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

Elementary properties

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

Elementary properties

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 <

A = {a

I n g e n e r a l , n e i t h e r t h e supremum n o r infimum e x i s t s . supremum d o e s e x i s t , we w i l l d e n o t e i t by V A o r

.

I f the

a and i f a a’ t h e infimum e x i s t s , we w i l l d e n o t e i t by A A o r Aaaa. V

By an o r d e r e d v e c t o r s p a c e E , we w i l l mean a v e c t o r s p a c e endowed w i t h an o r d e r which s a t i s f i e s t h e c o m p a t i b i l i t y conditions:

 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 <

then a

F o r e a c h a E E , we s e t la1 =

(-a)V 0

[lo] or [47].

aV(-a), a+

=

and

= aVO,

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

-(a/\O).

=

n Clai w i t h

F o r a more g e n e r a l f o r m u l a t i o n

( b u t which i s a c t u a l l y e q u i v a l e n t ) , c f .

a

=

s t r a i g h t forward.

(2.6)

/a

+

+

-

0;

(0)

a

A

a

(i)

a

=

a+ - a ;

(ii)

la1

(iii)

/-a/

(iv)

l X a / = / ~ / l a flu r a l l X ;

(v)

a

(Vi)

(a

=

a+ + a

= =

= 0 +

=

a

+

Va

-

;

la/;

i f and o n l y i f

/a1

=

0;

b)+ < a+ + b + , (a + b ) - < a-

+

b-,

b/ < /a1 + j b l .

The f o l l o w i n g i s due t o B i r k h o f f ( [ 9 ] ,

Chapter 1 4 ,

Theorem 8 ) .

(2.7) Theorem. -

(Birkhoff).

For a l l e l e m e n t s a , b , c of a Riesz

space, IcVa

-

cvbl

+

I a a - cr\bI

=

la - b l .

Riesz Spaces Proof.

U s i n g E x e r c i s e 7 and t h e d i s t r i b u t i v e l a w s , we

have (cVa - c v b )

(cVa)V ( c v b )

=

i n g t h e s e and a p p l y i n g ( 2 . 2 ) +

avb) - ( c

+

(cVa)A ( c v b ) = c v ( a v b ) -

-

cV ( a / \ b ) , and s i m i l a r l y , l a a - a b

(c

9

I

= c ( av

gives us

a/\b) = avb - a/\b

c V a - cvbl

la - b

=

) - a(a/\b). +

Add-

Iaa-cxbl=

( a g a i n by

Exercise 7 ) .

(2.8)

Corollary.

For a l l a , b , c o f a R i e s z s p a c e ,

< la - b l , IcVa - c v b l -

Iaa - abI < la - b l

A R i e s z s p a c e E i s Archimedean 0 = An(l/n)a.

every a , b € E ,

.

i f f o r every a € E + ,

A common ( e q u i v a l e n t ) d e f i n i t i o n i s t h a t f o r

if 0 < nb < a f o r a l l n , then b

R i e s z s p a c e s w i t h which

=

0.

A l l the

we w i l l work w i l l be Archimedean.

The s i m p l e s t example o f a non-Archimedean R i e s z s p a c e i s t h e p l a n e R'

ordered l e x i c o g r a p h i c a l l y : f o r a l l (al ,bl)

(a2,b2)tlR',

(al,bl)

5

,

( a 2 , b 2 ) i f (i) a l 5 a 2 o r ( i i ) a l = a 2

and b l 5 b 2 .

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

A l i n e a r subspace F o f a Riesz space E w i l l be c a l l e d a

Chapter 1

10

Riesz subspace of E i f i t i s closed under the l a t t i c e operat i o n s : a , b E F i m p l i e s avb, a/\bEF.

I t i s e a s i l y shown t h a t

f o r F t o be a R i e s z s u b s p a c e , i t i s s u f f i c i e n t t h a t a E F imply a+EF (Exercise 6 ) .

A Riesz subspace F i s i t s e l f a Riesz space

under t h e induced o r d e r , w i t h F

n E,

= F,.

The 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 o f R i e s z s u b spaces i s a Riesz subspace.

Consequently, f o r each s u b s e t A

o f E , t h e i n t e r s e c t i o n o f a l l t h e Riesz s u b s p a c e s c o n t a i n i n g A i s t h e s m a l l e s t Riesz subspace c o n t a i n i n g i t .

I t i s called

t h e R i e s z s u b s p a c e g e n e r a t e d by A . The sum o f two R i e s z s u b s p a c e s i s , i n g e n e r a l , n o t a Ri-esz s u b s p a c e ( c f . E x e r c i s e 1 2 ) . We t u r n t o R i e s z i d e a l s .

For a l i n e a r s u b s p a c e I o f a R i e s z s p a c e E , t h e f o l l o w -

(3.1)

ing a r e equivalent:

'1

2'

(i)

I i s a R i e s z s u b s p a c e , and

(ii)

0 < b < a€ I

a€I

i m p l i e s bE I ;

and ( b l < la1 imply bE I .

The p r o o f i s s t r a i g h t f o r w a r d .

I f I h a s t h e above p r o -

p e r t i e s , we w i l l c a l l i t a Riesz i d e a l o f E . ( 5 6 ) , t h e term "ideal"

As we w i l l s e e

i s j u s t i f i e d by t h e f a c t t h a t t h e R i e s z

i d e a l s a r e p r e c i s e l y t h e n u l l s p a c e s o f t h e homomorphisms of Riesz spaces. The 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 o f R i e s z

Riesz Spaces

11

i d e a l s i s a Riesz i d e a l , so a g a i n , f o r each s u b s e t A of E , t h e i n t e r s e c t i o n of a l l the Riesz i d e a l s c o n t a i n i n g A i s t h e I t i s c a l l e d t h e Riesz

s m a l l e s t Riesz i d e a l c o n t a i n i n g i t . i d e a l g e n e r a t e d by A .

This Riesz i d e a l has a simple c h a r a c t e r i z a t i o n :

( 3 . 2 ) The R i e s z i d e a l g e n e r a t e d by a s u b s e t A o f a R i e s z s p a c e

E c o n s i s t s of a l l bEE

satisfying:

m

5 z l X i [ a i l f o r some

Ibl

cm R +}. s e t { a l , - . - , a m } c A and { l l , - . - , ~

And t h e xi's c a n

be r e s t r i c t e d t o e l e m e n t s o f N .

Again, t h e 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 .

Unlike t h e

s i t u a t i o n w i t h R i e s z s u b s p a c e s , we have

(3.3) Theorem. Let { I } b e a c o l l e c t i o n of R i e s z i d e a l s o f a

a t h e s e t o f a l l f i n i t e sums R i e s z s p a c e E , and s e t I = C I c1 a' n z l a i , where a l , - * - , a n a r e a r b i t r a r y a ' s , n i s an a r b i t r a r y n a t u r a l number, and a i E Ia

(i

=

1

,

e

.

e

Then I i s a R i e s z

,n).

i

i d e a l , and I + = c a ( I a ) + .

n Suppose 0 < b i a = zlaiEI.

P roof. -

h e n c e , by t h e D e c o m p o s i t i o n Lemma ( 2 . 5 ) , b 0

i bi 2 ( a i ) ' -

t h i s gives

US

(i

=

b.EI. 1

l , . . ., n ) .

cli

(i

Since each I

,...

= 1

< b < zy(ai)+, Then 0 =

cybi, with

i s a Riesz i d e a l , "i , n ) , h e n c e bE 1. Thus

Chapter 1

12

c o n d i t i o n ( i i ) i n ( 3 . 1 ) 1' i s s a t i s f i e d . To v e r i f y c o n d i t i o n n .E I . We have 0 < a+ < z;(ai)+ ((vi) in ( i ) , c o n s i d e r a = c 1a 1 (2.6)), h e n c e , r e p e a t i n g t h e argument used a b o v e , we o b t a i n a + E I , which e s t a b l i s h e s ( i ) ( E x e r c i s e 6 ) .

For the f i n a l

e q u a l i t y i n t h e theorem, s e t b = a i n t h e f i r s t p a r t of t h e proof. QED

I n p a r t i c u l a r , t h e sum o f two Riesz i d e a l s i s a R i e s z ideal.

While t h i s i s n o t t r u e f o r R i e s z s u b s p a c e s , n o t e t h a t

t h e sum o f a R i e s z s u b s p a c e and a R i e s z i d e a l i s a Riesz s u b space (Exercise 1 3 ) .

a,bEE a r e d i s j o i n t i f IalAIbl = 0 . k i n t from b .

We a l s o s a y a is dis-

Properties of d i s j o i n t n e s s a r e c o l l e c t e d i n

E x e r c i s e s 5 , 8 , and 9 .

Two s u b s e t s A , B o f E a r e d i s j o i n t i f

e v e r y e l e m e n t o f A i s d i s j o i n t from e v e r y e l e m e n t o f B. 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 . 4 ) Let A , B be s u b s e t s o f a R i e s z s p a c e E , and H , I t h e R i e s z

i d e a l s which t h e y r e s p e c t i v e l y g e n e r a t e .

I f A and B a r e

d i s j o i n t , t h e n s o a r e H and I .

( 3 . 5 ) For two R i e s z i d e a l s H , I o f a R i e s z s p a c e E , t h e f o l l o w -

ing are equivalent:

'1

H and I a r e d i s j o i n t ;

13

Riesz Spaces

2'

a b = 0 f o r a l l a € H,,

'3

H

4'

H, (7 I + = 0 .

n

2'

Proof. we show 1' la1

I

=)

3'

=

0;

i m p l i e s 1' 4'

F i n a l l y , i f 4'

by t h e d e f i n i t i o n o f d i s j o i n t n e s s ; I f 1' h o l d s , t h e n f o r a € H

a 2'.

l a l ~ l a l= 0 , s o 3'

=

bE I + ;

holds.

'3

h o l d s , t h e n f o r a € H,,

n

I,

o f c o u r s e i m p l i e s 4'. a ~ b € H,

b € I,,

0 I,

= 0

QED

For any s u b s e t A o f E , we w i l l d e n o t e by Ad t h e s e t o f e l e m e n t s o f E d i s j o i n t from A , and we w i l l c a l l i t t h e d i s j o i n t I t i s e a s y t o v e r i f y t h a t Ad i s a R i e s z i d e a l .

of A.

t h a t i f H i s t h e R i e s z i d e a l g e n e r a t e d by A , t h e n Hd

Also =

Ad.

L e t E , F be two R i e s z s p a c e s , and d e n o t e t h e e l e m e n t s o f E by a ' s and t h o s e o f F by b ' s . E

x

F and d e f i n e ( a l , b l )

5

bl

b2.

5

Form t h e p r o d u c t v e c t o r s p a c e

( a 2 , b 2 ) i f and o n l y i f a l

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

x

5

a2,

F i s a Riesz s p a c e , w i t h

t h e o p e r a t i o n s V ,A g i v e n c o o r d i n a t e w i s e : ( a l , b l ) V ( a 2 , b 2 )

=

(alVa2,b1Vb2), and ( i i ) E and F a r e d i s j o i n t R i e s z i d e a l s o f E

x

F.

I f a R i e s z s p a c e i s t h e d i r e c t sum, i n t h e a l g e b r a i c s e n s e , o f two o f i t s R i e s z i d e a l s , t h e n t h e same two p r o p e r t i e s hold:

( 3 . 6 ) Theorem.

Suppose t h e R i e s z s p a c e E i s t h e d i r e c t sum

o f two R i e s z i d e a l s : E

=

I 3 H (that is, I

+

H = E, I

nH

= 0).

14

Chapter 1

For e v e r y a E E , b y a = aI av b

+

denote t h e r e s u l t i n g unique decomposition o f a

a[!.

Then f o r a l l a , b E E

aIVb I

+

aHVbH

a/\b = aIAbI

+

a# bH

=

*

In p a r t i c u l a r , a+

=

(a,)+ +

+,

-

a- = (aI)- + l a 1 = la11

aHI

+

*

In addition, a > 0 i f and o n l y i f 0 < a a ( a . - I ' H -

Proof.

Since I

nH

=

0 , I and H a r e d i s j o i n t ( 3 . 5 ) ,

h e n c e , b y E x e r c i s e 9 , a+ = ( a I ) +

+

(aH)+.

The r e s t o f t h e

theorem follows e a s i l y QED

(3.7) Corollary.

I f E = I EI H , I a n d H R i e s z i d e a l s , t h e n

H = I d a n d I = Hd .

Proof.

We h a v e H C I d

s i d e r bE I d , b > 0. bAbI = 0 , h e n c e b

=

.

Then a b

For t h e o p p o s i t e i n c l u s i o n , con=

0 f o r a l l a € I + , hence bI =

bHEH. QED

No te t h a t , b y t h e C o r o l l a r y , a R i e s z i d e a l I c a n h a v e most o n e co m pl em ent ar y ___

Riesz i d e a l , Id .

I n g e n e r a l , Id need

Riesz Spaces

15

n o t be complementary t o I , t h a t i s , we may have I

+

Id # E .

As an e x a m p l e , l e t E b e t h e R i e s z s p a c e o f c o n t i n u o u s f u n c t i o n s

on t h e r e a l i n t e r v a l

[ 0 , 1 ] , and I be t h e s e t o f f u n c t i o n s

which v a n i s h on [1/2,1].

Then I d c o n s i s t s o f t h o s e f u n c t i o n s

which v a n i s h on [0,1/2], and I + I d d o e s n o t c o n t a i n t h e c o n s t an t f u n c t i o n s .

5 4 . Order convergence

Consider a s e t S. set

a

(that is,

A mappingA->S

f

i n t o S of a d i r e c t e d

i s endowed w i t h an o r d e r < satisfying:

f o r a l l u l , a 2 € d , t h e r e e x i s t s a 3 € d s u c h t h a t a1,a2 < a ) w i l l be c a l l e d a n e t i n S.

For e a c h a€..$,

by sa and r e f e r t o f a s " t h e n e t { s , ) . t h e n o t a t i o n { s a } i n two d i s t i n c t ways:

we w i l l d e n o t e f ( a )

Warning:

We t h u s u s e

Standing a l o n e , {s,}

s i m p l y d e n o t e s an i n d e x e d s e t , w h i l e t h e p h r a s e " t h e n e t

{s,)"

i n d i c a t e s t h e i n d e x s e t i s d i r e c t e d (and t h e sa's a r e

not necessarily d i s t i n c t ) .

We t u r n t o a R i e s z s p a c e E . A n e t { a } i n E w i l l be c a l l e d a s c e n d i n g ( r e s p . d e s c e n d -

a

t o n i c n e t i s one which i s e i t h e r a s c e n d i n g o r d e s c e n d i n g .

aa

+

w i l l d e n o t e t h a t { a a } i s an a s c e n d i n g n e t , a n d a a + , t h a t i t i s

a descending n e t .

a and a,+a, t h a t i t i s a d e s c e n d i n g n e t w i t h a a' We e m p h a s i z e t h a t t h e s e n o t a t i o n s a l w a y s i m p l y t h a t

net with a = a

= A

a

cla

.

a cl + a w i l l d e n o t e t h a t { aa ) i s an a s c e n d i n g

V

Chapter 1

16

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

(4.1)

Given a a + a and b +b ( t h e same d i r e c t e d i n d e x s e t ) , t h e n a (i)

(-aa) +(-a) ;

(ii)

Xaa+Aa

(iii)

(aa

(iv)

(aolvba).f ( a v b ) ;

(v)

(aaAba) ( a / \ b ) .

+

for all

ba)+(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 <

whence a l s o 2b < a

t i n u i n g b y i n d u c t i o n , w e h a v e nb < a - c for a l l n I t follows b

=

c.

-

=

Con-

1,2,..

..

0.

QE D

Let E b e a n A r c h i m e d e a n R i e s z s p a c e , I: a R i e s z s u b s p a c e ,

(7.6)

and I t h e R i e s z i d e a l g e n e r a t e d b y F .

Then t h e f o l l o w i n g a r e

e qu i v a 1e n t : '1

F i s Dedekind d e n s e i n I ;

2'

F has a r b i t r a r i l y small elements i n I.

Proof.

We n e e d o n l y show 2'

and c o n s i d e r a € I .

Choose b E F

i m p l i e s 1'.

Assume 2'

such t h a t -b < a < b.

holds Then

a + b > 0 and b - a > 0 , h e n c e , by ( 7 . 5 ) , t h e r e e x i s t { c n } , a + b = V c and b - a = V d Write the acl B 6' l a s t o f t h e s e a - b = A ( - d ) . Then f r o m t h e f i r s t , Id,} c F s u c h t h a t

B

a A

=

B

v 0 c. Q

- b = V

(c

n o .

B

- b ) , and from t h e s e c o n d , a = A,(-d,)+

b=

(b - d o ) . QED

A s i s common w i t h c o n c e p t s o f c o m p l e t e n e s s , t h e r e i s a

s t a n d a r d c o m p l e t i o n t h e o r e m f o r Dedekind c o m p l e t e n e s s .

The

n e c e s s i t y o f Archimedeanness i n t h e following theorem f o l l o w s from ( 7 . 2 )

a n d t h e o b v i o u s f a c t t h a t a R i e s z s u b s p a c e o f an

Archimedean R i e s z s p a c e i s i t s e l f A r c h i m e d e a n .

Chapter 1

36

(7.7)

-___ Theorem.

(Nakano, J u d i n )

For e v e r y Archimedean R i e s z

s p a c e E , t h e r e e x i s t s a Dedekind c o m p l e t e R i e s z s p a c e

e,

unique

up t o R i e s z isomorphism, and a R i e s z isomorphism o f E o n t o a Dedekind d e n s e R i e s z s u b s p a c e o f

g.

For a p r o o f , s e e [ Z ] o r [ 3 6 ] . E i s c a l l e d t h e Dedekind c o m p l e t i o n o f E .

58.

Countability properties

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

c l o s e d i f i t i s c l o s e d i n E u n d e r o r d e r c o n v e r g e n c e o f sequences Clearly order closedness implies o-order c1osedness;but

i n A.

not conversely. The u n i o n o f a f i n i t e c o l l e c t i o n o f o - o r d e r c l o s e d s e t s and 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 a - o r d e r closed.

And ( 5 . 1 ) h o l d s h e r e : A R i e s z s u b s p a c e F i s a - o r d e r

c l o s e d i f and o n l y i f F,

i s o-order closed.

The a - o r d e r c l o s u r e o f a s e t A i s t h e i n t e r s e c t i o n o f a l l t h e a - 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 , hence i s t h e s m a l l e s t such s e t .

A g a i n , i n g e n e r a l , we c a n n o t o b t a i n t h e a - o r d e r

c l o s u r e o f A by s i m p l y a d j o i n i n g t o A t h e l i m i t s o f a l l 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 A. I f t h i s l a t t e r s e t i s i n s e r t e d i n p l a c e of A(') and t h e t e r m " a r b i t r a r y " r e p l a c e d by " c o u n t a b l y " r e s u l t i n g theorem h o l d s f o r t h e p r e s e n t c a s e .

,

i n (4.5)

then the

Similarly for

Riesz Spaces

37

( 5 . 2 ) and ( 5 . 3 ) w i t h " o r d e r c l o s u r e " r e p l a c e d by " a - o r d e r closure".

We w i l l c a l l a R i e s z i d e a l g e n e r a t e d by a s i n g l e e l e m e n t a p r i n c i p a l Riesz i d e a l . P r i n c i p a l

band i s d e f i n e d s i m i l a r l y .

From ( 3 . 2 ) , we h a v e :

( 8 . 1 ) The R i e s z i d e a l I g e n e r a t e d by an e l e m e n t a o f a R i e s z space E c o n s i s t s o f a l l b E E (n d e p e n d i n g on b ) .

Thus I

satisfying: =

/b/< n l a l f o r some n

lJnn[- la1 , l a ] 1 .

Combining t h i s w i t h ( 5 . 3 ) , i t i s e a s y t o show t h a t :

( 8 . 2 ) L e t I be t h e R i e s z i d e a l g e n e r a t e d by an e l e m e n t a . Then (i)

t h e o - o r d e r c l o s u r e of I c o i n c i d e s w i t h i t s o r d e r

c l o s u r e , hence i s t h e band H g e n e r a t e d by a ; ( i i ) f o r e v e r y bEI-I,

,

b = Vn(bA(nlal)).

A R i e s z s p a c e w i l l b e c a l l e d o-Dedekind c o m p l e t e i f e v e r y c o u n t a b l e s u b s e t which i s bounded above h a s a supremum and e v e r y one bounded below h a s an infimum.

of ( 7 . 2 )

Note t h a t t h e p r o o f

a c t u a l l y e s t a b l i s h e s t h e s t r o n g e r r e s u l t : a a-Dedekind

c o m p l e t e R i e s z s p a c e i s Archimedean. The R i e s z theorem ( 7 . 3 ) c a n n o t be e x p e c t e d t o h o l d f o r a

38

Chapter 1

o-Dedekind complete Ricsz s p a c e .

(8.3)

We h a v e t h e w e a k e r r e s u l t :

I n a 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 E, e v e r y p r i n c i p a l

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

-__ Proof.

Let 1 be t h e band g e n e r a t e d by a ; a n d , f o r sim-

p l i c i t y , we c a n t a k e a > 0.

We show c

=

v([O,b]

n

=

=

Vn(bA(na)).

I ) ; t h e remainder of the proof w i l l then

be t h e same a s f o r ( 7 . 3 ) .

(8.2), d

Given b E E + , s e t c

F o r e v e r y dE [ O , b ]

V,,(dA(na)) < vn(bA(na))

=

r-)

I , we h a v e , by

c.

QED

Notation.

Let E be a a-Dedekind complete R i e s z s p a c e ,

aEE, a n d I t h e b a n d g e n e r a t e d b y a.

By t h e a b o v e , I i s a p r o -

j e c t i o n b a n d , s o f o r e a c h b E E , w e h a v e t h e component b I o f b i n I , a n d f o r e a c h s u b s e t A o f E , we h a v e A I

=

projI(A).

H e n c e f o r t h , we w i l l a l s o d e n o t e t h e s e b y b a a n d Aa r e s p e c t i v e l y . Note t h a t i n t h i s n o t a t i o n , I

=

E

a

.

5 9 . R i e s z n o r m s a n d Ranach l a t t i c e s

We w i l l u s e t h e s y m b o l

[(1(1

t o d e n o t e a seminorm, s i m p l y

p o i n t i n g o u t when i t i s a c t u a l l y a norm. A s u b s e t A o f a Riesz s p a c e i s c a l l e d s o l i d i f a E A and

/b/

5 /a1 i m p i i e s bEA

-

e q u i v a l e n t l y , aEA

implies

Riesz Spaces

39

[-lal,lal] c A.

We r e c o r d t h e b a s i c p r o p e r t i e s o f s o l i d s e t s

i n Exercise 2 2 .

Note t h a t a R i e s z i d e a l c a n be d e f i n e d a s a

s o l i d l i n e a r subspace.

F o r a seminorm

(9.1)

.!I

on a R i e s z s p a c e E , t h e f o l l o w i n g

are equivalent; 1'

the unit ball of

2'

for a l l a,bEE,

[I .[I

is solid;

5 ;la\\.

Ibl < la1 i m p l i e s IIbll

The v e r i f i c a t i o n i s s i m p l e . A seminorm s a t i s f y i n g t h e a b o v e w i l l b e c a l l e d a R i e s z

s e_ m_ i n_ o r_ m , o r a R i e s z norm i f i t i s a c t u a l l y a norm. _ from

2O,

IIal] =

11

la(ll f o r a l l a E E .

From ( 2 . 8 ) ,

(9.2) If

II.11

Note t h a t ,

we h a v e

i s a R i e s z seminorm, t h e n f o r a l l a , b E E ,

11 av c

- bvcll <

/ l a c - bAcll <

a - bll

11 a

- bll

,

.

I t follows t h e l a t t i c e o p e r a t i o n s V , A a r e uniformly con-

t i n u o u s under

11 -11 .

Note t h e c o r o l l a r y t h a t E,

i s closed under

A t p r e s e n t , o u r i n t e r e s t i s R i e s z norms.

II.11. A R i e s z space

endowed w i t h a R i e s z norm w i l l b e c a l l e d a normed R i e s z s p a c e .

Chapter 1

40

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 .

( 9 . 3 ) A normed R i e s z s p a c e i s Archimedean.

Also

(9.4)

I n a normed R i e s z s p a c e , (i)

t h e norm c l o s u r e o f a R i e s z s u b s p a c e F i s a R i e s z

s u b s p a c e , and i t s p o s i t i v e cone i s t h e norm c l o s u r e o f F,; ( i i ) t h e norm c l o s u r e o f a R i e s z i d e a l 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 cone i s o b t a i n e d by a d j o i n i n g t o I + t h e (norm) l i m i t s o f i t s norm c o n v e r g e n t a s c e n d i n g s e q u e n c e s .

I n p a r t i c u l a r , a Riesz subspace F i s o n l y i f F,

norm c l o s e d

i f and

i s norm c l o s e d , and a R i e s z i d e a l I i s norm c l o s e d

i f and o n l y i f I + c o n t a i n s a l l t h e (norm) l i m i t s of i t s norm convergent ascending sequences I n a normed R i e s z s p a c e , o r d e r b o u n d e d n e s s i m p l i e s norm boundedness.

The c o n v e r s e i s f a l s e : i n t h e s e q u e n c e s p a c e R

1

t h e S c h a u d e r b a s e s I d n } , w h e r e e a c h dn i s t h e s e q u e n c e h a v i n g 1 i n t h e n t h p o s i t i o n and 0 e l s e w h e r e , i s norm bounded b u t n o t o r d e r bounded.

F u r t h e r m o r 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 a r e i n d e p e n d e n t o f e a c h o t h e r : i n R',

the

s e q u e n c e t l / n ) d n } norm c o n v e r g e s t o 0 , b u t i s n o t e v e n o r d e r b o u n d e d , s o c a n n o t b e o r d e r c o n v e r g e n t ; on t h e o t h e r h a n d , i n

,

Riesz Spaces

41

t h e sequence space c of convergent sequences, t h e sequence n { c l e k } ( n = 1 , 2 ; - * ) , w h e r e en i s t h e s e q u e n c e h a v i n g 1 i n t h e n t h p o s i t i o n and 0 e l s e w h e r e , o r d e r converges t o t h e element (l,l,l,.--) b u t i s n o t e v e n norm Cauchy.

F o r m o n o t o n i c n e t s , h o w e v e r , norm c o n v e r g e n c e i m p l i e s order convergence:

I n a normed R i e s z s p a c e , i f 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 , $ ~ )<

v e r g e s ; d e n o t e i t s l i m i t by f ( a ) .

is additive.

on E, l i m a( a

b,$a)

+

=

I t f o l l o w s it con-

CU.

The f u n c t i o n € t h u s d e f i n c d

In e f f e c t , f o r a , b E E + , f ( a

lim,[(a,$,)

+

(b,$,)l

+

= lim,(a,$,)

b) = +

l i m ( b , $ ) = f ( a ) + f ( b ) . And, a s i s e a s i l y v e r i f i e d , i t i s c1 a a l s o p o s i t i v e l y homogeneous. S o , a g a i n by E x e r c i s e 1 , f has a unique extension t o a1

o f E as 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 $. I t remains t o prove @

=

“,$y

t o show t h a t f o r e a c h a , ( a , $ )

2

4 2

(a,$a)

f o r a l l a: We h a v e

t h i s h o l d s from t h e above d e f i n i t i o n o f $ .

$iE b ,

Now s u p p o s e

@a f o r a l l n,we show ( 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 :

Riesz Space D u a l i t y

s u p a ( a , $cx)

<

f o r a l l aEE+,

such t h a t

then there e x i s t s $EEb

(a,$)

m

sup ( a , $ ) f o r a l l a€E+.

=

c1

CL

We g i v e a more d e t a i l e d d e s c r i p t i o n o f t h e R i e s z s p a c e structure of E

b

.

F i r s t , we record e x p l i c i t l y the following,

obtained i n the proof o f (10.1).

Combining t h i s w i t h E x e r c i s e 6 i n C h a p t e r 1 , we o b t a i n

(10.4)

For e v e r y 4 , $ € E b

and a E E + ,

In p a r t i c u l a r ,

(10.5)

F o r all + , $ E ( E

b

)+,

the following are equivalent

55

Chapter 2

56

'1

@A+ = 0 ,

f o r e v e r y aEE+ a n d

2'

with b

+

c

=

> 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,$,)

Riesz Space D u a l i t y

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 )<

co

3O

va$a e x i s t s i n E ~ ;

' 4

{$

=

c1

c1

f o r a l l a€E+;

} i s o r d e r e d bounded.

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

In E

b

,

o r d e r convergence i m p l i e s o(Eb,E)

convergence.

(10.11) closed.

Corollary 2.

b I n E , a .(Eb,E)

closed set is order

Chapter 2

58

In p a r t i c u l a r , a a ( E b , E ) c l o s e d Riesz i d e a l i s a l ~ a n d .

We c a n s a y m o r e :

( 1 0 . 1 2 ) The o ( E b , l l )

c l o s u r e o f a R i c s z i d c a l ,I o f ‘:1

band, and i t s p o s i t i v e cone i s t h e 0 ( L b , E )

is

closure of ,J+.

We w i l l p r o v e t h i s i n t h e n e x t 5 ( f o l l o w i n g ( 1 1 . 9 ) ) . The c o n v e r s e s o f ( 1 0 . 1 0 ) ,

( 1 0 . 1 1 ) , and ( 1 0 . 1 2 ) a r e f a l s e .

We p o s t p o n e a n e x a m p l e u n t i l we h a v e some c o n c r e t e E h ’ s .

b The p o s i t i v e c o n e ( E ) + o f E b i s D ( E b , E )

closed.

Tndeed,

w e c a n make a s t r o n g e r s t a t e m e n t :

(10.13)

h

€*.

(C ) + is closed i n

flence i t i s n o t o n l y “(F,’,l.)

b

c l o s e d i n I: ; i t i s a c t u a l l y ~ ( E ~ , Ec o) m p l e t e .

T h i s 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 o ( E b , E ) positive linear functional. [$I,$]

can be w r i t t e n ( 4

is closed i n E

*

.

+

Note t h a t , s i n c e e v e r y i n t e r v a l b

(E )+)

17

($ -

b (E ) + ) , i t follows i t

S i n c e i t i s a l s o bounded i n C

( 1 0 .1 4 ) Every i n t e r v a l o f E

and o f

b . is o(Eb,E)

*

compact.

,

we h a v e

Riesz Space D u a l i t y

F o r a s u b s e t A o f E , w e , s e t A'

=

b

I

(a,$)

0 for

=

.

a l l aEA}

( 1 0 . 1 5 ) For e v e r y R i e s z i d e a l I o f E ,

-__ Proof.

=

If

0 , s o $+€IL.

every s€T+, 0 < (a,$) Ricsz i d e a l .

IL i s a band o f E

I f $EIL, then f o r every a E I + , ( a , + + )

s u pb E , O , a l ( b , $ )

(10.12)

{$EE

59

2 (a,$)

=

b

.

=

0 < q < $ € I L , then f?r -

0, s o $ € I L .

Thus IL i s a

S i n c e IL i s o(Eb,E) c l o s e d , i t f o l l o w s from

t h a t it i s a band.

QED

(10.16)

Corollary.

-

For e v e r y R i e s z i d e a l I o f E , E

b

=

( I L ) d 0 11.

T h i s f o l l o w s from t h e R i e s z theorem ( 7 . 3 ) . S i n c e I' of

$IL

from I .

and ( I L ) d a r e d e t e r m i n e d by I , t h e c o m p u t a t i o n

f o r a n e l e m e n t ~p o f E b c a n b e c a l c u l a t e d ( I L1 Specifically,

and

( 1 0 . 1 7 ) Theorem.

Given a R i e s z i d e a l I , t h e n f o r e v e r y

d $ € ( ( I L ) ) + and aEE+,

Chapter 2

60

Proof. -__

Denote t h e r i g h t s i d e by f ( a ) .

f u n c t i o n f on E,.

We t h u s h a v e a

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

( c f . t h e p r o o f o f Lemma 1 i n ( l O . l ) , h e n c e h a s a u n i q u e e x t e n s i o n t o 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 E .

,

I t is c l e a r t h a t 0 c $ < 4 (a,$)

=

hence $ € ( I L )

We show $

d. In addition,

( a , + ) f o r a l l aEI,: hence f o r a l l a E I .

s e p a r a t i n g on ( I L ) d , we h a v e

+.

=

Since I i s

4.

$ =

QED

(10.18) Corollary.

(a,$)

0 f o r a l l c $ E ( I ' ) ~ and a E I d

=

Consider a c o l l e c t i o n { I

c1

1 o f R i e s z i d e a l s of E.

general theory of vector spaces,

(

I ' s a r e o ( E , E b ) c l o s e d , (nclIcl)' a

=

~

~

= 1n L y~( I a)) ' ,

By t h e

~a n d i f t h e

o(Eb,E) closure ( ~ ~ ( 1 ~ ) ~ ) .

For a f i n i t e c o l l e c t i o n , t h i s l a s t can be s t r e n g t h e n e d :

( 1 0 . 1 9 ) -~ Theorem.

For two R i e s z i d e a l s I , H o f E , ( I n H)L

Proof.

Consider $ € ( I

=

1'

+

H)',

+ El'.

a n d w e c a n assume

0>

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

following are equivalent: 1'

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

Riesz Space D u a l i t y

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.

Riesz Space D u a l i t y

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, <+ () a , + ) .

+iJ+, hence a+CJ

L

t u r n t h a t $ E ( J )'. I

Thus i n E b ,

.

it follows t h a t ( a + , + ) = 0 f o r a l l

Since (a',+)

2 (a,+)

> 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,

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

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

Riesz Space D u a l i t y

(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

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

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

Riesz Space D u a l i t y

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)

'<

(cf. (10.5)).

f

P

Consider $E(E ) + +

0

= $,

Since (Ta,p)

=

E

such t h a t ( T a , p ) =

( b , a ) , we c a n e v e n b y ( 1 1 . 6 ) , f i n d a p , O

(Tb,a)

and

bb

-, 0

, ; we

+

( a , p ) and ( T b , a )

=

f o r which ( T a , p )

=

0. QELJ

H e n c e f o r t h , we i d e n t i f y E w i t h i t s image i n E b b .

We

state this explicitly:

I f E b i s s e p a r a t i n g on E , t h e n u n d e r t h e b i l i n e a r f o r m ( a , $ ) on E

x

Warning.

Eb,

E i s a R i e s z s u b s p a c e o f Ebb - i n f a c t , o f E

bC .

While t h e c a n o n i c a l imbedding o f E i n Ebb 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 , suprema and i n f i m a o f f i n i t e s e t s , i t does n o t , i n g e n e r a l , p r e s e r v e suprema and infima of a r b i t r a r y s e t s .

Otherwise s t a t e d , it i s n o t

order continuous. E i s O(Ebb,Eb) d e n s e i n E b b ,

s i n c e i!: i s s e p a r a t i n g on E b ;

so a f o r t i o r i i t i s O(Ebb,Eb) d e n s e i n Ebc.

I t follows e a s i l y

Chapter 2

76

from t h i s and ( 1 2 . 8 ) t h a t t h e band o f Ebb g e n e r a t e d by E i s p r e c i s e l y Eb c .

A c t u a l l y , we have a much s t r o n g e r r e s u l t i n

( i i ) of (13.2) below. S i n c e E b c c o n t a i n s E , i t i s s e p a r a t i n g on E b , h e n c e , by t h e i n t r o d u c t i o n t o C h a p t e r 2 , E b c and E b 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 .

The f o l l o w i n g t h e o r e m s a y s

more: n o t o n l y i s E b c t h e s p a c e o f o r d e r c o n t i n u o u s l i n e a r b

i s i n t u r n t h e space o f o r d e r conbc 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 . Thus t h e f o l l o w i n g c a n b e f u n c t i o n a l s on E b , b u t E

c o n s i d e r e d t h e c e n t r a l d u a l i t y theorem f o r R ies z s p a c e s .

( 1 3 . 2 ) Theorem.

(Nakano).

Let E

b

b e s e p a r a t i n g on E , s o t h a t

t h e p a i r (EbC,Eb) 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 the other. (i)

Then each i s a c t u a l l y t h e space of order continuous 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 ; ( i i ) E c E b c and i t s o r d e r c l o s u r e i s a l l o f E

bc .

We w i l l n o t p r o v e ( 1 3 . 2 ) i n i t s f u l l g e n e r a l i t y .

We w i l l

e s t a b l i s h i t l a t e r f o r t h e two c a s e s i n w h i c h w e a r e i n t e r e s t e s :

E an L - s p a c e and E an M-space w i t h u n i t .

A R i e s z s p a c e F i s c a l l e d p e r f e c t i f F C i s s e p a r a t i n g on F , s o t h a t F c a n b e imbedded i n ( F c ) maps F o n t o Fcc (i)

(= (F')').

* , . and

t h i s imbedding a c t u a l l y

I n s h o r t , F i s p e r f e c t i f F = Fcc.

i n ( 1 3 . 2 ) above s a y s t h a t i f E b i s s e p a r a t i n g on E ,

t h e n Eb i s p e r f e c t .

Riesz Space D u a l i t y

77

We c l o s e t h i s 5 w i t h a t h e o r e m w h i c h we w i l l l a t e r n e e d .

W h i l e E b i s s e p a r a t i n g on E f o r a b r o a d r a n g e o f R i e s z s p a c e s , EC i s r a r e l y s o .

flowever, i f i t i s s e p a r a t i n g on E , t h e n we

have t h e c a n o n i c a l imbedding o f E i n (Ec)*.

I t c a n b e shown

h e r e a l s o , b y t h e same p r o o f a s was u s e d f o r ( 1 3 . 1 ) , t h a t t h i s c a n o n i c a l imbedding i s i n f a c t a Riesz isomorphism o f E i n t o Ecc.

I f E i s Dedekind c o m p l e t e , w e c a n s a y more:

Theorem. ( N a k a n o ) . ( 1 3 . 3 ) -~

I f E i s Dedekind complete and EC i s

s e p a r a t i n g on E , t h e n u n d e r i t s c a n o n i c a l i m b e d d i n g i n E C C , E

i s a R i e s z i d e a l whose o r d e r c l o s u r e i s a l l o f E c c .

Proof.

Lemma 1. F o r e v e r y d e c o m p o s i t i o n E c c (I

I b a n d s ) , E = (E 1, 2

n

11) 3 ( E

By E x e r c i s e 8 , E C

Proof. -___

(applying Exercise 8 again) ((Il)l)L

n ( ( I ~ ))' I ~n ) ~ I ~ .

= E

( ( I ~ ) =~

Corollary.

= E

n

= I1 0 I 2

12).

(I2Il 3 ( I l l l ,

=

E

=

and t h e r e f o r e

((I1)l)L 3 ((12)L)1*

Rut

0 I1 ( 1 2 . 7 ) ; and s i m i l a r l y ,

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

every aEE, aIEE.

Lemma 2 .

E i s Dedekind d e n s e i n t h e R i e s z i d e a l I o f Ecc

78

Chaptcr 2

g e n e r a t e d by 1 .

Proof.

Ry ( 7 . 6 ) , i t i s e n o u g h t o show t h a t i f b E I , b > 0,

then t h e r e e x i s t s a E E such t h a t 0 < a 5 b . b

so v

-

1> 0

c f o r some c E E .

h c = 0 , s o v h > O ( b - Xc)

Now A

Let II b e t h e b a n d o f E"

g e n e r a t e d hy (h

Straightforward computation gives C o r o l l a r y above, xc € E 11

h,

( b - r o c ) + z 0 f o r some

(b - ~ c ) + = b, so, finally,

h0 > 0 .

=

US

0

i

hclI

i

-

h.

hocl+. Ry t h e

so is t h e desired a.

Now I i s T k d e k i n d c o m p l e t e ( ( 1 0 . 1 ) a n d I l x e r c i s c 1 8 o f C h a p t e r l),

h e n c e , h y Lemma 2 ,

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 I:.

S i n c e E i s D e d e k i n d c o m p l e t e , we h a v e E T h a t t h e o r d e r c l o s u r e o f E i s Ec' c

t h a t i t i s a ( E c c , I 1 ) d e n s e i n I!

C

=

I.

f o l l o w s from t h e f a c t

combined w i t h ( 1 2 . 8 ) .

(11: r)

5 1 4 . 'The t r a n s p o s e o f a R i e s z homomorphism

Let E __ > F b e a l i n e a r m a p p i n g o f o n e R i e s z s p a c e i n t o By t h e t r a n s p o s e T t o f T , we w i l l , + Tt * mean t h e m a p p i n g E <-F d e f i n e d by +> another.

$€F $€F

* *

f o r t h e moment,

+

,

o r e q u i v a l e n t l y , by ( a , T t $ )

=

(Ta,$)

+ o T for e v e r y

f o r e v e r y aEE

.

b b (14.1) I f T i s o r d e r bounded, t h e n Tt(F ) c E

.

and

79

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

Proof.

Given $ € F b , t h c n f o r e v e r y o r d e r bounded s e t A

of I:, t sup / ( a , T ij~)l a€ A

(since T(A)

=

SUP

a€ A

l(Ta,$)l

i s o r d e r bounded).

<

~0

Thus Tt$EE b . QE D

I

E v e r y mapping E -->F o r d e r bounded.

which w e w i l l e n c o u n t e r w i l l be

S i n c e i t i s E b a n d Fb t h a t w e a r e i n t e r e s t e d i n ,

t h e term " t r a n s p o s e "

and n o t a t i o n Tt w i l l h e n c e f o r t h d e n o t e

t h e r e s t r i c t i o n o f t h e above t r a n s p o s e t o F

(14.2)

[i)

b

.

T p o s i t i v e implies Tt p o s i t i v e .

( i i ) I f Fb i s s e p a r a t i n g on F , t h e c o n v e r s e h o l d s : T

t

positive implies T positive.

b t I f I $ E ( F ) + , t h e n f o r e v e r y aEE,, ( a , T 11,) = t b 0, that is, T $E(E )+. ( i i ) I f a E E + , then f o r every

-__ Proof.

(Ta,$)

2

(i)

b $E(F )+, (Ta,$)

=

(a,Tt$) > 0 , hence, by [ l l . Z ) ,

TaEF,.

QE D we w i l l d e n o t e t h e s e t o f all b o r d e r bounded l i n e a r mappings o f E i n t o F by L ( E , F ) ( t h u s G i v e n two R i e s z s p a c e s E , F ,

Eb

=

Lb(E,IR)).

F b e i n g a v e c t o r s p a c e , Lb(E,F) i s a v e c t o r

s p a c e u n d e r t h e usual d e f i n i t i o n s o f a d d i t i o n a n d s c a l a r multiplication. The p r o o f o f ( 1 0 . 1 ) d e p e n d e d upon t h e f a c t t h a t lR i s

Chapter 2

80

Dedekind c o m p l e t e .

I f F i s D e d e k i n d c o m p l e t e , an a d a p t a t i o n o f

t h a t proof g i v e s us

b I f F i s Dedekind c o m p l e t e , L ( E , F ) i s a

( 1 4 . 3 ) Theorem. - _ _ c _

Dedekind c o m plete R i e s z s p a c e , w i t h t h e s e t o f p o s i t i v e l i n e a r mappings f o r p o s i t i v e c o n e .

(Note t h a t , a s a c o n s e q u e n c e , t h e s t a t e m e n t t h a t a l i n e a r mapping T i s p o s i t i v e c a n b e w r i t t e n T > 0.) Contained i n t h i s theorem i s t h e r e s u l t t h a t e v e r y o r d e r b o u n d e d l i n e a r mapping T i s t h e d i f f e r e n c e o f two p o s i t i v e l i n e a r mappings: T

=

.

T+ - T

U s i n g t h i s , we c a n s h a r p e n ( 1 4 . 1 )

considerably :

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

( 1 4 . 4 ) Theorem.

I

o r d e r bounded l i n e a r mapping E __ > F , t h e t r a n s p o s e E b <*

Fb

-__ Proof.

is order continuous.

