The Bidual of C(X)I

THE BIDUAL OF C(X) I NORTH-HOLLAND MATHEMATICS STUDIES 101 THE BIDUAL OF C(X) I Samuel KAPLAN Purdue University W...

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THE BIDUAL OF C(X) I

NORTH-HOLLAND MATHEMATICS STUDIES

101

THE BIDUAL OF C(X) I

Samuel KAPLAN Purdue University West Lafayette Indiana U.S.A.

1985

NORTH-HOLLAND -AMSTERDAM

0

NEW YORK

OXFORD

' Elsevier Science Publishers B V , 1985 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 of otherwise, without the prior permission of the copyright owner

ISBN: 044487631 6

Publishers:

ELSEVIER SCIENCE PUBLISHERS B.V P.O. BOX 1991 1000 BZ AMSTERDAM THE NETHERLANDS

Sole distributors for the U.S.A. and Canada:

ELSEVIER SCIENCE PUBLISHING COMPANY, INC 52 VAN DER B ILT AVENUE NEW YORK, N.Y. 10017 U.S.A.

L i b r a r y of Congres5 Cataloging i n Publication D a t a

&plan, Samuel, 1916The bidual of C(X) I.

(North-Holland mathematics studies ; 101) Bibliography: p. 1. Banach spaces. 2. Duality theory (Mathematics) 3. Embeddings (Mathematics) I. Title. 11. Title: Bidual of C.(X). I. 111. Series. 515.7'32 84-18665 QA322.2.K36 1985 ISBN 0-444-87631-6 PRINTED I N THE NETHERLANDS

PREFACE

T h e most commonly o c c u r r i n g , and p r o b a b l y t h e most i m p o r t a n t ,

n o n - r e f l e x i v e r e a l Banach s p a c e s ’ ( a l l o u r s p a c e s a r e o v e r t h e r e a l s ) a r e t h o s e o f t h e form C(X), X c o m p a c t , and t h o s e o f t h e 1 form L ( p ) , 1-1 a g i v e n m e a s u r e . For a n o n - r e f l e x i v e Banach s p a c e E , t h e r e a r e some n a t u r a l p r o b l e m s which do n o t a r i s e i n t h e r e f l e x i v e c a s e . such a r e : t o d e s c r i b e i t s b i d u a l E ” ,

Three

t o d e s c r i b e t h e imbedding

o f E i n E ” ; and - s i n c e E ” i s d e t e r m i n e d by E - t o s e e how

much o f t h e s t r u c t u r e o f

El’

can be o b t a i n e d from t h i s i m b e d d i n g . 1

For a s p a c e o f t h e form C(X) o r L ( p ) , t h e answer t o t h e f i r s t p r o b l e m i s w e l l known: t h e b i d u a l o f C(X) i s a C(Y), Y c o m p a c t , and t h a t o f L 1( p ) i s an L 1 (v) , v some m e a s u r e . A l s o t h e imbedd i n g o f L 1( p ) i n i t s b i d u a l i s w e l l known, and c a n b e d e s c r i b e d i n a r e l a t i v e l y s i m p l e and s a t i s f y i n g manner. The p r e s e n t work i s c o n c e r n e d w i t h t h e imbedding of a s p a c e C(X) i n i t s b i d u a l C“(X) and t h e s t u d y o f what p r o p e r t i e s of t h e l a t t e r c a n b e f o u n d s t a r t i n g from t h i s imbedding. my knowledge, l i t t l e h a s b e e n done i n t h i s d i r e c t i o n .

To

Or

r a t h e r , what we have i s a s c a t t e r e d body o f o c c a s i o n a l r e s u l t s ; t h e r e h a s been no s y s t e m a t i c a p p r o a c h t o t h e p r o b l e m . For a number o f y e a r s now, I have d e v o t e d m y s e l f t o s u c h a s y s t e m a t i c a p p r o a c h , and many o f t h e r e s u l t s have a p p e a r e d i n V

vi

Preface

a series of papers ( [ 2 2 ] - [ 2 8 ] ) .

My program i n t h i s monograph

i s t o p r e s e n t what I know, b o t h from my own work and t h a t of

o t h e r s , a s a u n i f i e d whole.

The c a n v a s i s a l a r g e o n e , and

t h e m a t e r i a l h e r e c o n s t i t u t e s o n l y t h e f i r s t volume. m a t i c i a n o f s t a n d i n g once remarked t o me t h a t C ' l ( X ) a gigantic C(Y),

A mathe-

was s i m p l y

much t o o c o m p l i c a t e d t o make g e n e r a l s t a t e -

A c t u a l l y , o f c o u r s e , many g e n e r a l s t a t e m e n t s

ments a b o u t .

Indeed, t h e b i d u a l of a general C ( X )

can be made a b o u t i t .

( i n common w i t h a l l m a t h e m a t i c a l o b j e c t s ) h a s an o r d e r l y and beautiful structure. C(X)

and i t s b i d u a l c a n be s t u d i e d t h r o u g h t h e i r r i n g

structure or, equivalently, t h e i r vector-space-cum-lattice structure.

We f o l l o w t h e l a t t e r r o u t e , examining them a s

Banach l a t t i c e s , t h a t i s , norm c o m p l e t e normed R i e s z s p a c e s . P a r t I i s an i n t r o d u c t i o n t o R i e s z s p a c e s , w i t h t h e c o n c e p t s and n o t a t i o n s w e w i l l u s e . ( s p a c e s o f t h e form C ( X ) ) ~'(p)).

P a r t I 1 d o e s t h e same f o r M l - s p a c e s and L - s p a c e s ( t h o s e o f t h e form

P a r t 111 c o n s i s t s of c l a s s i c a l m a t e r i a l on c ( x ) , i t s

d u a l , and b i d u a l

.

I n t h e r e m a i n d e r o f Volume I , we p r o c e e d w i t h t h e d e v e l o p ment o f o u r program.

I n p a r t I V , we s i n g l e o u t - o r o c c a s i o n -

a l l y d i s c o v e r - t h e s u b s p a c e s o f C l l ( X ) which seem t o be t h e most i m p o r t a n t .

The p r i n c i p a l one ( a f t e r C ( X )

t o be t h e s u b s p a c e U ( X )

i t s e l f ) seems

of "universally integrable" elements.

I t c o n t a i n s t h e s u b s p a c e s o f Bore1 e l e m e n t s and B a i r e e l e m e n t s , and s h a r e s many o f t h e i r p r o p e r t i e s .

In the context of C'l(X),

i t seems t o be a more n a t u r a l s u b s p a c e t h e n e i t h e r o f t h e l a t t e r

t o work w i t h .

Also p l a y i n g an i m p o r t a n t r o l e i s t h e norm

c l o s e d l i n e a r s u b s p a c e g e n e r a t e d by t h e " l o w e r s e m i c o n t i n u o u s "

Preface

vii

( e q u i v a l e n t l y , t h e "uppersemicontinuous")

elements.

I t seems

t o me w o r t h y o f more a t t e n t i o n t h a n h a s s o f a r been g i v e n t o i t . I n a n a t u r a l way, t h e d u a l Cl(X) o f C(X) i s t h e "home" o f a l l t h e s p a c e s L 1 (u),

p r u n n i n g t h r o u g h t h e Radon,

m e a s u r e s on X , and C"(X) t h e s e measures.

or regular,

i s t h e "home" o f t h e s p a c e s L m ( p ) f o r

Riemann i n t e g r a t i o n , t h e s u b j e c t o f P a r t V ,

t u r n s o u t t o be an i m p o r t a n t u n i f y i n g c o n c e p t i n s t u d y i n g t h e imbedding o f t h e s e L 1 ( p ) I s i n C'(X) and L m ( p ) ' s i n C v f ( X ) , I t i s c l o s e l y r e l a t e d t o what we c o n s i d e r an i m p o r t a n t c o n c e p t i n

R i e s z s p a c e s : t h e "Dedekind c l o s u r e " o f a s u b s e t .

This concept

a l s o i n c l u d e s Lebesgue i n t e g r a b i l i t y ( t o be d i s c u s s e d i n Volume 1 1 ) . I n P a r t V I , we examine t h e i d e a l o f " r a r e " e l e m e n t s ( c o r r e s p o n d i n g t o t h e nowhere d e n s e s e t s i n t o p o l o g y ) and i t s r e l a t i o n t o t h e Dedekind c o m p l e t i o n o f C(X). C'l(X) i s c l o s e l y c o n n e c t e d n o t o n l y w i t h i n t e g r a t i o n t h e o r y , but a l s o with general topology.

With e a c h Boolean a l g e b r a ,

t h e r e i s a s s o c i a t e d a t o p o l o g i c a l s p a c e , i t s S t o n e s p a c e , and t h e s t u d y o f Boolean a l g e b r a s h a s b e e n e n r i c h e d by t h e s t u d y of t h e t o p o l o g i c a l p r o p e r t i e s of t h e s e Stone spaces.

Parallel-

i n g t h i s , f o r e a c h C(X), Cll(X) seems t o p l a y t h e r o l e o f a " S t o n e s p a c e " f o r C(X).

As the reader w i l l see, there i s a

r e m a r k a b l e a r r a y 'of c o n c e p t s i n C " ( X )

analogous - i n f a c t ,

g e n e r a l i z i n g - t h e p r i n c i p a l concepts o f (general) topology. Some o f t h e s e a p p e a r i n C h a p t e r 1 0 , o t h e r s i n P a r t V I . The r e l a t i o n o f C " ( X )

t o the function spaces of ordinary

a n a l y s i s on X i s examined i n t h e monograph.

T h i s i s done

t h r o u g h t h e Isomorphism t h e o r e m , e s t a b l i s h e d i n 5 5 3 .

Although

t h e p r o o f i s r e l a t i v e l y s i m p l e , t h e t h e o r e m i s n o n - t r i v i a l , and

viii

Preface

h i g h l i g h t s t h e c e n t r a l r o l e of t h e Dini theorem i n t h e r e l a t i o n b e t w e e n C(X) a n d C"(X)

While t h e i n t r o d u c t o r y m a t e r i a l i n P a r t s I and I 1 i s s e l f c o n t a i n e d , o u r g o a l i s a l w a y s C"(X), two p a r t s a r e o f t e n o m i t t e d .

and t h e p r o o f s i n t h e s e

F o r an e n c y c l o p e d i c p r e s e n t a t i o n

o f R i e s z s p a c e s , w e r e f e r t h e r e a d e r t o Luxemburg and Zaanen [ 3 6 ] ; f o r a more l i m i t e d , b u t b e a u t i f u l l y w r i t t e n , p r e s e n t a t i o n ,

we r e f e r h e r o r him t o S c h a e f e r [ 4 7 ] ; a n d f o r a n e x t e n s i v e d e v e l o p m e n t o f t o p o l o g i c a l R i e s z s p a c e s , t o A l i p r a n t i s and Burkinshaw [ 2 ] .

F i n a l l y , f o r C(X) i t s e l f , t h e r e i s Semadeni [ 4 9 ] .

To k e e p t h e i n t r o d u c t o r y m a t e r i a l from becoming t o o d e n s e , I h a v e r e l e g a t e d some o f i t t o E x e r c i s e s i n t h e f i r s t t h r e e

chapters. I w a n t t o a c k n o w l e d g e h e r e some d e b t s .

To t h e D e p a r t m e n t o f

Mathematics of Purdue U n i v e r s i t y f o r providing, over t h e y e a r s , t h e a t m o s p h e r e a n d t h e a s s i s t a n c e n e e d e d t o c a r r y on t h e o r i g i n a l work a n d t h e work on t h i s monograph.

To t h e D e p a r t m e n t o f

M a t h e m a t i c s o f W e s t f i e l d C o l l e g e , U n i v e r s i t y o f London, w h e r e I carried

o u t a good p a r t o f t h e work a s a g u e s t o f t h e D e p a r t -

ment f o r more times t h a n I c a n remember.

To Owen B u r k i n s h a w

f o r k i n d l y p r o o f r e a d i n g t h e e n t i r e f i n a l copy.

And f i n a l l y , t o

E l i z a b e t h Young f o r t h e many h o u r s a n d t h e c a r e w h i c h s h e d e v o t e d t o t y p i n g t h i s volume.

Purdue U n i v e r s i t y July, 1984

TABLE OF CONTENTS

PREFACE

V

PART I BACKGROUND

CHAPTER 1

RIESZ SPACES

51

Ordered v e c t o r s p a c e s

2

52

Riesz spaces

5

53

R i e s z s u b s p a c e s and R i e s z i d e a l s

9

54

Order convergence

15

55

O r d e r c l o s u r e and b a n d s

20

56

R i e s z homomorphisms

26

57

Dedekind c o m p l e t e n e s s

31

58

Countabilit y p r o p e r t i e s

36

59

R i e s z norms and Banach l a t t i c e s

38

E xe r c i s e s

CHAPTER 2

44

RIESZ SPACE DUALITY

510 The s p a c e E b o f o r d e r bounded l i n e a r f u n c t i o n a 1s

52

9 1 1 E a s " p r e d u a l " o f Eb

62

ix

Contents

X

512

The s p a c e E C o f o r d e r c o n t i n u o u s l i n e a r functions

69 bb

74

513

The c a n o n i c a l i m b e d d i n g o f E i n E

514

The t r a n s p o s e o f a R i e s z homomorphism

78

515

The d u a l o f a Banach l a t t i c e

88

Exercises

89

PART I 1 L-SPACES AND Mll-SPACES

CHAPTER 3

§

16

MIL-SPACES AND L-SPACES

93

MI- s p a c e s

517

The components o f ll

100

918

MI-homomorphisms

108

519

L-spaces

109

520

The e x t r e m e p o i n t s o f K(E)

116

Exercises

CHAPTER 4

119

DUAL L-SPACES AND MIL-SPACES

521

The d u a l o f an MIL-normed s p a c e

120

522

1 ~ 1

a diversion

126

§23

The d u a l o f a n L - s p a c e

132

524

Mappings o f L - s p a c e s

138

(El

,E)

,

Contents

CHAPTER 5

xi

DUALITY RELATIONS BETWEEN AN L-SPACE AND ITS DUAL

525

T o p o l o g y on t h e d u a l o f an L - s p a c e

144

526

Dual b a n d s

146

527

B a s i c b a n d s i n t h e d u a l o f an L-space

147

528

Equi-order-continuity

152

529

Weak c o m p a c t n e s s

159

PART I 1 1 C (X)

CHAPTER 6

,

C (X)

,

C" (X) : THE FRAMEWORK

(C(X) ,X) - D U A L I T Y

The t o p o l o g y o f s i m p l e c o n v e r g e n c e

530

on C(X) 531

166

The d u a l i t y b e t w e e n t h e norm c l o s e d R i e s z i d e a l s o f C(X) and t h e c l o s e d

( o r open) s u b s e t s of X 532

169

The d u a l i t y b e t w e e n t h e MIL-subspaces o f C(X) and t h e u p p e r s e m i c o n t i n u o u s decomposition

933

CHAPTER 7

of X

The MIL-homomorphisms o f C(X)

(C(X)

,

176 181

C ' (X)) - D U A L I T Y

934

The i m b e d d i n g o f X i n C ' ( X )

186

535

Atomic a n d d i f f u s e Radon m e a s u r e s

188

xii

Contents

536

The v a g u e t o p o l o g y on C ' ( X )

193

537

Mapping d u a l i t y

199

CHAPTER 8

C"(X)

938

The i m b e d d i n g o f C(X) i n C " (

539

Some s i m p l e s e q u e n c e s p a c e s

8

206

208

PART IV THE STRUCTURE OF C"(X) : BEGINNINGS

THE FUNDAMENTAL SUBSPACES O F C"(X)

CHAPTER 9

5 40

Cll(X).

213

a n d C"(X)d

The b a s i c b a n d s a n d C"(X)

217

5 42

The 'Is emi c o n t i n u o u s " e 1emen t s

221

§43

The Up-down-up t h e o r e m

228

5 44

The s u b s p a c e U(X) o f u n i v e r s a l l y

§

41

0

integrable elements

2 31

945

The s u b s p a c e s s ( X ) a n d S(X)

234

546

Dedekind c l o s u r e s

2 35

547

The B o r e 1 s u b s p a c e Bo(X)

237

548

C(X)

238

CHAPTER 1 0

and C(X)

THE OPERATORS u AND L

549

The o p e r a t o r s u a n d L

248

550

The o p e r a t o r 6

256

Contents

551

&bands

552

Applications to general bands

CHAPTER 11

and u-bands

263 2 71

U(X)

553

The Isomorphism theorem

554

Immediate consequences o f the Isomorphism theorem

955

xiii

275

277

The universally measurable subsets of

x

289

§56 The Nakano completeness theorem

294

Appendix

297

PART V RIEMANN INTEGRATION

CHAPTER 12

THE RIEMANN SUBSPACE OF A BAND

557

The Dedekind closure of C(X),

303

558

A representation theorem

307

559

Examples of Riemann subspaces

310

CHAPTER 13

RIEMANN INTEGRABILITY

560

Riemann integrable elements

316

561

p-Riemann integrability

317

562

Riemann negligible elements

324

xiv

Contents 563

564

Riemann integrability and Riemann subspaces

32 7

Examples

331

PART VI THE RARE ELEMENTS

CHAPTER 1 4

THE LEAN ELEMENTS

565

Elementary properties

336

566

A "localization" theorem

342

CHAPTER 15

THE RARE ELEMENTS

567


348

568

A "localization" theorem

359

CHAPTER 1 6

THE DECOMPOSITION C'(X) = Ra(X)'

0

C(X)'

569

The decomposition o f C'(X)

36 3

570

The band Ra(X)d

365

571

The band Ra(X)

572

Examples

CHAPTER 1 7

ad

370 372

THE DEDEKIND COMPLETION OF C(X)

973

The Maxey representation

377

574

The Dilworth representation

378

Contents

xv

575

A third representation

38 3

576

A fourth representation

387

CHAPTER 1 8

THE MEAGER ELEMENTS

577


578

Sums o f positive elements o f a

389

Riesz space

390

579

Sums o f positive elements o f C " ( X )

395

580

The "localization" theorem

400

BIBLIOGRAPHY

410

This Page Intentionally Left Blank

PART I

BACKGROUND

1

CHAPTER 1

R I E S Z SPACES

Notation.

N

=

{ n , m , - . . } and IR

=

{ l , ~ , .} w i l l d e n o t e t h e

n a t u r a l numbers and t h e r e a l numbers r e s p e c t i v e l y . s p a c e s w i l l a l w a y s be o v e r

pi,

Our v e c t o r

w i t h t h e n e u t r a l element denoted

by 0 . A n i n d e x e d s e t {aalacl

w i l l g e n e r a l l y be w r i t t e n simply

{aaI .

5 1 . Ordered v e c t o r spaces

< on a s e t E , we mean, a s u s u a l , a b i n a r y By an o r d e r -

r e l a t i o n which s a t i s f i e s

(I)

a < a for a l l a€ E

(reflexivity) ;

(11)

a < b, b < c implies a < c

(transitivity) ;

(111)

a < b, b < a implies a

(anti-symmetry).

For a s e t A

=

=

b

{ a a ) i n E and bE E , A b.

A s u b s e t A o f E w i l l be

< b f o r some bE E . s a i d t o be bounded above i f A -

2

5 b

Any s u c h b-

Riesz Spaces

3

i n g e n e r a l , i t i s n o t u n i q u e - w i l l be c a l l e d an u p p e r bound of A .

C o r r e s p o n d i n g d e f i n i t i o n s h o l d f o r bounded below and

lower bound.

I f A i s bounded below and a b o v e , we w i l l s a y i t

i s o r d e r bounded. An e l e m e n t b of E i s c a l l e d t h e supremum of a s e t

3 if (i) b > A and ( i i ) c > A implies c > b. The a infimum o f A i s d e f i n e d i n t h e same way w i t h > r e p l a c e d by
0. -

We have i m m e d i a t e l y :

(1.1)

L e t E be an o r d e r e d v e c t o r s p a c e and { a J a s u b s e t o f E.

a

If V a e x i s t s , then a a ~ ( va ) = v ( A a a ) f o r a l l A > 0; (i)

a

cla

(ii)

-(Vaaa) = A a ( - a a ) ;

(iii) v a

cla

+

c =

+

c ) f o r a l l cE E .

More g e n e r a l l y , i f {b } i s a l s o a s u b s e t o f E and V b e x i s t s , B B B then ~aa

c1

+ vB bB

= V

cl,fdaa

+

be>.

a

Chapter 1

The n o t a t i o n

denotes t h a t a,B (over t h e r e s p e c t i v e index s e t s ) . V

~1

and

vary independently

The " d u a l " s t a t e m e n t t o ( l . l ) , o b t a i n e d by r e p l a c i n g A a l s o holds.

T h i s w i l l be t r u e t h r o u g h o u t t h e work.

V

by

For

e v e r y p r o p o s i t i o n i n v o l v i n g one o r more o f t h e s y m b o l s A,,

V,

_ _

t h e s t a t e m e n t o b t a i n e d by i n t e r c h a n g i n g t h e f i r s t

two w i t h e a c h o t h e r a n d t h e l a s t two w i t h e a c h o t h e r w i l l a l s o hold. A subset A of a vector space E i s a

cone

i f it s a t i s f i e s

the conditions:

(I)

a , b € A implies a + b€ A;

(11)

a € A i m p l i e s ha€ A f o r a l l

(111)

a, - a € A implies a

=

x

> 0; -

0.

A c o n e a l w a y s c o n t a i n s 0 , and c o n t a i n s n o o t h e r l i n e a r

subspace. An e l e m e n t a o f an o r d e r e d v e c t o r s p a c e E i s c a l l e d o-s i t i v e i f a > 0 (thus 0 i s positive). P

I t is immediate t h a t

t h e s e t of p o s i t i v e elements of E i s a cone.

I t is called the

p o s i t i v e cone o f E , and w i l l be d e n o t e d by E,.

I t is easily

verified that a < b i f a n d o n l y i f b - a € E,

s o t h e o r d e r i s known i f E, A i n a vector space E,

b

- a€

i s known.

( c f . E x e r c i s e 1) ,

Moreover, f o r any cone

t h e d e f i n i t i o n "a < b i f and o n l y i f

A" d e f i n e s a c o m p a t i b l e o r d e r o n E f o r w h i c h A i s

p r e c i s e l y E,.

Riesz Spaces

5

3 2 . Riesz s p a c e s

A R i e s z s p a c e , o r v e c t o r l a t t i c e , i s an o r d e r e d v e c t o r

s p a c e which i s a l a t t i c e u n d e r t h e o r d e r , t h a t i s , e v e r y f i n i t e { a l , . - . , an } h a s a supremum and an infimum i n E . We n . r e s p e c t i v e l y ; and, a s i s w i l l d e n o t e t h e s e by Vyai and A 1a 1 set A

=

customary, i f n

2 , a l s o by a1Va2 and alAa2.

=

n t h e v e r y d e f i n i t i o n , Vlai

Note t h a t , by n and ~~a~ a r e i n d e p e n d e n t o f t h e

o r d e r i n g o f t h e s u b s c r i p t s ; i n p a r t i c u l a r , a l v a 2 = a2Va1 and alAa2

=

a2Aal.

For two s u b s e t s A , B o f a R i e s z s p a c e E ( i n d e e d , o f any l a t t i c e ) , i f V A and V B e x i s t , t h e n (VA)V (VB)

= V (AIJ

B)

For t h r e e e l e m e n t s a l , a 2 , a 3 , t h i s g i v e s 3 u s ( a 1Va 2 ) v a , = a l v ( a 2Va 3 ) = v 1a 1. ’ s o we c a n w r i t e a 1Va 2v a 3 ( a s s o c i a t i v e law).

unambiguously.

And o f c o u r s e t h i s e x t e n d s t o any f i n i t e

number o f e l e m e n t s . The i d e n t i t i e s (1.1) w i l l be a p p l i e d r e p e a t e d l y t o a s e t c o n s i s t i n g o f two e l e m e n t s s o we s t a t e them f o r t h i s c a s e explicitly.

( 2 . 1 ) Given e l e m e n t s a , b o f a R i e s z s p a c e E , (i)

X(aVb) = ( 1 a ) v (Xb)

(ii)

-(aVb) = ( - a ) A ( - b )

(iii)

avb

+

c

=

(a + c)V (b

for a l l

+

x

> 0;

c).

This g i v e s u s immediately t h e s u p r i s i n g l y s t r o n g c o r o l l a r y :

Chapter 1

6

For a l l a,bE E ,

(2.2)

avb

avb - b

Pro of. -

=

+

a b

=

a + b.

(a - b)v 0

=

a - bAa ( t h i s l a s t e q u a l i t y

from E x e r c i s e 2 ) . QED

Note t h a t from ( i i ) a b o v e , f o r an o r d e r e d v e c t o r s p a c e t o be a R i e s z s p a c e i t i s s u f f i c i e n t t h a t t h e supremum e x i s t f o r every p a i r of elements. The f o l l o w i n g d i s t r i b u t i v e law i s e a s i l y v e r i f i e d : I f and V b e x i s t i n a Riesz s p a c e E , t h e n B B Vaaa (Vaaa)V ( V b ) = V (a vbB). In p a r t i c u l a r , B B a aV(V b ) = V ( a v b ) , a n d f o r t h r e e e l e m e n t s a , b , c ,

P a

av(bvc)

B

=

B

(avb)v ( a v c ) .

Not s o o b v i o u s i s a n o t h e r

( D i s t r i b u t i v e Law) Given a s u b s e t {a 1 o f a R i e s z s p a c e E , a i f Vaaa e x i s t s , t h e n (2.3)

bA(vaaa) = V a ( b h a a ) f o r a l l bE E .

Proof. Set a

=

Vaaa.

T h a t bAa

2

bAaa f o r a l l a i s c l e a r .

Suppose c > bAa f o r a l l a ; we h a v e t o show c > bAa. For e a c h U a, c > bAa = b + a - bva > b + a - bva. I t follows a a aa c 2 Va(b + aa - b v a ) = b + V a a a - b v a = b + a - b v a = h a . QED

7

Riesz Spaces

The " d u a l " law bv ( A a a a ) = A (bva ) a l s o h o l d s . F o r two c1 a e l e m e n t s , t h e laws r e d u c e t o bA(alVa2) = (bAal)V(bAa2) and bV(alAa2)

(bVal)A(bVa2).

=

By s t r a i g h t f o r w a r d a r g u m e n t from ( 2 . 3 ) , we c a n a l s o o b t a i n the apparently stronger

One o f t h e most u s e f u l p r o p e r t i e s o f a R i e s z s p a c e i s g i v e n by t h e

( D e c o m p o s i t i o n Lemma).

(2.5)

Given a R i e s z s p a c e E , i f

a,bl,b2EE+

and a

5

bl

with 0 < al

5

0

5

a2

Set al

=

a b l and a 2 = a

Proof. a - a/\bl

=

bl,

Ov(a

+

b 2 , t h e n a can be w r i t t e n a

5

=

al

+

b2 (al,a2 are not unique).

- b l ) 5 Ovb2

- al.

Then a 2

=

= b2 ( t h e second e q u a l i t y again

from E x e r c i s e 2 ) . QED

Given a , b E E

with a < b, the s e t {cla < c < b } w i l l be

d e n o t e d by [ a , b ] and c a l l e d an i n t e r v a l .

The D e c o m p o s i t i o n

Lemma c a n b e s t a t e d : For e v e r y b l , b 2 E E + ,

[O,bl

[O,blI

+

[O,b21.

+

b2] =

a2

8

Chapter 1

I t f o l l o w s f r o m t h e D e c o m p o s i t i o n Lemma, b y i n d u c t i o n , t h a t i f a , b l , . * - ,bnE E, 0 < a . < bi -

1 -

(i

-

,n).

1,.

=

Crib. 1 1'

and a
0;

b);

By an o r d e r bounded n e t , we w i l l mean one whose s e t o f v a l u e s i s o r d e r bounded. I f a a + a , t h e n , by r e p l a c i n g t h e n e t by a t e r m i n a l p a r t ,

,

i f n e c e s s a r y , we c a n assume i t h a s a f i r s t e l e m e n t a

and i s

"0

t h e r e f o r e an o r d e r bounded n e t ( s p e c i f i c a l l y , a for all a).

< a < a "0a -

And s i m i l a r l y f o r a d e s c e n d i n g n e t .

So hence-

f o r t h t h e n o t a t i o n s a + a and a + a w i l l always i m p l y t h a t we c1 a a r e d e a l i n g w i t h an o r d e r bounded n e t . Given a n e t { a 3 ( n o t n e c e s s a r i l y m o n o t o n i c ) , we w i l l s a y

"

t h a t t h e n e t {a } o r d e r converges t o a E E , i f t h e r e e x i s t n e t s a Ira}, { s 1 i n E , w i t h t h e same i n d e x s e t , s u c h t h a t ( i ) r a + a , a ( i i ) s + a , and ( i i i ) r < a < s f o r a l l a. We w i l l d e n o t e a- a a t h i s o r d e r c o n v e r g e n c e by a -+ a . a Note t h a t ( I ) from, t h e p r e c e d i n g p a r a g r a p h , t h e n o t a t i o n

"

a

-+

a net,

a always i m p l i e s t h a t we a r e d e a l i n g w i t h an o r d e r bounded (11) a + a and a + a a r e s p e c i a l c a s e s o f o r d e r c o n v e r g e n c e , a a

Riesz Spaces

17

a n d ( 1 1 1 ) f o r a n e t { a }, i f a = a f o r a l l a, t h e n a -t a . a a a For o r d e r c o n v e r g e n c e t o be a u s e f u l t o o l , l i m i t s m u s t b e

We v e r i f y t h i s .

a'

we 1 h a v e t o show a' = a . By d e f i n i t i o n , t h e r e e x i s t n e t s { r } . a 1 2 2 { s } s a t i s f y i n g t h e c o n d i t i o n s ( i ) , (ii), (iii) { s a } and I r a } , a We n o t e f i r s t t h a t f o r a l l f o r a' a n d a 2 r e s p e c t i v e l y . 1 In e f f e c t , choose Y E A 2 < s As a , f i E d ( t h e i n d e x s e t ) , r UV r a - a a' 1 1 1 1 2 such t h a t a , a < y; then r V r 2 < r V r 2 < a < s A s 2 < s A s ci a - y y - y - y y - a 0' 1 1 2 1 2 I t f o l l o w s t h a t a V a 2 = V ( r V r ) < A ( s A S ) = a1Aa2, whence

unique.

Suppose

a

2

a

a 1 = a2 .

a

a

-

c1

B

-+

B

and a

-t

U

a';

B

The a b o v e d e f i n i t i o n a n d d i s c u s s i o n c l e a r l y h o l d s i n a n y lattice.

In contrast, the following equivalent definition of

o r d e r convergence i s meaningful o n l y i n a Riesz s p a c e .

( 4 . 2 ) F o r a s e t { a } i n a R i e s z s p a c e E a n d aE E , a ing are equivalent: lo 2'

(a

-

a

a

-+

the follow-

a,

t h e r e e x i s t s a n e t { pc1 } i n E s u c h t h a t p c1 $ 0 a n d

a a ( 5 p,

f o r a l l a.

h o l d s , a n d f o r e v e r y a, s e t p

s - r a a' Then p a + 0 ( 4 . 1 ) , a n d F o r e v e r y a, r < a < s , r < a < s a- a a - a - u' Cc-versely, < a - a a 5 p a , t h a t i s , l a - aal 5 p a . hence pa assume 2' h o l d s . Then f o r e v e r y - p a -< a a - a 5 p a , h e n c e < a i a + p . T h u s a - p , a + pa a r e t h e d e s i r e d a pa- a a U r s a' a' QED Proof.

Assume 1'

c1

=

Chapter 1

18

From (4.1), we o b t a i n e a s i l y

3 b e two n e t s i n a R i e s z s p a c e E , w i t h t h e a b , then I f aa + a and b

( 4 . 3 ) Let { a , } , { b same i n d e x s e t .

c1

(i)

(-aa)

(ii)

laa

(iii)

a

(iv)

aavba

-+

avb,

(v)

a Ab a a

-+

aAb.

ci

-+

+

(-a),

Xa f o r a l l X ,

ba

+

-f

-f

a

+

b,

And, a s a c o r o l l a r y ,

(i)

a

a

+

a i f and o n l y i f ( a

(ii)

a

w

+

a i f and o n l y i f ( a ) +

(iii)

a

a

-+

a implies

(iv)

if a

(4.4)

a

-

w

a

+

a, b

a

laa/ +

-+

a) +

-+

0;

a + a n d (aa)- + a - ;

/at;

b , and a

< b

a-

a

f o r a l l a, t h e n

a < b.

T h e r e i s a p a r t i c u l a r k i n d o f m o n o t o n i c n e t w h i c h we w i l l o f t e n use.

F o l l o w i n g B o u r b a k i , we w i l l s a y t h a t A c E i s

f i l t e r i n g upward

-

i f a l , a 2 E ; A i m p l i e s alVa2EA.

d i r e c t e d s e t , h e n c e t h e i n j e c t i o n map

a n e t , an a s c e n d i n g o n e .

at+

A is then a

a of A into E is

This n e t w i l l be c a l l e d t h e

( c a n o n i c a l ) n e t a s s o c i a t e d w i t h A.

In general, we w i l l write

i t {aa} e v e n t h o u g h t h e i n d e x s e t {a} i s A i t s e l f .

I t is

19

R i e s z Spaces

clear that i f a

= VA,

then t h i s n e t order converges t o a .

A w i l l b e s a i d t o b e f i l t e r i n g downward i f a l , a Z f A i m p l i e s

a1Aa2EA.

A s above, t h e r e i s a c a n o n i c a l n e t a s s o c i a t e d w i t h A,

t h i s time a descending one.

And i f a

=

AA, then t h i s n e t order

converges t o a .

the set of a l l

G i v e n a s u b s e t A o f E , w e d e n o t e b y A(') limits i n E of o r d e r convergent n e t s o f A. that (i) A c A(1),

(ii) A c B i m p l i e s A(')

I t i s immediate

c B ( l ) , and

(iii)

( A I J B ) ( l ) = A(1) IJ B " ) .

In addition,

(4.5)

( i ) I f A is a l i n e a r subspace, then so i s A('); ( i i ) I f A i s a s u b l a t t i c e , t h e n s o i s A (1) ; ( i i i ) I f F i s a Riesz subspace, then so i s F ' l ) ,

I f I i s a Riesz i d e a l , then so is

(iv)

I'll,

with

and ( I ( ' ) ) +

i s t h e s e t o f suprema o f a r b i t r a r y s u b s e t s o f I + .

Proof.

(i) and ( i i ) a r e immediate; a l s o t h e f i r s t s t a t e -

ment i n ( i i i ) .

There e x i s t s a n e t {a } c F s u c h t h a t a

a € (F('))+. Then ( a a ] +

To p r o v e t h e l a s t s t a t e m e n t i n ( i i i ) , c o n s i d e r c1

+

a+

=

a, so a € (F+)(').

Thus ( F ( ' ) ) +

c1

a.

+

c ( F + )(1). ,

The o p p o s i t e i n c l u s i o n f o l l o w s f r o m t h e e a s i l y v e r i f i e d f a c t t h a t t h e l i m i t o f an o r d e r c o n v e r g e n t n e t o f p o s i t i v e e l e m e n t s

is positive. Now s u p p o s e I i s a R i e s z i d e a l . subspace.

We show t h a t 0 - b - a E I

By ( i i i ) , I ( ' )

i s a Riesz

i m p l i e s b e I(1)

,

hence

20

Chapter 1

i s a Riesz i d e a l .

I(') a

a

-f

a (by ( i i i ) a g a i n ) .

There e x i s t s a n e t { a } i n I + such t h a t a Then aaAb€ I,

f o r a l l a , and

I t remains t o prove t h e l a s t a A b -f a/ib = b . Thus b E I ( l ) . a statement in ( i v ) . I f a E E i s t h e supremum o f some s u b s e t o f

t h e n , by E x e r c i s e 1 4 , some n e t i n I + o r d e r c o n v e r g e s t o a ,

I,,

s o a € I'll.

C o n v e r s e l y , s u p p o s e a € ( I (1) ) + .

There e x i s t s

3

Since

n e t {a } i n I+ such t h a t aa a . Then aaAa -t a ~ =a a . a a Aa < a , i t f o l l o w s e a s i l y t h a t a = V a ( a a A a ) . a -f

QED

55. Order c l o s u r e and bands

A s u b s e t A of E w i l l be c a l l e d o r d e r c l o s e d i f i t i s

c l o s e d under o r d e r convergence i n E of n e t s i n A: f o r e v e r y n e t { a } i n A and a € E ,

+ a implies aEA. I t is easily verified a t h a t t h e union o f a f i n i t e c o l l e c t i o n of o r d e r c l o s e d s e t s and

a

a

t h e i n t e r s e c t 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 a r e o r d e r c l o s e d . We have a l r e a d y p o i n t e d o u t i n t h e p r o o f o f ( 4 . 5 )

t h a t E,

i s order closed. A useful property:

(5.1) A Riesz subspace F o f a Riesz space E i s o r d e r c l o s e d i f and o n l y i f F,

Proof. -____

i s order closed.

Suppose F i s o r d e r c l o s e d .

S i n c e F,

= F

n

E,,

i t i s t h e i n t e r s e c t i o n o f two o r d e r c l o s e d s e t s , h e n c e o r d e r

Riesz S p a c e s

closed.

C o n v e r s e l y , l e t F,

21

b e o r d e r c l o s e d , and s u p p o s e a n e t

{ a } i n I: o r d e r c o n v e r g e s t o a n e l e m e n t a . Then (a,)' CL + I t f o l l o w s a , a E F + , hence a E F . and ( a a ) - + a - .

+

a+

QED

Consider a subset A o f E .

The i n t e r s e c t i o n o f a l l t h e

o r d e r c l o s e d s e t s c o n t a i n i n g A i s an o r d e r c l o s e d s e t , hence t h e smallest one c o n t a i n i n g A.

We w i l l c a l l i t t h e o r d e r

c l o s u r e o f A. I n g e n e r a l A(1)

i s n o t o r d e r c l o s e d , hence i s a p r o p e r

s u b s e t o f t h e o r d e r c l o s u r e o f A. A(3)

and so on.

=

Let A ( 2 )

=

( A ( 1 ) )( 1 ) 9

I t may r e q u i r e a t r a n s f i n i t e

s e q u e n c e o f t h e s e i t e r a t i o n s t o o b t a i n t h e o r d e r c l o s u r e o f A. (Note, however, t h a t A i s o r d e r c l o s e d i f and o n l y i f A

=

A").)

I t i s c l e a r t h a t i f A c B , t h e n o r d e r closure A c o r d e r c l o s u r e B.

A l s o , f o r a n y two s u b s e t s A , B o f E ,

o r d e r c l o s u r e ( A ! J B ) = ( o r d e r c l o s u r e A) IJ ( o r d e r c l o s u r e B )

Combining ( 4 . 5 ) w i t h Z o r n ' s lemma, we c a n show

(5.2)

Theorem. (i)

Let E b e a R i e s z s p a c e .

I f A is a l i n e a r subspace, then so is i t s o r d e r

closure. (ii)

I f A i s a sublattice, then so i s i t s order closure

( i i i ) I f F i s a Riesz subspace and G i t s o r d e r c l o s u r e , t h e n G i s a R i e s z s u b s p a c e a n d G,

= o r d e r c l o s u r e F,.

Chapter 1

22

F o r a R i e s z i d e a l I , w e c a n do b e t t e r .

I(1) is order

c l o s e d , h e n c e i s t h e o r d e r c l o s u r e o f I (we t h u s h a v e n o n e e d o f Z o r n ' s lemma):

For a R i e s z i d e a l I o f a Riesz s p a c e E , I(1)

(5.3)

is order

c l o s e d , hence i s t h e o r d e r c l o s u r e o f I .

Proof.

By ( 5 . 1 ) a n d ( 4 . 5 ) , we n e e d o n l y show t h a t ( I ( 1 ) ) +

i s c l o s e d under t h e o p e r a t i o n o f a d j o i n i n g suprema o f a r b i t r a r y subsets of itself. idempotent.

Now t h i s o p e r a t i o n i s e a s i l y shown t o b e

Since (I('))+

i s o b t a i n e d by a p p l y i n g t h e o p e r a -

t i o n t o I + , we a r e t h r o u g h .

QED

We w i l l s e l d o m e n c o u n t e r o r d e r c l o s e d R i e s z s u b s p a c e s i n t h i s work. abound,

I n c o n t r a s t t o t h i s , o r d e r c l o s e d Riesz i d e a l s w i l l

The l a t t e r h a v e a name: a n o r d e r c l o s e d R i e s z i d e a l o f

a Riesz s p a c e E i s c a l l e d a

band

of E.

The r e m a i n d e r o f t h i s 5

i s devoted t o bands. Given a s u b s e t A o f a R i e s z s p a c e E , t h e i n t e r s e c t i o n o f a l l t h e bands c o n t a i n i n g A i s a band, hence t h e s m a l l e s t one

c o n t a i n i n g A. clearly I(1),

I t i s c a l l e d t h e band g e n e r a t e d by A.

It is

where I i s t h e Riesz i d e a l g e n e r a t e d by A.

We r e m a r k e d i n 53 t h a t f o r e v e r y s u b s e t A o f a R i e s z s p a c e , Aa i s a Riesz i d e a l .

In point of f a c t ,

Kiesz S p a c e s

(5.4)

23

For e v e r y s u b s e t A of a Riesz space E , A

Proof. -__

d . i s a band.

I t r e m a i n s t o show t h a t Ad i s o r d e r c l o s e d .

This

f o l l o w s f r o m ( 5 . 1 ) and ( i v ) o f E x e r c i s e 5 .

QED d d For a s u b s e t A of a R i e s z space E , w e w i l l denote (A ) s i m p l y by A dd

(5.5)

.

I t i s o b v i o u s t h a t A C A dd .

Let A b e a s u b s e t o f a R i e s z s p a c e E , I t h e R i e s z i d e a l

g e n e r a t e d by A ,

a n d H t h e b a n d g e n e r a t e d by A ( h e n c e t h e o r d e r

closure of I ) .

Then Ad

Id

=

=

H

d

a n d ( t h e r e f o r e ) Add

=

Idd

=

Hdd.

The v e r i f i c a t i o n i s s t r a i g h t f o r w a r d .

F o r an A r c h i m e d e a n

R i e s z s p a c e , w e c a n s a y more:

( 5 . 6 ) Theorem. -____

I f E i s an Archimedean Riesz s p a c e , t h e n f o r

every R i e s z i d e a l I , Idd i s i t s o r d e r c l o s u r e .

i f I i s a band, then Idd

Proof.

=

I.

Idd i s o r d e r c l o s e d ( 5 . 4 ) ,

of I i s contained i n I

dd

.

In particular,

so the order closure

For t h e o p p o s i t e i n c l u s i o n c o n s i d e r

24

bE ( I

Chapter I dd

< a < b } ; we show b = V A . ) + and s e t A = { a € I 1 0 -

Suppose A < c < b ; we h a v e t o show c = b . On t h e o n e h a n d , dd 0 < b - c < b , s o b - cE I on t h e o t h e r h a n d b - c E 1 d .

.

To

e s t a b l i s h t h i s , i t i s enough t o show t h a t i f dE I . s a t i s f i e s

0 < d < b - c , then d

=

0.

We have 0 c d < b - c < b , s o dEA,

so d < c , so 0 < 2d < c + (b - c) we h a v e 0 < nd < b for a l l n We t h u s h a v e b

- cE I d d

=

=

b.

1,2,...

P r o c e e d i n g by i n d u c t i o n ,

.

I t follows d

r) I d , whence b -

c

=

=

0.

0.

QED

( 5 . 7 ) C o r o l l a r y 1. A o f E , Add

I f E i s Archimedean, t h e n f o r e v e r y s u b s e t

i s t h e band g e n e r a t e d by A .

(5.8) Corollary 2.

For a R i e s z i d e a l I o f an Archimedean R i e s z

space E , the following a r e equivalent:

1'

the order closure of I i s E;

Z0

Id = 0.

For a counterexample t o ( 5 . 8 )

,

hence t o ( 5 . 6 )

,

when E i s

n o t A r c h i m e d e a n , l e t E b e t h e p l a n e IR2 o r d e r e d l e x i c o g r a p h i c a l l y ( 5 2 ) and H t h e y - a x i s . f o r e H~~

=

Then H i s a b a n d b u t Hd = 0 and t h e r e -

E.

We have s e e n ( 3 . 7 ) t h a t i f E then H

=

I d and I = H d .

=

I a H , I , H Riesz i d e a l s ,

T h u s , by ( 5 . 4 ) ,

I and H a r e b a n d s

( 3 . 7 ) can be s t a t e d a s f o l l o w s : A R ies z i d e a l I h as a

R i e s z Spaces

25

complementary Riesz i d e a l i f and o n l y i f E

I 3 Id; i n such

=

c a s e , ( i ) I d i s t h e o n l y c o m p l e m e n t a r y R i e s z i d e a l and ( i i ) I i s a band.

In g e n e r a l , even i f I is a band, I complementary t o I . i d e a l I i s a band.

+

Id

# E , so I d

is not

I n t h e example f o l l o w i n g ( 3 . 7 ) , t h e R i e s z However, i n an Archimedean R i e s z s p a c e ,

I d i s " a l m o s t " complementary t o I ( ( 5 . 9 ) b e l o w ) . We f i r s t a d o p t a n o t a t i o n .

I f two l i n e a r s u b s p a c e s F , G

of a vector space s a t i s f y F n G F

+

Thus f o r a R i e s z i d e a l I o f a R i e s z

G t o indicate this.

s p a c e , we w i l l a l w a y s d e n o t e I

(5.9)

Theorem.

0 , we w i l l w r i t e F 3 G f o r

=

+

I

d

by 1 3 I

d

.

I f E i s Archimedean, then

(Luxemburg-Zaanen).

f o r e v e r y Riesz i d e a l I of E , t h e o r d e r c l o s u r e o f I

Proof.

(I @ Id)d

=

Id

n

Idd

=

0 (Exercise 17).

Id i s E.

%,

Now

apply (5.8) QE D

Remark.

