CONVEX CONES
This Page Intentionally Left Blank
NORTH-HOLLAND MATHEMATICS STUDIES
56
Notas de Matematica (82) Edi...
67 downloads
1252 Views
16MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
CONVEX CONES
This Page Intentionally Left Blank
NORTH-HOLLAND MATHEMATICS STUDIES
56
Notas de Matematica (82) Editor: Leopoldo Nachbin Universidade Federal do Rio de Janeiro and University of Rochester
Convex Cones BENNO FUCHSSTEINER and WOLFGANG LUSKY University of Paderborn Germany
NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM
NEW YORK
OXFORD
North-Holland Publishing Company, I981
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 86290 0
Publishers: NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM NEW YORK OXFORD Sole distributors for the U.S.A. and Canada: ELSEVIER NORTH-HOLLAND, INC. 5 2 VANDERBILT AVENUE, NEW YORK, N.Y. 10017
Library of Congress Cataloging in Publication Data
F u c h s s t e i n e r , Benno. Convex cones. (North-Holland mathematics s t u d i e s ; 56) Bibliography: p . 1. Functions of r e a l v a r i a b l e s . 2 . Convex bodies. 3 . Cone. I. Lusky, Wolfgang, 194811. T i t l e . 111. S e r i e s .
PRINTED IN THE NETHERLANDS
PREFACE
The aim of t h i s book i s t o o u t l i n e an elementary theory o f l i n e a r funct i o n a l s on convex cones, b u t convex cones a r e here taken i n a s l i g h t l y more general way than usual, they need not be imbedded i n a vector space. In consequence, we do not have a general c a n c e l l a t i o n law f o r t h e a d d i t i o n . Typical examples f o r t h e cones we have in mind a r e 6 = R u I - -1 o r the upper semicontinuous 6- valued functions on some topological space. Accordingly, l i n e a r functionals on such cones a r e allowed t o a t t a i n values i n 6 instead of R . This g e n e r a l i t y has advantages w i t h respect t o extensions of l i n e a r f u n c t i o n a l s .
We believe t h a t , t o a l a r g e e x t e n t , t h e work of an a n a l y s t c o n s i s t s of a s k i l l f u l handling of i n e q u a l i t i e s . I n order t o t r e a t i n e q u a l i t i e s in their proper context, we combine t h e theory of l i n e a r f u n c t i o n a l s w i t h order t h e o r e t i c aspects i n s o f a r as we mostly deal with f u n c t i o n a l s being monotone w i t h respect t o s u i t a b l e order r e l a t i o n s . For those a p p l i c a t i o n s we have i n mind, monotonicity usually implies c o n t i n u i t y . Therefore the material of t h i s book i s presented i n an order t h e o r e t i c r a t h e r than topological s e t t i n g . Very few topological arguments a r e used in this book, b u t this i s n e i t h e r done t o c a r r y our viewpoint t o t h e extreme nor t o show disregard f o r t h e many beautiful arguments and techniques topology has contributed to analysis. We have adapted, and sometimes extended, many basic techniques which were developed in the study of compact convex s e t s . Accordingly, measure theory takes an important place throughout this book. For those readers who a r e not well acquainted w i t h t h a t f i e l d , we have gathered some elementary r e s u l t s and techniques i n t h e A p p e n d i x . V
vi
Preface
We want t o emphasize t h a t these notes a r e w r i t t e n f r o m a h i g h l y personal p o i n t o f view. The c h o i c e o f what t o i n c l u d e has been q u i t e s u b j e c t i v e , we c o n c e n t r a t e h e a v i l y on work done b y t h e f i r s t a u t h o r d u r i n g t h e l a s t years. While t h e i n t e r e s t o f b o t h authors l i e s more on t h e c o n c r e t e s i d e of a n a l y s i s , nevertheless we a r e aware o f t h e d e p l o r a b l e f a c t t h a t t h i s w i l l n o t always b e t r a n s p a r e n t t o t h e r e a d e r . To h e l p compensate f o r t h i s disadvantage, we have added many examples and a p p l i c a t i o n s from d i f f e r e n t areas. But, due t h e l a c k o f space and time,we have had t o o m i t many i m p o r t a n t examples, e s p e c i a l l y from e q u i l i b r i u m t h e o r y i n Mathematical Economics and from S t a t i s t i c a l Mechanics. The book i s m a i n l y addressed t o graduate s t u d e n t s . S i n c e no s p e c i f i c p r e r e q u i s i t e s a r e necessary, however, and no deeper knowledge o f any s p e c i f i c p a r t o f mathematics i s r e q u i r e d , these notes can as w e l l be read by every s t u d e n t . We have t r i e d t o p r e s e n t a l l arguments i n d e t a i l , although sometimes t h i s i s done r a t h e r b r i e f l y and t h i s i s e s p e c i a l l y t r u e i n the a p p l i c a t i o n sections. The book i s divided i n t o two chapters, which a r e t h e n s u b d i v i d e d i n t o two and s i x s e c t i o n s , r e s p e c t i v e l y . I n t h e f i r s t chapter, we p r e s e n t t h e b a s i c m a t e r i a l about l i n e a r f u n c t i o n a l s . We g i v e o u r f a v o u r i t e v e r s i o n o f t h e Hahn-Banach Theorem, and we o b t a i n i m p o r t a n t i n f o r m a t i o n about extens i o n and decomposition o f l i n e a r f u n c t i o n a l s . The second p a r t i s m a i n l y devoted t o t h e r e p r e s e n t a t i o n o f 1 i n e a r f u n c t i o n a l s by i n t e g r a l s . When r e f e r r i n g back t o items w i t h i n a chapter, t h e c h a p t e r number i s omi t t e d . We owe a g r e a t deal t o Mrs. Waltraud Bohmer f o r h e r e x c e l l e n t t y p i n g o f t h e manuscript. F i n a l l y , we thank t h e e d i t o r , Leopoldo Nachbin, and t h e p u b l i s h e r , E. F r e d r i k s s o n , f o r i n c l u d i n g t h i s volume i n t h e i r s e r i e s .
TABLE OF CONTENTS
PREFACE CHAPTER I
V
LINEAR FUNCTIONALS
Chapter I . 1.
1.1
1
The Sandwich Theorem
Semi groups
2
1.2
Cones
6
1.3
Some Consequences o f t h e Sandwich Theorem
11
1.4
Sum Theorem and F i n i t e Decomposition Theorem
14
1.5
Some elementary Appl ic a t i o n s
20
1.5.1
The Phragrnen-Lindelof
1.5.2
Bishop's General V e r s i o n o f t h e "Three
1.5.3
Seevers' P r o o f f o r t h e E x i s t e n c e o f
Principle
21
C i r c l es Theorem"
22
Jensen Measures 1.5.4
When i s a F u n c t i o n Dominated by an A r i t h m e t i c Mean o f o t h e r F u n c t i o n s ?
1.5.5
20
24
Decomposition w i t h r e s p e c t t o P o s i t i v e l y Independent F u n c t i o n a l s
26
1.5.6
The Jordan Decomposition
28
1.5.7
The Lemma o f Farkas
29
1.5.8
The S e p a r a t i o n Theorem
33
1.5.9
A f f i n e I n t e r p o s i t i o n and t h e M a z u r - O r l i c z Theorem
34
The Riesz-Konig Theorem
37
1.7
A S t r a s s e n - t y p e D i s i n t e g r a t i o n Theorem
42
1.8
Some Appl i c a t i o n s
49
1.6
1.9
1.8.1
An a b s t r a c t Flow Theorem
49
1.8.2
Flows i n networks
54
1.8.3
Supply-Demand Problems
58
A d d i t i o n a l Remarks and Comments vii
72
Table of Contents
viii
Chapter 1.2
Order Units and L a t t i c e Cones
Order U n i t Cones 2.2 The Kakutani-Krein-Stone-Yosida Theorem Order Complete Vector L a t t i c e s with Order Unit 2.3 L a t t i c e Cones 2.4 2.5 Riesz Property and F i n i t e Sum Property The Positive Dual Cone 2.6 Dual Orders and t h e Cartier-Fell -Meyer Theorem 2.7 Free L a t t i c e Cones 2.8 2.9 Simplicia1 Cones 2.10 Characters 2.11 Some Examples
2.1
87 89 93 96 101 106 113 118 122 130 139 149
2.11.1 Absorbing Functionals and Banach L a t t i c e s
149
2.11.2 AM- and AL- Cones
154
2.11.3 Characters o f AM-Cones
158
Kakutani's Characterization o f S u b l a t t i c e s o f C ( K ) - Spaces 2.11.5 Korovkin's Theorem
162
2.11.4
2.12
Remarks and Comments
CHAPTER I1
1.2 1.3 1.4 1.5
Countable Decomposition
177
Preliminaries The Main Decomposition Theorems Dini Cones Weak Dini Cones Remarks and Comments
178
Chapter 11.2 2.1 2.2 2.3 2.4 2.5 2.6
170
REPRESENTING MEASURES
Chapter I I . 1 1.1
167
Representing Measures
Decomposition Properties and Measure Theory Dini Cones and Representing Measures Weak Dini Cones and Signed Representing Measures Representing Measures on Weighted Cones Dirichlet States Elementary Examples and Applications
182 189 196 198 201 202 210 213 218 220 227
Table of Contents
2.6.1
The Riesz R e p r e s e n t a t i o n Theorem f o r non-Hausd o r f f Sets
2.6.2 2.6.3 2.6.4 2.6.5 2.7
Regul a r i t y and Support R e p r e s e n t a t i o n by Unbounded Measures Adapted Cones R e p r e s e n t a t i o n o f F u n c t i o n a l s w i t h Zero Mass
Remarks and Comments
Chapter II . 3
Boundaries
3.1
F i x p o i n t Boundaries, B a u e r ' s Maximum P r i n c i p l e and
3.2 3.3 3.4 3.5 3.6 3.7
More Boundaries
t h e K r e i n-Mi 1man Theorem Choquet's Theorem Maximal Measures The Choquet-Meyer Theorem D i n i Boundaries Remarks and Comments
Chapter 11.4
ix
I n t e g r a l R e p r e s e n t a t i o n o f Operators t a k i n g
227 2 28 232 237 241 245 251 252 258 26 7 274 281 287 303 305
Values i n an Order Complete Vector L a t t i c e Chapter 11.5
Generalized Hewitt-Nachbin Spaces
5.1
B a s i c D e f i n i t i o n s and t h e i r Meaning i n t h e
5.2 5.3 5.4 5.5 5.6
The F- C o m p a c t i f i c a t i o n
Classical S i t u a t i o n The F- R e a l c o m p a c t i f i c a t i o n F- Pseudocompactness Some Consequences Remarks and Comments
Chapter 11.6
6.1 6.2
Examples and A p p l i c a t i o n s
Completely Monotonic Functions
314 318 324 328 331 337 340 341
Kendal 1 ' s Theorem on I n f i n i t e l y D i v i s i b l e Completely Monotonic F u n c t i o n
6.3 6.4 6.5
313
M u l t i p l i c a t i v e Cones Banach Algebras and S p e c t r a l Theory The Bochner-Weil Theorem
345 349 352 362
Table of Contents
X
The L’evy-Khi n t c h i ne
6.7
Remarks and Comments
378
Measures and t h e Riesz R e p r e s e n t a t i o n Theorem
380
APPENDIX:
A 1
A A A A
(I-
Formul a
37 1
6.6
Algebras
380
2
Measures
383
3
The Riesz Representation Theorem
386
4
The Radon-Nikodym Theorem
393
5
Signed and L a t t i c e - v a l u e d Measures
396
REFERENCES
403
AUTHOR INDEX
423
SUBJECT INDEX
425
C H A P T E R
I
LINEAR FUNCTIONALS SECTION 1.1 THE SANDWICH THEOREM
E x t e n s i o n theorems f o r 1 i n e a r f u n c t i o n a l s b e l o n g t o t h e most fundamental t o o l s f o r l a r g e p a r t s o f modern a n a l y s i s , e s p e c i a l l y o f
functional
a n a l y s i s . The same i s t r u e f o r convex cones, where l i n e a r f u n c t i o n a l s p l a y an i m p o r t a n t r o l e . A s k i l l f u l 1 h a n d l i n g o f l i n e a r f u n c t i o n a l s t u r n s o u t t o be i n d i s p e n s a b l e i n t h i s area o f a n a l y s i s . Consequently we s t a r t t h i s book w i t h a s u i t a b l e v e r s i o n o f t h e Hahn-Banach theorem f o r semigroups and convex cones. We show t h a t between a s u b a d d i t i v e and a s u p e r a d d i t i v e f u n c t i o n a l t h e r e i s always an a d d i t i v e f u n c t i o n a l h a v i n g t h e same m o n o t o n i c i t y p r o p e r t i e s as t h e s u b a d d i t i v e f u n c t i o n a l (Sandwich Theorem). Since no a d d i t i o n a l work i s r e q u i r e d we immediately prove t h i s r e s u l t f o r maps t a k i n g values i n an o r d e r complete v e c t o r l a t t i c e . S i n c e we a r e d e a l i n g w i t h semigroups and cones i n s t e a d o f groups and v e c t o r spaces we have t o a d j o i n a smallest element t o t h e space where t h e f u n c t i o n a l s t a k e t h e i r values. We proceed g i v i n g two i m p o r t a n t consequences o f t h e Sandwich Theorem, namely t h e F i n i t e Decomposition Theorem and t h e Sum Theorem. The power o f these r e s u l t s i s i l l u s t r a t e d
a t a few s i m p l e a p p l i c a t i o n s ( s e c t i o n 1.5)
and l a t e r on we extend these theorems t o a more s o p h i s t i c a t e d s i t u a t i o n , where measures and i n t e g r a l s t a k e t h e p l a c e o f f i n i t e sums. The Sum Theorem leads t o a Strassen-type d i s i n t e g r a t i o n theorem and, f o r measures on compact spaces, t h e F i n i t e Decomposition Theorem
leads t o t h e Riesz-Konig Theorem
( a general and e f f e c t i v e v e r s i o n o f t h e Riesz R e p r e s e n t a t i o n theorem). Since t h e Riesz-Konig Theorem i s g e n e r a l i z e d l a t e r on t o a non-compact s i t u a t i o n we have postponed a p p l i c a t i o n s f o r t h i s r e s u l t u n t i l c h a p t e r 11. A l l t h e examples i n s e c t i o n 1.9 deal w i t h a p p l i c a t i o n s o f t h e D i s i n t e g r a t i o n Theorem t o networks and p r o b l ems i n Mathemati c a l Economi cs
. Finall y , i n
s e c t i o n 1.9 we mention a few f u r t h e r r e s u l t s and some r e l a t e d problems. Furthermore, we g i v e ( f r o m a h i g h l y personal v i e w p o i n t ) a s h o r t survey on 1
2
Linear Functionals
t h e h i s t o r y o f t h e Hahn-Banach theorem.
1.1 SEMIGROUPS
L e t S = ( S , t , < )be a preordered a b e l i a n semigroup always w i t h n e u t r a l element 0. By p r e o r d e r we mean a r e f l e x i v e and t r a n s i t i v e ( n o t n e c e s s a r i S . Furthermore we r e q u i r e c o m p a t i b i l i t y ~ ~ , s ~ ,t2 t E~ S, such t h a t
l y a n t i s y m m e t r i c ) r e l a t i o n < on
w i t h t h e semigroup s t r u c t u r e , i . e . whenever s1 < s2 and
t2 t h e n
tl
0 r E R
In the sequel we want t o deal with functions whose values are elements of I t turns o u t t h a t in a l l our definitions and proofs of the assertions in the f i r s t sections we need not change a single word i f we consider a t once the more general situation where a l l functions involved are allowed t o take values in R with order complete vector l a t t i c e R.
R .
So, from now on, l e t R be an order complete vector l a t t i c e . If the reader feels uncomfortable with the notion of general vector l a t t i c e s , he i s invited t o take R = IR
.
1.1.1.
Def i ni t i on :
i)
p : S fi i s subadditive, i f p ( 0 ) = 0 and p ( s + t ) I p(s) t p ( t ) for a l l s , t E S
ii)
q : S
iii)
p
: S
+
6 i s additive, i f
iv)
f :S
+
R i s monotone i f f ( s ) s f ( t ) whenever s
-.
+ fi i s superadditive i f q ( s + t ) 2 q ( s ) + q ( t ) for a l l
p
q(0) = 0 and s,t E S i s sub- and superadditive < t, s , t E S
Note, that the function which i s -- on S can neither be sub- nor superadditive since we required p(0) = 0 and q(0) = 0 in the preceding definition. On the other hand, define q : S + R by q ( s ) = if s E S 'I01 and q(0) = 0. Then q i s superadditive and even additive there i s no t E S with s + t = 0. Usually, we i f for every s E S'COI denote subadditive maps by p and superadditive maps by q . I t i s convenient t o use the symbol s for the pointwise order relation between functions on S , i . e . f 5 g i f f f ( s ) s g ( s ) for a l l s E S , where f,g: s + R .
-
Linear Functionak
4
1.1.2.
Sandwich Theorem:
Let p : -
S
be monotone and s u b a d d i t i v e -and l e t q : S + 6 -be superq s p. ---Then t h e r e i s a monotone a d d i t i v e p : S + R with
+
additive with
Remark
. q
, only
s u p e r a d d i t i v e and
q Ip
Note, t h a t we do n o t need monotony f o r
for
p.
Proof: In Q =
{p
pointwise
: S
R
+
I
o r d e r on
p
5
-
Q
S . C e r t a i n l y , q E Q, hence
pl
we c o n s i d e r t h e
* +. By Z o r n ' s
lemma
t h e r e i s a maximal c h a i n Q, i . e . a l i n e a r l y ordered subset o f Q, which .. D e f i n e ~ ( s )= s u p I p ( s ) l p E Q
i s maximal w i t h r e s p e c t t o i n c l u s i o n .
(1)
i s superadditive
p
We show t h a t
p
E Q with
p
p,
5
.
i s a maximal element o f p,
. Hence
would be a c h a i n c o n t a i n i n g
of
tj .
so E S
F i x an a r b i t r a r y a
= inf{
for
We c l a i m (1) - ( 5 ) and postpone t h e p r o o f s t o t h e n e x t page.
a l l s E S.
p 5 p
6
Q. Indeed, o t h e r w i s e t h e r e i s
for all
p
E
6
and thus
Q
U
{pol
properly. This contradicts the maximality
and d e f i n e
1 (p(n s o t t ) - p ( t ) ) l n €IN, t E S with u ( t ) + - - I
and
L(s) = sup{p(t)
t
0
s E S , where No
for all
(2)
u s u s p .
(3)
p
Thus,
= N U
with
t t m so < sl
IOl. We o b t a i n
i s monotone and s u p e r a d d i t i v e . p = p
be monotone.
since
m as ( m E No, t E S
p
,
i n view o f t h e m a x i m a l i t y of
We have
i s superadditive.
p
. I n particular,
p
must
5
The Sandwich Theorem
, we
Furthermore, u s i n g t h e d e f i n i t i o n o f Hence,
. Since
p ( s o ) = as
1
0
so E S
-
~ ( 5 )= as = i n f I T i ( p ( n s + t )
(5)
s
The map
+
conclude
s ;(so).
a
was a r b i t r a r y , we have
I
p(t))
s E S
for all
n E W, t E S l
.
i s subadditive.
as
Hence t h e theorem i s proved.
0
For r e a d e r s who a r e n o t t o o f a m i l i a r w i t h sub- and s u p e r a d d i t i v i t y we f i l l i n t h e d e t a i l s o f (1)
-
(5):
(1):
p ( 0 ) = 0 f o l l o w s from t h e d e f i n i t i o n . L e t
pl,p2
E Q
. Since
Because p
i s a chain, t h e r e i s
Q
s1,s2 E S . Consider with
E Q
p
pi
Ip ,
i = 1,2.
i s s u p e r a d d i t i v e , we have
P1(S1) + PZ(S2) 5 P(S1) + P ( S 2 ) I P(S1+S2) I !J(Sl+S2)
Hence, t a k i n g t h e suprema, we o b t a i n
+
p(sl)
.
p ( s 2 ) 5 p(sl+s2)
T h i s proves (1).
(2):
m = 0
Taking
We have m as 5 p(m so
+ t)
0
p(t)
I supIp(m
Ip ( s )
Take
+ so
m as
p(t)
for all
I m E INo, t E S
0
+
t)
for all
t sup{p(tl+t2)
hS1) +
u
+
I
(ml+m
with
m E INo, t E S w i t h
s E S
The monotony o f s1,s2 E S. Then
because
-
yields
m E INo
.
u 1.
and
t E S
with
. Hence,
P --m
i ( s ) = supIp(t)
(3):
;
i n the definition o f
(Recall, p
t
+
m so
g iff f(x) 2 g(x) for all x E X ) or with the pointwise order on some subset Y c X (i.e. f > g iff f(Y) 2 g(Y) for all y E Y ) then they are preordered cones.
8
Linear Functionals
ii) L e t T be a t o p o l o g i c a l space. Then t h e s e t s o f a l l I i - valued upper semicontinuous as w e l l as t h e l o w e r semicontinuous f u n c t i o n s a r e cones. iii) Vectorspaces iv)
( o v e r IR) w i t h t h e usual o p e r a t i o n s a r e cones.
< ) be an a b e l i a n semigroup and denote by (S, < ) * t h e monotone R- valued a d d i t i v e maps and t h e s u b a d d i t i v e
L e t S = (S,
t,
and (S, < )# maps r e s p e c t i v e l y . Then these s e t s a r e cones w i t h r e s p e c t t o t h e p o i n t w i s e
Rs
o p e r a t i o n s . I n f a c t t h e y a r e subcones o f
F a subset
G
c
F
i s s a i d t o be a subcone
. Here,
if
o f course, f o r a cone i s c l o s e d under a d d i -
G
t i o n and m u l t i p l i c a t i o n w i t h nonnegative r e a l numbers, i . e . t X2g2 E G
Xlgl
set T
x
Let al
t
tl,
g1,g2
S
o f a semigroup
whenever v)
whenever
. (Similarly,
if
tl t t2 E
whenever
al,a2
c
E A
F
x for
and
0 Ix 5 1
.
F. We endow
EatblaEA,bEBI
A = I x a l a E A3
A, B E Conv(F),
Conv(F)
=
T
i s s a i d t o be convex i f
L e t Conv(F) be t h e f a m i l y o f a l l nonempty convex subsets o f t h i s f a m i l y w i t h the f o l l o w i n g operations
A t B
a sub-
0 E T).
F be a cone. Then a subset A (l-x)a2 E A
t 0
i s c a l l e d a subsemigroup
and
t2 E T
x1 , x2
and
E G
x
L 0.
Then ( C l ) t o ( C 5 ) a r e e a s i l y v e r i f i e d , and
i s a cone under these o p e r a t i o n s . The n e u t r a l element o f t h i s
cone i s g i v e n by t h e s p e c i a l convex subset preordered cone i f we d e f i n e
A< B
iff
I01
c
F. Conv(F)
becomes a
A cB
Remark: The c o n v e x i t y o f t h e subsets i s necessary, o t h e r w i s e ( C 2 ) would n o t h o l d . ( v i ) A n i n t e r e s t i n g subcone o f Conv(F)
S(F) = I G G F 1 G
F I . For t h i s cone t h e c a n c e l l a t i o n l a w
i s a subcone o f
G t G 1 = G t G 2
*
i s g i v e n by
G1=G2
i s o b v i o u s l y n o t v a l i d . F o r example t a k e F = IRL and c o n s i d e r t h e f o l l o w i n g elements o f S(F) g i v e n by G1 = C(0,x) x E I R I ,
I
G2 = I(x,O)
Ix
E IRI
and
G = C(x,x)
1
x E I R I . Then
9
The Sandwich Theorem
G + G
3ut
+
G1
-IR 2 = G + G 2
1 -
G2. T h i s shows t h a t t h e r e a r e cones which a r e n o t sub-
cones o f vectorspaces s i n c e e v e r y subcone o f a v e c t o r space c e r t a i n l y satisfies (CL) (vi)
.
A r a t h e r odd example of
a cone i s t h e f o l l o w i n g . Consider
( t h e nonnegative f u n c t i o n s on some nonempty s e t X) multiplication and a d d i t i o n + by:
IR+ X
and d e f i n e s c a l a r -
0
for a l l fitg
A * f
=
fX
fu-g
=
f - g
and
h t 0
I: f,g E R
instead o f
hf
and
f
(Here we have chosen t h e symbols
x
+ g i n order t o prevent confusion w i t h the
usual m u l t i p l i c a t i o n and a d d i t i o n o f f u n c t i o n s ) . Then, i f we d e f i n e R:
f,
o
0' = 1,
becomes a cone under these o p e r a t i o n s . T h i s i s another example f o r a
cone which does n o t s a t i s f y t h e c a n c e l l a t i o n law.
1.2.3. Let
Definition: F
be a cone. A map
p : F
i s called functional.
+
A functional A E IR+, f E F.
p i s s a i d t o be homogeneous i f p ( x f ) = x p ( f ) f o r a l l A f u n c t i o n a l i s c a l l e d s u b l i n e a r ( s u p e r l i n e a r ) i f i t i s homogeneous and s u b a d d i t i v e ( s u p e r a d d i t i v e ) . A f u n c t i o n a l i s l i n e a r i f i t i s sub- and super1 inear.
1.2.4. Let -
Proposition: p be
(*) for a l l --
l i m sup
EJ.0
p ( ~ f )= 0
f E F. -Then e v e r y a d d i t i v e
(Recall t h a t
F and assume ~ that
s u b a d d i t i v e f u n c t i o n a l --on t h e cone
l i m sup
E J . 0
p I
p
i s linear. --
p ( ~ f )= i n f sup I p ( 6 f ) &
>O
I
E
t 61)
-
Linear Functionals
10
Proof: We have
mp
(1 m f)
=
n m
p(
since i s additive. Hence Consider some X 0 , x E IR,
*
rn
2 0
p ( f~)
such t h a t rn --t = p((x
- r n )f
1
f)
=
n p ( f ) for a l l
n,m E
IN
,f E
F
,
r ( f ) for a l l r E 4' r 2 0 . , and take an increasing sequence rn E Q ,
p(rf)
=
x ( r n + x in short). We have for f
) + u(rn f ) 5 P ( ( X - r n ) f
1
+
E F
rn ( f ) .
Taking the lim sup on b o t h sides we obtain by ( * ) t h a t
v ( X f ) s X p(f).
To prove the converse inequality we take a sequence rn E Q , rn
4 A.
Then
Certainly, every homogeneous functional p s a t i s f i e s (*). This enables us t o s t a t e the Sandwich Theorem now in terms of cones and functionals as a direct consequence of 1.1.2. and 1.2.4. This version of the Sandwich Theorem plays a central role in our theory. 1.2.5.
Let
Theorem:
g preordered cone and l e t p : F -, be monotone and sub1 inear, q : F + k superlinear with q s p. Then there i s 5 monotone linear p : F + R with q s p s p . F
Again, we emphasize t h a t the monotony of p does n o t mean a restriction. If no explicit preorder on F i s given, consider "equality". Then p becomes a monotone functional and 1.2.5. i s applicable. The notion of linearity we have adopted here indeed generalizes the usual definition of a linear map p : F + R , where F i s a vector space. To see this, l e t F be a vector space and take a functional p : F + R which i s linear according t o Definition 1.2.3. Since f E F, we obtain p((-x) f ) t p ( x f ) = ~ ( 0 = ) 0, x E IR, In particular p ( f ) > - m Thus our p ( ( - A ) f ) = - p ( f~ ) = ( - A ) p ( f ) . notion o f linearity coincides with the usual terminology in the vector space case. Let S be a group, i.e. an abelian semigroup with neutral element 0 such
.
11
The Sandwich Theorem
t h a t for every s E S there i s a n element -s E S vrith s t ( - s ) = 0. If U,V : S + R are additive maps with p I v on S then we can conclude p = v . Indeed, we have ~ ( s I ) ~ ( s ) and p ( - s ) s v ( - s ) , for a l l s E S. As before, we have p ( s ) t p ( - s ) = p ( s t ( - s ) ) = p ( 0 ) = 0, i . e . L I ( S ) = - u ( - s ) , and, similarly, v ( s ) = - v ( - s ) . Hence - ~ ( s )I - V ( S ) for a l l s E S. This yields p ( s ) = v ( s ) for a l l s E S . Thus, in the case of groups and additive maps, "domination means equality". Of course, here, the case of vector spaces and linear maps i s included.
SOME CONSEQUENCES OF THE SANDWICH THEOREM
1.3
Here we present some fundamental applications of the Sandwich Theorem. The theorems in t h i s and the next section are formulated in the context of semigroups. The corresponding assertions f o r cones are presented as corollaries (via Proposition 1 . 2 . 4 . ) However, we emphasize that the proofs for the semigroup case can be adapted immediately t o the case of cones. The reader merely has t o replace the words "sub- and superadditivity" by "sub- and super1 inearity", respectively. 1.3.1 Dominatina Extension Theorem: Let T -
be-a subsemigroup of the preordered abelian semigroup (S, < ) . Consider an additve map p : T + R and - a monotone subadditive p : S +
with l?t)
S
P(t)
for all t --
Then there i s a monotone additive ---L I S P 0" s. (by
'I
I LI
on T
I'
p
: S
+
+
R
with F(
we mean o f course ; ( t ) I
Proof: The map q : S
.
E T
R defined by
q(s) =
-
I LI
for a l l
p(t)
T
and -
t E T).
s E T
i(s)
if
-
otherwise
i s superadditive. The Sandwich Theorem 1.1.2. yields then a monotone additive p on S with q I LI I p , which means ;IU on T . o
12
Linear Functionals
The same argument as i n t h e preceding p r o o f , now w i t h an a p p l i c a t i o n o f Theorem 1.2.5 i n s t e a d o f t h e Sandwich Theorem 1.1.2.,
y i e l d s t h e cone v e r -
s i o n o f t h e Dominating Extension Theorem. On t h e o t h e r hand we can i n f e r t h i s v e r s i o n d i r e c t l y from t h e Dominating Extension Theorem by u s i n g P r o p o s i t i o n 1.2.4. Corollary: Let G -
(F, < ) . Consider -a linear G -and a monotone s u b l i n e a r p on F with
a subcone o f t h e p r e o r d e r e d cone
functional
i 2
Then t h e r e i s a monotone l i n e a r ----
p
p s p 0" F.
on F with -
I
on -
p
G and
The q u e s t i o n seems n a t u r a l whether i t i s always p o s s i b l e t o f i n d a monotone a d d i t i v e that is,
i n t h e p r e c e d i n g theorem which extends
p
G(t) = p ( t )
for all
i s a group. I n f a c t , i f T and hence ;(t) equivalent t o T
5
p(t)
and
;(t) = p ( t )
t E T.
t) I
for a l l
,
properly
T h i s i s c e r t a i n l y t h e case, i f
i s a group, t h e n
;(-
;
-t E T
for a l l
p(-t)
t E T.
f o r every t E T
t E T
which i s
I n p a r t i c u l a r , i n case t h a t
i s a vectorspace, t h e Dominating E x t e n s i o n Theorem i n c l u d e s t h e usual
v e r s i o n s o f t h e Hahn-Banach e x t e n s i o n theorem. The n e x t theorem p r o v i d e s us w i t h a necessary and s u f f i c i e n t c o n d i t i o n for
i
1.3.2.
t o admit a proper e x t e n s i o n . E x t e n s i o n Theorem:
canbe extended t o-a Then T, S, G , p be g i v e n as i n 1.3.1. on S ( i . e . 11 = u on T) with p 5 p on monotone a d d i t i v e p -
Let -
i f and o n l y i f ----
whenever
tl,t2 E T, s E S
and
tl < tp + s
.
S
'r
13
The Sandwich Theorem
Proof: (only i f part). then t ,t
1
;(tl) 2
= u(tl)
E T
and
(if part): Let for a l l
If
r E S.
p
5 p
u = u
i s a d d i t i v e and monotone and
I u ( t 2 + s ) = u ( t 2 ) + U(S) 5 ;(t2) + p ( s )
T
on
s E S,
for
tl< t2 + s .
+
$r) = i n f { i ( t )
p
p(s)
I
t E T, s E S
with
i s monotone, s u b a d d i t i v e and we have
From o u r assumption we o b t a i n
c 5 i;
on
r< t
p .
Hence
and t h e Sandwich Theorem y i e l d s a monotone a d d i t i v e u I U 5 u I = i on T. Now, n e c e s s a r i l y , u = ; on T
s)
i; I p on S .
T. On t h e o t h e r hand,
which f o l l o w s d i r e c t l y f r o m t h e d e f i n i t i o n o f
+ I on
=
i;
;, T
with
Again, these arguments c a r r y o v e r t o t h e case o f cones. A1 t e r n a t i v e l y , we can prove t h e cone v e r s i o n u s i n g P r o p o s i t i o n l . 2 . 4 . Corol 1a r y : Let -
G, F,
G,
Theorem 1.3.1. (i.e.
=
whenever
p
be g i v e n --as i n t h e c o r o l l a r y o f t h e Dominating p -can be extended t o-a monotone l i n e a r p
Then
u 0" G)
g1,g2 E G
with
,f
p
E F
I p ')" F ---i f and o n l y i f
and
_ .
91 < 92
Extension on F -
+ f -
As a consequence o f t h e E x t e n s i o n Theorem we o b t a i n a g e n e r a l i z a t i o n o f t h e w e l l known f a c t , t h a t t h e norm o f an element i s equal t o t h e maximum o f
1.3.3
x
x
o f a normed space
on t h e dual u n i t b a l l .
Norm Theorem:
Let S be-a preordered a b e l i a n semigroup. Consider a monotone s u b a d d i t i v e y p : S + R w i t h p ( n s ) = n p ( s j -f o r a l l s E 5 and n = 0,1,2, ... . Then p i s t h e p o i n t w i s e supremum o f all monotone a d d i t v e p 5 p and i s a t t a i n e d -a t some u , i . e . -for all f-o r e v e r y s E S t h i s supremum s E S -we have p ( s ) = maxIp(s) I p : S + R i s monotone, a d d i t i v e w i t h - p 5 p 0" S ) _ .
14
Linear Functionals
Proof: Let
s E S
and d e f i n e t h e subsemigroup
ii : T
L e t t h e a d d i t v e map n = 0,1,2,...
.
+
R
T = {ns
be d e f i n e d by
I
n= 0,1,2,
...I .
;(n s ) = p ( n s )
for all
Then t h e Dominating E x t e n s i o n Theorem p r o v i d e s us w i t h an
a d d i t i v e , monotone
!.I
:S
-,
R
such t h a t
!.I I p
T. Here t h e l a t t e r f a c t i m p l i e s e q u a l i t y on T
on
S
and
I !.I I
p
on
by d e f i n i t i o n o f
Corol 1a r y : Let -
p Then -p
the
&
p r e o r d e r e d c ~ n e F. monotone sub1 i n e a r f u n c t i o n a l fi i-s the pointwise ~ supremum o f a ll monotone li n e a r !.I I p
for every f E F
_.-
p ( f ) = max{U(f)
this supremum 3 a t t a i n e d
I 11
monotone, l i n e a r and
!.I
a t some
!.I
Ip
FI
,
and
i.e.
.
1 . 4 . SUM THEOREM AND FINITE DECOMPOSITION THEOREM As before, we f o r m u l a t e t h e theorems f o r b o t h cases, a b e l i a n semigroups and cones.
1.4.1.
Sum Theorem:
Let
be an a d d i t i v e 9 -on t h e
_.
!.I
-and l e t
pl,.
. .,p,
: S
+
!.I I p1 t
p r e o r d e r e d a b e l i a n semigroup
whenever
p2 t
*.. t
Pn
*
R with
uk
If, i n addition, -
!.I
...,un : S !.I,,
.
-,
I
that
r, t, sl,...,sk
< )
be monotone and s u b a d d i t i v e -such t h a t -
Then t h e r e a r e monotone, a d d i t i v e ul, --k = l,...,n, and !.I I !.I t !.I~t... t tone and i f we have ------
(S,
E S
such t h a t --
r < t t sk
, k = l , ...,n,
pk
,
i s mono--
The Sandwich Theorem
15
then we can assume
(**I
lJ =
+ u2 +...+ lJn
Proof: We consider
@
= S { l y * * * y n=l { ( s l ,
... , s n ) I
sk E S , k = l ,
..., n }
and denote
the diagonal by A @ ={(S,S,
...,S) I
s E
s}
@ i s a preordered abelian semigroup i f the preorder relation and the operations are defined pointwise, i .e.
+ tl’ ..., S n + t n ) n (s1,...,sn) < ( tl , . . . , t n ) i f f sk < t k for a l l k=l, ...,n ( s y . ., s n ) + (tl’“ . ’ t ) = (sl
.
i s a subsemigroup of 0 . We define an additive map : A @ -, R by (s,s,...,s) = ~ ( s ) , s E S . On @ we consider the monotone subadditive map p given by Obviously,
P(S”’..”Sn)
A @
=
Pl(S1) +
Pz(S2)
+...+
Pn(Sn),
S1’S2’“ ” S n
E
s.
We have I p On A @ in view of our assumptions. Hence we can apply the Dominating Extension Theorem 1.3.1. t o obtain a monotone additive map v : @ -, f? with 5 v on A @ and v 5 p on @ .I f , in addition, (*) holds, then the assumptions of the Extension Theorem 1.3.2. are s a t i s fied and we may even assume v = on A Q . Define p k ( s ) = V ( A s~ ) for a l l s E S , k = 1, ..., n , where s i s the k - t h component of A
i.e. a l l other components of A
A ~ S =(O,...,O,s,O,...,O), ~
Sand
~
S
are zero. The p k are monotone, additive and satisfy the required condi+ A,,S for a l l s E S. 0 tions since (s,s,...,s) = A ~ S ...+ Similarly as in the preceding section the cone case can be proven with the analogous arguments. On the other hand, using 1.4.1 and Proposition 1.2.4., we can derive the following corollary directly from the Sum Theorem.
16
Linear Functionals
Corol 1a r y : Let -
P
.,p,
ply..
&a
l i n e a r f u n c t i o n a l -on t h e p r e o r d e r e d cone (F, monotone sub1 i n e a r on F -such t h a t
and
P I
p1 t p2 t... t Pn
--Then t h e r e a r e monotone and
k = l,...,n,
linear
P I p
01
u I01 where t h e cone
operations a r e defined by
X
a(x,x) = ( ~ ~ h . 6f o) r
6 > 0,
o+o
*
= 0, O ( X , A) = 0,o
0 = 0
Y
0
+
(X,X)
+0
= (X,X)
=
(X,X).
X B R+ i f x i s i d e n t i f i e d w i t h p : X Q R, -, 6 and a s u p e r l i n e a r
may be considered as a subset o f
( x , l ) . NOW, d e f i n e a s u b l i n e a r q : X8R++6
= A 4(XL
P(X,X)
Then
-f
I
by
4
yields
f u n c t i o n a l i n between
p(0) = 0 and q(x,h)
= h ?(x
Y
q(0) = 0
.
q I p and t h e Sandwich Theorem g ves us a l i n e a r
.
Restriction o f that linear functional
to
X
gives the desired
h
.
o
Corollary: Let f and g be f u n c t i o n s X + 6 -such t h a t f i s concave ~ and g convex. Then t h e r e i s an a f f i n e f u n c t i o n h w i t h f I h I g i f and - -----
if -
only
-
f I g.
Remark :
If t h e f u n c t i o n f i s n o t d e f i n e d on a l l o f X t h e n we extend i t t o X b y p u t t i n g f ( x ) = - w whenever X i s o u t s i d e t h e domain o f f The theorem t h e n goes o v e r unchanged. As a consequence we o b t a i n : Every convex f u n c t i o n i s t h e p o i n t w i s e supremum o f t h e a f f i n e f u n c t i o n s i t dominates
.
O f course, one recovers t h e Sandwich Theorem from t h e preceding i n t e r -
p o s i t i o n theorem by r e q u i r i n g ready a cone. I f X
f ( 0 ) = g(0) = 0
i n case t h a t
X
i s al-
i s i n a d d i t i o n a v e c t o r space then, i n t h i s s i t u a t i o n ,
t h e i n t e r p o s i t i o n theorem i s due t o Mazur-Orlicz Q210 (see a l s o 208 1). Even i n s p e c i a l cases t h i s r e s u l t has proved t o be v e r y useful .
37
The Sandwich Theorem
One m i g h t ask under w h a t k i n d o f c o n d i t i o n a bounded a f f i n e f u n c t i o n between f
and
g
can be found. We l e a v e as an e x e r c i s e t h a t t h i s can be done i f
and o n l y i f t h e r e a r e constants
whenever
n,m E N
and
such t h a t t h e r e a r e
I n t h i s case we have
1.6
THE TIESZ
- K'ONIG
-
m
< C1 5 C2 < +
m
such t h a t
h o y . . . y ~ n y?lo'..., A, 2 0, xlY...,xny
xoy
ko E
X
C1 I h 5 C2
-
X ~ ~ . . . m~ E X
X
with
.
THEOREM
The aim of t h i s c h a p t e r i s t h e g e n e r a l i z a t i o n o f t h e F i n i t e Decomposition Theorem t o a measure t h e o r e t i c s i t u a t i o n . The r e s u l t which we o b t a i n h e r e i s a r a t h e r powerful g e n e r a l i z a t i o n o f t h e c l a s s i c a l Riesz R e p r e s e n t a t i o n Theorem. We would l i k e t o c a l l t h i s g e n e r a l i z a t i o n t h e Riesz-Konig Theorem s i n c e t h e u n d e r l y i n g i d e a appeared i n a paper o f Heinz Konig [139]. For those readers who a r e n o t f a m i l i a r w i t h t h e n o t i o n s and techniques o f elementary measure t h e o r y we have gathered t h e necessary m a t e r i a l i n t h e
Linear Functionals
38
appendix. F i r s t , some n o t a t i o n . L e t
X
be a s e t and l e t
lydenotes t h e c h a r a c t e r i s t i c f u n c t i o n o f As b e f o r e we denote by
supy
Y
Y
, i.e.
be a subset o f X. Then 1 if xEY ly(X) 0 otherwise
={
t h e f u n c t i o n a l which assigns t h e v a l u e
.
I f Y .t: 0 t h e n supy i s a subs u p y ( f ) = sup f ( y ) t o e v e r y f : X -, li YEY l i n e a r f u n c t i o n a l on t h e cone o f upperbounded f u n c t i o n s on X. By USC(n)
we denote t h e cone o f upper semicontinuous f u n c t i o n s on a t o p o l o g i c a l
.
space
n
1.6.1
Riesz-Konig Theorem:
Consider a cone f f
F and and assume t h a t by - a compact H a u s d o r f f space R ----,T a f u n c t i o n F + USC(n) --___i s g i v e n such t h a t , -for all w E R , -, ? ( w ) i s sublinear on F , Moreover l e t LI : F -,6 _--be l i n e a r such t h a t -
Then t h e r e i s a r e g u l a r B o r e l p r o b a b i l i t y measure
u(f) I
Proof:
R
d.r
for a l l f E F --
T
on n with -
.
The p r o o f i s a s u i t a b l e a d a p t a t i o n o f t h e p r o o f f o r t h e F i n i t e
Decomposition Theorem t o t h e p r e s e n t s i t u a t i o n . So, d e f i n e a s u b l i n e a r f u n c t i o n a l
on
USC(n)
for all for all
p
and a s u p e r l i n e a r f u n c t i o n a l
6
as f o l l o w s :
h E USC( a ) . Here w
E R
. Of
-f
Ih
i s meant p o i n t w i s e , i . e .
course, t h e cone
t h i s pointwise order r e l a t i o n
s
.
a l i n e a r monotone ( w i t h r e s p e c t t o 6 I v I p. Then
USC(n)
f ( w ) 5 h(w)
i s an o r d e r e d cone under
From t h e Sandwich Theorem we o b t a i n I) f u n c t i o n a l v on USC(n) such t h a t
t h e Riesz Representation Theorem y i e l d s a r e g u l a r B o r e l
39
The Sandwich Theorem
measure
T
on
R
with
v(h) = v
Since
and
'I
E C(n) c USC(n).