T

=

T+ - T - , s o T t = ( T + ) t - ( T - ) t .

> 0. o n l y prove t h e theorem f o r T -

By ( 1 4 . 2 ) , T

t

Hence w e n e e d > 0 , hence t o

show T t i s o r d e r c o n t i n u o u s , i t s u f f i c e s t o show t h a t + a + O

t F o r e v e r y aEE,, ( a , T $a) = t S i n c e {Tqa.}, l i k e { $ } , i s a d e s c e n d i n g ( T a , Q a . ) + 0 , by ( 1 0 . 9 ) . a t n e t , i t f b l l o w s a g a i n f r o m ( 1 0 . 9 ) t h a t Tq,+ 0 . i n Fb i m p l i e s

Tt+

a.

G 0 in E

b

.

QE D

Riesz Space D u a l i t y

81

E x a m i n a t i o n o f t h e a b o v e p r o o f shows t h a t f o r T p o s i t i v e , t h e c o n c l u s i o n h o l d s w i t h o u t t h e a s s u m p t i o n t h a t F be Dedekind complete.

(14.5) I f E

We r e c o r d t h i s .

~

> F i s a p o s i t i v e l i n e a r t r a n s f o r m a t i o n o f one

R i e s z s p a c e i n t o a n o t h e r , t h e n E <- Tt

Fb i s o r d e r c o n t i n u o u s .

Remark. I t c a n b e shown t h a t t h e c o n d i t i o n t h a t F b e ____ Dedekind c o m p l e t e c a n be e l i m i n a t e d f r o m ( 1 4 . 4 ) e n t i r e l y , n o t only for T positive.

The r e m a i n d e r o f t h i s 5 i s c o n c e r n e d w i t h t h e t r n s p o s e o f a R i e s z homomorphism.

We f i r s t c o n s i d e r two s p e c i a l c a s e s :

( i ) T i s a s u r j e c t i o n , and ( i i ) T i s an i n j e c t i o n .

-TI n

the

f i r s t c a s e , T h a s t h e f a c t o r i z a t i o n E L > E / T - l (O)->F,

-

w h e r e q i s t h e q u o t i e n t map a n d T a R i e s z i s o m o r p h i s m . In the T1 i s e c o n d c a s e , T h a s t h e f a c t o r i z a t i o n E ---> T(E) -> F, w h e r e i i s t h e c a n o n i c a l i n j e c t i o n o f T(E) i n t o F a n d T1 i s a

Riesz isomorphism o f E o n t o T ( E ) .

T h u s , i n e f f e c t , t h e two

s p e c i a l c a s e s r e d u c e t o t h e c a s e s ( a ) t h e q u o t i e n t map d e t e r m i n e d by a R i e s z i d e a l a n d ( b ) t h e c a n o n i c a l i n j e c t i o n o f a Riesz subspace.

( 1 4 . 6 ) Theorem.

-> E / I t h e q u o t i e n t map.

and E (E/I)

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

b

Then E b <- q t

Riesz i s o m o r p h i c a l l y o n t o I".

( E / I ) b maps

Chapter 2

82

Proof.

We c l e a r l y h a v e qt((E/I)b)

mapping i s o n t o , S i n c e qa

=

To show t h e

c 1'.

c o n s i d e r 4 i I L , a n d we c a n assume 4 > 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].

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

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 <- i t

Fb i s

interval preserving.

___ Proof.

i t i s o f c o u r s e t h e r e s t r i c t i o n map c$ k----->+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;

Riesz Space D u a l i t y

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 $ <

Since it i s one-one,

q , = p , o .

QED

Warning.

In general, i

t . i s n o t a R i e s z homomorphism.

For a Riesz i d e a l I o f F ,

((it)-')(O)

=

,'I

and i t i s n o t

h a r d t o show t h a t i n t h i s c a s e , i t i s a R i e s z homomorphism a n d m o r e o v e r maps ( I L ) d R i e s z i s o m o r p h i c a l l y o n t o i t ( F b ) .

Thus

(14.11) C o r o l l a r y 2.

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 F, b i s R i e s z isomorphic t o a R i e s z i d e a l of I .

Again, we i d e n t i f y t h e Riesz isomorphic spaces o b t a i n e d i n t h e above C o r o l l a r i e s :

I f a R i e s z s u b s p a c e E o f a R i e s z s p a c e F i s s e p a r a t i n g on b b F b , t h e n u n d e r t h e b i l i n e a r f o r m ( a , $ ) on E x F , F i s a R i e s z

Chapter 2

86

ideal of E

b

.

F o r e v e r y R i e s z i d e a l I o f F , u n d e r t h e b i l i n e a r form b ( a , @ ) on I x ( I L ) d , ( I L ) d i s a R i e s z i d e a l o f I .

We a r e now i n a p o s i t i o n t o d i s c u s s a g e n e r a l Riesz homomo r p h i s m .

(14.2)

(Kim) L e t E , F b e R i e s z s p a c e s , w i t h E

Theorem..

s e p a r a t i n g on E a n d Fb s e p a r a t i n g on F. mapping E ->

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

<-

Tt

t (T(E))b < A F b .

"t . I t f o l l o w s from t h e p r e c e d i n g r e s u l t s (and t h e f a c t t h a t T i s

a R i e s z isomorphism) t h a t Tt i s i n t e r v a l p r e s e r v i n g . Now assume T t i s i n t e r v a l p r e s e r v i n g , a n d c o n s i d e r a E E . b We show t h a t f o r e v e r y + E ( F ) + , ( T a + , $ )

f o l l o w t h a t Taf

=

(Ta)'.

=

( ( T a ) + , + ) ; it w i l l

Riesz Space D u a l i t y

87

(11.4)

(11.4) QE D

The a b o v e t h e o r e m a p p e a r e d i n [ 2 9 ] .

I t was p r o v e d

i n d e p e n d e n t l y b y Ando a n d L o t z f o r Banach l a t t i c e s ( c f .

[33]).

T i s a Riesz b homomorphism i f a n d o n l y i f f o r e v e r y R i e s z i d e a l J o f E , t t T t ( J ) i s a R i e s z i d e a l o f Eb a n d T ( J + ) = (T ( J ) ) + .

K i m a l s o gave t h e f o l l o w i n g c h a r a c t e r i z a t i o n :

The s e c o n d p a r t o f t h e

above p r o o f h o l d s v e r b a t i m w i t h

T a n d T L i n t e r c h a n g e d t o g i v e us

(14.13)

I f a l i n e a r m a p p i n g E __ 'I' > F o f o n e R i e s z s p a c e i n t o

another is interval preserving, then E

LTt-Fb

is a Riesz

homomorp h i s m .

Combining ( 1 4 . 1 2 ) ,

( 1 4 . 1 3 ) , and ( 1 4 . 4 ) , w e have

( 1 4 . 1 4 ) L e t E -> F b e a l i n e a r m a p p i n g 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 , a n d E b b ___ Ttt > Fb b i t s b i t r a n s p o s e Ttt

=

(Tt)t).

(that is,

88

Chapter 2

(i)

i f T i s a R i e s z homomorphism, t h e n T t t

i s an o r d e r

c o n t i n u o u s R i e s z homomorphism. ( i i ) I f T i s i n t e r v a l preserving, then Ttt

i s o r d e r con-

t i n u o u s and i n t e r v a l p r e s e r v i n g .

5 1 5 . The d u a l o f a Banach l a t t i c e

Notation.

norm d u a l by E ' ,

weak

L e t E b e a normed s p a c e .

a n d , a s c u s t o m a r y , we w i l l c a l l o ( E , E ' ) t h e

t o p o l o g y on E .

However, f o r o ( E ' , E ) , we w i l l a d o p t t h e

B o u r b a k i term, t h e v a g u e t o p o l o g y on E ' . for E

=

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

(They o n l y used i t

C(X) .)

I f E i s a normed R i e s z s p a c e , t h e n E ' c E

b

.

I n more

detail,

( 1 5 . 1 ) I f E i s a normed R i e s z s p a c e , t h e n (i)

E'

i s a v a g u e l y dense ( t h a t i s , a ( E ' , E ) dense) Riesz

i d e a l of Eb; ( i i ) under t h e induced R i e s z space s t r u c t u r e , E ' i s a Banach l a t t i c e .

Proof. -__

E v e r y o r d e r b o u n d e d s e t o f E i s norm b o u n d e d , s o

e v e r y e l e m e n t o f E ' i s o r d e r bounded. S i n c e E ' i s s e p a r a t i n g b on E , i t i s v a g u e l y d e n s e i n E . ( i i ) f o l l o w s from t h e f a c t t h a t the u n i t b a l l o f ' E ' i s the polar of the u n i t b a l l of E,

89

Riesz Space D u a l i t y

h e n c e , by E x e r c i s e 9 , i s s o l i d .

( 1 5 . 2 ) C o r o l l a r y 1.

I f E i s a normed R i e s z s p a c e , t h e n Eb i s

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

(15.3) C o r o l l a r y 2 .

I f E i s a normed R i e s z s p a c e , E '

is

Dedekind c o m p l e t e .

T h i s by E x e r c i s e 1 8 i n C h a p t e r 1. Combining ( 1 5 . 1 ) a n d ( 9 . 1 1 ) ,

( 1 5 . 4 ) I f E i s a Banach l a t t i c e , E '

=

E

b

.

EXE RC I SES

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 .

1.

L e t f b e a n a d d i t i v e a n d p o s i t i v e l y homogeneous f u n c t i o n

on E + : f ( a + b ) A > 0.

Then a

=

I--->

f(a)

+

f ( b ) , f(Aa)

=

Af(a) f o r a l l

f(a+) - f(a-) is linear functional

on E , a n d t h e o n l y o n e whose r e s t r i c t i o n t o E, with f .

coincides

Chapter 2

90

2.

Given a R i e s z i d e a l I o f E , t h e n f o r e v e r y @E(E

b

)+

and

a6 I<+,

3. I f E

h . i s s e p a r a t i n g on E , t h e n f o r e v e r y R i e s z i d e a l T

of E ,

4.

(i)

( ( P A L

If E

=

Id .

b . i s s e p a r a t i n g on E , t h e n f o r e v e r y R i e s z i d e a l

I of E ,

( 1 1 ) ~i s s e p a r a t i n g on T .

b ( i i ) For a R i e s z i d e a l ,J o f E , ( t J L ) d may b e 0 .

5.

(i)

F o r two R i e s z i d e a l s 1 , I I o f E , I

(Il)dj]

(Hqd

n

H = 0 implies

0.

=

h ( i i ) F o r two Riesz i d e a l s J , G o f E , i t i s p o s s i b l e t o d d h a v e J n G = 0 w h i l e (,J ) n ( G ) # 0 . L

L

6.

(i)

( 1 0 . 2 0 ) f a i l s t o h o l d i f E and Eb a r e i n t e r c h a n g e d (and of c o u r s e

changed from s u p e r s c r i p t t o

1

subscript). ( i i ) The above s t a t e m e n t h o l d s e v e n i f E i s

Dedekind

complete.

7 . Even i f E c i s s e p a r a t i n g on E , E C

n o t imply E

= I;

L

+

J

L

=

<J 0 G ( b a n d s ) d o e s

.

8 . I f Ec i s s e p a r a t i n g on E ,

t h e n Ec

=

and E i s D e d e k i n d c o m p l e t e ,

J 3 G does imply E = E

L

3 J

L

.

Riesz Space D u a l i t y

9.

(i)

91

If R c E i s s o l i d , t h e n i t s p o l a r A o - i n - E b i s s o l i d .

( i i ) I f A c E b i s s o l i d , t h e n AO-in-E i s s o l i d .

10

I f a s o l i d s e t A o f I:. i s a b s o r b i n g ( t h a t i s , IJn(nA) * b t h e n A'-in-E c E .

11.

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

(i)

b I f I: i s a norm c l o s e d R i e s z s u b s p a c e , t h e n E / F L F

b . i s o m e t r i c a l l y and R i e s z i s o m o r p h i c a l l y .

( i i ) H e n c e , i f I i s a norm c l o s e d R i e s z i d e a l , t h e n (IL)d= T

12.

=

b

.

I n a B a n a c h l a t t i c e , 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 l e band.

E),

=

PART I 1

L - SPACES AND MIL- SPACES

C(X)

i s an M I - s p a c e ( a n M - s p a c e w i t h u n i t ) , i t s d u a l C ' ( X )

i s an L-space, and i t s b i d u a l

Cll(X)

d e f i n e M-space a n d L - s p a c e b e l o w ) . C'l(X)

exist a t three levels.

i s a g a i n an M a - s p a c e (we

Thus t h e p r o p e r t i e s o f

A t t h e f i r s t l e v e l , are i t s p r o -

p e r t i e s a s an M I - s p a c e .

A t the next, deeper, level, are those

i t possesses a s a d u a l .

And a t t h e d e e p e s t l e v e l a r e t h o s e i t

possesses a s a b i d u a l . two l e v e l s .

I n P a r t I 1 h e r e , we d e a l w i t h t h e f i r s t

Our i n t e r e s t , o f c o u r s e , l i e s i n t h e t h i r d l e v e l ,

and w e w i l l b e o c c u p i e d w i t h t h o s e p r o p e r t i e s f o r t h e r e m a i n d e r of t h e work.

92

CHAPTER 3

M I - S P A C E S AND L-SPACES

The norms o f C ( X ) t h a t of

C l ( X )

and

Cl'(X)

a r e o f a s p e c i a l t y p e , and

of a t y p e " d u a l " t o t h i s .

In the p r e s e n t chapter,

we s t u d y t h e s e two t y p e s .

516 Mll-spaces

An e l e m e n t a > 0 f o r a R i e s z s p a c e E i s an o r d e r u n i t f o r E i f f o r every b E E , t h e r e e x i s t s X > 0 s u c h t h a t -Aa ib < l a e q u i v a l e n t l y , i f t h e R i e s z i d e a l g e n e r a t e d by a i s a l l o f E .

-

C l e a r l y , i f a i s an o r d e r u n i t , t h e n l a i s a l s o an o r d e r u n i t f ~ ? ^ every X > 0.

I n g e n e r a l , a R i e s z s p a c e d o e s n o t have an o r d e r

unit. I f we a r e working w i t h a p a r t i c u l a r o r d e r u n i t i n a R i e s z

s p a c e , we w i l l h a b i t u a l l y d e n o t e i t by 1.

An o r d e r u n i t II o f an Archimedean R i e s z s p a c e E d e f i n e s a c a n o n i c a l R i e s z norm ( c a l l e d an o r d e r u n i t norm) on E by jlaIj = inf{XI -

f o r a l l aEE.

xn -i a

We h a v e , i m m e d i a t e l y ,

93

i -

An}

94

Chapter 3

( 1 6 . 1 ) F o r a l l a , b E E , [ l a - b[I < b + €1.

i E and o n l y i f b - € 1 < a <

E

E q u i v a l e n t l y , IIaII < 1 i f and o n l y i f

la1 < ll

.

In

1 , and t h e u n i t b a l l B ( E ) o f E i s [ - n , R ] .

p a r t i c u l a r , \Ill;/

The norms d e f i n e d by two o r d e r u n i t s on a R i e s z s p a c e E are clearly equivalent. I t i s i m m e d i a t e from ( 9 . 1 ) R i e s z norm.

( 1 6 . 2 ) Let

t h a t a n o r d e r u n i t norm i s a

And t h e 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 .

11 I[ a

b e an o r d e r u n i t norm.

(1 avb(I

=

max(\I all

Then f o r a l l a , b E E + ,

, Ilhll J

.

A norm riith. t h i s p r o p e r t y ( w h e t h e r o r n o t it i s d e f i n e d

by a n o r d e r u n i t ) i s c a l l e d a n M-norm.

In keeping with t h i s

w e w i l l h e n c e f o r t h c a l l a n o r d e r u n i t norm

[I

an Ma-norm,

and we w i l l c a l l t h e ( u n i q u e ) o r d e r u n i t 1 d e f i n i n g i t s i m p l y the unit for

I[ . [I

( o r f o r E , i f t h e r e can be no c o n f u s i o n ) .

E v e r y R i e s z s p a c e w i t h an M-norm c a n b e i s o m e t r i c a l l y ( a n d

Riesz i s o m o r p h i c a l l y ) imbedded i n o n e w h i c h h a s a n o r d e r u n i t d e f i n i n g i t s norm.

Thus t h e R i e s z s p a c e s w i t h M-norms a r e

p r e c i s e l y t h e Riesz subspaces o f Riesz s p a c e s withMll-norms: o t h e r w i s e s t a t e d , e v e r y M-norm c a n b e c o n s i d e r e d i n d u c e d b y a n M I - norm.

I t i s t o be e x p e c t e d t h a t f o r an M R -norm, t h e r e s u l t s o f 559 and 1 5 can be s t r e n g t h e n e d .

To s t a r t w i t h , we h a v e

t r i v i a l l y t h a t u n d e r a n M-norm, o r d e r b o u n d e d n e s s and norm boundedness a r e e q u i v a l e n t .

Also

Ma-spaces and L-spaces

95

( 1 6 . 3 ) I n an MI-normed s p a c e , i f a b o u n d e d n e t {au} norm c o n verges, then it order converges.

Proof. -___

F o r s i m p l i c i t y , a s s u m e { a } norm c o n v e r g e s t o 0 . c1

F o r e a c h a, s e t ra

=

~ u p ~ , ~ I I a 'Then ~ ~ l .{r 1 i s a descending n e t

-

CL

o f r e a l n u m b e r s c o n v e r g i n g t o 0 , s o rclll+O. S i n c e la f o r a l l a, a a

+

0

I

<

-

r I c1

0 .

QED

(16.4)

C o r o l l a r y 1.

I n a n M I - normed s p a c e , e v e r y norm c o n -

v e r g e n t sequence o r d e r converges.

(16.5) C o r o l l a r y 2 .

I n a n MI-normed s p a c e , an q r d e r c l o s e d s e t

( o r e v e n a o - o r d e r c l o s e d o n e ) i s norm c l o s e d .

I n an MIL-normed s p a c e ) t h e R i e s z s u b s p a c e s c o n t a i n i n g t h e u n i t I have s t r o n g p r o p e r t i e s .

(16.6)

We g i v e two o f t h e m .

Let E b e a n MI-normed s p a c e a n d F a R i e s z s u b s p a c e c o n -

t a i n i n g the u n i t I o f E.

I f aEE i s i n t h e norm c l o s u r e o f F ,

t h e n i t i s t h e norm l i m i t ( h e n c e o r d e r l i m i t ) o f a n a s c e n d i n g sequence o f F , and a l s o o f a descending one.

Chapter 3

96

I t i s enough t o show t h a t f o r

Proof.

 0 , there exists

Choose c € F s a t i s f y i n g I I c - all 5

Then a - ( ~ / 2 ) l l< c < a + ( ~ / 2 ) 1 , h e n c e a - Ell < c -

( ~ / 2 ) 1< a , and c - ( ~ / 2 ) n i s t h e d e s i r e d b . QED

A g a i n , l e t F be a R i e s z s u b s p a c e c o n t a i n i n g I t .

Suppose

a E B ( E ) , t h e u n i t b a l l o f E , and t h a t { a 1 i s a n e t i n F s u c h c1

t h a t aa

-f

a.

V(-n)

Then (a,An)

+

a

also.

Thus a i s t h e l i m i t

o f a n o r d e r c o n v e r g e n t n e t o f t h e u n i t b a l l B(F) o f F .

Com-

b i n i n g t h i s w i t h Z o r n ' s lemma ( a s i n t h e p r o o f o f ( 5 . 2 ) , w e can e s t a b l i s h :

( 1 6 . 7 ) L e t E b e a n MIL-normed s p a c e , F a R i e s z s u b s p a c e c o n t a i n i n g t h e u n i t ll o f E , a n d G i t s o r d e r c l o s u r e .

Then B I G ) i s

t h e o r d e r c l o s u r e o f B(F).

As i n a normed a l g e b r a w i t h i d e n t i t y , w e h a v e :

( 1 6 . 9 ) Let E b e a n MIL-normed s p a c e . i d e a l , then inf a€ I

--Proof.

11 a

-

It]]

I f I i s a proper Riesz

= 1.

S u p p o s e IIa - lL\l = r < 1 f o r some a E I .

a > It - r n , h e n c e 0 < It < (1 - r ) - ' a , the hypothesis that I is proper.

Then

h e n c e IlEI, c o n t r a d i c t i n g

S i n c e O E I , t h e a b o v e infimum

M I - s p a c e s and L - s p a c e s

97

is attained. QED

(16.10) Corollary.

I n a n Ma-normed s p a c e , t h e norm c l o s u r e o f

a p r o p e r Riesz i d e a l i s a l s o p r o p e r .

The f o l l o w i n g i s o f c o u r s e what w e w o u l d w a n t .

( 1 6 . 1 1 ) Let E be a n MIL-normed s p a c e a n d I a p r o p e r norm c l o s e d Riesz i d e a l .

Then t h e q u o t i e n t norm on E / I i s a n MIl-norm w i t h

q I for unit.

--P r o o f . We show t h a t f o r e v e r y a E E / I , if

121 5 q n .

Suppose

some aEE s u c h t h a t q ( a ( 5 Xql.

( / a ( /5

11 all 5

1.

[I;//

5 1 i f and o n l y

Then f o r X > 1,

a

S i n c e t h i s h o l d s f o r a l l X > 1, 121

:q n ;

2 2 0.

Choose a n y aEE s u c h t h a t q a

Then q b

=

II;]l

5

Ilb;l

5

w e h a v e t o show

S i n c e t h e q u o t i e n t norm i s a R i e s z norm ( 9 . 8 ) ,

2 a l s o , hence

qa f o r

T h i s s a y s la1 5 X I , h e n c e J.21 =

A.

C o n v e r s e l y , s u p p o s e 121

=

=

5

qn.

IIa:] 2

1.

w e c a n assume

and s e t b = (av0)An.

5 1. QED

I f a R i e s z s p a c e w i t h a n M-norm i s norm c o m p l e t e , we w i l l

c a l l i t an M-space ( t h e common name i s ( A M ) - s p a c e ) .

And i f a

R i e s z s p a c e w i t h a n MI-norm i s norm c o m p l e t e , w e w i l l c a l l i t

Chapter 3

98

an MIL-space. A norm c l o s e d R i e s z s u b s p a c e o f a n M-space E i s i t s e l f an M-space.

We w i l l c a l l i t a n M - s u b s p a c e o f E .

F i n a l l y , i f an

M - s u b s p a c e o f a n M I - s p a c e E c o n t a i n s t h e u n i t IL o f E , we w i l l c a l l i t an MIL-subsnace o f E .

A s i s w e l l known, t h e B a i r e c l a s s e s i n f u n c t i o n t h e o r y a r e

each c l o s e d under uniform convergence.

This i s a property of

an MIL-space:

( 1 6 . 1 2 ) L e t E b e a n MIL-space, F a R i e s z s u b s p a c e c o n t a i n i n g t h e u n i t 1 o f E , and G t h e s e t o f e l e m e n t s which a r e l i m i t s o f Then F i s norm c l o s e d ( h e n c e

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 F.

an MIL- s u b s p a c e )

-__ Proof.

.

S u p p o s e a i s i n t h e norm c l o s u r e o f G , a n d f o r

s i m p l i c i t y , we c a n assume 0 < a <

(i)

n.

T h e r e e x i s t s { a n } (n = 0 , 1 , 2 , . - * ) i n G,

II anII 2

(n

l/zn

I n e f f e c t , by ( 1 6 . 6 ) ,

t h e r e i s a n a s c e n d i n g s e q u e n c e {cn3

i n G such t h a t a - (l/zn)Il i cn r e p l a c i n g e a c h cn by (c,)', = co a n d a n =

'n

0,1,2,...)

u n d e r norm c o n v e r g e n c e .

Cnan = a

S e t a.

=

such t h a t

-

5 a (n

= 0,1,2,-.

w e c a n assume cn

cn-l

(n

=

1,2,...).

1. 0

a ) ,

and,

f o r a l l n.

The s e q u e n c e

99

MIL-spaces a n d L - s p a c e s

t h e n has t h e above p r o p e r t i e s .

an

Now b y t h e d e f i n i t i o n o f G I f o r e a c h n , t h e r e e x i s t s a sequence {b

nm } ( m

=

i n I: w h i c h o r d e r c o n v e r g e s t o a n ;

1,2,...)

and w e can c l e a r l y t a k e 0 < bnm

brn = xn=Ob nm ( m

1,2,..-).

=

5

We show bm

F o r e a c h n , t h e s t a t e m e n t t h a t bnm

nm } (rn

t h e r e e x i s t s e q u e n c e s {p,,}, pnmJ-an9 qnm+an9 a n d qnm

2

qnm

9,

'

9

< (1/2")n Pnm -

m

cn,Oqnrn ( m

=

=

qm

m

= Cn,oq,rn

{q

brim

5

a , h e n c e aEG

-f

a

-f

=

Set

n

(as m

1,2;--)

-f

a)

. means

i n E such t h a t

Pnm f o r a l l m, a n d w e c a n t a k e

f o r a l l m.

S e t p,

m

=

~,,~p,,,

1,2,-.*).

(ii)

9,

5

f o r a l l m.

(l/2n)ll

5 bm 5 Pm f o r

m 5 Zm n z 0 bnm 5 cn,opnm

a l l m.

= p,,

and s i n c e t h e t h i r d

e x p r e s s i o n i s b m' w e h a v e ( i i ) .

q m +a

(iii)

--

We h a v e q1 5 q 2 5 .< a. Suppose c > q f o r a l l m. Then m N N c > c a for a l l N: i n e f f e c t , f o r every m > N , c 2 q,, Cn=oq,m; On N N h e n c e c 2 Vm(Cn=Oqnm) = Cn,0an. N Thus a = Vmqm, a n d we I t follows c > V (C a ) = a. N O n have ( i i i ) .

We h a v e p1

2

show c < a + (1/2

p2 N

2

> a. -

)ll f o r a l l N

Suppose c < p, = 0,1,2,...,

f o r a l l m ; we

whence c < a.

Fix N

Chapter 3

100

5 1 7 . The components o f II

Given a p o s i t i v e e l e m e n t a o f a R i e s z s p a c e E , eEE be c a l l e d a component o f a i f eA(a - e ) d e f i n i t i o n i s t h a t (ze)Aa

8

w i l l be d e n o t e d by

=

e.

=

0.

will

An e q u i v a l e n t

The s e t of a l l components o f a

(a).

I n t h i s 5 , w e a s s e m b l e t h e p r o p e r t i e s we w i l l need o f Z ( n ) ,

1 t h e u n i t f o r an MIL-norm on a R i e s z s p a c e E .

However, s i n c e

e v e r y a > 0 o f a R i e s z s p a c e i s an o r d e r u n i t f o r t h e R i e s z i d e a l which i t g e n e r a t e s , t h e r e s u l t s a r e c o m p l e t e l y g e n e r a l , t h a t i s , hold f o r

(17.1)

Z(n)

8

(a) f o r every a > 0 .

i s a s u b l a t t i c e of E.

Moreover, i t i s a Boolean

a l g e b r a w i t h 0 f o r i t s z e r o , 1 f o r i t s u n i t , and 1 - e f o r t h e complement o f e .

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 .

(17.2)

8 (l) i s

o r d e r c l o s e d (hence norm c l o s e d ) i n E .

Mll-spaces and L - s p a c e s

101

T h i s follows e a s i l y from t h e above and ( 4 . 3 ) .

Given a d e c o m p o s i t i o n E immediate t h e

=

1 1 3 I 2 (11,12 bands), i t i s

,

band components 111

1

% of

ll a r e a l s o com-

p o n e n t s o f I i n t h e s e n s e d e f i n e d above i n t h e p r e s e n t 5 .

Con-

v e r s e l y , s u p p o s e e 1 , e 2 a r e complementary components o f 1 ( t h a t i s , e 2 = ll - e 1) i n t h e p r e s e n t s e n s e , and l e t 11, I 2 be t h e R i e s z i d e a l s which t h e y g e n e r a t e . that E = I l @ I 2 with el

=

1111, e 2

=

I t i s a g a i n immediate

I I 2 . Thus an e l e m e n t o f

E i s a "component" o f ll i n one s e n s e i f and o n l y i f i t i s one

i n the other sense. Contained i n t h i s d i s c u s s i o n i s t h e r e s u l t t h a t every Riesz i d e a l g e n e r a t e d by a component o f It i s a p r o j e c t i o n band and e v e r y p r o j e c t i o n band i s a p r i n c i p a l R i e s z i d e a l .

(17.3)

More f u l l y :

For a R i e s z i d e a l I o f an Mll-normed s p a c e E , t h e

following a r e equivalent: 1'

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

2'

I i s t h e R i e s z i d e a l g e n e r a t e d by some component e o f I .

Moreover, (i) e

=

llI '.

( i i ) t h e norm on I i n d u c e d by t h e norm on E i s p r e c i s e l y t h e MI-norm d e f i n e d by e .

The c o n d i t i o n i n 2'

t h a t i t be a component o f It t h a t

generated I i s e s s e n t i a l .

I n g e n e r a l , n o t even a p r i n c i p a l

102

Chapter 3

band i s a p r o j e c t i o n band ( u n l e s s , o f c o u r s e , E i s 0-Dedeking complete ( 8 . 3 ) ) .

(17.4)

Corollary . The B o o l e a n a l g e b r a

(1) i s i s o m o r p h i c w i t h

t h e B o o l e a n a l g e b r a o f p r o j e c t i o n b a n d s o f E.

Remarks. -__

( i ) N o t e t h a t [jell = 1 f o r e v e r y component e o f ll.

( i i ) I t may h a p p e n t h a t t h e o n l y p r o j e c t i o n b a n d s o f E a r e 0 a n d E , s o t h e o n l y c o m p o n e n t s o f ll a r e 0 a n d It.

A f i n i t e subset {e1;*.,en} E r t i t i o n o f ll i f z nl e i

=

It.

o f g ( l l ) w i l l be c a l l e d a ( I t follows automatically that

the ei’s a r e mutually d i s j o i n t . )

A p a r t i t i o n { d l , ’ * - , d m }w i l l

b e c a l l e d a r efinement of { e l , . - . y e n } i f every d . < e . f o r 3 1 some i . Each e i i s t h e n c l e a r l y a sum o f d . ’ s . Given two I partitions {el;.-,en}, { d l , ‘ . , d m } o f ll, t h e s e t {eiAd. 3 3 ( i = l , . . . , n ; j = l,... m) i s a g a i n a p a r t i t i o n a n d i s a

.

r e f i n e m e n t o f t h e two g i v e n o n e s .

So by i n d u c t i o n , e v e r y

f i n i t e s e t o f p a r t i t i o n s h a s a common r e f i n e m e n t . Straightforward computation, using refinements, gives u s :

(17.5)

E v e r y e l e m e n t a o f 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 by

can be w r i t t e n i n t h e form a p a r t i t i o n o f ll.

c n1

i

1’

where { e

- ..

(1)

,en} i s a 1’ And e v e r y f i n i t e number o f e l e m e n t s o f t h i s =

e.

8

l i n e a r s u b s p a c e c a n a l l b e w r i t t e n i n t e r m s o f t h e same

M l l - spaces and L -spaces

103

partition.

The above representations are not unique.

If an element can

be written in terms of some partition, then it can be written in terms of any refinement of that partition.

(17.6) Given an MI-normed space E , the linear subspace generated by 8 (I) is a Riesz subspace (containing 1 of course).

Proof. Consider an element a of this subspace; we show a + is also an element.

a

=

c n11ie i' { e l , * - - ,ne} a partition

of I. By permuting the subscripts if necessary, we can assume 11,"',Xk

> 0 and Xk+l, . . . ,A,

5 0 (the cases 1 .1

and 1. < 0 for a l l n are trivial). 1 -

k Then zlXiei

> 0

-

for all n

= a+

(Exercise 8

in Chapter 1). QED

Remark. -- The lattice structure in this Riesz subspace is transparent: Consider elements a,b in it. By ( 1 7 . 5 ) ' they can n b = Cn1 ~ i e i ,{e1;--,e n 1 a partition of I. be written a = zlXiei, Then avb = ~n~ ( m a x ( A ~ , ~ ~ ) ) e ~ . Combining the above with ( 1 6 . 6 ) ' we obtain :

(17.7) Suppose a is in the norm closure of the linear subspace generated by%(l).

Then there exists a sequence {bkl

Y

Chapter 3

104

such t h a t : (i)

< a; b 1 - 0 , we c a n t a k e t h e b k ‘ s > 0

Z(l)

i s c a l l e d t o t a l i n E i f t h e l i n e a r subspace which i t

g e n e r a t e s i s norm d e n s e i n E . We p r e s e n t t h e c l a s s i c a l F r e u d e n t h a l s p e c t r a l t h e o r e m . F i r s t , two u s e f u l lemmas on c o m p o n e n t s .

We r e c a l l ( 8 . 3 ) t h a t

i n a u - 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 , t h e band g e n e r a t e d b y a ny e l e m e n t a i s a p r o j e c t i o n b a n d ; a l s o t h a t we d e n o t e t h e component i n t h i s band o f e a c h bEE by ba.

-__ Lemma.

(17.8)

Let E b e a-Dedekind complete.

If aEE+

(i)

and e

=

l a ,t h e n ae = a .

I t f o l l o w s a and

e g e n e r a t e t h e same b a n d . ( i i ) If aEE, e ad

=

-a

=

ll a+’

and d

=

ll - e , t h e n a e

=

a+ and

*

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 . Let E b e a - D e d e k i n d c o m p l e t e . we s e t

Given aEE, f o r e a c h A € R,

M I - s p a c e s and L - s p a c e s

105

We w i l l c a l l t h e e ( 1 ) ' s t h e s p e c t r a l e l e m e n t s o f a , and a {e,(h) I h E IR} t h e s p e c t r a l f a m i l y o f a . Note t h a t e a ( X ) = 1 f o r 1 <

-11

a [ ] and e a ( x )

( 1 7 . 9 ) Lemma.

=

Let E

- 11 a l l .

0 for

be

o-Dedekind c o m p l e t e .

),EX, d e n o t e e a ( x ) s i m p l y be e ( x ) , and s e t d ( h )

Given aEE =

and

I - e(1).

Then

P roof. -

(a

ae(x)

1. x e ( x ) ;

ad(1)

5

(a - Xn)e(x)

=

ld(X)'

(a - XI)' > 0 , by ( 1 7 . 8 ) .

Xn)e(A) = a e ( x > - x n e ( x > = a e ( x ) - 1 e C x ) .

x e ( x )> 0.

But

Thus a e ( l )

-

The s e c o n d i n e q u a l i t y f o l l o w s s i m i l a r l y from

( a - xn)d(X) = - ( a -

an)-

< 0. -

QED

( 1 7 . 1 0 ) Theorem.

(Freudenthal [18]).

I f an MI-space E i s

a-Dedekind complete, t h e n Z ( 1 ) i s t o t a l i n E.

P roof.

W e show i t c a n be a p p r o x i m a t e d i n

C o n s i d e r aEE.

t h e norm b y l i n e a r c o m b i n a t i o n s o f i t s s p e c t r a l e l e m e n t s . can assume 0 < a < I. F i x n , d e n o t e l / n by

E,

and s e t

We

Chapter 3

106

e k -- e a ( ( k dk

=

ek

dn

=

en '

-

-

1)~)

ek+l

I t i s i m m e d i a t e from ( 1 7 . 9 ) t h a t

Thus

n,e, d Ad k j

en ' =

0 for k # j ,

a = Cn a = ~ " a dk dk

'

=

Remark. -

I

[icn ( a dk - ( k

- i I E dk 1 1

o-Dedekind completeness i s , i n g e n e r a l , s t r i c t l y

stronger than the property t h a t % ( n ) i s t o t a l . t a k e f o r E t h e space c of convergent sequences

As a n e x a m p l e ,

107

MIL-spaces a n d L - s p a c e s

We r e c o r d a s i m p l e r e s u l t ( 1 7 . 1 2 )

w h i c h we w i l l n e e d l a t e r .

Let E be a o-Dedekind complete Mn-space, F a R i e s z subspace c o n t a i n i n g t h e u n i t ll o f E , a n d A t h e s e t o f a l l s p e c t r a l e l e m e n t s o f e l e m e n t s of F. verified,

Note f i r s t t h a t , as i s e a s i l y

f o r a n y aEE a n d a n y s p e c t r a l e l e m e n t e a ( x ) o f a , we

a (A) = e (a-AIL) ( 0 ) . H e n c e , i n t h e p r e s e n t c a s e , s i n c e ILEF, e v e r y e l e m e n t o f A c a n b e w r i t t e n i n t h e f o r m e , ( O ) , a E F .

have e

IL a+' tn,laEF+}.

But e a ( 0 ) A

=

=

T h u s , s i n c e F i s a R i e s z s u b s p a c e , we h a v e

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

(17.11)

L e t E b e a 0 - D e d e k i n d c o m p l e t e MIL-space,

subspace containing of elements o f F.

(and

n,

F a Riesz

and A t h e s e t o f a l l s p e c t r a l e l e m e n t s

Then 4 i s a s u b l a t t i c e o f & ( n ) c o n t a i n i n g ll

0).

(17.12) Corollary.

The l i n e a r s u b s p a c e G o f E g e n e r a t e d b y A

above i s a Riesz subspace.

Proof. For e a c h eEA, -~

G c o n t a i n s ll

- e.

I t f o l l o w s G con-

t a i n s t h e B o o l e a n s u b a l g e b r a o f e ( I L ) g e n e r a t e d b y A, h e n c e i s g e n e r a t e d bv t h e Boolean s u b a l g e b r a , hence i s a Riesz s p a c e (cf. the proof of (17.6)). QED

Chapter 3

108

§18.

M l l -hnmomorphisms

S i n c e u n d e r an Mll-norm, o r d e r boundedness and norm boundedn e s s a r e e q u i v a l e n t , we have

(18.1)

I f E , F a r e MIL-normed s p a c e s , t h e n f o r e v e r y l i n e a r

mapping E -> 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<,/[\aJII i s i n K J ,

IIall

1.

=

/' I

a G I a G 11 i s i n K G , a n d

Thus a € c o n v [ K ( J ) IJ K ( G ) ] .

!I aJII

+

11 aGll

=

I f e i t h e r component i s 0 ,

there i s nothing t o prove. QE D

When combined w i t h norm c o m p l e t n e s s , an L-norm d i s p l a y s a number o f r e m a r k a b l e p r o p e r t i e s .

Central t o these properties

i s t h e f o l l o w i n g g e n e r a l p r o p e r t y o f a n L-norm.

(19.3)

Theorem. I n a n L-normed s p a c e E , i f a n e t {aa 1 i s mono-

t o n i c a n d norm b o u n d e d , t h e n i t i s norm Cauchy.

Chapter 3

112

Proof.

For c o n c r e t e n e s s , l e t {a } be a n a s c e n d i n g n e t . c1

We c a n assume i t h a s a f i r s t e l e m e n t , w h i c h i n t u r n a l l o w s u s t o assume i t i s i n E,.

Set A

w e show t h e r e e x i s t s a.

such t h a t \ \ a a - a

Choose a.

such t h a t

a

=

11 3

,?

II a a

- a

-

E.

aO

ll a a 0 II 5 II aall 2

1, h e n c e

NOW c o n s i d e r

supallaclll.

a0

I

E

> 0;

c E for a l l a > cia. a 0 Then f o r a z a0, 1 - E <

5

II $1

-

II a a 0 II 5

E

*

QE D

I f a n L-normed s p a c e i s norm c o m p l e t e , we w i l l c a l l i t an

L - s p a c e ( t h e common name i s ( A L ) - s p a c e ) .

Thus an L - s p a c e i s a

Banach l a t t i c e whose norm i s a n L-norm.

(19.4) Theorem.

Proof.

An L - s p a c e i s D edeki nd c o m p l e t e .

By E x e r c i s e 2 5 i n C h a p t e r 1, i t i s enough t o show

t h a t i f { a } i s an a s c e n d i n g n e t i n E, w h i c h i s bounded a b o v e , a t h e n V a e x i s t s . { a } i s norm b o u n d e d , h e n c e , b y ( 1 9 . 3 ) , norm c1

ac1

Cauchy, h e n c e norm c o n v e r g e s t o some aEE.

Then a = V a ( 9 . 5 ) a a

and w e a r e t h r o u g h . QED

( 1 9 . 5 ) -~ Theorem.

I f E i s an L-space, t h e n f o r a monotonic n e t

i n E , o r d e r c o n v e r g e n c e a n d norm c o n v e r g e n c e a r e e q u i v a l e n t .

,

ME-spaces a n d L - s p a c e s

Proof. -~

113

We c a n c o n f i n e o u r s e l v e s t o an a s c e n d i n g n e t { a } U

And by (9.6), we n e e d o n l y show t h a t i f { a } i s o r d e r

i n E,.

U

c o n v e r g e n t , t h e n i t i s norm c o n v e r g e n t .

Assume a f a . N

It

t h a t { a } i s norm Cauchy, h e n c e norm c o n a v e r g e s t o some b E E . A f o r t i o r i , it order converges t o b , f o l l o w s from (19.3)

hence b

=

a.

QED

A seminorm

I/ - ; I

on a R i e s z s p a c e E i s o r d e r c o n t i n u o u s i f

f o r e v e r y n e t { a } i n E , a -f a i n p l i e s l i m [ l a - aclll = 0 . T t a a N i s e a s y t o s e e t h a t a R i e s z s e m i n o r m 11 .[I i s a r d e r c o n t i n u o u s i f a n d o n l y i f a 4 0 i m p l i e s l i m IIauII a. a

(19.6) Corollary.

I n (19.5) Let E

=

=

0.

I n an L - s p a c e E , t h e norm i s o r d e r c o n t i n u o u s

t h e m o n o t o n i c i t y c a n n o t be d i s p e n s e d w i t h .

1’ a n d f o r e a c h n , en

the nth position.

=

(O,... , O , l , O ; .

a

)

the 1 in

,

Then t h e s e q u e n c e { ( l / n ) e n } ( n

=

1,2;**)

norm c o n v e r g e s t o 0 b u t , n o t b e i n g e v e n o r d e r b o u n d e d , d o e s n o t order converge.

S i n c e , a s i s e a s i l y v e r i f i e d , k1 i s a n L - s p a c e ,

t h i s s u p p l i e s a c o u n t e r example. However, e v e r y Banach l a t t i c e h a s t h e w e a k e r p r o p e r t y ( 9 . 9 ) , hence ( 9 . 1 0 ) .

Combining t h i s w i t h ( 1 9 . 5 ) g i v e s u s t h a t a s u b -

s e t o f a n L - s p a c e i s o r d e r c l o s e d i f a n d o n l y i f i t i s norm closed.

More s h a r p l y ,

Chapter 3

114

( 1 9 . 7 ) -___ Theorem.

For a s u b s e t A o f an L - s p a c e , t h e f o l l o w i n g

sets coincide:

1'

t h e o r d e r c l o s u r e o f A;

2'

the a - o r d e r c l o s u r e of A;

3'

A(1).

4'

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 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 o f A;

5'

t h e norm c l o s u r e o f A .

P r o o f . The f i r s t s e t c o n t a i n s t h e s e c o n d and t h e t h i r d . -~ The l a t t e r two e a c h c o n t a i n t h e f o u r t h , w h i c h i n t u r n c o n t a i n s t h e f i f t h by ( 9 . 9 ) .

Finally, the f i f t h s e t contains the f i r s t

by ( 1 9 . 6 ) . QED

One c o n s e q u e n c e w o r t h n o t i n g i s t h a t i n a n 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 p r o j e c t i o n b a n d . A Riesz s p a c e E i s s a i d t o have t h e c o u n t a b l e sup p r o -

p e r t y (Fremlin) i f b = V A i n E i m p l i e s b

=

V,a,

f o r some

c o u n t a b l e s u b s e t l a n } o f A.

( 1 9 . 8 ) Theorem.

Proof.

An L - s p a c e E h a s t h e c o u n t a b l e s u p p r o p e r t y .

Suppose b = V A .

