and {1} measure 12 . We point out the fact that λ, as a Haar measure, is invariant with respect to the operation +. Further, λ(Us ) = 2−|s| , s ∈ {0, 1}
(σ, τ ) 7→ σ ∩ τ,
(σ, τ ) 7→ σ,
(σ, τ ) 7→ N \ σ,
are open continuous surjections. Proof. The proof follows by a straightforward verification. Lemma 10.21. The following assertions hold: (a) If Q ⊂ {0, 1}N is dense countable and A ⊂ {0, 1}N is comeager, then {σ ∈ A : (σ + Q) ∩ A dense in {0, 1}N } is comeager in {0, 1}N . (b) If Q ⊂ {0, 1}N is dense countable, f : {0, 1}N → R has the Baire property and f (σ + τ ) = f (σ), σ ∈ {0, 1}N , τ ∈ Q, then there exists a comeager B ⊂ {0, 1}N such that f is constant on B. Proof. For the proof of (a), we notice that τ + A is comeager in {0, 1}N for every τ ∈ Q, and hence \ B := A ∩ (A − τ ) τ ∈Q
is a comeager set as well. We claim that (σ + Q) ∩ A is dense for each σ ∈ B. Indeed, let σ ∈ B and V ⊂ {0, 1}N open be given. Since Q is dense, V − Q = {0, 1}N , and thus there exist v ∈ V and τ ∈ Q such that v − τ = σ. Hence v = σ + τ ∈ (σ + Q) ∩ A ∩ V. It follows that (σ + Q) ∩ A intersects any open set, and (a) is proved. To prove (b), let f : {0, 1}N → R be as in the hypothesis. Since f has the Baire property, we can find a comeager set A ⊂ {0, 1}N such that f |A is continuous (see Proposition A.61). Using (a) we find σ ∈ {0, 1}N such that (σ + Q) ∩ A is dense in {0, 1}N , and hence also in A. Since f is continuous on A and f (σ + τ ) = f (σ) for each τ ∈ Q, f is constant on A. This concludes the proof. Definition 10.22 (Q-invariance and tail sets). We say that a function f : {0, 1}N → R is Q-invariant, if f (σ) = f (τ ) whenever σ differs from τ at finitely many coordinates, σ, τ ∈ {0, 1}N . A set A ⊂ {0, 1}N is a tail set if cA is Q-invariant.
10.3 Baire measurability and boundedness of affine functions
325
Corollary 10.23. Let f : {0, 1}N → R have the Baire property and be Q-invariant. Then f is constant on a comeager subset of {0, 1}N . Proof. The proof follows from Lemma 10.21(b), because the set Q := {σ ∈ {0, 1}N : σ has finite support} is a countable dense set. Theorem 10.24 (0–1 laws). Let A ⊂ {0, 1}N be a tail set. (a) If A has the Baire property, then A is either meager or comeager. (b) If A is λ-measurable, then λ(A) is either 0 or 1. Proof. For the proof of (a), we notice that f := cA is Q-invariant because A is a tail set. By Corollary 10.23, f is constant on a comeager set. From this the assertion follows. For the proof of (b), assume that A is a λ-measurable tail set. We set A := {B ⊂ {0, 1}N : B is λ-measurable and λ(A ∩ B) = λ(A)λ(B)} N and show Q that A contains all λ-measurable Qn sets. Let B ⊂ {0, 1} be of the form ∞ B =Q F × k=n+1 {0, 1}, where F ⊂ Q k=1 {0, 1}. Since A is a tail set, we0 can write A = nk=1 {0, 1} × A0 , where A0 ⊂ ∞ k=n+1 {0, 1}. Then A ∩ B = F × A . We write 0 λ for the measure on {0, 1} assigning to each element 21 and compute
λ(A ∩ B) =
n O
λ0 (F ) ·
k=1
=
n O
∞ O
λ0 (A0 )
k=n+1 0
λ (F ) ·
k=1
∞ O k=n+1
0
λ ({0, 1}) ·
n O k=1
0
λ ({0, 1}) ·
∞ O
λ0 (A0 )
k=n+1
= λ(B)λ(A). Since any Q λ-measurable set B ⊂ {0,Q1}N can be approximated by the sets of the n form F × ∞ k=n+1 {0, 1}, where F ⊂ k=1 {0, 1} (see Proposition A.98), we get the required conclusion. Now, since A ∈ A, we obtain λ(A) = λ(A)λ(A), which concludes the proof. Definition 10.25 (Finitely additive and countably additive functions). A function µ : {0, 1}N → R is called finitely additive if µ is a finitely additive measure on N, when we view {0, 1}N as the family of all subsets of N. If µ is a countably additive measure on N, we say that µ is countably additive. Theorem 10.26. Let µ : {0, 1}N → R be finitely additive and have the Baire property in the restricted sense. Then µ is a continuous function and countably additive.
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10 Deeper results on function spaces and compact convex sets
Proof. We divide the proof in several steps. Let Q stand for the set {σ ∈ {0, 1}N : σ has finite support}. Step 1: The function µ|Q is uniformly continuous; that is, for every ε > 0 there exists k ∈ N such that |µ(α) − µ(β)| < ε whenever α|k = β|k , α, β ∈ Q. For the proof, we use the Baire property of µ to find a comeager set A ⊂ {0, 1}N such that f |A is continuous. Since \ A∩ (A − τ ) τ ∈Q
is comeager, we can find σ0 ∈ A such that σ0 + τ ∈ A for every τ ∈ Q. Let α, β ∈ Q be arbitrary. We set α0 := α ∩ (N \ β) ∩ (N \ σ0 ),
α1 := α ∩ (N \ β) ∩ σ0 ,
β0 := β ∩ (N \ α) ∩ (N \ σ0 ),
β1 := β ∩ (N \ α) ∩ σ0 .
Then a straightforward verification yields µ(α) − µ(β) = µ(α \ β) − µ(β \ α) = µ(σ0 + α0 ) − µ(σ0 + α1 ) − µ(σ0 + β0 ) + µ(σ0 + β1 ). Since σ0 + αi and σ0 + βi , i = 0, 1, are contained in A and µ|A is continuous at σ0 , the claim follows. Indeed, for a given ε > 0 we find k ∈ N such that |µ(σ) − µ(σ0 )| < ε for σ ∈ A with σ|k = σ0 |k . Then, for α, β ∈ Q satisfying α|k = β|k , we have αi |k = βi |k = 0|k ,
i = 0, 1,
and thus (σ0 + αi )|k = (σ0 + βi )|k = σ0 |k ,
i = 0, 1.
Hence |µ(α) − µ(β)| < 4ε. Let now ν : {0, 1}N → R be the continuous extension of µ|Q . Step 2: The function ν is countably additive. First we show that ν is finitely additive. Let σ, τ ∈ {0, 1}N have disjoint supports. We set σj := (σ(1), . . . , σ(j), 0, . . . ) and τj := (τ (1), . . . , τ (j), 0, . . . ), j ∈ N. Then ν(σ + τ ) = lim µ(σj + τj ) = lim µ(σj ) + µ(τj ) = ν(σ) + ν(τ ), j→∞
j→∞
and ν is finitely additive. If {σn } are elements of {0, 1}N with pairwise disjoint supports, from the continuity of ν we get ν
∞ X n=1
j j ∞ X X X σn = lim ν σn = lim ν(σn ) = ν(σn ), j→∞
and ν is countably additive.
n=1
j→∞
n=1
n=1
10.3 Baire measurability and boundedness of affine functions
327
We finish the proof by showing µ = ν on {0, 1}N . Fix σ0 ∈ {0, 1}N and let N ⊂ N be the support of σ0 . We identify {0, 1}N = {σ ∈ {0, 1}N : σ has support in N }. We have to verify that µ(σ0 ) = ν(σ0 ). To this end we prove the following assertion. Step 3: We have µ = ν on {0, 1}N . Let ω := µ − ν. According to the assumption, ω is a finitely additive function with the Baire property in the restricted sense on {0, 1}N that equals 0 on Q ∩ {0, 1}N . Hence ω is a Q-invariant function on {0, 1}N , and thus there exist a comeager set A ⊂ {0, 1}N and a ∈ R such that ω = a on A (see Lemma 10.21(b)). By Lemma 10.20, the set B := {(σ, τ ) ∈ {0, 1}N × {0, 1}N : σ, N \ σ, σ \ τ, σ ∩ τ ∈ A} is comeager in {0, 1}N × {0, 1}N . In particular, we can pick a point (σ, τ ) ∈ B. Then a = ω(σ) = ω(σ \ τ ) + ω(σ ∩ τ ) = 2a yields a = 0. Hence ω(σ0 ) = ω(N \ σ) + ω(σ) = 2a = 0, which is the desired conclusion. Definition 10.27 (Filters and ultrafilters). If X is a set, a system F of subsets of X is called a filter, if •
A1 ∩ A2 ∈ F whenever A1 , A2 ∈ F,
•
∅∈ / F,
•
A2 ∈ F whenever A1 ∈ F and A1 ⊂ A2 .
A filter F is called an ultrafilter (or a maximal filter) if for every filter F 0 containing F we have F = F 0 . T An ultrafilter U is called a free ultrafilter if U = ∅. It follows from Zorn’s lemma that every filter is contained in an ultrafilter and it is easy to see that there are free ultrafilters on N. Lemma 10.28. There exists an element f ∈ (`∞ )∗ whose restriction to (B`∞ , w∗ ) is not universally measurable. Proof. Consider the Cantor set {0, 1}N as a subset of (B`∞ , w∗ ) and let λ stand for the Haar measure on {0, 1}N . We select a free ultrafilter U on N and define f : `∞ → R as f (x) := limU x, x ∈ `∞ (equivalently, f (x) = x b(U), where x b is the continuous ˇ extension of x to the Cech–Stone compactification of N).
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10 Deeper results on function spaces and compact convex sets
In particular, for σ ∈ {0, 1}N we have ( 1, f (σ) = 0,
σ ∈ U, σ∈ / U.
Since f = cU on {0, 1}N , λ-measurability of f implies λ-measurability of U. Hence it suffices to show that U considered as a subset of {0, 1}N is not λ-measurable. To this end we notice that U is a tail set, that is, if σ differs from τ at finitely many coordinates, then σ ∈ U if and only if τ ∈ U. So if we assume that U is λ-measurable, Theorem 10.24 yields λ(U) ∈ {0, 1}. We consider now the mapping ϕ : {0, 1}N → {0, 1}N defined as ϕ(σ) := 1 − σ,
σ ∈ {0, 1}N ,
(equivalently, σ is mapped to its complement in N). Then ϕ is a homeomorphism of {0, 1}N onto itself and λ is ϕ-invariant, that is ϕ] λ = λ. It follows from the properties of ultrafilters that ϕ(U) ∪ U = {0, 1}N
and
ϕ(U) ∩ U = ∅.
Hence λ(U) = 21 , a contradiction. This concludes the proof.
10.3.B
Automatic boundedness of affine and convex functions
Definition 10.29 (Strong linear independency). A sequence {xn } in a locally convex 1 space is strongly linearly Pn independent if the following condition holds: if λ ∈ ` satisfies w-limn→∞ i=1 λi xi = 0, then λ = 0. Theorem 10.30. Let {xn } be a linearly independent sequence in a locally convex space E that converges weakly to 0. Then there exists a partition of N into sets N1 , N2 such that both sequences {xn }n∈N1 and {xn }n∈N2 are strongly linearly independent. Proof. Let {xn } be as in the hypothesis. We select a countable set {un : n ∈ N} of finitely supported elements in c0 that is dense in c0 . Let Q := {σ ∈ {0, 1}N : σ is finitely supported}. For σ ∈ {0, 1}N \ Q, we assign a sequence {pσ (k)}, where pσ (k) is the position of the k-th 1 in σ. For each n ∈ N, we define n Mn := σ ∈ {0, 1}N \ Q : there exists x∗ ∈ E ∗ such that for each k ∈ N 1o . we have |un (k) − x∗ (xpσ (k) )| ≤ n
10.3 Baire measurability and boundedness of affine functions
329
Claim. Every set Mn is dense and open in {0, 1}N \ Q. Proof of the claim. First we check that each Mn is open in {0, 1}N \ Q. We pick σ ∈ Mn and find x∗ ∈ E ∗ such that for each k ∈ N we have |un (k)−x∗ (xpσ (k) )| ≤ n1 . We find an index j0 ∈ N such that •
un (j) = 0 for j ≥ j0 ,
•
|x∗ (xj )| ≤
1 n
for j ≥ j0 .
Let U be the open set in {0, 1}N \ Q consisting of all elements of {0, 1}N \ Q that coincide with σ on the first pσ (j0 ) coordinates. Let τ ∈ {0, 1}N \ Q be such a sequence. Let k ∈ N be given. If k ≤ j0 , then pσ (k) = pτ (k) and thus |un (k) − x∗ (xpτ (k) )| = |un (k) − x∗ (xpσ (k) )| ≤
1 . n
If k > j0 , un (k) = 0 and pτ (k) ≥ k > j0 . Hence |un (k) − x∗ (xpτ (k) )| = |x∗ (xpτ (k) )| ≤
1 . n
Thus the chosen set U is contained in Mn . Next we prove that Mn is dense in {0, 1}N \ Q. Let σ ∈ {0, 1}N \ Q be given and let U be a neighborhood of σ consisting of all elements τ ∈ {0, 1}N \ Q that coincide with σ on the first m coordinates. We may assume that m is so large that un (k) = 0 for k ≥ m. We claim that there exists x∗ ∈ E ∗ such that un (k) = x∗ (xpσ (k) ),
k = 1, . . . , m.
(10.4)
Indeed, if this is not the case, then the mapping ϕ : E ∗ → Rm defined as ϕ(x∗ ) := (x∗ (xpσ (1) ), . . . , x∗ (xpσ (m) )),
x∗ ∈ E ∗ ,
satisfies ϕ(E ∗ ) 6= Rm . Hence there exist real numbers c1 , . . . , cm such that 0=
m X
ci x∗ (xpσ (i) ) = x∗
i=1
n X
ci xpσ (i) ,
x∗ ∈ E ∗ .
i=1
Pn
Since E ∗ separates points of E, i=1 ci xpσ (i) = 0, which contradicts the linear independence of the sequence {xn }. Let x∗ ∈ E ∗ satisfy (10.4). We find m0 > m such that •
m0 > pσ (m),
•
|x∗ (xj )| ≤
1 n
for j ≥ m0 .
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10 Deeper results on function spaces and compact convex sets
Let τ ∈ {0, 1}N \ Q be defined as follows: j = 1, . . . , pσ (m), σ(j), τ (j) := 0, j = pσ (m) + 1, . . . , m0 , 1, j > m0 . Then τ ∈ U ∩ Mn . Indeed, let k ∈ N be given. If k ≤ m, then pτ (k) = pσ (k), and thus by (10.4) |un (k) − x∗ (xpτ (k) )| = |un (k) − x∗ (xpσ (k) )| = 0. If k > m, then pτ (k) > m0 ≥ pσ (m), and hence |un (k) − x∗ (xpτ (k) )| = |x∗ (xpτ (k) )| ≤
1 . n
Thus τ ∈ Mn , as needed. Having observed that the sets Mn are dense and open, we can set M :=
∞ \
Mn
n=1
to get a comeager set in {0, 1}N \ Q. Since σ 7→ N \ σ, σ ∈ {0, 1}N , is a homeomorphism of {0, 1}N onto itself and {0, 1}N \ Q is comeager in {0, 1}N , the set N := M ∩ {N \ σ : σ ∈ M } is comeager in {0, 1}N , in particular it is nonempty. Let σ ∈ N . Then both σ and N \ σ are contained in M , and these two sets yield the required partition of N. Indeed, let us show that, for example, {xpσ (j) } is strongly linearly independent. We define a subspace of c0 as ∗ ∗ U := {{x∗ (xpσ (k) )}∞ k=1 : x ∈ E }.
Since σ ∈ M , the space U is dense in c0 . Let now λ ∈ `1 satisfy w- lim
j→∞
This means lim
j→∞
or equivalently,
j X
j X
λi xpσ (i) = 0.
i=1
λi x∗ (xpσ (i) ) = 0,
x∗ ∈ E ∗ ,
i=1 ∞ X
λi ui = 0,
u ∈ U.
i=1
Since U is norm dense in c0 and (c0 )∗ = `1 , λ = 0. This concludes the proof.
10.3 Baire measurability and boundedness of affine functions
331
Theorem 10.31. Let f be an affine function on a compact convex subset X of a locally convex space that has the Baire property in the restricted sense. Then f is bounded on X. Proof. Assume that f is not bounded. We translate the set X if necessary to achieve that 0 ∈ X and assume that f (0) = 0. Inductively we find a sequence {xn } in X such that xn+1 ∈ / span{x1 , . . . , xn } and 2n |f (xn )| ≥ 2 . (This is possible because f is bounded on F ∩ X for every finitedimensional subspace F of E.) Set xn x0n := , n ∈ N. n Since X is a bounded subset of E, x∗ (x0n ) → 0 for each x∗ ∈ E ∗ (see W. Rudin [403, Theorem 1.15 and Theorem 1.30]). By Theorem 10.30, we can find a subsequence {ynk } of {x0n } that is strongly independent. We relabel the sequence {ynk } as {yk }. Claim. The following assertions hold: (a) the mapping ϕ : {0, 1}N → X defined as X ϕ(σ) := σk 2−nk yk ,
σ ∈ {0, 1}N ,
k∈N
is a well-defined homeomorphism, (b) the function h : {0, 1}N → R defined as h(σ) := f (ϕ(σ)),
σ ∈ {0, 1}N ,
has the Baire property in the restricted sense on {0, 1}N . Proof of the claim. For the proof of (a), we notice that ϕ isP a well-defined mapping and ϕ({0, 1}N ) ⊂ X. Indeed, if σ ∈ {0, 1}N \ {0} and λ = k∈N σk 2−nk , then X σk 2−nk yk λ k∈N
is a well-defined point contained in X (see Theorem 2.29). Hence, ! X σk −nk 2 yk + (1 − λ)0 ϕ(σ) = λ λ k∈N
is well defined and contained in X as well. Further, for σ, τ ∈ {0, 1}N and x∗ ∈ E ∗ , we get |x∗ (ϕ(σ)) − x∗ (ϕ(τ ))| ≤ sup |x∗ (x)| x∈X
∞ X k=m
2−nk ,
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10 Deeper results on function spaces and compact convex sets
where m ∈ N is the least index such that σm 6= τm . Thus x∗ ◦ϕ is continuous for each x∗ ∈ E ∗ which gives that ϕ : {0, 1}N → X is continuous with respect to the weak topology of E. Since the weak topology of E coincides with the original topology on the compact set X, ϕ is continuous. Further, we check the injectivity of ϕ. Let σ, τ ∈ {0, 1}N satisfy ϕ(σ) = ϕ(τ ), that is, X X σk 2−nk yk = τk 2−nk yk . (10.5) k∈N
k∈N
Since the sequence {yk } is strongly independent, (10.5) yields that σ = τ . Hence, ϕ is injective and thus a homeomorphism. For the proof of (b), we have to verify that h has the Baire property on each compact set K ⊂ {0, 1}N . But this is obvious, since ϕ : K → X is a homeomorphic mapping. This concludes the proof of the claim. Now we are ready to finish the proof of Theorem 10.31. If x ∈ X and λ < 1, then f (λx) = f (λx + (1 − λ)0) = λf (x) + (1 − λ)f (0) = λf (x). Thus, for x, y ∈ X with x + y ∈ X, f (x + y) = 2f ( x+y 2 ) and consequently f (x + y) = 2f
x + y 2
1 1 f (x) + f (y) 2 2 = f (x) + f (y).
=2
Hence, if σ, τ ∈ {0, 1}N have disjoint supports, we get h(σ + τ ) = f (ϕ(σ + τ )) = f (ϕ(σ) + ϕ(τ )) = f (ϕ(σ)) + f (ϕ(τ )). Thus, the function h is finitely additive on {0, 1}N . By the claim, h has the Baire property in the restricted sense on {0, 1}N . By Theorem 10.26, the function h is countably additive. If ek = (0, . . . , 0, 1, 0, . . . ) denotes the element in {0, 1}N with digit 1 at k-th coordinate, k ∈ N, from the countable additivity of h we get ∞ X |h(ek )| < ∞. k=1
On the other hand, |h(ek )| = 2−nk |f (yk )| = 2−nk |f (n−1 k xnk )| ≥
2nk , nk
k ∈ N.
P Thus, the series ∞ k=1 h(ek ) cannot be absolutely convergent. This contradiction finishes the proof.
10.3 Baire measurability and boundedness of affine functions
333
Theorem 10.32. Let E = L1 (µ) for a σ-finite measure µ on a measurable space (X, Σ, µ) and let f be a linear functional on E ∗ (identified with L∞ (µ)) that has the Baire property in the restricted sense on RBE ∗ with respect to the w∗ -topology. Then there exists h ∈ L1 (µ) such that f (g) = X g(t)h(t) dµ(t), g ∈ L∞ (µ). Proof. Let f : E ∗ → R be as in the hypothesis. For a set A ∈ Σ, we set ν(A) := f (cA ). Then ν is a finitely additive signed measure on Σ. We want to prove that ν is countably additive. To this end, let {An : nS ∈ N} be a countable family of pairwise P∞ ∞ disjoint sets in Σ. Our aim is to show that ν( n=1 An ) = n=1 ν(An ). Let M := {n ∈ N : µ(An ) > 0}. Then ν(An ) = 0 for every n ∈ N \ M and cSn∈N An = cSn∈M An
µ-almost everywhere.
Thus ν
∞ [ n=1
X
[
An = f (cSn∈N An ) = f (cSn∈M An ) = ν
An
and
n∈M
ν(An ) =
X
ν(An ).
n∈M
n∈N
Hence we may assume that all the sets An are of strictly positive measure. Then the mapping ϕ : {0, 1}N → BE ∗ defined as ϕ : σ 7→
∞ X
σn cAn ,
σ ∈ {0, 1}N ,
n=1
is injective and continuous with respect to the w∗ -topology of BE ∗ . Thus h(σ) = f (ϕ(σ)), σ ∈ {0, 1}N , is finitely additive and has the Baire property in the restricted sense on {0, 1}N . By Theorem 10.26, h is countably additive. Thus ν
[ n∈N
X X X An = h en = f (cAn ) = ν(An ). n∈N
n∈N
n∈N
Since ν is absolutely continuous with respect to µ, the R Radon–Nikodym theorem provides a function h ∈ L1 (µ) such that ν(A) = A h dµ, A ∈ Σ. By Theorem 10.31, f is bounded on BE ∗ and thus continuous on E ∗ with respect to the norm topology. R Hence we can use the standard approximation techniques to deduce that f (g) = X gh dµ, g ∈ L∞ (µ). This concludes the proof. The Banach space `1 is also endowed with a natural order, precisely x ≤ y, x, y ∈ if x(n) ≤ y(n) for each n ∈ N. We recall that en is the n-th standard unit vector in `1 . `1 ,
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10 Deeper results on function spaces and compact convex sets
Notation 10.33 (Positive cone of `1 ). Let P stand for the positive cone in `1 , that is, P := {x ∈ `1 : x ≥ 0}. Apart from its usual norm topology, we consider on P a finer topology τ whose basis of neighborhoods for x ∈ P consists of the sets of the form V (x, ε) := {y ∈ P : y ≥ x, ky − xk ≤ ε}. Lemma 10.34. The cone P is a Baire space in the topology τ . Proof. Let {Vn } be a sequence of dense open sets in (P, τ ), and let V0 be a given nonempty τ -open set. Inductively we find points xn ∈ P and εn ∈ (0, 2−n ), n ≥ 0, such that V (x0 , ε0 ) ⊂ V0 and V (xn+1 , εn+1 ) ⊂ Vn+1 ∩ V (xn , εn ),
n ≥ 0.
Then the sequence {xn } satisfies xn ≤ xn+1 , n ≥ 0, and kxn − xn+1 k ≤ 2−n . It follows that the sequence {xn } converges to some x ∈ P . Since {xn } is increasing, x ∈ V0 ∩
∞ \
Vn ,
n=1
which concludes the proof. Lemma 10.35. Let f : (P, τ ) → R be a sublinear function with the Baire property. Then infn∈N f (en ) > −∞. Proof. Since the function f has the Baire property with respect to the Baire topology τ , there exists a τ -meager set M ⊂ P such that f |P \M is τ -continuous (see Lemma 10.34 and Proposition A.61). We pick x ∈ P \ M and find ε > 0 such that |f (y) − f (x)| < 1 whenever y ∈ V (x, ε) \ M . Let z ∈ P with kzk ≤ 1 be arbitrary. Since the mapping ϕ : P → P defined as ε ϕ(y) := y + z, y ∈ P, 2 is a homeomorphism of (P, τ ) onto (ϕ(P ), τ ), the set ϕ−1 (M ) is τ -meager. Since a nonempty open set in a Baire space is a Baire space again, we may select y ∈ V (x, 12 ε) \ (M ∪ ϕ−1 (M )). Then ε y + z ∈ V (x, ε) \ M, 2 and hence ε |f (y) − f (x)| < 1 and |f (y + z) − f (x)| ≤ 1. 2 By sublinearity, ε ε −2 ≤ f (y + z) − f (y) ≤ f (z). 2 2 4 Hence f (z) ≥ − ε . Since z with kzk ≤ 1 is arbitrary, we get f (en ) ≥ − 4ε , n ∈ N.
10.4 Embedding of `1
335
Lemma 10.36. Let f : X → R be a convex Borel function, where X := {λ ∈ P : kλk = 1} considered with the norm topology. Then f is lower bounded. Proof. Given f as in the hypothesis, we extend it on the whole P by setting ( 0, g(x) := x kxkf ( kxk ),
x = 0, x 6= 0,
x ∈ P.
x ) is a homeomorphism of P \{0} onto (0, ∞)×{λ ∈ Since the mapping x 7→ (kxk, kxk P : kλk = 1}, the function g is Borel on P . Moreover, g is sublinear. Hence the assertion follows from Lemma 10.35.
Theorem 10.37. Let f be a Borel convex function on a compact convex set X. Then f is lower bounded. Proof. Let f : X → R be as in the hypothesis. If we assume that f is not lower bounded, then there exist points xn ∈ X, n ∈ N, such that f (xn ) ≤ −n. Let Y := {λ ∈ P : kλk = 1} be considered with the w∗ -topology and ϕ : Y → X be defined as ∞ X ϕ(λ) := λn xn , λ ∈ Y. n=1
Then f ◦ ϕ is a convex Borel function on Y that is not lower bounded, a contradiction with Lemma 10.36. This concludes the proof.
10.4
Embedding of `1
Definition 10.38. If X is a set and {fn } a bounded sequence in `∞ (X), we say that {fn } is equivalent to `1 -basis (with constant C) if there exists C > 0 such that n
X
ci fi
`∞ (X)
≥C
i=1
n X
|ci |
i=1
for every finite sequence {ci }ni=1 of real numbers. Lemma 10.39. If {fn } is equivalent to `1 -basis, then span{fn : n ∈ N} is isomorphic to `1 via the mapping T : span{fn : n ∈ N} → `1 satisfying T fn = en (here {en } is the canonical basis of `1 ). Proof. The proof follows by a straightforward verification.
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10 Deeper results on function spaces and compact convex sets
Lemma 10.40. Let X be a set, α < β and {fn } be a bounded sequence of functions on X that satisfies the following condition: for every pair M, N of finite disjoint sets in N there exists x ∈ X satisfying x∈
\
{y ∈ X : fi (y) < α} ∩
i∈M
\
{y ∈ X : fi (y) > β}.
i∈N
Then {fn } is equivalent to `1 -basis with constant 21 (β − α). P Proof. Let c1 , . . . , cn be real numbers. Assume first that (α + β) ni=1 ci ≥ 0. We set M := {i : ci ≥ 0} and N := {i : ci < 0} and find x ∈ X such that
Then
n X
fi (x) > β,
i ∈ M,
fi (x) < α,
i ∈ N.
ci fi (x) =
i=1
X
ci fi (x) +
i∈M
≥β
X
≥ Hence
n
X
ci fi
`∞ (X)
ci + α
If (α + β) cases we get
i=1 ci
X
ci
i∈N n
n
i=1 n X
i=1
β−αX β+αX ci + |ci | 2 2 β−α 2
≥
i=1
Pn
ci fi (x)
i∈N
i∈M
=
X
n X
|ci |.
i=1 n
ci fi (x) ≥
β−αX |ci |. 2 i=1
i=1
< 0, we use the previous estimate for the numbers −ci . In both n
X
ci fi
`∞ (X)
i=1
n
≥
β−αX |ci |, 2 i=1
which concludes the proof. Lemma 10.41. Let E be a Banach space, K ⊂ E ∗ bounded, f ∈ BE ∗∗ and α < β. Assume that for every w∗ -open set U ⊂ E ∗ intersecting K there exist x∗ , y ∗ ∈ ∗ cow (U ∩ K) so that f (x∗ ) > β and f (y ∗ ) < α. Then E contains an isomorphic copy of `1 .
10.4 Embedding of `1
337
Proof. Let K be as in the hypothesis. We construct inductively points xn ∈ BE , n ∈ N, so that \ \ K∩ {z ∗ ∈ K : z ∗ (xn ) > β} ∩ {z ∗ ∈ K : z ∗ (xn ) < α} 6= ∅ (10.6) n∈M
n∈N
for any pair M, N of disjoint nonempty finite subsets of N. ∗ In the first step, we find x∗ , y ∗ ∈ cow K such that f (x∗ ) > β and f (y ∗ ) < α. Let x1 ∈ BE approximates f pointwise on x∗ , y ∗ ; precisely we need that x∗ (x1 ) > β and y ∗ (x1 ) < α. ∗ Since x∗ , y ∗ ∈ cow K, both sets {z ∗ ∈ K : z ∗ (x1 ) > β}
and {z ∗ ∈ K : z ∗ (x1 ) < α}
are nonempty. Assume now that the construction has been completed up to the n-th stage, that is, we have x1 , . . . , xn ∈ BE such that the set \ \ {z ∗ ∈ K : z ∗ (xi ) < α} {z ∗ ∈ K : z ∗ (xi ) > β} ∩ UM,N := i∈N
i∈M
intersects K for each pair of disjoint nonempty finite sets M, N in {1, . . . , n}. ∗ ∈ K ∩ UM,N and use the assumption For every such pair we select a point zM,N w∗ ∗ ∗ ∗ to find xM,N , yM,N ∈ co (K ∩ UM,N ) such that f (x∗M,N ) > β and f (yM,N ) < α. We find xn+1 ∈ BE satisfying x∗M,N (xn+1 ) > β Then
∗ and yM,N (xn+1 ) < α.
K ∩ UM,N ∩ {z ∗ ∈ K : z ∗ (xn+1 ) > β} 6= ∅ and K ∩ UM,N ∩ {z ∗ ∈ K : z ∗ (xn+1 ) < α} 6= ∅.
It follows that K∩
\
{z ∗ ∈ K : z ∗ (xi ) > β} ∩
i∈M
\
{z ∗ ∈ K : z ∗ (xi ) < α} 6= ∅
i∈N
for each pair of disjoint nonempty subsets of {1, . . . , n + 1}. This finishes the construction. By (10.6), {xn |K } satisfies the assumptions of Lemma 10.40. If C > 0 satisfies K ⊂ CBE ∗ , for any numbers c1 , . . . , cn we get n n
X
X
C ci xi E ≥ ci xi |K i=1
i=1
`∞ (K)
n
≥
β−αX |ci |. 2 i=1
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10 Deeper results on function spaces and compact convex sets
Thus
n
n
i=1
i=1
X
β−αX |ci | ≤ ci xi E , 2C which shows that the sequence {xn } is equivalent to `1 -basis and the proof is finished. Theorem 10.42. For a Banach space E the following assertions are equivalent: (i) E does not contain an isomorphic copy of `1 , (ii) f |BE∗ is universally measurable for any f ∈ E ∗∗ , (iii) for any f ∈ BE ∗∗ and µ ∈ M1 (BE ∗ ) there exists a sequence {xn } in BE such that xn → f on BE ∗ µ-almost everywhere, (iv) f |BE∗ satisfies the barycentric formula for any f ∈ E ∗∗ . Proof. For the proof of (i) =⇒ (iv), let f be an element of E ∗∗ . For a measure µ ∈ M1 (BE ∗ ) given, pick any numbers β > α. We set K := spt µ and find β 0 > α0 such that [α0 , β 0 ] ⊂ (α, β). By Lemma 10.41, there exists a w∗ -open set U intersecting K such that ∗
cow (U ∩ K) ⊂ {z ∗ ∈ K : f (z ∗ ) ≥ α0 } ⊂ {z ∗ ∈ K : f (z ∗ ) > α} w∗
co
∗
∗
0
∗
or
∗
(U ∩ K) ⊂ {z ∈ K : f (z ) ≤ β } ⊂ {z ∈ K : f (z ) < β}.
In both cases we get a w∗ -compact convex set L with strictly positive measure such that either f > α on L or f < β on L. Hence f satisfies the Haydon condition (Definition 4.13) and, by Theorem 4.19, f is strongly affine on BE ∗ . For the proof of (iv) =⇒ (iii) use Theorem 4.34. Since the implication (iii) =⇒ (ii) is obvious, we proceed to the last implication (ii) =⇒ (i). Assume that T : `1 → E is an isomorphic embedding satisfying kT −1 k ≤ 1. Then the dual operator T ∗ : E ∗ → `∞ is surjective and T ∗ (BE ∗ ) ⊃ B`∞ . Let g ∈ (`∞ )∗ be an element whose restriction to B`∞ is not universally measurable (see Lemma 10.28). Then f := g ◦ T ∗ is in E ∗∗ and it is not universally measurable by Corollary 5.27. This concludes the proof.
10.5
Metrizability of compact convex sets
In this section we will need notions and facts contained in Section A.4 and Subsection A.2.A. Definition 10.43 (Networks of topological spaces and distinguishable sets). For a topological space X, a family N of subsets of X is a network if for any open set U ⊂ X we have [ U = {N ∈ N : N ⊂ U }.
10.5 Metrizability of compact convex sets
339
A subset A of a completely regular space X is called distinguishable in X if there exists a separable metric space M and a continuous mapping f : X → M such that A = f −1 (f (A)). A completely regular space X is called distinguishable if X is distinguishable in ˇ its Cech–Stone compactification. Lemma 10.44. Let X be a completely regular space and ∆ := {(x, x) : x ∈ X} ⊂ X × X be the diagonal of X × X. Consider the following assertions: (i) (X × X) \ ∆ is Lindel¨of, (ii) there exists a countable family of continuous functions on X separating points of X, (iii) ∆ is a Gδ set in X × X, Then (i) =⇒ (ii) =⇒ (iii). Proof. Let X and ∆ be as in the statement and assume (i). For each x, y ∈ X, x 6= y, let fx,y be a continuous function on X such that fx,y (x) 6= fx,y (y). Then −1 −1 {fx,y (U ) × fx,y (V ) : (x, y) ∈ (X × X) \ ∆, U, V ⊂ R disjoint and open}
is a family of open subsets of (X ×X)\∆ that covers (X ×X)\∆. By the assumption, there exist countably many pairs (xn , yn ) ∈ (X × X) \ ∆ and open sets Un , Vn ⊂ R, n ∈ N, such that Un ∩ Vn = ∅ and (X × X) \ ∆ ⊂
∞ [
fx−1 (Un ) × fx−1 (Vn ) . n ,yn n ,yn
n=1
Then the family {fxn ,yn : n ∈ N} separates points of X. This proves (i) =⇒ (ii). For the proof of (ii) =⇒ (iii), let {fn : n ∈ N} be a countable family of continuous functions on X separating points of X. Then [ (X × X) \ ∆ = {fn−1 (U ) × fn−1 (V ) : U, V ⊂ R are disjoint closed intervals with rational endpoints} is an Fσ set. Thus ∆ is Gδ and the proof is complete. Lemma 10.45. Let X be a compact space and ∆ = {(x, x) : x ∈ X} ⊂ X × X be the diagonal of X × X. Then the following assertions are equivalent. (i) X is metrizable, (ii) (X × X) \ ∆ is Lindel¨of, (iii) there exists a countable family of continuous functions on X separating points of X, (iv) ∆ is a Gδ set in X × X.
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10 Deeper results on function spaces and compact convex sets
Proof. Obviously, (i) =⇒ (ii). By Lemma 10.44, (ii) =⇒ (iii) =⇒ (iv). Assume that ∆ is a Gδ subset of X × X. Then (X × X) \ ∆, as an Fσ subset of the Lindel¨of space X × X, is Lindel¨of. By Lemma 10.44, there exists a countable family {fn : n ∈ N} in C(X) separating points of X. We assume that kfn k ≤ 1, n ∈ N, and define ∞ X 1 |fn (x) − fn (y)|, x, y ∈ X. ρ(x, y) := 2n n=1
Then ρ is a continuous metric on X × X and thus ρ generates a weaker topology on X than the original one. By Proposition A.28, these topologies coincide. Lemma 10.46. Let X be a completely regular space. (a) A set A ⊂ X is distinguishable in X if and only if there exists a countable family {fn : n ∈ N} of continuous functions on X such that for every x ∈ A, y ∈ X \ A there exists n ∈ N such that fn (x) 6= fn (y). (b) The family of all distinguishable subsets of X is a σ-algebra. (c) The space X is distinguishable if and only if X is distinguishable in some compact space. (d) If X is distinguishable, then X is K-countably determined. (e) If X is compact and A ⊂ X Baire, then A is distinguishable in X. Proof. For the proof of (a), let f : X → M be a continuous mapping to a metric separable space M such that A = f −1 (f (A)). Let {gn : n ∈ N} be a countable family of continuous functions separating points of M . Then the family {gn ◦ f : n ∈ N} satisfies our requirement. Conversely, let {fn : n ∈ N} be as in (a). Then f : X → RN defined as f (x) = {fn (x)}, x ∈ X, is the required mapping with M = f (X). As the verification of (b) is routine, we proceed to the proof of (c). Y Assume that X is distinguishable in a compact space Y . By replacing Y with X if necessary we may assume that Y is a compactification of X. Let M be a separable metric space and f : Y → M be a continuous mapping ˇ such that X = f −1 (f (X)). Let βX be the Cech–Stone compactification of X and g : βX → Y be the continuous mapping which equals identity on X (see Theorem A.36). It is easy to verify that g ◦ f : βX → M distinguishes X in βX. To check (d), assume that X is distinguishable in βX. Let M be a separable metric space and f : βX → M be continuous such that X = f −1 (f (X)). Since f (βX) is compact, X = ϕ(f (X)), where ϕ : f (βX) → 2βX is defined as ϕ(y) := f −1 (y),
y ∈ βX.
Obviously, ϕ is a usco mapping. Hence X is K-countably determined.
10.5 Metrizability of compact convex sets
341
Finally, let A be a Baire subset of a compact space X. Let fn ∈ C(X), n ∈ N, be such that A is contained in the σ-algebra generated by {fn−1 (U ) : U ⊂ R open} (see Proposition A.48). Then {fn : n ∈ N} separates A from X \ A and A is distinguishable in X. Lemma 10.47. Let X, Y be completely regular spaces and ϕ : X → 2Y be a usco mapping. Then (a) graph of ϕ (that is, gr ϕ := {(x, y) ∈ X × Y : y ∈ ϕ(x)}) is a closed subset of X ×Y, (b) the projection πX : gr ϕ → X is a perfect mapping, (c) the mapping ψ : X → 2X×Y , x 7→ {x} × ϕ(x), is a usco mapping of X onto gr ϕ. Proof. For the proof of (a), let (x0 , y0 ) be a point of (X × Y ) \ gr ϕ. As ϕ(x0 ) is compact, we can find a neighborhood V of y0 such that V ∩ ϕ(x0 ) = ∅. By the upper semicontinuity of ϕ, U := {x ∈ X : ϕ(x) ⊂ Y \ V } is an open subset of X and x0 ∈ U . Thus U × V is an open neighborhood of (x0 , y0 ) disjoint from gr ϕ. −1 (x) = {x} × ϕ(x) is To verify (b) we first notice that πX is continuous and πX compact in X × Y for each x ∈ X. Let F ⊂ gr ϕ be a closed set. Then U := (X × Y ) \ F is an open subset of X × Y . −1 −1 Let x ∈ X satisfy πX (x) ⊂ U . By the compactness of πX (x), there exist open sets G1 ⊂ X and G2 ⊂ Y such that −1 πX (x) ⊂ G1 × G2 ⊂ U.
Since −1 {x ∈ X : πX (x) ⊂ G1 × G2 } = {x ∈ X : ϕ(x) ⊂ G2 } ∩ G1
is an open subset of X, −1 (x) ⊂ U } πX (F ) = πX ((X × Y ) \ U ) = X \ {x ∈ X : πX
is a closed set in X. Hence πX is perfect. As (c) is an immediate consequence of (b), the proof is finished. Lemma 10.48. Let X be a regular space. Then (a) {N : N ∈ N } is a network for each network N , (b) X is determined if and only if X admits a countable network.
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10 Deeper results on function spaces and compact convex sets
Proof. Let N be a network for a regular space X. Since for each x ∈ X and open set U containing x there exists an open set V such that x ∈ V ⊂ V ⊂ U , assertion (a) follows. For the proof of (b), we start with the assumption that X is determined. Let f : Y → X be a continuous surjection from a separable metric space Y . If B is a countable base of open sets in Y , then it easily follows that {f (B) : B ∈ B} is a countable network for X. Assume now that X has a countable network N = {Mn : n ∈ N}. Let {0, 1}N be the Cantor set identified with the family of all subsets of N as in Subsection 10.3.A. We set \ \ Mic , σ ∈ {0, 1}N . Nσ := Mi ∩ i∈σ
i∈σ /
We claim that Nσ is either empty or a singleton for each σ ∈ {0, 1}N . Indeed, let σ ∈ {0, 1}N and x, y ∈ X be distinct points. We find a neighborhood V of x such that y ∈ / V and n ∈ N such that x ∈ Mn ⊂ V . Then, x ∈ / Nσ if σn = 0, whereas y∈ / Nσ if σn = 1. Hence Nσ cannot contain both points x, y. We set Y := {σ ∈ {0, 1}N : Nσ 6= ∅}. and define f : Y → X as the mapping determined by the property f (σ) ∈ Nσ . Obviously, Y is a separable metric space and f is onto. To finish the proof we have to verify the continuity of f . Let σ ∈ Y and an open neighborhood U of f (σ) be given. We find an index n ∈ N such that f (σ) ∈ Mn ⊂ U. Then the set U := {τ ∈ {0, 1}N : τn = 1} is a neighborhood of σ and f (U ) ⊂ V . Thus f is continuous at σ and the proof is complete. Lemma 10.49. A determined compact space is metrizable. Proof. Let X be a determined compact space. By Lemma 10.48, there exists a countable network N for X consisting of closed sets. For any pair N, N 0 ∈ N of disjoint sets we choose a continuous function f(N,N 0 ) ∈ C(X) such that f(N,N 0 )
( 0 = 1
on N, on N 0 .
