Riesz spaces, Volume 2 (North-Holland Mathematical Library 30)

North-HollandMathematicalLibrary Board ofAdvisory Editors: M. Artin, H. Bass, J . Eells, W. Feit, P. J. Freyd, F. W. G...

Author: Adriaan Cornelis Zaanen | W. A. J. Luxemburg

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North-HollandMathematicalLibrary Board ofAdvisory Editors:

M. Artin, H. Bass, J . Eells, W. Feit, P. J. Freyd, F. W. Gehring, H. Halberstam, L. V. Hormander, M. Kac, J. H. B. Kemperman, H. A . Lauwerier, W. A. J. Luxemburg, F. P. Peterson, I. M. Singer and A . C . Zaanen

VOLUME 30

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD

RIESZ SPACES I1 A. C. ZAANEN

Leiden University, Leiden, The Netherlands

1983

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD

@

North-Holland Publishing Company, 1983

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 86626 4 Published by: NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD Sole distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 52, VANDERBILT AVENUE NEW YORK, N.Y. 10017

PRINTED IN THE NETHERLANDS

PREFACE

This i s Volume I1 of a r a t h e r d e t a i l e d work on Riesz s p a c e s ( v e c t o r l a t t i c e s ) . Volume I , w r i t t e n j o i n t l y w i t h W . A . J .

Luxemburg, appeared i n

1971. The c o n s i d e r a b l e t i m e i n t e r v a l between t h e appearance of t h e s e volumes

i s caused mainly by t h e f a c t t h a t i m p o r t a n t p a r t s of t h e t h e o r y were p u t i n t o a n e l e g a n t and more o r less f i n a l form o n l y v e r y r e c e n t l y . This i s t r u e f o r Chapter 13 on k e r n e l o p e r a t o r s ( i n t e g r a l o p e r a t o r s ) and s t i l l more f o r Chapter 18 on compact o p e r a t o r s . To mention one example, i t w a s unknown n o t s o l o n g ago t h a t , even i n a f a m i l i a r s p a c e l i k e a n L 2 - ~ p a c e , any p o s i t i v e o p e r a t o r m a j o r i z e d by a k e r n e l o p e r a t o r is i t s e l f a k e r n e l o p e r a t o r and any p o s i t i v e o p e r a t o r m a j o r i z e d by a compact o p e r a t o r i s i t s e l f compact. Chapters and s e c t i o n s i n t h e two volumes are numbered through. The p r e s e n t volume b e g i n s , t h e r e f o r e , w i t h Chapter 1 1 , s e c t i o n 76. T h i s does n o t imply t h a t e v e r y t h i n g i n Volume I i s needed f o r Volume 11. For t h e u n d e r s t a n d i n g of Chapters 12-20 i t i s s u f f i c i e n t t o b e f a m i l i a r w i t h Chapters 2-4 and p a r t of Chapter 6 ( F r e u d e n t h a l ' s s p e c t r a l theorem). Much of t h i s can a l s o b e found i n t h e f i r s t p a r t of Chapter I1 o f t h e book by H.H.

Schaefer ( [ I ] ,

1974) a s w e l l a s i n Chapters 1 and 4 o f t h e book by E . de Jonge and A.C.M.

van Rooij ( [ 1 ] , 1 9 7 7 ) .

,

Chapter 1 1 i s a n e x c e p t i o n ; i n t h i s c h a p t e r

w e d i s c u s s e x t e n s i o n s and g e n e r a l i z a t i o n s of what was proved i n Chapter 5 (and s e c t i o n 52 of Chapter 7) about prime i d e a l s i n d i s t r i b u t i v e l a t t i c e s and i n R i e s z s p a c e s . Chapter 12 c o n t a i n s a g e n e r a l i n t r o d u c t i o n t o o r d e r bounded o p e r a t o r s i n R i e s z s p a c e s . I n Chapter 13 t h i s m a t e r i a l i s immediately p u t t o u s e i n an i n v e s t i g a t i o n of k e r n e l o p e r a t o r s i n s p a c e s o f measurable f u n c t i o n s . T h i s c u l m i n a t e s i n A.V.

Buhvalov's 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 a n

Operator t o b e a k e r n e l o p e r a t o r . Chapters 14-16, mainly about normed Riesz s p a c e s ( i n p a r t i c u l a r Banach l a t t i c e s ) have t h e i r r o o t s i n a s e r i e s of n o t e s in t h e Proceedings o f t h e N e t h e r l a n d s Academy of S c i e n c e (1963-65;Notes

V

1-13

vi

PREFACE

by W . A . J . Luxemburg and A . C . Zaanen, Notes 14-16 by W . A . J . Chapter 17, based p a r t l y on c l a s s i c a l work of H.F.

Luxemburg). I n

Bohnenblust and S. Kaku-

tan5 around 1940, we d e a l with a b s t r a c t L -spaces, i n p a r t i c u l a r with ALP spaces and AM-spaces. Chapter 18 i s b u i l t around the r e c e n t r e s u l t s of P.G.

Dodds and D.H.

Fremlin on compact operators i n Banach l a t t i c e s . I n the

f i r s t p a r t of Chapter 19 we discuss O r l i c z spaces. In the p r o p e r t i e s of these spaces, g e n e r a l i z a t i o n s of the f a m i l i a r L -spaces, we recognize

P

b e a u t i f u l i l l u s t r a t i o n s of much t h a t was observed about a b s t r a c t normed Riesz spaces i n the e a r l i e r chapters. The second p a r t of Chapter 19 i s devot e d t o band-irreducible

operators ( i . e . ,

o p e r a t o r s t h a t leave no band in-

v a r i a n t except t h e n u l l band and the entkre space). Generalizations Of c l a s s i c a l eigenvalue theorems of 0. Perron, G. Frobenius (matrices with non-negative

e n t r i e s ) and R. Jentzsch ( p o s i t i v e k e r n e l o p e r a t o r s ) a r e de-

rived. These g e n e r a l i z a t i o n s (and even some s t i l l more g e n e r a l r e s u l t s ) a l s o occur i n the abovementioned book by H.H.

Schaefer; our p r e s e n t a t i o n

i s somewhat d i f f e r e n t because we have made an attempt t o keep a l l proofs f r e e of any r e p r e s e n t a t i o n methods. The attempt i s almost s u c c e s s f u l , with one exception i n Theorem 1 3 6 . 8 . A representation-free

(and Zorn-free) proof

of t h i s theorem i s d e s i r e d . F i n a l l y , Chapter 20 on orthomorphisms ( i . e . , o r d e r bounded operators t h a t a r e band preserving) contains r e s u l t s t h a t a r e mostly of very r e c e n t o r i g i n . Some o l d e r r e s u l t s t h a t were f i r s t proved

(1969-71) by m a n s of r e p r e s e n t a t i o n theorems a r e now presented i n a

representation- f r e e manner .

The p r e s e n t a t i o n of the m a t e r i a l i n some of the chapters has been s i g n i f i c a n t l y influenced by the work of s e v e r a l of my s t u d e n t s ( e i t h e r t h e s i s work o r otherwise). This holds f o r A.R. p a r t of Ch. and B.

1 9 ) , W.J.

Claas (Ch. 17), W.K.

Schep (Ch. 13 and the second

Vietsch (Ch. l a ) , C.B.

Huijsmans

de P a g t e r (Ch. 2 0 ) , E. de Jonge ( s e c t i o n s 119 and 133) and J.J. Grob-

l e r ( s e c t i o n 129). Furthermore, many d e t a i l s i n the t e x t come from conv e r s a t i o n s w i t h s t u d e n t s and colleagues. I n p a r t i c u l a r , the f r e q u e n t cont a c t s with W i m Luxemburg over a p e r i o d of many years have been very import a n t , a s e v i d e n t a l s o from the b i b l i o g r a p h i c a l r e f e r e n c e s . Several books dealing with Riesz spaces have been published since 1971. There i s some overlap, b u t not too much, of the p r e s e n t book w i t h the book on Banach l a t t i c e s and p o s i t i v e o p e r a t o r s by H.H. book by E. de Jonge-A.C.M.

Schaefer ( [ 1 ] , 1 9 7 4 ) .

The

van Rooij, very s u i t a b l e a s an i n t r o d u c t i o n t o

t h e theory, is from 1977. In the books by D.H.

Fremlin ( [ 1 ] , 1 9 7 4 )

and

v ii

PREFACE

C.D.

Aliprantis-0.

Burkinshaw ([ 11,1978) t o p o l o g i c a l R i e s z s p a c e s , more

g e n e r a l t h a n n o r m d o n e s , are i n v e s t i g a t e d . The books by H.E. 1974) and by J. Lindenstrauss-L.

T z a f r i r i ([11,1979),

Lacey (c11,

both about classical

Banach s p a c e s , a r e r e l a t e d . The same h o l d s f o r t h e book on k e r n e l o p e r a t o r s by M.A.

K r a s n o s e l s k i i , P.P.

(C11,1976).

Zabreiko, E . I .

P u s t y l n i k and P.E.

F i n a l l y , Yau-Chuen Wong and Kung-Fu Ng (C1],1973)

Sbolevskii discuss

p a r t i a l l y ordered topological vector spaces. Thanks ame due t o Sonja Wassenaar who p r e p a r e d t h e typed m a n u s c r i p t and t o t h e s t a f f a t t h e North-Holland

P u b l i s h i n g Company f o r t h e i r

cooperation. December 1982

A.C.

Zaanen

T ABLE OF CONTENTS

v

PREl' ACE

CHAPTER 11. PRill£ ll>EAL EXTENSlO'

6

16. Prime ideal separ ation 77. The hull-ke-rnel t opol ogy and the d\L&l hull-kernel t.opol ogy

19

78 . The unique priae ideal extension property 79 . Strongly order dense Rieu subs~cea

46

80. Extent iO'Il theorcas

S2

81 • PrUDe ideal extension and t he p rojection property 82. Hor.al and extrenally disconne c ted lattice•

57

32

63

89

CHAPTER 12. ORDER BOUNDED OPEAATORS

95

83, Order bounded opera t or s 84. Order cone i nuoua o pe rators

123

65 . The orde r dual of a Riesz sptlcre

131

86 . Integrals o n ideals o f

me~s ura!ble

f unction•

81. lntearah tmd singular: l i near function•h

88 . The largect. ideal on which every order bo~ded oper4tor ie nrder continuous 89. Annihilator• and weak topologies 90. The carrier of an order bounded linear luncliooal

141 14 6 158

169 176

91 . CO.plex lie•z spaces

187

92. Coaplox order bounded operators

201 209

CHAPTER 1 3. KEMtL OPEAATORS

93. Kernel oper,tors 94. The b4nd ot absol ute kernel ope rators 9S . The b ~nd guncrutcd by the kor nel operators ot f in i t e rank 96 . 8uhva l ov •s

theo r~

91 . Adjoint opera tors

ix

21 2 214

228 234 248

TABLE OF CONTENTS

98. Dunford'• theorc.

257

99.

262

Ctnerali~•d ca~ leaan opera~ors

CIIAP'tl:R 14. NORIUID R IESZ SPACES

279

100. Normed Riesz s p~ces 10 1. Banach lattices

28 1

102. The Banach dual

310

302

CHAPTER 15 . ORDER CONTlNUOUS NORMS

333

10). OrdQr continuous •lornas

335

104.

Meycr-Ni~berg ' s 1~

and disjoint sequence•

353

105. Ando'a thcoreQ and the order ~opology 1 0~. Order continuity and tbe ordeY dual

372 381

CIIAPTEI\ 16. EI

o ,

. We have

Po

. Write

w

=

kfl)

=

N n {PIw

EXERCISE 76.9. by

Q,

k(QU)

=

{ u p . Show that

and let {PIv n {PIu

. Since o,

w >

{uId

.

Let

,

{PIw

=

h'

N be

is dense in

be a (relative) neighbor-

i.e., w

{PIv n {PIu is not in

that there exists P E

is non-empty, we have

{uld

.

It follows, on

N not containing w ,

u > 0 be given in the Riesz space L

the set of all u-maximal ideals in L =

.

to show that this neighborhood contains a point of

inf(u,v)

inf(w,u)

so

account of i.e., P E

satisfying k(N) =

Po E {PIu

HINT: Let hood of

.

,

and denote

Show that the kernel

n (Q:Q€Qu) consists of all f E L such that either inf(lf1,u)

(i.e., f E {ujd) Hence, k(QU) Show that

=

Qu

.

I is x-maximal for some x

We use the same notations for prime ideals in a Riesz

proper prime ideals P u € L+

.

for some a E {a)

or

inf(lf1 , u )

is infinitely small with respect to

{ u } ~ holds for all u > 0 if and only if is dense in

{PIu

for every

u > 0

=

u

.

0

L is Archimedean.

if and only if

L

is

Archimedean. HINT: The intersection Q' = Q fl A, principal ideal A

is an ideal i n

A,

of an u-maximal ideal Q that does not contain u

.

and the Q'

is

14

PRIME IDEAL SEPARATION

u-maximal in AU

.

Ch. 11,9761

Indeed, if not, there is a properly larger ideal Q"

in

AU that is u-maximal. Hence, Q" is a prime ideal in A, , and so can be This extended (even uniquely by Theorem 52.2 (ii)) to a prime ideal in L prime ideal is properly larger than Q diction. Hence, Q'

is u-maximal in A, , i.e., Q'

AU

the unique extension, every u-maximal ideal

. Conversely, by

can be extended to a u-maximal ideal Q 0

5

v E Q

consists of the zeroelement and all

.

with respect to u Qu

inf ( I f I , u )

such that

is dense in

{PIv fl {PIu with

{PIu

f E AU

is a maximal ideal in

. By

Q fl AU

=

=

0 or

Q'

in Au

Theorem 52.2 (ii),

,

for all w E A,

Now, observe that

Q'

fl

that are infinitely small with

It follows that k(QU)

u (cf. Theorem 27.5).

f E L

all

.

inf(v,u) E Q'

if and only if

respect to

in L

inf(v,w) E Q'

holds if and only if

i.e., v E Q

. . Contra-

and does not contain u

inf ( 1 f 1 , u )

fl

=

Q consists of

is infinitely small

if and only if every relative open set

inf(v,u) > 0 contains at least one element of

'

QU

i.e., if and only if there i s at least one u-maximal ideal not containing the element

inf(v,u) > 0

follows that Qu k(Qu)

=

{uId

is dense in

for every

EXERCISE 76.10. by

Qu,v

. Note

Let

u > 0

that

{PIu

.

0 0 not in an ideal Q {vId c {uId

{,Id

.

Hence, since w

is not in

{uId

is not in

,

{vId

.

Since Q

EXERCISE 76.11.

is an ideal, Q

and

w

Q

. But

not in

not containing nor w

.

We generalize the preceding exercise. Let u,v

set of all u-maximal ideals in L

, and

that

then, by the

contains neither u

positive elements in the Archimedean space L

is

u,v

. Note

i.e., inf(w,u)

preceding exercise, there exists a v-maximal ideal Q inf(w,u)

that are

and

{u}~ there exists

that is v-maximal and does not contain u

, it follows that inf(w,u)

. Denote

in the Archimedean space L

. Show that

v-maximal and do not contain u

. It

{uId

is the set of all ideals in L

v-maximal. Equivalently, Q,,,

{PIu

is not in

for every u > 0 if and only if

the set of all ideals in L

dense in

inf(v,u)

denote by

that do not contain v

.

S

UIV

be the

Note that

S may be empty .(for example, if u I v ) . Show that u,v k(SU,v) = {inf(u,v)Id , where it is understood that the kernel of the empty

Ch. 1 1 , 5761

L

set is

15

PRIME IDEAL EXTENSION

. Show that

S

is dense in

u,v

and HINT: If u I v , then S u ,v therefore, that w = inf(u,v) > 0 , so

"'inf(u,v)

*

{PIinf (u,v) 0

0 and v t 0 in the Archimedean space L

,

R the set of all u-maximal ideals Q such that v is u,v contained in any ideal properly larger than Q (it is possible that v is

and denote by

already contained in Q ) . Show that k(R in

. The result that

{PIu

k(Ru,v)

=

U,V

{uld

) = {uId and R is dense u,v is due to A. Bigard and

K. Keimel (Cl1,Lemma 4 ) . R is the set of all u-maximal ideals Q such that u,v is contained in any ideal properly larger than Q , i.e.,

HINT: Note that w

=

sup(u,v)

R u,v

is the set of all ideals that are u-maximal as well as w-maximal. EXERCISE 76.13.

Show that

Let F

inf(lf1:fEF)

HINT: If

L

= 0

.

be a maximal dual ideal in the Riesz space I,

.

is Archimedean, this is trivial since n-lu E F+

for any

,... . In the general case, assume that u > 0 is a . There are two cases: u in F+ or u not in F+ . If u E F+ , then u = inf(v:vEF+) . But this is impossible since 1. 0 . It follows that u = 0 is the only lower bound of F+ in L+ . u

E F+ and n

= 1,2

lower bound of

F+

EXERCISE 76.14. different elements

Let X

be the distributive lattice consisting of five

(f3,x0,x1,x2,e) with

8

as smallest element, e

largest element, and furthermore xo < x 1 and page 16). Show that

w

=

as

xo < x2 (see the diagram,

{ e l is a prime ideal. Show that there are two

0 more proper prime ideals w I

and w 2

such that neither

w 1 = w2 nor Show that by adding yo,yl and y2 as in the diagram, X is 2 c w1 embedded in (obviously) the smallest Boolean algebra B(X) containing X

w

.

Determine the prime ideals in B(X) ideal in B(X)

~

and observe that, although X

each of the prime ideals in B(X)

.

is not an

is a unique extension of

PRIME IDEAL SEPARATION

16

.

one of the prime ideals in X wo

F

EXERCISE 76.15. Let

f E F

extra property that C

Note that

is not included in the extension of

= F+

, but

wo c w 1

the extension of

.

w1

be a dual ideal in the Riesz space implies af E F

. Define

is a cone in L

U {O}

Ch. 11,5761

f

05

g

to mean

that, with respect to this partial ordering, (i)

C

L with the

. Show that that g-f E C . Show

a # 0

for all

is a generating cone

L , (ii) L has the Riesz decomposition property, (iii) L is an anti-lattive (i.e., sup(f,g) exists if and only if f 20 g or f 0 5 g for

0 05 u 05 vl+v2 with

HINT: For (ii), let u = o

and

, v1

0

=

p and u

5

, v2

= 0

or

are all in F+

v +v -U 1 2

=

v1,v2 E C

. The cases

u = v1+v2 are trivial, s o assume that

. Then

).

that

u,vI,v2

inf(v +v -u,iu,ivl,iv2) E F+ I 2

v +v -p , so 1 2 u-p

5

.

(v,-p) + (v -p) 2

By the ordinary Riesz decomposition property in L

there exist elements

q 1,q2 such that

0 Hence, u 1

=

2

q 1 5 v -p

q,+ip

1

and

,05

u 2 = q2+:P

u = u1+u2 with

Since

~ , ~ ~ 2 ” J I -and UI

q2

ip

v2-u2

5

v2-p

5

and

u-p = q1+q2

satisfy uI

5

vI-ip and

fp

5

u 2 5 v2-iP

are all greater than or equal to

4P

,

all

Ch. 11,1761


these elements are in F+

. Hence

For (iii), bound of

assume that

u1 and

is equal to one of

u1

or

v

05

05

u1

v1 and

05

and

.

u2 0 5 v

0 C

are in the cone

u1,u2

, so ul

u2

0

17

and

u

05

v

.

v2

is an upper

Prove that, unless v

u2 0s u l

u2 (in which case

05

or

u1

05

u2 ),

there is a properly smaller upper bound, for example v-fp for p

=

.

inf(v-ul,v-u2) EXERCISE 76.16.

In Exercise 15.14 we presented an example of an

ordered vector space with the Riesz decomposition property, but not a Riesz space. Here is another example. Let

-

L

be the ordered vector space of all

real differentiable functions on the interval that

a

=

- m

and (or)

b

=

difference if we take the interval decomposition property, but

. It

(x:a<x

S

,

.

so

PRIME I U I ~ A LEXTENSION

Ch. 11,1771

THEOREM 77.2. The hull-kernel topology in

itself and a l l sets

that A

,

X I # X2

PROOF. I f

there e x i s t s a point i n h2

,

contains

X2 XI

b u t not

X2

are compact.

{XIx

. Similarly

XI

b u t not i n

XI

t h e topology i s a T -topology. 0

We prove now t h a t x E X

,

A

,

. If

X2

but not i n

is in

x

i f there i s a point i n

XI

s e t s of

XI

XI

or

but not i n

, i . e . , t h e set t h e o r e t i c complement of f X I x , , so ({X}x)c i s a neighborhood of X2 not

containing

. Hence,

is a T 0-topology such

A

there e x i s t s a point i n

({XIx)‘

t h e open s e t

21

i s compact. The s e t s

A

X2 not contained i n

{XIx

, being

a r e then a l s o compact. Note f i r s t t h a t t h e s e t s

form a base f o r t h e open sets i n t h e hk-topology;

closed sub(iXfx)c

,

f o r t h e compact-

ness proof i t i s s u f f i c i e n t , t h e r e f o r e , t o show t h a t any covering of

by

h

these base s e t s has a f i n i t e subcovering. L e t

be such a covering of Then t h e family any i n t e r s e c t i o n to

{XIz

for

Z

t h a t t h e subset easily that

D

,

A

and assume t h a t no f i n i t e subfamily covers

((XIy:y€D)

ny D

i s non-empty.

{XIyi

= y,

A

of

... A

X

A

.

has t h e f i n i t e i n t e r s e c t i o n property, i . e . ,

, SO

yn

Z

But t h i s i n t e r s e c t i o n i s equal =

y1

A

... A

yn > 0

. This

shows

has t h e f i n i t e i n t e r s e c t i o n property. It follows

i s included i n a t l e a s t one maximal lower s u b l a t t i c e

.

lo ’

Hence, i s then a maximal dual i d e a l , we have D C Xo E A Xo the i n t e r s e c t i o n n ({XIY:y€D) contains ho , s o t h i s i n t e r s e c t i o n i s

and s i n c e

non-empty.

Hence, A Let

This c o n t r a d i c t s

i s compact. Am

be t h e subset of

The r e l a t i v e hk-topology i n

XI,A2

Am

A

c o n s i s t i n g of a l l maximal dual i d e a l s .

i s c a l l e d t h e hk-topology i n

be maximal dual i d e a l s such t h a t

i s properly l a r g e r than

XI

and

X2

,

XI

so

# X2

. Then

X, U h 2

f i n i t e i n t e r s e c t i o n property. Hence, t h e r e e x i s t

Am

. Let

t h e union

X I U X2 does not have t h e

x 1 and

x2

in

X I U X2

22 xI I x2

such that

Let x1 E X I XI

and

{XIx1 X1

Ch. 11,8771

HULL-KERNEL TOPOLOGY x 1 E X, U X,

, we have x I E

XI

.

or x I E X p

since x 1 n x2 = e

cannot be in X I

Then x2

is a lower sublattice. Hence, we have x2 E X2 and {XIx2 are closed sets in the hk-topology of

and

{XIx2

, say.

. Since

. It X

follows that , containing

m X

X i respectively, such that the intersection of {XI and i s empty. Hence, {XIx1 is a closed set containing and not XI

containing X2 containing X I

. It follows that

the intersection of all closed sets

consists only of the point

hk-topology in Am

XI

itself, and so the

is a T -topology. Furthermore, the space Am 1

in its hk-topology; the proof is the same as above (note that Xo last proof is an element of A

is compact in the

). We have, therefore, the following theorem.

m

THEOREM 77.3. The hull-kernel topology i n

Am i s a compact

T -topology. 1

From Theorem 7 6 . 1 it follows immediately that the mapping

JI: X

+ w = X-X

is a one-one correspondence between the elements of A

R

the elements of by

J1

. The subset

onto the subset Rm

$({XIx)

of

mapped onto

{XIx

{oIX U { w )

hk-topology of A with

R

il

,

to

Y

of all maximal dual ideals is mapped

of all minimal prime ideals. The image

is exactly

{ w l X f l {wlY = {wlMY

Am

and

. Hence

{wIX

= {w}xvy

. The mapping

and

JI

{XIx

U {XIy = {XIxvy

{XIx fl {XIy = {XIxAy

is

onto

makes it possible to transplant the

R by defining the sets {wIX , x E X , together

to be a base for the closed sets. The topology generated thus in

is called the dual hull-kernel topology in il (briefly the dhk-topology). THEOREM 7 7 . 4 . The dual hu1l-ksrne1 topology i n

i s a compact , are then To-topoZogy. The s e t s { w l x , x E X , being closed subsets of dhk-compact. Furthermore, the dual hull-kernel topology i n the s e t Rm of il

a l l minimal prime ideaZs i s a compact T1-topoZogy, ueaker then the huZ1 kernel topology i n om

.