Let 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 supreme o f a l l f i n i t e s u b s e t s o f A.

Then,

Ma-spaces and L-spaces

115

by E x e r c i s e 2 6 i n C h a p t e r 1 , b i s i n t h e norm c l o s u r e of A

1’ h e n c e i n t h e norm c l o s u r e o f some c o u n t a b l e s u b s e t { a n ) o f A

I t f o l l o w s from ( 9 . 5 ) t h a t b

= V

a . n n

1’

S i n c e each an i s t h e

supremum o f a f i n i t e s u b s e t o f A , w e o b t a i n a c o u n t a b l e s u b s e t

o f A h a v i n g b f o r supremum. QE D

(19.9) C o r o l l a r y 1.

I n an L - s p a c e E , e v e r y o r d e r bounded s e t A

of mutually d i s j o i n t elements i s c o u n t a b l e , A

cnII anll

<

=

{ a n } , and

co *

P r o o f . We c a n assume A c E,. ____ there exists b

T h e n , by ( 1 9 . 8 ) , b

= VA.

{ a n } c A ; w e show A

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

.

{a,}

=

t i n c t from a l l t h e a n ’ s .

n a n f o r some

S u p p o s e t h e r e e x i s t s aEA

Then aAan

(Exercise 5 i n Chapter 1).

= V

=

dis-

0 f o r a l l n , h e n c e a/\b = 0

But t h e n a = 0 ( s i n c e aEA, h e n c e

< b ) , and w e h a v e a c o n t r a d i c t i o n . a -

n S i n c e t h e a n ’ s a r e mutually d i s j o i n t , xlai n ~ ~ ~ l a =i i11 lz l a i l j

n

= IIvlai[j

5 IIaII f o r a l l n .

= V

n hence 1a i ’

Thus CyIIan[I < \ I a [ I <m QED

A s t r o n g e r p r o p e r t y t h a n t h e above f o r a R i e s z s p a c e i s

t h a t e v e r y 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 (bounded o r n o t )

be c o u n t a b l e .

This i s c l e a r l y equivalent t o the property t h a t

every s e t of mutually d i s j o i n t bands be countable.

A Riesz

s p a c e w i t h t h i s p r o p e r t y w i l l b e s a i d t o b e a t most c o u n t a b l y

Chapter 3

116

decomnosable.

(19.10) C o r o l l a r y 2 .

F o r a b a n d ,J o f an L - s p a c e E , t h e f o l l o w -

ing are equivalent: lo

.J i s a p r i n c i p a l b a n d ;

2'

J i s a t most c o u n t a b l y d e c o m p o s a b l c ;

3'

,J i s c o u n t a b l y g e n e r a t e d .

-___ P r o o f . Assume ,J i s t h e b a n d g e n e r a t e d b y some a E E , ,

and

l e t {J } be a c o l l e c t i o n o f m u t u a l l y d i s j o i n t ( n o n - z e r o ) bands a i n J . For e a c h " , a J # 0 ; hence, by (19.91, is c o u n t a b l e . a A fortiori,

{Ja} i s countable.

Now a s s u m e J i s a t m o s t c o u n t a b l y d c c o m p o s a b l e , and l e t { a } be a maximal c o l l e c t i o n 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 J + . c1

Then J i s t h e b a n d g e n e r a t e d by { a a } . t h e band g e n e r a t e d b y a a .

i s countable.

'1

For e a c h a , l e t < J ab e

Then {J } i s c o u n t a b l e , whence { a c x } a

J i s thus countably generated.

T h a t 3'

implies

h o l d s f o r a g e n e r a l Banach l a t t i c e ( E x e r c i s e 2 3 i n C h a p t e r 1 ) . QED

§ 2 0 The e x t r e m e p o i n t s o f K ( E )

We w i l l d e n o t e b y e x t A t h e s e t o f e x t r e m e p o i n t s o f a s e t A i n a vector space.

In this 5,

w e c o l l e c t some p r o p e r t i e s

b l l - s p a c e s and L - s p a c e s

117

o f e x t K ( E ) , E a n L-normed s p a c e .

Let E be an L-normed s p a c e .

(20.1)

For e v e r y p r o j e c t i o n band

J of E ,

e x t K(E)

Proof. ___ aEK(E)

IIaII

=

n

=

[ e x t K ( J ) ] IJ [ e x t K(J

C o n s i d e r a € e x t K(E).

d

I].

If a

= 0 , then a E J , hence J S i m i l a r l y , i f aJ = 0 ,

J = K ( J ) , hence a € e x t K(J).

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 a € e x t K(E).

1.

Thus t h e l e f t s i d e o f t h e d e s i r e d e q u a l i t y i s c o n t a i n e d i n t h e right side. 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 a € e x t K ( J ) ; t o show a € e x t K ( E ) . +

K

=

1.

Suppose a

Then 0 < hb,

KC

=

Xb

+ KC,

b , c € K ( E ) , X,K

< a , s o Xb, K c E J ,

S i n c e a € exr. K ( J ) , i t f o l l o w s X

=

0 or

K

=

we h a v e

SO

2 0,

b,c€K(J).

0.

QED

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

s e t {Aa(AE IR}.

Thus lRa i s t h e ( o n e d i m e n s i o n a l ) l i n e a r s u b -

space generated by a .

118

Chapter 3

L e t E b e a n L-normed s p a c e .

(20.2)

Then f o r a € K ( E ) , t h e

following a r e equivalent: 1'

aE e x t K(E);

2'

lRa i s a R i e s z i d e a l o f E .

Assume a € e x t K ( E ) , a n d s u p p o s e 0 < b < a ; we

Proof.

If b = 0 o r b

show b = Ilblla. 0 < b < a.

w i t h (b/II bII

Set c

1E

=

K(E),

=

a , we a r e t h r o u g h .

a - b, so 0 < c < a also.

(c/ll

cII

1 E K(E),

and

[I bII

+

Suppose

Now

11 cII

=

[I a [ ] =

1.

S i n c e a i s a n e x t r e m e p o i n t o f K ( E ) , we must h a v e a = b / l l b [ l , whence b

=

Ilbll a .

C o n v e r s e l y , assume IRa i s a R i e s z i d e a l , and s u p p o s e a

=

Xb +

KC,

with b,c€K(E), b,c > 0, X +

n e s s , l e t X > 0. f o r some p € IR.

K

= 1.

For c o n c r e t e -

Then 0 < Xb < a , s o , b y a s s u m p t i o n , Xb T h i s g i v e s b = (li/X)a, and s i n c e

[I bII

=

pa

= 1, i t

follows b = a. QEP

(20.3)

Corollary.

The e l e m e n t s of K ( E )

are mutually d i s j o i n t .

I n c o n t r a s t t o 918, w e p o s t p o n e t h e c o n s i d e r a t i o n o f mapp i n g s o f an L - s p a c e t o 5 2 4 .

MI-s p a c e s and L - s p a c e s

119

EXERCISES

1. A R i e s z i d e a l o f a R i e s z s p a c e i s maximal i f i t i s p r o p e r

and c o n t a i n e d i n n o o t h e r p r o p e r R i e s z i d e a l .

I n an M a -

normed R i e s z s p a c e , a maximal R i e s z i d e a l i s norm c l o s e d . I t need n o t be o r d e r c l o s e d .

2.

Let E be a Riesz s p a c e . (i)

I f b i s a component o f a , a n d c i s a component o f b , t h e n c i s a component o f a .

( i i ) I f b , c a r e c o m p o n e n t s o f a , and c < b , then c i s a component o f b .

3. Let E be a n MI-normed s p a c e , and e , d € z ( I )

(i)

eAd

=

0 i m p l i e s e + d E & (I).

( i i ) ( e - d)',

( e - d ) - , le - d l € 6 ( I ) .

( i i i ) I f e # d , t h e n IIe - d/l

=

1.

CHAPTER 4

DUAL L- SPACES AND M I -SPACES

N o t a t i o n . As we w i l l s e e b e l o w , t h e d u a l o f an M I - s p a c e i s an L - s p a c e and t h e d u a l o f an L - s p a c e i s an M I - s p a c e . I n b o t h ~

c a s e s , t h e e l e m e n t o f t h e Mll-space w i l l a p p e a r i n t h e f i r s t p o s i t i o n i n t h e b i l i n e a r form (

a

,

.

)

giving the duality.

Since

o u r p r i m a r y s p a c e C(X) and i t s b i d u a l a r e b o t h M I - s p a c e s , t h i s i s i n accordance with t h e convention adopted a t the beginning o f 513.

5 2 1 . The d u a l o f an MI-normed s p a c e

Our f i r s t two p r o p o s i t i o n s h o l d f o r an M-normed s p a c e , and we s t a t e them s o .

( 2 1 . 1 ) Theorem.

I f E i s a n M-normed s p a c e , t h e n E j i s an

L-space.

Proof.

11 $11

+

[I $11.

Consider $ , $ C ( E ' ) + :

we h a v e t o show 114

I t i s enough t o show t h a t f o r e v e r y 120

E

+

> 0,

$11

=

Dual L-spaces a n d M a - s p a c e s

Since E '

121

i s an L - s p a c e , w e can d i s c u s s K ( E ' ) .

The f o l l o w -

ing i s a sharpening of (20.2).

(21.2)

Theorem.

L e t E b e a n M-normed s p a c e .

Then f o r $ € K ( E ' ) ,

the following are equivalent: io

4~ e x t K ( E ' ) ;

2'

IR$ i s a R i e s z i d e a l o f E ' ;

3'

$-l(O)

4'

$ i s a R i e s z homomorphism o f E i n t o R .

Proof. (20.2).

By

Conversely, 3'

i s a R i e s z i d e a l o f E ( h e n c e a maximal o n e ) ;

We e s t a b l i s h e d t h e e q u i v a l e n c e o f 1' a n d 2'

( R G ) ~= $ - ' ( O ) ,

2O,

h e n c e 2'

i m p l i e s '3

s i n c e IR$ i s o n e - d i m e n s i o n a l , IR$

i m p l i e s 2'

=

i m p l i e s .'4

We show 3'

By

the factorization E

3O,

(11.9).

((R$),)',

(10.15).

in

hence

'

t h e m a p p i n g E ->R h a s

L> E / + - ' ( O )

4 >lR.

.

q i s a R i e s z homomorphism; we show 4 i s a R i e s z i s o m o r p h i s m ,

w h i c h w i l l g i v e u s .'4 E/+-l(O)

i s p o s i t i v e (since $ i s ) , and, since

is one-dimensional, it is a b i j e c t i o n .

I t follows

Chapter 4

122

6 is

F i n a l l y , '4

a Kiesz i s o m o r p h i s m .

i m p l i e s '3

by ( 6 . 2 ) .

QED

We t u r n t o a n MI-normed s p a c e C . (18.1), E'

Note f i r s t t h a t f r o m

(we d o n ' t n e e d norm c o m p l e t e n e s s ) .

= Eb

( 2 1 . 3) I f E i s an M I - s p a c e , t h e n f o r e v c r y

(I,$). (Thus f o r

$ € ( E l ) + ,

$EK(E')

$ € ( E l ) + ,

i f and o n l y i f

11 $11 = (n,$) =

1.)

The p r o o f i s i m m e d i a t e .

( 2 1 . 4 ) C o r o l l a r y . For e v e r y

[I $11

=

(n,++)

+ € E l ,

+

(I,$-).

I t follows $ > 0 i f and o n l y i f

\I$\\

=

(n,$)

Remarks. 1. For an Ma-normed s p a c e , ( 2 1 . 1 ) i s o b v i o u s from the equality 2.

2'

I) $\I

=

( I , $ ) f o r $€(E')+.

F o r a n MI-normed s p a c e , we c a n s t r e n g t h e n ( 1 0 . 5 ) :

need only hold f o r a

=

a.

Let E be a n MI-normed s p a c e , a n d s u p p o s e T h i s s a y s 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 by

6 (n)

6 (I) i s

is total. norm d e n s e

i n E , h e n c e e a c h + € E l i s c o m p l e t e l y d e t e r m i n e d by i t s v a l u e s on g(I).

We now show t h a t a l s o $'

4 onG(n1.

i s determined by t h e v a l u e s o f

Dual L - s p a c e s and M a - s p a c e s

Let E b c a n MI-normed s p a c e w i t h

(21.5) each

@,€El

5

given

E

Then f o r

a n d e t t (I),

We c a n assume

Proof. (d,$+)

(ll) t o t a l .

123

II$II

=

1.

For every d < e , (d,$) 5

F o r t h e o p p o s i t e i n e q u a l i t y , w e show t h a t

(e,$+).

there exists d < e such t h a t ( d , @ ) > (e,$+)

> 0,

< a < e and ( a , @ ) > (e,@+)Choose aEE s u c h t h a t O -

-

2 ~ .

E.

T h e r e e x i s t s a p a r t i t i o n { e l , - . . , e n } o f ll a n d a

Lemma.

n l i n e a r combination z l l i e i

m

(i)

xlei

(ii)

o

=

such t h a t

e f o r some m < n,

5 Iqhiei 5 a , n ( i i i ) IIa - xlXieill 5 E . I n e f f e c t : By ( 1 7 . 7 ) , t h e r e i s a p a r t i t i o n o f ll a n d l i n e a r c o m b i n a t i o n s a t i s f y i n g ( i i ) and ( i i i ) . The comqon r e f i n e m e n t o f t h i s p a r t i t i o n and t h e p a r t i t i o n { e , n - e } i s t h e n t h e d e s i r e d one.

Note t h e 0 < 1. < 1 (i 1 -

=

l,a.+,n).

Chapter 4

124

I t follows

m

( ( a , $ > - C I X i ( e i ' @ ) l<

€

9

whence

zyxi(ei,$)

(a,$>

-

E

2

( e , @ + )-

2E

*

S u p p o s e ( e i , @ ) > 0 f o r i = l , . * . , k , where 1 i k i m, and (ei,+) < 0 f o r t h e remaining i ' s , i f t h e r e a r e any.

a l w a y s p e r m u t e l , . . . , n s o a s t o have t h i s .

5 z kl ( e i , $ )

=

(ei,$) < 0 f o r i = 1, * .

We a r e a s s u m i n g

m

t h a t ( e i , $ ) > 0 f o r a t l e a s t one i . )

k c,xi(ei,@)

(We c a n

< Then c l x i ( e i , $ ) k k ( z l e i , @ ) , s o zlei is t h e d e s i r e d d.

- ,n,

m

then clxi( e i , $ )

<

0 = (O,$),

If

so 0

i s t h e d e s i r e d d.

QEn

We r e c a l l ( 5 1 5 ) t h a t f o r a n y normed s p a c e E , we c a l l

o(E',E)

t h e v a g u e t o p o l o g y on

E l .

I f E i s an MIL-normed s p a c e , t h e n K(E')

of

( E l ) +

with t h e hyperplane {$CE1l(IL,$)

=

is the intersection l}.

The f i r s t o f

t h e s e i s v a g u e l y c l o s e d by ( 1 0 . 3 ) , a n d o f c o u r s e t h e s e c o n d i s I t f o l l o w s K(E')

also.

i s vaguely closed.

tained in the unit b a l l B ( E ' ) ,

Since it i s con-

w h i c h i s v a g u e l y c o m p a c t , we

obtain

( 2 1 . 6 ) I f E i s an MIL-normed s p a c e , K(E')

In a d d i t i o n ,

i s v a g u e l y compact.

Dual L - s p a c e s a n d MIL-spaces

(21.7)

125

I f E i s an MIL-normed s p a c e , e x t K ( E ' ) i s v a g u e l y

compact.

Proof.

We show e x t K ( E ' ) i s v a g u e l y c l o s e d i n K ( E ' ) .

Suppose a n e t {$ } i n e x t K ( E ' ) c o n v e r g e s v a g u e l y t o $ E K ( E ' ) . c1

We show $ i s a R i e s z homomorphism ( 2 1 . 2 ) .

Given a , b E E , t h e n ,

u s i n g (21.2), (avb,$)

=

l i m (avb,$ )

=

limcl max((a,+&

a

c1

(b,$a))

= max(1im ( a , $ ) , l i m a ( b , $ c l ) ) c1

=

a

max ( ( a , $ ) ,( b , > )

( 2 1 . 8 ) C o r o l l a r y 1.

I f E i s a n MI-normed s p a c e , e v e r y aEE

a t t a i n s i t s norm on e x t K ( E ' ) .

Proof.

aEE",

s o b y ( 2 3 . 2 ) b e l o w a n d t h e v a g u e com-

p a c t n e s s o f K ( E ' ) , a a t t a i n s i t s norm on K ( E ' ) .

I t follows

e a s i l y f r o m t h e K r e i n - M i l m a n t h e o r e m t h a t a a t t a i n s i t s norm on e x t K(E').

QED

(21.9) Corollary 2. i s s e p a r a t i n g on E .

I f E i s a n MIL-normed s p a c e , t h e n e x t K ( E ' )

Chapter 4

126

101

522.

( E ' ,E), a d i v e r s i o n

We h a v e s e e n t h a t f o r a n M I - s p a c e E , t h e norm on E '

s a t i s f i e s the identity

[I $11

(1,I @ ( ) .

=

Given a g e n e r a l R i e s z

s p a c e E , t h e r e i s a n i m p o r t a n t p o l a r t o p o l o g y on Eb t h a t i s d e f i n e d b y a g e n e r a l i z a t i o n o f t h e above i d e n t i t y , and whose p r o p e r t i e s a r e c o n s e q u e n t l y s t r i k i n g l y similar t o t h o s e o f an L-norm.

We t u r n a s i d e i n t h e p r e s e n t 5 t o s t u d y t h i s t o p o l o g y .

We n e e d some p r e p a r a t i o n .

A t o p o l o g y on a R i e s z s p a c e i s

called l o c a l l y s o l i d i f it has a b a s i s f o r t h e neighborhoods of 0 consisting of s o l i d sets ( 5 9 ) . called a locally solid space.

The R i e s z s p a c e i s t h e n

Under a l o c a l l y s o l i d t o p o l o g y ,

t h e o p e r a t i o n s a v b , a/\b a r e u n i f o r m l y c o n t i n u o u s ; and i f t h e topology i s Hausdorff, t h e p r o p e r t i e s (9.3) "sequences:

r e p l a c e d by " n e t s "

d e l e t e d from ( 9 . 7 ) ,

-

( 9 . 8 ) , with

i n ( 9 . 4 ) and " a - o r d e r c l o s e d "

a l l hold (the proofs there did not use the

convexness of the u n i t b a l l ) .

The r e a d e r c a n a l s o r e f e r t o [ Z ]

f o r a complete development o f t h e p r o p e r t i e s o f l o c a l l y s o l i d spaces. A t o p o l o g y i s b o t h l o c a l l y s o l i d a n d l o c a l l y convex i f i t

has a b a s i s f o r t h e neighborhoods of 0 c o n s i s t i n g of convex s o l i d s e t s - e q u i v a l e n t l y , i f i t c a n be g i v e n b y a f a m i l y o f R i e s z s emin or m s .

Such i s t h e t o p o l o g y we now t u r n t o .

L e t E b e a Riesz s p a c e . on E b b y Il$lla

=

(la1

,[$I).

Each aEE d e f i n e s a seminorm

11

a l l a

The t o p o l o g y d e t e r m i n e d by t h i s

f a m i l y o f seminorms ( a r u n n i n g t h r o u g h E ) w i l l b e d e n o t e d b y

101

( Eb ,E)

( N akano) .

the unit b a l l of

11 .!Ia

I t i s n o t h a r d t o s e e t h a t f o r e a c h aEE, i s p r e c i s e l y t h e p o l a r s e t [ - la1 , / a 1 1 .

Dual L - s p a c e s a n d M I - s p a c e s b ( E ,E) i s s i m p l y t h e p o l a r t o p o l o g y on E b d e f i n e d by

101

Thus

127

t h e i n t e r v a l s o f E.

for

More g e n e r a l l y ,

A c E , the family

.\I

,laE41

defines

a ( n o t n e c e s s a r i l y H a u s d o r f f ) t o p o l o g y , w h i c h we d e n o t e by

101

(Eb,A).

I f R i s t h e s o l i d h u l l of A, then c l e a r l y

I ~ I ( E ~ , =E )I O ~ ( bC , E + ) .

~ a l j ~ ~= ,I ~~ I)( E ~ , A ) I.n p a r t i c u l a r ,

Thus, i n s t u d y i n g t h e s e t o p o l o g i e s , w e can c o n f i n e o u r s e l v e s t o semi-norms o f t h e form To o u r k n o w l e d g e ,

II$\Ia,

101

a > 0.

(Eb,E) was f i r s t d e f i n e d by Dieudonng

I t s importance l i e s i n the f a c t t h a t i t i s t h e c o a r s e s t

[12].

which i s l o c a l l y

p o l a r t o p o l o g y on Eb ( f i n e r t h a n o ( E b , E ) ) We a l s o h a v e

solid.

(22.1)

Theorem.

b

101

complete under

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

(Dieudonng). (E , E ) .

-~ Proof. C o n s i d e r a Cauchy n e t

in

lOj(E

b

,E).

Then

i s a f o r t i o r i v a g u e l y Cauchy, h e n c e c o n v e r g e s i n E* t o some $ E E * .

We show f i r s t t h a t $ E E

t h a t f o r e a c h aEE,, that (a,

-

I)

0,

b.

I t i s e n o u g h t o show

0 i s b o u n d e d on [ - a , a ] . <

1 for a l l

c1

2

Then f o r b E [ - a , a ] a n d

u0.

0

a'a

0'

I(bAa) -

( W a) I

=

I(bAa

-

U

< (

IbI,I$a -

<

(a,14c1 -

<

1,

0

Choose a0 s u c h

>I 0, I ) 0

I) 0

128

Chapter 4

+ 1

D e n o t i n g t h i s l a s t q u a n t i t y by K , we have t h a t f o r e v e r y b E [ - a , a l , I ( b , $ > l = l i m a l ( b , + a ) l (K. We c o m p l e t e t h e p r o o f b y showing t h a t l i m f o r a l l aEE+.

C o n s i d e r aEE+ and

> 0.

E

]I$

c i a

Choose . a

-

$11

a = 0

such t h a t

> a0. for a l l a -

I t i s enough t o show t h a t f o r /(b,+a exists a

+)I 5 3

a.

~1

> a0 and bE [ - a , a ] , -

I $ a 1 c o n v e r g e s v a g u e l y t o +, t h e r e

2 ~ . Since

such t h a t I ( b , $

a1

-

$)I 5

E.

Then f o r a n y

a > a0,

A l l o u r a b o v e d i s c u s s i o n up t o ( b u t n o t i n c l u d i n g )

h o l d s w i t h E and Eb i n t e r c h a n g e d .

11 - [ ] +

11 all+

( la1

Every + € E b

(22.1)

defines a semi-

,I + / ) ,

and we t h u s o b t a i n t h e b t o p o l o g y 1u1 (E,A) d e f i n e d by e a c h s u b s e t A o f E In p a r t i c u l a r , b b we h a v e 1 0 1 (E,E ) . I f E i s s e p a r a t i n g on E , t h i s l a s t i s

norm

on E by

=

.

s i m p l y t h e t o p o l o g y i n d u c e d on E b y

1 0 I (Ebb,Eb).

129

Dual L - s p a c e s a n d M l l - s p a c e 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 .

( 2 2 . 2 ) Given a s u b s e t A o f E , 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 a l by A .

Then ~ O I ( E ~ , =A )/ ~ I ( E ~ , I ) .

I n t h e o p p o s i t e d i r e c t i o n , w e c a n s a y more:

( 2 2 . 3 ) Theorem.

Given a s u b s e t A o f E

i d e a l g e n e r a t e d by A. (i)

(n((E,A)

=

b

,

l e t J be t h e Riesz

Then

(a((E,J);

( i i ) 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 E w h i c h a r e

101

(E,A)

continuous i s p r e c i s e l y J .

P roof.

The p r o o f o f ( i ) i s e a s y ; w e show ( i i ) .

By ( i ) ,

w e c a n a s s u m e A = J, s o ( i i ) r e d u c e s t o (E,

101

That e v e r y element o f J i s

(E,J))'

101

(E,J)

= J.

continuous is c l e a r .

For

t h e c o n v e r s e , c o n s i d e r a l i n e a r f u n c t i o n a l I$ on E w h i c h i s IoI(E,J) continuous.

T h i s means t h e r e e x i s t s $ E J +

i s b o u n d e d on ( t h e n e i g h b o r h o o d o f 0 )

[-$,,$,lo.

s u c h t h a t I$

For s i m p l i c i t y ,

l(b,I$)l
assume s u p bE

But

[-$,$I0

[-$,$I i s c l o s e d i n E* ( 1 0 . 1 4 ) , hence and w e a r e t h r o u g h .

([-$,$I

0 0

)

in-^*= [-$ A QED

I,

Chapter 3

130

Remark. -__

I f E and E b a r e e x c h a n g e d i n ( 2 2 . 3 )

(and we t a k e

A i n ) , t h e n ( i i ) h o l d s i f J i s r e p l a c e d by t h e R i e s z i d e a l bb g e n e r a t e d by A i n E

.

The s i m i l a r i t y b e t w e e n

101

(Eb,E) and t h e L-norm t o p o l o g y on

t h e d u a l o f an MIL-normed s p a c e i s d i s p l a y e d i n t h e n e x t few r e s u l t s . Note f i r s t t h a t f o r e v e r y a € E , b E has t h e property:

114 +

$lia

=

f o r t h e seminorm

/14/la

II.II 4

t h e seminorm

II$IIa

on

And s i m i l a r l y

f o r a l l 4,$E(Eb ) + . b on E d e f i n e d b y 4 E E . +

11 * [ I a

Given a s u b s e t A o f a R i e s z s p a c e E , i f a m o n o t o n i c n e t

(22.4)

i n Eb i s

101

(Eb , E ) b o u n d e d , t h e n i t i s

101

(Eb , A )

Cauchy.

And

s i m i l a r l y f o r a s u b s e t A o f Eb and a m o n o t o n i c n e t i n E .

A l l t h a t n e e d s t o b e shown i s t h a t t h e n e t i s

f o r e a c h aEA, (19.3)

(22.5)

11 . \ l a - C a u c h y

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

.

For a m o n o t o n i c n e t i n E

b

,

l o l ( E b , E ) c o n v e r g e n c e and

o r d e r convergence a r e e q u i v a l e n t .

P roof.

That t o p o l o g i c a l convergence of a monotonic n e t

i m p l i e s i t s o r d e r c o n v e r g e n c e h o l d s f o r any l o c a l l y s o l i d

Dual L - s p a c e s a n d MIL-spaces

131

Hausdorff s p a c e : t h e p r o o f s o f (9.6) and (9.5) c a r r y o v e r t o The p r o o f o f t h e c o n v e r s e i m p l i c a t i o n i s i d e n t i -

such a space.

cal with that of (19.5).

QED

(22.6)

C o r o l l a r y 1.

I n Eb o r d e r convergence i m p l i e s

(01

(Eb,E)

convergences.

The p r o o f i s t h e same a s t h a t o f ( 1 9 . 6 ) . Combining ( 2 2 . 5 ) w i t h ( 9 . 4 )

(or rather, its adaptation to

a l o c a l l y s o l i d s p a c e ) , we h a v e

(22. 7) Corollary 2.

I n Eb t h e o r d e r c l o s u r e and

(01

(Eb,E)

c l o s u r e of a Riesz i d e a l c o i n c i d e .

We s h a r p e n ( 2 2 . 3 ) :

( 2 2 . 8 ) Theorem.

G i v e n a s u b s e t A of E b ,

the set of linear

f u n c t i o n a l s on E w h i c h a r e \ u I ( E , A ) c o n t i n u o u s o n e v e r y i n t e r v a l

o f E i s p r e c i s e l y t h e b a n d g e n e r a t e d by A .

Proof. ___

A g a i n w e c a n assume A i s a R i e s z i d e a l J o f E

b

.

Thus we h a v e t o show t h a t 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 E

Chapter 4

132

w h i c h a r e \ u \ ( E , J ) c o n t i n u o u s on e v e r y i n t e r v a l of E i s t h e b order closure of J - which, by ( 2 2 . 7 ) , i s t h e 1 0 1 (E , E ) I f J i s s e p a r a t i n g on E t h e n t h e G r o t h e n d i e c k

c l o s u r e of J .

c o m p l e t e n e s s theorem g i v e s t h e d e s i r e d r e s u l t ( c f . [ 4 5 ] , Chap.6, Theorem 2 ) .

I f J i s n o t s e p a r a t i n g on E , t h e n , b y a s t r a i g h t -

f o r w a r d a r g u m e n t , we c a n r e d u c e t h e t h e o r e m t o o n e on E/.J

L

and

again apply t h e Grothendieck theorem.

5 2 3 . The d u a l o f an L - s p a c e

We r e t u r n t o o u r main s t r e a m o f d e v e l o p m e n t .

(23.1)

I f E i s an L-normed s p a c e , t h e n E ' i s a

Theorem. -~

(Dedekind c o m p l e t e ) M I - s p a c e w i t h i t s u n i t I g i v e n by (n,a)

-__ Proof.

=

( [ a + ( [- Ila-\l

Consider

$ , $ € ( E l ) + ;

for a l l aEE.

we show

For c o n c r e t e n e s s , l e t t h e r i g h t s i d e b e

11 $ V t j [ l 5 II$II,

show

11$11

(I$([.

=

max((l$(l, [ I $ \ \ ) .

We n e e d o n l y

and f o r t h i s , i t i s enough t o show ( $ v t j , a ) <

f o r a l l aEK(E). F i x aEK(E).

II$I[

\I$V$[[

By (10.4), we have t o show ( $ , b )

for a l l b,c > 0 such t h a t b + c = a .

pair.

Then

+

($,c)

Let b , c be s u c h a

5

Dual L - s p a c e s a n d Mll-spaces

(4,b)

+

(UJ,C>

=

I [ bII

(4,

(IbII ll 4 < II !I !I @\I <

'3

=

= =

=

+

i s an M-space.

Thus E ' by n ( a )

(I1 blI II aII \I $11 [I $ II -

b p:r )

+

I] cII

133

(+,m ) C

II CII II $Ii + II C I I [I $11 II C I I ) II 441 +

Now d e f i n e t h e f u n c t i o n ll on E +

( l a ; ] . ll i s a d d i t i v e on E + b y t h e d e f i n i n g p r o p e r t y o f

t h e L-norm, a n d i s o f c o u r s e p o s i t i v e l y homogeneous.

Its

e x t e n s i o n t o a l l o f E i n t h e u s u a l manner i s t h e r e f o r e a n So w e c a n now w r i t e ( l L , a ) i n s t e a d o f n ( a ) f o r

element o f

E l .

a l l aEE.

Finally, given

aEE+, ( $ , a )

(11$(1

(la// 5

@ € [ E l ) + ,

I[ a / [ =

if

I]$ll -<

( & a ) ; hence

I#J

1, t h e n f o r e v e r y < 1.

Thus Il

is

the unit f o r E l . QE D

A g a i n , t h e r e m a r k a b l e p r o p e r t i e s o f a n L-norm a p p e a r when combined w i t h norm c o m p l e t e n e s s :

(23.2)

For a l i n e a r f u n c t i o n a l

$ o n an L-space E , t h e f o l l o w i n g

are equivalent:

'1

$ € E l ;

2O

@ab;

3'

$E.EC;

4'

+

i s norm c o n t i n u o u s on K ( E ) ;

5'

I$

i s bounded on K [ E ) .

Chapter 4

134

We a l r e a d y h a v e t h e e q u i v a l e n c e o f 1'

Proof.

I _

a n y Banach l a t t i c e ( 1 5 . 4 ) .

T h a t 1'

implies 3

0

t h e o r d e r c o n t i n u i t y o f t h e norm i n E ( 1 9 . 6 ) , from ( 6 . 0 ) .

i m p l i e s Z0,

0

1

and 2'

for

f o l l o w s from and t h a t 3'

o f c o u r s e i m p l i e s 4'

a n d 5'.

Since

the unit b a l l B(E)

o f E i s c o n t a i n e d i n t h e convex h u l l o f

K ( E ) IJ ( - K ( E ) ) ,

implies lo.

0

implies 1

5'

T t r e m a i n s t o show t h a t 4'

.

S u p p o s e $ @ E'.

Lemma.

F o r e a c h a E K ( E ) , nElN, a n d

__.

E

> 0, there exists

bEK(E) s u c h t h a t jlb - a ( / < Z E and l ( $ , b ) l > n.

We c a n assume

E

r: 1. -

4 i s discontinuous a t 0 (in the

n o r m ) , s o t h e r e e x i s t s c E E + s u c h t h a t IIcII Zn.

Set d = a + c.

5

I(@,c)l

E,

+

I(@,c)l

- ]($,a)/> Zn, a n d i t f o l l o w s by s t r a i g h t f o r w a r d com=

E,

l(@,d)l

2

I($,a)l

putation that b

Then [ I d - a [ [ <

F

d/[[dl[ has t h e d e s i r e d p r o p e r t i e s .

I t f o l l o w s from t h e Lemma t h a t t h e r e e x i s t s a norm Cauchy s e q u e n c e {a,} Since K(E)

i n K(E) s u c h t h a t

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

s e q u e n c e c o n v e r g e s t o some aEK(E). a on K ( E ) ,

> n (n = 1 , 2 , . * . ) .

-

Then 4 i s n o t c o n t i n u o u s a t

and we a r e t h r o u g h . (2ED

We e m p h a s i z e one o f t h e r e s u l t s c o n t a i n e d i n t h e a b o v e prob p o s i t i o n : F o r a n L - s p a c e E , E ' = E C = E . As i n d i c a t e d i n t h e p r o o f , t h i s i s a p r o p e r t y o f a l l Banach l a t t i c e s w i t h o r d e r

Dual L - s p a c e s a n d M I - s p a c e s

135

con t i n uo 11s no rm .

( 2 3 . 2 ) I f E i s an L - s p a c e , t h e n f o r $ € E l ,

P-r o o f .

We show t h a t f o r e a c h a E B ( E ) , t h e u n i t b a l l o f E ,

t h e r e e x i s t s bEK(E) such t h a t w r i t t e n i n t h e form a h +

=

K

jlajl

=

Xb -

l($,b)l

> [($,a)[.

a can be

where b , c E K ( E ) , X , K > 0 , and

KC,

(cf. the discussion a t the beginning of 519).

Assume, f o r c o n c r e t e n e s s , t h a t l($,a)l

l($,b)l

=

IV@,b)

<

hl($,b)l

=

(h

=

[I aII I(

<

I(

+

Then

+

+

KI($,C)I KI(+,b)l

K) I ( $ , b ) [

$,b)

$,b)

I($,c)(.

K($,C)I

-

< hlt$,b)l

2

I

I. QED

We d e n o t e b y A(K(E)) t h e s e t o f a l l bounded a f f i n e f u n c t i o n s on K ( E ) .

A(K(E)) i s c l e a r l y a v e c t o r s p a c e u n d e r p o i n t -

w i s e 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 .

We a l s o endow i t

w i t h t h e p o i n t w i s e g r d e r a n d t h e supremum norm.

(23.4)

4

Let E b e a n L-normed s p a c e .

t->$ I K ( E )

Then t h e r e s t r i c t i o n map

i s a l i n e a r b i j e c t i o n of E '

o n t o A(K(E)) w h i c h

Chapter 4

136

i s norm p r e s e r v i n g a n d maps

( E l ) +

onto (A(ti(E)))+.

I t follows

A(K(E)) i s an MIL-space a n d t h e mapping i s a R i e s z i s o m o r p h i s m .

Proof.

'The mapping i s c l e a r l y l i n e a r , p o s i t i v e , and

(by ( 2 3 . 3 ) ) n o r m - p r e s e r v i n g .

From t h i s l a s t , i t i s o n e - o n e .

E v e r y p o s i t i v e a f f i n e f u n c t i o n f on K(E) h a s an o b v i o u s ( u n i q u e ) e x t e n s i o n t o an a d d i t i v e , p o s i t i v e l y homogeneous f u n c t i o n on E + , h e n c e t o 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 4 on E ; a n d i f f i s bounded on K ( E ) , t h e n 4 i s bounded o n B ( E ) ,

so l i e s in

E l .

QED

We n e x t e s t a b l i s h t h e Nakano t h e o r e m ( 1 3 . 2 ) f o r t h e s p e c i a l case o f an L - s p a c e .

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

s h a r p e r form:

( 2 3 . 5 ) -~ Theorem.

For e v e r y L - s p a c e E , E

=

P r o o f . S i n c e E i s D e d e k i n d c o m p l e t e and E ' g i v e s u s t h a t E i s a Riesz i d e a l o f (El)' e q u a l t o (El)'.

(19.7)

=

E C , (13.3)

with order closure

a n d t h e norm c o m p l e t e n e s s o f E t h e n

g i v e u s t h a t E = (El)'. QED

Dual L - s p a c e s a n d MI1 - s p a c e s

137

A t r i v i a l a s p e c t o f t h e above t h e o r e m , b u t w o r t h r e c o r d i n g i s t h a t an L - s p a c e i s a band o f i t s b i d u a l . L - s p a c e , we a l s o h a v e E '

= EC ( 2 3 . 2 ) ,

S i n c e f o r an

w e a l s o have ( c f . t h e

d e f i n i t i o n following (13.2).

( 2 3 . 6 ) An L - s p a c e i s p e r f e c t

A s w e know, f o r a n y R i e s z s p a c e E ,

( i ) Eb i s Dedekind

bc . b c o m p l e t e , and ( i i ) E i s s e p a r a t i n g on E . F o r a n MIL-space F , b t h e s e two p r o p e r t i e s c h a r a c t e r i z e F a s a n E . S p e c i f i c a l l y ,

( 2 3 . 7 ) Theorem.

(Dixmier).

F o r an MIL-space F , t h e f o l l o w i n g

are equivalent:

'1

F i s t h e d u a l o f a Banach l a t t i c e ;

2'

F

3'

(i)

=

(F')'; F i s Dedekind c o m p l e t e , and

i s s e p a r a t i n g on F.

( i i ) F'

P roof. -

Assume 3'

i d e a l o f Fcc = ( F c ) ' .

holds. ((F')'

By ( 1 3 . 3 ) , F i s t h a n a R i e s z makes s e n s e , s i n c e , a s a b a n d o f

t h e Banach l a t t i c e F ' , FC i s i t s e l f a Banach l a t t i c e ) . f r o m ( 2 1 . 3 ) a n d ( 2 3 . 1 ) , IL(F) = IL(Fcc). a n d we h a v e 2', 1'

2'

i m p l i e s 1'.

But

I t f o l l o w s F = (F')',

F i n a l l y , a s we remarked above,

i m p l i e s 3'.

QED

Chapter 4

138

A c t u a l l y , G r o t h e n d i e c k h a s shown [19] t h a t 1'

above c a n

be r e p l a c e d by a c o n s i d e r a b l y w e a k e r s t a t e m e n t :

(23.8) (Grothendieck).

I n 1'

a b o v e , i t s u f f i c e s t o assume o n l y

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

I t then follows t h a t

t h i s p r e d u a l i s FC ( h e n c e a n L - s p a c e ) .

Remark. -

is essential.

I n ( 2 3 . 7 ) , t h e a s s u m p t i o n t h a t F i s an MIL-space The s p a c e c o o f s e q u e n c e s c o n v e r g i n g t o 0

s a t i s f i e s condition 3 bounded s e q u e n c e s .

0

,

but ((c0)')'

=

La, t h e s p a c e o f a l l

Note t h a t co i s even an M-space.

5 2 4 . Mappings o f L - s p a c e s

We h a v e s e e n ( 1 8 . 1 ) t h a t a l i n e a r mapping o f o n e MIL-normed s p a c e i n t o a n o t h e r i s norm c o n t i n u o u s i f a n d o n l y i f i t i s o r d e r bounded.

L - s p a c e s ( b u t n o t L-normed s p a c e s ) h a v e a much

stronger property (24.1). Henceforth, w e adopt t h e convention t h a t i f E , F a r e MIL-normed s p a c e s , a l i n e a r mapping o f E i n t o F w i l l a l w a y s b e r e p r e s e n t e d w i t h E on t h e

l e f t a n d F on t h e r i g h t : E-> T

F,

w h i l e i f E , F a r e L-normed s p a c e s , t h e mapping w i l l b e r e p r e m

s e n t e d w i t h F on t h e

l e f t and E on t h e r i g h t : F <- '

E

Dual L - s p a c e s a n d Mll-spaces

( 2 4 . 1 ) ____ 'Theorem.

139

E of one L-space

F o r a l i n e a r mapping F<---

i n t o another, the following are equivalent: 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;

3'

T i s order continuous.

P roof. -

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

Then t h e t r a n s p o s e

F ' __ Tt > E l i s a l s o norm c o n t i n u o u s , h e n c e o r d e r b o u n d e d ( 1 8 . 1 ) .

I t f o l l o w s by ( 1 4 . 4 ) t h e b i t r a n s p o s e F1<- T t t continuous.

Now E , F a r e b a n d s o f E " , F "

i s easy t o v e r i f y , TttlE TttlE

= T,

s o we h a v e .'3

El' i s o r d e r

respectively, hence, as

i s a l s o order continuous.

3'

i m p l i e s 2'

But

b y ( 6 . 0 ) , a n d 2'

i m p l i e s 1' by ( 9 . 1 1 ) . QED

F o r L - s p a c e s , o u r i n t e r e s t now l i e s i n i n t e r v a l p r e s e r v i n g maps.

( 2 4 . 2 ) Let F <- T

E b e a n i n t e r v a l p r e s e r v i n g l i n e a r map o f o n e

L-space i n t o a n o t h e r .

Then f o r e v e r y R i e s z i d e a l J c o n t a i n e d

i n TIE):

(i)

( T - I [ J ) ) + i s t h e p o s i t i v e cone o f a R i e s z i d e a l M

(ii)

T(M) = J ;

of E;

( i i i ) i f J i s a band, then so i s M.

140

Chapter 4

Note t h a t M i s o f n e c e s s i t y t h e l a r g e s t R i e s z i d e a l con t a i n e d i n T-'(J). Proof of (24.2).

For ( i ) , i t i s enough t o rem ark t h a t

since T is positive, ~ E ( T - ' ( J ) ) + implies [0,a1 c T-1 (J). ( i i ) f o l l o w s from ( i ) and t h e i d e n t i t y T(E)+

(T(E))+ (14.7).

=

( i i i ) f o l l o w s from t h e o r d e r c o n t i n u i t y of T ( 2 4 . 1 ) .

Finally,

QED

(24.3) C o r o l l a r y . Under t h e above h y p o t h e s i s , f o r e v e r y d e composition T(E)

J1 0 J2

=

o f T(E) i n t o b a n d s , t h e r e i s a u n i q u e l y d e t e r m i n e d d e c o m p o s i t i o n

E = G0 of E i n t o bands such t h a t T(GO)

Proof.

0 , T(G1)

=

L e t M1,M2

%,

Gla

G2

= J l , T(G2) = J 2 .

b e t h e b a n d s i n T -'(J1),

T-l(J2) given

by ( 2 4 . 2 ) .

MI

+

M 2 i s i t s e l f a b a n d ( E x e r c i s e 19 i n C h a p t e r l ) , s o w e

n e e d o n l y show t h a t (M1 e x i s t s aE(M1 (Ta)

+

0 < (Ta)J

Suppose n o t .

2 Ta and

Then t h e r e

Ta > 0 ( e l s e aEM1,M2), h e n c e

c a n n o t b o t h v a n i s h ; s a y (Ta) J2

1

M2)d = 0 .

M2)d, a > 0 .

a n d (Ta) J1

t

> 0.

Since

J1

T is interval preserving, there exists

141

Dual L - s p a c e s a n d M I - s p a c e s

. Rut t h e n b i b l l , c o n t r a d i c t i n g 1 This establishes ( i ) .

b E [ O , a ] s p c h t h a t 7%

aE(M1

+

M2)

d

.