I inf{v(h)
Ih
continuous w i t h
g I hl =
n
g dr
g E USC(n). A p p l i c a t i o n t o t h e c h a r a c t e r i s t i c f u n c t i o n s
-1,
Hence
n
i s monotone we have V(g)
for a l l
j h dT f o r a l l h
1,
yields:
T
(a)
v
= 1. Furthermore t h e monotony o f
i s p o s i t i v e . Therefore
T
obviously implies that
i s a p r o b a b i l i t y measure. The f o l l o w i n g
inequal it y completes o u r p r o o f : u(f) I E ( f ) I
"(7)
I
R
7
dr
,
f E F. 0
I t s h o u l d be observed t h a t t h e assumptions o f t h i s theorem a r e f u l f i l l e d
i n t h e s p e c i a l case when
n i s a s e t o f s u b l i n e a r f u n c t i o n a l s on
F
which i s compact under some t o p o l o g y such t h a t a l l t h e T , f E F , a r e upper semicontinuous. Here, o f course, by ^f we mean t h e e v a l u a t i o n : f(p) = p(f) set
R
for all
o f sub1 i n e a r
p€n
. Specialization
o f this situation t o a f i n i t e
f u n c t i o n a l s (equipped w i t h t h e d i s c r e t e t o p o l o g y )
recovers t h e F i n i t e Decomposition Theorem 1.4.3.: L e t pl, -
...,Pn
Proof:
Take
xi
be s u b l i n e a r -and l e t -
R ={pl,...,pnl
= ~ ( { p ~ } )i=l,...,n. ,
p
be l i n e a r on
and a p p l y 1.6.1. 0
F -such t h a t
Then p u t
Linear Functionals
40
1.6.2
Corollary:
Let -
X be topological Hausdorff space _.and l e t F be a subcone of USC(X). Assume n ___--t o be a compact subset of X -such t h a t :
supn(f)
=
supX(f) f o r a l l
f E F.
Then f o r every l i n e a r u on F with ---p(f)
supX(f) for a l l
I
t h e r e i s a Bore1 p r o b a b i l i t y measure ---p(f) 5
Proof:
Take
=
f
J f n
la
dT
T
f E F ,
on n with -
for all f --
E F.
and apply 1.6.1.
0
Later on, we shall use Corollary 1.6.2 i n connection w i t h boundaries ( t h e n i n t h e c o r o l l a r y will then be c a l l e d a sup-boundary o r even maxboundary s i n c e s u p n ( f ) = max f ( x ) f o r a l l f E USC(n)). XER
A very popular example f o r such a boundary i s given i n t h e following situation: Consider t h e Laplace equation
where f i s a continuous real-valued function on t h e closure G of some open, connected and bounded subset G of I R ~ , such t h a t t h e second p a r t i a l d e r i v a t i v e s o f f e x i s t and a r e continuous i n G. Such functions a r e c a l l e d harmonic. Let a G = E\G denote the topological boundary of G and l e t H ( E ) be the s e t of a l l harmonic functions on E Then
.
H(G) is not only a cone b u t a vector space. An important information about t h e harmonic functions i s contained in t h e following elementary result (of which we include a proof f o r the reason of completeness):
1.6.3
Maximum Princiole:
sup- ( f ) = max f ( t ) G
tEa G
for all
f E H(G).
The Sandwich Theorem
Proof:
If
f
i s c o n s t a n t , t h e n t h e a s s e r t i o n i s t r i v i a l . So, i t remains
t o deal w i t h t h e case when
to = (xl
0
0
, x2,.
.., x l
f
i s n o t c o n s t a n t . Assume t h a t t h e r e i s some
) E G such t h a t
f ( t o )> max f ( t )
t€aG
Then t a k e some (xl-xl)
0 2
I
p
41
> 0
p
.
such t h a t we have
for all
t = (x~,x~,...,x~) E
E .
And choose
E
> 0
so,
that
g(t) 5 f ( t )
we have
for all
t E
E.
But
proves t h a t t h e r e must be some
where
tl E G
g
a t t a i n s i t s maximum.
Elementary d i f f e r e n t i a l c a l c u l u s shows t h a t a necessary c o n d i t i o n f o r a maximum i s t h e f o l l o w i n g :
$2 Since
f
1
t=tl
5
0
for
i s harmonic we have a p p a r e n t l y
contradiction.
i = l,...,n
.
A g = 2~
> 0
NOW, i n t h i s c o n t e x t t h e meaning o f C o r o l l a r y 1.6.2 functions
f €
More p r e c i s e l y :
in
G
, hence
0
H(G)
i s t h a t t h e harmonic
a r e c o m p l e t e l y determined b y t h e i r values on
aG.
a
Linear Functionals
42
1.6.4
Proposition:
For every --
t
there - - i s a probability measure
G
E
f(t)
=
I
aG
f d~~
for all
0” a G -such t h a t
T~
f E H(G)
.
Proof: Use Coro for a l l f E H(G In case t h a t
G i s the open unit disc in
G = { ( r coscp, r sincp) 1 0
the measure
T~
(t d
=
sr
R2
< 1, 0 scp< 2 8 ) (polar coordinates)
( r , o ) ) i s the well-known ~Poisson measure, =
T t (cp)
P(r,o -cp)dm(cp)
where dm i s the Lebesgue measure on the unit c i r c l e and Poisson kernel ~-
P ( r , o ) i s the
n
P(r,o)
1 . 7 A STRASSEN
-
=
1 - rL 1 - 2 r c o s o t rz
TYPE DISINTEGRATION THEOREM
As in the case of the Finite Decomposition Theorem we can generalize the Sum Theorem to a measure theoretic situation. The resulting generalization i s very similar t o the Strassen disintegration theorem[304] which has proved t o be very useful in a couple o f applications. Although we can generalize some of the subsequent arguments t o the case o f 1 inear functionals attaining values in an order-complete vector l a t t i c e we shall r e s t r i c t our considerations t o li- valued 1 inear functionals. We do this in order t o make the results more transparent.
The Sandwich Theorem
We c o n s i d e r a measure space t o be
measurable f u n c t i o n s on
we denote
t h e convex cone o f
t
X
f dm
0 , We can assume A + 0 s i n c e otherwise
n n...n n . For 0 < h we define n 91 t l ( h ) = sup{p(k) 0 < k < h and t h e r e i s x > 0 with k < x c p ) Then v i s l i n e a r on the cone of a l l p o s i t i v e elements of F and can t h u s be extended t o a l i n e a r function on F which will be c a l l e d v again. The proof of t h e l i n e a r i t y of v follows from an elementary computation: Indeed, v i s s u p e r l i n e a r by d e f i n i t i o n . So i t remains t o show t h e suba d d i t i v i t y of u Therefore, l e t h l , h 2 z 0 and assume O < k< h l + h 2 . p E
.
.
Then d e f i n e k l and
k = kl
+
k2
=
inf(k,hl)
.
and
k2 = k
-
kl
. We have
From t h i s observation we derive
v(hl
0 0 and, f o r a l l O < h , 0 Iv(h) Iu(h) by d e f i n i t i o n of v and monotony o f u
From f + < ( l + c ) I , f o r every
E
> 0 , we get
Hence u ( f + ) = ~ ( 1 2) ( S I ( c p ) ) - l v ( w ) > 0 F i n a l l y , we claim t h a t
p =
(u(I))-'u
0
I v(1-f
+)
+
I u(1-f ) = 0
in view of t h e monotony of i s in
n , i = 1,..., n . gi
. .
v
.
96
Linear Functionals
We g e t
p ( f t ) = 1 from
quence o f
u(1) = v(f+)
v ( g 71) = 0, i = 1, ..., n.
and p ( g i )
-
p(g;)
i s a conse-
= 0
The l a t t e r f a c t f o l l o w s f r o m t h e ob-
h = g- , t h e u ( k ) must be equal t o zero f o r those j considered i n t h e d e f i n i t i o n o f v (because k = 0 i f j E A and
servation that, i f k
t
because p(inf{cp,gjl)
= 0
otherwise).
n i s compact. We have K(g+) K ( q ) =
since
~ ( g ' ) = s ~ p ( ~ ( g ) , O ) .T h i s shows t h a t Furthermore, we have
f o r a11
E
n
SEF
R
g
g E F which means
o
. Let
s
c
n be t h e s e t
i s a compact H a u s d o r f f space.
: F -,C(S) by @ ( f ) ( w ) = v ( f ) f o r a l l v E S and a l l f € F . i s c l e a r l y an o r d e r u n i t homomorphism. The p r e c e d i n g argument shows
Define
@
f E F with
that for K(f-) =
o
IS,l(f)
=
or
i s dense i n
IS,l(f)
K ( f - ) = 1 and
11 @ ( f ) I I
for all
u n i t isomorphism from
2.3
K(f-) = S
o
K
i s a l a t t i c e homomorphism.
K
K ( f t ) = 1 and
o f a l l l a t t i c e homomorphisms, i . e .
0
Hence t h e r e i s some
C(S)
= 1 there i s
K(ft) =
f E F.
F onto
o ,
K
i.e.
E S
IK(f)l
T h i s proves t h a t
@(F) i f
with
o
=
K ( f t ) = 1 and
1
. Hence
i s even an o r d e r
ISI/ i s a norm. F i n a l l y ,
o(F)
which i s a consequence o f t h e Stone Weierstrass theorem.
ORDER COMPLETE VECTOR LATTICES WITH ORDER U N I T
Recall t h a t a vector l a t t i c e
(F, G
L where
! means t h a t
i s uniquely determined.
6*
Here we have used the following notation: monotone 1 i near
_____L)
D
l a t t i c e cone homomorphism.
Of course, these diagrams are always assumed t o be commutative.
For convenience we sometimes write in the following f v g instead of sup(f,g), and we recall t h a t a l a t t i c e cone homomorphism 6* : G -+ L s a linear map with 6*(gl v g2) = 6*(g1) v 6 * ( g 2 ) for a l l g1,g2 E G
.
Before going any further into the details we 1 i ke t o adopt the follow notation :
-*
_____+
,
-
w
b
monotone sub1 inear monotone surjective sublinear or 1 i near, resp. i njecti ve l a t t i c e cone subhomomorphism,
124
Linear Functionals
where a l a t t i c e cone subhomomorphism rp between two l a t t i c e cones and ( L , < ) i s a map which s a t i s f i e s : cp(lg)=
m ( d Y (P(s1
f
92) < (P(91)
(P(SUP(91Y92)) = SUP((P(91)YV(92))
We use the symbol
>
a
f
and
(P(@
for a l l
(G$)
9Y91’92 E G ; A E R,
*
9 for a l a t t i c e cone isomorphism.
Before going into the construction of the free l a t t i c e cone, l e t us play the usual game. 2.8.2
Proposition:
( i ) ---Free l a t t i c e cones over (F, < ) are uniquely determined up t o l-a t t i c e cone isomorphisms. ( i i )& j ( G , j ) ( j the embedding F + G ) --__--be a free l a t t i c e cone over (F, < ) . Then a l l elements of G ---are of the form j ( f l ) v v j(fn),
...
with
n
E N,
fl,
...,f n E
F.
Proof: ( i ) Let (G,j) be a free l a t t i c e cone over I : G + G then completes the diagram
(F, < ) . The identity map
Because o f the uniqueness of this completion we conclude that I i s the only l a t t i c e cone homomorphism G + G with j = 1 0 j . Now, consider two free l a t t i c e cones ( G l , j l ) and ( G 2 , j 2 ) over ( F , < ) . Then look a t the diagrams jl j2
125
Order Units and Lattice Cones
and observe that
J1
3, = I
morphi sms
.
I G2
( i i ) Let
=
j,
...v
{j(fl)v
, sup(gl,g,)
E
i).
E Pi,
are l a t t i c e cone
j;
and
jf
j(fn)I n
(i.e.
G
= j;
j, = J,
and
, which implies t h a t
sub-lattice cone of g1,g2 E 7;
0
J,
and
j,
j , = J1
homomorphisms with and
*
= jy
0
j;
j,
.
Hence,
J
- I
1-
I+
are l a t t i c e cone iso-
fl,...,fn
E
FI , then
-
G
is a
G i s a subcone of G and for every Look a t the following diagram
Then according t o Definition 2.8.1 j * and 6* are uniquely determined. Replacing 6* by I makes the same diagram commutative, hence 6* = I . This proves t h a t I o j * = I , or G = G. 0 I
NOW, l e t us construct the free l a t t i c e cone f o r a given cone (F, < ). Recall that a s e t A c F i s called convex i f x f + (1-X)g E A whenever f,g E A and 0 I A I 1. For a given s e t B , the smallest convex s e t containing B i s denoted by < B > and i s called the convex hull of B . For a f i n i t e s e t I f l , ...,f n I c F the convex hull i s given by
--
n {
i =1
hi f i
I x1
2 0
,...,A n
2 0,
I n example 1.2.2(v) we have seen t h a t C o n v ( F ) = I A l A convex,
@
* A c FI
i s a cone i f one defines A+B=Ia+blaEA,bEBI A A = {xal a E A I
n
r xi i =1
=
11
.
126
Linear Functionals
I t i s r e l a t i v e l y t r i v i a l t o see t h a t
Convo(F) = { A i s a subcone o f We w r i t e a
ib.
A < B
IA
Conv(F). NOW, we endow t h i s cone w i t h a p r e o r d e r r e l a t i o n .
B
i
t h e r e i s some
b E B
such t h a t
i s t h e n a preordered cone under t h i s p r e o r d e r r e l a t i o n .
We c o n s i d e r two elements
we g e t
= sup n(A).
= TI
NOW, c o n s i d e r a second l a t t i c e cone subhomomorphism < a
A = E [ A ]
.
x1 ,... , A n
(by d e f i n i t i o n )
Put 6
=
n
then
t hi,
i =1
2 0
such t h a t
fi < hi I , i=l,...,n.
f i < 6 1 , i = I , ...,n . Hence,
( i i ) Again f o r an a r b i t r a r y
[A] E L F
take a f i n i t e s e t a
=
{fl,...,fn)
i n F such t h a t [ < a > ] = [ A ] . Since F i s negatively generated t h e r e a r e g l , . . .,gn < 0 such t h a t gi t f i < 0, i = l , . . . , n .
Put g
n
gi i =1 This shows t h a t = E,
Y
then g LF
t fi < 0
f o r i = l ,...,n.
i s negatively generated.
Hence *g t [ A 1 5
0
0
A t the end of t h i s s e c t i o n l e t us return t o Lemma 2.8.3. I t s content i s t h a t f o r a l a t t i c e cone G t h e r e i s a uniquely determined map *
.
129
Order Units and Lattice Cones
(mapping
n on n*) f r o m t h e cone o f s u b l i n e a r monotone maps
LF
i n t o t h e cone o f l a t t i c e cone subhomomorphisms
n
n*
=
2.8.6
o
+
F
+
G
such t h a t always
G
1 holds.
Proposition:
The map * i s sublinear. -Proof: For an a r b i t r a r y
[A1
that
[A] E L F
= [].
s u b l i n e a r monotone
a = {fl,
pick a f i n i t e set
...,f n l c F ...,n l
Then we have n*[A1 = s u p I n ( f i )
I
n : F
i s homogeneous.
+
G
. This
*
shows t h a t
i=l,
such
f o r any
Moreover t h i s f o r m u l a y i e l d s :
(n, + n2)* [ A 1
= sup{n,(fi)
t n2(fi)
< sup{n1( f i )
I
= n;
[ A ] + n;
I
...,
i=l, n l
..., .
i=l, n}
[A]
I i=l, ...,nl
+ sup{n2(fi)
Rather u s e f u l i s t h e f o l l o w i n g c r i t e r i o n 2.8.7
Theorem:
be monotone -l i n e a r maps F L e t u1 ,u2 value
(iii)
not -i s _
- m
there i s a ---
+
R
,
i.e.
here
a t t a i n e d . Then t h e f o l l o w i n g
A t 0
such t h a t
u1
=
R
= R
and t h e
I_-
equivalent:
x ( u l + p2)
.
Proof: (iii) + (i)
i s a consequence of t h e f a c t t h a t
(1-A) u l = x u z (i)
=+
(ii):R e c a l l , t h a t f o r
p*(*f
*
i s homogeneous, because
i n view o f ( i i i ) .
v *g) = max(u*(*f),
f,g E F we have i n general u*(*g)) = max(u(f),u(g)).
130
Linear Functionals
Application
t o our special case y i e l d s :
max(ul(f),vl(g))
+
max(v2(f), v 2 ( g ) ) = m a x ( ( q
which has ( i i ) as an immediate consequence. ( i i ) * ( i i i ) : P u t v = u1 t p 2 and assume t h a t v ( f ) and v ( g ) a r e both either
I 0
o r 2 0. Then v ( ) v ( f ) l g ) = v ( \ v ( q ) l f ) , and ( i i ) implies
! q ( l v ( f ) l g ) = q ( I v ( 9 ) l f ) . Hence,
whenever v ( f ) and v ( g ) have t h e same s i g n . I f v ( g ) and v ( f ) have opposite s i g n s , then say v ( f ) . And ( * ) applied t o this case gives
v ( f t g ) has the sign o f ,
Using t h e f a c t t h a t v i s a d d i t i v e , we see t h a t ( * ) is valid f o r a l l f , g E F. I f v = 0 then ( i i ) shows y1 = 0 and ( i i i ) i s t r i v i a l . So assume v(g) 0 f o r some g E F . I f v ( g ) < 0 then by ( i i ) we get pl(g) I 0 , and v ( g ) > 0 implies pl(g) 2 0. In any case
q(9) ~
v(9)
-
x
2 0
. Inserting
t h i s i n ( * ) gives the desired
result. 2.9
SIMPLICIAL CONES
Let (F,< ) and ( G , < ) be s t t i c e cones and denote by A ( G ) and t h e cones of 1 a t t i c e cone homomorphi sms , o r subhomo'morphisms S A (F,G) respectively, from F t o G ; i . e . A ( F , G ) = {u
SA(F,G) =
1 u ~p I
: F
p : F-GI
GI
.
Recall t h a t L denoted t h e canonical embedding o f F i n t o L F . Then we have t h e following remarkable property
Order Units and Lattice Cones
2.9.1
131
Theorem:
There i s a l i n e a r map u : S A ( F , G )
---__.-
S A ( L F ,G)
--f
-with the
following
properties: (i)
~ ( p o) L = p
(ii)
u
for all --
A(F,G)
into
p E SA(F,G) A(LF,G)
Proof: Consider
I* g i v e n by L F
-b
LF
( I
and d e f i n e
~ ( p =) p
o
I*. Then u
I i s t h e i d e n t i t y map from F onto F )
has t h e r e q u i r e d p r o p e r t i e s .
n
This property gives r i s e t o the f o l l o w i n g d e f i n i t i o n : 2.9.2
Def in i t i on:
A preordered cone (F,< ) i s c a l l e d s i m p l i c i a l i f t h e r e i s a l i n e a r map z : FT -+ ( L F ) : from t h e p o s i t i v e dual cone o f F i n t o t h e p o s i t i v e dual cone o f L F such t h a t z ( p ) o L = 11 f o r a l l LI E F: . R e c a l l , t h a t t h e p o s i t i v e dual cone monotone l i n e a r maps from
F
into
F: R
of (not
(F, < )
R),
was t h e cone o f a l l
and t h a t we assumed t h i s
cone t o be endowed w i t h i t s n a t u r a l o r d e r . So, a c t u a l l y , t h e n o t i o n " s i m p l i c i a l " depends on what k i n d o f image space Therefore, i f we have s e v e r a l
R
, we
R we have considered.
speak o f R- s i m p l i c i a l i n s t e a d o f
s i m p l i c i a l . F o r s i m p l i c i a l cones t h e map
z
i s u n i q u e l y determined
( t h e r e f o r e we may c a l l i t t h e s i m p l i c i a l map). T h i s f o l l o w s from t h e
f o l l o w i n g more general r e s u l t : 2.9.3 Theorem: L e t L -be t h e danonical embedding o f (F, < ) i n t o L F and l e t IT : FT (LF): z ( ~ o ) L = i.1 f o r a l l be a l i n e a r map w i t h p E F: -----+
.
132
Linear Functionals
-Then f o r a r b i t r a r y , with
v o
1
=
~3
*
u E F,
for all
cp
we have
E LF
t(p)(cp) =
s u p I ~ ( c p ) I v E (LF)*,
.
Proof: Consider an a r b i t r a r y element and l e t
P ( f i ),
i = 1, ...,n . Hence T(*fl v . . . v * f n ) > max(p(fl) ,...,p ( f n ) )
=
p*(*fl v . . . v * f n ) .
This proves the desired i n e q u a l i t y s i n c e a l l elements of k i nd.
L F are of this
0
Characters a r e closely r e l a t e d t o maximal monotone 1 i n e a r f u n c t i o n a l s . Where a monotone l i n e a r functional p : F + fi i s s a i d t o be maximal i f we have v = LI whenever v : F -,fi i s monotone l i n e a r with 1-1 5 v
.
Observe t h a t a general character p i s maximal s i n c e II < p* 2 v* We study the following s i t u a t i o n :
*v
.
implies
Let G be a l a t t i c e cone such t h a t t h e r e i s a s u r j e c t i v e l a t t i c e cone G . Furthermore we assume t h a t f o r every homomorphism y : L F monotone l i n e a r v : L F -, 6 t h e r e i s a monotone l i n e a r G : G + 6 w i t h w I o y. By 6 we denote the map y o 1 : F -,G .
0 we obtain p(f t x go) I p ( f ) t x p ( g o ) 5 0 . Hence f t x go E G and thus G
.
generates F The following lemma i s a variant of the statement of the Uni versa1 Property I . 2.6.4. 1.2.2. Lemma: Assume that --
G
unique linear --
generates F. : F - + R with
5
!.I
G IG =
(Note t h a t we require v ( g ) > -
: G !.I
.
for a l l
OD
+ IR
be linear. ---Then there i s a -~
g E G)
Proof: Let f E F and g E G with f t g E G. Then define ; ( f ) = v(ftg) - u ( g ) . This definition i s independent of the choice of g . Indeed, i f g1 E G with
f t g1 E G
!.I(f+g)
-
v(g) =
then f P(f t
t
g t g1 E G and we have
9 + 91)
-
!.I(g1)
-
v(9) = u ( f
Since ;1 i s certainly linear, we obtain t h a t unique linear extension of v 0
.
G
t
91)
-
"91).
i s the well defined
1.2.3 Lemma: Assume t h a t G generates F . Let v : F + R be linear, -and l e t -TI : F + 6 be sublinear. -Then the following are equivalent: i)
v(g)
5
n(g)
for a l l g E G --
.
183
Countable Decomposition
i i ) There i s a linear functional : F - * R -such t h a t and i ( g ) I 0 fo r a l l g E G . --
p I
+ n
Proof: Only i ) * i i ) requires a proof. Let p = n - LI. Then p i s sublinear and p ( g ) t 0 for a l l g E G. By the Dominating Extension Theorem I 1.3.1 there i s a linear functional v Ip with v ( g ) 2 0 f o r a l l g E G. Let f E F and g E G be such that f + g E G Then v ( f ) 2 - v ( g ) > Hence we may define i = v
-.
.
As before, l e t
F- = {f E F I p n ( f ) I 0
for a l l
- .
n
E NI
.
Partial Decomposition Theorem:
1.2.4
Let u : F Assume that F- generates F . ~1 owing are equi Val ent : Then the fol --
+ R
be F-- monotone -~ and l i n e a r . -
i s partially decomposable
i)
p
ii)
For every sequence ( f k ) in F- with --
sup(inf p n ( z f k ) ) > n m k=l
m
Proof: We only have t o combine Proposition 1.1.3 and Lemma 1.2.3.
0
Note t h a t in 1.2.3 and 1 . 2 . 4 as well as in the following theorem we For a function p : F + 6 we denote r e s t r i c t ourselves t o realvalued p by p i0 the fact t h a t !.I i s n o t dominated by 0 , i . e . there i s some f E F with p ( f ) > 0.
.
184
Representing Measures
1.2.5
DecomDosition Theorem:
Assume t h a t G c F- 2 a subcone g e n e r a t i n g F and t h a t f o r each n E N -such t h a t pn(g) < 0 . -Then t h e f o l l o w i n g are equivalent: t h e r e i s g E G --Every
i)
G-monotone l i n e a r
1.1
with -
1.1 i0
is -
-,R with
1.1 i0
is -
: F -tlR
p a r t i a l l y decomposable
ii)
Every
G- monotone l i n e a r
1.1 :
F
decomposable For e v e r y --
iii)
G- monotone l i n e a r
1.1
: F +IR OD
f o r every --
sequence
1 v(gk) >
gn E G w i t h
LI i0
with -
- k=l
-
and -
we have
Q)
-
Proof:
ii) i)i s , b y d e f i n i t i o n , e v i d e n t . i)3, iii)i s an immediate consequence o f t h e P a r t i a l Decomposition Theorem 1.2.4.
iii)-+ ii)F i x
1.1
as i n iii).P r o p o s i t i o n 1.1.3 g i v e s t h a t
decomposable w i t h r e s p e c t t o t h e
p
.
C a l l a sequence
"IG o f nonnegative r e a l numbers a r e p r e s e n t a t i o n i f
1.1
(tn)
IG
is
,n E
N
,
m
'
n = l tn P n j G
*
Observe t h a t i f
(tl,...yt~,t~+l,...)i s a r e p r e s e n t a t i o n t h e n
(tl,...,t~,O,O,O
e t c . ) i s a g a i n a r e p r e s e n t a t i o n . The reason f o r t h i s i s
that a l l the with
1 tn
Denote by
*0
p
"G
0. Observe f u r t h e r t h a t t h e r e i s a r e p r e s e n t a t i o n
5
since
1.1
IG
i s decomposable.
n t h e s e t o f a l l r e p r e s e n t a t i o n s . Consider i n n t h e p o i n t w i s e
order, i . e . f o r two r e p r e s e n t a t i o n s (t,)
5 (sn)
iff
( t n ) , (sn)
t, 5 s,
we w r i t e
for a l l
m E
N
.
Countable Decomposition
185
We claim: (1) There i s a representation which i s maximal with respect t o the pointwise order of n .
( 2 ) For every (t,) E n there are linear u, : G+R-= I r E R ~ r I O l with vn I pn,. for a l l n E N and with G m
Furthermore the un are maximal, i . e . i s linear with vn I p 5 p n I G
.
whenever
vn = p
p
F i r s t we show how these assertions imply t h a t IJ i s decomposable. Let (t,) E n be maximal and consider the vn as in ( 2 ) . Then the Dominating . Since the vn Extension Theorem gives us linear Gn 5 p, w i t h u I < "IG
-
are maximal, t h i s implies vn = u n l G and the 3, must be the unique extensions of
vn (given by Lemma 1 . 2 . 2 ) .
Furthermore
-
m
monotone functional -and l e t @ c F be i s a l i n e a r monotone p : F + 6 with p inf p ( f ) f€@
then define a s u p e r l i n e a r 6
with x f superlinear since 0 was downwards d i r e c t e d . I f 6(g) = supiA alX 1 0 , t h e r e i s
there i s al-
@
f E
Ig )
.
a
-
=
5
p
by
6 is indeed m
then define
I
p
Countable Decomposition
'(9)
1
=
yields
if
- m
g + O
if g = 0
0
a l i n e a r monotone
.
I n any case, t h e Sandwich Theorem I 1.2.5
,: F - 6
with
a = i n f 6(f) = i n f u(f) = i n f p(f).
f€Q
fE@
I p
6 5
. Then
0
fE@
I n t h e f o l l o w i n g c o r o l l a r y t o t h e Decomposition Theorem we show t h a t i n t h e case o f an o r d e r u n i t cone one can d r o p t h e r e s t r i c t i v e assumption t h a t a l l the
under c o n s i d e r a t i o n have t o a t t a i n values i n R
p
(instead o f
6 ) . 1.2.7 Let -
Corol 1a r v :
( F ,
- -1.
u
=
Clearly, t h i s r e s t r i c t i o n
; ( I ) = 1. D e f i n e a s u i t a b l e subcone f < 01. The cone
G
generates
F
U
G =
IF, of
t o t h e subcone i s r e a l v a l u e d , and F,
by
G t {a I] a E R I
since
-IE G. And, o b v i o u s l y , f o r each pn t h e r e is some g E G w i t h p , ( g ) < 0 , j u s t t a k e g = - I . Observe t h a t t r i v i a l l y e v e r y l i n e a r u I S
I
is
G- monotone. Now, e v e r y l i n e a r
I SI
on
FP can be
188
Representing Measures
extendedto a l i n e a r V(f) =
-w
linear
v
SI on a l l o f F by p u t t i n g
v I
f E F x Fp
for with
. Therefore
i s decomposable. Hence, 1.2.5 i m p l i e s t h a t
v $ 0
i s decomposable. So, c e r t a i n l y
(iii) * ( i) : As b e f o r e f i x holds f o r Then
. Let
F
i)i m p l i e s t h a t e v e r y G- monotone
,p
p I
LI
1 FLI
i s decomposable.
SI and c o n s i d e r G W
1 p(gk) >
g, E G such t h a t
k=l
. Assume
-w
. Put
that =
f,
iii)
m
1 gk.
k=l
i s decreasing and from iii)we o b t a i n
(f,)
m
m
W
Using t h e P a r t i a l Decomposition Theorem 1.2.4 we conclude t h a t
p
is
p a r t i a l l y decomposable. (ii) inf
m
* (iii):L e t fn E
and
= inf S ( f ) I m
p(fm)
F be decreasing. L e t
m
p 5
p :
S (Lemma 1.2.6).
I
F
+6
be l i n e a r w i t h
Then t h e r e a r e
tn 2 0
W
such t h a t
Since
p I
-p(I)
1 t p n = l n n'
= p(-I) I
Note t h a t
p n ( * I ) 5 SI(*I)
t tn p ( - I ) I - t tn n=l n=l
= i 1.
we o b t a i n
I tn I 1. n=l
Hence m
i n f SI(fm) m
= i n f u(fm) I
m
I
T h i s proves iii).
0
1 tn i n f pn(fm) I s u p ( i n f pn(fm)) I
n=l
i n f SI(fm) m
m
n
.
m
189
Countable Decomposition
DIN1 CONES
1.3
We want t o give some applications of the preceding section t o cones F(X) of upperbounded functions f : X -,li? containing a l l constant functions. ( X * 0 i s an arbitrary s e t ) . Hence t h r o u g h o u t t h i s section we assume t h a t F(X) i s an order unit cone whose order unit i s the function l X ,i . e . for all
f E F(X)
there i s
such that
X > 0
f
5
we consider the pointwise order with respect t o
X l X . Of course in
F(X)
x .
Definition:
1.3.1
i s called -Dini cone i f for every pointwise decreasing sequence
F(X) with
f n E F(X) ( ( f , ) t
(f,)
i n short) we have
supX(inf f n ) n a
=
inf supX(fn)
nEN
(By inf f n we mean the pointwise infimum).
n a
We call this condition the Dini
-
condition.
If F(X) = C ( K ) , the vector space of a l l continuous realvalued functions on a compact Hausdorff space K ,then, clearly, the Dini condition i s equivalent t o : Whenever f n E C ( K ) , (f,) t and (f,) converges pointwise t o a continuous function
f
on K then
(f,)
converges uniformly t o
f.
Dini's Lemma: of a l l upper semicontinuous functions on-a compact Hausdorff space i s- a- -Dini cone.
The cone U S C ( K ) --
1.3.2 Remark:
A max-stable order unit cone F(X) ( i , e . max(f,g) E F(X) whenever f , g E F ( X ) ) i s a Dini cone i f and only i f every decreasing sequence in
F(X)
converging pointwise t o zero converges uniformly to zero.
Proof: i ) Assume t h a t F(X)
i s a Dini cone and take f n E F ( X ) , (f,) t , which
Representing Measurss
190
converges p o i n t w i s e t o zero. The D i n i c o n d i t i o n i m p l i e s which c l e a r l y means t h a t
(f,)
i n f supX(fn) = 0 nEN
converges u n i f o r m l y t o zero.
ii) Assume t h a t every decreasing sequence i n F(X)
which converges p o i n t -
wise t o zero a l s o converges u n i f o r m l y t o zero
fn EF(X), (f,) c
Since
supX(inf fn) nEN
t h a t t h e assumption
i n f supX(fn) nEN
I
s u p X ( i n f fn) < i n f supX(fn) nEN nE N r E R
s u p X ( i n f fn) < r < i n f s u p X ( f n ) nEN nE N
- r,O)
for all
n E
i n f (supXfn) n a
F(X),
N , Then
- r
i n f (supxgn) = 0 nEN
= i n f (supX(fn
nEN
-
leads t o a contrasuch t h a t
.
i s max-stable and c o n t a i n s t h e c o n s t a n t s . We have by assumption on
.
i s always t r u e i t s u f f i c e s t o show
d i c t i o n . Therefore assume t h a t t h e r e i s
D e f i n e gn = max(fn
. Let
gn E F(X)
since
F(X)
gn J. 0 and t h e r e f o r e ,
. Since
r ) ) 5 i n f ( s u p X gn) nEN
weobtain t h e c o n t r a d i c t i o n i n f (supx fn) 5 r < i n f (supx f n ) nEN nEN
1.3.3.
Lemma:
a t t a i n s i t s maximum on L e t F(X) ---be a D i n i cone. -Then every f E F(X) -__-Proof: D e f i n e gn =
n(f
-
supX(f))
a t t a i n i t s maximum, t h e n
. Then
i n f gn(x) = nEN
gn E F(X), gn 2
-
0
for all
CJ,~
x E X.
.
If
.
f
X.
does n o t
Countable Decomposition
Hence
s u p X ( i n f gn) = n
-
03
, but
inf n
191
,a
supx(gn) = 0
contradiction. 0
The c o n d i t i o n o f Lemma 1.3.3 does n o t c h a r a c t e r i z e D i n i cones. The r e a d e r
w i l l f i n d a counterexample a f t e r Lemma 1.3.5. Let
VF(X)
be t h e max-stable cone generated by
I
V F ( X ) = {max(f l,...,fn)
fl
,..., fn E
F(X), i . e .
F(X), n E N I
.
Then we have 1.3.4
Lemma:
The r e. e q u i v a l e n t : -f o l l o w i n g a_ (i)
VF(X)
_i s _a -D i-n i cone
is_ a Dini _ --
(ii) F(X)
cone
(iii)For e v e r y sequence
fn I 0 i n F(X)
we _ -have
n m i n f supx ( IT f k ) = supx( I f k ) n k=l k=l
.
Proof: ( i i i ) a r e t r i v i a l . So i t remains t o show: ( i ) =b ( i i ) ( i i i ) + ( i ) : L e t F(X) be a D i n i cone and l e t (gn) t Define
13 = supx ( i n f 9,)
n
Hence i t remains t o prove we may assume
a I 13
supx(gn) Ia +
subsequence). Every
1 fn,...,f:n
and
E F(X)
gn
.
1
if
for all
Take a n u l t r a f i l t e r
decreasing sequence o f non empty s e t s :
n
such t h a t
a > -
i s o f the form
consequence o f t h e m a x i m a l i t y o f
. We
a = i n f supx(gn) n
n
always have
. Without
m
.
VF(X)
in
a 2 8.
loss o f q e n e r a l i t y
( o t h e r w i s e we go o v e r t o a
gn = max(f
o of
X
Xn = I x E X
1 ,,,..., fr;”
) where
containing the following
I gn(x)
o we f i n d a number
z a -
pn I kn
1
. As
f o r every
a
Representing Measures
192
pn 1 Yn = IX E X Ifn (x) 2 a -
E Q
.
T h i s f o l l o w s from t h e f a c t t h a t
and t h a t
h = ym E
m
x
n {Yi
u
xI
fi(x) 2 a -
=
Q
i s an u l t r a f i l t e r . Put
fn
n=l
kn
X,
j=l
. Then
Ii5
ml
cx E
I
1 fn = n ( f i n
-
a
- -)n1
pn 1 fn I gn Ia + n
because
f, 5 0
1
and
. Choose x,EX
(which i s nonempty). Then ( i i i ) p r o v i d e s us w i t h
such t h a t 1 pn (x,) IT Ti (fn n=l
-
OD
a
- -)1 '
-
1
t -ii ( p + n=l
E
-
a) >
--
n
pn i n f fn (x,) n
Combining t h i s i n e q u a l i t y w i t h m
2 i n f sup
r
for all
E
>0
X
u :C
- 1z
n
.
inf(hm(ym)) m
(XI
= tf :
x
+
tx E X R
1f
I f(x)
1
E )
-
1 2
we o b t a i n
n( xo ) I p 0 3 .
Since
1
t -= n=l n
we have
(i) o ( i i ) a f t e r Theorem 1.3.6.
be a t o p o l o g i c a l space. A non n e g a t i v e f u n c t i o n
t o -vanish a t i n f i n i t y i f Put
(h ) m
Ii n f g
We s h a l l g i v e a second p r o o f o f Lemma 1.3.4 Let
X
f : X
+
i s compact f o r a l l
i s said
R E
non negative, upper semicontinuous, vanishing a t i n f i n i t y }
.
> 0
.
Countable Decomposition
1.3.5
193
Lemma:
be a D i n i cone c o n s i s t i n g o f upper semicontinuous f u n c t i o n s on L e t F(X) ---t h e t o p o l o g i c a l space X . Then F(X) + U : C ( X ) __-i s a Dini cone. Proof: We can assume, t h a t hn = fn t cpn
(*)
B
, fn E =
F(X) F
i s max s t a b l e ( 1 . 3 . 4 ) . L e t
, cpn
€ U: C
(X)
Assume t h a t we have found i n t e g e r s functions be m kl
1
gn = max(fk ,a) + n > k,
such t h a t
h
t-
fkm+l
f
kmtl
(y) I a
hn h
kmtl
n
such t h a t t h e 5
m
f o r a l l elements y
kmt 1
I
...< km
are decreasing f o r a l l
considering the r e s t r i c t i o n s o f q u a l i t y together w i t h
,
1 = k 1 < k2
0
there i s a
g = f t b*
strongly
200
Representing Measures
"Strongly exposesl'means t h a t there i s some xo E X such t h a t g attains i t s supremum on X a t xo , and i f { x n ) i s any sequence in X with g(xn)
-,
g(xo)
then
II x,
-
xoll
+
0.
SECTION 11.2 REPRESENTING MEASURES
I n t h i s s e c t i o n we a p p l y t h e decomposition techniques, we have developed so f a r ,to t h e problem o f i n t e g r a l r e p r e s e n t a t i o n o f l i n e a r f u n c t i o n a l s . I n t h e f i r s t paragraph we show t h a t a s t a t e
on a v e c t o r l a t t i c e
LI
an i n t e g r a l r e p r e s e n t a t i o n i f and o n l y i f
LI
E(X)
admits
has t h e decomposition p r o -
p e r t y . T h i s f a c t immediately y i e l d s a proof of t h e D a n i e l l - S t o n e Theorem by means o f t h e Riesz R e p r e s e n t a t i o n Theorem. I n t h e second paragraph we p r e s e n t t h e c r u c i a l r e s u l t o f t h i s s e c t i o n ( R e p r e s e n t a t i o n Theorem 2.2.1). We show t h a t e v e r y s t a t e o f F(X) has an i n t e g r a l r e p r e s e n t a t i o n i f and o n l y i f F(X) i s a D i n i cone. L a t e r on we see t h a t t h i s r e s u l t g e n e r a l i z e s t h e c e l e b r a t e d Choquet Theorem. I n paragraph 2.3 we a p p l y t h e R e p r e s e n t a t i o n Theorem t o r e p r e s e n t a r b i t r a r y l i n e a r f u n c t i o n a l s b y s i g n e d measures. And i n 2.4 we s t u d y i n t e g r a l r e p r e s e n t a t i o n s o f l i n e a r f u n c t i o n a l s by means o f w e i g h t f u n c t i o n s . F i n a l l y , i n 2 . 5 we p r e s e n t a l o c a l v e r s i o n o f t h e R e p r e s e n t a t i o n Theorem. I n t h e a p p l i c a t i o n s s e c t i o n 2.6 we s t a r t b y g i v i n g a s l i g h t g e n e r a l i z a t i o n o f t h e Riesz-Konig Theorem. Then, b y means o f w e i g h t f u n c t i o n s , we e x t e n d some r e s u l t s t o t h e case o f r e p r e s e n t a t i o n b y unbounded measures. We e x t e n d t h e n o t i o n o f adaptedness i n such a way t h a t H e w i t t ' s Theorem about i n t e g r a l r e p r e s e n t a t i o n o f l i n e a r f u n c t i o n a l s on
C(X)
,
X
t o p o l o g i c a l space, i s
i n c l u d e d . And we conclude t h e paragraph by an i n v e s t i g a t i o n o f r e p r e s e n t a t i o n o f l i n e a r f u n c t i o n a l s (which a r e p o s i t i v e on t h e monotone f u n c t i o n s ) on c o n t i n u o u s l y o r d e r e d compact spaces. 201
202
Representing Measures
2.1
DECOMPOSITION PROPERTIES AND MEASURE THEORY
L e t us r e c a l l some elementary measure t h e o r e t i c a l f a c t s .
For a more
d e t a i l e d account o f measure t h e o r y t h e reader s h o u l d c o n s u l t t h e Appendix Y
cp :
-,Z
. Consider
Y be a s e t and 1 (pl(.X) = Icp- ( S ) 1 S E 11 i s t h e s m a l l e s t
a measure space
be a map. C l e a r l y
u- a l g e b r a i n
Y
such t h a t
This
T
on
cp
J
Z
by
f d-r =
Y
f
o
cpdr
cp-'(SZ).
=
n cp(Y) =
= 0 t h e n we may T(S) f o r a l l S E z
r(S1)
=
r ( S 2 ) whenever
S
f :
z +iT
.
Assume i n a d d i t i o n t h a t
Z
i s a compact H a u s d o r f f space, t h a t
o f t h e B a i r e subsets o f
Z
and t h a t
Then i n view o f t h e r e g u l a r i t y o f
2.1.1
.
Moreover we t h e n have
f o r a l l measurable
cp
i s z- mea-
~,(cp-l(S))
S E 1 and
cp-'(Z)
d e f i n i t i o n makes sense s i n c e we have
S 1 ,S 2 E z and cp-l(S,)
f
f o cp i s measurable whenever
s u r a b l e . I f T ( S ) = 0 whenever d e f i n e a measure
(Z,zy~), l e t
z consists
i s a p r o b a b i l i t y measure on
T
Z
.
(see A 3.5 Appendix), we have
T
Remark:
The f o l l o w i n g a r e e q u i v a l e n t : i)
T(S) = 0
for all
ii)
T(G) = 0
f o r a l l compact
iii) T(F) = 1 (Note t h a t
f o r a l l open
G c Z
is
G6
iff
cp(Y) n S = 4
with
S E Z
G
G6- s e t s
Fa- s e t s F F
c
cZ
= Z\G
Z with
Fa)
is
cp(Y) r l G = 4
with
F
3
q(Y)
.
.
o f upper bounded valued c o n s i d e r again a cone F(X) c o n s i s t i n g - 6-functions and suppose f u r t h e r m o r e t h a t F(X) c o n t a i n s t h e c o n s t a n t s
-L e_t us
.
L
2.1.2
Definition:
i)
A countable family
IYn
In
E
N)
c a l l e d countable c o v e r i n g o f
X
o f nonempty subsets o f if
X
=
U Yn
nEN
.
X is
203
RepresentingMeasures
Let EF be the smallest u- algebra on
ii)
X
in F ( X ) are measurable (see Appendix). Let linear with p 5 supX '
such t h a t a l l functions p
: F(X)
-+
Ir? be
has the decomposition property i f 1-1 i s decomposable with respect to (supy ) for a l l countable coverings C Y n I n E N1 of n
p
iii)
A positive IF- measure
on X
for u
T
on X
X.
i s called a representing measure
if
p ( f ) 5 J f d.r f o r a l l X
f E F(X)
.