( i ) Since I 3 Id i s a R i e s z i d e a l , t h e theorem

s a y s t h a t e v e r y e l e m e n t o f E+ i s t h e supremum o f some s u b s e t d of (I 3 I )+ ( i i ) Theorems ( 5 . 6 ) a n d ( 5 . 9 )

a r e contained i n a general

t h e o r e m o f Luxemburg a n d Zaanen ( [ 3 5 ] Theorem 2 2 . 1 0 ) , w h i c h s t a t e s t h a t t h e p r o p e r t y i n e a c h o f them i s , i n f a c t , a n e c e s s a r y and s u f f i c i e n t c o n d i t i o n f o r E t o be Archimedean.

Chapter 1

26

56. R i e s z homomorphisms

C o n s i d e r a l i n e a r mapping E ->F into another.

T o f one R i e s z s p a c e

Perhaps t h e weakest p r o p e r t y T can have i n v o l v -

i n g t h e Riesz space s t r u c t u r e i s o r d e r boundedness.

T w i l l he

s a i d t o b e o r d e r bounded i f i t c a r r i e s o r d e r bounded s e t s i n t o o r d e r bounded s e t s .

A stronger property is positivity.

T is

c a l l e d p o s i t i v e i f T(E+) c F + , o r , e q u i v a l e n t l y , i f i t p r e i m p l i e s Ta i Ta,. That a p o s i t i v e 2 1 l i n e a r mapping i s o r d e r b o u n d e d i s i m m e d i a t e . A stronger pro-

serves order: a

1 -

a

L

p e r t y s t i l l : T w i l l b e c a l l e d a R i e s z homomorphism i f i t p r e s e r v e s t h e l a t t i c e o p e r a t i o n s , t h a t i s , i f f o r a l l a , b E E:, T(avb)

=

(Ta)V(?'b) a n d T(ar\b)

=

(Ta)A(Tb).

A R i e s z homomorphism

clearly preserves order. A f o u r t h p r o p e r t y , s t r o n g e r t h a n o r d e r boundedness b u t

i n d e p e n d e n t o f t h e o t h e r two p r o p e r t i e s ,

is order continuity.

T w i l l be c a l l e d o r d e r continuous i f it p r e s e r v e s o r d e r con-

vergence: a

(6.0)

a.

+

a i m p l i e s Ta(,

+

Ta.

I f T i s o r d e r c o n t i n u o u s , t h e n i t i s o r d e r bounded

Proof. -__

I t i s e n o u g h t o show t h a t f o r a n i n t e r v a l o f t h e

f o r m [ O , a ] i n E , T ( [ O , a ] ) c [ - c , c ] f o r some c E F , .

As a set,

[ O , a ] i s f i l t e r i n g downward ( a l s o u p w a r d , b u t n o m a t t e r ) . L e t {a } b e t h e a s s o c i a t e d ( d e s c e n d i n g ) n e t ( c f . t h e d i s c u s s i o n c1

R i e s z Spaces

following ( 4 . 4 ) ) .

Then a,+O,

27

s o , b y h y p o t h e s e s , Ta

ci

+

s a y s t h e r e e x i s t s a n e t {b } i n F s u c h t h a t b 4 0 a n d ITa

"

CY

This

0.

a

I

< b

-

f o r a l l a. Now s i n c e t h e d i r e c t e d i n d e x s e t {a} i s a c t u a l l y t h e s e t

[ O , a l , i t h a s a f i r s t e l e m e n t a0 ( w h i c h i s a c t u a l l y a ) .

Since

f o r a l l Q, a n d t h u s { b c x }i s d e s c e n d i n g , i t f o l l o w s b F o f o n e R i e s z s p a c e i n t o another.

The f o l l o w i n g a r e e q u i v a l e n t :

1'

T i s a R i e s z homomorphism;

2'

T p r e s e r v e s one o f t h e l a t t i c e o p e r a t i o n s V , A ;

3'

~ ( a + )= ( ~ a ) +f o r a l l a E E :

4'

For a l l a , b E E ,

Proof.

Assume 4'

a/\b

0 i m p l i e s Ta/\Tb

=

For a , b E E ,

holds

( a - a/\b)A (b

=

a ~ b )= 0 ,

-

hence T ( a - a/\b)A T b - W b )

=

0.

Writing t h i s (Ta - T ( a / \ b ) ) A (Tb - T ( a / \ b ) ) = 0 ,

we h a v e Ta/\Tb

-

T(aAb)

=

0,

0.

iy.

Chapter I

28

w h i c h g i v e s u s 2'.

The r e m a i n d e r o f t h e p r o o f i s s t r a i g h t -

forward.

QED

(6.2)

I f E __ > F i s a R i e s z homomorphism, t h e n T(E) i s a

Riesz subspace o f F and T-l(O) i s a Riesz i d e a l of E .

The p r o o f i s s i m p l e . I f a l i n e a r mapping E

> F i s a b i j e c t i o n and b o t h T

~

and T - l p r e s e r v e t h e l a t t i c e o p e r a t i o n s , t h e n T w i l l be c a l l e d a R i e s z isomorphism.

For a l i n e a r b i j e c t i o n t o be a Riesz

isomorphism, i t s u f f i c e s t h a t T(E+)

The r e m a i n d e r o f t h i s

§

=

F,.

i s d e v o t e d t o two k i n d s o f R i e s z

homomorphisms w h i c h w i l l b e w i t h u s t h r o u g h o u t t h e w ork.

( 6 . 3 ) Theorem.

Let I b e a R i e s z i d e a l o f a Riesz s p a c e E , E / I

t h e q u o t i e n t v e c t o r s p a c e , and E -> E / I t h e q u o t i e n t map. Then (i)

q ( E + ) i s a c o n e , h e n c e d e f i n e s a n o r d e r on E / I .

Under t h i s o r d e r , (ii)

E/I i s a Riesz space;

(iii)

q i s a R i e s z homomorphism ;

(iv)

f o r every interval [a,b] of E , q([a,b]) = [qa,qb].

29

Riesz Spaces

*

We d e n o t e t h e e l e m e n t s o f E / I by % , b , ' * . .

P roof. -

n e e d o n l y show t h a t g , - S i E q ( E + ) i m p l i e s 2 t h a t qa

-5, s o

Z.

=

also -a

S i n c e qaE q ( E + ) , a +

jEE+

€ o r some j E 1 .

- a + j -> 0 , we have - i < a 5 j.

Z

iEE+

+

f o r some i E I .

Writing these a

q(-a) = i

+

2

0,

( E x e r c i s e 11) , s o

Thus a E I

-

Choose a s u c h

0.

=

( i ) We

0.

=

We n e x t show t h a t f o r a E E , qa'

=

(qa)',

which w i l l

e s t a b l i s h b o t h ( i i ) and ( i i i ) .

q i s p o s i t i v e by t h e v e r y

d e f i n i t i o n o f t h e o r d e r on E / I ,

s o qa'

b 2

qa,

b 2 0;

a + s u c h t h a t qb

Since

b

>

q a , bo + j

2

a f o r some j E 1 .

b.

t h a t qbO = since b

2

0, b o

+

i

bo + i v j > a , hence bo + iVj 2 a + .

2

q a , qa

+

2

-

0.

Suppose

We w i l l do t h i s by

qa'.

2

f i n d i n g an e l e m e n t b

-

'>

we have t o show

2

=

b.

Choose b o s u c h

0 f o r some i E I .

I t follows b o

And +

ivj

LO,

Thus b o + i v j i s t h e

d e s i r e d b. (iv) Since q i s p o s i t i v e , q ( [ a , b ] ) c [qa,qb].

For t h e

o p p o s i t e i n c l u s i o n , c o n s i d e r CE [ q a , q b ] and c h o o s e c E E t h a t qc qd

=

s.

=

such

Then, s e t t i n g d = ( c V a ) A b , we have dE [ a , b ] and

?. QED

Suppose E = I 3 H , I , H R i e s z i d e a l s ( h e n c e b a n d s ) . c a n o n i c a l mappinga+?al

o f E o n t o I i s a p r o j e c t i o n , t h a t i s , an

i d e m p o t e n t l i n e a r mapping o f E i n t o i t s e l f . by &I

.

The

We w i l l d e n o t e i t

S i n c e p r o j I e x i s t s whenever a R i e s z i d e a l I h a s a

complementary R i e s z i d e a l (and t h e r e f o r e b o t h a r e b a n d s ) , s u c h an I i s c a l l e d a F o j e c t i o n b a n d , and p r o j I i s c a l l e d a projection.

band

Chapter I

30

(6.4)

F o r a p r o j e c t i o n E-

Theorem.

> E on a R i e s z s p a c e ,

the following a r e equivalent: 1'

T i s a band p r o j e c t i o n ;

2'

( i ) T i s a R i e s z homomorphism, a n d < a f o r a l l aEE,. ( i i ) 0 5 Ta -

Proof. Assume 2'

T h a t 1'

h o l d s , and s e t H

The m a p p i n g E

i s t h e content of ( 3 . 6 ) .

T(E),

I

=

T-l(O).

Then E

=

H 3 I,

> E d e f i n e d b y S a = a - Ta i s a l s o a p r o j e c -

L

t i o n , w i t h S(E)

=

I , S - l ( O ) = I].

So we n e e d o n l y show S i s a

Note t h a t by i t s d e f i n i t i o n , S a l s o

R i e s z homomorphism.

satisfies

=

0

I i s a R i e s z i d e a l ; w e show FI i s a R i e s z i d e a l .

and, by ( 6 . 2 ) ,

(ii): 0 < Sa < a f o r aEE,

Then 0 < SWSb < ar\b (6.1)

implies 2

=

Now s u p p o s e ar\b

0 , hence S w S b = 0.

=

0.

I t f o l l o w s from

t h a t S i s a R i e s z homomorphism.

QED

( 6 . 5 ) A band p r o j e c t i o n p r e s e r v e s suprema and i n f i m a o f a r b i t rary sets.

A f o r t i o r i , it is order continuous.

If AA

Proof. A(projI(A))

=

=

0 i n E , t h e n , by 2'

( i i ) i n (6.4)

above,

0. QE D

Riesz Spaces

31

5 7 . Dedekind c o m p l e t e n e s s

A liiesz s p a c e w i l l b e c a l l e d D e d e k i n d c o m p l e t e i f e v e r y

s u b s e t w h i c h i s bounded a b o v e h a s a supremum i n E - e q u i v a l e n t l y , e v e r y s u b s e t b o u n d e d b e l o w h a s a n infimum i n E .

F o r l a t e r c o m p a r i s o n , we g i v e two a d d i t i o n a l d e f i n i t i o n s . An o r d e r e d p a i r ( A , B ) ~

o f s u b s e t s o f E w i l l be c a l l e d a D e d e k i n d

cut i f ( i ) A A implies c E R ) . property (that i s , c -

Note t h a t V A

=

A B i f e i t h e r e x i s t s ; and i n s u c h c a s e , i t l i e s

b o t h i n A and B , and i s i n f a c t t h e o n l y e l e m e n t i n t h e i r i n t e r -

section.

For a R i e s z space E , t h e following a r e e q u i v a l e n t :

(7.1)

1'

E i s Dedekind c o m p l e t e ;

2'

f o r every p a i r of subsets A , B such t h a t A < B, there

ic < B; e x i s t s c E E such t h a t A -

3'

f o r e v e r y Dedekind c u t ( A , B ) ,

A

n

B i s n o t empty.

The v e r i f i c a t i o n i s s i m p l e . Remark.

'3

above i s e s s e n t i a l l y Dedekind's o r i g i n a l

definition.

Note t h a t i f E i s Dedekind c o m p l e t e , t h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e e l e m e n t s o f E and t h e D e d e k i n d c u t s

32

Chapter I

o f E , g i v e n by c B

=

(A,B), where A

=

CaEEla 5 c } ,

2 c}.

{bEEIb

(7.2)

I+

A Dedekind c o m p l e t e R i e s z s p a c e E i s A r c h i m e d e a n .

Proof.

C o n s i d e r aEE+

5

b 5 ( 1 / 2 n ) a f o r a l l n , s o 2b

b

=

b

2 0 , t h i s g i v e s u s 2b

0.

and l e t b = A n ( l / n ) a ; w e show

b , whence b

=

=

(l/n)a f o r a l l n.

Since

0. QED

Judged by u s e f u l n e s s , t h e f o l l o w i n g i s p r o b a b l y t h e m o s t i m p o r t a n t p r o p e r t y o f a Dedekind c o m p l e t e R i e s z s p a c e [ 4 4 ] .

( 7 . 3 ) Theorem..

(F. Riesz)

I n a Dedekind c o m p l e t e R i e s z s p a c e

E , e v e r y band I i s a p r o j e c t i o n b a n d .

Proof.

We n e e d o n l y show I + I d s u f f i c e s t o show t h a t I + + ( I d ) + = E,.

c = V([O,b] Suppose 0

2

a

hence a

c

5 c , hence a

+

E , and f o r t h i s , i t

C o n s i d e r bEE,,

Since I i s a band, cEI;

I).

2

=

b - c, aEI. =

Then 0

5

and l e t

we show b - c E I d .

a + c

5

b , with a + c f I ,

0. QED

Note t h a t c

=

bI.

Exercise 2 1 c o n t a i n s t h e d e t a i l s o f t h e

Riesz Spaces

33

a c t u a l computation of bI i n various s i t u a t i o n s . Remark. -__

The R i e s z theorem c o n t a i n s a l l t h e d e c o m p o s i t i o n

theorems o f c l a s s i c a l i n t e g r a t i o n t h e o r y .

I t may have been

R i e s z ' d e s i r e t o show t h a t t h e s e d e c o m p o s i t i o n theorems were t h e same t h a t l e d him t o d e f i n e R i e s z s p a c e s .

I n a Dedekind c o m p l e t e R i e s z s p a c e E , t h e c o n c e p t s of l i m i t s u p e r i o r and l i m i t i n f e r i o r o f an o r d e r bounded n e t can

be d e f i n e d .

And t h e s e , i n t u r n , l e a d t o an a l t e r n a t e d e f i n i -

t i o n o f o r d e r convergence. Given an o r d e r bounded n e t { a c l ) i n E , we d e f i n e limsupclacl = /\cl(VB,ccaB)and l i m i n f c l a a c l e a r t h a t liminfaacl

5

limsup a

clcl

.

=

Vcl(~8,,ag).

I t is

I f (and o n l y i f ) t h e two a r e

equal, t h e n e t i s order convergent:

( 7 . 4 ) I f E i s Dedekind c o m p l e t e , t h e n f o r an o r d e r bounded s e t the following a r e equivalent:

{ a a } i n E and aEE,

2'

liminf a

=

a = limsup a

cla

cla

.

For e a c h a, s e t r

= A a and s = VB,aaB. Then a B>a - B cl r -f l i m i n f a a a , s a + l i m s u p a and r < a < s f o r a l l a. The c1 a a' a- a- a e q u i v a l e n c e o f 10 and 2' now f o l l o w s e a s i l y .

Proof.

QE D

We w i l l s a y a R i e s z s u b s p a c e F o f a R i e s z s p a c e E i s

Chapter 1

34

Dedekind d e n s e i n E i f e v e r y e l e m e n t o f E i s b o t h t h e supremum o f some s u b s e t o f F a n d t h e infimum o f some s u b s e t o f F .

As i s

o b v i o u s , i f F i s Dedekind d e n s e i n E , t h e R i e s z i d e a l g e n e r a t e d

by F i s a l l o f E ; h e n c e , g i v e n a R i e s z s u b s p a c e F , t o a s k a b o u t i t s Dedekind d e n s e n e s s makes s e n s e o n l y i n

t h e Riesz i d e a l

which i t g e n e r a t e s . N o t e t h a t i f E i s D e d e k i n d c o m p l e t e and F i s Dedekind dense i n E , t h e n t h e r e i s a one-one c o r r e s p o n d e n c e between t h e e l e m e n t s o f E and t h e Dedekind c u t s o f F . We w i l l s a y a R i e s z s u b s p a c e F o f a R i e s z s p a c e E

has

a r b i t r a r i l y small elements i n E i f f o r every a > 0 i n E , there e x i s t s bE F s u c h t h a t 0 < b < a.

(The common t e r m i n o l o g y i s

t h a t F i s o r d e r dense i n E . ) .

I f E i s Archimedean, t h e n f o r a R i e s z s u b s p a c e F , t h e

(7.5)

following are equivalent: 1'

F has a r b i t r a r i l y small elements;

2'

f o r e v e r y aEE,,

t h e r e e x i s t s a n e t Cb 1 i n F, c1

such

t h a t ba+a.

Proof.

We n e e d o n l y show t h a t 1'

h o l d s , and c o n s i d e r a > 0 .

Set A

enough t o show t h a t a = V A . c

3

0 such t h a t A < a

=

i m p l i e s 2'.

{bCF 10 < b < a } ; it i s

Suppose n o t .

- c.

Assume '1

Then t h e r e e x i s t s

We show t h a t b E F , 0 < b < c implies

b = 0 , which w i l l c o n t r a d i c t lo.

So c o n s i d e r b E F ,

0

5

b

5 c.

Since c

5

a , t h i s g i v e s us

35

Riesz Spaces

b < a , hence bE A , (a - c)

c

+

hence b < a - c.

a , whence 2 b € A ,

=

But t h e n 2b = b + b
A a n d b i s i n t h e norm

(9.5)

c l o s u r e o f A, t h e n b

P roof. A < c < b.

hence (Ib

-

= VA.

Suppose n o t .

Then t h e r e e x i s t s c s u c h t h a t

I t f o l l o w s t h a t f o r e v e r y aEA, b - a > b

all> - IIb - cI[ > 0 .

-

c > 0,

This contradicts the hypothesis

t h a t b i s i n t h e norm c l o s u r e o f A. QED

C o r o l l a r y 1.

(9.6)

I f a monotonic n e t {a } i n a normed R i e s z a

s p a c e norm c o n v e r g e s t o some e l e m e n t b , t h e n i t o r d e r c o n v e r g e s t o b.

Proof. (9.5),

For c o n c r e t e n e s s , assume { a } i s a s c e n d i n g . CL

i t i s e n o u g h t o show t h a t b > { a }. a = b. F o r e v e r y CL > a 0 , I[ b v a a - aa,l

'

bva aO

0

(Ib - ad/ ( 9 . 2 ) . - aJl

limdlbva aO

By

F i x a,; w e show =

[Ibvaa

-

0

S i n c e l i m IIb - aall = 0 , t h i s g i v e s a = 0 , h e n c e , b y t h e u n i q u e n e s s o f norm

42

Chapter 1

convergence, bva

= b. c10

Corollary 2.

(9.7)

I n a normed R i e s z s p a c e , e v e r y band -

i n d e e d , e v e r y o - o r d e r c l o s e d R i e s z i d e a l - i s norm c l o s e d .

Even f o r a m o n o t o n i c n e t , o r d e r c o n v e r g e n c e n e e d n o t i m pl y norm c o n v e r g e n c e : t h e exam pl e above i n t h e s p a c e c i s a n ascending sequence.

(9.8)

L e t E b e a normed Riesz s p a c e a n d I a norm c l o s e d R i e s z

ideal.

Then u n d e r t h e q u o t i e n t o r d e r a n d q u o t i e n t norm, E / I

i s a normed R i e s z s p a c e .

Proof.

Let B b e t h e u n i t b a l l o f E .

t h a t q(B) i s s o l d, t h a t i s , f o r every aEB, S i n c e 1qaI = q 1 a

We n e e d o n l y show [ I - q a l , \ q a [ J c q(B).

( ( 6 . 3 ) ( i i i ) , t h i s f o l l o w s from ( ( 6 . 3 ) ( i v ) ) . QED

A Banach l a t t i c e i s a normed R i e s z s p a c e w h i c h i s norm

complete.

As i s t o b e e x p e c t e d , t h e p r o p e r t i e s o f a normed

R i e s z s p a c e c a n g e n e r a l l y b e i m pr ove d upon f o r a Banach l a t t i c e . I t i s s t i l l t r u e o r d e r c o n v e r g e n c e and norm c o n v e r g e n c e

are independent.

However, we now h a v e :

Riesz Spaces

(9.9)

Theorem.

I n a Banach l a t t i c e E , i f a s e q u e n c e { a n ) norm

converges t o aEE,

t h e n some s u b s e q u e n c e o r d e r c o n v e r g e s t o a .

For s i m p l i c i t y , take a = 0.

Proof.

s e q u e n c e i f n e c e s s a r y , we c a n assume

We show an

43

+

0.

m For each n , {cnlaj

I}

And, b y t a k i n g a s u b -

anll < 1/2" (m

=

(n

=

1,2,

- - .).

n , n + 1 , 9 3 3 ) i s an

a s c e n d i n g Cauchy s e q u e n c e w i t h norms a l l < 1/2"'l;

hence i t

norm c o n v e r g e s t o some b n E E ,

Then bn+O,

by ( 9 . 6 ) , a n d l a n /

5 bn ( n

=

w i t h ]Ibnlj i 1/2"-l.

1 , 2 , - - - ) , s o we a r e t h r o u g h .

QE D

(9.10)

Corollary.

I n a Banach l a t t i c e , i f a s e t i s 0 - o r d e r

c l o s e d , t h e n i t i s norm c l o s e d .

F o r a g e n e r a l normed R i e s z s p a c e , w e were a b l e t o show t h i s only f o r Riesz ideal (9.7).

'

G i v e n A l i n e a r m a p p i n g E __ > F o f o n e normed R i e s z s p a c e i n t o a n o t h e r , we c a n a s k w h a t a r e t h e r e l a t i o n s b e t w e e n o r d e r c o n t i n u i t y a n d norm c o n t i n u i t y f o r T , o r ( s i n c e norm c o n t i n u i t y

i s e q u i v a l e n t t o norm b o u n d e d n e s s ) b e t w e e n o r d e r b o u n d e d n e s s a n d norm c o n t i n u i t y .

I n g e n e r a l , t h e r e i s none ( E x e r c i s e 2 4 ) .

However, i f E i s a Banach l a t t i c e , w e h a v e t h e

( 9 . 1 1 ) Theorem. ( C . mapping E

Birkhoff)

E v e r y o r d e r bounded l i n e a r

> F o f a B a n a c h l a t t i c e i n t o a normed R i e s z

Chapter 1

44

s p a c e i s norm c o n t i n u o u s .

Proof.

Assume T i s n o t norm c o n t i n u o u s .

e x i s t s a s e q u e n c e {a,)

1. n (n

IITanll (k

=

Then t h e r e

i n E s u c h t h a t l i m n ~ ~ a =n ~0 ~and By ( 9 . 9 ) , some s u b s e q u e n c e { a } "k

= 1 , 2 , ...).

order converges t o 0 , hence, i n p a r t i c u l a r , i s

1,2,...)

o r d e r bounded.

I t f o l l o w s {Ta } (k = 1 , 1 , . . . ) i s o r d e r nk

bounded i n F a n d t h e r e f o r e norm bounded. f a c t t h a t IITa "k

11 2

n k (k

=

This contradicts t h e

1,2,...). QED

(9.12)

Corollary

E v e r y p o s i t i v e l i n e a r mapping o f a Banach

l a t t i c e E i n t o a normed R i e s z s p a c e i s norm c o n t i n u o u s .

In

p a r t i c u l a r , t h i s h o l d s f o r e v e r y R i e s z homomorphism o f E .

Borkhoff's proof of (9.11)

(cf.

[ 9 ] ) was f o r a l i n e a r

functional, but i t c a r r i e s over verbatim.

EXERCISES

I n t h e s e e x e r c i s e s , E i s always a Riesz s p a c e , although some h o l d more g e n e r a l l y f o r a n o r d e r e d v e c t o r s p a c e and some

Riesz Spaces

45

for a lattice

1.

For a l l a , b , c E E , (i)

a 0;

(ii) a (-b);

(iii) if a < b and c < d, then a

+

c < b

+

d, avc < bvd,

and a c < bAd.

2.

For a l l a , b , c E E , c c - a b

=

-

avb

=

( c - a)A(C - b ) and

( c - a)V(c - b ) .

3.

For a l l aEE, a ( - a ) < 0 < aV(-a).

4.

Given a E E + , t h e n (i)

5.

f o r a l l b,cEE, av(b

+

c) < avb

+

avc;

( i i ) f o r a l l b,cEE, ah(b

+

c) < a b

+

a c .

(i)

a/\b = 0 i f and o n l y i f avb = a + b .

(ii)

a b

=

0 i f and o n l y i f &(Xb)

( i i i ) For a l l a , b , c E E + , i f a b

=

=

0 for all 1 > 0.

0 and a 0, a ( b

(i)

F o r a l l a , b E E , avb b - (b

- a)'.

=

=

0 , t h e n aA(b

+

c) = 0. c)

=

a + ( b - a ) + and ar\b

=

+

a h b + aAc.

Chapter 1

46

(ii)

For a l i n e a r s u b s p a c e F o f E t o b e a Riesz s u b s p a c e , it s u f f i c e s t h a t a E F i m p l i e s a+EF.

la - b l .

7.

avb - a/\b

8.

(Converse t o ( i ) i n ( 2 . 6 ) ) . Then b

9.

(i)

=

a+, c

a

=

G i v e n bAc = 0 , s e t a = b

+

b)-

a- + b - , and / a + b l

=

-

c.

.

I f a and b a r e d i s j o i n t , t h e n ( a

(a (ii)

=

I f a and b a r e d i s j o i n t and a

+

+

=

b)'

= a+ + b + ,

lal+Ibl.

b > 0 , then

a > O , b > O . ( i i i ) I f a l , . . . ,an a r e mutually d i s j o i n t and a l l d i s t i n c t from 0 , t h e n t h e y a r e l i n e a r l y i n d e p e n d e n t .

10.

lR i m p l i e s

11.

lim 1

I f E i s Archimedean, t h e n f o r e v e r y a E E ,

x c1 a

-f

cta

=

0 in

0.

I f I i s a Riesz i d e a l , then f o r a l l a , b E I ,

with a < b,

[a,b] c I.

12.

I n R',

l e t F1 b e t h e l i n e a r s u b s p a c e g e n e r a t e d b y t h e

element (1 ,l,O)

,

and F 2 t h e o n e g e n e r a t e d b y ( O , l , l ) .

Then F1 a n d F 2 a r e R i e s z s u b s p a c e s b u t F

13.

1

+

F2 i s not.

I f F i s a R i e s z s u b s p a c e and I a R i e s z i d e a l o f E , t h e n F + I i s a R i e s z s u b s p a c e , w i t h (F + I ) +

c F+

+

I.

Riesz Spaces

14.

Let A b e a s u b l a t t i c e o f E .

47

For a E E , . t h e f o l l o w i n g a r e

equivalent.

15.

'1

a i s t h e supremum o f some s u b s e t o f A ;

2'

t h e r e i s a n e t {a } i n A such t h a t a +a. c1.

c1

Let A , B be s u b s e t s of E. (i)

o r d e r c l o s u r e (AIJ B) IJ

(ii)

( o r d e r c l o s u r e A)

=

(order closure B).

O r d e r c l o s u r e ( A n B ) c ( o r d e r c l o s u r e A)

n

( o r d e r c l o s u r e B ) , and t h e i n c l u s i o n i s p r o p e r

in general. ( i i i ) If, i n ( i i ) , A , B a r e R i e s z i d e a l s , w e h a v e e q u a l i t y .

16.

(a)

F o r R i e s z i d e a l s 11,12,H o f E : (i)

H

n

(I1

(ii) if E (b)

18.

I1

Q

12, t h e n H

=

(fdnI,) 3 ( H n 1 2 ) .

12;

=

H I a n d i s a p r o j e c t i o n b a n d o f H.

If I a n d H a r e p r o j e c t i o n b a n d s , t h e n s o a r e 1 n I{ and I

17.

I1

n

+

=

H

n

I1

12)

If I i s a p r o j e c t i o n b a n d , t h e n f o r e v e r y R i e s z i d e a l H, H n I

(c)

=

+

+

H.

(I

d

n Hd .

(i)

For R i e s z i d e a l s I , H ,

(ii)

I f E i s Archimedean, t h e n , i n a d d i t i o n , d order closure (Id + H ) .

+

H)d

=

I

(InIj)d

If E i s D e d e k i n d c o m p l e t e , t h e n : (i)

e v e r y R i e s z i d e a l i s Dedekind c o m p l e t e ;

(ii)

e v e r y o r d e r c l o s e d R i e s z s u b s p a c e i s Dedekind complete.

=

48

19.

Chapter 1

I f E i s Dedekind c o m p l e t e , t h e n f o r e v e r y p a i r o f

(i)

bands I , H , I (ii)

+ H

i s a band.

The a b o v e n e e d n o t b e t r u e i f E i s n o t Dedekind complete.

20.

I f I i s a p r o j e c t i o n band, then f o r e v e r y a € E + \ aI

21.

=

V{bEIIO < b < a}.

Let E be Dedekind c o m p l e t e . (i)

Let I b e a R i e s z i d e a l a n d H i t s o r d e r c l o s u r e . Then f o r e a c h b E E + , h H

(ii)

=

V{aEIIO < a < b}.

Let H b e t h e b a n d g e n e r a t e d by a s u b s e t A . f o r e v e r y bEE,,

m

bH = V {bA ( c l n i / a i

( i i i ) Given a E E , t h e n f o r e a c h bEE,,

I ) I aiEA:

Then n i E N}.

ba = v n ( b A ( n l a l ) ) .

(We r e c a l l t h a t ba i s t h e component o f b i n t h e band g e n e r a t e d by a ( 5 8 ) (iv)

Let F be a R i e s z subspace and H t h e band g e n e r a t e d by F .

22.

.>

Then f o r e v e r y bEE,,

bH

=

V{bAa a € F + }

.

XE IR.

(i)

I f A i s s o l i d , t h e n XA i s s o l i d f o r a 1

(ii)

I f {A } i s a c o l l e c t i o n o f s o l i d s e t s , t h e n nclAa.

a.

and \J A

clcl

are solid.

( i i i ) Given A A C E , b y t h e s o l i d h u l l o f A , w e w i l l mean t h e i n t e r s e c t i o n o f a l l t h e s o l i d s e t s c o n t a i n i n g A. T h i s s o l i d h u l l i s p r e c i s e l y IJaEA

[- l a l , l a / ] .

(iv)

I f A i s s o l i d , i t s ~convex h u l l i s s o l i d .

(v)

I f A i s s o l i d , t h e n A c A+ - A+ ( A + = A

inclusion is proper i n general.

n

E+).

The

Riesz Spaces

23.

Let E b e a Banach l a t t i c e .

49

I f I i s t h e b a n d g e n e r a t e d by

a c o u n t a b l e s e t , then I i s a p r i n c i p a l band.

24.

(i)

The Banach l a t t i c e c o f c o n v e r g e n t s e q u e n c e s c a n b e written c

=

co

9 IRE

,

w h e r e co i s t h e s u b s p a c e o f

sequences c o n v e r g i n g t o 0 and U = ( l , l , l , * . * ) .

Let

P b e t h e p r o j e c t i o n o f c o n t o co d e f i n e d by t h i s decomposition.

Then, c o n s i d e r i n g c as t h e r a n g e

s p a c e , P i s norm c o n t i n u o u s b u t n o t e v e n o r d e r bounded ( h e n c e n o t o r d e r c o n t i n u o u s ) . (ii)

We d e n o t e , a s c u s t o m a r y , t h e s p a c e o f a l l f i n i t e s e q u e n c e s b y IR"), (Xl,X2,...),

and d e n o t e e a c h a €

IR")

by

it being understood t h a t only a f i n i t e

number o f t h e An's d i f f e r f r o m 0 . ( i l , A 2 , . . . , X n , - . - ) +>

Then t h e m a p p i n g

(X1,2x 2 , . . . , n ~ n ., . . ) i s o r d e r

c o n t i n u o u s b u t n o t norm c o n t i n u o u s .

25.

The f o l l o w i n g a r e e q u i v a l e n t : 1'

'2

E i s Dedekind complete; e v e r y a s c e n d i n g n e t i n E,

which i s bounded above

is order convergent; 3'

26.

e v e r y d e s c e n d i n g n e t i n E,

L e t E b e a normed R i e s z s p a c e .

i s order convergent.

G i v e n b = V A , l e t A1 b e

t h e s e t o b t a i n e d by a d j o i n i n g t o A t h e suprema o f a l l f i n i t e s u b s e t s o f A.

Then b i s i n t h e norm c l o s u r e o f A1.

CIIAPTIJR 2

RTESZ S P A C E D U A L I T Y

L e t L: be a v e c t o r s p a c c .

N o t a t i o n a n d t e r m i_ n o_ l og y~.

l i n e a r E u n c t i o n a l $ on E , t h e v a l u e o f 4 a t a E E by ( a , + ) .

We d e n o t e b y E

*

For a

w i l l be denoted

t h e v e c t o r space o f a l l l i n e a r func-

t i o n a l s on E ( a d d i t i o n a n d s c a l a r m u l t i p l i c a t i o n d e f i n e d p o i n t w i s e on E ) .

I t f o l l o w s from t h e v e r y d e f i n i t i o n o f l i n e a r func-

t i o n a l on t h e o n e h a n d , a n d t h e a b o v e d e f i n i t i o n o f a d d i t i o n and s c a l a r m u l t i p l i c a t i o n i n E b i l i n e a r f u n c t i o n on E E and E

*

E

*

E

=

is a

on t h e o t h e r , t h a t (.;)

.

a r e s e p a r a t i n g on each o t h e r : ( “ , G o )

i m p l i e s 40

aEE

x

*

*

0 , and ( a o , $ )

=

0 for a l l +€ E

*

D f o r a1 1

=

implies a0

=

*

i s c a n o n i c a l l y cndowed w i t h t h e t o p o l o g y a ( E , E ) o f

*

p o i n t w i s e c o n v e r g e n c e on E , a n d t h e terms “ c o n v e r g e n t i n E ”, “bounded i n E*“,

” c l o s e d i n E*“,

and s o f o r t h will always r e f e r

t o t h i s topology.

*

F u n c t i o n a l a n a l y s i s g e n e r a l l y works w i t h a p r o p e r subspace

of E , r a t h e r than E

*

itself.

*

F o r a s u b s p a c e F o f li , ( J ( F , € )

o f course coincides w i t h t h e topology induced by t h a t of E

*

.

Note a l s o t h a t E i s always s e p a r a t i n g on F , b u t F need n o t h e on E . The t o p o l o g y o f C

*

has three b a s i c properties.

50

0.

Riesz Space D u a l i t y

(1)

E* i s complete.

(11)

(Tihonov)

51

For e v e r y s u b s e t A o f E X , t h e f o l l o w i n g a r e

cquivalcnt: 1'

A i s compact;

2'

A i s c l o s e d and bounded.

Otherwise s t a t e d , C

*

has t h e IIeine-Rorel property.

( 1 1 1 ) For e v e r y l i n e a r s u b s p a c e I; o f E

*

,

t h e following are

equivalent: 1'

2

0

I: i s s e p a r a t i n g on

F i s dense i n E

*

E;

.

Consider a l i n e a r subspace F of E a l i n e a r f u n c t i o n a l @ +>

*

( a , @ ) on F .

.

Each a E E

defines

I f F is separating

on E , t h e n d i s t i n c t e l e m e n t s o f E d e f i n e d i s t i n c t l i n e a r f u n c t i o n a l s , and w e have a c a n o n i c a l imbedding o f E i n t o F

*

.

O t h e r w i s e s t a t e d , i f F i s s e p a r a t i n g on E , t h e n E a n d F a r e e a c h a s p a c e o f l i n e a r f u n c t i o n a l s on t h e o t h e r .

We a d o p t a c o n v e n t i o n f o r R .

so Chapter 1 a p p l i e s t o i t .

IR i s i t s e l f a R i e s z s p a c e ,

Iiowever, when d e a l i n g w i t h R, we

w i l l u s e t h e t e r m i n o l o g y common i n a n a l y s i s , r a t h e r t h a n t h e

52

Chapter 2

one w e have a d o p t e d f o r R i e s z s p a c e s .

F o r e x a m p l e , we w i l l

write sup 1 i n s t e a d o f v 1

.

o r d i n a r y convergence i n R .

I f { A } i s bounded t h e n t h i s

clcl

cla

"1

=

lim 1

w i l l denote

cla

c1

c l e a r l y c o i n c i d e s w i t h o r d e r convergence.

5 1 0 . The s p a c e E b o f o r d e r bounded l i n e a r f u n c t i o n a l s

Let E h e a R i e s z s p a c e and

+

a l i n e a r f u n c t i o n a l on E .

The d e f i n i t i o n s on l i n e a r mappings i n 5 6 a p p l y t o 4, r e s t a t e them h e r e f o r F =

IR.

+

i s o r d e r hounded i f i t i s

bounded on e v e r y o r d e r b o u n d e d s e t . n o n - n e g a t i v e on E + . max((a,+},(h,$})

and w e

I t is positive if it is

I t i s a R i e s z homomorphism i f ( a v b , $ )

f o r a l l a,bEE.

=

F i n a l l y , i t i s o r d e r con-

tinuous i f a

+ a implies (a,$} = lima(aa,$). a We w i l l d e n o t e t h e s e t o f o r d e r bounded l i n e a r f u n c t i o n a l s

on E by E

b

,

I t i s a l i n e a r subspace of E

positive linear functionals are a l l in E

h

*

.

.

Moreover, t h e Since they con-

s t i t u t e a c o n e , t h e y d e f i n e a n o r d e r on E b f o r w h i c h t h e c o n e

i s p r e c i s e l y ( Eh ) + .

(10.1)

Theorem.

(Riesz).

b b Under t h e o r d e r d e f i n e d by ( E ) + , E

i s a Dedekind c o m p l e t e R i e s z s p a c e .

Proof.

We show $+ e x i s t s i n Eb f o r e v e r y $ E E b

.

I t is

R i e s z Space D u a l i t y

e a s y t o show t h a t t h e n , f o r a l l + , J , E E

-~ Lemma 1.

by f ( a )

dE[O,b].

(J,

-

4)'

= +VJ,,

cE [ ( ) , a 1( c , + > *

+

For e v e r y cE [ O , a ]

b, so ( c , + ) + ( d , + )

I t follows f ( a ) + f ( b ) < f(a i f eE[O,a

+

f i s a d d i t i v e and p o s i t i v e l y homogeneous.

Consider a,bEE+. 0 < c + d < a

sup

=

,+

C o n s i d e r + E n b , and d e f i n e a

hence t h a t E b i s a R i e s z s p a c e . f u n c t i o n f on E,

b

53

+

d,+) < f(a

+

b).

For t h e o p p o s i t e i n e q u a l i t y ,

b).

+

(c

=

and d € [ O , b ] ,

b ] , t h e n by ( 2 . 5 ) ,

e

=

Hence ( e , + ) = ( c , + )

+

(d,+) < f(a) + f(b).

It

Thus f i s a d d i t i v e .

That

+

follows f ( a + b) < f(a)

f(b).

+

c

+

d , where c € [ o , a ] and

and 1 > 0 i s clear.

f ( x a ) = x f ( a ) f o r a l l aEE+

This e s t a b -

l i s h e s t h e Lemma. Applying E x e r c i s e 1 ( i n t h e p r e s e n t C h a p t e r ) , t h e r e i s a u n i q u e l i n e a r f u n c t i o n a l $ on E which c o i n c i d e s w i t h f on E, (hence i s p o s i t i v e ) ; w e show t h e o r d e r on E b , aEE+. wEEh

aEE,.,

5

a2

= $+.

IJJ

From t h e d e f i n i t i o n o f

i f and o n l y i f ( a , + , )

T h u s , by t h e v e r y d e f i n i t i o n o f satisfies w > 0, w hence w >

J,.

1. + ;

(a,w)

2

J,,

$ >

5

(a,+2) for a l l

+.

we show ( a , w ) > (a,$) ( c , ~ ) (c,+)

Suppose for a l l

f o r a l l c€[O,a].

Hence ( a , o ) > s u pc E [ o , a ] ( C 9 + )= f ( a ) = ( a , + ) . Thus E b i s a R i e s z s p a c e .

T h a t i t i s Dedekind c o m p l e t e

f o l l o w s from t h e f o l l o w i n g s t r o n g e r p r o p e r t y ( c f . E x e r c i s e 2 5 i n Chapter 1 ) :

54

Chapter 2

b I f an a s c e n d i n g n e t { $ 1 i n ( E ) + i s , ( E ~ , E ) c1 b b o u n d e d , t h e n $ + $ f o r some $ E E . Lemma 2 .

c1

For e a c h a E E + ,

i s a n a s c e n d i n g n e t i n lR f o r

{( a , @ , ) }

w h i c h , by t h e h y p o t h e s i s ,

~ u p , ( a , $ ~ )
( a , $ ) f o r a l l a i E + , hence

il,

$

Hut

for a l l aEE+.

2

@.

Given a E E + ,

Supu(a,$a)

=

(a,$) > (a,$,)

f o r a l l a , hence ( a , $ )

>

-

(a,@>.

T h i s e s t a b l i s h e s t h e Lemma a n d t h u s c o m p l e t e s t h e p r o o f

o f t h e Theorem.

QED

Remark.

I n p l a c e o f Lemma 2 , we c o u l d h a v e u s e d t h e

following equivalent property:

(10.2)

I f a s u b s e t {@a} o f E b i s f i l t e r i n g upward a n d s a t i s f i e s :


s u p a ( a , $cx)

0, there exist b,cE[O,a],

E

a , such t h a t (by@)

+

( c , + >2

E -

A l s o f r o m ( 1 0 . 3 ) - a n d t h e f a c t t h a t / b l 5 a i f and o n l y i f b = c - d with c,dE

[o , a ]

( 1 0 . 6 ) F o r e v e r y $EEb

a n d aEE+,

--

we h a v e

Whence ( o r by s t r a i g h t f o r w a r d c o m p u t a t i o n ) ,

(10.7)

F o r e v e r y aEE

and $ E E

I(a,+.)l

2

b

,

(la1

,Id)

*

An o b v i o u s p r o p e r t y o f E b *

(10.8)

Given a s e t { $ , ) . i n

f o r a l l aEE+, t h e n $ =

V$ ,.,

Eb and $ E E

b

,

i f ( a , $ ) = sup,(a,$,)


57

The o p p o s i t e i m p l i c a t i o n d o e s n o t h o l d , e v e n f o r a f i n i t e

s e t , w h i c h i s why w e n e e d ( 1 0 . 4 ) .

However, i t __ does hold i f

{$a} i s an a s c e n d i n g n e t o r a s e t w h i c h i s f i l t e r i n g u p w a r d . This i s contained i n the proof of (10.1).

More f u l l y , f r o m

that proof,

(10.9)

b i n (E ) + , t h e f o l l o w i n g a r e

F o r an a s c e n d i n g n e t

equivalent: 10

{ G a l i s a ( ~ b , ~c o)n v e r g e n t ;

2'

{$a} i s .(Eb,E)

3'

{$a} i s o r d e r c o n v e r g e n t ;

'4

{$a} i s o r d e r b o u n d e d .

bounded;

In another formulation, f o r a s e t

b c (E ) + which i s

f i l t e r i n g upward, t h e f o l l o w i n g a r e e q u i v a l e n t : t h e r e e x i s t s $€Eb

1'

(hence

such t h a t ( a , $ ) = sup ( a , $ ) c1

lima(a,$a))

f o r a l l aEE,;

2'

s u p a < a , @ c l )

0.

Riesz Space D u a l i t y We h a v e t o show $ € I L

4

,

E b = (IL) d 3 I',

HL.

+

61

s o i t i s enough t o show $

EHL.

so $ = $ By ( 1 0 . 1 7 )

(IL 1

II e a ch aEH+,

(Illd ,' f o r

+

O-< b-< a

Since every such b l i e s i n H , ( a , $ )

=

So t h e r i g h t s i d e

0.

i s 0 , a n d we a r e t h r o u g h .

If E = I

( 1 0 . 2 0 ) C o r o l l a r y 1.

E~

(i)

= HL

that H

+

I

T h a t HL =

H , I,H Riesz i d e a l s , then

3 11

( i i ) HL = I b and IL

P roof. -

%,

0

1'

b.

=

H

=

0 i s t r u e i n general vector spaces;

E f o l l o w s from ( 1 0 . 1 9 ) .

We t h u s h a v e ( i ) .

By

I b i n ( i i ) , we mean, more p r e c i s e l y , t h a t t h e b i l i n e a r b form ( , . ) on I x HL i n d u c e d by t h e c a n o n i c a l o n e on E X E

HL

=

-

d e f i n e s a R i e s z i s o m o r p h i s m o f HL o n t o I $

+-> $11 c l e a r l y

maps HL i n t o I

b

.

b

.

The mapping

The v e r i f i c a t i o n t h a t

t h i s mapping i s a R i e s z i s o m o r p h i s m ( i n t o ) i s s t r a i g h t f o r w a r d . b I t r e m a i n s t o show i t i s o n t o . Given $ € I , l e t $ b e t h e element o f Eb d e f i n e d by ( a , $ ) = ( a , $ ) f o r a l l a E I ( a , $ ) = 0 f o r a l l aEH.

and

Then $ € H I and c o i n c i d e s w i t h $ o n I ,

h e n c e i s c a r r i e d i n t o $ u n d e r t h e a b o v e mapping. QED

62

Chapter 2

(10.21) Corollary 2. b b a n d s o f E , t h e n .J

(Lotz [ 3 3 ] ) . +

I f ,J,G a r c a ( E b , E )

G i s a o(Eb,E)

closcd

c l o s e d band.

We w i l l p r o v e t h i s a t t h e e n d o f 511.

511. E a s " p r e d u a l " o f E

b

We e x a m i n e w h i c h p r o p e r t i e s o f 510 h o l d when we i n t e r b c h a n g e E and E b ( a n d o f c o u r s e r e p l a c e a ( E b , E ) by o ( E , E ) . b While E i s a l w a y s s e p a r a t i n g on E b , we n e e d n o t h a v e E s e p a r a t i n g on E . However, f o r t h e s p a c e s t o b e c o n s i d e r e d b i n t h i s work, E w i l l b e s e p a r a t i n g on E . Consequently, w e w i l l i n c l u d e t h i s a s s u m p t i o n i n many o f t h e t h e o r e m s b e l o w . E b , b e i n g Dedekind c o m p l e t e , i s A r c h i m e d e a n .