10.5 Metrizability of compact convex sets
343
Then {f(N,N 0 ) : N, N 0 ∈ N , N ∩ N 0 = ∅} is a countable family of continuous functions separating points of X. Thus X is metrizable by Lemma 10.45. Definition 10.50 (Hausdorff metric). Let X be a compact space with a compatible metric ρ that is bounded by 1. For a pair of sets A, B ⊂ X, let δ(A, B) := sup{dist(x, B) : x ∈ A}. Let K(X) denote the family of all compact subsets of X endowed with the following metric: if A = B = ∅, 0 dH (A, B) = 1 if exactly one of A, B is empty, δ(A, B) ∨ δ(B, A) if both A, B are nonempty. By Chapter II, Section 4.F in A.S. Kechris [262], dH is a metric on K(X) that turns it into a compact metric space. Moreover, it is compatible with the so-called Vietoris topology. Theorem 10.51. Let X be a compact convex set X with a determined boundary B. Then X is metrizable. Proof. Let N be a countable network of B (see Lemma 10.48) and let ε ∈ (0, 1) be fixed. For each finite family F ⊂ N , we set AF := {f ∈ BAc (X) : diam f (F ) ≤ ε, F ∈ F}. Let K([−1, 1]) denote the space of all compact subsets of [−1, 1] endowed with the Hausdorff metric dH (see Definition 10.50) and, for n ∈ N, let (K([−1, 1]))n be the product space of n copies of K([−1, 1]) with the maximum metric. If F = {F1 , . . . , Fn }, we define a mapping ϕF : AF → (K([−1, 1]))n as ϕF (f ) = (f (F1 ), . . . , f (Fn )),
f ∈ AF .
Since (K([−1, 1]))n is a compact metric space, ϕF (AF ) is its separable subspace. We select a countable set DF ⊂ AF such that ϕF (DF ) is dense in ϕF (AF ). Let [ D := {DF : F ⊂ N finite}. Step 1: For each f ∈ BAc (X) there exists a sequence {fn } of functions from D such that lim sup |fn (x) − f (x)| ≤ 2ε, x ∈ B. n→∞
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10 Deeper results on function spaces and compact convex sets
Given a function f ∈ BAc (X) , we find a finite open cover U of X such that diam f (U ) ≤ ε for each U ∈ U. We set N 0 := {N ∈ N : there exists U ∈ U such that N ⊂ U }. Let {F1 , F2 , . . . } be an enumeration of N 0 . We notice that diam f (Fi ) ≤ ε for each i ∈ N. By setting F n := {F1 , . . . , Fn },
n ∈ N,
we get f ∈ AF n for each n ∈ N. Thus for each n ∈ N we may use density of DF n in ϕF n (AF n ) to choose a function fn ∈ DF n such that dH (f (Fi ), fn (Fi )) < ε,
i = 1, . . . , n.
(10.7)
Let x ∈ B be an arbitrary point. We find a natural number n ∈ N such that x ∈ Fn . Let k ∈ N with n ≤ k be given. By (10.7) and the definition of the Hausdorff metric on K([−1, 1]), there exists y ∈ Fn such that |f (x) − fk (y)| < ε. Since diameters of f (Fn ) and fk (Fn ) are smaller or equal than ε, |fk (x) − f (x)| ≤ |fk (x) − fk (y)| + |fk (y) − f (x)| ≤ 2ε. Hence lim supn→∞ |fn (x) − f (x)| ≤ 2ε for each x ∈ B, concluding the proof. Step 2: For each f ∈ BAc (X) there existP a sequence {fn } of functions from D and a sequence {λn } of positive numbers with ∞ n=1 λn = 1 such that ∞ X
λn |fn (x) − f (x)| ≤ 3ε,
x ∈ X.
n=1
Given f ∈ BAc (X) , by Step 1 there exists a sequence {fn } in D such that −2ε ≤ lim inf(fn (x) − f (x)) ≤ lim sup(fn (x) − f (x)) ≤ 2ε, n→∞
x ∈ B.
n→∞
Corollary 3.75 yields −2ε ≤ lim inf(fn (x) − f (x)) ≤ lim sup(fn (x) − f (x)) ≤ 2ε, n→∞
x ∈ X.
n→∞
In other words, lim sup |fn (x) − f (x)| ≤ 2ε,
x ∈ X.
n→∞
A usePof the Simons inequality 3.74 provides a sequence {λn } of positive numbers with ∞ n=1 λn = 1 such that ∞ X n=1
λn |fn (x) − f (x)| ≤ 3ε,
x ∈ X.
10.5 Metrizability of compact convex sets
345
Step 3: For each f ∈ BAc (X) there Pnexist a finite family {f1 , . . . , fn } ⊂ D and a positive numbers {c1 , . . . , cn } with k=1 ck = 1 such that n X
ck |fk (x) − f (x)| ≤ 6ε,
x ∈ X.
k=1
If f ∈ BAc (X) , let {fk } and {λk } be as in Step 2. We find n ∈ N such that P −1 < 1 + ε and set λ := ∞ k=n λk satisfies (1 − λ) ck := (1 − λ)−1 λk ,
k = 1, . . . , n.
Then n X
ck |fk (x) − f (x)| = (1 − λ)−1
k=1
n X
λk |fk (x) − f (x)| ≤ 3ε(1 + ε) ≤ 6ε,
x ∈ X,
k=1
and Step 3 is proved. It follows that for every ε ∈ (0, 1) there exists a countable family Dε ⊂ BAc (X) such that BAc (X) ⊂ co Dε + 6εBAc (X) . This shows that BAc (X) is separable. Since Ac (X) separates points of X, X is metrizable by Lemma 10.45 and the proof is finished. Lemma 10.52. Let X be completely regular space such that • a countable set of continuous functions separates points of X, and X is distinguishable. Then X is separable metrizable. •
Proof. Let {gn : n ∈ N} be a family of continuous functions on βX that distinguishes X in βX. Let {fn : n ∈ N} be a family of continuous functions on X that separates points of X. We may assume that all functions fn are bounded. We extend the functions fn (and denote them likewise) to the whole βX. Let B be a countable base of open sets in R and let U := {fn−1 (B) : B ∈ B} ∪ {gn−1 (B) : B ∈ B}. We claim that {U ∩ X : U ∈ U} is a subbase of the topology of X. Indeed, let V ⊂ βX be an open neighborhood of x ∈ X. For each y ∈ βX we select sets Uy , Uy0 ∈ U such that x ∈ Uy , y ∈ Uy0 and Uy ∩ Uy0 = ∅. By the compactness of βX \ V there exist y1 , . . . , yn in βX \ V such that βX \ V ⊂
n [ i=1
Uy0 i .
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10 Deeper results on function spaces and compact convex sets
Thus x∈
n \
Uyi ⊂ V,
i=1
and {U ∩ X : U ∈ U} is a subbase of X. As any regular space with a countable base is metrizable (see R. Engelking [169], Theorem 4.4.7) and clearly separable, the proof is finished.
Lemma 10.53. Let X be completely regular space. Then the following assertions are equivalent. (i) X is determined, (ii) X is K-countably determined and admits a countable set of continuous functions separating points of X. Proof. Assume first that X is determined, that is, there exist a separable metric space M and a continuous surjective mapping f : M → X. Then f ×f : M ×M → X ×X defined as f (m, n) := (f (m), f (n)),
(m, n) ∈ M × M,
is a continuous surjection. Thus (X × X) \ ∆ = (f × f ) (f × f )−1 ((X × X) \ ∆) , as a continuous image of a Lindel¨of space, is Lindel¨of. By Lemma 10.44, X admits a countable family of continuous functions separating points of X. Conversely, let {fn : n ∈ N} be a family of continuous functions separating points of X and let ϕ : M → 2X be a usco mapping of a separable metric space M onto X. Let πM : M ×X → M and πX : M ×X → X denotes the canonical projections. We select a countable family {gn : n ∈ N} of continuous functions on M that separates points of M . Then the family {gn ◦ πM : n ∈ N} ∪ {fn ◦ πX : n ∈ N} separates points of gr ϕ.
(10.8)
10.5 Metrizability of compact convex sets
347
We claim that gr ϕ is distinguishable in β(gr ϕ). Indeed, we find a metric compactc of M and let ification M c π bM : β(gr ϕ) → M be the continuous extension of πM . To finish the reasoning of the claim we have to prove that gr ϕ = (b πM )−1 (b πM (gr ϕ)). (10.9) By Lemma 10.47(b), πM : gr ϕ → M is a perfect mapping. By Theorem A.37, c \ M. π bM (β(gr ϕ) \ gr ϕ) ⊂ M Hence (10.9) follows. Thus (10.8) along with (10.9) allows to use Lemma 10.52 and conclude that gr ϕ is a separable metrizable space. Hence X = πX (gr ϕ) is determined. Lemma 10.54. Let F be a countable family of bounded continuous functions on a topological space X. Then there exists a norm separable subspace G ⊂ C b (X) so that F ⊂ G, G contains constant functions and G is a lattice (with respect to the pointwise ordering). Proof. Given F as in the statement of the lemma, we may assume without loss of generality that 1 ∈ F. Then G := W(span F) − W(span F) is the desired space (recall that W(H) is the smallest min-stable cone generated by H; see Definition 3.10). Lemma 10.55. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Let F 1 ⊂ C b (ext X) and G ⊂ Kc (Y ) satisfy the following property: for every f ∈ F 1 there exist bounded sequences {un }, {vn } in G such that sup{(un ◦ ϕ)|ext X : n ∈ N} = f = inf{(−vn ◦ ϕ)|ext X : n ∈ N}. Then there exists a family F 2 ⊂ C b (ext Y ) so that f1 ∈ F 1 if and only if there exists f2 ∈ F 2 satisfying f1 = f2 ◦ ϕ on ϕ−1 (ext Y ) ∩ ext X. Proof. Given F 1 and G as in the hypothesis, for each f1 ∈ F 1 we find sequences {un }, {vn } according to the assumption and set f2 (y) := sup un (y),
y ∈ ext Y.
n∈N
Since ext Y ⊂ ϕ(ext X) and sup un (ϕ(x)) = inf −vn (ϕ(x)), n∈N
n∈N
x ∈ ext X,
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10 Deeper results on function spaces and compact convex sets
the function f2 satisfies f2 = sup un |ext Y = inf −vn |ext Y . n∈N
n∈N
Thus f2 is both upper and lower semicontinuous, and hence continuous on ext Y . Obviously, f2 is bounded. If x ∈ ext X ∩ ϕ−1 (ext Y ), then f2 (ϕ(x)) = sup un (ϕ(x)) = f1 (x). n∈N
Hence F 2 consisting of all such functions constructed for each f1 ∈ F 1 is the required family. Theorem 10.56. Let X be a compact convex set. Then the following assertions are equivalent. (i) X admits a continuous strictly convex function, (ii) ext X is a Baire subset of X, (iii) ext X is distinguishable in X, (iv) ext X is K-countably determined and countably many continuous functions on ext X separate points, (v) ext X is Lindel¨of and countable many continuous functions on ext X separate points, (vi) X is metrizable. Proof. We will prove (i) ⇐⇒ (vi) =⇒ (ii) =⇒ (iii) =⇒ (iv) =⇒ (v) =⇒ (iii) and (iv) =⇒ (vi). For the proof of (i) =⇒ (vi), let f be a strictly convex continuous function on X. Then ϕ : X × X → [0, ∞) defined as 1 x + y , ϕ(x, y) := (f (x) + f (y)) − f 2 2
(x, y) ∈ X × X,
is a positive continuous function and ∞ n \ 1o ∆= (x, y) ∈ X × X : ϕ(x, y) < . n n=1
Hence ∆ is a Gδ subset of X × X and X is metrizable by Lemma 10.45. The implication (vi) =⇒ (i) follows from Theorem 3.42. Further, (vi) =⇒ (ii) follows from Theorem 3.42 and Proposition 3.43. If ext X is a Baire subset of X, then ext X is distinguishable in X by Lemma 10.46(e). Hence (ii) =⇒ (iii).
10.5 Metrizability of compact convex sets
349
Assume now that (iii) is satisfied. By Lemma 10.46, ext X is a K-countably determined space. Define ϕ : X × X → X by the formula 1 ϕ(x, y) := (x + y), 2
(x, y) ∈ X × X.
It is easy to see that both ϕ−1 (ext X) and
ext X × ext X
are distinguishable in X × X. By Lemma 10.46(b), (ext X × ext X) \ ∆ = (ext X × ext X) ∩ ((X × X) \ ϕ−1 (ext X)) is a distinguishable set in X × X. It follows from Lemma 10.46(d) and Theorem A.111(e) that (ext X × ext X) \ ∆ is Lindel¨of. By Lemma 10.44, ext X admits a countable family of continuous functions separating points of ext X. Thus (iv) holds. For the proof of (iv) =⇒ (v), we just use Theorem A.111(e) to conclude that ext X is Lindel¨of. To close the chain of implications we have to prove (v) =⇒ (iii). To this end, assume that ext X is Lindel¨of and F 1 is a countable family of continuous functions on ext X that separates points of ext X. Obviously, we may assume that all functions in F 1 are bounded. Using Lemma 10.54 we enlarge F 1 (if necessary) to get a norm separable subspace of C b (ext X) that contains constant functions and is a lattice with respect to the pointwise ordering. Let F 2 ⊂ F 1 be a countable norm dense set in F 1 . Given f ∈ F 2 , ext X f ∗ = f on ext X by Corollary 3.23(c). Since ext X is Lindel¨of, for each f ∈ F 2 , according to Lemma A.54, there exist bounded sequences {fk }, {gk } in Kc (X) such that sup fk |ext X = f = inf (−gk |ext X ). k∈N
k∈N
Let G 1 consist of the elements of all these sequences for f running through F 2 . Claim. For every x1 ∈ ext X and x2 ∈ X \ ext X there exists g ∈ G 1 so that g(x1 ) 6= g(x2 ). Proof of the claim. We assume that g(x1 ) = g(x2 ) for each g ∈ G 1 . We find a Baire set A ⊂ X with x1 ∈ A ⊂ X \ {x2 } and a maximal measure µ ∈ Mx2 (X). We use Theorem 9.12, for maximal measures εx1 , µ, a Baire set A and a family G 1 , to get a continuous affine surjection ϕ : X → Y of X onto a metrizable compact convex set Y and a set G 2 ⊂ C(Y ) such that (a) ϕ] εx1 and ϕ] µ are maximal on Y ,
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10 Deeper results on function spaces and compact convex sets
(b) ϕ−1 (ϕ(A)) = A, (c) for each g1 ∈ G 1 there exists g2 ∈ G 2 such that g1 = g2 ◦ ϕ. We denote yi := ϕ(xi ), i = 1, 2. By (b), y1 6= y2 and by (a), y1 ∈ ext Y . We claim that families F 1 , G 2 satisfy the assumptions of Lemma 10.55. Indeed, if f ∈ F 2 , we get the required sequences from the choice of G 1 and property (c). Since F 2 is norm dense in F 1 , the claim easily follows. Thus we may use Lemma 10.55 to get a family F 3 ⊂ C b (ext Y ) so that f1 ∈ F 1 if and only if there exists f3 ∈ F 3 satisfying f1 = f3 ◦ ϕ on ϕ−1 (ext Y ) ∩ ext X. Since F 1 separates points of ext X, F 3 separates points of ext Y . Further, F 3 is a subspace of C b (ext Y ) that contains constant functions and is, moreover, a lattice. We define B := {f ∈ C b (ext Y ) : ϕ] µ(f ) = f (y1 )}. Then F 3 ⊂ B. To see this, let f3 ∈ F 3 be given. We find f1 ∈ F 1 so that f1 = f3 ◦ ϕ on −1 ϕ (ext Y ) ∩ ext X. By the definition of G 1 , there exists a bounded sequence {un } in G 1 such that supn∈N un = f1 on ext X. Using (c), we select a sequence {b un } in G 2 with un = u bn ◦ ϕ, n ∈ N. Then (ϕ] µ)(f3 ) ≥ (ϕ] µ)(sup u bn ) ≥ sup (ϕ] µ)(b un ) n∈N
n∈N
= sup µ(un ) ≥ sup un (x2 ) n∈N
n∈N
= sup un (x1 ) = f1 (x1 ) n∈N
= f3 (y1 ). Analogously we get (ϕ] µ)(f3 ) ≤ f3 (y1 ). Since ϕ] µ is maximal, it is carried by ext Y . HenceSwe may find an increasing sequence {Kn } of compact sets in ext Y such that (ϕ] µ)( ∞ n=1 Kn ) = 1 and y1 ∈ K1 . Now we may use the Stone–Weierstrass theorem for the restriction of B to each Kn to conclude that B = C b (ext Y ). Indeed, let f ∈ C b (ext Y ) and ε > 0 be given. We find Kn so that (ϕ] µ)(Kn ) > 1 − ε and a function g ∈ F 3 so that sup |f (y) − g(y)| < ε. y∈Kn
Since F 3 is a lattice containing constant functions, we may assume that kgk ≤ kf k. Then Z |(ϕ] µ)(f ) − f (y1 )| ≤ |f (t) − g(t)| d(ϕ] µ)(t) Kn
+ 2kf k(ϕ] µ)(ext Y \ Kn ) + |(ϕ] µ)(g) − g(y1 )| + |g(y1 ) − f (y1 )| ≤ ε(2 + 2kf k).
10.6 Continuous affine images
351
Thus ϕ] µ = εy1 . But this is impossible, because ϕ] µ ∈ My2 (Y ) and y1 6= y2 . This concludes the proof of the claim. The claim yields that ext X is distinguishable in ext X by means of functions from G 1 . This concludes the proof of (v) =⇒ (iii). To finish the proof we verify (iv) =⇒ (vi). If (iv) holds, Lemma 10.52 yields that ext X is a separable metrizable space. By Theorem 10.51, X is metrizable.
10.6
Continuous affine images
In this section we study the following problem. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a compact convex set Y . 1 • If ϕ(ext X) ⊂ ext Y , does it imply that ϕ (M1 ] max (X)) ⊂ Mmax (Y )? •
If ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, does it imply that ϕ] is injective on M1max (X)?
Theorem 10.57. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a compact convex set Y and let ext Y be a Lindel¨of space. (a) Then the following assertions are equivalent: (i) ϕ(ext X) ⊂ ext Y , (ii) ϕ] (M1max (X)) ⊂ M1max (Y ). (b) Further, the following assertions are equivalent: (i’) ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, (ii’) ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X). Proof. We first notice that the implications (ii) =⇒ (i) and (ii’) =⇒ (i’) are obvious. We start the proof of the converse implications by showing (i) =⇒ (ii). To this end, let µ ∈ M1max (X) be given. We fix an arbitrary closed set F ⊂ Y \ ext Y . Since ext Y is Lindel¨of, there exists a countable family of cozero sets {Un : n ∈ N} in Y such that ∞ [ ext Y ⊂ Un ⊂ Y \ F. n=1
S∞
ϕ−1 (
Then G := µ(G) = 1. Thus
n=1 Un )
is an Fσ set. By the assumptions, ext X ⊂ G and hence (ϕ] µ)
∞ [
Un = µ(G) = 1,
n=1
and hence µ(F ) = 0. Thus (ϕ] µ)∗ (Y \ ext Y ) = 0, and ϕ] µ ∈ M1max (Y ) by virtue of Proposition 3.80.
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10 Deeper results on function spaces and compact convex sets
We proceed with the proof of (i’) =⇒ (ii’). We start by proving ϕ(X \ ext X) ⊂ Y \ ext Y.
(10.10)
Indeed, given y ∈ ext Y ∩ ϕ(X), the set ϕ−1 (y) is a closed face. Since ϕ−1 (y) = co(ext ϕ−1 (y)) = co(ϕ−1 (y) ∩ ext X), the assumption yields that ϕ−1 (y) is a singleton. Hence (10.10) follows. Let µ ∈ M1max (X) be given. For any set F ⊂ X \ ext X, (10.10) gives ϕ(F ) ⊂ Y \ ext Y. This along with Proposition 3.80 and the first part of the proof yields (ϕ] µ)(ϕ(F )) = 0,
F ⊂ X \ ext X, F closed.
Hence µ(F ) ≤ µ(ϕ−1 (ϕ(F ))) = (ϕ] µ)(ϕ(F )) = 0,
F ⊂ X \ ext X, F closed,
and thus µ(ϕ−1 (ϕ(F ))) = µ(F ),
F ⊂ X, F closed.
(10.11)
If µ, ν ∈ M1max (X) are measures with ϕ] µ = ϕ] ν, then (10.11) yields µ(F ) = µ(ϕ−1 (ϕ(F ))) = (ϕ] µ)(ϕ(F )) = (ϕ] ν)(ϕ(F )) = ν(ϕ−1 (ϕ(F ))) = ν(F ) for any closed F ⊂ X. Hence µ = ν and ϕ] is injective on M1max (X). Remark 10.58. It can be easily verified that the mapping ϕ : X → Y is a homeomorphism of ext X onto ϕ(ext X) if ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X. Indeed, since ϕ(ext X) ⊂ ext Y
and
ϕ(X \ ext X) ⊂ Y \ ext Y,
it is not difficult to realize that ϕ(F ∩ ext X) = ϕ(F ) ∩ ext Y for any F ⊂ X. Hence ϕ : ext X → ext Y is a closed mapping (that is, ϕ maps closed sets to closed sets), and thus a homeomorphism on ext X. Hence we obtain that ext X is a Lindel¨of space if ext Y is Lindel¨of and ϕ is a surjective mapping with the properties as above. Theorem 10.59. Let ϕ : X → Y be a continuous affine mapping of a compact convex set X to a simplex Y . (a) Then the following assertions are equivalent:
10.6 Continuous affine images
353
(i) ϕ(ext X) ⊂ ext Y , (ii) ϕ] (M1max (X)) ⊂ M1max (Y ), (iii) ϕ(F ) is a face for each closed face F ⊂ X, (iv) ϕ(F ) is a closed extremal set for each closed extremal F ⊂ X. (b) Further, the following assertions are equivalent: (i’) ϕ(ext X) ⊂ ext Y and ϕ is injective on ext X, (ii’) ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X), (iii’) ϕ is a homeomorphism onto ϕ(X). Proof. For the proof of (a), we first verify (i) =⇒ (ii). To this end, let µ be a maximal probability measure on X. To show that ϕ] µ is maximal on Y , we use Mokobodzki’s maximality test 3.58. Let g be a continuous convex function on Y . Since Y is a simplex, g ∗ is an affine function (see Theorem 6.5). By the assumption , g ∗ ◦ ϕ = (g ◦ ϕ)∗
on ext X.
By Proposition 3.88, g ∗ ◦ ϕ ≤ (g ◦ ϕ)∗ on X. On the other hand, given x ∈ X, there exists a measure λ ∈ Mx (X) such that λ(g ◦ ϕ) = (g ◦ ϕ)∗ (x) (see Lemma 3.22). Then ϕ] λ ∈ Mϕ(x) (Y ) and (g ◦ ϕ)∗ (x) = λ(g ◦ ϕ) = (ϕ] λ)(g) ≤ g ∗ (ϕ(x)). Hence g ∗ ◦ ϕ = (g ◦ ϕ)∗ on X. Thus the equality (ϕ] µ)(g) = µ(g ◦ ϕ) = µ((g ◦ ϕ)∗ ) = µ(g ∗ ◦ ϕ) = (ϕ] µ)(g ∗ ) shows that ϕ] µ is a maximal measure on Y . We proceed with the proof by showing (ii) =⇒ (iii). Let F ⊂ X be a closed face. Since ϕ(F ) is obviously convex, we need to check its extremality. Let ν ∈ M1max (Y ) satisfy r(ν) ∈ ϕ(F ). We find a point x ∈ F with ϕ(x) = r(ν) and select a measure µ ∈ M1max (X) such that r(µ) = x. Since F is a closed face, µ ∈ M1 (F ). Then ϕ] µ is carried by ϕ(F ) and by the assumption, ϕ] µ is maximal. Since r(ϕ] µ) = ϕ(r(µ)) = r(ν) and Y is a simplex, ϕ] µ = ν. Hence ν ∈ M1 (ϕ(F )). Since a closed set is extremal if and only if it is a union of closed faces (see Proposition 2.69), we get (iii) =⇒ (iv). We proceed with the proof of (iv) =⇒ (i). But this is immediate, because a set {x} is extremal if and only if x ∈ ext X. This concludes the proof of (a).
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10 Deeper results on function spaces and compact convex sets
We start the proof of (b) by showing (i’) =⇒ (iii’). We know from (a) that ϕ(X) is a face of Y and hence a simplex. Since ext ϕ(X) = ϕ(X) ∩ ext Y , we may assume from now on that ϕ is a surjective mapping onto a simplex Y . By the assumption, ϕ(ext X) = ext Y
and
ϕ is injective on ext X.
Step 1: For any f ∈ Ac (X), the function g(y) := inf{f (x) : x ∈ ϕ−1 (y)},
y ∈ Y,
is lower semicontinuous and affine. It is easy to verify that g is lower semicontinuous and convex. Assume that g is not affine, that is, that there exist y1 , y2 ∈ Y , λ ∈ (0, 1) and ε > 0 such that λg(y1 ) + (1 − λ)g(y2 ) − g(λy1 + (1 − λ)y2 ) > ε > 0. By Proposition 4.9, we can find functions h1 , h2 ∈ Ac (Y ) so that h(yi ) − hi (yi ) < 31 ε and hi < g, i = 1, 2. Let h := h1 ∨ h2 and y := λy1 + (1 − λ)y2 . Since Y is a simplex, h∗ is affine (see Theorem 6.5) and hence the function f − ∗ h ◦ ϕ is lower semicontinuous affine and f − h∗ ◦ ϕ > 0 on ext X (since h < g and h∗ = h on ext Y ). By the Minimum principle 2.24, f − h∗ ◦ ϕ > 0 on X. On the other hand, 1 h∗ (y) > λh1 (y1 ) + (1 − λ)h2 (y2 ) > λg(y1 ) + (1 − λ)g(y2 ) − ε > g(y). 3 Thus there exists x ∈ ϕ−1 (y) such that f (x) < h∗ (y). Hence f (x) − h∗ (ϕ(x)) < 0, a contradiction. Step 2: The mapping ϕ is injective. Suppose that there exists a point y ∈ Y such that the fiber ϕ−1 (y) contains two different points, say x1 and x2 . Let f ∈ Ac (X) satisfy f (x1 ) > f (x2 ). Let g : Y → R be defined as above and let h := g ◦ ϕ − f . By the first step, h is upper semicontinuous and affine. Further, h = 0 on ext X by the assumption. By the Minimum principles 2.24 and 3.85, h = 0 on X. But h(x1 ) < 0, a contradiction. This finishes the proof of injectivity of ϕ, and thus ϕ is a homeomorphism. Since the remaining implications are obvious, the proof is finished. Theorem 10.60. Let F ⊂ H be function spaces on a compact space K, F be simplicial and Ac (H) = H. Then the following assertions are equivalent: (i) Ac (F) = H, (ii) ChF (K) = ChH (K).
10.6 Continuous affine images
355
Proof. If Ac (F) = H, Exercise 3.95 yields ChF (K) = ChAc (F ) (K) = ChH (K). Hence (i) =⇒ (ii). To prove (ii) =⇒ (i), suppose that ChF (K) = ChH (K). We know that Ac (F) ⊂ c A (H) = H. Let φ1 : K → S(H) and φ2 : K → S(Ac (F)) be the homeomorphic embedding from Definition 4.25. If ϕ : S(H) → S(Ac (F)) is the restriction mapping, then φ2 = ϕ ◦ φ1 . (10.12) By the assumption and Proposition 4.26(d), ϕ(ext S(H)) = ext S(Ac (F)) and ϕ is injective on ext S(H) by (10.12). Since S(Ac (F)) is a simplex (see Theorem 6.54), Theorem 10.59(b) implies that ϕ is a surjective homeomorphism. Hence Ac (F) = H, which finishes the proof. Notation 10.61. Let ϕ : X → Y be a continuous affine surjection of a compact convex set X onto a compact convex set Y . If f ∈ C(X), we denote as fϕ the function on Y defined as fϕ (y) := sup{f (x) : x ∈ ϕ−1 (y)},
y ∈ Y.
Lemma 10.62. Let ϕ be a continuous affine surjection of a compact convex set X onto a compact convex set Y and let f ∈ Ac (X). Then the following assertions hold: (a) fϕ ◦ ϕ ≥ f and fϕ is upper semicontinuous and concave. (b) fϕ = inf{h ∈ Ac (Y ) : h ◦ ϕ ≥ f }. Proof. Assertion (a) follows easily from the definition (see the proof of Theorem 10.59). To show (b), let h ∈ Ac (Y ) satisfy h ◦ ϕ ≥ f . Then fϕ (y) = max f (ϕ−1 (y)) ≤ h(y),
y ∈ Y.
Conversely, let y ∈ Y be fixed. Let p : Ac (X) → R be defined as p(g) = inf{h(y) : h ◦ ϕ ≥ g},
g ∈ Ac (X).
Then p is a sublinear functional on Ac (X) such that p(h ◦ ϕ) = h(y), h ∈ Ac (Y ). By the Hahn–Banach theorem we can find a linear functional s on Ac (X) such that s(f ) = p(f ) and s ≤ p on Ac (X). If g ∈ Ac (X) is negative, then s(g) ≤ p(g) ≤ 0. Further, s(1) ≤ p(1) and s(−1) ≤ p(−1). Since p(1) = 1, s(1) = p(1) = 1. Thus
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10 Deeper results on function spaces and compact convex sets
s is a positive continuous functional on Ac (X), and hence there exists x ∈ X with s(g) = g(x), g ∈ Ac (X) (see Proposition 4.31(a)). For any h ∈ Ac (Y ), h(ϕ(x)) = s(h ◦ ϕ) ≤ p(h ◦ ϕ) = h(y). By using −h instead of h we get h(ϕ(x)) = h(y), and thus y = ϕ(x). Hence fϕ (y) ≥ f (x) = s(f ) = p(f ), and the proof is finished. Lemma 10.63. Let ϕ be a continuous affine surjection of a compact convex set X onto a compact convex set Y . Then the following conditions are equivalent: (i) ϕ is open (that is, ϕ maps open sets to open sets), (ii) fϕ is continuous on Y for any f ∈ Ac (X), (iii) for any f1 , . . . , fd ∈ Ac (X), the set U := ϕ({x ∈ X : f1 (x) > 0, . . . , fd (x) > 0} is open in Y . Proof. Let ϕ be open and f ∈ Ac (X). For given y ∈ Y and c < fϕ (y) we find x ∈ ϕ−1 (y) with f (x) > c. Then W := ϕ({z ∈ X : f (z) > c}) is an open set in Y containing y and fϕ > c on W . Thus fϕ is lower semicontinuous. By Lemma 10.62(a), fϕ is continuous. This proves (i) =⇒ (ii). To show (ii) =⇒ (iii), we set V := {t ∈ Rd : ti > 0 for all i = 1, . . . , d}, d X S := λ ∈ Rd : λi ≥ 0 for all i = 1, . . . , d, λi = 1 , i=1 −1
d
Cy := {(f1 (x), . . . , fd (x)) ∈ R : x ∈ ϕ
(y)},
y ∈ Y.
By the definition, y ∈ U if and only if Cy ∩ V 6= ∅. If y ∈ / U , a separation argument shows that there exists λ ∈ Rd such that λ · t ≤ 0 for t ∈ Cy and λ · t > 0 for t ∈ V . We observe that λ is nontrivial and λi ≥ 0 for all i = 1, . . . , d, so that we may assume that λ ∈ S. It follows that [ Y \U = {y ∈ Y : sup λ · t ≤ 0} λ∈S
=
[ y∈Y: λ∈S
=
[ y∈Y: λ∈S
t∈Cy
sup x∈ϕ−1 (y) d X i=1
λi fi
d X
λi fi (x) ≤ 0
i=1
ϕ
(y) ≤ 0 .
10.6 Continuous affine images
357
Hence Y \ U is a projection on the first coordinate of the set d X C := (y, λ) ∈ Y × S : λi fi ϕ (y) ≤ 0 . i=1
By the assumption, fϕ is continuous on Y for each f ∈ Ac (X). Since kfϕ − gϕ k ≤ kf − gk, the mapping (f, y) 7→ fϕ (y) is continuous on Ac (X) × Y . Hence C is compact and, consequently, U is open. For the proof of (iii) =⇒ (i) we notice that the sets {x ∈ X : f1 (x) > 0, . . . , fd (x) > 0},
f1 , . . . , fd ∈ Ac (X), d ∈ N,
form a base of the topology of X. Lemma 10.64. If X is a Bauer simplex, then the sets of the form {x ∈ ext X : f (x) > a}, f ∈ Ac (X), a ∈ R, form a base of open sets in ext X. T Proof. If X is a Bauer simplex, ext X is a compact set. Thus the sets ni=1 {x ∈ ext X : fi (x) > 0}, where n ∈ N and the functions fi are continuous on ext X, form a base of the topology of ext X. Given functions f1 , . . . , fn in Ac (X), let f ∈ Ac (X) be such that f1 ∧ · · · ∧ fn = f on ext X (see Theorem 6.37). Then n \
{x ∈ ext X : fi (x) > 0} = {x ∈ ext X : f (x) > 0},
i=1
which concludes the proof. Lemma 10.65. Let X be a compact convex set, Y be a simplex and ϕ : X → Y be a continuous affine surjective mapping satisfying ϕ(ext X) = ext Y . Then fϕ is strongly affine on Y for any f ∈ Ac (X). Proof. Given a function f ∈ Ac (X) and y ∈ Y , let δy be the unique maximal measure in My (Y ). Let x ∈ ϕ−1 (y) be such that f (x) = fϕ (y) and let µx ∈ Mx (X) be a maximal measure. By Theorem 10.59(a), ϕ] µx is a maximal measure on Y and represents y. Hence δy = ϕ] µx and fϕ (y) = f (x) = µx (f ) ≤ µx (fϕ ◦ ϕ) = (ϕ] µx )(fϕ ) = δy (fϕ ). Since fϕ is upper semicontinuous and concave by Lemma 10.62(a), Corollary 4.8 gives fϕ (y) ≥ δy (f ) for any y ∈ Y . Hence δy (f ) = fϕ (y) for y ∈ Y . By Corollary 6.12, fϕ is strongly affine.
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10 Deeper results on function spaces and compact convex sets
Theorem 10.66. Let X, Y be Bauer simplices and ϕ : X → Y be a continuous affine surjective mapping satisfying ϕ(ext X) = ext Y . Then the following assertions are equivalent: (i) ϕ is open, (ii) ϕ : ext X → ext Y is open, (iii) fϕ |ext Y is continuous on ext Y for every f ∈ Ac (X). Proof. We start the proof by noticing that Lemma 10.65 yields that fϕ is strongly affine function for each f ∈ Ac (X). Since ext Y is a closed set, Corollary 5.32 gives that fϕ is continuous on X if and only if fϕ |ext X is continuous on ext X. Hence (i) ⇐⇒ (iii) by Lemma 10.63. Further, for any f ∈ Ac (X) and a ∈ R we have {y ∈ ext Y : fϕ (y) > a} = ϕ({x ∈ ext X : f (x) > a}).
(10.13)
Indeed, if fϕ (y) > a for some y ∈ ext Y , then max f (ϕ−1 (y)) > a. Since ϕ−1 (y) is a closed face in X (see Proposition 2.72(b)), there exists x ∈ ext ϕ−1 (y) = ϕ−1 (y) ∩ ext X such that f (x) = max f (ϕ−1 (y)) (cf. Proposition 2.64(b)). Hence y ∈ ϕ({x ∈ ext X : f (x) > a}). Since the reverse inclusion “⊃” is obvious, (10.13) follows. Thus (10.13) yields that fϕ |ext Y ∈ C(ext Y ) provided ϕ is open on ext X and f ∈ Ac (X). Conversely, if each fϕ |ext Y is continuous on ext Y , then ϕ is open on ext X due to Lemma 10.64. Hence (ii) ⇐⇒ (iii), which concludes the proof. Corollary 10.67. Let K, L be compact spaces and ϕ : K → L be a continuous surjection. Then ϕ is open if and only if ϕ] : M1 (K) → M1 (L) is open. Proof. Given ϕ : K → L as in the hypothesis, the mapping ϕ] : M1 (K) → M1 (L) is a continuous affine surjection of the Bauer simplex M1 (K) onto the Bauer simplex M1 (L) (see Proposition 6.38) satisfying the assumptions of Theorem 10.66. If we identify K and L with ext M1 (K) and ext M1 (L), respectively, then ϕ] = ϕ on ext M1 (K). Hence Theorem 10.66 yields the conclusion.
10.7 10.7.A
Several topological results on Choquet boundaries The Choquet boundary as a Baire space
Theorem 10.68. Let S be a function cone of continuous functions on a compact space K. Then ChS (K) is a Baire space.
10.7 Several topological results on Choquet boundaries
359
Proof. Let {Vn } be a sequence of open T∞sets in K such that each Vn ∩ChS (K) is dense in ChS (K). We need to show that n=1 Vn ∩ ChS (K) is dense in ChS (K). To this end, let V0 be an open set intersecting ChS (K). We select a point x0 ∈ V0 ∩ChS (K) and use Proposition 7.11(b) and Theorem 7.21 for the lower semicontinuous function cV0 to find f0 ∈ −S such that f0 (x0 ) > 0 and {x ∈ K : f0 (x) ≥ 0} ⊂ V0 . Inductively we construct functions fn ∈ −S and points xn ∈ ChS (K), n ∈ N, such that • f n+1 < fn , •
fn (xn ) > 0,
{x ∈ K : fn (x) ≥ 0} ⊂ Vn . Assume that the construction has been completed up to the n-th stage. By the inductive assumption and density of Vn+1 ∩ ChS (K) in ChS (K), there exists •
xn+1 ∈ {x ∈ K : fn (x) > 0} ∩ Vn+1 ∩ ChS (K). We apply Proposition 7.11(b) along with Theorem 7.21 to the function cVn+1 ∧ fn to find a function fn+1 ∈ −S with fn+1 (xn+1 ) > 0 and fn+1 < fn . This completes the construction. Let now f := infn∈N fn . Since the sequence {fn } is decreasing, f is an upper semicontinuous S-convex function. Moreover, by a compactness argument, {x ∈ K : f (x) ≥ 0} =
∞ \
{x ∈ K : fn (x) ≥ 0} 6= ∅.
n=0
By the Minimum principle 7.15(f), f attains its maximum at some point y ∈ ChS (K). Hence fn (y) ≥ 0, n ≥ 0, and thus y ∈ V0 ∩
∞ \
Vn ∩ ChS (K).
n=1
This finishes the proof.
10.7.B
Polish spaces as Choquet boundaries
Lemma 10.69. Let Y be a compact metric space and U be its open subset. Then for every ε > 0 there exists a family {gk : k ∈ N} of positive continuous functions on Y such that P∞ (a) k=1 gk = cU , (b) limk→∞ diam spt gk = 0, (c) diam spt gk < ε, k ∈ N.
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10 Deeper results on function spaces and compact convex sets
Proof. Given an open set U in a compact metric space Y , we find a countable open cover V = {Vk : k ∈ N} of U such that • diam V < ε, k ∈ N, k •
limk→∞ diam Vk = 0,
dist(Vk , Y \ U ) > 0. S To achieve this, we find compact sets Fn , n ∈ N, such that U = ∞ n=1 Fn . For each n ∈ N we find a finite open cover of Fn consisting of sets of diameter less than 21n ε that have positive distance from Y \ U . By enumerating all these families into a single sequence we get the required family V. We use Theorem 5.1.9 in [169] to find a family {gn : n ∈ N} of positive continuous functions on U such that P∞ • n=1 gn = cU , •
•
for every n ∈ N there exists kn ∈ N such that {x ∈ U : gn (x) > 0} ⊂ Vkn ,
the family W := {gn−1 ((0, 1]) : n ∈ N} is locally finite, that is, for each x ∈ U there exists a neighborhood intersecting only finitely many sets of W. Due to the properties of the cover {Vk : k ∈ N} we can extend functions gn by 0 outside U and get continuous functions on Y . We finish the proof by realizing that {gn : n ∈ N} satisfies the requirements. To verify the only non-obvious property (b), we show that, for each k ∈ N, Vk intersects only finitely many sets from W. Since W is locally finite, for each x ∈ Vk we can find an open neighborhood Ux of x intersecting only finitely many sets from W. By the compactness of Vk , we choose finitely many points x1 , . . . , xj in Vk such that Vk ⊂ Ux1 ∪ · · · ∪ Uxj . Hence Vk intersects only finitely many elements of W. From this the property (b) follows and the proof is finished. •
Theorem 10.70. Let P be a Polish space. Then there exists a metrizable simplex X such that P is homeomorphic to ext X. Proof. Let P be a Polish space and Y be its metric compactification. We fix a compatible metric ρ on T Y . By [169], Theorem 4.3.24, P is a Gδ subset of Y , and hence we may write P = ∞ n=0 Gn , where Y = G0 ⊃ G1 ⊃ · · · are open subsets of Y . For each n ≥ 0 we use Lemma 10.69 to find continuous functions {hnk }∞ k=1 on Y and a sequence of strictly positive numbers {βnk }∞ converging to 0 such that k=1 P∞ n • = c , h Gn k=1 k diam spt hkn ≤ 2−n βkn , k ∈ N. We set pnk := hnk cY \Gn+1 , •
Then
∞ X ∞ X n=0 k=1
k ∈ N, n ≥ 0.
hnk = cY \P .