I n particular i f we have equality for all bounded functions then T i s called s t r i c t l y representing.
f EF('X)
Note that for a representing measure T we always have T ( X ) = 1 since lX,-lXE F ( X ) . Indeed, then p ( f l X )= supX(* l X )= i 1 and therefore 1
J l Xd r = r ( X ) .
= p(lX) =
2.1.3
X
Examples:
Under the assumptions of the Riesz-Konig Theorem ( I 1.6.1) we obtain t h a t forsuch a cone F(X) every 1-1 has a representing measure (see also I 1 . 6 . 2 . ) . I n the Riesz-Konig Theorem the compactness of X was essential 1 as one sees from t h e following example: Let X = ,C I n E N 1 and consider the cone F(X) lim
n+
03
consisting of a l l bounded functions f : X
f(i)exists. Define
linear and
p
+
R
such t h a t
lim f(A) for a l l f E F(X) . Then p i s n-+ 03 5 supx b u t there i s no representing measure T tbr p on X . p(f)
=
I n the following section we want t o characterize those cones F(X) on which every linear functional p , with p I supx , has a representing measure. I t turns o u t t h a t these cones are j u s t the Dini cones. A t f i r s t we prove this result under additional restrictions thus obtaining a representation theorem which contains the Daniel1 - Stone Theorem as a special case. All proofs are based on the decomposition theorems of the previous chapter
204
Representing Measures
and the Riesz Representation Theorem (see I . 1.6 and Appendix).
Recall, a s t a t e p : F(X) + k i s a linear functional, monotone with respect t o the pointwise order in F ( X ) , such t h a t p 1. supx ( I 2 . 1 ) .
For the next proposition l e t us consider the following Situation: Let v : F X ) + g be a s t a t e . Consider a countable covering { Y n a)
max f , r )
b)
If
In
E N1
of
.
X
Furthermore assume:
F(X) whenever f E F ( X ) , r E R .
E
tn t 0 , n
E N
OD
, with It1
=
X tn I 1 n=l
and
00
JJ
1 tn supy t (1 - Itl) supx then there are states
I
n
n=l
,n
vn 1. supy
n
u
E N
, and
v
m
=
z tn v n
n=l
t
with (1 - Itl) v
.
Then we can formulate Proposition:
2.1.4
The following
are equivalent:
fi decomposable with
i) u
respect to
(supy ) n
i i ) For every positive decreasing sequence sup (inf sup ( f ) ) = 0 -we have
n
m
'n
.
(f,)
inf p(f,)
with
F(X) =
0
.
Proof: i)
4
i i ) : By assumption there are linear m
v
I
1 tn v n
n= 1
. Since
F(X)
pn
I sup
and
'n contains the constants, all
tn 2 0 such t h a t pn
must be states
205
Represenring Measures m
Now, t a k e
m
z
1 = ~ ( 1 )=
and
tn ~ ~ ( 1= ) z tn n=l n=l
(f,)
as i n i i ) . L e t
m
such t h a t for a l l
x
n=N+l n=l,
tn u,,(fl)
...,N.
E
take
m so l a r g e t h a t supy (f,) n
N
m
T h i s proves
Ie
Hence
0 Ii n f u ( f k ) I z tn un(fm) I t, supy (f,) k n= 1 n= 1 n I2 E
N E N
> 0 be a r b i t r a r y and t a k e
. Then
It
.
(Recall, i n f V(f,) m
N z
fmIfl, = 0
n=1
since
t,, I1)
+
OJ
tn vn(fl)
n=Nt1
I
.
> 0 was a r b i t r a r y .
E
* i):
ii)
Note t h e f o l l o w i n g two f a c t s :
(1) L e t for
p
u,,
and
v
be t h e s t a t e s d e s c r i b e d i n b ) . Assume t h a t ii)h o l d s
and assume t h a t I t l = z tn < 1. Then, s i n c e ii)deals o n l y w i t h
p o s i t i v e f u n c t i o n s , i i ) must a l s o h o l d f o r
(2)
Consider t h e subcone
i n s t e a d of
v
v
.
F-
F- = I f E F(X)
I supy
n
( f ) I0
for all
n 1
which p l a y s an e s s e n t i a l r o l e i n t h e P a r t i a l Decomposition Theorem. Then F- = I f E F(X) and
F-
generates
To prove i)l e t
1f
5 01
F(X).
(t,)
be as i n b ) . Assume t h a t t h e sequence
(t,)
is
maximal w i t h r e s p e c t t o componentwise o r d e r (compare t h e p r o o f o f 1.2.5). Such a maximal sequence i n e a s i l y found by i n d u c t i o n : t
1=
SUPIS
2 0
I
!J
I s supy
tn+l=Sup{s20jLlI
n
1
+
(1 - s)supxl
z t p p
k=l
'k
+ssupy
n
n+l
+ ( l - s - xtk)supXl. k=l
206
Representing Measures
According t o b ) we can f i n d s t a t e s
and
vn
v
m
If I t = 1 then, by d e f i n i t i o n , n n=1 m
I tn < 1 n=l i n f v(f,) m
. Hence,
i s a sequence as i n ii), we have
(f,)
if
LI i s decomposable. Therefore, assume
= 0 (consequence of ( 1 ) ) .
We claim,
v
i s p a r t i a l l y decomposable.
P r o o f o f t h i s c l a i m : We can a p p l y t h e P a r t i a l Decomposition Theorem 1.2.4 m
to Fix
F-
with
supy
n
instead o f
N > 0 a r b i t r a r i l y and p u t
Then
f,
2 0
(fm) +
and
m s u p ( i n f pn( 1 g k ) ) = k=l n m
s u p ( i n f pn(fm)) = 0 n m
m I gk) 5 k=l
-
m i n f v ( 1 gk) = m k=l
-
inf m
pn:
V(
f,
Let
= max(0,
.
-m
Hence t h e r e i s
with
m
1 gk
k=l
+
1 v(gk) >
k=l
and thus
i n f v(f,) m
.
= 0
n=l n*no
.
T h i s means
N . I n o t h e r words we have o b t a i n e d t h e c o n t r a d i c t i o n
since
N was a r b i t r a r y . So we can conclude,with v
the
i s p a r t i a l l y decomposable
(supy ) . n s > 0
m
1
01
,that
f,
and
no E N
such t h a t
v 5 s
supy
+ (l-s)supx
We then have lJ I
-
N ).
implies,by d e f i n i t i o n o f t h e
P a r t i a l Decomposition Theorem 1.2.4,that w i t h respect t o
gk E F-
tn supy
n
+ (s +
tn )supy 0
which c o n t r a d i c t s t h e m a x i m a l i t y o f
(tn
+ (1 -
m
z tn - s)supx
n= 1
0
.
207
Representing Measures
2.1.5
Corollary:
Assume t h a t F(X) i s max-stable --
-has t h e
SIP. Let
covering
i)
of
(Y,)
X
. Then
V F ( X ) = F(X))
(i.e.
be a s t a t e o f
LI
ii) I f (f,)
c
l i m LI (f,) wm
i n F(X) = 0
and t h a t --
F(X)
a countable
the following are equivalent: (supy ) n
decomposable w i t h r e s p e c t t o
LI
and c o n s i d e r
F(X)
and fmt 0
whenever
-f o_r_a l l
m
i n f supy (f,) m n
,then = 0
for all n E --
N
Proof: T h i s i s a d i r e c t consequence o f P r o p o s i t i o n 2.1.4 s i n c e 2.1.4.a) h o l d s and a repeated use o f I . 1.4.1. for
F(X) i n view o f t h e
2.1.6
SIP.
trivially
y i e l d s b ) s i n c e I . 1.4.1 ( * ) i s t r u e 0
Proposition:
be a ve c t o rl a t t i~ c e and c o n s i d e r a s t a t e p : E -* R . Then t h e r e i s a r e p r e s e n t i n g xF- measure T for v 0” X i f and o n l y
L e t E = F(X) i f LI -
has t h e decomposition p r o p e r t y
Remark : Note t h a t
i n t h i s case
T
E
i s s t r i c t l y representing since
i s in
p a r t i c u l a r a v e c t o r space (see t h e remarks a f t e r I. 1.2.5).
Proof: A t f i r s t we prove t h e n e c c e s s i t y :
If all
T
i s such a measure t h e n l e t
m E N
. Assume
(f,)
s u p ( i n f supy (f,)) n m n
+ =
in 0
F(X)
and
f,
for
2 0
f o r some s u i t a b l e
Yn
m
U Y = X . T h i s i m p l i e s f, n n= 1 f, dT = 0 l i m p(fm) = l i m W m wm X
1
+
0
p o i n t w i s e . Hence
b y t h e Monotone Convergence P r o p w t y
with
Representing Measures
208
( s e e Appendix
A 2).
Next, we prove the s u f f i c i e n c y : Let S = { p : F(X) -t R I p l a t t i c e homomorphism3 In I 2 . 2 . 2 we showed t h a t S i s a compact Hausdorff space i f we take the c o a r s e s t topology on S such t h a t the Gelfand transform 7 : S R ( i .e. Lf(p) = p ( f ) f o r a l l f E F(X). p E S ) i s continuous f o r a l l Since = {?If E E = F(X)3 i s dense i n C(S) with respect t o thesupnorm ( I 2.2.3) the Riesz Representation Theorem (Appendix) y i e l d s a unique probability measure ? on S which represents p i . e . p ( f ) = d.r S for all f E F(X).
.
-f
From this we want t o obtain a Let
x
E
cp :
X
xF- measure
T
on X which represents
p
S be the natural i n j e c t i o n , i . e . cp(x)(f) = f ( x ) f o r a l l f E F ( X ) . Nowy consider B o ( S ) , t h e u- algebra of a l l Baire
-t
X and
.
.
s e t s in S W e r e c a l l t h a t t h i s i s t h e u- algebra generated by C(S) Hence B o ( S ) is a l s o generated by E because of t h e density of E in
C(S). So we must have
xF s i n c e
c~’(B,(s)) =
Let Kn
(^focp)(x) =
f(x) for all
x E X, f E E
be a sequence o f compact subsets such t h a t
c S
We claim t h a t
m
;(
U
n=l
Kn)
=
m
U Kn n=l
3
. cp(X)
.
1.
Then we can define, according t o the introductory remarks of t h i s s e c t i o n and Remark 2.1.1, a measure T on XF by T(w-’(A)) = ; ( A ) f o r a l l A E Bo(S).
Proof of the claim: be such a sequence of compact subsets and define (K,) -1 Y n = cp (K,) c X . The decomposition property of p y i e l d s s t a t e s Let
vn
I
IJ
=
supy n
and
m
Z
n=l
A n pn
An L 0
. Another
for all
n E N with
W
1
n=l
An = 1
and
application of the Riesz Representation Theorem
:
209
Representing Measures
yields probability measures for a l l
f E F(X)
f E F(X)
.
This means :( p(f) =
and n
on Kn
T,,
E N ,
with
Hence p ( f ) =
pn(f)
S
^f
Thus , in view of the uniqueness of U
n= 1
-.
J f drn Kn
=
OD
z
d(
n=l
,;
A,
for a l l
)
T,,
OD
=
Z
n=l
Xn
T,,
.
K n ) = 1 and proves the claim. Therefore we have
J f d r for a l l X
f E F(X).
0
Theorem (Daniel1 - Stone)
2.1.7
F(X) a vector l a t t i c e and assume, Then the following equivalent:
Let E -
=
are
1-1
: F(X)
i)
There i s a unique representing IF-measure for
ii)
Whenever f, E F(X) , (f,)
J.
and
_.
A!
+
a state.
R
0” X
.
inf fm = 0 then lim v(f,) m n +a
=
0
.
Proof: 2.1.7 i s a combination of 2.1.5 and 2.1.6.
The uniqueness of the representing IF measure - for u can be seen as fol 1ows : Assume T~ and 72 are such measures. Then consider If E B d
zF(X) I
If X
d.rl =
I f dr2 1 X
. .
Obviously, this s e t i s a U- complete l a t t i c e and contains E Hence i t must be equal to B d zF X ) , since B d zF(X) i s the smallest u- complete l a t t i c e containing E
see Appendix A 1 ).
0
210
Representing Measures
Let
F(X)
66
be a g a i n a general cone o f upper bounded
and assume t h a t
F(X)
i s max-stable. L e t
l i n e a r f u n c t i o n a l . We c a l l
1.1
1.1
: F(X)
valued f u n c t i o n s be a monotone
+
! l i r J continuous i f i t s a t i s f i e s ii)o f
Theorem 2.1. 7, i . e . :
l i m v(fm)
0 whenever
Ill+=
f,
E F(X),
(f,)
4
and
i n f f, m
= 0
.
D I N 1 CONES AND REPRESENTING MEASURES
2.2
X
As b e f o r e we assume t h a t
6- valued 2.2.1
Representation Theorem:
The f o l l o w i n g i )
F(X) i s a cone of upper bounded
i s a s e t and
functions containing the constants.
F(X)
are e q u i v a l e n t :
i s a D i n i cone
ii) -Every s t a t e u : F(X)
. -+
iii)-Every l i n e a r f u n c t i o n a l
fi -has _ a representing 1.1 :
r e p r e s e n t i n g IF measure - on
F(X)
X
+
6
.
with
-
-
IF measure on 1.1 5
sup
X
.
has a X--
Proof:
iii)* ii) i s obvious. i i ) + i):L e t
F( X)
(f,)
J.
in
F(X)
. By
Lemma 1.2.6 t h e r e i s a s t a t e
1.1
of
with
I
i n f supX(fm) = i n f p ( f m ) 5 i n f fm d-c 5 s u p X ( i n f f m ) m m m X m where
T
i s a r e p r e s e n t i n g measure f o r
1.1
on
X ,which e x i s t s according
t o ii).The l a s t i n e q u a l i t y f o l l o w s f r o m t h e Monotone Convergence P r o p e r t y . On t h e o t h e r hand, Hence
s u p X ( i n f fm)5 i n f supX(fm) i s t r i v i a l . m m
s u p X ( i n f fm)= i n f s u p X ( f m ) which means t h a t m m
F(X)
i s a D i n i cone.
21 1
Represen ring Measures
i)
=$
o
Put
u : F(X)
iii): L e t =
I f
I
E VF(X)
with
r? be l i n e a r w i t h
bounded)
f
o
a v e c t o r l a t t i c e and be l i n e a r
+
i s a D i n i cone (Lemma 1.3.4).
ii Isupx
p IGIF(X)and
11
maximal, i . e . i f
v I sup x, v
I
. Assume
.
- o
E = o
and d e f i n e
.
supx
p 5
Then
E
ii is
furthermore, t h a t
l i n e a r , then
.
= u
is
: VF(X) -.fi
Let
G exists
Such
i n view o f Z o r n ' s Lemma t o g e t h e r w i t h t h e Dominating E x t e n s i o n Theorem 1.1.3.1 which a l s o shows t h a t
C must be monotone. v
can be extended u n i q u e l y t o a l i n e a r
@I'
ensures t h a t
>
;(f)
-
f E o
for all
a0
-
m
i
if
a E R
with
*
1 = u ( * lX) c G ( lX) 5 supX(* l X )=
l X )5 G(f).
< a = ;(a
1.3.6).
v
We c l a i m ,
G
*
n
and
then
alX I f
f
.
1 we have
has t h e decomposition p r o p e r t y .
X
. Then
with
hn t 0
there are states
m
m
un 5 supy
f E o
has t h e decomposition p r o p e r t y (Theorem
(Yn) be a c o u n t a b l e c o v e r i n g o f
Indeed, l e t
s i n c e t h e monotony o f
. Indeed,
i s bounded and t h e r e f o r e t h e r e i s a c o n s t a n t Since
E
on
1 A n = 1 and n= 1
I
n = l 'n 'n
*
use t h e Sandwich Theorem). Then t h e Dominating E x t e n s i o n Theorem 1.1.3.1 y i e l d s l i n e a r vn : E + R w i t h (To o b t a i n monotone
pn
'n l o lv n I o
vn
and
5
supy
for all
n
n. m
Then t h e m a x i m a l i t y o f all
n E N
.
=
Z
n= 1
and
An pn
Now t h e uniqueness o f t h e e x t e n s i o n s from
m
v =
implies
1 hn vn n=l
.
I n p a r t i c u l a r t h i s means t h a t
v
o
v
= unl
nl@
to
for
E yields
has t h e decomposition
property. E
i s a v e c t o r l a t t i c e and
applied t o EE- measure
E T
v
i s a s t a t e on
E
. Now,
P r o p o s i t i o n 2.1.6
and
u
shows iii).Indeed, P r o p o s i t i o n 2.1.6
on
X
with
v(f)
=
J f d-r f o r a l l f X
E E
yields a
, in
particular
212
Representing Measurns
f E 0. Note, t h a t , i f
for all
T h i s means
i n f fn = f .
n
u(f)
5
for all
2.2.2
;(f)
fn = max(f,-n)
zF = zD= zE . Therefore,
I i n f v(fn)
= inf
n
f E F(X)
f E F(X), t h e n
, which
1 fndr
n X
=
X
E @
and
we o b t a i n
f d.r
concludes t h e p r o o f .
Corollary:
be a -D i n i cone c o n s i s t i n g o f upper semicontinuous f u n c t i o n s on a t o p o l o g i c a l space X. Then every l i n e a r f u n c t i o n a l v : F(X) .* R w i t h p s supx has a r e p r e s e n t i n g zF-measure w h i c h can be u -on X --Let -
F(X)
extended t o t h e s m a l l e s t
u- a l g e b r a 1
c l o s e d compact subsets o f
Proof: Put
G(X) = F(X) t
U
X
c o n t a i n i n g IF -and a l l
X.
CL ( X ) .
Apply Theorem 2.2.1 t o
on
G(X)
Then (p
G(X)
i s a D i n i cone (Lemma 1.3.5).
can be extended t o a l i n e a r f u n c t i o n a l
on G(X) b y t h e Dominating Extension Theorem I 1.3.1). IIG measures -
T
.
This y i e l d s
xG c o n t a i n s a l l compact subsets o f X.
0
213
Representing Measures
2.3
WEAK D I N 1 CONES AND SIGNED REPRESENTING MEASURES
Throughout t h i s s e c t i o n f : X +R
F(X)
denotes a cone o f bounded f u n c t i o n s
n o t n e c e s s a r i l y c o n t a i n i n g t h e c o n s t a n t s . The aim o f t h i s
c h a p t e r i s t h e i n v e s t i g a t i o n o f s i g n e d measures on p r e s e n t i n g measures f o r
r(?)
t h e cone
and s i g n e d r e -
X
F ( X ) . The main t o o l s f o r t h i s i n v e s t i g a t i o n a r e
i n t r o d u c t e d i n 1.4 and a s i m p l e t r i c k c o n s i s t i n g i n t h e
t r a n s f e r between p o s i t i v e f i n i t e measures on ? = (X x (03) u ( X x { I ) ) and signed f i n i t e measures on X . We r e c a l l (Appendix Theorem A 5.2 o r Jordan Decomposition, s e c t i o n I 1.5.6.)
a s i g n e d measure
on
T
X
can
be w r i t t e n as
w i t h p o s i t i v e measures measures,
T+
variation
(TI
and
and
T-
T+
and
2.3.1
. S i n c e we
deal w i t h r e a l - v a l u e d
can be chosen t o be m u t u a l l y s i n g u l a r . The t o t a l
i s then
i s s a i d t o be f i n i t e i f
T
T-
Isl(X)
Ef = {g E E ( X )
xf
0)
I there
is
x
= u- algebra generated in
xE =
CI-
2 0
Xf
with by
algebra generated in X by
l g l S Af3
f-'Ef E(X)
=
{f-lhlh E E f l
.
.
For short we call a monotone linear functional p : E ( X ) -, R on a vector l a t t i c e E ( X ) Dini continuous i f p ( f n ) -, 0 whenever f n i s a n -,OD sequence in E ( X ) with f n J. 0 ( i . e . f n decreases pointwise t o zero).
Lemma 1: Let p : E ( X ) + R be a Dini continuous monotone linear functional. Then for every 0 5 f E E ( X ) there i s a Ifprobability measure mf such that --- I -
p(h) =
p(f) J
xf
( h f-')
dmf
for a l l
h
E
Ef
233
Representing Measures
Moreover, mf
is
unique i f
p(f)
0
Proof: I f p ( f ) = 0 then the assertion i s t r i v i a l . So assume p ( f ) > 0 and consider the vector l a t t i c e E; of a l l real valued linear functions on Xf given by the restrictions of f - 1 Ef . E; i s a n order unit vector
space (with 1 as order u n i t ) . And on xf
we define a s t a t e
E;
pf
by
where, of course, f a g stands f o r the extension t o X given by zero outside Xf pf i s Dini-continuous, since p was Dini continuous.
.
Hence by the Daniell-Stone Theorem there i s a unique probability measure representing the s t a t e pf NOW, by going back from E; t o Ef one gets
.
the desired result.
0
Let us turn our attention t o the u- completion (denoted by E u ( X ) ) o f E ( X ) . By that we mean the smallest u- complete vector l a t t i c e (with respect t o pointwise operations) which contains E ( X ) We recall t h a t u- complete means t h a t for every countable bounded s e t {fnln E N I the function dx) = sup f n ( x ) i s again an element of the vector l a t t i c e . n A s e t F i s called bounded i f there i s a n element g E E u ( X ) such t h a t If1 5 g for a l l f E F. E u ( X ) i s the intersection of a l l u- complete vector l a t t i c e s 3 E ( X ) I t i s a sublattice of the u- complete vector 1a t t i ce
.
.
E,(X)
where
IRX ‘E( X )
=
If E
X
R 1
i s the
E(X)
U-
I
there i s
g
E
EJX)
which means t h a t for every
such that If1 5 l g l l ,
complete vector l a t t i c e of a l l real
measurable functions. I n general we have E u ( X ) exercise t o show t h a t
E(X)
2 E,(X).
E,(X)
i f and only i f
f E Eu(X)
the function
=
EE(X)-
I t i s an easy Eu(X)
i s truncated,
Representing Measures
234
(f i s again in
A
l ) ( x ) = minCf(x),l)
Of course, i f
E,(X).
E(X)
i s truncated then
Eu(X)
is
truncated. Theorem 2:
Let
l~ :
Then -
p
: Eu(X)
E(X) R be a Dini continuous monotone linear functional. has t o-a- Dini continuous monotone linear - a unique extension -+
-+
IR
.
Proof: We f i r s t remark t h a t whenever u : E ( X ) -+R ( E ( X ) an arbitrary vector l a t t i c e ) i s monotone linear and Dini continuous then i t s Dini continuous extension t o E u ( X ) must be unique. I n f a c t , take two extensions v1 , v2 , then {f E
i,(x) I v l ( f )
hence equal to
=
v 2 ( f ) 1 i s a u- complete vector l a t t i c e
:
(Efl0
i(x),
Eu(X).
Now, we apply Lemma 1. If lJf
3
0 5 f E E ( X ) then we define the functional
+R
Then t h i s functional i s Dini continuous (Monotone Convergence Property) Therefore, i f 0 5 f l , f2 E E ( X ) , then and an extension of p IEf are Dini continuous extensions of p Hence Ufl, lJf2 IErnin(f1 ,f2) and LI coincide on { g E E,(X)I lgl 5 min(fl,f2)l llfl f2 Therefore, i f we take for g E E J X ) a n arbitrary f E E ( X ) with
.
.
.
l g l 5 f , we may define
235
Representing Measures
This functional does not depend on the choice of desired p r o p e r t i e s . 0
f
and i t has a l l t h e
Corollary 3 : Let -
E,(X) the
IJ
: E(X)
truncated. Then t h e r e i s a positive measure m ( w i t h r e s p e c t t o algebra generated by E ( X ) on X ) -such t h a t
is U-
-,R be monotone ~ l i n e a r and Dini continuous and assume ~ that -
v(f)
=
I X
f dm
for all
.
f E E(X)
Proof: of IJ t o E a ( X ) and consider Take the Dini continuous extension 2 = AEZ I t h e r e i s f E E & X ) such t h a t lA 5 f} E(X) Since E a ( X ) i s truncated we have i = t A E Z I1 E E u ( X ) } . E(X) A NOW, define f o r B E II E(X)
.
m(B)
=
sup{i;(lA) 1 A
E
with
Then m has a l l t h e desired p r o p e r t i e s .
A
= B3
.
0
This c o r o l l a r y e a s i l y leads t o t h e well known representation theorems f o r topological s i t u a t i o n s . Let us show this. On a Hausdorff topological space X we study t h e space C o ( X ) c o n s i s t i n g of a l l continuous functions f : X -,IR such t h a t If1 vanishes a t i n f i n i t y , i . e . the s e t s t x E X I I f ( x ) l L & I a r e compact f o r a l l E > 0 . Clearly, C o ( X ) i s a truncated vector l a t t i c e whose p o s i t i v e p a r t i s a subcone o f U C,+(X) .
-
Representing Measures
236
Theorem 4:
For every monotone linear v : C o ( X ) --
+
there i s a measure m (with -
R
respect t o the by C o ( X ) ) ~ - 0- algebra - generated p(f)
J f
=
dm -for a l l f E C o ( X )
-such t h a t
.
Proof: We only need t o prove t h a t suitable weight functions.
p
i s Dini continuous. This i s done by using
Consider a sequence f n in C o ( X )
with
f n +O.
Define W E C o ( X )
cp = f l
and a sequence
t
wn
E
W,(X)
=
1
n min(fl, -3 ) n=l n 1
with
Co(X)
cpn
+0
by:
otherwise Note,
cpn
w,,(xt)
+
Since
p
B u t the
i s indeed continuous since, by definition o f 0 for every net
xt
cp,
converginq t o a point x with
we have cp(x)
=
i s monotone we have
cpn
are uniformly decreasing t o zero since U C L ( X )
Dini cone (Lemma 1.3.5).
0
t R
is a
0
.
237
Representing Measures
Adapted Cones
2.6.4
Mokobozki - Sibony [2321 arld others (see also [721 and [255])introduced the concept of adapted cones. These cones came u p in potential theory. They have the property t h a t positive linear functionals always admit representing measures. Adapted cones are cones of nonnegative continuous functions on a locally compact topological space f u l f i l l i n g a suitable weight condition a t i n f i n i t y . We are going t o generalize the concept in such a way t h a t also nontopological situations and t h e i r representing measures are covered as we1 1 . be a truncated vector l a t t i c e of real-valued functions on some Let E ( X nonempty s e t X . E ( X ) i s said t o be adapted i f for every sequence f n E E ( X with f n J 0 there i s a 0 I Y E E ( X ) such t h a t f o r every there i s a 0 I cp E E ( X ) having the property t h a t cp-le,(x) converges uniformly t o zero, where E
> 0
cn(x)
As usual we p u t
f n ( x ) when f n ( x )
{
=
2
E
~(x)
otherwise and
= 0
0- (+-)
-1 (x) = +-
when ~ ( x =) 0.
cp
Lemma : Every monotone linear u : E ( X ) has - a representing measure.
+ R
on an adapted vector l a t t i c e E ( X ) -----
Proof: Consider arbitrary inf
nEN
E
> 0
u ( f n ) I E r (where r
fn E E(X)
and
with
fn
i s independent of
E R,
.
If we can prove
J
0
E
), then the Daniell-
Stone Theorem (section 2.6.3) yields the existence of a unique representing measure. I n order t o prove t h i s inequality we take the c p , ~ from the definition above, we define T = max(fl,cp,w) and we consider the subspace
i The functions
= {g E
c p , ~
E(X)
I
191 I A T
and a l l the
for some
x
E R,
I
.
f n are in t h i s subspace. Let
u
be the
Representing Measures
238
restriction o f
u
to
,?
and c o n s i d e r t h e s u b l i n e a r f u n c t i o n a l s on
g i v e n by:
Ix Ix
pn(g) = sup{T(x)-'g(x) qn(g) = s u p I T ( x ) - l g ( x )
i; i s monotone we have i;
Since
E X with
fn ( x ) I
E X with
fn ( x )
5 p max(pn,qn),
5 i n t o l i n e a r vn,nn w i t h
functional
I n particular then
6,
+
0.
(x)}
2 E Y (x)}
where
By t h e F i n i t e Decomposition Theorem we can decompose
E Y
p =
;(T)
fi = vn t
qn
the
From p n ( f n
-
E Y ) I
0
and nn I
and q,(f,-dn
(p)
we o b t a i n :
I 0
(adaptedness) we o b t a i n t h e d e s i r e d i n e q u a l i t y
Now, l e t
be a cone o f
F(X)
= p(T).
p 9., -1 vn and nn a r e monotone. P u t 6, = sup{(p (x)E,(x) I x E X I ,
vn I p pn
Since dn J. 0
adapted
.
real
f u n c t i o n s on
X
.
We c a l l
F(X)
i f t h e following c o n d i t i o n s a r e s a t i s f i e d :
i)
f o r every
f E F(X)
t h e r e i s some
ii) t h e t r u n c a t e d v e c t o r l a t t i c e
E
adapted, where generated means vector l a t t i c e o f functions
g E F(X)
generated by E
w i t h If1 I g F(X) i s
i s the smallest truncated,
X which c o n t a i n s F(X).
Theorem [ 1261 : Every monotone l i n e a r r e p r e s e n t i n g measure.
p
: F(X)
-,R ---on an adapted
cone
F(X)
-
has - a stri
RepresentingMeasures
239
Proof:
u can be extended t o a monotone l i n e a r
A l l we have t o show i s t h a t functional
on t h e t r u n c a t e d v e c t o r l a t t i c e
E(X)
generated by
F(X).
B u t t h i s i s easy. Because o f i ) i n t h e d e f i n i t i o n above
Ig
p(g) = i n f I u ( f ) defines a s u b l i n e a r f u n c t i o n a l on can be extended t o a l i n e a r g Io
since
If E F ( X ) I
E(X). Furthermore 2 p
on
E(X)
ii(g) 5 p(g) I~ ( 0 ) = 0
implies
.
. 6
u
p
Hence IF(X)' must be monotone =
0
Example 1 ( H e w i t t [1631 see a l s o [3171): ~
Let X
C(X) be t h e space o f continuous -r e a l valued f u n c t i o n s on X , where i s a normal t o p o l o g i c a l ___ space. _ Then e v e r y monotone l i n e a r functional _-
u :
-,R -can_ be
C(X)
r e p r e s e n t e d by
g
B a i r e measure. -____
Proof:
I t s u f f i c e s t o show t h a t C(X) fn J. 0, fn E C(X)
let
Yn = I x I f n ( X ) I T
n E C(X)
with
m
Z
n=l
every
, Zn
n
-2
x E X
~
E
> 0
and c o n s i d e r t h e d i s j o i n t c l o s e d s e t s
I x Ifn(x) 2
=
0 5 r n I 1 and
.
~~
g =
s
and
i s an adapted v e c t o r l a t t i c e . To show t h a t ,
T
1
E
. Take
Urysohn f u n c t i o n s = 0
nlYn
. The
function
-1
~+ f n( ) 1 i s o b v i o u s l y continuous and nowhere z e r o s i n c e i s i n some
Yn.
Put
cp = g
-1
and
Y
= 1. These f u n c t i o n s
have t h e d e s i r e d p r o p e r t i e s because o f
-1 cp
Example 2 (Mokobozky Let
X
-
E n = E n ' g I
f 5 c q
X
. We
such t h a t f o r e v e r y outside
0
-7
K
.
f ( x ) > 0. We c l a i m t h a t
P(X)
assume t h a t f o r e v e r y E
Furthermore we assume t h a t f o r every with
1
be a l o c a l l y compact H a u s d o r f f space and
q E P(X)
with
1
Sibony [2321 see a l s o [ 7 2 , p. 2831):
nonnegative f u n c t i o n s on a
"
m=n m
P(X)
> 0
a cone o f continuous f E P(X) t h e r e i s
t h e r e i s a compact s e t
x E X
there i s a function
i s an adapted cone i n t h e
K c X fEP(X)
Representing Measures
240
sense defined above. Clearly the property given above f o r P ( X ) a l s o holds f o r the p o s i t i v e cone of the truncated vector l a t t i c e generated by P ( X ) . For the proof of adaptedness we may t h e r e f o r e assume t h a t P ( X ) - P ( X ) = E ( X ) i s a truncated vector l a t t i c e . Now, take a sequence f n E P ( X ) with f n J. 0 . Then there i s some q E P ( X ) with
fl 5
E
q
o u t s i d e a s u i t a b l e compact
set (depending on E ) . P u t cp = f l t q and Y = q , then because of D i n i ' s Lemma on compact s e t s the functions cp and Y have t h e desired properties. Example 3:
Let us discuss a special case of example 2. Again X i s a l o c a l l y compact topological Hausdorff space and we denote by C a o ( X ) and C o ( X ) the complex-Val ued and real -Val ued continuous functions on X which vanish a t i n f i n i t y . We endow t h e s e cones w i t h t h e sup-norm. The cone of p o s i t i v e elements in C o ( X ) is c e r t a i n l y adapted. Hence, monotone f u n c t i o n a l s do have representing measures (Theorem 4 of section 2.6.3): W i t h Theorem 2.3.4 we obtain a s i m i l a r r e s u l t f o r bounded functions i n case t h a t p i s not monotone.
u For every l i n e a r functional -bounded ~ -
: C Co(X)
+
bounded complex-valued t i g h t measure such t h a t ~ - - v-p(f) =
I X
f dv f o r a l l
f E C Ro(X)
a t h-e r-e-i s a unique
.
Proof:
Because C
ao(X)
=
Co(X) + i Co(X)
functional on C eo(X)
from i t s r e s t r i c t i o n t o
r e s t r i c t our arguments t o C o ( X ) . functionals p r , p i by p(f) =
we e a s i l y recover every l i n e a r
pr(f)
+i
Co(X)
. Hence,
we can
Define bounded real-valued l i n e a r
pi(f)
for all
f E Co(X)
.
Since C o ( X ) i s a weak Dini cone these functionals have signed Baire representing measures mr and mi , respectively.
241
Representing Measures
Since, f o r example, r e p l a c i n g
mr
mr(A) = supImr(K)
by t h e measure
I K c A,
compact}
K
does n o t change t h e i n t e g r a l s f o r t h e f u n c t i o n s v a n i s h i n g a t i n f i n i t y we mr
can assume
mi
and
t o be t i g h t . Uniqueness i s easy t o see: E i t h e r
v i a t h e uniqueness a s s e r t i o n i n t h e D a n i e l 1
-
Stone Theorem o r f r o m t h e
f a c t t h a t d i f f e r e n t t i g h t measures l e a d t o d i f f e r e n t i n t e g r a l s .
2.6.5
0
R e p r e s e n t a t i o n o f F u n c t i o n a l s w i t h Zero Mass
We have seen t h a t o u r fundamental R e p r e s e n t a t i o n Theorem y i e l d s r e p r e s e n t i n g measures even i n f o r m a l l y more general s i t u a t i o n s (e.g.
signed
measures, measures on w e i g h t e d cones, unbounded measures). L e t us g i v e a n o t h e r exampl e o f t h i s k i n d .
L e t us s t a r t w i t h a s i m p l e o b s e r v a t i o n . We c o n s i d e r two cones
X
o f upper bounded f u n c t i o n s on cp : F(X)
{cp(f)
--t
If
G(Y).
E F(X)}
Denote by
.
and
cp(F)
F(X), G ( Y ) Y , r e s p e c t i v e l y , a n d a l i n e a r map
t h e subcone o f
G(Y)
given by
Lemma :
The f o l l o w i n g
are equivalent:
F? _ such _ -t h a t p ( f )
i )--For every l i n e a r
p
f E F(X)
p r o b a b i l i t y measure
U-
there i s
_ I -
2
: F(X)
a l g e b r a generated by cp(F)) ,(f)
ii) cp(F)
+
R
I
I Y
-+
T
4 supy(cp(f))
on Y
for all
( w i t h respect t o the
such t h a t
cp(f) d-r
for a l l
f E F(X)
i s a D i n i cone.
Proof:
i )=t. ii): Take an a r b i t r a r y s t a t e
w
of
cp(F)
t h e r e i s a s u i t a b l e r e p r e s e n t i n g measure
T
on
Hence,
v
and
T
i s a r e p r e s e n t i n g measure f o r
+ Y
R
.
Then, a c c o r d i n g t o i),
for
cp(F)
p =
+
R
v
o cp
.
must be a D i n i
242
Representing Measures
cone ( R e p r e s e n t a t i o n Theorem 2.2.1). ii)
3
i ) :L e t
6 : q(F)
+
assumption on state
on
v
q(F)
+
p r e s e n t i n g measure respect t o
6 5 supy
on
E F, cp(f) = g)
.
Because o f t h e
and t h e Sandwich Theorem g i v e s us a
v 2 6. Because o f ii)t h i s s t a t e has a r e -
with
R T
.
1-1
1f
6(g) = sup{p(f)
we have
1-1
be as i n i)and c o n s i d e r t h e s u p e r l i n e a r
F(X) - + R
1-1 :
g i v e n by
R
.
Y
And
T
has t h e r e q u i r e d p r o p e r t i e s w i t h
0
We g i v e two examples f o r s i t u a t i o n s where t h e lemma a p p l i e s .
Example 1: L e t now
F(X)
X which c o n t a i n s t h e
be a cone o f bounded f u n c t i o n s on
X
c o n s t a n t r e a l - v a l u e d f u n c t i o n s on
A linear
~ ( 1 =~ 0.) We r e c a l l t h a t
have mass zero i f -u ( f ) 5 supXlfl
.
for all
1-1
p : F(X) + R
i s said to
i s s a i d t o be normed i f
f E F(X).
Theorem: The f o l 1owing a r e equi Val e n t : i )
p : F(X) + R w_ i t h_mass - - - z e r o has a r e p r e s e n t i n g w i t h t o t a l v a r i a t i o n I T ~ I 1. -__
Every normed l i n e a r measure
on
T
X
For every normed l i n e a r
ii)
p o s i t i v e measure
v
0”
b r a generated by F(X))
p
: F(X)
X
x
X
with
-+
(rn,fn)
decreases -f o r a l l x,y inf nEN
-w_i t_h _mass _ _ _zero __
there i s a u-
(w i t h r e s p e c t -t o- _ t h_ e _p _ roduct such t h a t --
alge-
m( l X x ) 5
( f ( x ) -f(y))dm(x,y)
i i i ) -F o r e v e r y sequence
R
E R
x
for all --
f E F(X).
F(X) -such t h a t r n + f n ( X )
-
fn(Y)
E X we have --
sup (rn +fn(x) -fn(y)) x,y E X
= sup
inf (rn +fn(x) x,yEX nEN
-
fn(y)).
Proof: We c o n s i d e r
on Y
by
Y = X ^f(x,y)
x
X
= f(x)
and we c o n s t r u c t , f o r
-
f(y).
Let
G(Y) =
, a f u n c t i o n ?. F} , t h e n t h e map
f E F(X)
{? 1 f
E
243
RepresentingMeasures
f
-+
^f
i s l i n e a r from
F(X)
G(Y)
to
o n t o t h e same f u n c t i o n on
and two elements o f
F(X)
a r e mapped
i f and o n l y i f t h e y d i f f e r by a c o n s t a n t
Y
f u n c t i on. Hence, every
w i t h mass z e r o a n h i h i l a t e s t h e k e r n e l o f t h e map
IJ
and t h e r e must be a unique l i n e a r map
~ ( f= )u ( f ) c =
for all
(supX(f)
-
f E F(X).
If
G(Y)
v :
.+
such t h a t
R
+ i n f x ( f ) ) we g e t 1
Thus t h e Sandwich Theorem g i v e s us a s t a t e
. And
1
'IG(Y)
the equivalence ( i i )
lemma s i n c e ( i i i ) s t a t e s t h a t
(i)
=$
G(Y)
( i i ) : Take a norrned s t a t e
measure
a c c o r d i n g t o i)
T
and n e g a t i v e p a r t . hence,
-,^f
i s i n a d d i t i o n normed,then f o r
p
~ ( f =) u ( f ) = p ( f - c ) 5 ? I s u p X ( f ) - i n f X ( f ) } =
"'
f
. Then
'+(lX) = r - ( l X ) = A I
1
R
m = - ( ~ +
B
T
/2
G(Y)
+
supy(f) with
R
( i i i ) i s a consequence o f t h e
Q
has t o be a D i n i cone.
w i t h mass z e r o and a r e p r e s e n t i n g
decompose
must have mass z e r o
T
A
p
+
of
p
1
.
T
=
T+
-
T-
i n i t s positive
(F(X) contains t h e constants),
Then
( p r o d u c t measure)
-)
has t h e r e q u i r e d p r o p e r t y . (ii) + (i):
Take t h e marginalmeasures o f
p a r t s o f a measure on T ( A ) = m(A
x
X)
-
m(X
X x
m
as p o s i t i v e and n e g a t i v e
t o o b t a i n t h e r e p r e s e n t i n g measure A). 0
T
,
i.e.
Example 2: Take a t o p o l o g i c a l space on
X
.
We r e c a l l [ 2 4 l ] t h a t an o r d e r r e l a t i o n I
i s s a i d t o be continuous i f , whenever
neighbourhoods for all
X
U ( x ) , U(y)
of
x
and y
x 9y
, then there are
, respectively,such t h a t u 9 v
u E U ( x ) , v E U ( y ) . Such continuous o r d e r r e l a t i o n s a r i s e q u i t e
o f t e n . F o r example t a k e a p o i n t s e p a r a t i n g f a m i l y f u n c t i o n s on
X
and d e f i n e
F(X)
o f continuous
I t o be t h e s m a l l e s t o r d e r r e l a t i o n such
244
Representing Measures
t h a t the elements of f o r a l l f E F(X).
F(X) a r e monotone, i . e . x 5 y
f(x)
0
4
f(y)
In h i s pioneering work 12411 Nachbin has shown t h a t f o r completely regular spaces every continuous order is i n f a c t of t h i s kind.
.
NOW, l e t
X be compact with continuous order 5 Then by t h e fundamental Katetov - Nachbin Theorem ( a Tong - Katetov Theorem f o r normally ordered spaces, s e e [2411 o r [ 661) one can f i n d a monotone continuous function g with cp 2 g L Y whenever cp and Y a r e monotone and real valued w i t h cp lower semicontinuous and Y upper semicontinuous. This f a c t has the following consequence:
2 continuous real valued function 0" X , then t h e r e i s a mono-
Let f -
tone continuous function (*)
g
such t h a t --
1 I f + g I 5 7 supIf(x)
-
f ( y ) I x,y E
Proof: Define a monotone upper semicontinuous function semicontinuous function - ? by :
Put
A =
1
f ( x ) = sup{- f ( y ) I y
I XI
Ix
IZ I
Y(x)
= V
supX(f -
infi- f ( z )
7)
. Then -
V
f
t A L
-
^f - x
-
x
-f
Y
y
and a monotone lower
and by t h e Katetov
-
Nachbin Theorem there i s a monotone continuous function V Because - f 2 - f t -f we g e t f o r ( f t g ) : - x 5 ^ f t g s f t g
IY
IXI.
-
g i n between.
+ g s x .
1 Hence If t g l Ih supCf(x) - f ( y ) I x,y E X,y follows from t h e construction of ^f and y ) .
IX I
(the l a s t equality 0
As a consequence of t h i s i n e q u a l i t y and t h e lemma we obtain t h e following
i n t e r e s t i n g i n t e g r a l representation theorem ( a l s o due t o Nachbin [ 2 4 i l ) .
245
Representing Measures
Theorem : Let -
X
be a compact space w i t h continuous --o r d e r and l e t
l i n e a r functional on C(X)
such t h a t p ( g ) 2 0 --
LI
be-a_bounded __
f o r e v e r y monotone --
continuous f u n c t i o n . -Then t h e r e i s a p o s i t i v e -~ B o r e l measure v on
H = t ( x , y ) I y I XI with v(H) =
T IIp
II
and such t h a t ---
Proof: Observe t h a t
H
i s compact, map C ( X )
From t h e p o s i t i v i t y of
( * ) we o b t a i n
p(f) I
-,
C(H)
by
cp(f)(x,y)
=f(x) -f(y).
u on t h e monotone f u n c t i o n s and from t h e i n e q u a l i t y l l p II supH cp(f) f o r a l l f E C(X). S i n c e C(H)
1 7
i s o b v i o u s l y a D i n i cone (compactness o f
H ) t h e lemma g i v e s us a B a i r e
measure w i t h t h e d e s i r e d p r o p e r t i e s .