While E n e e d

n o t b e Dedekind c o m p l e t e , we do h a v e

( 1 1 . 1 ) I f Eb i s s e p a r a t i n g on E , t h e n E i s A r c h i m e d e a n .

Proof. -__ b

Suppose 0 < b < ( l / n ) a (n

every $ E ( E ) + (b,$)

=

0.

,

0

5

(b,@)

I t follows b

=

5 ( l / n ) ( a , $ ) (n

=

1,2;.*). =

1,2;-.),

Then f o r hence

0. QE n

The f o l l o w i n g s h o u l d b e compared w i t h t h e d e f i n i t i o n o f a

Riesz S p a c e D u a l i t y

63

positive linear functional.

(11.2)

Tf

E

b . i s s e p a r a t i n g on E , t h e n f o r a l l a E E , a E E +

if

b and o n l y i f ( a , + ) > 0 f o r 311 + € ( E ) + .

P r o o f . Suppose a#E+.

i d e a l g e n e r a t e d by a - .

# 0 ; l e t I be t h e Riesz

Then a

I _

We h a v e E b

=

d ( I L ) 3 1'

(lO.lh),so

c l e a r l y ( I L ) d i s s e p a r a t i n g on I .

I t follows there e x i s t s

s € ( ( I L ) d ) + such t h a t ( a - , $ ) > 0 .

S i n c e , by (10.181, ( a + , $ ) = O ,

we h a v e ( a , $ )

=

- ( a ,$I) < 0 . QED

(11.3)

Corollary.

E,

-

b is a(E,E ) closed in E.

We now t u r n t o t h e q u e s t i o n w h e t h e r t h e r e s u l t s i n 510 b c a r r y o v e r when w e i n t e r c h a n g e E a n d E . (10.1) of course does n o t c a r r y o v e r , s i n c e E n e e d n o t b e Dedekind c o m p l e t e . o f course (10.2)

( a l s o Lemma 2 i n ( 1 0 . 1 ) ) d o e s n o t c a r r y o v e r .

In contrast t o t h i s ,

(10.3) and i t s immediate consequences

a r e r e p l a c e d h e r e by s t r o n g e r r e s u l t s :

(11.4)

Then

Given a € E ,

b then f o r every + € ( E ) + ,

64

Chapter 2

M o r e o v e r , t h i s supremum i s a t t a i n e d . such t h a t ( a , + )

=

[O,@]

(a+,+).

(a+,+>2 (a+,$>

Proof.

T h e r e e x i s t s I)€

2

( a , + > f o r a l l $C LO,@],

so we

n e e d o n l y show t h e e x i s t e n c e o f a $ s a t i s f y i n g t h e l a s t Let I b e t h e R i e s z i d e a l g e n e r a t e d by a + .

equality.

g i v e s t h e decomposition Eb Then ( a , $ ) = ( a ' , + )

=

(1')

d

0 1'.

Set $

=

+

This

(1')d'

- ( a ,+) = ( a + , + ) = ( a + , + ) , where t h e s e c -

ond e q u a l i t y f o l l o w s f r o m ( 1 0 . 1 8 ) , a n d t h e l a s t f r o m t h e d e f i n i t i o n of

+. QED

b ( 1 1 . 5 ) For e v e r y a , b E E and + E ( E ) + , with p + u

=

there exist ~ , o € [ O , + l ,

4, s u c h t h a t (a"b,+)

=

(a,p)

(mb,+)

=

(a,o>

+

+

(b,o)

,

(b,D)

2

T h i s f o l l o w s f r o m ( 1 1 . 4 ) i n t h e same way t h a t ( 1 0 . 4 ) f o l l o w s from ( 1 0 . 3 ) .

(11.6)

I f E b i s s e p a r a t i n g on E , t h e n f o r a l l a , b E E + , t h e


a/\b

2'

b f o r e v e r y +E(E )+, t h e r e e x i s t p,crE[O,+],

=

0.

with


p +

such t h a t

= @,

0

(a,D)

P roof.

2'

65

T h a t 1'

=

(b,o) = 0.

i m p l i e s 2'

h o l d s , and s e t c = w b .

f o l l o w s from ( 1 1 . 5 ) .

b Then f o r e v e r y + E ( E ) + ,

Suppose letting

$ = p + u be t h e d e c o m p o s i t i o n p o s t u l a t e d by 2 O , we have

0 < ( c , @ )= ( c , p )

+

( c , ~ 5) ( a , ~ )+ ( b , o )

=

0.

I t follows c

=

0

QE D

( 1 0 . 8 ) c a r r i e s o v e r , b u t now n e e d s p r o v i n g .

Given a s e t { a } i n E and a f o r a l l + E ( E b ) + ,t h e n a = V a . a ,

L e t Eb b e s e p a r a t i n g on E .

(11.7)

aEE, i f ( a , @ ) = sup ( a , , $ )

P roof.

S i n c e f o r e v e r y a, ( a , + > > for a l l

i t f o l l o w s from ( 1 1 . 2 ) t h a t a > aa.

t h e r e e x i s t s bEE

such t h a t aa

5

Suppose a # vclacl.

b < a f o r a l l a.

@E(E~)+, Then

( E b ) + by b

i t s e l f i s s e p a r a t i n g on E , h e n c e t h e r e e x i s t s $E(E ) + s u c h t h a t (a

- b,$)

> 0.

Then ( a , @ ) > ( b , $ ) > sup,(aa,$)

=

(a,$)

and

we have a c o n t r a d i c t i o n . QE D

b

i s s e p a r a t i n g on E , t h e n f o r an b ascending n e t i n E , o ( E , E ) convergence i m p l i e s o r d e r con(11.8)

Corollary.

ve r g e n c e

If E

66

Chapter 2

( 1 0 . 9 ) d o e s n o t c a r r y o v e r ; e v e n f o r an a s c e n d i n g n e t , b o r d e r convergence d o e s n o t imply a ( E , E ) convergence: example i n t h e s equence s p a c e c p r e c e d i n g ( 9 . 5 ) , n s e q u e n c e {C e 1 (n = 1 , 2 , . I k

a

In t h e

the ascending

order converges t o the element

a )

b ( l , l , l , a - . ) b u t does n o t o(E,E ) converge t o it ( t h e r e e x i s t s b . $ € E w i t h v a l u e 1 on t h i s l a t t e r e l e m e n t a n d v a l u e 0 on all the elements of the sequence. I t follows of course t h a t (10.10), not carry over.

b F o r an e x a m p l e o f a o ( E , E ) c l o s e d R i e s z i d e a l

which i s n o t o r d e r c l o s e d , l e t E sequences.

( 1 0 . 1 1 ) a n d ( 1 0 . 1 2 ) do

=

c, the space of convergent

c i s a Banach l a t t i c e , a n d we w i l l s e e ( 1 5 . 7 ) t h a t

t h e r e f o r e E b i s a c t u a l l y t h e Banach s p a c e d u a l o f E .

It

f o l l o w s t h a t f o r l i n e a r s u b s p a c e s o f E , t h e norm c l o s u r e a n d b o(E,E ) closure coincide.

sequences converging t o 0.

Now c o n s i d e r c o , t h e s u b s p a c e o f co i s a R i e s z i d e a l w h i c h i s norm

b c l o s e d , hence o(E,E ) c l o s e d .

But i t s o r d e r c l o s u r e i s a l l

of E.

I t i s n o t t o b e e x p e c t e d t h a t ( 1 0 . 1 3 ) and (10.14) s h o u l d h h o l d w i t h E a n d Eh i n t e r c h a n g e d . (E ) + i s t h e s e t o f all 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 on E , b u t E + i s n o t t h e s e t o f a l l b t h o s e on E .

For a s u b s e t A o f E h , we w i l l d e n o t e t h e s e t { a E E l ( a , @ ) = O f o r a l l $CA}

by A L .

Only p a r t o f (10.15)

carries over:

( 1 1 . 9 ) F o r e v e r y R i e s z i d e a l .I o f E b ,

I n general, it i s n o t a band.

J

L

i s a R i e s z i d e a l of E.


Proof.

That J

67

i s a R i e s z i d e a l f o l l o w s by t h e a r g u m e n t

i n (10.15) using (l1.4),

To show i t n e e d n o t be a b a n d , t a k e

t h e R i e s z i d e a l co o f c a n d s e t <J b

u(E,E ) closed (cf. above), J

1

c

=

= 0'

( c0 )I.

S i n c e co i s

and w e h a v e shown

abovc

t h a t i t i s n o t a band. QED

We a r e now i n a p o s i t i o n t o p r o v e ( 1 0 . 1 2 ) . Consider a b R i e s z i d e a l <J o f E . I t s .(Eb,E) c l o s u r e i s (.J )' h e n c e , by 1

(11.9)

and ( 1 0 . 5 ) i s a b a n d .

I t r e m a i n s t o show t h a t ( ( J ) ' ) + L b The l a t t e r i s c o n t a i n e d i n ( E ) +

i s t h e 0 ( E b , E ) c l o s u r e o f J,.

by ( l O . l 3 ) , h e n c e i s c o n t a i n e d i n ( ( J ) ' ) + . L

For t h e o p p o s i t e

b ~ E ( E) + i s n o t i n t h e a ( ~ b , ~c l)o s u r e o f

i n c l u s i o n , suppose

< J + ; we show i t i s n o t i n ( J )" 1

J+ i s a c o n v e x c o n e , so by

t h e Hahn-Banach t h e o r e m , t h e r e e x i s t s a E E s u c h t h a t s ~ p ~ ~ ~ + (= a 0, 0 , it follows in

This completes t h e proof o f (10.12).

t h e a(Eb,E) closure of a R i e s z i d e a l i s order

c l o s e d , w h i l e (as we s h a l l s e e ) t h e o r d e r c l o s u r e n e e d n o t b e b o(E , E )

closed.

In E , exactly the opposite holds.

The

b

u(E,E ) c l o s u r e o f a Riesz i d e a l need n o t be o r d e r c l o s e d ( c f . t h e d i s c u s s i o n f o l l o w i n g ( 1 1 . 8 ) ) , b u t , a s w e now show, b its order closure is o(E,E ) closed.

( 1 1 . 1 0 ) I f Eb i s s e p a r a t i n g on E , t h e n e v e r y b a n d I o f E i s

Chapter 2

68

b a(E,E ) closed.

P roof. -

We show ( I L )

1

t a i n s some a E I d , a a fortiori (IL)

1

#

=

I.

Suppose n o t .

Then ( I L ) c o n 1

0 : E i s A r c h i m e d e a n , by (ll.l),

is also.

Now I i s a band o f ( I L )

our assumption, a proper band.

L

so

,

and by

I t f o l l o w s from ( 5 . 8 ) t h a t

t h e r e e x i s t s a E ( I L ) d i s j o i n t from I . 1

Now a v a n i s h e s on I L , and by (10.18), i t v a n i s h e s o n (IL)d.

a t h u s v a n i s h e s on a l l o f E b ,

h e n c e a = 0 , a n d we h a v e

a contradiction. QED

does c a r r y

(10.12)

o v e r on i n t e r c h a n g i n g E and E b ,

i f we

r e p l a c e "band" b y t h e w e a k e r " R i e s z i d e a l " :

(11.11) I f E

b

b i s s e p a r a t i n g on E , t h e n t h e a ( E , E ) c l o s u r e o f

a R i e s z i d e a l I i s a R i e s z i d e a l , and i t s p o s i t i v e c o n e i s t h e u ( E , Eb ) c l o s u r e o f I + .

The p r o o f i s t h e same a s t h a t u s e d a b o v e t o e s t a b l i s h (10.12).

The w e a k e r c o n c l u s i o n i s d u e t o t h e f a c t t h a t ( 1 1 . 9 )

i s weaker than ( 1 0 . 1 5 ) .

I f {J,} i s a c o l l e c t i o n o f R i e s z i d e a l s o f E b ,

f r o m t h e g e n e r a l t h e o r y o f v e c t o r s p a c e s , (CaJa)L =

then again

n,(J,)L,


a n d i f t h e J ' s a r e .(Eb,E) c1

closed,

69

(nc J1 a) L

= a(E,E

b

) closure

( C , ( J ~ ) ~l )l o.w e v e r , t h e t h e o r e m c o r r e s p o n d i n g t o ( 1 0 . 1 9 )

not

h o l d : g i v e n two b a n d s J , G o f E b ,

t h e n , i n g e n e r a l , <J

(JnGjL. F o L a n e x a m p l e , l e t E = c a g a i n , a n d s e t I J = ( ( c ~ ) ' ) ~ a, n d G = (c,)'.

show <J

1

+ G

1

# E.

J

L

=

(c,)~

J

n

=

G = 0 , s o (.JnG)L

=

co,

=

E:

0 , by E x e r c i s e 2 , and G

b since the l a t t e r is a(E,E ) closed.

does

Thus J + G L

= L

L

0 +co

CL#

+

We =

c6'

# E.

F i n a l l y , we give t h e proof of t h e Lotz theorem (10.21).

(J

+

G)l

.J + G .

=

JL n G L , h e n c e ( ( J + G ) L ) L

S i n c e (J

+

=

(.Jl)'

+

(Gl)'

(10.19)

=

G ) L i s a Riesz i d e a l o f E , by ( 1 1 . 9 ) ,

(10.15) g i v e s us t h e d e s i r e d r e s u l t . N o t e t h a t t h i s t h e o r e m d o e s n o t c a r r y o v e r t o E : t h e sum b b o f two o ( E , E ) c l o s e d R i e s z i d e a l s o f E n e e d n o t b e o ( E , E ) closed.

We p r e s e n t a n e x a m p l e i n E x e r c i s e 6 .

5 1 2 . The s p a c e E C o f o r d e r c o n t i n u o u s l i n e a r f u n c t i o n s

We r e c a l l ( 5 1 0 ) t h a t a l i n e a r f u n c t i o n a l space E i s o r d e r continuous i f a lima(acl,@).

c1

-t

@

on a Riesz

a implies ( a , $ ) =

We w i l l d e n o t e t h e s e t o f o r d e r c o n t i n u o u s l i n e a r

f u n c t i o n a l s on E by E C .

By ( 6 . 0 ) , E C

C

Eb.

(12.1) Given a R i e s z s p a c e E , t h e n f o r 4 E E are equivalent:

b

,

the following

Chapter 2

70

lo

@EEC;

2'

I$lEEc;

3'

f o r e v e r y n e t { a 1 o f I:,

Proof.

a 40 i m p l i e s ~

a

We show f i r s t t h a t 1'

h o l d s , and s u p p o s e a $ 0 b u t { ( a n , a

implies 3

I@I }

0

.

i (ma rY

a' 1 4 1 )

=

0.

Assume 1'

d o e s n o t c o n v e r g e t o 0.

Ry t a k i n g a s u b n e t i f n e c e s s a r y , we c a n assume t h e r e e x i s t s I

0 such t h a t ( a a ,

/@I)

> 1 for a l l

(1.

Then f o r e a c h a , t h e r e

e x i s t s b E [ - a a ] such t h a t ( b $) > 1 ( 1 0 . 6 ) . a a' a Y(. ' 0 ( s i n c e lbwl < a n ) , h e n c e , by 1 , l i m ( b , $ ) = 0 . a c1 a contradiction.

i m p l i e s 1'

The v e r i f i c a t i o n t h a t 3'

But h a

+

0

We t h u s h a v e

implies 2

0

a n d 2'

is straightforward.

qr: u

( 1 2 . 2 ) -___ 'Theorem.

___ Proof.

E C i s a band o f E

b

.

EC i s c l e a r l y a l i n e a r subspace.

g i v e s u s t h a t i t i s a Riesz i d e a l .

(12.1) then

Finally, that it is order

c l o s e d f o l l o w s from t h e

Lemma. ___

Given a n e t { @ } i n ( E c ) + , i f @ , i t @ E E b , t h e n @ € E C . ci

71


(b,$)

=

l i m ( b , $ ) u n i f o r m l y on [ - a , a ] . -_____

ct

Now s u p p o s e a 40 i n E ; we show l i I n B ( a R , ( P ) = 0 , w h i c h will B e s t a b l i s h t h e Lemma, a n d , w i t h i t , t h e Theorem. We c a n a s s u m e t h e n e t { a 1 h a s an i n i t i a l e l e m c n t , w h i c h we d e n o t e b y a o . B

agE[O,aO]

f o r a l l B,

s o , by t h e a b o v e r e s u l t , ( a , , @ )

l i m a ( a B , @ J u n i f o r m l y i n P.

Since a l s o l i m

B( , ‘I 4 ’@ CL )

=

=

0 for

a l l a , s t a n d a r d f u n c t i o n t h e o r y g i v e s u s t h a t l i m B ( a B , @ )= 0 .

QE D

E‘

h a s s t r o n g e r p r o p e r t i e s t h a n II

b

.

The f o l l o w i n g

i m p r o v e s on ( 1 1 . 9 ) .

(12.3)

F o r e v e r y R i e s z i d e a l .J o f E c ,

Proof. -__

.J

1

,I

1

i s a band o f E .

i s a Riesz i d e a l by (11.9).

That i t i s o r d e r

c l o s e d f o l l o w s f r o m t h e o r d e r c o n t i n u i t y o f t h e e l e m e n t s o f .J. QEI)

F o r t h e s i g n i f i c a n c e o f t h e n e x t two t h e o r e m s , s e e E x e r c i s e s 4 and 5 .

(12.4)

Theorem.

i d e a l ,J O f E‘,

I f E i s Archimedean, t h e n f o r e v e r y Riesz

t h e b a n d ( J L ) d o f E i s s e p a r a t i n g on J .

Chapter 2

72

C o n s i d e r $ E J , and s u p p o s e ( a , $ ) = 0 f o r a l l d a E ( J ) . Then ( a , $ ) = 0 f o r a l l a E ( . J ) %, J , h e n c e ( $ b e i n g L L L d o r d e r c o n t i n u o u s ) f o r a l l a i n t h e o r d e r c l o s u r e o f ( J ) %, .J . Proof. d

L

the latter is E.

By ( 5 . 9 ) ,

I t follows $

=

L

0.

QED

( 1 2 . 5 ) Theorem.

Let E b e Archimedean.

n

J , G of EC, i f J

( . J ~n ) ~

G = 0 , then (G )d c J L

=

0 , whence ( G ) d c J L

t h a t f o r every

a1

+

bl

=

d

1

c GL,whence

d We show t h a t f o r a E ( ( G L ) ) + a n d

g e n e r a t e d b y , s a y , $.

=

a n d (J )

Assume f i r s t t h a t G i s a p r i n c i p a l R i e s z i d e a l ,

Proof.

+A$

L

o.

=

$EJ+, (a,$)

F o r two R i e s z i d e a l s

E

5

> 0, (a,$)

0 , s o , by ( 1 0 . 5 ) ,

L

.

S p e c i f i c a l l y , we show

E.

there exist al,blEIO,a],

a , such t h a t b1,$)

+

(bl,$)

5

E/2.

In turn, there e x i s t a2,b2E[0,bl], a2

+

b2

=

bl,

such t h a t

C o n t i n u i n g i n t h i s f a s h i o n , we o b t a i n two s e q u e n c e s Can} {b,}

,

i n E+ w i t h t h e p r o p e r t i e s (i)

an

(ii)

(an,$>

+

bn

= +

(n

bn-l

O n , + )5

N o t e f i r s t t h a t bn+O.

E D n

=

2,3,...),

(n = 1 , 2 ; . - ) .

I n e f f e c t , s u p p o s e bEE

satisfies

0 < b < bn f o r a l l n . Then 0 < (b,$) < (bn,$) 5 ~ / f o2 r a~ l l n, whence ( b , + ) = 0 . I t follows b E G I . Since 0 < b < a, we


a l s o have bE(G ) d , h e n c e b

=

J.

73

0.

I t f o l l o w s xnan = a i n t h e s e n s e o f o r d e r c o n v e r g e n c e . Since $ i s order continuous, ( a , $ ) Thus ( G ) d c J 1

(12.3)

1

and ( 5 . 6 ) ) ,

.

zn(an,$) 5

=

I t f o l l o w s (.Jl)d

E.

c (Gl)dd

=

G

1

(by

and we have t h e t h e o r e m f o r G a p r i n c i p a l

I n t h e g e n e r a l c a s e , l e t { G 1 be t h e s e t o f a l l a d p r i n c i p a l Riesz i d e a l s i n G . Then (Jl) c (Ga)l f o r a l l a , Riesz i d e a l .

hence ( J ) d c na(Ga)l= (1 G )

a a 1

1

=

G

1

. QED

(12.6)

Corollary.

I f E i s Archimedean, t h e n for two R i e s z

ideals J , G of EC, i f J

G = 0 , t h e o r d e r c l o s u r e of J

1

+

G

1

is E.

T h i s f o l l o w s from t h e above and ( 5 . 9 ) . For t h e f o l l o w i n g , s e e Theorem 3 1 . 5 i n [ 3 5 ] .

( 1 2 . 7 ) Theorem.

I f a Riesz space E i s

(Luxemburg-Zaanen).

Archimedean, t h e n e v e r y band J o f E c i s o ( E C , E ) c l o s e d i n E C .

P roof. ( J )' 1

n

Set G = J

EC; s o EC = J 3 G.

G = 0 ; i t w i l l l o l l o w (J1)'

desired conclusion.

(J1)'

d

c ((Gl)d)l.

By ( 1 2 . 5 ) , Jl

n 3

We show

E C = J , which i s t h e (Gl)d, hence

S i n c e (G ) d i s s e p a r a t i n g on G ( 1 2 . 4 ) , 1

74

((C

Chapter 2

L

d ) )'

n

C = 0 , and we a r e t h r o u g h .

I f a Riesz spacc E i s Archimedean, t h e n f o r

(12.8) Corollary.

e v e r y Riesz i d e a l J o f E C , t h e o r d e r c l o s u r e of J and i t s o(EC,E) c l o s u r e i n EC c o i n c i d e .

§ 1 3 . The c a n o n i c a l i m b e d d i n g o f E i n E

We w i l l d e n o t e ( E b ) b s i m p l y by E b b ,

Assume Eb i s s e p a r a t i n g on E .

Notation.

( a , $ ) on E

x

a n d ( E b ) c by E b c .

Then w e h a v e t h e c a n o n i c a l i m -

b * b e d d i n g o f E i n (E ) : f o r e a c h aEE,

t h e l i n e a r f u n c t i o n a l $+->

bb

i t s image i n ( E b ) *

( a , $ ) on E

b

is

.

Thus f a r we h a v e worked w i t h t h e b i l i n e a r f o r m

Eb .

We a r e now i m b e d d i n g E i n t o E b b

.

In order

n o t t o c h a n g e ( a , $ ) t o ( $ , a ) , we a d o p t t h e c o n v e n t i o n t h a t t h e d u a l i t y between Eb and (Eb)* w i l l be d e n o t e d by a b i l i n e a r f o r m on ( E b ) * f o r $€Eb (@,$).

x

Eb

( i n s t e a d o f Eb

x

(E b ) *) .

Otherwise s t a t e d ,

,

t h e v a l u e of 0 a t $ w i l l b e w r i t t e n bb The same n o t a t i o n w i l l t h e n b e i n h e r i t e d f o r E . and Q E ( E b ) *

( 1 3 . 1 ) -___ Theorem.

Let E

b

b e s e p a r a t i n g on E .

The c a n o n i c a l

imbedding o f E i n t o (Eb)* i s a R i e s z isomorphism Ebc o f Ebb.

into

t h e band


75

T D e n o t e t h e c a n o n i c a l i m b e d d i n g b y E -->(E

Proof.

So T i s g i v e n b y ( T a , g )

=

( a , $ ) f o r a11 aEE

and $ E E

b * )

b

.

.

By

t h e v e r y d e f i n i t i o n o f o r d e r on E b , t h e l i n e a r f u n c t i o n a l Ta bb i s p o s i t i v e f o r e v e r y a E E + , hence l i e s i n E . I t f o l l o w s I n f a c t , T(E) c Eb c :

T(E) c E b b .

that i s the content of

(10.12).

We show t h a t ar\b

=

0 i n E i m p l i e s (Ta)A(Tb) = 0 i n E b

which w i l l complete t h e p r o o f . show t h e r e e x i s t p , a E [ O , $ ] , (Tb,a)

'
0.

qb i m p l i e s ( a , $ )

=

( b , $ ) , 4 determines a (positive)

l i n e a r f u n c t i o n a l ii, on E / T b y ( q a , + ) = ( a , $ ) f o r all a E E . t t h u s have a $ i ( E / I ) b such t h a t q $ = 4.

We

t t S i n c e q i s p o s i t i v e , s o i s q . And s i n c e q i s o n t o , (1 t i s o n e - o n e . Thus I L < q ( E / I ) b i s a p o s i t i v e b i j e c t i o n .

M o r e o v e r , from t h e a b o v e p r o o f t h a t q t i s o n t o , positive.

is also

(qt)-'

I t i s easily shown t h a t a b i j e c t i o n T s u c h t h a t

b o t h T and T - '

a r e p o s i t i v e i s a R i e s z i s o m o r p h i s m , s o we a r e

through.

QEn

H e n c e f o r t h , we i d e n t i f y '1

with (E/I)

b

.

We s t a t e t h i s

explicitly.

Given a R i e s z i d e a l I o f a R i e s z s p a c e E , t h e n u n d e r t h e b i l i n e a r form ( $ , $ )

we h a v e IL

on ( E / I )

x

'I

d e f i n e d by ( q a , $ ) = ( a , $ ) ,

(E/I)b.

To d e a l w i t h t h e c a n o n i c a l i n j e c t i o n o f a R i e s z s u b s p a c e ,

we n e e d some p r e l i m i n a r y r e s u l t s .

A l i n e a r mapping E

-> F o f o n e R i e s z s p a c e i n t o

a n o t h e r w i l l be c a l l e d i n t e r v a l p r e s e r v i n g i f (i)

T i s p o s i t i v e , and

(ii)

f o r every i n t e r v a l [a,b] of E , T([a,b])

=

[Ta,Tb].


83

Note ( 6 . 3 ) t h a t a q u o t i e n t map i s i n t e r v a l p r e s c r v i n g .

114.7)

If E

> F i s i n t e r v a l p r e s e r v i n g , t h e n f o r e v e r y Riesz

~

i d e a l T o f E , T ( I ) i s a Riesz i d e a l o f F and T(T+)

=

((T(T))+.

I n p a r t i c u l a r , T(E) i s a R i e s z i d e a l o f F.

If 0 < d < cET(I), then, since T i s interval

Proof. -__

preserving, dET(1).

We show t h a t i f c E T ( I ) , t h e n c ' + E T ( T ) ,

which w i l l e s t a b l i s h b o t h t h a t T ( 1 ) i s a Riesz i d e a l ( c f . ( 3 . 1 ) ) and t h a t T ( I + ) (Ta) Tb

=

+

5

Ta

+

,

=

((T(I))+.

c

=

T a f o r some a E T , s o 0 < c+

so t h e r e e x i s t s b E [ O , a + ]

=

such t h a t

c'.

QED

An e x t e n s i o n lemma

(14.8)

Lemma.

p o s i t i v e l i n e a r f u n c t i o n a l on F , (a,+) 5(a,$) Then

+

a n d @ i s o n e on E s u c h t h a t

for a l l aEE+.

can bc e x t e n d e d t o a p o s i t i v c l i n e a r f u n c t i o n a l

such t h a t

35

Proof.

~-

p(a)

Suppose $ i s a

Let E b e a R i e s z s u b s p a c e o f F .

= (a+,$)

on F

$.

D e f i n e t h e s u b l i n e a r f u n c t i o n a l p on F b y for a l l aEF.

Then p d o m i n a t e s

+

on E , s o , b y

Chapter 2

84

t h e Hahn-Banachtheorem, @ can h e e x t e n d e d t o a l i n e a r f u n c tional

T

on F w h i c h i s s t i l l d o m i n a t e d by p .

(a,$)

2

(a,?)

5 ~ ( a )= ( a , $ ) ,

p(a) = 0 , so

5 is so

positive.

For e v e r y a < 0,

Finally, f o r every a > 0,

T 5 +. QED

Remark. -

The above Lemma f i r s t a p p e a r e d i n o u r p a p e r [ 2 3 ]

( P r o p o s i t i o n ( 2 . 4 ) , t h e Lemma).

(14.9) Theorem. -

Let E be a Riesz s u b s p a c e o f a Riesz s p a c e F ,

a n d E __ > F t h e c a n o n i c a l i n j e c t i o n .

Then E b +IE.

C o n s i d e r a n i n t e r v a l [ 0 , $ ] o f Fb .

Since it is positive,

For the opposite inclusion, given

it([O, $1) c [O,it$].

@ € [ O , i t $ ] , t h e n by ( 1 4 . 8 ) , $ h a s e x t e n s i o n

t h a t TE[O,$].

Then @

=

5 to

a l l of F such

it$, a n d we a r e t h r o u g h . QED

( 1 4 . 1 0 ) C o r o l l a r y 1.

Given a Riesz subspace E o f a Riesz

s p a c e F; (i)

t h e r e s t r i c t i o n map

I+%-->

$ l E maps Fb o n t o a R i e s z

i d e a l o f Eb; (ii)

i f Fh i s s e p a r a t i n g on F (on E i s e n o u g h ) , t h i s

Riesz i d e a l i s o(Eb,E) dense i n Eb;


85

( i i i ) i f E i s s e p a r a t i n g on F b , t h e r e s t r i c t i o n map i s a R i e s z i s o m o r p h i s m o f Fb o n t o a R i e s z i d e a l o f E b .

( i i ) I f Fb i s s e p a r a t i n g on F , t h e n i t i s s e p a r a t i n g on E , h e n c e i t ( F b ) i s Proof.

( i ) follows from ( 1 4 . 9 )

s e p a r a t i n g on E .

( i i i ) i Li s o n e - o n e , o n t o , a n d p o s i t i v e , s o

i t i s e n o u g h t o show t h a t ( i t ) - '

(14.6)).

and ( 1 4 . 7 ) .

i s p o s i t i v e ( c f . the proof of

S u p p o s e i t q, > 0;we h a v e t o show q, > 0.

i t $+, s o i t J,

=

t i p f o r some P E [ O , $ + ] .

0 < it $

T

T i s a R i e s z homomorphism;

2'

T

P roof.

Then f o r a l i n e a r

F, the following are equivalent:

1'

t

b

i s interval preserving.

Assume T i s a R i e s z homomorphism.

Then T - l ( O )

i s a R i e s z i d e a l , and we h a v e t h e f a c t o r i z a t i o n

This gives, i n turn, the

Ttorization of T t :

t

Eb

< [4 fi/T-l(0))b

F , t h e f o l l o w i n g a r e e q u i v a l e n t :

'1

T i s norm c o n t i n u o u s ;

2'

T i s o r d e r bounded.

I n p a r t i c u l a r , e v e r y p o s i t i v e l i n e a r mapping i s norm cont i n u o u s , hence e v e r y R i e s z homomorphism.

A l s o , f o r an MILb normed s p a c e E ( w h e t h e r norm c o m p l e t e o r n o t ) , E ' = E

.

As u s u a l , i t i s R i e s z homomorphisms t h a t we a r e i n t e r e s t e d

in.

L e t E , F be MI-normed s p a c e s .

By an MI-homomorphism

of E

i n t o F , we w i l l mean a R i e s z homomorphism T w i t h t h e p r o p e r t y : Tll(E) = l l ( F ) .

We w i l l g e n e r a l l y u s e t h e symbol lL f o r t h e

u n i t s o f h o t h E and F ; t h u s t h e above e q u a l i t y w i l l be w r i t t e n T R = ll.

Note t h a t ( i n t h e u n i f o r m norm) [[ T(I = 1.

The term Mll-isomorphism i s c l e a r . t h a t an Ma-isomorphism i s an i s o m e t r y .

I t is easily verified By an imbedding o f one

Ma-normed s p a c e E i n t o a n o t h e r one F , we w i l l mean an Mll-isomorphism o f E o n t o a R i e s z s u b s p a c e o f F c o n t a i n i n g It.

( 1 8 . 2 ) An MIL-homomorphic image o f an Mll-space E i s an M I - s p a c e .

109

MIL-spaces a n d L - s p a c e s Proof.

Let E

-> F b e a n Ma-homomorphism.

a R i e s z subspace of F c o n t a i n i n g space.

We h a v e t o show

n,

a n d i s t h u s a n MIL-normed Consider a

T(E) i s norm c o m p l e t e .

norm Cauchy s e r i e s CnTan i n T ( E ) .

Then T(E) i s

By d i s c a r d i n g some terms a t

5 1/2"

t h e b e g i n n i n g a n d g r o u p i n g t h e r e s t , we c a n assume IITanll (n

=

1,2,...)

.

T h i s can be w r i t t e n

2

- (1/2")IL(F)

Tan

5

(1/2")Il(F).

For e a c h n , s e t b n = [anV ( ( - l / Z n )

Then IIbnjl

5 1 / 2 " a n d Tbn

=

n(E) 1 IA ( ( 1 / 2 n ) n(E) 1 .

Tan ( n = 1 , 2 , . . . )

.

From t h e f i r s t

o f t h e s e , I n b n i s norm C a u c h y , s o ( E b e i n g norm c o m p l e t e ) t h e r e

e x i s t s bEE

s u c h t h a t Cnbn

=

b i n t h e norm.

c o n t i n u o u s , we h a v e CnTan = CnTbn

S i n c e T i s norm

Tb.

=

QED

(18.3)

Corollary.

If E

-> F i s a n MIL-homomorphism o f o n e

MIL-space i n t o a n o t h e r , t h e n T ( E ) i s a n M I - s u b s p a c e o f F.

919 L - s p a c e s

Dual t o M-norms a r e L - n o r m s . R i e s z s p a c e E i s a d d i t i v e o n E,,

I f a R i e s z norm

11 . [ I

on a

that is,

= IIaII + [ l b [ [ f o r a l l a , b E E + ,

\ l a + b[I

i t w i l l b e c a l l e d a n L-norm.

A s s o c i a t e d w i t h a n L-norm

the set K

=

{aEE+I[Iall = 1 1 .

[I

is

110

Chapter 3

K i s c o n v e x a n d norm c l o s e d , a n d e v e r y a E E + c a l l be w r i t t e n u n i q u e l y i n t h e form a

Ah, b E K .

=

So

e v e r y aEE c a n b e w r i t t e n , u n i q u e l y , a ~b = a + , Kc = a

[la;].

= =

Xb -

I t follows w i t h b,cEK,

KC,

.

N o t e t h a t , w h i l e K i s d e t e r m i n e d by t h e norm l a t t e r i s , i n t u r n , d e t e r m i n e d by K . t h e n , from t h e a b o v e ,

11

/ a l I l = ~ ~ a ++i l11 a-11

=

1 a1 x +

=

a+

+

a-

[I . !I,

the

In e f f e c t , given aEE, =

Xb

Xc, h e n c e

+

a![

=

K.

Given a R i e s z s p a c e E , a c o n v e x s u b s e t K o f E + s u c h t h a t every a

0 has a unique r e p r e s e n t a t i o n a

a b a s e f o r E,. ~

Then, a s a b o v e , e v e r y a h a s t h e u n i q u e

representation a

Xb -

=

KC,

b,cCK, Xb

=

is e a s i l y v e r i f i e d t h a t t h e f u n c t i o n IIaII on E .

Xb, h E K , i s c a l l e d

=

a: =

= a

KC

X

+

K

, and i t i s an L-norm

Summing u p ,

( 1 9 . 1 ) Given a R i e s z s p a c e E , t h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e b a s e s o f E + a n d t h e L-norms on E .

11 - ! I ,

the correspondence base i s K

=

{aEE+I/Iall

e a c h b a s e K , t h e c o r r e s p o n d i n g L-norm i s a+

=

Xb, a

= KC,

[I a [ [ =

For e a c h L-norm

1 1 , and f o r

=

X

+ K,

where

b,cEK.

Because o f t h e

a b o v e , an L-norm i s a l s o c a l l e d a b a s e -

norm. Henceforth,

g i v e n an L-normed s p a c e E , i t w i l l b e c o n -

v e n i e n t t o d e n o t e t h e c o r r e s p o n d i n g b a s e by K ( E ) t h e b a s e o f E,.

and t o c a l l i t

We e m p h a s i z e t h a t t h i s n o t a t i o n a n d t e r m i n o l o g y

111

MIL-spaces a n d L - s p a c e s

w i l l a l w a y s mean t h a t w e h a v e a g i v e n f i x e d L-norm.

I t is

c l e a r t h a t f o r e v e r y R i e s z s u b s p a c e F o f E , t h e norm i n d u c e d on F i s a l s o on I,-norm a n d K ( F ) = K(E)

n

F.

Given a s u b s e t A o f a v e c t o r s p a c e , w e w i l l d e n o t e b y conv A t h e convex e n v e l o p e o f A.

(19.2)

E

=

Let E b e an L-normed s p a c e .

.J 3 G ,

For every decomposition

*J,G b a n d s , K(E)

Proof.

=

c o n v [ K ( J ) IJ K ( G ) ] .

Consider a E K ( E ) and w r i t e a

=

aJ

+

aG.

If neither

_ I

component i s 0 , we c a n w r i t e

Then a 0.

that

H e n c e , f o r s i m p l i c i t y , w e c a n assume

We c a n a l s o a s s u m e t h a t

I[ 011

=

1.

Chapter 5

146

a n d we c a n t a k e 0

Assume $ $ 0 i n E ' , 0.

w e h a v e t o show t h a t i n f a ( $ , , @ )

0.

=

5

5 n

for a l l a;

Suppose n o t .

Then t h e r e

e x i s t s X > 0 such t h a t

(i)

2X f o r a l l a .

($,,@)

For e a c h a , s e t em

~ ~ $ 0 A.p p l y i n g -

XI)'

8O,

I t ( $ c w . - X n ) + . By ( 1 7 . 9 1 , Xe, 5 $ , + O , we h a v e l i m , ( e a , @ )

< A l l + lL(,ixa)+ =

< 2X,

An

+

=

0.

Now $a

e a , h e n c e ($,,@)

=

hence

$,All

+

5 (An,@)+

= 0 , we c a n f i n d c1 s u c h t h a t

S i n c e lim,(e,,@)

(e,,@). ($,,@)

=

c o n t r a d i c t i n g ( i ) above. QED

Remark. -

On t h e o r i g i n a l s p a c e E ,

t h e g i v e n norm t o p o l o g y ,

so l o l ( E , E ' )

101

(E,E')

=

T(E,E')

o f f e r s n o t h i n g o f new

interest.

526. Dual bands

From ( 1 0 . 2 0 ) a n d E x e r c i s e 9 o f C h a p t e r 2 , we h a v e

( 2 6 . 1 ) L e t E b e an L-space (i)

If E

=

J1 @ J

E l

=

(J,P

(ii) If E'

=

2 (J1,J2 b a n d s ) , t h e n

3

(J1lL.

I1 3 I 2 (I 1 , I 2 bands), then

E = ( I 2 )L 3 ( I l l L .

=

An L - s p a c e a n d i t s Dual

147

O t h e r w i s e s t a t e d , e v e r y band J o f E d e t e r m i n e s u n i q u e d c c o m p o s i t i o n s o f b o t h E and

E

E l ;

.J 3 ,J

=

d

, E'

=

(JL)d 3 .JL;

and e v e r y band I o f E ' d e t e r m i n e s u n i q u e d e c o m p o s i t i o n s o f b o t h E ' and E : E '

=

Id, E

I

=

(I )d 3 I L

L

.

We w i l l c a l l (.JL) d

t h e band d u a l t o J o r t h e d u a l band o f J ; and w e w i l l c a l l (I )

d

t h e band d u a l t o I o r t h e d u a l band o f I .

1

-

T h e s e terms

a r e w e l l d e f i n e d o n l y i n t h e c o n t e s t o f a f i x e d L-space E and i t s dual E ' .

Each .J i s t h e d u a l b a n d o f i t s d u a l b a n d , a n d s i m i l a r l y f o r each I .

Thus w e c a n t a l k a b o u t a p a i r o f d u a l b a n d s ( I , , J ) ,

meaning t h a t I

=

( J L ) da n d ,J

=

(T. ) d . ( N o t e t h a t (,JL) d

=

( I , . J ) i s a p a i r o f d u a l b a n d s , t h e n c l e a r l y .J i s an b L - s p a c e and I i s i t s d u a l . Thus I = , J t = J c = J a n d J = I c If

(under t h e b i l i n e a r form following (cf.

(26.2)

(9.7),

(

a

, . ) on E ' x E ) .

We a l s o r e c o r d t h e

(12.7)).

I f E i s an L - s p a c e ,

c l o s e d and e v e r y band o f E '

t h e n e v e r y band o f E i s weakly

i s vaguely closed.

5 2 7 . B a s i c b a n d s i n t h e d u a l o f an L - s p a c e

Let E b e an L - s p a c e .

"small",

The p r i n c i p a l b a n d s o f E , b e i n g

can b e t h o u g h t of as t h e b u i l d i n g b l o c k s of E .

h a v e c h a r a c t e r i z e d them i n ( 1 9 . 1 0 ) .

In E ' ,

We

however, e v e r y

148

Chapter 5

band i s p r i n c i p a l - i s i n f a c t a p r i n c i p a l R i e s z i d e a l . b a n d s , t h e n , c a n be t h o u g h t o f a s " s m a l l " ? t h e bands d u a l t o p r i n c i p a l bands o f E . band a b a s i c band o f

What

The a n s w e r i s :

We w i l l c a l l s u c h a

E l .

We a d o p t a s p e c i a l n o t a t i o n f o r a b a s i c band o f

Given

E l .

a E E , l e t ,J h e t h e band o f E g e n e r a t e d by a , a n d I t h e band of E ' dual t o ,J.

Under t h e c o n v e n t i o n a d o p t e d a t t h e e n d o f 58,

f o r a n y bEE o r A c E , bJ i s d e n o t e d by h a , a n d an a b u s e o f l a n g u a g e , f o r a n y

or A C

$ € E l

n o t e d by $a a l s o , and A I b y A a .

E l ,

A 0 ; we h a v e t o show t h e r e e x i s t s a s u c h t h a t

consider E

is not order continuous.

a, hence

Then [ - c a , c a ] c E B ,

11 a,\] 5

> a. E f o r a l l fi QED

Let E b e a normed s p a c e ( a l t h o u g h t h e f o l l o w i n g c a n b e

154

Chapter 5

done more g e n e r a l l y ) .

F o r e a c h s u b s e t A o f E , t h e p o l a r A'

E l d e t e r m i n e s a g a u g e f u n c t i o n , w h i c h we w i l l d e n o t e by

I n g e n e r a l , it i s p o s s i b l e t o have

o f El.

\)

@[IA

=

m

in

\I - [ I A .

f o r some e l e m e n t s

As i s e a s i l y v e r i f i e d , t h e f o l l o w i n g a r e e q u i v a l e n t :

'1

11 $[IA

2'

A'

3'

A i s norm b o u n d e d .

2c

there is a $

i n A such t h a t :

;

for a l l

11;

I ( u m , a n ) 1 5 & f o r a l l m,n s u c h t h a t m > n .

To p r o d u c e t h e s e s e q u e n c e s , w e p r o c e e d i n d u c t i v e l y . Choose a1 a r b i t r a r i l y ; t h e n c h o o s e

al

1

5

such t h a t

"1

a n d II$lllA > $"+O,

26;

t h e n c h o o s e alEA s u c h ~ ( + l , a l ) >~

26.

Since

$ a V + l + ~ ) l& i t h a); h e n c e , s i n c e a l i s o r d e r c o n t i n u o u s ,

w e c a n c h o o s e a 2 > a1 t o s a t i s f y

Chapter 5

156

(the first of these, because also /all is order continuous).

l$,l

Now choose $ 2 such that

5 $a and 2

choose a2iA such that /($2,a2)l fashion, and setting wn

= $a

v $ ~(n

> ZE,

then

Proceeding in this

> ZE.

n+ 1

i1$2[[A

=

1 , 2 , * * * ) , we obtain

the desired sequences. Now IIwn - Wn+lllA 1 2 l(Wn I ( ~ ~ + ~ , a ~ >) l2~ -

E

= E.

-

wn+19an)l

3

\(wn,an)l

This contradicts 4'

-

and completes

the proof. QED

Remark. -

Contained in the above proof is the result that

a subset A of an L-space E is equi-order-continuous on E ' if and only if every countable subset of A is equi-order-continuous on E ' .

(28.3) Theorem. _____ In an L-space E, every weakly convergent

sequence is equi-order-continuous.

Proof. -

It is enough t o consider a sequence {an) which

converges weakly t o 0.

Denote the set {a,,)

+n $ 0 ;

=

(*)

we show limnl[$njlA

0 (28.2).

by A, and assume

Suppose not.

We can assume (for simplicity) that there exists

such that -

+n+l,an)l

>

E

for all n.

E >

0

An L - s p a c e and i t s Dual

157

In e f f e c t , by s u p p o s i t i o n , t h e r e e x i s t s

E

> 0 such t h a t

€ o r e a c h n o t h e r e i s a n n > no and an m s u c h t h a t l ( $ n , a m ) / > S e t no

2 ~ . We p r o c e e d b y i n d u c t i o n .

)I

and m1 s u c h t h a t I ( $ , , , a 1

1 a n d c h o o s e n1 > 1

=

The e l e m e n t s a l ; * * , a

> ZE.

ml

ml

s o limn($n,ai)

a r e o r d e r c o n t i n u o u s on E ' ,

=

0 (i

1,. . - , m , ) .

=

I t f o l l o w s t h e r e e x i s t n 2 > n l a n d m 2 > ml s u c h t h a t

I(

)I

$ n 2 9 aml

E

and I(+,

) ) > 2 ~ . Continuing i n t h i s

,a

2

m2

f a s h i o n , w e o b t a i n subsequences { $ nk ,a

k l($n

mk

>

I(

26,

$n

,amk)/ >

-

k

)I

k+l E

,a

mk

)I

< -

E

for a l l k.

},{a } such t h a t mk

f o r a l l k.

Then

Since t h e subsequences re-

'nk+ 1

t a i n t h e d e f i n i n g p r o p e r t i e s o f t h e o r i g i n a l s e q u e n c e s we have (*)

.