361
10.7 Several topological results on Choquet boundaries
We take into consideration only nonzero functions pnk and inductively find points xnk , ykn ∈ spt hnk ∩ P such that the family {xnk , ykn : k ∈ N, n ≥ 0} consists of pairwise distinct elements. We remark that this is possible because spt hnk ∩ P is infinite, whenever pnk 6= 0. We define a mapping γ : M(Y ) → M(Y ) as γ(µ) := µ −
Z ∞ X ∞ X n=0 k=1
Y
pnk dµ
1 (εxn + εykn ). 2 k
(10.14)
We define a subspace M of M(Y ) as M := {γ(µ) : µ ∈ M(Y \ P )}. Claim 1. The space M is closed (in the w∗ -topology). Proof of Claim 1. It is enough to show that the unit ball BM is compact (see Theorem A.7). Since BM is metrizable, it suffices to check the sequential compactness of BM . To this end, let {µi } be a sequence of measures in M(Y \P ) such that the sequence {γ(µi )} is contained in BM . We have to extract a w∗ -convergent subsequence whose limit is in BM . To this end, we use an estimate kµi k ≤ kγ(µi )k, the compactness of BM (Y \Gn+1 ) , n ≥ 0, and the diagonal method to find a subsequence (denoted again n ∈ as {γ(µi )}) such that the sequence {µi |Y \Gn+1 }∞ i=1 converges to a measure ν M(Y \ Gn+1 ), n ≥ 0. We set ν :=
∞ X
ν n |Gn \Gn+1 .
n=0
P∞ P∞
We notice that ν = n=0 k=1 pnk ν n . Further, kνk ≤ 1. To show this, we first observe that ∞ X ∞ Z ∞ X ∞ Z X X f pnk dµi ≤ |f |pnk d|µi | ≤ kµi k ≤ 1, n=0 k=1
Y
n=0 k=1
i ∈ N,
Y
and pnk µi = hnk cY \Gn+1 µi = hnk µi |Y \Gn+1 → hnk ν n = pnk ν n ,
k ∈ N, n ≥ 0.
Hence, for every f ∈ BC(Y ) , we obtain ∞ ∞ X ∞ X X |ν(f )| = (ν n |Gn \Gn+1 )(f ) = n=0
n=0 k=1
Z
f hnk dν n ≤ 1.
Y
(HerePwe use the following fact: If limi→∞ cji = cj and then ∞ j=1 |cj | ≤ 1.)
P∞
j=1 |cji |
≤ 1 for i ∈ N,
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10 Deeper results on function spaces and compact convex sets
Next we verify that γ(µi ) → γ(ν). To this end, we fix f ∈ C(Y ). For each λ ∈ M(Y \ P ) we denote Z 1 αkn (λ) := pnk (f − (f (xnk ) + f (ykn )) dλ, k ∈ N, n ≥ 0. 2 Y Then Z ∞ ∞ 1 XX γ(λ)(f ) = λ(f ) − pnk dλ (f (xnk ) + f (ykn )) 2 Y n=0 k=1 ∞ ∞ Z ∞ X ∞ Z X 1 XX ( pnk dλ)(f (xnk ) + f (ykn )) = pnk dλ − 2 Y Y =
n=0 k=1 ∞ X ∞ X
n=0 k=1
αkn (λ).
n=0 k=1
Let ε > 0 be given. We find δ > 0 such that |f (x) − f (y)| < ε provided ρ(x, y) < δ. Next we find N ∈ N such that 2−N < 21 δ and K ∈ N satisfying βkn < δ, n = 1, . . . , N , k ≥ K. Let A ⊂ N × N consists of all pairs (n, k), where n > N or n = 1, . . . , N and k > K. Then for any λ ∈ M(Y \ P ), the condition on diameters of the supports hnk yields Z N X K X X f dγ(λ) − αkn (λ) ≤ |αkn (λ)| Y
n=0 k=1
Z ≤ε
X
(n,k)∈A
pkn d|λ| ≤ εkλk.
Y (n,k)∈A
For fixed n ≥ 0, k ∈ N, Z 1 pnk (f − (f (xnk ) + f (ykn ))) dµi lim αkn (µi ) = lim i→∞ Y i→∞ 2 Z 1 = pnk (f − (f (xnk ) + f (ykn ))) dν n 2 Y Z 1 = pnk (f − (f (xnk ) + f (ykn ))) dν = αkn (ν). 2 Y P P∞ n n n Thus αk (µi ) → αk (ν) and since the series ∞ n=0 k=1 |αk (λ)| converges uniformly on BM(Y \P ) , an elementary limit procedure gives Z lim
i→∞ Y
f dγ(µi ) = lim
i→∞
∞ X ∞ X n=0 k=1
This concludes the proof of Claim 1.
αkn (µi ) =
∞ X ∞ X n=0 k=1
αkn (ν) =
Z f dγ(ν). Y
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10.7 Several topological results on Choquet boundaries
We proceed with the proof of the theorem by denoting π : M(Y ) → M(Y )/M the canonical quotient mapping and by setting X := π(M1 (Y )). Let ε : Y → M1 (Y ) denote the mapping assigning to each x ∈ Y the Dirac measure εx , and let ϕ := π ◦ ε. We notice the following fact. Claim 2. If x ∈ P , y ∈ Y and π(εx ) = π(εy ), then x = y. Proof of Claim 2. Assume that y ∈ Y \ {x} satisfies π(εy ) = π(εx ). Then εy − εx ∈ M , that is, there exists λ ∈ M(Y \ P ) such that ∞
εy − εx = γ(λ) = λ −
∞
1 XX 2 n=0 k=1
But the right-hand side has mass at most
1 2
Z Y
1 pnk dλ (εxnk + εykn ). 2
at each point of P , a contradiction.
It follows that ϕ : P → X is a homeomorphic embedding. Indeed, ϕ is injective by Claim 2 and, obviously, ϕ is continuous. Let F ⊂ Y be a closed set. Then for any x ∈ P and y ∈ F with ϕ(x) = ϕ(y), Claim 2 yields x = y. Hence ϕ(P ∩ F ) = ϕ(F ) ∩ ϕ(P ), and ϕ maps closed sets in P to closed sets in X. Hence ϕ is a homeomorphism. Claim 3. We have ext X ⊂ ϕ(P ). Proof of Claim 3. Let s ∈ ext X be given. Then π −1 (s) ∩ ext M1 (Y ) 6= ∅ by Proposition 2.72. Hence there exists x ∈ Y such that s = ϕ(x) (see Proposition 2.27). Assume that x ∈ / P , that is, x ∈ Gn \ Gn+1 for some n ≥ 0. Denote λ :=
Z ∞ X ∞ X n=0 k=1
pnk dεx
Y
1 (εxn + εykn ). 2 k
Then λ ∈ M1 (P ) and γ(εx ) = εx − λ. Hence 1 π(εx ) = π(λ) = 2
∞ X ∞ X
pnk (x)π(εxnk )
+
∞ X ∞ X
! pnk (x)π(εykn )
.
n=0 k=1
n=0 k=1
Since s is extreme, this implies ∞ X ∞ X
pnk (x)π(εxnk )
n=0 k=1
in other words,
∞ X ∞ X n=0 k=1
=
∞ X ∞ X
pnk (x)π(εykn ),
n=0 k=1
pnk (x)(εxnk − εykn ) ∈ M.
(10.15)
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10 Deeper results on function spaces and compact convex sets
Since this measure does not charge Y \ P , (10.15) holds only in case ∞ X ∞ X
pnk (x)(εxnk − εykn ) = 0.
n=0 k=1
As all points xnk , ykn , where pnk is nonzero, are different, this is impossible. Hence x ∈ P and Claim 3 is proved. To show that X is a simplex we verify the following fact. Claim 4. If Λ ∈ (Ac (X))⊥ satisfies |Λ|(X \ ϕ(P )) = 0, then Λ = 0. Proof of Claim 4. To verify this, let Λ be as in the hypothesis. We find λ ∈ M(Y ) with |λ|(Y \ P ) = 0 such that Λ = ϕ] λ and pick a continuous function g on Y such that g ∈ M⊥ . Then G(s) := µ(g),
s ∈ X, µ ∈ π −1 (s),
is an affine continuous function on X (see Corollary 5.27). Thus 0 = Λ(G) = (ϕ] λ)(G) = λ(G ◦ ϕ) = λ(g). Hence λ ∈ (M⊥ )⊥ = M . Since λ is carried by P , we get λ = 0, and consequently Λ = 0. Hence X is a simplex by Proposition 6.9. To finish the proof we verify the following fact. Claim 5. We have ϕ(P ) ⊂ ext X. To verify this, assume that ϕ(x) ∈ / ext X for some x ∈ P . Then there exists a measure µ ∈ M1 (X) carried by ϕ(P ) that represents ϕ(x) and is different from εϕ(x) . Then εϕ(x) − µ ∈ (Ac (X))⊥ and |εϕ(x) − µ|(X \ ϕ(P )) = 0. Thus µ = εϕ(x) by Claim 4. This contradiction concludes the proof.
10.7.C
K-countably determined boundaries
Theorem 10.71. Let X be a compact convex set such that ext X is K-countably deterT mined. Then ext X ∈ Π2 (Bos(X)); that is, ext X can be written as ∞ (F ∪ Vn ), n n=1 where Fn ⊂ X are closed and Vn ⊂ X open.
10.8 Convex Baire-one functions
365
Proof. If ext X is K-countably determined, there exists a countable family {Kn : n ∈ N} of compact sets in X such that for each x ∈ X \ ext X, y ∈ ext X, there exists n ∈ N withSy ∈ Kn ⊂ X \ {x} (see Theorem A.111(e)). For each F ⊂ N finite we set AF := n∈F Kn and 1 . BF := x ∈ X : exists µ ∈ Mx (X) such that µ(AF ) ≥ 2 Since AF is closed and the barycentric mapping is continuous, BF is closed as well. Further, AF ⊂ BF . We claim that \ ext X = AF ∪ (X \ BF ). F ⊂N,F finite
To this end, choose x ∈ ext X. If x ∈ BF for some finite F ⊂ N, then there exists µ ∈ Mx (X) with µ(AF ) ≥ 21 . Since µ = εx , x ∈ AF . Hence x ∈ AF ∪ (X \ BF ) for every F ⊂ N finite. Conversely, let x ∈ / ext X. We choose a maximal measure µ ∈ Mx (X). By the property of the family {Kn : n ∈ N}, there exists a set N ⊂ N such that [ Kn ⊂ X \ {x}. ext X ⊂ n∈N
S Since µ( n∈N Kn ) = 1 (see Theorem 3.81), there exists a finite set F ⊂ N so that µ(AF ) ≥ 21 . Since x ∈ / AF , we get x ∈ BF \ AF and thus x ∈ / AF ∪ (X \ BF ). This concludes the proof.
10.8
Convex Baire-one functions
We recall that a function f on a topological space X is fragmented if for any F ⊂ X and ε > 0 there exists a nonempty open set U ⊂ X such that diam f (U ∩ F ) < ε. Lemma 10.72. If f is a fragmented convex function on a compact convex set X, then f is lower bounded. Proof. Assume that there exist points xn ∈ X, n ∈ N, such that f (xn ) → −∞. We consider S := {λ ∈ `1 : λ ≥ 0, kλk = 1} with the w∗ -topology. Then S is a compact convex set and the mapping ϕ : S → X defined as ∞ X ϕ(λ) := r λn εxn , λ = {λn } ∈ S, n=1
is continuous and affine. Hence f ◦ ϕ is convex and fragmented on S. By Theorem A.127, f ◦ ϕ is a Baire-one function on S. Thus Theorem 10.37 yields that f ◦ ϕ
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10 Deeper results on function spaces and compact convex sets
is lower bounded. But f ◦ ϕ(en ) = f (xn ) → −∞, a contradiction (here {en } is the usual basis of `1 ). Lemma 10.73. Let f be a fragmented convex function on a compact convex set X and µ ∈ M+ (X) be a measure on X with kµk > 0. Then for every ε > 0 there exists a ν measure ν ∈ M+ (X) such that ν ≤ µ, kνk > 0 and ν(f ) ≥ kνk(f (r( kνk )) − ε). Proof. Let µ ∈ M+ (X) and ε > 0 be given. Set Y := co spt µ. By the assumption, there exists an open nonempty set U ⊂ Y so that diam f (U ) ≤ ε. Without loss of generality we may assume that inf f (U ) = 0 and sup f (U ) ≤ ε. By combining the Krein–Milman and Milman theorems 2.22 and 2.43, we get distinct points P y1 , . . . , ynP ∈ spt µ ∩ ext Y and coefficients λ1 , . . . , λn ∈ (0, 1) so that ni=1 λi = 1 and y := ni=1 λi yi ∈ U . We find a closed convex neighborhood V ⊂ U of y and disjoint open neighborhoods Ui of yi , i = 1, . . . , n, such that n X
λi z i ∈ V
whenever zi ∈ Ui .
(10.16)
i=1
Then µ(Ui ) > 0, i = 1, . . . , n. We set µi :=
µ|Ui , µ(Ui )
i = 1, . . . , n.
By convexity of f and (10.16), n X i=1
λi f (zi ) ≥ f
n X
λi zi ≥ 0,
zi ∈ Ui , i = 1, . . . , n.
(10.17)
i=1
Integrating (10.17) with respect to the probability measure µ1 ⊗ · · · ⊗ µn we obtain n X
λi µi (f ) ≥ 0.
i=1
P Let ν˜ := ni=1 λi µi and z be the barycenter of ν. ˜ From (10.16) we easily obtain that z ∈ V . Thus ν(f ˜ ) ≥ 0 ≥ sup f (U ) − ε ≥ f (z) − ε. We look for ν in the form ν = cν˜ with c > 0. To satisfy the requirement ν ≤ µ, we choose c ≤ min{µ(U1 ), . . . , µ(Un )}. Then the measure ν = cν˜ has all the required properties. Lemma 10.74. Let f be a fragmented convex function on a compact convex set X. P∞ P∞ Then n=1 λn f (x Pn ) ≥ f ( n=1 λn xn ) whenever numbers λn are positive, points xn are in X and ∞ n=1 λn = 1.
10.8 Convex Baire-one functions
367
Proof. Let P be the positive cone of `1 and S be its base, that is, P = {λ ∈ `1 : λ ≥ 0} and S = {λ ∈ P : kλk = 1} considered with the norm topology (see Notation 10.33). Without loss of generality we may assume that f ≥ 0 (see Lemma 10.72). Let λ0 ∈ S be given. We set αn := λ0n f (xn ),
n ∈ N.
Then {αn } ∈ P . We define g : P → R as ( 0, P g(λ) := kλkf ( ∞ n=1 Since λ 7→
λn kλk xn ),
λ = 0, λ 6= 0.
∞ X λn xn kλk n=1
is a continuous mapping of P \ {0} to X, g is a fragmented function on P \ {0}. Consequently, it is fragmented on P and thus g is a Baire-one function on P by Theorem A.127 (we recall that (P, k · k) is a complete metric space). Moreover, it is a sublinear function on P by the convexity of f . Let ϕ : {0, 1}N → R be defined as ϕ(σ) = g(σλ0 ),
σ ∈ {0, 1}N .
(Here σλ0 is the sequence {σn λ0n }.) Since σ 7→ σλ0 is continuous on {0, 1}N , ϕ is a Baire-one function on {0, 1}N . Set ∞ X N A := σ ∈ {0, 1} : ϕ(σ) ≤ σn αn . n=1
P∞
Since the mapping σ 7→ n=1 σn αn is continuous on {0, 1}N and ϕ is Baire-one, A is a Gδ set (see Theorem A.124). If σ ∈ {0, 1}N is finitely supported, say by a finite set J ⊂ N, then ϕ(σ) = g
X n∈J
∞ X X X 0 λ0n en ≤ λn g(en ) = λ0n f (xn ) = σn αn . n∈J
n∈J
n=1
Hence σ ∈ A and thus A is dense. The mapping σ 7→ N \ σ is a homeomorphism of {0, 1}N onto itself (see Subsection 10.3.A). Hence {N \ σ : σ ∈ A} is a dense Gδ set in {0, 1}N and we can find σ ∈ {0, 1}N such that both σ and N \ σ are in A. Then λ0 = σλ0 + (N \ σ)λ0 ,
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10 Deeper results on function spaces and compact convex sets
and thus ϕ(1) = g(λ0 ) ≤ g(σλ0 ) + g((N \ σ)λ0 ) = ϕ(σ) + ϕ(N \ σ) ≤
∞ X
σn αn +
n=1
∞ X
(1 − σn )αn =
n=1
∞ X
αn .
n=1
In other words, 0
g(λ ) = f
∞ X
λ0n xn
n=1
≤
∞ X
λ0n f (xn ).
n=1
This finishes the proof. Theorem 10.75. Let f be a fragmented convex function on a compact convex set X. Then f is lower bounded and satisfies the subbarycentric formula. Proof. Let f be a function satisfying our assumptions. Then f is lower bounded by Lemma 10.72 and universally measurable by Proposition A.118 and Theorem A.121. Let µ ∈ M1 (X) and ε > 0 be given. We are going to construct a transfinite sequence {µα }0≤α≤ω1 of positive measures on X which satisfies •
µ0 = µ,
•
µα+1 ≤ µα and either kµα+1 k < kµα k or µα = 0, R X f d(µα − µα+1 ) ≥ kµα − µα+1 k(f (xα ) − ε), where xα is the barycenter of µα −µα+1 kµα −µα+1 k , P µα = β<α (µβ − µβ+1 ) for α limit.
•
•
If α is an ordinal number and µα > 0, we use Lemma 10.73 to µα and obtain a measure να such that 0 < να ≤ µα . By setting µα+1 := µα − να we finish the inductive step for an isolated ordinal number. If α is a limit ordinal number and µβ have been defined for every β < α, we define µα ∈ M+ (X) by setting µα (g) := limβ<α µβ (g), g ∈ C(X). (It is not difficult to check that the right-hand side limit actually exists.) This finishes the construction. Let γ be the first ordinal number, for which µγ = 0. Since {kµα k}α<γ strictly decreases, γ is in fact a countable ordinal number. Thus we have µ=
X
(µα − µα+1 ),
α<γ
1 = kµk =
X α<γ
kµα − µα+1 k.
10.8 Convex Baire-one functions
369
Hence Z f dµ = X
XZ α<γ
≥
X
f d(µα − µα+1 )
X
kµα − µα+1 k(f (xα ) − ε)
α<γ
! =
X
kµα − µα+1 kf (xα )
−ε
α<γ
and r(µ) =
X
kµα − µα+1 kxα .
α<γ
Lemma 10.74 gives Z f dµ ≥ X
X
kµα − µα+1 kf (yα ) − ε ≥ f (r(µ)) − ε.
α<γ
As ε is arbitrary, µ(f ) ≥ f (r(µ)) which is the desired conclusion. Corollary 10.76 (Convex B1 -maximum principle). Let f be a Baire-one convex function on a compact convex set X. If f ≤ 0 on ext X, then f ≤ 0 on X. Proof. Since any Baire-one function is fragmented by Theorems A.124 and A.121, the preceding Theorem 10.75 asserts that any convex Baire-one function is Ac (X)convex. Thus B1 -maximum principle is a particular case of Theorem 3.86. Theorem 10.77. Let f be a Baire-one convex function on a compact convex set X. Then there exists a lower bounded sequence {fn } of continuous convex functions on X which converges to f . Proof. Since f is lower bounded, we can find lower bounded sequences {un } and {−ln } of upper semicontinuous functions such that un < f < ln , un % f and ln & f (see Theorem A.124). Let n ∈ N be fixed and x be an arbitrary point of X. Let µ ∈ Mx (Ac (X)) satisfy µ(ln ) = (ln )∗ (x) (see Lemma 3.21). Then (ln )∗ (x) = µ(ln ) > µ(f ) ≥ f (x) > un (x). Hence un < (ln )∗ . We use Corollary 3.49 along with Corollary 4.8 and find a convex continuous function fn with un < fn < ln . Obviously, the sequence {fn } is lower bounded and converges to the function f .
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10 Deeper results on function spaces and compact convex sets
10.9 10.9.A
Function spaces with continuous envelopes Stable compact convex sets
Definition 10.78 (Stable compact convex sets). Let X be a compact convex set. We say that X is stable if the mapping Φ 1 : X × X → X defined as 2
Φ 1 (x, y) := 2
x+y , 2
(x, y) ∈ X × X,
is open. Lemma 10.79. For a compact convex set X the following assertions are equivalent: (i) f ∗ is continuous for every f ∈ Kc (X), (ii)
ext Xf ∗
is continuous for every f ∈ Kc (X).
Proof. If f ∈ Kc (X), then h ∈ Ac (X) satisfies h ≥ f on X if and only if h ≥ f on ext X. Hence f ∗ = inf{h ∈ Ac (X) : h ≥ f } = inf{h ∈ Ac (X) : h ≥ f on ext X} = ext Xf ∗ . From this the conclusion follows. Lemma 10.80. Let X be a compact convex set such that f ∗ is continuous for every f ∈ Kc (X). Then (a) ext X is a closed set, and (b)
ext Xf ∗
is continuous on X for every f ∈ C(ext X).
Proof. By Theorem 3.24, ext X = {x ∈ X : f ∗ (x) = f (x) for any f ∈ Kc (X)}. Thus ext X is closed due to our assumption and (a) is proved. To show (b), let f ∈ C(ext X) be given and let ε > 0. Since ext Xf∗ = f on ext X (use Lemma 3.21), a compactness argument provides a function k ∈ Kc (X) such that f − ε < k < f on ext X. By Lemma 3.18(c), kext Xf ∗ − ext Xk ∗ k ≤ kf − k|ext X kC(ext X) < ε. Hence ext Xf ∗ , as a uniform limit of continuous functions, is continuous as well. Lemma 10.81. Let r : Mmax (X) ∩ M1 (X) → X be an open mapping. Then ext X is a closed set.
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371
Proof. Since the set D := {εx : x ∈ X} is a closed subset of M1 (X), the set Mmax (X) ∩ M1 (X) \ D is an open set in Mmax (X) ∩ M1 (X). Since the mapping r is open, the set X \ ext X = r Mmax (X) ∩ M1 (X) \ D
is also open. Lemma 10.82. Let ϕ be a mapping of a topological space X onto a topological space Y . Let Z ⊂ X be such that Z ∩ ϕ−1 (y) is dense in ϕ−1 (y) for every y ∈ Y . Then ϕ is open if and only if ϕ|Z is open. Proof. It is enough to show that, for an open set U ⊂ X, ϕ(U ) = ϕ(U ∩ Z). Let y ∈ ϕ(U ) be given. We find x ∈ U with ϕ(x) = y. By the assumption there exists z ∈ Z ∩ U ∩ ϕ−1 (y). Then ϕ(z) = y and z ∈ Z ∩ U . Hence y ∈ ϕ(Z ∩ U ). Lemma 10.83. Let λ be a Radon measure on a compact space K and let f be a λmeasurable function on K such that −1 ≤ f ≤ 1 λ-almost everywhere and λ(f ) = 0. Then there exists a sequence {fk } in C(K) such that −1 ≤ fk ≤ 1, λ(fk ) = 0 for any k ∈ N, and fk → f λ-almost everywhere. Proof. Let {gn } be a sequence in C(K) such that −1 ≤ gn ≤ 1, n ∈ N, and gn → f λ-almost everywhere (use Lusin’s theorem A.76). If λ(gn ) ≥ 0, choose αn > 0 such that hn := gn ∧ αn satisfies λ(hn ) = 0. If λ(gn ) < 0, choose βn < 0 such that hn := gn ∨ βn satisfies λ(hn ) = 0. Then λ(|gn − hn |) = |λ(gn − hn )| = |λ(gn )| → |λ(f )| = 0. Hence gn − hn → 0 in L1 (λ). Therefore, there exists a subsequence {gnk − hnk } converging to zero λ-almost everywhere (see W. Rudin [402], Theorem 3.12). Set fk := hnk , k ∈ N. Since gnk → f λ-almost everywhere, we get fk → f λ-almost everywhere. Theorem 10.84. Let K be a compact space. Then the space M1 (K) is a stable convex set. Consequently, any Bauer simplex is a stable convex set. Proof. Our aim is to prove that the mapping ϕ : M1 (K) × M1 (K) → M1 (K) defined as µ+ν ϕ(µ, ν) := , (µ, ν) ∈ M1 (K) × M1 (K), 2
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10 Deeper results on function spaces and compact convex sets
is open. Let n M := (µ, ν) ∈ M1 (K) × M1 (K) :
o dν dµ , are continuous . d(µ + ν) d(µ + ν)
If we show that M ∩ ϕ−1 (λ) is dense in ϕ−1 (λ) for any λ ∈ M1 (K), we can test by Lemma 10.82 the openness of ϕ just on the set M. To check this property, let λ ∈ M1 (K) be given. Let µ, ν ∈ M1 (K) be such that dµ λ = µ+ν 2 . Let f = d(µ+ν) . Then 0≤f ≤1
λ-almost everywhere
and
1 λ(f ) = . 2
Lemma 10.83 provides a sequence {fn } of continuous functions on K so that, for every n ∈ N, 0 ≤ fn ≤ 1, λ(fn ) =
1 and fn → f λ-almost everywhere. 2
Set µn := fn 2λ
and
νn := (1 − fn )2λ.
n and µn → µ, νn → ν. Thus the assumption of Then (µn , νn ) ∈ M, λ = µn +ν 2 Lemma 10.82 is satisfied for ϕ. Let (µ, ν) ∈ M be given and let {λi } be a net in M1 (K) converging to λ := µ+ν 2 . µi +νi It is enough to find measures (µi , νi ) ∈ M so that λi = 2 and µi → µ, νi → ν. Since (µ, ν) ∈ M, there exist continuous positive functions f, g on K such that
µ = f λ,
ν = gλ
and
f + g = 2.
We set µi := f λi and νi := gλi . Then µi + νi = 2λi and µi → µ, νi → ν. The reasoning is finished if we set ( µi :=
µi kµi k ,
kµi k ≥ 1,
µi + νi −
νi kνi k ,
kµi k < 1,
( µi + νi −
µi kµi k ,
kµi k ≥ 1,
and νi :=
νi kνi k ,
kµi k < 1.
If X is a Bauer simplex, then ext X is closed and X is affinely homeomorphic to M1 (ext X) by Proposition 6.39. Hence, X is a stable set according to the first part of the proof.
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373
Lemma 10.85. Let X be a compact convex set, x, y, z ∈ X and λ ∈ (0, 1) such that x = λy + (1 − λ)z. Assume that co(U ∪ V ) is a neighborhood of x for any neighborhoods U 3 y and V 3 z. If U and V are neighborhoods of y and z, respectively, then λU + (1 − λ)V is a neighborhood of x. Proof. Assume that there exist open sets U 3 y and V 3 z such that λU + (1 − λ)V is not a neighborhood of x. Then we can find a net {xi }i∈I converging to x such that xi ∈ / λU + (1 − λ)V for all i ∈ I. If U 0 , V 0 are arbitrary open neighborhoods of y and z, respectively, by the hypothesis there exists i0 ∈ I such that xi ∈ co(U 0 ∪ V 0 ) for all i0 ≤ i. Using this observation and varying U 0 and V 0 we easily construct a subnet {xj }j∈J of {xi }i∈I , nets {yj }j∈J , {zj }j∈J in X and a net {λj }j∈J in [0, 1] such that xj = λj yj + (1 − λj )zj , yj → y and zj → z. Then λj → λ. We modify the points yj , zj in the following way: If λj > λ, we set zj0 :=
1 − λj λj − λ yj + zj , 1−λ 1−λ
yj0 := yj ,
in the case λj ≤ λ, let yj0 :=
λj λ − λj yj + zj , λ λ
zj0 := zj .
Then xj = λyj0 + (1 − λ)zj0 ,
yj0 → y
and
zj0 → z.
For large indices j ∈ J, yj0 ∈ U and zj0 ∈ V . This contradicts our assumption xj ∈ / λU + (1 − λ)V . Theorem 10.86. Let X be a compact convex set. Then the following conditions are equivalent: (i) X is stable, (ii) Φλ : (x, y) 7→ λx + (1 − λ)y is an open mapping from X × X onto X for any λ ∈ [0, 1], (iii) Φ : (x, y, λ) 7→ λx + (1 − λy) is an open mapping from X × X × [0, 1] onto X, (iv) Int C is convex for every convex set C ⊂ X, (v) co G is open for any open set G ⊂ X, (vi) r : M1 (X) → X is open, (vii) r : Mmax (X) ∩ M1 (X) → X is open, (viii) f ∗ is continuous for any f ∈ C(X), (ix) f ∗ is continuous for any f ∈ Kc (X), (x)
ext Xf ∗
is continuous for any f ∈ Kc (X).
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10 Deeper results on function spaces and compact convex sets
Proof. We show (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) =⇒ (v) =⇒ (i), (ii) =⇒ (viii) =⇒ (ix) =⇒ (vii) =⇒ (i), (viii) ⇐⇒ (vi) and (ix) ⇐⇒ (x). (i) =⇒ (ii): If Φ 1 is open, Φλ is open for any dyadic rational number λ ∈ 2 [0, 1]. Let λ ∈ [0, 1] and y, z ∈ X be given. Set x := λy + (1 − λ)z. Given open neighborhoods U and V of y and z, respectively, we find a dyadic rational number α ∈ [0, 1] and a point z 0 ∈ V with x = αy + (1 − α)z 0 . Then co(U ∪ V ) ⊃ αU + (1 − α)V, where the set on the right-hand side is an open neighborhood of x, because Φα is open. Thus co(U ∪ V ) is a neighborhood of x, which allows to use Lemma 10.85 to infer that λU + (1 − λ)V is a neighborhood of x. (ii) =⇒ (iii): The proof follows from the equality [ Φ(U × V × I) = λU + (1 − λV ), λ∈I
where U, V are open sets in X and I is an open set in [0, 1]. (iii) =⇒ (iv): Let C be a convex set in X and x, y ∈ Int C, λ ∈ [0, 1] be given. Then λx + (1 − λ)y ∈ Φ(Int C × Int C × [0, 1]). Since the set on the right-hand side is open by (iii) and contained in C, we get that λx + (1 − λy) ∈ Int C. (iv) =⇒ (v): If G is an open set in X, then G ⊂ Int(co G) and thus by (iv) we get co G ⊂ Int(co G). Thus co G is open. (v) =⇒ (i): Let x = y+z 2 , x, y, z ∈ X, and U , V be convex neighborhoods of y, z, respectively. By (v), co(U ∪ V ) is a neighborhood of x, and thus Lemma 10.85 gives (i). (ii) =⇒ (viii): By induction it follows that for any n-tuple (λ1 , . . . , λn ) of numbers in [0, 1] with λ1 + · · · + λn = 1, the mapping Φ(λ1 ,...,λn ) : X n → X,
(x1 , . . . , xn ) 7→ λ1 x1 + · · · + λn xn ,
is open. Indeed, assuming the validity of the assertion for n − 1, let (λ1 , . . . , λn ) with λ1 + Pn−1 · · · + λn = 1 be given. If λ := i=1 λi and Φλ : X × X → X be given as Φλ (x, y) = λx + (1 − λ)y, then Φ(λ1 ,...,λn ) (x1 , . . . , xn ) = Φλ (Φ(λ−1 λ1 ,...,λ−1 λn−1 ) (x1 , . . . , xn−1 ), xn ). It follows that Φ(λ1 ,...,λn ) is open.
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375
Let f ∈ C(X) be given. Pick x ∈ X and ε > 0. By Lemma 3.21 and Proposition ε ∗ 4.3, there exists a molecular Pn measure µ ∈ Mx (X) such that µ(f ) > f (x) − 2 . We writePµ in the form µ = i=1 λi εxi , where x1 , . . . , xn ∈ X, λi are positive numbers and ni=1 λi = 1. For each i = 1, . . . , n we find an open neighborhood Ui of xi such that |f (y) − f (xi )| < 2ε for every y ∈ Ui . Due to the previous considerations, V := Φ(λ1 ,...,λn ) (U1 × · · · × Un ) is an open set P containing x. We pick y ∈ V and find points yi ∈ Ui , i = 1, . . . , n, such that y = ni=1 λi yi . Then f ∗ (y) ≥
=
n X i=1 n X
λi f ∗ (yi ) ≥
λi f (xi ) +
i=1
≥ µ(f ) −
n X
λi f ∗ (yi )
i=1 n X
λi (f (yi ) − f (xi ))
i=1 n εX
2
λi
i=1
≥ f ∗ (x) − ε. Thus f ∗ is lower semicontinuous on X. Since f ∗ is always upper semicontinuous, the proof of the implication is finished. (vi) ⇐⇒ (viii): We use Lemma 10.63. Let F be a continuous affine function on M1 (X). Then there exists a function f ∈ C(X) such that µ(f ) = F (µ), µ ∈ M1 (X). Hence we get Fr (x) = sup{µ(f ) : r(µ) = x, µ ∈ M1 (X)} = f ∗ (x),
x ∈ X.
Now the equivalence (vi) ⇐⇒ (vii) is precisely the equivalence (i) ⇐⇒ (ii) in Lemma 10.63. The implication (viii) =⇒ (ix) is obvious and (ix) ⇐⇒ (x) holds by Lemma 10.79. The next step is the proof of (ix) =⇒ (vii). By Lemma 10.80(a), the set ext X is closed. Thus maximal measures are precisely those measures which are carried by ext X (see Proposition 3.80), in other words Mmax (X) ∩ M1 (X) = M1 (ext X). We will use Lemma 10.63 again. If F is a continuous affine function on M1 (ext X), there exists a continuous function f on ext X such that µ(f ) = F (µ), µ ∈ M1 (ext X). Lemma 3.22 yields Fr (x) = sup{µ(f ) : r(µ) = x, µ ∈ M1 (ext X)} = ext Xf ∗ (x),
x ∈ X.
The function ext Xf ∗ is continuous on X by Lemma 10.80(b). Now Lemma 10.63 concludes the proof of the implication.
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10 Deeper results on function spaces and compact convex sets
(vii) =⇒ (v): By Lemma 10.81, ext X is closed and hence Mmax (X)∩M1 (X) = M1 (ext X). By Theorem 10.84, the set M1 (ext X) is stable. Thus, M1 (ext X) satisfies condition (v). Now, we are ready to verify (v) for X: Indeed, if G ⊂ X is open, then co(r−1 (G)) is open and (vii) implies that co G = r(co(r−1 (G))) is open as well.
10.9.B
CE-function spaces
Definition 10.87 (Continuous envelopes). A function space H on a compact space K is called a CE-function space if f ∗ is continuous for every f ∈ C(K). Recall that in Section 3.3 we defined [ M(H) := Mx (H) : x ∈ K and the H-barycenter mapping r : Mx (H) → K that assigns x to a measure µ ∈ Mx (H). Theorem 10.88. Let H be a function space on a compact space K. Then the Hbarycenter mapping r : M(H) → K is open if and only if H is a CE-function space. Proof. The proof follows the argument of Lemma 10.63. Suppose that the H-barycenter mapping r : M(H) → K is open. Since, given f ∈ C(K), r({µ ∈ M(H) : µ(f ) > 0}) = {x ∈ K : f ∗ (x) > 0} , we obtain that the set {x ∈ K : f ∗ (x) > 0} is open for every f ∈ C(K). For a ∈ R, the equality {x ∈ K : f ∗ (x) > a} = {x ∈ K : (f − a)∗ (x) > 0} yields that f ∗ is lower semicontinuous for every f ∈ C(K). Since f ∗ is always upper semicontinuous, the upper envelope f ∗ is continuous for every f ∈ C(K). For the proof of the converse implication, assume that f ∗ is continuous for every f ∈ C(K). It is sufficient to show that, given f1 , . . . , fd ∈ C(K) and U := {µ ∈ M(H) : µ(f1 ) > 0, . . . , µ(fd ) > 0} , the set K \ r(U ) is closed. Set n o V : = (t1 , . . . , td ) ∈ Rd : t1 > 0, . . . , td > 0 , d n o X S : = (t1 , . . . , td ) ∈ Rd : t1 ≥ 0, . . . , td ≥ 0, ti = 1 , i=1
C(x) : =
n
o (µ(f1 ), . . . , µ(fd )) ∈ Rd : µ ∈ Mx (H) ,
x ∈ K.
10.9 Function spaces with continuous envelopes
377
Then x ∈ / r(U ) if and only if C(x) ∩ V = ∅. For every x ∈ K, the set C(x), as a continuous affine image of a compact convex set Mx (H), is convex and compact in Rd . Since V is a cone, the Hahn–Banach separation theorem yields that the intersection V ∩ C(x) is empty if and only if there exists s ∈ S such that sup {s · c : c ∈ C(x)} ≤ 0 ≤ inf {s · t : t ∈ V } . Hence Lemma 3.21 gives
[ K \ r(U ) = x∈K:
sup
[ x∈K:
[ x∈K: s∈S
µ
sup
d X
µ∈Mx (H)
s∈S
=
si µ(fi ) ≤ 0
µ∈Mx (H) i=1
s∈S
=
d X
d X
si fi ≤ 0
i=1
∗ si fi (x) ≤ 0 .
i=1
Thus K \ r(U ) = πK
(x, s) ∈ K × S :
d X
∗ si fi (x) ≤ 0 ,
i=1
where πK : K × S → K denotes the projection onto K. By the assumption and Lemma 3.18(c), the mapping (x, f ) 7→ f ∗ (x) is continuous on K × C(K). Hence (x, s) 7→
d X
∗ si fi (x)
i=1
is continuous on K × S. Thus
{(x, s) ∈ K × S :
d X
∗ si fi (x) ≤ 0}
i=1
is a compact set in K × S and K \ r(U ) is closed, which finishes the proof.
378
10.10
10 Deeper results on function spaces and compact convex sets
Exercises
Exercise 10.89. Find a boundary of a compact convex set disjoint from the set of extreme points. Hint. Take an uncountable set Γ and let X be the unit ball of `∞ (Γ) endowed with the w∗ -topology. Then ext X = {x ∈ X : |x(γ)| = 1, γ ∈ Γ}. If we set B := {x ∈ X : |x(γ)| ∈ {0, 1}, x with countable support}, then B is a boundary disjoint from ext X (use Proposition 4.36). Exercise 10.90. Let X be a compact subset of an infinite-dimensional Banach space E. Prove that there exists a sequence {xn } of linearly independent vectors in E converging to 0 such that X ⊂ co{xn : n ∈ N}. Hint. We first propose the following observation. If E1 , E2 are subspaces of E with E1 ∩ E2 = {0} and E2 = span{e1 , . . . , ek } where {e1 , . . . , ek } are linearly independent vectors, then the vectors a1 + e1 , . . . , ak + ek are linearly independent for any a1 , . . . , ak ∈ E1 . To construct the required sequence {xn }, we find inductively sets Kn and Fn as follows. Let K1 := X and F˜1 ⊂ X be a finite 2−2 -net for K1 . If E1 := span F˜1 , let E2 be a subspace of E such that its dimension equals cardinality of F˜1 and E2 ∩E1 = {0}. We select a basis {e1 , . . . , ek } of E2 and form the set F1 in such a way that the i-the element of F1 is a little perturbation of i-the element of F˜1 in the direction of ei , i = 1, . . . , k, so that F1 is still a 2−2 -net for K1 and consists of linearly independent vectors. Assuming that the construction has been completed for n ∈ N, let Kn+1 := (Kn − Fn ) ∩ 2−2n UE (here UE is the open unit ball of E). Then for any element x ∈ Kn there exists y ∈ Kn+1 and z ∈ Fn such that y = x − z. We choose a finite 2−2(n+1) net F˜n+1 for Kn+1 . If now E1 := span{F1 ∪ · · · ∪ Fn ∪ F˜n+1 }, let E2 be a subspace of E such that its dimension equals cardinality of F˜n+1 and E2 ∩ E1 = {0}. We produce Fn+1 from F˜n+1 as in the first step of the construction to achieve that F1 ∪ · · · ∪ Fn+1 consists of linearly independent vectors, Fn+1 is a 2−2(n+1) -net for Kn+1 and Fn+1 ⊂ 2−2n UE . If x ∈ K1 is arbitrary, we can find x2 ∈ K2 and y1 ∈ F1 such that x2 = x − y1 . Then we find x3 ∈ K3 and y2 ∈ F2 such that x3 = x2 − y2 , hence x = x3 + y1 + y2 . By proceeding with this construction we get points xn ∈ Kn , n ≥ 2, and yn ∈ Fn , n ∈ N, such that x = xn+1 + y1 + · · · + yn , n ∈ N. Since xn+1 ∈ Kn+1 ⊂ 2−2(n+1) BE , we get x=
∞ X n=1
yn =
∞ X n=1
2−n (2n yn ).
10.10 Exercises
379
The required sequence {xn } is obtained by putting the elements of 2F1 , 22 F2 , 23 F3 , . . . into a single sequence. Exercise 10.91. Let {xn } be a strongly linearly independent sequence of vectors in a Banach space E that converges to 0. Prove that X := co{xn : n ∈ N} is a Bauer simplex with ext X = {0} ∪ {xn : n ∈ N}. Hint. By Theorem 2.43, ext X ⊂ {xn : n ∈ N} = {0} ∪ {xn : n ∈ N}. Assume that an , bn , n = 0, 1, . . . , are positive numbers such that a0 +
∞ X n=1
and a0 0 +
∞ X
an = b0 +
∞ X
bn = 1
n=1
an xn = b0 0 +
n=1
∞ X
bn xn .
n=1
P∞
Then n=1 (an −bn )xn = 0, and thus the strong linear independency yields an = bn , n ∈ N. Consequently, a0 = b0 . From this we get that X is a simplex and ext X = {0} ∪ {xn : n ∈ N}. Hence X is a Bauer simplex. Exercise 10.92. Let X be a compact convex subset of a Banach space E. Prove that there exist compact convex Bauer simplices X1 , X2 in E such that X ⊂ co(X1 ∪X2 ). Hint. If X is in Rd , the assertion is trivial because then we can find a d-simplex S ⊂ Rd with X ⊂ S. Otherwise we use Exercise 10.90 to find a linearly independent sequence {xn } in E tending to 0 such that X ⊂ co{xn : n ∈ N}. Using Theorem 10.30 we find a partition N1 , N2 of N into two disjoint infinite sets such that the sequences {xn }n∈N1 and {xn }n∈N2 are strongly linearly independent. By Exercise 10.91, the sets Xi := co{xn : n ∈ Ni },
i = 1, 2,
are Bauer simplices and, clearly, X ⊂ co(X1 ∪ X2 ). Exercise 10.93. Let F1 , F2 be nonempty disjoint faces of a compact convex set X such that X = co(F1 ∪ F2 ). Then F2 is the complementary face of F1 . Hint. Let F ⊂ X be a face disjoint from F1 and let x ∈ F \ F2 . Then there exist α ∈ (0, 1) and xi ∈ Fi , i = 1, 2, such that x = αx1 + (1 − α)x2 . Then x1 , x2 ∈ F , a contradiction with F ∩ F1 = ∅.