PROOF. We have only to show that in Rm

the dhk-topology is weaker

then the hk-topology, i.e., we have to show that every dhk-closed set i s hk-closed. Any dhk-closed set in Rm

is either

am

the sets f p I x

itself (and

Qm

.

is

{ u l x = {oIx fl Rm But are closed (as well as open) in the hk-topology of Rm

indeed hk-closed) or else an intersection of sets


Ch. 11,5773

(cf. Theorem 7.4(ii)),

23

any intersection of these sets is hk-closed. This

so

is the desired result. In Exercise 77.13 it will be indicated how to prove that the hk-topology and the dhk-topology in

{uIX

if the set of all

,

x E X

For any

the set theoretic complement of {uIx

will be indicated by

are identical if and onIy

sk,

is a Boolean algebra.

,

form a base for the open sets in the hk-topology of {ulx

,

{wIX = (w:x not in w )

. The sets

i.e., {oIX = (u:x€w)

Q

{wIX , x

,

E X

and the sets

together with the empty set, form a base for the open sets in the 61

dhk-topology of

.

The next theorem is due to J. Hashimoto ([11,1952,Theorem THEOREM 77.5. Let

and

S

be subsets of

T

X

4.2).

such that

i s non-

S

empty (T may be empty), and assume that the f a m i l y of s e t s

(

{w

lX, { w'1

:x€S ,y€T)

has the f i n i t e intersection property. Then the f m i Z y has a non-empty intersection. PROOF. Note first that for T

empty this is Theorem 6.8. In the

general case, we begin by observing that since {w),

n

{ w } , ~ = {w)xIAx2

I property. Let

F

,

the subset S of

X

is non-empty and

S

has the finite intersection

be the dual ideal generated by

all upper bounds of finite infima of elements of ideal generated by

T (where it is understood that

S (i.e., F

consists of

S ), and let

I = {el

if

I be the T is

empty). We show that F n I is empty. This is trivially true if If F n I is non-empty, there exist elements x l , y l ,...,y

m

E T

such that

XI A

and so

... A

Xn S y ,

V

...

V

ym

,

...,x

E S

and

I

=

{el

.

24

Ch. 1 1 , 9 7 7 1

HULL-KERNEL TOPOLOGY

contradicting the finite intersection hypothesis. Hence F n I is empty, so

by the prime ideal separation theorem there exists a prime ideal

such that

Ic

and F n wo

wo

=

0

. But

wO

then

This is the desired result. THEOREM 7 7 . 6 . Let

of

62

and T be subsets of

S

X

. Then any

subset A

of the f o m

is campact in the dhk-topology. For

S

non-empty, the set A is also

compact in the hk-topology. PROOF. We may assume that A open covering of A

with

2

is non-empty. Let

in the dhk-topology. Then A

n

U({W}~:Z€Z)

(n{wlz) =

be an

0 , i.e.,

non-empty. Hence, by the preceding theorem, there is a finite

intersection that is already empty (note that this finite intersection must contain at least one

{ w } ~). It follows that for appropriate '1

we have

=

(An{w},;)

0 ,

so

A c Uy{w)

. This shows that

2.

A

* * *

"n

is

dhk-compact. The proof that A

is hk-compact if

is non-empty is similar.

S

R is dhk-compact. is hk-compact.

COROLLARY 77.7. (i) Every hk-closed subset of R

(ii) Every dhk-closed proper subset of PROOF. (i) Every hk-closed subset of is of the form n ({W}~:~€T) with

R

is either the empty set or it

T non-empty. In both cases the set is

dhk-compact. (ii) Every dhk-closed subset A of the form A

=

n

:xES) with S

({w}

be a proper subset of preceding theorem, A

of

Ci

,

then A

=

62

is either D

non-empty. If A

D

is now hk-compact.

itself or else is also given to

is impossible. Hence, by the

25


Ch. 11,9771

The next theorem, which is of a topological nature, will be of great importance for the proof of the unique prime ideal extension from a distributive lattice

(X,e)

to the Boolean ring generated by

X

. We

shall

have to deal with a T -space possessing a base of open and closed sets. 0

Note that in this case the space is Hausdorff. Indeed, if x

and

y

are

different points in the space, then one of the points, say x , has a neighborhood V

occurring in the base and not containing y

V

complement of

is then an open neighborhood of

THEOREM 77.8.

Let

X

closed and compact subsets of

B(N)

(ii) B(N)

.

N of

be a Hausdorff space possessing a base

B(N)

open, closed and compact s e t s , and l e t

(i)

y

. The

be a coZlection of open,

such that

X

contains all s e t s of N , i s a BooZean ring with respect t o ordinary s e t t h e o r e t i c

union and intersection. Then the Stone representation space of the space

The mapping from each x E X

X

onto

51

i s homeomorphic t o X

x

.

,

PROOF. Given any xo E X

is a proper prime ideal in

consisting of a12 A E B(N)

B(N)

the set of all A E

.

B(N)

Conversely, given the proper prime ideal w the existence of a point

, i.e.,

.

i s homeomorphic t o X

giving r i s e t o the homeomorphism assigns t o

the prime ideal i n

containing the point

xo

B(N)

of a l l proper prime ideals i n B(N)

52

xo E X

such that

B(N)

not

not containing

in B(N) , we shall prove consists exactly of all

w

. For this purpose, let A = B(N)-Iu be the . Since A is a dual ideal it has the finite intersection property, i.e., fly Ai # 0 for A ,,...,A E h . Given AO,AI,...,A E A , the sets .A n A. (i=l,...,n) are relatively closed n subsets of the compact set A. with (A nA.) # 0 . It follows that 0 1 A E

B(N)

not containing xo

dual ideal corresponding to

n (A0nA:A€A)

w

is non-empty, i.e., S

=

n

(A:AEA)

is non-empty. We show that

S

consists of only one point. If not, there exist points xo,yo E S with

Xo

# yo

and

yo

(and so

. Hence, there exist disjoint neighborhoods respectively such that A 1 and

A 1 and Ag 1' =

are in

B(N)

(A:AEB(N),~~EA)

.

A2

A,

and A2

of

xO

are elements of the base

) . Now, let

N

26

Then A ' B

is non-empty (since A

, A 1 and A2 in

A'

3

B(N)

A

. The dual so

xo E S

= I7

.

A1

implies A E A '

, and finally A A'

is not in

,

E A'

is a dual ideal in implies xo E A

A U B E A'

,

so

the prime ideals A

The inclusion is proper, since A ' that has

but not in A

is in A '

X

or

(A:AEA) ,

element (namely, A 1 ) so

), the empty subset of

is prime,(kince

A E A

A c A'

satisfy

A'

B E A ' ) . Furthermore, since

or

it is evident that

E

. This shows already that

ideal A '

A E A'

1

implies A 1 fl A2 E A '

A'

implies B E A '

xo E B ,

A'

Ch. 11,8771


xo

and

contains an

as one of its points but not y

0 '

. Proper inclusion, however, is impossible,

since in a Boolean ring dual prime ideals are maximal. Hence, S

consists

of one point. It has been proved thus that there is a one-one correspondence between the prime ideals w A:AEB(N),x

{wlA

B(N)

and the points

w =

such that

not in A

is such that x

is not in A

. Hence

(w:AEw)

corresponds to (x:x not in A ) . But then This shows that the base sets to the base sets

x E X

, we have A E w if and only if the point x

Given now A E B(N) corresponding to

in

A(AEB(N))

{wIA

{wIA

corresponds to (x:xEA)

of the hk-topology in B

of the given topology in X

,

=

A

.

correspond

i.e., B

and X

are homeomorphic. We return to the situation that smallest element 8

and

B

X is a distributive lattice with

is the set of all proper prime ideals in X

Besides the hk-topology and the dhk-topology in B

.

we introduce the weakest

topology stronger than the hk-topology and the dhlc-topology. This is called the Boolean topology (briefly 5-topology) in B

. The family

is a subbase for the open sets of the 5-topology. Note that { u } ~= R

, so

4

and

$l are

=

Q,

and

included in the subbase. Note also that the

Ch. 11,5771


27

sets in the subbase are closed as well as open. The family of open and closed sets

is a base for the open sets in the 8-topology. There is a somewhat smaller

base. Indeed, since

we have

for every x

,

and so

is also a base. All sets in this base are open and closed. We prove that they are compact as well. It will be sufficient to show that every

{w}

is compact. For the proof (due to S . J . Bernau,C21,1972) we shall need Alexander's subbase theorem, stating that if the subset A

XO

of a topological

space has the property that every covering by sets of a subbase (for the open sets) has a finite subcovering, than A

is already compact (for a

proof of Alexander's subbase theorem, cf. for example the book by J.L. Kelley, General Topology, Ch. 5,Theorem 6 ) . In our case, let U B be a covering of

Then

{w),

0

{w}

x8

by sets of the subbase

n (n Bc) is empty. This intersection is of the form n{w}xn{wjY:xEs,yET

for certain subsets S

and

T of X

. Hence, this intersection being

empty, a finite intersection must already be empty by Theorem 77.5, i.e., {w),~

has a finite subcovering. The sets in the base

28

Ch. 11,5771


and, therefore, open, closed and compact. Furthermo

he 6- opology is

To

,

S2

with the 6-topology is now homeomorphic to the Stone representation

and hence is Hausdorff. According to the preceding theorem, the space

space of the Boolean ring of its own open, closed and compact subsets. There is a possibly smaller Boolean ring of open, closed and compact sets

N

containing

. The family of

N is a Boolean

all finite unions of sets of

{wlx

ring, the smallest Boolean ring containing all sets

.

is called the Boolean ring generated by the lattice of all

This family S

{wlx (for an

indication of the proof that S is indeed a Boolean ring, cf. Exercise

77.16). Note that all sets in S are open, closed and compact. Hence, il

s.

with the 6-topology is homeomorphic to the Stone representation space of Once more, we make some topological remarks. The non-empty subset Y

is called irreducibZe if for each pair of open

of the topological space X

, O2

such that 0 n Y and 0 n Y are non-empty it 1 2 follows that 0 f7 0 n Y is non-empty. In other words, Y is irreducible 1 2 if and only if for each pair of closed subsets FI,F2 of X such that subsets O 1

X

of

. Evidently, Y is irreducible if and only if for each finite family 0 1 , ...,0 of open subsets of X such that 0. n Y is non-empty for i = I, ...,n it follows it follows that Y c F 1 or Y c F2

Y c F 1 U F2

(nnI

that

0 . ) fl Y

is non-empty or, equivalently, if and only if for each

1

...,Fn

finite family F 1 ,

follows that Y c F.

of closed subsets of

X

for at least one value of

Each subset of X

such that Y c U p i i

=

1,. ..,n

consisting of one point is irreducible. If X

a Hausdorff space and Y

is an irreducible subset, then Y

it

.

is

consists of

one point. Every (relatively) open subset of an irreduaible set is also irreducible. If space X I and Y

Y 1 = $(Y) in X I

Q-](F2)

Q

is a continuous mapping from X is an irreducible subset of

is irreducible in X 1

such that Y 1 c F 1 U F2

included in Q-I(F,)

or

Q

that Y 1 c F 1 or Y 1 c F2

(F2)

, so

,

into the topological then its image

. For the proof, let FI,F2 be closed sets . Then the inverse images Q -1 (F1) and

are closed in X , and Y c $-'(F1) -1

X

,

since Y

U Q-](F2)

. Hence

Y

is

is irreducible. It follows

Y 1 is irreducible.

Ch. 11,5773

29


THEOREM 77.9. The folZaJing conditions f o r a subset

of a

Y

topological space are equivalent. (i) Y i s irreducible. (ii) The closure Y of Y

is irreducible. 0 i n X the intersection

(iii) For every open s e t

.

empty or dense i n Y (iv)

Every (reZatively) open subset of

is open and 0 n Y and 1 on^^. (iv) Y

0 0

0

n Y is e i t h e r

i s connected ( i . e . , i f

Y

0

n Y non-empty, there cannot e x i s t open 01,02 such t h a t n Y are non-empty, d i s j o i n t and with union equal t o 2

Every f i n i t e f a m i l y of non-empty (reZativelyl open subsets of

has a non-empty intersection. PROOF. The equivalence of (i) and (ii) follows by observing that for

any open 0 the intersections 0 Il Y and

0 n Y-

are simultaneously empty

or non-empty. Now consider condition (iii). To say that, for 0 open, 0 is dense in Y

meets every other non-empty relatively open subset of Y equivalent to (i).

Evidently, (iv) holds if Y

let (iv) hold and assume that Y 01,02

such that 0 n Y 1

Y

,

i.e., (iii) is

is irreducible. Conversely,

is not irreducible. Then there exist open

and 0 fl Y

1

2

are non-empty and disjoint. This

n Y is not connected, which contradicts (iv).

(0 UO )

would imply that

ll

is non-empty, is the same as saying that 0 n Y

0 f'l Y

if

2

Finally, (i) and (v) are evidently equivalent. The set theoretic union of a chain of irreducible subsets of irreducible. Hence, by Zorn's lemma, every irreducible subset of contained in a maximal irreducible set. Evidently (since Y

-

if and only if Y

X X

is is

is irreducible

is irreducible) every maximal irreducible set is closed.

We now consider irreducible subsets of

Q (with respect to the

hk-topology). THEOREM 77.10.

(i) The non-empty subset

and only i f the kernel (ii) The subset

of

k(Y)

Y

of

there e x i s t s a prime ideal ( i . e . , Y consists of a l l

Q

wo w

3

Y

Y

of

a i s irreducible i f

i s a proper prime ideal i n X

.

i s closed and irreducible if and only i f such t h a t Y i s the closure {w0}- of wo 1. Hence, Y i s a rnmirnal irreducible

wo

30

Ch. 11,1771


subset of

i f and on2y i f

0

Y

PROOF. (i) Let first k(Y) non-empty. Then x

and

it follows that x one wo E Y

, the

y

y

A

f o r some minima2 p r i m e idea2 uo

= {pol-

{wlX n Y

be prime. Let

and

y

A

,

is not in wo

A

y

is

are not i n

k(Y)

is not in k(Y)

Y non-empty intersection with such that x

y

A

since Y

Y

and y

y

{

~

= IwIx 3

be closed and irreducible. Then k(Y)

satisfying w

w

wo

. But

wo =

k(Y)

2

2

. It follows that

Y

Y

,

=

h(k(Y))

since Y

being in the hull of

consists of all w

Conversely, if

Y = {wo}-

and k(Y) = wo

is prime,

3

wo

for some prime ideal so

,

are not in k(Y) so

,

then

n~ {wly~ has~ a w E Y

k(Y)

.

is a prime ideal,

. Hence, all

is irreducible. Let us call this prime ideal wo

E Y satisfy

w

x

and

. Hence, there exists a prime ideal . It follows that x A y is not in

Y

is not in w

(ii) Let Y w

. If

n Y are non-empty, and

{wl

and

is irreducible.

be irreducible. In order to show that k(Y)

prime, it will be sufficient to show that if x {wlX n Y

least

i.e.,

and so this intersection is non-empty. It follows that Y

then x

n Y be Y is prime,

. This implies that, for at

is not in k(Y)

element x

Conversely, let Y

{w}

are not in k(Y) and s o , since k(Y)

.

is closed. Hence, every

, is

k(Y)

, i.e., Y = wo , then

{w,}

Y

a member of

-

.

is closed

Y is irreducible.

Note that since a prime ideal may contain several different minimal prime ideals, it may occur that two different maximal irreducible subsets of have a non-empty intersection. EXERCISE 77.11.

{ w } ~c {wIY

,

In the distributive lattice

i.e., { w } ~c {wIx

EXERCISE 77.12.

Let w 1

distributive lattice

(X,e)

the hk-closure of wI

,

of

w

,

and

(X,e)

if and only if x I y w2

,

.

show that

be proper prime ideals in the

. Show that

w 1 c w2

and also if and only if

if and only if w1

w2

is in

is in the dhk-closure

2 '

EXERCISE 77.13.

Let

am

in the distributive lattice minimal prime ideals

~1

be the collection of all minimal prime ideals (X,e)

. By

{pix

we denote the set of all

that do not contain x

. Show that the collection

Ch. 1 1 , § 7 7 1

31


{ v } ~ is a Boolean ring if and only if the hk-topology and the dhk-topology coincide on each { u l X Hence, the collection of a l l {u)x of all

.

is a Boolean algebra if and only if the topologies coincide on the whole of

Qm

. This result is due

to T.P. Speed (C11,1969).

HINT: Assume that the sets

{ u l x form a Boolean ring. Let {u}xo 0 of {u} is

fixed. We have to show that every hk-open subset (relatively) dhk-open. Any hk-open { u j Y T with

y,

{pix

the sets

for all

xo

5

with respect to

{plxo

T

xO

0 of this kind is a union of sets

. Note now that the complement of

is of the form

{ul YT { u I z T for some z E X , because

form a Boolean ring. Hence, the complement ' 0

respect to

{ulx

dhk-closed,

so

0

Oc

is the intersection of all

{u)zT

. Every

of {P),

0 with

x

5

xo

are dhk-compact by Theorem 7 7 . 4 , s o all sets are (relatively) compact for the dhk-topology in

The dhk-topology and the hk-topology in

{ulX , all

8 5 x 5 xo

{pix

xO

{ulx

Conversely, let the hk-topology and dhk-topology coincide in 5

{u1

0 is (relatively) dhk-open.

Note that all sets

{ulX , 8

is

T

is dhk-closed. Hence 0 is the intersection of

and a dhk-open set, so

, are

be

{pix

0

0

.

{ulx

0 are identical, so all sets

is arbitrary, it follows that xo are hk-compact, and so the collection of all {pix is a Boolean hk-compact. Since

ring by Corollary 8 . 3 . For the proof that the collection of all Boolean algebra if the topologies coincide on

Cim

, note

that

{pix

Qm

is a

is

compact in the dhk-topology (Theorem 7 7 . 4 ) . Hence, if the topologies coincide, then $2, all

{pix

is compact in the hk-topology, and so the collection of

is then a Boolean algebra (Theorem 8 . 4 ) .

EXERCISE 7 7 . 1 4 . Let X

be an infinite point set in which a topology

is introduced by defining that the open sets are the empty set and all subsets of

X possessing a finite complement. Show that X is an

irreducible Tl-space. EXERCISE 7 7 . 1 5 . Let X element 0

. Show that if

X

be a distributive lattice with smallest is linearly ordered, then Ci

is irreducible

in the hull-kernel topology. The converse does not hold; the lattice in Exercise 7 6 . 1 4 is a counterexample. EXERCISE 7 7 . 1 6 . Let and y

.

N be the family of all sets { w l x n {uly for

x

running through the distributive lattice X with smallest element

32

0

. Show that

s

the family

s

Boolean ring (i.e.,

s

Ch. 11,5781

THE UNIQUE PRIME IDEAL EXTENSION PROPERTY

of all finite unions of sets of

N

is a

is a distributive lattice with respect to set

theoretic union and intersection, the empty set is the smallest element of and the set theoretic difference of two sets in

s

is again in S ) .

HINT: It is easy to see that union and intersection of sets in S again i n

. Note now

S

of M

complement MC N

= {w}

N

{wIw E

fl

that if M = {wIx n {w}' with respect to

then N-M

=

N , and hence N-M E S

of two sets i n Then N-B

=

(N-Ni) E S

then A-B

=

U y (NI-B) E S 3

R

N n MC

.

,

are

then the set theoretic

is =

{wIx U { w ) so if Y' {Nn{wlX) U {Nn{wIy} is a union

Now let N E N

. It follows that

.

E N

if

and

A = Um N!

1

J

B

=

U y Ni E

E S and B E

s.

s ,

78. The unique prime ideal extension property

Let 8

,

(X',e)

and let

denote a distributive lattice X'

with smallest element

be a sublattice. It was proved in Theorem 51.8 that

(X,e)

every proper prime ideal w

in X

in X' (i.e.,

the extension is unique if

X'

. The

w'

fl

X

= w );

can be extended to a prime ideal w ' X

is an ideal in

converse is not true, that is to say, the extension can be unique

without X

being an ideal in X

.

The lattice in Exercise 76.14 is an

example; for this very simple lattice every proper prime ideal can be uniquely extended to a prime ideal in the Boolean ring generated by the lattice. This holds generally. We present the proof. THEOREM 78.1.

Every proper pfime idea2 i n the distributive l a t t i c e

has a unique extension t o a prime idea2 i n the Boozean r i n g

(X,e) generated by

X (i.e.,

B(X)

PROOF. As before, let R

B(X)

is the sma2Zest Boolean ring containing X

).

be the set of all proper prime ideals in X

.

As observed in the preceding section, the family of sets

is a base of open, closed and compact sets for the @-topology in R

. Note

now that, by the Stone representation, X is isomorphic to the family of all {wIx Hence, identifying the point x E X with {wlx for every x E X ,

.

Ch.

11,5781

we can identify each point of the Boolean ring of sets

33


the Stone representation space of explained there). Theorem 77.8,

B(X)

with a finite union

(cf. Exercise 77.16). It follows by Theorem 77.8 that

(wlx fl{wlY

B(X)

is homeomorphic to

Consider now a prime ideal P w1 E R

there exists a point

in

(in the sense

. According to

B(X)

such that the points of

P

are

-

{wjX n {w}'

exactly all finite unions of sets

that do not contain w 1 Conversely, any collection of all finite unions of this kind is the

.

collection of points of a proper prime ideal in B(X)

,

Given wo E Q

let P

be a prime ideal in B(X)

corresponds to a certain w 1 E Q

Then P

i.e., the points of

P

that extends

wo

are exactly all finite unions of sets

{ w ) ~ll

that do not contain w 1

. The points

with the points of

under the identification, i.e., with the points

wo

.

of

*

in the manner indicated above,

P n X = wo

correspond

({~}~:xEw~) On the other hand, these points are the points

exactly

{wlx f l {wly

of the special form

i.e., all

Idx

and so

- wo

w1

satisfying x E w 1

. This shows that

uniquely determined by

wo

-

. Hence

the prime extension P of

wo

is

It may be useful to note that even though in general the points of cannot be uniquely written as finite unions of sets

B(X)

that is needed in the proof is that every point of

e

IwIx n

10)

.

We return to the sithation that (X',&I)

lattice

.

If X

(X,e)

w'

of

X E

X

w

consists of all x' E X'

I' in X'

such that

w

. Indeed, I' l X c w

I' c

w'

X' A

x E w

, so

x' c w '

the following definition.

in X

satisfying x'

I' n X

C w

all

,

the prime extension

A

x E w

for every

in the following sense. Given

, the extension

w'

satisfies

implies that for x' E I' and x E X

. In view of

,

fl { w}'

is a sublattice of the larger

. The extension procedure is monotone

an ideal

{wIx

can be written as

, prime ideal exension i s unique.

i s an ideal in X'

Explicitly, given the proper prime ideal

X

we have

this monotonicity property we present

34


DEFINITION 78.2. The s u b l a t t i c e

Ch. 11,§781

of the d i s t r i b u t i v e l a t t i c e

(X,e)

(X',8) i s said t o have the monotone prime i d e a l extension property whenever f o r every i d e a l I' i n X ' and every proper prime idea2

I' n X c w

X' satisfying

that

there e x i s t s a prime extension

.