A s i s e a s i l y v e r i f i e d , t h e r e e x i s t s a (unique) p o s i t i v e l i n e a r mapping (n

=

1,2,-..).

1

Rm->

E'

( e n i s t h e e l e m e n t o f R"

p o s i t i o n and 0 e l s e w h e r e . ) T

such t h a t Ten

=

$n - $n+l

having 1 i n the n t h

T h i s g i v e s u s (a")'

0 and se-

and l($m,an)l

> E

Thus not only {an}, but every subsequence of (a,}

fails to be equi-order-continuous. But, by Smulian's theorem, some subsequence of {an} is weakly convergent and therefore, by (28.7), equi-order-continuous. We thus have a contradiction. QE D

Combining this theorem with (28.6), we have the

(29.3) Corollary 1. If a subset of an L-space is relatively weakly compact, then s o is its convex, solid hull.

An L - s p a c e and i t s Dual

161

I f a s u b s e t A o f an L - s p a c e E i s r e l a -

(29.4) Corollary 2.

t i v e l y weakly compact, t h e n e v e r y s e t B o f m u t u a l l y d i s j o i n t elements of A i s countable, B

=

{ a n } , and limnllanII

By ( 2 9 . 3 ) , we can assume A i s s o l i d .

P roof.

0.

=

To e s t a b l i s h

t h e p r o o f , i t i s enough t o show t h a t f o r e v e r y 1 > 0 , t h e r e i s o n l y a f i n i t e number o f e l e m e n t s o f B w i t h norm > 1.

Suppose

t h e r e e x i s t s A > 0 and an i n f i n i t e s e t {bn} c B s u c h t h a t IIb,ll

> 1 f o r a l l n. -

In E ' ,

second paragraph i n 527).

l l , , (n = 1 , 2 , . . - ) ( c f . t h e n The e n ' s a r e a l s o m u t u a l l y d i s j o i n t ,

s e t en

hence ( a s i s e a s i l y v e r i f i e d ) e n

= 0.

0.

I t f o l l o w s from ( 2 9 . 2 )

In p a r t i c u l a r , s i n c e A i s s o l i d ,

t h a t limn\lenllA = 0 . limn(en,/bnI)

+

=

But ( e n , l b n l ) = ( n , I b n l )

=

IIbnll

2

1 , s o we

have a c o n t r a d i c t i o n .

QED

(29.5)


E v e r y w e a k l y compact s u b s e t A o f an L-

s p a c e E i s c o n t a i n e d i n a p r i n c i p a l band o f E .

R e p l a c e A by i t s s o l i d h u l l A1.

Proof.

L e t { a n } be a

maximal s e t o f m u t u a l l y d i s j o i n t e l e m e n t s o f A1.

Then t h e

band J g e n e r a t e d by t h e s e t { a n } i s a p r i n c i p a l band ( 1 9 . 1 0 ) . That A 1 c

J f o l l o w s from t h e e a s i l y v e r i f i e d f a c t t h a t i f a

s o l i d s e t i s not contained i n J , then it contains a non-zero element of J

d

. QED

162

Chapter 5

A t o p o l o g y 7 on a R i e s z s p a c e i s c a l l e d Lebesgue i f o r d e r convergence i m p l i e s convergence i n

g.

An e q u i v a l e n t d e f i n i t i o n

f o r a l o c a l l y convex t o p o l o g y i s t h a t t h e f a m i l y o f d e f i n i n g seminorms a r e e a c h o r d e r c o n t i n u o u s . From ( 2 9 . 2 )

(and ( 2 9 . 3 ) ) , we h a v e

(29.6) Corollary 4. 7(E',E)

I f E i s an L - s p a c e , t h e Mackey t o p o l o g y

on E' i s a Lebesgue R i e s z t o p o l o g y .

Also

( 2 9 . 7 ) C o r o l l a r y 5.


?-(El

, E ) i s d e f i n e d by

t h e s e t o f a l l o r d e r c o n t i n u o u s R i e s z seminorms on

E l .

N o t e t h a t f o r an L - s p a c e E , s i n c e e v e r y i n t e r v a l o f E i s w e a k l y compact ( 2 9 . 1 ) - and c o n v e x - we h a v e :

(29.8) If E i s an L-space, then ( u )( E l , E ) c r ( E '

,E).

T h i s c a n a l s o b e s e e n from ( 2 9 . 1 ) seminorms

{I] - I I a I

aEE}

defining

IuI

and t h e f a c t t h a t t h e

(El ,E)

a r e a l l o r d e r con-

An L - s p a c e a n d i t s Dual

I n g e n e r a l , a n L - s p a c e E c o n t a i n s w e a k l y compact s e t s

tinuous.

n o t contained i n i n t e r v a l s , so T ( E ' , E )

lo/(E',E). of E '

16 3

i s s t r i c t l y f i n e r than

However, t h e two a l w a y s c o i n c i d e on t h e u n i t b a l l

((29.10) below).

We n e e d t h e f o l l o w i n g t h e o r e m .

I t was

p r o v e d o r i g i n a l l y b y Amemiya a n d Mori f o r a D e d e k i n d c o m p l e t e R i e s z s p a c e , a n d l a t e r , b y a n e l e g a n t p r o o f , shown t o h o l d f o r

a g e n e r a l Riesz s p a c e by A l i p r a n t i s and Burkinshaw ( [ Z ] , Theorem 1 2 . 9 ) .

We r e f e r t h e r e a d e r t o t h e i r p r o o f .

( 2 9 . 9 ) Theorem.

A l l t h e l i a u s d o r f f L e b e s g u e t o p o l o g i e s on a

R i e s z s p a c e E i n d u c e t h e same t o p o l o g y o n e v e r y o r d e r b o u n d e d subset of E.

Thus :

(29.10) C o r o l l a r y . T(E',E)


c o i n c i d e on t h e u n i t b a l l

then

101

[-ll,n] o f E ' .

(E',E)

and

PART I11

C (X) , C ' (X)

,

C" (X) : THE

164

FRAMEWORK

CHAPTER 6

(C(X) ,X) -DUALITY

X , Y w i l l d e n o t e compact H a u s d o r f f s p a c e s .

C(X) i s t h e s e t

o f r e a l c o n t i n u o u s f u n c t i o n s on X , a n d n ( X ) w i l l d e n o t e t h e f u n c t i o n o f c o n s t a n t v a l u e 1 on X .

Under t h e u s u a l p o i n t w i s e

d e f i n i t i o n s o f a d d i t i o n , s c a l a r m u l t i p l i c a t i o n , a n d o r d e r , C(X) i s a R i e s z s p a c e w i t h n(X) f o r a n o r d e r u n i t .

The MI-norm

d e t e r m i n e d by n(X) i s p r e c i s e l y t h e s t a n d a r d supremum n o r m , and u n d e r t h i s norm, C(X) i s c o m p l e t e . Thus C(X) i s a n M I - s p a c e .

I n more d e t a i l , f o r f , g E C ( X ) ,

f v g a n d fAg a r e g i v e n b y ( f V g ) ( x ) = m a x ( f ( x ) , g ( x ) ) a n d ( f A g ) ( x ) = m i n ( f ( x ) , g ( x ) ) f o r a l l xEX. f+(x)

=

( f ( x ) ) + , and f - ( x )

in particular, =

Ifl(x)

=

/f(x)I,

( f ( x ) ) - f o r a l l x€X.

Although X is n o t a v e c t o r s p a c e , i t i s u s e f u l (and s u g g e s t i v e ) t o t h i n k o f C(X) a s " d u a l " t o X.

This has been

done by v a r i o u s w r i t e r s , e s p e c i a l l y i n t h e c o n t e x t o f C a t e g o r y Theory ( c f .

[49],

523.2 or [4T).

We w i l l f o r m u l a t e t h e r e l a -

t i o n s b e t w e e n C(X) a n d X f r o m t h i s v i e w p o i n t .

Since X has a

n a t u r a l i m b e d d i n g i n t o t h e d u a l C'(X) o f C(X), C'(X) w i l l t h u s play the r o l e of "bidual"

t o X.

I n p a r t i c u l a r , we w i l l f e e l

free t o denote f ( x ) by ( f , x ) . X w i l l b e c a l l e d t h e K a k u t a n i - S t o n e s p a c e o f C(X).

165

Chapter 6

166

5 3 0 . The t o p o l o g y o f s i m p l e C o n v e r g e n c e o n C ( X )

X i s o f c o u r s e s e p a r a t i n g o n C ( X ) by v e r y d e f i n i t i o n : f # g means t h e r e e x i s t s xEX

such t h a t f ( x )

# g(x).

Dually,

C(X) i s s e p a r a t i n g on X , a n d i n d e e d , i n a v e r y s t r o n g s e n s e : (Urysohn) f o r e v e r y p a i r o f d i s j o i n t , c l o s e d , non-empty s u h s e t s Z1,Z2 o f X , ( i i ) f(x)

=

Remark. -___

t h e r e e x i s t s fEC(X) s a t i s f y i n g ( i ) 0 < f (n(X),

1 f o r a l l xEZ1,

(iii) f(x)

=

0 f o r a l l xEZZ.

I l e n c e f o r t h , we w i l l c a l l an f E C ( X ) w i t h t h e a b o v e

properties a Urysohn f u n c t i o n f o r t h e ( o r d e r e d ) p a i r (Z,,Z,). A s o n e would e x p e c t , we d e f i n e o ( C ( X ) , X )

a s t h e topology

on C ( x ) o f p o i n t w i s e c o n v e r g e n c e on X : a n e t { f } i n CL

C(X) c o n -

v e r g e s t o fEC(X) i n a ( C [ X ) , X ) i f f ( x ) = l i m f ( x ) f o r a l l xEX. CY.N

We w i l l c a l l i t by i t s common name: t h e t o p o l o g y o f s i m p l e convergence.

And we c a n d e f i n e o ( X , C ( X ) ) a s t h e t o p o l o g y o n X

o f p o i n t w i s e c o n v e r g e n c e o n C(X): a n e t { x } i n X c o n v e r g c s t o c1

xEX i n ~ ( y C, ( X ) )

if f ( x )

=

limolf(x,)

for a l l fEC(X).

Since

x

completely regular,o(X,C(X)) i s simply t h e o r i g i n a l topology on X .

We e x a m i n e o(C(X) , X ) . Norm c o n v e r g e n c e ( i n C ( X ) )

and s i m p l e c onvergence.

i m p l i e s b o t h o r d e r convergence

T h e r e i s no o t h e r i m p l i c a t i o n b e t w e e n

the t h r e e convergences: Let X = a m , t h e one-point (Alexandroff) c o m p a c t i f i c a t i o n o f N , and f o r e a c h n , l e t en b e t h e e l e m e n t o f C(X)

h a v i n g v a l u e 1 on n E N a n d v a l u e 0 e l s e w h e r e .

t h e s e q u e n c e {ne,}

Then

c o n v e r g e s t o 0 s i m p l y b u t n e i t h e r normwise n n o r i n t h e o r d e r , a n d t h e s e q u e n c e {c e . ) ( n = 1 , 2 , . * . ) o r d e r 1 1 c o n v e r g e s t o l(X) b u t d o e s n o t c o n v e r g e e i t h e r s i m p l y o r no r m w i s e .

is

16 7

(C(X) ,X)-Duality

For a monotonic n e t , however, w e have one a d d i t i o n a l implication:

( D i n i t h e o r e m ) F o r a m o n o t o n i c n e t { f } i n C(X) cr f E C (X), the following a r e equivalent : (30.1)

1'

{fci. 1 c o n v e r g e s t o f n o r m w i s e ;

2'

{fa} c o n v e r g e s t o f s i m p l y ;

We n e e d o n l y show 2'

Proof. -__

implies lo.

and

For c o n c r e t n e s s ,

assume { f } i s a d e s c e n d i n g n e t w i t h i n f f ( x ) = 0 f o r a l l xEX. a a Consider E > 0 . F o r e a c h xEX, c h o o s e a ( x ) s u c h t h a t (x)
0, the set {tET

1

If(t)

1

> E} i s

The f o l l o w i n g i s e a s i l y v e r i f i e d .

( 3 1 . 7 ) There i s a o n e - o n e c o r r e s p o n den ce between t h e f a m i l y o f

open s u b s e t s W of X and t h a t o f t h e norm c l o s e d R i e s z i d e a l s

H o f C(X): H

fEH}

W i f and o n l y i f W

=

{ x G X \ f ( x ) # 0 f o r some

i f and o n l y i f H i s t h e s e t o f c o n t i n u o u s f u n c t i o n s on

W which v a n i s h a t i n f i n i t y .

H e n c e f o r t h Z w i l l a l w a y s d e n o t e a c l o s e d s e t of

Notation.

X , a n d W an open s e t .

F o r a s u b s e t A o f C ( X ) , W(A) w i l l b e

t h e s e t X\Z(A). The s e t o f c o n t i n u o u s 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 on a l o c a l l y compact s p a c e T w i l l b e d e n o t e d by C w ( T ) .

Note

t h a t i t i s an M-space u n d e r t h e supremum norm, a n d i s an MEs p a c e i f and o n l y i f T i s c o m p a c t . U s i n g t h e above n o t a t i o n , we c a n s t a t e p a r t o f ( 3 1 . 7 ) a s

follows: I f X

= W IJ Z ,

W a n d Z c o m p l e m e n t a r y , t h e n ZL = Cw(W) .

A n o t h e r o b s e r v a t i o n on C w ( W ) . elements of C(X) of W.

Let Ho c o n s i s t o f t h e

e a c h h a v i n g f o r i t s s u p p o r t a comp.act s u b s e t

Then Ho i s a R i e s z i d e a l , i s c o n t a i n e d i n C w ( W ) ,

norm d e n s e i n t h e l a t t e r .

We n e e d o n l y v e r i f y t h e l a s t

and i s

(C(X) ,X) - D u a l i t y

statement.

173

Tn e f f e c t , s i n c e W i s c o m p l e t e l y r e g u l a r , Ho i s s o , by ( 3 0 . 4 )

s i m p l y d e n s e i n C'(W),

( o r t h e Dini Theorem), i s

norm d e n s e i n i t .

We c h a r a c t e r i z e v a r i o u s t y p e s o f R i e s z i d e a l s o f C(X) by p r o p e r t i e s of X . by i n t Q

.

We w i l l d e n o t e t h e i n t e r i o r o f a s e t Q i n X

Recall that a r e g u l a r c l o s e d s e t i s one which i s

t h e c l o s u r e o f an o p e n s e t , a n d a r e g u l a r o p e n s e t i s o n e which i s t h e i n t e r i o r o f a c l o s e d s e t .

We h a v e s e e n ( 3 1 . 2 ) t h a t f o r Q c X , QL i s a norm c l o s e d Riesz i d e a l .

F o r a n o p e n s e t W i n X , WL

(31.8)

___ Proof. WL

F o r an o p e n s e t , w e c a n s a y m o r e :

Let 2

X\W.

=

i s a b a n d o f C(X)

T h e n , as i s e a s i l y v e r i f i e d ,

(ZL)d, hence i s a band

=

QE D

For a Riesz i d e a l H o f C(X),

(31.9) C o r o l l a r y 1.

o r d e r c l o s u r e tI

P roof. simplicity, and H can

=

.'2

=

( i n t Z(H))'.

We c a n a s s u m e H i s norm c l o s e d . denote Z(H)

Also f o r

s i m p l y by Z a n d W(H) b y W.

So W

=

X\Z

S i n c e C(X) i s A r c h i m e d e a n , w h a t we w a n t t o p r o v e

b e s t a t e d : Hdd

=

( i n t Z)'.

A s we remarked i n ( 3 1 . 8 ) ,

2

Chapter 6

174

Hd

=

W"

(x\W)*

But t h e n Hd =

,

(w)" ,

s o , by t h e same r e m a r k , H d d

=

j i n t z)'.

=

QED

For a R i e s z i d e a l f1 o f C(X), t h e

(31.10) Corollary 2.

following are equivalent: lo

t h e o r d e r c l o s u r e o f H i s C(X);

'2

Z(H) h a s empty i n t e r i o r ( s o , b e i n g c l o s e d , i s nowhere

rare,

dense -

i n t h e Bourbaki terminology);

WIN) i s d e n s e i n X .

3'

The f o l l o w i n g a r e a l s o c o r o l l a r i e s , b u t we l i s t them a s theorems.

( 3 1 . 1 1 ) ____ Theorem.

F o r a norm c l o s e d R i e s z i d e a l H o f C ( X ) ,

the

following are equivalent:

1'

H i s a band;

'2

Z(H) i s a r e g u l a r c l o s e d s e t ;

'3

W ( H ) i s a r e g u l a r open s e t .

P roof. Z(H)

=

i f Z(H) H

=

I f H i s a band, t h e n , by (31.9) and (31.3),

m, so

Z(H) i s a r e g u l a r c l o s e d s e t .

Conversely,

i s a r e g u l a r c l o s e d s e t , t h e n , by ( 3 1 . 4 ) and ( 3 1 . 9 ) ,

(Z(H))'

=

order closure H. QED

(C(X) ,X) -Duality

175

I f a s e t i n X i s b o t h open and c l o s e d , w e w i l l c a l l i t clopen.

F o r a norm c l o s e d R i e s z i d e a l lI o f C ( X ) ,

( 3 1 . 1 2 ) Theorem.

the


H i s a p r o j e c t i o n band;

2'

Z(II) is clopen;

3'

W(t1)

-__ Proof..

i s clopen.

S u p p o s e H 3 Hd

=

C(X),

but Z(H)

some xEZ(H) i s i n t h e c l o s u r e o f W ( I I ) . t1 v a n i s h e s on

Conversely, i f X =

T t

v a n i s h e s on x , c o n t r a d i c t i n g ( l l ( X ) ) ( x ) Z1lJ

=

So

Then e v e r y e l e m e n t o f

x a n d e v e r y e l e m e n t o f Hd v a n i s h e s on x .

follows a l l of C(X)

C(X)

i s n o t open.

=

Z2, w i t h Z1,Z2 d i s j o i n t , t h e n c l e a r l y ,

( Z , P 3 (Z$. QED

Finally

,

( 3 1 . 1 3 ) Theorem.

For a R i e s z i d e a l H o f C ( X ) ,

are equivalent: 1' H i s a maximal R i e s z i d e a l ;

2'

Z(f1)

1.

consists of a single point.

the following

176

Chapter 6

-__ Proof.

Assume H i s m a x i m a l , a n d c h o o s e xEZ(f1).

a p r o p e r Riesz i d c a l o f C ( X )

w i t h 11.

{x}'

is

and c o n t a i n s [ I , hencc c o i n c i d e s

I t f o l l o w s from ( 3 1 . 3 ) t h a t Z ( l I )

=

{x}.

T h a t '2

i m p l i e s lo i s c l e a r . Qlill

Consider a c l o s e d s e t Z o f X. Cw(W), W

= X\Z.:

As we h a v e s e e n , ZL =

Tn a d d i t i o n , C(X)/ZL = C(Z) ( c f .

(33.6))

5 3 2 . The d u a l i t y b e t w e e n t h e M l l - s u b s p a c e s o f C ( X )

and t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n o f X

P a r a l l e l i n g 531, we c o l l e c t h e r e t h e o r e m s r e l a t i n g t h c M I - s u b s p a c e s o f C(X) t o t h e t o p o l o g y o f X. immediately a r i s e s .

A difficulty

E v e r y MIL-subspace F c o n t a i n s l l ( X ) ,

t h e a n n i h i l a t o r o f F i s always cmpty.

hence

A substitute for this

a n n i h i l a t o r i s t h e dccomposition of X i n t o t h e sets of cons t a n c y o f F.

(The a n n i h i l a t o r __ does e x i s t

-

in C'(X)!

And

i s closely r e l a t e d t o these s e t s of constancy.) By a d e c o m p o s i t i o n o f X , w e w i l l mean a f a m i l y x

=

{Za}

o f m u t u a l l y d i s j o i n t , c l o s e d , n o n - e m p t y s u b s e t s o f X whose

union i s X.

The d e c o m p o s i t i o n s o c c u r r i n g i n o u r s u b j e c t a r e

of a special kind.

A decompositionx

=

{Za} o f X i s c a l l e d

uppersemicontinuous i f f o r e v e r y c l o s e d s e t Z i n X , t h e union of a l l the Za's intersecting 2 i s also closed.

I t i s e a s i l y v e r i f i e d t h a t i f X s > Y i s a continuous mapping ( r e m e m b e r , X , Y a r e a l w a y s compact H a u s d o r f f ) , t h e n t h e

(C(X) , X ) - D u a l i t y

decomposition

2

=

{s

-1

(y) / y € s ( X ) }

177

of X i s uppersemicontinuous.

Conversely, every uppersemicontinuous decomposition

X i s d e t e r m i n e d b y a c o n t i n u o u s mapping o f X :

2

=

{Za} o f

In e f f e c t , l e t

X A> L b e t h e mapping w h i c h a s s i g n s t o e a c h xEX t h e Z t a i n i n g i t , and d e n o t e h y J f i n e d b y t h i s mapping q .

(Z,J)

cona t h e i n d u c t i v e t o p o l o g y on 5 d e -

The r e s u l t i n g t o p o l o g i c a l s p a c e

is called the q u o t i e n t s p a c e o f X d e t e r m i n e d by

c a l l e d t h e q u o t i e n t map, a n d ? very d e f i n i t i o n o f ? ,

the quotient topology.

is

By t h e

q i s continuous.

( 3 2 . 1 ) The q u o t i e n t s p a c e

Proof.

t ,q

(x,y) i s

compact H a u s d o r f f .

N o t e f i r s t t h a t an e q u i v a l e n t f o r m u l a t i o n o f t h e

u p p e r s e m i c o n t i n u i t y o f 5 i s t h a t f o r e v e r y open s e t W o f X , union o f a l l t h e 2 ' s c o n t a i n e d i n W i s a l s o open. c1

a s u b s e t Q o f X whole i f it i s a union o f Z ' s . whole i f Q

=

q

-1( q ( Q ) ) .

c1

the

Let u s c a l l

Q is then

T h u s , by t h e d e f i n i t i o n o f t h e

i n d u c t i v e t o p o l o g y , i f a n o p e n s e t W i n X i s w h o l e , t h e n q(W)

i s open i n

(2,y).

I t f o l l o w s t h a t t o show

(2,y)

is Hausdorff,

i t s u f f i c e s t o show t h a t d i s t i n c t Z ' s h a v e d i s t i n c t w h o l e (Y

neighborhoods i n X.

So c o n s i d e r Z

j o i n t open neighborhoods Wl,W2

"1

#

Za2,

and c h o o s e d i s -

o f Zal,Za r e s p e c t i v e 1y . 2

V1 b e t h e u n i o n o f a l l t h e Z ' s c o n t a i n e d i n

a

union of a l l those contained i n W2.

Since

2

Le t

W1, a n d V2 t h e i s uppersemicon-

t i n u o u s , V1 a n d V2 a r e o p e n , s o w e a r e t h r o u g h .

That

(t,7)

i s compact f o l l o w s from t h e c o n t i n u i t y o f q . QED

Chapter 6

178

mapping X

Returning to a continuous

__ >

resulting uppersemicontinuous decomposition

of X, form the quotient space ( X , y j . ical mapping ( 2 , J j

-

Y and the

2 = I S -1 ( y ) 1 y E s (X)}

Then s induces the canon-

2: Y such that the following diagram

commutes.

Then

(32.2)

2 is a homomorphism of

(X,r)onto Y.

The verification is simple.

We can now proceed with o u r examination of the MIl-subspaces of C(X).

Given a subset A of C ( X ) ,

a set Z in X will

be called set of constancy of A if (i) e v e r y f E A i s constant on 2 , and (ii) 2 is maximal with respect to this property.

If

A consists of a single element f, then t h e s e t s of constancy

are the sets { f - l ( X ) I X E I R l .

(32.3)

Given a subset A of C ( X ) ,

the set of sets of constancy

of A is an uppersemicontinuous decomposition of X.

(C(X) ,X)-Duality

Proof. -__

179

The family of continuous mappings { f l f E A 1 of X

into R defines a single continuous mapping X -> S IR

A

so

has the product topology). {

Then s(X)

IRA (as always,

is compact Hausdorff,

s - l(y) I y E s ( X ) 1 s an uppersemicontinuous decomposition of X.

But the sets s-l[y) are precisely the sets of constancy of A. QED

For any decomposition continuous), let F ( 2 )

=

x

of X (not necessarily uppersemi-

{fCC(X)lf

is constant on every

ZCX} .

Then, a s can b e verified b y straightforward computation, F(X) is an MIL-space.

(32.4) Given a subset A of C ( X ) ,

sets of constancy.

Then F(x)

let

b e the collection of its

is precisely the Ma-subspace I:

of C ( X ) generated by A.

____ Proof.

A s we have noted above, F(2)

and therefore contains F.

is an MIL-subspace

Conversely, consider fEF(x);

show f is in the norm closure of F, hence lies in F. [30.4),

we

By

it is enough t o show that f is in the simple closure

of F.

Lemma.

For every Z l , . . + , Z n € X ,a l l distinct, F contains

_ _ I

a Urysohn function for (Zl, lJnZ.) ( c f . the beginning of 5 3 0 ) . 2 1

Chapter 6

180

T h e r e e x i s t s R E F s u c h t h a t g(Z,) g(Z,)

tains A).

#

g ( Z z ) ( s i n c e I: c o n -

and g ( Z 2 ) a r e e a c h a s i n g l e r e a l n u m b e r , s o

by a d d i n g a n a p p r o p r i a t e m u l t i p l e o f I ( X ) t o g , i f n e c e s s a r y ,

w e c a n assume g ( Z 2 )

=

0.

Then,, h y m u l t i p l y i n g g b y an a p p r o -

p r i a t e s c a l a r , w e c a n assume g ( Z 1)

=

1.

The e l e m e n t (gVo)AI(X)

o f F i s now a Urysohn f u n c t i o n f o r ( Z l , Z 2 ) ; d e n o t e i t b y E ~ . S i m i l a r l y , F c o n t a i n s a Urysohn f u n c t i o n g i f o r e a c h p a i r (ZIJZi)

(i

=

3,...,n).

A;gi

i s t h e n a Urysohn f u n c t i o n f o r

( Z l , lJ;Zi). I t f o l l o w s f r o m t h e Lemma t h a t f o r e v e r y f i n i t e s u b s e t { x l , * - - , x n }o f X , t h e r e e x i s t s g E F s u c h t h a t g ( x i ) i

=

l,...,n.

=

f(xi)

€or

Thus f i s i n t h e s i m p l e c l o s u r e o f F .

OED

i t s sets of constancy a r e s i n g l e

I f A i s s e p a r a t i n g on X , points.

(32.5)

(32.4) then reduces t o

( W e i e r s t r a s s - S t o n e Theorem - l a t t i c e v e r s i o n ) .

A c C(X) i s s e p a r a t i n g on X ,

If

t h e n t h e M I - s u b s p a c e o f C(X)

g e n e r a t e d b y A i s C(X) i t s e l f .

( 3 2 . 4 ) s t a t e s t h a t i f we s t a r t w i t h an ME-subspace F o f C(X), t a k e t h e s e t 2 o f i t s s e t s o f c o n s t a n c y , t h e n t a k e F ( , ) , we a r r i v e b a c k a t F .

Dually,

(32.6) Every uppersemicontinuous decomposition

2

of X i s the


181

s e t o f s e t s o f c o n s t a n c y o f t h e MIL-subspace F ( X ) w h i c h i t

determines.

Proof.

We n e e d o n l y show t h e m a x i m a l i t y o f e a c h Z E

Specifically, consider

Zo€X

and x e Z o ; w e h a v e t o show t h e r e

e x i s t s fEF(z) such t h a t f ( x ) # f(Zo). space

(x,J)

determined b y x .

Denote b y Y t h e q u o t i e n t

Y i s compact H a u s d o r f f ( 3 2 . 1 ) ,

h e n c e t h e r e e x i s t s hEC(Y) s u c h t h a t h ( q x ) # h ( q ( Z o ) ) . f

=

hoq.

5.

Set

Then f i s a c o n t i n u o u s f u n c t i o n on X w h i c h i s c o n -

s t a n t on e v e r y Z € z , h e n c e i s i n

F(2)

- and s a t i s f i e s

f(x) # f(Zo).

QED

Summing u p ,

( 3 2 . 7 ) T h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e MIL-subs p a c e s F o f C(X) a n d t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n s of X.

F ;!

i f and o n l y i f

2

2

i s t h e s e t of sets of con-

s t a n c y o f F i f a n d o n l y i f F c o n s i s t s o f t h e e l e m e n t s o f C(X) c o n s t a n t on e v e r y Z E

2.

033. The MIL-homomorphisms o f C ( X )

For t h e following d i s c u s s i o n , it i s convenient t o use t h e notation

( f , x ) i n place of f(x)

(cf. the beginning of the

Chapter 6

182

Chapter).

Each xEX

d e f i n e s a f u n c t i o n on C(X) by f +>

(f,x),

and s i n c e C(X) i s s e p a r a t i n g on X , d i s t i n c t x ' s d e f i n e d i s t i n c t functions.

We c a n t h e r e f o r e i d e n t i f y e a c h x w i t h t h e f u n c t i o n

i t d e f i n e s , and h e n c e f o r t h w e w i l l do t h i s : x w i l l be c o n s i d e r e d a f u n c t i o n on C(X).

M o r e o v e r , by t h e v e r y d e f i n i t i o n

o f a d d i t i o n a n d s c a l a r m u l t i p l i c a t i o n i n C(X), x i s a c t u a l l y a linear functional.

Even m o r e , f r o m t h e d i s c u s s i o n o f t h e

b e g i n n i n g o f t h e C h a p t e r on t h e l a t t i c e o p e r a t i o n s i n C ( X ) , e v e r y xEX i s , i n f a c t , a n MI-homomorphism o f C(X) ( i n t o IR). And t h e y a r e t h e o n l y o n e s :

( 3 3 . 1 ) F o r a f u n c t i o n $ o n C(X), t h e f o l l o w i n g a r e e q u i v a l e n t

'1

$ i s a n MI-homomorphism o f C(X) i n t o R ;

2O

$EX.

P roof. -

I t r e m a i n s t o show t h a t ' 1

implies 2

0

.

We n o t e

f i r s t t h a t two MI-homomorphisms $1 , $ 2 o f C(X) i n t o 1R a r e i d e n t i c a l i f and o n l y i f (+1)-1(0) h o l d s f o r 4.

$-'lo) i s

=

Now assume 1'

($2)-1(0).

t h e n a R i e s z i d e a l H o f C(X), a n d

$ b e i n g a s i n g l e l i n e a r f u n c t i o n a l - i s a maximal o n e .

Z(H)

consists of a single point x (31.13).

Hence

But t h e n x -1( 0 )

=

H,

s o w e have $ = x . QED

C o n s i d e r a c o n t i n u o u s mapping X

f o s o f C(X) i n t o C ( Y ) .

We

We w i l l c a l l


183

t h i s mapping t h e t r a n s p o s e o f s , a n d d e n o t e i t b y s t . f o r a l l f E C ( X ) and yEY, ( s t f , y )

=

Thus,

(f,sy).

I t i s t r i v i a l t o v e r i f y t h a t C(X)

S

t

> C(Y)

Mn-

i s an

homomorphism.

We show t h a t , c o n v e r s e l y , f o r e a c h Mn-homomorph-

T i s m C(X) ->

C(Y), t h e r e e x i s t s a unique c o n t i n u o u s

X So ( f , s y )

=

(Tf,y) for all

i t i s immediate from t h i s i d e n t i t y t h a t f o r

e v e r y n e t { y a } i n Y a n d yoEY, i f ( g , y o ) = l i m a ( g , y a ) gEC(Y), t h e n ( f , s y o ) = l i m a ( f , s y a ) continuous.

That T

=

f o r a l l fEC(X).

i s a n MIL-subspace o f C(Y) ( 1 8 . 2 ) , a n d T - ’ ( O )

R i e s z i d e a l o f C(X).

for all Thus s i s

s t f o l l o w s f r o m t h e same i d e n t i t y .

As we know, f o r a n MIL-homomorphism C(X)

s(Y)

yoT

T

--A

C(Y), T(C(X))

i s a norm c l o s e d

And f o r a c o n t i n u o u s m a p p i n g X f l Z .

T i e t z e E x t e n s i o n Theorem.

__ > C ( Z )

is c l e a r l y t h e r e -

(33.6) then a l s o follows from t h e

CflAl'TER

7

(C(X) , C ' ( X ) ) - D U A I . T T Y

5 3 4 . The i m b e d d i n g o f X i n C 1 ( X )

The e l e m e n t s o f t h e d u a l C ' ( X ) o f C(X) a r e c a l l e d t h e Radon m e a s u r e s on X ; w e w i l l d e n o t e thcm by ~ i , v , u , p , u .

-

f e e l f r e e t o o m i t "Radon";

We w i l l

that i s , unless otherwise s t a t e d ,

" m e a s u r e : w i 11 a l w a y s mean Radon m e a s u r e . S i n c e C(Xj apply.

i s an ME-space, t h e r e s u l t s o f 5 5 2 1 , and 1 9

I n p a r t i c u l a r , C ' (X) i s a n L - s p a c e w i t h

IIuI[

=

( nfX) , p )

for a l l uEC'(X)+. Under ( f , x ) , e a c h xEX 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 on C(X) w i t h ( 1 , x )

=

1.

I t follows xEK(C'(X)j.

Combining t h i s

w i t h ( 3 3 . 1 ) and ( 2 1 . 2 ) , we o b t a i n :

( 3 4 . 1 ) -__ Theorem.

X = ext K(C'(X)).

K ( C ' ( X ) ) is o f t e n c a l l e d t h e s t a t e s p a c e o f C(X), and i t s e l e m e n t s a r e c a l l e d ____ states. e x t r e m e p o i n t s o f K(C1 ( X ) ) ,

The e l e m e n t s o f X , b e i n g t h e a r e t h e n c a l l e d t h e =re

186

s t a t e s on

(C(X) , C ' ( X ) ) - D u a l i t y

187

C(X).

The e l e m e n t s o f K ( C ' ( X ) )

are also called the probability

( K a d o n )_ m_ e a_ s u_ r e_ s on X.

-

A consequence o f (34.1) worth n o t i n g i s t h a t f o r X1,X

2 E X , x1

+

x 2 i m p l i e s xlAx2

=

0 in

c'(x)

(2'

i n (21.2)).

A s a c o n v e r s e t o ( 3 4 . 1 ) , we h a v e t h e f o l l o w i n g s p e c i a l c a s e o f K a k u t a n i ' s c l a s s i c t h e o r e m 1211.

( 3 4 . 2 ) ____ Theorem. ( K a k u t a n i ) .

Every MI-space E i s a C ( X ) .

S p e c i f i c a l l y , we c a n t a k e f o r X t h e s e t e x t K ( E ' ) endowed w i t h t h e vague t o p o l o g y .

Proof.

-I_.

E-

By ( 2 1 . 6 ) ,

C ( e x t K ( E ' ) ) by

e x t K(E') i s v a g u e l y compact. (Ta,$)

=

Define

( a , $ ) f o r a l l a E E and

$€ e x t K(E').

Then T i s a p o s i t i v e l i n e a r m a p p i n g w i t h T I =

Il(ext K ( E ' ) ) .

By (21.8), T i s an i s o m e t r y , h e n c e o n e - o n e .

We

show n e x t t h a t T p r e s e r v e s t h e l a t t i c e o p e r a t i o n s , h e n c e i s a n Mn-isomorphism o f E i n t o C ( e x t K ( E ' ) ) .

Given a , b E E , t h e n f o r

every $€ e x t K(E'), (T(aVb),$) = (avb,$) max((Ta,$), ( T b , $ ) )

=

(TaVTb,$).

=

max((a,$),(b,$))

(Here t h e s e c o n d e q u a l i t y

f o l l o w s f r o m t h e f a c t t h a t $ i s a R i e s z homomorphism ( 2 1 . 2 ) , and t h e l a s t from t h e d e f i n i t i o n o f o r d e r o n C ( e x t K ( E ' ) ) . ) F i n a l l y , T(E)

=

i s s e p a r a t i n g on e x t K ( E ' ) ,

t h e W e i e r s t r a s s - S t o n e Theorem ( 3 2 . 5 ) , T(E)

s i n c e E i s , s o , by = C(ext K(E')).

QE D

Chapter 7

188

With t h e i n t r o d u c t i o n o f C t ( X ) , w e c a n c o n s i d e r weak c o n v e r g e n c e on C(X).

The D i n i Theorem g i v e s u s an i m p o r t a n t p r o -

p e r t y o f t h i s convergence.

For l a t e r c o m p a r i s o n , we p r e s e n t i t

a s a s h a r p e n i n g o f t h e D i n i Theorem.

(34.3)

( D i n i Theorem).

F o r a m o n o t o n i c n e t {fa} i n C(X) and

fEC(X), t h e following a r e e q u i v a l e n t : 1'

{fa) c o n v e r g e s

2'

{fa} c o n v e r g e s t o f w e a k l y ;

3'

{fa] c o n v e r g e s t o f s i m p l y .

t o f normwise;

5 3 5 . Atomic a n d d i f f u s e Radon m e a s u r e s

S i n c e C t ( X ) i s an L - s p a c e , e v e r y norm c l o s e d R i e s z i d e a l i s a band (hence a p r o j e c t i o n b a n d ) .

I n p a r t i c u l a r , t h e band

o f C t ( X ) g e n e r a t e d b y X i s s i m p l y t h e norm c l o s e d R i e s z i d e a l which i t g e n e r a t e s .

I n f a c t , we h a v e m o r e :

( 3 5 . 1 ) The b a n d o f C'(X) g e n e r a t e d b y t h e s u b s e t X i s s i m p l y

t h e norm c l o s e d l i n e a r s u b s p a c e w h i c h X g e n e r a t e s .

-__ Proof,

I t i s e n o u g h t o show t h a t t h e l i n e a r s u b s p a c e

c x c p x i s a Riesz i d e a l . f o l l o w s from ( 3 . 3 ) .

S i n c e each1Rx i s a R i e s z i d e a l , t h i s

QED

(C(X) , C ' ( X ) ) - D u a l i t y

189

We w i l l c a l l t h e a b o v e b a n d t h e a t o m i c p a r t o f C ' ( X ) , a n d C ' ( X ) a has a s i m p l e , e v e n t r a n s p a r e n t ,

d e n o t e i t by C ' ( X ) a .

We l e a v e many o f t h e f o l l o w i n g s t a t e m e n t s t o t h e

structure.

reader's verification. a f t c r (34.1),

( O f major u s e f u l n e s s i s t h e f a c t , noted

t h a t the elements of X a r e mutually d i s j o i n t :

x1 # x 2 i m p l i e s x1/\x2

=

0 , h e n c e t h a t two s u b s e t s o f X w i t h

cinpty i n t e r s e c t i o n a r e d i s j o i n t i n t h e R i e s z s p a c e s e n s e . ) Let Q b e an a r b i t r a r y s u b s e t o f X , a n d J t h e b a n d w h i c h i t generates.

Then J i s s i m p l y t h e norm c l o s e d l i n e a r s u b s p a c e

g e n e r a t e d b y Q (same p r o o f as f o r ( 3 5 . 1 ) ) , a n d ,J

n

X

=

Q.

v e r s e l y t o t h i s l a s t i d e n t i t y , l e t J be a band o f C ' ( X ) a ,

set Q

J

=

n

X;

t h e n t h e band g e n e r a t e d by Q i s J .

Conand

Thus

(35.3) There i s a one-one correspondence between t h e bands J of a n d t h e s u b s e t s Q o f X : J

C'(X)a

n

Q = J

Q i f and o n l y i f

X i f and o n l y i f J i s t h e band g e n e r a t e d by Q .

We a l s o h a v e :

(35.3)

If X

g e n e r a t e d by versely,

Q , , Q,

= Q,IJ

Q,,Q,

n

=

respectively,

i f C'(X)a

J2 flX , t h e n Q,

0 Q,

=

Q,

8,

and J1,J2 a r e t h e bands

t h e n C'(X)a

J, 3 J 2 ( b a n d s ) , a n d Q, =

P, a n d X

=

Q, IJ Q,.

=

J 1 o J,.

=

J1

n

X, Q,

Con=

Chapter 7

190

Our n e x t g o a l i s t o g i v e a s i m p l e r e p r e s e n t a t i o n o f C ' ( X ) u F o r e a c h xEX,lRx i s a p r o j e c t i o n band o f C ' ( X ) ,

((35.4) below).

hence, given P E C ' (X) form Axx.

,

t h e component o f p i n t h i s band h a s t h e

Note t h a t t h u s , f o r e v e r y U E C ' ( X )

u n i q u e l y d e t e r m i n e d s e t o f numbers

X

, we

have t h e

I xEX},

Fix a s u b s e t Q o f X , and l e t G b e t h e l i n e a r s u b s p a c e o f C l ( X )

generated by Q.

( S o G i s t h e R i e s z i d e a l cxEQRx, a n d

E v e r y P E G h a s t h e form P

lies in C'(X)u.)

=

zyXx;xi I

(x1;

**,xn€Q).

F o r two e l e m e n t P = C n X x . and v 1 xi 1

=

z nl ~ X i ~ i

(we c a n c l e a r l y assume t h e y a r e l i n e a r c o m b i n a t i o n s o f t h e same e l e m e n t s o f Q ) , w e h a v e p + v

=

Cn(A 1 xi

+

K ~ . ) x Xu ~ ,= 1

f o l l o w i n g d e s c r i p t i o n o f t h e band J g e n e r a t e d by

Q i s now

e a s i l y e s t a b l i s h e d (remember, G i s a c t u a l l y a R i e s z i d a e a l )

( i ) E v e r y p E J l i e s i n t h e h a n d g e n e r a t e d b y some c o u n t a b l e s u b s e t {xn} o f Q , and i n t h e b a n d , i t h a s t h e u n i q u e r e p r e s e n t a t i o n p = ZnXxnxn i n t h e s e n s e o f norm c o n v e r g e n c e .

( i i ) Two e l e m e n t s p , v o f J c a n a l w a y s b e w r i t t e n i n t e r m s o f t h e same c o u n t a b l e s e t { x n } c Q : p = C A and w e h a v e p + v

=

Cn(A 'n

zn

max(X

,K

'n

'n

) x n , [IFiIj =

xn, v = C n ~ x n ~ n , 'n + K ) x n , Ap = Cn(AX )xn, pVv = 'n xn n

I.

C o n s e q u e n t l y , we c a n d e n o t e e a c h p E J b y 1-1 = CxEQAxx, i t being understood t h a t Ax = 0 f o r a l l but a countable s e t of The s e t o f c o e f f i c i e n t s {Xx)x€Q) each P E G .

i s uniquely determined f o r

I n t h i s n o t a t i o n , we h a v e

u

+ v = CX~q(Xx

+

KX)X,

X I S .

(C(X) , C ' ( X ) ) - D u a l i t y

191

Some g e n e r a l n o t a t i o n : G i v e n a s e t T, d e n o t e b y a l [ T ) s e t o f a l l r e a l f u n c t i o n s p on T s u c h t h a t Z t E T l p [ t )

1

o f <J o n t o V

1

(9).

1 Thus <J c a n b e i d e n t i f i e d w i t h R ( Q ) .

We

do t h i s :

(35.4) Theorem. -

Q.

Given a s u b s e t Q o f X , l e t ,J b e t h e b a n d

1x€Q X xx a s t h e f u n c t i o n on Q w i t h v a l u e Xx a t x , we h a v e LJ = a ' ( Q ) . We h a v e

g e n e r a t e d by

T h e n , c o n s i d e r i n g e a c h i~

We w i l l d e n o t e ( C I ( X ) a ) d i f f u s e p a r t o f C'(X)

d

(hence

by C ' [ X ) d ,

=

and c a l l i t t h e

the subscript d ) .

Its ele-

m e n t s a r e t h e d i f f u s e (Radon1 m e a s u r e s on X - a l s o c a l l e d t h e purely non-atomic ones.

The d e c o m p o s i t i o n C l ( X )

=

C'(X)

C ' (X)d g i v e s a s t a n d a r d r e p r e s e n t a t i o n o f e v e r y u E C ' {X)

a

3

as

t h e sum o f a u n i q u e l y d e t e r m i n e d a t o m i c m e a s u r e a n d a u n i q u e l y

192

Chapter 7

determined d i f f u s e one.

T h e s e two components o f p w i l l b e

d e n o t e d s i m p l y b y pa and p d ( i n s t e a d o f p

C ' (XIa

and 1-1 C ' (X) d l *

And f o r a s u b s e t A o f C 1 ( X ) , i t s images u n d e r p r o j e c t i o n i n t o C ' ( X ) u and C ' ( X ) d w i l l b e d e n o t e d by Au and A d . The f o l l o w i n g i s e a s i l y v e r i f i e d .

( 3 5 . 5 ) Let J be a Riesz i d e a l o f C ' ( X ) , and s e t Q

Then Ja i s c o n t a i n e d i n t h e

=

*J

n

X.

If J is a

band g e n e r a t e d by Q .

b a n d , t h e n J a c o i n c i d e s w i t h t h e b a n d g e n e r a t e d by Q .

For a Riesz i d e a l J o f C ' (X)

(35.6) Corollary.

,

the following

are equivalent:

'1

J c C ' (X)d;

2'

J

~

C'(X)a

X

g.=

i s a l w a y s s e p a r a t i n g on C(X) ( s i n c e X i s ) .

c o n t r a s t t o t h i s , C ' ( X ) d may c o n s i s t o n l y o f 0 :

Let X be t h e

A l e x a n d r o f f o n e - p o i n t c o m p a c t i f i c a t i o n o f N ; t h e n C(X) ( t h e s p a c e o f c o n v e r g e n t s e q u e n c e s ) a n d C'(X) = A t t h e o t h e r extreme, C'(X)d

In

1

=

c

(X) = C ' ( X ) u .

may a l s o b e s e p a r a t i n g on C ( X ) :

L e t X b e a r e a l i n t e r v a l ; t h e n t h e Lebesgue m e a s u r e on X i s d i f f u s e a n d , a s i s w e l l known, t h e b a n d i t g e n e r a t e s i s s e p a r a t i n g on C(X). Remark. -____

The s p a c e s X f o r w h i c h t h e r e e x i s t s a t l e a s t

o n e d i f f u s e m e a s u r e d i f f e r e n t from 0 c a n b e d e s c r i b e d p r e c i s e l y

(C(X) ,C' (X))-Duality

193

( ( 3 7 . 8 ) below).