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10 Deeper results on function spaces and compact convex sets
Exercise 10.94. Let X, Y be compact convex sets and ϕ : ext X → Y be a mapping. Prove that the following conditions are equivalent: (i) There exists a unique continuous affine extension ϕ b : X → Y of ϕ. (ii) There exists a continuous affine extension ϕ b : X → Y of ϕ. c (iii) For every g ∈ A (Y ), the function g ◦ ϕ satisfies conditions (i1) and (i3) of Theorem 6.32. Hint. Obviously, (i) =⇒ (ii). If ϕ b : X → Y is a continuous affine extension of ϕ, g◦ϕ b is a continuous affine extension of g ◦ ϕ. Hence (i1) and (i3) of Theorem 6.32 are satisfied for g ◦ ϕ by Theorem 6.32. Hence (ii) =⇒ (iii). Assume that (iii) holds. By Theorem 6.32, for any g ∈ Ac (Y ) there exists a unique extension g[ ◦ ϕ ∈ Ac (X) of g ◦ ϕ. Let T : Ac (Y ) → Ac (X) be defined as g 7→ g[ ◦ ϕ,
g ∈ Ac (Y ).
By the Minimum principle 2.24, T is a linear positive mapping such that T 1 = 1. Let φ1 : X → S(Ac (X)) and φ2 : Y → S(Ac (Y )) be the affine homeomorphisms given by Proposition 4.31. Then the dual operator T ∗ maps S(Ac (X)) to S(Ac (Y )) and thus we may define ϕ(x) b := (φ2 )−1 (T ∗ (φ1 (x))),
x ∈ X.
Then ϕ b is a continuous affine mapping extending ϕ. Indeed, we get from g(ϕ(x)) b = (T ∗ φ1 (x))(g) = T g(x) = g[ ◦ ϕ(x),
g ∈ Ac (Y ),
that g(ϕ(x)) b = g[ ◦ ϕ(x) = g(ϕ(x)),
g ∈ Ac (Y ), x ∈ ext X.
Hence ϕ b is an extension of ϕ. To finish the proof we have to verify the uniqueness of ϕ. b Let ψ be another continuous affine extension of ϕ. Then for any g ∈ Ac (Y ), g ◦ ψ ∈ Ac (X) and extends g ◦ ϕ. By the minimum principle, g ◦ ψ = g ◦ ϕ. b Hence g ◦ ψ = g ◦ ϕ b for every g ∈ Ac (Y ), and thus ψ = ϕ. b Exercise 10.95. Let Σ be a σ-algebra on a set X and νn : Σ → R, n ∈ N, be signed measures on Σ. Assume that ν(A) := limn→∞ νn (A) exists for every A ∈ Σ. Prove that ν is countably additive on Σ. Hint. Obviously, ν is finitely additive on Σ. Let Ai , i ∈ N, be disjoint sets from Σ. For n ∈ N, let fn : {0, 1}N → R be defined as [ fn (σ) = νn Ai , σ ∈ {0, 1}N . i∈spt σ
Then each fn is countably additive and continuous on {0, 1}N .
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381
Indeed, the countable additivity follows from the fact that νn is countably additive. To show its continuity, let σ k → σ in {0, 1}N and ε > 0 be given. Let i0 ∈ N be such S∞ that |νn |( i=i0 Ai ) < ε. Since σ k → σ on each coordinate, there exists k0 ∈ N such that σ k |i0 = σ|i0 for each k ≥ k0 . Let N := {1, . . . , i0 }. For k ≥ k0 we have |fn (σ k ) − fn (σ)| ≤ |fn (σ k cN ) − fn (σcN )| + |fn (σ k cN\N )| + |fn (σcN\N )| ≤ |νn |
∞ [
∞ [ Ai + |νn | Ai ≤ 2ε.
i=i0
Hence fn is continuous. If f (σ) := ν
i=i0
[
Ai ,
σ ∈ {0, 1}N ,
i∈spt σ
then f is finitely additive and it is a pointwise limit of the sequence {fn }. Hence f has the Baire property in the restricted sense (see Subsection A.2.D) and thus f is continuous and countably additive by Theorem 10.26. Hence ν is countably additive. Exercise 10.96. Prove that a compact convex set X with ext X countable is metrizable. Hint. Since any countable space has a countable network, the assertion follows from Lemma 10.48 and Theorem 10.51. Exercise 10.97. There exists a continuous affine surjection ϕ of a simplex X onto a compact convex set Y and a measure µ ∈ M1max (X) such that •
ϕ(ext X) = ext Y and ϕ is injective on ext X,
•
ϕ] µ ∈ / M1max (Y ).
Hint. Let H1 be the function space on the compact space K1 from Example 3.82 and let K2 be the quotient of K1 where all points of [0, 1] × {0} are identified with the point (0, 0) (see [169], Section 2.4). We write q : K1 → K2 for the quotient mapping and take H2 := {f ∈ C(K2 ) : 2f (0, 0) = f (t, −1) + f (t, 1), t ∈ [0, 1]}. Let X, Y be the state space of H1 , H2 , respectively, and φ1 , φ2 be the respective embeddings. Then ext X = φ1 (K1 \([0, 1]×{0})) and ext Y = φ2 (K2 \{(0, 0)}). We denote by ϕ : X → Y the restriction of the adjoint operator to the operator h 7→ h ◦ q, h ∈ H2 . Then X is a simplex and φ] λ ∈ M1 (X) is maximal for any continuous measure λ ∈ M1 ([0, 1] × {0}), even though φ] λ is carried by a compact set disjoint from ext X. (We recall that λ is continuous if λ({x}) = 0 for each x ∈ X.)
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10 Deeper results on function spaces and compact convex sets
Then ϕ(ext X) = ext Y and ϕ is even injective on ext X. On the other hand, if λ is any continuous probability measure on φ1 ([0, 1] × {0}), then λ is maximal on X, yet the measure ϕ] λ equals the Dirac measure at the point φ2 (0, 0), and hence ϕ] λ is not maximal. Exercise 10.98. There exists a continuous affine surjection ϕ of a metrizable simplex X onto a compact convex set Y such that • ϕ is injective on ext X, •
ϕ] (M1max (X)) ⊂ M1max (Y ) and ϕ] is injective on M1max (X),
•
ϕ is not injective on X.
Hint. Let K1 = {x1 , x2 , x3 , y1 , y2 , y3 } with the discrete topology and K2 be the quotient of K1 , if we identify y2 with x2 . Again, we denote by q : K1 → K2 the quotient mapping. Let H1 := {f ∈ C(K1 ) : 2f (x2 ) = f (x1 ) + f (x3 ), 2f (y2 ) = f (y1 ) + f (y3 )}
and
H2 := {f ∈ C(K2 ) : f (x1 ) + f (x3 ) = 2f (x2 ) = f (y1 ) + f (y3 )}. We take X, Y , φ1 , φ2 and ϕ : X → Y as above. Then X is a simplex, ext X = φ1 (K1 \ {x2 , y2 }), ext Y = φ2 (K2 \ {x2 }), ϕ : ext X → ext Y is a bijection, yet ϕ is not injective on X. Obviously, ϕ] maps injectively maximal measures to maximal measures. Exercise 10.99. There exists a continuous affine surjection ϕ of a simplex X onto a compact convex set Y such that • ϕ is injective on ext X, •
ϕ] (M1max (X)) ⊂ M1max (Y ),
•
ϕ] is not injective on M1max (X).
Hint. Let K1 = [0, 1] ∪ [2, 3] × {−1, 0, 1} endowed again with the “porcupine” topology as in Exercise 10.97 and let K2 be the quotient of K1 after identifying points (t + 2, 0) with (t, 0), t ∈ [0, 1]. Let H1 := {f ∈ C(K1 ) : 2f (t + i, 0) = f (t + i, −1) + f (t + i, 1), t ∈ [0, 1], i = 0, 2}, and H2 := {f ∈ C(K2 ) : 2f (t, 0) = f (t + i, −1) + f (t + i, 1), t ∈ [0, 1], i = 0, 2}, and let X, Y , φ1 , φ2 and ϕ be as above. Then ext X = φ1 (K1 \ (([0, 1] ∪ [2, 3]) × {0})), and ϕ maps injectively ext X onto ext Y .
ext Y = φ2 (K2 \ ([0, 1] × {0})),
10.10 Exercises
383
We claim that ϕ] (M1max (X)) ⊂ M1max (Y ). Indeed, a probability measure λ is maximal on X if and only if λ = (φ1 )] µ for some measure µ ∈ M1 (K1 ) whose restriction to [0, 1] ∪ [2, 3] × {0} is continuous. Similarly, any maximal probability measure on Y is of the form (φ2 )] µ for some measure µ ∈ M1 (K2 ) whose restriction to [0, 1] × {0} is continuous. From these observations the claim follows. Finally, if we take Lebesgue measure λ1 on [0, 1] × {0} and Lebesgue measure λ2 on [2, 3] × {0}, then ϕ] ((φ1 )] λ1 ) = ϕ] ((φ1 )] λ2 ). Hence ϕ] is not injective on M1max (X). Exercise 10.100. Let X be a compact convex set and r : M1 (X) → X be the barycentric mapping. If f ∈ C(X) is understood as an affine continuous function on M1 (X) via the formula µ 7→ µ(f ), µ ∈ M1 (X), show that fr = f ∗ (here fr is the function defined in Notation 10.61). Hint. Use Lemma 3.22. Exercise 10.101. If H is a Bauer simplicial space on K, then H is a CE-function space. Hint. The assertion follows immediately from Proposition 6.37. Exercise 10.102. If H is a CE-function space on a compact space K, prove that ChH (K) is closed. Hint. Let x ∈ K \ ChH (K). By Theorem 3.24, there is f ∈ C(K) such that f (x) < f ∗ (x). Then U := {y ∈ K : f ∗ (y) − f (y) > 0} is an open neighborhood of x and U ∩ ChH (K) = ∅. Exercise 10.103. Let H be a CE-function space on K. Prove that ChH (K) f ∗ is continuous for every f ∈ C(ChH (K)). Hint. Imitate the proof of Lemma 10.80. Exercise 10.104. Let K, L be compact spaces and ϕ : K → L be a continuous surjection. Prove that ϕ is open if and only if fϕ (y) := sup{f (x) : x ∈ ϕ−1 (y)},
y ∈ L,
is continuous on L for each f ∈ C(K). Hint. For necessity use the proof of Lemma 10.63. For sufficiency, asssume that ϕ is not open. Then there exist an open set U ⊂ K and x ∈ U such that ϕ(x) is not in the interior of ϕ(U ). Let f ∈ C(K) be positive such that f (x) > 0 and f = 0 on K \ U . Then the function fϕ is not continuous at ϕ(x).
384
10.11
10 Deeper results on function spaces and compact convex sets
Notes and comments
The idea of the proof of Theorem 10.2 goes back to R. Wittmann [474], where a more general result on the existence of a carrier in the sense of potential theory is established. A remarkable feature of the proof is that neither the Hahn–Banach theorem nor Bauer’s minimum principle are involved, thus Zorn’s lemma is not needed. A more general result than that of Theorem 10.2 is proved in H. Bauer [38]. A different approach yielding a very general existence result of the Shilov boundary containing results previously obtained can be found in R. Wittmann [475]. The relation to the Choquet boundary is studied there in detail. Wittmann’s method is based on the notion of so-called Shilov points. We also refer the reader to H. S. Bear [46]. The idea of I-envelopes can be traced to Section 5 in V. P. Fonf, J. Lindenstrauss and R. R. Phelps [179]. It was elaborated in V. P. Fonf and J. Lindenstrauss [178] where Theorem 10.7 is proved. Proposition 10.6 is taken from a paper by O. Kalenda [258] that along with [257] is devoted to a systematic study of I-envelopes. The notion of I-envelopes points to the following problem. Problem 10.105. Let E be a Banach space. Does there exist a topology τ (or even a locally convex topology τ ) on E ∗ such that I-env(A) = coτ (A) for any A ⊂ E ∗ ? It is not difficult to realize that I-env(A) is just the norm closed convex hull of A in the case when E ∗ is separable. Also, if E does not contain `1 , then it is not difficult to realize the existence of a desired locally convex topology on E ∗ . But in general the problem seems to be open. Theorem 10.8 is a consequence of a theorem by G. Rod´e contained in [390]. For the case of extreme points it was proved by R. Haydon [220]. We refer the reader to papers by G. Godefroy [198] and [199] for more information related to this theorem. Theorem 10.9 is a particular version of R. C. James’ theorem characterizing weakly compact sets in Banach spaces (see [244] and [243]). The presented proof follows the same arguments as the proof of Theorem 3.55 in [173]. The notion of angelicity is due to D. H. Fremlin; it is explicitly defined, for example, in J. Bourgain, D. H. Fremlin and M. Talagrand [80], Definition 3A or in B. Cascales and G. Godefroy [98]. Theorem 10.11 is a simpler variant of Theorem 462B in [183]; it follows the paper by J. D. Pryce [379]. Theorem 10.12 was proved by J. Bourgain and M. Talagrand in [81]; a different proof, which we follow, appeared in S. S. Khurana [268]. Theorem 10.12 provides a positive answer in a particular case to the so-called Boundary problem formulated by G. Godefroy in [198]. Let E be a Banach space and B ⊂ BE ∗ be a boundary of E, that is, any element x ∈ E attains its norm on B. Is it true that any bounded σ(E, B)-compact set A in E is also weakly compact? Positive answers were known for many particular cases: A is convex (see Proposition 5.1(β) in M. De Wilde [136] and also Corollary 1 on p. 100 in K. Floret [176]);
10.11 Notes and comments
385
B is w∗ -relatively sequentially compact in E ∗ (see B. Cascales and G. Vera [101], Corollary C); E = C(K), the space of continuous functions on a compact space K (see B. Cascales and G. Godefroy [98], Theorem 5); E = `1 (Γ) for a set Γ (see B. Cascales and R. Shvydkoy [100], Theorem 4.9]); E does not contain a subspace isomorphic to `1 ([0, 1]) (see B. Cascales, G. Manjabacas and G. Vera [99], Theorem D); E is an L1 -predual, (see J. Spurn´y [422], Theorem 1.1). This problem was solved in full generality by H. Pfitzner [373]. In the same paper he also proves that BE is angelic in the topology σ(E, B) for any boundary B ⊂ BE ∗ . For unbounded sets, an example constructed by W. Moors and E. Reznichenko [350], Section 4 (see also Section 4 in [422]) shows that the angelicity need not hold even for the topology σ(E, ext BE ∗ ). Several results on boundary topologies in L1 preduals can be found in W. B. Moors and J. Spurn´y [351]. Theorem 10.18 is proved in a paper A. J. Lazar [292] as a more general version of the classical Banach–Stone theorem (see for example Theorem 3.43 in [173]). The more precise description of isometries of spaces of affine continuous functions on simplices contained in Theorem 10.17 is given in the paper by T. S. S. R. K. Rao [385] and A. Curnock, J. Howroyd and N. C. Wong [128]. A. J. Ellis and W. S. So presented in [168] an example attributed to J. T. Chan showing that Theorem 10.18 does not hold for general compact convex set. An interesting question is whether a pair K1 , K2 of compact spaces is homeomorphic if the spaces C(K1 ) and C(K2 ) are isomorphic. It has turned out, as shown in a paper by A. Amir [14], that this is the case if there exists an isomorphism T : C(K1 ) → C(K2 ) such that kT k · kT −1 k < 2. H. B. Cohen showed in [122] that the bound 2 is the best possible. The case of C(K) spaces was further generalized for spaces of affine continuous functions on compact convex sets by C. H. Chu and H. B. Cohen [119]. They proved the following result: Let X, Y be compact convex sets such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k · kT −1 k < 2. If • X, Y are metrizable, or • ext X, ext Y are closed sets, then ext X is homeomorphic to ext Y . Another result of [119] is the following theorem: Let X, Y be simplices such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k · kT −1 k < 2. Then ext X is homeomorphic to ext Y . An example constructed in H. U. Hess [224] shows that the assumption on extreme points cannot be omitted. It seems to be an open problem whether the following assertion holds. Problem 10.106. Let X, Y be compact convex sets such that their extreme points are weak peak points and let T : Ac (X) → Ac (Y ) be an isomorphism with kT k·kT −1 k < 2. Is it true that ext X is homeomorphic to ext Y ?
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10 Deeper results on function spaces and compact convex sets
The results of Section 10.3 are based mainly upon papers by J. P. R. Christensen. Theorem 10.24(a) appears in [114], part (b) is standard and we include it just for a comparison (see, for example, Proposition 272O in [181]). The first version of Theorem 10.26 can be found in [114]; we follow the improved variant of [116], Chapter 5, and [456], Theorem 2.3.13. Theorem 10.30 and Exercise 10.92 are taken from J. P. R. Christensen [115]. Theorems 10.31 and 10.37 on automatic boundedness can be found in J. Spurn´y [430] and J. Saint Raymond [407]. Theorem 10.32 can be found in J. P. R. Christensen [116], Theorem 5.8, and [114], Theorem 4. A result related to Theorem 10.32 is due to M. Talagrand who proved in [442] that an affine function f : [0, 1]N → R measurable with respect to the sets with the Baire property is continuous. (We note that [0, 1]N is affinely homeomorphic to (B`∞ , w∗ ).) We do not know the answer to the following problem. Problem 10.107. Let f : X → R be an affine function on a compact convex set X that has the Baire property. Is f necessarily bounded? Section 10.4 follows the paper [220] by R. Haydon. We refer the reader to S. A. Argyros, G. Godefroy and H. P. Rosenthal [19] for information on spaces not containing `1 . Section 10.5 contains results of several authors. Theorem 10.51 can be found in [350] with a slightly different proof; for the case of extreme points it was proved by H. H. Corson [127] and a different proof was given by R. Haydon [219]. The main Theorem 10.56 is an amalgamation of results from papers by M. Herv´e [223], B. MacGibbon [333], J. E. Jayne [247] (see also [394], Section 5.10), G. Debs [141] and E. A. Reznichenko [387]. We remark that our proof of the result from [141] is different since it does not use the α-favorability of the set of extreme points of a compact convex set. We refer the reader to a paper by E. A. Reznichenko [387] where the following result (among others) was proved: Let X be a compact convex set such that •
X is a simplex, or
•
ext X is a Lindel¨of space.
Then w(X) = w(ext X) = nw(ext X). (We recall that w(Y ) and nw(Y ) are the weight and network weight, respectively, of a topological space Y .) It is an open problem whether the conclusion of this theorem holds without additional assumptions on the set X. Problem 10.108. Let X be a compact convex set. Is it true that w(X) = w(ext X) = nw(ext X)? We also mention the following problem from B. MacGibbon [333].
10.11 Notes and comments
387
Problem 10.109. Does there exist a nonmetrizable perfectly normal compact convex set? (We recall that a topological space is perfectly normal if it is normal and every open set is of type Fσ .) Descriptive structure of extreme and exposed points of convex sets was studied by many authors, we refer the reader to H. H. Corson [126], G. Choquet, H. H. Corson and V. Klee [112], J. E. Jayne and C. A. Rogers [248], P. Holick´y and V. Kom´ınek [233], P. Holick´y and M. Laczkovich [234] or P. Holick´y and T. Keleti [232]. The problem of transferring properties of a boundary of a compact convex set to the whole set is investigated in O. Kalenda and J. Spurn´y [260]. Theorem 10.57 is from M. Kaˇcena and J. Spurn´y [254]. It was stated in a more general version without a proof in S. Teleman [455]. However, Example 10.97 from C. J. K. Batty [32] shows that the result may fail in general. Theorem 10.59 and Theorem 10.60 are from D. A. Edwards and G. F. Vincent-Smith [161]; our proofs also use methods of E. A. Reznichenko [387]. An improvement can be found in D. A. Edwards [158], see also G. F. Vincent-Smith [461], A. de la Pradelle [134] and [135]. Theorem 10.66 and Corollary 10.67 are taken from J. Vesterstrøm [460]. Theorem 10.68 for the set of extreme points of a compact convex set is attributed in [5] to G. Choquet (unpublished result). The general result for Choquet boundaries is proved in D. A. Edwards [156] (see also Theorem I.5.13 in [5]). It can be found in G. Choquet [108], Theorem 27.9, that the set of extreme points of a compact convex set is even α-favorable. The result of Subsection 10.7.B can be found in G. Choquet [108], Theorem 29.9, and in R. Haydon [218]. Our exposition follows the paper [218] (see also [24], Section 3.6). P. R. Andeneas showed in [15] that every completely metrizable locally separable space is homeomorphic to ext X for a suitable simplex X. A related result is due to A. J. Lazar [294], who proved that any uncountable Polish space is homeomorphic to the set of extreme points of a closed convex body in `2 . A construction of simplices with a prescribed set of extreme points can be also found in P. J. Stacey [436]. A related result is due to D. Bensimon [51] and it reads as follows: Let M be a complete metric space of zero covering dimension. Then there exists a standard Choquet simplex X such that M is homeomorphic to ext X. Theorem 10.71 is taken from M. Talagrand [445]. We mention here another result of M. Talagrand [448]: There exists a simplex X such that •
ext X is K-analytic,
•
ext X is not contained in the smallest family of subsets of ext X containing compact sets and closed with respect to countable unions and intersections,
•
ext X is of type Kσδ in β(ext X),
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10 Deeper results on function spaces and compact convex sets
•
there exist an open set U ⊂ ext X and a point ω ∈ ext X such that ext X = U ∪ {ω},
•
ext X \ ext X is discrete.
Section 10.8 follows the paper J. Saint Raymond [408]. Subsection 10.9.A is taken from papers A. Clausing and S. Papadapoulou [121], S. Papadopoulou [369] and [368] and R. C. O’Brien [365]. We also refer the reader ˚ Lima [302], A. Clausing and G. M¨agerl [120] and to G. Debs [138] and [140], A. H. H¨ollein [237]. The results of Subsection 10.9.B follow the paper J. Vesterstrøm [460]. Exercise 10.90 is rather standard; we refer the reader to Lemma in J. P. R. Christensen [115] and H. H. Schaefer [409]. Exercise 10.92 is from [115]. Exercise 10.94 can be found in E. M. Alfsen [4], [2] and A. J. Lazar [292]. Exercise 10.95 is a scalar version of the Hahn–Vitali–Saks theorem (see, for example, N. Dunford and J. T. Schwartz [150], pp. 158–159). Examples of Exercises 10.97–10.99 are taken from [254].
Chapter 11
Continuous and measurable selectors
This chapter is devoted to continuous and Borel measurable selectors of multivalued mappings defined on compact convex sets. We start with the proof of the Lazar selection theorem 11.6 which is an affine analogue of the Michael selection theorem. Theorems 11.8 and 11.9 provide its further variants. First applications to affine retractions and extenders are contained in Theorem 11.13. Characterizations of separability of (Ac (X))∗ for the case of a simplex X is given in Theorem 11.14, the Michael selection theorem for compact spaces and the Dugundji theorem are presented as Theorems 11.17 and 11.18. Extension of Baire-α functions from compact subsets of extreme points is investigated in Theorem 11.23 as well as an application on extension of Baire-α functions from compact subsets of completely regular spaces. The last application shows that the mapping T from Definition 6.7 can be pointwise approximated by continuous mappings (see Theorem 11.27). The next section presents a general selection theorem and several of its applications. We start in Subsection 11.5.A with basic facts on measurability of multivalued mappings. The main selection result is contained in Theorem 11.35. Next, Subsection 11.5.C collects several applications of this result. Perhaps the most notable is Talagrand’s result on the existence of a Borel measurable mapping on a metrizable compact convex set X, that assigns to a point x ∈ X a maximal and simplicial measure m(x) representing x (see Theorem 11.41).
11.1
The Lazar selection theorem
Definition 11.1 (Affine mappings). A multivalued mapping ϕ : X → 2E from a convex set X to a vector space E is called affine if ϕ(x) is convex and αϕ(x) + (1 − α)ϕ(y) ⊂ ϕ(αx + (1 − α)y) whenever x, y ∈ X, α ∈ [0, 1]. Definition 11.2 (Semicontinuous mappings). A multivalued mapping ϕ : X → 2Y between topological spaces X and Y is termed lower semicontinuous if ϕ−1 (U ) := {x ∈ X : ϕ(x) ∩ U 6= ∅} is open in X for each open set U ⊂ Y . It is called upper semicontinuous if ϕ−1 (U ) := {x ∈ X : ϕ(x) ⊂ U } is open in X for each open set U ⊂ Y .
390
11 Continuous and measurable selectors d
Lemma 11.3. Let ϕ : X → 2R be a lower semicontinuous nonempty valued affine mapping on a simplex X and let U ⊂ Rd be a neighborhood of 0. Then there exists a continuous affine mapping h : X → Rd such that h(x) ∈ ϕ(x) + U for every x ∈ X. Proof. We proceed by induction on the dimension. If d = 1, let ϕ : X → 2R and U ⊂ R be as in the hypothesis. We find an open symmetric interval V ⊂ R such that 2V ⊂ U . Let g1 (x) := inf(ϕ(x) + V ),
x∈X
g2 (x) := sup(ϕ(x) + V ),
and
x ∈ X.
Since ϕ is lower semicontinuous and affine, g2 , −g1 are lower semicontinuous concave functions with g1 ≤ g2 . By the Edwards in-between theorem 6.6 there exists a continuous affine function h : X → R such that g1 ≤ h ≤ g2 . By the choice of V , h(x) ∈ ϕ(x) + U for every x ∈ X. d+1 Assume that the assertion holds for Rd . Let ϕ : X → 2R be a lower semicontinuous affine mapping and U be a neighborhood of 0. Let p and q be canonical projections of Rd+1 onto R and Rd , respectively. Let Up and Uq be open symmetric neighborhoods of 0 in R and Rd , respectively, such that Up × Uq ⊂ U . As p◦ϕ : X → 2R is lower semicontinuous and affine, we may apply the preceding argument to get a continuous affine function k : X → R such that k(x) ∈ (p ◦ ϕ)(x) + Up , Then the mapping from X to 2R
d+1
x ∈ X.
(11.1)
defined as
p−1 (k(x) + Up ),
x ∈ X,
is affine and lower semicontinuous. Moreover, ψ(x) := p−1 (k(x) + Up ) ∩ ϕ(x),
x ∈ X,
(11.2)
d+1
is a nonempty valued mapping from X to 2R . We claim that ψ is affine and lower semicontinuous. Since the intersection of a pair of affine mappings is again affine, we proceed to the lower semicontinuity of ψ. Let V ⊂ Rd+1 be an open nonempty set. Given a point x ∈ ψ−1 (V ), we find t ∈ ψ(x) ∩ V . Let W be an open ball around t such that W ⊂ p−1 (k(x) + Up ). Since the function k is continuous, we can find an open neighborhood G1 of x such that W ⊂ p−1 (k(y) + Up ) for each y ∈ G1 . As ϕ is lower semicontinuous and x ∈ ϕ−1 (W ), there exists an open set G2 containing x such that G2 ⊂ ϕ−1 (W ).
11.1 The Lazar selection theorem
391
We claim that G1 ∩ G2 ⊂ ψ−1 (V ). Indeed, if y ∈ G1 ∩ G2 , there is a point s ∈ ϕ(y) ∩ W . Since y ∈ G1 , s ∈ p−1 (k(y) + Up ). Then s ∈ p−1 (k(y) + Up ) ∩ ϕ(y) ∩ W ⊂ ψ(y) ∩ V, proving the claim. d Obviously, the mapping q ◦ ψ : X → 2R is also affine and lower semicontinuous. By the induction hypothesis, there exists an affine mapping l : X → Rd such that l(x) ∈ (q ◦ ψ)(x) + Uq ,
x ∈ X.
(11.3)
To finish the proof it is enough to verify that the mapping h : x 7→ (k(x), l(x)),
x ∈ X,
is the desired one. Obviously, h is continuous and affine. Let x ∈ X be given. By (11.3), l(x) = q(t) + uq for some t ∈ ψ(x) and uq ∈ Uq . As t ∈ ψ(x), p(t) = k(x) + up for some up ∈ Up . Thus (k(x), l(x)) = (p(t), q(t)) + (−up , uq ) = t + (−up , uq ) ∈ ϕ(x) + U. This finishes the inductive step as well as the proof. Lemma 11.4. Lex ϕ : X → 2E be an affine lower semicontinuous nonempty valued mapping from a simplex X to a locally convex space E and let U ⊂ E be a neighborhood of 0. Then there exists a continuous affine mapping h : X → E such that h(x) ∈ ϕ(x) + U for every x ∈ X. Proof. Let ϕ : X → 2E and U ⊂ E be as in the hypothesis. We may assume that U is an open convex balanced neighborhood of 0 in E. Since ϕ is lower semicontinuous, the family 1 n 1o , y ∈ E, Gy := ϕ−1 y − U = x ∈ X : y ∈ ϕ(x) + 2 2 is an open covering of X. We choose y1 , . . . , yd such that X⊂
d [
G yi .
(11.4)
i=1
By setting d n X 1 o d ψ(x) := λ = (λ1 , . . . , λn ) ∈ R : λi yi ∈ ϕ(x) + U , 2 i=1
we get an affine lower semicontinuous mapping with nonempty values.
x ∈ X,
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11 Continuous and measurable selectors
Indeed, (11.4) yields that ψ is nonempty valued, and, obviously, ψ is affine. Its lower semicontinuity follows from the observation that d n X o 1 λi yi − U 6= ∅ . ψ(x) = λ ∈ Rd : ϕ(x) ∩ 2 i=1
P Let W ⊂ Rd be an open convex balanced neighborhood of 0 such that di=1 wi yi ∈ 1 4 U whenever w = (w1 , . . . , wd ) ∈ W . Lemma 11.3 provides an affine continuous mapping h0 : X → Rd such that h0 (x) = (h1 (x), . . . , hd (x)) ∈ ψ(x) + W,
x ∈ X.
Then the mapping h : X → E defined as h(x) :=
d X
hi (x)yi ,
x ∈ X,
i=1
is the desired continuous affine mapping. Indeed, given x ∈ X, let λ ∈ ψ(x) and w ∈ W be such that h0 (x) = λ + w. Then d X i=1
hi (x)yi =
d X i=1
λ i yi +
d X i=1
1 1 wi yi ∈ ϕ(x) + U + U ⊂ ϕ(x) + U. 2 4
This finishes the proof. Lemma 11.5. Let X be a simplex, E be a locally convex space, ϕ : X → 2E be a lower semicontinuous affine multivalued mapping, h : X → E be a continuous mapping and U ⊂ E be a convex balanced neighborhood of 0. Then ψ(x) := ϕ(x) ∩ (h(x) + U ),
x ∈ X,
is a lower semicontinuous affine mapping provided ψ(x) is nonempty for each x ∈ X. Proof. Let V be an open subset of E and x ∈ ψ−1 (V ). We choose t ∈ ϕ(x) ∩ (h(x) + U ) ∩ V . Let u ∈ U be such that t = h(x) + u. We find a balanced open convex neighborhood W of 0 such that t + W ⊂ V and u + 2W ⊂ U . Let G1 , G2 be open neighborhoods of x such that G1 ⊂ ϕ−1 (t + W ) and
h(y) − h(x) ∈ W for y ∈ G2 .
Let y ∈ G1 ∩ G2 . We pick s ∈ ϕ(y) ∩ (t + W ) and let w ∈ W satisfy s = t + w. Then s = t + w = h(y) + u + w + h(x) − h(y) ∈ h(y) + u + w + W ⊂ h(y) + u + 2W ⊂ h(y) + U
11.1 The Lazar selection theorem
393
and s ∈ ϕ(y) ∩ (t + W ) ⊂ ϕ(y) ∩ V. Hence s ∈ ϕ(y) ∩ (h(y) + U ) ∩ V and G1 ∩ G2 ⊂ ψ−1 (V ). This concludes the proof. Theorem 11.6 (Lazar). Let ϕ : X → 2E be a lower semicontinuous affine mapping from a simplex X to a Fr´echet space E with nonempty closed values. Then there exists a continuous affine mapping h : X → E such that h(x) ∈ ϕ(x) for each x ∈ X. Proof. Given ϕ : X → 2E , let ρ be a translation invariant complete metric on E. We select a decreasing sequence {Un } of open convex balanced neighborhoods of 0 such that they form a base of neighborhoods of 0. We take p0n to be the Minkowski functional of the set Un and define pn := p01 + · · · + p0n . Then {pn } is an increasing sequence of pseudonorms on E that determines its topology. Moreover, if {yn } is a sequence in E which is Cauchy in every pseudonorm pn , it is convergent. For y ∈ E and r > 0 we set Un (y, r) := {z ∈ E : pn (y − z) < r}. By Lemma 11.4, there exists a continuous affine mapping h1 : X → E such that h1 (x) ∈ ϕ(x) + U1 (0, 2−1 ),
x ∈ X.
We inductively find continuous affine mappings hn : X → E such that hn+1 (x) ∈ ϕ(x) ∩ (hn (x) + Un (0, 2−n ) + Un (0, 2−n−1 ), x ∈ X, for each n ∈ N. For the inductive step, suppose that the mappings hk have been found for k = 1, . . . , n. Then we apply Lemma 11.4 to ϕ(x) ∩ (hn (x) + Un (0, 2−n )),
x ∈ X,
to get the desired mapping hn+1 (here we use Lemma 11.5). The completeness of E guarantees that the sequence {hn } converges uniformly to a function h : X → E. Since ϕ has closed values, h(x) ∈ ϕ(x) for each x ∈ X and the proof is complete. Theorem 11.7. Let X be a compact convex set. Then the following assertions are equivalent. (i) X is a simplex, (ii) any lower semicontinuous affine mapping ϕ : X → 2E with nonempty closed convex values in a Fr´echet space E admits a continuous affine selection.
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11 Continuous and measurable selectors
Proof. By Theorem 11.6, (i) =⇒ (ii). Conversely, assuming (ii), let f, −g ∈ C(X) be convex functions such that f ≤ g. Then ϕ : X → 2R defined as ϕ(x) := [f (x), g(x)],
x ∈ X,
is an affine lower semicontinuous function. According to the assumption, there exists a continuous affine function h ∈ Ac (X) such that f ≤ h ≤ g. By Theorem 6.6 and Theorem 6.56, X is a simplex. Theorem 11.8. Let F be a closed face of a simplex X, ϕ : X → 2E be an affine lower semicontinuous mapping to a Fr´echet space E with nonempty closed convex values and f : F → E be a continuous affine mapping with f (x) ∈ ϕ(x), x ∈ F . Then there exists a continuous affine selection g : X → E from ϕ such that g = f on F . Proof. Given the objects as in the hypotheses, let ψ : X → 2E be defined as ( f (x), x ∈ F, ψ(x) := ϕ(x), x ∈ X \ F. A routine verification shows that ψ is a lower semicontinuous affine mapping and thus admits a continuous affine selection g. Obviously, g satisfies the required conditions. Corollary 11.9. Let ϕ : X → 2K be a lower semicontinuous affine mapping from a simplex X to a metrizable compact convex set K. Then ϕ admits a continuous affine selection. Proof. By Proposition 2.45, we may assume that K is a compact convex subset of `2 . The conclusion now follows from Theorem 11.6.
11.2
Applications of the Lazar selection theorem
Lemma 11.10. Let H be a function space on a compact space K and let F be a metrizable subset of K. Then coH F is metrizable as well. Proof. If F is metrizable, M1 (F ) is metrizable as well. By Proposition 8.18, the barycentric mapping from Proposition 3.37 r : M1 (F ) ∩ M(H) → coH F is surjective. Since a continuous image of a compact metrizable space is metrizable (see Lemma A.34), the proof is complete. Lemma 11.11. Let H be a simplicial function space on a compact space K. Let F ⊂ ChH (K) be a compact set and let C = coH F . Then
11.2 Applications of the Lazar selection theorem
395
(a) δx (F ) = 1 for each x ∈ C, (b) C is a Choquet set, (c) the mapping x 7→ δx , x ∈ C, is continuous. Proof. Let F be as in the hypothesis and x ∈ C. By Proposition 8.18, there exists a measure µ ∈ Mx (H) carried by F . Then µ is H-maximal and µ = δx as H is simplicial. Since (b) follows from Proposition 8.31, we proceed to (c). Let {xα } be a net of points in C converging to x and assume that δxα 9 δx . By passing to a subnet if necessary we may assume that δxα → µ where µ 6= δx . Then µ ∈ Mx (H) and, moreover, µ is maximal (see Corollary 3.60). By the simpliciality of H, µ = δx , a contradiction. This concludes the proof. Definition 11.12 (Retractions and extenders). If F is a subset of a topological space K, a continuous mapping ρ : K → F is a retraction if ρ(x) = x for x ∈ F . The set F is called a retract. Let F be a subset of compact space K and H be a subspace of C(K). Let G := {h|F : h ∈ H}. A mapping L:G→H is a called an extender if L is a linear operator, and for each f ∈ H, • Lf = f on F , • (Lf )(K) ⊂ co f (F ). Theorem 11.13. Let X be a simplex. (a) If F is a closed metrizable face of X, then there exists an affine retraction ρ : X → F (that is, ρ : X → F is a continuous affine mapping such that ρ(x) = x, x ∈ F ). (b) If F is a closed metrizable face of X, then there exists an extender L : Ac (F ) → Ac (X). (c) Let K ⊂ ext X be a metrizable compact set. Then there exists an extender L : C(K) → Ac (X). Proof. For the proof of (a), let ϕ : X → 2F be defined as ( x, x ∈ F, ϕ(x) := F, x ∈ X \ F. Then ϕ satisfies the assumptions of Corollary 11.9 and thus ϕ admits a continuous affine selector. This selector is the required affine retraction of X onto F . Assertion (b) immediately follows from (a), since the mapping h 7→ h ◦ ρ, is the required extender.
h ∈ Ac (F ),
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11 Continuous and measurable selectors
For the proof of (c) we notice that F = co K is a closed metrizable face such that δx (K) = 1 for each x ∈ F and the mapping x 7→ δx , x ∈ F , is continuous (see Lemma 11.10 and Lemma 11.11). If ρ : X → F is a continuous affine retraction, the operator Lf (x) := δρ(x) (f ), x ∈ X, f ∈ C(K), is the required extender. This concludes the proof. Theorem 11.14. For a metrizable simplex X, the following assertions are equivalent: (i) ext X is uncountable, (ii) Ac (X) contains an isometric copy of C({0, 1}N ), (iii) Ac (X) contains an isometric copy of `1 , (iv) (Ac (X))∗ is nonseparable. Proof. Assume that ext X is uncountable. Since ext X is a Polish space (see Proposition 3.43), ext X contains a homeomorphic copy C of {0, 1}N . By Theorem 11.13(c), there exists an extender L : C(C) → Ac (X). Hence L(C(C)) is an isometric copy of C({0, 1}N ) contained in Ac (X). Thus (i) =⇒ (ii). Since C({0, 1}N ) contains an isometric copy of any separable Banach space (see [173], proof of Theorem 5.17), (ii) =⇒ (iii). Obviously, (iii) =⇒ (iv). Finally, assume that ext X is countable. Then Mbnd (X) is isometric to `1 (ext X) and the mapping π : Mbnd (X) → (Ac (X))∗ defined as π(µ) := µ|(Ac (X))∗ is easily seen to be an isometry (use Proposition 6.9). Hence (Ac (X))∗ is separable, which proves (iv) =⇒ (i). Theorem 11.15. Let H be a simplicial function space on a compact space K. Let F ⊂ K be a metrizable Choquet set and A := {h|F : h ∈ Ac (H)}. Then there exists an extender L : A → Ac (H). Proof. Assume that F is a Choquet set in K, X := S(Ac (H)) and φ : K → X is the embedding from Section 4.3. By Theorem 8.62, F is a P -set with respect to the function space Ac (H). As F is a Choquet set in K with respect to Ac (H) (see Lemma 8.16), by Theorem 8.39 there exists a closed face H ⊂ X such that φ(F ) = H ∩ φ(K). By Proposition 4.26(c) and Proposition 2.64, φ(F ) = H ∩ φ(K) ⊃ H ∩ ext X = ext H.
(11.5)
By the Milman theorem 2.43, co φ(F ) = H. By Lemma 11.10, H is metrizable.
11.2 Applications of the Lazar selection theorem
397
It follows from (11.5) that, if b h1 , b h2 ∈ Ac (X) satisfy b h1 = b h2 on φ(F ), then b h1 = c b h2 on H. Thus for any function h ∈ A there exists a unique function E1 h ∈ A (H) such that E1 h ◦ φ = h on F . Then E1 is a linear positive operator of norm 1. Since X is a simplex (see Theorem 6.54), Theorem 11.13 provides an extender E2 : Ac (H) → Ac (X). Then L : A → Ac (H) defined as Lh = (E2 ◦ E1 )(h) ◦ φ,
h ∈ A,
is the required extender. This concludes the proof. Theorem 11.16. Let H be a simplicial function space on a compact space K. Let F ⊂ ChH (K) be a metrizable compact set. Then there exists an extender L : C(F ) → Ac (H). Proof. Let X denote the simplex S(Ac (H)) and φ : K → X be the embedding from Section 4.3. If F ⊂ ChH (K) is a compact metrizable set, we use Theorem 11.13(c) to obtain an extender E : φ(F ) → Ac (X). Then Lf (x) := E(f ◦ φ−1 )(φ(x)),
x ∈ K, f ∈ C(F ),
is an extender from C(F ) into Ac (H). Theorem 11.17 (The Michael selection theorem). Let ϕ : K → 2E be a lower semicontinuous mapping from a compact space K to a Fr´echet space E such that ϕ has nonempty closed convex values. Then ϕ admits a continuous selection. Proof. Given ϕ : K → 2E as above, let ϕ b : M1 (K) → 2E be defined as ( ϕ(x), µ = εx , x ∈ K, ϕ(µ) b := E otherwise. It is straightforward to verify that ϕ b is a lower semicontinuous affine mapping with nonempty closed convex values. Theorem 11.6 provides an affine continuous selection fb : M1 (K) → E for ϕ. b Then f (x) := fb(εx ),
x ∈ K,
is a continuous selection for ϕ. Theorem 11.18 (Dugundji). Let Y be a metrizable compact subset of a completely regular space X and f : Y → E be a continuous mapping to a locally convex space E. Assume that co f (Y ) is a compact subset of E (this is automatically satisfied if E is a Fr´echet space). Then there exists a continuous mapping F : X → E such that F = f on Y and F (x) ⊂ co f (Y ), x ∈ X.