I' c w '

(X,e)

Note that if

has this property with respect to

w1,w2 are proper prime ideals in X exists for every prime extension w;

. As

satisfying w' c w; 1

satisfying w1 c w2 of

of

w'

,

such

(X',8)

and

then there

a prime extension

w1

in

w

w

w'

of

w

2

observed before, prime extension can be unique and

at the same time fail to be monotone (cf. Exercise 7 6 . 1 4 ) . For simplicity, assume now first that the ideal generated in the larger lattice X '

by the sublattice X

situation by saying the X in X '

prime ideal w ' w'

fl

X

=

by

@(w')

. As

X'

covers

intersects X

itself. We shall indicate this a consequence, every proper

now in a proper prime ideal. Indeed,

X would imply that the smallest ideal in X '

included in w ' i.e., X'

is X'

= w' = w'

. But this smallest ideal is

, which contradicts ll X , is therefore

X'

containing X

from R'

into R

, , defined

X' c w'

itself, s o

the hypothesis. The mapping

extension theorem then implies that

is

@

. The prime

ideal

is not only into, but also onto f2

@

Furthermore, we have

so

the mapping

@

is hk-continuous. We prove that if, in addition, X

the monotone prime ideal extension property, then $ THEOREM 78.3. I f

X covers

X'

extension property, then the mapping

and

X

has

is also hk-closed.

has t h e monotone prime idea2

$: w' -+ w '

n

X

from

R'

onto

R is

not only hk-continuous, b u t a l s o hk-closed. PROOF. Let F' c R' @(F')

be hk-closed. We have to show that the image

is also closed. Since F' = h(k(F'))

in R'

, we

have

.

Ch.


11,§781

Every w ' E F' so

the hull of

has, therefore, the property that $ ( w ' ) X

fl

in R

k(F')

0

h Xflk(F')

On the other hand, if

hull of

X

ll

3

$(F')

w

E R

w'

3

3

X fl k(F')

,

and

satisfies

. satisfies X fl k(F') c w

(i.e., w

in the

k ( F ' ) ) , it follows from the monotone prime ideal extension

property that there exists a prime ideal w ' and

35

k(F')

.

Hence w = $ ( w ' )

and

w'

in X' E F'

,

such that w ' ll X w

so

E $(F')

. This

= w

shows that

0

h Xnk(F') It follows that

$(F')

c $(F')

=

.

h(Xflk(F'))

,

so

$(F')

is hk-closed.

We now combine monotonicity and uniqueness of the extension. THEOREM 78.4. ?'he sub2attice (X',e)

(X,e)

of t h e d i s t r i b u t i v e Lattice

has t h e unique and monotone prime idea2 extension property i f and

o n l y if X

i s an idea2 i n X'

PROOF. If X

.

,

is an ideal in X'

the prime ideal extension is unique

and monotone, as already observed earlier in this section. For the proof of the converse, assume that X

has the unique and monotone prime ideal

extension property with respect to X'

. Then

X

property with respect to the ideal generated by just as well assume that X '

x'

X

in X'

is the ideal generated by

X = X'

), and we have to prove now that

is from R'

certainly has the same

. Hence, we may

X (i.e., X

. The mapping

$: w'

+

w'

covers fl X

onto R (as observed above), and by the uniqueness of the

extension the mapping is now one-one. Hence, since the extension is $

monotone, it follows from the preceding theorem that from R' SO

onto

52

. Given

its image $({u'}~~)

x' E X'

,

the set

{w'}~,

is open and compact in R

u ({wIy:yED)

. By

of the form

{wjX for some x E X

is a homeomorphism i s open and compact,

, and

hence of the form

the compactness this is equal to a finite union, and

. Hence

so

36


for some x E X

.

{w')~,

{w'}~

=

,

Ch. 11,9781

But then, by the unique extension, we have and

x'

so

.

x

=

It follows that X

=

.

X'

For later use it will be practical to have available a certain class of sublattices possessing the monotone prime ideal extension property. The following definition will be useful for this purpose. (X,e) of the d i s t r i b u t i v e l a t t i c e i s called a p-sublattice of X ' whenever f o r a l l x,y E X and y' E X' such t h a t y = x v y' and f o r e v e q proper prime i d e a l w ' i n X ' such t h a t y' E w ' there e x i s t s an element z E w ' n X s a t i s f y i n g DEFINITION 78.5. The s u b l a t t i c e

(x',e)

y = x v z . Note that every ideal X z =

y

A

y'

in

(X',8)

is a p-sublattice, since

satisfies the condition mentioned in the definition. It is also

easy to see that the lattice X

in Exercise 76.14 is not a p-sublattice of

.

the Boolean algebra B(X)

X of t h e d i s t r i b u t i v e l a t t i c e

THEOREM 78.6. Every p-sublattice

(X',e) has the montone prime i d e a l extension property. PROOF. Let

I'

be an ideal in X'

such that

I' ll X c w

of

satisfying

w'

o

generated in X'

by

. We have w'

I'

2

I'

and

w

and

a proper prime ideal in X

w

to show that there exists a prime extension

. To

this end, we consider the ideal J'

. We prove first that

J'

nX

= w

,

and

this will be the greatest part of the proof. By means of Zorn's lemma it follows from

I' Il X c w

I' 3 I' and 1; 0 A

y'

E 1;

property of

is prime. If not, there exist x',y' E X'

but

x'

and

y'

are not in

. Hence

which implies

1;

such that

I' nXc 0

x

A

y E 1; n X c w

against our hypothesis that x

=

X-w

. But

and y

.

. From the maximality

such that x

w

w

such that

it follows that there exist elements z ; , z ; E 1;

1;

in the set theoretic difference F y < y' v z;

in X'

is maximal with respect to the property that

We show that 1; x'

that there exists an ideal 1;

is prime,

are in F

= X-w

x' v z i

S

so

.

x E w

and x,y and

or y E w ,

This shows that

Ch. 11,§781

is prime. Assume now that J' fl X

1;

an element x x

=

=

in J' fl X

E I'

z'

. Since

in the ideal

Iw

y v y' v z'

=

y

Hence, since X

. Hence y' v z' . But

=

y v

E

z

w

.

y E w and x E w

It follows that

. Thus we

is not in w

= w ).

exists a prime ideal w '

X

fl w ' c w

w =

X n

w

. On

w'

.

,

is prime.

1;

zE1;)flXcw

z

so

and

w

E 1;

X

fl

x v y =

. This contradicts the hypothesis that = w .

is now a lower sublattice and F fl J'

is

By the prime ideal separation theorem there such that J ' c w '

in X'

It follows immediately that w ' prime extension of that

and

by =

find at last that J' fl X

The non-empty set F = X-w empty (since J' fl X

Hence

E 1;

z'

is

have

x v y

is a p-sublattice, there exists an element

suchthat x v y = y v z . B u t x

E I' implies

z'

, we

x E J'

generated in X'

for some y E w , so

y

5

. Then there

is not equal to w

but not in w

y' v z' for some y'

for some

37


3

I'

F

and

n

is empty.

w'

. It remains to prove that w ' . It follows from F fl w ' =

i.e., X fl w ' = w

the other hand, J' c w '

is a

0

implies w = X fl J' c X fl w ' .

This concludes the proof.

In theorem 7 8 . 4 it was proved that the sublattice (X,e)

of

(X',0)

has the unique and monotone prime ideal extension property if and only if

X

is an ideal in X'

. The last theorem shows that every p-sublattice has

automatically the monotone prime ideal extension property. The following statement is, therefore, an imediate consequence. THEOREM 78.7. The p-sublattice (X',0)

(X,0)

of t h e d i s t r i b u t i v e l a t t i c e

has the unique prime i d e a l extension property i f and only i f X

is an i d e a l i n X'

.

We recall (cf. section 76) that the distributive lattice of all principal ideals in the Riesz space L

is denoted by

X(L)

. As

observed

before, ideals, maximal ideals and prime ideals in L

correspond to

ideals, maximal ideals and prime ideals in X(L)

L

subspace of the Riesz space L' ideals in L and not in L' a subset of Of

X(L')

X(L')

,

, the so

generated by the elements of

L

. If X(L)

is a proper Riesz are principal

we cannot say immediately that X(L)

. Nevertheless, we

by identifying X(L)

elements of

shall consider X(L)

is

as a sublattice

with the set of all principal ideals in L'

.

38

Ch. 11,9781


The following theorem showsnow that p-sublattices occur frequently. Note first that if we say that (B,9) is a subring of theBooleanring (B',B), this means that elen?ent 9

(B,B)

is a sublattice of

(B',9)

and such that each initial segment

with the same smallest

(b:bEB,B 0 such that supCv,u')

u2 = sup(u,,v)

and u ,

. Then u?-

. Summarizing, we have

is

is

.

with u -v E L

argument as above shows that X(L)

.

is the principal ideal and

satisfies u 1

. Now, let

sup(AU,A

sup(v,u')

,

b

satisfy

it follows that u

still in L and generates AU still in L

a

for some proper prime ideal

same principal ideal. Hence, there exists a number u

x with

that

is a p-sublattice of B'

respectively. Since sup(Av,Au,)

generated by

a v b'

denote the principal ideals generated i n

and A,'

u'

ll

=

a v ac with

=

The last inequality implies that if w'

the

in this interval, let

xc be the complement (in the sense of the Boolean algebra) of

. Note now

,

arb,

. The

a v b'

is a Boolean algebra; for any

respect to b

is

and

0

5 u

2

-v

is a p-sublattice of

u'

X(L')

,

so

.

the same

Ch. 11,9781

39


For the cases considered in the last theorem it is now a simple matter to determine under which conditions prime ideal extension is unique. THEOREM 7 8 . 9 .

(B,B) of a Boolean ring

(i) A subring

unique prime i d e a l extension property i f and only i f we r e c a l l t h a t i n t h i s case the unique extension prime i d e a l

L

(ti) Let IL

be a Riesz subspace of t h e Riesz space by

t h e i d e a l generated i n L '

.

L

u ' E L'

and an element

a > 0

I n t h i s case, i f

P

,v E

L

The space

X(L) = X(1L)

i d e a l extension property i f and only i f means t h a t f o r any p a i r e x i s t s a number

B i s an i d e a l i n B' of a given proper

.

i s given by

B

in

w

w'

has the

(B',B)

satisfying

L

u E L+

L

. Denote

. Explicitly,

0 c u' 5 v

such t h a t

i s a proper prime i d e a l in

L'

by

has t h e unique prime

au 5

this

there u' 5 u

, it i s s u f f i c i e n t

.

for

the determination of the unique extension t o i n d i c a t e t h e unique extension P"

of

in

IL

P

.

. The

IL

to

PROOF. (i) (B,€i) theorem,

so

extension

P

i s simply t h e i d e a l generated by

P"

is a p-sublattice of

(B',6)

by the preceding

B has the unique prime extension property if and only if B

is an ideal in B ' (by Theorem 7 8 . 7 ) . (ii) The first part of the proof is similar as in part (i). For the determination of generated by

N

and

P"

A,

first that P"

must at least contain the ideal

. Furthermore, on account of

. Hence, for

is equal to

X(L) = X(1L)

,

the

must correspond with the same prime ideal in 0 5 u" E P"

for some u E P

,

, the so

N

u

principal ideal generated by is majorized by a multiple

. This implies that u is contained in the ideal generated by I L . Hence, P" is the ideal generated by P in I L . N

of u in

, note

in I L

P

prime ideals P x(L) = X(IL)

P"

P

We continue with our investigation of the situation that X ( L) = X ( I L ) . First a simple remark. If A

is a band in the Riesz space L

with

Projection property, then it is well-known (cf. Theorem 2 4 . 5 ) that the component u 1

in the band A

of any

u t 0

is given by

40

Ch. 11,5781


I be an ideal in L

Now, let

0

any

5

v E A

and A

the band generated by

is the supremum of all w E I satisfying 0

is obvious then that the component u l

7

=

of

u

in A

I (hence, 5

w

5

v )

. It

is now also given by

sup(w:w€I,o<w 0 element w E L+

A'

1.

is likewise a majorant in A

We have to show that A

that

I and the

be the component

occurring on the right has a majorant in the band

Then v = inf(z,u) v =

,

A'

observed above, we have

u' = sup v':v'EI,O5v' 0

in

IL

(namely,

1.

since L

is Dedekind complete. We prove that

0 < u'-w E IL

strictly between 0 and

assume instead that and

L'

,

u' E L

so

u'-w

IL

property, so

. It follows that

IL = L

.

is uniformly complete, but

is uniformly complete. Since IL

is an ideal in

IL

Hence, IL is uniformly complete and has the projection is Dedekind complete. The space L

of the Dedekind complete space exist v

and there would be no

. This is impossible, as shown

has the projection property, IL has the projection property

(by Theorem 2 5 . 2 ) .

and

IL

is now the Dedekind completion of L

is a Riesz subspace

such that for any u ' > 0

w in L satisfying 0

The case that L

IL

u E L+

that for any

Finally, we drop the condition that L

1L

,

is Dedekind

satisfying 0 < v i u '

in L

should have

above. Hence u ' = w

L'

so

u' E L

au ). Now, let

w

w

a > 0 and

. This shows in particular

there exists an element v > 0 v

so

is uniformly complete. Then L L

0 < au 5 u ' 5 u

.

has the projection property.

Assume now that, in addition, L

IL

E A

1

is a member of the set, and

u1

uniformly complete and has the projection property, so element u ' > 0 in

u

is an upper bound of the

is the supremum of the set. This shows that u '

u

, so 0 S

inf(u,w)

u1

< v

L

is Archimedean and

.

< u'

L'

i w

. Hence, by

in IL

there

Theorem 3 2 . 6 ,

is the Dedekind completion of

is of particular interest. In this case the ideal

IL is the space L'

itself. The following theorem is, therefore, an immediate consequence. THEOREM 7 8 . 1 1 . Let

.

L'

be t h e Dedekind compZetion o f t h e Archimedean

Then L has t h e unique prime i d e a l extension property W i t h Riesz space L respect t o L' if and only if X(L) = X(L') , i.e., if and onZy if for any


42

o

i

u ' E L'

that

au

5

there e x i s t s a number u' 5 u

. In

t h i s case

The condition that for any have

au

5

u'

5

u

o

a >

and an element

Ch. 11,1781

L+ such

u E

has a ls o t h e p r o j e c t i o n property.

L

u' > 0

for some u E L+

in the Dedekind completion L'

we

is the condition, due to J.J. Masterson,

that occurs in Exercise 37.16. In this exercise we introduced for each ideal A

in L

the ideal A#

ideal in L'

generated by

is of the form A#

for some A

Exercise 37.16 that every ideal in L' if and only if L

in L'

A

.

In general not every

in L

. It was

is of the form A#

shown now in

for some ideal

A

in L

L

is Dedekind complete (i.e., L = L' ), Masterson's condition is trivially

satisfied. If L

satisfies Masterson's condition. Of course, if

satisfies Masterson's condition, then L has the

projection property, as shown by the last theorem. We show that Masterson's condition is strictly between Dedekind completeness and the projection property.

Let

THEOREM 78.12.

Dedekind compZetion to

L denote an Archimedean R i e s z space w i t h

. Then Masterson's

L'

condition f o r

L' I i s s t r i c t l y between Dedekind completeness of

property for

.

L

PROOF. We first present a space L

L

L

(with respect

and the p r o j e c t i o n

satisfying Masterson's condition

without being Dedekind complete. Let L be the Riesz space of all bounded real functions f on the interval C 0 , l l

such that f has an at most

countable range. The Dedekind completion L' is the space of all bounded

.

real functions on C 0 , l l

This shows already that L

complete. For the proof that L

the bounded function u ' ( x ) t 0 on identically zero and

0

2

is not Dedekind

satisfies Masterson's condition, consider

u'(x) < 1

C0,lI

. We may assume that . For n

holds for all x

=

u'

2,3,

is not

...

,

let

and let Em

=

. ...)

(x:u'(x)=O)

of the sets E

...)

(n=2,3,

(n=2,3,

and

u(x)

=

not identically zero and observing that

Since u'

is not identically zero, one at least

2n-l t (n-l)-'

5

2u

holds

=

n

. . The last inequality follows by for n = 2,3, ... . This shows that

0 for x E Em

u 2 u'

-1

for x E E n Then u E L+ , the function u is

is not empty. Now, let u(x)

L

Ch. 11,0781

43


satisfies Masterson's condition. Next, we present a space L with the projection property, but not satisfying Masterson's condition. Let L be the Riesz space of all real ...)

sequences f = (f(l),f(2), L'

is the sequence space I 1

u' =

There is no u E L+

with finite range. The Dedekind completion

,

so

,...) E L' .

satisfying au

S

u' 5 u

for some a > 0

, and

L

so

does not satisfy Masterson's condition. On the other hand, as proved in Theorem 25.1,

part (iii) of the proof, L has the projection property.

We shall return to the situation that L

L'

is a Riesz subspace of

in the next section. The present section will be concluded by presenting a necessary and sufficient condition for a sublattice to have the unique prime ideal extension property.

(X,e)

THEOREM 78.13. The s u b l a t t i c e

of the d i s t r i b u t i v e Zattice

(XI,@) has t h e unique prime i d e a l extension property w i t h respect t o i f and onZy i f t h e BooZean ring generated by X'

ring generated by PROOF. Let B' sublattice of

. By

B be the subring of

B'

X'

. Then

generated by

X X

is a

. In

consist of all finite suprema of (relative) differences

. Evidently, B

of elements of X

X i s an i d e a l i n the BooZean

be the Boolean ring generated by

. Let

B'

other words, let B X

.

X'

is exactly the Boolean ring generated by

Theorem 78.1 the lattices X

and

X'

have the unique prime ideal

extension property with respect to B and B' respectively. It follows that X has the unique prime ideal extension property with respect to X' if and only if

B has this property with respect to B'

is an ideal in B'

.

,

i.e., if and only if B

Most of the results in the present section were proved by W.A.J.

Luxem-

burg in the paper (C51,1973). EXERCISE 78.14. In Theorem 78.10 it was proved that if subspace of

L'

such that for every

0 _< u' E L'

L is a Riesz

there exists

E L+ and

44

satisfying au < u' 0

implies the projection property for L

. This is no

the hypothesis a little weaker. Show that if 0< v

S

u < w

property but

,

in L

to to

longer true if we make

then it is very well possible that L'

without using that X(L)

P to

extension of

,

X(1L)

=

P in IL . This was proved . It can be proved directly, i.e.,

generated by

that if the extension is unique, then the

IL must be the ideal P"

generated by

Assume, for the proof, that the unique extension P#

. Then there exists an element

larger than P" P"

has the projection

In Theorem 78.9 it was proved that if prime extension

IL is the ideal P"

there exists a prime ideal Q

so

satisfying

is unique, then the prime extension of any prime ideal P

L'

after proving first that X(L) = X(1L)

, and , but

L'


L has not.

EXERCISE 78.15. from L

L

there exist v,w E L+

such that for every 0 < u' E L'

P"

Ch. 11,9781


is not in Q

u'

unique prime extension

u' > 0

P

in IL

, but

in P#

Q nL

3

P"

.

3

not in

includes

,

L = P

Q n L satisfies (QnL)#

of

P

is properly

in IL such that Q

. It follows that

(QnL)#

of

P#

so

the

. Note

(QnL)# = Q by the uniqueness, and derive a contradiction. Note also, for the extension of Q n L , that Q n L is indeed a proper prime

now that

ideal in L

.

EXERCISE 78.16. proper subring R

We recall that if

,

then R'

R'

is a commutative ring with a

is said to be integrally dependent upon R

whenever for every x' E R' there exists a polynomial n- 1 p(x) = xn + a x + + . a with coefficients an-l,...,ao in R such n- 1 that p(x') = 0 Such subrings R of R' are in some sense analogous to

...

.

the p-sublattices introduced in Definition 78.5.

Prove the following

statements. (i)

If w '

is a proper prime ideal in R'

proper prime ideal in R (ii) such that

If w

.

is a proper prime ideal in R

I' n R c w

satisfies J' Il R c w

,

.

then the ideal J'

a proper prime ideal in R such that

extension w '

such that

I' c

w'

.

and

then w = w ' I'

generated by

(iii) (Monotone prime ideal extension). If w

,

I'

fl

R

n R is a

an ideal in R'

I' and

I'

is an ideal in

c w

,

then w

w

R'

and

has a prime

Ch. 11,1781

Let the sets R

(iv) R'

45


and

of all proper prime ideals in R and

0'

respectively be provided with the hull-kernel topology. The mapping fl R

$: w ' + w '

is now from R'

,

onto Cl

and

is hk-continuous and

$

hk-closed. HINT: For (ii) and (iii), let w ' w'

3

I' such that

w'

fl R

w'

fl

R

is included in w = w

.

be an ideal in R'

satisfying

is maximal with respect to the property that

w'

. Then, by

It follows that

Exercise 52.14(i),

, so

J' c w '

w'

.

n Rcw

J'

is prime and

A second proof is obtained by considering the quotient ring

Let $: R'

-f

S

be the canonical homomorphism from R'

is integrally dependent upon $(R)

. Hence, by

in $(R)

and

in S

such that w"

I' and

3

Let R and

that prime extension from R x'E

R'

.

w ' fl R = w

to

fl $(R)

R'

be as in the preceding exercise. Show is unique if and only if for each

there exists a finite set

...,x

{xl,

}

of elements in R

containing x'

contains

and conversely (i.e., the radical of the ideal generated by {xl,

...,xn)

such

{x,,

x'

...,x

-+

w'

fl R

).

, we

is now a homeomorphism. Hence, for x' E R'

By the compactness of

{P')xl

...,xn

there exist xl,

have

such that

U~IP'I , so any P' satisfying x' E P' also satisfies i: {x1,...,x 1 c P , and conversely. n For the converse, assume that Pi # Pi , but $(Pi) = $(Pi) It may

I P ' I ~ ~=

.

be assumed that there exists an element x' E R' {P'IX, but that $(Pi)

such that P i

...,x

=

Uy{P')

but not of

. xi

$(P;)

It follows that

...,xn 1

{xI,

. This contradicts

$(Pi)

is in

in R

Pi is not. By hypothesis there exist x l ,

is a subset of

= $(Pi)

.

1,

equals the

HINT: Assume first that prime extension is unique, so the mapping $: w '

.

R'

that every proper prime ideal in R' radical of the ideal generated by

= $(w)

is a prime ideal in

It is not difficult to prove now that w ' = $-'(w-)

EXERCISE 78.17.

S

is a proper prime ideal

$(w)

N

satisfying w '

S = R'/I'.

. Then

the Cohen-Seidenberg theorem (cf. again Exercise

52.14), there exists a prime ideal w R'

onto S

such

46

Ch. 11,1791

STRONGLY ORDER DENSE RIESZ SUBSPACES

75. Strongly order dense Riesz subspaces According to the definition in section 21 the ideal A space L

is said to be order dense in L

equal to L dense in L

if the band generated by

A

. In an Archimedean space L , therefore, the ideal A is whenever Add = L . This definition can be extended to an of the Archimedean space L

arbitrary non-empty subset D said to be order dense i n

L

if

Ddd

=

L

, i.e.,

D

e ( s o {e}d

= {O}

is

is

. By way of example, if the

Archimedean space L possesses a weak unit subset of

in the Riesz

), then any

L containing e is order dense. F o r an ideal this definition of

order denseness agrees quite well with our intuitive ideas of denseness, but not for an arbitrary subset. We shall introduce, therefore, a different notion, the notion of strong order denseness for a linear subspace of an Archimedean Riesz space. For an ideal strong order denseness will be the same as order denseness, but for a more general linear subspace strong order denseness is a much stronger requirement than only order denseness. The definition follows now. The linear subspace D is called strongly order dense i n

of the Archimedean space L

L if, for every

u >

0 in L

,

the

intersection {u}dd n D does not consist only of the zero element. Here we restrict ourselves to the case that D


one of the main results will be that the Riesz subspace D order dense in L

if and only if for every u > 0 in L

element v

such that 0 < v

in D

THEOREM 79.1.

5

u

, then

, and

there exists an

holds.

If the linear subspace

strongly order dense i n L

L

is strongly

D o f the Archimedean space

D i s order dense i n

L

L is

, but the converse

i s not always true (not even i f

D is a Riesz subspace). An ideaZ, however, is strongly order dense i f and only i f the ideal i s order dense. PROOF. Let D be strongly order dense in L {uldd

nD

+ COI

for every

d If Ddd # L would hold, then D # {O} in L

satisfying 0 < u E Dd

{ddd 0 D L.

= {O}

. But

o

< u E L+

, so

so

.

there would exist an element u

then we have

. Contradiction. Hence

,

Ddd = L

{ujdd

,

C

i.e., D

Dd

,

so

is order dense in

Ch. 11,5793

47


TO show that order denseness (for a Riesz subspace) does not always

imply strong order denseness, let L functions on [0,11 Then D

, and

be the Riesz space of all real

let D be the Riesz subspace of all constants.

is order dense, but not strongly order dense.