5 3 6 . The v a g u e t o p o l o g y on C ' ( X )

0 on C ' ( X )

A linear functional

l i e s i n C(X)

i f i t i s v a g u e l y c o n t i n u o u s on C ' ( X ) ;

i f and o n l y

a n d i n d e e d , by t h e

G r o t h e n d i e c k c o m p l e t e n e s s t h e o r e m , i t s u f f i c e s f o r Q t o be

We c a n s a y m o r e :

v a g u e l y c o n t i n u o u s on t h e u n i t b a l l B ( C ' ( X ) ) .

( 3 6 . 1 ) -~ Theorem.

A l i n e a r f u n c t i o n a l 0 on C ' ( X )

lies i n C(X)

i f a n d o n l y i f 0 i s v a g u e l y c o n t i n u o u s on K ( C ' ( X ) ) .

Proof.

We n e e d o n l y show t h a t i f

@ i sv a g u e l y c o n t i n u o u s

on K ( C ' ( X ) ) , t h e n i t i s v a g u e l y c o n t i n u o u s on B ( C ' ( X ) ) .

a n e t {pa} i n B ( C ' ( X ) )

J(@,p) - (@,pa)l

A,@-

pa +

c o n v e r g e s v a g u e l y t o p , b u t ( 0 , ~ )#

Then w e c a n a s s u m e t h e r e

l i m a ( G,p,).

2

E

for a l l a.

exists

=

K

), = {O

a

> 0 such t h a t

For e a c h a , w e c a n w r i t e

w h e r e P , u ~ E K ( C ' ( X ) ) , X,K,, > 0 , and a 1 (cf.519). Now t h e real i n t e r v a l [ 0 , 1 ] i s compact.

i s v a g u e l y c o m p a c t , s o b y t a k i n g an a p p r o p r i a t e

s u b n e t , w e c a n assume t h e r e e x i s t X , K €

and

E

KaUas

a and K ( C ' ( X ) )

that

Suppose

limaha,

K

=

lim

K

c1

a'

} converges vaguely t o

{pa} O.

On t h e o t h e r h a n d ,

@

such

converges vaguely t o p, On t h e o n e h a n d i t f o l l o w s

t h a t { p } c o n v e r g e s v a g u e l y t o AD c1

IR a n d p , o E K ( C ' ( X ) )

KU,

whence Xp

-

KO

= p

b e i n g v a g u e l y c o n t i n u o u s on K ( C ' ( X ) ) ,

. we

Chapter 7

194

S i n c c X i s s e p a r a t i n g on C(X), t h e l i n e a r s u b s p a c e g e n e r a t -

A f o r t i o r i , C ' ( X ) r L i s also.

ed by X i s vaguely dense i n C ' ( X ) .

More s h a r p l y , b y t h e Krein-Milrnan t h e o r e m , t h e c o n v e x h u l l o f X i s v a g u e l y d e n s e i n K(C' ( X ) ) .

I t f o l l o w s K(C'(X)=)

is also.

We w i l l n e e d t h e f o l l o w i n g f u r t h e r s h a r p e n i n g .

( 3 6 . 2 ) F o r two R i e s z i d e a l s J1,J2 o f C ' ( X ) ,

the following arc

equivalent: 1'

< J 2 i s c o n t a i n e d i n t h e v a g u e c l o s u r e o f J1;

2'

( J 2 ) + i s c o n t a i n e d i n t h e v a g u e c l o s u r e o f (.J1)+.

3'

K ( J 2 ) i s c o n t a i n e d i n t h e vague c l o s u r e o f K ( J 1 ) .

P roof. -

2'

T h a t 1'

i m p l i e s 2'

f o l l o w s f r o m (10.12).

Assume

By 2O, some n e t {pa} i n (J1)+

h o l d s and c o n s i d e r p E K ( J , ) . 1

converges vaguely t o P.

Then

IIuil

=

(n(X),u)

lirna[lpal[ , s o , t a k i n g a s u b n e t , w e c a n assume

lima(ll(X),pa)

IIuaI[ #

Thus 2'

i m p l i e s 3.

T h a t 3'

=

0 f o r a l l a.

Then {vc13 i s a n e t i n K(.J1)

For e a c h a, s e t v U = c o n v e r g e s v a g u e l y t o p.

=

and 0

implies 1

i s immediate.

QED

(36.3) Corollary.

For a R i e s z i d e a l J o f C ' ( X ) ,

the following

(C(X) , C ' (X)) -Duality

195

are equivalent:

'1

,J i s v a g u e l y c l o s e d ;

.7 0

%J+ i s v a g u e l y c l o s e d ;

0

3

4'

R(J)

is vaguely closed;

K(J)

i s vaguely closed.

By t h e d i s c u s s i o n i n 5 4 1 0 , 11, t h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e v a g u e l y c l o s e d b a n d s tJ o f C ' ( X )

and

t h e ( w e a k l y c l o s e d , h e n c e ) norm c l o s e d R i e s z i d e a l s I I o f i f and o n l y i f J

JH

C'(X):

Now c o n s i d e r a s u b s e t Q o f X . l i n e a r subspace of C ' ( X ) i d e a l of C(X)

(31.2),

n

X

=

Z(QL)

=

1

.

But ' Q

i s a Riesz

so QLL i s , i n f a c t , the vaguely closed

Q (31.3).

c l o s e d band of C ' ( X ) ,

then J

X,

Two e a s y c o n s e q u e n c e s a r e t h a t

L

HL i s p r e c i s e l y t h e v a g u e l y

a n d [ i i ) i f ,J i s a v a g u e l y

c l o s e d band g e n e r a t e d by Z(H),

n

<J

QLL i s t h e vaguely c l o s e d

g e n e r a t e d by Q .

( i ) f o r a Riesz i d e a l H of C ( X ) ,

J

=

Moreover, i f Q i s a c l o s e d s u b s e t o f

___ b a n d g e n e r a t e d by Q .

QLL

i f and o n l y i f I1

fIL

=

=

(JnX)'

- equivalently,

Z(J ) L

Summing u p ,

X.

There i s a t h r e e - w a y correspondence between

( 3 6 . 4 ) Theorem.

t h e c l o s e d s u b s e t s 2 o f X , t h e norm c l o s e d R i e s z i d e a l s H o f C(X),

and t h e v a g u e l y c l o s e d bands J o f C ' ( X ) .

Z,H,

a n d .J

c o r r e s p o n d t o e a c h o t h e r i f and o n l y i f t h e f o l l o w i n g h o l d :

J

n

(i)

Z

=

Z(H)

=

(ii)

H

=

ZL

=

J .

(iii) J

=

€1'

=

t h e v a g u e l y c l o s e d band g e n e r a t e d b y 2 .

X;

1'

=

Chapter 7

196

Remark 1. -

In ( i i ) , w e c o u l d have w r i t t e n Z

1

.

i t depends

on w h e t h e r we t h i n k o f X a s t h e " p r e d u a l " o f C(X) o r a s a s u b -

s e t o f Cl(X). Remark 2 . b e emp ty : l e t J

C1(X)d f o r a real i n t e r v a l .

=

a bo v e t h e o r e m , s i n c e f o r __ any X , Ct(X)d

n

X =

We saw i n 535 t h a t f o r two s u b s e t s Q l , Q 2 bands J1,J2 which t h e y g e n e r a t e , J2

=

I n d e e d , by t h e

@,w e

have t h a t

Qln

Q,

=

o f X and t h e

fl i f and o n l y i f

F o r c l o s e d s u b s e t s o f X , w e c a n s a y more:

0.

(36.5) Let Z1,Z2 they generate.

b e c l o s e d s u b s e t s o f X and J 1 , J 2 t h e b a n d s Then t h e f o l l o w i n g a r e e q u i v a l e n t :

io

z1 n z 2

2'

J~

3'

(vague c l o s u r e J1)n

n J~

P roof. -

=

+;

= 0;

( v a g u e c l o s u r e J,)

We n e e d o n l y show t h a t 1'

Urysohn f u n c t i o n f o r t h e p a i r ( Z l , Z 2 )

=

0.

i m p l i e s 3'. ( c f . 530).

n(X) - f i s a Urysohn f u n c t i o n f o r ( Z 2 , Z l ) .

Let f be a Then g

=

assume 1-1 > 0 , h e n c e i t s u f f i c e s t o show t h a t ( n ( X ) , p ) The v a g u e c l o s u r e o f J1 i s ( Z 1 ) l L ; s i n c e g€(Z1)', follows ( g , u ) +

(g91-I)

=

0.

=

0.

S i m i l a r l y , ( f , p ) = 0.

=

Now s u p p o s e p l i e s

i n t h e v a g u e c l o s u r e s o f b o t h J1 a n d J2; w e show p

(f3.1-I)

X may

i s v a g u e l y c l o s e d i f and o n l y i f C1(X ), = 0 .

C '( X) ,

Jln

n

I f a band J i s n o t v a g u e l y c l o s e d , J

We c a n

0.

= 0.

it

Then (n(X),.cl)

=

QED

197

(C(X) ,C' (X))-Duality

A non-empty s u b s e t Q o f a t o p o l o g i c a l s p a c e i s c a l l e d

d e n s e - i n - i t s e l f i f , a s a s p a c e i n i t s own r i g h t , i t h a s no If Qis d e n s e - i n - i t s e l f ,

isolated point.

let ,J b e t h e v a g u e l y c l o s e d

Given a s u b s e t A o f C ' ( X ) , band g e n e r a t e d by A .

Then

8.

then so i s

. J / T X w i l l be c a l l e d t h e s u p p o r t

I f A c o n s i s t s o f a s i n g l e e l e m e n t 1~., w e w i l l c a l l .J

of A. -

X

t h e s u p p o r t o f p.

(36.6)

Theorem.

--

For e v e r y non-empty s u b s e t A o f C ' ( X ) , ,

~

the

support of A i s d e n s e - i n - i t s e l f .

___ Proof.

Let J b e t h e v a g u e l y c l o s e d b a n d g e n e r a t e d by A ,

X h a s an i s o l a t e d p o i n t x.

and s u p p o s e J

is a closed set.

Then Z

=

(,JnX)\{x 1

Let J1 b e t h e v a g u e l y c l o s e d b a n d g e n e r a t e d

by Z .

J = J g R x . 1

Note f i r s t t h a t w r i t t e n J1

%,

Rx.

Jln

Rx

=

0 by ( 3 6 . 5 ) ,

s o J1 + R x c a n b e

S i n c e Rx ( b e i n g o n e - d i m e n s i o n a l ) i s a l s o

v a g u e l y c l o s e d , i t f o l l o w s f r o m t h e L o t z Theorem ( 1 0 . 2 1 ) t h a t J1 3 Wx i s v a g u e l y c l o s e d .

I t i s i m m e d i a t e t h a t J1 3 R x c < J .

The o p p o s i t e i n c l u s i o n f o l l o w s f r o m t h e f a c t t h a t J l 3 R x c o n t a i n s Z IJ

{x)

and i s vaguely closed.

This e s t a b l i s h e s (*).

Chapter 7

198

Now A c C ' ( X ) d ,

h e n c e A c { x } d , hence by (*)

,

A c .Jl.

T h i s c o n t r a d i c t s t h e a s s u m p t i o n t h a t t h e v a g u e l y c l o s e d hand g e n e r a t e d by A i s . J . QED

G i v e n a c l o s e d s u b s e t Z o f X, l e t W

=

v a g u e l y c l o s e d band g e n e r a t e d by Z , and I1

s o b y s t a n d a r d Banach s p a c e t h e o r y , ,J 11'.

=

=

X\Z,

<J b e t h e

Z ' .

Then .J

(C(X)/II)

'

=

HL,

and C ' ( X ) / . J

S i n c e ,Jd c a n b e i d e n t i f i e d w i t h C ' ( X ) / , J , t h e l a s t i d e n t i t y

can be w r i t t e n J d = 1 1 ' . Now H = Cw(W) and C[X)/II

=

C(Z)

( c f . § 3 1 ) , s o t h e above

can b e w r i t t e n :

( 3 6 . 7 ) -____ Theorem.

L e t 2 b e a c l o s e d s u b s e t o f X, W

-J t h e v a g u e l y c l o s e d b a n d g e n e r a t e d b y

Z.

=

X\Z,.and

Then

.J = C ' ( Z ) ,

Jd

=

(Cw(W))'.

We r e s t a t e o n e o f t h e s e for e m p h a s i s .

Corollary. (36.8) -J

=

=

Given a v a g u e l y c l o s e d b a n d (J o f C'(X),

C'(JnX),

T h e r e i s u n d o u b t e d l y a body o f r e s u l t s f o r t h e Mll-subs p a c e s o f C(X) p a r a l l e l i n g t h o s e w e h a v e g i v e n f o r norm c l o s e d

( C ( X ) ,C' ( X ) ) -Duali t y

Riesz i d e a l s .

199

However, w e h a v e l i t t l e t o o f f e r i n t h i s d i r e c -

One p r o b l e m : f o r an M I - s u b s p a c e F , w e know no r e a s o n a b l e

tion.

d e s c r i p t i o n o f F'. R e c a l l t h a t i n t h e d u a l i t y b e t w e e n C(X) a n d X, t h e r o l e played by Z ( H )

f o r a norm c l o s e d R i e s z i d e a l H was t a k e n f o r F ,

We n o t e h e r e t h a t t h e s e

by t h e s e t o f s e t s o f c o n s t a n c y o f F .

s c t s of constancy a r e simply the i n t e r s e c t i o n s with X of the t r a n s l a t e s o f F'.

5 3 7 . Mapping d u a l i t y

The d i s c u s s i o n i n 5 2 4 on a n bin-homomorphism a n d i t s t r a n s -

We e x p a n d on t h e r e s u l t s

pose a p p l i e s t o t h e p a i r (C(X),C'(X)). given there.

Let C(X)

__ > C(Y) h e a norm c o n t i n u o u s

o r d e r bounded - l i n e a r mapping. vaguely c o n t i n u o u s .

Then C ' ( X )

d e t e r m i n e d b y i t s v a l u e s on Y . mapping C ' ( X )

C " ( Y ) .

For t h e r e m a i n d e r

s , T , S w i l l correspond w i t h each o t h e r a s above.

We s h a r p e n t h e f i r s t h a l f o f ( 2 4 . 4 ) .

( 3 7 . 2 ) The f o l l o w i n g a r e e q u i v a l e n t : lo

T i s a Markov m a p p i n g ;

2'

S(K(C' Y)j)

3O

s(Y) c K(C' (X 1 .

Also (24.5)

= KlC'

(XI);

(by combining i t w i t h ( 3 3 . 2 ) ) .

( 3 7 . 3 ) The f o l l o w i n g a r e e q u i v a l e n t :

'1 2'

T i s a n MIL-homomorphism;

S (i) i s i n t e r v a l p r e s e r v i n g a n d ( i i ) maps K ( C ' ( Y ) ) i n t o K ( C ' ( X ) ) ;

(C(X) ,C'(X))-Duality

3O

201

x.

s(Y) c

I f S s a t i s f i e s t h e a b o v e , t h e n , from ( 2 4 . 7 ) , i f G i s a band o f C ' ( Y ) , S(G)

The f a c t t h a t S i s

i s a band o f C ' ( X ) .

v a g u e l y c o n t i n u o u s g i v e s u s more:

(37.4) I f s , T , S s a t i s f y (37.3), then f o r every vaguely closed band G o f C ' ( Y ) , S(G)

Proof.

i s a l s o a vaguely c l o s e d band.

is

By ( 3 6 . 3 ) i t s u f f i c e s t o show t h a t K ( S ( G ) )

vaguely c l o s e d .

By ( 2 4 . 7 ) ,

K(S(G))

S(K(G));

=

is

since K(G)

v a g u e l y compact and S i s v a g u e l y c o n t i n u o u s , S ( K ( G ) )

is also

v a g u e l y compact. QED

( 3 7 . 5 ) Theorem. H

=

T

-1

Suppose s , T , S s a t i s f y ( 3 7 . 3 ) .

( 0 ) , and J = S ( C ' ( Y ) ) .

e a c h o t h e r by ( 3 6 . 4 ) : Z = Z ( H )

Set Z

=

s(Y),

Then Z,H, and J c o r r e s p o n d t o = J

n

X, H

=

ZL

=

Jl, and

J = HL = t h e v a g u e l y c l o s e d band g e n e r a t e d by Z .

T h i s f o l l o w s from ( 3 3 . 3 ) ,

(36.4),

( 3 7 . 3 ) , and ( 3 7 . 4 ) .

The f o l l o w i n g r e s u l t c o n t a i n e d i n t h e above i s w o r t h noting.

I f C(X)

[Tt(C'(y>)l

?i

X.

-> C(Y)

i s an MI-homomorphism, t h e n Tt(Y)

=

Chapter 7

202

( 3 7 . 6 ) C o r o l l a r y 1.

L e t C(X) ___ > C(Y) be an MIL-homomorphism.

Then (i)

T i s a s u r j e c t i o n i f and o n l y i f T t i s an i n j e c t i o n .

( i i ) T i s a n i n j e c t i o n i f and o n l y i f T t

(37.7) Corollary 2 .

L e t C(X) ->

T

is a surjection.

C(Y) b e an MI-homomorphism,

Then and s e t J = T t ( C ' ( Y ) ) . t ( i ) T [ C ' ( Y ) ] = Ja; U

( i i ) T~ [ C ' ( Y ) ~ 3] J ~ .

Proof.

( i ) f o l l o w s e a s i l y from t h e comment f o l l o w i n g

( 3 7 . 5 ) above a n d t h e f a c t t h a t T t i v e elements.

t

p r e s e r v e s t h e norms o f p o s i -

( i i ) f o l l o w s from ( i ) . QED

( 3 7 . 8 ) C o r o l l a r y 3.

L e t C(X)

and s e t J = T t ( C ' ( Y ) ) .

->C(Y) be an MI-homomorphism,

I f T i s s u r j e c t i v e , T t maps CI(Y),

and

C 1 ( Y ) d e a c h R i e s z i s o m o r p h i c a l l y and i s o m e t r i c a l l y o n t o <Ju and J d respectively.

We showed i n ( 3 6 . 6 ) t h a t a n e c e s s a r y c o n d i t i o n f o r C ' ( X ) t o c o n t a i n non-zero d i f f u s e measures i s t h a t X c o n t a i n a s u b s e t which i s d e n s e - i n - i t s e l f .

We show t h i s c o n d i t i o n i s a l s o s u f -

ficeint. We w i l l u s e t h e f o l l o w i n g s t a n d a r d p r o p e r t y o f compact

(C(X) ,C' (X)) -Duality

203

spaces.

Lemma.

~f

Y

C

X i s~ a c o n t i n u o u s s u r j e c t i o n , t h e n t h e r e

e x i s t s a c l o s e d s u b s e t 2 o f X w h i c h i s mapped o n t o Y w h i l e no p r o p e r c l o s e d s u b s e t o f Z i s mapped o n t o Y .

The f o l l o w i n g t h e o r e m i s d u e t o W . and Semadeni ( c f .

Theorem.

(37.9)

-

Rudin and t o P e l c z y n s k i

[49] 519.7).


X c3ntains a dense-in-itself subset;

1'

there i s a continuous s u r j e c t i o n of X onto a r e a l

2'

interval ; 3O

c'(x)d #

Proof.

0.

Assume X c o n t a i n s a s e t Q d e n s e - i n - i t s e l f .

Choose

d i s t i n c t p o i n t s xo,xl of Q y then d i s j o i n t closed neighborhoods of xOyx1 r e s p e c t i v e l y .

Vo,V1

Choose d i s t i n c t p o i n t s x o o , x o l

o f Q i n t h e i n t e r i o r o f V o and x l o , x l l

of Q i n t h e i n t e r i o r

Then c h o o s e d i s j o i n t c l o s e d n e i g h b o r h o o d s V o o , V o l

o f V1.

x o o , x o l r e s p e c t i v e l y i n V o and VloYVl1 i n Vl.

of xlo,xll

of

respectively

C o n t i n u i n g t h i s p r o c e s s i n d u c t i v e l y , we o b t a i n , f o r

(ij = 0 or 1 for j = l,...,n); each n , Zn closed s e t s V. i l . . . in d e n o t e t h e i r union by Zn. Finally, set Z =

nnZn.

F o r e a c h x E Z , t h e r e i s a u n i q u e s e q u e n c e { i n )(n

=

1,2,..*),

204

Chapter 7

w i t h in

=

.

0 o r 1 f o r a l l n , such t h a t xEnnVili2...i n

Let s x be t h e e l e m e n t i n t h e r e a l i n t e r v a l [ 0 , 1 ] w h i c h h a s t h e dyadic representation 0 . i i 1 2 mapping o f Z o n t o [ 0 , 1 ] ,

- - . in

* - . .

x t->s x i s t h e n a

and i s c l e a r l y continuous.

e x t e n s i o n t h e o r e m now g i v e s u s 2

0

The T i e t z e

. <s

Now assume t h e r e i s a c o n t i n u o u s s u r j e c t i o n Y X onto t a r e a l i n t e r v a l Y. Then C(Y) __ >C(X) i s an M l l i s o m o r p h i s m tt ( i n t o ) , and t h e r e f o r e C ' ( Y ) C'(X) i s s u r j e c t i v e . It

<s

f o l l o w s from ( 3 7 . 7 ) t h a t C ' ( Y ) d c st t ( C ' ( X ) d ) . ( i t c o n t a i n s t h e L e b e s g u e m e a s u r e on Y ) , C I ( X ) d h a v e 3'.

T h a t 3'

Since C'(Y)d# 0

# 0.

We t h u s

i m p l i e s 1' was shown i n ( 3 6 . 6 ) .

QED

Combining t h e a b o v e w i t h ( 3 6 . 7 ) , we o b t a i n e a s i l y :

( 3 7 . 1 0 ) C o r o l l a r y 1.

Let Z b e a c l o s e d s u b s e t o f X a n d J t h e

v a g u e l y c l o s e d band which i t g e n e r a t e s .

Then t h e f o l l o w i n g a r e

equivalent:

1'

'2

Z contains a dense-in-itself subset; Jd # 0 .

Corollary 2. (37.11) -


1'

X is dense-in-itself;

2'

C ' (X)d i s v a g u e l y d e n s e i n C ' ( X ) ;

3'

c ~ ( x > , i s s e p a r a t i n g on

c(x).

(C(X) , C ' (X))-Duality

205

If X contains no dense-in-itself subset, it is called dispersed or scattered..

-

The equivalence of '1

and 3'

(37.9) can be stated as follows.

(37.12) Corollary 3.

The following are equivalent:

1'

X is dispersed;

2O

c'(x)d

=

0.

in

CHAPTER 8 C"

The i m b e d d i n g o f C(X)

538.

S i n c e C'(X)

(X)

i s an L - s p a c e , t h e r e s u l t s i n P a r t I1 on t h e

d u a l o f an L-space a l l a p p l y t o C"(X). existence of C(X). consider C(X) do s o .

Since

C l ( X )

The new f a c t o r i s t h e

i s s e p a r a t i n g on C ( X ) , w e c a n

as a R i e s z s u b s p a c e of

C1l(X)

(5131, and w e w i l l

We h a v e m o r e :

( 3 8 . 1 ) The u n i t

C(X)

i n C1'(X)

n(X)

of C(X)

i s a l s o t h e u n i t o f C"(X).

Thus

i s an MIL-subspace o f C " ( X ) .

T h i s f o l l o w s from ( 2 1 . 3 ) and ( 2 3 . 1 ) .

As w e p o i n t e d o u t i n 513, w h i l e t h e i m b e d d i n g o f C ( X )

C"(X)

in

p r e s e r v e s t h e suprema and i n f i m a o f f i n i t e s e t s ( t h e

l a t t i c e o p e r a t i o n s ) , it does n o t , i n g e n e r a l , p r e s e r v e t h o s e of i n f i n i t e s e t s .

(Thus i t i s n o t o r d e r c o n t i n u o u s . )

Hence-

f o r t h , t h e n o t a t i o n f = V A w i l l a l w a y s mean t h a t f i s t h e supremum o f A i n C " ( X ) ,

e v e n when A a n d f a r e b o t h i n C ( X ) . 206

C" (X)

And s i m i l a r l y f o r f

-f

c1

f.

207

I f we w a n t t o i n d i c a t e t h a t t h e s e

r e l a t i o n s h o l d i n C(X), w e w i l l w r i t e " f

''f

a

+

=

V A - i n C(X)" a n d

f i n C(X)".

( 3 8 . 2 ) Under t h e r e s t r i c t i o n o f t h e form ( * , . ) on C"(X)

t o C"(X)

x

K ( C ' ( X ) ) , Cf'(X)

=

x

C'(X)

A ( K ( C ' ( X ) ) ) , t h e s p a c e o f bounded

a f f i n e f u n c t i o n s on K ( C l ( X ) ) , a n d C(X) i s t h e s u b s p a c e o f vaguely continuous ones.

The f i r s t s t a t e m e n t was e s t a b l i s h e d i n ( 2 3 . 4 ) ,

the second,

i n (36.1).

S i n c e C"(X) i s a n M a - s p a c e , norm b o u n d e d n e s s i n Cl'(X) a n d o r d e r b o u n d e d n e s s i n C"(X)

are equivalent.

Thus

w e can

s i m p l y r e f e r t o a s e t ( o r a l i n e a r mapping) a s "bounded",

w e w i l l do s o .

and

( A l l t h i s of course a l s o h o l d s i n C(X).)

We r e c o r d a f u r t h e r s h a r p e n i n g o f t h e D i n i Theorem.

In

s p i t e o f t h e c o n v e n t i o n on o r d e r c o n v e r g e n c e a b o v e , w e a c t u a l l y do u s e t h e p h r a s e " i n Cf1(X)" i n 4'

below.

This i s t o emphasize

the contrast with the other three statements.

(38.3)

( D i n i Theorem).

For a bounded monotonic n e t { f } i n

C(X) a n d f E C ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t :

1'

I f a } c o n v e r g e s t o f normwise;

'2

{f } converges t o f weakly;

'3

{ f a ) converges t o f simply;

c1

c1

208

Chapter 8

' 4

{ f a } o r d e r c o n v e r g e s t o f i n C"(X).

P-r o o f .

The new s t a t e m e n t i s 4'.

T h a t 1'

implies 4

0

And, s i n c e e v e r y e l e m e n t o f C ' (X) i s o r d e r

f o l l o w s from ( 9 . 6 ) .

i m p l i e s .'2

c o n t i n u o u s on c l l ( x ) , 4'

QED

5 3 9 . Some s i m p l e s e q u e n c e s p a c e s

Some s e q u e n c e s p a c e s

w i l l r e c u r t h r o u g h o u t t h e work.

To

a v o i d u n n e c e s s a r y r e p e t i t i o n , w e c o l l e c t t h e m a t e r i a l on t h e s e here.

I t i s w e l l known, a n d i n any c a s e , e a s i l y v e r i f i e d .

Throughout t h i s g , T i s a f i x e d s e t . k"(T)

i s t h e s p a c e o f bounded f u n c t i o n s on T endowed w i t h

p o i n t w i s e a d d i t i o n , s c a l a r m u l t i p l i c a t i o n , and o r d e r , and w i t h t h e supremum norm.

I t i s a D e d e k i n d complete MI-space, w i t h t h e

f u n c t i o n n ( T ) €or u n i t . T R g e n e r a t e d by I l ( T ) .

Note t h a t i t i s t h e Riesz i d e a l o f

c o ( T ) w i l l d e n o t e t h e s u b s e t o f km(T) c o n s i s t i n g o f t h o s e f u n c t i o n s f which "vanish a t i n f i n i t y " :

set { t E T l If(t)I > i d e a l o f k"(T).

E }

is finite.

f o r every

E

> 0 , the

c o ( T ) i s a norm c l o s e d R i e s z

Note t h a t f o r e v e r y f E c o ( T ) , f ( t ) # 0 f o r

only a countable s e t of t ' s . c(T) i s t h e l i n e a r subspace co(T) I t t u r n s o u t t o be a n MIL-subspace.

+

W

n (T)

of km(T).

C"

(X)

209

1

We h a v e a l r e a d y d e f i n e d XO. (T) ( c f .

535).

We e m p h a s i z e t h a t t h e a b o v e s p a c e s a r e i n d e p e n d e n t o f a n y

t o p o l o g y w i t h w h i c h T may b e endowed.

F o r e x a m p l e , when we

1 1 d i s c u s s e d il ( T ) i n 535, t h e n , a s f a r a s il ( X ) was c o n c e r n e d , X

was s i m p l y a d i s c r e t e s e t . C'(X)a.

( N ot e t h a t t h i s h o l d s a l s o f o r

The t o p o l o g y o f X d o e s n o t e n t e r i n t o i t s p r o p e r t i e s

a s an L - s p a c e .

Under two d i f f e r e n t t o p o l o g i e s on X

t h a t X i s c o m pact - t h e r e s u l t i n g C ' ( X ) , ' s

-

such

w i l l be i d e n t i c a l .

Where t h e y w i l l d i f f e r i s i n t h e v a g u e t o p o l o g i e s . )

We r e c o r d t h e d u a l i t y p r o p e r t i e s o f o u r s p a c e s . w i t h t h e M- s p ace ( b u t n o t a n M I - s p a c e )

(39.1)

il 1( T )

=

( c o ( T ) ) ' = (c0(T))'

We s t a r t

co(T).

= ( c , ( T ) ) ~ under the

b i l i n e a r form , p ) = 'tETf ( t ) p ( t ) 1 f o r f E c o ( T ) , pE R ( T ) . (

(39.2)

am(T)

=

(al(T))' = (al(T)IC

=

(al(T)Ib under t h e

=

(c0(T))".

b i l i n e a r form ( f9p) = CtETf(t)p(t)

f o r fERm(T), PER' (TI.

Thus Rm(T)

And t h e

o r i g i n a l i m b e d d i n g o f co(T) i n am(T) c o i n c i d e s w i t h i t s c a n o n i c a l imbedding i n i t s b i d u a l .

Chapter 8

210

U n l i k e c o ( T ) , c ( T ) i s a n MIL-space h e n c e c a n b c r e p r e s e n t e d

as a C ( X ) .

The a p p r o p r i a t e X i s t h e c l a s s i c a l A l e x a n d r o f f o n e -

p o i n t c o m p a c t i f i c a t i o n a T o f T , T b e i n g c o n s i d e r e d t o have t h e d i s c r e t e - h e n c e l o c a l l y compact - t o p o l o g y . C ( n T ) , and ( c ( T ) ) '

=

C'(aT).

Thus c ( T )

=

Now aT i s d i s p e r s e d , h e n c e

C ' ( C X T )=~ 0 ( 3 7 . 1 2 ) , a n d we h a v e C ' ( a T )

=

C ' ( ~ L T )= ~ gl[aT).

L e t u s d e n o t e t h e new p o i n t i n u T b y t,:

aT

=

T IJ

{t,}

Each t E T i s a n o r d e r c o n t i n u o u s l i n e a r f u n c t i o n a l o n C('T), 1 1 w h i l e tm i s n o t , s o we h a v e C ' ( a T ) = 9. ( T ) 3 W t , , with k (T) ( C ( a ~ )'.

.

=

Summing up :

From ( i i ) and ( 3 9 . 2 ) ,

(39.4)

km(T) c a n a l s o b e w r i t t e n a s a C ( X ) .

The a p p r o p r i a t e X

i s t h e S t o n e - C e c h c o m p a c t i f i c a t i o n RT o f T , T b e i n g c o n s i d e r e d

Let u s d e n o t e t h e p a r t B-T

t o have t h e d i s c r e t e topology. Z:

RT = T I J

Z (Z i s a c l o s e d s e t i n PT, a n d T an o p e n o n e ) .

I t i s e a s i l y v e r i f i e d t h a t ZL ( 3 6 . 7 ) , C ' ( @ T ) = L1(T)

=

3 C'(Z).

co(T).

Hence, by ( 3 9 . 1 ) and

Summing u p ,

by

C" (X)

211

( i i i ) h e r e f o l l o w s from ( 3 9 . 2 ) a n d ( 2 3 . 5 ) .

A s i s c u s t o m a r y , w e w i l l d e n o t e R"(N),

1

R (N) s i m p l y b y Rm,

c o , c , and R

customary t o s a y t h a t c

0

c o ( N ) , c(IN), a n d

1

.

A word o f w a r n i n g : I t i s 1 a n d c h a v e t h e same d u a l s , v i z R .

In t h e p r e s e n t work, t h i s i s an i n c o r r e c t s t a t e m e n t . k1

=

.9

1

(N), b u t c'

= k

1

(&).

W h i l e IN a n d aW

(c0)'

=

h a v e t h e same

c a r d i n a l i t y , t h e p o i n t t m i n aW h a s d i f f e r e n t p r o p e r t i e s t h a n t h e p o i n t s i n IN.

As we p o i n t e d o u t a b o v e , t h e p o i n t s o f N a r e

o r d e r c o n t i n u o u s on c , w h i l e t w i s n o t . N o t e t h a t , a s a c o n s e q u e n c e , we m u s t a l s o d i s t i n g u i s h between Rm

=

R"(N)

a n d R"(aN).

PART I V

THE STRUCTURE OF C ” (X) : BEGINNINGS

212

CIIAPTER 9

THE FUNDAMENTAL SUBSPACES OF C" (X)

5 4 0 . C"(X)a a n d C"(X)d

The d e c o m p o s i t i o n C ' (X)

C ' (X)a 3 C ' ( X ) d

=

i n t o the atomic

and d i f f u s e p a r t s o f C ' ( X ) g i v e s u s t h e d e c o m p o s i t i o n C"(X) = C'l(X)a 0 C"(X)d, w h e r e C"(X)a

[C' ( X ) d ] L a n d C'l(X)d

=

=

By t h e t e r m i n o l o g y a d o p t e d i n 5 2 6 , C1'(X)a and

[C'(X)a]L.

C ' ( X ) a a r e d u a l b a n d s , w h i c h i s why w e u s e t h e n o t a t i o n C"(X)a; and s i m i l a r l y f o r c " ( x ) d and C ' ( X ) d .

We e x a m i n e t h e b a n d C"(X)a. C''(X)a = [ C ' ( X ) a ] ' , we h a v e C ' ( X ) a every fEC"(X)a,f

Remark.

=

a"(X).

Specifically, for

Note t h a t i n t h e p r e s e n t w o r k , w h i l e c(X) a n d =

C"(X)a,

1( X ) , b e i n g t h e p r e d u a l

i s never contained i n the l a t t e r .

The i d e n t i t y o f t h e d u a l p a i r (C"(X)a, (k"(X),a

1

R (X), and

Cx,-XX,xEC' (X)a, ( f , p ) = EXEX( f , x ) X x ( 3 9 . 2 ) .

co(X) a r e c o n t a i n e d i n iz"(X) o f a"(X),

=

=

i s t h e f u n c t i o n on X w i t h t h e v a l u e ( f , x ) on

e a c h xEX; s o f o r i-1 _ I _

S i n c e C'(X)a

1

C1(X)a) w i t h

(X)) means t h a t t h e p r o p e r t i e s o f C ' ( X ) a a r e w e l l

known, a n d e v e n t r a n s p a r e n t .

We r e c o r d two t h a t we w i l l u s e

frequently.

213

Chapter 9

214

(40.1)

F o r A c C"(X)a a n d fEC"(X),

,

the following are cqui-

valent:

'1

f

2'

(f,x)

=

VA; supgEA(g,x) f o r a l l

=

EX.

I t i s i m m e d i a t e t h a t t h e t o p o l o g y on C"(X)u i n d u c e d by

a(C"(X) , C ' ( X ) ) c o i n c i d e s w i t h o ( C " ( X ) a ,

C ' (X)u).

'Thus w e can

s i m p l y t a l k a b o u t t h e v a g u e t o p o l o g y on C''(X)a w i t h o u t ambiguity.

(40.2) Corollary.

F o r a bounded n e t { f } i n C"(X)a a n d c1

f E C " (X)a, t h e f o l l o w i n g a r e e q u i v a l e n t : 1'

{f } o r d e r converges t o f ;

2'

{fa} c o n v e r g e s v a g u e l y t o f ;

3'

(f,x)

c1

=

lim (f

u

x) f o r a l l

a,

XEX.

We a d o p t t h e same c o n v e n t i o n f o r c o m p o n e n t s i n C"(X)u and

C''(X)d t h a t we d i d i n 5 3 5 f o r components i n C t ( X ) u and C'(X),: p r o j a and p r o j d w i l l d e n o t e t h e b a n d p r o j e c t i o n s o n t o C"(X), a n d C"(X)d r e s p e c t i v e l y ; f o r f E C " ( X ) , f a and f d w i l l b e t h e c o r r e s p o n d i n g c o m p o n e n t s ; a n d f o r A c C''(X), we w i l l d e n o t e p r o j a ( A ) and p r o j d ( A ) by Au and A d . Now c o n s i d e r t h e image C(X)u o f C(X) u n d e r p r o j , .

Since

X i s s e p a r a t i n g on C(X), p r o j a c l e a r l y maps C(X) M l - i s o m o r p h i c -

a l l y o n t o C(X)u.

Even m o r e , s i n c e e a c h fEC(X) i s d e f i n e d by

S u b s p a c e s o f C" ( X )

i t s v a l u e s on X , X,

215

i t f o l l o w s t h a t i f R € C ( X ) ~ i s c o n t i n u o u s on

t h e n t h e r e e x i s t s fEC(X) s u c h t h a t f u

=

g.

Otherwise s t a t e d

C(X)a c o n s i s t s p r e c i s e l y o f t h o s e e l e m e n t s o f C"(X)a w h i c h a r e c o n t i n u o u s on X .

i n g e n e r a l , C(X)a # C ( X ) ! !

Nevertheless,

Given f E C ( X ) , f a i s t h e e l e m e n t o f C"(X)u w h i c h a g r e e s w i t h f on e v e r y xEX, s o " l o o k s l i k e " f .

But f a d o e s n o t c o i n c i d e w i t h

f o n C ' ( X ) d i n g e n e r a l ; fa v a n i s h e s on C ' ( X ) d w h i l e f n e e d n o t .

A s t r i k i n g e x a m p l e i s f u r n i s h e d b y n(X)

for X t h e r e a l i n t e r v a l

[0,1].

Il(X)a h a s t h e c o n s t a n t v a l u e 1 on

C"(X)d;

in particular,

=

1 but ( l ( X ) a , u )

=

X b u t v a n i s h e s on

f o r t h e Lebesgue m e a s u r e

u on X ,

(l(X),p)

0.

W h i l e C(X)a i s M a - i s o m o r p h i c w i t h C ( X ) , t h e i m b e d d i n g o f C(X)a i n C t ' ( X ) a

i s i n a d e q u a t e f o r t e l l i n g u s a b o u t t h e imbedd( w h i c h i s e s s e n t i a l l y w h a t t h i s work i s

i n g o f C(X) i n C ' l ( X )

However, w e r e c o r d h e r e a p r o p e r t y f o r w h i c h i t

about).

Remember t h a t V A i n '1

adequate. i n C"(X)

(40.3)

b e l o w means t h e supremum

.

For A c C(X)

1'

f

2'

f a = VA,.

= VA;

_Proof. implies 2

0

and f E C ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t :

.

S i n c e b a n d p r o j e c t i o n p r e s e r v e s s u p r e m a , '1 Assume '2

holds.

By a d j o i n i n g s u p r e m a o f f i n i t e

s u b s e t s o f A t o A , we c a n r e d u c e t h e p r o b l e m t o t h e f o l l o w i n g :

Let I f c 1 } b e a n a s c e n d i n g n e t i n C ( X )

and f € C ( X ) : i f (fa)=l.fu

216

Chapter 9

(in C1l(X)u), show that fa+f (in Ctt(X)).

This follows f r o m

(40.2) and the Dini Theorem (38.3). QED We have been discussing C(X)a.

What about C(X)

The latter may vary from 0 to all of C(X).

n

C”(X)a?

Specifically, from

(37.11):

(40.4) Theorem. The following are equivalent: 1’

X is dense-in-itself;

zo

c(x) n

C ~ I ( X )=~ 0 .

And from (37.12):

(40.5) Theorem. The following are equivalent: 1’

X is dispersed;

2O

C(X) c cyX)a.

C’(X)a C(X)

n

is always separating on C(X),

Ct‘(X)d

=

0.

so we always have

Combining this with (40.4), we have that

for X dense-in-itself, C(X)

has only 0 in common with both

Ct’(X)u and C t t ( X ) d .

A final property of C(X),.

Recall (cf.141

that f o r a

S u b s p a c e s o f C"(X)

217

s u b s e t A o f a R i e s z s p a c e E , A(1) d e n o t e s t h e s e t o f a l l l i m i t s o f n e t s i n A which a r e o r d e r c o n v e r g e n t i n E .

( 4 0 . 6 ) Every e l e m e n t o f C1l(X)a i s t h e l i m i t o f an o r d e r c o n -

v e r g e n t n e t i n c ( x ) ~ : [ C ( X ) ~ I ( ' )= C ~ I ( X ) ~ .

Proof.

C o n s i d e r f€C"(X),,

The collection.&!

and we c a n assume 0 < f < ll(X)u.

= { a } of a l l f i n i t e s u b s e t s of X i s a d i r e c t e d

s e t under i n c l u s i o n .

We w i l l o b t a i n a n e t i n C(X)u, i n d e x e d by

A , which o r d e r c o n v e r g e s t o f .

Given a

=

{ x l , - + . , x n ) , choose

gaEC(X) s u c h t h a t 0 < g (i = l , * * - , n ) . a -< n ( X ) and g a ( x i ) = f ( x i ) Then ( f , x ) = l i m a ( g a , x ) = l i m a ( ( g a ) a , x ) f o r e v e r y xEX. It f o l l o w s from ( 4 0 . 2 )

t h a t (ga)a

-t

f.

QED

As an i l l u s t r a t i o n o f o u r e a r l i e r s t a t e m e n t t h a t t h e

imbedding o f C(X)u i n C 1 l ( X ) u d o e s n o t t e l l u s a b o u t t h e imbeddi n g o f C(X) i n Cll(X), we w i l l s e e l a t e r (548) t h a t , i n c o n t r a s t t o t h e above r e s u l t , C(X)(')

i s a p r o p e r s u b s e t o f C"(X).

5 4 1 . The b a s i c bands and Cg(X)

We f i r s t i n t r o d u c e t h e B a i r e e l e m e n t s o f C"(X) b r i e f l y , i n o r d e r t o e s t a b l i s h t h e i m p o r t a n t theorem ( 4 1 . 3 ) .

The a - o r d e r

c l o s u r e o f C(X) i n Cl'(X) w i l l be c a l l e d t h e B a i r e s u b s p a c e o f

218

Chapter 9

C"(X),

a n d d e n o t e d by B a ( X ) .

B a i r e e l e m e n t s o f C"(X).

-

ordinal

0 c

c(

< fi,{>

I t s e l e m e n t s w i l l be c a l l e d t h e

S e t Ba(X)O

=

C(X), and f o r each

t h e f i r s t uncountable o r d i n a l , l e t

Ba(X)a b c t h e s e t o f a l l l i m i t s o f o r d e r c o n v e r g e n t s e q u e n c e s i n IJ

=

--

Ou f o r a€C(X)'

( a - u)EC(X)'

and u€C(X)',

-C(X)'.

=

This g i v e s i n t u r n t h a t ( a - mu)€C(X)'

Consequently,

, and ( u - a)€C(X)'. and ( u - R A U ) € C ( X ) ~ .

The f i r s t o f t h e s e , f o r e x a m p l e , f o l l o w s from t h e i d e n t i t y R

- mu

=

( a - u)VO and t h e f a c t t h a t C(X)'

For s i m p l i c i t y , we w i l l w r i t e C(X)" similarly for C ( X ) ' ~ ,c and C(X)uu = C(X)',

XIuu,

C(X R'

and C(X)uR.

and C(X)"

i s a sublattice. f o r (C(X)')&, While C(X)"

and =

a r e new i n g e n e r a l .

C(X)'

S u b s p a c e s of C t t ( X )

(42.5)

C ( x p

In e f f e c t , - ( C ( X ) ' ) ~

Since C(X) c C(X)u

=

=

225

-C(X)RU.

(-c(x)')'

=

and C ( X ) c C ( X ) ' ,

((-C(X)p)Q

=

(C(Xp)Q.

we a l s o have

C(X)Q c C(Xp',

(42.6)

c ( X p c C(X)RU.

The argument u s e d f o r ( 4 2 . 4 ) g i v e s u s t h a t C ( X ) I u C ( X ) U R a r e norm c l o s e d a l s o .

and

Unexpectedly perhaps, t h e y a r e ,

i n f a c t , a-order closed:

( 4 2 . 7 ) -~ Theorem. (i)

C ( X ) R U i s c l o s e d under t h e o p e r a t i o n s of t a k i n g

i n f i m a o f a r b i t r a r y s u b s e t s and suprema o f c o u n t a b l e s u b s e t s . ( i i ) C(X)"

i s c l o s e d under t h e o p e r a t i o n s of t a k i n g

suprema o f a r b i t r a r y s u b s e t s and i n f i m a o f c o u n t a b l e s u b s e t s .

Proof.

We show C ( X ) R U i s c l o s e d u n d e r t h e o p e r a t i o n o f

t a k i n g suprema o f c o u n t a b l e s u b s e t s . ifn} c C(X)lU.

fl 5 f 2 nl(X)

Suppose f = v n f n ,

S i n c e C ( X ) R U i s a s u b l a t t i c e , w e c a n assume

z - - -Let .

A

=

> f f o r sonte n ) .

EESC(X)'~1 R > f j

(A i s n o t empty, s i n c e

We show f = h A , h e n c e l i e s i n C ( X ) Q u .

226

Chapter 9

S i n c e C ( X ) k i s a l s o a s u b l a t t i c e , A i s f i l t e r i n g downward, s o i t i s enough by ( 1 0 . 9 ) ,

-___ Lemma.

t o prove t h e

F o r e a c h u E C ' ( X ) + and

E

> 0, there exists LEA

such t h a t ( t , l J >

+

2E.