Now s e t G o

=

M1

=

n

(Ta).,

M2, G1

=

M1

n

(Go)

d

,

G2 =

M2

n

(Go)

d

. QED

Go above i s , i n g c n e r a l , n o t a l l o f T

-1

largest Riesz ideal contained i n the latter.

(0).

I t is the

In f a c t , it i s

n o t h a r d t o see t h a t T - l ( O ) b e i n g a R i e s z i d e a l i s e q u i v a l e n t t o o u r T b e i n g a R i e s z homomorphism.

A l i n e a r mapping E __ > F o f o n e Mll-normed s p a c e i n t o

a n o t h e r i s c a l l e d a Markov mapping i f ( i ) T i s p o s i t i v e a n d ( i i ) Tll = ll.

N o t e t h a t e v e r y MIL-homomorphism i s a Markov

mapping. The f o l l o w i n g i s a n e a s y c o n s e q u e n c e o f ( 2 1 . 2 )

and ( 2 3 . 1 ) .

( 2 4 . 4 ) F o r a l i n e a r mapping C __ > F o f o n e Ma-normed s p a c e into another, the following are equivalent: 1'

T i s a Markov m a p p i n g ;

2'

T ~ ( K ( F ~c) )K ( E ~ ) .

And f o r a l i n e a r mapping F<-

E o f one L-space i n t o a n o t h e r ,

the following are equivalent: lo

T(K(E)) c K (F);

2i

T t i s a Markov m a p p i n g .

Combining t h e f i r s t p a r t w i t h ( 1 4 . 1 2 ) , we h a v e

Chapter 4

142

( 2 4 . 5 ) Theorem. _-

For a l i n e a r mapping E ->

T

I: o f o n e bill-

normed s p a c e i n t o a n o t h e r , t h e f o l l o w i n g a r c e q u i v a l e n t : 1'

T i s an M I

2'

T

t

homomorphism;

( i ) i s i n t e r v a l p r e s e r v i n g , and ( i i ) maps K(F')

i n t o K(E').

And c o m b i n i n g t h e s e c o n d p a r t w i t h ( 1 4 . 1 3 ) ,

( 2 4 . 6 ) 'Theorem. -

F o r a l i n e a r mapping

F:

<-

T

E o f one I,-spnce

into another, the following are equivalent: 1'

T ( i ) i s i n t e r v a l p r e s e r v i n g , and

( i i ) maps K(E) 2'

T t i s a n MI-homomorphism.

-__ Proof.

Suppose 2

i n t o K(F);

0

A c t u a l l y ( 1 4 . 1 3 ) o n l y g i v e s u s t h a t 1' holds.

Then b y ( 2 4 . 5 ) ,

interval preserving.

i m p l i e s 2'. Ttt t h e mapping F"<E" i s

S i n c e E , F a r e bands i n E",F"

it f o l l o w s t h a t T t t ( E i s a l s o i n t e r v a l p r e s e r v i n g .

l a s t i s T , s o w e have ( i ) i n lo.

T h a t 2'

respectively, But t h i s

i m p l i e s ( i i ) i n 1'

f o l l o w s from ( 2 4 . 4 ) .

QED

A s i s t o b e e x p e c t e d , a mapping w i t h t h e p r o p e r t i e s ( i i ) above h a s r a t h e r s a t i s f y i n g p r o p e r t i e s :

(i),

Dual L - s p a c e s a n d M I - s p a c e s

(24.7)

Let F <-

1

143

E be a l i n e a r mapping o f o n e L - s p a c e i n t o

a n o t h e r w h i c h i s i n t e r v a l p r e s e r v i n g a n d maps K(E) i n t o K ( F ) . Then (i)

T(ext K ( E ) ) c e x t K(F);

and f o r e v e r y b a n d J o f E , (ii)

T(,J) i s a band o f F ,

( i i i ) T(K(J)) = K(T(J)). I n p a r t i c u l a r , T(E) i s a b a n d o f F.

-__ Proof.

( i ) f o l l o w s from ( 2 0 . 2 )

c o n s i d e r a band J o f E .

and ( 1 4 . 7 ) .

To show ( i i ) ,

By ( 1 4 . J ) , T ( J ) i s a R i e s z i d e a l o f F ;

we h a v e t o show i t i s o r d e r c l o s e d . Then by ( 1 9 * 8 ) and ( 1 9 * 5 ) ,

=

Suppose b = V A , A c ( T ( J ) ) + .

Cnbn ( i n t h e

where

[ b n l c ( T ( < J ) ) + . F o r e a c h n , t h e r e e x i s t s anE.J+

such t h a t

Tan = bn ( ( 1 4 . 7 ) a g a i n ) , a n d , s i n c e T c l e a r l y p r e s e r v e s t h e norms o f p o s i t i v e e l e m e n t s ,

[I anll

=

\ I bnII .

I t f o l l o w s Znan i s

a norm Cauchy s e r i e s i n J , s o Can = a i n t h e norm f o r some a E J .

S i n c e T i s norm c o n t i n u o u s , Ta = b , a n d w e h a v e ( i i ) . f o l l o w s from T ( J + )

=

(iii)

( T ( J ) ) + (14.7) and t h e f a c t t h a t T p r e -

s e r v e s t h e norms o f p o s i t i v e e l e m e n t s . QE n

Remark. with (18.3).

The l a s t s t a t e m e n t i n ( 2 4 . 7 )

s h o u l d b e compared

CfIAPTER 5

D U A L I T Y RELATIONS BETWEEN AN L-SPACE AND ITS DUAL

As w e s t a t e d a t t h e b e g i n n i n g o f P a r t 11, i n o r d e r t o

s t u d y t h e p r o p e r t i e s o f C"(X)

as a b i d u a l , w e must f i r s t b e

c l e a r a b o u t what p r o p e r t i e s d e r i v e from t h e f a c t t h a t i t i s a dual.

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

L e t E be an L - s p a c e . which w i l l concern u s . lol(E',E),

T h e r e a r e f o u r t o p o l o g i e s on E '

These a r e t h e vague t o p o l o g y a ( E ' ,E)

,

t h e Mackey t o p o l o g y T ( E ' , E ) , a n d t h e norm t o p o l o g y ,

l i s t e d i n order of increasing fineness.

The o n l y p a r t o f t h i s

t h a t n e e d s showing i s t h a t -c(E',E) i s f i n e r t h a n Since E i s a Riesz ideal of

El'

El',

(E',E).

( 2 3 . 5 ) , t h i s f o l l o w s from ( 2 2 . 3 ) .

Our i n t e r e s t a t p r e s e n t l i e s i n l o l ( E ' , E ) . a c t u a l l y a band o f

101

Since E i s

t h e condition given i n (22.3) f o r a

l i n e a r f u n c t i o n a l Q, on E ' t o l i e i n E

- t h a t 0 be l o l ( E ' , E )

c o n t i n u o u s on E ' - c a n b e r e p l a c e d b y t h e w e a k e r c o n d i t i o n t h a t Q, b e of E '

(cf.

l a l ( E ' , E ) c o n t i n u o u s on t h e u n i t b a l l B ( E ' ) = [ - I l , n ] (22.8)).

I n d e e d , as t h e f o l l o w i n g theorem shows, i f

w e know t h a t Q E E " , t h e n i t s u f f i c e s f o r 0 t o b e 144

lal(E',E)

145

An I . - s p a c e a n d i t s D u a l

c o n t i n u o u s on t h e p r o p e r s u b s e t

6 (1) o f

B(E').

&(a).

We r e c a l l t h a t e a l w a y s d e n o t e s an e l e m e n t o f

Note

a l s o , from t h e Dedekind c o m p l e t e n e s s o f E ' and t h e f a c t t h a t

e(1) i s o r d e r c l o s e d i n E ' , t h a t a n e t { e a } i n & ( l l ) o r d e r c o n verges i n

E ( 1 ) i f and o n l y i f i t o r d e r converges i n E ' .

( 2 5 . 1 ) -___ Theorem. tional

@

on E ' ,

I f E i s an L - s p a c e , t h e n f o r a l i n e a r f u n c the following are equivalent:

loO E E ;

is order continuous;

2' 3'

0 is I ~ I ( E I , E c o) n t i n u o u s ;

4'

i s I ~ I ( E I , E )c o n t i n u o u s on B ( E ' ) ;

5'

@ E E " a n d @ i s o r d e r c o n t i n u o u s o n 6 (1);

6'

oEE" a n d

Q, i s

101

(E',E)

7'

0EE" and

@

is

101

( E ' , E ) c o n t i n u o u s a t 0 on C$

8'

4 E E " and e $ 0 i m p l i e s l i m ( e c1

Proof. equivalent.

c o n t i n u o u s o n 6 (I);

c1

6O,

0) = 0 .

a'

We a l r e a d y h a v e t h a t t h e f i r s t f o u r p r o p e r t i e s a r e By t h e r e m a r k p r e c e d i n g t h e p r o p o s i t i o n , 2'

i m p l i e s S o , w h i c h i n t u r n i m p l i e s 8'. implies

(I);

w h i c h i m p l i e s 7';

A l s o , o f c o u r s e , '4

and, by (22.6),

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 8'

Assume cp s a t i s f i e s 8'.

7'

i m p l i e s 8'.

implies 2

0

I t f o l l o w s from (21.5)

. and t h e

same k i n d o f a r g u m e n t a s was u s e d i n t h e p r o o f o f ( 1 2 . 1 ) 101 a l s o s a t i s f i e s 8'.

that

o

> 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<, by A a .

By

$ I w i l l be d e -

Even t h o u g h a l i e s i n E , n o t

El, t h i s d o e s n o t l e a d t o c o n f u s i o n .

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

n o t a t i o n i s t h a t l a t e r a w i l l be a m e a s u r e p , a n d t h e n J a w i l l be L

1

(u)

a n d 1, w i l l h e t h e d u a l s p a c e L m ( p ) .

b o t h b a n d s J,I a r e d e t e r m i n e d by a . n o t a t i o n , n o t o n l y do we h a v e J

=

Ea,

In any c a s e ,

Note t h a t u n d e r o u r but also I

=

We d e v e l o p t h e p r o p e r t i e s o f b a s i c b a n d s . C o n s i d e r aEE a n d t h e p a i r o f d u a l b a n d s t e r m i n e d by a .

(522).

(\I-

11

a l l a

((Ea)L)d, the restriction of

an L-norm.

I(

Now a a l s o d e f i n e s t h e seminorm

Since the n u l l space of =

de-

a

i s c l e a r l y (E,)'

[I . ] l a

E l )

and

is a n

to

Thus ( E l ) a i s endowed w i t h an MIL-norm

t h e c a n o n i c a l MI-norm o f

on E '

a n d a n L-norm

s -

11

11 - \ I a

N o w , a s w e know, t h e band E a i s a n L - s p a c e and b e i n g i t s d u a l b a n d , i s t h e n i t s d u a l s p a c e ( w i t h laf o r t h e unit).

Consequently, i n t h e remainder of t h i s 5 , we w i l l t h i n k

o f them t h i s way, and w r i t e (( simply

a ( (

s i m p l y ((

-11

and ((

.\Ia

11 -!Ia.

The t o p o l o g y on

d e f i n e d by t h e L-norm

11 - [ I a

is, in

149

An L - s p a c e a n d i t s Dual

g e n e r a l , s t r i c t l y c o a r s e r t h a n t h a t d e f i n e d by t h e Ma-norm

!I.:/. Il+lla

I n e f f e c t , w e c a n assume IIaII =

.;I,

under

]I

( l ~ $ l , l a l )2

[/+{I

IIaII

=

;]+:I.

Since ( E ' ) a i s complete

i t f o l l o w s t h a t i n g e n e r a l , i t i s n o t complete under

s o i s n o t an L - s p a c e .

a i l a ,

1 ; then f o r every

=

N e v e r t h e l e s s , we show t h a t i t

p o s s e s s e s many o f t h e i m p o r t a n t p r o p e r t i e s o f a n L - s p a c e .

i s o f c o u r s e Dedekind c o m p l e t e ( E x e r c i s e 1 8 i n Chapter 1 ) .

And s i n c e a i s o r d e r c o n t i n u o u s on

i s an o r d e r c o n t i n u o u s norm.

Moreover, p a r t i a l l y compensating

f o r the lack of Il-[la-completeness, the u n i t b z l l Ii,-complete:

In e f f e c t ,

*;Ia [-na,lla] i s

[-lla,ILa] i s v a g u e l y c o m p a c t ( 1 0 . 1 4 ) ,

s o t h e B o u r b a k i t h e o r e m ( [ 4 5 ] Chap. 6 , C o r . t o P r o p . 3) g i v e s the desired result.

The p r o o f o f ( 1 9 . 5 ) now c a r r i e s o v e r

v e r b a t i m , a n d we h a v e :

(27.1) in

Let E b e a n L - s p a c e a n d a E E . o r d e r convergence and

I1

Then f o r a m o n o t o n i c n e t

Ila-convergence are e q u i v a l e n t .

A s a s u b s t i t u t e f o r ( 9 . 9 ) , w e have

(27.2)

Let E b e a n L-space a n d a € E .

s e q u e n c e { I $ ~ }i n

[I . ; l a - c o n v e r g e s

I f an o r d e r bounded t o $€

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 $.

The p r o o f i s e s s e n t i a l l y t h e same a s t h a t f o r ( 9 . 9 ) .

150

Chapter 5

(27.2)

i n t u r n g i v e s us t h e f o l l o w i n g s u b s t i t u t e f o r ( 1 9 . 7 ) .

( 2 7 . 3 ) L e t E b e an L - s p a c e a n d a E E .

Then f p r an o r d e r b o u n d e d

the following sets coincide.

subset A of 1'

the order closure of A;

2'

t h e o-order c l o s u r e of A;

3'

A(1).

'4

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 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 o f A. 5'

the

11

*

;closure

o f A.

(19.8) and ( t h e r e f o r e )

(19.9) c a r r y over (indeed, i n a

s t r o n g f o r m ) , and p a r t o f (19.10) w i t h " p r i n c i p a l band" r e p l a c e d b y " b a s i c band".

(27.4)

Theorem.

We combine a l l t h i s i n

Let E b e a n L - s p a c e .

Then f o r a b a n d I o f E ' ,

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

I i s a b a s i c band;

2'

I h a s t h e c o u n t a b l e sup p r o p e r t y ;

3'

e v e r y s e t o f m u t u a l l y d i s j o i n t c o m p o n e n t s o f 1, i s

countable;

' 4

I i s a t most c o u n t a b l y decomposable.

P roof.

The p r o o f o f ( 1 9 . 8 )

(with (19.5) r e p l a c e d by

An L - s p a c e a n d i t s Dual

0

(27.1)) gives us that 1 g i v e s u s t h a t 2' (cf.

implies 2

implies 3

0

.

3'

And t h e p r o o f o f ( 1 9 . 9 )

and 4 '

F i n a l l y , s u p p o s e 4'

(17.3)).

.

0

151

are clearly equivalent

holds.

Then t h e band ,J o f

E d u a l t o T i s a l s o a t most c o u n t a b l y d e c o m p o s a b l e ; h e n c e by ( 1 9 . 1 0 ) , J i s a p r i n c i p a l band. 0

is 1

Thus I i s a b a s i c b a n d , w h i c h

. QED

Note t h a t ( 1 9 . 9 )

c a r r i e s o v e r i n f u l l f o r c e t o 3'

above:

f o r e v e r y c o u n t a b l e , m u t u a l l y d i s j o i n t s e t o f components { e n } o f n ( E i ) , CnIIenlla

i

a.

Given a n L - s p a c e E , (E',E),

(25.1) applies not only t o the pair

b u t a l s o t o e v e r y p a i r ( T , < J ) o f d u a l b a n d s , s i n c e .J

i s an L - s p a c e and I = %J'.

I f J i s a p r i n c i p a l band o f E , w e

c a n a d d two more e q u i v a l e n t c o n d i t i o n s :

(27.5)

Let E b e an L - s p a c e and aEE.

t i o n a l 0 on

t h e f o l l o w i n g are e q u i v a l e n t :

1'

@EEa;

2'

0 i s o - o r d e r c o n t i n u o u s on

3'

0 is

Proof. on E ' ,

Then f o r a l i n e a r f u n c -

11.

; ] , - c o n t i n u o u s on

[-n,,na].

A s f o r every element of E ,

h e n c e on

f o l l o w s from ( 2 7 . 2 ) .

s o 1'

i m p l i e s 2'.

Suppose 0 i s

0

i s order continuous T h a t 2'

1) .\I a - c o n t i n u o u s

i m p l i e s 3' on [ - n a , J l a ] .

Chapter 5

152

Now

101

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

((E'),,Ea)

so, a fortiori, follows

I[

e l l a ,

l~l((E')~,E~)-continuous on [ - l a , l a ] .

Q, i s

T t

i s I~I((E')~,E~)-continuous on e v e r y i n t e r v a l o f

$I

b

S i n c e Ea i s a band i n = (E,)";

,

=

(Ea)', so

now c f . t h e comment f o l l o w i n g

((E1)a)b

=

(23.5)),

(22.8) a p p l i e s t o g i v e u s OEE,.

Thus 3'

implics lo.

QED

-~ Remark.

Actually, t h e topology

t o p o l o g y d e f i n e d by

\I.I[ a

101 ( ( E 1 I a , E a )

c o i n c i d e on [ - l l a , l a ] ( c f .

and t h e (29.8)

below).

Given a R i e s z s p a c e F , l e t us d e n o t e b y Foc 0 - 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 F.

I t f o l l o w s by

t h e same a r g u m e n t s a s were u s e d f o r FC t h a t F5' Fb.

And c l e a r l y FC c Fac c I:b .

s e t o f Fac.

the set of

i s a band o f

I n g e n e r a l FC i s a p r o p e r s u b -

We h a v e s e e n t h a t f o r an L - s p a c e E , E '

h e n c e , from t h e a b o v e , E '

=

EC = EoC

band o f E a n d I i t s d u a l b a n d i n E l . L - s p a c e , we h a v e I = J C a l s o h a v e , by ( 2 7 . 5 ) ,

=

Jac.

that J

=

=

Eb.

=

E C = Eb ,

Now l e t J b e a

J b e i n g i t s e l f an

For J a p r i n c i p a l band, w e Ic

=

Iac.

528 E q u i - o r d e r - c o n t i n u i t y

In

5528, 2 9 , w e d e v e l o p t h e m a t e r i a l w e w i l l n e e d on

weak c o m p a c t n e s s .

Our t r e a t m e n t , t h e c h a r a c t e r i z a t i o n o f weak

c o m p a c t n e s s by e q u i - o r d e r - c o n t i n u i t y , stems f r o m Nakano.

For

An L - s p a c e and i t s D u a l

153

e x t e n s i v e d i s c u s s i o n s o f t h e t o p i c , we r e f e r t h e r e a d e r t o [ 1 7 ] a n d [ 2 ] ( c f . a l s o [27]).

/I - 1 1

Let

(28.1)

b e a ( n o t n e c e s s a r i l y R i e s z ) seminorm on a R i e s z

s p a c e F , and B i t s u n i t b a l l .

Then t h e f o l l o w i n g a r e e q u i -

valent:

[I

1' (19.6)

i s order continuous (cf. the d e f i n i t i o n preceding

1; o o

i n F implies lima(sup

2'

a

3'

a

'4

a a+ O i n F i m p l i e s l i m a ( s u p

5'

a GO in F implies [-a

-+

CY -f

c1

i n F implies [ - l a

c1

Proof.

1'

i m p l i e s 2':

a

a

a' a

I

Ilb[l)

0;

=

Ibl+,l , l a a l l c B f o r some a ;

I b I'I

1c

Suppose 2

aa

IIlb[l)

=

0;

B f o r some a.

0

f a i l s t o hold.

Then

t h e r e i s a n e t { a } i n F and a n E > 0 s u c h t h a t a 0 while a a f o r e v e r y a, t h e r e e x i s t s b w i t h Ib 1 < ( a a ( a n d (1 ball > E . a a -+

Since b

+

a

I] .]I

0 , i t follows

c l e a r l y i m p l i e s 3'

and

4O,

I t r e m a i n s t o show t h a t 5' E

IIaB[l 5

for a l l 6 > a.

c a + 0 a n d J a a l-L c

a

a n d e a c h o f t h e l a t t e r i m p l i e s 5'. i m p l i e s 1'.

B

11

Suppose aa

+

0 , and

There e x i s t s a n e t {ca} such t h a t

f o r a l l a.

Since (l/E)ca+O, t h e n , by So,

[ - ( ~ / E ) c ~ (, l / ~ ) ac] c B f o r some a.

h e n c e [lc

2'

> 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.

<

-

E

for a l l B > 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 .

<

m

for a l l ~ E E ' ;

i s absorbing i n E '

I n s u c h c a s e ( s i n c e A'

(El

=

lJn(nAo));

i s convex and v a g u e l y c l o s e d ) ,

a seminorm on E ' w i t h A'

aEE i s o r d e r c o n t i n u o u s on

E l ,

A s we know, e v e r y i s an

F o r a s u b s e t A o f E , we w i l l

.\IA

say t h a t A i s equi-order-continuous o n 2 i f

(1. \ I A

11 . ; l a

and ( t h e r e f o r e )

o r d e r c o n t i n u o u s seminorm on E l .

(Note t h a t

is

for its unit ball.

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

continuous.

11 * \ IA

is order

i s t h e n a u t o m a t i c a l l y a seminorm

a n d t h e r e f o r e A i s norm b o u n d e d . ) We g i v e some c h a r a c t e r i z a t i o n s o f e q u i - o r d e r - c o n t i n u i t y i n L-spaces.

(28.2)

Theorem.

- c _ _

For a s u b s e t A o f E ,

Let E b e an L - s p a c e .

the following are equivalent: 1'

A i s e q u i - o r d e r - c o n t i n u o u s on

2'

+

3'

+,+O

' 4

e v e r y bounded monotone n e t i n E '

5'

a i n~ E ~ ' implies l i m

[IQ, a

a

i n E' i m p l i e s l i m n \ l + n ( l A

E l ;

=

0;

=

0;

is

e v e r y b o u n d e d monotone s e q u e n c e i n E '

11

*

[IA-Cauchy;

is

11

- IlA-Cauchy

An L - s p a c e and i t s Dual

Proof.

1'

-~

uity.

2

0

exists

o f c o u r s e i m p l i e s 3'.

\IA

- $

<

.

-

a

an

<

> E for a l l n. -

"n+ 1 plete) 4

Assume 3'

+ $ f o r some

@ € E l ,

<

h o l d s , and suppose

11 -IIA-Cauchy.

{$"I i n E ' i s n o t

> 0 and a1 < a 2

E

by t h e d e f i n i t i o n e q u i - o r d e r - c o n t i n -

i m p l i e s 2'

a descending n e t

155

-

*

such t h a t

a

But ( E '

h e n c e , hy

Then t h e r e

b e i n g D e d e k i n d com-

2O,

(In a n d we h a v e a c o n t r a d i c t i o n .

Thus 3'

implies 4

0

.

-

$!IA

4'

of

0,

=

c o u r s e i m p l i c s 5'. Assume 5'

holds.

We show t h a t c o n d i t i o n 4'

s a t i s f i e d , t h a t i s , i f $a+O, then l i m (sup a

of (28.1) i s

O'

11$!A)

l+I+,l

S u p p o s e a n e t { $ } s a t i s f i e s cp + O b u t n o t t h e c o n c l u s i o n .

"

CY

Then t h e r e e x i s t s c > 0 s u c h t h a t f o r e v e r y f o r which

(*)

I$

<

-

[@al

but

/I

> 2 ~ .

T h e r e e x i s t s e q u e n c e s t u n } i n E ' a n d {a,} (i) (ii)

ul

3

w2

I(wn,an)l

(iii)

,... > 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")'

<--- T~

Since

Ell.

t i. s v a g u e l y c o n t i n u o u s , a n d { a n } c o n v e r g e s v a g u e l y t o 0 i n m

El',

{Ttan} c o n v e r g e s v a g u e l y t o 0 i n (a ) ' .

h e n c e , by ( 2 3 . 5 ) ,

R

1 i s a band i n ( a " ) ' ,

t c o n s i d e r t h e component (T a n )

Now R m

=

,

(a')'

so f o r each n , we can

o f T t q n . I t f o l l o w s from t h e R

t c l a s s i c a l P h i l l i p s Lemma [ 4 2 ] t h a t t h e s e q u e n c e { ( T a n ) c o n v e r g e s t o 0 n o r m w i s e . Thus f o r l a r g e e n o u g h n , whence

I(

e n , ( T t a n ) 1)

I 5

E.

But

(1

R1 (Ttan)

}

11

I!

ZE,

158

Chapter 5

a n d we h a v e a c o n t r a d i c t i o n .

QED

Given a s e t B c o n t a i n i n g t h e o r i g i n i n a R i e s z s p a c e , t h e union of a l l t h e s o l i d sets contained i n B i s again a s o l i d s e t , hence t h e l a r g e s t one i n B. F o r t h e f o l l o w i n g p r o p o s i t i o n , n o t e t h a t i f any s e t B i n a n Ar ch imed e an R i e s z s p a c e F s a t i s f i e s S o i n ( 2 8 . 1 ) , t h e n B i s absorbing i n F.

(28.4) (28.1),

I f a c o n v e x s e t B i n a R i e s z s p a c e s a t i s f i e s 5'

in

t h e n t h e l a r g e s t s o l i d s e t B1 c o n t a i n e d i n B i s a l s o

c o n v e x and s a t i s f i e s p r o p e r t y S o i n ( 2 8 . 1 ) .

Hence B1 i s

a b s o r b i n g an d i t s gauge f u n c t i o n i s a n o r d e r c o n t i n u o u s R i e s z seminorm.

I f i n a d d i t i o n , B i s t h e u n i t b a l l o f i t s gauge

f u n c t i o n , t h e n B1 i s t h e u n i t b a l l o f

%

gauge f u n c t i o n .

The p r o o f i s s t r a i g h t f o r w a r d .

C o r o l l a r y 1. (28. 5) -

F o r e v e r y o r d e r c o n t i n u o u s seminorm

a R i e s z s p a c e F , t h e s m a l l e s t R i e s z seminorm

2 II.II

11 [I *

on

e x i s t s and

i s a l s o order continuous.

Corollary 2 . (28.6) -

I f a s u b s e t A o f a normed R i e s z s p a c e E i s

An L-space and its Dual

159

equi-order-continuous on El, then the norm closed, convex, solid hull A1 of A is also equi-order-continuous.

It is clear that (A,)'

is the largest solid set in ,'A

so

(28.4) applies.

(28.7) _____ Theorem. Let E be an L-space.

I f a subset A of E is

equi-order-continuous, then the convex solid hull of A hence also A - is relatively weakly compact.

Proof. By ( 2 8 . 6 ) , we can assume that A itself is convex and solid.

So we need only show that (Ao)'

We achieve this by showing that (Ao)' consider @€(Ao)O-in-(E')*;

(A0)O-in-(E')*.

=

we have to show QEE.

is sufficient, by ( 2 3 . 5 ) , to show that on El.

is weakly compact.

@

So

For this, it

is order continuous

This follows from the fact that 0 is bounded on A',

which is the unit ball of an order continuous norm. QED

5 2 9 . Weak compactness

We can now describe the weakly compact subsets of an L-space. We first record.

160

Chapter 5

(29.1) Theorem. -

Intervals of an L-space E are weakly compact.

This follows from (10.14) and the fact that E is a Riesz ideal of E" = ( E l ) b

.

(29.2) Theorem. (Nakano).

F o r a subset A of an L-space E, the

following are equivalent: '1

A is equi-order-continuous on

'2

A is relatively weakly compact.

-__ Proof.

We have already proved '1

E l .

implies 2'

(28.7).

For

the converse, suppose A is relatively weakly compact but not equi-order-continuous. quences {$n} in for all n.

E l ,

By (28.6), there exist

{an} in A such that $,+O

E

> 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)

C o r o l l a r y 3.

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.

I f E i s an L - s p a c e ,

?-(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)

I f E i s an L - s p a c 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) <

E.

Since f

cr (XI

i s c o n t i n u o u s , t h e r e e x i s t s an o p e n

n e i g h b o r h o o d W(x) o f x s u c h t h a t f

(XI

(Y) <

E

f o r a l l yiW(x).

Now X i s c o m p a c t , s o t h e r e e x i s t x l , - - ,*x n s u c h t h a t { W ( x l ) , . - . , W ( x n ) } c o v e r s X. Then a > a ( € ) implies fa(y) <

Choose a(€) 2 w ( x , ) , . * . , a ( x n ) . E

f o r a l l yEX.

QED

I n common w i t h norm c o n v e r g e n c e a n d o r d e r c o n v e r g e n c e , simple convergence h a s t h e f o l l o w i n g e a s i l y v e r i f i e d , p r o p e r t y

(30.2)

I f { f } a n d {p } c o n v e r g e s i m p l y t o f a n d g r e s p e c t i v e l y , c1

t h e n {fcrVg,),

a

{fclAgcl},

Ifc1 + g a l , a n d {Afcl}

o f IR) c o n v e r g e s i m p l y t o f v g , f A g , f

+

(A

any element

g , a n d Af r e s p e c t i v e l y .

Chapter 6

168

Otherwise s t a t e d , t h e l a t t i c e o p e r a t i o n s and v e c t o r s p a c e o p e r a t i o n s a r e continuous under simple convergence.

F o r a R i e s z s u b s p a c e F o f C(X) t o b e s i m p l y

(30.3) Corollary.

c l o s e d , i t s u f f i c e s t h a t F+ be s i m p l y c l o s e d .

i s simply c l o s e d .

Note a l s o t h a t C ( X ) +

The f o l l o w i n g i s t h e c r u c i a l p a r t o f M . H .

Stone's proof

of t h e Stone-Weierstrass theorem.

( 3 0 . 4 ) _____ Theorem.

(Stone).

F o r a s u b l a t t i c e A o f C(X), t h e

following are equivalent:

'1

A i s simply closed;

2'

A i s norm c l o s e d .

We n e e d o n l y show t h a t 2'

P roof.

i m p l i e s 1'.

Assume A i s

norm c l o s e d , a n d c o n s i d e r f i n t h e s i m p l e c l o s u r e o f A . E

> 0.

We show t h e r e e x i s t s gEA

such t h a t

i A

( i ) F o r e a c h xoEX, t h e r e e x i s t s g

\I g

- fll

Fix

5 3 ~ .

such t h a t

xO

(XI

2

f ( x ) - 3~

f o r a l l xEX,

gXO

gx (x,)

< f(xo)

+

E *

0

In e f f e c t , s i n c e f i s i n t h e simple c l o s u r e of A, then f o r

169

(C(X) , X ) - D u a l i t y e a c h xEX, t h e r e e x i s t s g ( g x .(XI

-

<

f(x)

-

E ;

(xo)

€ A such t h a t J g xO

- f(xo)

< r-, -

xO

and from t h e s e c o n d o f t h e s e i n e q u a i t i e s ,

0

t h e r e e x i s t s a n o p e n n e i g h b o r h o o d W x) o f x s u c h t h a t ( y ) 2 f ( y ) - 3~ f o r a l l y 6 W ( x ) . X b e i n g compact, t h e r e xox e x i s t { x l , . . - , x n } s u c h t h a t IW(x,), * . , W ( x n ) } c o v e r s X. Set g

Since x

was a r b i t r a r y i n ( i ) , w e now d r o p t h e s u b s c r i p t .

0

We t h u s h a v e { g x l x E X } c A s u c h t h a t f o r e v e r y x , gx 3 ~ l l ( X ) and gx(x) < f(x) +

c o v e r s X.

Set g

=

m

tliTlgx

i 3 ~ l l ( X ;) s o we a r e t h r o u g h .

+

3~ f o r a l l y E V ( x ) .

, x m } s u c h t h a t {V(x,)

.

f -

For e a c h x , choose a n e i g h b o r -

E.

hood V(x) o f x s u c h t h a t g x ( y ) < f(y) Again t h e r e e x i s t s { x , , . .

2

, . . * ,V(xm) 1

< f + Then g E A a n d f - 3 ~ n ( X )5 g -

QED

(30.5)

Corollary.

(i)

Norm c l o s e d R i e s z i d e a l s o f C ( X )

a r e simply closed.

( i i ) M I - s u b s p a c e s o f C(X) a r e s i m p l y c l o s e d .

531. 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 of C(X)

and t h e c l o s e d ( o r open) s u b s e t s o f X

F o r e v e r y s u b s e t Q o f X , we s e t QL

=

{fEC(X)] f(x)

=

0 for all

a n d f o r e v e r y s u b s e t A o f C ( X ) , we s e t

~€41;

Chapter 6

170

Z(A) Z(A)

{x€X I f ( x )

=

=

0 f o r a l l fEA). (In l i n e with our d u a l i t y

i s c a l l e d t h e z e r o - s e t o f A.

a p p r o a c h , w e would p r e f e r t o d e n o t e Z ( A )

b y A&, b u t Z ( R ) i s

the established notation.) We h a v e i m m e d i a t e l y :

( 3 1 . 1 ) For e v e r y A c C ( X )

,

Z(A) i s a c l o s e d s u b s e t o f X .

In the opposite d i r e c t i o n , given

Q

c X,

it i s e a s i l y

v e r i f i e d t h a t QL i s a R i e s z i d e a l a n d s i m p l y c l o s e d , s o a f o r t i o r i norm c l o s e d .

( 3 1 . 2 ) For e v e r y

Thus:

Q c X,

Q'

Now i t i s c l e a r t h a t

i s a norm c l o s e d R i e s z i d e a l .

(0)' =

QL

M o r e o v e r , by t h e Urysohn t h e o r e m , t h i s property.

( 3 1 . 3 ) For e v e r y

Dually ,

0is

We t h u s h a v e :

Q c X, Z(QL)

=

(q

0.

the closure of

Q i n X).

the largest set with

171

(C(X) , X ) - D u a l i t y ( 3 1 . 4 ) For e v e r y s u b s e t A o f C(X),

i s t h e norm c l o s e d

(Z(A))'

Riesz i d e a l H g e n e r a t e d by A.

H c (Z(A))'

-__ Proof.

by ( 3 1 . 2 ) .

clusion consider f € ( Z ( A ) ) I , B

=

{ g E H 10 c: g 5 f } .

i s f i l t e r i n g upward. we show f ( x )

=

H e n c e , b y t h e D i n i Theorem ( 3 0 . 1 1 , i f

I f f(x)

=

it w i l l follow t h a t

0 , w e are through.

c a n assume h ( x ) > f ( x ) . =

Suppose

I t f o l l o w s t h e r e e x i s t s hEII, w i t h

h(x) > 0 ( t h e r e e x i s t s kEAwith k(x) # 0 ; s e t h

t h a t g(x)

Set

B i s n o t empty ( i t c o n t a i n s 0 ) and i t

Then xeZ(A).

f(x) > 0.

a n d we c a n assume f > 0 .

s u p g E B g ( x ) , f o r e v e r y xEX,

Consider x .

fEfl.

For t h e o p p o s i t e con-

Then g

=

=

l k l ) , and w e

hnf i s an element of B such

f(x).

QED

In particular,

(31.5) For e v e r y Riesz i d e a l I o f C ( X ) ,

(Z(1))'

i s t h e norm

closure of I.

Summing u p ,

(31.6) There i s a one-one correspondence between t h e f a m i l y o f c l o s e d s u b s e t s 2 o f X a n d t h a t o f norm c l o s e d R i e s z i d e a l s II of C(X):

H <->

Z i f and o n l y i f Z = Z(H)

i f and o n l y i f H

=

ZL.

Chapter 6

172

Given H <-->

Z,

s e t W = X\Z:

Thus 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 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 open s e t s o f X .

T h i s corresp o n d en ce can be g iv en w i t h -

out reference t o the closed s e t s .

First a definition.

A func-

t i o n f on a l o c a l l y compact s p a c e T i s s a i d t o v a n i s h a t i n f i n i t y i f f o r every -~ compact.

E

> 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

following are equivalent: 1'

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

(C(X) ,X) - D u a l i t y

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 <-

S

Y.

For each fEC(X),

f o s i s a c o n t i n u o u s f u n c t i o n on Y, s o an e l e m e n t o f C(Y). t h u s h a v e a mapping f

I--->

f o s o f C(X) i n t o C ( Y ) .

We

We w i l l c a l l

(C(X) ,X) - D u a l i t y

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
( 3 3 . 2 ) Theorem.

X,Y,

=

(Stone).

mapping

T:

Given two c o m p a c t H a u s d o r f f s p a c e s

t h e r e i s a one-one correspondence between t h e c o n t i n u o u s

m a p p i n g s s o f Y i n t o X a n d t h e Ma-homomorphisms T o f C(X) i n t o C(Y).

s i f a n d o n l y i f (f,sy)

T <->

= (Tf,y)

for a l l ffC(X)

a n d yEY.

Proof.

-~

I t r e m a i n s t o show t h e e x i s t e n c e o f s , g i v e n T .

F o r e a c h yEY, yoT i s an M ~ - h o m o m o r p h i s m o f C(X) i n t o R , h e n c e , by (32.1),

an element o f X.

o f Y i n t o X; d e n o t e i t b y s .

f E C ( X ) a n d yEY.

We t h u s h a v e a m a p p i n g y k--> 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 <---1 i s a c l o s e d s u b s e t o f X and { s (x) ( x € s ( Y ) } i s an

S

Y,

184

Chapter 6

uppersemicontinuous decomposition o f Y.

Y b e a c o n t i n u o u s mapping a n d T

( 3 3 . 3 ) L e t X <s (i) (ii)

T-'(O) -1 {s

=

(s(Y))'and

(x) ( x E s ( Y ) l

s(Y)

=

=

st .

Then

Z(T-'(O));

i s t h e s e t of s e t s of constancy of

T ( C ( X ) ) , a n d T(C(X)) i s t h e s e t o f e l e m e n t s o f C(Y) c o n s t a n t -1 on e v e r y s (x),

Otherwise s t a t e d , T - l ( O ) and s(Y) c o r r e s p o n d t o e a c h o t h e r u n d e r ( 3 1 . 6 ) , and T(C(X)) and { s each o t h e r under (32.7).

-1

(x) I x € s ( Y ) }

correspond t o

The p r o o f o f ( 3 3 . 3 ) i s s t r a i g h t -

forward.

(33.4) Corollary. (i)

s i s a s u r j e c t i o n i f a n d o n l y i f T i s an i n j e c t i o n .

( i i ) s i s an i n j e c t i o n i f and o n l y i f T i s a s u r j e c t i o n .

We a p p l y t h e a b o v e t o t h e m a t e r i a l o f 5 § 3 1 , 3 2 .

Let F b e

an MIL-subspace o f C ( X ) a n d 2 i t s s e t o f s e t s o f c o n s t a n c y , and d e n o t e t h e q u o t i e n t s p a c e

( Z J ) by

Y.

Let Y

q u o t i e n t map.

Then an a d a p t a t i o n o f t h e a r g u m e n t f o r ( 3 2 . 6 ) , t p l u s ( 3 3 . 4 ) a b o v e , g i v e s u s t h a t C(Y) %> C ( X ) i s an MILi s o m o r p h i s m o n t o F.

do t h i s :

Thus F c a n b e i d e n t i f i e d w i t h C(Y).

We

(C(X) , X ) - D u a l i t y

Let F b e an M I - s u b s p a c e o f C ( X ) ,

(33.5)

constancy o f F, Y q u o t i e n t map.

=

(2,y) t h e

185

2

t h e s e t of s e t s of

q u o t i e n t s p a c e , and Y

Then u n d e r t h e d e f i n i t i o n ( f , q y )

=

(f,y) for a l l

f E F and qyEY, w e h a v e F = C ( Y ) .

Now l e t Z b e a c l o s e d s u b s e t o f X a n d H Let X <-

S

ZL ( s o Z

=

=

Z(H)).

t Z b e t h e i n j e c t i o n map.

MI-homomorphism o n t o . ~

Then C ( X )

onto C(Z).

Thus

We do t h i s :

C(X)/H

can be i d e n t i f i e d w i t h C ( Z ) .

(33.6)

For e v e r y c l o s e d s u b s e t Z of X , =

i s an

Factoring it:

t " we o b t a i n a n M a - i s o m o r p h i s m (s ) o f C ( X ) / H

definition (qf,z)

>c(z)

~

C(X)/ZL

( f , z ) for a l l qf€C(X)/Z'

=

C(Z)

under the

and z E Z .

L

Remark. The mapping C ( X ) s t r i c t i o n map f k--> 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

<

the

m.

n (The sum d e n o t e s t h e supremum o f a l l f i n i t e sums C 1 I p ( t i ) I . ) 1 I t i s immediate t h a t f o r e v e r y p C (T), p ( t ) # 0 f o r only a countable set of t ' s .

Under 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 , t h e norm IIp1I

?,!

1

CtETIp(t)

=

( T ) i s a R i e s z s p a c e ; a n d endowed w i t h

1,

i t i s an L-space.

Henceforth, the

1 symbol R ( T ) w i l l a l w a y s mean t h i s L - s p a c e .

We r e t u r n t o o u r s u b s e t Q o f X a n d t h e b a n d J w h i c h i t generates.

X x i n J , d e f i n e t h e f u n c t i o n p' x€Q x Then, from t h e p r e c e d i n g d i s c u s s i o n Xx (x€Q).

For e a c h

on Q by p'(x)

=

u

=

C

p' i s c l e a r l y an i s o m e t r i c R i e s z i s o m o r p h i s m t h e mapping p t-->

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

J<-->H

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 )

<--

SL

cl(\lr)

is

i s completely

Moreover, e v e r y c o n t i n u o u s

Y , where C'(X) h a s t h e vague t o p o l o g y , i s

t t h e r e s t r i c t i o n t o Y of such a transpose T : C(X) ->

<- T~

equivalently,

S i n c e t h e l i n e a r s u b s p a c e o f C'(Y) g e n e -

r a t e d by Y i s v a g u e l y d e n s e i n C 1 ( Y ) , T t

S

-

t CCY) by ( s f , y )

=

(f,sy).

In effect, define

I t is easily verified

t h a t s t i s l i n e a r a n d norm c o n t i n u o u s . We t h e n h a v e tt tt S C1(Y), with s J Y = s , s o s t .i s t h e d e s i r e d T . C ' (X)
is straightforward.

The r e s t

Chapter 7

200

There i s a three-way correspondence between

( 3 7 . 1 ) Theorem.

t h e norm c o n t i n u o u s l i n e a r m a p p i n g s C(X) __ > C ( Y ) , c o n t i n u o u s l i n e a r m a p p i n g s C ' (Xj <--S

<-

C'(X)

the vaguely

C' (Yj , a n d t h e m a p p i n g s

Y c o n t i n u o u s w i t h r e s p e c t t o t h e t o p o l o g y o f Y and T , S , and s c o r r e s p o n d t o e a c h

t h e vague topology o f C t ( X ) .

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 :

tt ;

(i)

s

=

T~

(ii)

s

=

Sly;

(iiij T

=

st = s ~ / c ( x ) .

In ( i i i ) , S

of t h i s 5,

t

=

s

t d e n o t e s Ctt(X) S > 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).

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

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) -

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

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

=

--

O
Ba(X)B.

' J O <-a < Q

*

Then C ( X ) = Ba(X)O c B a ( X ) l c . . - c Ba(X) (Y

Ba(X)a

=

Ba(X).

The s u b s p a c e s Ba(X)

(1

(0 < cx c (?)

w i l l be c a l l e d t h e B a i r e c l a s s e s o f C " ( X ) .

( 4 1 . 1 ) Theorem.

E v e r y B a i r e c l a s s i s an MIL-subspace o f C"(X).

Ba(X) i s a 0 - o r d e r c l o s e d ME-subspace o f C'l(X).

The f i r s t s t a t e m e n t f o l l o w s f r o m ( 1 6 . 1 2 ) ; (5.2),

t h e s e c o n d , from

t h e d i s c u s s i o n i n 5 8 , and ( 9 . 1 0 ) .