Finally, let I be an order dense ideal in L

. Since

given in L

I is an ideal and u E Idd = L

element v E I satisfying 0 < v that 0 < v E {uIdd

n I , so

2 u

, and let u > 0 be , there exists an

(cf. Theorem 19.3(iii)).

{uIdd Il I # { O )

.

It follows

I is

This shows that

strongly order dense. For the main theorem in this section, let L be a Riesz subspace of the Archimedean Riesz space L' (the same notations as in the preceding section). For any subset D of L'

,

respect to

L'

Dd

element f

, we

will be denoted by shall write fdd

the disjoint complement of

.

Sometimes, if D

D with

consists of one

. The following theorem

instead of

is essentially due to A. Bigard (C11,1972,Lemma 2). THEOREM 79.2. Given the Riesz subspace

space

L' (i)

have

L of the Archimedean Riesz

, the following conditions are equivalent. L i s strongly dense i n L' , i . e . , for every

Iu'Idd

n L # {Ol

satisfying

0

v

(iii) Every

5

.

u' > 0

(ii) For every u'

.

u' t 0

in

L'

in

1

(

there e z i s t s an element

B'

of a l l bands

B'

in

L'

-+

v

B = B'

n

.

L

A U , ll L

#

in L

i s a one-one mapping from the s e t

onto the s e t of a l l bands

B in L

.

B'

-+

B = B'

i s a l a t t i c e isomorphism between the Boolean algebras o f a l l bands i n L respectively. The inverse mapping i s given by

the bands bands

B'

and

B' Il L and

C'

are d i s j o i n t compZements i n

C' f l L

we

, where A,

{O)

Furthermore, i f these conditions hold, then the mapping and

L'

,

(iv) For every u ' > 0 i n L' we have denotes the idea2 generated by u ' i n L

The mapping

in

L' s a t i s f i e s

u' = sup v:vEL,O 0

B' L'

=

Tidd

n

L'

. Finally,

i f and on2y i f the

are d i s j o i n t complements i n L

.

L

48

*

PROOF. (i)

, we

(ii). Given u' > 0 in L'

by hypothesis. Let 0 

0 ,

il

so

L # {O) since

such that

so

let 0 < w E zdd fl L

. Finally, let

Then q > 0 , because otherwise it would follow from

0 5 inf(w,z)

of

{u'Idd

s p does not hold, i.e.,

If follows that =

have

inf(u',p)

L' is Archimedean there exists a natural number n

q

Ch. 11,9793


since q

5

inf(nw,np) = nq = 0

5

contradicts 0 < w E zdd

inf(w,p)

=

inf(w,np)

and w

and

. Furthermore, q

are members of

p

L

is a member

. Note now

that

so

(q-n inf(u',p))+ q

so

-1

v = n q

5

=

0

. In other words, we have

n inf(u',p)

5

nu' ,

satisfies v E L

and

0< v

5

u'

. This shows that

condition

(ii) is satisfied. (ii)

*

element v

(iii). Given u' > 0 in L' in L

,

satisfying 0 < v < u'

there exists by hypothesis an

. Hence

u'

is an upper bound of

the non-empty set

).

Mu, = (v:vEL,O 0 holds for at least one

contained in B;

. Contradiction. Using

follows easily that for any band

. To show that the mapping I7 L # Bi I7 L . If B; # Bi

contains an element f ' h'

, it

n L is a band in L ,

is strongly

for all d E D , but on the other hand

L

that

n L and

but not in B; n L

.

.

{h'ldd n L # { O } The set d (B;) n L , i.e., the set is This shows that B; n L # B; n L

.

50

Ch. 11,9791


It has to be shown next that the mapping B' +B'n L all bands in L form

B

=

B' fl L

,

i.e., we must prove that every band for some band

B'

. Then

bands in L '

, and

Theorem 19.3(ii)). and

C

B IC

in L

is a subset of

C'

. It follows that C , so

is evidently included in B' fl L

In addition, it follows that complements in L' B'

C'

@

, we

@

C

in L

order dense in L' , i.e., B' if

B'

B

B' fl L and

=

and

C'

, which C'

is order dense in L

,

then u and

B'

. On

the other hand dd holds for B' = B

C'

=

.

Cdd are disjoint

0< v

would be disjoint from u'

5

, and

is impossible. Hence, B'

then v @

C'

would is

are disjoint complements. Conversely,

,

then it is evident that

.

If

u t 0 in L dd = B

is disjoint (in L ' ) from B'

u I (B'eC')

i.e., B

holds in L ' (cf.

fl L = B

L are disjoint bands in L

,

, let

C' = Cdd are

is a subset of

B' n L c B

such that

and

is of the in L

B

B' ll L must be included in

are disjoint complements in L' C = C' fl

is disjoint from B fB C dd and from C ' = C , so

L

fl

some u' > 0 in L'

in L

in L

and

B' I C'

. Hence, B'

B' = Bdd and

. Indeed, if

could take v

be disjoint from B

Bdd

=

implies that

the disjoint complement (in L ) of B

B'

(B'flL) I C , since B'

Hence

B

. Given the band

in L'

C be its disjoint complement in L

is onto the set of

. This implies

C

u

=

0

, and

so

B

@

.

are disjoint complements in L

inf(B',B') = B' f l Bi and sup(B',B') = 1 2 I 1 2 hold in the Boolean algebra structure for bands Bi

C

Finally, we recall that =

B;

((Bi)dfl(B;)d)d

in L' (cf. Theorem 22.7).

The mapping

B'

+

and

B' ll L preserves inter-

sections and disjoint complements, s o the mapping preserves suprema and infima, and hence it also preserves the operation of taking the complement in the sense of the Boolean algebra. Since the mapping is one-one, this shows that the mapping is a lattice isomorphism between the Boolean algebras of all bands in L' COROLLARY 79.3. L'

and

space

L' L"

and

L respectively.

(i) If L i s a strongly order dense Riesz subspace of

i s a strongly order dense Riesz subspace of the Archimedean

, then L i s strongly order dense i n

L"

.

(ii) If L i s a strongly order dense Riesz subspace of the Archimedean space L' and B' i s an ideal i n L' , then B = B' f l L i s strongly order dense i n B' .

PROOF. (i) Given u" > 0 in L" , there exists an element u' in L' u' 5 u" Corresponding to u' , there exists an element u in L such that 0 0 in B '

.

there exists an

L satisfying 0 < v 5 u' Given u ' > 0 in B ' , there exists an element v in L satisfying 0 < v 5 u ' (since L is in B '

element v

strongly order dense in L' ) . From 0 < v < u ' E B'

,

v E B'

v E B'

so

n

L

=

B

. This is the desired

it follows that

result.

Given the Riesz subspace L of the Riesz space L' , we shal

say that

L has the unique band extension property with respect to L' whenever, for every band

B

B' l l L = B

that

in L

.

, there

exists a unique band

B'

in L'

such

For an Archimedean space this property is equivalent to

strong order denseness of

L

.

in L'

THEOREM 79.4. If L i s a Riesz subspace o f the Archimedean space

then L has the unique band extension property with respect t o L

only i f

i s strongly order dense i n L'

.

PROOF. It is evident from the preceding results that if

L

,

L'

i f and

L'

is strongly

order dense in L' , then L has the unique band extension property. Given the band

B in L , the band B '

=

For the converse, assume that

Bdd is the unique extension. L has the unique band extension property.

I01 in L has the band IO} in L' as its unique extension. Hence, given any band B' # { O } in L' , we must have B ' n L # 101 , since The band

otherwise B'

,

L'

the band

{u'jdd

nL#

would also be an extension of {u'jdd {O}

. This shows

{O}

. Given now u' . Hence

>

0 in

satisfies {u'Idd # { O }

in L'

L

that

is strongly order dense in L'

.

EXERCISE 79.5. Let L be a Riesz subspace of the Archimedean space L' Show that if L

is included in the Dedekind completion L#


.

L# = (L')#

L

L'

of

In particular, show that if it is given that L# = ( L ' ) #


the Archimedean space L'

, then

are not necessarily equal (and

.

so

L

the Dedekind completions L# L'

(c,)

.

,

.

then

is strongly order dense in (L')#

and

is not necessarily included in L#

HINT: For a counterexample in the converse direction, let L' =

, then

and the Dedekind completions satisfy

In the converse direction, show that if

L

L

=

em

1. and

52

Ch. 1 1 , 5 8 ~ 1

EXTENSION THEOREMS

be the Riesz space of all real functions on

EXERCISE 7 9 . 6 . Let L' the interval [0,1] Show that

for every band is equal to

and let L

be the Riesz subspace of all constants.

.

is not strongly order dense in L'

L

in L' , except when B '

B'

.

{O}

EXERCISE 79.7. Let

In particular, show that

, the

L'

=

L' be the Riesz space of all real Lebesgue

measurable functions on the interval [ O , l l

,

and let L

subspace of all continuous functions. Show that L dense in L'

intersection B' n L

be the Riesz

is not strongly order

.

HINT: Similarly as in Exercise 18.14 one can easily construct a function u'(x)

2

0 in L' , not identically zero and such that any satisfying 0

continuous v(x)

v(x)

5

5

u'(x)

is identically zero.

80. Extension theorems In the present section we shall again assume that

.

space of the Riesz space L' by

L

that

By

L

is a Riesz sub-

IL we denote the ideal generated in L'

. Since all results in this section will be

about the various ways

L can be embedded in IL , there is hardly any l o s s of generality in

assuming immediately that L' = IL denote by

.

I in L , we shall now

For any ideal

I' the ideal generated by

I in L'

. Note

that for

I

=

L

this

notation is consistent. Before stating any specific theorems, we make one general remark. Given the prime ideal Q is a prime ideal in L

in L'

,

it is evident that Q n L

. Conversely, given the prime

ideal P

in L , there

is (in view.of the prime ideal extension theorem) at least one prime ideal

Q

in L'

such that Q n L

prime extension Q

of

P

=

P'

Also, even in the case that

extension of

P

.

THEOREM 80.1.

Let

L'

P

=

IL

equivalent. (i) For every prime ideal (ii)

.

It is not necessary, however, that the

is exactly the ideal P'

L'

.

P in L'.

is prime, P' may not be the only prime

. The following P in

conditions are now

L , the ideal

For every minimal prime ideal M

minimal prime ideal i n

generated by

in

L

P'

i s prime in

, the ideal

M'

is a

L'

.


Ch. ll,§8Ol

(iii) For every minimal prime i d e a l i n L'

i n L , the i d e a l

M

.

53

I f these conditions are s a t i s f i e d , then t h e mapping

M

one mapping from the s e t of a l l minimal prime i d e a l s i n L of a l l minimal prime i d e a l s i n L' . PROOF. (i)

M' Q

*

(ii). Given the minimal prime ideal M

is prime by hypothesis. For the proof that M' is a prime ideal in L '

included in M' fl L

ideal in L

is now an ideal in L' in L'

M'

including M

=

M , and

including M

, we

Q

must have

3

.

M'

i s a one-

onto t h e s e t

,

in L

Q fl L

the ideal

MI

is a prime

. Hence, since

Q fl L = M

so

and since

M'

is minimal, assume that

. Then

satisfying Q c M'

+

i s prime

M'

Q

is the smallest ideal

It follows that

Q

=

M' , i.e.,

is a minimal prime ideal. (ii)

*

(iii). Evident.

6;;)

(i). Given the prime ideal P

prime ideal in L

such that M

prime, s o on account of

L , let M be a minimal

in

is included in P

P' 3 M '

. By

hypothesis M'

is

P' is prime.

the ideal

Now assume these conditions satisfied. It is evident that the mapping M so

II. = M' f l L , so M I = M2 ), 2 all that remains to be proved is that every minimal prime ideal Q in

-t

M'

is one-one (indeed, M'

1

L ' is of the form M'

=

Mi

implies M'

1

ll

for some minimal prime ideal M

the minimal prime ideal Q minimal prime ideal in L

in L '

. We prove

properly included in Q , then M'

. Hence, let

that Q

=

M'

. Indeed, if

M'

is a were

would be a prime ideal (by condition

(iii)) properly included in the minimal prime ideal Q Hence Q = M'

in L

be given. Evidently, M = Q fl L

.

. This is impossible.

It is a natural question now whether there are some simple conditions for L P'

and

in L'

L'

implying that for every prime ideal P

in L

the ideal

is indeed prime. In the following theorem we present some

conditions of this kind. THEOREM 8 0 . 2 . Besides assuming t h a t

and Zet

L

be strongly order dense i n

p r o j e c t i o n property. Then P'

L ' = IL

, l e t L' be Archimedean

. Furthermore,

let i s prime f o r every prime i d e a l L'

L

have the

P

in

L

.

54

EXTENSION THEOREMS

Ch. Il,§801

PROOF. The proof is analogous to Masterson's proof in Exercise 37.17 for the particular case that L'

,

be a prime ideal in L

and let

is the Dedekind completion of inf(u',v')

. For notational convenience, we

or v' E P '

proved that u' E P'

denote the principal ideals generated by respectively. Note that D I E ,

E

The bands

and

. On account of

L

d D n L cP

we have

or

S

w

E Dd

1

w

. Since n

L

, so

n

L

P

shall D

and

E c Dd , in particular v' E Dd

.

by Theorem

. Assume that

d D

n

L c P holds. We

holds, as follows. Take w E L

has the projection property, we have w = w

and w2 E Ddd

L

in L' by

v'

. Let

to be

Ddd fl L are disjoint complements in

L c P

Ddd

prove that in this case v' E P ' v'

so

and

u'

and Ddd are disjoint complements in L'

Dd

79.2 the intersections Dd n L

L

. It has

in L'

= 0

. It follows then from

satisfying

+ w2

with

v' 5 w = w1+w2 with

v',wl in Dd and w2 in Ddd that v' 5 w 1 (indeed, writing s = (v'- wli+ for brevity, it follows from s 2 w2 that inf(s,w 2 ) = s . and w2 E Ddd , so inf(s,w ) = 0 , i.e., s = 0 so v' S w 1 ) . But s E D 2 d This shows that v' E P ' If Hence v' 5 w E D n L c P c P ' Ddd u'

nL

.

1

.

holds, we prove similarly that u' E P'

c P

.

holds. Thus we have

E P' or v' E P' , which is the desired result. In the next theorem we present some conditions

only be satisfied if

L

Dedekind completion of

L'

and

L

,

coincide. For the case that L'

THEOREM 80.3. Besides assuming that

(ii)

an element

u1,u2 E L+

1

L'

2

such that

u

=

=

IL

(iii) L

=

L'

.

is the

such t h a t

, l e t L' be Archimedean.

u;,u; E (L')+

u1 + u2 and

(Interval density property). I f fo E L

strong that they can

the theorem i s due to J. Quinn.

The follozJing conditions are now equivalent. (i) I f 0 5 u < u'+u' with u E L+ and elements

so

f' 5 f o 5 g'

f' < g'

.

ui 5 u!

i n L'

for

, there e x i s t i = 1,2

.

, there e x i s t s

.

Note t h a t i n (i) we have written u 0

PROOF. Assume t h a t t o prove t h a t

Then

has t h e unique i d e a l extension

L

if x(L)

i f and only

there e x i s t s an element

0 < u' E L'

f o r some

.

IL

=

L'

property w i t h respect t o every

Ch. I l , § 8 O l

X(L)

=

u

=

, i.e.,

X(L')

E L+ such t h a t

for

au 5 u' 5 u

has t h e unique i d e a l e x t e n s i o n p r o p e r t y . One way

L

i s by o b s e r v i n g t h a t unique i d e a l e x t e n s i o n

X(L')

i m p l i e s unique prime i d e a l e x t e n s i o n , and so

X(L)

=

by Theorem 78.9.

X(L')

A proof which makes no u s e of Theorem 78.9 can b e given a s follows. Given

,

0 < u' E L ' and l e t

U'

A'

ideal

A,'

let

A = A

Il

b e t h e p r i n c i p a l i d e a l g e n e r a t e d by

. By

L

A

generated by

in

so t h e r e e x i s t s an element u E A

,

L'

u E A

u E A' = AU,

implies that

A,'

t h e unique e x t e n s i o n

A,,

so

,

so u

2

. It

= A'

such t h a t

I

in

follows t h a t

J Il L = I

. Given

h y p o t h e s i s an element

u E L+

f o l l o w s from ctu 5 u ' that

u' 5 u

that

I' c J

u' E I '

L

. Hence

J

=

, there

nL

R' =

u

I'

,

, and

= I

in

w'

w' + w'

L'

L'

=

IL

X(L)

=

is besides

E A',

J

X(L') I'

.

no

i s an i d e a l i n

t h e r e e x i s t s by

au 5 u ' 5 u

f o r some

a > 0

. It

J c I'

, and

. Show t h a t

. It

denote by

is evident

R'

the s e t

and hk-closed.

Bedides

.

nL

i s hk-continuous

w e may i n t r o d u c e i n R' t h e T -topology, h a v i n g a l l sets P as a b a s e f o r t h e open s e t s . E v i d e n t l y t h e T -topology i s ({oi}u:uE~+) P weaker t h a n t h e hk-topology. Show t h a t t h e t o p o l o g i e s a r e i d e n t i c a l i f and t h e hk-topology,

only i f

X(L) = x(L')

.

L'

so i t f o l l o w s now from

.

(~w'lu:u€L+)

Show t h a t t h e mapping

J

have proved t h u s t h a t

EXERCISE 80.5. Assume a g a i n t h a t of a l l proper prime i d e a l s

in

u' 2 0

such t h a t u € J

that

. We

any

u'

.

o t h e r i d e a l e x t e n s i o n . Assume, f o r t h i s purpose, t h a t satisfying

L' ,

u' 5 u . On t h e o t h e r hand -1 a u' f o r some a > 0

For t h e proof i n t h e converse d i r e c t i o n , assume t h a t We have t o show t h a t , f o r any i d e a l

in

u'

i s then equal t o t h e

Ch. 11,1811


57

81. Prime i d e a l e x t e n s i o n and t h e p r o j e c t i o n p r o p e r t y Throughout t h e p r e s e n t s e c t i o n we assume as our b a s i c h y p o t h e s i s t h a t L' = I L

, L'

L

i s Archimedean and

Theorem 80.2 i t has been proved t h a t i f , i n a d d i t i o n , L p r o p e r t y , then

P'

i s prime f o r every prime i d e a l

. In

L'

i s s t r o n g l y o r d e r dense i n

has t h e p r o j e c t i o n

P

in

. It

L

may be

asked whether t h e r e i s a r e s u l t i n t h e converse d i r e c t i o n . The answer i s a f f i r m a t i v e . It w i l l be one of t h e main theorems i n t h e p r e s e n t s e c t i o n t h a t i f , b e s i d e s t h e b a s i c h y p o t h e s i s , we assume t h a t p r o p e r t y , then only i f

i s prime i n

P'

due t o K.K.

We f i r s t make a simple remark. S i n c e

L'

,

u

E L+ has t h e p r o p e r t y t h a t

element

, this

L

L' i s

main r e s u l t i s

i s Archimedean and

L'

L

B

Bdd

i f and only i f

L ' (where, a s b e f o r e , Bdd

t h e second d i s j o i n t complements of if

L i f and

is

L

i t follows from Theorem 79.2 t h a t t h e bands

a r e d i s j o i n t complements i n

C

a r e d i s j o i n t complements i n

u = v+w

in

Kutty and J . Quinn ([11,1972).

s t r o n g l y o r d e r dense i n and

P

has t h e p r o j e c t i o n p r o p e r t y . For t h e s p e c i a l c a s e t h a t

L

t h e Dedekind completion o f t h e Archimedean space

B

has t h e p r o j e c t i o n

L'

f o r every prime i d e a l

L'

and

and

C r i n t h e space

u = v+w w i t h

i s a l s o t h e (unique) decomposition of dd v E Bdd and an element w E C

.

v E B

u

in

Cdd

denote

L ' ). Hence,

and

L'

and

Cdd

,

w E C

then

as a sum of an

THEOREM 81.1. Assume, besides the basic hypothesis mentioned above,

that

L'

has the projection property, and l e t

B

be a band i n L

. Then

i s a projection band in L if and only i f B ' = Bdd holds. I t follows that L has the projection property i f and o n l y i f B ' = Bdd hoZds for

B

every band

B

in L

.

PROOF. Assume f i r s t t h a t

B'

C

Bdd

satisfies

. Given

u' E B '

Observe now t h a t

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

(since

L'

0 < u'

E Bdd , l e t

z

L ) and

z

L

u' 2 0

. Since Bdd

in

satisfy

E L+

h a s an o r d e r p r o j e c t i o n on t h e band

z

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

Bdd

B

i s e v i d e n t , we have t o prove only t h a t any

u' < z

B (since

B

. is

h a s a l s o an o r d e r p r o j e c t i o n on t h e band

has t h e p r o j e c t i o n p r o p e r t y ) . According t o t h e remark

h e d i a t e l y preceding t h e p r e s e n t theorem, t h e s e p r o j e c t i o n s a r e t h e same, i.e.,

pBz = PBddZ

. on

u' < PBddz, and s o by an element of

account of

u' 5 PBz B

. It

. Note

u' < z

and

now t h a t

follows t h a t

u'

u ' E gdd

P8z E B

E B'

. This

, so

i t follows t h a t u'

i s majorized

is the desired result.

58

11,9811

PRIME IDEAL EXTENSION AND PROJECTION PROPERTY Conversely, assume t h a t

We have t o show t h a t every

. Given

w IB

and

u = v+w

with

L ' ) . I t f o l l o w s from

0 5 v I z

satisfying and

u

v E Bdd

and

satisfying

w E Bd ( s i n c e

B' = B

u = v+w

is certainly so t h a t

0 5 v E Bdd = B '

. We

L

i s of t h e form

u E L+

, it

u E L+

v E Bdd

i s a band i n

B

with

t h a t t h e r e e x i s t s an element

s h a l l prove now t h a t

n

,

and t h e proof w i l l b e complete. For t h e proof of

t h i s w i l l imply then t h a t t h a t i t f o l l o w s from

v

is in

,w E

z E B

,

L

Bd

z E B

. Since

inf(z,u) = v

,

, note

v E B

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

Bdd

L

L = B

.

can b e w r i t t e n a s

u

are i n

inf(z,u) = v

dd

z

so

that

.

i n f ( z , u ) = inf(z,v+w) = i n f ( z , v ) = v This concludes t h e proof.

THEOREM 81.2. Assume, besides the basic hypothesis, t h a t

i s prime i n L'

projection property. Then P' L

i f and only if

L

P'


L

i s prime f o r every prime i d e a l

P'

t o prove, t h e r e f o r e , t h a t i f prime f o r every prime i d e a l

in

L

in

P

L

in

i s properly l a r g e r than

Bdd Bdd

and l e t

is not i n

such t h a t

u'

v = v;+v;

with

we d e r i v e from

i s not i n

B'

Note t h a t B'

not

n

,

v'

1

B'

is n o t i n

L = B c B ' ) , so v;

v E L+

, but

L

. Then

P = Q

n

B'

, so

have only

P

in

. Now

,

L

is

P'

f a i l s t o have t h e p n o j e c t i o n

B

in

L

take

v E L+

.

such u' > 0

t h e r e e x i s t s an element

L

and

,

v ; E Bdd

,

L (because o t h e r w i s e i s likewise not i n

t h e r e exists a prime i d e a l vi

in

satisfying

u' 5 v,

v ' E Bd Taking components i n Bdd, 2 t h a t u' 5 v ' , s o v i i s n o t i n B' (because u' 1 i s an i d e a l i n L' ). R e c a p i t u l a t i n g , w e have

u' I v and

B'

v i E Bdd

v = v'+v' with 1 2

Bdd

. We

i s prime f o r any prime i d e a l

p r o p e r t y . By t h e p r e c e d i n g theorem t h e r e e x i s t s t h e n a band that

P

has the projection P

h a s t h e p r o j e c t i o n p r o p e r t y . Assume, on t h e c o n t r a r y , t h a t

L

then

has the


PROOF. I t was proved i n Theorem 80.2 t h a t i f property, then

L'

in

L'

i s prime i n

L

Q

v; E Bd

vi L

, v;

not i n

.