F o r s i m p l i c i t y o f n o t a t i o n , we assume 0 < f
0 , w > 0 , and p.hv

t h e r e e x i s t s W € C ( X ) ~s u c h t h a t w

2

0 , ( w , p ) = 0, ( w , w )

=

0,

> 0.


Proof.

I[ u[[ =

We c a n a s s u m e

I _

I I v / [ = 1.

229

We w i l l d e f i n e , b y

i n d u c t i o n , a s e q u e n c e I f n ] i n C(X)+ s a t i s f y i n g (i)

> f2 >...9 fl -

(ii)

(fn,p>

5

(n = 1 , 2 , . . - ) ,

l/zn+l

(iii) (fn,u> > 1/2

+

(n

1/2"+'

=

1,2,...)

Choose f l , g l E C ( X ) + s u c h t h a t fl

+

(f,,!J> (cf.

(10.5)).

fl,

.,fn-l

g1 = -t

m i ,

(gp") I 1 / 2

Then f l h a s t h e t h r e e p r o p e r t i e s .

have been chosen.

fn

+

(fn,u)

gn +

=

Assume

Choose f n , g n E C ( X ) +

such t h a t

fn-y

(gn,") 5

Then ( f n , v ) = ( f n - l , v ) 1/2 + 1/2"+l,

2

- ( g n , v ) 2. 1 / 2

+ l/Zn

- 1/2"+'

and t h u s f n h a s t h e t h r e e p r o p e r t i e s .

=

w = A f n n

i s now t h e d e s i r e d e l e m e n t o f C(X)u. QED

(43.2)

(Up-down-up T h e o r e m ) . C(X)2UR = C(X)"R"

= C"

(X) .

P ro o f . The c o r e o f t h e t h e o r e m i s c o n t a i n e d i n t h e

f o l l o w i n g Lemma ( c f . 5 4 1 f o r a d i s c u s s i o n o f C:(X)).

P r o o f o f t h e Lemma.

.

2 30

Chapter 9

( I ) I t i s e n o u g h t o show

c (X I n e f f e c t , s u p p o s e t h i s i n c l u s i o n h o l d s ; we show t h a t

I i s a Dedekind

t h e n , f o r e v e r y b a s i c band I , I, c C(X)RU. complete MI-space w i t h I ( X ) ,

f o r u n i t , hence, by the

F r e u d e n t h a l Theorem ( 1 7 . 1 0 ) , I + i s t h e norm c l o s u r e o f t h e s e t of p o s i t i v e l i n e a r combinations o f elements of&(ll(X))

n

I.

S i n c e C(X)Qu i s a norm c l o s e d w e d g e , t h i s e s t a b l i s h e s ( I ) . CL(X); w e h a v e t o show e E C ( X ) e U .

So c o n s i d e r e E t ( n ( X ) )

e l i e s i n a b a s i c b a n d , s o e = ll(X)v f o r some v E C ' ( X ) , a n d we c a n t a k e p € C ' (X)+ w i t h

Il(X),

EC(X)9'u

11 1111

1.

=

The s t a t e m e n t t h a t

i s e q u i v a l e n t t o : n(X) - Il(X),IEC(X)S.

We show

t h i s l a s t statement.

(11) F o r e v e r y v > 0 s u c h t h a t

ponent e(v) of

n(X)

,IAV

= 0,

t h e r e e x i s t s a com-

,

s u c h t h a t e(v)EC(X)"

(e(v),u) = 0,

( e ( v ) , v > > 0.

Let w be 'the u s c e l e m e n t o b t a i n e d i n ( 4 3 . 1 ) , and s e t e ( v ) = I ( X ) w ( t h e c o m p o n e n t o f n(X) i n t h e b a n d g e n e r a t e d b y w ) Then e ( v )

=

vn[(nw)An(X)],

hence l i e s i n C(X)UR, and c l e a r l y

has t h e desired properties. We c o m p l e t e t h e p r o o f o f t h e Lemma b y s h o w i n g t h a t n ( x ) - n ( x l v = v C e ( v ) Iv > 0 ,

,IAV

D e n o t e t h e supremum on t h e r i g h t b y d . d 5 n(X) - n(X),.

Suppose ( l t ( X )

=

01. dAlt(X)p = 0 , h e n c e

- n(x)u)

- d > 0.

Let I b e

t h e band i t g e n e r a t e s and J t h e band o f Cl(X) d u a l t o I . Choose v E J , v > 0 .

Then pAv

=

0 , s o we have a c o r r e s p o n d i n g

2 31

S u b s p a c e s o f C" (X)

e(u).

Rut t h e n e ( v ) A d

=

0 , contradicting the definition of d.

We c a n now e s t a b l i s h t h e t h e o r e m ( 4 3 . 2 ) .

Every element

o f C"(X)+ i s t h e supremum o f some s u b s e t o f C:(X)+.

t h e Lcmma, C"(X)+ c C(X) fEC(X)"'. i n C(X)

lies in

.

Hence b y

C o n s i d e r a n y f E C " ( X ) : w e show

N o t e f i r s t t h a t f o r e v e r y n , nn(X) a n d -nll(X) l i e

,

2

Choose n s u c h t h a t f

h e n c e i n C(X)'"'.

Then ( s i n c e C(X)'up c(x)'~'.

0 u'

-nn(X).

i s a wedge) f + n n ( x ) > 0, hence l i e s i n

[Ience, f i n a l l y , f

=

(f

+

nn(x))

+

(-nn(x)) also

c(x)'~'. QEII

The Up-down-up t h e o r e m i s c o n t a i n e d i n ___ Note. [25].

my p a p e r

P e d e r s o n ' s g e n e r a l Up-down-up t h e o r e m f o r C * - a l g e b r a s

appeared i n 1972 [41].

F r e m l i n p u b l i s h e d h i s Up-down-up-down

t h e o r e m f o r p e r f e c t R i e s z s p a c e s i n 1 9 6 7 [16].

5 4 4 . The s u b s p a c e U ( X )

of universally

integrable elements

We now h a v e

C ( X p c C(X)RU c C(X)RU'

CIX)

=

= C"(X)

C ( x p c C(XpR

C

.

C(XpRU

Of t h e s e , t h e i n t e r m e d i a t e members C(X)'

a n d C(X)Ru o f t h e t o p

232

Chapter 9

row a r e o n l y w e d g e s , n o t l i n e a r s u b s p a c e s , and s i m i l a r l y f o r t h e i n t e r m e d i a t e members C(X)u a n d C(X)" Now C(X)u

=

-C(X)',

s o C(X)'

n

l a r g e s t one c o n t a i n e d i n C (X) s i m i l a r l y f o r C(X)'' subspaces.

2!

(lC(X)U'.

C(X)'

o f t h e b o t t o m row.

i s a l i n e a r subspace, the

( e q u i v a l e n t l y , i n C(X)');

and

They a r e e a c h , i n f a c t M I -

We t h u s h a v e t h e c h a i n o f Ma- s u b s p a c e s :

c(x) c

c ( x l Rn

n C ( X ) ~ c'

c(xiu c C ( X ) ' ~

Have w e f o u n d new MIL-subspaces?

(44.1) Theorem. -

C(X)'

n

C(X)'

cll(x).

The f i r s t one i s n o t new:

= C(X).

P ro o f . We n e e d o n l y show t h e l e f t s i d e i s c o n t a i n e d i n C(X)

*

Lemma. ( i ) E v e r y f€C(X)'

i s vaguely lowersemicontinuous

on K ( C ' ( X ) ) . ( i i ) Every f€C(X)'

i s v a g u e l y u p p e r s e m i c o n t i n u o u s on

K(C'CX1).

I n e f f e c t , t h e r e i s a n e t { f } i n C(X) s u c h t h a t f t f . c1

I t follows ( f , p ) = s u p ( f a , p ) f o r a l l pEK(C'(X)). f

Is

0.

c1

Since the

a r e v a g u e l y c o n t i n u o u s on K ( C ' ( X ) ) , t h i s g i v e s u s ( i ) ,

( i i ) i s proved s i m i l a r l y . I t f o l l o w s f r o m t h e Lemma t h a t e v e r y f€C(X)'

n

v a g u e l y c o n t i n u o u s on K ( C ' ( X ) ) , h e n c e by ( 3 6 . 1 ) , l i e s

C(X)u i s

Subspaces of C"(X)

233

in C ( X ) . QED

Our c h a i n o f MIL-subspaces i s now r e d u c e d t o

The m i d d l e member

new, i n g e n e r a l d i s t i n c t from b o t h C(X)

and C r f ( X ) . We s h a l l s e e t h a t i t can be i d e n t i f i e d w i t h t h e f a m i l y o f f u n c t i o n s on C which a r e i n t e g r a b l e w i t h r e s p e c t t o e v e r y Radon m e a s u r e .

( T h i s i s a l r e a d y foreshadowed by t h e f a c t

t h a t an e l e m e n t o f C"(X) supremum o f

USC

l i e s i n i t i f and o n l y i f i t i s a

e l e m e n t s and an infimum o f Rsc o n e s . )

Con-

s e q u e n t l y , we w i l l c a l l i t s e l e m e n t s t h e u n i v e r s a l l y i n t e g r a b l e e l e m e n t s o f C"(X),

( 4 4 . 2 ) -____ Theorem.

and w e w i l l d e n o t e i t by U(X).

U(X)

i s a a - o r d e r c l o s e d MI-subspace o f C"(X).

The a - o r d e r c l o s e d n e s s f o l l o w s from ( 4 2 . 8 ) .

U(X) i s p e r h a p s t h e most i m p o r t a n t MI-subspace o f C"(X) a f t e r C(X)

itself.

I t w i l l come t o d o m i n a t e o u r work.

2 34

Chapter 9

5 4 5 . The s u b s p a c e s s ( X ) and S(X)

We a r e s t i l l l o o k i n g f o r MIL-subspaces. a R i e s z s u b s p a c e o f C"(X) t i o n , u s i n g C(X)'

= -C(X)'

(42.1).

C(X)'

- C(X)'

is

A s t r a i g h t f o r w a r d computa-

a n d t h e f a c t t h a t b o t h c o n t a i n IL(X).

gives u s :

(45.1)

The f o l l o w i n g R i e s z s u b s p a c e s c o i n c i d e : C(X) R - C(X) 9

C(XIu - C ( x I u , (C(Xl'l+

- (C(X)')+,

a n d (C(X)')+

-

We w i l l d e n o t e t h i s common s u b s p a c e by s ( X ) .

(C(X)u>+.

While i t i s

a R i e s z s u b s p a c e c o n t a i n i n g IL(X), i t i s , i n g e n e r a l , n o t norm c l o s e d , s o i s n o t an MI-subspace. Appendix t o C h a p t e r 1 1 . )

(We g i v e an e x a m p l e i n t h e

I t s norm closure

g

an MIL-subspace.

We w i l l d e n o t e i t by S ( X ) . We now h a v e t h e c h a i n o f M a - s u b s p a c e s :

C(X) c S(X) c U(X) c C'l(X)

I n g e n e r a l , S(X) i s d i s t i n c t f r o m b o t h C(X) a n d U(X), s o we h a v e i s o l a t e d a s e c o n d new M I - s u b s p a c e .

C(X)RU - C(X)'U s p a c e o f Cl'(X).

(= C(X)U'

- C(X)UR) i s a l s o a R i e s z s u b -

I t c o n t a i n s U(X> o f c o u r s e , a n d a r e a s o n a b l e

c o n j e c t u r e m i g h t be t h a t i t i s d i s t i n c t from C"(X). n o t know w h e t h e r o r n o t t h i s i s t r u e .

,

We do

2 35

Subspaces o f C"(X)

5 4 6 . Dedekind c l o s u r e s

B e c a u s e o f i t s l a t e r i m p o r t a n c e , we s i n g l e o u t a n o p e r a t i o n w h i c h we h a v e u s e d a b o v e .

Given a s u b s e t A o f a R i e s z

s p a c e E , b y t h e D e d e k i n d c l o s u r e o f A i n E , we w i l l mean t h e s e t o f e l e m e n t s o f E w h i c h a r e e a c h b o t h t h e supremum o f some s u b s e t o f A a n d t h e infimum o f some s u b s e t o f A .

Otherwise R s t a t e d , t h e Dedekind c l o s u r e o f A i n E i s t h e s e t A AU. The D e d e k i n d c l o s u r e o f A o f c o u r s e c o n t a i n s A .

If it

c o i n c i d e s w i t h A , we w i l l s a y t h a t A i s D e d e k i n d c l o s e d i n E .

We l i s t some e a s i l y v e r i f i e d p r o p e r t i e s . c l o s u r e o f a s e t i s Dedekind c l o s e d .

The D e d e k i n d c l o s u r e o f

a Riesz subspace i s a g a i n a R i e s z s u b s p a c e s . a u t o m a t i c a l l y Dedekind c l o s e d .

The D e d e k i n d

A Riesz ideal is

If a subset A of E i s , a s an

o r d e r e d s e t , D e d e k i n d c o m p l e t e , t h e n A i s Dedekind c l o s e d i n E . The c o n v e r s e i s f a l s e ( c f .

( 4 6 . 3 ) b e l o w a n d t h e Remark f o l l o w -

ing i t ) .

( 4 6 . 1 ) I f a l i n e a r s u b s p a c e F o f C''(X)

contains ll(X),

D e d e k i n d c l o s u r e o f F i s norm c l o s e d .

I t follows t h a t

(i)

then the

F a n d i t s norm c l o s u r e h a v e t h e same D e d e k i n d c l o s u r e ,

and ( i i ) i f F i s D e d e k i n d c l o s e d , t h e n i t i s norm c l o s e d .

Proof.

An e x a m i n a t i o n o f t h e p r o o f o f ( 1 6 . 6 ) shows t h a t

2 36

Chapter 9

F t h e r e need o n l y b e a l i n e a r s u b s p ace c o n t a i n i n g

I (we do n o t

need t h e sequ e n c e s o b t a i n e d t o b e m o n o to n ic) . QED

(46.2) Corollary.

I f a R i e s z subspace of C"(X)

contains IL(X),

t h e n i t s Dedekind c l o s u r e i s a Dedekind c l o s e d M I - s u b s p a c e .

I n o u r s e a r c h f o r new M I - s u b s p a c e s , what a b o u t t h e Dedekind c l o s u r e s o f o u r t h r e e M I - s u b s p a c e s C ( X ) , U(X)?

The r e s u l t s a r e n e g a t i v e .

(44.1) t h a t C(X)

(46.3) C(X)

Remark.

S ( X ) , and

We h a v e a l r e a d y shown i n

i s Dedekind c l o s e d .

We r e s t a t e i t f o r m a l l y :

i s Dedekind c l o s e d i n C l ' ( X ) .

In general, C(X)

i s f a r from b e i n g Dedekind

complete.

( 4 6 . 4 ) The Dedekind c l o s u r e o f S ( X )

i n C''(X)

i s U(X)

(so U(X)

i s Dedekind c l o s e d i n C " ( X ) ) .

Proof.

By ( 4 6 . 1 ) , i t i s enough t o show t h a t t h e Dedekind

closure of s ( X ) U(X)

and t h e

i s U(X).

T h i s f o l l o w s from t h e d e f i n i t i o n o f

S u b s p a c e s o f Cll(X)

-____ Lemma.

s(x)'

237

= c ( x ) ' ~ , s ( ~ ) ' = c(x)".

C(X)' c s ( X ) , s o C(X)RUc S ( X ) ~ . For t h e o p p o s i t e i n -

c(x)'

c l u s i o n , s ( ~ )= (42.6)

=

C(X)",

-

c(x)'

=

+ c(x)'

c(x)'

c c(x)RU + c(x)RU

s o S ( X ) ~c C(X)RU. QE D

5 4 7 . The B o r e l s u b s p a c e Bo(X)

What do we o b t a i n i f we r e p e a t t h e o p e r a t i o n s t h a t we have c a r r i e d o u t i n t h i s c h a p t e r , b u t s t a r t i n g t h i s t i m e w i t h S(X) instead of C ( X ) ?

(47.1)

(i)

On t h e w h o l e , l i t t l e t h a t i s new.

S(X)'

= U(X)

R

= C(X)''.

( i i ) s ( x ) ~= u ( x ) ~= c ( x ) ' ~ .

The v e r i f i c a t i o n i s s i m p l e .

The 0 - o r d e r c l o s u r e o f S ( X ) , however, d o e s g i v e u s somet h i n g new.

We w i l l c a l l t h i s a - o r d e r c l o s u r e t h e B o r e l s u b -

=ace - o f C"(X),

and d e n o t e i t b y Bo(X).

c a l l e d t h e B o r e l e l e m e n t s o f C"(X). d e f i n e t h e c l a s s e s Bo(X),

( 4 1 . 1 ) g i v e s us:

(0

5

c1

I t s e l e m e n t s w i l l be

A s w i t h Ba(X), we c a n

2 a).

The argument u s e d f o r

238

Chapter 9

( 4 7 . 1 ) Bo(X) i s a 0 - o r d e r c l o s e d MIL-subspace o f C'l(X). Bo(X),

Every

i s a n MIL-subspace.

Note t h a t ( i ) Bo(X) i s a l s o t h e 0 - o r d e r c l o s u r e o f s ( X ) , and ( i i ) Ba(X) c Bo(X) c U ( X ) .

( i ) f o l l o w s f r o m ( 1 6 . 5 ) , and

( i i ) from C(X) c S(X) c U(X) t o g e t h e r w i t h t h e f a c t t h a t U(X)

is 0-order closed (44.2).

548. C(X) ")

and C(X) ( 2 )

I n o u r s e a r c h f o r new MIL-subspaces, w e h a v e s o f a r i g n o r e d an o b v i o u s s e t o f c a n d i d a t e s : c C ( X ) ( 1 ) , C ( X ) ( 2 ) , . . . }

(cf. §54,5).

One c o n s e q u e n c e o f t h e Up-down-up t h e o r e m ( 4 3 . 2 ) i s t h a t C(X)(3)

=

C"(X).

So we n e e d o n l y e x a m i n e C(X)(')

A g a i n , we h a v e n o t h i n g new. U(X)

and C ( X ) ( 2 ) .

We show i n t h i s 5 t h a t C ( X ) ( l )

( a l l r o a d s l e a d t o U[X)) and C(X)(')

ment on t h e i d e n t i t y g i v e n a b o v e ) .

=

C"(X)

=

(an improve-

Both p r o o f s a r e n o n t r i v i a l ,

and we f i r s t r e c o r d some g e n e r a l t h e o r e m s i n f u n c t i o n a l analysis.

The f i r s t i s t h e Hahn-Banach t h e o r e m i n t h e f o r m we

w i l l use i t .

(48.1)

(Hahn-Banach T h e o r e m ) .

L e t E b e a l o c a l l y convex s p a c e .

I f t h e convex s u b s e t s A , B o f E h a v e t h e p r o p e r t y t h a t t h e c l o s u r e of A - B does n o t c o n t a i n 0 , then t h e r e e x i s t s a cont i n u o u s l i n e a r f u n c t i o n a l @ on E s u c h t h a t

Subspaces o f C"(X)

2 39

Let K b e a compact convex s u b s e t o f a l o c a l l y

( 4 8 . 2 ) Lemma.

convex s p a c e E , and KO a c l o s e d convex s u b s e t o f K .

Suppose

f i s a c o n v e x l o w e r s e m i c o n t i n u o u s f u n c t i o n on K , a n d

(i)

( i i ) g i s a c o n c a v e u p p e r s e m i c o n t i n u o u s f u n c t i o n on K O s u c h t h a t g ( a ) < f ( a ) f o r a l l aEKo.

Then t h e r e e x i s t s a n

a f f i n e c o n t i n u o u s f u n c t i o n h on K s u c h t h a t h ( a ) > g ( a ) f o r a l l aEKo,

(i)

( i i ) h ( a ) < f ( a ) f o r a l l aEK.

P roof.

Form t h e p r o d u c t s p a c e E

x

IR ( w i t h e l e m e n t s ( a , ) , ) ) ,

and s e t G = I(a,1)(aEKo F = t(a,?,)la€K,

7

x x

f(a13.

I t i s e a s i l y v e r i f i e d t h a t G and F a r e convex s e t s .

We show

S u p p o s e a n e t { ( a ?,a)} a' i n F - G c o n v e r g e s t o 0 ( i n t h e t o p o l o g y o f E x lR). F o r e a c h

t h a t 0 i s n o t i n t h e c l o s u r e of F - G.

a. (aa,Xa)

a n d ( c ,T I E G . a ,K a) E F a c 1 K i s compact, s o , t a k i n g a s u b n e t i f n e c e s s a r y , w e c a n assume =

(ba7Ka) - ( c a 7 n a ) , where (b

{ b a } c o n v e r g e s t o bEK.

s o a c t u a l l y b€Ko.

I t follows { c } a l s o converges t o b , c1

Choose r , s E IR s u c h t h a t g ( b ) < r < s < f ( b ) .

Then e v e n t u a l l y g ( c ) < r a n d f ( b ) > s , h e n c e K - IT > f ( b a ) a a a a This c o n t r a d i c t s t h e assumption t h a t t h e g(c,) > s - r > 0. n e t {la}

} converges t o 0 inlR. a I t f o l l o w s from ( 4 8 . 1 ) t h a t t h e r e e x i s t s a c o n t i n u o u s =

{

K

~ -

linear functional

on E

x

lR, a n d t E R , s u c h t h a t

240

Chapter 9

(i)

( (a,

SUP

(a, X)EG Now ( E

x

XI ,Q)

IR)'

< t
0 : i n e f f e c t , s u p p o s e u aEKo, u f ( a ) (a,$)

+

5

0; then f o r every

5 ug(a) hence ( ( a , f ( a ) ) , Q > = ( ( a , f ( a ) ) , ( $ , u ) )

uf(a) 5 ( a , $ )

+

ug(a)

=

=

((a,g(a)),O), contradicting ( i ) .

F o r s i m p l i c i t y , assume u = 1 .

Then ( i ) becomes

In p a r t i c u l a r , (a,+)

+

g(a) < t < ( b , $ )

+

f(b)

f o r a l l aEKo, bEK. The f u n c t i o n h on K d e f i n e d by h ( b ) = t - ( b , + ) now h a s the desired properties. QED

Let K b e a compact c o n v e x s e t i n a l o c a l l y

(48.3) Corollary.

convex s p a c e .

Then f o r a f u n c t i o n f o n K , t h e f o l l o w i n g a r e

equivalent:

'1

f i s t h e p o i n t w i s e supremum o f a s e t A o f c o n t i n u o u s

a f f i n e f u n c t i o n s on K ; f i s a lowersemicontinuous convex f u n c t i o n .

2'

Assume 1' h o l d s .

P roof. -

Then f i s c l e a r l y l o w e r s e m i -

c o n t i n u o u s ; w e show i t i s c o n v e x . and b i n K , X and show f ( c ) Kg(b)

5

Xf(a)

5 Xf(a)

+

Suppose c

K

non-negative, and X +

+

Kf(b).

K

Xa

= =

+

Kb, w i t h a

We h a v e t o

1.

F o r e v e r y g 6 A , g ( c ) = Xg(a)

K f ( b ) ; h e n c e SupgEAg(C)

5 lf(a)

+

Kf(b).

+


241

S i n c e t h e l e f t s i d e i s f ( c ) , we are through. Now as s um e 2'

Consider anEK

holds.

a n d 1 < f ( a o ) ; w e show

t h e r e e x i s t s a c o n t i n u o u s a f f i n e f u n c t i o n h on K s u c h t h a t h ( a ) < f ( a ) f o r a l l aEK a n d h ( a o ) > 1. t h e p r e s e n t K , KO c o n s i s t o f t h e p o i n t a

I n ( 4 8 . 2 ) , l e t K be 0'

f be t h e given f ,

a nd g b e t h e f u n c t i o n on K O d e f i n e d by g ( a o )

=

1.

Then t h e

conditions there are s a t i s f i e d , so the h given there i s the d e s i r e d one. QED

Remark.

The a b o v e o f c o u r s e h o l d s w i t h "supremum"

p l a c e d b y " inf i m um " ,

" convex" b y " c o n c a v e " ,

re-

and "lowersemi-

c o n t i n u o u s " by " u p p e r s e m i c o n t i n u o u s " .

t u r n t o C'l(X).

We

The f o l l o w i n g i s e a s i l y d e r i v e d f r o m

( 4 8 . 3) ( a n d ( 3 8 . 2 ) ) .

For f E C " ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t :

( 4 8 . 4 ) Theorem. lo

f EC ( X ) R ;

2'

f i s v a g u e l y l o w e r s e m i c o n t i n u o u s on K ( C ' (XI).

We w i l l n e e d t h e f o l l o w i n g s p e c i a l c a s e o f t h e i n s e r t i o n

theorem o f D.A. Edwards [ 1 5 ] .

( 4 8 . 5 ) Theorem. -___F

(Edwards).

We u s e h i s p r o o f .

F o r e v e r y p a i r U E C ( X ) ~ ,R€C(X)

R

242

Chapter 9

such t h a t u < R , t h e r e e x i s t s fEC(X) s u c h t h a t u < f < k.

Proof.

We c o n s i d e r o n l y t h e v a g u e t o p o l o g y on K(C'(X)),

w e w i l l o m i t t h e m o d i f i e r "vague". K(C'(X))

and

so

A l s o we w i l l w r i t e K f o r

f o r ll(X).

I t s u f f i c e s t o e s t a b l i s h the following:

( I ) T h e r e e x i s t s a s e q u e n c e {f,}

(n

=

0,1,2,...) i n C(X) s u c h

that

u - 2-nn

-

f n -< R

+

z-"n

(n

=

0,1,2;.*)

(n

=

1,2,...).

and < fn f n - 1 - 2-"n -

5 fn-l

+ z-"n

Let u s show f i r s t t h a t t h e t h e o r e m f o l l o w s from ( I ) .

t h e s e c o n d s e t o f i n e q u a l i t i e s , t h e s e q u e n c e {€,I Cauchy, h e n c e norm c o n v e r g e s t o some f € C ( X ) .

By

i s norm

T h e n , by t h e

f i r s t set of i n e q u a l i t i e s ,

u - 2-nn < f < R

+

f o r a l l n.

I t follows u < f < R.

We p r o c e e d t o e s t a b l i s h ( I ) , o r r a t h e r t h e s t r o n g e r property:

We w i l l c h o o s e f o t o s a t i s f y p r o p e r t y ( i i ) b e l o w , a n d t h e

r e m a i n i n g f ' s by i n d u c t i o n t o s a t i s f y ( 1 1 ) . n

A c t u a l l y , we w i l l

Subspaces of C”(X) only derive f1 from f O . same

243

‘The derivation of fn from fn-l is the

.

Note first that, by ( 4 8 . 4 ) , for every

ic

> 0, u

uppersemicontinuous affine function on K and R

f

-

K is I ~ an

Kn is a lower-

semicontinuous one, and that

Set

K

=

foEC(X)

1 and a p p l y (48.2) with K O = K (also (38.2)).

We obtain

such that

We proceed to derive fl.

Proof of (iii):

In effect, by (ii), ( f o Since f 0 - (u -

n)

-

(u - ll) ,p) > 0 for all p E K . .

is an ilsc element, it is lowersemicontinuous

244

Chapter 9

on K.

(fO (U

-

-

But K is compact, so there exists 0

-

Since

for all p E K .

x

< 1

such that

I t follows that f o -

An. k - u >

0, we also have R - ( u

-

It)

2 n

Writ-

> XI. -

ing these two inequalities in the form -1 ( u - z n) + xn < fo + 2-'n, -1 < a + z-ln, ( u - 2 n) + An we obtain the first inequality in ( * ) . Again by (ii), ( ( a .

1) - fO,p) > 0 for all p E K .

+

(a

+

f0 is again lowersemicontinuous on K , so there exists 0 < such that ( (a.

(a

+

lL)

-

+

nj) - fO,p) >

for a l l p E K .

ll) < 1 -

It follows that

This gives us the first of the following;

f0 2 A l l .

the second is trivial.

xn

(fO

-

2-ln)

(fO

-

2 - l +~ I n < f0

+

R(uVf).

Another, useful, corollary:

(49.7)

fAg = 0 implies R(f)Au(g)

=

0.

Of the equalities and inequalities obtained so f a r , the binomial ones can be sharpened when one of the elements lies in C ( X ) .

252

Chapter 1 0

I f g € C ( X j , t h e n f o r every fEC"(X),

(49.8)

e(f

+

u(f

g ) = E(f)

+

p,

u(f)

+

g.

+ g)

=

T h i s f o l l o w s from ( 4 9 . 1 ) .

Proof. ___

The f i r s t a n d f o u r t h o f t h e s e a r e s p e c i a l c a s e s o f

By a p p l y i n g ( 4 9 . 5 ) a n d ( 4 9 . 3 ) ,

( 4 9 . 4 ) ; we show t h e s e c o n d . together with k(g) L(f)vp

=

=

g = u ( g ) , we h a v e L ( f V g ) < Q(f)Vu(g) =

k(f)vk(g', < a ( f v g ) , which g i v e s u s e q u a l i t y . QE D

S e t t i n g g = 0 i n ( 4 9 . 9 ) , we o b t a i n

(49.10)

(U(f))+

= U(f+)

(E(f>)+

=

Q(f+)

(u(f))-

=

L(f-j

(k(f))-

=

U(f-)

The O p e r a t o r s u a n d P,

253

Whence,

(49.11)

P roof.

u(f+)vu(f-)

u(f)

=

U(f+) - e ( f - ) .

a(f)

=

a(f')

lu(f)l =

ll(f+)

+

a(f-1

= u[f')va(f-).

la(f)j

PJf+)

+

u(f-)

=

+ u(f-)

I

=

=

a(f')vu(f-).

[u(f+)va(f-)]

u ( f + v f - ) = u ( If

h o l d i n g by ( 4 9 . 4 ) . a(f')

=

lu(f) IV l a ( f ) =

u(f-).

-

I),

v

[a(f+)vu(f-)l

t h e second l a s t e q u a l i t y

For t h e i n e q u a l i t y , k ( l f l ) = G(f+ (a(f))-

(g,(f))++

t h e t h i r d term by u ( f + )

+

=

=

la(f)

1,

Q(f-) - a(lf1)

+

f-)
0 and 6 ( f ) = 0 i f and o n l y i f fEC(X).

Also t h a t 6 ( f ) i s a usc element.

From t h e i d e n t i t i e s

k ( - f ) = - u ( f ) and u ( - f ) = - k ( f ) , w e have 6 ( - f ) more

=

6(f).

Even

The O p e r a t o r s u a n d II

257

The v e r i f i c a t i o n i s s t r a i g h t f o r w a r d .

Proof. [u(f)

+

a(g)l

6 ( f ) - 6(g)

- [il(f)

+

=

[u(f) - a ( f ) ] -

u(g)l 5 u(f

+

g) -

[ ~ ( g )- a ( g ) ]

a(f

+

g)

=

6(f

= +

8).

The o t h e r i n e q u a l i t i e s a r e o b t a i n e d i n t h e same way. QED

I n t e r c h a n g i n g f and g , and u s i n g 6 ( - h ) the

-___ Proof.

a(fvg) = u(fvg) - a(fvg)

=

6 ( h ) , we obtain

258

Chapter 10

QED

We s h a r p e n o n e o f t h e i n e q u a l i t i e s i n ( 5 0 . 2 ) .

(50.5)

S(f

+

g)

5

6(f'Jg)

+

6(fAg) < S(f)

S e t t i n g g = 0 , we o b t a i n :

+

6(g).

The O p e r a t o r s u a n d EL

(50.6)


F o r e v e r y f€C"{X), 6(f)

Setting f

=

259

f+, g

=

=

6(f+)

+

6(f-).

f - , and a p p l y i n g C o r o l l a r y 1, w e

obtain :

(50.7) Corollary 2.

For e v e r y f€C"(X), S(jfl)

C o r o l l a r y 3. (50.8) -

5 6(f).

For e v e r y f € C " . ( X ) , 6(f) < 2u(lfl)

Proof.

Since f + , f - > 0 , we have 6 ( f f ) < u(f+)

and & ( f - ) < u(f-) < u(lf1).

5

u(lf()

Now a p p l y ( 5 0 . 6 ) . QED

( 5 0 . 9 ) I f gEC(X), t h e n f o r e v e r y f € C " ( X ) , S(f

+

g) = S(f).

This i s immediate from (49.8). t h a t f o r e v e r y fEC"(X), 6 ( f ) = s ( l L ( X )

A u s e f u l consequence i s

- f).

Chapter 1 0

260

For a l l f , g E C " ( X ) ,

(50.10)

I)&(f)

- 6 k ) I I 5 211 f

- gll.

Thus t h e o p e r a t i o n & ( - ) i s norm c o n t i n u o u s .

T h i s follows from ( 4 9 . 1 6 .

Under R i e s z homomorphisms, R i e s z s u b s p a c e s show up a s i m a g e s o f R i e s z s u b s p a c e s , and R i e s z i d e a l s a s i n v e r s e images of R i e s z i d e a l s .

I t may b e w o r t h n o t i n g , t h e r e f o r e , t h a t u n d e r

t h e mapping C"(X)---> 6

C"(X),

t h e i n v e r s e images o f R i e s z i d e a l s

a r e Riesz subspaces (6 i s o f course n o t l i n e a r ) :

(50.11) Theorem. -

I f I i s a norm c l o s e d R i e s z i d e a l o f C"(X),

then &-'(I) (i)

i s a n MJl-subspace o f C l ' ( X ) ,

(ii)

contains C(X),

and

( i i i ) i s closed under t h e operations u ( - ) , & ( * ) ,

Proof.

&

(

a

+

.

g ) E I an d , by ( 5 0 . 4 1 ,

Thus & - l ( I ) i s a R i e s z s u b s p a c e .

T h a t i t i s norm

c l o s e d f o l l o w s f r o m t h e norm c o n t i n u i t y o f & ( * )

(50.10).

F i n a l l y , t h a t & - l ( I ) c o n t a i n s 1 f o l l o w s from ( i i ) , w h i c h i s clear.

)

I f s ( f ) E I , t h e n by ( S O . l ) , & ( X f ) E I ; and i f

& ( f ) , 6 ( g ) E I , t h e n , by ( 5 0 . 2 1 , 6 ( f G(fVg)EI.

and

( i i i ) f o l l o w s from t h e e a s i l y v e r i f i e d i n e q u a l i t i e s :

0 < & ( U ( f ) , & ( & ( f ) ) , & ( & ( f5) )s / f ) .

QED

The O p e r a t o r s u a n d R

Kemark. ___

261

( i i ) and ( i i i ) c l e a r l y h o l d even i f I i s n o t

norm c l o s e d . For I a b a n d , w e c a n s a y m o r e :

(50.12)

I f I i s a band o f

Cll(X),

then 6

-1

( I ) i s Dedekind c l o s e d

i n C"(X).

We p r o v e a s t r o n g e r r e s u l t : i f A , B c 6 - ' ( 1 ) , and V A = f

=

A B "modulo I " , t h e n f E A ; ' ( I ) .

A

Precisely,

I f A,B c 6-'(1),

( 5 0 . 1 3 ) Let I b e a band o f C " ( X ] .

5 f 5 B,

A

5

f

5

B,

and (AB - VA)EI, t h e n f E 6 - ' ( 1 ) .

Proof.

We w i l l u s e t h e o b v i o u s

F o r e v e r y kEC"(X), Lemma. -

Now s e t g show ( u ( h )

= VA

and h

= AB.

the following a r e equivalent:

By h y p o t h e s i s

(h - g ) E I .

k(g))EI; s i n c e k(g) < f < u(h), it w i l l follow

-

u ( f ) - p,(f)CI, which i s t h e d e s i r e d r e s u l t . Fix v E I

1

,v

0.

Since 6-'(I)

i s a s u b l a t t i c e , we can

assume A i s f i l t e r i n g u p w a r d s a n d B downwards, h e n c e w e c a n r e p l a c e them by n e t s { g a l a n d { h g } s u c h t h a t g U f g and h U + h .

We

Chapter 10

262

Then { k ( g ) } i s a l s o a n a s c e n d i n g n e t a n d { u ( h ) } a d e s c e n d i n g

8

c1

net, so k(gcl)+k < k(g) and uChS)+u > u(h1. A p p l y i n g t h e Lemma a n d t h e o r d e r c o n t i n u i t y o f v , we h a v e

(!L,v>

=

lima(a(ga),u>

=

lima

= (g,v>

-

and. s i m i l a r l y ,

(u,v>

=

(h,v>

*

Thus

QED

As a n a p p l i c a t i o n o f t h e o p e r a t i o n 6('),

we present a

t h e o r e m o f K r i p k e a n d Holmes [ 3 0 ] ( t h e y d i d i t i n t h e c o n t e x t of ordinary function theoryJ.

(50.14)

F o r e a c h f E C " ( X ) , t h e d i s t a n c e o f f f r o m C(X)

(1/2)11 6 ( f )

11,

Proof. -__

and t h i s d i s t a n c e i s a t t a i n e d .

For e v e r y g E C ( X ) , 6 ( g ) = 0 , hence, by ( 5 0 . 1 0 ) ,

/ I f - gll > (1/2)1[ ~ ( f ) l [ . We show t h e r e e x i s t s gEC(X)

11 f

is

- gjl < (1/2)11 6 ( f ) l l .

D e n o t e (1/2)\1 6 ( f )

11

by r .

such t h a t

Then

The O p e r a t o r s u a n d i! 6(f)
0.

f I , i t f o l l o w s from t h e Lemma i n ( 5 1 . 3 ) t h a t f

V{hEH 1 0 < h < f}.

Since

=

T h i s i s f i l t e r i n g upward, h e n c e , by

D i n i ' s t h e o r e m ( 3 8 . 3 1 , f i s i n t h e norm c l o s u r e o f H , h e n c e i n H. QED

Some a d d i t i o n a l c h a r a c t e r i z a t i o n s o f an R-band:

(51.6) Let I b e a band o f C ' l ( X ) ,

and H = C ( X )

n

I.

J i t s d u a l band i n C l ( X ) ,


The O p e r a t o r s

and R

11

1'

I i s an R - b a n d ;

2'

I i s t h e vague c l o s u r e o f H ;

3'

1 1 i s s e p a r a t i n g on J ;

4O

I

=

267

Ill ;

J.

5'

I

6'

I

i s vaguely closed;

1

= G1

f o r some v a g u e l y c l o s e d b a n d

By ( 5 1 . 5 ) a n d ( 2 6 . 2 ) , 2' -__f'roof. ( 5 1 . 3 ) , hence t o lo.

S u p p o s e 2'

(1IJ.)'

=

I

(IJ.)'

1

.

=

I.

Thus 4'

( H ' )

=

S u p p o s e 5'

=

holds.

Conversely,

=

1

5'

(GL)J.

i f 4'

T h i s s a y s t h a t 2'

(HL)'.

holds.

holds.

then I

i s e q u i v a l e n t t o 3'.

I t f o l l o w s from ( 1 0 . 1 5 )

=

(I

1

holds then I

4'

holds.

iJ. n c ( x )

o f c o u r s e i m p l i e s 6'. =

and (26.2) t h a t =

i m p l i e s 5'.

T h e n , i n t h e d u a l i t y b e t w e e n C(X) and

C ~ ( X ) , ( ( I ~ =) ~I )~~ . But ( I J . )J .

So 4'

J ' , 2'

in

i s e q u i v a l e n t t o 3'

T h i n k i n g o f i t a s a s u b s e t o f C(X), t h i s can b e

w r i t t e n (HI)' tl'

=

of Cl(X).

Then, t h i n k i n g o f H as a s u b s e t o f C " ( X ) ,

holds.

I.

=

Since I

C,

G,

s o 5'

=

Finally,

I

n

c(x) = H.

i f 6'

holds,

holds. QE D

A b a n d I o f C"(X) w i l l b e c a l l e d a u - b a n d i f n ( X ) , usc element.

I f two c o m p o n e n t s e , d o f E(X) s a t i s f y e

+

is a d =

lt(X), t h e n o n e i s a u s c e l e m e n t i f a n d o n l y i f t h e o t h e r i s a n

~ s ecl e m e n t .

I t follows t h a t f o r a b a n d I o f C"(X),

u-band ( r e s p . an k-band)

a u-band).

i f a n d o n l y i f I d i s an k - b a n d ( r e s p .

And i t f o l l o w s f r o m t h i s , i n t u r n , t h a t t h e

c h a r a c t e r i z a t i o n s o f k-bands have "dual"

u- b a n d s

.

I is a

characterizations f o r

Chapter 1 0

268

Whereas f o r an k - b a n d I , C(X)

I plays t h e central r o l e ,

f o r a u - b a n d I , i t i s C(X), t h a t p l a y s t h e c e n t r a l r o l e . can s a y t h i s a n o t h e r way. w i t h C(X)/(C(X)

n

Id).

We

Note t h a t C(X)I c a n b e i d e n t i f i e d

T h u s , where t h e r e i s a o n e - o n e c o r r e s -

pondence b e t w e e n t h e k - b a n d s o f C1'(X) and t h e norm c l o s e d R i e s z i d e a l s o f C(X), we w i l l s e e t h a t t h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e u - b a n d s o f C"(X)

and t h e q u o t i e n t

s p a c e s o f C(X) ( w i t h r e s p e c t t o norm c l o s e d R i e s z i d e a l s ) . One c h a r a c t e r i z a t i o n o f an k - b a n d i s t h a t i t i s t h e s m a l l e s t band o f Cl'(X) c o n t a i n i n g some R i e s z i d e a l o f C(X). T h a t i s t h e c o n t e n t o f 2'

i n (51.3).

Correspondingly a u-band

can b e c h a r a c t e r i z e d a s t h e l a r g e s t band o f C'l(X) " c o n t a i n i n g " some q u o t i e n t s p a c e o f C ( X ) - t h e word " c o n t a i n i n g " b e i n g appropriately defined.

(51.7)

T h a t i s t h e c o n t e n t o f '3

below.

L e t I b e a band o f C'l(X) and J i t s d u a l band i n C ' ( X ) .


I i s a u-band;

2'

J i s vaguely closed;

3'

For e v e r y band I1 o f C"(X), i f (i)

I1

3

I , and

o n t o C(X), i s a b i I1 j e c t i o n ( h e n c e a n M I - i s o m o r p h i s m ) , t h e n I1 = I . ( i i ) t h e p r o j e c t i o n o f C(X)

P roof.

The e q u i v a l e n c e o f 1' and 2'

t h a t o f l o and 5' above b a n d ) .

i n (51.6)

i s a restatement of

( w i t h I t h e r e r e p l a c e d by I d , I t h e

We p r o v e t h e e q u i v a l e n c e o f 1' a n d 3'.

Suppose

The O p e r a t o r s u and

269

,?.,

we h a v e two b a n d s 1,11 w i t h I c 11, s o t h a t ( I 1 ) always have ((C(X), ) I 1 C(X),)

a n d C(X)

n

=

C(X),

( t h a t i s , C(X)

( I , ) d c C(X)

n

o n t o C(X),

w e show ( I l l d

i s o n e - o n e (so C(X)

n

I1 (I,)d

=

h o l d s a n d I1 s a t i s f i e s t h e h y p o t h e s e s o f

=

I d , whence I 1 = I .

h e n c e , by ( 5 1 . 3 ) ,

(Il)d

Conversely, suppose 3

=

n

C(X)

C(X)

o n t o C(X),

=

C(X)

n

Id c

h o l d s ; w e show I i s a u - b a n d .

n

To

t h e band g e n e r a t e d by

d (so i t s d i s j o i n t i s I1). I

(Il)d,

By ( * ) , C(X)

Id.

f i t o u r n o t a t i o n , d e n o t e by

0

onto

0

0

C(X)

We

Id'

Now s u p p o s e 1

(Il)d,

.

And i t i s c l e a r t h a t

Id.

and C(X), a r e M a - i s o m o r p h i c ) i f a n d o n l y i f C(X)

3';

d

projects

I1

n

c I

I1

( * ) t h e p r o j e c t i o n o f C(X)

C(XI

d

By t h e v e r y d e f i n i t i o n o f

I d , s o , by ( * ) , t h e p r o j e c t i o n o f

i s one-one.

I t f o l l o w s f r o m 3'

that I

=

I1.

I1 Sincc (I,)d

i s a n ,?.,-band, I l i s a u - b a n d , s o we a r e t h r o u g h .

QED

We g i v e c h a r a c t e r i z a t i o n s o f k - b a n d s a n d u - b a n d s i n terms

o f t h e i r d u a l b a n d s i n C'(X).

(51.8)

Let I b e a b a n d o f C"(X)


I i s a n R-band;

2'

J = (c(x)

I n such c a s e , I

=

n 11'. (C(X)

n

I)".

a n d J i t s d u a l b a n d i n C'(X).

C h a p t e r 10

270

___ Proof.

Set H

=

C(X)

n

( I L ) d , i t f o l l o w s f r o m '4

0

I.

Suppose 1

i n ( 5 1 . 6 ) t h a t .J

holds.

S i n c e ,J

CI(X)/HL

=

=

=

H'.

C o n v e r s e l y , i f J = H I , t h e n fl i s s e p a r a t i n g on LJ, h e n c e , b y 3'

i n ( 5 1 . 6 ) , I i s an R - b a n d . QE D

Remark. -

The a b o v e i s a c t u a l l y ( 3 6 . 7 ) r e s t a t e d .

a n d J i t s d u a l b a n d i n C'CX).

Let I b e a band o f C " ( X )

(51.9)


'1

I i s a u-band;

2'

J

=

(C(X)I)'.

In such case, I

Proof.

=

(C(X)I)I'.

S e t II

I _ _

(C(X)I)l t o <J

=

HL.

n

I

d

.

Then C(X),

=

Thus we n e e d o n l y show t h a t 1'

Since J

HL.

=

C(X)

=

=

d [I )

I'

t h e i d e n t i t y ,J

=

C[X)/H, h e n c e

is equivalent IIL

is equivalent

t o I d b e i n g an k - b a n d ( 5 1 . 6 ) , s o w e a r e t h r o u g h .

QED

Given a b a n d I o f C'l(X), l e t <J b e i t s d u a l b a n d i n C ' ( X ) , and Q

=

J

n

X.