T h i s B a i r e measure can b e extended D t o a B o r e l measure by t h e usual procedure.
F o r a g e n e r a l i z a t i o n o f t h e p r e c e d i n g theorem t h e r e a d e r s h o u l d c o n s u l t [ 1661.
2.7
REMARKS AND COMMENTS
S e c t i o n 2.1:
The i n t e r r e l a t i o n between decomposition p r o p e r t i e s and
ineasure theory, d e s c r i b e d i n t h i s s e c t i o n , a r e t a k e n f r o m [1201 (see a l s o [1191,[1231,[1241).
An e s s e n t i a l t o o l i s P r o p o s i t i o n 2.1.4;
we l i k e t o
observe t h a t t h e u n d e r l y i n g s i t u a t i o n o f t e n occurs. F o r example, when F(X)
s a t i s f i e s a m i l d l a t t i c e c o n d i t i o n and
LI
i s maximal ( t h u s y i e l d i n g
b ) as a consequence o f t h e F i n i t e Sum Theorem). The p r o o f o f P r o p o s i t i o n 2.1.6
l o o k s r a t h e r complicated. I n f a c t , one can
g i v e a s i m p l e p r o o f b y use o f t h e D a n i e l l - S t o n e Theorem (as i t was done i n [1203). B u t we wanted t o g i v e a p r o o f o f t h e D a n i e l l - S t o n e Theorem as w e l l , so, we had t o proceed d i f f e r e n t l y . O f course, 2.1.7
i s somewhat weaker
t h a n t h e usual v e r s i o n o f t h e D a n i e l l - S t o n e Theorem s i n c e we assumed t h e functions i n
E(X)
t o be bounded. T h i s r e s t r i c t i o n w i l l be a b o l i s h e d i n
246
Represenring Measures
s e c t i o n 2.6.3 where the r e s u l t i s extended t o the case of unbounded functions. In f a c t , t h e theorem we a r e presenting t h e r e i s s l i g h t l y stronger than the usual Daniell-Stone Theorem, i n s o f a r as we obtain umonotone extensions without S t o n e ' s condition. The Daniell-Stone Theorem, in i t s o r i g i n s , goes back t o D a n i e l l ' s 1917 paper [ 831 ( s e e a l s o [3031).
I t seems t o be an i n t e r e s t i n g question, whether we can base a proof o f t h e Daniell-Stone Theorem e n t i r e l y on decomposition techniques, thus avoiding the use o f t h e Riesz Representation Theorem completely. In f a c t , t h i s i s possible; we sketch the procedure: Let E ( X ) be, with respect t o pointwise operations, a vector l a t t i c e and assume t h a t l X i s an order u n i t . Consider a decomposable s t a t e LI Define f o r A
c
.
X:
Call a sequence A n , n E N , o f pairwise d i s j o i n t s e t s a d i s j o i n t covering of A i f u An = A Define
.
nEN
m(A)
=
inf{
50
iii(An)
E
n=l
Then pick o u t subsets o f =
{A
c
X
I m(A)
Now, one can prove t h a t
1
X
by:
t
m(X\A)
5
is a
(A,)
=
d i s j o i n t covering of A 3
11
.
.
a- algebra
2
Z
E(X)
and t h a t m
is a
a d d i t i v e measure on which represents ii . ( I f t h e reader t r i e s t o prove these a s s e r t i o n s he should make extensive use of t h e FSP f o r (E(X),s)*).
a-
Represenring Measures
247
Section 2.2: Again, the material i s taken from [1201. A different proof of the Representation Theorem 2.2.1 was given by M. Neumann. His essential tools were Choquet's Theorem and Simons' Convergence Lemma. In [1201 one finds a condition for the existence of s t r i c t representing measures. This condition i s f a l s e and does n o t ensure the existence of s t r i c t represent i ng measures. Section 2.3: This material appeared in [1211. I n section 11.4 we are transferring the Representation Theorem t o a vector-Val ued s i tation, and i t seems interesting t o remark t h a t this procedure f a i l s for Theorem 2.3.4 ( a t least in the non-weakly-u-distributive case). Another open problem i s the uniqueness of signed representing measures. Whereas, in the case of positive representing measures, one can easily obtain results 1 i ke the Choquet-Meyer uniqueness theorem, no similar results for the signed case are knwon to the authors. Probably, one has t o work with a decomposition property being similar t o the one f u l f i l l e d by the preduals of abstracts L- spaces ( [2101,[2111). The Weighted Representation Theorem appeared, as a corollary, Section 2.4: in [120]. Frequently weight functions are considered in modern analysis. For example, see the extensive work a b o u t the weighted spaces C V o ( X ) and CV(X), where V i s a so-called Nachbin family ( [ 371,[ 771,[1441 [1961,[2421 [257 1 and 13061). Observe t h a t Lemma 2.5.3 i s a special case o f the abstract Section 2.5 Di sintegrat o n Theorem I 1.7.1. We included a separate and simple proof in order t o save the reader from going t h r o u g h the details of section I 1.7. The material i s taken from [1241. Theorem 2.5.7 can be considered as an integral representation theorem for those s t a t e s behaving locally simplicial. Of course, i n this case the representing measure i s unique. This section was mainly written t o demonstrate the wide Section 2.6: range of applications for the Representation Theorem. For this reason some of the proofs are rather brief. For example, the reader should observe that, although the proof of the uniqueness in Theorem 2 (section 2.6.3) i s correct, one needs some additional work t o verify the crucial assertion t h a t If E c , ( X ) I v l ( f ) = v2(f)l i s a vector l a t t i c e .
248
Representing Measures
The u- completeness i s no problem, b u t t o prove t h a t t h i s space i s a l a t t i c e one certainly needs transfinite induction (which, of course, proceeds in the obvious fashion). The intrinsic reason for that space t o be a l a t t i c e i s t h a t i ( X ) already i s a l a t t i c e . I n Lemma 1 (section 2.6.2) we work with the Stone-Czech compactification although we do n o t assume t h a t the space under consideration i s completely regular. Since some textbooks, unfortunately, define the Stone-Czech compactification only f o r completely regular spaces an additional explanation seems appropriate. Of course, by i3 X we mean the linear functionals on C B ( X ) which are given by the u l t r a f i l t e r s on X . Then everything which i s known for the Stone-Czech compactification goes through, except, t h a t X + B X need n o t be injective. (Another way o f defining B X i s via the structure space of C B ( X ) given by the KakutaniKrein-Stone-Yosida Theorem).
Section 2.6 deals w i t h several versions, and generalizations, of the fundamental Riesz Representation Theorem (which was originally proved by F. Riesz 1262 3 in 1909 f o r the case X = [0,11, see also [2651). This theorem has caught the attention of many outstanding mathematicians; they have genralized the result and reformulated the proof countless times. The case of compact subsets of Rn was settled by J. Radon [ 2591, for general metrizable compact sets the theorem i s due t o S . Saks I2821 and S. Banach (1937). For the general case of structure spaces of M- spaces Kakutani [1811 gave the proof. For non-Hausdorff compact s e t s , see A. Markov [2261 and A.D. Alexandrov t 1 1 (compare our Theorem 2.6.1). The C o ( X ) - case (see our Theorem 4 in section 2.6.3) i s due to several authors, independently. This case i s related t o the integral representation of the dual spaces with respect t o the s t r i c t topology (introduced by R.C. Buck [ 6 0 1, see also, for example, [1651 and t2901). For pseudocompact X the integral representation of monotone 1 inear functionals C ( X ) + R was f i r s t given by I . Glicksberg 11411. Hewitt's result [1631,[3171 ( o u r example 1 in section 2.6.4) i s usually stated in a more general form: Every bounded linear functional : C ( X ) -+ R can be represented by an integral, where bounded means t h a t for every 0 5 f E C ( X ) the s e t { p ( g ) I - f I g I f , g E C ( X ) I has t o be bounded. Of course, monotone functionals are bounded and a bounded functional can
248
Represenring Measures
be decomposed i n t o t h e d i f f e r e n c e o f two monotone ones. I f t h e r e a d e r i s i n t e r e s t e d i n e l e g a n t and s h o r t p r o o f s o f t h e Riesz R e p r e s e n t a t i o n Theorem he s h o u l d c o n s u l t [3161 o r [1381. G e n e r a l i z a t i o n s t o s e t - v a l u e d measures a r e t r e a t e d i n [2781. F o r more i n f o r m a t i o n about t h e s i t u a t i o n cansidered i n Example 2 ( s e c t i o n 2.6.5)
t h e r e a d e r s h o u l d have a l o o k i n t o Nachbin's b e a u t i f u l
book on t o p o l o g y and o r d e r [2411. Choquet's Theorem, which i n a c e r t a i n way i s a l s o a g e n e r a l i z a t i o n o f t h e Riesz R e p r e s e n t a t i o n Theorem,will
be t r e a t e d e x t e n s i v e l y i n t h e f o l l o w i n g
sections. There a r e many analoga o f t h e Riesz R e p r e s e n t a t i o n Theorem f o r t h e v e c t o r valued case. A p a r t f r o m t h e r e s u l t we g i v e i n s e c t i o n 11.4 t h i s a s p e c t
w i l l n o t be t r e a t e d i n t h i s book. For t h e s e m a t t e r s t h e i n t e r e s t e d r e a d e r s h o u l d c o n s u l t t h e e x c e l l e n t monograph o f D i e s t e l and Uhl [ 861.
Other R e p r e s e n t a t i o n Theorems:
L e t us b r i e f l y mention some o t h e r i n t e -
g r a l r e p r e s e n t a t i o n theorems f o r non-compact s e t s . A most i m p o r t a n t development i n t h i s f i e l d was i n i t i a t e d b y E d g a r ' s theorem. I n [ 9 3 1 he proved t h a t e v e r y p o i n t x i n a s e p a r a b l e RNP- s e t X ( s e e d e f i n i t i o n on page 198) i s t h e b a r y c e n t e r o f a p r o b a b i l i t y measure mx o n t h e u n i v e r s a l l y measurable s e t s i n X such t h a t t h e extreme p o i n t s o f X have
mx- measure 1. The p r o o f . amazingly s i m p l y , works v i a C h a t t e r j i ' s m a r t i n g a l e convergence theorem and von Neumann's s e l e c t i o n theorem. The use o f von Neumann's s e l e c t i o n theorem i s t h e reason t h a t t h e p r o o f cannot be t r a n s f e r r e d t o non-separable RNP- s e t s (where t h e theorem f a i l s ) . A f o r e r u n n e r o f t h i s theorem was g i v e n by Choquet [ 7 0 1. A non-separable analogue i s a l s o g i v e n b y Edgar . I n [ 9 4 ] he shows t h a t e v e r y element x
i n an
RNP- s e t
X
can be r e p r e s e n t e d by a t i g h t p r o b a b i l i t y measure
mx, which i s maximal w i t h r e s p e c t t o t h e d i l a t i o n o r d e r ( e q u i v a l e n t [951
t o b e i n g maximal, among t h e t i g h t measures, w i t h r e s p e c t t o t h e Choquet o r d e r ) . I n t h e s e p a r a b l e case these maximal t i g h t measures a r e supported be t h e extreme p o i n t s - which does n o t h o l d i n t h e non-separable case
-
i n f a c t t h e n t h e y o n l y vanish on t h e s e t s which a r e movable b y d i l a t i o n s .
250
Representing Measures
An equivalent non-separable version of Edgar's theorem i s g i v e n by P. Mankiewicz [225]. The c h a r a c t e r i z a t i o n of RNP- s e t s which a r e simplexes by means o f uniqueness of maximal measures is the work of R.D. Bourgin and G.A. Edgar [571. These r e s u l t s have been extendedfurther by the impressive work of E . G . F . Thomas ([307]to [311] on conuclear cones. Let us mention a special case of h i s r e s u l t s : Let S be a complete bounded closed convex Souslin subset of a l o c a l l y convex Hausdorff space. (Recall t h a t a s e t i s Souslin i f i t i s t h e continuous image of a separable completely metrizable space). Then, i f S has the Radon-Nikodym property, every point in S i s t h e barycenter of a Radon probability measure concetrated on t h e extreme points ( i . e . S has t h e i n t e g r a l representation property). This author shows furthermore t h a t the i n t e g r a l representation property characterizes t h e Radon-Nikodymproperty i n the following way: S has t h e RNP i f and only i f a l l closed convex subsets of SN do have the integral representation property. Actually a1 1 these r e s u l t s carry o v e r [ g i i l t o the case when t h e cone generated by S i s replaced by a socalled conuclear cone r with RNP which has t h e property t h a t closed convex h u l l s of compact subsets of r a r e again compact: (Another, b u t very r e l a t e d , 1 ine of non-compact Choquet theorems f o r bounded, weakely-closed, convex subsets of t i g h t measures was given by von Weizsacker[3251, [326] and von Weizsacker-Wink1 e r [327] ( s e e a l s o [331] ) . They show t h a t , roughly speaking, these s e t s do have t h e i n t e g r a l repres e n t a t i o n property.
SECTION 11.3 BOUNDARIES
This section i s dedicated t o the investigation of the so-called geometric situation. We show t h a t , in case of a compact convex s e t , the restrictions o f the affine continuous functions, as well as the restrictions of the upper semicontinuous convex functions, t o the extreme points are constituting a Dini cone. (Actually, we present this fact slightly more general since we replace the compact convex s e t by an arbitrary s t a t e space of an order unit cone, b u t t h i s generalization i s not essential). For this case the Representation Theorem then yields Choquet's celebrated theorem in the generalization of Bishop - de Leeuw. We s t a r t by an investigation of boundaries. We show t h a t there i s a minimal f i x p o i n t boundary for the so-called exposed maps. This immediately leads t o Bauer's Maximum Principle (and, of course, t o the Krein-Milman Theorem). I n section 3.2 we characterize the minimal fixpoint boundary in terms of separation properties; i t turns o u t t h a t t h i s boundary can actually be smaller than the Choquet boundary. This leads us t o the definition of five different boundary notions, and we investigate the relations between these boundaries. Since we did n o t need the notion of maximal measure for the proof of Choquet's Theorem (section 3.2) we append some information a b o u t the Choquet order in section 3.4. I n section 3.5 we t r e a t the Choquet-Meyer uniqueness theorem as well as the characterization of Bauer simplices ( t h i s i s done as an application of the material gathered in chapter 1 . 2 ) . Finally, in 3.6 we investigate in what kind of situation a Maximum Principle forces a cone t o be a Dini cone. Relevant results are obtained via Simons' Convergence Lemma. As a byproduct we get information a b o u t situations when the Choquet boundary i s the minimal max-boundary. This aspect of minimal boundaries will be further treated in chapter 11.5. 251
252
3.1
RepresentingMeasures
FIXPOINT BOUNDARIES, BAUER'S MAXIMUM PRINCIPLE AND THE KREINMILMAN THEOREM
I n t h e f o l l o w i n g we c o n s i d e r a f a m i l y o f upperbounded f u n c t i o n s on some nonempty s e t
X
and we s t u d y subsets
of
Y
X
i n which a l l f u n c t i o n s
X- suprema. I n p a r t i c u l a r we a r e
under c o n s i d e r a t i o n a l r e a d y a t t a i n t h e i r
i n t e r e s t e d i n " s m a l l " Y. The reason f o r t h i s i s , t h a t i n most cases t h e r e 1evant p r o p e r t i e s o f t h e f u n c t i o n s under c o n s i d e r a t i o n a r e c o m p l e t e l y
determined by t h e i r r e s t r i c t i o n s on Y , hence t h e r e i s some economical advantage i n making Y as small as p o s s i b l e . T h i s i d e a i s used w i t h g r e a t success i n d e a l i n g w i t h d i f f e r e n t i a l equations a d m i t t i n g a maximum p r i n c i p l e (as we have seen i n c h a p t e r 1.1.6). We s t a r t w i t h t h e d e f i n i t i o n o f t h e c r u c i a l n o t i o n Def in i t i o n :
3.1.1 Let
X
be a nonempty s e t and l e t F(X)
functions
f :X
f o r every
f E F
I n general even
+
6 .
A subset
, there X
Y
c
X
be a f a m i l y o f upperbounded i s c a l l e d a boundary f o r
i s a corresponding
i s n o t a boundary f o r
f e a t u r e s o f Dini cones i s t h a t i n t h i s case As we have seen i n 1.1.6.3 u n i t c i r c l e i n R2
with
y E Y
F(X) if,
f(y) = supX(f)
.
F(X). One o f t h e i m p o r t a n t happens t o be a boundary.
X
the, l e t us say, t o p o l o g i c a l boundary o f t h e
i s a boundary f o r t h e c o r r e s p o n d i n g harmonic f u n c t i o n s .
Another example f o r a boundary i s
K
i n case o f
C(K)
.
K ,
f o r compact
a r e s u l t which a l s o h o l d s t r u e f o r pseudocompact K T h i s l i s t c o u l d be c o n t i n u e d a l m o s t i n d e f i n i t e l y . We s h a l l deal w i t h some o t h e r i m p o r t a n t examples i n t h e n e x t s e c t i o n s . I n t h i s s e c t i o n we a r e g o i n g t o i n v e s t i a a t e a p a r t i c u l a r boundary b y means o f exposed selfmaps
3.1.2
i) L e t called
y : X
-t
X
.
Definition:
r r-
be a f a m i l y o f f u n c t i o n s i n v a r i a n t i f y(Y) c Y
y : X
for all
+
X
.A
Y c X.
i s a s i n g l e t o n we speak o f y- i n v a r i a n t i n s t e a d o f
subset
Y c X
I n case t h a t
is
r = Iyl
{yl- i n v a r i a n t .
253
Boundaries
ii) L e t
f : X
nuous f u n c t i o n s y : X -, X
= Ix
-,
r? .Then we c a l l a f a m i l y
F(X)- exposed
v a r i a n t subset
af K
F ( X ) c o n s i s t o f upper s e m i c o n t i -
be a compact space and l e t
X
E K
K c X
I f(x)
t h e s e t where
f
r
o f functions
i f f o r e v e r y nonempty c l o s e d compact
and e v e r y
= supK(f)3
f E F(X)
,is
r-
again
r-in-
the set i n v a r i a n t . Note t h a t
af K ,
K- supremum, i s nonempty s i n c e e v e r y upper
attains i t s
semicontinuous f u n c t i o n a t t a i n s i t s maximum on a compact s e t . L e t us g i v e an i m p o r t a n t example f o r t h i s s i t u a t i o n . 3.1.3
Example (Geometric S i t u a t i o n ) :
K o f a l o c a l l y convex Haus-
Consider a nonempty compact convex s u b s e t d o r f f v e c t o r space
x x+
(1-x)y E K
-,6
E
.
R e c a l l t h a t ,K
whenever
i s s a i d t o be convex i f
x
x,y E K, 0 5
5 1
,
and t h a t a f u n c t i o n
f ( x x + ( 1 - x ) y ) 5 x f ( x ) + (1-x)f(y) Conv(K) we denote t h e cone o f upper semicontinuous convex f u n c t i o n s f : K -+ r? . NOW, f o r z E E , d e f i n e f : K
whenever
i s c a l l e d convex i f
x,y E K, 0 5
5 1
We c l a i m t h a t
r
x+z, x - z E K
x+z
if
x
otherwise
= IyzIz
E E3
r- i n v a r i a n t compact k
We have t o show y z ( x ) E yz(x) = x
. By
by
yz : K j K
a
x
is
c K
afk
Conv(K)
-
and a r b i t r a r y
. If
x +z
or
exposed. To see t h i s we t a k e f E Conv(K), z E E, x E afK. x-z
and t h i s i s by assumption an element o f
i s not i n
af
k
K
then
. I n case
that I
x+z since that
as w e l l as x - z
a r e elements of
K
we know t h a t
yz(x) = x + z E K
and k i s r- i n v a r i a n t . Now t h e c o n v e x i t y o f f 1 1 f ( x ) I f(x+z) + f ( x - z ) , hence x + z E a f K because x E
k
f ( x ) = supl((f).
I n any case y z ( x ) E a f K
.
implies
Representing Measures
254
P r o p o s i t i o n [129]:
3.1.4
Consider a f a m i l y
F(K)
on a c l o s e d compact __--Then t h e s e t --of -
r
o f p o i n t s e p a r a t i n g upper semicontinuous f u n c t i o n s
set
, -and l e t r be an x ~for all y E r)
K
{x E K I y ( x )
F(K)
-
exposedfamily.
I=
.
i s-a boundary f o r F(X) -
Proof: Fix
fo E F(K)
. Since
r
is
F(K)- exposed
a
f0
K
must be
r-
invariant.
Furthermore t h i s s e t i s closed, hence compact. Thus
K =
{i c
a K 10
i s c l o s e d and compact and r - i n v a r i a n t )
f0
must be nonempty. By Z o r n ' s lemma t h e r e i s a minimal element Again, s i n c e
KO i s
r-
Hence, by m i n i m a l i h y o f I n o t h e r words,all KO = I x o l
Obviously, a
r-
maximum (because
K
.
.
af K O = KO f o r a l l f E F ( K ) .
KO we g e t
a r e c o n s t a n t on
i s a singleton since
a common f i x p o i n t
f E F(K), af K O € K
invariant,we have, f o r
f E F(K)
KO i n
F(K)
.
KO
This y i e l d s t h a t
was assumed t o be p o i n t s e p a r a t i n g .
i n v a r i a n t s i n g l e t o n must be a f i x p o i n t . So we have found xo
, where
Ixol c a
f0
t h e a r b i t r a r i l y chosen
K).
fo a t t a i n s i t s
0
Using t h i s r e s u l t i t i s now obvious how t o make boundaries s m a l l e r , j u s t by making
r
larger.
To t h i s end we n o t e t h a t t h e u n i o n o f an a r b i t r a r y
c o l l e c t i o n o f F(K) - exposed f a m i l i e s i s a g a i n F(K)- exposed. Hence t h e r e i s a unique maximal F(K)- exposed f a m i l y , namely rmax = UIr Replacing
r
Ir
is
i n P r o p o s i t i o n 3.1.4
F(K)- exposed) by
.
rmaxwe g e t a minimal f i x p o i n t
boundary. The f o l l o w i n g theorem g i v e s a complete c h a r a c t e r i z a t i o n of t h i s minimal f i x p o i n t boundary i n terms o f p r o p e r t i e s w i t h r e s p e c t t o
F(K).
255
Boundaries
3.1.5
Theorem [1291:
L e t F(K) be a f a m i l y o f p o i n t - s e p a r a t i n g upper semicontinuous f u n c t i o n s on a _ compact c l o s_ e d s_ et K . 5 max(K) -be t h e c o l l e c t i o n -o f those _ ~ _ such t h a t f o r each compact,closed w i t h {XI 5 R c K x E K ----______ t h e r e i s some
f E F(K)
with -
f ( x ) = sup-(f) > i n f - ( f ) . K K
(*)
Then max(K) -----i s t h e s e t o f a l l common f i x p o i n t s o f rmax. I n particular f o r F(K). max(K) i s-a boundary Proof: x E max(K) and p u t
Let
KO = Assume since is
n {k
c K
compact closed, rmax - invariant, x E
{ x ) t: KO. Then t h e r e i s x E max(K).
rmax-i n v a r i a n t s i n c e
2
mality o f rmax
KO
.
Hence
f E F(K)
with
rmaxi s
{XI = KO and x
KO
k~ .
( f ) > infK (f) 0
t h e o t h e r hand,
F ( X ) - exposed and
we have
KO
f ( x ) = sup
. On
x E af KO 8 KO
We i n f e r
v a r i a n t . S i n c e . B KO ~
Y
Ik
KO
is
af KO rmaxin-
a c o n t r a d i c t i o n t o t h e minimust be a f i x p o i n t f o r a l l
*
Now c o n s i d e r x
E K
with y(x) = x
for all
Then t h e r e i s a compact, c l o s e d subset such t h a t f o r each be c o n s t a n t on
KO
f E F(K)
.
KO c K
we have e i t h e r
F o r each
y E KO
y
define
E rmax.Assume x B max(K). containing
x
f ( x ) < supK ( f ) 0
yy
: K
--t
K
.
.
with
{XI t: KO,
or f must
by y y ( x ) = y
Then r i s L e t r = {yyl y E KO} and y ( z ) = z i f z E K \ { x I Y F(K)- exposed. Indeed, l e t k c K be compact, c l o s e d and r- i n v a r i a n t . f E F(K).
Let Hence all
KO
c
y E KO
k
Assume, x E af
k.
This implies,
.
Hence
-
af K
f
is
. Then
y = y ( x ) E k f o r a l l y E KO. Y i s c o n s t a n t on KO , i . e . y E af k f o r
c ?i
r-
invariant. If
-,
x B af K then, t r i v i a l l y
256
Representing Measures
( a I?) = af I? f o r a l l y E KO . We have r c rmax s i n c e rmax i s t h e Y f maximal F(K)- exposed f a m i l y . Because y ( x ) = y x f o r y E KO\ {XI , Y x cannot be f i x p o i n t f o r a l l y E rmax, a c o n t r a d i c t i o n .
y
*
We conclude
x E max(K).
0
Two o t h e r consequences o f 3.1.4 a r e t h e c e l e b r a t e d Krein-Milman Theorem and Bauer's Maximum P r i n c i p l e . L e t us come back t o t h e Geometric S i t u a t i o n o f Example 3.1.3.
Again
H a u s d o r f f v e c t o r space x = y = z
whenever
K
E
.
i s a compact convex s e t i n a l o c a l l y convex Recall t h a t
x = x y t (1-x)z
x E K
with
we denote t h e s e t o f a l l extreme p o i n t s o f rates the
PO
nts o f
K
since f o r
a continuous l i n e a r f u n c t i o n a l obviously
i s c a l l e d extreme p o i n t i f
y,z E K, 0 < K
.
Note t h a t
x,y E K w i t h
x
+
u on E w i t h p ( x )
x
< 1
.
ex(K)
By
Conv(K)
sepa-
t h e r e i s always
y
~ ( y ), and
E Conv(K) (compare Remark 2.5.8).
p
H. Bauer's Maximum P r i n c i p l e :
3.1.6
f o r Conv(K) , i . e . -e v e r y upper semicontinuous ex(K) -i s_ a boundary a t_ t a_i n_s _i t s K- supremum on ex(K) convex f u n c t i o n on K _
.
Proof: Take t h e s e t
r
is
r
= Iy,lz
E
El. which we considered i n 3.1.3. We know t h a t
Conv(K)- exposed, hence i t s common f i x p o i n t s a r e a boundary (3.1.4).
But, o b v i o u s l y ,
ex(K)
.
r
i s e x a c t l y t h e s e t o f a l l common f i x p o i n t s o f 0
Let
n
z
i=1 n
X
be a subset o f a v e c t o r space
Xi xi
with
n E N
, xl,
... ,xn
E X
E
, then and
a v e c t o r o f t h e form
x1
,..., xn
2 0
such t h a t
xi = 1 i s c a l l e d , as u s u a l l y , a convex combination o f X . i=1 By co(X) we denote t h e s e t o f convex combinations and , i n case t h a t
E
257
Boundaries
-
i s a topological vector space , co(X) stands for the closure o f co(X). O f course, co(X) and are the convex hull o f X and the closed convex hull respectively.
z(X)
3.1.7
Krein-Milman Theorem:
convex -be a -compact ~ convex ~ subset _ of_a locally _ - Hausdorff ~ -vector K = =(ex K ) , i . e . K i s the closed convex hull o f its space. Then -------extreme points. ~Let K
Proof:
.-.
.
K = =(ex K ) 3 ex(K) i s certainly a compact convex subset o f K The s e t k must be nonempty since every f E Conv(K) attains i t s K- supremum on i?, according to Bauer's Maximum Principle. Assume there i s some xo E K \ K . Since K x k i s open in K and E i s locally convex, there
i s an open convex A c K x k containing xo
. Then Hahn-Banach
Separation
Theorem (1.5.8) applied t o A and B = k yields a continuous linear functional u such t h a t u ( A ) n p ( B ) = 0 . Without loss o f generality assume p ( a ) > u ( b ) for a l l a E A, b E B (otherwise take - p ) Hence supK(u) 2 supA(p) > supB(p) since p a t t a i n s i t s supremum on
.
the compact s e t B v l K E Conv(K)
.
. This
IB
contradicts 3.1.6 since ex 0
K c k and
Represenring Measures
258
3.2
MORE BOUNDARIES
In the l a s t section we have mainly considered the minimal fixpoint boundary and we have characterized t h i s boundary in terms of a separation property with respect t o compact subsets. This separation property ( i n the definition of max(K)) can be modified in several ways thus leading t o different subsets of a compact s e t . In the important cases these modified sets are again boundaries and, under additional assumptions, there are many relations between these boundaries. Another way of picking out special subsets as boundaries can be given in terms of characterizations o f the support o f representing measures. This concept i s mainly due to Heinz Bauer and has become a very useful tool in several parts of analysis
.
We s t a r t by giving a l i s t of the different subsets under consideration. Let K denote a compact Hausdorff space and l e t F(K) consist of upper semicontinuous 6- valued functions on K We assume (for convenience) that the elements of F ( K ) do separate the points of K , b u t i t i s n o t assumed, a t least n o t for the moment, t h a t F(K) i s a cone of functions. The u- algebra under consideration will be the family of Borel sets and by a representing measure for xo E X we mean a -,regular Borel probability measure T such t h a t
.
f(xo) 5 By
S U ~ ~ ( Tw )e
I
f
X
dT
for a l l
f E F(X)
.
denote the support of the measure
T
.
For completeness we also include the definition o f max(K)
3.2.1
i)
.
Definition max(K) i s the s e t of those x E K such that for each closed k c K with {XIc k there i s an f E F(K) with
*
f ( x ) = sup-(f) > i n f - ( f ) K K
.
strong maximum max(K) i s called the s e t of ~ _ _ _ _points. -
259
Boundaries
ii)
Max(K) (called the s e t of ~maximum points) i s the s e t of those x E K such that for each nonempty compact I? c K with x 13 k there i s some f E F(K) with
w
f ( x ) > sup-(f). K
i i i ) Max(K) i s the s e t of those x E K such that for every compact, nonempty k c K with x B k there i s some f E F(K) with f ( x ) 2 sup-(f) and K iv)
Let
0 > B > a
. Then
Max
a,B
f(x) > inf-(f) K (K)
.
i s the s e t of those x E K
such t h a t for every compact, nonempty k c K with there i s some f E F(K) with f I 0 and
x B k ,
f ( x ) 2 B > a 2 sup-(f) K
v)
C h ( K ) (called the Choquet boundary) i s the s e t of those x E K such t h a t the Dirac measure 5, a t x i s the only representing measure for x.
vi )
N
C h ( K ) (called the -weak Choquet boundary) i s the s e t of those x E K such that for every representing measure T of x we have x E S U ~ ~ ( T )
.
3.2.2
(i)
ProPosi tion:
MaxaYB(K) c Max(K)
c
N
N
Max(K) and C h ( K ) c C h ( K )
( i i ) max(K) c C h ( K ) n $x(K) ( i i i ) Ch(K)
U %(K)
N
t
Ch(K)
( i v ) Max(K) contained in every closed boundary in the closure of max(K). ----
of
K
, inparticular
Represen ring Measures
260
Proof: ( i ) i s an immediate consequence o f t h e d e f i n i t i o n s . ( i i ) Consider x E max(K). Let
k
be as i n D e f i n i t i o n 3.2.1 iii)and p u t
t h e i n e q u a l i t y o f D e f i n i t i o n 3.2.1 i ) AJ
to
k
^K
=
ku
{XI Then a p p l y
t o obtain the inequality
o f 3.2.1 iii).Hence, x E Max(K). Now, l e t
be a r e p r e s e n t i n g measure f o r
T
We have t o prove
s u p p ( ~ )= Cxl
x.
which c l e a r l y i m p l i e s
T =
6
X
. Assume
{XI and c o n s i d e r k = S U ~ ~ ( Tu ) Cx3 (which cannot be a s i n g l e f s a t i s f y i n g t h e i n e q u a l i t y o f 3.2.1 i). Put ^K I y E s u p p ( ~ )I f ( y ) = f ( x ) l , t h e n t h i s i n e q u a l i t y i m p l i e s t h e contradiction: SUPP('I)
t o n ) . Then t a k e an
Hence s u p p ( ~ ) = {XI and (iii)
I 4
Ch(K) c Ch(K)
Now, l e t
N
x E Max(K)
x E Ch(K).
i s a g a i n an immediate consequence o f D e f i n i t i o n 3.2.1. and l e t
'I
be a r e p r e s e n t i n g measure f o r
We have t o prove x E s u p p ( ~ ) . So, assume
-
K = supp(~)
.
By 3.2.1 iii)t h e r e i s some
f(x
t sup-(f) K
and
( i v ) Take
x E Max(K)
t h e n 3.2.1 i i ) contradiction to
and l e t
g i v e s some
k
f E F(K)
f(x) >
with
.
J f dr (a c o n t r a d c t i o n ) . R
be a c l o s e d boundary o f
f E F(K)
K.
If x C K
f ( x ) z sup-(f). This i s i n K b e i n g a boundary. The second p a r t o f t h e a s s e r t i o n
f o l l o w s t h e n from Theorem 3.1.5.
with
.
x C s u p p ( ~ ) and p u t
f(x) > inf-(f) K
These two i n e q u a l t i e s c l e a r l y i m p l y
x
0
26 1
Boundaries
3.2.3
Theorem t1311:
L e t F(K) + F(K)
Then
c F(K),
i.e.
f
+ g E F(K) whenever f,g E F(K).
CWK) c Max(K).
Proof:
x d Max(K). Then b y d e f i n i t i o n t h e r e is a compact c X which does x such t h a t f ( x ) I s u p - ( f ) f o r a l l f E F(K). By t h e K Riesz-Konig Theorem I 1 . 6 . 1 t h e r e i s a p r o b a b i l i t y measure T on such Let
not contain that
f(x) I
K
f dr
for all
f E F ( K ) . We can assume t h a t
because o t h e r w i s e we can r e p l a c e
T
x
3.2.4
and
N
x d Ch(K)
because
x
i s regular,
b y a r e g u l a r measure h a v i n g t h e same
i n t e g r a l s f o r continuous f u n c t i o n s . Hence, for
T
B k
3
T
i s a r e p r e s e n t i n g measure
supp(~).
0
Corollary:
L e t F(K)
t F(K) c F ( K ) .
i) There i s
fi s m a l l e s t
c l o s e d boundary of K
boundarv.
ii) --A l l the sets
d
, --called the Shilov N
max(K), Max(K), Max(K), Ch(K), Ch(K)
t h e i r closures a r-e-a_ l l_equal to_ t h e S h i l o v boundary ___ IL
a r e boundaries,
.
r\r
i i i ) max(K) c Ch(K) c Max(K) = Max(K) = Ch(K).
Proof: ( i i i ) f o l l o w s from t h e i n c l u s i o n s g i v e n i n 3.2.2 and 3.2.3. Since max(K) i s a boundary (Theorem 3.1.5) a l l t h e o t h e r s e t s must t h e n be boundaries, and from 3.2.2 ( i v ) we g e t t h a t t h e c l o s u r e o f Max(K) i s t h e s m a l l e s t c l o s e d boundary. E v e r y t h i n g e l s e f o l l o w s t h e n f r o m t h e f a c t t h a t t h e s e t s under c o n s i d e r a t i o n a r e boundaries and subsets o f
Max(K).
262
Representing Measures
Before we a r e going t o c h a r a c t e r i z e t h e Choquet boundary in terms of separation properties we remark t h a t conditions on t h e representing measures in t h e d e f i n i t i o n of the Choquet boundary can be formally weakened. Remark:
3.2.5
Let M,
denote the s e t of p r o b a b i l i t y measures on
Then
Ch(K) = i x E K
I TIXI
for all
> 0
K representing T
x.
E Mx3
Proof: I t i s t r i v i a l t h a t the Choquet boundary is contained i n t h e s e t on t h e r i g h t s i d e . NOW, assume t h a t f o r every T E Mx we have TIXI > 0 . We claim t h a t then
-
-( T - T
T =
l-T{XI
~ 1 x 1= 1 f o r a l l {XI
E M,
M,
i s an element of
)
Hence a c o n t r a d i c t on.
3.2.6
T
.
Indeed, i f
with
~ 1 x 1< 1 then
;{XI = 0.
0
Theorem:
be-a-cone containing a l l constant functions -and l e t a < p Let F(K) Then C h ( K ) = Max ( K ) .
< 0
.
a,!3
Proof: Let x E Max
(K)
x d
k
.
If
T
E
M,
T(K)
with
of
K with
f s 0 and f ( x ) 2 p > a t s u p - ( f ) .
K
then B
Hence
k
and consider a nonempty compact subset
a,D Let f E F(K)
5
< f(x)
aB
< 1
5
K
f dr
r(k) a
f o r every compact
. k
c K
with
x B
.
By regula-
Boundaries
r i t y of
we get then
T
Now, assume x Assume
E
Ch(K)
263
~ 1 x 12 1 - B/a > 0
.
for a l l
f E F ( K ) . The Sum Theorem yields linear K
and f(X)
p2 I supK
aB
5
pl,v2
: F(K)
B a) v2(f)
u l ( f ) + (1 -
respectively such that
K
Then B/a
T~
t
then
>
(1 - B/a)
sup-(g) K
T~
-,
with
pi(f)
I
f
i=1,2,
dTi,
i s a n element in Mx
T~
, T ~
on i;,
for a l l
f E F(K).
which i s n o t equal t o g
E
F(K)
with
x
E
Max
(K)
B ' ( l - a) s u P K ( g ) * Put
f E F(K) , f
and therefore
x B R.
.
for a l l f E F(K)
the Dirac measure 6, , a contradiction. Hence there i s g(')
Ch(K).
such t h a t
By the Riesz-Konig Theorem there are probability measures and
E
;supi;,(f)+ (1 - B/,)supK(f).
f(x)
p1 I sup-
x
and l e t k c K be non empty compact with
(*)
I
yields
And 3.2.5
Max
I0
a,B
, f(x)
2
B > a 1 supl((f)
( K ) 3 Ch(K)
.
. Hence
a,P
We conclude t h i s section with a discussion of three examples. If F(K) i s N N a cone we have proved (3.2.4) max(K) c C h ( K ) c C h ( K ) = Max(K) = Max(K) c closure of (max(K)). We show that i n general a l l preceding inclusions are proper.
264
Representing Measures
Exampl es :
I 1I n
i ) Let K
Let F(K)
If : K + R
=
.
Clearly, 1 = lim f(Ti) n +-
E N) U {O}
I f(0)
=
i s a vectorspace of continuous functions. We have 0 B Ch(K) s i n c e , "X 1- 6 , 6, Dirac measure a t Ti 1 , we obtain 'I = n = l 2"
F(K) with
f ( 0 ) = "X 1-n f(,)1 = f dr n=l 2 K
.
measure on K with
f dp
for all
Xi 2 0
with
I
f(0) =
K
On t h e other hand, l e t
m
P =
.X X i di , f o r s u i t a b l e
1 =o
Dirac measure a t 0 )
. Fix
m ,N E
by
1
fm,N (E)
and f
m,N
Ch(K)
6,
*
1
2m-N)
x xi
i =o
=
1
be a p r o b a b i l i t y
W e have
.
, and define f
c N
, n
=
is t h e
( 6,
m,N
E
F(K)
m
0
for m * n s N
1
for n > N
n+ m
Hence, i f = A.
=
(zm -
I
,m
N
.
f E F(K) m
p
( 0 ) = lim f m , N);( 1 = 1. W e obtain
1=
P
K i s compact. "1 1-n f ( 1 x)} . n=l 2
N
I fm y N dp + m ,
=
x0
t
m
(2m-2m-N)~mt 1 i=Nt1
for all
Am = 2 - m ( l - ~ 0 )
.
(1 - Xo)r and 0 E supp(p) ry rv C h ( K ) = Max(K) = Max(K) t
m E N
Thus 0
xi
.
.
We conclude, rJ
E Ch(K)
and
.
i i ) Let K be as i n i ) . Define now F(K) = I f : K
T r i v i a l l y , we have t h a t
+
R
1 f(o)
I 1I n
=
lim f ( n1) = 1 ( f ( 1 ) w-
E N)
t
1 f(-2))3
,
is the smallest boundary of
K with
265
Boundaries
respect t o But
F(K). Hence
f4
0 d Ch(X)
since
Let
0 C supp(
I an E R
iii) L e t co = {(a,) co
must be t h e c l o s u r e o f
K
1 + '2 b2)
b1
,n
.
, with l i m
E N
1 1 . II w i t h
be t h e sequence whose
an = 01
n+
i s a Banach space i f we c o n s i d e r en E co
max(K).
n'th
.
II ( a n ) l l = sup l a n l nEN
1
component i s equal t o
II enII
and a l l o t h e r components a r e zero. C l e a r l y ,
1 for all n
n
= -
.
. Note
m
t h a t we may w r i t e a l l elements o f
co
as
(a,)
=
z n an en where t h e
n=l
s e r i e s converges i n t h e norm t o p o l o g y . D e f i n e t h e f u n c t i o n m
Pm : co
+
P (
by
R
z n an e n )
= am
n=l
. Clearly,
i s continuous w i t h
Pm
r e s p e c t t o t h e rIOrTntoplOgy . L e t us c o n s i d e r t h e f u n c t i o n defined by
f(n) = n
[rl
=
- [ 2I-
sup{n E N U I03 [ n 5 r3
For every with
mo, N E
f ( m ) = mo
N
Let
1
n E N3
F(K)
functions on
, there
r 2 0
. n > 1
for all
m E N ,m
is
,n
E N
, and:
2 N
, and
- n-" ( x ~ ( +~ en) )
IPN(xn)I I 1 f o r a l l
.
for
f(n) < n
x1 = el
xn = n-'(nn+t)el
K = ixn
N
.
L e t us d e f i n e by i n d u c t i o n
We have c l e a r l y
+
, where
By elementary a n a l y s i s we have
(*I
f :N
We have
xn
+
x1
for all
N > 1 and n E
and consequently,
n = 2,3, N
.
K
...
Consider i s compact.
c o n s i s t o f the r e s t r i c t i o n s o f a l l realvalued l i n e a r continuous co. C l e a r l y
f
a p r o b a b i l i t y measure on
K
P
nl K f u n c t i o n s separate t h e p o i n t s o f
E F(K)
with
K
. We f(xl)
n E N , and t h e s e
for all claim, =
K
x1 E Ch(K) : L e t
f dT
for all
T
f E F(K).
be
266
Representing Measures m
Hence
T =
1
i=1 measures a t xi
di,
Xi
1
. Assume
be an index such t h a t
0 = f(Xl)
,
A. 2 0
m
Z
i=1
AN = sup Xi
Ai
N-(Ntl)~N
-
i=N
a c o n t r a d i c t i o n . Thus
there i s
g E F(K) N
such t h a t
m
t e o E U
2 N
such t h a t
f(Xi)
where
f = -P
=
1 -N N AN t
di
NIK
are the Dirac
xi
* 0 . Let
N > 1
E F(K). Hence
W
k=Nt1
m
1 lk k-k 2 N-(Ntl)~N k=Nt1
x1 E Ch(K).
with
i s a neighbourhood of mO
,
.Define
i>1
f dT = Z
2
x
1
=
X
mo E
1
i > 1 such t h a t
there i s
m
is
A. =
g(xl)
x
for all
mO
.
) en
sup Xi > 0 irN+1
Hence
-t
I n o t h e r words, t h e r e
U = {y E colg(xl)
> g(y))
0 we have
m t N , N s u f f i c i e n t l y large. (*) yields
f ( m ) = mo
and t h e r e f o r e we a r r i v e a t a c o n t r a d i c t i o n :
(Here we used t h e same n o t a t i o n f o r
,
x1 B max(K). Indeed, o t h e r w i s e
But
. Since
mO
- (Ntl)-N.N-l
= supK(g) > i n f K ( g ) .