398
11 Continuous and measurable selectors
ˇ Proof. If X is as in the theorem, we consider its Cech–Stone compactification βX. 1 1 Then F := M (Y ) is a metrizable closed face of M (βX). Moreover, the function fb: F → E defined as Z µ 7→
f (y) dµ(y),
µ ∈ F,
Y
is continuous (we consider the Pettis integral). We remark that fb is well defined by the assumption on co f (Y ). By Corollary 11.13, there exists a continuous affine retraction r : M1 (βX) → F . Then F (x) := fb(r(εx )), x ∈ βX, is a continuous extension of f with the required properties. This concludes the proof.
Theorem 11.19 (Borsuk–Dugundji). Let X be a completely regular space and Y its metrizable compact subset. Then there exists a linear extender L : C(Y ) → C b (X). ˇ Proof. We consider the Cech–Stone compactification βX of X and let H := C(βX). c Then H = A (H) and ChH (βX) = βX. Thus the extender from Theorem 11.16 is the desired one.
11.3
The weak Dirichlet problem for Baire functions
Let H be a function space on a compact space K. We recall the inductive definition of functions of H-affine classes from Section 5.6. For any countable ordinal α and a family of functions H we define H0,b := H and having Hβ,b , β < α, already defined for an ordinal number α ∈ (0, ω1 ), we define H Sα,b to be the space of all pointwise limits of bounded sequences of functions from β<α Hβ,b . We assume in Lemmas 11.20–11.22 that X is a simplex. Lemma 11.20. Let H := {fn : n ∈ N} be a bounded family of continuous affine functions on a closed face F ⊂ X. Then there exists a family A = {hn : n ∈ N} ⊂ Ac (X) and a mapping r : X → F such that r(x) = x on F and hn (x) = fn (r(x)) for all n ∈ N and all x ∈ X. Proof. Let H be as in the hypothesis. Without loss of generality we may assume that fn (F ) ⊂ [0, 1] for all n ∈ N. We consider a mapping ϕ : F → [0, 1]N (the space [0, 1]N is considered with the pointwise topology) defined as x 7→ {fn (x)}n∈N ,
x ∈ F.
11.3 The weak Dirichlet problem for Baire functions
399
Then ϕ is continuous and affine on F . We define a multivalued mapping Φ : X → N 2[0,1] as ( ϕ(x), x ∈ F, Φ(x) := ϕ(F ), x ∈ X \ F. Then Φ has nonempty convex compact values and Φ is affine. (Here we use the assumption that F is a face.) Since Φ = ϕ on the closed set F and Φ = ϕ(F ) on the complement of F , it is easy to verify that Φ is even lower semicontinuous. By Theorem 11.6, there exists an affine continuous mapping ψ : X → [0, 1]N such that ψ(x) ∈ Φ(x) for all x ∈ X. Set hn := πn ◦ ψ, n ∈ N, where πn : [0, 1]N → R is the projection to the n-th coordinate. Clearly, each hn is a continuous affine function on X. Since ψ = ϕ on F , hn |F = fn |F for all n ∈ N. If x ∈ / F , ψ(x) = ϕ(y) for some y ∈ F . If we denote this y as r(x) and set r(x) := x on F , we obtain the desired mapping r, which concludes the proof. Lemma 11.21. Let H, F , A and r be as in Lemma 11.20. Then, for every f ∈ Hα,b , there exists a function h ∈ Aα,b such that h(x) = f (r(x)) for all x ∈ X. Proof. If α = 0, that is, f = fn for some n ∈ N, take h := hn . Suppose that the lemma has been verified for all β < α where α ∈ (0, ω1 ). Pick f ∈ Hα,b . Then there exist functions fn ∈ Hαn ,b , n ∈ N, where αn < α, such that {fn } is bounded and fn → f on F . By the induction hypothesis, for every n ∈ N we are able to select hn ∈ Aαn ,b such that hn (x) = fn (r(x)) for every x ∈ X. Since fn (r(x)) → f (r(x)), the function h := limn→∞ hn is well defined, h ∈ Aα,b and h(x) = f (r(x)) for all x ∈ X. This finishes the proof. Lemma 11.22. Let F be a bounded family of continuous functions on a compact set K ⊂ ext X nad let T be the operator from Definition 6.7. Set F := co K. (a) The family H := {T f |F : f ∈ F} consists of continuous affine functions on F . (b) If f ∈ F α,b , then T f |F ∈ Hα,b . Proof. Assertion (a) follows from Lemma 11.11 and (b) can be obtained by a straightforward use of transfinite induction. Theorem 11.23. Let X be a simplex and K ⊂ ext X be compact. Then for any bounded function f on K of Baire class α there exists a function h ∈ Aα (X) such that h = f on K and h(X) ⊂ co f (K). Proof. By setting F := co K we obtain a closed face in X (see Lemma 11.11). Given a function f ∈ B bα (K), we find a countable family F := {fn : n ∈ N} ⊂ C(K) such that f ∈ F α,b and fn (K) ⊂ co f (K) for every n ∈ N (for the existence of such a set use transfinite induction).
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11 Continuous and measurable selectors
Define gn := T fn |F , n ∈ N. By Lemma 11.22(b), the family H := {gn : n ∈ N} consists of continuous affine functions on F . Moreover, the values of every function in H are contained in co f (K). Lemma 11.20 provides a family A = {hn : n ∈ N} of continuous affine functions on X and a mapping r : X → F such that r(x) = x on F and hn (x) = gn (r(x)) for each x ∈ X and n ∈ N. Since T f |F ∈ Hα,b due to Lemma 11.22(b), an application of Lemma 11.21 finishes the proof. Theorem 11.24. Let H be a simplicial function space on a compact space K and F ⊂ ChH (K) be compact. Then for every f ∈ B bα (F ) there exists a function h ∈ (Ac (H))α,b such that f = h on K and h(K) ⊂ co f (F ). Proof. Let X denote the simplex S(Ac (H)) and φ : K → X be the embedding from Section 4.3. Given f ∈ B bα (F ), Theorem 11.23 provides a function h ∈ Aα (X) such that h = f ◦ φ−1 on F and h(X) ⊂ co f (F ). Then h ◦ φ ∈ (Ac (H))α,b and h = f on F . This finishes the proof. Corollary 11.25. Let K be a compact subset of a completely regular space X. Then for every bounded function f on K of Baire class α there exists a bounded function h on X of Baire class α such that f = h on K and h(X) ⊂ co f (K). ˇ Proof. We consider the Cech–Stone compactification βX of X and let H := C(βX). c Then H = A (H) is simplicial and ChH (βX) = βX. By Theorem 11.24, any function f ∈ B bα (F ) can be extended to a function h ∈ Hα,b = B bα (βX) such that h(βX) ⊂ co f (F ). This concludes the proof.
11.4
Pointwise approximation of maximal measures
Let H be a simplicial function space on a metrizable compact space K and X be the state space S(Ac (H)). By Theorem 6.54, X is a simplex. Let π be the quotient mapping from Definition 4.25. Theorem 11.26. There exists a sequence {γn } of continuous affine mappings, γn : X → M1 (K), such that π(γn (s)) → s for every s ∈ X. Proof. Let {hk : k ∈ N} be a dense subset of Ac (H). For any n ∈ N we define a mapping ϕn on S(Ac (H)) whose values are subsets of M1 (K) as n n \ 1o ϕn (s) := µ ∈ M1 (K) : |µ(hk ) − s(hk )| < , n
s ∈ X.
k=1
By (4.9), for each s ∈ X there exists µ ∈ M1 (K) such that π(µ) = s, and then µ ∈ ϕn (s). Hence all maps ϕn have nonempty values. It is straightforward to verify that ϕn is affine, and so also ϕn is affine. (Here ϕn is the mapping which assigns to each s ∈ X the closure of ϕn (s) in M1 (K).)
11.4 Pointwise approximation of maximal measures
401
We show that ϕn and ϕn are lower semicontinuous. To this end, let n ∈ N and let V be an open subset of M1 (K). If s ∈ (ϕn )−1 (V ) = {s ∈ X : ϕn (s) ∩ V 6= ∅}, then there exists a measure µ ∈ V such that |µ(hk ) − s(hk )| <
1 n
for any k = 1, . . . , n. There exists an open neighborhood U of s such that for any t ∈ U and k = 1, . . . , n we have |µ(hk ) − t(hk )| <
1 . n
Hence µ ∈ ϕn (t) and we see that the set (ϕn )−1 (V ) is open. Since {s ∈ X : ϕn (s) ∩ V 6= ∅} = {s ∈ X : ϕn (s) ∩ V 6= ∅}, the mapping ϕn is also lower semicontinuous. By Theorem 11.9, for any n ∈ N there exists a continuous affine mapping γn : X → M1 (K) such that γn (s) ∈ ϕn (s) for any s ∈ X. If s ∈ X, then obviously γn (s)(hk ) → s(hk ) for each k ∈ N. Since the set {hk : k ∈ N} is dense in Ac (H), it immediately follows that γn (s)(h) → s(h) for any h ∈ Ac (H). This verifies that π(γn (s)) → s. Theorem 11.27. There exists a sequence of positive linear continuous operators Tn : C(K) → Ac (H) such that Tn (f )(x) → δx (f ) for any f ∈ C(K) and any x ∈ K. Proof. As in Section 4.3, let φ be the evaluation mapping from K into X. We may apply Theorem 11.26 to obtain a sequence of continuous affine mappings γn : X → M1 (K) such that π(γn (s)) → s for each s ∈ X. If n ∈ N and f ∈ C(K), set τn (f )(s) := γn (s)(f ), Tn f (x) := τn (f )(φ(x)),
s ∈ X, x ∈ K.
Obviously, {Tn } is a sequence of positive linear continuous operators on C(K) with values in C(K). By (4.10), Tn f ∈ Ac (H) for each f ∈ C(K). Fix x ∈ ChH (K). We claim that γn (φx ) → εx . (11.6) Otherwise we would have γnk (φx ) → µ 6= εx for a subsequence, in view of the compactness of M1 (K). Then we would obtain π(µ) = lim π(γnk (φx )) = φx . k→∞
402
11 Continuous and measurable selectors
Since εx is the only H-representing measure for x, it would follow that µ = εx . This contradiction proves (11.6). Given f ∈ C(K), we see that, for any x ∈ ChH (K), Tn f (x) = τn (f )(φx ) = γn (φx )(f ) → εx (f ) = f (x). Now, let x ∈ K be arbitrary. Since in metrizable case maximal measures are carried by ChH (K) (see Theorem 3.79), the Lebesgue dominated convergence theorem assures that Z Z f (y)dδx (y) = δx (f ) Tn f (y)dδx (y) → Tn f (x) = ChH (K)
ChH (K)
for any f ∈ C(K), which finishes the proof of the theorem.
11.5 11.5.A
Measurable selectors Multivalued mappings
Definition 11.28 (Measurable mappings). Let S be a family of sets in a set X and ϕ : X → 2Y be a mapping from X to a topological space Y . We say that ϕ is S -lower semimeasurable if ϕ−1 (U ) := {x ∈ X : ϕ(x) ∩ U 6= ∅} ∈ S for each open U ⊂ Y . If ϕ−1 (U ) := {x ∈ X : ϕ(x) ⊂ U } ∈ S for each open U ⊂ Y , ϕ is called S -upper semimeasurable. The mapping ϕ is S -measurable if ϕ is both S -upper and S -lower measurable and S -semimeasurable if ϕ is S -upper measurable or S -lower measurable. In the sequel we use the notation from Definition 5.2 for a family S of sets in a set X. Lemma 11.29. Let S be an algebra and let ϕ : X → 2Y be a mapping from X to a metrizable space Y . (a) If ϕ is S -upper semimeasurable, then ϕ is Σ2 (S )-lower semimeasurable. (b) If ϕ is S -lower semimeasurable and has compact values, then ϕ is Σ2 (S )-upper semimeasurable. (c) If ϕ1 , . . . , ϕn : X → 2Y are S -lower measurable and Y is separable, then the n mapping ψ(x) := ϕ1 (x) × · · · × ϕn (x), x ∈ X, from X to 2Y , is Σ2 (S )-lower semimeasurable.
11.5 Measurable selectors
403
Proof. We select a compatible metric on Y . For the proof of (a), let S ϕ : X → 2Y be S -upper measurable and let U ⊂ Y be an open set. We write U = ∞ n=1 Fn , where Fn ⊂ Y are closed. Then ∞ [
ϕ−1 (U ) =
{x ∈ X : ϕ(x) ∩ Fn 6= ∅}
=
n=1 ∞ [
X \ ϕ−1 (Y \ Fn )
n=1
is in Σ2 (S ). To verify (b), let ϕ : X → 2Y be S -lower semimeasurable with compact values S F , where and let U ⊂ Y be an open set. Again, we write U = ∞ n=1 n n 1o Fn := y ∈ U : dist(y, Y \ Fn ) ≥ , n ∈ N. n If K ⊂ Y is compact, then K ⊂ U if and only if K ⊂ Fn for some n ∈ N. Hence ϕ−1 (U ) =
∞ [
ϕ−1 (Fn ) =
n=1
∞ [
(X \ ϕ−1 (Y \ Fn )) ,
n=1
and ϕ is Σ2 (S )-upper semimeasurable. To verify (c), we first notice that, by the separability of Y , it is enough to check that ψ−1 (U ) ∈ Σ2 (S ) for any set of the form U = U1 × · · · × Un , where U1 , . . . , Un are open in Y . But this is immediate, because ψ−1 (U ) =
n \
{x ∈ X : ϕi (x) ∩ Ui 6= ∅} ∈ S .
i=1
This concludes the proof. Lemma 11.30. Let Y be a metrizable separable space, S be an algebra and ϕn : X → 2Y , n ∈ N, be Σ2 (S )-upper semimeasurable mappings with closed values such that ϕ1 has compact values. Then ϕ(x) :=
∞ \
ϕn (x),
x ∈ X,
n=1
is a Σ2 (S )-upper semimeasurable mapping. Proof. We select a compatible metric on Y and a countable base B of open sets in Y that is stable with respect to finite unions. We fix an open set U ⊂ Y . Then [ −1 (ϕ1 ∩ ϕ2 )−1 (U ) = (ϕ−1 (11.7) 1 (V ∪ U ) ∩ ϕ2 (Int(Y \ V ) ∪ U )). V ⊂Y open
404
11 Continuous and measurable selectors
To verify this, let ϕ1 (x) ∩ ϕ2 (x) ⊂ U . Since ϕ1 (x) \ U is compact, dist(ϕ1 (x) \ U, ϕ2 (x) \ U ) > 0. Let V ⊂ Y be an open set such that ϕ1 (x) \ U ⊂ V and dist(V , ϕ2 (x) \ U ) > 0. Then ϕ1 (x) ⊂ V ∪ U and ϕ2 (x) ⊂ Int(Y \ V ) ∪ U. Hence we get “⊂” in (11.7). Since “⊃” is obvious, (11.7) follows. If ϕ1 (x) ⊂ V ∪ U for an open set V ⊂ Y , then by the compactness of ϕ1 (x) \ U there exists a set B ∈ B such that B ⊂ V and ϕ1 (x) ⊂ B ∪ U . If ϕ2 (x) ⊂ Int(Y \ V ) ∪ U , then ϕ2 (x) ⊂ Int(Y \ B) ∪ U . Thus [ −1 (ϕ1 ∩ ϕ2 )−1 (U ) = (ϕ−1 1 (B ∪ U ) ∩ ϕ2 (Int(Y \ B) ∪ U )) B∈B
is in Σ2 (S ). It follows that ϕ1 ∩ ϕ2 is Σ2 (S )-upper semimeasurable. By induction we prove that ϕ1 ∩ · · · ∩ ϕn is Σ2 (S )-upper semimeasurable, n ∈ N. Given an open set U ⊂ Y , using a compactness argument, we obtain that ϕ−1 (U ) =
[
(ϕ1 ∩ · · · ∩ ϕn )−1 (U )
n∈N
is in Σ2 (S ). Hence ϕ is Σ2 (S )-upper semimeasurable. Lemma 11.31 (Reduction lemma). If S is an algebra and {An : n ∈ N} is a cover of X consisting of sets from Σ2 (S ), then there exists a disjoint partition {Bn : n ∈ N} of X consisting of sets from Σ2 (S ) such that Bn ⊂ An , n ∈ N. S Proof. The proof is analogous to that of Lemma 5.4(f). Let An = ∞ k=1 Ank , where Ank ∈ S , n, k ∈ N. We enumerate the family {Ank : n, k ∈ N} as {Bj0 } and define B1 := B10 ,
0 Bj := Bj0 \ (B10 ∪ · · · ∪ Bj−1 ),
j ≥ 2.
Then the sets Bj are in S . We set A01 :=
[
{Bj : Bj ⊂ A1 }
A0n :=
[
{Bj : Bj ⊂ An , Bj * A01 ∪ · · · ∪ A0n−1 },
and n ≥ 2.
Then A0n ⊂ An and A0n ∈ SΣ2 (S ), n ∈ N. Moreover, {A0n : n ∈ N} is a disjoint family whose union equals ∞ n=1 An .
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Lemma 11.32. Let (Y, ρ) be a separable metric space, S be an algebra in X and ϕ : X → 2Y be a Σ2 (S )-upper semimeasurable mapping with compact values. Then {x ∈ X : diam ϕ(x) < ε} ∈ Σ2 (S ) for each ε > 0. Proof. We select a countable base B of open sets in Y that is stable with respect to finite unions. We claim that [ {x ∈ X : diam ϕ(x) < ε} = ϕ−1 (B). (11.8) B∈B,diam B<ε
Indeed, if diam ϕ(x) < ε, we pick an open set V ⊃ ϕ(x) with diam V < ε. Since B is stable with respect to finite unions, by the compactness of ϕ(x) we find B ∈ B such that ϕ(x) ⊂ B ⊂ V . Since the reverse inclusion is obvious, (11.8) holds. Using Σ2 (S )-upper semimeasurability of ϕ we conclude the proof. Lemma 11.33. Assume that S is an algebra and ϕ : X → 2Y is a Σ2 (S )-upper semimeasurable mapping with closed values, where Y is a metrizable compact convex set. Then ψ(x) := co ϕ(x), x ∈ X, is a Σ2 (S )-upper semimeasurable mapping. Proof. We start the proof by choosing a countable dense set {fn : n ∈ N} in Ac (Y ). For each n ∈ N we define ψn (x) := {y ∈ Y : fn (y) ≤ max fn (ϕ(x))},
x ∈ X.
Then each ψn is a Σ2 (S )-upper semimeasurable mapping from X to 2Y . Indeed, if U ⊂ Y is an open set, we denote α := min fn (Y \ U ). Then ψn (x) ⊂ U
⇐⇒
Y \ U ⊂ Y \ ψn (x)
⇐⇒
α > max fn (ϕ(x))
⇐⇒
ϕ(x) ⊂ fn−1 ((−∞, α)).
Hence ψn−1 (U ) ∈ Σ2 (S ). It follows from the Hahn–Banach theorem and density of {fn : n ∈ N} in Ac (Y ) that ∞ \ ψ(x) = ψn (x), x ∈ X. n=1
By Lemma 11.30, ψ is Σ2 (S )-upper semimeasurable.
406
11.5.B
11 Continuous and measurable selectors
Selection theorem
Let Y be a topological space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Then for any compact K ⊂ Y , S K := {s|K : s ∈ S} is a function cone of continuous functions as defined in Chapter 7. Lemma 11.34. Let Y be a separable metrizable space and S ⊂ C(Y ) be a set. Then S is separable in the topology of the uniform convergence on compact subsets of Y . Proof. We fix a countable base {Vn : n ∈ N} of Y that is closed with respect to finite unions. For any n, m ∈ N with Vn ∩ Vm = ∅ we select a continuous function fn,m on Y such that ( 0 on Vm , fn,m = 1 on Vn . Let B consist of all products of finitely many just constructed functions and let A :=
n X
qi fi : qi ∈ Q, fi ∈ B, n ∈ N .
i=1
Then for any compact set K ⊂ Y , {f |K : f ∈ A} is dense in C(K) by the Stone– Weierstrass theorem A.29. We enumerate A as {gq : q ∈ N} and for any n, p, q ∈ N we find a function fnpq ∈ S such that 1 sup |fnpq (x) − gq (x)| < , p x∈Vn provided it exists. Otherwise we set fnpq := 0. We claim that the family {fnpq : n, p, q ∈ N} is dense in S in the topology of uniform convergence on compact sets. Indeed, let K ⊂ Y be compact, f ∈ S be arbitrary and ε > 0. We find p ∈ N with p2 < ε and select q ∈ N such that kf − gq kC(K) < p1 . Since the base {Vn : n ∈ N} is stable with respect to finite unions, an easy compactness argument provides n ∈ N such that supx∈Vn |f (x) − gq (x)| < p1 . Let fnpq ∈ S be the function chosen for the triple (n, p, q). Then kf − fnpq kC(K) ≤ sup |f (x) − fnpq (x)| < x∈Vn
2 < ε. p
This concludes the proof. Theorem 11.35. Let S be an algebra of sets in a set X, Y be a separable metrizable space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Let ϕ : X → 2Y be a Σ2 (S )-measurable mapping with nonempty compact values. Then there exists a Σ2 (S )-measurable function f : X → Y such that f (x) ∈ ChS|ϕ(x) (ϕ(x)), x ∈ X.
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407
Proof. We fix a compatible metric on Y . Using Lemma 11.34, we select a countable set Q := {vj : j ∈ N} ⊂ S that is dense in S in the topology of uniform convergence on compact sets. Let v0 ∈ S be an arbitrary strictly negative function on Y . We set Tj := {y ∈ Y : vj ≤ 0} and
Sj := {y ∈ Y : vj < 0},
j ≥ 0.
By induction, we construct mappings αn : X → N ∪ {0}, n ≥ 0, such that (a) αn is Σ2 (S )-measurable, n ≥ 0, (b) ϕ(x) ∩ Sαn (x) 6= ∅, x ∈ X and n ≥ 0, (c) vαn (x) > vαn−1 (x) on ϕ(x), x ∈ X and n ∈ N, (d) diam(ϕ(x) ∩ Tαn (x) ) < n1 , x ∈ X and n ∈ N. We start the construction by setting α0 (x) := 0, x ∈ X. Then the required property (a) is obvious and (b) is satisfied by the choice of v0 . Assume now that the construction has been completed for all k = 0, . . . , n − 1 for some n ∈ N. For each j ≥ 0 we define Aj := x ∈ X : ϕ(x) ∩ Sj 6= ∅ , Bj := x ∈ X : vαn−1 (x) < vj on ϕ(x) , 1 Cj := x ∈ X : diam(ϕ(x) ∩ Tj ) < . n It follows from Lemma 11.30 and Lemma 11.32 that both Aj and Cj are in Σ2 (S ). Further, the set [ Bj = ({x ∈ X : αn−1 (x) = m} ∩ {x ∈ X : vj − vm > 0 on ϕ(x)}) m≥0
=
[
−1 αn−1 (m) ∩ {x ∈ X : ϕ(x) ⊂ (vj − vm )−1 ((0, ∞))}
m≥0
is in Σ2 (S ) by the induction hypothesis and Σ2 (S )-upper semimeasurability of ϕ. Hence Dj := Aj ∩ Bj ∩ Cj ∈ Σ2 (S ), j ≥ 0. S We claim that X = ∞ j=0 Dj . Indeed, let x ∈ X be arbitrary. We denote u := vαn−1 (x) and use the induction hypothesis to realize that ϕ(x) ∩ {y ∈ Y : u(y) < 0} 6= ∅. By the Minimum principle 7.15(f) there exists y0 ∈ ChS|ϕ(x) (ϕ(x)) such that u(y0 ) < 1 ) such that U (y0 , r) ⊂ {y ∈ Y : u(y) < 0} and find a con0. We pick r ∈ (0, 2n tinuous function g on Y with g(y0 ) = u(y0 ) whose support is contained in U (y0 , r). Then for f := u ∨ g we get from Theorem 7.21 and Proposition 7.11(b) f (y0 ) = inf{s(y0 ) : s ∈ S, s > f on ϕ(x)}.
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11 Continuous and measurable selectors
Since Q is dense in S with respect to the topology of uniform convergence on compact sets, we can find j ≥ 0 such that vj (y0 ) < 0
and
vj > f on ϕ(x).
(11.9)
It follows that ϕ(x) ∩ Tj ⊂ ϕ(x) ∩ spt g ⊂ U (y0 , r).
(11.10)
By (11.9) and (11.10), we get that x ∈ Dj as required. Using Lemma 11.31 we obtain a countable partition {Ej : j ≥ 0} of X consisting of sets from Σ2 (S ) such that Ej ⊂ Dj , j ≥ 0. We define αn : X → N ∪ {0} as αn (x) := m,
x ∈ Em .
Since αn is constant on each set from the partition {Ej : j ≥ 0}, αn is Σ2 (S )measurable. It follows from the inclusions Ej ⊂ Dj , j ≥ 0, that αn satisfies all the required conditions (b), (c), (d). This finishes the construction. We set ϕn (x) := Tαn (x) , x ∈ X, and use (b), (c) and (d) to deduce that f (x) := ϕ(x) ∩
∞ \
ϕn (x),
x ∈ X,
n=1
is a well-defined (single-valued) mapping from X to Y . Further, for x ∈ X we denote y := f (x) and show that y ∈ ChS|ϕ(x) (ϕ(x)). By (c), the function v(z) := sup vαn (x) (z), z ∈ ϕ(x), n∈N
is a lower semicontinuous S|ϕ(x) -concave function. Further, since {y} = ϕ(x) ∩ T∞ n=1 Tαn (x) , we get v(y) ≤ 0
and v > 0 on ϕ(x) \ {y}.
By Theorem 7.15(c), {y} is S|ϕ(x) -extremal, and thus y ∈ ChS|ϕ(x) (ϕ(x)). Since each mapping αn is Σ2 (S )-measurable, the mappings x 7→ ϕn (x), x ∈ X, are Σ2 (S )-upper semimeasurable. By Lemma 11.30, x 7→ f (x), x ∈ X, is Σ2 (S )-upper semimeasurable. Since f is single-valued, it is Σ2 (S )-measurable. This finishes the proof. Corollary 11.36. Let S be an algebra of sets in a set X, Y be a separable metrizable space and S ⊂ C(Y ) be a convex cone that separates points of Y and contains constant functions. Let ϕ : X → 2Y be an S -semimeasurable mapping with nonempty compact values. Then there exists a Σ2 (S )-measurable function f : X → Y such that f (x) ∈ ChS|ϕ(x) (ϕ(x)), x ∈ X. (11.11)
11.5 Measurable selectors
409
In particular, if X is a metrizable space and ϕ : X → 2Y with nonempty compact values is upper or lower semicontinuous, then there exists a mapping f ∈ Baf1 (X, Y ) satisfying (11.11) (see Definition 5.18). Proof. The first part follows from Lemma 11.29 and Theorem 11.35. If ϕ is upper or lower semicontinuous, then ϕ is Σ2 (Bas(X))-measurable, where Bas(X) is the algebra from Definition 5.13. An application of the first part yields a Σ2 (Bas(X))-measurable selection f satisfying (11.11). But this means that f is in Baf1 (X, Y ).
11.5.C
Applications of the selection theorem
We recall that, given a family of functions F, W(F) denotes the min-stable cone generated by F (see Definition 3.10). Theorem 11.37. Let E be a metrizable separable locally convex space and S be an algebra of sets in a set X. Let ϕ : X → 2E be a Σ2 (S )-semimeasurable mapping with nonempty compact convex values. Then there exists a Σ2 (S )-measurable mapping f : X → E such that f (x) ∈ ext ϕ(x),
x ∈ X.
Proof. We apply Corollary 11.36 to the separable metrizable space E and the cone S := W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}). Then ChS|ϕ(x) (ϕ(x)) = ext ϕ(x) (see Proposition 7.6(c) and Exercise 7.63) and hence the assertion follows. Theorem 11.38. Let E be a separable metrizable locally convex space and S be ∗ ∗ an algebra of sets in a set X. Let ϕ : X → 2(E ,w ) be an S -lower semimeasurable mapping with nonempty convex w∗ -compact values. Then there exists a Σ2 (S )measurable mapping f : X → (E ∗ , w∗ ) such that f (x) ∈ ext ϕ(x),
x ∈ X.
Proof. Let {Un } be a basis of neighborhoods of 0 in E and let Yn be the polar of Un , that is, Yn := {x∗ ∈ E ∗ : |x∗ (x)| ≤ 1 for each x ∈ Un }, n ∈ N. We set
( ϕ−1 (Y1 ), n = 1, Xn := −1 −1 ϕ (Yn ) \ ϕ (Yn−1 ), n ≥ 2.
Then {Xn : n ∈ N} is a partition of X consisting of sets from S and ϕ|Xn : ∗ Xn → 2(Yn ,w ) is an S -lower semimeasurable mapping for each n ∈ N. For each n ∈ N, (Yn , w∗ ) is a compact metrizable space and we can apply Corollary 11.36 to
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11 Continuous and measurable selectors
the cone W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) to obtain a Σ2 (S )-measurable selection fn : Xn → (Yn , w∗ ) such that fn (x) ∈ ext ϕ(x), x ∈ X. Then the mapping f (x) := fn (x),
x ∈ Xn , n ∈ N,
is the desired selection. This concludes the proof. Definition 11.39 (Convex cone S on M+ (X)). If E is a locally convex space, we consider the convex cone S := W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) on E. If X is a compact convex set in E, any function s ∈ W({x∗ + c : x∗ ∈ E ∗ , c ∈ R}) can be regarded as a continuous function on M+ (X) via the identification with µ 7→ µ(s), µ ∈ M+ (X). Hence we can view S as a convex cone of continuous functions on M+ (X) that separates points of M+ (X) and contains constant functions. If X ⊂ E is a compact convex set and x ∈ X, then the set Mx (X) of all probability Radon measures on X representing x is a compact convex set (in the w∗ -topology). Moreover, it is an ordered compact convex set (see Section 7.5) and the cone S|Mx (X) is dense in the cone −Lc (Mx (X)) (here Lc (Mx (X)) is the cone of all continuous affine isotone functions on Mx (X); see Definition 7.41). Hence by Theorem 7.54(d), ChS|Mx (X) (Mx (X)) = Ch−Lc (Mx (X)) (Mx (X)) = ext((Mx (X))max ). (Here (Mx (X))max is the set of all maximal elements in Mx (X).) Hence µ ∈ ext(Mx (X))max is both a maximal and an extreme point of Mx (X) (see Theorem 7.61). Theorem 11.40. Let S be an algebra in a set X and K be a metrizable compact convex subset of a locally convex space E. Let ϕ : X → 2K be an S -upper semimeasurable mapping with nonempty compact convex values and f : X → K be an S -measurable mapping with f (x) ∈ ϕ(x) for x ∈ X. Then there exists a Σ3 (S )-measurable mapping m : X → M1 (K) such that m(x) ∈ ext Mf (x) (ϕ(x)) max , x ∈ X. Proof. We recall that r : M1 (K) → K is the barycentric mapping. Let S be the cone on M+ (X) as above. 1 We define a multivalued mapping M : X → 2M (K) as M (x) := Mf (x) (ϕ(x)),
x ∈ X.
We claim that M is a Σ2 (S )-upper semimeasurable mapping.
11.5 Measurable selectors
411
Indeed, we notice that M (x) = M1 (ϕ(x)) ∩ r−1 (f (x)),
x ∈ X,
and the mappings M1 (x) := M1 (ϕ(x)),
M2 (x) := r−1 (f (x)),
x ∈ X,
are Σ2 (S )-upper semimeasurable. To verify this, let ε : K → M1 (K) be the homeomorphic embedding assigning to y ∈ K the Dirac measure εy at y. Then, by Corollary 2.28, M1 (x) = co{εy : y ∈ ϕ(x)} = co(ε(ϕ(x)), x ∈ X. Thus by Lemma 11.33, M1 is Σ2 (S )-upper semimeasurable. Concerning M2 , its Σ2 (S )-upper semimeasurability follows from the following equality (M2 )−1 (F ) = f −1 (r(F )) ∈ Σ2 (S ) valid for any closed set F ⊂ M1 (K) (we recall that F , and consequently r(F ), is a compact set). Hence Lemma 11.29 and Lemma 11.30 imply the Σ2 (S )-upper semimeasurability of M . Since ∆3 (S ) is an algebra of sets and Σ2 (∆3 (S )) = Σ3 (S ) by Proposition 5.4(e), Corollary 11.36 provides a Σ3 (S )-measurable selector m : X → M1 (K) such that m(x) ∈ ChS|M (x) (M (x)) = ext Mf (x) (ϕ(x)) max , x ∈ X. The proof is complete. Theorem 11.41. Let X be a metrizable compact convex set. Then there exists a mapping m ∈ Baf1 (X, M1 (X)) such that m(x) ∈ ext((Mx (X))max ),
x ∈ X.
Proof. We use Corollary 11.36, where Y = M1 (X), S is the convex cone from Definition 11.39 and ϕ : X → 2Y assigns to x ∈ X the set Mx (X). The role of the algebra S is played by the algebra Bas(X) generated by closed sets in X (see Definition 5.13). If r : Y → X is the barycentric mapping and F ⊂ Y is a closed set, ϕ−1 (F ) = r(F ) is closed in X. Hence ϕ is a Bas(X)-lower semimeasurable mapping. By Corollary 11.36, there exists a Σ2 (Bas(X))-measurable function m : X → Y such that m(x) ∈ ChS|ϕ(x) (ϕ(x)) = ext((Mx (X))max ), Hence m is the required mapping by Theorem 5.19.
x ∈ X.
412
11.6
11 Continuous and measurable selectors
Exercises
Exercise 11.42. Let K be the one-point compactification of an uncountable discrete set A and X := M1 (K). Then ext X is uncountable and Ac (X) does not contain `1 . Hint. Since ext X = {εx : x ∈ K}, ext X is uncountable. Further, Ac (X) is isometric to C(K) and C(K) is isomorphic to c0 (A) (we recall that c0 (A) denotes the space of all continuous functions on the discrete space A that vanish at infinity, see Subsection A.3.B). Thus it suffices to show that c0 (A) does not contain `1 . Let S : `1 → c0 (A) be an isomorphism and let B ⊂ A be countable such that it contains supports of all vectors Sen , n ∈ N (here {en : n ∈ N} denotes the usual basis in `1 ). Then any vector in S(`1 ) is supported by B, and thus S(`1 ) is an isomorphic copy of `1 in c0 , which is impossible. ˇ Exercise 11.43. Let βN be the Cech–Stone compactification of N and F := βN \ N. Prove that there exists no bounded linear operator E : C(F ) → C(βN) such that Ef = f on F for each f ∈ C(F ). Hint. Assume that E is such an operator and let P f := f − E(f |F ),
f ∈ C(βN).
If we identify `∞ with C(βN) and c0 with {f ∈ C(βN) : f = 0 on F }, then P is a bounded linear projection of `∞ onto c0 . But such a projection does not exist (see Theorem 5.15 in [173]). Exercise 11.44. Let X be a simplex such that ext X is countable. Then X is metrizable and every strongly affine function is in A1 (X). Hint. By Exercise 10.96, X is metrizable. If f is a strongly affine function on X, it is bounded by Lemma 4.5 and f |ext X is a Baire-one function due to Theorem A.124. If h ∈ A1 (X) is an extension of f |ext X provided by Theorem 6.49, f = h by the minimum principle (all maximal measures are carried by ext X). that ext X is uncountable. Then Exercise 11.45. Let X be a metrizable simplex such S for any α < ω1 there exists a function f ∈ Aα (X) \ β<α Aβ (X). Hint. Since ext X is an uncountable Polish space (see Proposition 3.43 and Theorem A.114), it contains a compact perfect subset F ⊂ ext X (see Corollary 6.5 of [262]). Using Theorem 22.4 in [262] we find a bounded function f ∈ B bα (F ) that is not of any lower Baire class. Let h ∈ Aα (X) be an extension provided by Theorem 11.23. Then h is not of any lower affine class. Exercise 11.46. If H is a simplicial function space on metrizable compact space K, α ∈ (0, ω0 ) and f ∈ B bα (K), then T f ∈ (Ac (H))α+1,b (see Definitions 6.7 and 5.35).
11.6 Exercises
413
Hint. If f is continuous on K, T f ∈ (Ac (H))⊥⊥ ∩ B b1 (K) by Theorem 6.8. By Proposition 5.42, T f ∈ (Ac (H))1,b . The assertion for all finite ordinals now follows by induction. Exercise 11.47. If H is a simplicial function space on a metrizable compact space K, α ∈ [ω0 , ω1 ) and f ∈ B bα (K), then T f ∈ (Ac (H))α,b . Hint. If f ∈ B bω0 (K), then there exists a bounded sequence {fn } converging to f such that fn ∈ B bαn (K), where each αn < ω0 . By Exercise 11.46, T fn ∈ (Ac (H))αn +1,b . Since T fn → T f , we get T f ∈ (Ac (H))ω0 ,b . Now the assertion follows by transfinite induction. Exercise 11.48. If H is a simplicial function space on a compact space K, α ∈ [0, ω1 ) and f ∈ B bα (K), then T f ∈ (Ac (H))α+1,b , c
T f ∈ (A (H))α,b ,
α < ω0 , α ≥ ω0 .
Hint. If f ∈ B bα (K) is given, we find a bounded countable family F of continuous functions on K such that f ∈ F α,b (use transfinite induction). We apply Theorem 9.12 to the function space Ac (H) and obtain a metrizable compact space K 0 , a function space H0 on K 0 , F 0 ⊂ C(K 0 ) and a continuous surjection ϕ : K → K 0 such that •
h0 ◦ ϕ ∈ Ac (H) for any h0 ∈ H0 ,
•
for every g ∈ F there exists g 0 ∈ F 0 such that g = g 0 ◦ ϕ,
•
H0 is simplicial and Ac (H0 ) = H0 .
It follows that there exists a function f 0 : K 0 → R such that f = f 0 ◦ ϕ. By Theorem 5.26(c), f 0 ∈ B bα (K 0 ). Using Exercises 11.46 and 11.47 we get that T f 0 ∈ (Ac (H0 ))β,b = (H0 )β,b , where β is either α + 1 (if α is finite) or α (if α ≥ ω0 ). Since h0 ◦ ϕ ∈ Ac (H) for any h0 ∈ H0 , the assertion follows. Exercise 11.49. Let X be a simplex and α ∈ [0, ω1 ). Then Abar (X) ∩ B α (X) ⊂ Aα+1 (X),
α < ω0 ,
Abar (X) ∩ B α (X) = Aα (X),
α ≥ ω0 .
Hint. If f ∈ Abar (X), then T f = f . Hence the assertion follows from Exercise 11.48 (we recall that Ac (Ac (X)) = Ac (X), see Proposition 4.2).
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Exercise 11.50. Let X be a compact space, {fn } be a bounded increasing sequence of continuous functions and f := limn→∞ fn be discontinuous. Then there exists a subsequence {fnk } such that gk := fnk+1 − fnk , k ∈ N, form a sequence equivalent to the standard basis of c0 (that is, there exists C > 0 such that for every real numbers c1 , . . . , cn we have n
X
C −1 k(c1 , . . . , cn )k∞ ≤ ck gk ≤ Ck(c1 , . . . , cn )k∞ ). k=1
Hint. Without loss of generality we may assume that 0 < f1 ≤ · · · ≤ fn ≤ · · · ≤ 1. Let x ∈ X and c, η > 0 be such that f (x) < c < c + η < lim sup f (y). y→x
P 1 Let εk > 0, k ∈ N, be chosen such that ∞ k=1 εk < 12 η. Inductively we find points xk ∈ X, indices nk ∈ N and neighborhoods Uk 3 x such that, for k ∈ N, (a) nk+1 > nk , (b) Uk+1 ⊂ Uk , (c) x1 = x and xk+1 ∈ Uk , (d) diam fnk (Uk ) < εk , (e) fnk+1 (xk+1 ) > c + η, (f) fnk (z) > f (z) − εk for z ∈ {x1 , x2 , . . . , xk }. We start the construction by setting x1 := x and finding n1 ∈ N such that fn1 satisfies (f) for z = x1 . Then we select a neighborhood U1 3 x with (d). Assume that the construction has been completed up to the k-th stage. Let xk+1 ∈ Uk be chosen in such a way that f (xk+1 ) > c + η. Then we find nk+1 > nk such that (e) and fnk+1 (z) > f (z)−εk+1 for z ∈ {x1 , x2 , . . . , xk+1 } are satisfied. We finish the construction by finding a neighborhood Uk+1 of x with (b) and diam fnk+1 (Uk+1 ) < εk+1 . Fix i ∈ N. If k < i, then (d) gives fnk+1 (xi+1 ) − fnk (xi+1 ) ≤ |fnk+1 (xi+1 ) − fnk+1 (x)| + |fnk+1 (x) − fnk (x)| + |fnk (x) − fnk (xi+1 )| < εk+1 + εk + εk . (11.12) For k > i, (f) yields fnk+1 (xi+1 ) − fnk (xi+1 ) ≤ f (xi+1 ) − fnk (xi+1 ) < εk .
(11.13)
11.6 Exercises
415
For k = i we have by (e) and (d) fni+1 (xi+1 ) − fni (xi+1 ) > c + η − (fni (x) + fni (xi+1 ) − fni (x)) > c + η − c − εi = η − ε i .
(11.14)
We claim that gk := fnk+1 − fnk , k ∈ N, form a sequence equivalent to the standard basis of c0 . To see this, let c1 , . . . , cn be real numbers and let i ∈ {1, . . . , n} be an index such that |ci | = k(c1 , . . . , cn )k∞ . Then, for any x ∈ X, n n X X ck gk (x) ≤ |ck |(fn
k+1
k=1
(x) − fnk (x))
k=1
≤ |ci |
n X
(fnk+1 (x) − fnk (x)) ≤ |ci |(f (x) − f1 (x))
k=1
≤ k(c1 , . . . , cn )k∞ . Hence
n
X
ck gk ≤ k(c1 , . . . , cn )k∞ . k=1
To show the estimate from below, assume that |ci | = ci . Then (11.12), (11.13) and (11.14) give n X
ck (fnk+1 (xi+1 ) − fnk (xi+1 ))
k=1
≥ ci (fni+1 (xi+1 ) − fni (xi+1 )) − |ci |
X
(fnk+1 (xi+1 ) − fnk (xi+1 ))
k∈{1,...,n}\{i}
≥ ci η − εi −
X
(εk+1 + 2εk )
k∈N\{i}
≥ ci η −
∞ X
3εk
k=1
3 ≥ ci η. 4 Hence
n
X
3
ck gk ≥ ηk(c1 , . . . , cn )k∞ 4 k=1
in this case.