B'

would be i n

. Since

such t h a t

Q

vi

is not i n

contains

€5'

and s o , by h y p o t h e s i s , t h e

but

Ch. I l , § S l l

corresponding ideal P'

is prime in L '

not in P'

v; E P'

. Indeed, if

2

v;

, we

get then that v'

v-v' > v

=

1

2 -

inf(w,V,)

-

. We

shall prove now that v;

is

would hold, there would exist (by the

P' ) an element w E P

definition of v

59


2

inf(w,v)

satisfying w v; 2

0

2

. Since also

v;

, so

.

On account of vi E Bdd the last inequalities imply that

-

v

inf(w,v) E gdd

, so

We have also w E P c Q v - inf(w,v) E Q v; E Q

in P'

= B c Q

inf(w,v) E Q

,

that v E Q

and so

. , and

vi E Q

hence it follows from (since 0

is a statement contradicting the defini-tionof Q

not in P' P' c Q

nL

, so

. The element since v'1

vi

is likewise not in P'

but neither vi

nor v;

It follows that L

vi

v ) . But

S

v;

Q ), vi

is not

. Indeed, we have

is not in Q (by the definition of

either. Thus we arrive at the situation that

5

. Hence,

inf(vi,v;)

= 0

is

,

. This yields a contradiction.

is a member of P'

has the projection property. For a completely different

proof we refer to Exercise 81.7. We shall now consider the situation that L space and L'

the Dedekind completion of L

is an Archimedean Riesz

. The conditions in the basic

hypothesis are then automatically satisfied (i.e., L' Archimedean and L

=

IL

, L'

is

is strongly order dense in L' ) and, moreover, L'

has

the projection property. Hence, Theorem 81.1 and Theorem 81.2 in the beginning of this section become very simple statements. Before discussing these statements, we prove a simple but remarkable theorem. THEOREM 81.3.

space L and compZetion of

If L'

i s the Dedekind eomp2etion of the Arehimedean

I i s an arbitrary idea2 in

I

.

PROOF. Note first that

L' , s o

I'

L

, then

I'

is the Dedekind

is an ideal in the Dedekind complete space

exist v and

I is a Riesz subspace of 1'. there w in I such that 0 < v s u' s w (cf. Theorem 3 2 . 6 ) .

Hence, let u'

P

I'

is Dedekind complete. Evidently

The proof will be complete if we show that for any u' > 0 in I' 0 in I' be given. By the definition of

I'

, there

exists

60

Ch. 11,§811

PRIME IDEAL EXTENSION AND PROJECTION PROPERTY

an element w E I such that

u' 5 w

. Also,

u'

since we have

E L'

and

L' is the Dedekind completion of L , there exists an element v E L+ that 0 < v 2 u ' It follows that v E I' Il L = I Hence 0 < v 2 u '

.

with

v

in I

and w

.

. This is

such w

2

the desired result.

We state and prove the final theorem in this section. THEOREM 81.4. Let

The band

(i)

B

i . e . , i f and only i f

that band

L' be the Dedekind completion of t h e Archimedean

.

L

Riesz space

in

i s a projection band i f and only i f

L

Bdd i s t h e Dedekind comp2etion of

B

. It

B'

follows

L

has the p r o j e c t i o n property i f and only i f i t i s true f o r every

B

in L

(ii)

that

L

and only i f

B

Bdd i s the Dedekind compZetion of

i s a prime idea2 i n L'

P'

dd B ,

=

.


in L

P

i f

has t h e p r o j e c t i o n property.

(iii) The following conditions are equivaZent. (a) L s a t i s f i e s Masterson's'condition i . e . ,

i n L' that

u > 0

there e x i s t s an element (YU 2

u'

5

.

u

(b) For every prime i d e a l

Q

i n L'

Q fl L

such t h a t

=

(c) Every idea2

J

the Dedekind completion of

J

i n L and a number

P (and a c t u a l l y

n

L

.

u' > 0

a > 0

such

there i s only one prime i d e a l

in L

P

i n L'

f o r every

Q

satisfies

i s then equal t o

, i.e.,

J = (JnL)'

P' ). J

is

PROOF. (i) and (ii) follow from Theorem 81.1 and Theorem 8 1 . 2 respectively. The equivalence of (a) and (b) in (iii) was proved already in For the equivalence of (a) and (c), note that by Theorem 8 0 . 4

Theorem 78.11. L

if

has the unique ideal extension property with respect to L

= (JllL)'

EXERCISE 81.5.

holds for every ideal J L' = IL

Assume that

and

L'

(i)

J = (JnL)'

(ii)

X(L)

(iii) L (iv)

Q

=

X(L')

L

and

.

.

in L' L' have the same principal ideals.

holds for every ideal J

, i.e.,

in L '

is Archimedean. Show that

the following conditions are equivalent.

.

if and only

satisfies Masterson's condition, i.e., (a) is equivalent to the

statement that J

L'

L'

has the unique prime ideal extension property with respect to =

(QflL)'

holds for every prime ideal Q

in L '

.

Ch. 11,5813

61

PRIME IDEAL EXTENSION (ii), (iii) follows from Theorem 78.9

HINT: The equivalence of (i),

and Theorem 80.4. It is evident that (i) implies (iv). To show that (iv) implies (iii), let P be prime in L and let Q be an arbitrary prime extension of

P

determined by

P

. Then .

Q

EXERCISE 81.6. Assume that and

L

= P'

(QflL)'

=

L'

by (iv), so

, L'

IL

=

Q

is uniquely

has the projection property

is strongly order dense in L' (note that these conditions are

satisfied if

L'

L ) . Show that the following

is the Dedekind completion of

conditions are equivalent. (i)

L


(ii)

For every minimal prime ideal M

(iii) Every minimal prime ideal N (iv)

Every band

in L

E

L'

in

in L'

satisfies E

HINT: Condition (i) is equivalent to P' ideal P

the ideal M'

is prime

is then a minimal prime ideal in L' ).

in L' (and actally M'

satisfies N (EflL)'

=

.

being prime for every prime

and this is equivalent to (ii) by Theorem 80.1. For the

in L

equivalence proof of (ii) and (iii), assume that (ii) holds, and let

. Then

a minimal prime ideal in L' easily seen that N fl L

N fl L

and this will imply by (ii) that M'

, which

included in N

(NflL)'

. By

that M

=

,

is minimal).

it follows from (ii) that

is included in N

.

and

N

is

is prime in

.

N = (NflL)'

be a minimal prime ideal

It follows from (iii) that N = (NflL)'

in L'

, so N

For the proof that (i) implies (iv), let (i) hold and let

. Then

B = E rl L

is a band in L (since L'

is strongly order dense in L' , cf. Theorem 79.2), condition (i). But

Bdd

=

(EflL)dd

=

=

M'

such

. This

. Conversely, assume that

in L

=

Bdd is a band in L'

E

=

E

from (iv) that

(EflL)'

, i.e., Bdd

(iv) holds, and let

such that E fl L = B

= B'

, and

E

be a band

is Archimedean and dd so B' = B by

E (cf. again Theorem 79.2),

E = B' = (EflL)'

. Than

properly

(NflL)'

is minimal, so

,

is prime, so (ii) is satisfied.

shows that M' in L'

be

Since N fl L

Theorem 52.3 there exists a minimal prime ideal N N fl L

N

and it is

is prime in L' with M'

Conversely, assume that (iii) holds, and let M in L

,

properly included in N fl L

in L

is impossible since N

a minimal prime ideal in L

. But

is prime in L

is even a minimal prime ideal in L (since other-

wise there is a minimal prime ideal M

L'

.

(NflL)'

=

L

so

B be a band

. It follows

this is equivalent to (i).

Ch. 11,1811

PRIME IDEAL EXTENSION AND PROJECTION PROPERTY

62

EXERCISE 81.7. In Theorem 81.2 it was assumed that

, L'

IL

=

has

L is strongly order dense in L' , and it was

the projection property and is prime i n

proved that P'

L'

for every prime ideal P

L'

in L

if and

only if L has the projection property. Since it was shown already in Theorem 80.2 that P'

in L

is prime for every prime ideal P

if L

has

the projection property, the proof in Theorem 8 1 . 2 was restricted to showing that L has the projection property if P

in L

Assume

. There is a

P'

mapping

+

is prime for every prime ideal

completely different proof for this last statement.

P

prime for every prime

IT:M

P'

. Note

first that by Theorem 80.1 the

is a one-one mapping of the set M

M'

onto the set M'

prime ideals in L

of all minimal

.

of all minimal prime ideals in L'

Note, furthermore, that in L' every prime ideal contains a unique minimal prime ideal (since L'

M'

. Under

set

has the projection property; cf. Theorem 37.ll),

. Now consider the hull-kernel

the same holds in L

the mapping

{M'IU

so

inverse map

-1

is continuous. Thus,

of the compact space continuous on

the image of any open base set

IT

{MIu

morphic mapping from

-1

{MIu

u E L+

onto

{M'IU

,

therefore,

. Note

follows from the hypothesis that L ' (and hence also in

IT

{MIu

,

so

IT

is also

induces a homeo-

that the compactness of

has the projection property

By Exercise 37.14 the hull-kernel topology in M'

{M'IU ) is extremally disconnected, s o by the homeo-

morphism the same holds in {MIu

is the

is a continuous one-one mapping

IT

onto the Hausdorff space

. For every

(cf. again Theorem 37.11).

of

{MIu

so

and

maps open sets onto open sets, which implies that the

IT

IT

topologies in M

{MIu

,

i.e., the closure of every open subset

is open. Furthermore, {MIu

Combining now the compactness of

is compact by the homeomorphism.

{MIu

for every

u 2 0 with the fact that

every prime ideal in L contains a unique minimal prime ideal, we may conclude already that L has the principal projection property (cf. Theorem 37.II). Once having shown that

L

has the principal projection property,

Theorem 31.3 can be applied, i.e., L will have the projection property if it can be shown that the lattice of all principal bands in L complete. Denote by

BU

is equivalent to

, and

{MIu c {MIv (cf. Theorem 35.5).

BU c Bv in the lattice of all principal bands we have {MIU

u E L+

the band generated by

is open and closed, we have

1. c BU

B UT

is Dedekind

note that

Assume now that

. Then, since

Ch. 11,§821

and the set A

, being

well as closed. Since A =

{MI

B

C

uT

If

is compact, it follows from Theorem 37.2 that

{MIu

,

. Note

that

{MI

uT

c{M)

for all

uo

implies

T

B is an upper bound of the set of all uo uO Bv is another upper bound, then {MI C {MIv for all T , s o

A c {MIv

for all

{MIu , is open as

the closure of an open subset of

for some uo E L+

uO B

63


,

T

so

.

i.e., {MIU c {MIv 0

supremum of the set of all

UT

,

It follows that BuO c Bv

.

B

UT

so

B uO

BU

T

.

is the

82. Normal and extremally disconnected lattices.

Let

(X,e,e)

be a distributive lattice with the smallest element 0

and largest element e (in the terminology of Chapter 1, e

is therefore

the unit of X ) . In this lattice we can introduce, besides the given ordering, the inverse ordering, also called the duaZ ordering. Then X with respect to the dual ordering a distributive lattice with element and 0 as largest element. Evidently, x v y original ordering are the same as x A y and x v y

e

is

as smallest

and x A y in the with respect to

the dual ordering. It follows immediately that a proper prime ideal for one of these orderings is a prime dual ideal for the other ordering, and conversely. Hence, a maximal ideal for one of the orderings is a maximal dual ideal for the other ordering.

It will be evident that the above duality relation can be used to derive dual results from known results. A s a first example, let us consider Theorem 51.8 (ii), the prime ideal extension theorem from an ideal in to the whole of ideal in

(X,O)

(X,0)

. We

and let w

recall the explicit formulation. Let be a proper prime ideal in I in X

exists a proper prime ideal w ' W'

is uniquely determined by

w

satisfying w '

. Explictly, we have

fl

I

(X,€J)

I be an

. Then there . The ideal

= w

Note that this is equivalent to the statement that every prime dual ideal A in I can uniquely be extended to a prime dual ideal A ' in X

. Explicitly,

Ch. 11,1823

NORMAL AND EXTREMALLY DISCONNECTED LATTICES

64

f o r some aEI) =

= (X:fiX,XA&h

(x:x€X,x>b

f o r some bEh

For t h e l a s t e q u a l i t y , n o t e t h a t i f x 2 b

for

b = x

.

E X

a

A

x

a

A

f o r some

E X

Conversely, i f

x t b E

a E I then

x

, then A

b = b E X.

Now, f o r the p a r t i c u l a r case t h a t the l a t t i c e has a l a r g e s t element I

is a principal ideal

I n s t e a d of

I = Ce,x

I

(x:8<x<xo)

=

, we

,

e

and

l e t us d u a l i z e t h i s r e s u l t .

1 0

= (x:8<x<x )

x

) . It follows t h a t i f (with r e s p e c t t o the o r i g i n a l

get then a subset of t h e form

0 Y = Cx , e l = (x:xo<x<e) Note t h a t with r e s p e c t t o the o r i g i n a l ordering 0 Y i s again a l a t t i c e (but now x i s t h e s m a l l e s t element), and with 0 r e s p e c t t o t h e dual ordering Y i s a p r i n c i p a l i d e a l i n X ( t h e p r i n c i p a l i d e a l generated by

0

i s a prime dual i d e a l i n

ordering) h extension

.

X'

to

X

,

A'

and

Equivalently, every proper prime i d e a l extension

w'

to

X

,

and

= (x:x€X,x are equivalent. (i)

X

i s completely normal. containing a prime dual ideal i s prime

(ii) Any dual ideal i n X

itself. (iii) The f a m i l y of a l l prime dual i d e a l s containing a given prime

dual ideal i s linearly ordered by inclusion. (iv) The f a m i l y of a l l prime ideals contained i n a given proper prime ideal i s linearly ordered by inclusion. (v)

For any prime dual ideal

A

and f o r any

x

not i n X

there

e x i s t s a unique prime dual ideal A 1 such t h a t X I 3 A and X I i s a maximal prime dual ideal with respect t o the property of not containing (vi)

For any proper prime ideal

w

and f o r any

a unique prime ideal w1 such t h a t w 1 c w and ideal wiht respect t o the property of containing

w1

x

.

x E w

x

there e x i s t s

i s a minimal pr*e

.


Ch. 11,§821

PROOF. (i)

=s

dual ideal in X xo v yo E F arbitrary

Let X be completely normal, and let F

(ii).

E X

yo E F

or

and set wo

=

(x:€I<x<w )

=

be a

to show that

. For this purpose, choose an .

xo v yo v zo

normal, any sublattice of the form sublattice Y

. We have

containing the prime dual ideal A

implies xo E F zo

79

Since X

is completely

(x:w,<x~wo) is normal, i.e., the

has the property that, with respect to the dual

0

ordering, every principal ideal in the sublattice is extremally disconnected. In other words, Y

(x:€I<xrw )

=

is completely extremally dis-

0

connected with respect to the dual ordering. Note now that the intersections

Y fl F and Y that Y

F

Il

fl

X

are not empty (since

zo

and Y n A

is an ideal in Y

E Y n

A c

Y n F ); note also

is a prime ideal in Y with

respect to the dual ordering. Hence, with respect to the dual ordering, Y is completely extremally disconnected and Y n F

is an ideal containing

. It follows from Theorem 82.15 that Y f l F . Note now that xo and yo are members of Y

the prime ideal Y n A a prime ideal in Y

with respect to the dual ordering, it is given that X Hence xo E Y n F (ii)

*

or yo E Y

(iii).

XI

Let

prime dual ideal

A

the other. Then X

, and

,

y

o

and, ~

~

and

A2

be prime dual

assume that neither of

deals containing the

A ,A2

x1 n

is prime by hypothesis (ii). Also, since neither of x1,x2 E X

is contained in i2 , SO

X2 is contained

AI,A2

such that x 1 is in X I

but not

.

in X 2 and x2 is in A 2 but not in A 1 Then x 1 v x2 is in x 1 n so (since X I n X is prime) we have x I E X I n A 2 or x2 E X I n A 2

-

2

This contradicts the defintion of prime dual ideals containing (iii) (iii)

*

A

. Hence, the family of

x 1 and x2

A

be a prime dual ideal, and let x

respect to the property of not containing x

. Let

kind. By hypothesis (iii) we have either A

cX

follows from the x-maximality property that

A

x

-

x2,

all

is linearly ordered by inclusion.

There exists at least one prime dual ideal containing A

dual ideal A ,

.

(iv). These are equivalent by duality :(cf. Theorem 76.1).

(v). Let

containing A

1

XI

or = A2

(vi)

(vi).

.

and maximal with and

A

be not in A

2

A2

cXI

be of this

,

so it

. Hence, the prime

and maximal with respect to not containing

is unique. (v)

n

as desired.

is contained in the dual ideal

in the other, there exist

~

n F , since Y n F is a prime ideal. It

or yo E F

xo E F

follows that

O

is then

These are equivalent by duality.

* (i). It is sufficient to show that, for an arbitrarily chosen

~

.

80

Ch. 1 1 , § 8 2 1


xo E X

,

the sublattice Y

the proper prime ideal

w

(x:x <x)

=

0-

is normal. In other words, given

in Y , we have to show that

unique minimal prime ideal in Y

w

contains a

. Hence, let the proper prime ideal

w

in

Y be given, and let p 1 and p 2 be minimal prime ideals in Y , both of these contained in w By Lemma 8 2 . 1 (and by the remarks preceding the

.

lemma) the prime ideals w , u l u',pi

and

pi

to X

.

and

p2

in Y

have unique prime extensions

Explicitly,

x:xEX,xcb for some bEw and similarly for p i

and

pi

. Evidently, xo E

p i t w1

. We

show that

is a prime ideal, minimal with respect to the property of containing

pi

xo

.

Indeed, if

included in p i included in p 1

11;

,

is a prime ideal in X

then p; fl Y

,

containing xo

and properly

is a non-empty prime ideal in Y

properly

(that the inclusion is proper follows from the explicit

extension formula). This is impossible since

p1

is a minimal prime ideal

. Hence, p i is minimal with respect to containing xo . The same holds for p i . By hypothesis (vi) the prime ideal contained in w ' and minimal with respect to containing xo is unique, s o p ; = p i . Hence in Y

p l = p;

ny

= p1

2

n

i.e., the given prime ideal w

Y

= p

2 '

in Y

contains a unique minimal prime

ideal. We observed already that several of the conditions for normality in Theorem 8 2 . 5 are mentioned by A.A. Montecro ( C 1 1 , 1 9 5 4 ) in a paper without detailed proofs. In this paper are also mentioned conditions for completely normal lattices (Theorem 8 2 . 1 7 ) and for completely extremally disconnected lattices (Theorem 8 2 . 1 5 ) as well as the theorem that a topological space is normal or extremally disconnected if and only if the lattice of its closed subsets is so (Theorem 8 2 . 1 0 ) . There is another condition for complete normality, due to W.H. Cornish ([I 1 , 1 9 7 2 ) . The condition extends the normality condition saying that

X

=

FI 1

V p2

holds for any pair of different minimal prime ideals.


Gh. 11,9821

THEOREM 82.18. The d i s t r i b u t i v e l a t t i c e i f and only i f

ideals

X

v w2

= w1

(X,O) i s completely normai!

hoZds f o r any pair of incomparable prime

( i . e . , prime ideals

w1,w2

81

w1,w2 such that neither o f them i s

included i n the other). PROOF. Let X be completely normal and let u1,w2 be incomparable prime ideals in X w1 v w2

X

.

. Assume now

Then

w 1 v w2

that the ideal

w 1 v w2

satisfies

is included in a proper prime ideal

w

0 '

By condition (iv) for complete normality in the preceding theorem the prime ideals contained in wo and

w2

are linearly ordered by inclusion, s o (since w

are included in wo ) we have

diction. Hence w, v w 2

X

=

.

w1 c

w2

or

. Contra-

w2 c w 1

The converse is also easy. It is given now that w 1 v w2 incomparable wl,w2

,

and we have to show that X

i.e., we have to show that if

w1

and

in the proper prime ideal wo , then

w2

w

0

=

X

. Contradiction.

for

are prime ideals contained

w 1 c w2

. But

X

is completely normal,

or

w2 c w 1

then X

. Assume this

and incomparable

to be false. Then there exists a proper prime ideal wo prime ideals w1,w2 contained in wo

=

1

= w

= wo '

1

so

There is no connection between normality and the property to be extremally disconnected (cf. Exercise 82.20).

However, if

as well as completely extremally disconnected, then X normal. Similarly, if normal, then X

(X,O,e)

(X,e)

is normal

is completely

is extremally disconnected and completely

is completely extremally disconnected.

THEOREM 82.19.

(i) If the d i s t r i b u t i v e l a t t i c e

(X,e)

i s normal as

well as completely extremally disconnected, then X i s completely normal. (ii) S i m i Z a r l y , i f

(X,e)

has a largest element

is is completely

e and X

extremally disconnected as well as completely normal, then X extrema l l y disconnected. PROOF. (if Let

(X,O)

be normal (i.e., every proper prime ideal

contains a unique minimal prime ideal) and completely extremally disconnected (i.e., the proper prime ideals containing a fixed proper prime ideal are linearly ordered by inclusion). We shall prove that X

is

completely normal (i.e., the prime ideals contained in a fixed proper prime

82


Ch. 11,1821

ideal are linearly ordered by inclusion). For this purpose, let w ,w 1

let u0

prime ideals contained in the proper prime ideal wo , and

.

unique minimal prime ideal contained in wo

be

2

be the

It follows immediately that

is also the unique minimal prime ideal contained in w 1 , and similarly

p0

. Hence, w 1 and w2 are proper prime ideals containing the prime . By hypothesis, the prime ideals containing uo are linearly ordered by inclusion, s o or w2 c w 1 . This is the desired result. w1 c w2 for w2

ideal uo

(ii) The proof is similar. A s observed earlier, any Riesz space

connected (equivalently, the lattice X(L)

L

is completely extremally dis-

of all principal ideals in L

is completely extremally disconnected). Hence, if

L

is normal (i.e., if

any proper prime ideal contains a unique minimal prime ideal), then L

is

completely normal. In this case, therefore, we have the situation that if the proper prime ideal ideals containing in wo

, is

wo

wo

is given, then the family of all proper prime

, as well as the family of all prime ideals contained

linearly ordered by inclusion. In other words, the union of

these two families is a chain of proper prime ideals such that member of the chain. Given the proper prime ideals wo corresponding chains are identical if chains are disjoint if set Q

wo

and

wh

wo

and

wh

and

are incomparable. It follows that the

principal projection property (i.e., for any u E L+ {uId

0 '

of all proper prime ideals is the disjoint union of all such chains

minimal prime ideal in the chain. We recall that if =

the

are comparable and the

Q ) of the

Each chain is the closure (in the hull-kernel topology of

L

is a

wo

0'

0

{uIdd ), then L

L

has the quasi

we have

is normal (by Lemma 3 7 . 1 0 ) , and s o

L

is

completely normal in this case. Not every Riesz space is normal; it was indicated in Exercise 3 7 . 1 3 that every maximal ideal in the space C([O,ll) of all real continuous functions on C 0 , l l

contains infinitely many

different minimal prime ideals. EXERCISE 82.20. Show that the distributive lattice

(X,e)

in Exercise

7 6 . 1 4 is completely normal but not extremally disconnected. Hence, with

respect to the dual ordering, X

is completely extremally disconnected but

not normal. Show that by taking the smallest element of X obtains a lattice X'

away, one

that is completely normal as well as extremally

disconnected. Actually, there are only two proper prime ideals in X'

, and

both of these are minimal as well as maximal. Show also that by appropriately

Ch. 11,8821

83


adding one extra element to X

, one

can obtain a lattice X"

that is nor-

mal as well as extremally disconnected, but neither completely normal nor completely extremallydisconnected.Finally, show thatby adding to X' its "reflection" with respect to its largest element, one obtains a lattice X"' that is neither normal nor extremally disconnected. Note that all possible combinations occur. EXERCISE 82.21. Show that every subset of the topological space E

is

normal (with respect to the relative topology in the subset) if and only if the lattice X

of closed subsets of

E

is completely normal. Formulate

and prove the dual result. HINT: We indicate how to prove that complete normality of that the relative topology in the arbitrary subset E" Let FY = F1 fl E" =

of

E

X

implies

is normal.