> g(x
g(xl)
A k k-k PN(Xf(k))
g E F(K)
and i t s e x t e n s i o n t o
*
So, we have indeed max(K) c Ch(K).
The same argument as b e f o r e shows t h a t i n f a c t
Ch(K) = K
.
co).
267
Boundaries
3.3 CHOQUET'S THEOREM I n t h i s section we deal with representation of points of compact convex sets by means of integrals over the extreme points. We do t h a t a t several levels of abstraction, although a l l these results are - more or less consequences of Choquet ' s fundamental representation theorem in case of the geometric situation (described in 3.1).
I n f a c t a l l theorems of this section can be obtained as corollaries of Theorem 3.3.6. B u t in order t o i l l u s t r a t e different viewpoint we give for a1 1 theorems different approaches.
We recall the geometric situation: K i s a nonempty compact convex subset of a locally convex Hausdorff vector space E . Again Conv(K) denotes the i- valued upper semicontinuous convex functions on K and A ( K ) stands for the subcone of affine continuous functions. A function cp i s said t o be affine i f
x,y E K , 0 5 X 5 1
for all of K
.
.
And by
ex(K) we denote the extreme points
The importance of Bauer's Maximum Principle (or the Krein Milman Theorem) comes from the f a c t that i t allows t o identify A ( K ) and A ( K ) lex(K) (restrictions t o ex(K)): 3.3.1
Lemma:
For every --
h E A ( K ) lex(K) there i s a unique
The map h
+
-_.
6 ----i s linear and we have supK 6
6 E A ( K ) with =
~up,,(~)h
.
lex(K)
= h.
Proof: Assume there are h l , h 2 E A ( K )
such t h a t their restrictions t o ex(K)
are equal. By Bauer's Maximum Principle we get supK(hl - h 2 ) = ~ ~ p ~ ~- h(2 ) ~= )0 .( The h ~same holds for supK(h2- h l ) . Hence h l
= h2
. So,
the extension t o
K
i s unique and the map h
+
6
268
Representing Measures
must be l i n e a r s i n c e i t s i n v e r s e ( r e s t r i c t i o n map) i s l i n e a r . The r e s t o f t h e a s s e r t i o n f o l l o w s a g a i n f r o m t h e Maximum P r i n c i p l e .
0
The n e x t o b s e r v a t i o n i s i m p o r t a n t f o r t h e subsequent theorems: 3.3.2
Lemma:
Conv(K)lex(K)
a D i n i cone.
Proof: P r o v i n g t h e D i n i c o n d i t i o n d i r e c t l y i s n o t t h a t easy. B u t i n Lemma 1.3.4 we have a f o r m a l l y weaker c o n d i t i o n . L e t Conv( K )
l W K ) . Then
take
(pn E Conv(K)
From Bauer's Maximum P r i n c i p l e we g e t of
%
be a sequence i n
having the
(pn
K we g e t ( f r o m D i n i ' s Lemma) some m ...
5 0
as r e s t r i c t i o n s .
'pn
. Because o f
6 0
t h e compactness
x E K with W
.
Y(x) = i n f supK( I: gn) where Y = I: The f u n c t i o n Y i s a g a i n m n=l n=l an element o f Conv(K) hence ( B a u e r ' s Maximum P r i n c i p l e ) x can be assumed t o be an element o f Lemma 1.3.4
3.3.3
ex(K). T h i s shows t h a t c o n d i t i o n ( i i i ) o f
i s fulfilled.
Choquet's Theorem:
F o r e v e r y x E K t h e r e i s a p r o b a b i l i t y measure mx --
on ex(K) -
(with
a l g e b_ r a making a l l elements o f A(K) r e s p e c t t o t h e s m a l l e s t u- ~ --_ _ lex(K) measurablel such t h a t h(x) =
I
ex(K)
h dmx -f o r a l l h E A(K)
.
Proof: l e t 6 be t h e e x t e n s i o n t o K (Lemma 3.3.1). I N K ) The map h + 6 ( x ) i s o b v i o u s l y a s t a t e o f A(K) B u t t h i s must be lex(K) a D i n i cone s i n c e i t i s a subcone o f a D i n i cone (Lemma 3.3.2). By t h e For
h E A(K)
.
R e p r e s e n t a t i o n Theorem (2.2.1)
t h e r e i s t h e n a measure mx
with
269
Boundaries
for all
( o r for all
h E A(K)lex(K)
Since A ( K ) i s a vector space this measure s t r i c t l y represents x Inserting for h the function 1 one gets t h a t mx(ex(K)) = 1 .
1
h E A(K)).
. 0
.
Choquet's original theorem was stated for metrizable K I n t h i s case the Baire and the Borel sets coincide and - since ex(K) i s then a Borel subset of K 12501- the measure mx can then be considered as a Borel measure supported by the extreme points. The non-metrizable case was f i r s t proved by Bishop and de Leeuw ([441 . Using the Sandwich Theorem we immediately can transfer the result t o Conv(K) instead of A( K ) : 3.3.4
Theorem:
x
For every --
E K
there i s a probability measure mx
on
ex(K) (with -
respect to the u- ~~algebra making a l l elements o f Conv(K) ~ -smallest measurable) -such that
4x1
I
I
cp dmx
ex(K)
for a l l
cp E
Conv(K)
.
Proof: Define a superlinear 6 : C o n v ( K ) J e x ( K ) 6(cp) =
supI;p(x) I % E Conv(K) with
(p
lex(K)
From Bauer's Maximum Principle we get 6
Theorem gives us a s t a t e
2 6
on
6
+
= cp)
sup ex( K ) Conv(K) lex(K)
. and the Sandwich
I
Conv(K)lex(K) i s a Dini cone (3.3.2),hence the Representation Theorem (2.2.1) yields a representing measure mx for v By definition of 6 this measure
.
f u l f i l l s the desired inequality.
0
*
270
Representing Measures
Now, i t i s quite natural t o ask i f these theorems can be transferred t o s t a t e spaces of order unit cones. F i r s t we have t o find a suitable boundary which can replace the extreme points of a compact convex s e t . Fortunately, the extreme points of the s t a t e space can take this role. This i s not completely t r i v i a l since we cannot always consider the s t a t e space as a compact convex subset of a locally convex Hausdorff vector space. This i s so, because s t a t e s are allowed t o attain the value .
-
3.3.5
Theorem:
Let ( F , < , I ) ----be an order unit cone. Consider a sequence f n @ F such that u ( f n ) i s decreasing --for every s t a t e p . --_Then. there i s a character v with inf v ( f n ) = inf S I ( f n ) nEN
nEN
where SI i s the -order unit functional. Proof: Consider on the compact s t a t e space n the cone of continuous functions F. For W,P E n we define functions n + R by
^F given by the Gelfand transforms of yP
Yw
The s e t r
P
i f there i s a
u
otherwise
I
= { y p y wp,w E
0
n l i s obviously
x
I 1 with p s x p + ( 1 - x ) ~
?- exposed.
The s e t K of states ( n o t necessarily characters) v f u l f i l l i n g the equality of 3 . 3 . 5 i s nonempty (Lemma 1.2.6) and compact since i t i s the s e t where the upper semicontinuous function inf ^fn attains i t s maximum. K
i s r- invariant since
Tn
nEN
i s decreasing. Therefore
.
r l K is FIK -
exposed. Hence, there i s a r- fixpoint in K By definition of the t h i s must be an extreme point and the assertion i s proved since ex-
yP ,w
treme points are characters ( I .2.10.6).
0
27 1
Boundaries
3.3.6
Theorem:
Let ( F , < , I ) be an order u n i t cone, ~denote by char(F) the characters of F . ---Then for every s t a t e P of F there i s a probability measure -
m
P
on char(F)
-
(with respect t o the
U-
algebra generated -by the
restrictions t o char(F) of the Gelfand transforms of F ) -such t h a t for all --
f E F.
Proof: Theorem 3 . 3 . 5 states t h a t char(F) i s a boundary of the s t a t e space for Exactly as in the proof Of 3 . 3 . 4 we then show the Gelfand transforms t h a t for a s t a t e p of F there i s a s t a t e i; o f ? [char(F ) such t h a t F(p) I for a l l T E ^F . Another immediate consequence of Theorem 3 . 3 . 5 i s t h a t F l c h a r ( F ) i s a
.
-
Dini cone. Hence by the Representation Theorem we find a representing measure m for This measure does the j o b . 0
.
3.3.7
Theorem:
Let K _ be -a_ compact space and F(K) a_ family _ _of- point separating _ _ Hausdorff -upper semicontinuous functions on K Then for every x E K there i s-a~ _ __ probability measure T on the Choquet boundary C h ( K ) (with to ~ _ respect _ _ _ the u- algebra generated by the restrictions -of the elements of F ( K ) to -~ C h ( K ) ) -such that
.
_ .
-_.
f(x) 5
I
Ch(K)
f d-r -for all f E F(K).
Reprssenting Measurns
272
Proof: Obviously we can replace
F(K)
by the cone generated by
F(K) and the
constant functions, t h a t changes n e i t h e r the Choquet boundary nor the inequality f o r
T
We embed the s e t p(x)(f) = f(x) p(Ch(K)) from 3.3.6.
. So,
K
l e t us assume t h a t
F(K)
i s an order u n i t cone.
n o f F(K) by
i n t h e s t a t e space
: K
p
-,n
with
f E F(K) (canonical embedding). We c l a i m t h a t
for all
i s equal t o t h e s e t o f characters. Then the a s s e r t i o n f o l l o w s
Proof o f the claim: L e t 1.1
be a character. By the Riesz-Konig Theorem there i s a r e g u l a r
representing measure
for
T
and s p l i t up the support
1.1
on
s u p p ( ~ )o f
1.1 5
T ( S 1 ) U l t T(s2)1.I2
u 2 ( f ) < u ( f ) . Hence p
+
1.1~
and
T
. We p
f E F(K)
:
v1,u2
If T(s2) > 0 then we have s t a t e s
such t h a t
. Take an a r b i t r a r y
K
given by
have, by d e f i n i t i o n o f
p2
,
cannot be an extreme p o i n t o f t h e
s t a t e space, which i s a c o n t r a d i c t i o n t o t h e assumption t h a t character. Therefore T ( S ~ =) 0 and T i s supported by S1
p
is a
. Since
f
was a r b i t r a r y we have
and t h i s i n t e r s e c t i o n must be nonempty and f o r every P(X) 2 x E k
1.1
.
.
x E
k
we have
Since a character i s maximal we then g e t p ( x ) = p f o r a l l can only c o n s i s t o f one p o i n t because t h e F(K) were assumed
273
Boundaries
t o be p o i n t s e p a r a t i n g . And f o r t h i s p o i n t we have proved t h a t i t i s t h e s u p p o r t o f any r e p r e s e n t i n g measure, hence
F o r t h e second p a r t t a k e an p1 3p2
and
0
v 1 !J we have v = p On a vector space every s t a t e ( i n f a c t every linear functional) i s t r i v i a l l y maximal.
.
.
We know (Theorem 3 . 3 . 6 ) t h a t for every s t a t e p there i s a representing measure m f o r !J which i s carried by the characters of the s t a t e space. P
Representing measure means t h a t m p(f) I
I 'i dmll
for a l l
P
i s a probability measure with
f E F
,
And "carried by the characters" means t h a t i t i s a measure on the characters
endowed with the smallest U- algebra such t h a t a l l the f E are Gelfand transforms of F) ; t h i s i s equivalent t o saying measurable (? t h a t there i s a Baire probability measure T~ on the s t a t e space n which gives, with respect to ? , the same integrals as m and which has the !J property that i t vanishes on any Baire s e t having empty intersection with
282
Representing Measures
t h e s e t o f c h a r a c t e r s (Remark 2.1.1). Furthermore we know (Theorem 3.4.7) on t h e s t a t e space
t h a t a B a i r e p r o b a b i l i t y measure
'I
i s maximal w i t h r e s p e c t t o t h e Choquet o r d e r i n g
R
(meaning maximal as a l i n e a r f u n c t i o n a l w i t h r e s p e c t t o t h e
VFB
-
point-
w i s e o r d e r ) i f and o n l y i f t h e r e i s a measure c a r r i e d b y t h e c h a r a c t e r s which has t h e same i n t e g r a l s w i t h r e s p e c t t o t h e f u n c t i o n s i n
VFB
(or
because o f Observation 3.4.1,
or
C(R) V? s i n c e t h i s c o n t a i n s VFB)
.
Now, l e t us combine a l l t h i s knowlegde w i t h Theorem I . 2.9.11. 3.5.1
Theorem:
Let F be-a s i m p l i c i a 1 ---o r d e r u n i t cone and c o n s i d e r an -
R- valued maximal
o f F. Then t h e r e i s a unique p r o b a b i l i t y measure -
state
P
by t h e --
c h a r a c t e r s --o f the state
J
P(f) =
7
dmu
m
P -
carried
space -such t h a t -for all
f E F
.
Proof:
.
F: Now, c o n s i d e r as G t h e F i n G v i a t h e Gelfand t r a n s form. Then a l l t h e requirements o f S i t u a t i o n I . 2.9.9 a r e o b v i o u s l y f u l -
Since
i s R- valued i t i s an element o f
1.1
and map b y T : F
V':
cone
-,G
t h e cone
f i l l e d . Theorem I . 2.9.11 g i v e s a unique l i n e a r map u(;)
o
(*) for a l l
T =
i
* -,GT
with
and
u(;)(g)
G
u : F,
E Ff
= sup(v(g)
. Hence
I
u(U)
v E G,* is
with
voT =
V? = G- p o i n t w i s e maximal i f
is
maximal. T h e r e f o r e i t s s t r i c t r e p r e s e n t i n g measure on t h e s t a t e space must be maximal i n t h e Choquet o r d e r i n g and t h e e x i s t e n c e o f t h e s t r i c t r e p r e s e n t i n g measure in
'c1
(*)
f o l l o w s from t h e p r e c e d i n g remarks.
a l s o y i e l d s t h e uniqueness o f
would be dominated b y m
P
m
because o f
u
The p r o p e r t y
s i n c e any o t h e r such measure
(*); hence i t must be equal t o
because i t corresponds t o a maximal measure
.
fi
P
m
u
Boundaries
283
u
A c t u a l l y we have proved a l i t t l e b i t more s i n c e t h e map
we used i n t h e
proof i s additive: 3.5.2
Remark:
Consider t h e same s i t u a t i o n as i n 3.5.1 and assume t h a t t h e s e t o f maximal R- valued s t a t e s i s a convex s e t . Then t h e map
When
F
u
+
m
U
i s affine, i.e.
R- v a l u e d and
i s a v e c t o r space t h e n e v e r y s t a t e i s a u t o m a t i c a l l y
maximal; thus t h i s s i t u a t i o n i s much more t r a n s p a r e n t : 3.5.3
Theorem: are equibe an order unit vector space. Then t h e f o l l o w i n g --
L e t (E, < , I )
valent:
i)
E i s simplicia1
ii) Ef
-has t h e FSP o r , equivalently,
( E T - E*,)
i s 2 vector l a t t i c e ~~
w i t h~ r e s p ect t o t he o rder g i v en b y t h e p o s i t i v e cone ET -
iii) --F o r e v e r y two p r o b a b i l i t y measures there i s ---
iv)
v)
an
ml,m2
.
the characters c a r r i e d by -
f E E such that
Every s t a t e u of --
E has a unique r e p r e~ s e n t i n g measure m v ca_ r r i e_ dby- t h e c h a r a c t e r s . _ There i s an a f f i n e
.___---
u
+
m v from t h e s t a t e space i n t o t h e
p r o b a b i l i t y measures _ c a r_ r i e_ db y- t h e c h a r a c t e r s such t h a t ~ ( f )= J f dmU
for all
f E F -and s t a t e s
P
.
Representing Measures
284
Proof: i s t h e c o n t e n t o f Theorems 1.2.9.5 and 1.2.9.7. (i) Y (ii)
--
(i) * (iv)
f o l l o w s f r o m o u r l a s t theorem.
(iv)
i s a consequence o f Remark 3.5.2.
(v)
(i): Consider t h e embedding T : E + G = C(R) g i v e n by t h e Gelfand (v) transform. Then ( v ) says t h a t t h e r e i s an a d d i t i v e map u : ET -,GT w i t h u(1-1)o T = p f o r a l l Theorem 1.2.9.8.
1-1 E
. Hence,
Er
E
i s simplicia1 by v i r t u e o f
Since e v e r y s t a t e has a r e p r e s e n t i n g measure ( s t r i c t s i n c e (iii) Y (iv): we a r e i n a v e c t o r space s i t u a t i o n ) uniqueness o f t h a t r e p r e s e n t i n g measure must o b v i o u s l y be e q u i v a l e n t t o t h e p r o p e r t y t h a t f separates these measures. n A d d i t i o n a l i n f o r m a t i o n i s o b t a i n e d when t h e s i m p l i c i a 1 map i s c l o s e d . We s t a r t b y making a few remarks on elementary topology. We have endowed t h e s t a t e space
nF
o f an o r d e r u n i t cone w i t h t h e c o a r s e s t t o p o l o g y such
t h a t , f o r e v e r y f E F , t h e Gelfand t r a n s f o r m i s continuous. T h i s means t h a t a f u n c t i o n @ : nF1 + nF2 i s continuous i f and o n l y i f 7 o is continuous f o r e v e r y f E F2. can c o n s i d e r t h e cone R
nF
O f course, i n s t e a d o f t h e s t a t e space one = i r w l r 1 0 , w E nF3 generated by t h e s t a t e
space. F o r a v e c t o r space t h i s cone i s equal t o a l s o h o l d s i f t h e elements o f
f E F t h e r e i s some a E R
(F,< )
* . This
F a r e bounded f r o m below, i . e . such t h a t
aI
< f. We endow
same topology. I t should be observed t h a t a f f i n e maps T :
R
equality f o r every
nF w i t h t h e
nF
1
+
n
F2
between s t a t e spaces correspond u n i q u e l y t o l i n e a r maps + R nF between t h e generated cones ( t h i s e x t e n s i o n , o r 1 2 r e s t r i c t i o n , we denote b y t h e same symbol). A l i n e a r map T :
T : R
nF
i s continuous i f and o n l y i f i t s r e s t r i c t i o n T
RRF1
lnFl
t o t h e base R
-+ F1
i s continuous. F i n a l l y we remark t h a t we endow t h e bounded measures ( o r t h e base g i v e n b y t h e p r o b a b i l i t y measures) on
nF 1
w i t h t h e weak-star
F2
285
Boundaries
-
C(nF ) 1 all
VFB
i s sup-norm dense i n
t h i s t o p o l o g y i s t h e same as t h e c o a r s e s t t o p o l o g y such t h a t , f o r
f E F
3.5.4
-
-
C(nF ) . S i n c e VFB 1
topology w i t h respect t o
, the
functions
m
+
1f
dm
a r e continuous.
Theorem:
Consider -a g a i n a s i m p l i c i a l ---o r d e r u n i t v e c t o r space E are equivalent:
. -Then t h e
following
i s c l~ o s e d in the s t at e space. (i) -The _ _s -e t o f c h a r a c t e r s (ii) The sup-norm c l o s u r e o f
a t t i ce w i t h respect to -i s a -v e c-t o r l~
t h e p o i n t w i s e o r d e r on t h e c h a r a c t e r s . (iii)
The 9 -
u
a s s i g n i n g --t o each s t a t e
p
i t s unique maximal measure
_ .
(------maximal i n t h e Choquet o r d e r ) i s continuous.
(iv)
The s i m p l i c i a l
map
1 : E:
+
g
(L E )
continuous.
Proof:
a n be t h e c h a r a c t e r s o f E . T h i s i s a compact s u b s e t o f t h e s t a t e space. The map p : 2 + c(a a ) g i v e n by t h e r e s t r i c t i o n s t o a R i s monotone. And because o f (i) * (ii):
Let
t h i s map i s i n j e c t i v e and we o n l y have t o p r o v e t h a t p ( E ) i s a sup-norm dense subspace o f c ( a s2 ) . We know t h a t p ( E ) separates t h e p r o b a b i l i t y measures on
as1
(Theorem 3.5.3 ( i i i ) ) .
c(a
norm continuous l i n e a r f u n c t i o n a l s on i n C(a n )
(ii) ( i v ) :
Hence
by t h e Hahn-Banach Theorem
0 ) . Thus
(I
^E.
E*, t h e r e i s a unique such t h a t p(p)(?) = p(f)
p E
p(i)
must be dense
e : E + Clos(E) o f
E
into
An easy e x e r c i s e shows t h a t f o r each s t a t e in for all
~(1.1)
LE
(2);
Clos f E E
i s continuous and l i n e a r . Now, if Clos o f t h e f r e e l a t t i c e cone
separates t h e sup-
1.5.8).
Consider t h e canonical embedding
t h e sup-norm c l o s u r e o f
p(2)
(E)
and t h a t t h i s map
p
i s a l a t t i c e then by d e f i n i t i o n
t h e r e i s a unique l a t t i c e cone homomorphism
286
E*
Representing Measures
: LE
+
Clos(E)
such t h a t
commutes. Then c o n s i d e r t h e map z : E: T h i s map i s l i n e a r w i t h
X(p)
0
L =
for all
p
g i v e n by
(LE)
+
. Hence
p
p
+
p(p)o c*.
IT must be t h e
s i m p l i c i a l map. F o r i t s c o n t i n u i t y we have t o show t h a t f o r a r b i t r a r y
W E LE p
t h e map
-,p ( p ) ( t * ( q ) )
(iv)
*
(iii):
E*
: LE
E*
o
L
-,V
= E
p + z ( p ) ( ~ )
Consider t h e embedding
: E
+
V
2
and l e t
i s a f f i n e (Remark 3.5.2 and Theorem 3.4.7) we have p; ) and u must be continuous s i n c e I: i s continuous. a
(iii) * (i):R e c a l l t h a t i f c h a r a c t e r (Lemma 3.4.4). c l o s e d because a Theorems 3.5.3
E
i s continuous.
p
be t h e unique l a t t i c e cone homomorphism such t h a t
. Since
u ( p ) o e* = ~ (
i s continuous. B u t t h i s i s equal t o
and i s c l e a r l y continuous s i n c e
~ ( p )=
dV t h e n
Hence a-1{6ylv E
i s continuous.
QI
p =
u
and
p
must be a
i s t h e s e t o f c h a r a c t e r s and
0
and 3.5.4 are, o f course, minor m o d i f i c a t i o n s o f r e s u l t s
well-known f o r t h e geometric s i t u a t i o n when A(K) a r e t h e a f f i n e c o n t i n uous R- valued f u n c t i o n s on a compact convex subset K o f a l o c a l l y convex H a u s d o r f f v e c t o r space. The f u n c t i m
t h e n serves as o r d e r u n i t lK and an easy e x e r c i s e (Hahn-Banach Theorem) t h e n shows t h a t t h e s t a t e s of A(K) are exactly given by t h e p o i n t evaluations i n K K then i s c a l l e d
.
a siniplex ( o r r a t h e r a Choquet-simplex) i f A(K) i s s i m p l i c i a l . Because of t h e f a c t t h a t t h e n t h e cone generated by K i s a l a t t i c e cone t h i s
n o t i o n r e a l l y i s an a p p r o p r i a t e g e n e r a l i z a t i o n o f t h e w e l l known d e f i n i t i o n of a simplex i n R"
.
I n t h i s s i u t a t i o n the characters are obviously given
by t h e extreme p o i n t s of
K
. When t h e
extreme p o i n t s a r e c l o s e d t h e n
K
i s c a l l e d a Bauer-simplex (because o f t h e p i o n e e r i n g work o f Heinz Bauer i n combining p o t e n t i a l t h e o r y and t h e t h e o r y o f compact convex s e t s ) .
287
Boundaries
3.6
DIN1 BOUNDARIES
I n t h i s section we investigate two problems. The f i r s t one i s motivated by the observation (Lemma 1.3.3) t h a t X i s a boundary whenever F(X) i s a Dini cone. We look for conditions such t h a t F(X) becomes automatically a Dini cone whenever i t has X as a boundary. An essential tool for finding meaningful conditions i s Simons Convergence Lemma (see 12971 and [2441 ,[2681). This lemma i s very powerful in several areas of functional analysis, i t i s particularly useful in context with the proof of James 'Theorem (see [298 I and [1771 12583 ,[lo61 ). The second problem looks very different from the f i r s t one, b u t they are related with respect t o the techniques which are involved. We have already seen (section 3.2) t h a t a compact convex subset may have a boundary for the affine functions smaller t h a n the s e t of extreme points (Choquet boundary). We ask under w h a t kind of conditions such a behaviour i s prohibited, in other words: When are the characters the smallest boundary of the s t a t e space (with respect t o the Gelfand transforms) of an order unit cone? W e s t a r t with Simons Convergence Lemma ( i n the version of M. Neumann [2441). 1
Consider C, , the s e t of sequences y all
n
= (y,,y2,...)
m
E
and I ~ I =1 we
t y n f n . This s e r i e s always converges in
n=l
we can write: (1)
m
m 1
n=l
yn(fn - c ) ,
i s a common upper bound f o r the
fn.
for
N , be a sequence of uni-
formly upper bounded functions on some nonempty s e t X denote the function
with yn 2 0
since
288
3.6.1
Representing Measures
Simons Convergence Lemma:
Assume t h a t -supx< y,f
>
= lim
m
(supx
E ynfn) n=l
W m
for a l l --
1
y E Lt
. Then,if
Y c X
_i s _a
> I
boundary f o r I < y,f
y E
L,),1
we have -s u p y ( l i m sup fn) 2
n
(**I 2
-)a
m
infIsupX( z
n= 1
Xnfn)
I
m
J t i y
E N , An t 0
m
An =
II
n=l
1)
Remark: C o n d i t i o n ( * ) i s q u i t e o f t e n a u t o m a t i c a l l y f u l f i l l e d . F o r example when t h e
.
This follows a r e upper semicontinuous f u n c t i o n s on a compact s e t X fn immediately w i t h (1) f r o m D i n i ' s Lemma. Also,when t h e fn a r e u n i f o r m l y bounded ,then ( * ) h o l d s . P r o o f o f 3.6.1.: L e t a be t h e i n f on t h e r i g h t s i d e o f ( * * ) . 6 > 0 there are y E Y
can show t h a t f o r each ntm
g, E Km = I such t h a t
Z
i=m g,(y)
and thus, s i n c e
I
Xifi
ntm
z
i=m
t a - 2 6. 6
Xi
= 1, Xi
t 0, i=m,mtl,
n
.
g,
E
(a- 6(lth)
...,ntm,
-
+ m
E N,
n
E NI
-
2 6
. We can c l e a r l y
. Let
p be an upper bound f o r t h e 6 > 0 a r b i t r a r i l y and l e t
px)(l-A)-'
Km by i n d u c t i o n , such t h a t
m
l i m sup f n ( y ) 2 a n+m
s u p y ( l i m sup fn) 2 a
r e s t r i c t o u r s e l v e s t o t h e case a > - m fn. We t h e n have - m < a I p < m F i x
Define
and , f o r a l l
Because,then we have
was a r b i t r a r y ,
1 > h > 0 be so t h a t
The theorem f o l l o w s i f we
t a - 2 6
.
289
Boundaries
.
+ 6 2-m A m We c l a i m t h a t t h e r e i s y E Y
such t h a t
g,(y)
ta
-2
6 for a l l
(1t
x ) -1(g,tx
m 2 1.
Proof o f t h e claim: Define h, =
m
E xi-' i=l
gi
and p u t
ho = 0
. Since
m E N , we have
for a l l
I f we m u l t i p l y
the l a t t e r i n e q u a l i t y by
I
x
(Note, t h a t always supX(hm) > We d e r i v e
+ supX(hmtl)
supX(hm-l)
x-m(supX(h,l)
-
-
since
supX(hm)) 2
Because o f ( * ) we have s u p X ( l i m hm) = l i m supX(hm) m+m
.
(ltx)
we o b t a i n
+ 6 2-"(1+ h ) A". a >
- -).
gmtl)E
I(,,
Representing Measures
290
The boundary property of We conclude
Y yields
y E Y
such t h a t h(y) = s u p X ( h ) .
hence
t (l-x)-'(a-6(1tA)
m E
for a l l
N
.
-
AD) 2 a
-
2 6
0
For our considerations we only need a special version of the Convergence Lemma,dealing with the situation when the f n are decreasing. I n t h i s case the inequality has a simple form: 3.6.2
Corollary:
Let the --
f n be as in 3.6.1
pointwise decreasing.
Then
and assume furthermore that the f n are
supy inf f n t inf supx f n n n This inequality shows t h a t the Convergence Lemma must have nice applications for Dini cones. Let us f i r s t briefly recall the notation we adopted in the context of Dini cones. We assume that F = F(X) i s a cone o f upper bounded functions on a nonempty s e t X containing a l l constant functions. F i s endowed with the pointwise order on X and the function lX serves as order unit. VF i s
Boundaries
the max-stable cone generated by and
functions in VF(X), sup- norm.
5
29 1
VFB consists of the bounded
F(X).
denotes i t s closure with respect t o the
Let us introduce two additional closures. By F ( X ) we denote the cone consisting o f a l l
< where
y
m
Y,cp>
and
E L:
=
t Ynfn
n= 1
cp = (f,)
i s a uniformly upper bounded sequence in
F ( X ) . And by F(X) we denote the cone of a l l uniformly converging sums 1 gn n€N
, where the
g n are elements of
F v R
F(X)
Imax(f,r) I f E F, r E R3
=
def
.
was called a Dini cone if inf supX(fn)= supx inf (f,)
n a
nEN
for a l l
(f,)
+
in F(X) , where
+
(f,)
means t h a t the sequence i s
pointwise decreasing. We are going t o call a subset Y
c
X
a Dini boundary for F(X)
if
inf supX(fn)= supy inf (f,) nEN
for a l l sequences say that
p
nEN
(f,)
+
in F ( X ) .
If
has a representing measure u(f) I
I
Y
f d-r
for a l l
p T
: F(X) -ti i s linear,then we
0”
Y
f E F(X) ,
if
292
where
Representing Meagtres
must be a p o s i t i v e measure w i t h respect t o the
T
generated by the r e s t r i c t i o n s t o
Y
o f the elements o f
u- algebra F(X).
An example o f a D i n i boundary i s the s e t o f characters f o r the Gelfand transform o f an order u n i t cone (Theorem 3.3.5).
One o f our problems w i l l
be t o f i n d o u t i f a l l D i n i boundaries are o f t h a t kind. This example shows, furthermore ,that D i n i boundaries and subsets c a r r y i n g representing measures f o r a l l s t a t e s are i n t i m a t e l y connected.
3.6.3
Improved Representation Theorem t1291:
Consider
Y cX
F(X)
Y
, then
&a
-i s-a
t h e f o l l o w i n g a r e equivalent:
D i n i cone and
boundary f o r
_ .
F(X)
6
i s a D i n i boundary f o r F(X)
For every s t a t e ----
of
For every s t a t e ---
1.1
l i m sup n +m whenever Let -
F(X)
t he r-e i s a representing -measure on Y -
o f F(X) we have --
p ( f n ) I sup
fiY
( l i m sup f n ( y ) ) n+m
is an upper -
(f,)
bounded sequence i n F(X).
fn E F(X) w i t h fn I 0 m
supy(
z fn) >
n=l
-
E p(fn) > n=l
If i n f supx(
mEN
IJ
-- . m I fn) > -
n=l
.Then
03
whenever ---there i s a s t a t e
(viii)
-i s-a
boundary f o r F(X)
Y i s-a boundary for Y
Y
m
o f F(X) w i t h -
293
Boundaries
then supy( Z. f n ) > n= 1 whenever
(f,)
&2
m
sequence
fi
with
F(X)
fn
5 0
-for a l l
n E N .
Proof: ( i ) * ( i v ) : Since F(X) i s a Dini cone,formula (1) (preceding 3.6.1) shows t h a t ( * ) of 3.6.1 i s a tomatically f u l f i l l e d . Now, take a decreasing sequence (f,) + in F(X) , then from 3.6.2 we obtain supy inf f n 5 nf supx f n . n n Since the inequality in the other direction i s t r i v i a l the cone must have Y as a Dini boundary. ( i v ) 4 ( v ) : When Y i s a Dini boundary i t must be a boundary f or F(X) (exactly the same argument as in Lemma 1.3.3). B u t then we can use a routine argument: Take a s t a t e IJ of F ( X ) . Define a superlinear 6 on F ( X ) l y ( r e s t r i c t i o n s t o Y ) by:
Since Y
i s a boundary we have 6
get a s t a t e
G on F ( X ) l y with G p(f) 5
Now, since Y
I
; ( f l y ) for a l l
i s a Dini boundary of
supy t 6.
. By
the Sandwich Theorem we then
B u t t h i s means
f E F(X). F(X)
the cone F ( X ) J y must be a
i ) ) . From the Representation Theorem 2.2.1 Dini cone (Lemma 1.3.4 i i i ) we then get a representing measure on Y for n This clearly i s a representing measure for LI
.
.
( v ) * ( v i ) : i s an immediate consequence o f the Monotone Convergence
Property ( i n the version of Fatou's Lemma) applied t o the representing measures on Y .
294
Representing Measures
-
(vi)* ( v i i ) + ( v i i i ) i s t r i v i a l .
( v i i i ) ( v ) : Apply ( v i i i ) t o the constant sequence t o s e e t h a t , f o r a l l f , we have
fn
=
(f
-
supX(f))
Then take an a r b i t r a r y s t a t e u of F(X) and consider on F ( X ) l y ( r e s t r i c t i o n s t o Y ) the superlinear functional
From ( 2 ) we g e t 6
of
I supy.
Hence the Sandwich Theorem gives us a s t a t e ;I
F(X)\y j(f
Iy
) > p(f) -
for all
f E F(X).
We claim t h a t ( v i i i ) implies t h a t F(X)
i s a Dini cone. Hence
1Y
has a
representing measure (Representation Theorem) which i s then a l s o a representing measure on Y f o r u Thus ( v ) i s proved s i n c e p was a r b i t r a r y .
.
Proof o f the claim: We use an argument we already used in t h e proof of Lemma 1.3.4. Take an a r b i t r a r y sequence cpn I 0 i n F(X) and p u t m m a = i n f supy ( z cp,) and 0 = supy ( t cpn) m n= 1 n= 1
.
We need t o prove a IB s i n c e a t 0 i s t r i v i a l . I n order t o do t h a t , we take numbers N n such t h a t Nn 1h e r e g = t a I s u p Y( g n) $ a + - , w cpn. n k=l 1 1 Then we consider 0 t f n = n ( g n - a - -). Since cpn 5 0 , we o b t a i n , by n d e f i n i t i o n of a , for every m E N and every E > 0 some y, E Y with 1 f n ( y m ) 2 - -2 - E f o r a l l n 5 m. Hence n m i n f s u p y ( t f n ) t - "t 1-2 > - m m n=l n=l n
"
.
295
Boundaries
Now, ( v i i i ) g i v e s us some y
z fn(y ) >
n=1
-
E Y
.
m
i n f gn(y ) 5 p , n
This inequality,together w i t h
forall
n1 ( p + t - a ) > - w
Z
with
implies
E > O .
n=l Hence (v) for
B 2 a
.
* (i):Actually,we prove t h e statement t h a t Y F ( X ) . T h i s statement i s , f o r m a l l y , s t r o n g e r t h a n
state o f
F(X)
and l e t
a r e p r e s e n t i n g measure 1-1
( i ) . Let
be i t s r e s t r i c t i o n t o t h e subcone
F(X). Then
ji i s c l e a r l y a r e p r e s e n t i n g measure f o r
for
T
u be a
(Monotone Convergence P r o p e r t y and f o r m u l a (1) b e f o r e 3.6.1).
state o f
F(X) has a r e p r e s e n t i n g measure on
R e p r e s e n t a t i o n Theorem ( h e r e ( i i ) +. ( i ) )
Y
vergence P r o p e r t y t h a t (iii)
=S
(ii)
(ii )
*
(iv);
Y
.
So e v e r y
And e x a c t l y as i n t h e
one shows w i t h t h e Monotone Con-
F(X).
must be a D i n i boundary f o r
is trivial Let
be a r b i t r a r y i n
(f,) J.
a = i n f supx fn n
We have t o prove
p 2 a
= max(fn,a-I).
Then
'pn
i s a D i n i boundary
p
and
,for a < y,(p > E
=
F(X). P u t
supy i n f f n ' n -
w
F(X)
.
I n o r d e r t o do t h a t ,define whenever
y
E Li
since
the
'pn
a r e c o n s t i t u t i n g a sequence which i s u n i f o r m l y bounded. NOW, t h e i n e q u a l i t y coming o u t o f Simons' Convergence Lemma has t h e f o l l o w i n g f o r m supy i n f n since the
'pn
r i g h t hand s i d e
'pn
2
i n f supx n
'pn
a r e decreasing. The l e f t hand s i d e i s L a
, hence
p
L a
.
5 max(p,a-I)
and t h e
296
Represenring Measures
( v ) * ( i i i ) : Assume t h a t every s t a t e of F(X) has a representing measure on Y . We then know t h a t every s t a t e of V F has a representing measure
on Y (2.2.1 and 1.3.4). We claim that every s t a t e of has a representing measure. Then ( i i i ) i s an immediate consequence of 2.2.1 and 1.3.3 (applied t o Proof of the claim: Take an arbitrary s t a t e i; of VFB,
q).
take a maximal s t a t e l.~ 2 i; , r e s t r i c t i t t o VFB and extend i t t o a maximal s t a t e i,i on VF Then the corresponding representing measure for G on Y i s clearly a representing measure for l.~ Since both p and
.
I vFB .
the integral are sup-norm continuous we immediately have a representing measure on Y for IJ 0
.
3.6.4
Remark:
I t i s quite obvious that whenever one of the conditions of Theorem 3.6.3 i s replaced by the corresponding condition for VF, VFB or VFB (instead
-
of
F(X))
then we obtain an equivalent condition. The reasons for that are
i)
the Dini property remains valid i f we replace F(X) by VF,
ii)
any element of i n VFB ,
iii)
states of
VF
i s the infimum of a decreasing sequence
VFB are sup-norm continuous and therefore they
(as we1 1 as integrals with respect t o probabil i t y measures) have unique extensions t o the sup-norm closure.
Now some examples. 3.6.5
Examples:
( i ) Let X be a pseudo-compact topological space, i.e. X i s completely regular such that every bounded continous real valued function attains i t s maximum on X (see 1921 or [140]). Then every s t a t e on the (bounded) continuous real-valued functions has a Baire representing measure on X. (One should observethat every continuous R- valued function on a pseudocompact space i s automatically bounded 1320, p.341). This example i s due t o G1 icksberg [ 1411 see a1 so [3171 ,[1181.
297
Boundaries
(ii)
X
Let
be a convex closed and bounded ( n o t necassarily compact)
subset o f a l o c a l l y convex t o p o l o g i c a l Hausdorff vector space and consider t h e continuous, r e a l valued bounded and convex functions ConvB(X) on X. Then, i f Y c X i s a boundary f o r ConvB(X),there i s , f o r every, x E X a representing measure on (iii)
Let
Y.
be a convex bounded p o l i s h subset o f a l o c a l l y convex t o -
X
pological Hausdorff vector space
and consider the weakly
E
r e a l valued,bounded and convex f u n c t i o n s
*
t p Conv ~
o(E¶E )
(X)
* (E) Conv u(E,E )
a t t a i n s i t s maximum on X
X
then
on
continuous,
X
.
I f every
i s weakly
compact. Proof: ( i ) and ( i i ) are,obviously,immediate
consequences o f Theorem 3.6.4.
Only t h e p r o o f o f ( i i i ) needs a d d i t i o n a l remarks. Let
F(X)
be t h e r e s t r i c t i o n s o f
. All
n o f F(X)
space
(E*+R)
to
we have t o prove i s
and consider the s t a t e
X
n
since
= X
x E X with
i n i t s canonical topology. Clearly, here we i d e n t i f y a l l 6, E n
, where
extend
p
t o a state o f
representing measure
Convo(E,E*) on
T
generated by Convu(E,E*) to
Now, l e t p E n ¶ t h e n we can
= f ( x ) , f E F(X).
6,(f)
(X)
and from Theorem 3.6.4 we g e t a
( w i t h respect t o
X
n i s compact
(X) ,or,equivalently,by
1
, the
u- algebra
the r e s t r i c t i o n s o f
X ) . By t h e Hahn-Banach Theoremyand t h e f a c t t h a t
E*
has a countable
X
base ( s i n c e i t i s polish),we obtain,that every closed convex subset o f belongs t o
.
1
Hence every open convex subset belongs t o t and 1
must t h e r e f o r e be t h e ure
( s e c t i o n 2.6.2). T =
where
xn
2 0
convex h u l l s
c X
T
T
i s a t i g h t meas-
as
Tn
1 An = 1 and where
Kn, Kn
u- algebra o f Bore1 sets. Thus
This means,that we can w r i t e
n=l
with
compact supports
X
Since
o f the
X
Kn
T,
are p r o b a b i l i t y measures w i t h
i s complete and contains t h e closed
, these
closed convex h u l l s must be
298
Representing Measures
E**
a g a i n compact. A p p l i c a t i o n o f t h e Hahn-Banach Theorem i n r e s p e c t t o t h e weak*topology) y i e l d s t h a t t h e s t a t e by t h e i n t e g r a l w i t h r e s p e c t t o
i s an element
T,
(with
on
11,
xn
F(X)
-
.
[2501
( T h i s i s e x a c t l y t h e argument which was used i n t h e p r o o f o f P r o p o s i t i o n 1 . 2 ) . Hence m
lJ =
where t h e since
p
n=l
xn E X.
AnXn
3
The s e t
was a r b i t r a r y .
X
i s complete, thus
11
E X
and
X = R
0
L e t us now i n v e s t i g a t e under what k i n d o f c o n d i t i o n s t h e Choquet boundary i s a minimal boundary. L e t F ( X ) , F(X) t h a t a subset
Y c X
3.6.6
be as b e f o r e . R e c a l l
o f a t o p o l o g i c a l space i s c a l l e d L i n d e l o f i f e v e r y
Y
open c o v e r i n g o f
and
c o n t a i n s a c o u n t a b l e sub-covering.
Theorem:
L e t X be a compact nonempty H a u s d o r f f --space and l e t F(X) be a cone o f uppersemicontinuous 6- valued f u n c t i o n s which ~contains a l l constant t h e f o l l o w i n g asserfunctions and separates t h e p o i n t s o f X . Consider t i o_ ns f or a -
subset
Y c X
.
For t h e Choquet boundary CH(X) ( --w i t h respect t o (i) --Ch(X) c Y (ii)
Y
f o r F(X) -i s-a boundary -
(iii)Every s t a t e of (iv)
F(X)
F ( X ) ) we have --
(o r equivalently f o r F(X)
has measure on - a r e p r e s e n t i n g ~-
or -
Y
s u p X ( i n f fn) = supy ( i n f fn) f o r every decreasing sequence n a nEN
i n F(X) Z
(v)
Ch(X)
(vi)
Ch(X) c Z
-We have:
c
(i)
4
f o r every --
Fo- subset
-
Z
c
-
X
with -
Y cZ
f o r e v e r y L i n d e l o f set Z c X with Y --
(ii) o (iii)
(iv)
4
(v)
(vi).
c
q)
Z.
(f,)
299
Boundaries
Proof: ( i ) + (ii):I t i s c l e a r from t h e d e f i n i t i o n s o f l i k e w i s e t h e Choquet boundary o f
X
Ch(X)
that
F(X),
w i t h respect t o
Ch(X)
F(X) o r
is
5
Hence (ii)f o l l o w s i n view o f C o r o l l a r y 3.2.4. ( i v ) i n c o n n e c t i o n w i t h D i n i ' s Lemma and Lemma 1.3.4, t e l l s us t h a t o ( i v ) f o l l o w from Theorem 3.6.3. i s a D i n i cone. Thus ( i i ) * (iii)
-
(iv)
(v):
Assume t h e r e i s
x B Z.