416
11 Continuous and measurable selectors
If i ∈ {1, . . . , n} is such that −ci = k(c1 , . . . , cn )k∞ , then we apply the inequalities above to (−c1 , . . . , −cn ) and get n n
X
X
3 3
ck gk = (−ck )gk ≥ ηk(−c1 , . . . , −cn )k∞ = ηk(c1 , . . . , cn )k∞ . 4 4 k=1
k=1
Exercise 11.51. Let X be a metrizable compact convex set and f be an affine lower semicontinuous function on X that is not continuous. Then there exists a sequence {gn } of functions in Ac (X) that is equivalent to the standard basis of c0 . Hint. Without loss of generality we may assume that 0 < f ≤ 1 (see Lemma 4.20). By Proposition 4.12, Proposition A.53 and Lemma A.54 there exists an increasing sequence {fn } of affine continuous functions converging to f . By Exercise 11.50 we can extract a subsequence {fnk } such that gk = fnk+1 − fnk form a sequence equivalent to the standard basis of c0 . Exercise 11.52. Prove that the space c of convergent sequences endowed with the sup-norm is isomorphic to c0 . Hint. Consider the operator T : c → c0 which maps (x1 , x2 , . . . ) 7→ ( lim xn , x1 − lim xn , x2 − lim xn , . . . ), n→∞
n→∞
n→∞
(x1 , x2 , . . . ) ∈ c.
Exercise 11.53. Let X be an infinite-dimensional metrizable simplex. Then Ac (X) contains an isomorphic copy of c0 . Hint. Assume first that ext X is closed. Then ext X contains infinitely many distinct points xn , n ∈ N, and x such that xn → x. Let F := {x} ∪ {xn : n ∈ N}. Then C(F ) is isometric to the space c of convergent sequences which is isomorphic to c0 (see Exercise 11.52). By Theorem 11.13(c) there exists an extender L : C(F ) → Ac (X). Then T (C(F )) is isomorphic to c0 . If ext X is not closed, there exists a concave continuous function f such that f∗ is not continuous (see Theorem 6.37). Then f∗ is an affine (see Theorem 6.5) lower semicontinuous function that is not continuous. Hence Exercise 11.51 finishes the argument.
11.7
Notes and comments
The main selection Theorem 11.6, as well as its consequences Theorems 11.7, 11.8 and Corollary 11.9, is proved in A. J. Lazar [293]. Our proof follows the papers
11.7 Notes and comments
417
Ch. L´eger [299] and A. J. Lazar and J. Lindenstrauss [296], where a selection theorem for L1 -preduals is proved. Theorems 11.13, 11.15 and 11.16 are from A. J. Lazar [293]. Theorem 11.14 is a particular version of A. J. Lazar and J. Lindenstrauss [296], Theorem 2.3, and E. H. Lacey and P. D. Morris [291], Theorem 1.1. Theorem 11.17 is an affine version on simplices of the Michael selection theorem (see E. Michael [341] and [342]). Theorem 11.18 can be found in J. Dugundji [149] who proved there the following result by topological methods: If A is a closed subset of a metrizable space X and f : A → E is a continuous mapping to a locally convex space E, then there exists a continuous mapping F : X → E extending f such that F (X) ⊂ co f (A). R. Arens showed in [17] that X can be replaced by a paracompact space if E is a Banach space. Theorem 11.19 is a variant of results from J. Dugundji [149] and K. Borsuk [77]. An application to potential theory is given in J. Lukeˇs and J. Kol´aˇr [276] and D. Werner [465] (see also P. Harmand, D. Werner and W. Werner [216], Chapter II, pp. 96–98). Section 11.3 follows the paper J. Spurn´y [428] and the results of Section 11.4 can be found in J. Lukeˇs, J. Mal´y, I. Netuka, M. Smrˇcka and J. Spurn´y [319]. Section 11.5 follows ideas from an unpublished paper by G. Debs [137] (see also [139]). Theorem 11.41 is proved in M. Talagrand [443]. The existence of a Borel measurable selection assigning to a point x of a compact convex set X a maximal measure µ ∈ Mx (X) was established by M. Rao [384]. If X is a simplex, the mapping x 7→ δx has the property that T f (x) := δx (f ), x ∈ X, is Borel for every f ∈ C(X) (see Theorem 6.8). We do not know the answer to the following question. Problem 11.54. Let X be a compact convex set. Does there exist a mapping x 7→ µx , x ∈ X such that µx ∈ Mx (X) is maximal and x 7→ µx (f ) is Borel for each f ∈ C(X)? The question of the existence of Borel measurable selectors is a widely investigated area; we refer the reader for example to K. Kuratowski and A. Maitra [286], K. Kuratowski and C. Ryll-Nardzewski [287], Ch. Castaign and M. Valadier [102], D. H. Fremlin [180], R. W. Hansell [207] and [208], J. Kaniewski and R. Pol [261] and J. Spurn´y and M. Zelen´y [433]. Exercises 11.48 and 11.49 are proved in M. Capon [96]. The paper by M. Kaˇcena and J. Spurn´y [253] shows that there exists a metrizable simplex X such that T (B bα (X)) \ B bα (X) 6= ∅ for every α ∈ [0, ω0 ). If U is a bounded open set in Rd and H(U ) is the function space from Definition 13.26, Theorem 1.3 of [253] shows that the shift of classes ceases to exist for H(U ) at the second stage. Exercise 11.49 prompts a question whether the shift of finite classes is essential. This problem was answered in J. Spurn´y [420] by the following result.
418
11 Continuous and measurable selectors
There exist a metrizable simplex X and a strongly affine function f : X → R of the second Baire class that is not of the second affine class. Problem 11.55. Given n ≥ 3, does there exist a metrizable simplex X and a strongly affine function f ∈ B n (X) that is not of affine class n? Exercise 11.50 is a very particular version of techniques described in H. P. Rosenthal [398]; we also refer the reader to S. A. Argyros, G. Godefroy and H. P. Rosenthal [19].
Chapter 12
Constructions of function spaces
The goal of this chapter is to present methods of how to construct new function spaces from given ones. More precisely, we will deal with products of function spaces and inverse limits of function spaces. The key feature of both of these constructions is that they preserve simpliciality. We begin with products of function spaces, so after the basic definitions and auxiliary results we come to the results describing Choquet boundaries and maximal measures of products of function spaces (see Theorem 12.10 and 12.13). In order to show that the product preserves simpliciality, we need to prove in Subsection 12.1.C several technical results on peaked partitions of unity and approximations in product spaces. These results are essential for the proof of Theorem 12.21, showing that the product of simplicial spaces is simplicial. This method also opens a way for a construction of tensor products of compact convex sets which we indicate in Exercises 12.85–12.87. Inverse limits of function spaces and compact convex sets are studied in Subsections 12.2.B and 12.2.C. Again, the principal results are the preservation of simpliciality (see Theorem 12.34), its consequence to compact convex simplices (see Corollary 12.35) and a description of the Choquet boundary of the inverse limit (see Theorem 12.36). The general results are applied in the framework of compact convex sets to get in Theorem 12.40 and Corollary 12.42 that Ac (X) is an L∞,1+ε -space provided X is a simplex. Perhaps the most important result is Theorem 12.45 that asserts that any metrizable simplex is the inverse limit of a suitable sequence of finite-dimensional simplices. A much simpler method yields that any simplex is the inverse limit of a system of metrizable simplices (see Theorem 12.47). A fundamental object in the theory of metrizable simplices is the Poulsen simplex. Its construction and properties are presented in Subsection 12.3.A. Theorem 12.60 summarizes its remarkable properties. Further we show in Subsection 12.3.B the existence of a nonmetrizable simplex and a strongly affine function on it that is not determined by its values on the set of extreme points. Finally, a striking example due to M. Talagrand on functions of the second affine class is constructed in Subsection 12.3.C.
420
12.1 12.1.A
12 Constructions of function spaces
Products of function spaces Definitions and basic properties
Q If {Ei }i∈I is a family of topological spaces, i∈I Ei denotes their cartesian Q product with If J ⊂ I, the natural Q the usual product topology. Q Q projection of i∈I Ei onto E is denoted as π . If A ⊂ E and z ∈ i i J i∈J i∈I i∈I\J Ei , then n o Y πJz (A) := x ∈ Ei : (x, z) ∈ A . i∈J
If f :
Q
i∈I
Ei → R is a function and y ∈
Q
πJy (f )(x) := f (x, y),
i∈I\J
Ei , we define
x∈
Y
Ei .
i∈J
Q
Q In case f (x, y1 ) = f (x, y2 ) for any x ∈ i∈J Ei and y1 , y2 ∈ i∈I\J Ei , we write πJ (f ) instead of πJy (f ). If J = {j}, we write πj (f ) instead (f ). N of π{j}Q If fi : Ei → R are functions, i = 1, . . . , n, we write ni=1 fi : ni=1 Ei → R for the function ! n n O Y fi (x1 , . . . , xn ) := f1 (x1 ) · · · fn (xn ), (x1 , . . . , xn ) ∈ Ei . i=1
i=1
Definition 12.1 (Products of function Qspaces). Let Hi be a function space on a compact space K , i ∈ I. We set K := i i∈I Ki and define the algebraic tensor product J H as the linear span of the set i i∈I ( ) O Q hi ⊗ c i∈I\J Ki : J ⊂ I finite, hi ∈ Hi for i ∈ J . i∈J
N The multiaffine product i∈I Hi is defined as n o O Y Hi := f ∈ C(K) : πjy (f ) ∈ Hj for any j ∈ I, y ∈ Ki . i∈I
i∈I\{j}
S
We recall that {Jγ }γ∈Γ is a partition of a set I if γ∈Γ Jγ = I and Jα ∩ Jβ = ∅ for distinct and {Jγ }γ∈Γ is a partition of Q α, β ∈ Γ. If {Ki }i∈I are compact spaces Q Q I, then i∈I Ki is naturally homeomorphic to γ∈Γ ( i∈Jγ Ki ). This provides an Q Q Q identification of C( i∈I Ki ) with C( γ∈Γ ( i∈Jγ Ki )). Proposition 12.2. Let Hi be a function space on a compact space Ki , i ∈ I. Let {Jγ }γ∈Γ be a partition of I. Then the following assertions hold: J N (a) i∈I Hi ⊂ i∈I Hi ,
12.1 Products of function spaces
(b)
J
(c)
N
i∈I i∈I
Hi =
J
Hi =
N
421
J
i∈Jγ Hi ), N γ∈Γ ( i∈Jγ Hi ). γ∈Γ (
Proof. The proof follows by a straightforward verification. Unless stated otherwise, Hi are N function spaces on compact spaces Ki , i ∈ I, and H denotes the multiaffine product i∈I Hi . Notation 12.3. If J ⊂ I, we write HJ for the space of functions from H that depends only on coordinates from J; precisely, HJ := {h ∈ H : x, y ∈ K, πJ (x) = πJ (y) =⇒ h(x) = h(y)}. The space Hf of functions depending on finitely many coordinates is defined as Hf := {h ∈ H : exists J ⊂ I finite such that h ∈ HJ }. Q The function space πJ (H) on i∈J Ki is defined as n o Y Ki : f ⊗ cQi∈I\J Ki ∈ H . πJ (H) := f ∈ C i∈J
Proposition 12.4. The following assertions hold: (a) if I1 ⊂ I2 ⊂ I and h ∈ HI1 , then h ∈ HI2 , N (b) πJ (H) = i∈J Hi , (c) if h ∈ HJ , then h = πJ (h) ⊗ cQi∈I\J Ki , (d) if µ ∈ M+ (K) and h ∈ HJ , then µ(h) = ((πJ )] µ)(πJ (h)), (e) Hf is dense in H. Proof. We start the proof by noticing that assertions (a)–(c) are easy to verify from the definitions. If µ ∈ M+ (K) and h ∈ HJ , then ((πJ )] µ)(πJ (h)) = µ(πJ (h) ◦ πJ ) = µ(πJ (h) ⊗ cQi∈I\J Ki ) = µ(h). This proves (d). For the proof of (e), let h ∈ H and ε > 0 be given. By the compactness of K we can select a finite open cover {V1 , . . . , Vm } of K consisting of standard open basic sets such that diam h(Vi ) < ε, i = 1, . . . , m. We set J := {i ∈ I : there exists l ∈ {1, . . . , m} such that πi (Vl ) 6= Ki } Q and select y ∈ i∈I\J Ki . We define h0 : K → R as h0 (x) := h(πJ (x), y),
x ∈ K.
422
12 Constructions of function spaces
Then h0 ∈ Hf . If z ∈ K is given, we find Vl containing z. Then (πJ (z), y) ∈ Vl as well, and thus |h0 (z) − h(z)| = |h(πJ (z), y) − h(z)| < ε. Thus Hf is dense in H. Lemma 12.5. Let Hi be a function space on a compact space Ki , i = 1, 2. Then Ac (H1 ) Ac (H2 ) ⊂ Ac (H1 ⊗ H2 ). Proof. Let ai ∈ Ac (Hi ), i = 1, 2. We show that a1 ⊗ a2 ∈ Ac (H1 ⊗ H2 ). We assume first that both functions a1 , a2 are positive, and fix (x1 , x2 ) ∈ K1 × K2 and ε > 0. We find δ > 0 such that δ(a1 (x1 ) + a2 (x2 ) + δ) < ε, and choose functions hi ∈ Hi , i = 1, 2, such that, for i = 1, 2, we have ai ≤ hi
and hi (xi ) < ai (xi ) + δ.
(This is possible by Corollary 3.23(a).) Then h1 ⊗ h2 ∈ H1 ⊗ H2 , and a1 (x1 )a2 (x2 ) ≤ h1 (x1 )h2 (x2 ) < (a1 (x1 ) + δ)(a2 (x2 ) + δ) = a1 (x1 )a2 (x2 ) + δ(a1 (x1 ) + a2 (x2 ) + δ) < a1 (x1 )a2 (x2 ) + ε. Hence (a1 ⊗ a2 )∗ = a1 ⊗ a2 . Let now a1 be positive and a2 be arbitrary. Obviously, (a1 ⊗ c)∗ = a1 ⊗ c for any c ∈ R. Using the first step we get a1 ⊗ a2 ≤ (a1 ⊗ a2 )∗ = (a1 ⊗ (a2 + ka2 k − ka2 k))∗ = (a1 ⊗ (a2 + ka2 k) − a1 ⊗ ka2 k)∗ ≤ (a1 ⊗ (a2 + ka2 k))∗ + (a1 ⊗ (−ka2 k))∗ = a1 ⊗ (a2 + ka2 k) + (a1 ⊗ (−ka2 k)) = a1 ⊗ a2 . For the lower envelope we have (a1 ⊗ a2 )∗ = −(a1 ⊗ (−a2 ))∗ = −(a1 ⊗ (−a2 )) = a1 ⊗ a2 . Hence a1 ⊗ a2 ∈ Ac (H1 ⊗ H2 ). Finally, let a1 , a2 be arbitrary. Then a1 ⊗ a2 = (a1 + ka1 k) ⊗ a2 − ka1 k ⊗ a2 ∈ Ac (H1 ⊗ H2 ). This concludes the proof.
12.1 Products of function spaces
423
Proposition 12.6. The following assertions hold: (a) If x ∈ K, µ ∈ Mx (H) and J ⊂ I, then (πJ )] µ ∈ MπJ (x) (πJ (H)). In particular, (πi )] µ ∈ Mxi (Hi ), i ∈ I. N (b) If x = (xi )i∈I ∈ K and µi ∈ Mxi (Hi ), then i∈I µi ∈ Mx (H). N (c) Ac (H) ⊂ i∈I Ac (Hi ). (d) H = Ac (H) if and only if Hi = Ac (Hi ) for every i ∈ I. Proof. To verify (a), let x ∈ K, J ⊂ I and µ ∈ Mx (H) be given. For h ∈ πJ (H), we set h0 := h ⊗ cQi∈I\J Ki . Then h0 ∈ H, and hence ((πJ )] µ)(h) = µ(h ◦ πJ ) = µ(h0 ) = h0 (x) = h(πJ (x)). To prove (b), let µi ∈ Mxi (Hi ) for i ∈ I and h ∈ H. We denote µ := We first handle the case I = {1, . . . , n}. Then Z Z µ(h) = ··· h(y1 , . . . , yn ) dµn (yn ) · · · dµ1 (y1 ). K1
N
i∈I
µi .
Kn
Since the function yn 7→ h(y1 , . . . , yn ) is in Hn , Z h(y1 , . . . , yn ) dµn (yn ) = h(y1 , . . . , yn−1 , xn ),
(y1 , . . . , yn−1 ) ∈
Kn
n−1 Y
Ki .
i=1
Proceeding by induction, µ(h) = h(x1 , . . . , xn ). If I is an arbitrary index set, J ⊂ I is finite and g ∈ HJ , then by the previous part and Proposition 12.4(b),(d), O µ(g) = ((πJ )] µ)(πJ (g)) = ( µi )(πJ (g)) = πJ (g)(πJ (x)) = g(x). i∈J
Since Hf is dense in H (see Proposition 12.4(e)), µ ∈ MxQ (H). For the proof of (c), let h ∈ Ac (H), j ∈ I and y ∈ i∈I\{j} Ki be given. We have to show that πjy (h) ∈ AcN (Hj ). For any xj ∈ Kj and µj ∈ Mxj (Hj ), we set µ := µj ⊗ εy . Since µ = µj ⊗ i∈I\{j} εyi , we get µ ∈ M(xj ,y) (H) by the previous part. Hence πjy (h) = h(xj , y) = µ(h) = (µj ⊗ εy )(h) Z Z = h(s, t) d(µj ⊗ εy )(s, t) Q Kj
Z = Kj
Thus h ∈
N
i∈I
Ac (Hi ).
i∈I\{j}
Ki
h(s, y) dµj (s) = µj (πjy (h)).
424
12 Constructions of function spaces
To show (d), let Hi = Ac (Hi ) for each i ∈ I. Then by (c), H ⊂ Ac (H) ⊂
O
Ac (Hi ) =
i∈I
O
Hi = H.
i∈I
Conversely, let H = Ac (H) and j ∈ I beQgiven. We pick a function fj ∈ Ac (Hj ) and show that fj ∈ Hj . We set K 0 := i∈I\{j} Ki and f := fj ⊗ cK 0 . Then f ∈ Ac (H). Indeed, for x ∈ K and µ ∈ Mx (H) we get (πj )] µ ∈ Mxj (Hj ) by (a). Thus f (x) = fj (xj ) = ((πj )] µ)(fj ) = µ(fj ◦ πj ) = µ(f ). Hence f ∈ Ac (H) = H and fj = πj (f ) ∈ Hj . This concludes the proof.
12.1.B
Maximal measures and extremal sets
In this section, Hi is a function space on aN compact space Ki , i ∈ I (or i = 1, 2, respectively). IfQ H is the multiaffine product i∈I Hi , we denote by K the underlying compact space i∈I Ki . If H is a function Sspace on a compact space K and x ∈ K, as in Theorem 8.32 we denote Fx (H) := ν∈Mx (H) spt ν. Lemma 12.7. Let Hi be a function space on a compact space Ki , i = 1, 2, and let ϕ : K1 → K2 be a continuous mapping such that Fϕ(x) (H2 ) ⊂ ϕ(Fx (H1 )) for each x ∈ ChH1 (K1 ). Then ϕ] µ is H2 -maximal for any H1 -maximal measure µ ∈ M+ (K1 ). Proof. Let µ ∈ M+ (K1 ) be H1 -maximal. Then µ is carried by ChH1 (K1 ) by Proposition 3.64. For any function g ∈ C(K2 ) we know by Theorem 8.32 that g ◦ ϕ is constant on Fx (H1 ) for µ-almost all x ∈ K1 . Hence there exists an Fσ set A ⊂ K1 such that µ(K \ A) = 0 and g is constant on ϕ(Fx (H1 )) for all x ∈ A. By the assumption, g is constant on Fϕ(x) (H2 ) for all x ∈ A. Then (ϕ] µ)(K2 \ ϕ(A)) = 0 and g is constant on Fy (H2 ) for each y ∈ ϕ(A). Hence ϕ] µ is H2 -maximal by Theorem 8.32. 12.8. Let H = Proposition Q F (H ). i i∈I xi
N
i∈I
Hi . Let x = (xi )i∈I ∈ K. Then Fx (H) =
Proof. Let µ ∈ Fx (H) be given. By Lemma 12.6(a), (πi )] µ ∈ Mxi (Hi ) for any i ∈ I. By Lemma A.100(a), πi (spt µ) = spt(πi )] µ ⊂ Fxi (Hi ), Thus spt µ ⊂
Q
i∈I
Fxi (Hi ), which gives Fx (H) ⊂
Q
i∈I
i ∈ I. Fxi (Hi ).
12.1 Products of function spaces
425
N Conversely, let µi ∈ Mxi (Hi ) for i ∈ I. Then i∈I µi ∈ Mx (H) by Proposition 12.6(b), and thus Lemma A.100(b) yields Y O spt µi = spt( µi ) ⊂ Fx (H). i∈I
Q
i∈I
Fxi (Hi ) ⊂ Fx (H), which finishes the proof. N N Proposition 12.9. Let J ⊂ I, H = i∈I Hi and G := i∈J Hi . Q (a) Let E ⊂ K be a closed H-extremal set and y ∈ i∈I\J Ki . Then πJy (E) is either empty or G-extremal. Hence
i∈I
(b) If E ⊂ K is a closed H-extremal set, then πJ (E) is G-extremal. Q (c) If Ei ⊂ Ki are closed Hi -extremal, i ∈ I, then i∈I Ei is a closed H-extremal set. Proof. For the proof of (a), assume that πJy (E) is a nonempty set that is not Gextremal. Then there exist x ∈ πJy (E) and µ ∈ Mx (G) such that spt µ is not contained in πJy (E). We claim that the measure µ ⊗ εy ∈ M(x,y) (H). Indeed, by Propositions 12.6(b) and 12.2(c), O O µ ⊗ εy ∈ M(x,y) G ⊗ Hi = M(x,y) Hi . i∈I\J
i∈I
Hence µ ⊗ εy is a measure H-representing (x, y) ∈ E, but µ ⊗ εy is not carried by E. Hence we have arrived at a contradiction with the H-extremality of E. Q To show (b), let x ∈ πJ (E) and µ ∈ My x (G) be given. Then there exists y ∈ i∈I\J Ki such that (x, y) ∈ E. By (a), πJ (E) is G-extremal, and hence spt µ ⊂ y πJ (E) ⊂ πJ (E). We start the proof of (c) by showing the assertion for the case I = {1, 2}. Let Ei ⊂ Ki be closed Hi -extremal sets, i = 1, 2, and E := E1 × E2 . Let (x1 , x2 ) ∈ E and µ ∈ M(x1 ,x2 ) (H1 ⊗ H2 ). Since K \ E = ((K1 \ E1 ) × K2 ) ∪ (K1 × (K2 \ E2 )), we get µ(K \ E) ≤ µ((K1 \ E1 ) × K2 ) + µ(K1 × (K2 \ E2 )) = ((π1 )] µ)(K1 \ E1 ) + ((π2 )] )(K2 \ E2 ). By Proposition 12.6(a), (πi )] µ ∈ Mxi (Hi ) and thus ((πi )] µ)(KQ i \ Ei ) = 0, Ni = 1, 2. Hence µ(K \E) = 0 and E is H1 ⊗H2 -extremal. By induction, i∈I Ei is i∈I Hi extremal provided I is finite. Q Let now I be an arbitrary set and E := i∈I Ei . Assume that µ(K \ E) > 0 for some x ∈ E and µ ∈ Mx (H). We find f ∈ C(K) such that f = 0 on E and µ(f ) > 0. Let ε > 0 be arbitrary.
426
12 Constructions of function spaces
Q By Proposition 12.4(e), there exist a finite set J ⊂ I and g ∈ C( i∈J Ki ) such Q that, for K 0 := i∈I\J Ki , we have kf − g ⊗ cK 0 kC(K) < ε. Then (πJ )] µ ∈ MπJ (x) (πJ (H)) and, by the Q first part of the argument, πJ (x) is contained in the πJ (H)-extremal set EJ := i∈J Q Ei . Hence spt(πJ )] µ ⊂ Ej . Further, |g| < ε on EJ . Indeed, we pick t ∈ i∈I\J Ei . Then for any s ∈ EJ , (s, t) ∈ E. Hence |g(s)| = |(g ⊗ cK 0 )(s, t) − f (s, t)| < ε. Using this we obtain Z |µ(g ⊗ cK 0 )| = |((πJ )] µ)(g)| ≤
|g| d((πJ )] µ) < ε. EJ
Finally, 0 < |µ(f )| ≤ |µ(f − g ⊗ cK 0 )| + |µ(g ⊗ cK 0 )| < 2ε. Since ε is arbitrary, we get a contradiction. N Q Theorem 12.10. Let H = i∈I Hi . Then ChH (K) = i∈I ChHi (Ki ). Proof. The assertion follows from Proposition 12.9, because we have the following chain of equivalences for x = (xi )i∈I ∈ K: x ∈ ChH (K)
⇐⇒
{x} is H-extremal
⇐⇒
{xi } is Hi -extremal for each i ∈ I
⇐⇒
xi ∈ ChHi (Ki ) for each i ∈ I.
N N Proposition 12.11. Let J ⊂ I, H = i∈I Hi and G := i∈J Hi . (a) If µ ∈ M+ (K) is H-maximal, then (πJ )] µ is G-maximal. (b) If µ, ν ∈ M+ (K) with µ ≺H ν, then (πJ )] µ ≺G (πJ )] ν. Proof. Lemma 12.7 yields that for a verification of (a) it is enough to show that FπJ (x) (G) ⊂ πJ (Fx (H)) for any x ∈ K. But this follows from Proposition 12.8, which gives Y Y FπJ (x) (G) = Fxi (Hi ) = πJ Fxi (Hi ) = πJ (Fx (H)). i∈J
i∈I
Q To prove (b), let µ, ν ∈ M+ (K) be given. We write K 0 := i∈I\J Ki . For an arbitrary f ∈ −W(G), the function f ⊗ cK 0 is contained in −W(H). Hence ((πJ )] µ)(f ) = µ(f ⊗ cK 0 ) ≤ ν(f ⊗ cK 0 ) = ((πJ )] ν)(f ). Hence (πJ )] µ ≺G (πJ )] ν by Proposition 3.56.
427
12.1 Products of function spaces
N Proposition 12.12. Let H := H1 H2 and µ ∈ M+ (K1 × K2 ). If (πi )] µ is Hi maximal, i = 1, 2, then µ is H-maximal. In particular, if µi ∈ M+ (Ki ) is Hi -maximal, i = 1, 2, then µ1 ⊗µ2 is H-maximal. Proof. Let h ∈ H be arbitrary. We define T : K1 → C(K2 ) by assigning x1 7→ π2x1 (h). Then T (K1 ), as a compact subset of the Banach space C(K2 ), is norm separable. Let {hn : n ∈ N} ⊂ T (K1 ) be a dense countable set. Since (π2 )] µ is H2 -maximal, Theorem 8.32 yields that the set Fn := {x2 ∈ K2 : hn = hn (x2 ) on Fx2 (H2 )} T satisfies ((π2 )] µ)(K2 \ Fn ) = 0 for each n ∈ N. Then the set E2 := n∈N Fn is of full (π2 )] µ-measure and each hn is constant on Fx2 (H2 ) for every x2 ∈ E. Since {hn : n ∈ N} is dense in T (K1 ), we get h(x1 , y) = h(x1 , x2 ),
(x1 , x2 ) ∈ K1 × E2 , y ∈ Fx2 (H2 ).
(12.1)
Analogously we find a set E1 ⊂ K1 such that ((π1 )] µ)(K1 \ E1 ) = 0 and h(y, x2 ) = h(x1 , x2 ),
(x1 , x2 ) ∈ E1 × K2 , y ∈ Fx1 (H1 ).
(12.2)
Combining (12.1) and (12.2) we get h(x1 , x2 ) = h(y1 , y2 ),
(x1 , x2 ) ∈ E1 × E2 , y1 ∈ Fx1 (H1 ), y2 ∈ Fx2 (H2 ).
Since µ(K \ (E1 × E2 )) ≤ µ((K1 \ E1 ) × K2 ) + µ(K1 × (K2 \ E2 )) = ((π1 )] µ)(K1 \ E1 ) + ((π2 )] µ)(K2 \ E2 ) = 0, we get that h is constant on F(x1 ,x2 ) (H) = Fx1 (H1 ) × Fx2 (H2 ) for µ-almost all (x1 , x2 ) ∈ K1 ×K2 (here we use Proposition 12.8). By Theorem 8.32, µ is H-maximal. This concludes the proof. N + Theorem 12.13. Let H := i∈I Hi and µ ∈ M (K). If (πi )] µ is Hi -maximal, i ∈ I, then µ is H-maximal. N In particular, if µi ∈ M1 (Ki ) are Hi -maximal measures, i ∈ I, then i∈I µi is H-maximal. Proof. Let µ ∈ M+ (K) be as in the hypothesis. Step 1. We first prove the assertion by induction for finite index sets I. Assume that it holds for all finite sets with n elements and let I = J ∪ {i0 }, where J has n elements. For each i ∈ J, (πi )] ((πJ )] µ) = (πi )] µ
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12 Constructions of function spaces
N is Hi -maximal. By Nthe inductive hypothesis, (πJ )] µ is i∈J Hi -maximal. By Proposition 12.12, µ is i∈J Hi ⊗ Hi0 -maximal. In other words, µ is H-maximal. Step 2. Let I be an arbitrary index set. We select an H-maximal measure ν with µ ≺H ν. By Proposition 12.11, for any finite set J ⊂ I, we have (πJ )] µ ≺πJ (H) (πJ )] ν. Since (πi )] ((πJ )] µ) = (πi )] µ for i ∈ J, the first part of the proof yields that (πJ )] µ is πJ (H)-maximal. Hence (πJ )] µ = (πJ )] ν. We get that µ(E) = ((πJ )] µ)(E) = ((πJ )] ν)(E) = ν(E) Q for any set E := i∈I Ei , where Ei ⊂ Ki is Borel, i ∈ I, and Ei 6= Ki only for indices in a finite set J. Hence, by Proposition A.99, µ = ν. We conclude that µ is H-maximal.
12.1.C
Partitions of unity and approximation in products of function spaces
Theorem 12.14. Let H be a simplicial space on a compact space K and −f, g be lower semicontinuous H-concave functions on K such that f ≤ g. Let F ⊂ K be a closed H-extremal set, h ∈ C(F ) satisfy f |F ≤ h ≤ g|F and h(x) = µ(h),
x ∈ F, µ ∈ Mx (H).
Then there exists a function a ∈ Ac (H) such that f ≤ a ≤ g and a = h on F . Proof. By setting ( h f := f 0
on F, on K \ F,
and
( h g := g 0
on F, on K \ F,
we obtain functions f 0 ∈ Kusc (H) and g 0 ∈ S lsc (H) with f 0 ≤ g 0 . By the Edwards in-between theorem 6.6 there exists a function a ∈ Ac (H) such that f 0 ≤ a ≤ g 0 . Then a satisfies the required properties. Definition 12.15 (Partition and peaked partition of unity). PA family {h1 , . . . , hn } of positive functions in H is called a partition of unity if ni=1 hi = 1. If moreover khi k = 1, i = 1, . . . , n, then {h1 , . . . , hn } is called a peaked partition of unity. Lemma 12.16. Assume that H is a simplicial function space on a compact space K, {f1 , . . . , fn }, {g1 , . . . , gn } are families of functions in Ac (H) and x1 , . . . , xp are pairwise distinct points in ChH (K). Let φ1 , . . . , φm be functions on ChH (K) and {αij : i = 1, . . . , n, j = 1, . . . , m} and {βjl : j = 1, . . . , m, l = 1, . . . , p} be families of real numbers such that
12.1 Products of function spaces
(a) fi |ChH (K) ≤
Pm
j=1 αij φj
429
≤ gi |ChH (K) , i = 1, . . . , n,
(b) φj (xl ) = βjl , j = 1, . . . , m, l = 1, . . . , p. Then there exists a family {ψ1 , . . . , ψm } of functions in Ac (H) such that P (c) fi ≤ m j=1 αij ψj ≤ gi on K, i = 1, . . . , n, (d) ψj (xl ) = βjl , j = 1, . . . , m, l = 1, . . . , p. Proof. We proceed by induction on m. Step 1. Assume that m = 1 and we are given fi |ChH (K) ≤ αi1 φ1 ≤ gi |ChH (K) , φ1 (xl ) = β1l ,
i = 1, . . . , n,
l = 1, . . . , p.
(12.3)
Without loss of generality we may assume that all numbers αi1 ≥ 0 (if αi1 < 0 for some i = 1, . . . , n, we would consider −gi |ChH (K) ≤ (−αi1 )φ1 ≤ −fi |ChH (K) for this index i in (12.3)). We set n 1 o fi : i ∈ {1, . . . , n} with αi1 6= 0 , A1 := αi1 o n 1 gi : i ∈ {1, . . . , n} with αi1 6= 0 . B 1 := αi1 By (12.3) and the Minimum principle 3.16, any function from A1 is on K smaller than any function from B 1 . We set ( max{f : f ∈ A1 } a1 := min{β1l : l = 1, . . . , p} ( min{g : g ∈ B 1 } b1 := max{β1l : l = 1, . . . , p}
if A1 = 6 ∅, if A1 = ∅, if B 1 = 6 ∅, if B 1 = ∅.
Then −a1 , b1 are continuous H-concave functions satisfying a1 ≤ b1
and
a1 (xl ) ≤ β1l ≤ b1 (xl ),
l = 1, . . . , p.
Using Theorem 12.14, we find a function ψ1 ∈ Ac (H) such that a1 ≤ ψ1 ≤ b1 and ψ1 (xl ) = β1l , l = 1, . . . , p. Obviously, fi ≤ αi1 ψ1 ≤ gi whenever αi1 6= 0. Otherwise fi |ChH (K) ≤ 0 ≤ gi |ChH (K) , and thus fi ≤ αi1 ψ1 ≤ gi again by the Minimum principle 3.16. This finishes the first step of the proof.
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12 Constructions of function spaces
Step 2. Assume now that the assertion holds for m − 1 and we have given functions {φ1 , . . . , φm } satisfying the assumptions (a) and (b). We rewrite (a) as fi |ChH (K) −
m−1 X
αij φj ≤ αim φm ≤ gi |ChH (K) −
j=1
m−1 X
αij φj ,
i = 1, . . . , n.
j=1
As in the first step we may assume that αim ≥ 0 for every i = 1, . . . , n. Then we obtain for i ∈ {1, . . . , n} with αim 6= 0, m−1 m−1 X αij X αij 1 1 fi |ChH (K) − φj ≤ φm ≤ gi |ChH (K) − φj . αim αim αim αim j=1
(12.4)
j=1
Hence for any r, s with αrm 6= 0 6= αsm we obtain m−1 m−1 X αsj X αrj 1 1 gs |ChH (K) − φj . fr |ChH (K) − φj ≤ αrm αrm αsm αsm
(12.5)
j=1
j=1
By interchanging r with s we get from (12.5) that m−1 X αrj αsj 1 1 1 1 gs ≤ φj ≤ fs fr − − gr − αrm αsm αrm αsm αrm αsm
(12.6)
j=1
on ChH (K). For αim = 0 we get for the functions {φ1 , . . . , φm−1 } fi |ChH (K) ≤
m−1 X
αij φj ≤ gi |ChH (K) .
(12.7)
j=1
We apply the induction hypothesis to (12.6) and (12.7), where we consider the families 1 1 fr − gs ∪ {fi }, αrm αsm
1 1 gr − fs ∪ {gi }, αrm αsm
where r, s are such that αrm 6= 0 6= αsm and i satisfies αim = 0. It yields the existence of a family {ψ1 , . . . , ψm−1 } ⊂ Ac (H) such that ψj (xl ) = βjl ,
j = 1, . . . , m − 1, l = 1, . . . , p,
m−1 m−1 X αsj X αrj 1 1 gs − ψj , fr − ψj ≤ αrm αrm αsm αsm
(12.8)
αrm 6= 0 6= αsm , (12.9)
j=1
j=1
fi ≤
m−1 X i=1
αij ψj ≤ gi ,
αim = 0.
(12.10)
12.1 Products of function spaces
431
We set Am :=
m−1 X αij 1 fi − ψj : i ∈ {1, . . . , n} with αim 6= 0 , αim αim j=1
B m :=
m−1 X αij 1 gi − ψj : i ∈ {1, . . . , n} with αim 6= 0 . αim αim j=1
By (12.9), any function from Am is smaller than any define ( max{f : f ∈ Am } am := min{βml : l = 1, . . . , p} ( min{g : g ∈ B m } bm := max{βml : l = 1, . . . , p}
function from B m . Again we if Am = 6 ∅, if Am = ∅, if B m = 6 ∅, if B m = ∅.
Then −am , bm are continuous H-concave functions such that am ≤ bm and am (xl ) ≤ βml ≤ bm (xl ), l = 1, . . . , p (here we use (b), (12.4) and (12.8)). Using Theorem 12.14, we get a function ψm ∈ Ac (H) such that am ≤ ψm ≤ bm
and ψm (xl ) = βml ,
l = 1, . . . , p.
(12.11)
Then {ψ1 , . . . , ψm } is the required family of functions (for αim 6= 0 we get (c) from (12.11) and the definition of Am , B m , for αim = 0 we obtain (c) from (12.10)). This concludes the proof. Proposition 12.17. Let H be a simplicial function space on a compact space K, {f1 , . . . , fn } ⊂ Ac (H) and ε > 0. Assume that φ1 , . . . , φm are positive functions on ChH (K), {x1 , . . . , xm } is a family of distinct points in ChH (K) and {αij : i = 1, . . . , n, j = 1, . . . , m} is a family of real numbers such that Pm (a) j=1 φj = 1 on ChH (K), (b) for j, k = 1, . . . , m, we have ( 1, j = k, φj (xk ) = 0, j 6= k, Pm (c) |fi (x) − j=1 αij φj (x)| ≤ ε for x ∈ ChH (K) and i = 1, . . . , n. Then there exists a partition of unity {ψ1 , . . . , ψm } in Ac (H) such that (d) for j, k = 1, . . . , m, we have ( 1, j = k, ψj (xk ) = 0, j 6= k, Pm (e) |fi (x) − j=1 αij ψj (x)| ≤ ε for x ∈ K and i = 1, . . . , n.
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12 Constructions of function spaces
Proof. Using (a) and (c) we get for any i = 1, . . . , n fi − ε − αim ≤
m X
αij φj − αim
j=1
=
m X
αij φj − αim
m−1 X
φj
j=1
j=1
=
m X
(αij − αim )φj
j=1
≤ fi + ε − αim on ChH (K). Positive functions φ1 , . . . , φm−1 satisfy P (f) fi −ε−αim ≤ m−1 j=1 (αij −αim )φj ≤ fi −αim +ε on ChH (K) for i = 1, . . . , n, Pm−1 (g) 0 ≤ j=1 φj ≤ 1, (h) 0 ≤ φj ≤ 1, j = 1, . . . , m − 1, (i) for j = 1, . . . , m − 1, l = 1, . . . , m, we have ( 1, j = l, φj (xl ) = 0, j= 6 l. Using Lemma 12.16 we find functions ψ1 , . . . , ψm−1 in Ac (H) such that (f)–(i) hold with φj replaced by ψj (in Lemma 12.16, we consider families {fi − αim − ε}ni=1 ∪ {0} ∪ {0}m−1 j=1 ,
{fi − αim + ε}ni=1 ∪ {1} ∪ {1}m−1 j=1
and n + m inequalities given by (f), (g) and (h)). We denote the properties of the functions ψj as (f∗ )–(i∗ ). We set ψm := 1 −
m−1 X
ψj .
j=1
It follows from (g∗ ) and (h∗ ) that ψj ≥ 0, j = 1, . . . , m, and thus {ψ1 , . . . , ψm } is a partition of unity. From (i∗ ) and ψm (xl ) = 1 −
m−1 X j=1
( 1, ψj (xl ) = 0,
m = l, m 6= l,
we obtain (d). Finally, by rewriting (f∗ ) we get (e). This concludes the proof.
12.1 Products of function spaces
433
Proposition 12.18. Let Hi be a function space on a compact space Ki , i = 1, 2, and let H1 be simplicial. Assume that {f1 , . . . , fn } is a family in Ac (H1 ) ⊗ H2 and ε > 0. Then there exist a partition of unity {ψ1 , . . . , ψm } in Ac (H1 ), {x1 , . . . , xm } ⊂ ChH1 (K1 ) and {gij : i = 1, . . . , n, j = 1, . . . , m} ⊂ H2 such that (a) for 1 ≤ j, l ≤ m we have ( 1, j = l, ψj (xl ) = 0, j 6= l, (b) kfi −
Pm
j=1 ψj
⊗ gij k < ε for i = 1, . . . , n.
Proof. Let (H2 )n stand for the cartesian product of n copies of H2 endowed with the maximum norm, that is, g = (g(1), . . . , g(n)) ∈ (H2 )n .
kgk∞ = max kg(i)kH2 , i=1,...,n
Let pi : (H2 )n → H2 , i = 1, . . . , n, be the projection on the i-th coordinate. Let f : K1 → (H2 )n be a continuous mapping defined as f (x) := (π2x (f1 ), . . . , π2x (fn )),
x ∈ K1 .
(The mapping f is continuous by Proposition A.35.) For each g ∈ (H2 )n we set ε Ug := x ∈ K1 : kg − f (x)k∞ < , 3
(12.12)
and choose g1 , . . . , gp ∈ (H2 )n such that K1 ⊂
p [
Ugj .
j=1
Let Vgj := Ugj ∩ ChH1 (K1 ),
j = 1, . . . , p.
We set V 0 := ∅ and inductively define families V j , j = 1, . . . , p, as follows. Let ( V j−1 ∪ {Vgj } if V j−1 ∪ {Vgj +1 , . . . , Vgp } does not cover ChH1 (K1 ), V j := V j−1 otherwise. After relabeling we can assume that V p = {Vg1 , . . . , Vgm } for some m ∈ {1, . . . , p}. The family V p is an open cover of ChH1 (K1 ) such that for any l = 1, . . . , m there exists [ x l ∈ V gl \ Vgj . j∈{1,...,m}\{l}
434
12 Constructions of function spaces
We denote C := {g1 , . . . , gp } − co{g1 , . . . , gm } and
D := C + U 0,
ε . 3
For any i = 1, . . . , n, the set pi (C) is a compact subset of H2 . By Theorem A.32 and the compactness of K2 , there exist points {ξi1 , . . . , ξiqi } in K2 and their open neighborhoods {Wi1 , . . . , Wiqi } such that K2 ⊂
qi [
Wir
and
r=1
ε diam h(Wir ) < , 3
r = 1, . . . , qi , h ∈ pi (C).