Cons i d e r a b a s i c b a n d C"(X)

u

of C"(X)

( t h a t i s , lJEC'(X)

i s t h e band o f C"(X) d u a l t o t h e band C ' ( X ) (cf. 1-1 1-1 R e c a l l t h a t f o r f E C " ( X ) , t h e component o f f i n C"(X)

and C"(X) 527)).

i s d e n o t e d by f

1'

u'

and f o r A c C"(X),

u n d e r t h e p r o j e c t i o n i n t o C"(X)

P

t h e image o f A i n C"(X)

i s d e n o t e d by A

1'

. LJ

1 C ' ( X ) l - l c a n b e i d e n t i f i e d w i t h t h e s p a c e L (p) ____ Remark.

as t h e d u a l o f C'(X) 1-1' LJ' Thus C ' ( X ) i s t h e "home" o f

( t h e Radon-Nikodym T h e o r e m ) , s o C"(X)

can be i d e n t i f i e d w i t h L m ( p ) . 1 t h e s p a c e s L ( p ) , a n d C"(X) t h a t o f t h e s p a c e s L m ( u ) ,

1-1 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 . We saw i n ( 4 0 . 6 )

t h a t C"(X)a = [ C ( X ) a ] ( l ) .

b a n d , we h a v e a n e v e n s t r o n g e r r e s u l t

For a b a s i c

Subspaces of Ct1(X) (41.2) Theorem.

Given uEC'(X),

every fEC"(X)

an order convergent sequence in C(X)

Proof.

C(X),,

it is u ( C " ( X )

11

P

is thc limit of

.

is separating on C'(X)

u (since C(X) is), so

)-dense in C"(X)

It follows from the

C'(X)

Ll '

219

I'

1J'

discussion in the opening paragraph of 5 2 5 that C(X)

101

(C'l(X)lJ,C' (X)l-l)-densein C"(X),,.

in C 1 ' ( X )

1'

under the norm

A fortiori, it is dense

1-1'

and we can assume 0 5 f 5 n(X)

By the above, there is a sequence {gn} c C(X)p -

gn;I11

limnllf - g i

n

=

0.

is also

(1 .{I 1-1 .

Now consider f€Cl'(X)

limnll f

li

Then,

n(X)l-lilU = 0.

]I - [ I p

L1

.

such that

being a Riesz norm,

(27.2) now gives the desired

result. QED

We can now establish the theorem mentioned at the beginning o f the 5 : the canonical band projection of C"(X) Cll(X),, maps the second Baire class Ba(X)2

onto

onto all of C1l(X)u.

This is the classical theorem that for every U , each element of

Lm(p) -

which is an equivalence class of u-integrable

functions - contains a function of the second Baire class.

(41.3)

Proof.

Consider fEC"(X),,

.

By (41. 2) above, there is a

sequence ifn? in C(X) such that (fn),,

-+

f.

We can assume

Chapter 9

220

{ f n } i s bounded: i n e f f e c t , l e t

=

IIfll; s o - x I ( X )

< f < P -

All(X),, ; t h e n we n e e d o n l y r e p l a c e e a c h f n by [fnV ( - x l l ( X ) ) Now s e t g

(All(X)).

=

limsupn f n

=

AnVmZnfm.

Then g E B a ( X ) 2

and - 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 suprema and i n f i m a

-

gP = limsupn(fn),, = f .

QED

By E x e r c i s e 23 o f C h a p t e r 1 , t h e u n i o n o f a c o u n t a b l e s e t of p r i n c i p a l bands o f Cl(X) i s c o n t a i n e d i n a p r i n c i p a l b a n d . I t follows t h e union of a countable s e t of b a s i c bands o f

i s c o n t a i n e d i n a b a s i c band.

C"(X)

One c o n s e q u e n c e i s t h a t

t h e u n i o n o f a l l t h e b a s i c b a n d s o f Ctf(X) i s a l s o t h e i r ( a l g e b r a i c ) sum, h e n c e a R i e s z i d e a l .

( 4 1 . 4 ) C:(X)

We d e n o t e i t by C;(X).

i s a 8 - o r d e r c l o s e d R i e s z i d e a l o f C"(X) whose

o r d e r c l o s u r e i s C"(X).

T h a t C''(X) remarks.

i s o - o r d e r c l o s e d f o l l o w s from t h e a b o v e

Since it i s a Riesz i d e a l , the statement t h a t i t s

o r d e r c l o s u r e i s Ctt(X) i s e q u i v a l e n t t o t h e ( s t r o n g ) s t a t e m e n t t h a t e v e r y fEC"(X)+ below.

i s t h e supremum o f t h e e l e m e n t s o f C:(X)

T h i s f o l l o w s e a s i l y from t h e f a c t t h a t e v e r y band o f

Cll(X) c o n t a i n s a b a s i c b a n d .

We c a n d e s c r i b e C:(X)

n

C f t ( X ) a (= [C:(X)la)

S i n c e e v e r y e l e m e n t o f C l ( X ) a i s o f t h e form P

precisely. =

znXxnxn, a

band o f Ctt(X)u i s b a s i c i f and o n l y i f i t c o n s i s t s o f a l l t h e

S u b s p a c e s o f Ct'(X)

elements o f C"(X)a

which v a n i s h o f f some c o u n t a b l e s u b s e t o f X .

Thus t h e b a s i c bands o f C"(X)a able subset of X I .

(41.5)

Ck(X)

221

a r e t h e b a n d s {L"(Q)

IQ a count-

I t follows:

C't(X)a c o n s i s t s o f a l l fEC"(X)a

such t h a t

(f,x) # 0 on o n l y a c o u n t a b l e s e t o f x ' s ( d e p e n d i n g on f ) .

(41.6)

Corollary.

C:(X)

n

C't(X)a i s t h e o - o r d e r c l o s u r e o f

co (XI

5 4 2 , The " s e m i c o n t i n u o u s " e l e m e n t s

I n e x p l o r i n g how t h e s t r u c t u r e o f C'l(X) i s d e t e r m i n e d by t h e MIL-subspace C(X), o u r f i r s t a p p r o a c h , t h e one t o be f o l lowed i n t h e r e m a i n d e r o f t h i s c h a p t e r , w i l l be t o l o o k f o r " l a y e r s " o f C"(X)

"between"

s i b l e " from C(X). o f such kind. C(X)

C(X) and C"(X) which a r e " a c c e s -

Our p r i n c i p a l i n t e r e s t w i l l b e MIL-subspaces

The t o p o l g i e s on

Cll(X) a r e o f no h e l p i n t h i s .

i s c l o s e d u n d e r t h e norm t o p o l o g y o f Cl'(X), and u n d e r

o(Cll(X) , C ' ( X ) ) , d e n s e i n C"(X).

101 (Clt(X) , C ' ( X ) ) ,

and -r(Cl'(X) , C ' ( X ) ) , i t i s

Thus t h e norm t o p o l o g y i s t o o f i n e and t h e

o t h e r t h r e e a r e too coarse. I n c o n t r a s t t o t h i s , t h e use of o r d e r convergence e n a b l e s

u s t o u n c o v e r v a r i o u s s u b s p a c e s o f t h e k i n d we a r e s e e k i n g .

Chapter 9

222

(An e x c e p t i o n t o t h e c o n c l u s i o n o f t h e a b o v e p a r a g r a p h i s s u p p l i e d by s e q u e n t i a l convergence i n t h e above t o p o l o g i e s .

I t e n a b l e s u s t o d i s c o v e r and s t u d y t h e v a r i o u s R a i r e c l a s s e s . tiowever, t h e same r e s u l t c a n b e o b t a i n e d u s i n g a - o r d e r c o n vergence.)

A c t u a l l y , t h e f i r s t s u b s e t s w e examine a r e n o t l i n e a r s u b -

We n e e d some p r e l i m i n a r i e s .

spaces.

A subset P of a vector

s p a c e i s c a l l e d a wedge i f i t i s c l o s e d u n d e r a d d i t i o n a n d Note t h a t a c o n e

m u l t i p l i c a t i o n by n o n - n e g a t i v e s c a l a r s . (cf.

§ 1 ) i s a wedge w h i c h c o n t a i n s no l i n e a r s u b s p a c e o t h e r

than 0.

I f P i s a wedge, t h e n P

(-P) and P

-

P are both linear

s u b s p a c e s , t h e f o r m e r t h e l a r g e s t o n e c o n t a i n e d i n P and t h e l a t t e r t h e s m a l l e s t one c o n t a i n i n g P.

(So P

- P is the linear

s u b s p a c e g e n e r a t e d by P . )

(42.1)

I f a wedge P i n a R i e s z s p a c e E i s a l s o a s u b l a t t i c e o f

E, then P t h a t aVbEP

- P is a Riesz subspace o f E.

If fact, it suffices

f o r a l l a,bEP.

T h i s f o l l o w s from ( a - b)'

= avb

- b.

Given a s u b s e t A o f a R i e s z s p a c e E , w e s e t

A'

='A

=

CaEE

{aEE

1 1

a = V B f o r some B c A);

a = A B f o r some B c A ) .

Subspaces of C " ( X )

Note that (i) (A',)' A c A'',

and (iii) - A '

=

= A'

and (Au)'

223

=

A',

(ii) A c A'

and

(-AIu.

(42.2) If P is a wedge in a Riesz space - in particular, if P is a linear subspace

-

than P p and Pu are wedges.

The verification is straightforward. If A is a subset of a Riesz space E, then, as we noted in (i) above, 'A rema.

is closed under the operation of taking sup-

In general, it is not closed under the operation of tak-

ing infima.

And similar statements hold for AU with "suprema"

and "infima" interchanged. lattice, then A '

However, by ( 2 . 4 ) , if A is a sub~

is closed under the operation of taking

infima of finite subsets and AU, under the operation of taking infima of finite subsets.

We thus have

(42.3) If A is a sublattice of a Riesz space E, then so are A R

and AU.

We now turn to C " ( X ) .

Consider the wedges C ( X ) '

We will call the elements of C(X)' of C(X)'

the u s c ones, of C"(X).

and C(X)'.

the ~ s elements, c and those The notations " R s c "

and " u s c "

are of course suggestive of "lowersemicontinuous" and "uppersemicontinuous" respectively, but they are for these words.

=

abbreviations

224

Chapter 9

By t h e p r e c e d i n g , C(X)' l a t t i c e s o f C"(X).

(42.4)

C(X)'

P roof. -

and C(X)'

a r e wedges and s u b -

In a d d i t i o n ,

and C(X)'

a r e norm c l o s e d .

C o n s i d e r f i n t h e n o r m - c l o s u r e o f C(X)'.

each n = 1 , 2 , . * . ,

c h o o s e fn€C(X)'

For

< l/n. s u c h t h a t ( I f n - f[l -

Then f - (l/n)n(X) < f -

< f

n -

+

(l/n)n(X),

hence < f n - (l/n)n(X) < f , f - (2/n)n(x) -

hence, i n t u r n , f = NOW

v n [fn

- n ( x ) c c ( x ) c c(x)',

-

(l/n)n(X)].

s o [fn - ( I / ~ ) R ( X ) I : C ( X ) ~f o r a l l n .

I t f o l l o w s f€C(X)R. QED

S i n c e C(X) = - C ( X ) , we have C(X>u 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 ><

(f,u>

+

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 <

uEC'(X)+

and c > 0 .

n(X).

Fix

We n o t e f i r s t

In e f f e c t , t h e r e i s a descending n e t i n C(X)'

o r d e r con-

v e r g i n g t o f n , a n d we c a n as s um e i t s members a r e a l l < n(X). E v e n t u a l l y , some member o f t h e n e t w i l l h a v e

these three

properties. Now s e t Ln

=

V5=lti

[n

( i ) an d ( i i ) a r e c l e a r . n

=

1, s i n c e fil

=

ki.

=

1,2,.'-).

Then

Also ( i i i ) and ( i v ) hold f o r

Assume ( i v ) h o l d s f o r n .

We show t h a t

Subspaces o f C"(X)

then ( i i i ) and ( i v ) hold f o r n

1.

+

227

This w i l l e s t a b l i s h the

four propert i es. (Rn+l'l-l)

=

(RnVRn+yd

=

(R;+]d

<

(R;+lW)

+

+

< ( fn+],U)

+

(fn+]d)

+

=

Thus ( i i i ) h o l d s f o r n

+

- R;+$+,!d

((an

( (Rn - f n + l ) + , " ) E

(1 - P ) E

+

(2 - 2 - n ) E *

1.

so

Thus ( i v ) h o l d s f o r n + 1.

n IIn '. w e show R s a t i s f i e s t h e Lemma, w h i c h w i l l complete t h e proof. L€C(X)' , and by (ii) , R 2 f. Thus Now s e t R

LEA.

= V

now (f,v)

=

limn(fn,v),

(a,~)

=

limn(Rn,u).

and

S i n c e , by ( i i i ) , (an,v) 5 (fn,v)

we h a v e

+

2s f o r a l l n ,

Chapter 9

228

Applying ( 7 . 4 ) ,

we have t h e

Corollary.

C ( X ) R U and C(X)"

(42.8)

are o-order closed.

5 4 3 . The Up-down-up theorem

C o n s i d e r t h e two c h a i n s :

C(X)

C(X)R c C(X)RU c C(X)RUR c

* '

C ( x p c C(X)Ut c c(x)'Ru

-

-

= c

* *

Wedges a t t h e same s t a g e i n t h e two c h a i n s a r e t h e n e g a t i v e s of e a c h o t h e r , and t h e r e a r e v a r i o u s c r o s s i n c l u s i o n s ( c f . ( 4 2 . 6 ) , f o r example). chains.

A n a t u r a l q u e s t i o n i s : how l o n g a r e t h e s e

We show t h e y a r e q u i t e s h o r t .

F i r s t , a p r o p o s i t i o n which we w i l

n e e d , and which i s o f

interest in itself.

Suppose we have p vEC'(X) s u c h t h a t FI > 0 ,

v > 0 , and

A s we know, t h e r e e x i s t s fEC"(X)+

~JAV=

0.

t h a t ( f , p ) = 0 , ( f , w ) > 0 (11.6).

such

I n f a c t , we c a n choose f t o

be U S C :

( 4 3 . 1 ) Given p,wEC'(X)

s u c h t h a t p > 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.

S u b s p a c e s o f C"(X)

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

<

g(a)l

> 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 <

E'

=

x

i n f ( ( b , K ) ,o). (b,K)EF

IR, s o

0 =

($,u) with $EE', uEW.

Note t h a t u > 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).

+

S u b s p a c e s o f C"(X)

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 <

x

( u - ll),,)

n)

>

-

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

+

<

t,

2-ln,

+

2-ln.

+

We thus obtain the second inequality in ( * ) . (*),

(For the

in

we need only take the smaller of the two x's obtained

above. ) The two inequalities in ( * ) now give us ( u - 2-ln)v(f0

-

2-'n)

<

(fO

+

2-'n)r\(a

+ 2-ln)

-

xn,

which gives us (iii). Finally, applying (48.2) and (38.2) to (iii), we ontain an flEC(X) satisfying ( 1 1 ) .

QED With the Edwards theorem, we can establish our identity for c(x)(~).

S u b s p a c e s o f C"(X)

(48.52)Theorem

___ Proof.

C(X)

U(X).

=

Suppose f € C ( X ) ( l ) .

I f c l } i n C(X) s u c h t h a t f

245

=

This says t h e r e e x i s t s a n e t

V,ba2afp

f

= A V

a B>cc R -

(cf.

(7.4)).

Thus f i s a supremum o f u s c e l e m e n t s a n d a n i n f i m u m o f e s c e l e m e n t s , h e n c e l i e s i n U(X). For t h e o p p o s i t e i n c l u s i o n , s uppose f E U ( X ) . e x i s t n e t s {ua}i n C(X)u a n d {La? i n C(X)

R

Then t h e r e

(we a s s u m e , a s we

c a n , t h a t t h e y h a v e t h e same d i r e c t e d i n d e x s y s t e m ) s u c h t h a t

u +f and k n + f . c1

element f

N

A p p l y i n g ( 4 8 . 4 ) , we o b t a i n , f o r e a c h a , a n

o f C(X) s u c h t h a t u,'

fa 5 La.

Then f ,

+

f , so

f E c (X) (I).

QED

I t r e m a i n s t o show t h a t C ( X ) ( 2 )

=

C"(X).

S i n c e C(X)(')

=

U(X), w e s t a t e t h i s i n t h e f o r m

Theorem. U(X)"'

(48.6)

=

C1l(X)

I n f a c t , w e e s t a b l i s h a much s t r o n g e r r e s u l t ,

(Ba(X)2)(1)

C" (X) :

(48.7) Theorem. -

E v e r y fEC"(X)

v e r g e n t n e t i n Ba(X)2.

i s t h e l i m i t o f an o r d e r con-

=

Chapter 9

246

P roof.

L e t 3 be t h e s e t o f b a s i c bands o f C"(X).

It is

Now c o n s i d e r f€Cll(X),

a n d we

a d i r e c t e d s e t under inclusion. < f < I(X). c a n assume 0 -

By ( 4 1 . 3 ) , f o r e a c h I € $ , we c a n

c h o o s e g ( I ) E Ba(XI2 s u c h t h a t g ( I ) I

=

f I ; a n d we c a n c l e a r l y

take g(1) t o s a t i s f y 0 < g(1) < n(X). {g(I)IIEJ] Let gI

=

g

i s a n e t ; w e show i t o r d e r c o n v e r g e s t o f . =

liminfIag(T)

hI = f I f o r a l l I € J .

(f,p) f o r a l l

Fix Io.

Since3 is directed,

and h

I t w i l l f o l l o w ( g , p)

uEC'r(X), h e n c e g

=

h

We show

1imsuprE3g(I).

=

=

=

( h , 11)

=

f.

For e v e r y I € J s u c h t h a t I 2 I o , g ( I ) I

=

fI :

0

in effect, g(III

(g(I)I)Io

=

=

.

(fI)Io = f I

0

I t follows

0

0

(cf. (6.5))

= fI

The p r o o f t h a t h I 0

i s t h e same 0

We r e c o r d t h e M I - s u b s p a c e s o f C"(X)

which we have s i n g l e d

out i n t h i s chapter:

We a l s o h a v e t h e v a r i o u s B a i r e c l a s s e s a n d B o r e 1 c l a s s e s .

247

Suhspaces of C"(X)

We also record the fact that we have established Nakano's theorem (13.2) for the special case

o f C(X).

As is to be

expected, the statement here is considerably stronger than that of the general case.

(48.8)

(i)

C'(X)

=

(C1l(X))c.

(ii)

c"(x)

=

(c'(x))~.

(iii) The order closure of C(X) C(X) (2)

=

Cf'(X).

is C " ( X ) ;

indeed

CHAPTER 1 0

THE OPERATORS u and R

549. The o p e r a t o r s u and R

T h e r e i s a p e r v a s i v e a n a l o g y t h r o u g h o u t t h i s work between t h e e l e m e n t s o f C"(X) and t h e s u b s e t s o f a t o t a l l y d i s c o n n e c t e d compact H a u s d o r f f s p a c e [ i n f a c t a h y p e r s t o n i a n o n e ) . elements correspond

The u s c

t o c l o s e d s e t s , t h e Rsc e l e m e n t s t o open

s e t s , and ( t h e r e f o r e ) t h e e l e m e n t s o f C(X) t o c l o p e n ( b o t h c l o s e d and open) s e t s .

The Edwards Theorem ( 4 8 . 5 ) c o r r e s p o n d s t o

the property t h a t t h e clopen s e t s c o n s t i t u t e a b a s i s f o r t h e topology.

The e l e m e n t s o f U(X) c o r r e s p o n d t o t h e s e t s m e a s u r -

a b l e w i t h r e s p e c t t o e v e r y r e g u l a r m e a s u r e , t h e e l e m e n t s o f Bo(X) t o t h e Bore1 s e t s , t h e e l e m e n t s o f Ba(X)l t o t h e s e t s which a r e s i m u l t a n e o u s l y a n F,

and a G g ,

and s o o n .

"Analogy" i s t o o weak

a word; t h e p r o p e r t i e s o f C"(X) a r e a c t u a l l y g e n e r a l i z a t i o n s o f the topological properties. C.A.

Akermann [l] h a s u s e d t h i s a p p r o a c h f o r C * - a l g e b r a s .

That c a s e i s more d i f f i c u l t , and he d e f i n e d h i s " t o p o l o g i c a l " properties only f o r projections. For e a c h fEC"(X), we d e f i n e

248

The O p e r a t o r s u and k

249

Thus 2 ( f ) i s t h e l a r g e s t k s c e l e m e n t b e l o w f , and u ( f ) i s t h e s m a l l e s t u s c e l e m e n t above i t .

I n t h e above a n a l o g y , t h e y

c o r r e s p o n d t o t h e i n t e r i o r and t h e c l o s u r e o f a s e t , r e s p e c t i v e ly .

By t h e v e r y d e f i n i t i o n , k ( f )

=

f is equivalent to the state-

ment t h a t f i s a n Q s c e l e m e n t , and f = u ( f ) t o t h e s t a t e m e n t t h a t f i s a usc element.

And by ( 4 4 . 1 ) ,

2(f)

=

f = u(f) is

I t i s immediate

e q u i v a l e n t t o t h e s t a t e m e n t t h a t fEC(X).

< g implies k(f) < k ( g ) and u ( f ) < u(g), that k(xf) that f -

A k ( f ) and u ( X f )

=

Au(f) f o r a l l X > 0 , and t h a t k ( - f )

=

=

-u(f).

We p r o c e e d t o d e v e l o p t h e p r o p e r t i e s o f o u r two o p e r a t o r s . The v e r i f i c a t i o n o f many o f t h e s e r e q u i r e s o n l y r o u t i n e c a l c u l a t i o n , and w i l l be o m i t t e d .

(49.1) L(f)

+

Proof.

k(g) I Q ( f

+

+

g) i s t h e l a r g e s t s u c h below f

k(-g) < k(f);

+

2) < u(f)

+

u(g).

i s an ksc element,

g, since klf) + k(g)

k(f) + k(g) 5 k(f + g),

ity.

g ) 5 9>(f) + u ( g ) < u(f

For t h e f i r s t i n e q u a l i t y , Q ( f ) < f and k ( g ) < g,

so k ( f ) + k ( g ) 5 f and k ( f

+

+

g, i t f o l l o w s

By t h e f i r s t i n e q u a l i t y , k ( f

since Q(-g)

=

+

g)

+

- u ( g ) , w e have t h e second i n e q u a l -

The o t h e r two f o l l o w b y symmetry. QED

Chapter 10

250

The verification of the following is straightforward.

For a finite set, the last inequality in the first line a nd the first inequality in the second line become equalities

(49.4)

We also have

The Operators u and R -~ Proof.

251

The first inequality in the first row is immediate

(cf. (49.3)); we show the second. u(g) - Q(f)Au(g)

=

(u(g)

-

From the first equality, u(g)

a(f) -

+

> (g - f)' -

=

g - fAg.

f ) A u ( g ) is a usc element

(cf. the discussion preceding (42.4)),

hence we obtain u(g)

-

, we have < u(fAg).

QED

In particular,

(49.6) Corollary. f E C"(X)

For an t s c element R , a usc element u , and

, RAu(f) < u(RAf), uVR(f) > 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-) <

and - r e p l a c i n g

5

lu(f)l. QE D

T h i s f o l l o w s from t h e f a c t t h a t i f Kn(x) < f < xll(X) , t h e n Kn(x) < a(f)

( 4 9 . 1 5 ) C o r o l l a r y 1.

< -

f

< U(f) -

<

[I f\I

For f E C " ( X ) + ,

xn(x).

=

[lu(f)[l.

Combining t h i s w i t h ( 4 9 . 1 3 ) , we h a v e

(49.16) Corollary 2 .

II u ( f )

- u ( g ) ;I <

jl a ( f )

- R(g)

I1 f

- R[I

*

11

Thus t h e o p e r a t i o n s u(-) a n d a ( * ) a r e norm c o n t i n u o u s .

We w i l l n e e d t h e f a c t t h a t Z ( n ( X ) ) 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 s u ( * ) a n d I,(.):

Proof.

(Zu(e))An(x)

=

u(Ze)An(X)

=

u [ ( Z e ) ~ n ( x ) ]= u ( e )

'l'hc Operators

11

(the second equality from (49.9)), so u(e) n(X)

(cf. §17).

255

and R

is a component o f

Similarly for e ( c ) .

QED

Notation. -~ a(u(f)),

We can consider combined operations like

a(a(f)),

u(k(u(f))),

expressions simply au(f),

and so on. a a ( f ) , uau(f),

We will write such

. - ..

We note immedi-

ately that the operations a ( . ) and u ( . ) are idempotent: aa(f)

=

k(f),

uu(f)

=

u(f).

Also, it can be shown by simple

examples that the two operations do not commute: ku(f)

# uk(f),

in general. We record two properties we will use.

The verification is simple. and ua(f)

Note that, in general, ku(f)

are not comparable orderwise.

(49.19) The operations a u ( . ) and u a ( * ) are idempotent: every fEC"(X)

(i)

,

auau(f)

=

au(f),

(ii) u a u a ( f )

=

ua(f).

Proof. -

for

In the first inequality of (49.18) above, replace

256

Chapter 1 0

f by u ( f ) ;

t h i s g i v e s us k u ( f ) < kuku(f).

< ku(f). inequality there, kuku(f) -

From t h e l a s t

Thus we h a v e ( i ) .

(ii) is

shown s i m i l a r l y . QED

The p o s s i b l e o p e r a t i o n s o b t a i n a b l e a s c o m b i n a t i o n s o f

a ( . ) and u ( . ) t h u s r e d u c e s t o s i x : a ( . ) , u ( * ) , k u ( * ) , u k ( - ) , a u k ( * ) , and u k u ( * ) .

5 5 0 . The o p e r a t o r 6

F o r e a c h f € C ” . ( X ) , we d e f i n e

S(f)

=

u(f) - a(f).

6 ( f ) corresponds t o t h e f r o n t i e r of a set i n topology, and,

i n function theory, t o the function giving the s a l t u s of a function a t every point. I t i s immediate t h a t 6 ( 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)

C o r o l l a r y 1.

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) <

11 6 ( f ) i l n ( X )

=

2rn(X).

263

We w r i t e t h i s

u ( f ) - rn(X) < e(f) Then by t h e Edwards Theorem ( 4 8 . 5 ) ,

+

rn(x).

t h e r e e x i s t s gEC(X) s u c h

that (i)

u ( f ) - rn(X) < g

5

a(f)

+

rl(X).

I t follows f

- r m )

2

g < f

+

rn(x),

t h a t i s , \If - gII 5 r .

QE D

-~ Remark.

From t h e two p a r t s o f t h e p r o o f , f o r e v e r y

gEC(X) s a t i s f y i n g ( i ) , wc h a v e \If - g [ l

=I

r (not j u s t < r).

i s e a s y t o v e r i f y t h a t , c o n v e r s e l y , e v e r y gEC(X) w i t h ! I f =

It

- g!I

r satisfics (i).

951. R - b a n d s and u - b a n d s

A n a t u r a l p r o b l e m i n t h e s t u d y o f Cl'(X) i s t h e c l a s s i f i c a -

t i o n o f i t s bands.

Thus f a r we h a v e e n c o u n t e r e d t h e p a i r o f

c o m p l e m e n t a r y b a n d s C't(X)a and C''(X)d, a n d t h e b a s i c b a n d s { C " ( X ) l l j ~ E C ' ( X ) } . I n t h e p r e s e n t 5 , w e l o o k a t two a d d i t i o n a l , relatively

simple types.

We f i r s t r e c o r d some s i m p l e p r o p e r t i e s o f a g e n e r a l b a n d I o f C"(X).

I i s i t s e l f an

MIL-space, w i t h n(X),

( b u t n o t e t h a t w i t h one e x c e p t i o n - I

=

MI-subspace, s i n c e it does n o t c o n t a i n

n(X));

p r o j e c t i o n p r o j I o f C"(X)

C"(X)

for unit

- I i s n o t an

and t h e c a n o n i c a l

onto I i s an o r d e r continuous

Chapter 10

264

MIL-homomorphism.

Also ( c f .

(18. 3 ) ,

( 6 . 5 ) , and ( 4 8 . 8 ) ) :

( 5 1 . 1 ) F o r e v e r y band I o f C"(X),

i s an MIL-subspace o f I ,

(i) C(X),

( i i ) t h e o r d e r c l o s u r e o f C(X),

Remark.

- c _ _

form ( c f .

For p a r t i c u l a r bands,

(40.6),

(51.2) Corollary.

is I.

( i i ) may t a k e a s t r o n g e r

(41.2)).

For t h e o r d e r c l o s u r e o f a s u b s e t A o f a

band I t o be I , i t s u f f i c e s t h a t t h i s o r d e r c l o s u r e c o n t a i n C(X)I *

I n c o n t r a s t t o ( i i ) a b o v e , t h e o r d e r c l o s u r e o f C(X)

n

I is,

i n g e n e r a l , a p r o p e r s u b s e t o f I (we d e s c r i b e i t i n ( 5 2 . 1 ) ) . However, a s we now show, t h e l a t t e r o r d e r c l o s u r e i s s t i l l a band.

( 5 1 . 3) Theorem.

F o r e v e r y Riesz i d e a l H o f C(X)

,

the order

c l o s u r e o f H i s a band o f C1'(X) ( h e n c e t h e band g e n e r a t e d b y H ) .

P r o o f . Let I1 b e t h e b a n d o f C " ( X )

g e n e r a t e d by H; w e show

The O p e r a t o r s u and k

265

I1 i s i n f a c t t h e o r d e r c l o s u r e o f H

Lemma.

Let I.

For e v e r y f E C ( X ) + ,

=

V{h€I-Il~Oc h < f}.

be t h e R i e s z i d e a l o f C " ( X )

I1 i s t h e o r d e r c l o s u r e o f I o , s o f T

g e n e r a t e d by H . =

V{gEIo 1 0 < g

5

Then f}.

But f o r e v e r y s u c h g , t h e r e e x i s t s hEH s u c h t h a t g < h i f : choose kEH s u c h t h a t g c k , and s e t h

=

hf.

This e s t a b l i s h e s

t h e Lemma, and t h e Theorem f o l l o w s from ( 5 1 . 2 ) .

QED

We t u r n t o t h e f i r s t of o u r s p e c i a l t y p e s o f b a n d . I of C"(X)

(51.4)

i s an Rsc e l e m e n t .

w i l l be c a l l e d a n & - b a n d i f n ( X ) ,

Let I be a band o f C l ' ( X ) ,

and s e t H

A band

=

I.

C(X)

The

following a r e equivalent: 1'

I i s an & - b a n d ;

2'

I i s t h e band o f C " ( X )

3'

g e n e r a t e d by H;

I i s t h e o r d e r c l o s u r e of H .

P roof.

Assume '1

holds.

Then n ( X ) ,

i s t h e supremum o f a

s u b s e t o f C ( X ) which i s c l e a r l y i n I , hence i n H . e a s i l y from t h i s .

T h a t 2'

i n t h e proof o f (51.3).

i m p l i e s 1'

2'

follows

f o l l o w s from t h e Lemma

The e q u i v a l e n c e o f 2'

and '3

is the

Chapter 1 0

266

content of (51.3).

(51.5) C o r o l l a r y . There i s a one-one co r r es p o n d en ce between t h e R-bands I o f C " ( X )

and 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 i f and o n l y if H

I <->

=

C(X)

n

I i f and o n l y i f I i s t h e

generated by H (equivalently, t h e order c l o s u r e

band o f C " ( X ) of H).

P r o o f . The --

o n l y s t a t e m e n t remain in g t o be proved i s t h a t

i f I i s t h e b a n d g e n e r a t e d by a norm c l o s e d R i e s z i d e a l H o f

C(X),

f

=

then C(X)

I

=

H.

Consider f E C ( X )

n

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

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 ) .

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

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)

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

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)

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

'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

The O p e r a t o r s u a n d R

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

The O p e r a t o r s u a n d R

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.

<

g.

For all Rl,L2EC(X)

R

,

(Rl)u

5

(R2)u

implies

5 L2' By the definition of C(X)',

the Lemma can be stated:

every pair of subsets A,B o f C(X) which are bounded a b o v e , V A u -< VBu implies V A < VB.

We show this. 275

For

2 76

C h a p t e r 11

Assume V A VgEB(faAg,) VgEB(fAg)

.

For e v e r y f E A , . f < V B hence f = aa' a I t fOllOWS f r o m ( 4 0 . 3 ) t h a t f = = v g € B ( f A g )U .

-

< VB

a -

u

Since t h i s holds f o r a l l f E A , V A < VB.

VB.

Lemma 2 . For a l l t e C ( X ) ' -___

u

< -

u

and uEC(X)',

u -

R

implies

a

a

In e f f e c t ,

(a.

-

u)&

=

-

u

> 0. u -

S i n c e I? - u i s a n k s c

e l e m e n t , i t f o l l o w s f r o m Lemma 1 t h a t Q - u > 0. Now c o n s i d e r a n y f , g E U ( X ) s u c h t h a t f a 5 g,. some A c C(X)u a n d g

=

A B f o r some B c C(X)

R

f

=

(cf. 544).

vA f o r Then

= VA a n d ga = A B ~ . S i n c e , by s u p p o s i t i o n , f a 5 g a , t h i s a a g i v e s u s A < B * h e n c e , by Lemma 2 , A - B; h e n c e , f i n a l l y , a - a' f / g.

f

QED

(53.2)

Corollary.

Given fEU(X), f i s c o m p l e t e l y d e t e r m i n e d by

i t s v a l u e s on X .

Proof. know f .

so f

By t h e I s o m o r p h i s m t h e o r e m i f w e know f

(We m u s t know t h a t f E U ( X ) ! ! ) .

a'

t h e n we

Now C''(X)a = k"(X),

i s c o m p l e t e l y d e t e r m i n e d by t h e s e t o f v a l u e s a C ( f a , x ) I x E X 3 But ( f , x ) = (f , x ) f o r a l l x E X . . Thus f i s a d e t e r m i n e d by t h e s e t o f v a l u e s { ( f , x ) I x E X } ,

We e m p h a s i z e a g a i n t h a t , a s f a r a s we know, i t i s o n l y U(X) whose e l e m e n t s a r e d e t e r m i n e d by t h e i r v a l u e s on X . may b e

-

It

i n some s t i l l t o b e d e f i n e d s e n s e - t h a t U(X) may b e

t h e l a r g e s t s u b s p a c e o f Ctt(X) c o n t a i n i n g C(X) f o r w h i c h t h i s holds.

We s t a t e t h i s i n a d i f f e r e n t way.

The I s o m o r p h i s m

t h e o r e m g i v e s u s an i d e n t i f i c a t i o n o f U(X) w i t h a s u b s p a c e o f ilm(X),

t h a t i s , w i t h a s e t o f f u n c t i o n s on X .

( B u t remember:

U(X) i s r e a l l y d i s t i n c t f r o m t h i s f u n c t i o n s p a c e . )

I t may b e

t h a t U(X) i s t h e l a r g e s t s u b s p a c e o f C1'(X) w h i c h c o n t a i n s C(X) and whose e l e m e n t s c a n b e i d e n t i f i e d i n a " n a t u r a l " way w i t h f u n c t i o n s on X .

554.

Immediate c o n s e q u e n c e s o f t h e Isomorphism theorem

The u - o r d e r c l o s e d n e s s o f U(X) and t h e o r d e r c o n t i n u i t y o f band p r o j e c t i o n g i v e s u s :

(54.1)

U(X)a

is o-order closed.

Whence,

( 5 4 . 2 ) Given i f n } c U(X) a n d f € U ( X ) , t h e f o l l o w i n g a r e e q u i valent:

C h a p t e r 11

278

1'

{f,}

order converges t o f i n U ( X ) ;

2'

{f,}

order converges t o f i n Clt(X) ;

0

3

{fn} converges t o f vaguely;

4'

{f,}

5'

{ ( fn ) u } o r d e r c o n v e r g e s t o f a .

i s bounded a n d ( f , x )

Proof. t h a t U(X)

That lo i m p l i e s 2

0

=

limn(fn,x) €or a l l x E X :

f o l l o w s e a s i l y from t h e f a c t T h a t 2'

i s u-order closed i n Ctt(X).

implies 3

(40.2)

0

.

Since ( (fn)a,x)

g i v e s u s t h a t 4'

i m p l i e s 5'.

=

( f n , x ) and ( f , , x ) F i n a l l y , assume 5

of

3'

f o l l o w s from t h e o r d e r c o n t i n u i t y o f e v e r y u E C ' ( X ) . course implies 4

0

0

=

holds.

~ ~ I)s o~m o) r p. h i s m t h e o r e m a n d t h e Then f a = A ~ ( V ~ > ~ ( The -

o r d e r closedness o f U(X) f

=

limsup f 11

Thus f n

-t

n

.

gives us t h a t f

=

An(VmZnfm).

(f,x),

0-

Thus

A s i m i l a r argument g i v e s us t h a t f = l i m i n f n f n .

f.

QED

Contained i n t h e Isomorphism theorem i s t h e f o l l o w i n g :

( 5 4 . 3 ) The p r o j e c t i o n o f C " ( X )

o n t o C"(X

and B o ( X )

MIl-isomorphically i n t o C ' t ( X ) u .

and C ( X ) u

l a t t i c e i s o m o r p h i c a l l y i n t o C"

We h a v e a l r e a d y s e e n t h i s f o r C ( X ) then characterized C(X)a:

i n 540.

M o r e o v e r , we

279

(54.4) For g E C " ( X ) u , lo

gEC(x)u;

'2

g is

the following are equivalent:

continuous on X.

Using this, we can a l s o characterize (C(X) R )a and (C(X)u)a:

(54.5) For g E C " ( X ) , ,

the following are equivalent:

'1

g i (CCX)'

2'

g is lowersemicontinuous on X.

ja ;

Similarly for (C(X)u)a,

Proof.

with "uppersemicontinuous" in '2

Given L € C ( X ) ' ,

then by the Lemma in the proof of

(44.1), R is lowersemicontinuous on X. all x € X , , i t follows R, versely, assume 2'

Since (R,,x)

=

( L , x ) for

is a l s o lowersemicontinuous on X.

holds.

Con-

Then by standard function theory

(and ( 4 0 . 1 ) ) , g is the supremum in Ctl(X)a of the set B of elements of C"(X)a

B

=

A d , A c C(X).

below i t which are continuous on X. p, = V A then satisfies p,

a

By ( 5 4 . 4 ) ,

= g.

QED

We recall (48.4) that an element f of C t l ( X ) lies in C ( X )

1

if and only if f is vaguely lowersemicontinuous on K ( C ' ( X ) ) . It is not sufficient that f be lowersemicontinuous on X. ever, for fEU(X), the latter is sufficient:

How-

Chapter 11

280 (54.6) Corollary 1. lo

f€C(X)R;

2’

f is lowersemicontinuous on X;

3’

fa is lowersemicontinuous on X.

0

Proof.

I t remains only to show that 3

pose 3’ holds. that R, f

For f € U ( X ) , the following are equivalent:

fa.

=

implies 1’.

Then, by ( 5 4 . 5 ) , there exists L€C(X)’

Supsuch

It follows from the Isomorphism theorem that

= J?,.

QED

For a second corollary, we recall some function theory. We are interested in C ” ( X ) a , ~ we ~ state it in terms of functions on X.

Consider a bounded function g on X.

function ^g on X as follows: neighborhoods of x ; then g ( x )

g determines a

let {V } be the family of all c1

=

infa(sup

g(y)).

We denote

va

limsup It is easily verified that 2 is g(y). Y+X an uppersemicontinuous function above g , and is the smallest this by ^g(x)

one. of g.

=

We will call it the qpersemicontinuous upper envelope The lowersemicontinuous lower envelope of g is defined

s imi larl y

.

We then have

(54.7) Corollary 2.

For f E U ( X ) , u(f),

tinuous upper envelope of f a , and il(f), tinuous lower envelope.

is the uppersemiconis the lowersemicon-

281

T h i s f o l l o w s e a s i l y from ( 5 4 . 6 ) and t h e Isomorphism The o r em. The c o n d i i o n fEU(X) c a n n o t b e d r o p p e d : i n t e r v a l and f

=

ll(X)d

.

Let X b e a r e a l

Then f a = 0 , w h i c h i s i t s own u p p e r -

semicontinuous upper envelope, but it i s n o t hard t o v e r i f y that u(f)

=

ll(X),

hence u ( f ) a

=

Il(X)a

#

I n ( 5 4 . 5 ) , we c h a r a c t e r i z e d (C(X)')a

0.

a n d (C(X)')a

in

We c a n do t h e same f o r s ( X ) , ,

C " ( X ) a , u s i n g o n l y C"(X)a.

S(X)a,

a n d Ro(X),.

Ba(X),,

(54.8) For e v e r y g€C"(X),,

the following are equivalent:

lo

gEs(XIa :

2'

g i s t h e d i f f e r e n c e o f two ( b o u n d e d ) l o w e r s e m i c o n -

tinuous functions.

S i n c e t h e Mll-isomorphism o f U(X) o n t o U(X)a i s a n i s o metry, t h i s gives, i n turn:

(54.9)

F o r e v e r y gEC"(X).

, the following are equivalent:

lo

gES(X),:

2'

g i s i n 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 ( b o u n d e d ) 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 s on X.

By a B a i r e f u n c t i o n on X , we mean o n e d e f i n e d i n t h e

282

Chapter 11

standard way in function theory, starting from continuous functions (that is, by (54.4), the elements of C(X)a).

Similarly

for a Borel function.

(54.10) For gEC"(X)a 1'

2

0

gEBa(X)a

, the following are equivalent:

:

g is a H a i r e function on X .

(54.11) For g E C " ( X ) a

, the following are equivalent:

'1

gEBo(X):

2'

g is a Borel function on X.

(54.8), (54.9), (54.10), and ( 5 4 . 1 1 ) each have a corollary paralleling (54.6).

We leave their formulations to the reader.

What about a l s o characterizing I J ( X ) a

in C"(X),

using only

We shall see below that this fails (54.20).

C(X),?

Given an

element of C"(X)cc, we cannot, with the above restriction, determine whether o r not it lies in

U(aa.

Before showing this, we have to look at U ( X ) (thus far, we have been considering U ( X ) a ) . Note that, by the I s o m o r p h i s m theorem,

(54.12)

n

C"(X)cc

U(X)

In c o n t r a s t t o t h i s (and u n l i k e t h e c a s e w i t h C ( X ) ) , always i n t c r s c c t s C"(X)

U

i n more t h a n 0 :

F o r e v e r y xoEX, t h e e l e m e n t f o o f C ' f [ X ) u

( 5 4 . 1 3 ) Thcorcm.

w h i c h h a s v a l u e 1 on x

0

and

I)

on t h e r e s t o f X i s a u s c e l e m e n t

( h e n c e l i e s i n U(X)).

--__ Proof.

fo

=

u.

Set A

{fEC(X)+I ( f , x o ) > l } and u

A i s f i l t e r i n g downward.

t i n u o u s on C " ( X ) (u,xo)

=

=

=

A A ; we show

Since each xiY

i s order con-

- and X i s a normal s p a c e - it f o l l o w s

1 and ( u , x )

0 for a l l

=

x # xo.

T h u s ua

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 u v a n i s h e s on

fo.

=

Cl(X),

We and

thus coincides with ua. C o n s i d e r U E C ' ( X ) ~ , a n d we c a n t a k e p > 0. that 0 <

E

E

such

< 1 / 2 ; we show ( u , ~ )< 2 ~ . ui/\x0 = 0 , h e n c e t h e r e

exist f,gEC(X)+, f (g,xo) <

Take

+

&

g

=

n(X),

such t h a t ( f , p ) <

Then ( f , x o > > (1

-

E),

SO

E:

and

(1/(1 -

~))feA.

I t follows

C o r o l l a r y 1. (54.14) -

c o ( X ) + c C(X)'.

Hence co(X) c s ( X )

T h i s f o l l o w s f r o m t h e a b o v e a n d t h e f a c t t h a t C(X)u i s a

C h a p t e r 11

284

norm c l o s e d wedge. Combining ( 5 4 . 1 4 ) w i t h ( 4 1 . 6 ) a n d t h e f a c t t h a t U(X) i s 0 - o r d e r c l o s e d , we h a v e

(54.15) Corollary 2 . .