,
and a l l
U Kn, Kn 5 Kn+l n= 1
Z =
We have
x B Kn
Hence
x E Ch(X) such t h a t
m
for all
for all
n E N
IY
a r e compact.
Kn
n.
By Theorem 3.2.6 t h e r e a r e
'pn
E F(X)
such t h a t
3 2 supK (wn) n Put
F(X)
'pn
and
I0
for all
.
n E N
n
1 '4( ,then we have i n c o n t r a d i c t i o n t o ( i v ) : k=l
fn =
m
supx i n f fn 2 n
-1
Z
n=l
n
> - 3 2 supy i n f fn n
.
Hence ( v ) must be t r u e . (v)
-
(vi):
x ECh(X) with
x
Let Z c X
B U(z)
such t h a t
x
be a L i n d e l o f s e t such t h a t
B
f o r each
Z
Y
z E Z.
zn E Z
Fo- s e t c o n t a i n i n g
course , c o n t r a d i c t s ( v ) . ( v i ) + v) i s t r i v i a l since sets.
0
.
Z
I f t h e r e were
U(z)
By t h e L i n d e l o f p r o p e r t y we c o u l d s e l e c t m
a c o u n t a b l e number o f elements s e t wou d be an
c
we would o b t a i n open neighbourhoods
Y
with
Z
c
U n=l
n(Z,) . The
but not containing
Fo- subsets o f t h e compact
x X
latter
. This,
of
are Lindelof
300
Representing Measures
(ii) ( i ) i n the preceding theorem does n o t h o l d i n general. The f o l l o wing example, due t o Simons [ 297,p. 7061 ,may i l l u s t r a t e t h i s . This example shows furthermore t h a t t h e r e are i n f a c t boundaries which are d i s j o i n t from the Choquet boundary.
3.6.7
Example:
.e""(n) = { f l f : R
L e t n be an uncountable s e t and consider Endow
k?(a)
-,R , f
bounded).
w i t h the weak-star topology w i t h respect t o
k?(n) = I d r p : n -,R such t h a t
E I v ( w ) l < -1.
That i s the coarsest
WE61
topology such t h a t a l l t h e functions
r(n) 3 f
+
1 cp(W)f(W), l p E e'(n)
=
WEQ
a r e continuous.
i s then compact. Consider F = Convc(X) , the continuous convex f u n c t i o n s on X I t i s easy t o see t h a t the extreme p o i n t s o f X are
.
ex(X) = { g E X I g ( W ) =
*
1 for all
w E
nl
.
And t h i s i s o f course the Choquet boundary Ch(X) ( w i t h respect t o F = Convc(X)). F i x x E X , then we f i n d , f o r every h E Convc(X) ( i n f a c t f o r every continuous f u n c t i o n ) , countably many
%
E
1
l (n) such t h a t f o r every y
E X
( J u s t consider the compact subsets For each
n we f i n d f i n i t e l y many
< Ij yx > '4( i n
=
-
m
else
- m
defines an R- valued monotone linear functional which has by ( i i ) a representing measure T . And application of the Monotone Convergence Property ( i . e . the f a c t t h a t the integral d.r i s u- monotone) X yields easily
-
supx inf f n n
2
inf n
p(fn)
.
Hence F(X) must be a Dini cone since the decreasing sequence was chosen arbitrarily 0
.
This vector valued representation theorem has some well known results as consequences. First of a l l an analogue of the Riesz Representation Theorem due t o J.D. Maitland Wright: 4.3 Corollary [340]: Consider C ( K ) , the continuous real valued functions on some compact space K -and l e t R be an order complete vector l a t t i c e . Then every monotone : C ( K ) -,R has a n R- valued representing measure (with respect linear
-Proof: C(K)
i s a Dini cone and
'C( K )
=
Bo(K)
(Baire s e t s ) .
0
312
Note t h a t
Representing Measures
K
i s n o t required t o be Hausdorff.
One should observe that from t his theorem one easily recovers the existence of the Loomis-Sikorski homomorphism (Theorem I 2.3.3) by p u t t i n g C ( K ) = C(S) = R, S extremally disconnected. Another consequence i s a vector valued Choquet Theorem:
4.4 Corollary [1341: K be-a compact --______convex subset of a locally convex Hausdorff vector space, A ( K ) the realvalued continuous affine functions on K and let R be an order complete vector l a t t i c e . Then for every monotone linear ---~ ---
Let -
I.I
: A(K)
-+
R
there e x i s t s an R- -~ valued measure T (with - respect -t o the --ex(K) n Bo(K)) ---on the extreme points ex(K) -such t h a t
u- algebra
Proof:
A ( K ) I ex( K ) i s isometrically order isomorphic t o A ( K ) ( c . f . Lemma 3.3.1) 0 and furthermore i t i s a Dini cone (Lemma 3.3.2). For weakly u- dist ri buti ve R t h i s re sult was already obtained by G.F. Vincent-Smith [3191.
SECTION 11.5 GENERALIZED HEWITT - NACHBIN SPACES
In t h i s chapter we t r e a t y a s a n application of the preceding methods, some topological aspects of the theory of continuous functions in a generalized way.Using the notions of characters and Dini cones,which we have discussed in preceding sections,it i s possible t o transfer classical notions such as Stone - Czech - compactification, realcompactifi cation and pseudocompactness from the situation of a completely regular topological space X t o a general s e t where a given order unit cone F ,consisting of bounded functions, takes overthe place of the continuous, or bounded continuous,functions on X in the classical situation. I n the f i r s t section we introduce the notion of F- compactification and F- realcompactification. We show t h a t , indeed, these definitions include the classical definitions of Stone - Czech - and realcompactification. Our
generalizations allow us t o describe these compactifications easily e.g. by means of decomposition properties of s t a t e s or integral representations of s t a t e s and characters. So, many theorems in the following a r e j u s t recollections of what we have proved so f a r , which, in t h i s context, i s seen from a different angle and obtains a new meaning. There i s one new aspect: Here we also discuss f i l t e r convergence properties. Finally, we consider some applications. Forexample, a generalization of G1 icksberg's integral representation theorem ( o r Alexandrov - G1 icksberg theorem [317]) ,for pseudocompact spaces which i s , ultimately, contained already in Theorem I1 2 . 2 . 1 and Theorem I1 3.3.6 and, therefore, sheds a new l i g h t on these theorems. 313
314
5.1
Representing Measures
BASIC DEFINITIONS AND THEIR MEANING IN THE CLASSICAL SITUATION
Throughout t h i s chapter l e t X a nonempty --s e t and l e t F be an order unit cone consisting of bounded realvalued functions on X , which separate -always denotes the s t a t e space of F . the points of X . R -----2 . 6 . 3 ) t o general
We want t o extend the notion of Dini continuity (see s t a t e s u on F . 5.1.1 Let
Definition: p
be a s t a t e on F.
p
i s called Dini continuous
whenever f n E F i s pointwise decreasing such t h a t
if
inf f n n
inf u ( f n ) I 0 n I 0
.
This definition indeed includes the former definition of Dini continuity since we have 5.1.2 i)
Lemma: p
i s Dini continuous ---i f and only i f -inf p ( f n ) n
5
supX(inf f n ) n
for a l l pointwise decreasing sequences f n
_.-
ii)
If
F i s max-stable,then p p ( f n ) -,0 for a l l sequences
E F.
i s Dini continuous -i f and
fn E
on’ly if F pointwise decreasing -t o zero.
Proof: i ) The sufficiency i s obvious. To prove the necessity assume inf p ( f n ) > a > supX(inf f n ) for some a E R and some decreasing n n sequence f n E F. Then, with g n = f n - a , we have, g n i s decreasing, inf g n I n
0
but
inf n
p(g,)
> 0
, a contradiction.
i i ) We only need t o prove the sufficiency of t h i s condition. Indeed, l e t f n E F be a r b i t r a r i l y decreasing such t h a t inf f n I 0 . n
315
Generalized Hewitt-NachbinSpaces
Then p(max(f,,O))
+
0 by assumption.
A Dini continuous character of F
5.1.3
i)
is called Dini character.
Definition:
Put p XF v XF p XF
= {p
E n
= Ip E
n
I I
p
character) and
p
Dini character3
.
is called the F- compactification of X ,
v
XF is called the
F- realcompactification (or F- Hewitt-Nachbin-completion). Moreover, X is defined to be F- compact, if p XF c X . Similarly, X is defined to be F- realcompact (and sometimes called F- Hewitt-Nachbinspace) if v XF c X . ii)
X is called F- pseudocompact if every element of the sup-norm closure VF of VF attains its maximum on X . (Recall VF = {max(fl ,..., fn) I fl,...,fn E F, n E N1).
-
Rema rk :
If we take F to be the continuous, bounded realvalued functions'on a completely regular topological space X , denoted by CB(X) , then p XF is just the classical Stone-Czech-compactification. Indeed, recall that the Stone-Czech-compactification p X of a completely regular space X is the set of all multiplicative nonzero linear functionals on CB(X) endowed with the weak-star topology. This means that p X coincides with the set of all lattice homomorphisms S on CB(X) (see 1 . 2 . 2 ) . Indeed, by the Kakutani-Krein-Stone-Yosida Theorem the sup-norm closed vector lattice CB(X) is isometrically isomorphic to C(S). So any multiplicative nonzero linear functional p on CB(X) corresponds to a multiplicative linear functional on C(S) which must be the Dirac functional at some element in S . Therefore p E S . Conversely, by the same theorem, every p E S is multiplicative on CB(X). Clearly, S is the extreme point set by Theorem of the state space of CB(X). Therefore, B X = B X CB(X) I 2.10.6 ).
316
Represenring Measures
S i m i l a r l y , the realcompactification o f a completely regular Hausdorff space
X,
v X
a l s on CB(X) on
i s t h e s e t o f a l l m u l t i p l i c a t i v e nonzero l i n e a r f u n c t i o n which can be extended t o m u l t i p l i c a t i v e l i n e a r f u n c t i o n a l s
C(X), t h e space o f a l l continuous r e a l v a l u e d f u n c t i o n s on
general we have,
*
vX c
BX
X.
In
(examples see below). Note, t h a t if
E p X w i t h p ( f ) E f ( X ) f o r a l l f E CB(X) t h e n p can be represented by a zero-one measure m on X , i . e . m(A) E 1 0 , l ) f o r a l l Bore1 s e t s A . T h i s i s a consequence o f z e r o - s e t f i l t e r t h e o r y ([1401).Thereforey i n t h i s case, p can be extended t o a m u l t i p l i c a t i v e l i n e a r f u n c t i o n a l on C(X), thus p E v X . Conversely, i t i s w e l l known t h a t i f p i s a m u l t i p
p l i c a t i v e l i n e a r f u n c t i o n a l on
C(X)
then
p ( f ) E f(X)
f E CB(X)
for all
(see Theorem 5.3.1 below). As general r e f e r e n c e s f o r t h e c l a s s i c a l s i t u a t i o n we r e f e r t o [ 1 4 0 ] ,[3201and [3241. 5.1.4 Let X true: -
Proposition:
-be_ a
completely r e g u l a r H a u s d o r f f space. Then t h e f o l l o w i n g
are
ii)
Proof:
i) f o l l o w s d i r e c t l y f r o m t h e preceding remarks. ii)Consider
p
E v X cB(x)
c pXcB(x) = p X
and a l a t t i c e cone homomorphism on
. Hence
p
i s multiplicative
CB(X). We c l a i m t h a t
p(f) E f(X)
a l l f E CB(X) from which, b y t h e p r e c e d i n g remarks, we o b t a i n and t h e n v X CB(X) *
p
for
E vX
f E CB(X) a r b i t r a r i l y and assume p ( f ) B f ( X ) . Consider t h e
So t a k e sequence
fn = max(1 - n ( f - p ( f ) ) 2 ,0)
which i s t h e n p o i n t w i s e decreasing t o
zero. Hence we o b t a i n t h e c o n t r a d i c t i o n
since
p
was D i n i continuous. Here we used t h e f a c t t h a t
p
i s a multi-
317
Generalized Hewitt-Machbin Spaces
plicative l a t t i c e cone homomorphism on C B ( X ) . Therefore p ( f ) E f(X). Conversely, l e t
p E
uX
. Consider
a sequence
decreasing t o zero,and assume inf u ( f n ) n Kn = I T E B X
I
T ( f n ) 5 pa 1.
=
f n E C B ( X ) , pointwise
a > 0
i s compact and
Kn
consists of the Dirac functionals
bX
at x
E X.
. ?
Consider c
U Kn nEN
Furthermore
, where p
B
U Kn. nEN
Therefore, by Urysohn'slema, we find a function g E C ( B X ) with g(p) > g ( T ) for a l l T E U Kn Indeed, consider g n E C ( p X ) with nEN
.
m
.
1 9,. ' 'nIKn = o for a l l n , a n d p u t 9 = n x= l 2n Consequently, with g ( x ) = b x ( g ) for a l l x E X , we have
0
5
gn
~ ( 6 =)
1
, gn(ll)
g(p)
B g(2)
5
fact t h a t
p
= 1
g(X).
By the preceding remarks t h i s contradicts the i s extendable t o a multiplicative linear functional on C ( X ) . =
The notion of F- pseudocompactness clearly coincides with the classical one i f F= C B ( X ) . Recall t h a t there are indeed non-compact completely regular Hausdorff spaces X which are nevertheless C B ( X ) - pseudocompact. An easy example ([1401) i s X = {a I a ordinal number with a < w13 endowed with the order topology; here
w1
i s the f i r s t uncountable ordinal. The pseudo-
compactness here follows easily from the fact t h a t every sequence in X has a cluster point. B u t for general F we have already more examples of F- pseudocompact s e t s a t hand. I t follows e.g. from 1.3.3 t h a t X i s F- pseudocompact whenever F i s a sup-norm closed max-stable Dini cone. We have already seen that F being a Dini cone i s equivalent t o X being F- pseudocompact (Improved Representation Theorem 3.6.3 with Y = X ) . Let K be a compact convex subset of a locally convex Hausdorff space and denote by ex(K) i t s extreme p o i n t s e t . Another interpretation of Bauer's Maximum Principle (11. 3.1.6) i s t h a t ex K i s A(K),exK-and ConvB(K)lex Kpseudocompact. Here A ( K ) = {f:K + R I f affine, continuous), ConvB(K) = tf:K + R [ f convex, uppersemicontinuous, boundedl.This i s a n easy consequence o f Theorem 5 . 4 . 1 below.
fact
318
Representing Measures
5.2 THE F- COMPACTIFICATION We want t o s t u d y f i l t e r p r o p e r t i e s o f t h e s e t s s u p p o r t i n g c e r t a i n l i n e a r f u n c t i o n a l s . L e t us f i r s t i n t r o d u c e d i f f e r e n t f a m i l i e s o f subsets o f
X
which we s h a l l discuss i n t h i s c o n t e x t . Let
be a s t a t e o f
p
subsets
Z(p)
i s defined t o be t h e f a m i l y o f a l l
such t h a t f o r e v e r y a < p < 0
Z c X
, and
f I0
F. Then
p(f)
2 p > a 2 sup
there i s
(X L Z ) ( 0
i s c a l l e d t h e s e t o f s t r o n g domination f o r
Z(p)
f E F with
p
.
Furthermore, we c o n s i d e r
x I
D(p) =
CY c
N(u)
{Y c X
I
supy]
P 5
p $ SUP
(XLY)}
U(P) i s t h e s e t of domination, for
u
.
*
i s c a l l e d t h e complementary s e t
N(p)
Recall, by Face(p) we mean t h e c o l l e c t i o n o f a l l t h o s e s t a t e s such t h a t t h e r e a r e 0 < l,l I x v + ( l - A ) v o . We c l e a r l y have
i f and only i f
I 1 and a s t a t e
Face(v) c Face(p)
Furthermore, o b v i o u s l y , F
x
p
whenever
of
v0
v E Face(p)
Let i)
n be a r b i t r a r y . --Then we have:
Z(u) c N ( I J ) n D ( v )
ii) Z ( P ) c Z ( V ) iii)
F
R
of
Face(p) = { p l (see Remarks 2.5.2 and 2.5.8).
Lemma: p E
of
(see 11. 2.5.2).
i s an extreme p o i n t o f t h e s t a t e space
A t f i r s t we prove some t e c h n i c a l f a c t s concerning
5.2.1
u
F with
for all
Z1 n Z2 E N(v)
v E Face(u)
for all
Z1,Z2 E
Z(p)
P ( u ) , N(u) and Z(v).
319
Generalized Hewitt-NachbinSpaces
Proof: i)
follows from the definition. We claim that
Z ( p ) c N(p)
whenever Z E
This proves
Z(p).
Indeed, l e t Z E there are s t a t e s
Z(p) c D ( p ) .
By the Finite Decomposition Theorem I , p 2 and 0 I x I 1 with and
(1-x)p2
p I A p1 t
According to the definition of Z(p),with a obtain f E F , f I 0 such t h a t
Hence, in view of
Therefore
0 I
p I
i i ) Let v
A
Hence iii)
I
A p
- n
2
= - 1
we
sup ( X \ Z ) ( f )
=
. Consider f E F, f
x
z,))
=
I
(l-X)vo
0 , with
p > a t sup ( X \ Z ) ( f )
Z
E
Z(V).
we would have
N(p)
"'Pyl
w t
some real numbers a < B
( f ) which implies
, b u t Z1 n Z2 6
sup ( x \ ( z l n
since n was arbit.rary.
0
Then we have N 5 X
(l-h)vo(f) 2 p(f) t (X\Z)
n.
D(u).
Z(p).
and there i s
v ( f ) 2 p > a t sup If
(f)
(X\Z)
and 0
1, p
,
Face(u) and Z E
E
v(f) t
x
A
=
(l-A)supZ(f) 2 p ( f ) t
supz and hence Z
f o r some v0 E n
Then a
f I 0
1 s p(f) 5
This implies
+
(f)
A sup(x\z)
-
1.4.4
Z(p). pl
supz
p I
u
y2
-
0.
Representing Measures
320
with
Y i = XxZi,
we would o b t a i n
0 Ix I1 w i t h
p I
Then c o n s i d e r
By t h e F i n i t e Decomposition Theorem 1.1.4.4
i=1,2.
x
a,P
supy with
1
+ (l-x)supy
2
.
a c 2 p < p < 0 and
flyf2 E F, fl I 0, f 2 I 0,
such t h a t p ( f i ) 1 p > a 2 sup
'i
(fi)
,i=
1,2
.
We a r r i v e a t t h e c o n t r a d i c t i o n 2B I
p(fl
+ f 2 ) I x supy (fl) 1
+ (l-x)supy (f2) 2
5 a
.
Z1 n Z2 E N ( p )
Hence
NOW, we f o r m u l a t e t h e main c h a r a c t e r i z a t i o n o f t h e F- c o m p a c t i f i c a t i o n which i s , p a r t i a l l y , a r e c o l l e c t i o n o f former theorems.
5.2.2
Theorem:
The f o l l o w i n g are equivalent:
_.
i)
P
ii)
p
E P XF
i s a character of F
iii) p has t o VF ---and f o r every f i n i t e - -a unique dominated e x t e n s i o n c o v e r i n g Y1,...,Yn o f X t h e r e i s some k I n such t h a t ----
iv)
p
can be extended t o-a c h a r a c t e r o f VF
v)
p
can be extended ---t o a l a t t i c e cone homomorphism VF --
maximal s t a t e -vi)
p
vii)
P
viii) p
+R
being a
E ex n can be extended t o- an nt_ o f- t h e s t a t e space aVF of VF -- _extreme _ _ _ p_o i_ i s maximal and Z(P) = N ( p )
321
GeneralizedHewitr-Nachbin Spaces
If
ix)
X
Z(p)
&
identified, b-y evaluation, with a subset of n , then i s a f i l t e r base of a _ f i l_ t e_ r -in n converging 2 p .
Proof: i)
o
ii)
i s j u s t the definition of i s Theorem I
i i ) o iv) ii)
o
-
v)
.
2.10.10 i )
i s Corollary I 2.10.4 together with Theorem I
2.10.10.
i s Theorem I 2.10.6
i i ) o vi) ii)
D XF
v i i ) i s Theorem I
2.10.6 together with Theorem I
2.10.10.
i i ) =$ i i i ) : The uniqueness of the dominated extension follows d i r e c t l y from the definition of a character (Definition I 2.10.1). Let Y 1 , . . . , Y n be a f i n i t e covering of X . From the Finite Decomposition Theorem I
n z
i=l
A. = 1
1.4.4 we obtain s t a t e s n 1 such t h a t p I x. i=l
Since, by v i ) , p 5 supy . k
p
E ex n
pi xipi
, there i s
k
and real numbers and In
p . I supy 1
i
such t h a t
xi
L
0
for all
p =
with i
.
u k , hence
i i i ) =$ v ) : Let fi be the unique dominated extension t o VF . Then fi clearly i s maximal. Another consequence of i i i ) , the uniqueness, the covering property and the Dominating Extension Theorem I 1.3.1 i s t h a t ;5 supy f o r some k I n , whenever Y1, ...,Y n i s a f i n i t e covering of k X . Finally, because p can be extended t o the vector space plF = F - F which i s a subspace of VF-VF , which means there i s an extension
p > a
and p u t
t (1-x)supx (X\Z) would o b t a i n s t a t e s u1,p2 w i t h p I x p l t ( l - x ) p 2
have
u1 5
p I
sup
SUP(^,^),
p 2 I supx
x
p a-l s i n c e by t h e Sum Theorem I
Indeed, f i x
Z E Z(p).
. T h i s would
imply
pl
= p
=
,
f o r some
Then
f
N(p).
1.4.1 we
because
p
p I sup
was (X\Z)
Thus we have
g E F. Put
i s w e l l defined,
u(f) > Therefore, viii) base.
Z E
cannot
where
assumed t o be an extreme p o i n t . Hence we would a r r i v e a t which c o n t r a d i c t s
. We
> a
Z E Z(p)
*
ix): v i i i ) L e t G 1 Z(u)
From Z(p) =
N(p)
=
f E F, f I 0, and
sup ( X \ Z ) ( f )
and hence
N(p)
*
= Z(u).
and Lemma 5.2.1 i i i ) show t h a t Z(p) i s a f i l t e r be an u l t r a f i l t e r converging t o some i; E n
we d e r i v e
.
GeneralizedHewitt-Nachbin Spaces
p I
supy f o r a l l
323
Y E G
s i n c e otherwise t h e r e i s Y E G with p Ir s u p y and hence X Y E N ( p ) = Z ( p ) c G which c o n t r a d i c t s t h e f i l t e r p r o p e r t y o f G . Therefore p 5 , and , by maximal i t y of p , p = 11 . Hence Z(p) has only one accumulation p o i n t , namely p . That i s , Z ( p ) converges t o p s i n c e n i s compact. i x ) * vi):
Let u Z(U)
Hence
v I SUPy
E
Face(p). Then we have, by Lemma 5.2.1,
= Z(V)
for all
v I p
c Wv).
Y E
Z(p)
for all
and thus
v 5 lim Z ( p ) = p
. We obtain
v E Face(p)
and t h e r e f o r e , by d e f i n i t i o n of Face(p), v = p which means Hence p E ex n . This completes t h e proof of Theorem 5.2.2.
Face(u)
=
{PI.
324
5.3
RepresentingMeasures
THE F- REALCOMPACTI FICATION
I n t h i s section we turn t o various characterizations of the F- realcompactification u XF Recall t h a t ZF i s the smallest u- algebra on X such
.
that a l l elements in F are measurable. Let measure m i s said t o represent p i f p ( f )
p
5
E n. Recall, a IF f d m for a l l f E F. X
If equality holds for a l l f E F , then m i s called s t r i c t l y representing. Let E be the sup-norm closure of VF - VF. Note, that every s t a t e of VF has a unique extension t o a s t a t e of VF - VF a n d , furthermore, t o a s t a t e of E . (Recall, VF consists of bounded functions on X ) . By definition, a l l states are continuous with respect t o the sup-norm. Let p be a character of F . Then by Theorem 5.2.2 and the preceding remark, p has a unique extension t o a character of E Occasionally we denote this extension again by p , because no confusion can a r i s e . Since E i s a vector l a t t i c e , p on E necessarily i s a l a t t i c e homomorphism (see I 2 . 2 . )
.
Let Y c X and assume 1-1 I supy on dominated extension of the character we obtain p 5 supy on E .
F. I n view of the uniqueness of the p
t o E and the Sandwich Theorem
5.3.1 Theorem 11273: Let be-a character o f F. -Then the following --a r e equivalent: - p -
i)
p
i s Oini continuous
ii)
For every sequence f n E F with - fn there i s x E X with r f n ( x ) > -nEN
iii)
p
iv)
p
has -
a ( s t r i c t l y ) representing
I 0
and
1 p(fn) > nEN
-
OJ
- O D
ZF-
measure
has fi ( s t r i c t l y ) representing zF- measure m which i s a 10,l)- measure, i . e . m ( A ) = 1 or m ( A ) = 0 -for a l l A E IF
325
Generalized Hewitt-Nachbin Spaces
with -
Far
vi)
p I supy
Z
any sequence no nemp ty
the i n t e r s e c t i o n
E Z(p)
n
fl
Czn I
f E E
vii)
u(f) E f(X)
viii)
F o r e v e r y f E E -t h e r e i s x E X with u ( f ) If ( x ) -u , as d c h a r a c t e r o f E , -i s D i n i continuous.
ix)
n E NI
whenever
Proof:
i)=9 ii)i s an easy consequence o f t h e D i n i c o n t i n u i t y o f u 2.5.7 iii) ii)t o g e t h e r w i t h 2.5.8. ii)a iii) i s Theorem (Recall,
iii)* i v ) : If
p
as a c h a r a c t e r i s an extreme p o i n t o f Let
m
and
R
. i s maximal).
p
be t h e r e p r e s e n t i n g measure o f iii)and l e t
A E IF.
0 < m ( A ) < 1 t h e n we have
m = m ( A ) ml where
ml
+
(1-m(A))
i s t h e measure w i t h
m2
ml(B)
and
=
m2
i s d e f i n e d by
m(A) m2(B) = p
iv)
m((x'A)n B, m(X A)
i s an extreme p o i n t o f 4
v):
B E IF . This contradicts the f a t t that
for all
By Theorem
t h e s t a t e space o f 2.5.7.
E = VF
we know t h a t
- VF
.
has t h e decomposition
p
m
property, i . e . there are such t h a t
A
with
n 2 0
= 1
t A, n=l
and s t a t e s
m
p I
Since
p
t xn n=l
and
pn
i s an extreme p o i n t of R p
=
un
I supy
n
pn
5 supy
we have f o r some
n
.
n
for all
n.
pn
of
F
326
v)
Representing Measures
c;)
f o l l o w s , by c o n s i d e r i n g complements, d i r e c t l y from t h e d e f i n i t i o n
vi)
of
N(p)
v)
* vii):
and t h e f a c t t h a t Put
Yn = { x E X
I f(x)
Zn = I x E X
N(p) =
I p(f)>. I f
t
supy
n
2
1
p(f)
(Theorem 5.2.2). p(f)3
and
B f(X)
then
t
X =
U (YnU Zn). nEN
n such t h a t
Hence, by v), t h e r e i s some p 5
I f(x)
Z(U)
or
p I supz
n
.
The second i n e q u a l i t y l e a d s t o t h e c o n t r a d i c t i o n supz ( f ) 4- n 5 p ( f ) I supz ( f ) n n 1
Hence we have
LI I sup
maximality o f
p
p(f)
1
vii) viii)
1
there i s
Hence Thus
is
Y-,
monotone, 1.e.
ix):
Let
a
with
m I: n=l
fn E E
f 2
f
+
on
p(f)
u(f) E f(X).
+
be p o i n t w i s e decreasing t o zero and assume,
< a = i n f v(fn) n
1 min(fn,a)
E E
2"
fn(x) 2 a f o r a l l p(fn)
, since
is trivial.
o We have
and consequently, by t h e Sandwich Theorem and
'n
u ( f ) , a c o n t r a d i c t i o n . Thus
.t
viii)
*
,p
*
0
n
,a
.
i x ) * i) i s t r i v i a l .
D
. and by v i i i ) t h e r e i s
contradiction since
x E X
with
i n f f, = 0. n
Y,
Generalized Hewin-NachbinSpaces
327
Example: Theorem 5.3.1 and Proposition 5.1.4 provide us with examples for realcompact Hausdorff spaces X which are n o t compact. E . g . take X = N . Then the multiplicative li ne a r functionals on C ( X ) differ ent from zero ( i . e . the Dini characters on C B ( X ) ) a re , according t o Theorem 5.3.1, j u s t the characters of C B ( X ) which a re s t r i c t l y represented by a { O , l l - measure on N . These measures, of course, a re the Dirac measures of the elements in N . Hence , we obtain the well-known f a c t N = v N . Let us return t o Theorem 5.3.1. As remarked previously every character of F can be extended uniquely t o a character of E = sup-norm closure of VF - VF. Furthermore, by Theorem 5.3.1 , i f p i s a character of F then p i s a Dini character i f and only i f i t s unique extension t o E i s a Dini character of E . On the other hand, there may e xi st a Dini continuous character T of E such t h a t T i s n o t maximal while T i s always maximal on the vector space
E.
I e. we have
v XE
3
v
XF
in general
(here we have identi-
fied the Din characters of F with t h e i r unique extensions t o For the same reason we obtain B XE 2 B XF in general. We can identify the elements of
E).
X , by evaluation, with elements of
Hence, as a subset of the compact Hausdorff space p XE
the s e t X
P XE. is
completely regular. Nevertheless, the classical realcompactification v X of X i s d i f f ere nt from w X E . Indeed, by Theorem I 2.2.3 there i s a n order unit isomorphism E C(B XE).
+
C ( p X E ) , hence we may identify E
with
C ( p X E ) , in general, i s only isometrically l a t t i c e isomorphic t o
a subspace of
C B ( X ) . Hence v x = w xC B ( X )
2
v
XE
in general.
Representing Measures
328
5.4
F- PSEUDOCOMPACTNESS
I n this section we give characterizations of F- pseudocompact sets in terms of Dini-cones and representing measures. Then we study the connection between F- pseudo, F- real- and F- compactness. As special cases we obtain several cl assi cal resul ts concerning pseudocompactness and the real -and Stone-Czech- compactification of completely regular Hausdorff spaces.
5.4.1
Theorem [1271:
The following are equivalent:
g
F- pseudocompact
i)
X
ii)
p XF = v XF
iii) F
& 5 -Dini cone
xF- ~measure on
Every s t a t e o f F has - a representing
iv)
X
.
Recall, F i s a Dini cone i f and only i f supX(inf f n ) n
=
for every decreasing sequence
inf supX(fn) n (f,)
in F
.
Proof: i i i ) co iv) i s Theorem 2.2.1. iv) + i i ) follows a t once from Theorem 5.3.1 i i i ) i ) . i i ) i ) : According t o o u r remark in the introduction o f section 5.3, everystate o f VF has a unique extension t o a s t a t e of Therefore we can identify P,
.
Since we know by Theorem 5.3.1 t h a t every character continuous i f and only if i t s unique extension t o E we obtain with the obvious identifications v XF = v XVF -- u x w
.
p
of F i s Dini i s Dini continuous
329
Generalized Hewitt-Nachbin Spaces
Now, by assumption, 3.3.5
there i s
v
X z
=
x~ .
@
= VXE
LI E BXvT
Consider
by Theorem
with
= vXF
By Theorem 5.3.1 v i i ) there i s x E X Hence X i s F- pseudocompact.
. Then
f E
with
f(x) = u(f) = supX(f).
i ) + i i i ) : I n view of Lemma 1.3.4 i t suffices t o prove t h a t VF cone. Let h n be a decreasing sequence in VF. We always have
i s a Dini
inf s u p X ( h n ) 2 supX(inf h n ) .
n
n
Hence we have t o prove the converse inequality. To t h i s end we may clearly assume a = inf s u p ( h ) > - 03 .
X n
n
Let 6 < a be arbitrary and p u t
g n = max(hn,6). Then g n
i s decreasing
-
and uniformly bounded. X i s a boundary f o r VF by assumption Therefore we may apply Simons' Convergence Lemma 3.6.1 to prove
m
( * ) inf{supX( I.
n= 1
I
A,,
gn)
supX(lim sup 9,)
n
I xi
.
m
z xn
t 0 , i = l, . . . , m ,
n=l
i).
= 11 I
'
+-
Here we consider the pointwise lim sup, i . e lim sup g n ( x ) = inf sup{g,(x)
n
for
x E X
since the
n
-,-
. Hence
m t n3
the right-hand side of ( * ) i s equal t o
g n are decreasing. The left-hand
inf supx(gn) = a since 6 < a
n
We obtain
I
supX(inf g n ) 2 a
n
side
0-f
supX(inf 9,)
n
( * ) i s equal t o
.
. This
implies, since 6 < a
was a r b i t r a r y ,
Representing Measures
330
that s u p X ( i n f hn) 2 a = i n f supX(hn). n n Hence V F
i s a D i n i cone.
o
Theorem 5.4.1 c o n t a i n s t h e well-known Alexandrov ([317l),which
G l i c k s b e r g Theorem X
,
has an i n t e g r a l r e p r e s e n t a t i o n by a measure on
X
s t a t e s t h a t , f o r a c o m p l e t e l y r e g u l a r H a u s d o r f f space
e v e r y s t a t e on
C(BX)
i f and o n l y i f
X
Moreover,if
-
i s pseudocompact.
K
i s a compact convex subsetof a l o c a l l y convex H a u s d o r f f space, t h e n t h e s e t o f a l l extreme p o i n t s o f K, ex K , i s Convg(K) , e x Kand A(K)
I
(Recall
Convg( K
ex
-
pseudocompact.
I
ex K =
K
1
f : K
+
, f upper semicontinuous, convex, bounded3
A(K) l e x K = { f
ex K
If : K
+
R
,
,
a f f i n e continuous))
T h i s i s a consequence o f Theorem 5.4.1
iii)* i), Lemma 3.3.2
and t h e
f a c t t h a t a subcone of a D i n i - c o n e i s a Dini-cone. We conclude t h i s s e c t i o n w i t h a c o r o l l a r y which c l a r i f i e s t h e c o n n e c t i o n between
F- r e a l -
and
F- pseudocompactness and
F- compactness. I n t h e
c l a s s i c a l s i t u a t i o n i t i s a well-known t o p o l o g i c a l r e s u l t ([1401): 5.4.2
Corollary:
The f o l l o w i n g are equivalent:
_.
i) X & F- pseudocompact ii) X i s F- compact
and
F- realcompact
Proof: i)
4
ii):
compact ii)
=,
.
i):
p XF c X
By Theorem 5.4.1 we have
By d e f i n i t i o n ,
, every
X
c h a r a c t e r of
a
XF = v XF c X. Hence
X
is
F-
i s F- realcompact. Furthermore, here, s i n c e F
is the Dirac functional o f a p o i n t o f
hence i t i s D i n i continuous. Therefore,
13 XF
= v XF
.
0
X,
GeneralizedHewitt-NachbinSpaces
33 1
SOME CONSEQUENCES
5.5
n ,
I n t h i s s e c t i o n we t u r n o u r a t t e n t i o n t o subsets o f t h e s t a t e space
elements o f
X
X
. Here,
u pXF
X
i n p a r t i c u l a r t o subsets o f
as u s u a l l y , we i d e n t i f y t h e
w i t h t h e i r corresponding D i r a c f u n c t i o n a l s , t h a t i s t o say,
n
i s regarded as a s u b s e t o f
. As
such, c l e a r l y
X
i s a topological
space c a r r y i n g t h e t o p o l o g y i n h e r i t e d f r o m t h e t o p o l o g y o f t h e s t a t e space.
A t f i r s t we want t o s t u d y subsets F l y - realcompact whenever ly
. Where,
in
c l e a r l y ,F
.
to Y
F
X
X
is
of
X
which a r e
F- compact o r
-
Fly Or F- realcompact, r e s p e c t i v e -
i s t h e cone o f t h e r e s t r i c t i o n s o f a l l elements
IY
Note, t h a t , b y Theorem 5.2.2 that i f
Y
X
ix),
i s dense i n
X U p XF
. We
point out
F- compact i t i s n o t n e c e s s a r i l y compact o r even c l o s e d as
is
a t o p o l o g i c a l subspace o f
R
.
Example: Take
X = N
and d e f i n e
F = If : X + R I f
-
lim n-t
bounded w i t h
f(n) =
n , since
bn, n E N
, i.e.
p XF = X
t
.
f(2)l
n c l e a r l y are a l l Dirac
Here t h e extreme p o i n t s o f t h e s t a t e space functionals
$ f(1)
. However,
p XF
i s n o t closed i n
1i m n+ 0
T h i s example a l s o shows t h a t
A subset
Y c X
X U p XF i s , i n g e n e r a l , n o t c l o s e d i n
i s c a l l e d sup-boundary i f we have
supy(f) = supX(f)
for all
f E F
Note t h a t sup-boundary n o t n e c e s s a r i l y means t h a t
Y
i s a boundary.
Remark: Let
Y c X
Define
cp :
be a sup-boundary and
aY be t h e s t a t e space o f
nY + R by
rp(p)(f) = p ( f
IY
)
for all
p
E
fiY
and
f E F.
F
IY
*
n
.
Representing Measures
332 tp
i s continuous and i n j e c t i v e . We c l a i m :
(1)
I f v E n then there i s
(2)
q ( P yF
IY
P xF
)
p E
sly
with
~ ( p 2 ) v
.
f E F,y
.
*
P r o o f o f (1): D e f i n e 6 5 supy
by
6 ( f ) = supIv(g) (To o b t a i n
6 5 supy
Ig
E
F, gIy
5 fl
for all
we need t h e assumption t h a t
Y
be a sup-boundary).
6 i s s u p e r l i n e r r and hence t h e Sandwich Theorem p r o v i d e s us w i t h t h e d e s i r e d s t a t e p E n w i t h S 5 p l s u p y whence rp(!.~) 2 v . T h i s proves (1). Y ( 2 ) , ( 3 ) a r e easy consequences of Theorems 5.2.2 iii)and 5.3.1 v)
5.5.1
Proposition:
be a sup-boundary of
Let Y -
i)
If
i s closed i n X
Y
Fly
-
ii) If Y
is
X
and
. --Then we have X
F- compact t h e n Y
compact. is an Fa- -subset o f X and X i s F- realcompact, -
Fly
( R e c a l l , an
-
Fa- subset i s a c o u n t a b l e union o f c l o s e d subsets o f X )
i):We have Y = q-l(i(Y)) i : Y
then
Y
realcompact.
Proof: -
where
is
+
X
,
i s t h e embedding, and, b y ( 2 ) ,
.
333
Generalized Hewitt-NachbinSpaces
Cp(YUPYF Since, b y Theorem 5.2.2 Y u p YF
) c x u g x F = x .
ix),
i s dense i n
Y
we o b t a i n
Y U 13 Y
IY
. Hence
= Y
IY
IY
Y
ii):Take t h e f u n c t i o n
is
F
compact.
I y-
as b e f o r e . S i n c e
cp
cp :
Y u w YF
subsets o f there i s
u
Y
w YF
.
IY
no E N w i t h
definitions o f
Z(p)
p 5
and
u
Let
5.5.2
w YF
.
that the restriction o f
IY
(Here we closed
be a r b i t r a r y . By Theorem 5.3.1 v ) IY T h i s means by Theorem 5.2.2 and t h e
f i l t e r b a s e o f a f i l t e r converging t o
. Hence we have
.
is
E w YF
supy
N(p)
X
-*
IY continuous we o b t a i n t h a t Y i s a Fa- subset o f Y U w YF OD IY used t h a t X i s F- realcompact). So assume Y = U Yn , Yn n=l
c Y
p
.
. Since
Z(V) t o Y
is a
i s c l o s e d we o b t a i n
Y
D
ProDos it ion :
The f o l l o w i n g
equivalent:
i) p E p X F ' w X F
ii)There i s an
Fa- s u b s e t
Y
o f X u p XF with
Y
3
X
,but
p
B Y.
Proof: ii) i ) : The same argument as i n t h e l a s t p a r t o f t h e p r o o f o f P r o p o s i t i o n 5.5.1 i i ) ( i . e . u l t i m a t e l y Theorem 5.3.1 v))shows t h a t Y 3 X U w XF . T h i s y i e l d s i). i)+ ii): By Theorem 5.3.1 v ) t h e r e i s a c o u n t a b l e c o v e r i n g such t h a t
p $
supy
OD
Hence
p
B
U n=l
n
for all
n E N
Ync X
of
X
.
in where in denotes
the closure o f
Yn
in
X U p XF. 0
334
Representing Measures
Reformulation of Proposition 5.5.1 and 5.5.2 well known r e s u l t s : 5.5.3
Theorem:
The following i)
y i e l d s generalizations of
X
&
are equivalent:
F- realcompact.
i i ) F o r every with X --
iii) X
p
c Y
i s the --
--t h e r e is
E 13 X F \ X
and -
p
B Y
an Fo- subset Y
. Fo- ~subsets of
intersection o f a l l
X
u
of 3! XF
X
u
p XF
containing X.
We c l o s e t h i s section by proving t h a t some topological spaces a r e automatically F- realcompact. Recall, a topological space X i s c a l l e d a Lindelof space i f f o r any open covering of X t h e r e i s always a countable subcovering. 5.5.4
Theorem:
Let X be endowed with a topology such t h a t X & 5 Lindelof space and i s F- realcompact. a l l f E F a r e upper semicontinuous. Then X I
Proof: Let
v
p
E v XF
and assume
1-1
B X
. Consider
F = I
7 IY
E Z(p)l
, where
is the closure of Y under t h e given topology. In view of t h e d e f i n i tions of Z(1-1) and N(u) and Theorem 5.2.2, F i s a f i l t e r base of Z(p). (Here we use t h a t a l l f E F are uppersemicontinuous). 1-1 B X implies n F = 0 t h e r e f o r e { X \ ! 1 Y E Z ( p ) > is an open covering of X which contains a countable subcovering { X \ Y n I n E N) i n view of t h e Lindelof property. We a r i i v e a t Theorem 5.3.1 v i ) .
n yn
n€N
= 4
vn E
Z
, which c o n t r a d i c t s 0
335
GeneralizedHewitt-NachbinSpaces
We conclude t h i s s e c t i o n w i t h a s l i g h t g e n e r a l i z a t i o n o f a w e l l known r e s u l t due t o Choquet [ 72 1, p. 146: Again, l e t
K
be a compact convex
s u b s e t o f a l o c a l l y convex H a u s d o r f f space. Then, under t h e g i v e n t o p o l o g y , i s a B a i r e space. F o r g e n e r a l i z a t i o n s t o cones, see D.A.
ex K
R e c a l l , a B a i r e space
X
i s a t o p o l o g i c a l space w i t h t h e p r o p e r t y t h a t
t h e i n t e r s e c t i o n o f every sequence dense i n
X
Edwards [991.
(P,)
.
X
o f dense open subsets o f
is
The B a i r e space p r o p e r t y indeed remains t r u e f o r a l l F- compact s e t s . To t h i s end l e t
X
be a g a i n an a r b i t r a r y s e t and c o n s i d e r a cone
F
as
above, c o n s i s t i n g o f bounded r e a l v a l u e d f u n c t i o n s on X h a v i n g lX as o r d e r unit. R e c a l l , by Theorem 5.2.2. , D X F = e x 0 , where R i s t h e s t a t e space o f F. 5.5.5
Theorem:
If a x ,
i s endowed the ~w i t h s t a t e space o f F , t h e n BXF ---
r es t r i c t i o n -o f t h e topology on n is a -_--
,the
B a i r e space.
Proof: The p r o o f i s e s s e n t i a l l y t h a t o f [ 8 8
, p.3951.
We can, by e v a l u a t i o n , r e g a r d t h e elements o f
F
T h e r e f o r e i t i s no l o s s o f g e n e r a l i t y t o assume t h a t p E
9XF c R = X
t h e elements
Indeed, by d e f i n i t i o n o f f I 0 such t h a t
Put
Uf
= -JC(
I x E X I f(x)
Y E
Z(p)
>c(
.