In g e n e r a l , I cannot be r e c a p t u r e d from

i n d e e d , Q may e v e n b y e m p t y .

Q

-

I f I i s a u - b a n d o r an R - b a n d ,

c a n b e r e c a p t u r e d f r o m Q: however, i t -

(51.10).

I f I i s a u - b a n d , w i t h d u a l b a n d J , t h e n <J

n

X is a


c l o s e d s e t Z o f X , C(X),

=

J

C(Z),

271

C ' ( Z ) , and I =

=

C'l(Z).

T h i s i s e a s i l y v e r i f i e d from t h e m a t e r i a l i n t h e p r e s e n t section (cf.

(51.11)

(36. 7 ) ) .

I f I i s an f - b a n d , w i t h d u a l b a n d . J , t h e n J

I

o p e n s e t W o f X a n d C(X)

n

X i s an

Cw(W).

=

This also is e a s i l y verified.

-__ Remark.

I n § 5 6 , we g i v e w h a t i s p r o b a b l y t h e l a r g e s t

f a m i l y o f bands1 o f C"(X) r e c a p t u r e d from J

w i t h t h e p r o p e r t y t h a t each I can be

X (J t h e b a n d o f

C l ( X )

dual t o I ) .

9 52. A p p l i c a t i o n s t o g e n e r a l bands

Given a band I o f C " ( X ) ,

t h e n , i n g e n e r a l , Il(X),

n e i t h e r an Rsc n o r u s c e l e m e n t . g e n e r a t e d by k(Il(X),) k-band i n I .

is

IIowever, t h e R i e s z i d e a l

i s a n R-band,

and i s c l e a r l y t h e l a r g e s t

And t h e R i e s z i d e a l g e n e r a t e d b y u ( l l ( X ) , )

u-band, and i s c l e a r l y t h e smallest u-band c o n t a i n i n g I .

is a

By

a n a b u s e o f n o t a t i o n , we w i l l d e n o t e t h e f i r s t b a n d b y R ( 1 ) and t h e s e c o n d by u ( 1 ) . From t h e d e c o m p o s i t i o n

Cll(X)

= 1 3 Id,

w e have

272

Chapter 1 0

Ctt(X)

=

(Q(1))

u(1) 3 a(Id). =

u(1

d

O t h e r w i s e s t a t e d , ( ~ ( 1 ) =) ~k ( I d )

(and

1).

We h a v e i m m e d i a t e l y :

( 5 2 . 1 ) For e v e r y b a n d I o f C " ( X ) , a ( I )

i s t h e b a n d g e n e r a t e d by

~ ( xn ) I.

For e v e r y band I o f C t l ( X ) , ( a ( I ) ) L i s t h e

(52.2) Corollary.

vague c l o s u r e o f I

Remark,

L

.

C l e a r l y C(X)

n

k(1)

=

C(X)

s m a l l e s t band f o r which t h i s i s t r u e .

n

I

and k ( I ) i s t h e

I t f o l l o w s a l l t h e bands

b e t w e e n I a n d ~ ( 1 ) h a v e t h e same i n t e r s e c t i o n w i t h C(X), and ~ ( 1 )i s t h e s m a l l e s t b a n d f o r w h i c h t h i s i s t r u e .

For t h e bands between I and u ( I ) , t h e r e i s a c o r r e s p o n d i n g statement: projection

t h e y a l l h a v e t h e "same"

image o f C(X) u n d e r b a n d

- t h e word "same" now m e a n i n g " M I - i s o m o r p h i c " .

We

s t a t e t h i s precisely.

(52.3)

F o r e v e r y b a n d I o f C"(X),the p r o j e c t i o n o f u ( 1 ) o n t o I m p s

u ( I)

M I - i s o m o r p h i c a l l y o n t o C(X),.

Moreover, u ( I ) i s t h e

l a r g e s t band f o r which t h i s i s t r u e .

T h i s f o l l o w s f r o m t h e a b o v e Remark a n d ( * ) i n t h e p r o o f


273

of (51.7).

And from ( 5 2 . 2 ) ,

w e have:

( 5 2 . 4 ) Let J b e t h e b a n d d u a l t o t h e b a n d I o f C"(X).

Then

t h e vague c l o s u r e o f 3 i s t h e band d u a l t o u ( 1 ) .

We c a n now d e s c r i b e (C(X) a r b i t r a r y b a n d I o f C"(X).

n

I ) ' and (C(X),)'

From ( 5 1 . 8 ) a n d ( 5 1 . 9 ) , we h a v e

t h a t (C(X)

n

(C(X)I)'

<J i f a n d o n l y i f I i s a u - b a n d .

=

I)'

=

f o r an

J i f a n d o n l y i f I i s an R - b a n d , a n d

From t h e f i r s t o f t h e s e , we h a v e i m m e d i a t e l y :

( 5 2 . 5 ) F o r e v e r y b a n d I o f C"(X), t o a(11,

(c(x) n

SO

I ) I I

=

(C(X)

n

I ) ' i s t h e band dual

~ ( 1 ) .

The c o r r e s p o n d i n g s t a t e m e n t f o r C(X),

i s less d i r e c t .

( 5 2 . 6 ) L e t I be a b a n d o f C''(X) a n d J i t s d u a l b a n d i n C ' ( X ) .

Then ( C ( X ) , ) '

so (C(X),)"

can be i d e n t i f i e d w i t h t h e vague c l o s u r e o f J.,

can be i d e n t i f i e d w i t h u ( 1 ) .

Denote t h e v a g u e c l o s u r e o f J b y J1. t h e b i l i n e a r form

( ( . , - ) ) on C(X),

x

We h a v e t o d e f i n e

J1 g i v i n g t h e d u a l i t y

2 74

Chapter 1 0

Note f i r s t t h a t i f I i s a u - b a n d , s o t h a t .J i t s c l f i s v a g u e l y c l o s e d , t h e n f o r fCC(X), and l i E t J , ( ( f , ] ~ ) ) = b e i n g t h e c a n o n i c a l b i l i n e a r form on C " ( X ) c a u s e I and .J a r e d u a l b a n d s .

(f,Ll),

XC1(X).

the l a t t e r This be-

Now l e t I b e a g e n e r a l b a n d , and

Denote t h e r e s t r i c p r o h , u ( I ) the p r o j e c t i o n of u(1) onto I. tion of proj s i m p l y by p . So p i s an M l l I , u ( I ) to c ( X )U ( I ) i s o m o r p h i s m o f C(X) o n t o C(X),. Then, f o r f E C ( X ) I and Ll(I)

uEJ1,

the desired valuc ((f,ii)) ((f,lJ))

i s g i v e n by

(P-'(f),lJ)

=

-

And t h e i m b e d d i n g o f f i n t o t h e b i d u a l o f C(X), i s g i v e n by

f

L-, p-l

(f).

I n s t u d y i n g a p a i r o f d u a l b a n d s ( I , J ) , t h e r e i s o f t e n no l o s s i n r e p l a c i n g C'(X) b y t h e v a g u e c l o s u r e ,JI o f <J a n d

( t h e r e f o r e ) C''(X) b y u ( 1 ) . ourselves t o t h e support

O t h e r w i s e s t a t e d , we c a n c o n f i n e

.J1n X

o f J i n s t e a d o f a l l o f X.

We

will do t h i s i n p r a c t i c e b y s i m p l y a s s u m i n g t h a t J i s v a g u e l y d e n s e i n C'(X)

- e q u i v a l e n t l y , t h a t u(1) = C"(X).

553. The Isomorphism theorem

We pointed o u t in 540 that the projection of

Clt(X)u

maps C(X)

bin-isomorphically onto C ( X )

0.

Cl'[X)

onto

. We show now

that, in fact:

(53.11 (Isomorphism theorem). C"(X)u

maps U(X)

Proof. on U(X).

The projection o f C " ( X )

onto

MI-isomorphically onto U(X)a.

We need o n l y show that the projection is one-one

And for this, it is enough to show that for a l l

f,gEU(X), fu

5 g, implies f

-___ Lemma 1.

0 ; we show p i s n o t

Now i n t h e c l a s s i c a l example a b o v e ,

X c o u l d have been any compact s p a c e and p any s t r i c t l y p o s i -

t i v e element o f C1(X)d

.

So we a r e t h r o u g h .

QED

Remark 1.

I t i s n o t d i f f i c u l t t o show t h a t

all

the order

c o n t i n u o u s l i n e a r f u n c t i o n a l on U(X), n o t o n l y t h o s e i n C"(X), l i e i n C ' ( X ) a - o t h e r w i s e s t a t e d , U(X)' Remark 2 .

= C'(X)a.

( 5 4 . 2 3 ) i s a s p e c i a l c a s e o f a g e n e r a l theorem

by M a s t e r s o n [ 5 2 ] .

355. The u n i v e r s a l l y m e a s u r a b l e s u b s e t s o f X

L e t ( 1 , J ) be a p a i r o f d u a l b a n d s . vague c l o s u r e o f J by J1, call J

n

X t h e s e a t of J

be no c o n f u s i o n .

(u(I),J1)

Then, d e n o t i n g t h e

a r e dual bands.

and a l s o t h e s e a t o f L .

Note t h a t J

We w i l l There can

X = { x E X [ . ( f , x ) # 0 f o r some

f E I } ; s o i t c a n be d e f i n e d i n d i f f e r e n t l y u s i n g e i t h e r J o r I . S i m i l a r l y , t h e support of J , be c a l l e d t h e s u p p o r t o f I .

Jln

X,

( c f . 536) w i l l a l s o

Thus t h e s u p p o r t o f J i s s i m p l y

t h e s e a t o f J1 - e q u i v a l e n t l y , t h e s u p p o r t o f I i s s i m p l y t h e s e a t of u ( 1 ) .

C h a p t e r 11

290

We e x t e n d t h e a b o v e t o a r b i t r a r y s u b s e t s . A o f C 1 ( X ) , l e t $J b e t h e b a n d g e n e r a t e d b y A .

Given a s u b s e t Then by t h e --s e a t

w e w i l l mean t h e s e a t o f J , a n d s i m i l a r l y f o r t h e s u p p o r t

__ of A

o f A.

I n p a r t i c u l a r , we c a n t a l k a b o u t t h e s e a t a n d s u p p o r t o f

a s i n g l e element

of Cl(X).

Similarly

f o r t h e s e a t and s u p -

p o r t o f a s u b s e t o f C'l(X), a n d i n p a r t i c u l a r o f a s i n g l e e l e men t

. F o r a band J o f C l ( X ) , w e c l e a r l y h a v e J

.

Note a l s o t h a t f o r a

=

JU

n

X.

t h e s e a t of I i s t h e s e a t

E q u i v a l e n t l y , f o r a band I o f C ' ( X ) , of Ia

.OX

R i e s z i d e a l G o f C f ( X ) , i f we

d e n o t e i t s o r d e r c l o s u r e b y .J, t h e n J

n

X

=

G

X.

The s u p p o r t o f a s e t ( i n e i t h e r Cl(X) o r C"(X))

is, i n

g e n e r a l , n o t t h e c l o s u r e o f i t s s e a t : t h e s e a t of C1(X)d i s ~

always empty, b u t i f X i s d e n s e - i n - i t s e l f , t h e n t h e s u p p o r t o f C1(X)d i s a l l o f X.

We w i l l s i n g l e o u t b e l o w a f a m i l y o f b a n d s

o f C"(X) w i t h t h e p r o p e r t y t h a t t h e s u p p o r t o f e a c h i s t h e closure of its seat.

D i s t i n c t b a n d s o f C l ( X ) c a n h a v e t h e same s e a t , a n d s i m i l a r l y f o r d i s t i n c t b a n d s o f C"(X). band o f Cl(X) o r o f C"(X)

Thus, i n g e n e r a l , a

c a n n o t be r e c a p t u r e d from i t s s e a t .

I f w e know t h a t a b a n d J o f C 1 ( X ) i s v a g u e l y c l o s e d , t h e n w e

can r e c a p t u r e it from i t s s e a t Z u s u a l , ZL = Z'-in-C(X)

(so

know Jd i s v a g u e l y c l o s e d ,

J

= J

n

= C1(Z)).

X ; J = (Z')',

I t follows t h a t i f we

we c a n a l s o r e c a p t u r e J f r o m i t s

s e a t (which i n t h i s c a s e i s an open s e t ) . know t h a t a b a n d I o f C"(X)

where, a s

E q u i v a l e n t l y , i f we

i s a u - b a n d - o r a n R-band - t h e n

we can r e c a p t u r e i t f r o m i t s s e a t .

We w i l l d e s c r i b e b e l n w w h a t

i s p r o b a b l y t h e l a r g e s t f a m i l y o f b a n d s o f C"(X) w h i c h c a n b e

291

r e c a p t u r e d from t h e i r s e a t s .

We now s i n g l e o u t a p a r t i c u l a r f a m i l y o f s u b s e t s o f X , "universally measureable ones.

I n t h i s 5 and t h e f o l l o w i n g o n e ,

w i l l be d e n o t e d s i m p l y by

z(ll(X))

ways d e n o t e a n e l e m e n t o f t . either C"(X) For

o r C'(X))

e € 6 ,C(e)

the

t,

and t h e l e t t e r e w i l l a l -

Also t h e s e a t o f a s e t A ( i n

w i l l b e d e n o t e d b y Q(A).

is c l e a r l y the set { x E X I(e,x)

1).

=

I t is

e a s i l y v e r i f i e d t h a t t h e m a p p i n g e +-> Q(e) ( t h a t i s , t h e r e s t r i c t i o n of Q(.)

6

t o g ) i s a B o o l e a n a l g e b r a homomorphism o f

o n t o t h e Boolean a l g e b r a o f a l l s u b s e t s of X.

is not one-one. this. write

However i t i s one-one

Note f i r s t t h a t

6

n

C"(X),

=

&,

OT,

=

6 n

In general, it

we record

C"(X),;

(n(X),);

we w i l l

6,.

( 5 5 . 1 ) The r e s t r i c t i o n o f t h e o p e r a t o r Q ( * ) t o z , hence i s a Boolean a l g e b r a isomorphism o f 6 ,

is one-one,

w i t h t h e Boolean

a l g e b r a of a l l s u b s e t s o f X.

F o r e a c h s u b s e t (! o f X , w e w i l l c a l l t h e u n i q u e t h a t Q(e)

=

eE2,

such

Q t h e c h a r a c t e r i s t i c e l e m e n t of Q i n C 1 ' ( X ) a . These

-

two B o o l e a n a l g e b r a s , t h a t o f a l l s u b s e t s o f X a n d t h a t o f t h e i r c h a r a c t e r i s t i c e l e m e n t s i n C 1 l ( X ) a a r e o b j e c t s commonly dealt with in analysis. p a i r o f Boolean a l g e b r a s :

We a r e more i n t e r e s t e d i n a d i f f e r e n t

292

C h a p t e r 11

n

( 5 5 . 2 ) The r e s t r i c t i o n o f t h e o p e r a t o r Q ( . ) t o

U(X)

is also

o n e - o n e ( b u t n o t o n t o ) , hence i s a Boolean a l g e b r a isomorphism w i t h a Boolean s u b a l g e b r a o f t h a t of a l l t h e s u b s e t s o f X .

Th s f o l l o w s from Q(e)

=

Q ( e a ) , t h e Isomorphism t h e o r e m ,

and ( 5 5 1 ) .

We w i l l c a l l t h e s e t s CQ(e) IeE

U(X) 1 t h e u n i v e r s a l l y

8

m easurable s u b s e t s o f X , and f o r e v e r y s u c h Q c a l l e t h e c h a r a c t e r i s t i c element o f Q .

=

Q ( e ) , we w i l l

We emphasize t h a t t h i s

t e r m , " t h e c h a r a c t e r i s t i c e l e m e n t o f Q" w i l l o n l y be used when we know t h a t Q i s a u n i v e r s a l l y m e a s u r a b l e s e t , and t h a t t h e n t h e term r e f e r s t o a unique element o f

u(x).

Given Q c X , i f

we d o n ' t know t h a t i t i s u n i v e r s a l l y m e a s u r a b l e ( o r know t h a t i t i s n o t ) , t h e n we can o n l y r e f e r t o i t s " c h a r a c t e r i s t i c e l e -

ment i n C 1 r ( X ) a r r .

Even f o r a u n i v e r s a l l y m e a s u r a b l e s e t , we c a n

of c o u r s e s t i l l r e f e r t o i t s c h a r a c t e r i s t i c e l e m e n t i n C 1 ' ( X ) a . I n g e n e r a l , t h i s w i l l be d i s t i n c t from i t s c h a r a c t e r i s t i c e l e ment, a l t h o u g h , i n some c a s e s , t h e two w i l l c o i n c i d e ( f o r example, when Q c o n s i s t s o f a s i n g l e e l e m e n t x ) . Some immed a t e p r o p e r t i e s o f t h e f a m i l y o f u n i v e r s a l l y measurable s e t s

I t c o u l d e q u a l l y w e l l have been d e f i n e d a s

t h e family of s e a t s of elements o f lJ(X),

{Q(f) l f C U ( X ) I .

c l o s e d u n d e r suprema and i n f i m a o f c o u n t a b l e s e t s . e of

8 0

U(X)

I t is

An e l e m e n t

i s a u s c e l e m e n t ( r e s p . an llsc e l e m e n t ) i f and

o n l y i f Q(e) i s a c l o s e d s e t ( r e s p . an open s e t ) . I n p a r t i c u l a r eE8

,nU ( X )

over,

l i e s i n C(X)

i f and o n l y i f Q(e) i s c l o p e n .

More-

293

(55.3)

F o r e v e r y eE

8 n

(i) Q(u(e))=

U(X),

91e),

( i i ) Qlk(e)) = i n t e r i o r Q(e), ( i i i ) Q(s(e)) = frontier Q(e).

T h i s f o l l o w s from (54.7) and o r d i n a r y f u n c t i o n t h e o r y .

We s i n g l e o u t a n o t h e r f a m i l y o f b a n d s o f C"(X). c l u d e s t h e R-bands and t h e u - b a n d s . c a l l e d a U-band i f l l ( X ) I E U(X).

a l s o a U-band.

n 1~1.

IU(X

It is clear that collection

In particular,

( 5 5 . 4 ) F o r a b a n d I o f C"(X) 1'

I i s a U-band;

2O

U(X), c U(X);

(55.5)

=

=

n(X)

-

2 n

Jl(X),,

If follows e a s i l y t h a t u(x)

t h i s h o l d s , t h e n IL(X),€U(X),

30 u ( x )

I

[u(x)

Corollary.

n

I]

u(x),

=

u

X)

n

U(X).

hence I d i s

[u(x)

=

I.

s o I i s a U-band.

,

w i l l be

A b a n d I o f C"(X)

of U-bands i s a Boolean a l g e b r a i s o m o r p h i c t o

F o r e v e r y U-band I , n(X)

I t in-

n

11

3

Moreover, i f Thus

the following are equivalent:

B [u(x)

n rdl.

The m a p p i n g I t->U(X)

n

I i s an isomorph-

i s m o f t h e Boolean a l g e b r a ' o f U-bands o n t o t h a t o f t h e p r o j e c t i o n bands o f U ( X ) .

C h a p t e r 11

294

I t i s c l e a r t h a t t h e mapping I

6

-

2

Q(1)

i s an i s o m o r p h i s m

o f t h e B o o l e a n a l g e b r a o f U-bands o n t o t h a t o f t h e u n i v e r s a l l y measurable s u b s e t s of X.

Thus a U-band i s d e t e r m i n e d by i t s

M o r e o v e r , from ( 5 5 . 3 ) , t h e s u p p o r t o f a U-band i s t h e

seat.

closure of i t s s e a t .

A s we s t a t e d e a r l i e r , t h e U-bands may

c o n s t i t u t e t h e l a r g e s t f a m i l y o f b a n d s i n C'l(X) w i t h t h e s e two p r o p e r t i e s .

5 5 6 . The Nakano c o m p l e t e n e s s t h e o r e m

We g i v e t h e c l a s s i c a l c h a r a c t e r i z a t i o n o f Dedekind comp l e t e n e s s f o r C(X) i n t e r m s o f t h e t o p o l o g y o f X ( 5 6 . 3 ) .

( 5 6 . 1 ) The f o l l o w i n g a r e e q u i v a l e n t :

'1

C(X) i s Dedekind c o m p l e t e ;

2'

f o r every t s c element t , U ( L ) E C ( X ) ;

3'

f o r every usc element u , L ( U ) E C ( X ) .

Proof.

Assume C(X) i s Dedekind c o m p l e t e a n d c o n s i d e r a n

ksc element il. show f

=

u(Q).

hence f > u(a). {hEC(X)lh > t}.

1'

i m p l i e s 2'.

Let A = { g E C ( X ) ( g < a } and f = V A - i n - C ( X ) ; we f

2 u(a):

in effect, f > A, hence f > VA = t,

For t h e o p p o s i t e i n e q u a l i t y , l e t B = Then B > A , s o f u(k).

Thus '2

Set R

VA; w e show u ( k )

=

=

VA-in-

Suppose fEC(X1, f > A; then f > VA 0

implies 1

.

a n d 3'

2'

=

II,

a r e clearly e q u i -

valent.

QED

We n e x t show t h a t i t s u f f i c e s f o r '2 f o r elements of

(we a r e s t i l l d e n o t i n g

or 3 C$

0

above t o h o l d

(Il(X)) b y

&

and

r e s e r v i n g t h e l e t t e r e t o denote an element o f z ) .

( 5 6 . 2 ) The f o l l o w i n g a r e e q u i v a l e n t : 1'


4'

for every

element e i n

6,

u(e)Ec(~)

SO

f o r every usc element e i n

E,

a(e)Ec(x)

RSC

____ P r o o f . We n e e d o n l y show 4'

i m p l i e s 2'

h o l d s a n d c o n s i d e r a n Rsc e l e m e n t I ? , 0 < R < n

above.

Assume 4'

n(X).

For each

1 , 2 , * * * , set

=

n

n

w h e r e t h e e i ( k / n ) ' s a r e t h e s p e c t r a l e l e m e n t s o f I? d e f i n e d i n 517.

Rn

i s c l e a r l y a n Rsc e l e m e n t .

By t h e p r o o f o f t h e

Freudenthal theorem ( 1 7 . 1 0 ) , we have limn+m[l R by ( 4 9 . 1 6 ) ,

(n

=

limn,,[lu(R)

- u(tn)[l = 0.

Qn[I

=

0 , hence,

We show u ( k n ) E C ( X )

1,2;..); it w i l l follow t h a t u(a)EC(X). n R n = ~ ~ = ~ ( k / n ) e ~ ( k h/ enn )c e, by (49.4), u(R,)

=

C h a p t e r 11

296

VE=l(k/n)u(eQ(k/n)). (k

=

l,...,n),

S i n c e , by a s s u m p t i o n ,

u(eQ(k/n))EC(X)

u(R,)EC(X). QED

Combining t h e above w i t h ( 5 5 . 3 ) i t ) , we have ( c f .

(56.3)

(Nakano)

(and t h e remark p r e c e d i n g

[541):


1'


6'

f o r e v e r y open s e t W i n X , W i s open.

-

We h a v e b e e n d e a l i n g w i t h t h e s e t components o f n(X) i n C l t ( X ) .

t

=

8

(n(X)) o f

all t h e

Note i n t h e f o l l o w i n g c o r o l l a r y ,

we a r e d e a l i n g o n l y w i t h t h o s e i n C(X).

(56.4) Corollary.


1'


7'

8

(n(x))-in-c(x) is ( a ) t o t a l i n C(X), and ( b ) Dedekind c o m p l e t e .

Proof. (56.2). assume 7'

T h a t 1'

i m p l i e s 7'(a)

That i t implies 7 O ( b )

is contained i n the proof of

f o l l o w s from ( 1 7 . 2 ) .

h o l d s ; w e show t h a t t h e n 4'

Conversely,

( i n 56.2)) holds.

We n o t e f i r s t t h e f o l l o w i n g c o r o l l a r y o f ( 1 7 . 7 ) .

Lemma 1 .

If;(Il(X))-in-C(X)

i s t o t a l i n C(X), t h e n e v e r y

fEC(X)+ i s i n t h e norm c l o s u r e o f t h e s e t o f e l e m e n t s below i t o f t h e form Xe, where e €

6

( n ( X ) ) - i n - C ( X ) and

> 0.

I t f o l l o w s f i s t h e supremum ( i n C t ’ ( X ) ) o f t h i s s e t o f elements.

Lemma 2 . elements of

Every e € i $ which i s Rsc i s t h e supremum o f t h e

t (n(X))-in-C(X)

below i t .

By d e f i n i t i o n , e i s t h e supremum o f a l l t h e e l e m e n t s o f C ( X ) + below i t .

Let i e a ) be t h e s e t o f a l l e l e m e n t s o f

6

(n(X))-

in-C(X) which have t h e f o l l o w i n g p r o p e r t y : t h e r e e x i s t f E C ( X ) + below e and A > 0 s u c h t h a t Xea 5 f < e. u s i n g t h e comment a f t e r Lemma 1 , t h a t e

I t is easily verified, =

V e

clcl

.

Lemma 2 g i v e s u s t h a t a l s o e v e r y e € z which i s infimum o f t h e e l e m e n t s o f $- ( l l ( X ) ) - i n - C ( X ) above i t .

USC

is the

T h a t 4’

h o l d s c a n now be p r o v e d by e x a c t l y t h e same argument used t o p r o v e 2’

(56.1). QED

Appendix

We p r e s e n t h e r e t h e example p r o m i s e d i n 9 4 5 t o show t h a t t h e Riesz subspace s(X) of C ” ( X ) (45.1),

s(X)

=

n e e d n o t be norm c l o s e d .

(C(X)u)+ - (C(X)’)+,

and i t i s i n t h i s form

By

Chapter 11 that we will treat it. We have to exhibit an element o f C"[X) which is not in (C(X)')+

(C(X)u)+

-

& in its norm closure. Since

but

U(X)

is

norm closed, this element will still be in U(X), and thus we can work within LJ(X).

It follows from the Isomorphism theorem This means we can

that, equivalently, we can work in U(X)=.

confine ourselves to bounded functions on X and use the tools and terminology of ordinary function theory (in particular, cf. (54.6)). Specifically, let X be a real (compact) interval. produce a sequence If,]

We will

of bounded, non-negative functions on

X and a function f on X such that:

for each k, fk = uk - vk, uk and vk non-negative

(I)

uppersemicontinuous functions on X; limkll f - fk[I

(11)

=

0;

(111) f cannot be expressed in the form f

=

u - v, u and

v non-negative uppersemicontinuous functions on X.

Lemma. -

Let X be the real interval [a,b].

1 > 0, there exists a non-negative g on X with

F o r every

I[ gI[

=

1 such

that: (i)

g can be written in the form g

=

u - v, u and v non-

negative uppersemicontinuous functions on X; (ii) for every such representation of g, I[u/l 2 1.

Proof.

I _

To avoid burdensome details, we establish the

Lemma for 1 = 2; it will be clear that the same procedure

( c o n s i d e r a b l y l e n g t h e n e d ) w i l l work f o r a r b i t r a r y n , h e n c e f o r any

> 0.

Let { x n } (n

1 , 2 , . . - ) be a descending sequence of d i s t i n c t

=

p o i n t s i n t h e open i n t e r v a l ( a , b ) which c o n v e r g e s t o e a c h n , l c t {xnm} (m

=

For

3.

1,2,...) be a d e s c e n d i n g s e q u e n c e o f

d i s t i n c t p o i n t s i n t h e o p e n i n t e r v a l ( X ~ , X ~w-h i~c h) c o n v e r g e s to x n,m

( t a k e xo

n

1,2,.-.

=

=

b).

We d e f i n e g b y g ( a )

n

=

=

0 otherwise.

=

2 , u(xnm)

=

1 for

=

1 f o r n,m

=

1,2;-*,and

F i n a l l y d e f i n e v by v ( a ) = v ( x n )

=

1 for

Then u and v a r e u p p e r -

and v ( x ) = 0 o t h e r w i s e .

1,2,...,

g(xnm)

and g ( x ) = 0 o t h e r w i s e .

Now d e f i n e u by u ( a ) u(x)

=

semicontinuous and u - v = g , which e s t a b l i s h e s ( i ) . To show ( i i ) , a s s u m e g = u - v , w i t h u a n d v n o n - n e g a t i v e uppersemicontinuous.

Since v > 0 , t h e c o n d i t i o n g(xnm)

g i v e s us u(xnm) > 1 f o r n,m

=

1,2,-..

.

Since u i s uppersemi-

continuous, t h i s g i v e s u s , i n t u r n , t h a t u(xn) > 1 for n The c o n d i t i o n g ( x n )

1,2;.*.

n = 1,2,.

-

9

.

-

=

0 then gives us v(x ) > 1 f o r

n S i n c e v i s u p p e r s e m i c o n t i n u o u s , we o b t a i n v ( a ) > 1.

Applying t h e c o n d i t i o n g ( a )

u(a)

=

1

=

=

1, w e o b t a i n , f i n a l l y , t h a t

2.

QED

We p r o c e e d t o c o n s t r u c t t h e e x a m p l e .

l e t X be t h e r e a l i n t e r v a l [0,1]. al > b 2 > a 2 > * ” >

bk > ak > . . -

For c o n c r e t e n e s s ,

Choose a s e q u e n c e 1 > b l >

converging t o 0.

For each k ,

l e t Z k be t h e c l o s e d i n t e r v a l [ a k , b k ] a n d gk t h e f u n c t i o n on 2 Z k g i v e n by t h e Lemma w i t h = k . D e f i n e hk ( k = 1 , 2 , - - . ) o n X , byhk(x) = (l/k)gk(x) k d e f i n e f k = Clhi ( k

on Zk, h k ( x ) =

1,2,...)

=

0 elsewhere.

and f = cyhi

Finally

(norm c o n v e r g e n c e ) .

Chapter 1 1

300

Then { f k } and f have t h e desired properties.

Remark.

A study o f t h e problem o f t h e n o r m closure o f

s ( X ) o n a real interval h a s been carried out b y Ryan [ 4 6 ] .

PART V

RIEMANN INTEGRATION The a p p r o a c h t o Riemann i n t e g r a t i o n p r e s e n t e d h e r e stems from t h e f o l l o w i n g p a p e r s : C a r a t h e o d o r y [ l l ] , Loomis [ 3 2 ] , Bauer [ S ] , and Semadeni [ 4 8 ] .

The r e a d e r i s r e f e r r e d t o them

for further references.

301

CHAPTER 1 2

THE RIEMANN SUBSPACE O F A BAND

I n 5 4 0 , we d i s c u s s e d C ( X j a , a n d i n 5 4 1 , C ( X ) , band C"(X)

u'

pEC'(X).

We now t u r n t o C(X),

€or a b a s i c

f o r a g e n e r a l band

I of C"(X). N o t e f i r s t t h a t t h e o r d e r c l o s u r e o f C(X),

is a l l of I .

I n s h a r p e r f o r m t h e Up-down-up t h e o r e m h o l d s f o r C(X), i n I . R T h i s f o l l o w s f r o m t h e e a s i l y v e r i f i e d f a c t t h a t (C(X) ) I = ( C ( X ) I ) p . , (C(X)'')I

= (C(X)I)ue,

and s o o n .

Note a l s o t h a t

p r o j I d o e s n o t p r e s e r v e D e d e k i n d c l o s u r e : w h i l e C(X) i s ( 4 6 . 3 ) , C(X), i s , i n g e n e r a l , n o t

D e d e k i n d c l o s e d i n Cl'(X) Dedekind c l o s e d i n I .

E x a m i n a t i o n o f i t s Dedekind c l o s u r e w i l l

l e a d u s t o Riemann i n t e g r a t i o n . C o n s i d e r a c o n c r e t e case: I J

=

=

L m ( p ) and i t s d u a l band

1 L ( p ) , f o r p L e b e s g u e m e a s u r e on a r e a l i n t e r v a l .

them s i m p l y by La and L

1

.

L e t C b e t h e image i n Lm o f t h e c o n C(X) , ) . By a s t a n d a r d t h e o r e m L 1 t h e s e t o f l i n e a r f u n c t i o n a l s on L

tinuous functions (that is C on t h e t o p o l o g y

o(L1,C),

which are a(L1,C) which a r e o(L1,C)

We d e n o t e

=

continuous is C i t s e l f .

What a b o u t t h o s e 1 c o n t i n u o u s on t h e u n i t b a l l B(L ) ? T h i s i s no

l a r g e r ; G r o t h e n d i e c k ' s completeness theorem gives us t h a t i t i s s t i l l C.

What a b o u t t h e s e t o f t h o s e t h a t a r e 0(L1,C) 1

o u s on K ( L ) ?

continu-

T h i s s e t i s l a r g e r ; w e show t h a t i t i s t h e

Dedekind c l o s u r e o f C i n Lm. 302

Riemann Subspace of a Band

303

Now, contained implicitly in a theorem of Caratheodory [ll] is the fact that this Dedekind closure is precisely the image R in La of the Riemann integrable functions.

We thus have

a topological characterization of R: R consists of those elements of Lm which are a(L1,C) continuous on K ( L 1 ) . In this chapter,we establish o u r results not for the above concrete case but for a general band.

§ 5 7 . The Dedekind closure of C ( X ) ,

We first record a theorem from function theory - which we have met earlier ($54)

-

and a sharpening of it for convex

sets.

(57.1) Let Q be a subset of a topological space T, and h a bounded function on Q. -

h(t)

We define =

limsup h(q) qE Q

on the closure

by

.

q+t Then

(i)

h is uppersemicontinuous on Q;

(ii) if h is uppersemicontinuous on Q , then E(q)

=

h(q)

for q E Q .

Again, we will call 5 the uppersemicontinuous upper envelope of h.

304

Chapter 1 2

A f u n c t i o n f on a c o n v e x s e t Q ( i n some v e c t o r s p a c e ) i s

concave ( r e s p . convex) i f t h e f o l l o w i n g h o l d s . q 1 , q 2 E Q and X 1 , X 2

2

Xlf(q1)

(resp. f ( l p l

(57.2)

+

X2f(q2)

0 such t h a t

Proof.

x2

=

1 , f(X1ql

x2q2) 5 xlf(ql)

+

+

x2q2)

=

1.

XIF;(pl)

+

A2F(p2).

Set p

=

Alpl

+

and

x1,x2 2

X2p2.

0 such t h a t

We h a v e t o show h ( p ) >

I t i s enough t o show t h e f o l l o w i n g ( q ' s

always denote elements of 9).

F o r e v e r y q1 E p l + N a n d q 2 E p 2 + N

,

X1ql

+

X2q2Ep +N;

hence

The d e s i r e d i n e q u a l i t y f o l l o w s d i r e c t l y from t h i s .

Our g o a l i s ( 5 7 . 5 1 b e l o w .

1

X2f(q2))*

I f h i s c o n c a v e , t h e n s o i s f;.

Consider p1,p2En

X2

+

+

+

I n t h e above t h e o r e m , l e t T b e a l o c a l l y convex s p a c e

and Q a c o n v e x s u b s e t .

X1

xl

For e v e r y

Riemann S u b s p a c e o f a Band

( 5 7 . 3 ) Lemma 1.

Let I b e a b a n d o f C"(X), w i t h d u a l b a n d J .

F o r a bounded f u n c t i o n h on K ( J ) ,

'1

305

the following a r e equivalent:

h i s t h e p o i n t w i s e infimum on K ( J )

o f some s u b s e t o f

C(X),; h i s concax'e a n d u p p e r s e m i c o n t i n u o u s on K ( J )

2'

with res-

p e c t t o u (J,C(X) I ) .

T h a t '1

Proof.

i m p l i e s 2'

i s c l e a r , s i n c e every element

i s a f f i n e a n d u ( J , C ( X ) d c o n t i n u o u s on K ( J ) .

o f C(X),

To show

t h e c o n v e r s e , w e n o t e f i r s t t h a t C ~ ( J , C ( X ) ~c o) i n c i d e s w i t h t h e r e s t r i c t i o n t o .J o f t h e v a g u e t o p o l o g y u ( C ' (X) , C ( X ) ) ; c a n work w i t h t h e l a t t e r .

s o we

S e c o n d l y , b y t h e comment a t t h e e n d

o f 8 5 2 , we c a n , f o r s i m p l i c i t y , assume .J i s v a l u e l y d e n s e i n C'(X).

K(J)

i s then vaguely dense i n K ( C ' ( X ) ) ,

uppersemicontinuous upper envelope K(C'(X)).

so t h e

o f h i s d e f i n e d on a l l o f

-

By ( 5 7 . 1 ) and ( 5 7 . 2 ) , h i s c o n c a v e a n d v a g u e l y u p p e r -

s e m i c o n t i n u o u s , a n d c o i n c i d e s w i t h h on K ( J ) . and (38.2) g i v e s u s t h a t

o f some s u b s e t o f C(X).

Applying (48.3)

i s t h e p o i n t w i s e infimum on K ( C ' ( X ) )

'1

follows immediately.

QE D

For a monotonic n e t i n I , o r d e r convergence is e a s i l y s e e n t o be e q u i v a l e n t t o p o i n t w i s e c o n v e r g e n c e on K(J)

(cf. (10.9)).

Combining t h i s w i t h ( 5 7 . 3 ) , w e o b t a i n :

( 5 7 . 4 ) Lemma 2 .

Let I b e a b a n d o f C"(X), w i t h d u a l b a n d J .

For f E I , , t h e f o l l o w i n g a r e e q u i v a l e n t :

Chapter 1 2

306

l o f E (C(X) 2'

y;

f i s u p p e r s e m i c o n t i n u o u s on K ( J ) w i t h r e s p e c t t o

O(J,C(X)~).

I n t h e two Lemmas, we c a n o f c o u r s e r e p l a c e " u p p e r s e m i c o n t i n u o u s " by " low ers em icontinuous " ,

"concave"

by "convex",

and ( C ( X) I ) u by (C(x),)'.

( 5 7 . 5 ) Theorem.

L e t I b e a band o f Cf'(X ), w i t h d u a l b a n d < J .

For a l i n e a r f u n c t i o n a l $ on J , t h e f o l l o w i n g a r e e q u i v a l e n t . 1'

$ i s i n t h e D edeki nd c l o s u r e o f C(X ), ;

2'

$ i s a ( J , C ( X ) I ) - c o n t i n u o u s on K ( J ) .

Proof.

N ot e t h a t i f

element of I .

s a t i s f i e s ZO,

In effect, u(J,C(X),)

t h e n i t must b e a n

i s c o a r s e r t h a n t h e norm

topology o f J , s o (23.2) g i v e s t h e d e s i r e d conclusion.

The

t h e o r e m f o l l o w s f r om Lemma 2 and t h e comment f o l l o w i n g i t .

UED

H e n c e f o r t h , we w i l l c a l l t h e D edeki nd c l o s u r e o f C ( X ) , Riemann s u b s p a c e o f I , a n d d e n o t e i t b y R ( 1 ) .

the

The t e r m i n o l o g y

o f c o u r s e comes f r om t h e C a r a t h e o d o r y t h e o r e m r e f e r r e d t o i n the introduction to this chapter. Remark. above theorem.

o ( J , C ( X ) I ) c a n be r e p l a c e d by o(C'(X),

I t then reads:

C(X))

in the

The Riemann s u b s p a c e o f a band

Riemann S u b s p a c e o f a Band

o f C"(X)

307

c o n s i s t s o f t h e l i n e a r f u n c t i o n a l s on tJ w h i c h a r e

v a g u e l y c o n t i n u o u s on K ( J ) .

By i t s d e f i n i t i o n , R(1) (C(X),)'

=

(C(X)')I

Warning!

(C(X),)un

=

and (C(X)I)'

=

(C(X)7)2.

(C(X)'),,

While f E R ( 1 ) i f and o n l y i f f

Now

s o we have:

=

uI

=

RI f o r

some usc e l e m e n t u a n d some Rsc e l e m e n t t , w e may n o t b e a b l e t o choose u and R s o t h a t R

i

u.

We g i v e a n e x a m p l e i n 563.

558. A r e p r e s e n t a t i o n theorem

F o r e v e r y b a n d I o f C"(X),

C(X),

i s an M I - s u b s p a c e o f I

w h i c h i s s e p a r a t i n g on t h e d u a l b a n d J .

This represents a q u i t e

general situation:

( 5 8 . 1 ) -___ Theorem.

Let E b e an L-space, E '

i t s d u a l , and F an

M I - s u b s p a c e o f E ' w h i c h i s s e p a r a t i n g on E . compact s p a c e X, d e n s e i n C'(X)

Then t h e r e e x i s t s a

a band J o f C'(X) - which w e c a n t a k e v a g u e l y

-

a n d an i s o m o r p h i s m o f E o n t o J s u c h t h a t E '

i s M I - i s o m o r p h i c t o t h e band I d u a l t o J and F i s Ma-isomorphic

308

Chapter 1 2

t o C(X),.

F i s an MIL-space, s o i s M I - i s o m o r p h i c t o C(X) f o r

Proof.

some X c o m p a c t .

L e t C(X) __ To > F b e t h i s isomorphism, F i > E t

the canonical i n j e c t i o n o f F i n t o

E l ,

and s e t T = i o T o .

'

C(X) -> E l i s an ME-isomorphism o f C(X) i n t o C o n s i d e r C t ( X ) 0.

Proof.

We show the equivalent statement: for every Rsc

< 0. element R , R E C " ( X ) ~ implies R -

so R+

=

V{f€C(X)j

T h u s R'

is also an ilsc element,

But, by ( 4 0 . 4 ) ,

0 < f < a'}.

consists only of 0.

il+

=

the latter set

0.

We proceed with our example.

Let X be the real interval

[ 0 , 1 ] , u the characteristic element of the closed set [ 0 , 1 / 2 ] ,

and R the characteristic element of the interval [O,l/Z). (Remember, we mean the characteristic elements in U(X); so u is a usc element and

k

an Rsc element (cf. § 5 5 ) . )

Then

Chapter 1 2

314

( b y ( 5 4 . 1 3 ) ) , h e n c e ud

u - REC"(X)a

I t r e m a i n s t o show u d & C(X)d.

that fd (u,x)

=

ud = R d .

Suppose t h e r e e x i s t s f E C ( X ) s u c h

We show R < f < u , whence ( k , x )

f o r a l l xEX, which i s i m p o s s i b l e .

f - REC"(X)a

f - u = ( u ( f ) , p ) f o r e v e r y ~ E J .

The p r o o f i s i m m e d i a t e .

We n e x t g i v e some c h a r a c t e r i z a t i o n s o f x ( p ) .

For s i m p l i c -

i t y , we t a k e p > 0.

(61.3)

Given u € C ' ( X ) + , t h e n f o r f E C " ( X ) , t h e f o l l o w i n g a r e

equivalent:

1'

f i s p-Riemann i n t e g r a b l e ;

h>f

g 0 , the following a r e equivalent:

1'

f is lean;

2'

f

=

u ( g ) - g f o r some ~ E c ~ * ( x ) ;

3'

f

=

h - R(h) f o r some h € C " ( X ) ;

4O

f < 6(f).

336

The Lean E l e m e n t s

Proof. ___ u(f)

S u p p o s e 1' h o l d s .

p(f)

=

S(f).

u(f) < s(f)

=

u(f)

-

-

=

0.

so f

=

f - k(f).

T h a t 3'

Finally, i f f

t ( u ( g ) - g) 0

plies 1

5

Then k ( f ) = 0 , s o f < u(f) =

holds.

S u p p o s e 4'

e(f) < u ( f ) , so u ( f )

Thus 1' h o l d s .

a(f)

i n 3'.

Thus 4'

=

1'

i m p l i e s 3':

i m p l i e s 2'

u(g)

-

337

holds.

Then

u(f) - Elf), so

=

i n effect, k(f)

=

0,

f o l l o w s by s e t t i n g g

=

-11

g , t h e n by ( 4 9 . 2 ) ,

u ( g ) - u ( g ) = 0 , whence k ( f )

=

0 < k(f) =

Thus 2'

0.

im-

. QED

I f f i s l e a n , t h e n s o are f + and f - .

The c o n v e r s e i s f a l s e .

For o n e e x a m p le, l e t X be a r e a l i n t e r v a l and g a n d h t h e c h a r a c t e r i s t i c elements o f t h e sets o f r a t i o n a l p o i n t s and i r rational points respectively h are lean.

(555)).

( r em em be r, g , h E I J ( X )

Suppose k(g) > 0 ; t h e n e(g),

g and

i s a lowersemicon-

t i n u o u s f u n c t i o n o n X w hi ch i s > 0 a n d v a n i s h e s o n t h e i r r a t i o n a l points, an impossibility. Then f +

=

g and f -

=

Similarly for k(h).

11, b o t h l e a n , w h i l e If

I

=

Set f

=

g - h.

R ( X ) , which i s n o t

lean. F o r a n e x a m p l e removed f r o m f u n c t i o n i m a g e r y , l e t X b e d e n s e - i n - i t s e l f and f f - = l.(X)d, f

=

l(X)u

- n(X)d.

Then f +

=

n(X),

and

b o t h l e a n ( c f . t h e comment f o l l o w i n g ( 4 0 . 5 ) ) , w h i l e

n(x).

=

Remark.

N ot e t h a t i n t h e s e two e x a m p l e s , 6 ( f )

T h u s, i n c o n t r a s t w i t h ( 6 5 . 1 ) ,

(65.2)

=

2.n(X).

6 ( f ) is, i n general, not lean.

If f i s l e a n , t h e n k ( f ) < 0 < u(f).

C h a p t e r 14

338

Proof.

f f and f - a r e l e a n , h e n c e , by ( 4 9 . 1 1 ) ,

u(f+) - a(f-)

=

u(f+) > 0 , and L ( f )

=

a(f')

u(f)

=

- u(f-) = -u(f-)

.(

0.