C z x)

n

c l l ( ~c ) ~~ ( x ) .

Thus U(X) c o n t a i n s a 1 t h e e l e m e n t s o f C"(X)a w h i c h v a n i s h

o n a l l b u t a c o u n t a b l e number o f

X I S .

In addition:

( 5 4 . 1 6 ) C o r o l l a r y 3. a b l e number o f

Proof.

X I S ,

fa

I f fEU(X)

v a n i s h e s on a l l b u t a c o u n t -

t h e n fEC".(X)u.

a l s o v a n i s h e s on a l l b u t a c o u n t a b l e number o f

x's, s o b y t h e p r e c e d i n g r e m a r k , f a E U ( X ) .

i t f o l l o w s from

t h e Isomorphism theorem t h a t f = f a .

QED

The n e x t c o r o l l a r y s t a t e s t h a t t h e p r o p e r t y ( 4 0 . 1 ) o f C''(X)u h o l d s a l s o f o r U(X)u a n d f o r U(X)

n

C"(X)u.

(54.17) C o r o l l a r y 4 . F o r B c U ( X ) u a n d gEU(X)a, are equivalent: 1'

g = VB;

the following

285 '2 0

3

g

VB-in-U(X)u;

=

(g,x)

suphEB(h,x) for all xEX.

=

Similarly, with U(X)u

replaced by U(X)

_-Proof. We need only show that

2'

n

C"(X)u

implies .'3

This follows

easily from (54.13).

(54.18) Corollary 5. gEU(X)u

, ga

+

QED

For a bounded net {gal in U(X)u

g in U(X)u

placed by U(X)

implies g

a

+

and

g. Similarly with U(X)u re-

C''(X)u.

Combining (54.17) with the Isomorphism theorem we have the

(54.19) Theorem.

For A c U ( X ) and fEU(X), the following are

equivalent: '1

f

2'

(f,x)

=

VA-in-U(X); =

supkcA(k,x)

for all XEX.

We return to the question of characterizing U(X)u starting from C ( X ) u .

in C " ( X ) u

In line with the characterizations ob-

tained earlier in this section, the characterization should be that U(X)u

is the Dedekind closure of S ( X ) ~ (cf. (54.8) and

540), o r , equivalently, that U ( X ) u consists of those elements of

C ' l ( X ) u

which are each both the infimum o f some subset of

286

C h a p t e r 11

( C ( X ) u ) e (= ( C ( X ) R ) u b y ( 5 4 . 5 ) ) a n d t h e supremum o f some s u b s e t We show f i r s t t h a t t h i s D e d e k i n d

of (C(X)u)u (= ( C ( X ) u ) u ) .

c l o s u r e i s a l l o f Cll(X)u.

( 5 4 . 2 0 ) E v e r y fEC"-(X)u

a n d t h e supremum o f some s u b s e t o f (C(X) R

s u b s e t o f (C(X)')u

Proof. g,

i s s i m u l t a n e o u s l y t h e infimum o f some

For e a c h xEX, l e t

b e t h e e l e m e n t o f Cll(X)u w i t h v a l u e 1 on x a n d 0 e l s e w h e r e ,

and s e t u

X

=

uX i s a

(f,x)gx.

immediate t h a t f

=

VxExux.

ll(X)u - (1 - ( f , x ) ) g x .

USC

e l e m e n t (54.13), a n d i t i s

A g a i n , f o r e a c h xEX, s e t hx

Then h,E(C(X)')),

and f

=

=

AxEXhx. QED

We s h a l l s e e l a t e r t h a t U(X)u i s , i n g e n e r a l , a p r o p e r s u b s p a c e o f C"(X)u.

Thus t h e c o n j e c t u r e f a i l s , a n d we a p p a r e n t -

l y c a n n o t c h a r a c t e r i z e U(X)u s t a r t i n g f r o m C(X)u. B e f o r e c o n t i n u i n g t h e d i s c u s s i o n , we n o t e some c o n s e q u e n c e s of (54.20).

(54.21)

C o r o l l a r y 1.

s(X),,

S ( X ) u , a n d U(X)), a l l h a v e Ct'(X)a

f o r t h e i r Dedekind c l o s u r e s .

S i n c e Cll(X)u i s D e d e k i n d c o m p l e t e , t h e a b o v e a n d t h e Isomorphism theorem g i v e us

The D e d e k i n d c o m p l e t i o n s o f s ( X ) , S(X),

( 5 4 . 2 2 ) Theorem. a n d U(X)

can a l l be i d e n t i f i e d w i t h C"(X),.

Note t h a t , i n c o n t r a s t t o ( 5 3 . 2 0 ) , C ( X ) a closed.

Bo(X),

i s Dedekind

I n e f f e c t , i f f i s i n t h e Dedekind c l o s u r e o f C ( X ) a ,

t h e n i t i s e a s i l y s e e n t o b e c o n t i n u o u s on X , h e n c e l i e s i n C(XIa.

We e x a m i n e ( 5 4 . 2 0 )

g determines t h e s e t s A

=

S e t g,

morphism t h e o r e m , A < B. and ( g * ) ,

g

=

so gEU(X),

(s*1,

=

.

=

=

Consider gEC"(X)a

=

f

=

=

By t h e I s o -

g = AB,.

V A and g*

(VN,

= AB.

= g,

= VA,

Then g,

i g* -

and s i m i l a r l y

t h e n t h i s common e l e m e n t l i e s i n U(X),

,

that is, g

=

fa for

g*, as i s e a s i l y v e r i f i e d using t h e I s o -

Summing u p : a n e l e m e n t g o f C ' l ( X ) a

morphism t h e o r e m .

.

< g } and B = a -

And c o n v e r s e l y , i f g E U ( X ) ,

f E U ( X ) , t h e n g,

U(X),

((g*)=

= g*,

I f g,

f o r (g*),).

{ U € C ( X ) Iu ~

S o , b y ( 5 4 . 2 0 1 , VA,

3 g}.

{REC(X)'~~,

more c l o s e l y .

i f and o n l y i f g,

=

lies in

g*.

We w i l l s e e l a t e r t h a t g i v e n gEC"(X),

=

a"(X),

t h e n g,

*

and g * h a v e t h e f o l l o w i n g p r o p e r t y : F o r e v e r y p € C ' ( X ) + , ( g , p ) j*gdp and ( g , , p )

=

=

j * g d u , w h e r e I*gdu i s t h e u p p e r i n t e g r a l o f

g w i t h r e s p e c t t o p a n d /*gdu i s t h e l o w e r i n t e g r a l .

R e t u r n i n g t o o u r d i s c u s s i o n on n o t b e i n g a b l e t o c h a r a c t e r -

ize U(X)a

from C(X),,

t h e r e i s an a d d i t i o n a l message i n ( 5 4 . 2 0 ) :

While, by t h e Isomorphism t h e o r e m , w e can o b t a i n t h e i n t r i n s i c p r o p e r t i e s o f U(X) embedding o f U ( X )

by s t u d y i n g U ( X ) u i n C"(Xj

, we c a n n o t l e a r n a b o u t t h e

from t h a t of U ( X ) a

i n Cl'(X)u.

For

C h a p t e r 11

288

e x a m p l e , U(X) i s Dedekind c l o s e d i n C"(X) w h i l e U(X)u i s n o t Dedekind c l o s e d i n C"(X)u. L e t X be a r e a l i n -

We g i v e a s e c o n d , c l a s s i c a l , e x a m p l e . t e r v a l , p t h e Lebesgue m e a s u r e on X , a n d x t i o n of a l l f i n i t e subsets of X. under i n c l u s i o n .

For e a c h

=

{ Z } be t h e c o l l e c -

Note t h a t x i s a d i r e c t e d s e t

Z€z,

l e t f Z be t h e elemen t o f

C1l(X)a h a v i n g v a l u e 1 a t e a c h p o i n t o f Z and v a l u e 0 e l s e w h e r e Then { f , } i s an a s c e n d i n g n e t i n U(X) ( 5 4 . 1 3 ) , s u p Z ( f Z , x ) = 1 = ( I l ( X ) , x ) f o r a l l xEX. o r d e r converge t o l(X) (l(X),p)

=

1.

However,

If,} does n o t

since ( f , , ~ ) = 0 f o r a l l Z€x, while

( I f , } o r d e r c o n v e r g e s t o Il(X)=.)

A f i n a l p r o p e r t y o f U(X).

o u s o n Cll(X).

and we h a v e

Every p € C ' ( X ) i s o r d e r c o n t i n u -

S i n c e U(X) i s a - o r d e r c l o s e d i n C"(X), i t f o l l o w s

e v e r y pEC'(X) i s a - o r d e r c o n t i n u o u s on U(X), c o n s i d e r e d a R i e s z space i n i t s e l f .

I n g e n e r a l , a u€C'(X) i s n o t o r d e r c o n t i n u o u s

on U(X) ( c o n s i d e r e d a s a R i e s z s p a c e i n i t s e l f ] . We show t h a t t h e u ' s w h i c h a r e o r d e r c o n t i n u o u s on U(X) a r e p r e c i s e l y t h e e l e m e n t s o f C ' (X)u.

( 5 4 . 2 3 ) Theorem.

For uEC'-(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

lJEC'(X)a ;

2'

p i s o r d e r c o n t i n u o u s on

Proof.

U(X).

C o n s i d e r p E C ' ( X ) a , 1-1 > 0.

To show p i s o r d e r c o n -

t i n u o u s on U(X), i t i s enough t o show t h a t f a + O

i n U(X) i m p l i e s

0 = l i m (fa,p). a

By t h e Isomorphism t h e o r e m , ( f a ) a + O i n U(X)a,

Hence by ( 5 4 . 1 7 ) ,

(fcl)a+O.

it follows 0

Cll(X)a,

=

Since p i s order continuous on

l i m ( ( f ) ,p) a Cla

Thus 1'

lim(fN ,p).

=

i m p l i e s 2'.

,

Conversely, consider p € C ' ( X ) d o r d e r c o n t i n u o u s on U(X).

1-1 > 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'

C(X) i s Dedekind c o m p l e t e ;

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):

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 ;

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.

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 ;

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 ) <--by ( 2 4 . 7 ) ,

Tt(E)

Tt

El'.

E l .

E i s a b a n d o f E"

i s a band J of Cl(X).

Thus

( 5 2 3 ) , hence

Since F is separating

on E , Tt maps E o n e - o n e , h e n c e Tt I E i s an i s o m e t r i c R i e s z i s o morphism o f E o n t o J .

And s i n c e E i s s e p a r a t i n g on F , .J i s

s e p a r a t i n g on C ( X ) , h e n c e v a g u e l y d e n s e i n C t ( X ) . For s i m p l i t S E. Let I b e t h e band o f C''(X) c i t y d e n o t e T I E b y S: J
Then I __ St > E '

s an ME-isomorphism o f I o n t o

i t r e m a i n s t o show t h a t S t f I = T f f o r e v e r y f € C ( X ) . aEE, ( S t f I , a )

=

( f I , S a ) = ( f Sa)

=

E l ;

For e v e r y

( f , T t a ) = ( T f , a ) , s o we a r e

through. QE D

Combining t h i s t h e o r e m w i t h ( 5 7 . 5 ) , we o b t a i n t h e

(58.2)

C o r o l l a r y 1.

Under t h e h y p o t h e s e s o f t h e a b o v e t h e o r e m ,

t h e Dedekind c l o s u r e o f F c o n s i s t s o f t h o s e l i n e a r f u n c t i o n a l s on E w h i c h a r e c r ( E , F )

(58.3) Corollary 2.

ing are equivalent:

c o n t i n u o u s on K ( E ) .

Under t h e h y p o t h e s e s o f ( 5 8 . 1 ) , t h e f o l l o w -

Riemann S u b s p a c e o f a Band

'1

E'

309

F" ( w i t h t h e g i v e n i m b e d d i n g o f F i n E' t h e c a n o n -

=

i c a l imbedding o f F i n i t s b i d u a l ) ; 2'

E

3'

K(E)

=

Proof.

F'

(under t h e b i l i n e a r form

(

*

,

a

)

on E '

E);

x

i s ~ ( E , F )c o m p a c t .

Assume 1'

holds: E'

=

F"

=

(F')

'.

Since both E

and F ' a r e L - s p a c e s , i t f o l l o w s f r o m ( 2 3 . 5 ) t h a t E

=

F'.

=

(El)'

And s i n c e E i s c o m p l e t e l y d e t e r m i n e d b y i t s v a l u e s on F , i t i s e a s i l y v e r i f i e d t h a t t h e b i l i n e a r f o r m ( . , - ) on E ' restricted to F

x

course implies 3

0

E , when

x

E , g i v e s an isomorphism o f E o n t o F ' .

.

Assume 3'

~ ( J , C ( X ) ~ ) - c o m p a c th,e n c e J

holds. =

C'(X)

of

2'

is

Then i n ( 5 8 . 1 ) , K ( J ) ( 3 6 . 3 ) , hence I

=

C"(X).

I t f o l l o w s E ' = F".

QED

This gives us, i n turn:

(58.4)

Theorem.

The d u a l E ' o f a n L - s p a c e E i s t h e b i d u a l o f

an ME-space i f a n d o n l y i f E ' c o n t a i n s a n ME-subspace F s u c h that (i)

F i s s e p a r a t i n g on E , a n d

( i i ) K(E)

Remark.

i s a(E,F) compact.

By c o m b i n i n g t h e a b o v e w i t h D i x m i e r ' s t h e o r e m ,

we o b t a i n c o n d i t i o n s f o r a n ME-space t o b e t h e b i d u a l o f an ME-space w i t h o u t a s s u m i n g i n a d v a n c e t h a t i s i s a d u a l s p a c e .

Chapter 1 2

310

§ 5 9 . Examples o f Riemann s u b s p a c e s

A s e x a m p l e s , w e w i l l d e s c r i b e t h e Riemann s u b s p a c e s o f t h e f o l l o w i n g b a n d s o f Clf(X): a b a s i c band C"(X)

a U-band ( i n lJ' p a r t i c u l a r a u - b a n d o r a n 2 - b a n d ) , a n d t h e two p a r t i c u l a r b a n d s Crl(X) a n d Cl'(X)a. A b a s i c band I

f u r n i s h e s t h e example from which 1-I we h a v e t a k e n t h e name "Riemann s u b s p a c e " . We show i n t h e =

C'l(X)

n e x t c h a p t e r t h a t R(C"(X) j e c t i o n o f C"(X)

) i s t h e image u n d e r p r o j

LJ

!J

(the pro-

o n t o C'l(X) ) o f t h e s e t o f e l e m e n t s o f C " ( X )

v

w h i c h a r e "Riemann i n t e g r a b l e " w i t h r e s p e c t t o 1-1. For I R(C"(X))

=

=

C1'(X) i t s e l f , we a l r e a d y h a v e t h e a n s w e r ( 4 6 . 3 ) :

C(X).

Remark. -

We w i l l s e e i n t h e n e x t c h a p t e r t h a t t h i s

g e n e r a l i z e s t h e c l a s s i c a l t h e o r e m t h a t a f u n c t i o n on a r e a l i n t e r v a l i s Riemann 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 r e g u l a r measure - e q u i v a l e n t l y R i e m a n n - S t i e l t j e s 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 f u n c t i o n o f bounded v a r i a t i o n - i f and o n l y i f it i s continuous ( c f . 564). F o r I a U - b a n d , we h a v e a s i m p l e c h a r a c t e r i z a t i o n o f R ( 1 ) . I t s h o u l d be compared w i t h t h a t o f a g e n e r a l b a n d i n ( 5 7 . 5 ) .

( 5 9 . 1 ) Let I b e a U-band, w i t h d u a l b a n d J .

ment f o f 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 : 1'

fER(J);

2'

(i)

f E U ( X ) , and

( i i ) f i s c o n t i n u o u s on J

X.

Then f o r an e l e

Riemann S u b s p a c e o f a Rand 0

Proof.

Suppose 1

holds.

Then ( i ) f o l l o w s f r o m ( 5 7 . 6 )

and ( 5 5 . 4 ) , a n d ( i i ) f o l l o w s f r o m ( 5 7 . 5 ) . v e r s e l y suppose 2

0

311

Thus 2'

holds.

Con-

holds.

( * ) T h e r e e x i s t a u s c e l e m e n t u a n d a n Rsc e l e m e n t R s u c h t h a t =

(UIIa

( R I ) a = fa.

S i n c e f i s c o n t i n u o u s on . J n X ,

fa i s also.

Hence t h e r e

e x i s t an 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 h and a l o w e r s e m i c o n t i n u -

o u s o n e g on X w h i c h c o i n c i d e w i t h f f o r some u s c e l e m e n t u , a n d g = R a

n

v a l u e s on <J

n

X),

X ; s o uI a

( u ~ =) (~k I ) a

=

h

=

ua

€ o r some k s c e l e m e n t .t.

Then u and R c o i n c i d e w i t h f on J I a v a n i s h on X \ ( J

on J n X ( 5 7 . 1 ) .

a

X.

Since the elements of

they a r e completely determined by t h e i r =

=

fa.

This can be w r i t t e n

RIU

f a , and w e have ( * ) .

Now u I , k I E U ( X )

( ( 5 5 . 4 ) a g a i n ) , s o from ( * ) t h e I s o -

morphism t h e o r e m , a n d Z o ( i )

,

uI

= kI

= f.

Thus fER(.I)

. QED

For emphasis, w e s i n g l e o u t t h e f o l l o w i n g , c o n t a i n e d i n t h e above.

(59.2) C o r o l l a r y .

F o r a U-band I , R ( I ) c U(X).

N o t e t h a t f r o m ( 5 1 . 1 0 ) a n d ( 4 6 . 3 ) , we h a v e

31 2

Chapter 1 2

( 5 9 . 3 ) For a u-band I , R(1)

=

C(X),.

T h i s does n o t h o l d f o r an R - b a n d : Take f o r X terval [0,1].

Let G be t h e ( o n e - d i m e n s i o n a l ) band o f C ' ( X )

g e n e r a t e d by t h e l e f t e n d p o i n t x C"(X).

the real in-

0 o f X , and s e t I = (;I i n

=

Now l e t u be t h e u s c e l e m e n t d e f i n e d by u(0)

=

1,

u(x)

=

sin(l/x)

0 < x < 1,

and R be t h e R s c e l e m e n t d e f i n e d by R(0) = -1,

L(x)

=

0 < x < 1.

sin(l/x)

(Remember t h a t an e l e m e n t o f U ( X ) by i t s v a l u e s on X.)

Set f

=

can be c o m p l e t e l y s p e c i f i e d

(1/2)(u + L ) .

Then fEU(X)

l i e s i n I , s i n c e i t v a n i s h e s on x = 0 , hence on G .

and

Moreover,

i t i s c o n t i n u o u s on t h e h a l f - o p e n i n t e r v a l ( 0 , 1 ] , which i s

Gdn X.

I t f o l l o w s from ( 5 9 . 1 ) t h a t f E R ( 1 ) .

Suppose f = gI f o r some gEC(X). (0,1],

We show f(? C(X),.

Then g c o i n c i d e s w i t h f on

s i n c e no e l e m e n t o f C(X) can have t h e v a l u e s s i n ( l / x )

on ( 0 , 1 ] , we have a c o n t r a d i c t i o n .

For I = C " ( X ) u ,

we a l r e a d y a l s o have R(1) ( c f . t h e comment

f o l l o w i n g ( 5 4 . 2 2 ) ) : R(C"(X)u) = C ( X ) u .

More g e n e r a l l y :

( 5 9 . 4 ) For e v e r y band I o f Ct'(X)u, R(1) c o n s i s t s o f t h o s e e l e ments o f I which a r e c o n t i n u o u s on t h e s e a t Q ( 1 ) o f I .

Riemann Subspace of a Band

-~ Proof.

some f € I }

31 3

Recall that Q ( 1 ) is the set {xEXl (f,x) # 0 for

( § 5 5 ) , so I = k " ( Q ( 1 ) ) .

The proposition then follows

from the fact that a function on any Q

c

X is continuous on Q

if and only if it is the restriction to Q of some lowersemicontinuous funcion on X and of some uppersemicontinuous function on X. QE D

We de not have a description of R(C"(X)d)

(but cf. (65.5)).

We give an example to show that it is properly larger than C(X),

.

( S o , unlike C(X)a,

C(X)d

is not Dedekind closed.)

First, a

(59.5) -__ Lemma.

If X is dense-in-itself, then for every usc ele-

ment u , U€C"(X)~ implies u > 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 <-O .

,

hence u ~ € R ( C " ( X ) ~ ) .

= Rd,

s o , b y t h e Lemma, f

-

5 (f,x) 5

( f - R ) d = 0, so

R > 0.

Similarly,

CIIAL'TER

13

RTEbIANN INTEGRABILITY

C o n s i d e r t h e c l a s s i c a l Lebesgue t h e o r e m : a bounded f u n c t i o n o n a r e a l i n t e r v a l i s Ricmann i n t e g r a b l c i f a n d o n l y i f i t s s e t o f p o i n t s o f d i s c o n t i n u i t y h a s (Lebesgue) measure z e r o . While q u i t e a t t r a c t i v e , t h e t h e o r e m h a s an u n s a t i s f a c t o r y f e a t u r e : i t r e q u i r e s f i r s t d e f i n i n g "Lebesgue measure z e r o " ,

a

c o n c c p t o u t s i d e t h e domain o f Riemann i n t e g r a t i o n . T h i s u n s a t i s f y i n g f e a t u r e can be e l i m i n a t e d . s u b j e c t o f m e a s u r e on o u r r e a l i n t e r v a l . Lebcsgue m e a s ure.

I-lerc m e a s u r e means

C o r r e s p o n d i n g t o "measure"

i n t h e same way

t h a t "Riemann i n t e g r a l " c o r r e s p o n d s t o " L e b e s g u e have t h e concept of " c o n t c n t " .

Consider t h e

integral",

we

The t h e o r e m f o r s e t s c o r r e s p o n d -

i n g t o the Lebesgue theorem above r e a d s : a s e t h a s c o n t e n t ( i s q u a d r a b l e ) i f and o n l y i f i t s f r o n t i e r h a s measure z e r o .

But

t h e f r o n t i e r o f a s e t i s c l o s e d , and f o r a c l o s e d s e t , h a v i n g measure z e r o i s e q u i v a l e n t t o having c o n t e n t z e r o .

Thus t h e

theorem can he s t a t e d : a s e t h a s c o n t e n t i f and o n l y i f i t s f r o n t i e r has content zero. The way t o e l i m i n a t e m e a s u r e f r o m L c b e s g u e ' s t h e o r e m i s

now c l e a r .

We r e p l a c e t h e s e t o f p o i n t s o f d i s c o n t i n u i t y o f a

f u n c t i o n f by t h e s a l t u s f u n c t i o n o f f , which g i v e s t h e p o i n t s

of discontinuity.

The s a l t u s f u n c t i o n i s a l w a y s p o s i t i v e a n d

uppersemicontinuous,

and f o r s u c h a f u n c t i o n , h a v i n g Lebesgue 31 5

Chapter 1 3

316

i n t e g r a l z e r o i s e q u i v a l e n t t o h a v i n g Riemann i n t e g r a l z e r o . Thus t h e L e b e s g u e t h e o r e m b e c o m e s : a bounded f u n c t i o n i s Riemann i n t e g r a b l e i f a n d o n l y i f i t s s a l t u s f u n c t i o n h a s Riemann i n t e g r a l zero. T h i s w i l l b e o u r a p p r o a c h t o Riemann i n t e g r a b i l i t y , u s i n g t h e f a c t t h a t f o r an element f o f C " ( X ) ,

6(f) is the gencraliza-

tion of the saltus function of a function.

560. Riemann i n t e g r a b l e e l e m e n t s

We f i r s t d e f i n e Riemann i n t e g r a b i l i t y w i t h r e s p e c t t o a band J o f C l ( X ) .

I n 5 6 1 , we e x a m i n e i t w i t h r e s p e c t t o a

s i n g l e e l e m e n t 1-1 o f C ' ( X ) . Given a b a n d J o f C l ( X ) , an

e l e m e n t f o f C"(X) w i l l b e

s a i d t o b e Riemann i n t e g r a b l e w i t h r e s p e c t t o J i f 6 ( f ) E t J L . We w i l l d e n o t e t h e s e t o f s u c h e l e m e n t s b y

B(J).

Thus X ( J )

=

6 - l (.JL). From ( S O - l l ) , we h a v e :

(60.1)

F o r e v e r y band J o f C ' (X),

3 (J)

(i)

i s an Ma-subspace o f C"(X),

(ii)

c o n t a i n s C(X),

( i i i ) i s closed under t h e operations u ( - ) , a ( - ) ,

a(.).

I t i s e a s y t o s e e from t h e d e f i n i t i o n t h a t i f f C % ( J ) and

Riemann I n t e g r a b i l i t y

a(f) 5

g 5 u ( f ) , t h e n gE.2 (J).

31 7

A c t u a l l y , from ( 5 0 . 1 3 1 , we have

a much s t r o n g e r r e s u l t :

(60.2)

If A,B c X(J), A

In p a r t i c u l a r ,

a(J)

5 f 5 B , and (AA

- VB)EJL,

then f € x ( J ) .

i s Dedekind c l o s e d .

561.

u-Riemann i n t e g r a b i l i t y

Given p E C ' ( X ) , an e l e m e n t f o f C''(X) w i l l be s a i d t o be Riemann i n t e g r a b l e w i t h r e s p e c t t o IJ,o r 1-1-Riemann i n t e g r a b l e , if ( 6 ( f ) ,

14) =

0.

( 6 1 . 1 ) Given p E C ' ( X ) , t h e n f o r e v e r y f E C " ( X ) , t h e f o l l o w i n g a r e equivalent :

'1

f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o 1-1;

2'

f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o C ' ( X )

Fi'

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 a b l e f a c t t h a t f o r gEC"(X)+, g E ( C J ( X ) I . I y i f and o n l y i f ( 8 , 11-11)

=

0.

Thus t h e s e t o f 1-1-Riemann i n t e g r a b l e e l e m e n t s o f C"(X) precisely by

a (1-1).

a(C'(X) ) .

u

is

We w i l l a l s o d e n o t e t h i s (MI-)s u b s p a c e

Some c h a r a c t e r i z a t i o n s o f

2 (J):

31 8

Chapter 13

( 6 1 . 2 ) Given a b a n d J o f C l ( X ) , t h e n f o r f E C " ( X ) , t h e following are equivalent: 1'

f i s Riemann i n t e g r a b l e w i t h r e s p e c t t o J ;

2'

f i s Riemann 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 u E L J :

3'

t h e r e e x i s t a u s c e l e m e n t u a n d an k s c e l e m e n t a. s u c h

that R < f 5 u and u - L E J ' :

'4

( ~ ( f ) , u )= ( f , p > = ( 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
t h e r e e x i s t s u b s e t s A , B o f C(X) s u c h t h a t A < f < B and ( f , ~ = )

'4

SUP

€5A

(g,p)

=

inf ( ~ , L O ; hEB .

t h e r e e x i s t s e q u e n c e s { g n l , { h n } i n C(X) s u c h t h a t

The p r o o f i s s t r a i g h t f o r w a r d .

319

Riemann I n t e g r a b i l i t y

I n t h e c l a s s i c a l t h e o r y o f Riemann i n t e g r a t i o n on a E u c l i d e a n s p a c e , Rieman i n t e g r a b i l i t y i s d e f i n e d b y o n e o f t h e properties 2

0

,

3 O , o r 4'

above, but using a d i s t i n g u i s h e d

family of "step functions" i n place of continuous functions. Among t h e p r o p e r t i e s p o s s e s s e d b y t h e s e s t e p f u n c t i o n s a r e t h e following.

Each s t e p f u n c t i o n i s a f i n i t e l i n e a r c o m b i n a t i o n

(I)

of f u n c t i o n s which can be o b t a i n e d as s p e c t r a l e l e m e n t s o f continuous functions ( c f . 517). ( 1 1 ) Each s t e p f u n c t i o n i s Riemann i n t e g r a b l e .

In ( 6 1 . 5 ) we generalize the classical procedure t o t h e p r e s e n t c a s e i n t h e f o r m o f a c o n d i t i o n f o r p-Riemann i n t e g r a b i l i t y i n terms of s p e c t r a l elements o f continuous f u n c t i o n s . F i r s t , h o w e v e r , we m u s t a n s w e r t h e f o l l o w i n g q u e s t i o n .

Are t h e

s p e c t r a l e l e m e n t s o f c o n t i n u o u s f u n c t i o n s Riemann i n t e g r a b l e ( b y o u r d e f i n i t i o n ) ; a n d i f n o t , a r e s u f f i c i e n t l y many Riemann integrable t o enable us t o carry out the classical process? The q u e s t i o n was a n s w e r e d b y Loomis

[32].

Indeed, he proved a

more g e n e r a l r e s u l t , w h i c h r e d u c e s , i n o u r c o n t e x t , t o t h e

f o 1l o w i n g :

(61.4) Theorem (Loomis).

Given p E C ' { X ) ,

then f o r every fEC"(X),

the following are equivalent:

'1

f i s p-Riemann i n t e g r a b l e ;

2'

a l l b u t a c o u n t a b l e number o f t h e s p e c t r a l e l e m e n t s o f

f a r e p-Riemann i n t e g r a b l e .

Chapter 1 3

320

Proof.

We c a n assume p > 0 , IIui1

=

1, and 0 < f < n(X).

F o r t h e r e m a i n d e r o f t h e p r o o f , we w i l l d e n o t e n(X)

s i m p l y b y I]

and e a c h s p e c t r a l e l e m e n t e f ( X ) s i m p l y b y e ( 1 ) . Assume 1' h o l d s . (6(e(x)),p) e s t a b l i s h 2'.

2

E

We show t h a t f o r e v e r y E > 0 ,

f o r o n l y a f i n i t e number o f A ' s , w h i c h w i l l S p e c i f i c a l l y , we p r o v e t h e

Then

We show t h e e q u i v a l e n t i n e q u a l i t y

Note f i r s t t h a t f A X n n - f A A n - l n we w o r k w i t h t h i s l a t t e r f o r m .

= (fAXnn -

An-ln)'

;

Riemann I n t e g r a b i l i t y

This i s t h e f i r s t i n e q u a l i t y i n ( i i ) .

T h i s i s t h e second i n e q u a l i t y i n ( i i ) , s o we have ( i i ) , and

with it, ( i ) .

I t follows f r o m ( i ) t h a t f o r n = l,.-.,m,

321

Chapter 1 3

32 2

(Here we have used t h e a s s u m p t i o n t h a t f - hence e a c h g n i s Riemann i n t e g r a b l e ; c f . 4'

i n (61.2).)

Then

Since l E = l ( 6 ( e ( ~ n ) ) , 2 ~ )m E , t h i s g i v e s u s m < 1 / ~ and , comp l e t e s t h e p r o o f o f t h e Lemma.

Now assume 2'

holds.

I t i s c l e a r t h a t t h e proof of

F r e u d e n t h a l ' s s p e c t r a l theorem (17.10) can be m o d i f i e d ( f o r o u r f ) s o t h a t t h e s p e c t r a l e l e m e n t s a r e u-Riemann i n t e g r a b l e . Thus f i s i n t h e norm c l o s u r e o f j ; ! ( u ) .

Since

x(u)

i s norm

c l o s e d , f E j;! ( p ) .

QED

We can now g i v e o u r g e n e r a l i z a t i o n o f t h e c l a s s i c a l cond i t i o n f o r Riemann i n t e g r a b i l i t y .

( 6 1 . 5 ) Given p E C ' ( X ) + , t h e n f o r f € C t t ( X ) , t h e f o l l o w i n g a r e equivalent:

' 1

f i s u-Riemann i n t e g r a b l e ;

2'

t h e r e e x i s t s e q u e n c e s { g n } , {h,}

satisfying:

Riemann I n t e g r a b i l i t y

323

t h e g n ' s and h n ' s a r e l i n e a r combinations o f

(i)

u-

Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s

of C ( X ) ; gl 5 g 2

(ii)

2 * *-* f<-< * . * I

( i i i ) supn(gn,v)

Proof.

(f,v)

S u p p o s e 1' h o l d s .

t o s a t i s f y 4'

(*)

=

h2 < h1 , =

infn(hn,p).

Choose s e q u e n c e s { g ; } , { h , l }

of (61.3).

For e a c h n , t h e r e e x i s t g n , h n , l i n e a r combinations o f p -

Riemann i n t e g r a b l e s p e c t r a l e l e m e n t s o f e l e m e n t s o f C ( X ) , s u c h t h a t gn 5 g;

,

hn

2

hA

and

[ I g n - gA[19 IIhn - hA[I

5

l/n.

I n e f f e c t , gn c a n b e o b t a i n e d b y t h e m o d i f i c a t i o n o f t h e proof of (17.10)

p o i n t e d o u t a t t h e end of t h e p r e c e d i n g p r o o f ,

a n d s i m i l a r l y f o r hn ( t h e a p p r o x i m a t i n g s t e p e l e m e n t s i n ( 1 7 . 1 0 ) can b e chosen above t h e g i v e n a ) . Applying (17.12) and ( 6 0 . 1 ) , w e can assume, i n a d d i t i o n , t h a t t h e s e q u e n c e {g,} descending.

They t h u s s a t i s f y ( i ) a n d ( i i ) o f 2'.

f o l l o w s f r o m 4' T h a t 2'

i s a s c e n d i n g and t h e sequence {hn} i s

i n ( 6 1 . 3 ) a n d t h e norm c o n t i n u i t y o f

i m p l i e s 1'

f o l l o w s f r o m 5'

(iii) p.

i n (61.3). QED

Remark. p-Riemann i n t e g r a b i l i t y i n o r d i n a r y a n a l y s i s i s d e f i n e d f o r f u n c t i o n s on X C''(X)a.

- i n our context, f o r elements of

I t i s easy t o v e r i f y t h a t the space of functions

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

Chapter 13

324

2 (u),

precisely

562. Riemann n e g l i g i b l e e l e m e n t s

We r e t u r n t o a g e n e r a l band J o f C ' ( X ) . R(J)

n J'

We w i l l c a l l

t h e s e t o f Riemann n e g l i g i b l e ( w i t h r e s p e c t t o J )

e l e m e n t s , an d w e w i l l d e n o t e i t b y f X ( J 1 . w i l l also write

,TX

I f J = C'(X )

I.r'

we

(1-1).

( 6 2 . 1 ) Given a b a n d J o f C ' ( X ) , t h e n f o r f € C " ( X ) , t h e f o l l o w i n g are equivalent: 1'

fCh%(J);

2'

u ( f ) ,k(f)EJ1:

'3

u(/fl)EJ'.

S u p p o s e 1' h o l d s , t h a t i s , 6 ( f ) E J '

Proof. f - 6(f), f

w e h a v e 2'. 3'

holds.

+

6(f)EJL. T h a t 2'

h'?, (J)

Then

Since f - 6cf) < tcf) < u(f) < f + 6(f),

i m p l i e s '3

f o l l o w s from (49.12).

Th en, a g a i n by ( 4 9 . 1 2 1 ,

u ( f ) , t ( f ) E J L , hence G(f)EJ'. fE

and fEJ'.

(u(f)I,

T h a t fEJ'

Suppose

l t ( f ) IEJL, hence

is trivial.

Thus

. QED

(62.2)

C o r o l l a r y 1.

Given a b a n d J o f C ' ( X ) ,

t h e n f o r fEC"(X),

Riemann Integrability fE Jvx (J) if and only if

1 f( -<

325

u for some usc element u in J L .

I t follows , f $ ( J ) is the Riesz ideal of C " ( X )

generated by the

usc elements in (J'),.

Proof.

Suppose g € l X ( J ) and 0 5 If1 5 /gl. Then u(lf/) 2

~ ( / g / ) .Since,

by '3

hence, again b y '3 ideal.

above,u(jgj)EJL, it follows u(lfl)EJL.

above, fcJY*x(J).

Thus J v x ( J ) is a Riesz

The rest of the Corollary is clear.

Remark. Note that, since h?$?(J)= X ( J )

n

QED JL,

,qx(J) is norm

closed.

It follows from (62.1) that Riemann negligibility can be defined first, and then used to define Riemann integrability. Specifically, define g to be Riemann negligible if u(lgl)EJL, then (since 6(f) is a positive usc element) define f to be Riemann integrable if 6 ( f ) is Riemann negligible.

(As

we

pointed out in the introduction to the present chapter, this eliminates the mention of Lebesgue negligibility in Lebesgue's characterization of Riemann integrability.

By definition X ( J )

=

6-'(JL).

Contained in the above dis-

cussion is the fact that actually X ( J )

=

&-lWX(J)).

This

gives us:

(62.3) Given a band J of C' (X), let H be the Riesz ideal

C h a p t e r 15

326

g e n e r a t e d Oy Sa(J)). Then (i)

S-'(H)

=

g(H

1

) , a n d H i s t h e s m a l l e s t Riesz i d e a l of

Clt(X) w i t h t h i s p r o p e r t y ; ( i i ) jZ(HI)

=

J ( J ) , and H

1

i s t h e l a r g e s t b a n d o f C1(X )

with t h i s property

The b a n d H 1 g e n e r a t e d by S ( ( R ( J ) )

(that is, the order clos-

u r e o f t h e above H) i s o f c o u r s e t h e smallest band o f C"(X) s u c h t h a t S-'(H1) t h a n J'

=

T h i s band may b e s t r i c t l y s m a l l e r

X(J).

- equivalently,

H

I

may b e s t r i c t l y l a r g e r t h a n J .

an example, l e t X b e a f i n i t e s e t .

Then Cll(X)

=

As

C(X), s o

S ( C t l ( X ) ) = 0 , s o f o r e v e r y band J o f Cll(X), H = 0 , s o H1

=

0.

We e m p h a s i z e ( i i ) a b o v e . : f o r e v e r y band J o f C l ( X ) , t h e r e

i s a l a r g e s t b a n d ( i t c o n t a i n s J ) w i t h t h e same Riemann i n t e g r a b l e e l e m e n t s i n C"(X).

We c l o s e t h i s 9 w i t h a n i n t e r e s t i n g r e s u l t .

(62.4)

Giv e n a band J o f C t ( X ) , e v e r y Riemann i n t e g r a b l e e l e -

ment o f C1l(X)d i s Riemann n e g l i g i b l e :

g(J) n

Proof.

C"(X)d

c&?,(J).

Consider f €$(J)

we show f E J'. a ( f ) = 6 ( f ) EJ'.

By ( 5 4 . 1 2 ) ,

n C 'l (X )d,

L(f)

=

and we c a n t a k e f > 0;

0 , hence u ( f )

Thus 0 < f < u ( f ) € J',

=

u(f) -

whence f EJ'. QED

327

Riemann I n t e g r a b i l i t y

563. Riemann i n t e g r a b i l i t y and Riemann s u b s p a c e s

Let J be a band o f C'(X) Then

J(J),

=

a n d I its d u a l band i n

Cll(X).

J ( J ) / J ~ % ( J ) A. r e a s o n a b l e e x p e c t a t i o n i s t h a t I n ( 6 3 . 2 ) , we l i s t

R ( J ) , = R(1) t h e Riemann s u b s p a c e o f I . some common b a n d s f o r which t h i s i s t r u e . p r i n c i p a l b a n d s {C'(X)

Among them a r e t h e

l ~ € C ' ( X ) l , which a r e t h e o n e s d e a l t w i t h

LJ i n ordinary integration theory.

We w i l l s e e below, however,

t h a t f o r a g e n e r a l band o f C'(X),

t h e e x p e c t a t i o n need n o t be

born o u t . First,we record:

( 6 3 . 1 ) For e v e r y band J o f C'(X),

w i t h d u a l band I ,

nJ),= R ( I ) Proof. f - R(f)EJ'.

Suppose f E J ( J ) .

Then 6 ( f ) E J ' ,

I t follows a ( f ) , = f I

=

so u ( f ) - f ,

u ( f ) I , whence f I E R ( I )

(57.6). QED

( 6 3 . 2 ) L e t J b e a band of Cl(X),

w i t h d u a l band I .

f i e s any o f t h e f o l l o w i n g c o n d i t i o n s , t h e n X ( J ) ,

If J satis-

= R(1):

( i ) J i s a p r i n c i p a l band; (ii) J i s vaguely c l o s e d - e q u i v a l e n t l y , I i s a u-band;

Chapter 1 3

328

( i i i ) Jd i s v a g u e l y c l o s e d - e q u i v a l e n t l y , I i s a n R-band; (iv)

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 ,

U(1)

=

Cll(X).

I n e a c h c a s e , i t o n l y r e m a i n s t o show t h a t

Proof. R(1) c a ( J ) , .

Assume ( i ) h o l d s , t h a t i s , J

=

C'(X)

f o r some u E C ' ( X ) .

I-r

By ( 2 7 . 4 ) , t h e r e e x i s t s e q u e n c e s { ( g ) I , n P I ( h n ) l - r l i n C(X) s u c h t h a t ( g ) + f , ( h ) + f . LJ n1-1 n1-1

Consider f E R ( 1 ) .

To show t h i s , we r e p l a c e Cgn3,thn3 by s e q u e n c e s

C1711> k i t h t h e d e s i r e d p r o p e r t y . -

g1

h e prmceed i n d u c t i l e l y .

Set

=

gl,

then s e t hl

= hlvgl.

We h a v e (gl)l-l

= (gl)p,

and (E1I-r 1 = (hlll-lv(Fl)l-l = ( h l ) l - l v ( g l ) u = ( h l l V .

-

-

g l , - ..,gm, El,..-,En

Ti

{En},

< * - *

n Set

have been chosen s o t h a t

of course,

Assume

El z - -+ -gn* <<

5Kl a n d (Ek)l-l = =

( g k ) l - l , (Ek)l-l = ( h ) ( k = l , . . . , n ) k1-1 (gn+lVzn)AEn. C o n t i n u i n g i n t h i s f a s h i o n , we

f -<

.

obtain (*), N o w s e t R = Vn g n' u = An hn '

T h e n R-< u a n d R = f = u . 1-1 1-1

To e s t a b l i s h t h a t f E J ? j ( J ) I , we n e e d o n l y show t h a t uE%'(J). T h i s f o l l o w s from u - I! EJ'

( s i n c e (u

-

a) I

=

0 ) a n d 3'

in

(61.2). Assume ( i i ) h o l d s .

Then by ( 5 9 . 5 ) a n d ( 6 0

C(X), c 2 ( J ) , . Assume ( i i i ) h o l d s , a n d c o n s i d e r f E R ( 1 ) .

We c a n assume

Riemann I n t e g r a b i l i t y

We show t h a t i n t h i s c a s e , n o t o n l y i s f i n

0 < f < ll(X)I.

but is actually i n

je(J),,

329

J ( J ) . By ( 5 7 . 6 ) , t h e r e e x i s t a n Rsc 2 f 5 uI.

element k and a u s c element u such t h a t k I (uV0)A

l l ( X ) ; t h e n (u,) I = f a l s o .

S e t El

= ( k V 0 ) A U ( X ) I (by

a s s u m p t i o n , l l ( X ) l i s a n Rsc e l e m e n t ) ; t h e n ( R 1 ) I

R I E I , s o we a c t u a l l y have R1 We t h u s have R1 - R1 € I d =

, ' J

=

I t follows

hence f € X ( J ) .

Assume ( i v ) h o l d s .

Then i t f o l l o w s e a s i l y from t h e f a c t

t h a t p r o j , mgps C ( X ) MU-isomorph c a l l y o n t o C(X), again, R(I) c

But

f also.