For any h ∈ pi (C) we find yh ∈ K2 and r ∈ {1, . . . , qi } such that yh ∈ Wir and |h(yh )| = khk. Hence khk −
ε < max |h(ξir )| ≤ khk, 3 r=1,...,qi
h ∈ pi (C).
Since pi (D) ⊂ pi (C) + U (0, 3ε ), we obtain khk −
2ε < max |h(ξir )| ≤ khk, r=1,...,qi 3
h ∈ pi (D).
(12.13)
We define functionals Γir : (H2 )n → R, i = 1, . . . , n, r = 1, . . . , qi , as Γir (g) := g(i)(ξir ),
g = (g(1), . . . , g(n)) ∈ (H2 )n .
Then equations (12.13) can be rewritten as khk∞ −
2ε < max |Γir (h)| ≤ khk∞ , i=1,...,n 3
h ∈ D.
(12.14)
r=1,...,qi
We define functions φj (x) on ChH1 (K1 ) as ( 1, j = min{l : x ∈ Vgl , l = 1, . . . , m}, φj (x) := 0, j 6= min{l : x ∈ Vgl , l = 1, . . . , m},
j = 1, . . . , m.
Then φj ≥ 0,
m X j=1
φj = 1,
( 1, φj (xl ) = 0,
j = l, j 6= l,
j, l = 1, . . . , m.
Further, for each x ∈ ChH1 (K1 ), there exists a unique index jk such that φjk (x) 6= 0. Hence, from (12.12) we get ε kf (x) − gjk k∞ < . 3
12.1 Products of function spaces
435
This can be rewritten as m X
ε
f (x) − φj (x)gj ∞ < , 3
x ∈ ChH1 (K1 ).
(12.15)
j=1
Since {Vg1 , . . . , Vgm } is a cover of ChH1 (K1 ), we see from (12.12) that ε f (x) ∈ {g1 , . . . , gm } + U (0, ), 3 Further
Pm
j=1 φj (x)gj
x ∈ ChH1 (K1 ).
∈ co{g1 , . . . , gm }, and thus f (x) −
m X
φj (x)gj ∈ D.
j=1
Using (12.14) and (12.15) we obtain m m X X Γir (f (x)) − = Γ f (x) − φ (x)g φ (x)Γ (g ) j j j ir j ir j=1 j=1 m X
ε φj (x)gj ∞ < , ≤ f (x) − 3
(12.16)
j=1
i = 1, . . . , n, r = 1, . . . , qi , x ∈ ChH1 (K1 ). We apply Proposition 12.17 to the family of functions {Γir ◦ f : i = 1, . . . , n, r = 1, . . . , qi } ⊂ Ac (H1 ) and to the family of numbers {Γir (gj ) : i = 1, . . . , n, r = 1, . . . , qi , j = 1, . . . , m}, and get a partition of unity {ψ1 , . . . , ψm } ⊂ Ac (H1 ) such that m X ε Γir (f (x)) − ≤ , ψ (x)Γ (g ) j ir j 3 j=1 i = 1, . . . , n, r = 1, . . . , qi , x ∈ ChH1 (K1 ), ( 1, j = l, ψj (xl ) = j, l = 1, . . . , m. 0, j= 6 l, Since {Ug1 , . . . , Ugp } is a cover of K1 , from (12.12) we infer f (x) ∈ {g1 , . . . , gp } + U 0,
ε , 3
x ∈ K1 .
(12.17)
(12.18)
436 Since
12 Constructions of function spaces
Pm
j=1 ψj (x)gj
∈ co{g1 , . . . , gm } for each x ∈ K1 , we obtain f (x) −
m X
ψj (x)gj ∈ D,
x ∈ K1 .
j=1
From (12.14) and (12.17) we get m X 2ε kf (x) − ψj (x)gj k∞ − < max Γir f (x) − ψj (x)gj i=1,...,n 3 j=1 j=1 r=1,...,qi m X ψj (x)Γir (gj ) = max Γir (f (x)) − i=1,...,n j=1 r=1,...,q m X
i
ε ≤ , 3 Hence
x ∈ K1 .
m X
f (x) − ψj (x)gj ∞ < ε,
x ∈ K1 .
(12.19)
j=1
To obtain the required family of functions {gij : i = 1, . . . , n, j = 1, . . . , m} in H2 , we define gij := pi (gj ), i = 1, . . . , n, j = 1, . . . , m. Property (a) then follows from (12.18) and (b) from (12.19). This finishes the proof.
12.1.D
Products of simplicial spaces
Proposition 12.19. Let Hi be function spaces on compact spaces Ki , i = 1, 2, and let H1 be simplicial. Then Ac (H1 ⊗ H2 ) = Ac (H1 ) ⊗ Ac (H2 ). Proof. It follows from Proposition 12.18 that any function in Ac (H1 ) ⊗ Ac (H2 ) can be uniformly approximated by functions from Ac (H1 ) Ac (H2 ). Using this fact, by Lemma 12.5 and Proposition 12.6(c) we obtain Ac (H1 ) ⊗ Ac (H2 ) ⊂ Ac (H1 ) Ac (H2 ) ⊂ Ac (H1 ⊗ H2 ) = Ac (H1 ⊗ H2 ) ⊂ Ac (H1 ) ⊗ Ac (H2 ), which finishes the proof. Theorem 12.20. Let Hi be simplicial function spaces on compact spaces Ki , i = 1, 2. Then H1 ⊗ H2 is simplicial.
12.1 Products of function spaces
437
Proof. Let H := H1 ⊗ H2 . We show that Ac (H) has the weak Riesz interpolation property (see Theorem 6.16). To this end, let a, b, c, d be functions from Ac (H) such that a ∨ b < c ∧ d. We choose ε > 0 such that c ∧ d − a ∨ b > ε. Since Ac (H) ⊂ Ac (H1 ) ⊗ Ac (H2 ) by Proposition 12.6(c), we may use Proposition 12.18 to find a partition of unity {ψ1 , . . . , ψm } ⊂ Ac (H1 ), points x1 , . . . , xm in ChH1 (K1 ) and a family of functions {aj , bj , cj , dj : j = 1, . . . , m} ⊂ Ac (H2 ) such that ( 1, j = l, ψj (xl ) = j, l = 1, . . . , m, (12.20) 0, j 6= l, and for the functions a0 :=
m X
ψj ⊗ aj ,
b0 :=
j=1 0
c :=
m X
m X
ψj ⊗ bj ,
j=1
ψj ⊗ cj ,
0
d :=
j=1
m X
(12.21) ψj ⊗ dj
j=1
we have a ∨ b < a0 ∨ b0 < c0 ∧ d0 < c ∧ d.
(12.22)
Then also π2x (a0 ) ∨ π2x (b0 ) < π2x (c0 ) ∧ π2x (d0 ),
x ∈ K1 .
(12.23)
For any j = 1, . . . , m we obtain from (12.20) and (12.21) that x
π2 j (a0 ) = aj ,
x
π2 j (b0 ) = bj ,
x
π2 j (c0 ) = cj ,
x
π2 j (d0 ) = dj .
From (12.23) it follows that aj ∨ bj < cj ∧ dj ,
j = 1, . . . , m.
Since Ac (H2 ) has the weak Riesz interpolation property (see Theorem 6.16), there exists a function hj ∈ Ac (H2 ) with aj ∨ bj < hj < cj ∧ dj , Then the function h :=
m X
j = 1, . . . , m.
(12.24)
ψj ⊗ hj
j=1
belongs to Ac (H1 ) ⊗ Ac (H2 ) = Ac (H) (see Proposition 12.19). Since the functions ψj are positive, (12.24) and (12.22) yield a ∨ b < h < c ∧ d. This concludes the proof.
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12 Constructions of function spaces
Theorem 12.21. Let Hi be simplicial function spaces on compact spaces Ki , i ∈ I. Then N (a) i∈I Hi is simplicial, N (b) Ac (H) = i∈I Ac (Hi ), N Q (c) δx = i∈I δxi for x = (xi )i∈I ∈ i∈I Ki . Proof. Step 1. First we verify by induction that (a) and (b) hold for a finite set I. From Proposition 12.19 and Theorem 12.20, we know their validity in case I has two elements. Assume now that they hold for any index set containing n indices, and let I = J ∪ {i0 }, where J has n elements. Using induction hypothesis, Propositions 12.2(c) and 12.19, we get that the space O O Hi = H i ⊗ H i0 i∈I
i∈J
is simplicial, and Ac (H) = Ac
O
O Hi ⊗ Ac (Hi0 ) H i ⊗ H i 0 = Ac i∈J
i∈J
=
O
Ac (Hi ) ⊗ Ac (Hi0 ) =
i∈J
O
Ac (Hi ).
i∈I
NStep 2.c Let now Ic be an arbitrary index set and H := i∈I A (Hi ) = A (H). Let J ⊂ I be finite and ! O Ac (Hi ) h∈ i∈I
N
i∈I
Hi . We first show that
J
(that is, h depends only on coordinates from J). By the first part of the proof, O O πJ (h) ∈ Ac (Hi ) = Ac Hi . i∈J
i∈J
Using Lemma 12.5, Propositions 12.4(c) and 12.2(c) we get O O h = πJ (h) ⊗ cQi∈I\J Ki ∈ Ac Hi ⊗ Hi = Ac (H). i∈J
i∈I\J
N N By Proposition 12.4(e), ( i∈I Ac (Hi ))f is dense in i∈I Ac (Hi ). It follows that !f O i∈I
c
A (Hi ) =
O i∈I
c
A (Hi )
⊂ Ac (H) = Ac (H).
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439
Since the second inclusion follows from Proposition 12.6(c), we get the required N c c equality A (H) = i∈I A (Hi ). N c Let a, b, c, d ∈ Ac (H). Then a, b, c, d ∈ i∈I A (Hi ), and therefore Proposition 12.4(e) yields the existence of a finite set J ⊂ I and functions ! O 0 0 0 0 c a ,b ,c ,d ∈ A (Hi ) i∈I
J
such that a ∨ b < a0 ∨ b0 < c0 ∧ d0 < c ∧ d. Then πJ (a0 ) ∨ πJ (b0 ) < πJ (c0 ) ∧ πJ (d0 ). Since
N
i∈J
Hi is simplicial, there exists a function O O h0 ∈ Ac Hi = Ac (Hi ) i∈J
i∈J
such that πJ (a0 ) ∨ πJ (b0 ) < h0 < πJ (c0 ) ∧ πJ (d0 ). Then the function h := h0 ⊗ cQi∈I\J Ki belongs to
N
i∈I
Ac (Hi ) = Ac (H) and satisfies a ∨ b < h < c ∧ d.
This finishes the proof of (a) and (b). Q To verify (c), let x = (xi )N i∈I be a point in i∈I Ki . By Proposition 12.6(b) and Theorem 12.13, the measure i∈I δxi is H-maximal and H-represents x. Since H is simplicial, O δx = δ xi i∈I
as needed. This concludes the proof. Theorem 12.22. Let Hi be function spaces on compact spaces Ki , i ∈ I, such that N i∈I Hi is simplicial. Then Hi is simplicial for each i ∈ I. Proof. We Q fix j ∈ I and choose aj , bj , cj , dj ∈ Ac (Hj ) such that aj ∨ bj < cj ∧ dj . 0 Let K := i∈I\{j} Ki . Then the functions a := aj ⊗ cK 0 ,
b := bj ⊗ cK 0 ,
c := cj ⊗ cK 0 ,
are in Ac (H) (see Lemma 12.5), and a ∨ b < c ∧ d.
d := dj ⊗ cK 0
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12 Constructions of function spaces
By the simpliciality of Ac (H) we find a function h ∈ Ac (H) with a∨b < h < c∧d. We select an arbitrary y ∈ K 0 . Proposition 12.6(c) yields πjy (h) ∈ Ac (Hj ). Then aj ∨ bj < πjy (h) < cj ∧ dj , and Ac (Hj ) has the weak Riesz interpolation property. Hence Hj is simplicial by Theorem 6.16.
12.2 12.2.A
Inverse limits of function spaces Admissible mappings
Definition 12.23 (Admissible mapping). Let Hi be a function space on a compact space Ki , i = 1, 2, and let ϕ : K1 → K2 be a continuous surjective mapping. It is said to be an admissible mapping of (K1 , H1 ) to (K2 , H2 ), if h2 ◦ ϕ ∈ H1 for any h2 ∈ H2 . If ϕ : (K1 , H1 ) → (K2 , H2 ) is an admissible mapping, H2 is naturally isometrically embedded in H1 via the mapping h2 7→ h2 ◦ ϕ. If S(Hi ) denote the respective state spaces of Hi , i = 1, 2, then the restriction mapping ψ : S(H1 ) → S(H2 ) is a continuous affine mapping. By the Hahn–Banach theorem, it is surjective. Another way of describing ψ is by the formula ψ(s1 )(h2 ) = s1 (h2 ◦ ϕ),
h2 ∈ H2 , s1 ∈ S(H1 ).
If φi : Ki → S(Hi ) denote the respective homeomorphic embedding of Ki into S(Hi ), i = 1, 2, we obtain that φ2 ◦ ϕ = ψ ◦ φ1 .
(12.25)
Proposition 12.24. Let ϕ : (K1 , H1 ) → (K2 , H2 ) be an admissible mapping. Then the following assertions hold. (a) If µ ∈ Mx (H1 ) for x ∈ K1 , then ϕ] µ ∈ Mϕ(x) (H2 ). (b) If g is a continuous H2 -convex function, then g ◦ ϕ is H1 -convex. (c) If µ ≺H1 ν for µ, ν ∈ M+ (K1 ), then ϕ] µ ≺H2 ϕ] ν. (d) ϕ(ChH1 (K1 )) ⊃ ChH2 (K2 ). (e) ϕ] (Mmax (H1 )) ⊃ Mmax (H2 ). Proof. To check (a), let x ∈ K1 and µ ∈ Mx (H1 ) be given. Then ϕ] µ(h2 ) = µ(h2 ◦ ϕ) = h2 (ϕ(x)), and thus ϕ] µ ∈ Mϕ(x) (H2 ).
h ∈ H2 ,
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441
By an analogous straightforward argument we get (b). For the proof of (c), let µ ≺H1 ν for µ, ν ∈ M+ (K1 ) be given. For any g ∈ Kc (H2 ), we obtain ϕ] µ(g) = µ(g ◦ ϕ) ≤ ν(g ◦ ϕ) = ϕ] ν(g). Hence ϕ] µ ≺H2 ϕ] ν as required. Now let ψ : S(H1 ) → S(H2 ) be the restriction mapping. By Proposition 2.72(c), ψ(ext S(H1 )) ⊃ ext S(H2 ). In view of (12.25) and Proposition 4.26(d), −1 ChH2 (K2 ) = φ−1 2 (ext S(H2 )) ⊂ φ2 (ψ(φ1 (ChH1 (K1 )))) = ϕ(ChH1 (K1 )).
Analogously we proceed with the proof of (e). By Proposition 7.49, ψ(Mmax (S(H1 ))) ⊃ Mmax (S(H2 )). Since µ ∈ M+ (Ki ) is Hi -maximal if and only if (φi )] µ is Ac (S(Hi ))-maximal, i = 1, 2, (see Proposition 4.28(d)) and ψ] ◦ (φ1 )] = (φ2 )] ◦ ϕ] , we conclude that ϕ] (Mmax (H1 )) ⊃ Mmax (H2 ). Proposition 12.25. Let ϕ : (K1 , H1 ) → (K2 , H2 ) be an admissible mapping and let H2 be simplicial. Then the following assertions are equivalent: (i) ϕ(ChH1 (K1 )) = ChH2 (K2 ), (ii) (g ◦ ϕ)∗ = g ∗ ◦ ϕ for any continuous H2 -convex function g on K2 , (iii) ϕ] (Mmax (K1 )) = Mmax (K2 ). Proof. We start by showing (i) =⇒ (ii). Let g ∈ Kc (H2 ) be given. Proposition 12.24(a) and Lemma 3.22 imply (g ◦ ϕ)∗ ≤ g ∗ ◦ ϕ. On the other hand, g ∗ is H2 -affine by Theorem 6.5, and hence g ∗ ◦ ϕ is H1 -affine (use Proposition 12.24(a)). By the assumption, g ∗ ◦ ϕ = (g ◦ ϕ)∗ on ChH1 (K1 ), and hence the minimum principle of Proposition 3.88 gives g ∗ ◦ ϕ ≤ (g ◦ ϕ)∗ on K1 . Hence g ∗ ◦ ϕ = (g ◦ ϕ)∗ . To show (ii) =⇒ (iii), let µ ∈ Mmax (H1 ) be given. For any g ∈ Kc (H2 ), ϕ] µ(g) = µ(g ◦ ϕ) = µ((g ◦ ϕ)∗ ) = µ(g ∗ ◦ ϕ) = ϕ] µ(g ∗ ), and ϕ] µ is H2 -maximal by Theorem 3.58. Finally, (iii) =⇒ (i) is obvious.
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12 Constructions of function spaces
12.2.B
Construction of inverse limits
Definition 12.26 (Inverse system and limit of function spaces). We assume that I is an up-directed set. Let (Ki , Hi ) be a function space, i ∈ I. We say that ((Ki , Hi ), pij )i,j∈I is an inverse system of function spaces if •
pij : (Kj , Hj ) → (Ki , Hi ) is an admissible mapping, i ≤ j,
•
pii is identity, i ∈ I,
•
pij ◦ pjk = pik , i ≤ j ≤ k.
If I = N, we say that the system is an inverse sequence. Let K := lim Ki be the inverse limit of the system (Ki , pij )i,j∈I , that is ←
K=
(xi )i∈I ∈
Y
Ki : pij (xj ) = xi , i ≤ j, i, j ∈ I
i∈I
and let πi : K → Ki be the i-th projection. The inverse limit lim((Ki , Hi ), pij )i,j∈I of this inverse system is the function space ←−
(K, H), where H :=
[
{h ◦ πi |K : h ∈ Hi }.
i∈I
If every Ki is nonempty, it follows from Theorem 3.2.13 in [169] that K is a nonempty compact space because each pij is surjective. Further, every πi is surjective and pij ◦ πj = πi , i ≤ j. Definition 12.27 (Inverse system of compact convex sets). Let I be an up-directed set and (Xi , pij )i,j∈I be a system of compact convex sets and mappings such that •
pij : Xj → Xi is a continuous affine surjection, i ≤ j,
•
pii is identity, i ∈ I,
•
pij ◦ pjk = pik , i ≤ j ≤ k.
Then ((Xi , Ac (Xi )), pij )i,j∈I is an inverse system of function spaces as defined in Definition 12.26. We call it an inverse family of compact convex sets and the compact convex set X := lim(Xi , pij )i,j∈I is its inverse limit. Sometimes we write briefly ←− X = lim Xi . ← Further, πi : X → Xi is an affine continuous surjection for each i ∈ I. Proposition 12.28. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) := lim((Ki , Hi ), pij )i,j∈I be its inverse limit. Then ←−
(a) H is a well-defined function space, (b) every πi : (K, H) → (Ki , Hi ) is an admissible mapping.
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443
Proof. To show that H is a function space, we first notice that H contains constant functions. If x, y ∈ K differs at a coordinate i ∈ I, let h ∈ Hi be such that h(xi ) 6= h(yi ). Then h ◦ πi ∈ H separates the points x and y. Let f ◦ πi , g ◦ πj ∈ H, where i, j ∈ I, f ∈ Hi and g ∈ Hj . We select k ∈ I such that i, j ≤ k. Then f ◦ πi + g ◦ πj = f ◦ pik ◦ πk + g ◦ pjk ◦ πk ∈ H, because f ◦ pik , g ◦ pjk ∈ Hk and Hk is a vector space. An analogous verification shows that cf ∈ H whenever c ∈ R and f ∈ H. Thus H is a vector space. Since (b) follows by the definition, the proof is finished. Proposition 12.29. If ((Ki , Hi ), pij )i,j∈I is an inverse system of function spaces and (K, H) is its inverse limit, we consider the family (S(Hi ), ψij )i,j∈I , where ψij : S(Hj ) → S(Hi ) are the restriction mappings. (a) The family (S(Hi ), ψij )i,j∈I is an inverse system of compact convex sets. (b) S(H) is affinely homeomorphic to lim S(Hi ). ←−
Proof. The system (S(Hi ), ψij )i,j∈I is inverse by a straightforward verification. For the proof of (b), let X := lim S(Hi ) be the limit of the system (S(Hi ), ψij )i,j∈I . Let ←
Φ : H → Ac (S(H)) be the mapping from Definition 4.25, that is, Φ(s)(h) = s(h),
s ∈ S(H), h ∈ H.
We define the required affine homeomorphism ϕ : X → S(H) as ϕ(s)(h ◦ πj ) = sj (h),
s = (si )i∈I ∈ X, h ∈ Hj for some j ∈ I.
Obviously, ϕ is a well-defined affine continuous mapping. We show that ϕ is injective. Let s, t ∈ X be distinct points, that is, they differ at a coordinate j ∈ I. Let h ∈ Hj be such that sj (h) 6= tj (h). Then ϕ(s)(h ◦ πj ) = sj (h) 6= tj (h) = ϕ(t)(h ◦ πj ), and ϕ(s) 6= ϕ(t). In order to prove surjectivity of ϕ it suffices to show that ϕ(X) ⊃ ext S(H) (use the Krein–Milman theorem 2.22). Assume that this is not the case and hence there exists t ∈ ext S(H) \ ϕ(X). Using Proposition 2.41 and density of Φ(H) in Ac (S(H)) (see Proposition 4.26(b)) we find a function h ∈ H such that Φ(h) > 0 on ϕ(X) and Φ(h)(t) < 0. Let h = h0 ◦ πj for some h0 ∈ Hj and j ∈ J. Then 0 ≤ Φ(h)(ϕ(s)) = sj (h0 ),
s ∈ X.
Hence sj (h0 ) ≥ 0 for each sj ∈ S(Hj ). It follows that h0 , and consequently h, is a positive function. Since Φ preserves order (see Proposition 4.26(e)), Φ(h) ≥ 0, which is a contradiction. This concludes the proof.
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12 Constructions of function spaces
Proposition 12.30. Let (Xi , pij )i,j∈I be an inverse system of compact convex sets and let X := lim Xi be its limit. Let (X, H) be the inverse limit of the family ←
((Xi , Ac (Xi )), pij )i,j∈I , where every (Xi , Ac (Xi )) is regarded as a function space. Then (a) H = Ac (X), (b) S(H) is affinely homeomorphic to X. Proof. Given the objects as in the hypothesis, we notice that H = {f ◦ πi : f ∈ Ac (Xi ), i ∈ I} is contained in Ac (X). By Proposition 12.28(a), H contains constants and separates points of X. Thus Exercise 4.49 yields H = Ac (X) and (a) holds. By Proposition 4.31(a), X is affinely homeomorphic to S(Ac (X)) = S(H) = S(H). This concludes the proof. Theorem 12.31. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) be its inverse limit. Let µ ∈ M1 (lim Ki ) be a measure such that (πi )] µ is ← Hi -maximal for any i ∈ I. Then µ is H-maximal. Proof. Let K := lim Ki and let ν ∈ M1 (K) satisfy µ ≺H ν. Proposition 12.28(b) ←
and Proposition 12.24(c) yield (πi )] µ ≺Hi (πi )] ν for each i ∈ I. Since (πi )] µ is Hi -maximal, (πi )] ν = (πi )] µ. Thus Lemma A.103 yields ν = lim(πi )] ν = lim(πi )] µ = µ. ←
←
This concludes the proof. Proposition 12.32. Let ((Ki , Hi ), pij )i,j∈I be an inverse system of function spaces and (K, H) be its inverse limit. (a) Let x = (xi )i∈I ∈ K and µ ∈ M1 (K). Then µ ∈ Mx (H) if and only if ((πi )] µ, pij )i,j∈I is an inverse system of measures (see Definition A.101) with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. (b) Let x = (xi )i∈I ∈ K and let (µi , pij )i,j∈I be an inverse system of measures with µi ∈ Mxi (Hi ). Then µ := lim µi is an H-representing measure for x. ←−
Proof. For the proof of (a), we first assume that µ ∈ Mx (H). Propositions 12.28(b) and 12.24(a) yield that (πi )] µ ∈ Mxi (Hi ) for any i ∈ I. Lemma A.103 implies that ((πi )] µ, (pij )] )i,j∈I is an inverse system. Conversely, let ((πi )] µ, pij )i,j∈I be an inverse system of measures with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. Let h ∈ Hj for some j ∈ I. Then µ(h ◦ πj ) = ((πj )] µ)(h) = h(xj ) = (h ◦ πj )(x). Thus µ ∈ Mx (H).
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445
To check (b), Definition A.101 provides a measure µ ∈ M1 (K) such that (πi )] µ = µi , i ∈ I. By (a), µ ∈ Mx (H), and we are done.
Inverse limits of simplicial function spaces
12.2.C
We again consider an inverse system ((Ki , Hi ), pij )i,j∈I and its inverse limit (K, H). S Proposition 12.33. If every Hi is simplicial, then i∈I {h◦πi : h ∈ Ac (Hi )} is dense in Ac (H). Proof. Let a ∈ Ac (H) and ε > 0. By Corollary 3.23 and an easy compactness argument, there exist functions −f, g ∈ W(H) such that a − ε < f ≤ a ≤ g < a + ε. We write f = (f1 ◦ πi1 ) ∨ · · · ∨ (fn ◦ πin ), g = (g1 ◦ πj1 ) ∧ · · · ∧ (gm ◦ πjm ),
fk ∈ Hik , ik ∈ I, k = 1, . . . , n, gk ∈ Hjk , jk ∈ I, k = 1, . . . , m.
Let i ∈ I be greater than all the indices i1 , . . . , in , j1 , . . . , jm . Then (f1 ◦ pi1 i ) ∨ · · · ∨ (fn ◦ pin i ) ≤ (g1 ◦ pj1 i ) ∧ · · · ∧ (gm ◦ pjm i ), and the former function belongs to Kc (Hi ), the latter to S c (Hi ). Using the simpliciality of Ac (Hi ), we find a function h ∈ Ac (Hi ) such that (f1 ◦ pi1 i ) ∨ · · · ∨ (fn ◦ pin i ) ≤ h ≤ (g1 ◦ pj1 i ) ∧ · · · ∧ (gm ◦ pjm i ). Then a − ε < f ≤ h ◦ πi ≤ g < a + ε. Since πi is an admissible mapping, Proposition 12.24(b) yields h ◦ πi ∈ Ac (H). This concludes the proof. Theorem 12.34. If every Hi is simplicial, then H is simplicial. Proof. We show that Ac (H) has the weak Riesz interpolation property. Assume that a1 , . . . , a4 ∈ Ac (H) satisfy a1 ∨ a2 < a3 ∧ a4 . By Proposition 12.33 we may assume that ak = bk ◦ πik , k = 1, . . . , 4, where ik ∈ I and bk ∈ Ac (Hik ). Let i ∈ I be greater than indices i1 , . . . , i4 . As above we employ the simpliciality of Hi to find a function h ∈ Ac (Hi ) such that (b1 ◦ pi1 i ) ∨ (b2 ◦ pi2 i ) < h < (b3 ◦ pi3 i ) ∧ (b4 ◦ pi4 i ). Then h ◦ πi ∈ Ac (H) (use Proposition 12.24(b)) and a1 ∨ a2 < h ◦ πi < a3 ∧ a4 . This finishes the proof.
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12 Constructions of function spaces
Corollary 12.35. Let (Xi , pij )i,j∈I be an inverse family of Choquet simplices. Then its inverse limit is a Choquet simplex. Proof. Let X := lim Xi be the inverse limit and let (X, H) be the inverse limit of ←
the family ((Xi , Ac (Xi )), pij )i,j∈I whose elements are regarded as function spaces. By Theorem 12.34, H is a simplicial function space. Proposition 12.30 yields H = Ac (X), and thus Ac (X) is simplicial as well. But this means exactly that X is a simplex. Theorem 12.36. Let x = (xi )i∈I be a point of K. (a) If xi ∈ ChHi (Ki ) for every i ∈ I, then x ∈ ChH (K). (b) Assume that each Hi is simplicial and pij (ChHj (Kj )) ⊂ ChHi (Ki ), i ≤ j. Then x ∈ ChH (K) if and only if xi ∈ ChHi (Ki ) for every i ∈ I. Proof. For the proof of (a), let x ∈ K satisfy xi ∈ ChHi (Ki ) for each i ∈ I. Let µ ∈ Mx (H) be given. By Proposition 12.32, ((πi )] µ, pij )i,j∈I is an inverse system of measures with (πi )] µ ∈ Mxi (Hi ) for i ∈ I. Then (πi )] µ = εxi for each i ∈ I and Lemma A.103 gives µ = lim(πi )] µ = lim εxi = εx . ←
←
Hence x ∈ ChH (K). To verify (b), let x ∈ ChH (K) be given and let the assumption of (b) be satisfied. We fix i ∈ I, and select a function f ∈ Kc (Hi ) along with ε > 0. Since x ∈ ChH (K), there exists h ∈ H such that f ◦ πi < h and (f ◦ πi )(x) < h(x) < (f ◦ πi )(x) + ε. Without loss of generality we may assume that h = g ◦πj for some h ∈ Hj and j ≥ i. Then (f ◦ pij )(xj ) = (f ◦ pij ◦ πj )(x) = (f ◦ πi )(x) < (h ◦ πj )(x) < (f ◦ pij )(xj ) + ε gives (f ◦ pij )(xj ) < h(xj ) < (f ◦ pij )(xj ) + ε. This along with Proposition 12.25 yields (f ∗ ◦ pij )(xj ) = (f ◦ pij )∗ (xj ) < h(xj ) < (f ◦ pij )(xj ) + ε. Hence f ∗ (xi ) < f (xi ) + ε. Since ε > 0 is arbitrary, f ∗ (xi ) = f (xi ) for any f ∈ Kc (Hi ). By Theorem 3.24, xi ∈ ChHi (Ki ).
12.2 Inverse limits of function spaces
12.2.D
447
Structure of simplices
∞ n Definition 12.37 (Admissible basis of `∞ n ). We recall that `n is the space R considered with the supremum norm and pointwise ordering (that is, x ≤ y if and only if y −x has positive coordinates). Let {e1 , . . . , en } be the canonical basis of `∞ n . A basis {a1 , . . . , an } of `∞ is called admissible if a = T e , i = 1, . . . , n, for some isometry i i n ∞ . An admissible basis is positive if the operator T above is positive. T : `∞ → ` n n
Lemma 12.38. (a) Let {a1 , . . . , an } be an admissible basis of `∞ n . Then there exists a permutation π : {1, . . . , n} → {1, . . . , n} and signs i ∈ {−1, 1}, i = 1, . . . , n, such that ai = i eπ(i) , i = 1, . . . , n. If, moreover, {a1 , . . . , an } is positive, all signs i are positive. ∞ (b) Let m > n, T : `∞ n → `m be an isometry and let {u1 , . . . , un } be an admissible ∞ basis of `n . Then there exist an admissible basis
{v1 , . . . , vm } ⊂ `∞ m and a family of numbers {ai,j : i = 1, . . . , n, j = n + 1, . . . , m} such that • •
P T ui = vi + m j=n+1 ai,j vj , i = 1, . . . , n, Pn i=1 |ai,j | ≤ 1, j = n + 1, . . . , m.
Further, for each k = n + 1, . . . , m, the space Fk := span{T u1 , . . . , T un , vn+1 , . . . , vk } is isometric to `∞ k . (c) If T in (b) is positive and {u1 , . . . , un } is an admissible positive basis, all numbers ai,j , i = 1, . . . , n, j = n + 1, . . . , m, and the basis {v1 , . . . , vm } can be chosen to be positive. ∞ (d) Let F ⊂ `∞ m be a subspace isometric to `n , n < m. Then there exist subspaces ∞ Fn+1 , . . . , Fm−1 of `∞ m such that F ⊂ Fn+1 ⊂ · · · ⊂ Fm−1 ⊂ `m and Fi is ∞ isometric to `i , i = n + 1, . . . , m − 1.
Proof. For the proof of (a), let ai := T ei , i = 1, . . . , n, for some isometry T : `∞ n → ∞ . Since e ∈ ext B ∞ , T e ∈ B ∞ , and `∞ . Let e stand for the vector (1, 1, . . . , 1) ∈ ` `n `n n n thus there exist signs i ∈ {−1, 1}, i = 1, . . . , n, such that T e = (1 , . . . , n ). ∞ Let S : `∞ n → `n be the isometry that maps ei to i ei , i = 1, . . . , n. Then ST is a positive isometry of `∞ n onto itself.
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12 Constructions of function spaces
Since e = ST (e1 + · · · + en ),
kST ei k = 1,
and ST ei ≥ 0,
i = 1, . . . , n,
it follows that ST ei = eπ(i) , i = 1, . . . , n, for some permutation π : {1, . . . , n} → {1, . . . , n}. Hence T ei = i eπ(i) as needed. ∞ For the proof of (b), let T : `∞ n → `m be an isometry and {u1 , . . . , un } be an ∞ admissible basis of `n . We write T ui in coordinates with respect to the canonical basis {e1 , . . . , em } of `∞ m as T ui = (ci,1 , . . . , ci,m ),
i = 1, . . . , n.
For any choice of signs 1 , . . . , n we have n n n X
X X i ci,1 , . . . , i ci,m . i ui = 1 = T i=1
i=1
i=1
Pn
Hence i=1 |ci,j | ≤ 1 for each j = 1, . . . , m. For any i = 1, . . . , n, we can find an index π(i) ∈ {1, . . . , m} and i ∈ {−1, 1} ∞ such that i ci,π(i) = 1. Hence cl,π(i) = 0 for all l ∈ {1, . . . , n}\{i}. Let S : `∞ m → `m be the isometry that maps eπ(i) to i ei , i = 1, . . . , n, and the remaining vectors onto {en+1 , . . . , em }. P Then S defines the required admissible basis {v1 , . . . , vm }. We write T ui = vi + m j=n+1 ai,j vj , i = 1, . . . , n. Let k ∈ {n + 1, . . . , m} be fixed. It is easy to observe that the basis k k X X T u1 − a1,j vj , . . . , T un − an,j vj , vn+1 , . . . , vk j=n+1
j=n+1
is isometric to the usual basis (e1 , . . . , ek ) of `∞ k and spans Fk . This concludes the proof of (b). To verify (c), we follow the proof of (b) and notice that T ui are positive vectors. Hence the same argument as in (b) provides the required positive basis {v1 , . . . , vm } and numbers ai,j , i = 1, . . . , n and j = n + 1, . . . , m. ∞ To prove (d), let F ⊂ `∞ m be isometric to `n for some n < m. We find an ad∞ missible basis {v1 , . . . , vm } of `m from (b). Now, the spaces Fn+1 , . . . , Fm from (b) satisfy our requirements. This finishes the proof. Lemma 12.39. Let X be a compact convex set. c (a) Let {e1 , . . . , en } be the canonical basis of `∞ n , {f1 , . . . , fn } ⊂ A (X) be a peaked partition of unity and let E be the linear span of {f1 , . . . , fn }. Then the mapping T : E → `∞ n defined as T
n X
ai fi := (a1 , . . . , an ),
(a1 , . . . , an ) ∈ `∞ n ,
i=1
is a positive isometry satisfying T fi = ei , i = 1, . . . , n.
12.2 Inverse limits of function spaces
449
Further, the mapping ϕ : X → Rn defined as ϕ(x) := (f1 (x), . . . , fn (x)),
x ∈ X,
is an affine continuous surjection onto the (n − 1)-simplex co{e1 , . . . , en }. (b) Let E ⊂ Ac (X) be isometric to `∞ n and 1 ∈ E. Then there exist a positive isometry T : E → `∞ and a peaked partition of unity {f1 , . . . , fn } ⊂ Ac (X) n such that E = span{f1 , . . . , fn } and T fi = ei , i = 1, . . . , n. (c) Let {e1 , . . . , ep } ⊂ Ac (X) be a peaked partition of unity and Ep be the linear span of {e1 , . . . , ep }. If Ep+r ⊃ Ep is a subspace of Ac (X) isometric to `∞ p+r , then there exist positive numbers {a1,m , . . . , am,m },
m = p, . . . , p + r − 1,
and peaked partitions of unity {e1,m , . . . , em,m } ⊂ Ep+r ,
m = p + 1, . . . , p + r,
such that •
ei = ei,p+1 + ai,p ep+1,p+1 , i = 1, . . . , p,
•
ei,m = ei,m+1 + ai,m em+1,m+1 , i = 1, . . . , m, m = p + 1, . . . , p + r − 1, Pm i=1 ai,m = 1, m = p, . . . , p + r − 1.
•
Proof. For the proof of (a), let T and {f1 , . . . , fn } be as in the hypothesis. We find points {x1 , . . . , xn } such that fi (xi ) = 1, i = 1, . . . , n. Obviously, T is positive. For a given (a1 , . . . , an ) ∈ `∞ n , let i0 be an index satisfying |ai0 | = k(a1 , . . . , an )k. Then n n X X ai fi (x) ≤ |ai0 | fi (x) = k(a1 , . . . , an )k, i=1
On the other hand,
x ∈ X.
i=1 n X
ai fi (xi0 ) = ai0 ,
i=1
and hence T is an isometry. Obviously, T ei = fi for each i = 1, . . . , n. Further, ϕ(xi ) = ei , i = 1, . . . , n. For any x ∈ X, ϕ(x) =
n X
fi (x)ϕ(xi ),
i=1
and thus ϕ(X) ⊂ co{e1 , . . . , en }. This proves (a). To verify (b), let T : E → `∞ n be an isometry. Since 1 is an extreme point of BE , each coordinate of T 1 equals either 1 or −1. By composing T with an isometry of `∞ n
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12 Constructions of function spaces
onto `∞ n if necessary we may assume that T 1 = (1, . . . , 1) =: e. Then T is positive and the functions fi := T −1 ei , i = 1, . . . , n, form the required partition of unity. Indeed, n n X X 1 = T −1 (e) = T −1 ei = fi , i=1
i=1
and all functions fi are positive. To prove (c), let Ep ⊂ Ep+r be given. Let T : Ep → `∞ p be a positive isometry given by (a). Using (b), we find a positive isometry S : Ep+r → `∞ p+r . Let I : Ep → −1 ∞ ∞ Ep+r denote the inclusion. Then SIT : `p → `p+r is an isometry into. Using ∞ Lemma 12.38(d) we find a space Fp+1 ⊂ `∞ p+r isometric with `p+1 such that ∞ SIT −1 (`∞ p ) ⊂ Fp+1 ⊂ `p+r .
Let J : Fp+1 → `∞ p+1 be a positive isometry. Since the set {T e1 , . . . , T ep } is the ∞ canonical basis of `p , using Lemma 12.38(c) we can find a positive admissible basis {v1 , . . . , vp+1 } of `∞ p+1 and positive numbers {a1 , . . . , ap } such that JSIei = vi + ai vp+1 ,
i = 1, . . . , p,
(12.26)
P and pi=1 ai ≤ 1. ∞ Let E := (JS)−1 (`∞ p+1 ) and V : E → `p+1 denote the restriction of JS. Since 1 ∈ E and V is a positive isometry, V 1 = (1, . . . , 1) ∈ `∞ p+1 . We set ei,p+1 := V −1 vi ,
i = 1, . . . , p + 1,
and ai,p := ai ,
i = 1, . . . , p.
Then {e1,p+1 , . . . , ep+1,p+1 } is a peaked partition of unity in Ep+r . Indeed, all the functions are obviously positive and have norm one. Since V 1 = (1, . . . , 1) =
p+1 X
vi ,
i=1
we obtain 1=
p+1 X
ei,p+1 .
i=1
We claim that
Pp
i=1 ai,p
= 1. Indeed, from (12.26) we get
ei = ei,p+1 + ai,p ep+1,p+1 , Hence 1=
p X i=1
ei,p+1 + ep+1,p+1
i = 1, . . . , p. p X i=1
ai,p .
12.2 Inverse limits of function spaces
451
If x ∈ X is a point where ep+1,p+1 (x) = 1, we obtain 1=
p X
ai,p .
i=1
Now the rest of the argument follows by an application of the this procedure to the family {e1,m , . . . , em,m } for all m = p + 1, . . . , p + r − 1. This concludes the proof. Theorem 12.40. Let X be a simplex and E ⊂ Ac (X) be a finite-dimensional space c isometric to `∞ n . Let {f1 , . . . , fk } ⊂ A (X) and ε > 0. Then there exist m ∈ N and a finite-dimensional subspace F ⊃ E of Ac (X) such that F is isometric to `∞ m and dist(fi , F ) < ε, i = 1, . . . , k. Proof. Given the objects as in the theorem, we first notice that without loss of generality we may assume that k = 1. Let f ∈ Ac (X), ε > 0 and E be given. Using the continuity of f and the compactness of X we can find a partition {U1 , . . . , Um } of ext X consisting of nonempty sets such that diam f (Uj ) < ε, j = 1, . . . , m. We select points xj ∈ Uj and set φj := cUj , αj := f (xj ), j = 1, . . . , m. Then m X f (x) − αj φj (x) ≤ ε,
x ∈ ext X.
j=1
By Lemma 12.17, there exists a peaked partition of unity {ψ1 , . . . , ψm } in Ac (X) such that m X f (x) − αj ψj (x) ≤ ε, x ∈ X. j=1
Then F := span{ψ1 , . . . , ψm } satisfies the required conditions by Lemma 12.39(a). Definition 12.41 (L∞ -spaces). We recall that a Banach space E is termed an L∞,λ space if for any finite-dimensional space A ⊂ E there exists a finite-dimensional subspace F ⊃ A of E whose Banach–Mazur distance from some `∞ n is smaller or equal than λ where n = dim F (that is, for every ε > 0 there exists an isomorphism −1 k < λ + ε). T : F → `∞ n such that kT kkT Corollary 12.42. If X is a simplex, then Ac (X) is an L∞,1+ε -space for every ε > 0. Proof. Let E ⊂ Ac (X) be a space of dimension n ∈ N and ε > 0. We choose ε0 ∈ (0, n) such that n ε0 < ε. n − ε0
452
12 Constructions of function spaces
Using Theorem 5.6(i) of [173], we find a basis {e1 , . . . , en } of E consisting of unit vectors and functionals {e∗1 , . . . , e∗n } ⊂ (Ac (X))∗ of norm 1 such that ( 1, i = j, ∗ ei (ej ) = 1 ≤ i, j ≤ n. 0, i 6= j, Using Theorem 12.40 we find a finite-dimensional space F 0 ⊂ Ac (X) isometric to `∞ m 0 for some m ∈ N such that dist(ei , F 0 ) < nε2 , i = 1, . . . , n. We select {f1 , . . . , fn } ⊂ 0 F 0 such that dist(ei , fi ) < nε2 , i = 1, . . . , n, and set G := span{f1 , . . . , fn }. If S : E → G is defined as n n X X ci ei := ci fi , c1 , . . . , cn ∈ R, S i=1
i=1
then S is an isomorphism satisfying ke − Sek ≤
n X
|ci |kei − fi k ≤
n ε0 ε0 X ∗ |e (e)| kek, ≤ i n2 n
e=
i=1
i=1
It follows that kS −1 k ≤
1 1−
ε0 n
n X
ci ei ∈ E.
i=1
.