(with F,,F2 closed) be relatively

E-such that ; ' F

and F" are disjoint, i.e., 2 F1 n F2 has no points in common with E" By the complete normality

closed subsets of Fo

F ; = F2 fl E"

and

of

X

.

there exist closed sets F;,F;

.

such that F; fl F 1

F;

=

Il

F2

= Fo

Fi U F' = E The set theoretic complements 01,02 of F;,F; are 2 open and disjoint, and the intersections with the complement : F of FO satisfy Fi fl FE c 0; fl : F for i = 1,2 Hence, the intersections with and

.

certainly satisfy Fi fl E" c 0; n E"

the smaller set E"

In other words, the sets 0; = 0. fl E" (i=1,2) 1

the relative topology of E"

1

, and

=

1,2

.

are disjoint and open in

we have FY c 0; 1

for i

for i

1

= 1,2

. This

shows that the topology in E" is normal. EXERCISE 82.22. (i) Let the distributive lattice

{pix = The lattice

(p:p~n~,xnot in p

).

is called dLsjunctive if x # y

implies {plx #

(X,9) is disjunctive if and only if x # y

Show that {XId

(X,8)

be the set of all minimal prime ideals in Om (X,9) and, exactly a s several times before, let

# {y}d

.

I v } ~.

implies

(ii) Let there be given a base for the closed sets in the topological space E

. Show that

E

is Hausdorff if and only if, for any different

points x1,x2 in E , there exist sets FI,F2 in the base such that xi is not in F. for i

=

1,2 and such that F 1 U F2

=

E

.


84

( i i i ) Show t h a t t h e d i s j u n c t i v e l a t t i c e

{pix

x E X, t h e s e t

i f , f o r each

Ch. 11,5823

i s normal i f and only

(X,9)

i s Hausdorff i n i t s d u a l h u l l - k e r n e l

topology. d i f and only i f {xId = {y} Y For ( i i i ) , assume f i r s t t h a t a l l { u I X a r e Hausdorff i n t h e

HINT: For ( i ) , n o t e t h a t by Theorem 6.5. dual hull-kernel X

{vIX

= {p}

topology. We have t o prove t h a t any p r o p e r prime i d e a l i n

c o n t a i n s a unique minimal prime i d e a l . Hence, l e t

prime i d e a l i n now t h a t

ul

u2

for

and l e t

x E X b e such t h a t

w

u2

and

p1

. Assume . Then

. The

{pix

sets

form a b a s e f o r t h e c l o s e d s e t s i n t h e dual h u l l - k e r n e l

{uIx . Hence, {uIz , u2

i s not i n

I n o t h e r words, we have

there exist s e t s

{u}

is not i n

and

E

z1

pl

{t~}

, z2 E

. On

z2

p2

d i s j u n c t i v e p r o p e r t y ) . I t f o l l o w s from

zi E

E w

way

i s chosen. C o n t r a d i c t i o n . Hence, w

z1

and

ui

{ u I Z 2 such t h a t

and Iulz

U IuIZ2 =

1

i = 1,2

for

c w

{vIx

.

(by t h e

z 1 v z2 = x

i s not i n

z 1 v z2 = x

t h e o t h e r hand

z1 v z2

x

be a given proper

w

i s not i n

x

a r e d i f f e r e n t p o i n t s i n t h e Hausdorff space z S x

topology of

u1

,

c o n t a i n s d i f f e r e n t minimal prime i d e a l s

w

and

{uIZ

X

that

i n view of t h e

w

c o n t a i n s only one minimal

ideal. Conversely, assume t h a t

{uIX i s

n o t Hausdorff i n i t s d u a l h u l l - k e r n e l

different have

Cul

i s normal and f o r some

X

z1

ul,u2 U {u}

i n {uIx 22 p

such t h a t f o r a l l

. In

# [ulx

the s e t

x E X

topology. Then t h e r e e x i s t and

z1 E pl

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

E u 2 we

z2

z 1 v z2

.

i n the ideal

I n view of t h e and u 2 we have { u L l V z 2 # {uIX d i s j u n c t i v e p r o p e r t y t h i s shows t h a t x i s n o t i n I , so t h e r e e x i s t s a I

g e n e r a t e d by

prime i d e a l

such t h a t

w

and

I c w

x

is not i n

w

. Hence,w

proper prime i d e a l c o n t a i n i n g d i f f e r e n t minimal prime i d e a l s

ul

is a and

u2

.

Contradiction. EXERCISE 82.23.

Let

t h e f a m i l y of a l l s e t s

(X,9)

{xIdd

o r d e r e d by i n c l u s i o n . Show t h a t element

feldd

=

{el

. Show

be a d i s t r i b u t i v e l a t t i c e , and l e t for A

x E X

,

A

be

these s e t s being p a r t i a l l y

i s a d i s t r i b u t i v e l a t t i c e with smallest

t h a t t h e l a t t i c e A i s d i s j u n c t i v e . Formulate

and prove t h e corresponding r e s u l t f o r a Riesz space. HINT: Note t h a t

t h e set of a l l

{uIx

A

, partially

(for

orderdd by i n c l u s i o n , i s isomorphic t o

x E X ) , p a r t i a l l y o r d e r e d by i n c l u s i o n . This

f o l l o w s from Theorem 6.5 ( i i ) , according t o which {xIdd c {yIdd

a r e e q u i v a l e n t (and hence

IwIx

=

{ulx

c

I U ) ~and

{u)

and = {yIdd

Ch. 11,8821

85


{uIx

are equivalent). The set of all smallest element

{pie

is a distributive lattive with

equal to the empty set (cf. section 6 ) . Hence A

{eldd

is likewise a distributive lattice with Note that the supremum and infimum of

=

{e}

as smallest element.

{xvyIdd and dd { x ~ y } respectively. ~ ~ The disjoint complement of the given element {XI

in A all

is the set of all {y}dd {yIdd

such that

{xlId # {x21d

, and

satisfying y

y E {xld

. Hence, if

{xlIdd and

so

IxIdd and

complements. It follws that A

by

. By

y

i.e., the set of

,

then

{xIdd c I

{yIdd

=

{x}dd

. Show that

(X,e)

I is a

implies y E I

. We

is a d-ideal (cf. Theorem 5.4

be the same lattice as in the preceding exercise. Prime

ideals in X will be denoted by in A

,

in the distributive lattice

recall that any minimal prime ideal in X Let A

= 0

{x,fdd # {x2}dd

{x2Idd have different disjoint

x E I implies

d-ideal if and only if x E I and (iv)).

x

are

is disjunctive.

EXERCISE 82.24. (i) The ideal I is called a d-ideal if

A

{yldd

w

(exactly as before), prime ideals

a d-prime ideal is meant a prime ideal which is also

a d-ideal. Show that the mapping

is a one-one mapping of the set of A l l prime ideals in A all d-prime ideals in X

with minimal prime ideals in X

, and

(ii) Minimal prime ideals in X

,

the set

{uIx

correspond

conversely. Formulate and prove

the corresponding result for Riesz spaces. each x € X

onto the set of

such that minimal prime ideals in A

are denoted by

u

. Show that, for

is Hausdorff i n its dual hull-kernel topology

if and only if every d-prime ideal in X

contains a unique minimal prime

ideal. Formulate and prove the corresponding result for Riesz spaces. HINT: For (ii), note that every d-prime ideal in X

contains a unique

minimal prime ideal if and only if the disjunctive lattice A

is normal.

Apply Exercise 82.22. EXERCISE 82.25. Show that the following conditions for the Riesz space

L

are equivalent.

(i) topology*

For every u E L+

the set

Cu}u

is compact in the hull-kernel

86

Ch. 11,9821

NORMAL AND EXTREMALLY DISCONNECTED LATTICES (ii)

The lattice of all

lattice of all

is a Boolean ring. Equivalently, the

{uldd is a Boolean ring.

(iii) L has theM-complementation property, i.e., for all u,v E L+ satisfying 0 2 v dd dd Iv+v,l = {u}

2

.

u

there exists an,ielement v2 I v

such that

L(iv) For every u E L+ the hull-kernel topology and the dual hullkernel topology coincide on {p}u (v) Every proper d-prime ideal in L is a minimal prime ideal.

.

HINT: The conditions (i), (ii), (iii) are equivalent by Corollary 37.3; the conditions (ii) and (iv) are equivalent by Exercise 77.13. Note that (v) is equivalent to the condition that every proper prime ideal in A

is a

minimal prime ideal. It follows from Theorem 8.5 and Theorem 8.6 that this is equivalent to (ii). EXERCISE 82.26. element of x

if

i.e., 2

.

8 A

Let

Given x E X

be a distributive lattice with smallest

(X,8)

, the

is the largest element disjoint to x ,

x ~ x * = eand x* y = 8

The lattice

implies y

(X,e)

x*

2

. Evidently

(i) 5

e = 8*

Show that

x . (ii)

Show that

complements, then p

(xVy)* = x* A

(iii) Show that and only if x** (iv)

y

x

x

(X,8)

be pseudocomplemented.

y*

A

x**

5

and

. Show that

x* = x***

. Hence, if

p

and

q

for every are pseudo

is a pseudo complement. A

y

A

=

.

y

(xAy)** = x**

if and only if

8

5 z* A

x**

if and only if x**

y**

y = 8

A

y**

A

.

, and 2 z*

ahso if

. Use this

Let x E X be given. Show that the smallest pseudo complement

such that z* t x

z*

and

q

y** = 0

Show that

to prove that (v)

A

is uniquely determined.

is the largest element of X

implies x* 2 y* ; furthermore, x

y

x*

is called pseudocompZemented if every x E X has a

pseudo complement. In the following, let x

is called pseudo complement

element x* E X

satisfies z*

=

. Use this

x**

are pseudo complements, the supremum of x

pseudo complements i s

(xvy)**

(vi) Let D(X) = (x:X*=e)

. Furthermore implies y E D(X) .

x

A

y E D(X)

y

t

x

= =

(x*AY*)*

.

(x:x**=e)

x v x* E D(X)

(vii) Given the proper prime ideal w

to prove that if x

and y

. Show that

for every in X

,

x

in the set of

x,y E D(X)

implies

and x E D(X)

the following

,


Ch. 11,§821

conditions are equivalent: (a) w

,

is not in w

x E w

(c)

Note that if X

87

is minimal, (b) x E w

,

implies x** E w

implies that

w f l D(X)

(d)

x*

is empty.

is the lattice of all ideals in a distributive lattice

with smallest element or in a Riesz space, then Id is the pseudo

I for any I E X

complement of

HINT: For (ii), write x* Similarly x* so

I

(xvy)*

x**

A

y = 8

first that x** x** w

x**

A

y**

I z*

A

y**

I

w

is minimal

generated by

. Hence

y = t3

z A

y*

A

. For (iii),

. For y

A

=r

y

5 z*

A

y

,

A

x

A

y

A

.

= t3

implies y I x* x

so

z

(xVy)* , i.e.,

z I

= t3

and then that x**

, so

x*

y**

z = 8

A

z = d'

A

.

=

,

x***

It follows

. Hence

=)

X-w

and

,

x (note that

(d)

8

is not in F ; otherwise 0

and hence q I x* E w

proper since x E w-w, (c)

A

z I

, so

z A (xvy) = t3

note that x

= t3

A z

. Then

= z

. Now write ( ~ y ) l * = w , s o x A y w . Then = (ay)** . For (vii) we first prove (a) (b). Assume that and x E w as well as x* E w . Let F be the dual ideal

for some q E X-w

=r

y*

(iv), let x

exists a prime ideal w 1 (b)

A

.

disjoint to F

. This

, which , so w1

q

A

x

is impossible). There

. The

c w

is impossible since w

is easy. For (d)

=

inclusion is

is minimal.

* (a), assume that (d) holds and

w

is not

minimal. Then there exists a prime ideal w 1 c w with an element x E w-w 1 Then x* E w 1 c w , s o x v x* E w But then x v x* E D(X) n w , which contradicts

.

(a).

EXERCISE 82.27.

The pseudocomplemented lattice

lattice if x* v x** = e lattice X*

V

(X,e)

y* = e

for every x E X

is a Stone lattice if and only if x

, i.e.,

if and only if X

(X,e)

. Show that A

y

= t3

is normal. Hence, the pseudocom-

(X,8)

prime ideal in X

contains a unique minimal prime ideal.

x

(X,e)

be a pseudocomplemented lattice. Show that

is a Stone lattice if and only if

...,xn XI A ... A Xl,

Let

(xIA..Ax ) * = xr v n

...

V

x* n

. Equivalently, X is a Stone lattice if and only 8 implies x; v ... v x* = e . n

in X x = n

implies

is a Stone lattice if and only if every proper

plemented lattice

EXERCISE 82.28.

is called a Stone

the pseudocomplemented

for all if

HINT: It is sufficient to show that in a Stone lattice we have (XAy)* = x* v y* Theorem 82.5.

. This can be

proved similarly as (iii)

* (iv) in

CHAPTER 12

ORDER BOUNDED OPERATORS

L

In section 83 we assume that

and M

are Riesz spaces, with M

complete. The linear operator T from L Tf 2 0 holds i n if

M

into M

for all f t 0 in L

and

Dedekind

is called positive if

T i s called order bounded

T maps order bounded sets (in L ) onto order bounded sets (in M ) .

We recall that a subset D exist

f,g E L

such that

of f

5

L d

5

is said to be order bounded if there

g

for all d E D

. Any

positive

T is order hounded if and only if T = T 1-T 2 TI and T2 positive (Jordan decomposition). The vector space of all

operator is order bounded and with

order bounded operators (from L

Lb

. For

TI and

T2 in Lb

positive. This makes

Lb

into M ) is denoted by

we write T

1

5

Lb(L,M)

, briefly

T2 whenever T2-TI is

an ordered vector space with the set of all

positive operators as positive cone. We shall prove that

Lb is a Dedekind

complete Riesz space with respect to this ordering (L.V. Kantorovitch,l936; is the space of real

H. Freudenthal,l936). In the special case that M numbers,

is 'exactly the space L"

Lb

of all order bounded linear

functionals on L

.

theorem that L"

is a Dedekind complete Riesz space goes back to F. Riesz

The space L"

is called the order dual of

(1928) already. Denoting the null operator by T+ = sup(T,8)

of any

+

T u for all u E L+ There exist dua Tf+

=

T E

Lb

. The

the positive part

satisfies

0

sup Tv:o -u-v 0 -

[-u-v,u'+vl]

. But

0

then D

< 0 -

so

=

. It follows already that

D

is included in [-w,wl

is included in for w

From these remarks it is evident that the algebraic bounded subsets D1 and D2 DEFINITION 8 3 . 2 . Let

=

.

u+v+u'+v'

sum

of

D,+D2

is again bounded.

L be as before and l e t

be a (not necessari-

M

l y Dedekind complete) Riesz space. The operator T from L i n t o M i s called order bounded i f the image under T of every bounded subset of L i s a bounded subset o f

. In

other words, T i s order bounded whenever the image of every s y m e t r i c order interval i n L i s a bounded subset of

M

. Equivalently,

M

T i s order bounded whenever, f o r every

u

E L+ , the s e t

(ITf 1 :-ur-fr-u)

i s a bounded subset of

M+

. Evidently,

bounded operators from L i n t o M of all operators from

L(L,M)

L

the s e t

. The space

#

L

operator T from L i n t o M and T2 p o s i t i v e .

. I n particular,

into M

a l l order bounded linear functionals on

algebraic duaL

of a l l order

Lb(L,M)

i s a linear subspace of the space

L"

the s e t

of

L"

L i s a linear subspace of the

i s called the order duaL o f

i s called regular i f

L

. The

T = T -T with 1 2

TI

It is obvious that every regular operator is order bounded. In the decomposition theorem which follows now it is shown that for M same. From here on we assume again that M

Dedekind

into M ) are the

complete regular and order bounded operators (from L

is Dedekind complete.

THEOREM 8 3 . 3 . (i) Any positive operator from

L

into M

i s order

bounded. (ii) (Jordan decomposition). The operator

T from

L into M

order bounded i f and o n l y i f there e x i s t positive operators (from L i n t o M

such t h a t

)

PROOF. (i) Let more, let u € L+

T = T -T 1

2

T2

'

T be a positive operator from L

and -u 2 f s u

TI and

is

. Then

-Tu

5

Tf

2

into M

Tu, so

. Further.

iTf/ < Tu

100

Ch. 12,§831


This shows that

Tu

is an upper bound of the set

.

(ITf I :-uSf5u)

Hence, T is order bounded. are order bounded, T

TI and T2 positive, then T I and T2 T is order bounded. For the converse, assume that

T = TI-T2 with

(ii) If

so

, and

is a bounded subset of M+ bounded subset of M T(U)

exists in M

, which

,

the set

so

(Tv:OSvSu)

is certainly a

implies that

0 is Dedekind complete. The element TO = 0 is one

since M

t 0

T(U)

the set

= sup Tv:O s p(f) for a l l Then there e x i s t s an operator T from V i n t o M such t h a t for a l l f E YJ and Tf 2 p(f) f o r a l l f E V . that

p(f+g)

s p(f)+p(g)

for a l l

the linear f E Vl Tf = t(f)

.

Ch.

12,1831


111

PROOF. Almost t h e same a s f o r t h e c l a s s i c a l Hahn-Banach theorem.

COROLLARY 83.14. Let

be an element i n t h e p o s i t i v e cone of the L and l e t vo be a p o s i t i v e element i n the

uo

Dedekind complete Riesz space pyincipal ideal T

operator

= 0

and Tf

in

A(u ) 0

L

for all f

. The

erists a positive Tug = vo

i n t o i t s e Z f ) such that

.

A(u )

complement of

0

0 i vo 5 aOuO f o r an a p p r o p r i a t e number

p ( f ) = a f + for 0 a l l f E L , i s s u b l i n e a r . On t h e l i n e a r subspace of a l l m u l t i p l e s of uQ we d e f i n e tfXu ) = X v Then t(huo) i 0 = p(Xuo) f o r X i 0 and 0 0 ' t(Xu ) = Avo 5 Xaouo = p(AuO) f o r X > 0 , s o t can be extended t o an 0

operator For

T

u E L+

mapping

L

i n the d i s j o i n t

PROOF. Note f i r s t t h a t

a. > 0

. Then there

uo

generated by

( i . e . , T maps

in

L

p from L

such t h a t

we have

T(-u)

i s p o s i t i v e . Replacing restriction of

A(uo)

.

TIA(uO)

Tug = vo

and

0

, so

p(-u)

S

=

Tf 5 p ( f )

Tu

for a l l

. This

0

5

f E L

shows t h a t

.

T

by t h e minimal p o s i t i v e extension of t h e

T

, we

i n t o i t s e l f , defined by

can make

T

vanish on t h e d i s j o i n t complement

i s a l i n e a r subspace Vl L (but not n e c e s s a r i l y a Riesz subspace of L ) such

I n t h e next theorem i t w i l l be assumed t h a t of t h e Riesz space t h a t f o r any

f E L

t h e r e e x i s t s an element

g E V,

satisfying

Note t h a t t h i s i s equivalent t o assuming t h a t f o r any gl

and

g2

in

Vl

satisfying

t h a t t h e i d e a l generated by

g,

5

f 5 g2

, and

t h a t t h e converse i m p l i c a t i o n i s f a l s e ; t a k e f o r continuous f u n c t i o n s on

f i g

L

. Note,

incidentally,

t h e space of a l l r e a l

L

vanishing a t t h e o r i g i n and f o r

[-1,11

.

there e x i s t

t h e condition implies

i s t h e whole space

Vl

f E L

the

V,

space of a l l odd f u n c t i o n s . THEOREM 83.15

(L.V.

Kartorovitch C21,1937/. Let

subspace of the Riesz space element

g E V1

L

such t h a t for any

satisfying f i g

, and l e t

t

from V I i n t o the Dedekind complete Riesz space p o s i t i v e means, therefore, that there e x i s t s a p o s i t i v e i.e.,

Tf = t ( f )

forcall

operator f E VI

V1

f E L

be a Zinear there e x i s t s an

be a positive operator M ( t o say that

t

is

f o r a l l f E L+ n v1 ). Then from L i n t o M uhich estends t

t(f) 2 0

.

T

,

112


Ch. 12,9831

f E L+ be given. The set of all g E V 1 satisfying g 2 f

PROOF. Let

t(g) t 0 holds for each

is non-empty by hypothesis and

It follows then from the Dedekind completeness of M

g

in this set.

that

P(f

.

exists in M+ 0

f

5

Note already that p(f)

fg in L

I all of

5

p(f+g)

=

L

implies 0

by defining p(f)

pI(f+g)+}

< -

p(f ++g+ )

f,g E L

. Furthermore it

p(af)

ap(f)

M

=

. We 5

p(f)

for a 2 0

. Hence, p

t(f)

=

p(f)

f+

+

p(g+) (af')

,

so

.

t(f)

on V 1

. Then

f E L

p(f)

+

af+

f E L+

Il

p

V1

a t 0 that

on V1

i.e.,

it is useful to observe

, but

5 f+ 5

by taking the infimum for all

.

into

this is evident, since

f E V1

It follows then from f

to

for all

p(g)

for real

is a sublinear mapping from L

by

p(f+)

5

=

f+ has then a

g E V,

that

g E V 1 with

g t f+

, we

p(f)

theorem there exists therefore an operator T

is an extension of

T

is positive. For

Tf t 0

and

. This concludes the proof that t is dominated the sublinear mapping p . In view of the generalized Hahn-Banach

obtain

so

= =

is dominated by

. For

f E L+ n V 1

for

in M.. We extend p

is not necessarily a member of V 1

in V 1

g

t(g)

5

t(f)

p(f2)

in this case. For an arbitrary

first that majorant

t

= 5

for arbitrary

p(f+)

f E V1

for all

p(f+)

follows from

shall prove now that

t(f)

t(f)

=

p(f,)

2

t

and

f E L+

T

T

from L

is still dominated by

into M p

such that

. We show that

we have

. This concludes the proof.

COROLLARY 83.16. If K

i s a Riesz subspace of

L

such t h a t the i d e a l

i s the whole space L , then any p o s i t i v e operator from K i n t o the Dedekind complete Riesz space M has a p o s i t i v e l i n e a r extension

generated by to

L

K

. I n particular,

l i n e a r extension t o Riesz space and

L

L

any p o s i t i v e linear functionaZ on K has a p o s i t i v e

. This holds

i n particular i f

i s the Dedekind completion of

K

K

.

i s an Archimedean

We have derived the Kantorovitch extension theorem from the generalized Hahn-Banach theorem. It is possible to give

an independent proof of the

Ch. 12,5831

113

ORDER BOUNDED OPRATORS

Kantorovitch theorem which is adapted to the special situation and which of course makes essential use of Zorn's lemma (cf. Theorem X. 3 . 1 in the book by B.Z. Vulikh c 1 1 ) . We continue with still another extension theorem, based again on the Hahn-Banach theorem. We shall deal again with the extension of a positive operator

t

subspace K

from a Riesz

to the whole space L

present situation the ideal generated by

It is assumed now, however, that

K

, but

is not assumed to be

in the

L itself.

is dominated in a spezial manner by a

t

sublinear mapping. THEOREM 83.17.