= f.

f < u l , w i t h ( u , ) ~= f . I 1

=

S e t u1 =

XCJ)

(not j u s t

B(J

(52.3) that,

, I)

QED

The r e m a i n d e r o f t h i s s e c t i - n i s d e v o t e d t o a n example t o show t h a t f o r a g e n e r a l band J o f C l ( X ) , w i t h d u a l band I ,

X(J), #

For c l a r i t y , we f i r s t c o n s t r u c t a n example i n

R(1).

o r d i n a r y f u n c t i o n t h e o r y , t h e n show t h a t we a c t u a l l y h a v e a n example i n C ' l ( X ) . Let w = t h e f i r s t i n f i n i t e o r d i n a l ; o1 = t h e f i r s t u n c o u n t a b l e o r d i n a l ;

N

=

fi

= the s e t of ordinals

t h e s e t o f o r d i n a l s { B 16 < w ) ;

{ B I B 5 01;

IN1 = t h e s e t o f o r d i n a l s t a l a < w l l ;

-

N1

=

t h e s e t o f o r d i n a l s Icxlci

We e n d o w m and

ml

5

with t h e order topology.

p a c t Hausdorff. Let X =

N1

Z =

m1

Q

WEjl

=

all.

fl

x x

X

(product topology);

w) IJ (wl W)

x

m);

U ( ~ 1W)

They a r e t h e n com-

330

So

Chapter 13

Z is a closed subset of X, and Q c Z. Consider the function f on Q with value 1 on the points of and 0 on the points of w1

N1x w

x

IN.

istic function of the closed set W, istic function of the open set X \ ( w ,

Let u be the characterw,

x x

and

m).

il

the character-

Then u is upper-

semicontinuous, R is lower semicontinuous, and

Note that we do not have

2 u ; we show there exist no

R

R lower-

semicontinuous and u uppersemicontinuous such that R < u and ( * ) is satisfied.

u(wl,n)

=

Suppose such a pair R,u exists.

0 for all n E N .

Now

Since u is uppersemicontinuous, i t

follows that limsupa+w u(a,n) < 0 for all n E N . Hence, since 1 N1 is uncountable, there exists a. such u(a,n) < 0 for all c1L c1 0

and nEl?i.

By assumption, R a o , which conthis gives in turn that R ( a , w ) -

tradicts the definition of R . The above construction has been carried out in identify this with C r r ( X ) a . (37.12).

X is dispersed, so C " ( X )

=

It f o l l o w s u above is a usc element of C " ( X )

an Rsc element. R l Q = RI,

Let I be the band R"(Q)

so by ( * ) , f E R ( I ) ,

.

Then ulQ

=

Cll(X). and k u I and

but it is easily shown, using the

above argument, that there is no gEC"(X) such that ( k ( g ) ) I (u(g)),

= f.

Now

k"(X).

=

Riemann I n t e g r a b i l i t y

331

5 6 4 . Examples

Consider t h e c a s e J

=

C(X)

u'

and we c a n t a k e p

2

We

0.

have no s p e c i a l comment t o make a b o u t x ( p ) beyond t h e s t a t e m e n t i n ( 6 3 . 2 ) , b u t we t a k e t h i s o c c a s i o n t o remark on s t a n d a r d We r e l a t e i t t o o u r

Riemann i n t e g r a t i o n i n f u n c t i o n t h e o r y .

a p p r o a c h by - a s u s u a l - i d e n t i f y i n g L"(X)

with CII(X)u

" f u n c t i o n " will mean an e l e m e n t o f C l f ( X ) a ) .

I t i s e a s y t o show

(SO

t h a t t h e s e t o f f u n c t i o n s which a r e Riemann i n t e g r a b l e ( w i t h respect to

u)

i n the standard sense i s precisely

X ( U ) ~ ,and

the

s e t o f t h o s e which a r e Riemann n e g l i g i b l e i n t h e s t a n d a r d s e n s e is precisely

y x ( ~ =)$ ~ X(p)

CII(X)u.

And r e m a r k a b l y , by ( 6 2 . 4 ) .

Thus, by ( 6 3 , 2 ( i ) ) , t h e s p a c e o f Riemann i n t e g r a b l e f u n c t i o n s modulo t h e s p a c e o f Riemann n e g l i g i b l e o n e s c a n be i d e n t i f i e d w i t h t h e Riemann s u b s p a c e

) - otherwise s t a t e d , LJ This i s t h e Caratheodory

R(C"(X)

w i t h t h e Riemann s u b s p a c e o f L m ( p ) . theorem. We l o o k a t o t h e r e x a m p l e s .

(64.2)

X ( C ' (X)) = C(x) = .R(C"(X)).

332

Chapter 1 3 Proof.

( C * ( X ) ) ' = 0 , s o I ( c ~ ( x ) )= 6

-1

(0) = c(x).

We

have a l r e a d y n o t e d t h e s e c o n d e q u a l i t y i n 5 5 9 ( c f . ( 4 6 . 3 ) ) . QED

We have remarked e a r l i e r ( 5 5 9 ) t h a t t h i s g e n e r a l i z e s t h e c l a s s i c a l theorem f o r a r e a l i n t e r v a l : a f u n c t i o n f i s RiemannS t i e l t j e s 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 f u n c t i o n o f bounded v a r i a t i o n i f and o n l y i f f i s c o n t i n u o u s .

The above i s a s t r o n g s t a t e m e n t .

For f EC"(X) t o be

Riemann 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 a t o m i c Radon m e a s u r e , f must l i e i n C(X); i t i s n o t enough t h a t i t be c o n t i n u o u s on X .

The o n l y r e s u l t we have t o o f f e r a t p r e s e n t on x ( C l ( X ) d ) i s a c h a r a c t e r i z a t i o n due t o C . Goffman f o r t h e c a s e where X i s

m e t r i c and d e n s e - i n - i t s e l f ( o r a l c o m m u n i c a t i o n ) .

( 6 4 . 4 ) Lemma.

First, a

I f X i s a compact m e t r i c s p a c e , a p o s i t i v e u s c

e l e m e n t u l i e s i n Cii(X)u i f and o n l y i f ( u , x ) = 0 f o r a l l b u t

a c o u n t a b l e number of

XIS.

Riemann I n t e g r a b il i t y

Proof.

333

The s u f f i c i e n c y f o l l o w s from ( 5 4 . 1 6 ) .

Comversely,

s u p p o s e U E C " ( X ) ~and t h a t ( u , x ) > 0 f o r a n u n c o u n t a b l e number Then t h e r e e x i s t s 1 > 0 and a c l o s e d u n c o u n t a b l e s u b -

of x ' s .

> 1 €or a l l xEZ 0'

s e t Zo o f X ,

such t h a t ( u , x )

take X

S i n c e X i s compact m e t r i c , Z o c o n t a i n s a c l o s e d

=

1.

d e n s e - i n - i t s e l f subset 2.

For s i m p l i c i t y ,

L e t e be t h e c h a r a c t e r i s t i c e l e m e n t

Then ( e , x ) < ( u , x ) f o r a l l x , s o t h e Isomorphism theorem

of Z .

gives us e < u. L e t I be t h e band o f C"(X) g e n e r a t e d by e , and J i t s d u a l band i n C'(X).

Then J 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 ( c f . (51.10) and ( 3 6 . 7 ) ) , h e n c e , b y ( 3 7 . 1 0 ) , c o n t a i n s a d i f f u s e measure ~ > 0 . But t h e n ( e , u ) > 0 , s o ( u , ~ )> 0 , c o n t r a d i c t i n g u EC'l(X)u. QED

S i n c e f o r e v e r y f EC"(X), 6 ( f ) i s a p o s i t i v e u s c e l e m e n t , t h e sbove g i v e s us o u r v e r s i o n o f Goffman's t h e o r e m :

( 6 4 . 5 ) I f X i s compact m e t r i c , t h e n f o r f E C " ( X ) , i f and o n l y i f ( 6 ( f ) , x )

=

f Ex(C'(X)d)

0 f o r a l l b u t a c o u n t a b l e number o f

X I S .

Goffman's s t a t e m e n t i s a s f o l l o w s :

(64.6)

(Goffman).

A bounded r e a l f u n c t i o n on a c l o s e d i n t e r v a l

i s Riemann 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 p u r e l y n o n a t o m i c

3 34

Chapter 13

measure if and only if its set of points of discontinuity is countable.

PART V I

THE RARE ELEMENTS I n t o p o l o g ? , , iiiiport;iiit r o l c s a r c p l a y e d deiisc s e t s a n d t h e s e t s o f E i r s t c a t e g o r y

ti),

-

t h e no\
i n R o u r b a k i ' s tcriiiPlaying s i m i lar

i n o l o g ) , , t h e r a r e s e t s a n d t h c incagcr s e t s .

r o l c s i n the s t u d y of' C " ( S ) , we h n v c t h e " r a r c c l c i n e i i t s " and

t l i c "mcagcr c l e m c n t s " .

lJn1 i k e t h e s i t u a t i o n

iii

t o p o l o g y , the

more i n i p o r t n i i t r o l e Iicre sccms t o hc t h a t o f t h e r a r e c l c m c n t s , a n d most

or

P a r t 1'1 i s J e v o t c d t o t h e s e .

We f i r s t e x a m i n e t h e c l e m e n t s o f C " ( X ) t h c s e t s w i t h empty i n t e r i o r i n t o p o l o g y . "1 e:in".

335

corresponding to \lie c a l l s u c h c l c m c n t s

CHAPTER 1 4 THE LEAN ELEMENTS

565. Elementary p r o p e r t i e s

An e l e m e n t f o f C t ' ( X ) w i l l be c a l l e d

lean

i f a ( If

1)

=

0.

If1 i s t h e n a l s o l e a n a n d , i n f a c t , s o i s e v e r y e l e m e n t of t h e

R i e s z i d e a l g e n e r a t e d by f .

Thus a n a l t e r n a t e d e f i n i t i o n i s

t h a t f i s l e a n i f t h e R i e s z i d e a l which i t g e n e r a t e s i n t e r s e c t s C(X) only i n 0.

There i s a n abundance o f l e a n e l e m e n t s .

We show t h a t € o r

u ( f ) - f and f - a ( € ) a r e l e a n .

every f EC"(X),

l e a n e l e m e n t i s o f t h i s form.

Note

Moreover, e v e r y

t h a t the following proposi-

t i o n c o r r e s p o n d s t o s t a n d a r d t o p o l o g i c a l ones on s e t s w i t h empty i n t e r i o r [ 31,581

.

( 6 5 . 1 ) For f € C l ' ( X ) ,

f > 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

<

k(f

+

0, then f

=

+

t(g)

u(f).

=

g) < u(f)A(f

+ g) < L(f)

g i s lean.

+

Hence g)

+

<

0, whence

=

0.

QED

(65.5) Corollary 1.

Suppose f,g

0 are lean.

< R and fAR Rsc element R such that g -

Proof.

By (49.7), u(f)AR

=

=

If there is an

0, then f

0 , so u(f)Ag

=

+

g is lean.

0 and (65.4)

applies.

Let I be a u-band or an R-band.

(65.6) Corollary 2 . f EC"(X),

i f fI and f

Proof.

I

For every

are lean, then f is lean.

For concreteness, let I be a u-band, and we can

Chapter 1 4

340 a s su me I f 1
II

II

USC,

a n d If

=

Ifl

1

and If1

=

If I

I

dl,

s o , by t h e

1

a r e l e a n . We show u ( I f 1 ) A l f d I I i t w i l l f o l l o w from ( 6 5 . 4 ) t h a t I f 1 i s l e a n . I f I I .Y(X),,

which i s

If

If

=

0;

= 0.

so, a fortiori, u(lflI)Alfl

I

QED

I n t u i t i v e l y , a l e a n e l e m e n t s h o u l d be " n e g l i g i b l e " .

We

n e x t e x p l o r e t h e c o n s e q u e n c e s o f two e l e m e n t s o f C1l(X) d i f f e r i n g by a l e a n e l e m e n t .

Proof. is lean.

0 5 ( L ( f ) - u ( g ) ) + 5 ( f - g)',

hence (L(f) - u ( g ) ) +

S i n c e i t i s a n ~ s ecl e m e n t , i t f o l l o w s ( a ( f ) - u ( g ) ) + = 0.

QED

A f o r t i o r i , i f f - g is lean, then L(f) < u(g).

This gives

us :

( 6 5 . 8 ) C o r o l l a r y 1.

2,(P

f

I f h i s l e a n , t h e n f o r e v e r y gE C"(X ),

h) 5 u ( g > .

(65.9) Corollary 2 .

If h i s l e a n , then f o r every usc element u,

a(u

i h)

<

u,

The Lean Elements

and for every Rsc element u(R

t

341

k,

h) > k.

In particular, for every gEC(X), R(g

?

h)<- g < u(g

(65.10) Corollary 3.

II s

+

*

11).

If h is lean, then for every gEC(X), hll L

I1 gII -

This last follows from the preceding and (49.14).

We will be interested in knowing when a given f EC"(X) differs from a usc element or an ksc element by a lean element. Now for every f, f ially lean.

<

u(f),

so (f - u(f))+

We can expect that if u(f)

0, hence is triv-

=

is replaced by a

smaller usc element u , then (f - u ) + will no longer be lean How much smaller must u be?

We have a simple answer.

(65.11) Theorem, F o r every f EC"(X),

(f

-

uk(f))+

is lean, and

uk(f) is the smallest usc element for which this is true.

Proof.

o

<

(f

-

ua(f))+

<

(f

-

.t(f))+

=

Since

f - a(f).

this last is lean (65.1), it follows (f - uk(f))+

is lean.

Now

suppose u is a usc element such that (f - u ) + is lean; we show u > u&(f).

It is enough to show that u 2 R(f).

k((f

-

u)')

=

0,

34 2

SO

Chapter 1 4

I,(f

-

U)

< 0 (49.11), -

SO

a(f) - u < 0 (49.2).

Thus u > ?(f).

QED

Remark.

D u a l l y , u s i n g t h e f a c t t h a t u ( f ) - f i s l e a n , we

have: (ku(f)

-

f)'

i s l e a n , and L(f) i s t h e l a r g e s t fsc element

f o r wh ich t h i s i s t r u e .

A l l t h e s t a t e m e n t s i n P a r t VI h a v e

similar dual s t a t e m e n t s . ( 6 5 . 1 1 ) h a s a c o r r e s p o n d i n g t h e o r e m i n t o p o l o g y : G i ven a s e t A i n a t o p o l o g i c a l s p a c e T , l e t I: t h e i n t e r i o r of A.

=

Int,t h e c l o s u r e o f

Then A \ F h a s empty i n t e r i o r , a n d I: i s t h e

s m a l l e s t c l o s e d s e t f o r w hi ch t h i s i s t r u e . i n t h e n e x t 5 , we exam i ne t h i s f u r t h e r .

For t h e d i s c u s s i o n

Let u s say t h a t A has

empty i n t e r i o r a t a p o i n t t o f T i f t h e r e e x i s t s a n e i g h b o r h o o d

W o f t s u c h t h a t A n W h a s empty i n t e r i o r .

Then t h e s e t F

a b o v e i s p r e c i s e l y t h e s e t o f p o i n t s o f T a t w h i c h A d o e s n__ ot h a v e empty i n t e r i o r . (A

n

Thus A h a s a u n i q u e d e c o m p o s i t i o n A

=

i n t o t h e s e t o f i t s p o i n t s a t which i t does n o t

F ) IJ ( A \ F )

h a v e empty i n t e r i o r a n d a s e t w i t h empty i n t e r i o r .

(65.11)

g i v e s u s a c o r r e s p o n d i n g decomposition o f e a c h f EC"(X) : f = fAuk(f)

+

(f

-

uk(f))+.

For a n e l e m e n t o f & ( I L ( X ) ) ,

d e c o m p o s i t i o n i s i n t o two e l e m e n t s o f E ( l l ( X ) ) r e p l a c e d by

V,

and

+

the

can be

making t h e a n a l o g u e c l e a r e r .

566. A ' l l o c a l i z a t i o n " t h e o r e m

The f o l l o w i n g " l o c a l i z a t i o n " and e a s i l y v e r i f i e d ( c f .

theorem i n topology i s obvious

[31], S8,VII):

I f a s e t A h a s empty

34 3

The Lean E l e m e n t s

i n t e r i o r a t e v e r y p o i n t , t h e n A h a s empty i n t e r i o r .

The

( ( 6 6 . 3 ) bel ow ) r e q u i r e s more

correspond-ing theorem f o r C"(X) work t o p r o v e .

( 6 6 . 1 ) L e m m a 1.

I f fAg i s l e a n , t h e n s o a r e f + A g , f - A g ,

lf/Ag,

IflAIgl, and f - .

Proof.

5

-IfAg/

fAg < f+Ag < I fl A g

<

IflAlgl 5 /fAgl.

S i n c e t h e s e t o f l e a n e l e m e n t s i s s o l i d , i t f o l l o w s f+A g,

/flAg,

f - = 1 - f - 1 , s o , by w hat w e h a v e j u s t

and IflAlgj a r e l e a n .

shown, t o p r o v e t h a t f - A g i s l e a n , i t s u f f i c e s t o show t h a t (-f-)Ag i s lean.

(-f-)Ag

=

i s l e a n s i n c e fAg i s l e a n .

o

so

<

f-

=

(fA0)Ag

=

(fAg)AO

=

- ( f A g ) - , w hi ch

Finally, f > fAg, s o - f < -(fAg),

(-f)+ < (-(fAg))+

=

(fAg)-;

s i n c e (fAg)- i s l e a n ,

it follows f - i s lean. QED

No te t h a t f A g l e a n d o e s n o t i m p l y t h a t f + i s l e a n : l e t f

=

n(X)

and g

that f is lean. the

=

0.

I t f o l l o w s t h a t fAg l e a n d o e s n o t i m p l y

The f o l l o w i n g Lemma p r e s e n t s a c a s e i n w h i c h

implication does hold.

Lemma (66.2) - 2 . G i ven a n Rsc e l e m e n t k , l e t I b e t h e band w h i c h it generates.

Then f o r f E I , i f f A k i s l e a n , f i s l e a n .

Chapter 1 4

344

Proof. (49.4)

Case I , R > 0 : By ( 6 6 . 1 ) ,

f a ( 1fI)AR = R(lflAR)

=

0.

IfIAR i s l e a n , h e n c e

Since t h e o n l y element o f I

d i s j o i n t from R i s 0 , i t f o l l o w s R ( l f 1 ) = 0 . < 0: Case 11, R -

Thus f i s l e a n .

By t a k i n g n e g a t i v e s , t h i s c a s e c a n be

writ t e n :

Given a u s c e l e m e n t u > 0 , l e t I be t h e band which i t

(*)

generates.

Then f o r g € 1 , i f gvu i s l e a n , g i s l e a n .

We w i l l p r o v e t h i s a t t h e end of 567. We p r o c e e d t o p r o v e ( 6 6 . 2 ) f o r t h e c a s e o f a g e n e r a l k s c L e t H be t h e band g e n e r a t e d by R'.

element R .

Then H i s a n

R-band, s o by ( 6 5 . 6 ) , i t i s enough t o show t h a t f H ahd f

are

H lean. fHAR+

=

fHARH = (fA!L)H, which i s l e a n by t h e comment p r e -

ceding (65.3).

S i n c e '.9

i s a n Rsc e l e m e n t , Case I above g i v e s

us t h a t f H i s lean. f dA(-R-) = f dAR = (fAR) d , w h i c h , a g a i n , i s l e a n by H H H H t h e comment p r e c e d i n g ( 6 5 . 3 ) . Now - R - = 9,110, h e n c e i s a n Rsc e l e m e n t , and s i n c e R ed by - k - ,

i s c l e a r l y c o n t a i n e d i n t h e band g e n e r a t H Case I 1 above g i v e s u s t h a t R is lean. H QED

We a r e now i n a p o s i t i o n t o e s t a b l i s h o u r l o c a l i z a t i o n theorem.

( 6 6 . 3 ) Theorem. Then f o r f E C " ( X ) ,

Let

{aa]

be a c o l l e c t i o n o f Rsc e l e m e n t s .

fARalean f o r a l l a i m p l i e s t h a t V a ( f A R a )

is

The Lean E l e m e n t s

34 5

lean.

We p r o v e t h e f o l l o w i n g t h e o r e m , which i s e a s i l y shown t o be e q u i v a l e n t t o t h e above.

L e t { L a } b e a c o l l e c t i o n o f Rsc e l e m e n t s a n d R = v,Lu.

If

FARcx i s l e a n f o r a l l a , t h e n f A k i s l e a n .

Proof.

< k ; s o we For s i m p l i c i t y , we c a n assume t h a t f -

have t o show t h a t f i s l e a n . Case I , L > 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 <- R <- O ,

fARa 0

hence, t a k i n g n e g a t i v e s , 0 < -f < -(fARa ) , which i s l e a n . 0

Now remove t h e r e s t r i c t i o n of p o s i t i v e n e s s o r n e g a t i v e n e s s on R .

L e t I b e t h e band g e n e r a t e d by R + .

By ( 6 5 . 6 ) , we n e e d

are lean. I We r e d u c e t h e c a s e o f f I t o Case I a b o v e . R + = v w ( R a ) + and

o n l y show t h a t f I and f

I -< R I = R + , s o i t s u f f i c e s t o show t h a t f I A ( R a ) + i s l e a n f o r a l l a. (Ra)+ 5 R + , so (Ra)+ = ((La)+),, so fIA(La)+ = f

fIA((ka)+)I

=

( f A ( L a ) + ) , , which i s l e a n by ( 6 6 . 1 ) a n d t h e

comment p r e c e d i n g ( 6 5 . 3 )

.

We r e d u c e t h e c a s e o f f

I

t o Case I 1 a b o v e .

-1

= LAO

=

34 6

Chapter 1 4

Va(LaAO) = V c l ( - ( ~ c l ) -and ) ~ fId 5

=

- f

, so

i t s u f f i c e s to

A(-(Lcl)-) i s l e a n f o r a l l a. f I d = c - f - ) d I I ( f + E I ) , s o we have t o show t h a t ( - f - ) d A ( - ( t 1 - 1 i s l e a n . I

show t h a t f

f i r s t that (-f-)A(-(ila)-) =

(fAo)A(LaAo)

Note

= (fAka)Ao = - ( f A k , t ) - ,

S i n c e - f - 5 ( - f - ) d , we have ( - f - ) A ( - ( k a ) - ) < I ( - f - ) d ~ ( - ( k a ) - ) 0 , h e n c e , by t h e p r e c e d i n g s t a t e m e n t , I ( - f - ) d ~ ( - ( ~ c l ) i-s) l e a n . I QED

which i s l e a n .

We l e a v e t h e v e r i f i c a t i o n o f t h e f o l l o w i n g t o t h e r e a d e r .

C o r o l l a r y 1. (66.4) -

Let {I,} be a c o l l e c t i o n o f il-bands and I

be t h e band w i t h ll(X)I

=

V ~ I ( X )( ~ s o I i s a l s o an il-band).

If

c1

fI

i s l e a n f o r a l l a, t h e n f I i s l e a n . c1

(66.5) Corollary 2 .

For e a c h f EC"(X), t h e r e e x i s t s a l a r g e s t

R-band I f o r which f I i s l e a n .

Analogous t o ( 6 6 . 3 ) , we h a v e :

( 6 6 . 6 ) Theorem..

u =

A

c1

ua'

Let Iua} be a c o l l e c t i o n o f u s c e l e m e n t s and

Then f o r f E C " ( X ) ,

t h a t (f - u ) + i s lean.

( f - ua)+ l e a n f o r a l l a i m p l i e s

The Lean E l e m e n t s

Remark.

A t f i r s t glance,

347

( 6 6 . 6 ) a n d ( 6 6 . 3 ) seem t o b e t h e

same t h e o r e m , a n d i t i s t r u e t h a t t h e y r e d u c e t o t h e same t h e o -

rem i n t o p o l o g y . Proof o f

-

k(f),

u)')

5

fAu

~ +

t h e y do n o t c o i n c i d e .

By ( 6 5 . 7 ) , 9 ( f a ) 5 ua f o r a l l a.

(66.6).

follows L(f) < ~

E((f

However, i n C " ( X ) ,

=

(f

uu, h e( n c e~ 2 ( f ) 5 f A u . -

whence 2 ( ( f - u ) '

u)+ =

=

Then n ( f )

f , whence t ( f )

+

k((f

It

+ -

u)')

5

0.

QED

Remark.

Note t h a t a l s o ( 6 6 . 5 ) a n d ( 6 5 . 1 1 ) 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 .

CHAPTER 1 5

THE RARE ELEMENTS

§ 6 7 . Elementary p r o p e r t i e s

An e l e m e n t f o f C"(X) w i l l be c a l l e d r a r e i f u ( If ~

that is, i f ku(lf1)

=

0.

1)

i s lean,

A rare element i s of course l e a n .

As i s

We w i l l d e n o t e t h e s e t o f r a r e e l e m e n t s by Ra(X).

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 Ra(X) a r e much s t r o n g e r a n d more s a t i s f y i n g t h a n t h o s e o f t h e s e t o f l e a n e l e m e n t s .

To

b e g i n w i t h , w h i l e t h e l a t t e r s e t i s n o t even a l i n e a r s u b s p a c e , we h a v e :

(67.1)

Ra(X) i s a norm c l o s e d R i e s z i d e a l .

S u ppos e f , g E R a( X ) .

Proof.

ku(lf

+

gl) 5 Rullfl <

Then

atuofl)

< %u(lfl) =

hence !Lu(\f

+

lgl)

+

+

+

u(lgl)l

(49.1)

u(lgl)

(49.1)

u(lgl),

gl) 5 ku(lg/)

=

0.

Thus ( f

+

g) E R a ( X ) .

c l e a r t h a t i f f E R a( X ) , t h e n Xf E Ra(X) f o r a l l A ER. 34 8

I t is So Ra(X)

The Rare Elements is a linear subspace. g E lia(X)

ideal.

That f E Ra(X)

349

and ( g l

follows from the definition. Finally, the operation

(49.16)), hence Ra(X)

So

<

If

1

implies

Ra(X) is a Riesz

u ( . ) is norm continuous (iterate

is norm closed. QED

As is easily seen, Ra(X) is the Riesz ideal generated by

the positive lean usc elements, hence by all the lean usc elements (cf. the comment following (65.2)). Some additional characterizations:

(67.2) For f EC"(X), '1

f E Ra(X);

2'

f',f-

3O

Lu(f)

the following are equivalent:

E Ra(X); =

0 = uR(f);

< 0 < uL(f). 4O Lu(f) -

Proof. If 2'

That l o and 2'

are equivalent is contained in (67.1)

holds, then (by iterating (49.10))Lu(f)

uL(f-)

=

0, and similarly for u!,(f).

course implies 40 . (ku(f))'

=

Finally assume ' 4

0 and Lu(f-)

=

(uk(f))-

=

=

ku(f')

Thus 3' holds. holds. 0.

-

'3

Then k u ( f ' )

Thus 2'

of =

holds. QED

' 4

above should be compared with (65.2).

In that case, the

inequalities were only necessary; here they are a l s o sufficient.

Chapter 1 5

350

( 6 7 . 3 ) Ra(X) 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 s u ( * ) , a ( . ) ,

(therefore) 6( * )

Proof.

and

.

S u ppos e f E R a( X ) .

Then i t h a s p r o p e r t y 4'

we show u ( f ) a n d k ( f ) a l s o h a v e t h i s p r o p e r t y . k(f) < 0 , and u a ( u ( f ) ) > ua(f) > 0.

in (67.2);

au(u(f))

=

Similarly for k(f).

QED

We h a v e s e e n t h a t f,g l e a n d o e s n o t i m p l y t h a t f + g , f v g , fAg a r e l e a n .

(67.4)

However,

I f f i s l e a n a n d g i s r a r e , t h e n f + g , f v g , a n d fAg a r e

lean.

(67.5) Corollary.

I f a R i e s z i d e a l I o f C'l(X) i s c o n t a i n e d i n

t h e s e t o f l e a n e l e m e n t s , t h e n I + Ra(X) i s a l s o .

I t follows

t h a t i f a R i e s z i d e a l i s maximal w i t h r e s p e c t t o b e i n g c o n t a i n e d i n t h e set of l e a n elements

,

t h e n i t c o n t a i n s Ra(X ).

The Rare Elements

Again, i f two e l e m e n t s o f

C'l(X)

351

d i f f e r o n l y be a r a r e e l e -

ment, we c a n s a y more t h a n we c a n i f t h e y d i f f e r by a l e a n element.

( 6 7 . 6 ) Given f , g E C " ( X ) , i f ( f - g ) + i s r a r e , t h e n ( u ( f ) - u ( g ) ) +

and ( P ( f ) - Q ( g ) ) + a r e r a r e .

Proof.

By ( 4 9 . 2 ) , ( u ( f ) - u ( g ) ) + , ( R ( f ) And by ( 4 9 . 1 0 ) ,

(u(f - g))'.

(u(f - g))'

=

R(g))+

-

2

u ( ( f - g)'),

which

i s r a r e by h y p o t h e s i s .

( 6 7 . 7 ) Given f , g E C 1 ' ( 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'

(u(f)

u(g)j+ is rare;

2'

(u(f) - u(g))+ i s lean;

3'

Ru(f) 5 Q u ( g ) ;

4O

uRu(f) 5 u R u ( g ) .

-

And 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 : 11.

(a(f)

21.

( a ( f ) - . ~ ( g ) ) +i s l e a n ;

3'.

u Q ( f ) 5 uR(g);

4'.

QuR(f) < RuR(g).

Proof. 4'.

T h a t 2'

-

~ ( g ) ) +i s r a r e ;

We o f c o u r s e have t h a t 1' i m p l i e s 3'

i m p l i e s 2'

f o l l o w s from ( 6 5 . 7 ) .

and '3

implies

F i n a l l y , suppose

352 4'

Chapter 15

holds; we show ( u ( f )

( u ( f ) - u(g))+

=

-

u(g))'

is rare.

[ u ( f ) - uRu(f) + u R u ( f ) - uRu(g)

+

uRu(g) - u ( g ) ] +

< ( u ( f ) - u R u ( f ) ) + + ( u R u ( f ) - uEu(g))+

+

(u!Lu(g) - u(g))+.

The second term in this last line vanishes by supposition, and uRu(g) 5 u ( g ) , so the last term vanishes.

We thus have:

This last is lean by (65.1), hence, being a usc element, is rare. The equivalence o f l', 2 ' , 3', 4 ' i s proved similarly. QED

Combining (67.6) and (67.7):

(67.8) I f ( f - g ) + is rare, then the eight properties in (67.7) hold.

I f we interchange f and g in the above three propositions,

then combine the statements thus obtained with the original ones, we have

(67.9)

I f f - g is rare, then the following s i x properties

-

The Rare Elements

353

o f which t h e f i r s t t h r e e a r e e q u i v a l e n t and t h e l a s t t h r e e a r e

equivalent - a l l hold: (i)

u ( f ) - u(g) is rare;

(ii)

!,u(f)

iu(g);

=

(iii) uau(f)

utu(g);

=

(iv)

e(f) - a(g) is rare;

(v)

uk(f) = uL(g);

(vi)

kua(f) = k u k ( g ) .

(67.10)

I f r i s r a r e , t h e n f o r e v e r y g E Cl'(X),

C o r o l l a r y 1.

ku(g

2

r ) = ku(g),

uL(g t r ) = uk ( g ) .

I f r i s r a r e , then f o r every usc elementu,

(67.11) Corollary 2.

ku(u

?

r) = k(u),

and f o r e v e r y ksc e l e m e n t L, uk(k

2

r)

=

u(L).

In p a r t i c u l a r , f o r every g E C ( X ) , ku(g ? r ) = g = u k ( g

We c a n a l s o s t r e n g t h e n ( 6 5 . 1 1 )

(67.12)

Theorem.

r).

(Note t h a t u k u ( f ) > uL(f)).

For every fEC"(X), (f - uku(f))'

i s r a r e , and

u L u ( f ) i s t h e s m a l l e s t u s c e l e m e n t f o r which t h i s i s t r u e .

Chapter 1 5

354

Proof.

To show t h a t ( f - u L u ( f ) ) + i s r a r e , i t s u f f i c e s t o

show t h a t ( f - k . u ( f ) ) + i s r a r e .

- ku(f))+

0 < (f

k u ( f ) ) + = u ( f ) - k u ( f ) , which i s l e a n ( 6 5 . 1 ) ,

< (u(f)

-

hence ( u ( f ) - n u ( f )

being a usc element) r a r e . Now suppose u i s a u s c e l e m e n t s u c h t h a t ( f - u ) + i s r a r c , we show t h a t u > 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<

-

u )<

0

(49.2),

0.

QED

I f f E Ra(X), t h e n 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 c o n t a i n e d i n Ra(X) ( s i n c e Ra(X) i s i t s e l f a R i e s z i d e a l ) . general, the band g e n e r a t e d b y f i s n o t . f E Ra(X) does n o t imply l l ( X ) f E Ra(X).

In

Otherwise s t a t e d ,

A s an example, l e t X be

a r e a l i n t e r v a l , { r n } t h e s e t o f r a t i o n a l p o i n t s of X , and

u t h e u s c e l e m e n t d e f i n e d by ( u , r I 1 ) (u,x)

=

=

l / n (n = 1 , 2 , . - . ) and

I t i s easy t o v e r i f y t h a t u i s r a r e b u t

0 otherwise.

R ( X ) u (which i s 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 { r n l ) i s n o t r a r e - i n d e e d u(ll(X)u) = n ( X ) . We do have t h e f o l l o w i n g :

( 6 7 . 1 3 ) For f EC"(X), f

1'

f E Ra(X);

2'

n(x) (f -

Proof.

(xi 1

3

0 , t h e following a r e equivalent:

+ E Ra(X) f o r a l l

A > 0.

Suppose 1' h o l d s , and c o n s i d e r 1 > 0 .

Let

The Rare Elements

e

=

n(X)

(XI 1

(f-

0 < e < (l/X)€€

+.

Then, by ( 1 7 . 9 ) ,

(ii)

< f.

e -

Thus

Ra(X).

C o n v e r s e l y , s u p p o s e 2' 0 < f < n(X).

Xe < f

355

h o l d s , a n d , f o r s i m p l i c i t y , assume

Consider X > 0 .

( f - h n ( X ) ) + E Ra(X).

t h e l a t t e r i s i n Ra(X) b y a s s u m p t i o n . ( f - hll(X))+ < ll(X);

f < n ( X ) , hence

the d e s i r e d i n e q u a l i t y then follows.

I t f o l l o w s from ( i ) and ( i i ) t h a t f i s i n t h e norm c l o s u r e o f Ra(X).

S i n c e Ra(X) i s norm c l o s e d , f E Ra(X). QED

While t h e band g e n e r a t e d by a n e l e m e n t o f Ra(X) need n o t be c o n t a i n e d i n Ra(X), i t i s contained i n t h e s e t of l e a n elements. T h i s 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 t h e o r e m , which i s t h e c l a s s i c C a t e g o r y theorem f o r compact s p a c e s .

(67.14)

(Category theorem).

I f { f n } c Ra(X)+ and f = V n f n ,

then f i s l e a n .

Proof.

I t s u f f i c e s t o show t h a t V n u ( f n ) i s l e a n , s o , f o r

Chapter 1 5

356

s i m p l i c i t y , we can t a k e t u n } c Ra(X)+ and f pose f i s n o t l e a n .

=

vnun.

Now s u p -

Then t h e r e e x i s t s g EC(X) s u c h t h a t 0 < g < f.

Again f o r s i m p l i c i t y , we c a n assume IIgll > 1 . S e t vn

=

unAg ( n

=

Then t h e v n t s a r e u s c e l e -

1,2,..-).

m e n t s , Cvnl c Ra(X)+, and g

=

Vnvn.

h n } c C(X)+ s a t i s f y i n g :

. . .) ; for all n

We remark f i r s t t h a t , s i n c e IIgll > 1, ( g - l ( X ) )

Hence,

0.

s i n c e ( g - n ( X ) ) E C(X), no l e a n e l e m e n t c a n dominate i t . We o b t a i n t h e h n ' s i n -

We p r o c e e d t o e s t a b l i s h t h e Lemma. ductively.

By t h e above r e m a r k , v 1 i- ( g - IL(X)).

Now v1 i s

t h e infimum o f a l l t h e e l e m e n t s o f C ( X ) between i t and g ; hence t h e r e e x i s t s hlEC(X)

such t h a t

v1 hl

5

h l 5 g9 (g -

n(x) 1 -

That i s , hl s a t i s f i e s t h e Lemma. Assume h l , - . - , h n have b e e n c h o s e n t o s a t i s f y t h e Lemma.

To e s t a b l i s h t h i s , we show t h e s t r o n g e r i n e q u a l i t y : Vn+l

(g

(g - l ( X )

- n(X>) - hn)

Suppose hn

+

Vn+l

>

(g - n ( X ) ) .

hn

+

Write t h i s

t h e l e f t s i d e i s i n C(X) and t h e 5 v ~ + ~Here .

r i g h t s i d e i s l e a n , hence ( g - n ( X )

- hn)

5 0 , o r hn 2 Cg

- 1(X)).

T h i s c o n t r a d i c t s t h e i n d u c t i o n h y p o t h e s i s t h a t hn s a t i s f i e s ( i i i )

357

The R a r e E l e m e n t s

i n t h e Lemma.

I t f o l l o w s f r om ( * ) ( a s i n t h e c h o i c e o f h l ) t h a t t h e r e exists

EC(X) such t h a t

hnvvn+1 5 h n + l 5 g , hn+l b

s

-

n(x)),

w hic h e s t a b l i s h e s t h e L e m m a . But t h e n h n + g , s o

by D i n i ' s t h e o r e m , limnllg - hnll

=

0.

A ga in t h i s c o n t r a d i c t s ( i i i ) i n t h e L e m m a .

QED

We h a v e s e e n ( 6 5 . 1 ) t h a t u ( f ) l e a n ( a l t h o u g h tE(f)

,

-

f and f

i n general, is not).

-

t ( f ) are always

If o n e o f t h e s e i s

r a r e , we c a n s a y much m o r e :

(67.15) For f E C " ( X ) ,

t h e following are e q u i v a l e n t :

1'

u(f)

2'

f - t ( f ) i s rare;

3'

6(f) is rare;

4'

6(f) is lean.

P roof. -

-

f is rare;

If one of u ( f )

-

f and f - k ( f ) i s rare, then,

s i n c e t h e o t h e r i s l e a n , i t f o l l o w s from ( 6 7 . 4 ) t h a t 6 ( f ) i s l e a n , hence (being a usc element) rare. imply

4O,

w h i c h i m p l i e s 3'.

3'

Thus 1' a n d 2'

o f c o u r s e i m p l i e s '1

each

a n d 2'.

QED

A t r i v i a l c o r o l l a r y i s t h a t f o r a u s c e l e m e n t u a n d a n Lsc

Chapter 1 5

358

element R ,

6 ( u ) and 6 ( k ) a r e r a r e .

For e v e r y f E S(X) , 6 ( f ) E Ra(X).

( 6 7 . 1 6 ) Theorem.

Proof.

We have a s t r o n g e r r e s u l t :

Suppose f i r s t t h a t f E s ( X ) , t h a t i s , f

where u1 and u 2 a r e u s c e l e m e n t s . 6(ul)

Then, by ( S O . Z ) ,

6(u2) E Ra(X) ( c f . t h e above r e m a r k ) .

+

holds f o r s(X).

=

u

1 - u2

0 < 6(f) <

Thus t h e theorem

That i t h o l d s f o r i t s norm c l o s u r e S ( X ) f o l l o w s

from t h e f a c t s t h a t t h e o p e r a t i o n 6(.)

i s norm c o n t i n u o u s ( 5 0 . 1 0 )

and R a ( X ) i s norm c l o s e d . QED

( 6 7 . 1 7 ) C o r o l l a r y 1, 1'

f is rare;

2'

f i s lean;

3'

u(f) is lean;

4'

k(f)

Proof. u(f) = f Thus '3

+

For f E S ( 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 :

is lean.

'1

o f c o u r s e i m p l i e s 2'.

Suppose 2'

holds.

Then

( u ( f ) - f ) , hence i s l e a n by ( 6 7 . 1 6 ) above and ( 6 7 . 4 ) .

holds.

hence i s r a r e .

Suppose 3'

holds.

> 0 by ( 6 5 . 2 ) , Then u ( f ) -

Since S ( f ) i s a l s o r a r e ( 6 7 . 1 6 ) ,

L(f)

=

u(f) - 6(f) is rare.

k(f)

5

0 by ( 6 5 . 2 ) ,

F i n a l l y , suppose 4'

hence - k ( f )

therefore rare, so k(f) is rare.

i t follows

holds.

Then

i s a p o s i t i v e l e a n usc element,

Since f - k ( f ) i s a l s o r a r e ,

The R a r e E l e m e n t s

by ( 6 7 . 1 6 ) , i t f o l l o w s f

=

L(f)

+

(f

-

359

!,(f)) i s r a r e . QED

For a g e n e r a l e l e m e n t f o f C t l ( X ) ,

a u ( f ) a n d u a ( f ) h a v e no

particular order relation, that is, a l l possible order relations occur.

However:

(67.18) C o r o l l a r y 2 .

Proof.

For f E S ( X ) ,

QU(f)

-

uf,(f).

u ( f ) - t ( f ) is l e a n , by (67.16),

so a ( u ( f ) - k(f)) = O .

Applying (49.2) g i v e s us a u ( f ) - uL(f) < 0.

QED

We c l o s e t h i s 5 w i t h t h e p r o m i s e d p r o o f o f ( * ) i n ( 6 6 . 2 ) . 0 < u - gvu, hence u i s l e a n , and t h e r e f o r e (being a p o s i t i v e U

S

~e l e m e n t )

rare.

I t f o l l o w s from t h e C a t e g o r y theorem (67.14)

t h a t I i s contained i n the s e t of lean elements.

Thus g i s l e a n .

QED

568. A " l o c a l i z a t i o n " t h e o r e m

The p r o p e r t i e s e s t a b l i s h e d i n 566 f o r l e a n e l e m e n t s h o l d , w i t h one e x c e p t i o n , f o r r a r e elements.

Note t h a t w e h a v e b e e n

f o r c e d t o change t h e o r d e r of development.

360

Chapter 1 5

(68.1) Theorem. f o r fEC"(X), fAR,

L e t {La}

b e a c o l l e c t i o n o f Rsc e l e m e n t s .

rare f o r a l l

c1

i m p l i e s t h a t Va(fARa)

Then

i s rare.

A g a i n , we p r o v e t h e f o l l o w i n g e q u i v a l e n t t h e o r e m i n s t e a d .

L e t {La} b e a c o l l e c t i o n o f Rsc e l e m e n t s , a n d g,

= VIJP,il.

I f fAk,x i s r a r e f o r a l l u , t h e n fnil i s r a r e .

P roof.

Again, f o r s i m p l i c i t y , w e c a n assume t h a t f < 8 . ~

We show i ? u ( f ) < 0 < uR(f)

(cf. (67.2)).

Ru(f) < 0: W e f i r s t p r o v e t h a t i t s u f f i c e s t o show P u ( f ) A Rc 0.. Assume Ru(f)AR < 0. au(f)Au(R) < 0. comes R u ( f ) < 0.

Then u ( k u ( f ) A L ) < 0 , s o by ( 4 9 . 6 ) ,

But f < R , so Ru(f) < u(k), so t h i s last beNow t o p r o v e k u ( f ) A k < 0. ku(f)AR

(the l a s t inequality

Ru(f)A(VaRa) = Vcx(Ru(f)Aila) < Vclku(fARa) f o l l o w i n g from ( 4 9 . 4 )

and ( 4 9 . 6 ) ) .

=

S i n c e LU(fAka) < 0 for

a l l a , Vaku(fARm) 5 0 , a n d we a r e t h r o u g h . uR(f)

0 : Choose cto a r b i t r a r i l y .

Then u k ( f ) > v R ( f A R c l )-> 0 . 0

QED

A g a i n we h a v e t h e c o r o l l a r i e s :

( 6 8 . 2 ) C o r o l l a r y 1.

Let

{Ia} b e a c o l l e c t i o n o f R - b a n d s a n d I

t h e b a n d w i t h I l ( X ) I = Vall(X)I

a implies t h a t f I i s rare.

.

rare f o r a l l a

Then f I

a

The R a t e Elements

(68.3) Corollary 2.