I . Then we o b t a i n
i s a neighbourhood o f
p
which converges t o
(Theorem 5.2.2
p
a neighbourhood base o f
compact subset o f
R
and
1~.
Since
.
~1
there i s
p
E Uf,
.
X = n. F o r any
a r e neighbourhoods o f
0 2 B >
z ( 1 - 1 ) ~f o r
n
as f u n c t i o n s on 1-1
.
f E F with
C1c Y
. Hence
Y
i s a filterbasis o f a f i l t e r
Z(p)
ix))
t h e elements o f
(closure o f
Z(p)
form
Representing Measures
336
Since 2 ( p ) i s a neighbourhood base o f 0 t fl,f2 E F, E R , such t h a t
V E U ~
lYa?
g
there are
E IR, f 3 E F
oxF
To show t h a t Pn
subsets
nu
I . !
f242
with
i s a B a i r e space c o n s i d e r a sequence o f dense open
o f 9XF.
Note t h a t d e n s i t y of
Pn
means t h a t
Pn
c o n s t r u c t a decreasing sequence
n
n + l * an+l
pxF c
u
F fn,%’
n pXF
fn,S,
0 , whenever
n U
nonempty and r e l a t i v e l y open. F i x such a s e t
0 * F
we deduce
E BXF,
. We
U
Ff
n y*n
We c l a i m t h a t
i s compact we o b t a i n
W
n 3XF
8
P1 n... n Pnn U
c
(
n
i=1
W
convex and
n
n pXF n\W
is
3.3.5
n...n
O X F c P1
.
n U
Pn
for a l l
n
, i.e
03
n
i=1
P
i s dense i n 8XF
0 n o t e t h a t , by c o n s t r u c t i o n o f W , W i s compact i s convex. L e t
L e t nW be t h e s t a t e space o f
W
* 0 .
a
n’ n
n.
U was a r b i t r a r y . Thus ?XF i s a B a i r e space.
To show
Since
n=l
n U i s non-empty. T h i s i m p l i e s t h a t
Pi)
since
n Ff
W:=
for all
0
Once we know this,we have W 03
use i n d u c t i o n t o
such that
m
Since
U cpXF i s
nw D
, by
p
is
F
Iw
p
FIw
.
F
-
W I
ex fiw
.
F
i s a c l o s e d sup-boundary o f
compact, i . e .
( a p p l i e d t o a c o n s t a n t sequence) pE
+
.
-compact and W
P r o p o s i t i o n 5.5.1,
an extreme p o i n t
be t h e r e s t r i c t i o n map. IW Then we have W c o w o P c n
: F
,
ex ow
.
ex fiw c W
* 0 . Hence
W
F
IW’
By Theorem
possesses
337
Generalized Hewitt-NachbinSpaces
W
has, by c o n s t r u c t i o n , t h e p r o p e r t y t h a t , w i t h
n , we have
p2 E
p2 E
maximal s t a t e o f F
p
=
x
+ (l-x)v2 ,
v1
Assume
p !f
ex R
and one e n d p o i n t Let
=
fixF
, vl,
. Then
v1 1.1 =
A v2
+
n ,
dn n
I
there i s
x
0 5
x
v1
5 1
E W
> 0
i s not i n W
or
v2
such t h a t
1.1 E
i n t ( L n n)
.
G2 E W w i t h
p =
a 3
2
.
t (l-a)vl}
and, b y compactness, t h e r e i s
E W
v2
.
( l - ~ ) v ~
be an extreme p o i n t o f p E
*
E W whenever
A two-dimensional computation shows t h a t , b y m i n i m a l i t y o f that
FIw
.
t h e r e i s a 1i n e d
of
d W we o b t a i n
is a
1.1
i s maximal regarded as a s t a t e of
1.1
v2 E
vl,
p2 L p l y
i m p l i e s t h a t t h e extreme p o i n t
i s convex, we have
A = i n f { a E [O,l]
Since with
, since
n\W
Note t h a t , s i n c e
. This
W
and
E W
p1
n
ex nw and t h a t
. Essential W
x , v2
must
arguments f o r t h i s computation a r e
n\W
as w e l l as
a r e convex. 0
5.6
REMARKS AND COMMENTS
A l l t h e m a t e r i a l , e x c e p t 5.5.5,
of t h i s s e c t i o n i s t a k e n f r o m 11271. We
l i k e t o emphasize t h a t t h e analogy between t h e t h e o r y o f c o n t i n u o u s f u n c t i o n s and, more o r l e s s , Choquet t h e o r y , which we have presented h e r e goes f a r beyond t h e few examples we gave. Most o f t h e m a t e r i a l concerning t h e s i t u a t i o n which we c a l l e d t h e c l a s s i c a l s i t u a t i o n goes back t o H e w i t t ' s fundamental paper 11621 o f 1948. By t h e same t i m e s i m i l a r r e s u l t s were found b y
L. Nachbin, a l t h o u g h h i s i n t e r e s t
i n t h e m a t t e r was a c o m p l e t e l y d i f f e r e n t one. He was m a i n l y i n t e r e s t e d i n spaces which admit complete u n i f o r m s t r u c t u r e s . Nachbin's r e s u l t s 12391 were n o t p u b l i s h e d b e f o r e 1954. Because o f these o u t s t a n d i n g c o n t r i b u t i o n s realcompact spaces a r e sometimes c a l l e d H e w i t t
- Nachbin spaces. The i n t e r -
r e l a t i o n between complete u n i f o r m s t r u c t u r e s and realcompact spaces a r e expressed by t h e f o l l o w i n g a s s e r t i o n ( compare S h i r o t a l2921) :
338
Representing Measures
A completely regular space such t h a t every closed d i s c r e t e subspace has
nonmeasurable c a r d i n a l i t y admits a complete uniform s t r u c t u r e i f and only i f i t is realcompact. The original d e f i n i t i o n o f realcompact spaces is d i f f e r e n t from t h e one we gave f o r t h e generalized s i t u a t i o n . S t a r t w i t h a completely regular TI- space X and a continuous map f : X + R and denote by f * : 13 X + R U I-) the canonical extension of f t o a map from t h e StoneCzech compactification to t h e A l e x a n d r o v - c o m p a c t i f i c a c t i o n of R . Denote by v f X the s e t p XxI y I f * ( y ) = -I. Then v X was o r i g i n a l l y defined as
UX =
n
{vf X If E
c(x)}.
Since we d i d not admit unbounded functions i t i s not completely obvious how t h i s d e f i n i t i o n c a r r i e s o v e r t o t h e generalized s i t u a t i o n . B u t i t can I f does not a t t a i n i t s be done. Consider f o r example- v"F = i f E maximum on X I and define f o r f E v"F v f X = {x E p XF I f ( x ) < supx f 1. Then, in f a c t , v XF =
n Ivf X If E VFI N
This i s l e f t as an exercise f o r the reader. Hewitt proved t h a t X i s realcompact i f and only i f every real zero-set u l t r a f i l t e r i s f i x e d . Recall, t h a t a zero-set f i l t e r i s a f i l t e r having a f i l t e r base consisting of zero s e t s of continuous functions, and such a f i l t e r i s s a i d t o be real i f every countable i n t e r s e c t i o n i s nonempty. This c h a r a c t e r i z a t i o n corresponds t o t h e f a c t t h a t X 3 v X F i f and only i f every u l t r a f i l t e r being contained i n some
Z(p),
p E
B X F , and which
has t h e countable i n t e r s e c t i o n property, has t o be fixed. A condition which can be easilyobtained from Theorem 5.3.1. As a general observation one finds t h a t t h e properties of zero-set f i l t e r s a r e r e f l e c t e d in t h e corresponding properties of t h e f i l t e r s Z ( p ) , p E p X F
.
Other characterizations o f realcompact spaces a r e t h e following 13241 due t o Katetov [1831, Mrowka [2341, Frolik [1121, Wenjen [3281 and, of course, imp1 i ci t e l y t o Hewi tt [ 1621 :
339
Generalized Hewitr-Nachbin Spaces
Let
X
be a c o m p l e t e l y r e g u l a r
T1- space. Then t h e f o l l o w i n g a r e e q u i -
Val e n t :
1)
x
2)
For every
i s realcompact
and 3)
E BX\X
f(x) > 0
For every
t h e r e i s some
for all
For every
x E 8X\X
t h e r e i s some
5)
X =
6)
X
2 E 3X\X
E A
that
n{BlpX i s the
2
B
X
3
X
f(2) = 0
,B
intersection o f
f E C(X)
which has no
u {XI.
t h e r e i s some
X n A =
and
with
x E X.
continuous e x t e n s i o n t o
4)
f E CB(X)
0
.
i s an
G6- s u b s e t A o f pX
such
Fu- s e t } .
u- compact subspaces o f gX
.
The r e a d e r i s c e r t a i n l y a b l e t o p r o v e these equivalences and t o t r a n s f e r them t o t h e general s i t u a t i o n . Then he can o b t a i n p r o o f s f o r t h e g e n e r a l i z e d statement
v i a an a p p l i c a t i o n o f t h e r e s u l t s i n s e c t i o n 5.
Condition 4) i s p a r t i c u l a r l y i n t e r e s t i n g since i t i l l u s t r a t e s t h e i n t e r r e l a t i o n between realcompact spaces and L i n d e l o f spaces. A space
X
is
L i n d e l o f i f and o n l y statement 4 ) h o l d s f o r e v e r y c o m p a c t i f i c a c t i o n ( i n s t e a d o f gX )
. This
r e s u l t i s due t o Mrowka
[2351.
Another d e f i n i t i o n o f realcompact spaces i s t h e f o l l o w i n g : . X compact i f and o n l y i f
m
X
i s real-
i s homeomorphic t o a c l o s e d s u b s e t o f
Rm
some c a r d i n a l number.
F o r more m a t e r i a l about u n i v e r s a l p r o p e r t i e s and p r o d u c t s o f g e n e r a l i z e d r e a l compact
spaces t h e r e a d e r i s r e f e r r e d t o [ 127 I .
F o r i n f o r m a t i o n about pseudocompact spaces ( c l a s s i c a l s i t u a t i o n ) t h e r e a d e r i s r e f e r r e d t o H e w i t t ' s o r i g i n a l paper [162] and subsequent work
([1401~[1411,[3171~[324I).
SECTION 11.6 EXAMPLES AND APPLICATIONS
Most " r e a l " a p p l i c a t i o n s o f i n t e g r a l r e p r e s e n t a t i o n theorems deal w i t h s i t u a t i o n s where t h e extreme p o i n t s o f t h e s t a t e space a r e e i t h e r compact o r n o t f a r from b e i n g compact. I n f a c t , one o f t h e few e x c e p t i o n s i s t h e Poulsen simplex, which p l a y s a r o l e i n s t a t i s t i c a l mechanics ([376] ,[377]
,
[ 581). T h i s simplex (which i s unique i n t h e m e t r i z a b l e case [2131,[2211)
has a dense extreme p o i n t s e t . S i n c e we c o u l d n o t go deeper i n t o s t a t i s t i c a l mechanics, which i n i t s e l f i s convex-cone-techniques,
an i m p o r t a n t area f o r a p p l i c a t i o n s o f
we o m i t t h i s example. I n s t e a d o f t h a t , we a r e pre-
s e n t i n g s e v e r a l examples which i n p r i n c i p l e c o u l d be t r e a t e d w i t h t h e RieszKonig Theorem, t o g e t h e r w i t h some a d d i t i o n a l techniques. Nevertheless, we hope t h a t by these examples we can evoke t h e i m p r e s s i o n t h a t i n t e g r a l r e p r e s e n t a t i o n s a r e an i m p o r t a n t t o o l i n several areas o f mathematics. The examples are: B e r n s t e i n ' s i n t e g r a l r e p r e s e n t a t i o n o f c o m p l e t e l y monot o n i c f u n c t i o n s , K e n d a l l ' s theorem on i n f i n i t e l y d i v i s i b l e c o m p l e t e l y monotonic f u n c t i o n s , t h e m u l t i p l i c a t i v i t y o f extreme s t a t e s i n case of mu1 t i p l i c a t i v e cones and t h e a p p l i c a t i o n s t o t h e Gelfand Representation and t h e GelfanrkNaimark Theorem, t h e Bochner-Weil Theorem i n harmonic a n a l y s i s and, f i n a l l y , t h e Levy-Khintchine formula. 340
341
Examples and Applications
6.1
COMPLETELY MONOTONIC FUNCTIONS
Let E be t h e vector space of a l l i n f i n i t e l y o f t e n d i f f e r e n t i a b l e functions f : I 0,+ m [ + R on t h e open h a l f - l i n e . We endow E with t h e topology of uniform convergence o f a1 1 d e r i v a t i v e s ( including t h e zero order d e r i v a t i v e ) on compact subsets of t h e open half l i n e . With t h i s topology E becomes a l o c a l l y convex Hausdorff space. Clearly the given topology i s generated by t h e following family of seminorms Ip,,, I m E N , n E No} , where
A function f E E i s c a l l e d completely monotonic i f ( - l ) nf ( n ) 2 0 f o r a l l n = O,l, ... . Examples a r e the functions e x p ( - a x ) , xqa , where a 2 0 . Note t h a t w i t h t h e usual pointwise operations, the completely monotonic functions form a cone. The function x - ~ shows t h a t completely monotonic functions a r e not n e c e s s a r i l y bounded. B u t i f such a function i s bounded then c l e a r l y i t s l i m i t lim f ( x ) e x i s t s . We want t o prove an
i n t e g r a l representation f o r Indeed, roughly speaking, f exp(-ax) for different a X = { f E E I f completely X i s convex. W e must show,
x+o+
a r b i t r a r y completely monotonic functions f . i s t h e "sum" of functions of t h e type . A t f i r s t , we t r e a t t h e bounded case. P u t monotonic, f ( x ) I1 for a l l x >'O} . C l e a r l y , t h a t X i s compact under the given topology.
Lemma : X
subset of _i s_a compact --
E.
Proof: Let R be the Alexandrov compactification o f R be mutually d i s j o i n t copies of I O,m [ . Introduce a topology i n
and l e t
Rn, n=O,l,
u Rn by defining a subset t o be open
nEN
only i f a l l i t s i n t e r s e c t i o n s w i t h t h e Rn
a r e open. The union
is then a l o c a l l y compact space. We may i d e n t i f y t h e functions
...
,
i f and U
nENo
Rn
f E X with
342
Representing Measures
:
u
Rn
i=
{
?. I
^f
+
R
'.
d e f i n e d by
f(n)
=
I Rn
.. .
n = 0,1,2,
I f we endow
w i t h t h e topology o f u n i f o r m convergence on a l l compact
f E XI
u Rn , t h i s i d e n t i f i c a t i o n becomes a homeomorphism. We show,
subsets o f
i
i s equicontinuous. Then, by A s c o l i ' s Theorem,
must be c o n d i t i o n a l l y
compact w i t h r e s p e c t t o t h e topology o f compact convergence. C l e a r l y , i s c l o s e d i n t h e space o f a l l f u n c t i o n s from t h e topology o f compact convergence. Hence To show t h e e q u i c o n t i n u i t y o f
r > 0
For each that
and
i
= 1for all
r > 0
n and a l l
f o r some
;? r < ro < r w i t h
Since
(-1)
x 2 r
for all
to
and
X
?
R w i t h respect t o must be compact.
there i s a constant
for all
T h i s f o l l o w s e a s i l y by i n d u c t i 0 n . n = 0 U(r,O)
Rn
,it s u f f i c e s t o p r o v e t h e f o l l o w i n g :
n = 0,1,2,...,
(-l)n f ( ' ) ( x ) I U(r,n)
2
u
-X
. Assume
x 2 r
and
U(r,n)
such
f E X.
i s c l e a r s i n c e we may t a k e
t h a t we have a l r e a d y found
U(r,n)
r > 0. Then by t h e mean v a l u e theorem we o b t a i n
(ro) = 2 (f(n) (r)
f(n+l)
f(n+l)
-
An)({))
I
i s non-increasing we have
and
. And
f E X
we can d e f i n e
2
U ( r , n t l ) = F(U(r,n)tU($,n))
Pro pos it i on : F o r each --
t h e r e i s & p o s i t i v e Bore1 measure
f E X
T
on
with
R,
T(R+) I 1 such t h a t
j
f(x) =
exp(-ax) dr(a)
for a l l x > o
.
R+
Proof: Put X
n
Denote by
Conv(X)
= { exp(-ax) l a E
, hence
t h e continuous convex f u n c t i o n s
R+I U
compact. We show,
and d e f i n e yr : X
+
X
{
01.
r > 0
+
R.
n i s a c l o s e d subset o f
Clearly,
n i s a boundary f o r
as f o l l o w s : I f
X
and
Conv(X). L e t f E X\{Ol
r E b{Ol put
343
Examples and Applications
yr(f)(x)
= f(r)-'f(r+x)
r < 0
x > 0. Define
for all
x > 0.
If
f E X.
Note, t h a t even i n t h e l a t t e r case
define yr(f)(x) = f(x)
-
yr(0)(x) = 0
f(x-r)
+
for all
f(x)f(-r)
for all
i s again completely
yr(f)
monotonic. L e t us c o n s i d e r t h e f i x p o i n t boundary i n t r o d u c e d i n c h a p t e r
I 1 3.1. K c X
be a compact
. Let
5
say
r
The s e t
r
T h i s shows t h a t
r-
Conv(X)
-
exposed. Indeed, l e t
i n v a r i a n t s e t and f i x some element i n
E a K 5
ylrl(f), y - l r l ( f )
r
of
f
Conv(X),
f E aE K . Then we have
r
and t h e r e f o r e
necessarily s a t i s f y
u
exp(-a x ), a E R U I - -1
functions
exp(-ax)
boundary f o r with
is
is
Conv(X)-
f(x+y) = f ( x ) f ( y )
x,y > 0. I t i s w e l l known t h a t , among t h e c o n t i n u o u s f u n c t i o n s ,
for all only
01
be a r b i t r a r y and c o n s i d e r
0
exposed. The f i x p o i n t s
with
+k
functional w i t h
x > 0.
p f I supx
{m}
bX E
Conv(X)
f E X. Then, o f course,
7 on n w i t h on
is a
-bX E Conv(X). And
E Conv(X)
5
uf(f bX) I
i s a linear
I * b x ( w ) d ;(u)
for
wER
n n a t u r a l l y defines a p r o b a b i l i t y
whose r e s t r i c t i o n t o
i n t e g r a l i n e q u a l i t i e s now i m p l y
X. Hence R
be t h e f u n c t i o n s
f E X. By C o r o l l a r y I 1.612 we o b t a i n .
for all
The measure R+ U
let
p f ( s ) = ~ ( f )f o r a l l
with
t h i s e q u a l i t y . Only t h o s e
a E [O,m] a r e elements o f
for all
a p r o b a b i l i t y measure
measure on
, satisfy
{a)
x > 0
Conv(X). F o r
bx(f) = f(x)
uf : Conv(X)
all
Ir E R,r
= {yr
f(x) =
j
R+
we c a l l
. The
T
preceding
exp(-ax)dT(a) f o r a l l
x
0.
7
R+ 0
Theorem (Berns t e i n [ 36
I,
A
+
function
-i s_
f :
3 O,=[
see a1 so [ 2501 ) : c o m p l e t e l y monotonic i f and o n l y i f t h e r e
R
a p o s i t i v e Bore1 measure
f(x) =
J R+
exp(-ax)dT(a)
T
0"
for _ all _
R+
such t h a t
x >
o . The measure
T
i s unique.
Representing Measures
344
Proof: The s u f f i c i e n c y i s s t r a i g h t f o r w a r d , i t remains t o show t h e n e c e s s i t y . I f i s completely monotonic and for a l l R,
. Then
x > 0
1>
fc E X
E
then put
> 0
. There
f
fE(x) = f ( e ) - l f ( x + c )
i s a p o s i t i v e B o r e l measure
T~
on
with
(*)
J
f(x+e) = f ( e )
exp(- a ( x t e ) ) e x p ( a e ) d-rE(a)
R+
for all
x > 0, i n view o f t h e p r e c e d i n g P r o p o s i t i o n . L e t
measure on
dT
=
Rt
i.e. T
I g(a)f(e)
exp(a&)d-re(a)
for all
g E Co(Rt)
Rt
f o r a l l continuous i s independent o f
I
be t h e B o r e l
d e f i n e d by
R,
Ig
T
E
g : R+
+R
. Indeed,
with
t a k e any
l i m g ( r ) = 0 . Because o f ( * ) r1 > el > 0 Then we have
e x p ( - a n ) f ( & )e x p ( a & ) d T E ( a ) =
.
f(n) =
R+
I
R+
exp(-an)
f(E1)
exp(a&, )dTe ( a ) 1
The subspace generated b y t h e f u n c t i o n s
for all
exp(-an),
n E N
.
n E N , i s dense i n
w i t h r e s p e c t t o t h e sup-norm which i s a consequence o f W e i e r s t r a s s ' theorem. T h i s shows, T i s independent o f e , f ( x ) = e x p ( - a x ) d r ( a )
Co(Rt)
I
R
for all
x > 0
and
'I
i s t h e unique measure w i t h t h i s p r o p e r t y .
0
345
Examples and Applications
6.2
KENDALL’S THEOREM ON INFINITELY DIVISIBLE COMPLETELY MONOTONIC FUNCTION
L e t cp : 1 0,-
be a bounded c o m p l e t e l y monotonic f u n c t i o n .
R
+
c a l l e d i n f i n i t e l y d i v i s i b l e if f o r e v e r y
Y with
monotonic f u n c t i o n
y n = cp
n E N
. Later
cp
is
there i s a completely
on we s h a l l see, i n t h e c o n t e x t
o f t h e L6vy-Khintchine formula, t h e reason f o r t h e importance o f t h i s n o t i o n . A t t h e moment we a r e g o i n g t o g i v e a c h a r a c t e r i z a t i o n o f t h e i n f i n i t e l y d i v i s i b l e c o m p l e t e l y monotonic f u n c t i o n s . T h i s c h a r a c t e r i z a t i o n Kendall [1931 ( s e e a l s o [ 2 5 0 1 ) .
i s due t o D.G. Theorem: Let -
cp
be a bounded e l y d i v i s i b l e c o m p l e t e l y monotonic f u n c t i o n . - i n f i n i t~
Then t h e r e i s a p o s i t i v e -Bore1 measure ----
on [O,ml such t h a t -
T
1 - exp(-a x 1 d r ( a ) [o,-] 1 - exp(-a)
}
for all --
x E lo,-[
.
B e f o r e we p r o v e t h e theorem we d i s c u s s some b a s i c p r o p e r t i e s o f i n f i n i t e l y d i v i s i b l e c o m p l e t e l y monotonic f u n c t i o n s . L e t monotonic f u n c t i o n s that
X
P,,~,
m E N
with
cp
cp(x) 5 1
.
X
be t h e s e t o f c o m p l e t e l y
I n t h e l a s t s e c t i o n we proved
i s compact w i t h r e s p e c t t o t h e t o p o l o g y generated b y t h e seminorms
elements o f
,n
.
E No
By
K
we denote t h e s e t of i n f i n i t e l y d i v i s i b l e
.
X
Lemma :
i)
K
iii)
I f cp E K -
iv)
If
_i s _a
cp
E
compact s u_ bset o f
x and
element o f ~v)
I f cp E K to -
cp
.
and -
K
.
a > 0 p
2 0
X
then
. cpa E K.
then t h e f u n c t i o n --
then t h e function --
1/ n exp(n(cp ( x )
exp(p(cp(x)
- 1)) is an
- 1)) converges p o i n t w i s e
346
Representing Measures
Proof:
i)- i v ) a r e s t r a i g h t f o r w a r d v e r i f i c a t i o n s .
F o r ( v ) we use t h e f a c t t h a t
exp(n(a
l/n
- 1)) -, a as n
for all
a > O . o r cp(x) = 1, f o r some
I t remains t o prove v i ) . I f cp(x) = 0
,
x > 0
then by complete m o n o t o n i c i t y , t o g e t h e r w i t h a s i m p l e c a l c u l a t i o n o r w i t h B e r n s t e i n ' s Theorem, we have cp cases j u s t t a k e
= cp(x) cp(xta)
al(x)
= n(l
-
= y e = cp
o r cp
. Assume
cp(a)
l/n
)
/ cp(a)
for all
x
, respectively.
For these
l/n
-
x > 0
.
K by
. I n order
1 - cp(a)
Then cpn E
1
y1 E
x > 0 cp(x)
and cpn(x) =
=
now, 0 < cp(x) < 1 f o r a l l
e = 1. Then we can d e f i n e
A t f i r s t , take
'n
5,
= 0
cp(xta) 1/n
to find
put
51
l/n
and
exp(n(cp(x)
l/n
- l))exp(-pn)
= exp(n(cp(x+a)
for all The f i r s t t h r e e e x p o n e n t i a l s converge t o
11n
n E N
-
l))exp(pn(cpn(x) and
x > 0
-
1))
.
cp(x), @ ( a ) and cp(xta). Hence
t h e f o r t h e x p o n e n t i a l a l s o converges p o i n t w i s e t o a f u n c t i o n
E~
which i s
e a s i l y seen t o be c o m p l e t e l y monotonic. (We may go o v e r t o a s u i t a b l e subsequence which converges i n t h e g i v e n t o p o l o g y on compactness o f with
e
5, = 5,
=
1 cp'-'
K).
. Now
51
assume
and
K
i n view o f t h e
i s c l e a r l y i n f i n i t e l y d i v i s i b l e and s a t i s f i e s v i ) 0
0, be t h e function w i t h
Q E C
Fix
Since
= Q(x)
for a l l
for all
LI
f E Conv(C). We have
-
for all
on
x > 0.
with
R
p
uQ 5 supc
( * bX) 5
Q
The measure
p r o b a b i l i t y measure Q(x) = uQ(dX) =
on
T
dX d?
n
n
r
= {T
a,@
I
Qa(x)dT(a)
[ O ,-I
I a E lO,-[,p Q(x)
Ta,,(Q)(x)
1
I
‘I
=
+ +
E R}
T
common f i x p o i n t s o f By d e f i n i t i o n of t h e
s o l v e t h e equations:
r
is
r
( * 6,)d;
p(Q(l+a)
-
Q(a))
-
Q(a))
x > 0.
n i s a boundary. D e f i n e T
a,a
: C
if lB(Q(l+a)
-,
-
C
by Q(a))l < 1
otherwise.
a,@( Q ) E C
e a s i l y checks t h a t
Q E C.
for all
t o obtain a
for all
o f functions
p(Q(x+a)
Q(X)
I n f a c t , we have
by
n defines n a t u r a l l y a
T h i s proves t h e Theorem. I t remains t o show t h a t a family
6
-,
such t h a t we have
[Op] =
on
hX E Conv(C).
Conv(C)
Q ’
n i s compact we can a p p l y C o r o l l a r y I 1.6.2
p r o b a b i l i t y measure
*
We have
Q E C.
and d e f i n e t h e l i n e a r f u n c t i o n a l
LI ( f ) = f ( Q )
Q
6,(Q)
i n view o f iii)and v i ) o f t h e lemma. One
Con(C)
-
exposed. T h e r e f o r e t h e s e t o f a l l
must be a boundary ( P r o p o s i t i o n 3.1.4). T
a,@
the fixpoints are the functions
Q
in
C
which
348
Representing Measures
(1)
Q(x)(Q(lta)
Define
- Q ( a ) f o r a l l a,x
-
1 for all
Q ( x ) R(a) = Q ( x t a )
-
Q(a)
-
x > 0
.
Q ( x ) R ( l ) = R(x)
Then (1) i m p l i e s
Q(x) f o r a l l
.
a,x > 0
for all
.
x > 0
Define
H(x) =
.
> 0
a = 1
From ( 2 ) we i n f e r w i t h
(3)
= Q(xta)
- Q(x)
R(x) = Q ( 1 t x )
(2)
- Q(a))
-Q(x) =
Q(Xt1)
l t R ( X )
.
= 1 t R(l)Q(X)
We o b t a i n H ( x t a ) = 1t R ( 1 ) Q ( x t a ) = 1t R ( 1 ) Q ( x ) t R ( 1 ) Q ( a ) t R ( l ) * Q ( x ) Q ( a ) = H ( x ) H ( a ) for a l l
a,x > 0. Here t h e second e q u a l i t y f o l l o w s from ( 2 ) and ( 3 ) .
Hence H(x) = exp(8 x )
,B
E
[-my-[
, and
therefore
~ ( x )= ( e x p ( B x ) - 1) ~ ( 1 1 - l f o r a l l With
Q ( l ) = 1 we o b t a i n
t h a t both, R(l) > 0
R(l)
and
B
R(l)
, are
t
x > 0.
1 = exp(B). The monotony o f
Q
p o s i t i v e o r negative. I f B > 0
implies and
t h e n cp w i t h
cp(x) = exp(- Q ( x ) ) = exp{
-
exp(g x ) - 1 exp(8) 1
for all
-
x > 0
i s n o t completely monotonic. Hence we o b t a i n t h e f u n c t i o n s =
-1 -1
exp(-ax ) exp(- a)
for all
Q,
with
x > 0, a E [ o p ] ,
Since these f u n c t i o n s i n f a c t s o l v e (1) t h e y a r e t h e o n l y s o l u t i o n s ( i n c )
o f ( 1 ) . T h i s means
n i s a boundary.
0
349
Examples and Applications
6.3
MULTIPLICATIVE CONES
Throughout t h i s s e c t i o n we assume t h a t ( F , + I ) i s an order u n i t cone such t h a t F = F, + R I , i . e . f o r every f E F t h e r e i s some h 2 0 such f + A I . The cone i s s a i d t o be m u l t i p l i c a t i v e i f we have defined that 0 on F, such t h a t an a s s o c i a t i v e d i s t r i b u t i v e mu t i p l i c a t i o n
-
f = xf
f .(AI) = (XI)
for all
f E F,
.
This ensures t h a t t h e r e l a t i o n s one usually expects from an algebra do hold ( w i t h R, instead of R ) .
The m u l t i p l i c a t i o n i s s a i d t o be monotone i f , f o r a l l f
4
h, g
iI ,
we have f g < h g
and
f , g , h E F,
with
f + h g < h + f g .
And t h e m u l t i p l i c a t i o n i s s a i d t o be weakly monotone i f , f o r a l l
f , g , h E F,
with
f p(x),the
}
x n
n (7)
1
n=l
Of course, these functions f u l f i l l the usual relations, l i k e =
x I- x
and
exp(ln(1
X - x)) =
I -
X
,
where exp(z) i s defined t o be: exp(z) =
(5)
zn z n!
a
n=O
*
Now, we consider the fol owing Situation 1: that
Let N be a real normed algebra and P
i)
I E P , -I !k P
ii)
II xII I t x E P
iii)
P
*
for all
x
c N
be a cone such
E N
P = { x y Ix,y E P ) c P
Remarks : We introduce in
-
N
the order given by
x s y
y-XEP.
Then i ) and i i ) guarantee t h a t I unit functional by SI and I S I ]
P
, i.e.
i s an order unit
. We
denote the order
denotes the order unit norm
354
Representing Measures
I S I I = maxISI(x),
(6)
SI(-x)l
=infIx>O
IXI -x
and XI t x E P I .
E P
Because o f ii) we have
(7)
ISII(x)
I IIxII
Now, c o n s i d e r a r b i t r a r y
- X Z = (y-x)z
Then y.2
E P
x,y,z E P
x E N.
for all
and
with
x Iy
and
z II .
x t y z - ( y t x z ) = (x-y)(I-z)
E P.
Hence i i i ) i s e q u i v a l e n t t o t h e a s s e r t i o n t h a t t h e m u l t i p l i c a t i o n i s monotone. Define
%(N)
t o be t h e nonzero m u l t i p l i c a t i v e l i n e a r f u n c t i o n a l s
v : N - r R .
Theorem I n case o f S i t u a t i o n 1 we have
i) SI(x)
Iv
= supIv(x)
ii) 6-'p(x)
I SI(x)
I . The sup i s a t t a i n e d a t some
E %(N
Ip ( x )
x E P, where
for a l l
v E %(N
6=supIp(y)lyEP,I-y
PI.
Proof: The lemma i n t h e p r e c e d i n g s e c t i o n shows t h a t e v e r y c h a r a c t e r i s m u l t i p l i c a t i v e on
P
, hence
contained i n +(N)
N
on
and
(&mark i n 6 . 3 ) . So t h e c h a r a c t e r s a r e
i ) i s an immediate consequence o f Theorem 3 . 3 . 5
( a p p l i e d t o c o n s t a n t sequences). A s s e r t i o n i )y i e l d s = (sI(x))2
S+X)
f o r x E P. F o r x E N we o b t a i n by ii)o f S i t u a t i o n 1 t h a t SI(X.X) 2 SI(X) 2 S i n c e we have I S I ] 6 II II we immediately o b t a i n
.
ISII I
p
( f r o m t h e c o n s t r u c t i o n o f t h e s p e c t r a l r a d i u s norm), I f
6 =
m
t h e theorem i s proved. So, l e t us assume Then f o r e v e r y p ( x ) I6 x
.
x
Since
6
< O D
> SI(x)
x
. Note,
that
we have
> SI(x)
P(X) 2 6 S I ( X ) .
0 I
6 t 1 since X
2
. Hence
I
p(1) = X
p(,)
1. Take I 6
was a r b i t r a r i l y chosen t h i s y i e l d s 0
or
x 2 0.
355
Examples and Applications
, we d i s t i n g u i s h between r e a l 1 in e a r i t y and complex 1 i n e a r i t y . R e c a l l a map u between
Since from now on we a l s o deal w i t h v e c t o r spaces o v e r
N1
two v e c t o r spaces
and
N2
t
,if
i s called real-linear
u(a x + B y ) = a u ( x ) + B ~ ( y )f o r a l l and r e a l numbers
.
p
a,B
.
a,p
and complex numbers
E N1
X,Y
if
i s c a l l e d complex-linear,
*
A l l o u r examples deal w i t h i n v o l u t i o n s . An i n v o l u t i o n
:
N
+
N
is a
r e a l -1 i n e a r map such t h a t
for a l l
A(N)
By
N
x,y E
(X*)*
= x
(xyy
= y* x*
(A xf
=
N
and
x x* A
,
x denotes
E K . (i.e.
t h e c o n j u g a t e complex number
K = R
or
K
t )
=
we denote t h e s e t o f nonzero m u l t i p l i c a t i v e l i n e a r f u n c t i o n a l s
A ( N ) a r e supposed t o be r e a l - l i n e a r , i f K = R and c o m p l e x - l i n e a r i f K = 0 . I n case o f a normed a l g e b r a N w i t h i n v o l u t i o n an element I, E A(N) s a i d t o be symmetric i f +
K.The elements o f
-
v ( x * ) = I,(x)
for all
x E N.
By A ~ ( N ) we denote t h e symmetric elements o f Example 1 ( Consider
A
*-
A(N).
algebras)
t o be a
*
- algebra w i t h u n i t
I. A
*-
complex Banach a1 gebra w i t h continuous i n v o l u t i o n . By ,A, r e a l normed a1 gebra o f s e l f a d j o i n t elements
Asa = IX E A Obviously, we have
A = As,
I X*
=
@ i ,A,
algebra i s a we denote t h e
XI .
. Let
P denote t h e c l o s e d cone
of
x
356
Representing Measures
-
P = Ix
generated by
i s closed. Hence
Ix
with
E As,}.
S i n c e t h e i n v o l u t i o n i s continuous
As,
P c Asa. Let x 4
Observation 1: y E As,
2
.
x = y
II I - x I I < 1. Then t h e r e i s
be s e l f a d j o i n t w i t h
Proof: Recall t h a t f o r
I 1 a l l < 1 we d e f i n e
with
a E As,
by the
f o l l o w i n g a b s o l u t e l y converging s e r i e s
Clearly,
For
d
z = I
m i s s e l f a d j o i n t . From t h e t r i a n g l e i n e q u a l i t y we o b t a i n :
- dr-a’,
we have
must be s e l f a d j o i n t and assertion. Observation 2:
z
f o r some
E As,
Let
.
a E As,
and
E
Put a = I
. Then
> 0
Proof: Apply Observation 1 t o
x = I t ( ( Ia l l +
Now, d e f i n e an o r d e r r e l a t i o n i n As,
a Ib That means
a
E)-’
Ix
101
-
( I l a II +
.
i s an o r d e r u n i t of
c -
As,
x E P with
.
and we have SI I II II
Proof: O b s e r v a t i o n 2 shows t h a t , f o r a l l
E )
I +a = z 4
0
Lemma 1:
I
Hence, y =
t o obtain the
x
by
i f and o n l y i f t h e r e i s some
P = Ix E As,
.
I I z II < 1 and z E AS,
m= y2 .
a E Asa
, we
have
.
a t x = b.
357
Examples and Applications
-
a 5 ( I l a II + &)I .
T h i s c l e a r l y proves t h e c l a i m .
0
II II of
L e t us c a l l t h e norm
monotone if II II
A
i s monotone on
P.
Lemma 2:
The f o l l o w i n g are equivalent: II II
i)
ii) SI(x)
% monotone = 11 x l l
for all
x 2
o
Proof:
SI
Since
i s monotone w i t h r e s p e c t t o t h e g i v e n o r d e r ii)i m p l i e s i ) .
II II
Now, l e t
II II I p
be monotone, t h e n
(restriction t o the positive
cone) i s a monotone s u b l i n e a r f u n c t i o n a l . Hence, f o r e v e r y f i x e d 0
9
u : P
x E P, t h e r e i s a monotone l i n e a r f u n c t i o n a l
II
p 5 I1
~ ( x =) I I x I l (Theorem I 1.3.3).
and
, we
II I l l s 1
and
.
ASa = P- R, I
state o f
SI(X) The i n e q u a l i t y
2
R
Because o f
~ ( 1 )= I I I I I = 1. So,
obtain
+
p
with x 5 IIx I I I
can be extended t o a
From Lemma 2.1.2 we d e r i v e
U(X)
= IIXII
.
i s a consequence o f Lemma 1.
IIxII 2 SI(x)
c
Another u s e f u l o b s e r v a t i o n i s : Lemma 3:
fi
The m u l t i p l i c a t i o n -
& monotone ---i f and o n l y
,A,
if
A i s commutative.
Proof:
P
-
P
c
P
i s equivalent t o the assertion t h a t the multiplication i s i s commutative, t h e n x x* y a y * = ( x y ) . ( x y ) * .
-
monotone. I f A
T h i s c l e a r l y imples Since
,A,
ASa = P
,A,
c ,A,
-
I
Ix
if
P
{A
P
P
c
( I
E R,}
.
P
P . Hence, P
c
P
*
P c P.
i s an o r d e r u n i t ) we have
. Therefore
we g e t , f o r a r b i t r a r y
Representing Measures
358
, that
x,y E Asa
x-y
i s s e l f a d j o i n t . This implies
,
( x . y ) = (X.Y)* = y*x* = y * x and
x
must commute w i t h
Therefore
A
Example 2 Let
A
. Hence,
y
As,
i s a commutative algebra.
must be commutative. (C*- a l g e b r a s )
be a
algebra w i t h u n i t .
C*-
A
C*-
*
algebra i s a
- algebra
such t h a t
II X ~ * I I
=
II~II
for a l l
II x I I
I n particular t h i s implies
2
=
x E A.
2 I I x II and
I I x I1 = IIx* II
for all
x E A. Remark : Let A
be a commutative
C*-algebra then, f o r
a,b E As,
the following
are true: 2 2 2 I I a t b I l t I I a II
i)
I1 I- a 2 II I 1
ii) If II a2q1 5 1 then I I a 2 t b 2I I < : l
iii) I f
1>
Let
iv)
> 0
E
1 1 1 - a 2 - b2I I s l
then
, then
t h e r e i s y E As,
such t h a t
c
I t
a 2 t b 2 = y2.
Proof:
i) I I a2 t b 2 I I = II ( a + i b ) ( a - i b ) l I = I l a t i b l l I I a - i b l l = l l a t i b l l 2
.
From t h e t r i a n g l e i n e q u a l i t y we d e r i v e 2 IIatiblI= IIatibll+ IIa-ibllt 2 Ilall. Hence 2 2 2 2 I I a t b I12 I I a l I = I l a II
ii) L e t (ltc) I
t > 0
-
a
2
, t h e n from Observation 2 we g e t some c
= c2
( I t &= )
And
E+
0
.
. Hence, (1tc)llIii
E As,
such t h a t
we o b t a i n from i ) : =
I I2 t~c 2 112 I I C 2 11 = I I ( l + t ) I - a 2 I I .
proves t h e a s s e r t i o n
359
Examples and Applications
iii) L e t
E
> 0 , then f r o m O b s e r v a t i o n 2 we g e t some ( 1 + ~I)- a 2 - b 2 = c2
(1tE) t I I ( l + c ) I Again,
E
-+
-
.
such t h a t
From i)and ii)we o b t a i n :
a 2 I I = I I b2 t c 2 l l t I I c 2 I I = l I ( l + ~ ) I - 2a - b 2 I I .
proves t h e a s s e r t i o n .
0
iv)
S i n c e t h e norm i s homogeneous we can assume
Put
x =
E
c E Asa
I1 a 2 + b2 11s 1 - E
I + ( a 2 + b 2 ) , t h e n a c c o r d i n g t o iii)we have
And t h e a s s e r t i o n f o l l o w s from O b s e r v a t i o n 1
.
II I - x I I
.
I (1-E).
o
Lemma 4:
If A
C*- a l g e b r a . t h e n t h e norm i s monotone.
-i s_ a commutative
Proof: E As,
Consider al,...,am
Put
a =
x
,@
a' i
yields that for and
p
I + D = x2
and
= 1b
6,p
2 0
. Then
2
bn E As,
bly...,
. Assertion
6,p
-+
0
gives the desired r e s u l t .
Now, c o n s i d e r on
such t h a t
A
2
. 0
A ~ ( A ) t h e s t a t e space t o p o l o g y . Then
And f o r commutative
61+ a = y
i) o f t h e Remark y i e l d s
II ( 6 + p ) I + a + @ l l t II 6 I + a I I And
have t o show
( v ) o f t h e Remark and i n d u c t i o n
y,x E As,
there are assertion
. We
aS(A)
i s compact.
we o b t a i n t h e f o l l o w i n g i m p o r t a n t r e p r e s e n t a t i o n .
Gel fand-Naimark Theorem: Let -
A
-be_
a commutative
C*- ___-algebra w i t h u n i t . Let A
=
symmetric nonzero m u l t i p l i c a t i v e l i n e a r f u n c t i o n a l s on A
x E A
+
i
E C a(A)
, -given
i s o m e t r i c a l g e b r a isomorphism
by
from
i ( ~= )~ A
onto
AS(A)
.Then
( x ) -for all u E A
C,(A).
be the
is an
Representing Measures
360
Remark: Because o f
A
=
,
x E A
A ~ ( A ) we have, f o r a l l
/\
z
x* = x
(where t h e b a r
I = 1A *
denotes complex c o n j u g a t i o n ) . Furthermore P r o o f o f t h e theorem: Consider
Asa
. We
have t o prove t h a t
isomorphism f r o m ,A,
onto
x
+
x
i s an i s o m e t r i c a l g e b r a
Consider i n ,A,
C R(A).
the order
i n t r o d u c e d i n Example 1. Then
i s an o r d e r u n i t cone (Lemma 1) (Asa,s,I) w i t h monotone m u l t i p l i c a t i o n (Lemma 3 ) . A s t r a i g h t f o r w a r d computation shows ( A s a ) = AS(A)
*
From Lemma 2 and Observation 3 we o b t a i n The theorem o f t h i s s e c t i o n y i e l d s t h e n
for all
. This
x 2 0
I I X II for a l l
x E As,
= II x II f o r a l l
SI(x)
x 2 0.
implies Sup{
=
1 i(u) I v
E A)
( t o see t h a t c o n s i d e r
x2 2 0).