Let

be a sublinear mapping from the Riesz space

p

L

i n t o the Dedekind complete space

M such t h a t p i s absolute andmonotone ( i . e . , p(f) = p(lf1) f o r a l l f E L and 0 2 f 5 g i n L implies p(f) S p(g) i n M 1. Furthermore, l e t t be a p o s i t i v e operator from the Riesz subspace K of L i n t o M such t h a t lt(f)l 5 p(f) holds f o r all f E K . Then there e x i s t s a p o s i t i v e operator T f r a L i n t o M such t h a t T extends t and 1Tfl S T(lf1) S p(f) holds f o r a l l f E L . PROOF. The mapping for all

f E L

,

pl from L

into M

is sublinear just as well as

, defined by p,(f) p , and we have

= p(ff)

for all f E K . Hence, by the Hahn-Banach theorem, there exists an operator T from L into M which extends t and such that Tf S p 1 (f) holds for all f E L . It follows that for f E L+ we have T(-f) so

Tf 2 0

S

, which

pl(-f)

=

p (-f)

shows that

T

+I

=

p(0)

= 0

,

is positive. Finally, using now that T is

positive, we get ITfl 5 T(lf1)

5

pl((fl)

=

p(lf/) = p(f)

for every

f E L

As an application of Corollary 83.16 we shall prove now that if a Riesz subspace of the Dedekind complete space M , then M

. L

is

contains the L , but not in a unique manner in general. This result is due to A.I. Veksler ([11,1969). Precisely stated, we have the

Dedekind completion L-

of

1 I4

ORDER BOUNDED OPEPATORS

Ch. 12,1831

following theorem. THEOREM 83.18. If L

space

i s a Riesz subspace o f the Dedekind compZete L1 of M such t h a t L 1 i s

, then there e x i s t s a Riesz subspace

M

Riesz isomorphic t o the Dedekind compZetion

LA of

t be the identity mapping on L

PROOF. Let operator T

,

into M

operator from L

from L-

into M

u- > 0

Indeed, for any satisfying 0 0 in M

,

are strictly positive disjoint elements in L^

and v-

and v-

such that

. We

i s a one-one mapping, we make the following

Before proving now that T remark. If u1

t

Tu- 2 Tu = u > 0 in M

so

i s a positive

t

there exists an element u

in L-

,

u-

2

. Then

by Corollary 83.16 there exists a positive

so

is strictly positive, i.e., u- > 0 in LA

then u-

under an isomorphism

L

L invariant.

t h a t leaves

and v

. Then, in general, u

v-

Z

be disjoint, but it is possible to choose u

and v

and v

and v

in L

will not

in such a manner that

these elements are incomparable. Indeed, if the choice is made completely

0< u < v

arbitrarily, it may happen that Archimedean, the inequality nu

v

2

Take the largest natural number n

,

satisfies v 1 2 v-

= inf(vl+nu,v-)

Hence, the elements u but

v1 2 u

Tu-

2

Tv-

and

, we

Tu- and

choose u

are incomparable. But

u

Tv-

and

Tu

5

We prove now that T

=

vI in L

does not hold, so

immediately that

does not hold for all n u

I v1

inf!v,\v-)

v

5

v-

2

Tu^-Tv^ # 0

Tu-

.

For the proof that

T

5

u-

and

v

Z

1

,

v-

are incomparable. It follows

as indicated, Tv-

5 Tv,

,

so

Tu

=

u

and

Tv

I

=

v

1

so we get a contradiction.

and

Tv-

.

f- < 0 implies

f- has positive and negative parts

respectively that are strictly positive. Then u=

.

= v-

is one-one, that is to say, Tf- # 0 for f^ # 0

. Hence, assume now that

positive and disjoint, so Tf-

... .

= 1,2,

we have

In the beginning it was shown already that f- > 0 or Tf- # 0

is

holds. Then v 1 = v-nu

satisfy 0 < u

and v,

L

are incomparable, because if for example

v,

Tu-

nu

for which

since on account of

inf(v,,v-)

. In this case, since

and v^

u- and

are strictly

are incomparable. It follows that

is a Riesz isomorphism it remains to show that

Ch. 12,1837 -1

is also positive (cf. Theorem 18.5).

the inverse operator T We have to show that Tf- < 0

would imply u-

and

115


v-

of

f- Z 0 . It is impossible that f- < . Hence, assume that the positive and

f- are strictly positive,

incomparable. But

Tf-

It follows that f-

0 implies TuA

Z

.

0

Z

so

t Tv-

Tu-

,

so

Tf-

Z

0.

0 since this negative parts

Tv-

and

Let

are

we get a contradiction.

It has been proved thus that L^ is Riesz isomorphic to the image T(L-)

L-

of

in M

As before, let

To each by

f E L

f-(T)

=

i.e., f-

L

and M

be Riesz spaces with

we assign a mapping b

is an operator, and if

(f+)-

f-

from

and

f

positive operators, so

(f-)-

into M

Lb(Lb(L,M),M)

, defined

is a linear mapping,

is a positive element, then the

. In other words,

.

f E L

we have

i s an order bounded

f-

It is now also evident that the mapping y : f + finto

Dedekind complete. into M

is a positive operator. For an arbitrary

operator from Lb(L,M)

from L

M

Lb(L,M)

for all T E L (L,M) . Obviously f-

Tf

corresponding f-

with

.

is a positive operator

LEMMA 83.19 (C.T. ~ e k e r , C l l , 1 9 7 7 ) . The operator

y

, as defined above,

i s a Riesz homomorphism. PROOF. It is sufficient to prove that y(f+) = (y(f))+ Equivalently, it is sufficient to prove that (f+)- = (f-)'

Given any

f E L

, we

f E L

f E L

. .

have to prove therefore that

holds for all T E Lb(L,M) For T

for all for all

positive we have

, and

it is sufficient to do this for positive T.

116

Ch. 12,1831


by Namioka’s formula in Theorem 83.9, and

=

0

sup Sf:BSS 0 in M

0 in L

.

is A

TI

is order closed.

= TI

(B)

be the

is ru-closed

is also ru-closed.

is an ideal in L ; explicitly,

Hence, since A

is a ru-closed ideal, A

is a a-ideal by hypothesis. Note

now that

I = M(C0,ll)

. Let

i s properly included i n

u E L+

E L

k

.

may be regarded as a Riesz

of a l l ( r e a l ) Lebesgue measurable

122


, provided

C0,ll

f u n c t i o n s on

Ch. 12,1833

t h a t any measurable f u n c t i o n almost e q u a l

t o a continuous f u n c t i o n is i d e n t i f i e d w i t h t h e continuous f u n c t i o n . According t o Theorem 8 3 . 1 8 t h e Dedekind completion in

. Show t h a t

M

M

i s p r o p e r l y contained i n

C*

HINT: There e x i s t s a sequence

u

0

C

in

C

of

C^

.

C

i s contained

such t h a t t h e pointwise

l i m i t of t h e sequence i s s t r i c t l y p o s i t i v e on a s e t of p o s i t i v e measure f o r example, p a r t ( i ) of E x e r c i s e 1 8 . 1 4 ) .

(cf.,

L

EXERCISE 8 3 . 3 1 . Let

be a Riesz space and l e t

complete R i e s z space. Furthermore, l e t (i)

Du

Let

be t h e s u b s e t of

DU = (,C;lTuil Show t h a t

T € L (L,M) b by

:n€N, u=Cnu. I 1 , a l l uiEL

T ,T

1

2

E Lb(L,M)

be t h e s u b s e t of

EU

M

and l e t

J. (TlhT2)(u)

E

HINT: sequence

(i) If (wij:i=l

w j i ij than o r eqaul t o

and

v

CITuil so

= C

CITl(ui)

S

sup D

U

u = C;ui

such t h a t

, defined

TI

. and

CCITw..I 1J

2 iTl(u)

. For

so any upper bound of Hence

sup D

U

2

IT1 (u)

( i i ) Note t h a t

Du

.

T2

are positive

by

.

. = Cmv

( a l l ui,v.

l j ,...,n ; j = l , ..., m) f o r every j . Then = ITl(u)

.

u E L+

+> .

EU = ( E p I ~ i ~ T 2 ~ i : n €,u=Cyui,all N uiEL+) Show t h a t

be a Dedekind

, defined

ITl(u> = sup Du ; p r e c i s e l y , DU 4 ITI(u)

( i i ) Now, l e t and l e t

M

M

, so DU , t h e set

every

f

J such t h a t

E L+ ), t h e r e e x i s t s a double

f o r every i u. = C . W . . 1 mJ 1 J C ITv. I a r e both l e s s 1 J i s upwards d i r e c t e d . Since C;ITuil

DU

and

i s bounded above by

satisfying

If1

5

u

i s an upper bound of t h e set

ITI(u)

we have

(Tf: If

I~u)

.

,

Ch. 12,1841

123


It is evident from these formulas that EU is directed downwards, because the set DU in part (i) is directed upwards.

84. Order continuous operators

In the present section it will be assumed that L is an arbitrary is a Dedekind complete Riesz space. The important particular case that M is the space of all real numbers will be discussed Riesz space and M

in more detail in section 85. In accordance with the notations introduced in the preceding section, we shall denote the set of all order bounded operators by

. In Theorem 83.4

Lb(L,M)

it was proved that

Lb(L,M)

is a

Dedekind complete Riesz space with the positive cone consisting of all positive operators from L

into M

.

The operator T E L,,(L,M) is called sequentially order continuous fo-order continuous) if it follows from u C 0 in L that inflTu I = 0 n n in M Similarly, T E &(L,M) is called order continuous if inf ITu I = 0 Obviously, if T holds for any downwards directed system u C 0 in L

.

.

is positive, then T is o-order continuous (or order continuous) whenever u

C 0 implies Tun

C

0 (or u

C

0 implies Tu

C

0 1. We shall prove

that T is o-order continuous if and only if order convergence of fn to f in L implies order convergence of Tfn to Tf in M ; this justifies the terminology. The terminology is not uniform. Soviet mathematicians call a o-order continuous operator an ( o ) - l i n e m operator and order continuous operators are called completeZy linear. In the Nakano terminology o-order continuous and order continuous linear functionals are called

continuous and universally continuous linear functionals respectively. Luxemburg and Zaanen have called these functionals integrals and normal integrals respectively. The sets of all order continuous and all o-order continuous operators from L into M will be denoted by Ln(L,M) and Lc(L,M)

respectively. As long as L and M

.

h.1, and Lc Evidently, we have Ln that Ln and Lc are bands in Lb '

c

Lc

remain fixed, we write c

. It will be

proved now

124

ORDER CONTINUOUS OPERATORS

Ch. 12,1841

Lc if and only if T+ E Lc and T- E L , +c i . e . , if and only if IT1 E Lc . S i m i l a r l y , T E Ln if and only if T E Ln aizd T- E Ln , i.e., if and unZy if IT] E fn . LEMMA 84.1. We have T E

Ln ; the proof for Lc is quite Ln implies T+ E Ln . Let T E Ln and

PROOF. We present the proof for similar. We show first that T E u

C 0

. We have

in L

to show that

inf T+u + T

purpose that there exists an element u E L T

. Take

v E L+

0

such that

5

. Then

v Iu

0

=

. We may

such that 0

0 5 v-inf(v,uT) = inf(v,u) - inf(v,u ) 5 u-u

so

in view of formula ( I )

assume for this 5

u

for all

5 u

T '

in Theorem 83.6 we get

v-inf(v,u )

-I>

.

+

I T (u-u,)

This implies

o Since T E

so

i T + U ~I

li

T inf(v,u

Ln and inf(v,u

T

) J. 0

>I

.

+ T+~-T~

, we

have

0 I inf T+u .

does not possesss any atoms

p

(we recall that an atom is by definition a set E

of positive measure

possessing as measurable subsets only sets of measure zero or sets

El that the set theoretic difference E-El is of measure zero). Lebesgue

such

measure in the real line is an example. As on earlier occasions, let M(r)

=

M(')(X,p)

be the Riesz space of all real pmeasurable tind

everywhere finitevalued functions on X (with identification of

u-almost

u-almost

everywhere equal functions, and with p-almost everywhere pointwise ordering). We shall prove that every positive linear functional on M(r) is identically zero, i.e., {M(r)}" We note first that if and

u E {M(r))+

.

{e)

=

is a positive linear functional on M(r)

$

satisfies $(uxE)

=

a > 0 for some measurable set C c X

...

and if furthermore the measurable subsets E (n=l,Z, ) of E satisfy n Indeed, if we should have En 4 E , then $(ux ) > 0 for some n En $(uxE ) = 0 for all n , then F = E-E satisfies ~ ( U X ) = a for all n n Fn n , and hence the function

.

satisfies $(u,)

2

ka for k

Note that uo E {M(r)}+

= 1,2,

...

, which

leads to a contradiction.

follows from the fact that for each x E X

only

finitely many terms of CT ux differ from zero. A s a first consequence Fn of what has been proved we find that if ,$ is a positive linear functional on M(r) such that +(u) > 0 for some u E {M(r)}+ , then $(uxE) > 0 for some set E

of finite measure. Now, since 1-1

is possible to decompose E

does not have atoms, it

into disjoint sets El and

F1

of equal

,

132

THE ORDER DUAL OF A RIESZ SPACE

measure

11.1(E) ; for one of these, say

Similarly, E l has a subset E2 E

of measure

$(qE ) > 0 1 such that $(uxE ) > 0

iu(E)

(En:n=1,2,

...)

, descending to a set of measure zero, and such that

for all n

and

.

, we must have

El

Proceeding by induction, we obtain a sequence of

Ch. 12,5851

. Then

for k

$(ul) 2 k

a

2

.

of subsets =

$(ux

En

) > 0

. This is impossible, and hence there does

= 1,2,..

not exist any positive linear functional on

N(r)

different from the null

functional. EXAMPLE 85.2. (The space L ; 0 0 and

p

does not possess any atoms. Furthermore, let p

fixed real number satisfying 0 u-measurable functions f

on

p < 1

0

(cf. also Exercise 7 1 . 9 ) . be a positive linear functional on

satisfies f'(x)

.

> 0

L It will be proved P that @ is the null functional, and for this purpose we first show that there exists a finite constant C > 0 such that @(u) 2 C holds for all Now, let $

Ch. 12,1851

u

I

E L+P satisfying X updp

sequence

I,

u : d p

and

series

zm1 n-2/P

and

uo = 1

so

I

= 1

...)

(un(x):n=1,2,

= 1

133


.

If no such C exists, there exists a

of non-negative functions such that

.

$(un) > n3Ip for all n u

n

n-'Ip

u

of the

. But

satisfies uo E L P

$ ( u o ) 2 +(n-"P

holds for all n

The partial sums sk

satisfy now

un) > n 1 IP

. This leads to a contradiction. Hence there exists a

constant C > 0 such that

X

.

It follows that if 0 5 u 4 u E L , then $(un) I. $ ( u ) for all u € L+ P P There exists a sequence Assume now that $ ( u ) > 0 for some un€ L+ P of step functions (s (x):n=l,Z, ...) , each s vanishing outside a set n of finite measure, such that 0 5 s 4 u , and so (by the result n established above) there is a step function s ( x ) 2 0 such that $ ( s ) > 0

.

But then $(x,)

> 0

be a subset of

E

for some set E c X

such that p(E ) = :a

satisfying p(E) = a < and

$(x

)

2

m

. Let

i$(xE) ; since

1 El does not have any atoms there exists at least one E l having these properties. Generally, for n = 2,3, , let En be a subset of En-]

...

that p(E ) = 2 - n ~ and m n v = C l vn Since

.

$(x

En

) t 2-" $(xE)

. We set

vn

=

Zn

xEn

the series In 1- v E L I n P

for k

=

I, :v

dp

El

such

and

converges, and similarly as above it follows that

. On the other hand we have

... . This shows that

l,2,

the null functional.

$(xE)

>

0 is impossible, and

so

.

LI

A

V =

.

Q is

134

Ch. 1 2 , 9 8 5 1

THE ORDER DUAL OF A RIESZ SPACE

Note that in these examples the situation is different as soon as 1-1 has at least one atom, because then there exists a nonzero positive linear functional. Given the order bounded linear functional $ relatively uniformly to zero, we have

0

5

u

< EnuO n

for some uo E L+

...)

(~,:n=1,2,

0

$

$(un)

+

5 $+(U n )

-

c

.

$+(U0)

E

in L+

0 as

n

+

-.

L

,

converging Indeed, since

0 , we have

C

E

,

C 0

In the converse direction, we shall prove now that

is a linear functional on

L

...)

($(u ) : n = l , 2 , n (un:n=l , 2 , . .) in L+

such that

bounded for every decreasing sequence relatively uniformly

on the Riesz space

and for an appropriate sequence

of positive numbers satisfying

and similarly for $ if

..)

(un:n=l , 2 , .

and given the decreasing sequence

to zero, then $

.

is

converging

is order bounded. Precisely, we

have the following theorem. THEOREM 85.3. The Zinear functional

sup1 $(un)

I

satisfying

0 for all x E X . Assume now that L"

=

L-

E >

$(un) 2

For each

x

,

so

n 2 nx

0

for all

n

,

and let

there exists a natural number

n

converges on

X

w(x)

= Cm v ( x ) 1 n

such that v (x) = 0 for

. This implies that

$(w)

exists as a finite number. On the other hand

SO

$(vn) 2 $(un)

-

$E 2 $E

impossible. Hence $(un)

.I 0

for all n for u

J.

, which

implies $(w)

0 , and s o

L"

= Lz

.

=

m.

This is

Li and f o r any subset E of X , l e t @ ( x E ) Evidently A i s now a non-negative o-additive measure on

Let X(E) =

0

2 @

.

E L"

=

t h e c o l l e c t i o n of a l l s u b s e t s of if

A(En)

k

> 0

en

=

X

. The measure

i s a d i s j o i n t union of s u b s e t s of

U y En

f o r only f i n i t e l y many En

Ch. 12,1871

INTEGRALS AND SINGULAR LINEAR FUNCTIONALS

148

.

n

for a l l

has t h e property t h a t A(En) > 0

holds

Indeed, i f n o t , we may assume t h a t

, and

n

A

, then

X

then the f u n c t i o n

, equal

f

to

a

U y En , s a t i s f i e s $ ( f ) 2 CnE1 k an - 1 .a = k n i s impossible. I t follows i n p a r t i c u l a r t h a t t h e r e a r e only

and zero o u t s i d e

. This

, say x I , ...,x P X-Up{x 1 , any a t most 1 i

f i n i t e l y many p o i n t s i n Hence, w r i t i n g

X1 =

f o r which X({xi}) ' countable subset of

X

A-measure zero. There a r e now two p o s s i b l e cases, e i t h e r A(XI)

.

> 0

If

XI

A(X,)

,

= 0

$(f) = 0

then

and vanishing o u t s i d e

X1

. Since

f o r any

f E L

f 2 0

any

A(XI)

f

XI

vanishing o u t s i d e

f o r any

f E L

. Note

XI

0

=

. i s of or

t h a t i s bounded on $(f) = 0

. Hence

already t h a t i n t h i s case

> 0

can be approximated from

below by a sequence of bounded f u n c t i o n s , i t follows t h a t any

-1

on n for a l l

for

i s not only an i n t e g r a l

@

but even a normal i n t e g r a l . If

A(Xl)

E

1

, we

> 0

t o t h e measure

show f i r s t t h a t

A (we r e c a l l t h a t

A(EI) = 0

satisfies either

c E

not contain an atom, we have

X

or

contains an atom with r e s p e c t

X

E c X,

i s an atom i f

A (E) > 0

A(E-E ) = 0 ) . Indeed, i f

= El U E'

1 1 of p o s i t i v e measure. S i m i l a r l y , E i = E2 U E;

1

with

and

El

with

E2

and

and any does

XI

d i s j o i n t and

Ei E;

disjoint

and of p o s i t i v e measure. Proceeding i n t h i s manner, we o b t a i n a d i s j o i n t sequence

(En:n=l ,2,.

TJ; En c X,

. This,

..)

, each

En

of p o s i t i v e measure, such t h a t

however, is impossible (as observed above). Hence

contains an atom. S i m i l a r l y a s above it follows now t h a t t h e number of d i f f e r e n t atoms i n

XI

i s f i n i t e , say

X(S) = 0

not contain any atom, s o and c a l l i n g

A1 U S

x i.e.,

X

=

A]

, we

.(upi)

,

again

(upi})

. But

A ],...,A then

have now

A

xi

.

Then S = XI-Uy Ai does P U S i s s t i l l an atom, 1

i s t h e d i s j o i n t union of a f i n i t e number of atoms, some of which

c o n s i s t of one point and the o t h e r ones c o n s i s t of an uncountable number of

Ch. 12,9871

149


points. Let A

be one of the uncountable atoms. For the investigation of the

given functional $ real function on

on A we may assume that

X vanishing outside A

Restricting ourselves to points union of the sets

,

(x:f(x)a)

of these sets is of positive measure (since A example, that = A(A)

= 1

(x:f(x)>a)

. This set

$(f)

the set A and

. Let

X(A) = 1

. We write

= CY

is the disjoint

, and

(x:f(x)=a)

.

, where B 1

; n = 2,3,

Bn = (x:a+(n-l)-'tf(x)>a+n-l)

This contradicts $(f) = a

and

... . =

n0 '

is of measure zero.

(x:f(x)>a)

is of measure zero. Then

1

h(x:f(x)=a f

. Hence,

=

= (x:f(x)>a+l)

Only one of the sets B is of positive measure, say for n n so A(B ) = 1 . But then "0

It follows that

only one

is an atom). Assume, for

is of positive measure, i.e., X(x:f(x)>a)

is of the form Uy Bn

Similarly, (x:f(x)O) is n

.

of Lebesgue measure at most ascending as

n

2/m

for every n

increases, the set

(x:fm(x)>O)

and s o , since Fmn

is

is of Lebesgue measure at

2/m . In combination with f (rn) = 1 f o r all n this shows that (at m least for m > 2 ) the function fm is not continuous, i.e., fm is not an

most

element of

L

. Once again on

account of

fm(rn) = 1

evident that the least upper bound of the sequence

for all n

...)

(vm1,vm2,

it i s

exists

in L and this least upper bound is in fact the function identically one.

.

4 1 in L Hence v mn Assume now that :L

# I01

.

. Then there exists a positive integral

4

on L such that $ ( I ) = 1 In view of vmn t n l , there exists a natural number n ~ ( I V ~) < 2-m-1 A s observed above, the set such that m m (X:vmn(x)>0) is of Lebesgue measure at most 2/m for every n , in

.

154

INTEGRALS AND SINGULAR LINEAR FUNCTIONALS

. Finally, let

particular for n = n m wm

Ch. 12,1871

inf,,v1nl,v2n2,...,v f mnm

=

.

Then w C and (x:w (x)>O) is of measure at most 2/m Hence w C 0 m m m (this holds even pointwise almost everywhere), which implies $(wm) C 0

.

But

m so

of

m

...

$(l-wm) < 2-2 + $(1)

= 1

.

+ 2-m-l

4

on account

We thus get a contradiction. It follows that L"= L"

.

The result that every positive linear functional on the space C(C0,Il) is singular may seem surprising at first since it follows among other things that the familiar Riemann integral on

C(C0,ll)

is a singular linear

functional, although the Riemann integral has the property that if (un:n=l,2, ...)

x E [O,ll

,

is a sequence in c(C0,ll)

then

+

un(x)dx

0

+

satSsfying un(x)

0

for every

(note that by Dini's theorem u

converges uniformly). This seems to contradict the result above. Observe, however, that u

n

C 0 in C(C0,ll)

wise. It is evident that but

u

n

J. 0

in C(C0,ll)

the sequence of all u

n

u (x) C 0 pointn u (x) C 0 pointwise implies u 0 in C(cO,ll), n n

is not the same as

+

does not always imply that the pointwise limit of is identically zero (remember that we had

vmn l.n

in the example above, but the pointwise limit of this eequence is not identically one).

This explains why a non-negative linear functional $

+

satisfying $(u ) 0 if n integral in the Riesz space sense.

c(C0,ll)

We return to the situation that L any

$ E L"

the null i d e a l

The null ideal of and instead of an ideal in L

$

N(4)

,

N($)

+

u (x) n

of

$

0

on

pointwise is not yet an

is an arbitrary Riesz space. For is defined by

is sometimes also called the absolute kernel of $ , we sometimes write N

$ '

It is evident that N($)

contained in the n u l l space (or k e r n e l ) of

9

, which

is is by

Ch. 1 2 , § 8 7 1


d e f i n i t i o n t h e l i n e a r subspace of a l l i s a normal i n t e g r a l , then

The d i s j o i n t complement

(N($))d

of

0

F i n a l l y , note t h a t i f

A

then

E L"

$

i s included i n

observation t h a t f o r

and

. This

N($)

f E A

if

Ad = { O ]

A

(equivalent t o {A]

included i n

Add

Archimedean, then

C($) f

i s c a l l e d the carrier of

. The ,

E L

$

Riesz seminorm

p$

i s a Riesz norm on

$ ;

'

.

C($)

i s i d e n t i c a l l y zero on the i d e a l

$

A,

follows immediately from t h e

in

L

i s s a i d t o be quasi o r d e r dense

Add = L ) and

A

generated by

, any

. If

$(f) = 0

we have

We r e c a l l t h a t . t h e i d e a l i f t h e band

N($)

by

$

\ $ ! ( i f \ ) for a l l

p (f) =

satisfying

i s a band. Note t h a t

N($)

we s h a l l denote t h e c a r r i e r of defined by

f E L

155

i s s a i d t o be order dense

A

. Since

{A] = L

satisfies

{A]

order dense i d e a l i s quasi order dense. I f

{A] = Add

f o r every i d e a l

is

is

L

A (by Theorem 2 2 . 3 ) , so

order dense i d e a l s and q u a s i o r d e r dense i d e a l s a r e now t h e same. It w i l l be proved now t h a t every i n t e g r a l i s normal on a q u a s i order dense i d e a l .