QED

We n o t e t h e u s e f u l c o r o l l a r y t h a t i f a u s c e l e m e n t u i s l e a n , > 0 , and i f an Esc e l e m e n t R i s l e a n , t h e n R < 0. then u -

(Note

that (59.5) i s a corollary of t h i s . ) Note a l s o t h a t t h e c o n c l u s i o n o f ( 6 5 . 2 ) i s n o t a s u f f i c i e n t c o n d i t i o n f o r f t o be l e a n , a s t h e examples p r e c e d i n g ( 6 5 . 2 ) show. These examples a l s o show t h a t t h e s e t o f l e a n e l e m e n t s i s not c l o s e d under a d d i t i o n o r t h e l a t t i c e o p e r a t i o n s i s n e i t h e r a l i n e a r subspace nor a s u b l a t t i c e of

V,A.

C'l(X).

So i t

I t con-

t a i n s w i t h e a c h e l e m e n t t h e R i e s z i d e a l g e n e r a t e d by t h a t c l e ment.

I t i s t h u s a union of Riesz i d e a l s , a c t u a l l y t h e union

o f a l l t h e R i e s z i d e a l s i n t e r s e c t i n g C(X) o n l y i n 0 .

I t is

t h u s a s o l i d s e t , and c o u l d be d e f i n e d a s t h e l a r g e s t s o l i d s e t i n t e r s e c t i n g C(X) o n l y i n 0 . An o b v i o u s , b u t u s e f u l , consequence o f t h e s o l i d n e s s i s t h a t i f f i s l e a n , t h e n f o r e v e r y band I o f Cl'(X), f I i s a l s o lean. Another p r o p e r t y :

( 6 5 . 3 ) The s e t o f l e a n e l e m e n t s i s norm c l o s e d .

T h i s f o l l o w s from t h e e a s i l y v e r i f i e d f a c t t h a t i f a s o l i d

s e t i n t e r s e c t s C(X) o n l y i n 0 , t h e n s o does i t s norm c l o s u r e .

The Lean Elements

339

The examples preceding (65.2) actually show that the sum of two lean elements f,g need not be lean even if f A g

=

0.

ever, under "stronger" disjointness of f and g, their sum

How-

%

lean.

> 0 are lean and u ( f ) A g (65.4) If f,g -

Proof.

By (49.1), k(f

+

g)

-

u(f)

(using Exercise 4 in Chapter 1) R(f u(f)Af

R(f

+

+

g)

u(f)Ag =

=

f.

Thus 0

.


If h i s l e a n , then f o r every usc element u,

a(u

i h)

k.

In particular, for every gEC(X), R(g

?

h) 0 and f i s i n t h e band g e n e r a t e d by Lemma 2 , i t i s enough t o show t h a t If \ A l l i s l e a n .

:

By

il = V a i l a

v t r ( R w ) + , and by Lemma 1 , I f \ A ( L a ) + i s l e a n f o r e v e r y a. f o l l o w s L ( l f [ ) A ( e w ) += 0 f o r a l l a ( 4 9 . 4 ) , whence L( whence ( a g a i n by ( 4 9 . 4 ) )

=

It

f()AL

=

0,

IfjAR i s l e a n .

Case 11, L < 0 : Choose a n a r b i t r a r y cto.

5 f ui,u(f). k u ( ( f - u)')

=

I t i s enough t o show u z ku(f).

0 , s o Lu(f - u ) < 0 , s o L(u(f)

whence ( a g a i n by ( 4 9 . 2 ) ) k u ( f )

-

u
0:

( 6 8 . 5 ) Given a n Rsc e l e m e n t il > 0 , l e t I b e t h e band which i t generates.

P roof. -

Then f o r f € 1 , i f fAR i s r a r e , f i s r a r e .

By ( 6 8 . 4 ) , ) f l A L i s r a r e , s o f o r s i m p l i c i t y , we c a n

assume f > 0.

For e v e r y n E N , 0 < fAnR < n ( f A k ) , s o fAna i s

rare.

=

Since f

Vn(fAnL), i t f o l l o w s from ( 6 8 . 1 ) t h a t f i s r a r e QED

The a b o v e d o e s n o t h o l d € o r a g e n e r a l Lsc e l e m e n t R : be a r e a l i n t e r v a l and { r n } t h e r a t i o n a l p o i n t s o f X .

Let X

L e t f be

36 2

Chapter 1 5

t h e c h a r a c t e r i s t i c element of t h e s e t { r n l , and u t h e u s c e l e ment d e f i n e d by ( u , r n )

wise.

Finally, l e t R

by R a n d fAL

= =

l / n (n

-u.

=

and ( u , x )

=

0 other-

Then f i s i n t h e b a n d g e n e r a t e d But u ( f ) = n ( X ) , s o f i s n o t

R , w hi ch i s r a r e .

=

1,2,...)

rare. A s with l e a n elements ( c f .

( 6 6 . 6 ) ) , t h e f o l l o w i n g theorem

i s analogous t o (68.1), and while they a r e a c t u a l l y d i f f e r e n t , t h e y r e d u c e t o t h e same t h e o r e m i n t o p o l o g y .

( 6 8 . 6 ) Theorem.

u

= Aauu.

Then f o r f E C " ( X ) ,

t h a t ( f - u)'

Proof. Aauu =

u.

Let {uul be a c o l l e c t i o n of usc elements, and (f

-

ua)+ r a r e f o r a l l a i m p l i e s

i s rare.

By ( 6 7 . 8 ) , R u ( f ) Thus ( f - u)'

5 ua f o r a l l a, h e n c e Ru(f)

< ( f - Ru(f))'

> 0 .

QED

363

Chapter 1 6

364

( 6 9 . 2 ) Theorem.

( R a ( X ) L ) d = Ra(X)'.

We p r o v e t h e more g e n e r a l t h e o r e m :

( 6 9 . 3 ) I f I i s a norm c l o s e d R i e s z i d e a l o f

then ( I l ) d

Cl'(X),

=

IC.

Proof.

We show f i r s t t h a t ( I ) d i s a R i e s z i d e a l o f 1'. 1

Note t h a t , s i n c e Cl(X) i s a band o f C 1 " ( X ) ,

n

[ (11) d - i n - c l l l (X)]

Cl(X),

and t h u s ( I ) d 1

(II) d

=

i s a band o f

Combining t h i s w i t h ( 1 4 . 1 1 ) , we have t h a t

(IL)d-in-C1'l;X).

( I L ) d i s a Riesz i d e a l o f Ib.

Now e a c h u E ( I ) d I

i s o r d e r con-

t i n u o u s on Cll(X), hence - s i n c e I i s a R i e s z i d e a l - on I . Thus ( I L ) d i s a R i e s z i d e a l o f 1'. I t r e m a i n s t o show t h a t ( I ) d i s a l l o f 1'. I

t h a t , by ( 1 5 . 4 ) and ( 2 1 . 1 ) ,

Ib

=

Note f i r s t

I ' and s o i s a n L - s p a c e .

t h e imbedding o f ( I ) d i n 1 ' i s c l e a r l y norm p r e s e r v i n g . I

Now Since

i s norm c o m p l e t e , i t i s norm c l o s e d i n I f , h e n c e , by

(19.7), order closed.

i s a band o f 1'.

Thus

Finally,

( I l l d i s s e p a r a t i n g on I , h e n c e v a g u e l y d e n s e i n 1'; t h e Luxemburg-Zaanen theorem ( 1 2 . 7 ) ,

h e n c e , by

( I ) d = 1'. I

QED

We c a n t h u s w r i t e o u r d e c o m p o s i t i o n o f C 1 ( X )

C'(X) = Ra(X)' t i o n of X, X

0 =

[X

C(X)'.

i n t h e form

By ( 2 0 . 1 ) , t h i s g i v e s us t h e decomposi-

n Ra(X)']

IJ [X

n

C(X)'].

So e v e r y xEX

is

C'(X) e i t h e r i n Ra(X)'

=

o r C(X)'.

Ra(X)'@

C(X)'

365

We c a n s h a r p e n t h i s :

( 6 9 . 4 ) O f x € X i s an i s o l a t e d p o i n t , i t l i e s i n C(X)';

i f not,

it l i e s in R ~ ( x ) ' .

Proof.

Let u b e t h e c h a r a c t e r i s t i c e l e m e n t o f x .

x is not isolated.

Suppose

Every f€C(X) s u c h t h a t 0 < f < u vanishes

on X\x ( s i n c e u d o e s ) , h e n c e , b e i n g c o n t i n u o u s on X , a l s o v a n i s h e s on x ; i t i s t h e r e f o r e 0 . is isolated.

Thus u E R a ( X ) .

Suppose x

Then u i s c o n t i n u o u s on x, h e n c e ( b y t h e d i s -

c u s s i o n following (40.2)

and t h e Isomorphism theorem) ufC(X)

I t f o l l o w s t h a t Wu c C ( X ) .

.

But IRu i s i n f a c t a b a n d , s o t h i s

i n c l u s i o n a c t u a l l y g i v e s u s t h a t i t i s d i s j o i n t from Ra(X). Thus uERa(X)d X

=

[Ra(X)']'

,

a n d t h e r e f o r e v a n i s h e s on

Ra(X)'.

QED

5 7 0 . The b a n d Ra(X)

d

Ra(X) also d e t e r m i n e s a d e c o m p o s i t i o n o f C"(X): Ra(X)dd 3 Ra(X)d.

=

Ra(X)dd i s t h e band g e n e r a t e d by Ra(X), or

equivalently, i t s order closure. Ra(X)d = (Ra(X)')'.

C"(X)

I t i s a l s o (C(X)')',

O t h e r w i s e s t a t e d , ( R a ( X ) d d , Ra(X)')

p a i r o f d u a l b a n d s , and (Ra(X)d,C(X)c) decomposition can be w r i t t e n :

is also.

while is a

So t h e a b o v e

366

Chapter 1 6

(70.1)

C''(X)

Ra(X)"

=

%,

C(X)".

We examine Ra(X)d i n more d e t a i l .

Our f i r s t aim i s t o

show t h a t C(X)

i s Dedekind d e n s e i n R a ( X ) d , t h a t i s , t h e Ra (XI Dedekind c l o s u r e R(Ra(X) d ) o f C(X) i s a l l o f Ra(X)d ( 7 0 . 3 ) . Ra (XI

( 7 0 . 2 ) Theorem.

F o r e v e r y f€IJ(X) Jl ( f )

Proof.

Ra(X)d

= f

,

Ra(X)

d = u(f)

We show t h e s e c o n d i d e n t i t y .

Ra(X)d

Set I

=

Ra(X) d .

Note f i r s t t h a t f o r an Q s c e l e m e n t Q , ( u ( k ) - . f ) € R a ( X ) , h e n c e u(Q), = R,.

Now c o n s i d e r f E U ( X ) .

elements such t h a t

?,!

c1

+f.

Choose a n e t

{aa)

o f Ilsc

Then ( Q a ) , + f I , s o b y t h e above

identity, u(a ) +fI.

a 1

Now f < u(f) < U(Q) for a l l c1

u ( ~ , ) , + f , , and t h u s f I

( 7 0 . 3 ) C o r o l l a r y 1.

Proof. _-

=

(L,

h e n c e f I 5 ~ ( f 5) ~

u(f),.

R(Ra(X)d) = Ra(X)

d

.

The Up-down-up t h e o r e m h o l d s Ra(X)d' f o r F i n Ra(X)d ( c f . t h e i n t r o d u c t i o n t o C h a p t e r 1 2 ) , s o FRuQ

=

Ra(X)

S e t F = C(X)

d

.

d S i n c e R(Ra(X) ) = FR

FU, we t h u s n e e d o n l y

C'(X)'

show t h a t Feu'

=

FR

=

R a ( X t 3 C(X)'

36 7

FU.

=

We f i r s t show F R

=

FU.

Consider f E F

'.

Then, by t h e

above mentioned d i s c u s s i o n i n t h c i n t r o d u c t i o n t o Chapter 1 2 , t h e r e i s a n e s c e l e m e n t R s u c h t h a t p. ( 7 0 . 2 1 , U(')

=

f.

Thus F E F U .

= f , whence, by Ra (XI T h i s g i v e s u s F' c F u ;

Ra(X)d t h e o p p o s i t e i n c l u s i o n i s shown t h e same way. F'

FU g i v e s u s FRu

=

Fa.'

=

F

=

Fuu

=

FU

=

F i!

,

The i d e n t i t y

h e n c e , f i n a l l y , F 'uk

-

a. .

Remark. -___

Contained i n t h e above a r e t h e i d e n t i t i e s

S i n c e R z ~ ( X )i s~ D e d e k i n d c o m p l e t e , we h a v e t h e

(70.4) Corollary 2 . C(X)

Ra(X)d i s t h e D e d e k i n d c o m p l e t i o n o f

Ra(XId'

(70.5) p r o j

maps s u p r e m a a n d i n f i m a i n C(X) i n t o s u p r e m a Ra (XI and i n f i m a i n Ra(X)d: f o r e v e r y A c C(X), f = AA-in-C(X) implies f Ra (XI A

(A

Proof. =

(So, a f o r t i o r i , f

Ra(X)

Ra(X)d -

d) -in-C(X) Ra(X)

u

d).

= A(A

Ra(X)d')

C o n s i d e r A c C(X) s u c h t h a t AA-in-C(X)

AA is r a r e .

Since proj

=

0.

Then

preserves infima i n C"(X),

Ra ( X I

36 8

Chapter 16

t h i s gives us

d)

A(A

=

Ra (XI

u Ra (X)

d

=

0.

QE D

d , c o n s i d e r e d a s a mapping o f C ( X ) Ra (XI d . i n t o Ra(X) i s o r d e r c o n t i n u o u s , and t h e r e f o r e i s a l s o o r d c r (70.6) Corollary.

proj

c o n t i n u o u s c o n s i d e r e d a s a mapping o f C(X)

o n t o C(X) Ra(X) d '

(70.7)

Theorem.

-

C(X)'

=

[C(X)

dIc. Ra (XI

C(X)

0 , and s u p p o s e f,+O

p

P r o o f . C o n s i d e r pEC(X)',

in

I t f o l l o w s e a s i l y f r o m (70.6) t h a t t h e n f,+O

in

Ra(X)d'

R a ( X ) d , whence i n f , ( f a , p )

=

0.

Thus U E [ C ( X )

C o n v e r s e l y , c o n s i d e r $E[C(X)

dlc.

Ra(X)d

IC.

~t f o l l o w s from

Ra (XI (70.6) t h a t t h e l i n e a r f u n c t i o n a l $oproj c o n t i n u o u s , h e n c e an e l e m e n t o f C(X)',

on C ( X ) i s o r d e r Ra (XI and c o i n c i d e s w i t h on

QED

Remark.

The a b o v e t h e o r e m a n d ( 5 4 . 2 3 ) a r e s p e c i a l c a s e s

of a g e n e r a l theorem o f J . J . Masterson [ 5 2 ] , which s t a t e s t h a t an Archimedean R i e s z s p a c e a n d i t s D e d e k i n d c o m p l e t i o n h a v e t h e same o r d e r c o n t i n u o u s l i n e a r f u n c t i o n a l s t h e comment p r e c e d i n g ( 2 6 . 2 ) ) .

(cf.

( 5 4 . 2 2 ) and

C'(X)

Ra(X)'

=

3 C(X)'

369

The f o l l o w i n g a r e o f c o u r s e e q u i v a l e n t : 1'

c(x)'

20 c(x)

i s S e p a r a t i n g on c ( x ) ;

n

R ~ ( x = ) 0~; ~

t h e mapping o f C ( X ) o n t o C ( X ) Ra (XI i s an Ml-i s o m o r p h i s m . 3'

g i v e n by p r o j Ra(X)d

Combining t h i s w i t h (70.4), w e h a v e :

( 7 0 . 8 ) 'Theorem.

I f C(X)'

i s s e p a r a t i n g o n C ( X ) , t h e n Ra(X) d

i s t h e D e d e k i n d c o m p l e t i o n o f C(X).

Remark,

I n g e n e r a l , t h e D e d e k i n d c o m p l e t i o n o f C(X) c a n

b e r e a l i z e d i n a n a t u r a l way o n l y a s a q u o t i e n t s p a c e o f C'l(X), not as a subspace (cf. Chapterl7). s e p a r a t i n g on C(X)

-

The p r e s e n t c a s e - C(X)'

i s p r o b a b l y t h e o n l y one i n which t h e r e

e x i s t s a n a t u r a l , u n i q u e l y d e t e r m i n e d , s u b s p a c e o f C"(X)

which

c a n b e i d e n t i f i e d w i t h t h e d e s i r e d Dedekind c o m p l e t i o n .

(Using

t h e Axion o f C h o i c e , V e k s l e r

[ 5 3 ] shows t h a t t h e r e e x i s t i n

C"(X) many c o p i e s o f t h e D e d e k i n d c o m p l e t i o n o f C(X) c o n t a i n i n g C(X) i n i t s g i v e n i m b e d d i n g i n C"(X).) We c o m p l e t e t h i s 5 w i t h t h e c a s e t h a t C(X) i s D e d e k i n d complete.

(70.9)

I f C(X) i s D e d e k i n d c o m p l e t e , t h e n p r o j

maps C(X)

Ra ( X I o n t o Ra(X)

~

d

.

Chapter 16

370

Proof.

C o n s i d e r fERa(X)

d

;

we h a v e t o show f E C ( X ) Ra(Xld'

By t h e Remark f o l l o w i n g ( 7 0 . 3 ) , f ment R , h e n c e b y ( 7 0 . 2 ) , f

=

R

f o r some ~ s ecl e -

Ra (XI

But b y t h e h y p o t h e s i s

= u(k)

Ra ( X ) d m and ( 5 6 . 1 ) , u ( k ) E C ( X ) , s o we a r e t h r o u g h .

(>En

(70.10) Corollary.

I f C(X) i s D e d e k i n d c o m p l e t e and C(X)'

s e p a r a t i n g on C ( X ) , d o n t o Ra(X)

then p r o j

is

maps C(X) M I L - i s o m o r p h i c a l l y Ra (XI

.

T h i s f o l l o w s from ( 7 0 . 9 ) and t h e e q u i v a l e n c e s p r e c e d i n g (70.8). Remarks. -___

(1) I f t h e two c o n d i t i o n s o f ( 7 0 . 1 0 )

fied, X is called hyperstonian.

are s a t i s -

( 2 ) Note t h a t ( 7 0 . 1 0 ) i s a

p a r t i c u l a r c a s e o f Nakano's theorem ( 1 3 . 2 ) .

5 7 1 . The b a n d Ra(X)

dd

An i m m e d i a t e q u e s t i o n i s w h e t h e r Ra(X)dd c o n t a i n s a l l t h e lean elements.

We s h a l l s e e i n t h e n e x t 5 t h a t t h i s n e e d n o t

be s o . For e v e r y fERa.(!),

u ( f ) a n d k ( f ) a r e a l s o i n Ra(X) ( 6 7 . 3 ) .

Does t h i s p r o p e r t y h o l d f o r Ra(X)dd? However, b y ( 7 0 . 2 )

,

Again t h e a n s w e r i s n o .

C'(X)

=

Ra(X)'

0 C(XlC

371

F o r f E U ( X ) , i f f E R a ( X ) d d , t h e n u ( f ) , a ( f ) E R a ( X ) dd .

(71.1)

T h i s i m m e d i a t e l y e x t e n d s t o all f i n t h e R i e s z i d e a l g e n e r a t e d b y [J(X)

(71.2)

n

Ra(X)dd.

Indeed:

L e t I b e t h e R i e s z i d e a l g e n e r a t e d b y U(X)

n

Ra(X) dd .

Then f o r f € C " ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t : lo

fEI;

Z0

t(f),u(f)EI;

3'

1( f ) ,u(f) ERa(X) d d .

___ Proof. A l s o 2'

a n d 3'

2'

a r e c l e a r l y e q u i v a l e n t ( k ( f ) , u ( f ) ELJ(X)).

c l e a r l y i m p l i e s .'1

We show 1'

s i d e r f E I , a n d assume f i r s t t h a t f > 0.

i m p l i e s 3'.

Con-

Then t h e r e e x i s t s

g E U ( X ) n Ra(X)dd s u c h t h a t 0 < f < g , whence 0 < k(f)

u(g).

Since

gEIJ(X),

u(g)ERa(X)dd

(71.1),

0 , t h e n , by ( 7 1 . 1 ) ,

u(f)€Ra(X)dd,

hence u ( f ) i s l e a n , hence u ( f ) i s r a r e , hence f

is rare,

QED

5 7 2 . Examples

How l a r g e and how small c a n e a c h o f t h e b a n d s Ra(X)', C(X)',

Ra(X)dd, and Ra(X)d b e ? S i n c e e v e r y i n f i n i t e compact s p a c e h a s a t l e a s t one non (69.4) gives us:

isolated point,

(72.1)

Ra(X)'

= 0 i f and o n l y i f

X is a finite set.

A fort-

i o r i , t h e same h o l d s f o r Ra(X).

However, Ra(X)' i n f i n i t e : L e t X = OlN

may b e o n l y o n e - d i m e n s i o n a l , e v e n f o r X =

{1,2,...,n;*.,w~,

t h e Alex an d r o f f onec , Cl(X) = R 1( X ) ,

p o i n t c o m p a c t i f i c a t i o n o f N.

Then C(X)

and C"(X)

Let u b e t h e c h a r a c t e r i s t i c e l e -

ment o f

=

W.

Ru = Ra(X).

k"(X)

(cf. 539).

Then Ra(X)

=

=

n u , Ra(X)d = k m ( N ) , and Ra(X) dd

C o r r e s p o n d i n g l y , Ra(X)'

I n c o n t r a s t t o Ea(X)',

= lRw

we h a v e C(X)'

b u t i t may b e 0 f o r X i n f i n i t e .

and C(X)'

= L!

1

=

(IN).

# 0 for X finite,

Consider t h e c l a s s i c a l

Ct(X)

Theorem.

=

Ra(X)'

3 C(X)'

373

I f X i s d e n s e - i n - i t s e l f and

(Szpilrajn [51]).

s e p a r a b l e , t h e n f o r e v e r y n o n - z e r o r e g u l a r m e a s u r e 1-1, t h e r e exists

3

nowhere d e n s e s u b s e t Z s u c h t h a t p ( Z ) > 0 .

I n o u r c o n t e x t t h i s becomes:

(Szpilrajn).

(72.2)

I f X i s d e n s e - i n - i t s e l f and s e p a r a b l e ,

t h e n f o r e v e r y p E C t , ( X ) , p > 0 , t h e r e e x i s t s fERa(X)+ s u c h t h a t ( f , u ) > 0.

Hence

(72.3) Corollary.

I f X i s d e n s e - i n - i t s e l f and s e p a r a b l e , t h e n

c(x>c = 0.

I n p a r t i c u l a r , C(X)' C(X)'

=

0 , t h e n Ra[X)'

=

=

0 for X a real interval.

Cl[X),

and t h e r e f o r e Ra(X)dd

When =

C'I(X).

Thus i n t h i s c a s e Ra(X) i s o r d e r d e n s e i n C"(X). A c o m p l e t e l y d i f f e r e n t e x a m p l e o f a s p a c e X w i t h C(X)'

is X

=

BN\N:.

Dieudonn6 h a s shown t h a t i t h a s i n f a c t a much

stronger property (cf.

[ 1 4 ] , Lemma 8 ) .

A t t h e o t h e r e x t r e m e C(X)'

X

=

= 0

& i s an example.

c a n b e s e p a r a t i n g on C(X),

A n o t h e r e x a m p l e i s o b t a i n e d by t a k i n g

374

Chapter 1 6

f o r C(Y) t h e b i d u a l C"(X),

X any compact s p a c e .

C l ( X ) , w h i c h i s s e p a r a t i n g on C"(X).

(L"(p))c

= L

=

A more i n t e r e s t i n g e x a m p l e

1-1 t h e L e b e s g u e m e a s u r e on some r e a l i n -

i s t h e MU-space L a ( , ) ,

terval.

Here C(Y)'

1

(p),

w h i c h i s s e p a r a t i n g on L " ( p ) .

We c o n c l u d e w i t h some r e s u l t s on t h e v a g u e c l o s u r e s o f Ra(X)'

o r e q u i v a l e n t l y , on u(Ra(X) d d ) a n d u ( R a ( X ) d ) .

a n d C(X)',

Let W be t h e s e t o f i s o l a t e d p o i n t s o f X , and 2

=

X\W.

Then ( c f . ( 3 6 . 7 ) a n d t h e d i s c u s s i o n f o l l o w i n g ( 3 1 . 7 ) ) ZL

s o C1(X)

=

C 1 ( Z ) 0 al(W)

( 6 9 . 4 ) , W c C(X)'

a n d C"(X)

=

C"(Z)

o

=

co(W),

A l s o , by

a"(W).

a n d Z c Ra(X)'.

(72.4) Let W be t h e s e t o f i s o l a t e d p o i n t s o f X , and Z

=

X\W.

Then (i)

t h e v a g u e c l o s u r e o f Ra(X)d i s C l ( Z ) ;

(ii)

u(Ra(X)

dd

( i i i ) a(Ra(X) d )

Proof.

=

a"(W).

( i ) and ( i i ) are e q u i v a l e n t by t h e p r e c e d i n g

identities (cf. Z c Ra(X)',

) = C"(Z);

( 5 2 . 4 ) ) ; and a l s o ( i i ) and ( i i i ) .

Since

and C'(Z) i s t h e v a g u e l y c l o s e d band g e n e r a t e d by Z ,

i t f o l l o w s C ' ( Z ) i s c o n t a i n e d i n t h e v a g u e c l o s u r e o f Ra(X)'. 1 F o r t h e o p p o s i t e i n c l u s i o n , a (W) c C(X)', h e n c e Ra(X)' c [al(W)ld i n C'(Z).

=

C l ( Z ) , h e n c e t h e v a g u e c l o s u r e o f Ra(X)'

is contained

Thus ( i ) h o l d s , a n d we a r e t h r o u g h . QED

Note t h a t

1(W) = ( C ( X ) c ) a a n d ( t h e r e f o r e ) L"(W)

=(Ra(X)d) a .

C'(X)

Ra(X)'

=

This gives us (11) below.

3 C(X)'

375

include (I), which we have re-

We

marked earlier, €or comparison.

(72.5) Corollary 1 . [ I ) The following statements are equivalent: lo

c(x)c

2'

Ra(X)'

3'

Ra(X)dd

0;

=

C'(X);

=

C"(X)

=

- otherwise stated,

Ra(X)

is

order dense in C"(X).

(11) The following weaker statements are equivalent: lo

(c(x)c)a

'2

Ra(XlC is vaguely dense in c'(x);

3'

u ( R ~ ( X ) ~ ~ )= C"(X).

=

0;

We single out the following for emphasis.


If X is dense-in-itself, then U(REI(X)~~)

C" (X) .

Other consequences of (72.4):

=

376

C h a p t e r 16

-__ Proof.

(a) f o l l o w s from (72.4

( i i i ) and ( 5 2 . 1 ) .

fol-

(b)

lows from ( 7 2 . 4 ) , ( 3 3 . 6 ) , a n d ( 5 1 . 1 0 ) .

Q E 13

F i n a l l y , a n e n l i g h t e n i n g e x a m p l e i s s u p p l i e d b y C(X)

L m ( u ) , 1-1 t h e L e b e s g u e m e a s u r e on some r e a l i n t e r v a l . a l r e a d y r e m a r k e d t h a t C(X)'

= L

1

(p),

Cl(X).

d

E q u i v a l e n t l y , u(Ra(X) )

=

We h a v e

hence i s s e p a r a t i n g on

C(X), a n d t h u s i s v a g u e l y d e n s e i n C l ( X ) . l a t e d p o i n t s , s o , b y ( 7 2 . 6 ) , Ra(X)'

=

But X h a s no i s o -

i s a l s o vaguely dense i n

u(Ra(X)

dd

) = C"(X).

We c a n now a n s w e r ( i n t h e n e g a t i v e ) t h e two q u e s t i o n s r a i s e d a t the beginning of 571. Ra(X) d # 0 w h i l e k(Ra(X) d ) lean.

=

I n t h e above e x a m p l e ,

0 , hence a l l i t s elements a r e

We t h u s h a v e l e a n e l e m e n t s n o t c o n t a i n e d i n Ra(X)

And s i n c e u(n (X)

dd) R a ( X ) dd such t h a t u ( f ) E , Rn(X) .

=

dd

.

l l ( X ) , w e h a v e an e l e m e n t fERa(X)

dd

CHAPTER 1 7

TfIE D E D E K T N D C O M P L E T I O N OF C ( X )

5 7 3 . The Maxey r e p r e s e n t a t i o n

By a n a b u s e o f l a n g u a g e , we w i l l c a l l a R i e s z i d e a l I o f

C"(X) I

n

lean

i f a l l of i t s elements are lean - equivalently, i f

c ( x ) = 0.

( 7 3 . 1 ) Theorem.

i d e a l I , C"(X)/I

Proof.

(Maxey [ 3 7 ] ) .

F o r e v e r y maximal l e a n R i e s z

i s t h e D e d e k i n d c o m p l e t i o n o f C(X).

T i s norm c l o s e d , b y ( 6 5 . 3 ) ,

s o C"(X)/I

i s an

M I - s p a c e w i t h qIl(X) f o r t h e u n i t (q t h e q u o t i e n t m a p ) , a n d q i s an MI-homomorphism.

I t f o l l o w s q maps C(X) M I L - i s o m o r p h i c a l l y

onto q(C(X)).

(*)

q(C(X)) i s Dedekind d e n s e i n C"(X)/I.

By ( 7 . 6 ) , w e n e e d o n l y show t h a t q ( C ( X ) ) h a s a r b i t r a r i l y

small e l e m e n t s i n C " ( X ) / I .

Consider fEC"(X)/T,

t o show t h e r e e x i s t s g E q ( C ( X ) ) s u c h t h a t 0 < 377

> 0 ; we h a v e < If.

Otherwise

378

Chapter 1 7

s t a t e d , i f we d e n o t e by H t h e Riesz i d e a l o f C"(X)/T by f , t h e n we h a v e t o show t h a t H

n

generated

q(C(X)) # 0 .

c o n t a i n s I p r o p e r l y , h e n c e , by h y p o t h e s i s , i t i s n o t

q-l(H)

l e a n : q - l ( ~ )n C ( X )

+

0.

I t f o l l o w s 11 n q ( c ( x ) )

+

0.

T h i s e s t a b l i s h e s ( * ) ; we c o m p l e t e t h e p r o o f b y s h o w i n g t h a t C"(X)/I C"(X)/I;

(cf.

i s Dedekind c o m p l e t e .

w e show t h e r e e x i s t s ?EC"(X)/I

(7.1)).

Similarly, s e t

-

A1

5

-

B1,

(A,B)

Let

Set

gl

b e a Dedekind c u t i n

i n q ( C ( X ) ) , and n q ( C ( X ) ) , a n d B1

il =

A1

=

=

hence - q b e i n g an isomorphism

fEC"(X) s u c h t h a t A1

5 f

< B1.

Then

il 5

e l e m e n t o f A i s a supremum o f e l e m e n t s o f

ii

i s an infimum o f e l e m e n t s o f

G1,

q-'(i,) -1 q (B1)

n

A1

qf
u(VclR(ua)); show t h a t u = U ( V ~ ( )u) . We o f c o u r s e h a v e u a a we show t h a t f o r e v e r y u s c e l e m e n t v s u c h t h a t v > u(V,R(u,)), Now s e t u =

u(V

For e v e r y a,

w e have v > u.

(ua i s r e g u l a r ) .

v > ~ ( u , ) , hence v > uR(ua) = ua

hence v > u(V u ) I t follows v > v u acl a a'

=

u.

QED

( 7 4 . 4 ) C o r o l l a r y 1. (i)

ul, u 2 r e g u l a r u s c e l e m e n t s i m p l i e s ulvu2 i s r e g u l a r .

( i i ) R1,

R2 regular

RSC

elements implies k 1 ~ k 2 i s regular.

Dedekind Completion o f C(X)

381

( 7 4 . 5 ) C o r o l l a r y 2 . Under t h e h y p o t h e s i s o f ( 7 4 . 3 ) : ( i ) v,u(fa)

- v,a(fa)

i s rare;

( i i ) A,u(fa)

- ~,l(f,)

i s rare.

Proof.

we p r o v e ( i ) .

V,u(fa)

v u k ( f a ) 5 u(v,f,)

- VaR(fa)

proof o f ( 7 4 . 3 ) .

Now a p p l y ( 6 7 . 1 6 ) .

5 U ( Vc lfc l ) , s o 0

= 6(vaa(f,)),

< V,u(f,)

-

t h i s l a s t by t h e

QED

We w i l l d e n o t e t h e s e t o f r e g u l a r u s c e l e m e n t s by C(X)Ur and t h e s e t o f r e g u l a r Rsc e l e m e n t s by C(X)". The Dedekind c u t s o f C ( X ) with t h e

a r e i n one-one correspondence

r e g u l a r p a i r s i n C(X), h e n c e w i t h t h e r e g u l a r u s c e l e -

ments and w i t h t h e r e g u l a r Rsc e l e m e n t s .

Since t h e r e i s a

o n e - o n e c o r r e s p o n d e n c e between t h e Dedekind c u t s o f C(X) and t h e e l e m e n t s o f i t s Dedekind c o m p l e t i o n ? ( X ) , i t f o l l o w s t h e r e i s one between t h e r e g u l a r u s c e l e m e n t s and t h e e l e m e n t s o f

t ( X ) ; and s i m i l a r l y f o r t h e r e g u l a r Q s c e l e m e n t s .

Thus, b y

t h e Maxey t h e o r e m :

( 7 4 . 6 ) For a maximal l e a n R i e s z i d e a l I o f C'l(X), t h e q u o t i e n t map C"(X) %>

C"(X)/I

i s a b i j e c t i o n of each of the follow-

i n g s e t s onto C"(X)/I:

(i)

c(x)"

(ii) c(x)'~.

As a c o r o l l a r y , we have ( c f .

[13]):

Chapter 1 7

382

( 7 4 . 7 ) Theorem.

(Dilworth).

C(X)"

a n d C(X)"

a r e each iso-

m o r p h i c a s l a t t i c e s t o t h e Dedekind c o m p l e t i o n o f C ( X ) .

(So,

i n p a r t i c u l a r , t h e y a r e Dedekind c o m p l e t e . )

We w i l l n e e d t h e f o l l o w i n g g e n e r a l i z a t i o n o f ( 7 4 . 6 ) . A c t u a l l y , we w i l l n e e d i t o n l y f o r I more g e n e r a l l y . c o n t a i n s Ra(X)

(74.8)

R a ( X ) , b u t we s t a t e i t

=

R e c a l l t h a t e v e r y maximal l e a n R i e s z i d e a l (67.5).

L e t I be a l e a n R i e s z i d e a l c o n t a i n i n g Ra(X), and A modulo I .

a n e q u i v a l e n c e c l a s s o f C"(X)

I f A c o n t a i n s an k s c

element o r a usc element, then:

(i)

A c o n t a i n s e x a c t l y one r e g u l a r p a i r

(ii)

f o r e v e r y ksc e l e m e n t R i n A , u ( a )

every usc element u i n A,

USC

Proof.

=

uo, and f o r

~ ( u )= Q,;

( i i i ) k0 i s t h e l a r g e s t

smallest

(ko,uo);

RSC

e l e m e n t i n A , a n d uo i s t h e

element.

( i ) S u p p o s e A c o n t a i n s a n Rsc e l e m e n t R .

u ( k ) - RERa(X)

again, Ru(k)EA.

Then

( 6 7 . 1 6 ) , hence u ( 2 ) E A ; and a p p l y i n g (67.16) S e t uo

=

u ( k ) , R,

=

a u ( ~ ) . (k0,uo) i s a

r e g u l a r p a i r ; t h a t i t i s u n i q u e w i l l f o l l o w from ( i i ) . ( i i ) C o n s i d e r two E S C e l e m e n t s (u(k23 - k 2 )

+

(E2 - k l )

+

(kl

1 , ~ 2i n A .

- u(al))ERa(X)

+

u ( R ~ )- u ( a , ) = I

+

Ra(X)

I t f o l l o w s from (67.17) and (67.9) t h a t uku(k2) = u k u ( k l ) .

=

I. But

D e d e k i n d C o m p l e t i o n o f C(X)

and uRu(R1)

uilu(k2) = u(k,)

same u 0 .

=

u(Rl).

Thus R 1

and

a2

give the

S i m i l a r l y , two u s c e l e m e n t s i n A g i v e t h e same L o .

( i i i ) Given a n y Lsc e l e m e n t R i n A , R f i k(uo)

38 3

=

5 u(R)

=

uo, so

S i m i l a r l y , f o r any u s c element u i n A , u

Po.

2 uo. QED

Contained i n t h e above and t h e comment f o l l o w i n g i t

-

i t can a l s o be s e e n from (65.2)

-

is the r e s u l t that the only

lean r e g u l a r element i s 0. Finally,

i f ( a n d o n l y i f ) C(X) i s D e d e k i n d c o m p l e t e , t h e n

t h e o n l y r e g u l a r e l e m e n t s a r e t h e e l e m e n t s o f C(X):

( 7 4 . 9 ) The f o l l o w i n g a r e e q u i v a l e n t : 1'

C(X) i s D e d e k i n d c o m p l e t e ;

2O

C(xpr

=

C(X);

3O

C(X)Q'

=

C(X).

This is contained i n the r e s u l t s of t h i s be s e e n from (56.1)

§.

I t can a l s o

(and ( 7 4 . 2 ) ) .

575. A t h i r d r e p r e s e n t a t i o n

By t h e Maxey t h e o r e m ( 7 3 . 1 ) , t h e D e d e k i n d c o m p l e t i o n A

C(X) o f C(X) i s a n Mn-homomorphic image o f C"(X), w i t h C(X) mapped M X - i s o m o r p h i c a l l y ,

s u c h t h a t t h e D e d e k i n d c l o s u r e of

Chapter 1 7

384

A

( t h e image o f ) C ( X ) i s i t s D e d e k i n d c o m p l e t i o n .

C(X)

is

c l e a r l y t h e s m a l l e s t MI-homomorphic image o f C1'(X) w i t h t h i s property.

5 , we p r e s e n t w h a t i s p r o b a b l y t h e

In the present

l a r g e s t Mn-homomorphic image o f C1'( X)

w i t h t h e same p r o p e r t y .

We e x t e n d o u r term "Riemann s u b s p a c e " ( 5 5 7 ) t o q u o t i e n t spaces of

Cl'(X).

Given a norm c l o s e d Riesz i d e a l I o f C"(X),

t h e n by t h e Riemann s u b s p a c e o f C 1 I ( X ) / I , we w i l l mean t h e Dedekind c l o s u r e o f q ( C ( X ) ) ( q t h e q u o t i e n t map); we w i l l d e n o t e i t by R ( C " ( X ) / I ) .

For a band o f Crl(X), t h e d e f i n i t i o n r e d u c e s

t o t h e former one. The blaxey t h e o r e m c a n now be s t a t e d a s f o l l o w s : I f I i s a maximal l e a n R i e s z i d e a l , t h e n (i)

R(C"(X)/I)

( i i ) R(C"(X)/I)

We show t h a t f o r I

=

i s Dedekind c o m p l e t e , and =

C"(X)/I.

Ra(X), ( i ) s t i l l h o l d s ( ( 7 5 . 3 ) b e l o w ) .

I n t h e f o l l o w i n g two lemmas, q i s t h e q u o t i e n t map o f C'I(X) o n t o C"(X)/Ra(X).

In general, q is not order continu-

o u s ; however, i t h a s a p a r t i a l o r d e r c o n t i n u i t y :

(75.1)

Lemma 1.

( i ) Given a s e t {u } o f u s c e l e m e n t s , c1

.

u = A u i m p l i e s qu = A q u a a a a ( i i ) Given a s e t { i } o f Rsc e l e m e n t s , a i = v i implies q i = V q i acl a a

.

P-r o o f . -

We p r o v e ( i ) .

S i n c e q p r e s e r v e s o r d e r , qu

5 qua

D e d e k i n d C o m p l e t i o n o f C(X)

< qua f o r a l l a ; S u p p o s e t h a t f o r some f E C " ( X ) , q f -

for a l l

we show q f < qu.

The a s s u m p t i o n t h a t q f < qu

u s ( f - u )'€Ra(X)

for a l l a.

c1

(f

-

385

u)+ERa(X),

a

for a l l a gives

I t f o l l o w s from (68.6) t h a t

hence q f 5 qu. QED

(75.2)

Lemma 2 .

The f o l l o w i n g c o i n c i d e :

lo

q(c(x)ur);

2O

q (C (XI R r ) ;

3O

q(C(XIU);

4O

q(C(X)5;

5'

R [ C" (X) /Ra (XI ] .

Proof.

(74.8),

That l o ,

2O,

3O,

a n d '4

c o i n c i d e f o l l o w s from

s o i t s u f f i c e s t o show t h a t q ( C ( X ) u ) = R [ C " ( X ) / R a ( X ) ] . c h o o s e A , B c C(X) s u c h t h a t V A

Given u€C(X)',

Then b y Lemma 1, A q ( B )

quE R[C" (X) / Ra (X) ] Conversely,

=

qu

=

q(R ( u ) )

= Vq(A)

=

,

a(u), AB

=

u.

hence

. if

?

= Aq(A),

A c C(X), t h e n , s e t t i n g u = A A

a n d a p p l y i n g Lemma 1 a g a i n , w e h a v e q u =

?.

Thus ? € q ( C ( X ) u ) . QED

( 7 5 . 3 ) Theorem.

R[C"(X)/Ra(X)]

i s Dedekind c o m p l e t e , h e n c e i s

t h e D e d e k i n d c o m p l e t i o n o f C(X).

Proof.

S i n c e t h e q u o t i e n t map q maps C(X)ur o n e - o n e

Chapter 1 7

386

and o r d e r p r e s e r v i n g o n t o R[C"(X)/Ra(X)]

( ( 7 4 . 8 ) and ( 7 5 . 2 ) ) ,

t h i s f o l l o w s from ( 7 4 . 7 ) . Q E 1)

i s D e d e k i n d c o m p l e t e , i t i s Dedekind

S i n c e R[C"(X)/Ra(X)] c l o s e d i n C"(X)/Ra(X).

i t h a s a much s t r o n g e r

Actually,

property:

((75.4)

I f a subset

o f R[C"(X)/Ra(X)]

( s a y ) , t h e n V F-in-C'l(X)/Ra(X) a R[C"(X)/Ra(X)].

Proof.

-~

-

fa

= q(ka)

k

< AIL(X).

a-

e x i s t s and l i e s i n

{fa} i lq(lL(X)) f o r some A.

f o r some

RSC

element k

Set R = V R cta

.

i s bounded a b o v e

c1

(75.2),

Now f o r e a c h a, a n d we c a n assume

Then, by ( 7 5 . 1 ) ,

qk

=

V

cta

. QED

Whence, p e r h a p s s u r p r i s i n g l y ,

( 7 5 . 5 ) Theorem.

Proof. C"(X)/Ra(X).

R[C"(X)/Ra(X)]

Suppose

i s o r d e r c l o s e d i n C''(X)/Ra(X).

{ f a ] c R[C"(X)/Ra(X)],

By ( 7 5 . 4 ) , f o r e v e r y

a,

e x i s t s and b e l o n g s t o R[C"(X)/Ra(X)].

qct =

Since

'from ( 7 5 . 4 ) a g a i n t h a t f€R[C"(X)/Ra(X)].

0 , we h a v e aAb - aAc
1,

and

-

ua,

Chapter 18

406 hl

=

P1

h

=

(Pa

c1

-

v B < a p I+

c1

@

It follows f r o m (78.5) that p+ = Ch

ci

.

> 1.

Thus, if we show that

{ g c x } , {ha} satisfy the conditions of ( 8 0 . 4 ) , i t will f o l l o w

that p + is meager. Cg,

exists: in effect, by (78.5) again, 7 g M

= VR(l =

k.

Combining this with (79.5) gives us C g M 5 u ( C k ( g a ) ) ; thus (ii) is satisfied. a l l a.

q,

Since 0 < h (x 5 p,

=

(p - 4,)'.

ha is mcager f o r

It remains to show that 0 < h M 5 pa €or each a.

are positive.)

QED

The Meager E l e m e n t s

(80.9) Thcorcm. ti

407

For e v e r y f E C " ( X ) , t h e r e e x i s t s a u s c e l e m e n t

such t h a t (i)

~ i r ( f

The Bidual of C(X)I

The bidual of C(X) I

The Fall of the House of Zeus

The Mystery of the Trail of Terror

The Theory of the Decomposition of Azomethane

The Location of the treasury of Atreus

The Theology of the Gospel of Mark

The Fall of the House of Usher

The Sailor of the Seas of Fate

The Fall of the House of Usher

Encyclopedia of the Wars of the Roses

The Fall of the House of Usher

The Lord of the Rings The Return of the

The Rise of the Image, the Fall of the Word

The Sailor of the Seas of Fate

Fall of the House of Usher, The

The Manual of the Warrior of Light

The Capture of the Earl of Glencrae

The Beginning of the Age of Mammals

The End of The Novel of Love

The Sailor of the Seas of Fate

The Foundations of the Origin of Species

The Wreck of The Rivers of Stars

The Theology of the Book of Isaiah

Conquest of the Planet of the Apes

The Fall of the House of Usher

The End of the History of Art?

The End of the Novel of Love

The History of the Siege of Lisbon

The Harrowing Of The Dragon Of Horasbreath

The Bidual of C(X)I

The bidual of C(X) I

The Fall of the House of Zeus

The Mystery of the Trail of Terror

The Theory of the Decomposition of Azomethane

The Location of the treasury of Atreus

The Theology of the Gospel of Mark

The Fall of the House of Usher

The Sailor of the Seas of Fate

The Fall of the House of Usher

Encyclopedia of the Wars of the Roses

The Fall of the House of Usher

The Lord of the Rings The Return of the

The Rise of the Image, the Fall of the Word

The Sailor of the Seas of Fate

Fall of the House of Usher, The

The Manual of the Warrior of Light

The Capture of the Earl of Glencrae

The Beginning of the Age of Mammals

The End of The Novel of Love

The Sailor of the Seas of Fate

The Foundations of the Origin of Species

The Wreck of The Rivers of Stars

The Theology of the Book of Isaiah

Conquest of the Planet of the Apes

The Fall of the House of Usher

The End of the History of Art?

The End of the Novel of Love

The History of the Siege of Lisbon

The Harrowing Of The Dragon Of Horasbreath

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