X i s an isometry. Obviously t h i s i s o m e t r y must be an a l g e b r a isomorphism s i n c e t h e elements o f A a r e m u l t i p l i c a t i v e and l i n e a r .
So
x
+
That t h e i s o m e t r y i s o n t o i s an immediate consequence o f t h e StoneWeierstrass Theorem.
Example 3
0
(Gelfand r e p r e s e n t a t i o n o f commutative a l g e b r a s ) .
We s t a r t w i t h a commutative complex normed a l g e b r a
I . L e t us f i r s t embed A T h a t i s a commutative morphism J : A morphism
3
+
: a(A)
construction o f
B
t(A)
in
the free
C*-
A
w i t h u n i t element
a l g e b r a generated by
C*- a l g e b r a such t h a t f o r e v e r y nonzero i n t o a C*- a l g e b r a B t h e r e i s a unique
-,B
II 211
with
=
1.
**-
homohomo-
L e t us f i r s t d e s c r i b e t h e
a(A). Consider f o r a moment A
as a v e c t o r space w i t h
( i n s t e a d o f t ) and denote by A Q A t h e a l g e b r a i c t e n s o r p r o d u c t ( a g a i n w i t h r e s p e c t t o t h e r e a l numbers). r e s p e c t t o t h e r e a l numbers
R
A.
361
Examples and Applications
D e f i n e m u l t i p l i c a t i o n , norm and i n v o l u t i o n i n
x
(
i
A @ A
Then
ai 8 bi)*
=
f bi 8 ai 1
A 8 A
by:
.
i s a normed r e a l a l g e b r a w i t h i n v o l u t i o n and u n i t
I 8 I.
Take t h e q u o t i e n t a l g e b r a w i t h r e s p e c t t o t h e i d e a l generated by
.
{ (ai) 8 b t a 8 ( i b ) la,b E A1
Since t h i s i d e a l i s i n v a r i a n t under t h e i n v o l u t i o n we have a c a n o n i c a l l y d e f i n e d i n v o l u t i o n i n t h e q u o t i e n t . O f course, t h e d i f f e r e n c e between
A Q A
and t h e q u o t i e n t i s t h a t we can now i d e n t i f y
.
(ia) Q b = - a Q ( i b )
T h i s means t h a t we can c o n s i d e r t h e q u o t i e n t a l g e b r a as a complex a l g e b r a such t h a t t h e i n v o l u t i o n
*
satisfies
( x ( a 8 b ) ) * = x(b Q a ) .
NOW, t a k e as
t(A)
the completion o f t h i s algebra w i t h respect t o t h e
s p e c t r a l r a d i u s norm. By abuse o f n o t a t i o n we can now i d e n t i f y As(a(A)) = A(A)* Observe, t h a t t h e u n i v e r s a l p r o p e r t y o f w i t h t h e s p e c t r a l r a d i u s norm of
implies that
A
endowed
can be c o n s i d e r e d as a complex subalgebra t(A) yields:
And a p p l i c a t i o n o f t h e Gelfand-Naimark Theorem f o r
t(A).
-Gelfand
p
t(A)
Representation ( c f . [ 275
F o_ r e_ very
2 i f II a II =
a E A llalI2
-we have for all
m e t r i c a l g e b r a isomorphism -~
1)
p(a) = s u p { l v ( a ) l a E A between
, Jhe~ p A
IU E =
A(A)). & p a r t i c u l a r ,
II II and t h e r e i s an i s o -
and a subalgebra
of
C(A(A)).
362
6.5
Representing Measures
THE BOCHNER
-
WEIL THEOREM
We rephrase some o f t h e r e s u l t s o f s e c t i o n s 6.3 and 6.4 f o r t h e case o f without unit). commutative i n v o l u t i v e normed algebras ( perhaps Let
N
be such an algebra, i . e . a commutative complex normed a l g e b r a w i t h
involution x By
Nsa
+
x*
.
We assume t h e i n v o l u t i o n t o be continuous.
we denote t h e r e a l a l g e b r a o f s e l f a d j o i n t elements i n
Nsa = C X E N I X = x * }
u : N
A functional x E N
all
+
.
a i s s a i d t o be symmetric i f u ( x * )
. The symmetric
c o m p l e x - l i n e a r f u n c t i o n a l s on
u n i q u e l y t o t h e r e a l - l i n e a r f u n c t i o n a l s on symmetric m u l t i p l i c a t i v e
If
i s continuous w i t h
p
C(V)
Ns,
=
N
u(x)
correspond
N , i.e. t h e
.
u i s called positive i f
= 1 t h e n we c a l l i t a s t a t e .
f o r a p o s i t i v e f u n c t i o n a l we have
for
AS(N) a r e t h e non-zero
Nsa.
c o m p l e x - l i n e a r f u n c t i o n a l s on
m u l t i p l i c a t i v e r e a l - l i n e a r f u n c t i o n a l s on
A symmetric c o m p l e x - l i n e a r f u n c t i o n a l
N:
p(xx*) 2 0
for all
Note t h a t
x E N.
Theorem 1:
i)
If
measure
&
u T
continuous and p o s i t i v e t h e n t h e r e i s a p o s i t i v e f i n i t e
on A,(N) -
such t h a t --
ii) _ If
N -----i s a Banach a l g e b r a t h e n e v e r y p o s i t i v e f u n c t i o n a l _ i s -autom a t i c a l l y continuous. iii)
Every extreme p o i n t -~
of t h e s e t o f s t a t e s i s m u l t i p l i c a t i v e . -
-
_
-
I
_
_
_
363
Examples and Applications
Proof:
As described i n 6.3 we adjoin a u n i t I t o N and obtain t h e algebra NI. By B we denote the Banach algebra given by the completion of N I . The convolution i s extended t o N I by putting I* = I , and t o B by c o n t i n u i t y . From s e c t i o n 6.4 we know t h a t Bsa ( s e l f a d j o i n t elements of B ) c o n s t i t u t e s an order u n i t cone, with I as order u n i t . Since B i s commutative the m u l t i p l i c a t i o n i n BSa i s monotone. Now, extend
p t o N I by putting p(1) = C ( p ) . ( I f NI i s not a Banach algebra then extend i t f u r t h e r t o B by continuity).We claim t h a t p i s p o s i t i v e on B , i . e . ~ ( y y*) 2 0 f o r a l l y E B . Since p i s l i n e a r and continuous i t s u f f i c e s t o prove t h a t f o r y = I - x , x E N. In f a c t
-
Therefore p is monotone, hence continuous with respect t o t h e order u n i t norm. Since t h e order u n i t norm i s equal t o the s p e c t r a l radius norm ,IJ i s continuous with respect t o t h e norm. This proves i i ) because we needed c o n t i n u i t y only f o r t h e case when N was not a Banach algebra. a r r i v e d a t t h e s i t u a t i o n considered i n s e c t i o n 6.3 a s s e r t i o n Since we i ) follows from t h e theorem i n t h a t s e c t i o n and i i i ) i s an immediate 0 consequence of t h e preceding lemma. We i l l u s t r a t e this r e s u l t f o r a concrete s i t u a t i o n . Let ( G , - ) be a l o c a l l y compact abelian group. Locally compact a b e l i a n groups have a very rich s t r u c t u r e concerning d u a l i t y (Pontryagin d u a l i t y theorem) and i n v a r i a n t measures (Haar measure, s e e 12401). For a general introduction i n t o t h e theory of these groups t h e reader i s r e f e r r e d t o any textbook on harmonic a n a l y s i s ( 1 i ke [219 I , [148 1 , [184 1 o r [273 I ) . Let us r e c a l l some of t h e basic notions. A m u l t i p l i c a t i v e map 6 denotes the s e t 3 : G -+ t z E t I I z I = 1) i s c a l l e d a character on G o f a l l characters and i s endowed with t h e topology of uniform convergence i s then l o c a l l y compact and i t i s an abelian on compact subsets of G . group with respect t o pointwise m u l t i p l i c a t i o n . Take an element s E G , then s E --t i ( s ) defines a c h a r a c t e r on . Hence, t h e r e i s a
.
364
RepresentingMeasures
canonical homomorphism G
-
which i s continuous a c c o r d i n g t o A s c o l i ' s
+
1emma. One o f t h e most i m p o r t a n t p r o p e r t i e s o f
m
sure
, that
i s a p o s i t i v e nonzero, r e g u l a r Borel' measure m which i s
, i.e.
translation invariant
A c G. Consider
i s t h a t i t admits a -Haar mea-
G
C ao(G)
, the
m(A.s) = m(A)
s E G
for all
and B o r e l s e t s
space o f complex-valued continuous f u n c t i o n s on
G
which vanish a t i n f i n i t y . Endow C ao(G) w i t h t h e sup-norm t o p o l o g y and denote by MB(G) t h e space of continuous l i n e a r f u n c t i o n a l s on C a,(G). Example 3 i n s e c t i o n 2.6.4 sho\:s t h a t f o r every unique bounded complex-valued t i g h t measure iC
IG
x(f) = Thus we can i d e n t i f y
MB(G)
.
G
measures on
f di
S p e c i a l elements o f
MB(G)
For
L 1 (G)
x,y E MB(G)
I G
L 1(dm)
. The
define
we d e f i n e c o n v o l u t i o n
*
y) =
.
E C U,(G)
a r e o b t a i n e d when t h e Haar measure
. Similarly,we
f d(x
J
s ,tEG
f(S*t)
L 2 (G)
.
x*y
and i n v o l u t i o n x* by
L2( G )
even an i d e a l o f
. L1(G)
i s a H i l b e r t space.
dx(s)dy(t)
(MB(G),*,*)
i n v o l u t i v e complex Banach a l g e b r a w i t h u n i t element G)
m is
subspace o b t a i n e d ' i n t h i s way i s
where t h e b a r denotes complex c o n j u g a t i o n . t h e u n i t of
such t h a t
G
w i t h t h e space o f bounded complex-valued t i g h t
m u l t i p l i e d by a d e n s i t y cp E denoted by
,f
x E MB(G) t h e r e i s a on
i s a commutative be
( D i r a c measure a t
i s a complex Banach subalgebra ( w i t h o u t u n i t ) ,
(MB(G),*,*)
.
365
Examples and Applications
For
x E MB(G)
we d e f i n e bounded, continuous, complex-valued f u n c t i o n s
and
Fx
by:
6
on
(F,)(s)
=
sEG
,
S(S) dx(s)
E
G
These f u n c t i o n s a r e c a l l e d t h e F o u r i e r - t r a n s f o r m o f x , r e s p e c t i v e l y . When
cotransform o f
complex-valued f u n c t i o n s on
i) a r e
FX
x
and t h e F o u r i e r
-
C aB(E) ( t h e bounded continuous
endowed w i t h t h e sup-norm and t h e i n -
v o l u t i o n g i v e n b y complex c o n j u g a t i o n , t h e n
C LB(g)
c o n s t i t u t e s a complex
i n v o l u t i v e Banach a l g e b r a w i t h r e s p e c t t o p o i n t w i s e m u l t i p l i c a t i o n . The importance o f t h e F o u r i e r - t r a n s f o r m stems from t h e f a c t t h a t Banach a l g e b r a homomorphism f r o m
MB(G)
into
F
is a
C OB(6). T h i s homomorphism
i s symmetric w i t h r e s p e c t t o t h e i n v o l u t i o n s , i . e . F
u
-
*
= F
u
for all
p E
MB(G)
.
Observation 1: Every
3 E 6 ~-~ d e f i n e s an element (*)
And f o r e v e r y ---
v,(x)
v,
= (Fx)(S')
1
v E A ( L (G))
S
1 E A ~ ( L( G ) )
for a l l
t h e r e i s some
Hence
C
1
= A ( L (GI)
= A,(L
by
1 x E L (G) S E
.
G -such t h a t
v = v-
S
.
1( G ) ) .
Sketch o f t h e P r o o f : R e c a l l t h a t t h e F o u r i e r t r a n s f o r m i s a symmetric Banach a l g e b r a homomorphism. T h i s immediately i m p l i e s t h e f i r s t p a r t .
366
Representing Measures
Now t a k e
1
v E A ( L (G))
Im1-l v ( x
SO(S) =
So E
Then
(1)
s,
. I f one
x
*
y =
1 x E L (G)
and
*
~ ( x + ) 0
.
s E G
Define f o r
bS)
keeps i n mind t h a t
tEG
*
x
bt d y ( t )
v, = v
t h e n one e a s i l y sees
with
R e c a l l t h a t a Banach a l g e b r a separates t h e p o i n t s o f
A
.
.
for all 0
A
i s s a i d t o be semisimple [275 I
if A(A)
Lemma 1:
1 L (G)
i s semisimple
Proof: I n view o f Theorem 1 i ) i t s u f f i c e s t o show t h a t f o r a g i v e n p o s i t i v e 1 x * x * E L (G) t h e r e i s a p o s i t i v e l i n e a r f u n c t i o n a l p such
element that
~ ( *x x*) > 0
F o r y E L 1( G )
.
Such a f u n c t i o n a l i s e a s i l y found:
consider the l i n e a r operator
.
T : L 2 (G) + L 2 (G) g i v e n by Y then t h i s o p e r a t o r i s i n -
2 T cp = y Y cp f o r a l l cp E L (G) If y P 0 Y j e c t i v e . Take now cp + 0 and c o n s i d e r t h e l i n e a r f u n c t i o n a l
P
1
: L (G)
g i v e n by ~ ( y )=
< q,Tycp>
(
< ,>
scalar product i n
L'(G)).
T h i s f u n c t i o n a l i s indeed p o s i t i v e s i n c e
Observation 2:
i) E
separates t h e p o i n t s o f
G
, i . e . t h e canonical map G
+
is
+
t
Examples and Applications
367
in
G)
inj e c t i ve . ii)
6
The l i n e a r h u l l o f
C a ( G ) ( c o n t i n u o u s complex f u n c t i o n s on
i s , w i t h r e s p e c t t o u n i f o r m convergence on compact subsets, dense i n
c
D(G)
.
Proof: i ) i s a consequence o f Lemma 1 and t h e f a c t i i ) Stone-Weierstrass Theorem. 0
6
1
= A(L ( G ) )
(Observation 1 ) .
Another a p p l i c a t i o n o f Theorem 1 i s t h e c h a r a c t e r i z a t i o n o f F o u r i e r - t r a n s forms o f p o s i t i v e f i n i t e measures. L e t us b e g i n w i t h a d e f i n i t i o n :
A function
i s s a i d t o be o f p o s i t i v e t.ype on
f E C ag(G)
f i n i t e s e t o f numbers we have
E a
cl,...,cn n
G
i f f o r every
s ~ ~ . . .E ~G s ~
and elements
ci Ck f ( s i S i l ) 2 0
1
i,k = l O b s e r v a t i o n 3:
f E
i ) Every
6 _ is _ of
p o s i t i v e -t y p e on
G
ii)
. -finite tight
The F o u r i e r t r a n s f o r m of e very p o s i t i v e -~ G _ is _ o f positive t y pe on 6 . iii)If f _ is _ o f p o s i t i v e -t y p e on G then f ( e ) iv)
f E C aB(G)
is _ of _
(*) f o r every --
I
p o s i t i v e ----t y p e i f and o n l y i f f d(x
*
x*) 2 0
x E MB(G).
Sketch o f Proof: i)
direct verification.
ii)Take
cl,.
..,cn
E t
2 (f(s)l
and
fl,. ..,gn
- . Then
E G
measure
T
-for a l l s
on E G
.
Representing Measures
368
n
X
ck
Ci
i, k = l
FT(Ti
5;')
=
SEG
(
X i,k
ck si
Ci
5;')
J I t ci Ci\ 2 ( s ) d.r(s)
=
dT(S)
(S)
2 0
.
s€G
iii) Take
c1 = 1
, c 2 --
a and
. Then
s1 = e, s2 = s
the definition o f
p o s i ti ve t y p e y i e l ds (a
Taking i(f(s) type) Q =
6
t
1) f ( e ) t a f ( s )
a = 1 yields that
-
f(s-'))
- Ifol , i f
(Note, t h a t
f(s-'). f(s)
f(s)
af(s-l) z
+
o
.
i s r e a l , a = i shows t h a t
f(s) t f(s-l)
i s real.
. Hence fo =
t
f(e) 2 0
, since
f
i s o f positive
NOW, e x p l o i t i n g t h e same i n e q u a l i t y w i t h 0, proves t h a t f ( e ) 2 I f ( s ) I
.
i v ) Note t h a t f o r sums o f D i r a c measures ( * ) i s e q u i v a l e n t t o t h e c o n d i t i o n r e q u i r e d i n t h e d e f i n i t i o n o f p o s i t i v e type. Then a p p r o x i m a t i o n o f f i n i t e measures by sums o f D i r a c measures y i e l d s t h e equivalence o f p o s i t i v e t y p e and ( * )
.
0
The f o l l o w i n g theorem i s due t o Bochner 150 1 f o r t h e case t o H e r g l o t z [1601 f o r
G = Z
, and
G = R " , due
t o A. Weil [3221 f o r t h e general case.
Theorem 2:
. -Then t h e
Let f E
C aB(G)
i)
is o f p o s i t i v e type on -
f
following are equivalent: G
ii) There ---- i s a unique p o s i t i v e measure f(s) =
I, S ( s )
ZEG
d.r ( S )
T
E MB(G)
for a l l
-such t h a t
s E G.
369
Examples and Applications
Proof:
ii)* i )L e t x
be a sum o f D i r a c measures on
G
.
Elementary computation
yields:
I
*
f ( s ) d(x
I
x*)(s) =
G
S ( s ) S ( t ) d x ( s ) dx*(t)dT(S)
&G2
and t h i s q u a n t i t y i s Hence
f
since
1 0
i s assumed t o be p o s i t i v e .
T
i s o f p o s i t i v e type.
i)* ii) S i n c e f
6
functional
Denote by
p
on
i s bounded ( O b s e r v a t i o n 3 iii)we can d e f i n e a l i n e a r MB(G)
by
the r e s t r i c t i o n t o
1
L (G)
. We
claim t h a t
p
i s positive i n
t h e sense o f Theorem 1. I n f a c t w i t h O b s e r v a t i o n 3 i v ) we o b t a i n : U(X +
-
X*
x
*
x*) = 5
- i((6, ;(he)
-
X)
=
*
(de
f(e)
Hence, Theorem 1 p r o v i d e s a p o s i t i v e measure U(X)
=
I
V(X)
dr(v)
-
X)*)
6
1
= A(L ( G ) )
;(be
*
he)
. T
1 on A(L ( G ) ) such t h a t for all
VW1(G)) With
+
( O b s e r v a t i o n 1) we can r e w r i t e :
1 x E L (G)
.
37 0
Representing Measures
i s a B a i r e measure on
6
(since a l l
measurable). T h e r e f o r e we can extend
T
t o a r e g u l a r Bore1 measure, i . e .
Note, t h a t
T
S have t o be p we
a t i g h t measure. Combining t h e formula above w i t h t h e d e f i n i t i o n o f have:
Hence
1,
f(s) =
Uniqueness o f
T
?EG
S ( s ) dT(2)
for all
s E G.
i s an immediate consequence o f Observation 2 i i ) .
Note, t h a t Theorem 2 s t a t e s t h a t t h e f u n c t i o n s o f p o s i t i v e t y p e on the restrictions t o
G
( c o n s i d e r e d as a subgroup o f
.
6
)
G
are
o f the Fourier
Since t h e elements o f MB(6) a r e transforms o f p o s i t i v e measures on ^G d i f f e r e n c e s o f p o s i t i v e measures t h i s means t h a t t h e r e s t r i c t i o n map t o G o f t h e F o u r i e r transforms o f
i s b i j e c t i v e . NOW, c o n s i d e r
MB(g)
G
6 .
as subgroup o f
subgroup. Assume t h a t t h e r e i s an element x o E E\G symmetric neighbourhood xo U2 = I x o s t
I s,t
U
of
E U3 c Z \ G
e E
. Then
o f the characteristic functions o f CP E L2('G)
n L 1(t)
f: U
and
xo E E L G
. T h-i s
P o n t r y a g i n D u a l i t y Theorem: G = ~~
lX u2
p
E L2(g)
E.
proves
. Then
cp =
cp
+
lU * lxou2
0,
I t i s e a s i l y shown t h a t
G.
d i c t s t h e f a c t t h a t t h e r e s t r i c t i o n map t o can be no element
a compact
take the convolution
vanishes on
must be t h e F o u r i e r transform o f some
. Take
such t h a t
0
and cp
This i s then a closed
G
n
L1(c)
. But
cp
t h i s contra-
i s b i j e c t i v e . Hence, t h e r e
37 1
Examples and Applications
6.6
- KHINTCHINE FORMULA
THE LEVY
Again
MB(G)
denotes t h e bounded complex-valued
l o c a l l y compact a b e l i a n group measures i n MB(G).
. By
G
MB,(G)
i g h t measures on t h e
we denote t h e p o s i t i v e
Q u i t e o f t e n ,in p r o b a b i l i t y t h e o r y ,one has t o deal
w i t h i n f i n i t e l y d i v i s i b l e elements of t h a t f o r every given
t h e r e i s some
n E N
*
LI = v
*. . .
v
MB+(G)
*v
v
i.e.
E MB,(G)
p
such
with
E MB,(G)
(n-times).
F o r example, c o n s i d e r t h e p r o b a b i l i t y d i s t r i b u t i o n o f a random v a r i a b l e which, f o r a r b i t r a r y
, can
n
be w r i t t e n as t h e sum o f
n
independent
random v a r i a b l e s . T h i s i s t h e s i t u a t i o n which leads t o t h e c e n t r a l l i m i t theorem. Other examples a r e g i ven by t r a n s 1 a t i on
- in v a r i a n t Markov-semi -
) . S i n c e t h e s e examples p l a y an i m p o r t a n t r o l e i n
groups ( s e e [1421,[27]
appl ied p r o b a b i 1 it y t h e o r y
t h e c h a r a c t e r i z a t i o n o f in f i n i t e l y d i v i s i b l e
measures i s o f general i n t e r e s t . I n o r d e r t o a v o i d t e c h n i c a l i t i e s we r e s t r i c t o u r c o n s i d e r a t i o n s t o t h e s p e c i a l case
G =(R,+).
B u t a l l t h e arguments can b e t r a n s f e r r e d t o a more
general s i t u a t i o n . Those readers who a r e i n t e r e s t e d i n t h e general t h e o r y of convol u t i o n semi groups a r e r e f e r r e d t o [158 1,1171 1
81 1 , see a1 so
[1781 and [1421. Recall where
if
G = R
x E R
transform
.
t h e n a1 1 c h a r a c t e r s a r e o f t h e form
Hence we can i d e n t i f y o f a measure
F
lJ
F,(x)
=
R
p €
MB(R)
fi w i t h
, we
ei t x d p ( x )
,
t
+
e
-i tx
and,for t h e F o u r i e r -
R
o b t a i n t h e w e l l known f o r m u l a x E R
=
6 .
Consider t h e f o l l o w i n g s i t u a t i o n : Fix
N E N~
and l e t
f,goygly...ygN~l
f u n c t i o n s such t h a t :
(1)
lim It1
f(t) = 1 + -
: IR + R
be bounded
continuous
372
Representing Measures
(2)
f(t) > 0
(3)
f
t
for all
+
0
N-times continuously d i f f e r e n t i a b l e a t 0 with
is
and f ( " ) ( O )
f")(O)
> 0
(4)
g0(O)
g1(0) =
(5)
t h e functions g k a r e (N-k)gi") ( 0 )
=
=
* * .
k
< m
for 0 5 m < N
0
gN-,(0)
1 -< m 5 N
0 for
lim S U P It g k ( t ) l It1 + -
(6)
=
=
1
times d i f f e r e n t i a b l e a t 0 with
-
k
for k = OY1s2,...,N-1.
Theorem: Let i)
: R +
Whenever p1
ii)
be continuous. -Then t h e following a r e equivalent:
R
f tF
p E MB(R)
P
2-pf
has compact ~support and pl,p2
-,
E K,
-such t h a t
then
There a r e a polynomial P ( t ) of degree < N a p o s i t i v e regular Bore1 measure T W\{Ol and a t 0 w i t h a + ~ ( b I 0 ) I ) 1 -representation --holds f o r a l l t E R: such t h a t t h e following integral ---
u(t)
=
A d +
P(t)
t
f(N)(O)
(For N = 0 t h e sum term of t h e integrand vanishes). Note t h a t the integrand i n t h e i n t e g r a l representation of u i s a bounded continuous function on R ( a consequence of T a y l o r ' s formula together with the assumptions on f and go,...,gN-l) .
Examples and Applications
373
B e f o r e we p r o v e t h e theorem we mention a n o t h e r o b s e r v a t i o n which can be proved by elementary c a l c u l a t i o n : Remark: Let
MB(R)
p E
have compact s u p p o r t . Then t h e F o u r i e r t r a n s f o r m
J
F (x) =
u
R
eixt
dp(t)
i s i n f i n i t e l y o f t e n d i f f e r e n t i a b l e and t h e moments o f
are given by
LI
P r o o f o f t h e Theorem:
(ii) + (i):L e t
x
> 0
. Since
must h o l d f o r
have compact s u p p o r t such t h a t
p
the f i r s t
(N-1) d e r i v a t i v e s o f
. Therefore
Fp
the f i r s t
(N-1)
IF I 5
u
xf
f o r some
v a n i s h a t zero, t h e same
f
moments o f
vanish
p
(remark above). A p p l i c a t i o n o f F u b i n i ‘ s theorem t o t h e i n t e g r a l representation yields:
where
= T
T~
+ a 6
0
and where
FU(c’) f ( E ) - l
i s extended c o n t i n u o u s l y
i n t o t h e o r i g i n ( 1 ’ H o p i t a l ’ s lemma). Th s f o r m u l a immed a t e l y y i e l d s t h e required inequalities.
(i) (ii): measures p
Consider t h e v e c t o r space such t h a t
(7)
p
(8)
F ( 0 ) = F (1)( 0 ) =
E c MB(R)
cons s t i n g o f those
has compact s u p p o r t and t h e F o u r i e r t r a n s f o r m
u
P
moments a r e z e r o
*.*
= F(N-l)(0)
u
= 0
F
v
i s r e a l valued
, i.e. the f i r s t
(N-1)
374
Representing Measures
(9)
.
FP vanishes a t i n f i n i t y
Recall, t h a t t h e F o u r i e r t r a n s f o r m q u i t e o f t e n vanishes a t i n f i n i t y , f o r example i f !J has a continuous densi t y ( R i emann-lebesgue Lemma). Observe t h a t i f f o r a continuous f u n c t i o n v we have t h a t
Iv
!R
then
v
0
dp =
for all
1-1 E
must be a polynomial o f degree
, the
Now consider
R*
each element
r o f IR*
E
< N
.
one p o i n t c o m p a c t i f i c a t i o n o f a linear functional
on
cpr
R
,
and a s s i g n t o
E :
r+O,m
r = m
1F !J( N ) ( 0 )
One e a s i l y checks t h a t
r
+
if
cpr(p)
r = O
i s continuous on R*
f o r every
!J
E
The c o n d i t i o n imposed i n ( i ) guarantees
I u d!J I
R
sup CP,(!J)
for a l l
rER*
!J
E E
.
The Riesz-Konig Theorem g i v e s us a r e g u l a r Bore1 p r o b a b i l i t y measure on IR*
such t h a t
I u d!J = I R*
IR Define p
a = 0(0)
and
E vanish and s i n c e
T = u
cpr(u) d u ( r )
IR\
cp,(!~)
I01
.
p E
E
.
Since t h e f i r s t (N-1) moments o f
i s equal t o
moment we o b t a i n v i a F u b i n i ' s theorem
for a l l
f(N)(0)-l
times t h e N-th
a
E
.
Examples and Applications
J u
R
du =
J v
R
dp
for all
p E
375
E ,
where
B u t v is a continuous function ( T a y l o r ' s formula). F i n a l l y , s i n c e J ( u - v ) d u = 0 f o r a l l p E E the functions
u and v
R
must be equal u p t o a polynomial of degree < N . The condition a + ~ ( R \ { 0 1 ) I 1 i s f u l f i l l e d because T was a p r o b a b i l i t y measure. Let us i l l u s t r a t e this r e s u l t f o r concrete cases. F i r s t observe t h a t we have proved again Bochner's Theorem i n t h e following fomulation: Bochner's Theorem: A continuous function u -
: R +R i s t h e Fourier of-a p o s i t i v e - transform ~ T on R w i t h T ( R ) 5 1 i f and only i f f o r every regular Borel measure ~-~ -----complex Borel measure p 0" R with compact support -and w i t h p1 2 F 2 - p 2 (p, , p 2 E R+) -we have: u
Proof: Take N
=
0 , f = 1, go = l .
0
I t i s not a t a l l d i f f i c u l t t o transform t h i s condition i n t o the positivetype-condi t i o n .
376
Representins Measures
Another s p e c i a l case i s : The Levy
-
Let u : R
Khintchine
hu
g continuous f u n c t i o n . Then,for every
A 2 0, e i s t h e F o u ri e r t r a n s f o~ rm _ o f -some p o s i t i v e f i n i t e r e g u l a r B o r e l measure on R -i f and only i f t h e r e a r e a 2 0 , E R and a p o s i t i v e -f i n i t e Borel measure T JIJ R \ I O ) such t h a t -+
R
Formula:
_
_
_
_
_
I
_
_
_
~
-
_.-
for all --
t E IR
Proof:
A s t r a i g h t f o r w a r d c a l c u l a t i o n shows t h a t f o r any u ( t ) , which has an i n t e g r a l r e p r e s e n t a t i o n as above, t h e f u n c t i o n s e x u(t), x 1 0 , a r e indeed F o u r i e r t r a n s f o r m s o f p o s i t i v e f i n i t e B o r e l measures o n R . N = 2
put
,
f ( t ) = ( l + t 2 ) -t2 I , g o ( t ) = 1 and
x
Assume t h a t f o r every
t 0
the function
exu
gl(t)
=
( 1 + t 2 -1
.
i s the Fourier transform
of some p o s i t i v e f i n i t e B o r e l measure on R. We c l a i m t h a t t h e r e i s some
0
0
such t h a t
.
M2 FT s f s M I F Now, c o n s i d e r plf
t F
1-1
3 = (pl MI
M 2 0
where
t - p2 f T
-
. Hence,
p
,p2
p1
1-1)
t 0
. Since
that
The F o u r i e r -
sin t 7 ).
( F T ) ( t ) = (1 We can f i n d
[-1,+1].
u
and p1
M1 FT 2 F
.
Fy t 0
E MB(R)
u
w i t h compact s u p p o r t such t h a t we o b t a i n f o r t h e measure
I n a d d i t i o n we have
F- 5 M f
f o r some
we o b t a i n f r o m ( * )
i s independent o f
= ( p 2 M 1 ~t u )
-
u
. For
t h e o t h e r i n e q u a l i t y we p u t
and o b t a i n w i t h ( * )
J u d u
So, t h e c l a i m i s proved.
) a1
if
-
cp
i s open
A r a t h e r good-natured o b j e c t i s Bd x ( X ) = (X,x)
{flf
: X +IR
i s bounded and
an a r b i t r a r y measurable space.
(I, Bor(lR))- measurable},
T h i s i s n o t o n l y a v e c t o r space, b u t
i n a d d i t i o n a v e c t o r l a t t i c e (under p o i n t w i s e suprema and i n f i m a ) which is
u- complete. Here
whenever
u- complete means t h a t
fn E B d I ( X ) , n E N, such t h a t
pleteness o f Instead o f respectively
BdI(X) BdB(X)
.
sup fn belongs t o nm
sup fn i s bounded. The nm
r e f l e c t s t h e p r o p e r t y r e q u i r e d i n 1.1 and
BdB, (X), we w r i t e
Bor(X)
BdI(X) u-com-
(iii).
and B a i r e ( X ) ,
Measures and the Riesz Representation Theorem
A 2
MEASURES
2.1
Definition:
Let (X,E) be a measurable space. A map m : X measure i f m(D) = 0 and i f (i) m(A U 8 ) t m ( A
(ii)
If
fi
B)
=
for all
m ( A ) t m(B)
An E X, n E N , such t h a t
+
An+1
2
An
383
i s called
uI+-)
IRt
A,B E
for a l l
x n E N
then we have always sup m ( A n )
n a
=
m(
U An) nEN
.
Sometimes the measures defined A measure i s c a l l e d f i n i t e i f m ( X ) < above a r e c a l l e d p o s i t i v e measures t o s e t them a p a r t from signed measures. 03.
Signed measures a r e differences of positive measures. The tripe1 (X,r,m) i s usually s a i d t o be a ~measure space. P o s i t i v e f i n i t e measures w i t h m ( X ) = 1 a r e c a l l e d p r o b a b i l i t y measures, and (X,Z,m) i s then s a i d t o be a p r o b a b i l i t y space. An obvious consequence of i .e.
2 . 1 ( i ) i s t h a t p o s i t i v e measures a r e monotone,
m ( A ) I m(B)
whenever
A,B E II w i t h
A
c
B.
And .:om 2 . ( i i ) t h e important property follows t h a t measures a r e 0- a d d i t i v e , t h a t means : m(U{An
which
In
E Nl)
I 1
n a
m(An),
when An E E, n E N
,.
implies e q u a l i t y i n c a s e t h a t t h e An a r e pairwise d i s j o i n t .
A s e t N E x with m ( N ) = 0 i s c a l l e d an m-nullset and two functions f, g on X having d i f f e r e n t values only on an m-nullset a r e s a i d t o be equal m- almost everywhere ( f = g m- a . e . ) .
Appendix
384
(X,x,m)
Let
Then a measurable f u n c t i o n measure m
?\
For
Then of
m
from
X
to
X
~p :
x
--t
. This
X
a measurable space.
can be used t o t r a n s f e r t h e
can be done i n t h e f o l l o w i n g way.
define
E
m
(i,i) be
be a measure space and l e t
. This
i s a measure on
under 10
.
measure i s c a l l e d t h e -~ image measure
The most i m p o r t a n t f e a t u r e of measures i s t h a t t h e y enable us t o d e f i n e i n t e g r a l s i n a v e r y n a t u r a l way. Roughly speaking, i n t e g r a l s a r e p o s i t i v e l i n e a r f u n c t i o n a l s w i t h s t r o n g convergence p r o p e r t i e s .
(X,r,m) be a measure space. For s i m p l i c i t y we assume t h a t m i s a f i n i t e measure. To d e f i n e t h e i n t e g r a l we s t a r t w i t h measurable step-
Let
f u n c t i o n s . These a r e f u n c t i o n s o f t h e f o l l o w i n g t y p e :
n s = E a k=l where
1 Ak
1 Ak
,
akEW,
AkEE,nEH,
denotes t h e c h a r a c t e r i s t i c f u n c t i o n o f
Ak, For a such a step-
f u n c t i o n an i n t e g r a l i s g i v e n by
X
s dm
n E ak m(Ak) k=l
=
One can e a s i l y show t h a t t h i s d e f i n i t i o n makes sense,i.e. s,
another r e p r e s e n t a t i o n o f
s
=
m
Now, f o r a r b i t r a r y
J
X
f dm =
m E B . 1 , B . E IR, 6 . E j=1 J Bj J J
x ,
then
n
f E Bd 1 (X)
sup[
i f there i s
we d e f i n e
J s dm I f X
2 s,s
measurable s t e p - f u n c t i o n } .
385
Measures and the R i e v Representation Theorem
The i n t e g r a l o f
f
B E II of
on a subset
J f d m = B
J f . l B d m X
X
i s d e f i n e d as
.
An easy c a l c u l a t i o n y i e l d s t h a t by
-,
f
j f d m X
a l i n e a r f u n c t i o n a l on
Bd Z ( X )
i s g i v e n . Furthermore t h i s f u n c t i o n a l i s
monotone, i.e.
1f X
dm 5
X
f
dm
for
f,? E B d I ( X ) w i t h
f I
f .
B u t i n c o n t r a s t t o o r d i n a r y l i n e a r f u n c t i o n a l s t h e i n t e g r a l has t h e f o l l o w i n g s t r o n g convergence p r o p e r t y : Monotone convergence p r o p e r t y : (i.e.
f n+l < for a l l n) - fn -
L e t fn, n E N, be-a d e c r e a s i n g sequence n a
bounded, t h e n ~inf n€N
j X
fn dm
is
i n B d z ( X ) -such t h a t i n f (f,) -
=
I
X
i n f (f,) n€N
T h i s convergence p r o p e r t y r e f l e c t s t h e
0-
dm.
a d d i t i v i t y o f t h e measure m.
Only f o r convenience we have s t a t e d t h i s p r o p e r t y f o r bounded f u n c t i o n s .
A s i m p l e e x e r c i s e shows t h a t i t goes o v e r unchanged t o unbounded f u n c t i o n s . V a r i a t i o n s o f t h e above r e s u l t can be found i n c o u n t l e s s forms i n t h e 1 i t e r a t u r e (Lebesgue Dominated Convergence Theorem, F a t o u ' s Lemma, Lemma o f Beppo -Levi
, etc. ) .
NOW, l e t us c o n s i d e r a monotone, r e a l v a l u e d l i n e a r f u n c t i o n a l B d Z (X)
h a v i n g t h e monotone convergence p r o p e r t y , i . e . i n f n€N
u on u(fn) =
386
Appendix
p ( i n f fn) whenever n€N
5 fn
f,+l
for a l l
n E N and
i n f fn i s bounded. nm
One c o u l d have t h e i d e a t o c a l l such a l i n e a r f u n c t i o n a l an a b s t r a c t i n t e g r a l . B u t t h i s does n o t l e a d t o a n y t h i n g new, s i n c e from an a b s t r a c t integral
for
p :
A E 1
Bdx(X) + R we can e a s i l y o b t a i n a measure by d e f i n i n g
. And -
If X
Finally, a t r i v i a l
- we
no wonder dm =
p(f)
- but useful
should be mentioned. L e t
have then for a l l
-
be a measure space and l e t
measurable space. Then g i v e n a measurable f u n c t i o n
J f dm
=
where
A 3
J
X
dm
focp
i s t h e image measure o f
--
f o r m u l a f o r i n t e g r a l s o f image measures
(X,f,m)
X
.
f E Bz(x)
forall
m under
cp
cp : X + X
(XJ)
be a
we have
f EBd;(i),
.
THE RIESZ REPRESENTATION THEOREM
w i l l be a nonempty compact Hausdorff space and p : C(X) + I R i s a monotone l i n e a r f u n c t i o n a l on t h e r e a l v a l u e d continuous f u n c t i o n s on X Here "monotone" i s always used w i t h r e s p e c t t o t h e p o i n t w i s e o r d e r o f C(X). For t h e e x t e n s i o n o f p t o an i n t e g r a l we need t h e knowledge o f t h e f o l l o w i n g s i m p l e f a c t ( p r o o f l e f t t o t h e r e a d e r ) :
Throughout t h i s s u b s e c t i o n X
.
3.1
Remark:
Let Y
semicontinuous. I f Y < i s a continuous
beupper semicontinuous on (o
g with
(i.e.
X
~ ( x < ) q ( x ) for a l l
Y 5 g < cp
.
and l e t cp be l o w e r x E X)
then there
C e r t a i n l y , t h e reader has observed immediately t h a t , because of t h e compact-
387
Measures and the Riesz Representation Theorem
ness of X , t h i s i s a s p e c i a l case o f t h e famous Tong - K a t e t o v Theorem. We s t a t e d i t i n t h i s s p e c i a l form o n l y t o enable everybody t o f i n d t h e simple proof f o r himself. Now, we extend
G
By d e f i n i t i o n
;:
t o a functional
p
USC(X)
6
-*
on t h e cone o f upper
X. T h i s i s done b y
semicontinuous f u n c t i o n s on
i s s u b l i n e a r and monotone ( w i t h r e s p e c t t o p o i n t w i s e
o r d e r ) . Furthermore,
c o i n c i d e s on
LI
d i f f i c u l t t o see i s t h e f a c t t h a t
C(X)
with
.
p
Somewhat more
G i s indeed l i n e a r .
3.2
Lemma:
(i)
G
i s linear
( i i)
i
has t h e monotone converqence p r o p e r t y USC(X) , i.e. when n E N , jsi p o i n t w i s e d e c r e a s i n g sequence in USC(X) then
,Y,
~
i i ( y n ) = G ( i n f yn) n a
inf na
Proof:
. i s s u b l i n e a r , so i t remains t o
(i) We a l r e a d y observed t h a t
i s s u p e r a d d i t i v e . F o r t h i s purpose we t a k e a r b i t r a r y
prove t h a t I Y ~ , YE ~ USC(X).
the function
Then f o r
cp =
g
Using Remark 3 . 1
-
yll
g E C(X)
+
E
with
g 2
yll
+
y2
and
i s l o w e r semicontinuous w i t h
we f i n d a continuous
h with
h2
cp>
t
> Y2
cp y12
> 0
. We
. put
h 1 -- g - h t t r ~ a ~n d g e t
Using t h e f a c t t h a t
LI
i s a d d i t i v e on
C(X),
t h i s yields:
NOW, t a k i n g on t h e r i g h t hand s i d e t h e infimum o v e r
g
and
t
we have
Appendix
388
Hence,
must be s u p e r a d d i t i v e .
(ii) The i n e q u a l i t y
i s t r i v i a l , since w i t h g 1 Y = i n f Y, n€N some
no E N
i s monotone. Now, we t a k e an a r b i t r a r y g E C(X) by compactness of X , we find, f o r E > 0
. Then
such t h a t
Taking t h e i n f over
gt
g and
E
> Y
E
.
,
Hence
on t h e r i g h t s i d e we g e t t h e d e s i r e d
inequal ity
I n t h e c o n s t r u c t i o n o f C we have approached t h e upper semicontinuous f u n c t i o n s f r o m above by continuous f u n c t i o n s w i t h o u t l o o s i n g a- a d d i t i v i t y . Now, we c o n t i n u e t h i s process by approaching t h e bounded f u n c t i o n s f r o m below by upper semicontinuous f u n c t i o n s . By d o i n g t h i s we l o o s e u- a d d i t i v i t y as w e l l as l i n e a r i t y . Hence we have t o r e s t r i c t t h e con-
s t r u c t e d f u n c t i o n a l t o a s u i t a b l e subspace where a1 1 d e s i r e d p r o p e r t i e s remain v a l i d . F o r t u n a t e l y , i t t u r n s o u t t h a t
Bor(X)
i s such a s u i t a b l e
subspace. L e t us d e s c r i b e t h i s procedure i n g r e a t e r d e t a i l . We denote t h e space o f bounded f u n c t i o n s on functionals f o r
b E Bd(X) ;(b)
X
by
by:
= SUP{;(Y)
I \y
E USC(X),
5 b)
Bd(X). We d e f i n e
389
Measures and rhe Riesz Representation Theorem
and
-
G(b) =
;(-
b ) = i n f { - ;(P)
Both f u n c t i o n a l s a r e monotone and d e f i n i t i o n s u p e r l i n e a r whereas G super1 in e a r i t y o f
;
IY
E USC(X),
-
\y
2 b}
.
( c a l l e d t h e l o w e r i n t e g r a l ) i s by
(upper i n t e g r a l ) i s s u b l i n e a r . The
imp1 i e s :
Hence we have -
(3.1)
v
ll2ll.
Obviously,
equals
; on
from t h e d e f i n i t i o n o f
.
USC(X). And, s i n c e
C ( X ) c USC(X), we o b t a i n
;(Y) s ;(Y) f o r a l l
that
Y E USC(X).
Hence, we have
;(Y) as a consequence o f
=
G(Y)
for all
Y E USC(X)
(3.2)
(3.1).
L e t us t u r n o u r a t t e n t i o n t o
U-
a d d i t i v i t y . The b e s t we can do i s gathered
i n t h e f o l l o w i n g obvious 3.3
Let
Remark: bn E Bd(X), n E N
supremum
, be
-
b
=
sup bn n a
(ii) ;(b)
s
sup L ( b n ) n a
a p o i n t w i s e i n c r e a s i n g sequence w i t h bounded
Then
.
I f we have i n a d d i t i o n
G(bn) =