For e a c h f E C " ( X ) ,

361

there exists a largest

t - b a n d I f o r which f I i s r a r e .

Again ( 6 8 . 3 ) and ( 6 7 . 1 2 ) r e d u c e t o t h e same t h e o -

Remark.

rem i n t o p o l o g y . ( 6 6 . 1 ) h o l d s w i t h " l e a n " r e p l a c e d by " r a r e " .

We r e c o r d

the f a c t :

( 6 8 . 4 ) I f fAg i s r a r e , t h e n s o a r e f+Ag. f - A g ,

IflAg,

If I A I g l ,

and f - .

The p r o o f i s t h e same. (66.2) a l s o c a r r i e s over t o r a r e n e s s , but only f o r L > 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))'

<

<

u ( f ) - k u ( f ) E Ra(X)

( t h i s l a s t by ( 6 7 . 1 6 ) . QED

CHAPTER 16 THE DECOMPOSITION C1(X)

=

Ra(X)'

0

C(X)'

569. The decomposition of C'(X)

Ra(X) Ra(X)L.

determines the decomposition C1(X)

=

(Ra(X)

I

)

d 0

Each of these summands has a simple characteristization.

(69.1) Theorem.

Proof. -

Ra(X)L

=

C(S)'.

c C(X)':

Ra(X)I

and suppose fa+O in C(X).

In

ffect, consider I,rERa(X) , p > 0 , I

Then u

U is order continuous on C"(X),

=

Since

~~f~ is in Ra(X).

( u , ~ , r ) = infa( f,,p)

=

Thus

0.

1-1 is order continuous on C(X).

For the opposite

every U € (Ra(X)I) Ra(X)

, v

> 0, fails to be order continuous on C(X).

is separating on (Ra(X)

ment uERa(X)+ C(X)

d

inclusion, it is enough to show that

1

)d, so there exists a usc ele-

such that ( u , ~ , r ) > 0.

Choose a net { f a } in

such that faJ.u. Then fw.'cO in C(X),

but lima(fa,v)

=

(u,I,r> > 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),

<

u(f) 5

so

k ( f ) , u ( f ) ERa(X)dd.

QE D

I f C(X)'

i s s e p a r a t i n g on C ( X ) ,

t h e n C(X)

s o t h e e l e m e n t s o f Ra(X)dd a r e a l l l e a n .

n

R E L ( X )=~ ~ 0,

T h i s h a s a consequence

t h a t e v e r y e l e m e n t o f U(X) i n Ra(X)dd i s r a r e :

(71.3)

Ra(X).

I f C(X)'

i s s e p a r a t i n g on C ( X ) , t h e n U(X)

(Hence a l s o t h e R i e s z i d e a l I o f ( 7 2 . 1 )

n

Ra(XIdd c

above.)

372

Chapter 1 6 Proof.

I f f € U ( X ) n Ra(X)dd, f > 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.

(72.6) Corollary 2 .

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 <

A1

2

<

-

=

-

,.

of

<

such t h a t

B1.

s,.

-

n

C(X).

C(X). Choose Since every

and e v e r y element

i t f o l l o w s ;Z < qf <

B.

QED

(73.2) Corollary.

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 f o r . e v e r y

maximal l e a n R i e s z i d e a l I , Ct'(X)

Remark.

=

C(X)

@

I (algebraically!).

A l i p r a n t i s a n d B u r k i n s h a w h a v e shown t h a t M a x e y ' s

t h e o r e m h o l d s f o r a g e n e r a l R i e s z s u b s p a c e o f a Dedekind comp l e t e R i e s z s p a c e ([2],

Theorem 2 . 1 8 ) .

574. The D i l w o r t h r e p r e s e n t a t i o n

( 7 4 . 1 ) Given a n Rsc e l e m e n t k a n d a

USC

element u, t h e following

Dedekind Completion of C(X)

379

are equivalent: '1

a

=

there exists a Dedekind cut ( A , B )

of C(X)

such that

VA, u = AB; 2O

u(n)

=

u, t ( u )

=

R.

The verification is simple.

A pair ( a , u ) satisfying the above conditions will be called a regular pair. called a regular -__

The element t of the pair will be

P,sc element, and the element u, a regular usc

element . Dilworth calls the corresponding functions (he works with functions, not elements o f C"(X))

normal, but we want to

.~

emphasize the relation of our elements to the regular open sets and closed sets in topology.

(74.2) For a usc element u , the following are equivalent: '1

u is a regular usc element;

2O

u

=

3'

u

= U(R)

ut(u);

for some

And for an Rsc element

LSC

element R .

R , the following are equivalent:

'1

R is a regular R S C element;

2O

R = Ru(R);

3'

R = a(u)

___ Proof.

for some usc element u.

In each statement, 1' and '2

equivalent, and '2

implies 3'.

That 3'

are easily seen to be implies

2'

follows from

Chapter 1 7

380

t h e f a c t t h a t t h e o p e r a t i o n s u a ( . ) a n d f,u(.)

a r e idempotent.

QED

W ith t h e c o r r e s p o n d e n c e o f t h e r e g u l a r e l e m e n t s t o r e g u l a r sets i n a t o p o l o g i c a l s p a c e , w e a l s o have c o r r e s p o n d i n g theorems.

For t h e f o l l o w i n g c f .

(74.3)

( [ 1 3 ] , Theorem 4 . 2 ) .

I f { f } i s a bounded s e t i n C"(X), cl

t h e elements o f which

a r e a l l regular usc elements o r a l l regular (i)

RSC

elements, then

f ) is a regular usc element, a a ( i i ) R ( A f ) i s a r e g u l a r Rsc e l e m e n t . a a u(V

Proof.

F o r c o n c r e t e n e s s , assume t h a t f ' s a r e r e g u l a r usc a

e l e m e n t s : {ua}.

A

u

is

USC,

so ( i i ) f o l l o w s from ( 7 4 . 2 ) .

a a u ) . S i n c e v ~ ( u , ) i s Rsc, i t i s enough t o acl a > 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)].

-f

f"

in

-.

.f - i n - C " ( X ) / R a ( X )

-

c1

Get+?,

i t follows QED

Dedekind Completion of

Comb n i n g ( 7 5 . 2 ) ,

(75.5), a n d ( 7 4 . 9

387

,

w e have t h e

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 q ( C ( X ) )

(75.6) C o r o l l a r y .

i s o r d e r c l o s e d i n C"(X)/Rs(X).

576. A f o u r t h r e p r e s e n t a t i o n

By ( 5 0 . 1 1 ) , s - l ( R a ( X ) ) i s a n Mll-subspace o f C"(X)

which and a ( . ) .

c o n t a i n s C(X) a n d 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 s u ( .

I t corresponds t o t h e family of s e t s i n a topological space which have nowhere d e n s e f r o n t i e r s . Corresponding t o a s t a n d a r d theorem i n topology

w e have :

( 7 6 . 1 ) The f o l l o w i n g c o i n c i d e :

'1

6 - l (Ra(X))

2'

S(X)

'3

C(X)u

+

Ra(X)

4'

C(X)'

+

Ra(X)

5'

C(X)ur

%,

6'

C(X)"

0 Ra(X).

Proof.

Ra(X)

+

Ra(X)

T h a t S o a n d 6'

comment p r e c e d i n g ( 7 4 . 9 ) .

a r e d i r e c t sums f o l l o w s f r o m t h e Now 6' c 4'

c '2

a n d 5'

c '3

c .'2

Chapter 1 7

388

c l o f o l l o w s from ( 6 7 . 1 6 ) a n d ( 6 7 . 3 ) .

T h a t 2'

p r o v e t h a t 1' c S o , easily verified, Then f Ru(f)

= +

6'.

Suppose 6(f)ERa.(X).

If - u R ( f )

I 5

-

Lu(f))EC(X)"

+

Since, as i s

6 ( f ) , it follows f

u R ( f ) + ( f - u R ( f ) E C ( X ) U r + Ra(X). (f

I t remains t o

-

u!L(f)€Ra(X).

Similarly, f =

Ra(X). QED

We h a v e s e e n ( 6 7 . 1 8 ) t h a t f o r f € S ( X ) , R u ( f ) < u!L(f). ( 6 7 . 2 ) , t h i s a l s o h o l d s f o r fERa(X). S(X)

+

Ra(X)

(cf.

(67.10),

Indeed, it holds f o r

f o r example).

(76.2) For f E s-l(Ra(X)), ku(f)

5

By

We r e c o r d i t :

uR(f)

From ( 7 5 . 2 ) a n d ( 7 6 . 1 ) w e h a v e :

( 7 6 . 3 ) Let q b e t h e q u o t i e n t mapping o f C"(X)

o n t o C"(X)/Ra(X).

Then q [ S - l (Ra(X) ) ]

=

R [ C " (X) /Ra(X)

1.

Whence :

(76.4)

6-I(Ra(X))/Ra(X)

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).

CHAPTER 1 8

THE MEAGER ELEMENTS

5 7 7 . Elementary p r o p e r t i e s

The a - o r d e r c l o s u r e o f Ra(X) w i l l be d e n o t e d b y Me(X), and i t s e l e m e n t s w i l l be c a l l e d meager.

A meager e l e m e n t i s

t h e g e n e r a l i z a t i o n t o C"(X) o f a m e a g e r , o r f i r s t c a t e g o r y , s e t i n topology. We have i m m e d i a t e l y t h a t Me(X) i s a a - o r d e r c l o s e d R i e s z ideal.

Moreover, i t i s a l e a n i d e a l .

Category theorem ( 6 7 . 1 4 ) .

(77.1)

(Category theorem).

This i s simply the

We r e s t a t e i t :

E v e r y meager e l e m e n t i s l e a n .

S i n c e Me(X) i s a - o r d e r c l o s e d , we h a v e , u n l i k e t h e c a s e w i t h Ra(X):

(77.2)

fEMe(X) i m p l i e s n(X),EMe(X).

389

Chapter 18

390

O t h e r w i s e s t a t e d , Me(X) c o n t a i n s , w i t h e a c h e l e m e n t , t h e band g e n e r a t e d b y t h a t e l e m e n t . U n l i k e R a ( X ) , Me(X) i s t i o n s u ( * ) , Q(+),

c l o s e d under any of t h e o p e r n -

I n t h e f i r s t example f o l l o w i n g ( 6 5 . 1 ) ,

6(.).

g i s meager w h i l e u ( g )

not

=

n(X).

However, we n o t o n l y h a v e

Me(X) c Ra(X)dd ( o b v i o u s ) , b u t :

(77.3)

For a l l fEMe(X), u ( f ) , a . ( f ) E R a ( X )

-~ Proof. By ( 7 1 . 2 )

, we

dd

n e e d o n l y p r o v e t h a t Me(X) i s c o n -

t a i n e d 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 U(X) fEMe(X), f > 0 , w e h a v e f -

.

= V

g n n'

n

Ra(X)dd.

{gn} c Ra(X)+.

Given

Then

0 < f < ~~u(g~)€Ll(X s o) ,we a r e t h r o u g h . QE n

Note, f i n a l l y , t h e c o r o l l a r y o f (71.3) ;

(77.4)

I f C(X)'

i s s e p a r a t i n g on C ( X ) , t h e n Me(X)

=

Ra(X).

(Thus Ra(X) i s o - o r d e r c l o s e d . )

5 7 8 . Sums o f p o s i t i v e e l e m e n t s o f a R i e s z s p a c e

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

The Meager Elements

391

the meager elements the "localization" theorems that were obtained for the lean elements and the rare elements.

We know

of no direct method o f doing this, and do not actually reach

our goal until 5 8 0 .

5578, 79 develop the machinery we will use.

r

Given a set {a), the s y m b o l

will always denote a

The collection of all r's is a directed set

Cinite subset.

under the order defined by inclusion. Let E be a Riesz space and {aa} a set of positive elements.

For each

r

=

non-empty, set a,

{al;*-,an},

=

Zna 1 a.' 7

and for 'I the empty set, set a r arl

5

=

Then

0.

r l c r2 implies

ar , and thus {ar}allT is an ascending net.

If a

= Vrar

2

(hence aT+a), we will write a

=

Caaa, or simply a

=

Caa, and

call a the __ sum of the a a l s . We emphasize that we use the

2 0 for all

symbol Caa only if a a

01

(and the above supremum

exists).

(78.1)

(78.2)

If a

If a

=

=

Zaa and b

=

Cba,

then

a + b

=

I(aa

+

Caa, b a

C b a , and b a 5 aa for a l l a , then

= -

ba).

b

=

C(aa

-

ba).

We will often use this last in the form: Caa uaa

-

ba).

-

Zba

=:

Chapter 18

392

(78.3)

L e t E b e Dedekind c o m p l e t e .

Suppose a

decomposed i n t o two d i s j o i n t s e t s : { a } (i) b

=

CaB a n d c

=

za

=

=

La

and

CL

{ o } IJ {y}.

{CL}

is

Then

e x i s t , and

Y

( i i ) a = b + c.

C o n s i d e r two s e t s { a ) , { B I .

As a g r e e d ,

r

w i l l always

d e n o t e a f i n i t e s u b s e t o f t h e f i r s t s e t ; and i n t h e f o l l o w i n g ,

A w i l l always d e n o t e a f i n i t e s u b s e t o f t h e second s e t . usual,

{a}

x

{B} is the set of a l l ordered p a i r s (n,R).

As

It

i s c l e a r t h a t i n t h e d i r e c t e d s e t of a l l f i n i t e s u b s e t s o f

{a]

t h o s e o f t h e form T x A

{B},

x

Consider a s e t { a

(a,B)

1 of p o s i t i v e elements of

i s , t h e s e t i s i n d e x e d by {a} write aaB f o r a

are cofinal.

x

(8)).

E (that

For s i m p l i c i t y , w e w i l l

Thus Lacla w i l l mean 1

( a ,Kla(a,BI * From t h e a b o v e p a r a g r a p h , i t i s c l e a r t h a t t h e f o l l o w i n g e q u a l (u,B)*

i t y h o l d s (when e i t h e r s i d e e x i s t s ) .

(78.4)

a

ct

=

L e t E b e Dedekind c o m p l e t e .

"aaB,

then a = C C a c1

V

=

Ca

aB'

B aB

=

CaaB

=

CBCaaclB-

The f i r s t e q u a l i t y i s o u r h y p o t h e s i s .

Proof.

a

I f a =Cctaa, and f o r e a c h n ,

We show

By ( * ) a b o v e , i t s u f f i c e s t o show t h a t a =

a l l T,AarxA'

The Meager E l e m e n t s

a

a

rxA = ‘crEr,BEA

r

=

5

393

uB

BE a aa B

‘&EraN

< a.

-

Since t h i s holds f o r a l l

r,A,

the opposite inequality,

f i x To.

V

> va l l r , aa r x a -

a11

Vall

, a ar x n

<

a.

To e s t a b l i s h

Then

a

A r0xn

a(ZatroCBcnaafi)

=

‘ail

=

‘acro(Vall

=

‘atroaa

=

ar

a‘BEaaaB)

. 0

> V a rarxA -

I t follows Vall

r r

= a .

I t r e m a i n s t o show t h a t 1

= z z a a a a@‘ Note f i r s t < a < a f o r a l l r , hence

, hence ar x B - r a e x i s t s - otherwise s t a t e d , zNaaBe x i s t s f o r each

t h a t f o r e a c h 6, V

(a,BIaclB

a l l rarxg

a

D e n o t e t h i s sum by b show t h a t bA

5

< a

aB

-

B

.

We show C b e x i s t s . B B

I t i s enough t o

a f o r a l l A. ’A

=

‘B€AbB

= C B E n ( V a l r~a r x B ) =

“all

r

=

‘all a

rarxA

<

Denoting obtain b

(‘BcaarxB)

(cf. proof above).

b by b , and c o p y i n g t h e p r o o f o f a B B =

za

aB’

=

Ca

aB’

we

a.

Chapter 18

394

Now l e t {a} b e t h e s e t o f a l l o r d i n a l s < a f i x e d o r d i n a l 0.

Let E b e Dedekind c o m p l e t e , and s u p p o s e z a each a, ~

n

exists.

~ exi< s t s , a~n d i ta i s e~a s y t o s e e t h a t V

Then f o r a
a

p

1=

c an -

(78.5)

L e t E b e Dedekind c o m p l e t e , and c o n s i d e r a bounded s c t

{ a n } i n E + , w h e r e now {a} i s t h e s e t o f o r d i n a l s o r d i n a l 0.

< a fixed

Set bl

=

b

=

a

a

1’ (a - V c1

a )+ B
f o r a > 1.

Then

Proof.

We u s e i n d u c t i o n on 0 .

for 0 a finite Case I ,

ca
Assume i t h o l d s f o r e v e r y 0 < Oo.

ordinal.

oo =

h a s an i m m e d i a t e p r e d e c e s s o r n o : Then

Ca
+

0

= v a < a aa

ba 0

+

0

=

The e q u a l i t y c l e a r l y h o l d s

ba

0

“a
0

- v n < n0 a n ) +

aa)vaa

= 0

0

= vn
Case 11, 0, i s a l i m i t o r d i n a l : preceding the proposition,

T h e n , by t h e o b s e r v a t i o n

The Meager Ilcments

=

57!).

v,L
Q 1: I1

Sums of positive elements of C T T ( X )

Suppose we have { f l y ) , { p r Uc ) C"(X)+

for a11

(1.

and I f l l

5

395

such that En

5

u(pa)

Tf YpLx exists, then Cu(pcL) exists, hence 7 f n exists Tu(~i,,).

We will need an answer to the question:

under what circumstances do

hie

have I f c x - u ( Q Q ?

In general, Y u ( p rt ) and u ( T p 1 are not comparable. example where Zu(pci)

An

u ( T p i , ) is obtained by taking for X thc

real interval [ O , I ] , for p l thc characteristic clcment of the hali-open interval [ 0 , 1 / 2 ) ,

for p 2 that of (1/2,1], and then

examining the sum p

An example where C L I ( ~ ~ <) i ~ ( ~ pis~ ~ )

+ p2.

obtained by taking for X a real interval, letting {rn? be the rational points in X , taking for pn the characteristic element of rn (n

=

1,2,.'-), and then considering Fpn.

We return to the question raised above.

The conditions

in the following proposition may secm arbitrary, but they will be what we need €or our "localization" theorem.

(79.1) Suppose { p a } is a subset of C " ( X ) +

If

such that Cp,

exists.

Chapter 1 8

396

6 ( p ) E R a ( X ) f o r all a , and a ( i i ) { a } i s a s e t o f p o s i t i v e RSC elements s a t i s f y i n g

(i)

U

R

< u ( p ) f o r all a , t h e n

u -

a

CR

(Y

< Ll(J:p(J.

-

We f i r s t e s t a b l i s h t h e t h e o r e m f o r a f i n i t e s c t

Proof.. {pl,---,pn}.

i i ( ~ p ~ ) ] we + ; h a v e t o show t

Sct R

=

[zpi

-

0 < a.

=

[cai

- U(ZPi)l+

CU(Pi)

<

= C(U(Pi)

=

0.

- 7Pi -

Pi)

C6(Pi)

<

E Ra(X). I t f o l l o w s 9“

=

0.

We t u r n t o t h e g e n e r a l c a s e . proved,

‘r

f o r e v e r y f i n i t e s u b s e t T o f {a}.

< u(p )

r

-

Ry t h e f i n i t e c a s e j u s t

( 7 9 . 2 ) C o r o l l a r y 1.

Let { R

a

1 , {ma 1 b e s e t s o f & s c e l e m e n t s

such t h a t 0 c -

and s u p p o s e C R (i) (ii)

< R

a -

< u(m,)

a -

f o r a l l u,

( o r e q u i v a l e n t l y , Cma) e x i s t s .

< u(cma), a (ka- mu) E R a ( X ) .

CR

c

c1

m

Hence

Then

The Meager E l e m e n t s

39 7

( i ) f o l l o w s from ( 7 9 . 1 ) and ( 6 7 . 1 6 ) .

Proof.

0 < C ( Q u - mu)

For ( i i ) ,

- Cmu

=

CRu

<

u(Cmul

=

6 (Cma)

-

Cmc,

E Ra(X). QED

The f o l l o w i n g c a n be t h o u g h t o f a s a t h i r d " l o c a l i z a t i o n " t h e o r e m f o r r a r e n e s s ( w i t h ( 6 8 . 1 ) and ( 6 8 . 6 ) ) .

(79.3)

Corollary 2 .

L e t { g a l b e a s e t o f Rsc e l e m e n t s a n d { r u 3

a s e t of r a r e elements such t h a t 0 < ru

5 Ru

f o r a l l u.

Then i f I R u e x i s t s , C r u i s r a r e .

Proof.

For e a c h a , s e t

We show f i r s t t h a t { R u } ,

ma = ( Q a - u ( r a ) ) + .

{mu] s a t i s f y t h e h y p o t h e s e s o f ( 7 9 . 2 ) .

< mu F o r e a c h a, i t i s c l e a r t h a t mu i s Rsc and 0 -

u(R,)

2

u(m,).

R u = RaAu(ra)

+

(La- u ( r a ) ) +

=

5 Q u ; we show

QuAu(ru)

+

ma.

Hence

u( R u)

=

uR(R a )

=

uR(RaAu(ru)

=

uR(mu)

+

ma) (67.10)

< u(ma).

Now 0

5

ru

5

R u - ma f o r a l l a , h e n c e , by ( i i ) i n C o r o l l a r y 1,

CraERa(X). QED

Chapter 18

398 For a general set {r

c1

}

of rare elements, Trcy,i f i t exists,

need not be rare (even if the set is countablc).

We will need the following stronger form of (79.3).

(79.4) Corollary 3. Suppose {ga}

(i)

c C"(X)+

satisfies:

~ g ,exists;

(ii) cg, 5 u ( z k ( g a ) ) . If{r 3 is a set o f rare elements such that a

€or all a,

O < r

a 5 ga

Then zr

0.

is rare.

Proof. a(g,)

+

F o r each a , set h

a

= ga-!2,(g,).

Then 0 < r < a -

ha, so by the Decomposition Lemma ( 2 . 5 1 ,

where 0 < p

< a(g,)

a-

and 0 < (la 5 ha.

so it suffices to show that Cp,

rare follows from (79.3).

r

a = p, + q,, Then crC, = Cp, + C ( l a ,

and C q a are rare.

That Cpa is

We show Ch, is rare, hence Cq,

is

rare.

(78.2) (ii j

Finally, as a consequence of (78.5):

(79.5) Let { k } be a bounded set of positive Rsc elements, where 0.

The Meager E l e m e n t s

ttu}

399

i s t h e s e t o f o r d i n a l s < a f i x e d o r d i n a l 0.

Set

and

Then

We f i r s t show t h a t x m

Proof. -__

LY

<

-

~p(g,l

f o l l o w t h a t u(rm ) < u(jye(gn)) < u(Vka). a (m

being esc) m

i:ga

= V J R b~y

< k(ga),

a -

hence z m

m

5V

R ~ ;i t w i l l

,5 g a ,

< zt(ga).

a-

hence

And z k ( g a ) <

(78.5).

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 u ( v k ) < u(1m ) . We a a u s e i n d u c t i o n on 0. The i n e q u a l i t y h o l d s f o r 0 = 2 , i n f a c t , VJR, =

=

ml

=

zma.

Assume t h e i n e q u a l i t y h o l d s f o r e v e r y

0 < 0,.

Case I , 0, h a s a n i m m e d i a t e p r e d e c e s s o r a o .. (78.3) (49.1)

400

Chapter 18

QED

580. The " l o c a l i z a t i o n " t h e o r e m

A s w i t h l e a n e l e m e n t s ( 6 6 . 1 ) and r a r e e l e m e n t s ( 6 8 have :

( 8 0 . 1 ) I f fAg i s m e a g e r , t h e n s o a r e f + A g , f - A g ,

IflAg,

and f - .

The p r o o f i s t h e same. ( 6 8 . 5 ) a l s o c a r r i e s o v e r t o Me(X), and i n d e e d w i t h t h e R S C element R > 0 r e p l a c e d by a g e n e r a l g > 0:

The Meager E l e m e n t s

401

( 8 0 . 2 ) Given g > 0 , l e t I b e t h e band which i t g e n e r a t e s .

Then

f o r f E I , i f fAg i s meager, f i s meager.

The a r g u m e n t i s t h e o n e u s e d f o r ( 6 8 . S ) , w i t h a n a p p e a l t o t h e a - o r d e r c l o s e d n e s s o f Me(X) a t t h e e n d .

We n e x t c a r r y o v e r ( 7 9 . 3 ) t o Me(X).

(80.3)

Let { k } b e a s e t o f ksc e l e m e n t s a n d {h } c1

i'L

R

s e t o f mea

ger ones such t h a t <

hn

5

i a f o r a l l a.

Then i f C i a e x i s t s , Cha i s m e a g e r .

F o r e a c h a, ha

Proof.

-~

r a r e elements.

=

cnran, { r a n } a s e t of p o s i t i v e

T h e n , b y ( 7 8 . 4 ) , Chc,

= C

aC n r an

=

'nCaran*

For

< h 5 i a , h e n c e , b y ( 7 9 . 3 ) , c a r a n e x i s t s and an i s r a r e ; d e n o t e i t by r n . Then xh = c r n E M e ( X ) .

each n , 0 < r

c1

QED

(80.3)

i n t u r n g i v e s t h e s t r o n g e r form:

(80.4) C o r o l l a r y . (i)

S u p p o s e {ga} c C " ( X ) +

satisfies:

cga exists

( i i ) cg

< u(ca(g,)).

a I f { h } i s a s e t o f meager e l e m e n t s s u c h t h a t a

402

Chaptcr 18

0 < ha

2

ga for a l l a ,

t h e n 1 1 1 ~ i s meager.

The p r o o f i s t h e same a s f o r ( 7 9 . 4 ) . We c a n now e s t a b l i s h t h e " l o c a l i z a t i o n " t h e o r e m .

Notc t h a t

i t c o r r e s p o n d s t o a c l a s s i c a l t h e o r e m o f B a n a c h ' s i n topology,

and, l i k e h i s theorem, u s e s w e l l - o r d e r i n g .

(80.5)

Theorem. -

Then f o r f E C " ( X ) ,

L e t { a } b e a c o l l e c t i o n o f Rsc e l e m e n t s . c1

fARa m e a g e r f o r a l l a i m p l i e s

V

a

is

(fAg,)

meager.

As b e f o r e , we prove t h e f o l l o w i n g e q u i v a l e n t theorem instead.

L e t { a } b e a c o l l e c t i o n o f Rsc e l e m e n t s , a n d R a I f f A R a i s meager f o r a l l a , t h e n fAR i s meager.

P roof. -

< R. F o r s i m p l i c i t y , w e c a n assume f -

Lemma. -

The t h e o r e m h o l d s f o r {La} c C " ( X ) +

Proof.

For e a c h B, s e t f a = f A R a ;

thus f

= VaRa.

and f E C " ( X ) + .

=

Vf,.

We

T h e Meagcr E l e m e n t s

well-order t h e index s e t

{ ~ t } ,

403

s o , f o r s i m p l i c i t y , we c a n assume

i t i s t h e s e t o f a l l o r d i n a l s < an o r d i n a l 0.

hl 11

Set

fl

=

(fa

f )+ ci > 1. B {ha} s a t i s f y t h e c o n d i t i o n s o f ( 8 0 . 4 ) ,

cx

=

We show t h a t { g } , cx h e n c e xh i s m e a g e r .

v

-

R
S i n c e , by ( 7 8 . 5 ) ,

CY

zh

= ri

Vfu

f, t h i s w i l l

=

c o m p l e t e t h e p r o o f o f t h e Lemma. By ( 7 8 . 5 ) a n d ( 7 9 . 5 ) ,

v9

< u(VE
h

< f

a-

CY

, so h

a

%l

I t r e m a i n s t o show t h a t 0 < hn 5 g o .

i s meager.

h 0 = [ f a - V B
O <

=

[ f A k C Y - V B < wf A P F ] +

=

[fAka

fA

-

Now, i n any R i e s z s p a c e , f o r a ( b

-

c)'.

=

(Va<,kB) 1'.

3,

b, c > 0 , we h a v e aAb - aAc <

Thus 0 < h a -< [ f A ( k , =

fA(ka

=

fAga

<

ga.

-

-

Vll<,kB)+]+ VB
We p r o c e e d t o p r o v e t h e t h e o r e m . Case I , K > 0 and f l i e s i n t h e band g e n e r a t e d by

p :

We show If / A k i s m e a g e r ; i t w i l l f o l l o w f r o m ( 8 0 . 2 ) t h a t meager. (80.1)

By h y p o t h e s i s , f o r e v e r y a , fAka i s m e a g e r .

twice, we have t h a t

hence V P meager.

+ =

a

I+

=

R.

lf1ALa+

i s meager.

Now

is

Applying

V i a

I t f o l l o w s f r o m t h e Lemma t h a t

1fI

= k,

lflAP is

404

Chapter 18

Then f A k a -< f < k < 0 , whence,

- 0 : Fix a .

Case 11,

,Q <.

Thus - f , and t h e r e -

taking negatives, 0 < -f < -(fAta)€Me(X). f o r e f , i s meager. Case 111, p,+,

0: L c t I b e t h e band

a n d R - b o t h d i f f e r from

a r e b o t h meager. I R+ = vkCi+ and We r e d u c e t h i s t o Case I .

g e n e r a t e d b y k f ; we show f I a n d f f I i s meager: < a

f

I -

I

a l l a.

= g,+.

I t r e m a i n s t o show t h a t f I A k

fIALo;+

= fIA(ka'l1

=

+ . (y.

i s meager f o r

B u t , by ( 8 0 . 1 1 ,

(fAkaC)I.

fAY,'

i s m e a g e r , s o ( f A 2 C Y + ) Ii s m e a g e r , s o w e a r e t h r o u g h .

We r e d u c e t h i s t o Case I T :

i s meager:

V(-ta

)

=

-a

,

fId

5 R

and f I f

A(-a,

= -k

I

.

- ) i s meager.

I t r e m a i n s t o show t h a t , g i v e n a, T h i s f o l l o w s from:

Id

(-f-)A(-La-)

=

-(fAka)-,

w h i c h i s m e a g e r , s o we h a v e ( i ) .

( i i ) , we need o n l y v e r i f y t h a t - f

< f d.

-

For

We l e a v e t h i s t o t h e

I

reader. QED

(80.6)

C o r o l l a r y 1.

I t h e b a n d w i t h n(X),

Let 11 } b e a c o l l e c t i o n o f R - b a n d s , a n d c1

.

= Vall(X)I

meager f o r a l l

Then f I

c1

CY

ct

i m p l i e s t h a t f I i s meager.

(80.7)

C o r o l l a r y 2 . For e a c h f € C " ( X ) ,

there exists a largest

The Meager E l e m e n t s

405

R-band T s u c h t h a t f I i s m e a g e r .

( 6 6 . 6 ) and ( 6 8 . 6 1 , w e h a v e t h e " o t h e r " " l o c a l i z a t i o n "

As i n

theorem:

Let { u } b e a b o u n d e d s e t o f u s c e l e m e n t s , a Then f o r f E C " ( X ) , ( f - ua)+ m e a g e r f o r a l l a

( 8 0 . 8 ) Theorem. and u

.

= AU

a implies t h a t (f

u ) + i s meager.

-

< n(X) f o r a l l a. a A l s o , b y w e l l - o r d e r i n g t h e i n d e x s e t { a ) , we c a n a s s u m e {a} i s

We c a n a s s u m e f < n(X)

Proof.

and u

t h e s e t o f a l l o r d i n a l s < some o r d i n a l 0. Set R Then Aqa (p

-

qa)'

=

=

0,

n(X) - u , p

(p

-

4,)'

f - u , and q

= =

a

=

a n d p+

( f - u,)',

u

a =

-

u f o r e a c h a.

(f - u)'.

Thus

i s m e a g e r f o r a l l a , a n d we h a v e t o show t h a t p+ i s

meager.

For each

U,

set

Note t h a t f o r e v e r y a, P . ~ = h e n c e i s a n Rsc e l e m e n t .

(n(x)

-

U)

A l s o t h a t II

-

(ua -

= V P . ~a n d

n(x)

LI) =

p+

=

Vp,.

Now s e t

a > 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<) u

(ii)

( C - u ) + i s m e a g e r , and

-

uKu(f),

( i i i ) u i s t h e s m a l l e s t usc e l e m e n t f o r w h i c h ( i i ) h o l d s .

r h i s f o l l o w s f r o m t h e a b o v e Theorem t o g e t h e r w i t h ( 6 5 . 1 1 ) h'ote t h a t w h i l e i n t h e two f o r m c r t h e o r e m s , we

and ( 6 7 . 1 2 ) .

were a b l e t o s p e c i f y t h c u s c e l c m e n t p r e c i s e l y , i n t h e p r e s e n t o n e , we h a v e o n l y e s t a b l i s h e d i t s e x i s t e n c e . Remarh. -__

Again a s w i t h l c a n n e s s a n d r a r e n e s s ,

(80.5)

and

( 8 0 . 8 ) 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 .

In t o p o l o g y , m e a g e r s e t s p l a y a more i m p o r t a n t r o l e t h a n r a r e ones.

I n C " ( X ) , t h e r a r e o n e s seem t o h e more i m p o r t a n t .

The f o l l o w i n g d i s c u s s i o n t h r o w s some l i g h t on t h i s .

To make

t h e c o m p a r i s o n s more i n t u i t i v e , we p r e s e n t t h e d i s c u s s i o n i n t e r m s o f o r d i n a r y f u n c t i o n t h e o r y - more e x a c t l y , b o u n d e d f u n c t i o n s w i t h t h e supremum norm

--

w i t h no r e f e r e n c e t o C " ( X).

S o , b y u ( f ) , we w i l l mean t h e u p p e r s e m i c o n t i n u o u s u p p e r e n v e l o p e o f f , by 6 ( f ) ,

"rare",

"meager",

t h e s a l t u s f u n c t i o n o f f , and so on. "Riemann i n t e g r a b l e " ,

The t e r m s

a n d s o on a r e m o d i f i e d

i n t h e o b v i o u s way. In working w i t h s u b s e t s o f a s e t X , i f a s e t i d e a l o r s e t r i n g i s n o t l a r g e enough f o r o u r p u r p o s e s , w e u s u a l l y t a k e i t s U-closure.

When d e a l i n g w i t h f u n c t i o n s , h o w e v e r , i f a R i c s z

i d e a l o r Riesz subspace i s n o t l a r g e enough, t h e n between t h e i d e a l ( r e s p . subspace) and i t s 0 - o r d e r c l o s u r e , t h e r e l i e s i t s norm c l o s u r e , a n d w e o f t e n f i n d t h i s l a s t t o b e t h e c o r r e c t

Chapter 18

408

enlargment.

And i f o u r i d e a l o r s u b s p a c e i s a l r e a d y norm

c l o s e d , t h e n i t may t u r n o u t t o be p e r f e c t l y a d e q u a t e , w h e r e i t s correspondent

i n s e t t h e o r y was n o t .

I n d e e d we may o b t a i n a

s h a r p e r theorem than i s p o s s i b l e i n s e t theory.

We g i v e two

examples. Example 1. f u n c t i o n on X .

L e t X b e com pact a n d .9 a l o w e r s e m i c o n t i n u o u s Then 6 ( k ) i s r a r e .

Compare t h i s w i t h t h e

s t a n d a r d theorem: t h e s e t o f p o i n t s of d i s c o n t i n u i t y o f R i s meager. Even t h o u g h a f u n c t i o n - s a y a p o s i t i v e o n e - i s r a r e , t h e s e t IxEXlf(x) > 0 ) i s , i n g e n e r a l , o n l y meager.

(In Riesz

s p a c e c o n c e p t s , f E R a( X ) i m p l i e s o n l y t h a t lLfEMe(X) . )

Other-

w i s e s t a t e d , t h e above s e t c a n n o t d e t e r m i n e w h e t h e r f i s r a r e o r o n l y meager. Example 2 .

L e t X be a r e a l i n t e r v a l , and c o n s i d e r t h e

c l a s s i c Le b e s gue t h e o r e m : a bounded f u n c t i o n f on X i s Riemann

i n t e g r a b l e i f and o n l y i f i t s s e t o f p o i n t s o f d i s c o n t i n u i t y h a s Le b e s g u e m e a s u r e 0 .

I n f u n c t i o n t e r m s : f i s Riemann i n t e g r -

a b l e i f an d o n l y i f 6 ( f ) h a s L ebes gu e i n t e g r a l 0 . can immediately sharpen t h e theorem.

But now we

6 ( f ) i s a p o s i t i v e upper-

s e m i c o n t i n u o u s f u n c t i o n , a n d f o r s u c h f u n c t i o n s , t h e L ebesgue i n t e g r a l v a n i s h i n g i s e q u i v a l e n t t o t h e Riemann i n t e g r a l ( e x i s t i n g and) v a n i s h i n g .

Thus t h e t h e o r e m becom es: a bounded f u n c -

t i o n f i s Riemann i n t e g r a b l e i f and o n l y i f S ( f ) h a s Riemann

--

integral 0. ( No te t h a t t h i s c l o s e l y p a r a l l e l s t h e s e t t h e o r e m ( o r d e f i n i t i o n ) : a s e t h a s J o r d a n c o n t e n t i f and o n l y i f i t s f r o n t i e r has Jordan content 0.)

The Meager E l e m e n t s

409

The n e e d f o r L e b e s g u e ' s t h e o r e m i n t h e f o r m h e g a v e i t i s e x p l a i n e d b y t h e d i s c u s s i o n i n Example 1 .

The f u n c t i o n s h a v i n g

Riemann i n t e g r a l 0 f o r m a norm c l o s e d i d e a l ; t h o s e h a v i n g Lebesgue i n t e g r a l 0 form a o - o r d e r c l o s e d i d e a l .

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45. Robertson, A.P. and Robertson, Wendy, Topological Vector Spaces, Cambridge Univ. Press, 1964. 46. Ryan, F.D., On the uniform closure of the linear space generated by the non-negative uppersemicontinuous functions, Purdue Univ. Dissertation, 1965. 47. Schaefer, H.H., Banach lattices and Positive Operators, Springer-Verlag, 1974. 48. Semadeni, Z., Functions with sets of points of discontinuity belonging to a fixed ideal, Fund. Math. 5 2 (1963), 25-39.

414

Bib1 iography

, Banach Spaces of Continuous Functions, Polish

49.

Scientific Publishers, Warsaw, 1971. 50. Stone, M . H . ,

Notes on integration I-IV, Proc. Nat. Acad.

Sci. U . S . A . 35 (19A9),

34 (1948), 336-342, 447-455, 483-490, and 50-58.

51. Szpilrajn, E., Remarques sur les fonctions complstemcnt additives d'ensembles, Fund. Math. 2 2 (1934), 3 0 . 3 - 3 1 1 . 52. Masterson, J.J., Dual space of a vector lattice and its cutcompletion, Trans. A.M.S.

128 (1967), 515-522.

53. Veksler, A . I . , On a new construction o f Dedckind completion of vector lattices and of Il-groups with division, Sibirskii Mat. Zhur. 10 (1969), 1206-1213. (Translated by Consultants Bureau, 1970). 54. Nakano, H., Uber das System aller stetigen Funktionen auf einem topologischen Raum, P r o c . Imp. Acad. Tokyo 17 (1941), 308-310. 55. Stone, M.H., Boundedness properties in function lattices, Can. J. Math. 1 (1949), 176-186.

INDEX OF SYMBOLS

19

166

13 23

21s

59

237

66 38

a

+

aAb

165

192,214,

92

192,214

189

222

191

222

92

135

213

8

213

8

218

8

220

7

172

5

208

5

208

50

conv A

111

15 15

256

1s

39 2

16 415

Symbols

416

E+

4

169

c*

50

170

Eb

52

291

EC

69

E

bb

74

2

$c

74

E'

88

306

38

384

28

316

5

100

348

ext A

116

Ea E/ I

2 34

r

391

2 34 11

int Q

173

391 50

110

166 126

79

127

191 208

78

249

87

223

144

255 2 33

Me (X)

389

249 255

2 324

223

Symbo 1s

172

165

Y

165

172 170

3

VA

3

claa

AA

3

Aaaa

Fa

n

G

3

25 93 165

38

417

I N D E X OF TERMINOLOGY

a r b i t r a r i l y small elements Archimedean atomic

9

189

Baire classes

218

B a i r e element

218

Baire function

281

Baire subspace

217

Banach l a t t i c e

42

band

22

band, b a s i c band

148

g e n e r a t e d by A

band, p r i n c i p a l

22

37

29

band, p r o j e c t i o n base

34

110

Bore1 e l e m e n t

237

Bore1 f u n c t i o n

282

Bore1 s u b s p ace

237

bounded above

2

bounded b e l o w

3

bounded

207

418

Terminology

c h a r a c t e r i s t i c c l e m e n t i n C" c h a r a c t e r i s t i c clcment clopen s e t component cone

CI

291

292

175 100, 101

3

C o u n t a b l e Suli p r o p e r t y

114

c o u n t a b l y decomposable ( a t most)

D e c o m p o s i t i o n lemma decomposition of X Dedekind c l o s u r e

7

176 233

Dedekind c l o s e d

235

Dedekind c o m p l e t e

31

Dedekind c o m p l e t i o n Dedekind c u t

31

Dedekind d e n s e

34

dense-in-itself diffuse

36

197

191

disjoint dispersed

1 2 , 13 205

d u a l bands

147

equi-order-continuous

f i l t e r i n g upward f i l t e r i n g downward

hyperstonian

370

18 19

154

115

419

Terminology

420

108

imbedding infimum

3

interval

7

interval preserving

82

Kakutani-Stone space

165

109

L-norm L-space

112

R-band 2 6 5 lean

336, 377

Lebesgue t o p o l o g y

162

l i m i t inferior

33

l i m i t superior

33

locally solid l o w e r bound

126 3

lowersemicontinuous lower envelope

M-norm M-space

94

97

M-subspace

98

MIL- homomorphism MIL-isomorphism Mll-norm MIL-space

108

94 98

MI-subspace

98

Mackey t o p o l o g y Markov mapping meager

108

389

144

141

280

Terminology

net ~

15

, ascending

15

__ , d e s c e n d i n g

15

__ , m o n o t o n i c

15

net associated with A norm, b a s e

116

__ , o r d e r u n i t

93

normed R i e s z s p a c e

order closed

39

3 , 1 6 , 26

o r d e r bounded

20

order closure

21

order continuous

26, 113

16

o r d e r convergence order unit

partition

18

93

102

p o s i t i v e cone

4

p o s i t i v e element

4

p o s i t i v e mapping

26

p r o b a b i l i t y Radon m e a s u r e s p r o j e c t i o n , band

29

quotient space of X

Radon m e a s u r e rare

186

348

refinement

102

regular element

379

177

187

421

rerminology

422

regular pair regular set

379 173

Riemann i n t e g r a b l e

316, 3 1 7

Riemann n e g l i g i b l e

324

Riemann s u b s p a c e

3 0 6 , 384

R i e s z homomorphism Riesz i d e a l

26

10

, principal

-

R i e s z isomorphism R i e s z norm

28

39

K i e s z seminorm Riesz space

39

5

, perfect

-

Riesz s u b s p a c e

seat

37

76

10

2 8 9 , 290

separating

50

s e t of constancy

178

simple convergence spectral state

166

10.5

186

,

pure

186

s t a t e space

186

support supremum

1 9 7 , 2 8 9 , 290 3

a-Dedekind complete n-order closed a-order closure

36 36

37

Terminology transpose

78, 1 8 3

U-band

293

u-band

267

unit

94

unive rs a 11 y rneasurab1e

29 2

universally integrable

233

u p p e r bound

3

uppersemicontinuous decomposition uppersemicont i n u o u s upper envelope Urysohn function

vague topology

166

88

vanish at infinity

wedge

227

zero set

170

172

176

280, 30.3

423

This Page Intentionally Left Blank

The bidual of C(X) I

The Care of the Self: of the History of Sexuality

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 Care of the Self: of the History of Sexuality

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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