(12.27)
Since G has dimension n, Theorem 5.6(ii) in [173] yields the existence of a projection P : F 0 → G with kP k ≤ n. We set H := ker P and F := E ⊕ H. Then F contains E and the operator T : F → F 0 defined as T f := Se + h,
f = e + h ∈ E ⊕ H,
satisfies by (12.27) 0 kf k − kT f k ≤ kf − T f k = kSe − ek ≤ ε kek n ε0 −1 ε0 1 ≤ kS kkSek ≤ kP (Se + h)k n n 1 − εn0 n kT f k. ≤ ε0 n − ε0 By our choice of ε0 , kf k − kT f k ≤ εkT f k and thus
kT k ≤ (1 − ε)−1 This finishes the proof.
and kT −1 k ≤ 1 + ε.
12.2 Inverse limits of function spaces
453
Remark 12.43. It is known that a Banach space E is an L∞,1+ε -space for every ε > 0 if and only if E ∗ is isometric to L1 (X, Σ, µ) for a suitable measure space (X, Σ, µ) (see p. 59 in [251]). Hence Corollary 12.42 provides another proof of Theorem 6.25. Lemma 12.44. Let E ⊂ Ac (X) be isometric to `∞ n and let 1 ∈ E. Then the set Sn := {e∗ ∈ E ∗ : e∗ ≥ 0, e∗ (1) = 1} is affinely homeomorphic to an (n − 1)-simplex. Proof. We start the proof by using Lemma 12.39(b) to find a peaked partition of unity {f1 , . . . , fn } ⊂ Ac (X) such that E = span{f1 , . . . , fn }. Let ψ : Sn → Rn be defined as ψ(e∗ ) := (e∗ (f1 ), . . . , e∗ (fn )), e∗ ∈ Sn . Then ψ is an affine homeomorphism of Sn to Rn . By Lemma 12.39(a) it is enough to show that ψ(Sn ) = ϕ(X) (ϕ is defined as in Lemma 12.39(a)). Let φ : X → Sn be the evaluation mapping defined as φ(x)(f ) := f (x), f ∈ E, x ∈ X. Obviously, for any e∗ ∈ Sn , ψ(e∗ ) has positive coordinates whose sum equals 1, hence ψ(Sn ) ⊂ ϕ(X). Conversely, if xi is a point of X with fi (xi ) = 1, then ψ(φ(xi )) = ei . Hence ψ(Sn ) ⊃ ext ϕ(X), and the conclusion follows. Theorem 12.45. Let X be a metrizable simplex. Then the following assertions hold. (a) There exists an increasing family {En } of finite-dimensional subspaces of Ac (X) S ∞ c such that E1 = span{1}, En is isometric to `∞ n and A (X) = n=1 En . (b) There exists an inverse sequence (Xn , pnm )n,m∈N such that each Xn is an (n−1)simplex and X = lim Xn . ←
Proof. For the proof of (a), let {fk : k ∈ N} be a dense subset of BAc (X) . We inductively use Theorem 12.40 to find an increasing sequence {nk } of natural numbers and spaces Enk ⊂ Ac (X) such that • n = 1 and E n1 = span{1}, 1 •
Enk is isometric to `∞ nk , k ∈ N,
•
Enk ⊂ Enk+1 , k ∈ N,
dist(fi , Enk ) < 2−k , i = 1, . . . , k, k ∈ N. By Lemma 12.39(a),(c), for each k ∈ N we can find spaces •
Enk ⊂ Enk +1 ⊂ · · · ⊂ Enk+1 −1 ⊂ Enk+1 such that each Ej is isometric to `∞ j , j = nk + 1, . . . , nk+1 − 1. Hence the proof of (a) is finished. To verify (b), let {En } be subspaces given by (a). We set Xn := Sn , n ∈ N, where Sn is the compact convex set from Lemma 12.44, and define pnm : Xm → Xn ,
454
12 Constructions of function spaces
n ≤ m, to be the restriction. By Lemma 12.44, the system (Xn , pnm )n,m∈N is an inverse sequence of (n − 1)-simplices. It is a routine verification to show that the mapping ψ : S(Ac (X)) → lim Xn defined as ←
ψ(s) := (s|En )n∈N ,
s ∈ S(Ac (X)),
is an affine homeomorphism. Since X is affinely homeomorphic to S(Ac (X)) (see Proposition 4.31), we are done. Corollary 12.46. For any metrizable simplex X,Sthere exists an increasing sequence {Fn } of finite-dimensional faces of X such that ∞ n=1 Fn is dense in X. Proof. Let {En } be a sequence of spaces in Ac (X) isometric to `∞ n such that E1 = c span{1} and their union is dense in A (X) (see Theorem 12.45). Let n ∈ N be fixed. By Lemma 12.39 there exists a peaked partition of unity {f1 , . . . , fn } such that En is its linear span and T : E → `∞ n defined as T fi := ei , i = 1, . . . , n, is a positive isometry. Let xi ∈ X be points such that fi (xi ) = 1,
i = 1, . . . , n.
Without loss of generality we may assume that all points xi are extreme. Then the set Fn := co{x1 , . . . , xn } is a finite-dimensional face of X. S We claim that {Fn } is the required sequence. Indeed, set F := ∞ n=1 Fn and assume the existence of ∞ [ Fn . x∈X\ n=1
Using the Hahn–Banach theorem we find a function f ∈ Ac (X) and δ > 0 such that 0 < f < 1 and m < m + δ < f (x),
where m := max f (F ).
Let n ∈ N and g ∈ En be such that 0 < g < 1 and kf − gk < 41 δ. If g = for some numbers a1 , . . . , an , then k(a1 , . . . , an )k`∞ = kgk. n Since
1 g ≤ δ + m on F, 4
for any i = 1, . . . , n we get 1 0 ≤ ai = g(xi ) ≤ m + δ. 4
Pn
i=1 ai fi
(12.28)
12.3 Several examples
455
By (12.28), kgk ≤ m + 41 δ. Hence 1 1 m + δ < f (x) ≤ δ + g(x) ≤ m + δ, 4 2 a contradiction. This finishes the proof. Theorem 12.47. Every simplex is the inverse limit of an inverse system of metrizable simplices. Proof. If X is a simplex, let E denote the family of all norm separable closed subspaces of Ac (X) that contain 1 and possess the weak Riesz interpolation property. By Theorem 9.12, this family is up-directed, that is, for any E1 , E2 ∈ E there exists S E ∈ E with E1 ∪ E2 ⊂ E. Moreover, {E : E ∈ E} is dense in Ac (X). For any E ∈ E, let XE := {e∗ ∈ E ∗ : e∗ ≥ 0, e∗ (1) = 1}. For any E, F ∈ E with E ⊂ F , let pE,F : XF → XE be the restriction mapping. Since each E ∈ E possesses the weak Riesz interpolation property, (XE , pE,F )E,F ∈E is an inverse system of metrizable simplices (as in Exercise 6.91). To finish the proof it is enough to realize that the mapping ψ : S(Ac (X)) → lim XE defined as ←
ψ(s) := (s|E )E∈E ,
s ∈ S(Ac (X)),
is an affine homeomorphism.
12.3 12.3.A
Several examples The Poulsen simplex
Lemma 12.48. Let X be a simplex in a locally convex space E and z ∈ E \ span X. Then co(X ∪ {z}) is a simplex and ext co(X ∪ {z}) = ext X ∪ {z}. Proof. We start the proof by noticing that z is an extreme point of co(X ∪ {z}). Further, any point x in co(X ∪{z}) can be uniquely decomposed as x = αy+(1−α)z for some y ∈ X and α ∈ [0, 1]. Hence there exists a unique maximal measure µ ∈ Mx (co(X ∪ {z})), namely µ = αδy + (1 − α)εz , where δy is the unique maximal measure for y on X. Thus co(X ∪ {z}) is a simplex and ext co(X ∪ {z}) = ext X ∪ {z}. Lemma 12.49. Let X be a simplex in a Banach space E. Then there exists a simplex Y ⊂ E ⊕ `2 such that X is a closed face of Y and ext Y is dense in Y .
456
12 Constructions of function spaces
Proof. Let X be a simplex in a Banach space E and let {en : n ∈ N} be the canonical basis of `2 . We identify x ∈ E with (x, 0) ∈ E ⊕ `2 and y ∈ `2 with (0, y) ∈ E ⊕ `2 . Let En := span{ek : k = 1, . . . , n} ⊂ `2 and let πn : E ⊕ `2 → E ⊕ En be the projection. Let Y0 := X. We construct inductively simplices Yn ⊂ E ⊕ `2 , points zn ∈ E ⊕ `2 and εn > 0, n ∈ N, such that (a) Yn = co(Yn−1 ∪ {zn }), n ∈ N, (b) Yn ⊂ E ⊕ En , n ∈ N, (c) ext Yn ⊂ ext Ym , n ≤ m, (d) πn (Ym ) = Yn and kπn (x) − xk ≤ 2−n+1 , x ∈ Ym , n ≤ m, (e) for each k ∈ N there exists n ≥ k such that dist(y, ext Yn ) < εk , y ∈ Yk , and εn+1 < 21 εk , (f) εn+1 ≤ εn , n ∈ N. To start the construction, we choose y1 ∈ Y0 and set ε1 := 2−1 , z1 := y1 + 2−1 e1 , Y1 := co(Y0 ∪ {z1 }). Assume now that the construction has been completed up to the n–th stage. We denote Mn (ε) := {x ∈ Yn : dist(x, {z1 , . . . , zn }) ≥ ε},
ε > 0.
Now we distinguish two cases. If Mn (εn ) 6= ∅, we choose yn+1 ∈ Mn (εn ) and set εn+1 = εn . Otherwise we find εn+1 ∈ (0, 12 εn ) such that Mn (εn+1 ) 6= ∅ and choose yn+1 ∈ Mn (εn+1 ). In both cases we set zn+1 := yn+1 + 2−n−1 en+1 , Yn+1 := co(Yn ∪ {zn+1 }). This finishes the construction. Properties (a)–(d) are obviously satisfied. Given k ∈ N, we find m ≥ k such that 2−m+3 < εk and notice that by (d) and the construction, kπm (zj ) − πm (zi )k ≥ εj − 2−m+2 ,
m ≤ i < j.
By the compactness of Ym we observe that infi,j≥m kπm (zj ) − πm (zi )k = 0, and thus there must exist j ∈ N with εj < εk . Hence we find n ≥ k such that the (n + 1)-th step of the construction follows the second case, which verifies property (e). S Let Y := ∞ n=1 Yn . Then Y is convex and compact (it follows from (d)).
12.3 Several examples
457
Further, ext Yn ⊂ ext Y . Indeed, let x ∈ ext Yn equal αy1 + (1 − α)y2 for some y1 , y2 ∈ Y . Then x = απm (y1 ) + (1 − α)πm (y2 ),
n ≤ m,
and thus πm (y1 ) = πm (y2 ) = x for each m ≥ n. Hence y1 = y2 = x. S Thus ext Y is dense in Y by (e). It also follows from (d) that ∞ n=1 {f ◦ πn : f ∈ Ac (Yn )} is dense in Ac (Y ). Since each space in this union has the Riesz interpolation property, Ac (Y ) has the weak Riesz interpolation property and hence Y is a simplex (see Theorem 6.16). Since X is obviously a face of Y , the proof is finished. Corollary 12.50. There exists a simplex X ⊂ `2 such that ext X is dense in X. Proof. An immediate consequence of Theorem 12.49. Lemma 12.51. Le ϕ : X → Y be a continuous affine surjection of a compact convex set onto an n-simplex Y . Then ϕ is open. Proof. Let {e0 , . . . , en } be affinely independent vectors in some space Rd and ϕ : X → co{e0 , . . . , en } be a continuous affine surjection. Let U ⊂ X be a closed neighborhood of x ∈ X and let y = ϕ(x). For each ei we find xi ∈ X such that ϕ(xi ) = ei . Let α ∈ (0, 1) be such that αx + (1 − α)xi ∈ U , i = 0, . . . , n. Then V := co{αy + (1 − α)ei : i = 0, . . . , n} is a neighborhood of y contained in ϕ(U ). Hence ϕ is open. Lemma 12.52. Let F be a closed face of a metrizable simplex X and {e1 , . . . , en } ⊂ Ac (X) be a peaked partition of unity such that each ei attains its maximum on F . Then there exists an affine continuous retraction g : X → F such that ei = ei ◦ g, i = 1, . . . , n. Proof. Let {x1 , . . . , xn } ⊂ F be such that ei (xi ) = 1, i = 1, . . . , n. Let π : X → co{x1 , . . . , xn } be defined as π(x) :=
n X
ei (x)xi ,
x ∈ X.
i=1
Then ei = ei ◦ π for all i = 1, . . . , n. We define ϕ : X → 2F by the formula ϕ(x) := π −1 (π(x)) ∩ F,
x ∈ X.
458
12 Constructions of function spaces
Then ϕ is an affine mapping with nonempty closed convex values. Since the set co{x1 , . . . , xn } is an (n − 1)-simplex (see Lemma 6.68 and Exercise 6.85), the mapping π : F → co{x1 , . . . , xn } is open by Lemma 12.51. Hence for any open set U ⊂ X, {x ∈ X : ϕ(x) ∩ U 6= ∅} = {x ∈ X : π −1 (π(x)) ∩ F ∩ U 6= ∅} = π −1 (π(F ∩ U )), and thus ϕ is lower semicontinuous. If f (x) = x for x ∈ F , then f is a continuous affine mapping with f (x) ∈ ϕ(x), x ∈ F . By Theorem 11.8, there exists a continuous affine selection g : X → F such that g(x) = x on F . Then π ◦ g = π and hence ei ◦ g = ei ◦ π ◦ g = ei ◦ π = ei ,
i = 1, . . . , n.
This concludes the proof. Lemma 12.53. Let F be a closed proper face of a metrizable simplex X with ext X = X. Let {e1 , . . . , en } ⊂ Ac (X) be a peaked partition of unity, {f1 , . . . , fn+1 } ⊂ Ac (F ) be a partition of unity, and let {a1 , . . . , an } be a set of positive numbers such that ei |F = fi + ai fn+1 , i = 1, . . . , n. Then, for each ε > 0, there exists a peaked partition of unity {g1 , . . . , gn+1 } ⊂ Ac (X) such that • g | i F = fi , i = 1, . . . , n + 1, •
kei − (gi + ai gn+1 )k < ε, i = 1, . . . , n.
Proof. Let xi ∈ ext X be such that ei (xi ) = 1, i = 1, . . . , n. We find a point yn+1 ∈ ext X \ F such that n X ei (yn+1 ) − ei aj xj < ε,
i = 1, . . . , n.
(12.29)
j=1
This is possible because ext X \ F is dense in X (see Lemma 2.90). For j = 1, . . . , n we set yj := xj if xj ∈ / F , otherwise we find yj ∈ ext X \ F such that |ei (xj ) − ei (yj )| < ε,
i = 1, . . . , n.
(12.30)
Let H := co(F ∪ {y1 , . . . , yn+1 }). Then H is a closed face of X and by Lemma 12.52 there exists a continuous affine retraction ϕ : X → H such that ei ◦ ϕ = ei , i = 1, . . . , n. For i = 1, . . . , n + 1 we find gei ∈ Ac (H) defined by conditions gei = fi on F and ( 0, i 6= j, gei (yj ) = j = 1, . . . , n + 1, 1, i = j,
12.3 Several examples
459
(use Theorem 6.6), and define gi := gei ◦ ϕ,
i = 1, . . . , n.
Then {g1 , . . . , gn+1 } is a peaked partition of unity and gi = fi on F . Since ei ◦ ϕ = ei and gi ◦ ϕ = gi , we get kei − (gi + ai gn+1 )k = kei ◦ ϕ − (gi ◦ ϕ + ai gn+1 ◦ ϕ)k =
max
j=1,...,n+1
|ei (yj ) − (gi + ai gn+1 )(yj )| < ε.
(Use (12.30) for j = 1, . . . , n and (12.29) for j = n + 1.) This finishes the proof. Theorem 12.54. Let X1 , X2 be metrizable simplices with ext Xi = Xi , i = 1, 2. Let Hi ⊂ Xi be a proper closed face, i = 1, 2, and let ϕ : H2 → H1 be an affine homeomorphism. Then ϕ can be extended to an affine homeomorphism of X2 onto X1 . Proof. Let {xn : n ∈ N} and {yn : n ∈ N} be dense subsets in the unit spheres of Ac (X1 ) and Ac (X2 ), respectively, such that x1 = y1 = 1. We construct inductively peaked partitions of unity {ej1,m , . . . , ejm,m } ⊂ Ac (X1 ),
j j {f1,m , . . . , fm,m } ⊂ Ac (X2 ),
m ≤ j, m ∈ N,
and sets of positive numbers {a1,m , . . . , am,m }, such that Pm (1) i=1 ai,m = 1, (2)
m ∈ N,
m ∈ N,
eji,m = eji,m+1 + ai,m ejm+1,m+1 , j j j fi,m = fi,m+1 + ai,m fm+1,m+1 ,
(3)
j eji,m (ϕ(t)) = fi,m (t),
(4)
−j keji,m − ej+1 i,m k < 2 , j j+1 kfi,m − fi,m k < 2−j ,
m ∈ N, 1 ≤ i ≤ m, j ≥ m + 1,
t ∈ H2 , m ∈ N, 1 ≤ i ≤ m, j ≥ m,
m ∈ N, 1 ≤ i ≤ m, j ≥ m.
m (5) If Em := span{em 1,m , . . . , em,m }, then for each n ∈ N there exists m ∈ N such that dist(xk , Em ) < 2−n , 1 ≤ k ≤ n.
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12 Constructions of function spaces
m , . . . , f m }, then for each n ∈ N there exists m ∈ N such (6) If Fm := span{f1,m m,m that dist(yk , Fm ) < 2−n , 1 ≤ k ≤ n. 1 } := {1}. We start the construction by setting {e11,1 } := {1} and {f1,1 j j Assume that {ej1,m , . . . , ejm,m } and {f1,m , . . . , fm,m } have been constructed for 1 ≤ m ≤ j ≤ p such that (1), (3), (4) hold for m ≤ j ≤ p, (2) holds for m + 1 ≤ j ≤ p, and (5) and (6) hold for some n ∈ N. By Theorem 12.40, there exists a finite-dimensional space Ep+r ⊂ Ac (X1 ) isometric to `∞ p+r such that Ep ⊂ Ep+r and
dist(xk , Ep+r ) < 2−n−1 ,
1 ≤ k ≤ n + 1.
By Lemma 12.39, there exist sets of positive numbers {a1,m , . . . , am,m },
p ≤ m ≤ p + r − 1,
and peaked partitions of unity {g1,m , . . . , gm,m } ⊂ Ep+r , such that Pm •
• •
i=1 ai,m = 1, p ≤ m ≤ p + r − p ei,p = gi,p+1 + ai,p gp+1,p+1 , 1 ≤ i
p + 1 ≤ m ≤ p + r,
1, ≤ p,
gi,m = gi,m+1 + ai,m gm+1,m+1 , 1 ≤ i ≤ m, p + 1 ≤ m ≤ p + r − 1.
For every 1 ≤ m ≤ p + r, p + 1 ≤ j ≤ p + r and 1 ≤ i ≤ m we set ( epi,m , 1 ≤ m ≤ p, eji,m := gi,m , p + 1 ≤ m ≤ p + r. Then (1), (2) and (4) hold for the constructed functions eji,m and (5) holds for n + 1. We now transfer the constructed objects to Ac (X2 ). Since p+1 {ep+1 1,p+1 ◦ ϕ, . . . , ep+1,p+1 ◦ ϕ}
is a partition of unity on H2 and p fi,p (t) = epi,p (ϕ(t)) = gi,p+1 (ϕ(t)) + ai,p gp+1,p+1 (ϕ(t)) p+1 = ep+1 i,p+1 (ϕ(t)) + ai,p ep+1,p+1 (ϕ(t)),
t ∈ H2 , 1 ≤ i ≤ p,
Lemma 12.53 provides for any ε > 0 a peaked partition of unity {h1,p+1 , . . . , hp+1,p+1 } ⊂ Ac (X2 ) such that •
hi,p+1 (t) = ep+1 i,p+1 (ϕ(t)), t ∈ H2 , 1 ≤ i ≤ p + 1,
461
12.3 Several examples •
p kfi,p − (hi,p+1 + ai,p hp+1,p+1 )k < ε, 1 ≤ i ≤ p.
We take ε < 2−2(p+1) and define p+1 fi,p+1 := hi,p+1 ,
1 ≤ i ≤ p + 1,
p+1 p+1 p+1 fi,p := fi,p+1 + ai,p fp+1,p+1 ,
1 ≤ i ≤ p,
.. . p+1 p+1 p+1 fi,l := fi,l+1 + ai,l fl+1,l+1 ,
1 ≤ i ≤ l,
.. . p+1 p+1 p+1 f1,1 := f1,2 + a1,1 f2,2 .
Then p+1 p kfi,k − fi,k k < 2−p ,
1 ≤ k ≤ p, 1 ≤ i ≤ k.
Similarly we construct m m m m m {f1,m , . . . , fm,m }, . . . , {f1,2 , f2,2 }, {f1,1 },
p + 2 ≤ m ≤ p + r.
Then (4) holds for j ≤ p + r − 1. We interchange the roles of Ac (X1 ) and Ac (X2 ), that is, we find by Theorem 12.40 a space Fp+r+s ⊂ Ac (X2 ) isometric to `∞ p+r+s such that Fp+r ⊂ Fp+r+s and dist(yk , Fp+r+s ) < 2−(n+1) ,
1 ≤ k ≤ n + 1.
j Now we proceed with the construction as above and find vectors fi,m and then eji,m . Then (1)–(4) hold for relevant indices and (6) for n + 1. This completes the inductive construction. We set
ei,m := lim eji,m , j→∞
j fi,m := lim fi,m , j→∞
1 ≤ i ≤ m, m ∈ N.
It follows from (4), (5) and (6) that span{ei,m : 1 ≤ i ≤ m, m ∈ N}
and
span{fi,m : 1 ≤ i ≤ m, m ∈ N}
are dense in Ac (X1 ) and Ac (X2 ), respectively. Further, by (2), ei,m = ei,m+1 + ai,m em+1,m+1 , fi,m = fi,m+1 + ai,m fm+1,m+1 ,
m ∈ N, 1 ≤ i ≤ m,
(12.31)
and by (3) ei,m (ϕ(t)) = fi,m (t),
t ∈ H2 , m ∈ N, 1 ≤ i ≤ m.
(12.32)
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12 Constructions of function spaces
Hence {e1,m , . . . , em,m } and
{f1,m , . . . , fm,m },
m ∈ N,
are peaked partitions of unity, in particular, they are linearly independent. It follows from (12.31) that the equalities T ei,m = fi,m ,
1 ≤ i ≤ m, m ∈ N,
j define an isometry T : Ac (X1 ) → Ac (X2 ). Since ej1,1 = 1 and f1,1 = 1 for each j ∈ N, we have e1,1 = 1 and f1,1 = 1, and thus T 1 = 1. Hence T is positive. Let φi : Xi → S(Ac (Xi )) be the affine homeomorphism from Proposition 4.31, i = 1, 2. As in the proof of Theorem 10.14, the adjoint operator T ∗ : (Ac (X2 )∗ →(Ac (X1 ))∗ defines an affine homeomorphism of S(Ac (X2 )) onto S(Ac (X1 )). Since
T e(t) = e(ϕ(t)),
t ∈ H2 , e ∈ Ac (X1 ),
the mapping ∗ φ−1 1 ◦ T ◦ φ2
is an affine homeomorphism of X2 onto X1 that extends ϕ. This finishes the proof. Corollary 12.55. Let X1 , X2 be metrizable simplices with ext Xi = Xi , i = 1, 2. Then X1 is affinely homeomorphic to X2 . Proof. We pick points xi ∈ ext Xi , i = 1, 2. Then the faces Fi := {xi }, i = 1, 2, are affinely homeomorphic and thus the conclusion follows from Theorem 12.54. Definition 12.56. (Poulsen simplex) A metrizable simplex, whose set of extreme points is dense, is called the Poulsen simplex. Lemma 12.57. Let F be a closed face of a simplex X and x1 , x2 be distinct points in the complementary face F 0 . Then there exists a positive function f ∈ Ac (X) such that f = 0 on F and f (x1 ) 6= f (x2 ). Proof. Given the objects as in the hypothesis, we find a function g ∈ Ac (X) such that 0 ≤ g < 21 and δ := g(x1 ) − g(x2 ) > 0. Then k := cF ∨ g is a convex upper semicontinuous function. We claim that k ∗ = k = g on F 0 . Indeed, let z ∈ F 0 be arbitrary. We find a measure ν ∈ Mz (X) such that ν(k) = ∗ k (z) (see Lemma 3.21). We find a maximal measure µ with ν ≺ µ and observe, using Proposition 3.56, that k ∗ (z) = ν(k) ≤ µ(k) ≤ k ∗ (z).
12.3 Several examples
463
Since µ is maximal, Lemma 8.13 and Theorem 8.39 yield µ(F 0 ) = 1. Hence Z ∗ k(t) dµ(t) = µ(g) = g(z). k (z) = µ(k) = F0
We select a function h ∈ Ac (X) such that k ≤ h and h(xi ) < g(xi ) + 12 δ, i = 1, 2. By Theorem 6.6, there exists a function a ∈ Ac (X) such that k ≤ a ≤ 1 ∧ h. Then the function f := 1 − a satisfies our requirements. Proposition 12.58. Let F be a closed face of a simplex X and let A := {f ∈ Ac (X) : 0 ≤ f ≤ 1, f |F = 0}. Let ϕ : X → RA be defined as ϕ(x) := (f (x))f ∈A , x ∈ X, and let Y := ϕ(X). (a) The set Y is a simplex and ϕ(F ) is an extreme point of Y . (b) The mapping ϕ|F 0 is injective and ϕ(F 0 ) is the complementary face of ϕ(F ). (c) If X is the Poulsen simplex, then Y is the Poulsen simplex. Proof. For f ∈ A, let πf : Y → R be the restriction on the coordinate determined by f . Then πf ∈ Ac (Y ). It follows that for any function f ∈ Ac (X) constant on F there exists a function g ∈ Ac (Y ) such that f = g ◦ ϕ. For the proof of (a) we notice that ϕ(F ) = 0 is an extreme point of Y . Indeed, if αϕ(x)+(1−α)ϕ(y) = 0 for some x, y ∈ X and α ∈ (0, 1), then f (αx+(1−α)y) = 0 for each f ∈ A, and thus αx + (1 − α)y ∈ F (use Corollary 8.64). Hence x, y ∈ F and ϕ(x) = ϕ(y) = 0. Further, let g1 , −g2 ∈ Kc (Y ) be such that 0 ≤ g1 ≤ g2 ≤ 1. By Theorem 6.6, there exists a function f ∈ A such that g1 ◦ ϕ ≤ f ≤ g2 ◦ ϕ on X. Then g1 ≤ πf ≤ g2 and Y is a simplex by Theorem 6.6. To verify (b), we notice that ϕ is injective on F 0 by Lemma 12.57. The second assertion requires to show that (ϕ(F ))0 = ϕ(F 0 ). In other words, we need to prove that c∗ϕ(F ) (t) = 0
⇐⇒
t ∈ ϕ(F 0 ),
t ∈ Y.
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12 Constructions of function spaces
Assume first that c∗ϕ(F ) (t) = 0, that is, t ∈ (ϕ(F ))0 . Let x ∈ X be such that ϕ(x) = t. For any ε > 0 there exists a function g ∈ Ac (Y ) such that g ≥ cϕ(F ) and g(t) < ε. Thus f := g ◦ ϕ ∈ Ac (X) satisfies f ≥ cF and f (x) < ε. Hence c∗F (x) = 0 and x ∈ F 0 . Thus t ∈ ϕ(F 0 ). Conversely, let x ∈ F 0 be given. For any ε > 0 we use Theorem 6.6 to find a function f ∈ Ac (X) such that f = 1 on F and f (x) < ε. For the function g ∈ Ac (Y ) satisfying f = g ◦ ϕ we get g ≥ cϕ(F ) and g(ϕ(x)) < ε. Thus c∗ϕ(F ) (ϕ(x)) = 0 and ϕ(x) ∈ (ϕ(F ))0 . Finally, let X be the Poulsen simplex. Since ϕ : F 0 → ϕ(F 0 ) is an affine bijection and (ϕ(F ))0 = ϕ(F 0 ), ϕ(ext X ∩ F 0 ) = ϕ(ext F 0 ) = ext ϕ(F 0 ) = ext Y ∩ (ϕ(F ))0 . Further, F as a closed proper face of X has empty interior (see Lemma 2.90), and thus ext X ∩ F 0 = ext F 0 is dense in X. Hence ext ϕ(F 0 ) is dense in Y and Y is the Poulsen simplex. This finishes the proof. Theorem 12.59. Let Fi , i = 1, 2, be closed faces of the Poulsen simplex and let Fi0 be their complementary faces, i = 1, 2. Then F10 is affinely homeomorphic to F20 . Proof. For each i = 1, 2, let ϕi : Xi → Yi be as in Proposition 12.58. Then the singletons {yi } := ϕi (Fi ), i = 1, 2, are obviously affinely homeomorphic, and thus by Theorem 12.54 there exists an affine homeomorphism ϕ : Y1 → Y2 sending ϕ1 (F1 ) onto ϕ2 (F2 ). By Proposition 12.58, ϕ(ϕ1 (F10 )) = ϕ({y1 }0 ) = {y2 }0 = ϕ2 (F20 ). 0 0 Thus ϕ−1 2 ◦ ϕ ◦ ϕ1 is an affine homeomorphism of F1 onto F2 .
Theorem 12.60 (Properties of the Poulsen simplex). Let S stand for the Poulsen simplex. (a) Any metrizable simplex is affinely homeomorphic to a face of S. (b) Let X be a metrizable simplex with the following properties: •
Any metrizable simplex is affinely homeomorphic to a face of X.
•
If F1 , F2 are faces of X with dim F1 = dim F2 < ∞, then there exists an affine homeomorphism ϕ of X onto itself such that ϕ(F1 ) = F2 .
Then X is affinely homeomorphic to S.
12.3 Several examples
465
(c) If K ⊂ ext S is a compact set, then its interior (relative to ext S) is empty. (d) Every Polish space is homeomorphic to a closed subset of ext S. (e) If K1 , K2 are homeomorphic compact subsets of ext S, then there exists an affine homeomorphism ϕ of S onto itself such that ϕ(K1 ) = K2 . (f) The set ext S is arcwise connected (that is, for any points x1 , x2 ∈ ext S there exists a homeomorphic copy of [0, 1] contained in ext S that joins x1 with x2 ). Proof. To prove (a), let X be a metrizable simplex. By Proposition 2.45 we may assume that X is a norm compact subset of `2 . By Lemma 12.49, there exists a compact simplex Y ⊂ `2 ⊕ `2 such that X is a closed face of Y . Due to Corollary 12.55, Y is affinely homeomorphic to S. To verify (b), let X possess the prescribed property. By the assumption, X contains a face X0 affinely homeomorphic to the Poulsen simplex S. If F ⊂ X is an arbitrary finite-dimensional closed face, there exist a closed face of the same dimension F0 ⊂ X0 and an affine homeomorphism ϕ : X → X such that ϕ(F ) = F0 . Then ϕ(F ) = F0 ⊂ ext X0 ⊂ ext X. Hence F ⊂ ext X. Since finite-dimensional faces are dense in X by Corollary 12.46, X = ext X. Thus X is affinely homeomorphic to S by Corollary 12.55. To show (c), let K ⊂ ext S be a compact set. Assume that its interior relative to ext S is nonempty, that is, that there exists an open set U ⊂ S intersecting K such that U ∩ ext S ⊂ K. Since ext S is dense in S, U ∩ ext S is dense in U . Thus F := co K is a face in S (see Exercise 6.84) that is proper (any point in ext S \ K belongs to S \ F ). Since F contains U , we obtain a contradiction with Lemma 2.90. If P is a Polish space (a separable completely metrizable topological space), then there exists a metrizable simplex X such that ext X is homeomorphic to P (see Theorem 10.70). By (a), X can be realized as a closed face of S, which proves (d). If ϕ : K1 → K2 is a homeomorphism between compact subsets of ext S, we can extend it to an affine homeomorphism ψ : co K1 → co K2 . Indeed, co Ki , i = 1, 2, are closed faces with closed sets of extreme points, hence they are Bauer simplices (see Lemma 11.11). By Proposition 6.39, co Ki is affinely homeomorphic to M1 (Ki ), i = 1, 2, and the claim follows. Theorem 12.54 thus implies assertion (e). To prove (f), let x1 , x2 ∈ ext S be distinct points. By (d), ext S contains a homeomorphic copy l of [0, 1] joining a pair of points y1 , y2 ∈ ext S. By (e) we can find an affine homeomorphism ϕ : S → S such that ϕ(yi ) = xi , i = 1, 2. Hence ϕ(l) is a simple arc joining x1 and x2 .
12.3.B
A big simplicial space
Construction 12.61. Let H1 be a simplicial function space on a compact space K1 such that H1 = Ac (H1 ). Let K2 be the compact space constructed from K1 as in
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12 Constructions of function spaces
Definition 6.13, where B = ChH1 (K1 ), Lb = [0, 1] and µb is Lebesgue measure on [0, 1] for each b ∈ B. If t ∈ [0, 1], we write (b, t) for a point in Lb . We consider the space K1 as a subset of K2 . Let p : K2 → K1 be the natural projection, that is, ( y, y ∈ K1 , p(y) = x, y ∈ Lx for some x ∈ B. Let H2 := {f ∈ C(K2 ) : f |K1 ∈ H1 , f (x) = µx (f ) for x ∈ ChH1 (K1 )}. Then H1 can be considered to be a subspace of H2 if we identify f ∈ H1 with f ◦ p ∈ H2 . Lemma 12.62. Let (K1 , H1 ) be a simplicial function space satisfying H1 = Ac (H1 ) and let (H2 , K2 ) be as in Construction 12.61. (a) Then H2 is a function space on the compact space K2 with H2 = Ac (H2 ) and ChH2 (K2 ) = K2 \ K1 . (b) The mapping p : K2 → K1 is admissible and p(ChH2 (K2 )) = ChH1 (K1 ). (c) If µ ∈ M1 (K2 ) is such that p] µ is continuous, then µ(K2 \ K1 ) = 0. (d) If µ ∈ M1 (K1 ) is H1 -maximal and continuous, then it is H2 -maximal. (e) The function space H2 is simplicial. (f) For any x ∈ K1 , the unique H2 -maximal measure H-representing x is continuous. Proof. We start the proof of (a) by noticing that K2 is a (Hausdorff) compact space and H2 is a function space. Indeed, 1 ∈ H2 , the space H2 separates points of K1 and any pair of points in K2 \ K1 not contained in the same Lb . Given x ∈ ChH1 (K1 ), let R1 g ∈ C([0, 1]) be a monotone function satisfying 0 g = 0. Then ( g on Lx , h := 0 otherwise, is in H2 and separates points of Lx . If f ∈ Ac (H2 ) is given, f |K1 ∈ Ac (H1 ) = H1 and f (x) = µx (f ) for any x ∈ ChH1 (K1 ). Thus f ∈ H2 . Obviously, any point of K1 has an H2 -representing measure distinct from the Dirac measure, and thus ChH2 (K2 ) ⊂ K2 \ K1 . On the other hand, it is easy to construct an H2 -exposing function for any point in K2 \ K1 . Hence ChH2 (K2 ) = K2 \ K1 . Since (b) is obvious, we proceed to the proof of (c). Let µ ∈ M1 (K2 ) be such that p] µ is continuous on K1 . Assume that µ(K2 \ K1 ) > 0. Since K2 \ K1 is a
12.3 Several examples
467
disjoint union of open sets Lx , x ∈ ChH1 (K1 ), there exists x ∈ ChH1 (K1 ) such that µ(Lx ) > 0. Then p] µ({x}) = µ(Lx ) > 0, a contradiction with the assumption. To prove (d), let µ ∈ M1 (K1 ) be H1 -maximal and continuous. Let ν ∈ M1 (K2 ) be any measure with µ ≺H2 ν. By (b) and Proposition 12.24(c), µ ≺H1 p] ν. Since µ is H1 -maximal, µ = p] ν. By the continuity of µ and (c), ν is carried by K1 , and hence µ = ν. 1 The next step P∞ measure, P∞is to prove (e). Let µ ∈ M (K1 ) be a discrete H1 -maximal that is, µ = n=1 an εxn , where the numbers an are strictly positive, P∞ n=1 an = 1 and xn are distinct points in ChH1 (K1 ). We claim that ν := n=1 an µxn is the unique H2 -maximal measure with µ − ν ∈ H⊥ . 2 To see this, let ν 0 ∈ M1 (K2 ) be an H2 -maximal measure with µ − ν 0 ∈ H⊥ 2 . Then 0 is H -maximal (see Proposition 12.25). Thus p ν 0 = µ by p] ν 0 − µ ∈ H ⊥ and p ν 1 ] ] 1 Proposition 6.9. From this and from the fact that ν 0 does not charge anyPpoint in K1 (by Construction 12.61 and H2 -maximality of ν 0 ), we obtain that ν 0 = ∞ n=1 an λxn for some probability measures λxn on Lxn . Fix n ∈ N. If g ∈ C([0, 1]) is a continuous R1 function with 0 g = 0, then ( g on Lxn , h := 0 otherwise, is in H2 and ν 0 (h) = µ(h) = 0. Hence λxn is a multiple of µxn . Since they are both probability measures, λxn = µxn . Thus ν 0 = ν. It follows that any point x ∈ K1 has a unique H2 -maximal measure H2 -representing x. Indeed, let δx bePthe unique H1 -maximal measure H1 -representing x. We decompose δx = (δx )c + P∞ n=1 an εxn , where (δx )c is the continuous part of δx , the numbers an are positive, ∞ n=1 an ≤ 1 and the points xn are distinct elements of ChH1 (K1 ) (see Proposition A.78). Then we get that δ := (δx )c +
∞ X
a n µ xn
n=1
is the unique H2 -maximal measure H2 -representing x. To see this, let µ be an H2 -maximal measure with µ − δ ∈ H⊥ 2 . Then p] µ is H1 -maximal and p] µ − δx ∈ H⊥ 1 . Thus (δx )c +
∞ X
an εxn = p] µ = µ|K1 + p] (µK2 \K1 ).
n=1
By (c), p] (µ|K2 \K1 ) is discrete. Hence µ|K1 = (δx )c . Since ∞ X n=1
an µxn − µ|K2 \K1 ∈ H⊥ 2,
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12 Constructions of function spaces
it follows that
∞ X
an εxn − µ|K2 \K1 ∈ H⊥ 2.
n=1
P∞
Hence µ|K2 \K1 = n=1 an µxn by the preceding argument. Thus µ = δ. Hence H2 is simplicial, which finishes the proof of (e). Finally, given x ∈ K1 , we repeat the argument from the proof of (e) to deduce that the unique H2 -maximal measure H2 -representing x is continuous. This concludes the proof. Construction 12.63. We inductively construct simplicial function spaces (Hα , Kα ) together with admissible mappings pβα : (Kα , Hα ) → (Kβ , Hβ ) and homeomorphic embeddings iβα : Kβ → Kα for all ordinals β ≤ α ≤ ω1 . Let K0 := {0} and H0 := C(K0 ). Suppose that α ∈ (0, ω1 ) and (Kβ , Hβ ) along with the respective mappings have been constructed for all β < α. If α is an isolated ordinal, say α = γ + 1, we use Construction 12.61 for the space (Kγ , Hγ ) to get (Kα , Hα ). Let pγα be the mapping p : Kα → Kγ from Construction 12.61 and iγα : Kγ → Kα be the identity mapping. Further we set pβα := pβγ ◦ pγα
and
iβα := iγα ◦ iβγ ,
β < α,
and pαα be the identity. If α is a limit ordinal, let (Kα , G α ) be the inverse limit of ((Kβ , Hβ ), pβγ )β,γ<α . We set (Kα , Hα ) := (Kα , G α ) (the closure is meant to be in C(Kα )). If πβ : Kα → Kβ , β < α, is the projection, we define pβα := πβ ,
β < α.
For any β < α, let iβα : Kβ → Kα be the natural homeomorphic embedding, that is, iβα (x) = (p0β (x), p1β (x), . . . , x, iβ,β+1 (x), . . . ),
x ∈ Kβ .
Finally, let iαα be the identity mapping. For any α < ω1 , we denote by πα the mapping pαω1 : Kω1 → Kα and by iα : Kα → Kω1 the embedding iαω1 . Lemma 12.64. Let (K, H) := (Kω1 , Hω1 ). Then the following assertions hold. (a) The function space Hα is simplicial and Hα = Ac (Hα ), α ∈ [0, ω1 ]. (b) If x ∈ iα (Kα ), α < ω1 , then its unique H-maximal measure is carried by iα+1 (Kα+1 ). (c) For any continuous measure µ ∈ M1 (K), there exists ordinal α < ω1 such that µ(iα (Kα )) = 1.
12.3 Several examples
(d) ChH (K) = K \
S
469
α<ω1 iα (Kα ).
(e) The function ( 0, f (x) := 1,
x ∈ ChH (K), x ∈ K \ ChH (K),
is universally measurable and belongs to H⊥⊥ . Proof. We begin by proving (a) by transfinite induction. Obviously, (K0 , H0 ) is a simplicial function space and Ac (H0 ) = H0 . Assume that (a) has been verified for all β < α for some α ≤ ω1 . If α is isolated, say α = γ + 1, then Hα is simplicial and Ac (Hα ) = Hα by Lemma