THEOREM 87.6. For every

order dense. For every

$

E L,,:

$

nomal integral i s the quasi order dense ideal PROOF. Let

N

E Lc,sn

$

we may assume t h a t of

4

satisfies

p o s i t i v e on

. For

c(4)

=

, and

on

(01

. The

r e s t r i c t i o n of

with

$

E L"

and

$ E L"

. This

$

sn n i s t h e n u l l f u n c t i o n a l . Since $

4

i s i d e n t i c a l l y zero on

C($)

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

is s t r i c t l y

tr) C(4)

$

C($)

C($)

by Theorem

t h e minimal p o s i t i v e l i n e a r extension

of t h i s r e s t r i c t i o n i s a normal i n t e g r a l on

L

i s q u a s i o r d e r dense

N($)

s o t h i s i s a normal i n t e g r a l on

84.4. Hence, by Theorem 8 3 . 7 ( i i ) , $

acts as a

.

N($c,sn)

t h e proof t h a t

i s quasi

$

i s p o s i t i v e . We have t o show t h a t t h e c a r r i e r

$

C(4)

N($)

the n u l l ideal

the largest idea2 on which

f:L

. On

C($)

= $

L

. We

implies t h a t holds on

C($)

t h e o t h e r hand

. Hence

C($) =

thus g e t

JI E L" n

$ {O}

,

n

0

0

for some sequence

, then

f(x) = 0 for all other x E ( 0 , l )

linear functional Q

on L

...)

(xn:n=1,2,

there exists a positive

such that the restriction of

Q

to the

is not an integral. It follows now that La

principal ideal Af

(0,l)

in

of all f vanishing outside an at most countable set

(xn:n=1,2,

.

consists

...)

with

Then La = Lan, lim f(x ) = 0 , the set (xn:n=1,2,...) depending upon f n an 0 So Lan is order dense in L This implies (L ) = Lin by Theorem

.

88.11 (i). Hence La = Lan and a 0 (L ) = = L" # Li sn

(Lan)'

.

=

L"

sn

' but LL # Li and

EXERCISE 88.17. Show that La consists of all f E L condition that if and

.

(un:n=l ,2,. . )

is any sequence in L

T is any positive operator in

Lb '

satisfying the

such that un

then T(inf (un, I f I ) )

J.

0

.

J.

0

LARGEST IDEAL OF ORDER CONTINUITY

168

EXERCISE 88.18.

Show t h a t i f

t h e band generated by in

and

L

,

B

EXERCISE 88.19. Let

L

T

E Lb

A c La

0

. Show t h a t

B

A

be an i d e a l i n

,

La

TIA

and

then

A

.

, then

, the

L

. If

Ad = (La)d

La

C

by ( i i ) , s o

(La)'

It follows from

A

t

there exists

w > 0

not i n

, there 0

5

La

that

in

Ad

T

Ad

3

Lb

know t h a t

E Lb

with

on

A

. But

then again

equality

Ad = (La)d

i n Example 85.1. Then ideal i n

.

L

TI =

(B n :n=l,2,

...

. For

Bo

with

f T I nE Bn

that

fnn

f o r every

l Ln

-+

T

f

Lb

does not

Ln

.

follows from ( i ) t h a t i f 0 A c Ls Hence

.

i s always t r u e .

s e p a r a t e s t h e p o i n t s of

Ad

i s properly l a r g e r ,

i s not i n such t h a t

(La)d

u

=

. Since

. If TIAd

L

necessary, r e p l a c e

. Then

T

w

inf(w,v) > 0


Tu > 0

.

L

T

vanishes on

is

.

, so by t h e Add

,

E Ls , which implies Tu = 0 (since u E La Lb does not s e p a r a t e t h e p o i n t s of L , t h e

any

and

, so

E Lb f

L have t h e p r i n c i p a l p r o j e c t i o n property. I f

,

we have a unique decomposition f = f m + %'n d f i n E Bo ll Bn , and i t follows from Theorem 3 0 . 5 ( i i )

f E Bo

I f [ 2 Ifinl c 0

. This

. We

=

.

. Hence,

if

f E La

, then

holds i n p a r t i c u l a r f o r any sequence

Tfln TI

+

0

of

B 4 Bf , where Bf i s t h e p r i n c i p a l band n ask whether t h e converse holds. Let La be t h e s e t of

p r o j e c t i o n bands s a t i s f y i n g generated by

.

LC

) i s a sequence of p r o j e c t i o n bands s a t i s f y i n g Bn 4 Bo i s a l s o a p r o j e c t i o n band, we s h a l l say t h a t t h e sequence T

Bo

exhausts

0

does not n e c e s s a r i l y hold. Take L = M(r)(X,u) as a 0 L" = {O) , so L = L = L Take f o r A any

EXERCISE 88.20. Let

where

T

1. Contradiction. I f

. If

w

0 5 v E La

minimal p o s i t i v e l i n e a r extension of SO

(La)d

such t h a t

exists

. We

u E La ll Ad

there e x i s t s

= OLS

. It

c Ao

Assume now t h a t we know i n a d d i t i o n t h a t

Hence

A

l a s t e q u a l i t y does not n e c e s s a r i l y hold.

and T vanishes on A , then T E Ls , and s o 0 c A c Ls . I n t h e converse d i r e c t i o n Ls c (La)'

(La)d

=

and ( i i ) f o r any

Formulate and prove t h e corresponding statement f o r

(La)'

ll Ln

{A)

is

f o r example, s a t i s f i e s these c o n d i t i o n s ) . Show t h a t 0 a 0 A = (L ) = L Show t h a t i f i t i s given i n a d d i t i o n t h a t


HINT: A

0

{B}

i s an i d e a l

A

h a s an extension which i s a member of

,

s e p a r a t e t h e p o i n t s of

T E Lb

and

if

such t h a t ( i ) d i f f e r e n t

L

have d i f f e r e n t r e s t r i c t i o n s t o

the r e s t r i c t i o n

(the ideal

Lb

A

Lb

i s an i d e a l i n

B

OIB} =

i s t h e band generated by

{A)

members of

then

12,1881

Ch.

Ch. 12,§8Q]

of a l l Bf

f E L

such t h a t

Tfin

HINT: W e have t o show t h a t

Given

8 2

, we

T E Lb

. Take

any

-pn

,

3

La

. Let

. Show t h a t 0

Tu C 0 " + pn = (6f-un)

together i s

pn

, we

B

have

E v i d e n t l y , t h e component of u -6f is n (6f)nn-Pn

La

La c La

and w r i t e

band g e n e r a t e d by a l l t h e p

holds f o r every sequence

have t o show t h a t

6 > 0

band g e n e r a t e d by

0

-t

. Evidently

T E Lb

and f o r every

Tf > 0

169


6f-u

n

, so

4 Bf

in

Bf

is

Bn

_
O , s o

so

B

inf

infyu = O . n

EXERCISE 88.21.

Lan

in

d n n

n

Bf

Tu n

satisfies

.

0 5 (un)in 5 f ' Hence nn < T(6f) = 6Tf This h o l d s f o r a l l

.

Formulate and prove t h e corresponding s t a t e m e n t f o r

.

89. A n n i h i l a t o r s and weak t o p o l o g i e s For t h e f o l l o w i n g we need some elementary f a c t s about a n n i h i l a t o r s and weak t o p o l o g i e s . L e t dual

W# ( i . e , , W#

Furthermore, l e t fixed

.

W

be a r e a l o r complex v e c t o r space w i t h a l g e b r a i c

i s t h e v e c t o r space of a l l l i n e a r f u n c t i o n a l s on

W'

For any non-empty

b e a l i n e a r subspace of subset

A

of

W#

. We

keep

t h e annihilator

W

A'

W

of

W ).

and

W'

A

is

OB

of

d e f i n e d by Ao = (+EW':b(f)=O

and f o r any non-empty

i s d e f i n e d by

subset

f o r a l l fE*) B

of W'

,

t h e inverse annihilator

B

ANNIHILATORS AND WEAK TOPOLOGIES

170

Ch. 12,9893

Compare this with the formulas in ( 1 ) of section 88, where we had W and W'

W'

N

L

=

and W

= {O}

. It is evident that

Ao

B1 c B2 c W'

and

implies OB

throughout this section that '(W')

=

OB

and

respectively. Note that A

1

1

C

=I

A2

OB

{O)

2

,

(r

are linear subspaces of 0 0 W implies A 1 3 A2 2 Wo =

3

'(W')

w .

A

It is of some importance to observe that A 0 0 W For the proof, note first that A C (A )

0

.

In the converse direction, substituting B Ao c {o(Ao)}o

. We

=

{O(A0)Io

=

implies Ao in B

Ao

shall assume

separates the points

i.e., W'

of C

L

=

. Hence there is equality. Similarly, OB

for any 0 0 0 2 { (A )I

.

(OB)' we obtain 0 0 0

C

=

{ ( B)

1 for any

BcW'.

We now recall that if X and, for each u E {a} mapping from X system

, Ya

into Y a

,

is a non-empty point set, {a) an index set is a non-empty topological space and

then the weak topology in X

is the weakest topology i n

(Fa:aE{a})

X which makes all Fa

continuous. The system of all inverse images Fi'(Oa) through the open sets in Y the set each a Yu

in

,

Ou

(i.e., a

we restrict Oa

runs

to run only through a subbase for the topology in

.

Let W

Oa

is variable, and for each fixed a

and W'

is still a subbase for the weak topology

be the same as introduced above. The weak topology in

generated by the members of W' i s denoted by

W

, where

is again variable), is a subbase for the weak topology. If for

the system of all F - l ( O ) a a

x

Fa a generated by the

o(W,W')

and the weak to-

pology in w' generated by the members of W is denoted by o(W',W)

o(W,W')

. Hence,

is the weakest topolony in W such that every $ EW' is a continuous

(real or comp1ex)functionon W . Similarly, o(W',W)

is the weakest topologyin

W' such that every f E W i s a continuous (real or complex) function on W'

.

Werecall thatthevalue of the function f at the point $ E W' is $(f) The inverse images $

-1

(0)

, with

$

running through W'

and

.

0

through the open subsets of the space of real or complex numbers, form a subbase for o(W,W')

. For the complex case, the open circles

( z : I z-z

0

ICE)

form a subbase for the topology in the complex plane, so the inverse images $

-1

(z:lz-z

0

ICE)

form a subbase for

(f :FEW, I $ (f)-zol

<E)

,

o(W,W')

,

i.e., the sets


Ch. 12,5891

depending upon the parameters

$,zo and

E

,

171

form a subbase for u(W,W')

.

The finite intersections

form a base for S

. Let S be one of these base sets, and fo E S . Then I$,(fo)-zkl < holds for

o(W,W')

is not empty. Let

k = l , . ..,n

the circle

E~

. Writing ck

-

I$

k

assume that

(f )-zkl = 6k O

for k = I ,

...,n

,

,

(z: Iz-zk I < E k )

(z: (z-$k(fo) 1 ~ 6 ~ is ) included in

so

It follows that the open neighborhood

with

6

=

min(61,...,6

the union of sets

) , is contained in the given set n of this kind, i.e., S = U(Uf:fES)

Uf the system of all sets

is a base for

parameters by

. Note

o(W,W')

. Then

. This

S

shows that

>

0

. We

shall denote the set

. Finally, we observe that the same result holds in

U(fo;$l,...,$n;6)

the real case.

(i) Addition and mZtipZication by (real o r compzex) o(W,W') In other oords, the space W equipped with the topoZogy O(W,~') i s a topoZogica2 vector space. THEOREM 89.1.

constants are continuous i n (ii)

The

o(W,W')

a Zinear subspace of

is

that a base set of this kind depends on the

E W' and 6

fo E W ;

S

W

(iii) The topoZogy

.

cZosure of any Zinear subspace o f

.

o(W,W')

i s Hausdorff.

W

i s again


172

Ch. 12,1891

PROOF. (i) For the addition it is sufficient to observe that if fl+fi = f3

in W

and the neighborhood

U(f3;$1,...,$n;6)

is given, then

is a member of this neighborhood as soon as

f; E U(fl;$l,...,$n;i6) f;+f; and f; E U(f2;+1,...,$n;i6) For the multiplication, if f = a f 1 0 0 and V 1 = U(fl;$l,...,$n;6) , we have to prove the existence of a neighborhood

.

vo

of

fo and a neighbordood

af E V 1

f E Vo

for all

f E Vo

For

af-fl that

=

af-a f

l$k(af)-$k(fl)l (ii) Denote the

0 0

a l f + a2f2 E AA

point of

. By

,

=

(a-a )f 0

0

0

. We may

assume that

0 < 6 < 1

and from

+ a (f-f ) + (a-ao)(f-fo) 0 0

holds for k

< 6

o(W,W')

We have to show that if

in the complex plane such that

a

it follows from 0 < 6' < 6

a E Co

and

of

a E Co

. Let

V1

in the defintion of

Co

and all

...,n

I,

=

and

so

af E V 1

fl,f2 E A-

al,a2

and

i.e., every neighborhood

U

.

by

closure of the linear subspace A

.

A-

are constants, then of

a f + a2f2 I 1

part (i) there exist neighborhoods

contains a

U 1 and U2

of

f

and

a f' + a f' E U for all f; E U 1 and all 1 1 2 2 f2 are members of A , the neighborhoods u1 each contain at least one point of A Hence, choosing f; in

f2 respectively such that f; E U2 and

U2

U1

A

n

. Since

and

f l and

in U2 n A

f;

, we

.

have

a f' + a f' E U l A 1 1

desired result. (iii) W' exists

$

separates the points of

E W'

such that

neighborhoods U(f1;$;6)

$(fl) # and

W

$(f2)

U(f ; $ ; 6 ) 2

2 2

. This

is the

. Hence, if f l # f2 in W ,there . Let 6 = ~l$(fl)-$(f2)l . The are now disjoint,

so

o(W,W')

is Hausdorf f. THEOREM 98.2. (i) For any non-empty subset

if

A = '(Ao)

.

(iii) For any linear subspace

is

'(Ao)

.

of W'

B

the inverse

OB

i s o(W,W') closed. (ii) The linear subspace A of

annihilator

A

W

is o(W,W')

of W

the

closed if and only

o(W,W')

closure of

A

Ch. 12,8891

173


PROOF. (i) If OB =

0

B

$o , then

consists of one point

,

(f:$ 0 (f)=O)

{$ } = 0

and all we have to prove is that the complement be a point in the complement,

open. Let

(f:$o(f)#O)

fo U(fo;$o,~~$o(fo)~) is an open neighborhood of fo

is an interior point of the complement of

of

OB

is

If

so

. Hence,

OB

so

the complement

of

satisfies A

W

=

0

is closed by the result in (i).

Conversely, assuming that A

closed, we have to show that A

=

,

in '(Ao) fo

,

OB

, then

W'

is an intersection of closed sets. This shows that

.

set

fo disjoint from

is an arbitrary non-empty subset of

(ii) If the linear subspace A

than A

. The

~(w,w*) open.

B

OB

u(W,W')

is

$O(fo) # 0

so

0

0

(A )

. Since

A

either, i.e., $ (f ) # 0

Hence, let

0

fo be given such that

fo

. From

does not exist any

,

0

then A

is o(W,W')

fo

is not in A $,

for at least one

is not in A

there is a neighborhood V = U(fo;$l,...,$n;G) disjoint from A

is closed.

is not properly larger

In other words, we have to show that if

is not in '(Ao)

0

(A )

is trivially included

'(Ao)

it is sufficient to prove that

OB

of

. Since

A

,

then

E Ao

.

is closed,

fo such that V

is

the disjointness it follows in particular that there

g E A

satisfying $,(g)

=

$,(fo)

for k

= 1,

...,n

simultaneously. We introduce now a linear mapping complex number space

where $,, ...,$,

En

, defined

En

, and

no point

@(A)

g E A

from W

are the same elements of

definition of the neighborhood of

@

V

into n-dimensional

by

. The

W'

as occurring in the

image @ ( A )

is a proper subspace of

is mapped onto the point

En

is a linear subspace

because (as observed above)

($l(fo),...,$n(fo))

of

En

.

A s follows from simple linear algebra facts, there exists now a linear

subspace L $(A)

of

En

of dimension n-1 (a hyperplane) such that L

but does not contain the point

(fo),...,$n(fo))

contains

. Denoting the


174

coordinates in En L C

be

C i = l ckzk = 0

ck $ k ( g ) = 0

$,(g)

by

Ch. 12,1891

...,zn , let the equation of the hyperplane

zI,

. Since

Q(A)

, we

is included in L

have

for every g E A , s o the point $, = Z ckOk E W' satisfies for every g E A , i.e., 4' E A' Furthermore, since

= 0

.

is not contained in L , we have C c $ (f ) # 0 , s o (@l(fO),...,$n(fO)) k k 0 @'(f0) # 0 Hence we have now that @'(f0) # 0 for at least one $, E Ao

.

.

This is the desired result. The reader who is familiar with the theory of locally convex vector spaces will easily see another proof. Since A

, there

in the locally convex topology u(W,W')

. Observing now

# 0

is a member of

W'

(iii) Let A '(Ao) A-

by

S

,

exists a

that every

for brevity. Then A

W with closure A

and denote

is included in the closed set S

is the smallest closed set including A

and

continuous linear functional

o(W,W') the proof is complete.

be a linear subspace of

o(W,W')

for all g E A

= 0

such that @,(g)

continuous linear functional $' $,(f0)

is a closed linear subspace

,

A c A- c S

so

. But

. It follows

that

-

The sets A

and

S

are closed linear subspaces (note that we use here that

the closure of a linear subspace is a linear subspace), so by part (ii) the

-

second and third terms in ( I ) are A be rewritten as

S c A- c S

and

respectively. Hence ( I ) can

S

. This shows that

A- = S

There exist similar theorems for o(W',W)

,

-

i.e., A

.

= 0 (A 0 )

. One might

as for o(W,W')

repeat the proofs. It is more illuminating, however, to observe that the theorems for a(W',W)

are in a certain sense the same as for o(W,W')

that no further proofs are necessary. Explicitly, observe that every

,

so

f E W

can be regarded as a linear functional on W' ; the value of the functional f at the point

$

E W'

is the number @(f)

. Different

functionals, since (due to our hypothesis that W'

of W ) for f, # f2 there exists at least one $ E W' $(fl) # $(f2)

. It follows that

W

if we look at the points

functionals on W' ). Then the topology o(W',W) and, for any non-empty subset B

such that

can be regarded as a linear subspace W"

of the space of all linear functionals on W' (i.e., W" but we use the notation W"

f give different


of

W'

, the

is the same as W f of

W

is the same as

,

as linear u(W',W")

inverse annihilator OB

in W

Ch. 12,9891

175


. The followinn theorem follows now imme-

is the same as the annihilator Bo in W" diately.

THEOREM 89.3 (i) The system of a l l f i n i t e intersections

$o E W' ; f l , ...,f n E

depending upon the parameters base f o r t h e

.

topology i n W'

o(W',W>

w

and

6 > 0

, is a

(ii) Addition and multiplication by constants are continuous

operations i n the Hausdorff topology the annihilator

of

W

W'

is o(W',W)

w'

B of

o(W',W)

the

closure of

o(w',w)

any non-empty subset

closed. The linear subspace

closed if and only if B

=

B is

(OB)' (OB)'

. For .

P

,

is the Banach dual

then o(W,W')

usually called the weak topology in W = W

weak s t a r topology in W' closed sets in W

=

P

=

and

o(W

w*)

P' P

P

o(W',W)

and

is what is = a(w*,W ) P P

is the norm

that fail to be closed in the weak topology (unless

P

is of finite dimension), it is true that every norm closed linear sub-

W

P

space of of

W

W

is closed in the weak topology, i.e., any linear subspace

P

is weakly closed if and only if the subspace is norm closed. For

P

the proof, it is sufficient to show that the linear subspace A closed if A Since A

fo

is norm closed. Let

element $o E WE joint from A

, so

fo

is an interior point of the complement of

. Also,

0

(A )

closure of Let W

A

if A

of W

# 0

. It

fo

dis-

A (in the

P

is norm closed if and only if

is an arbitrary linear subspace of W the norm P '

is equal to )'A(' and W'

.

be as before, and let Z' be a linear subspace of W'.

It is not difficult to see that the topology o(Z',W)

in 2'

. It follows that the

is the

relative topology in Z' of

o(W',W)

closure of any subset D of

Z' is the intersection of Z' and the B is a linear subspace of

u(W',W)

closure of

.

is weakly closed. It is an immediate consequence

that the linear subspace A =

be a point i n the complement of A

= 0 for all f E A and $,(f0) is a weakly open neighborhood of

such that $o(f)

weak topology). Hence A 0

is weakly

is norm closed, there exists (by the Hahn-Banach theorem) an

follows that U(fo;$o;~l$o(fo)l)

A

B of

p

. Although in general there are many

W*

A

any linear subspace

is a normed vector space with respect to the norm

If W = W if W'

is

Ao

. For

o(W',W)

D

. In particular, if

o(Z',W) Z'

,

176

CARRIER OF A LINEAR FUNCTIONAL

then the o(Z',W)

B

closure of

0

is equal to

(

B) 0 fl Z'

. Hence,

Z' of any subset A

denoting the annihilator with respect to 0

AO , we have A" = A f l 2 ' , and for the o(Z',W) subspace B of Z' we can write ( 0 B) 0 = (0B) 0 course the o(Z',W)

Ch. 12,5901

of W

by

closure of the linear Z' Note that of

.

n

topology is Hausdorff, but it may happen that although

distinct members of W

act differently in W'

the smaller space Z' (i.e., '(W')

=

I01

they act

, but

identically on

it may be that

# I03

'(Z')

).

We conclude with the following theorem which will be of use later (cf. Theorem 106.2). THEOREM 8 9 . 4 (i) Let

A 1 and

A2

be linear subspaces o f

only the n u l l element i n comon and such that o f W' , i . e . , (AIbA2)0 = I03 Then Ao and

.

1

element i n comon. If, i n addition, A 1 and

0 A1

0

0

separates the points of W

A2

(ii) Similarly, l e t

B 1 and

w , i.e.,

' ( B ~ ~ B ~ )= C O I

. Then

then

OB1 e

0

,

f E $ = 0

so

o

$(f) =

. This shows

n

0

A2

for all

that :A

A]

so

=

A 01

0

(A1) and A2

@

A20

=

):A('

, then f E A] 0

and A2

Assume now that, in addition, A1 0

and

B2 separates the points o f

PROOF. (i) If $ E A l

a(W,W') = {O}

.

W

closed, then W', having


B I 0 B2 OB2 have only the null

OBI and

element i n common. If, i n addition, B, 0

are

be linear subspaces o f

B2

, having

A2

0 0 , i . e . , o2(AI@A2)

only the null element i n common and such that of

@

A2d A

W

separates the points have only the n u l l

A

B2

are

, i.e.,

o(W',W) 0

closed,

0

( B l b B2)0 =

$(f) = 0 for all

I01

f E A1

.

and all

8 , i.e., $ E (A 1@A 2lo = I O I , so have only the null element in comon.

and A2

. Then

are

o(W,W')

closed, so

separates the points of W

(ii) Similarly.

90. The carrier of an order bounded linear functional

Before investigating further the properties of the null ideal and bhe carrier of an order bounded linear functional, we prove a certain decomposition lemma with some corollaries.

Ch. 12,5901

177


Let

LEMMA 90.1.

and

P

be Riesz seminorms on the Riesz space L

A

L is a A-Cauchy

such that every monotone order bounded sequence in sequence. Furthermore, l e t

c1 W

p(un)

izn}

0

E

u E L

, there

(b)

A(wn)

(c)

u

n

5 z

n

PROOF. Let

n

4

u

5

,

n

5

and

,

=

n

,

... um

+

ZY;

(. =

z

n

C

n

C 0 as

that p(zn)

-.

and it follows from +

-u.

for all n

E

. Furthermore, since

(x -u. ) + x. 5 x. for n i,m 1 1 for all n Ogserving that u 5 x xn-yn 5 zn n n ' yn+zn for all n By the Riesz decomposition property there have

xn-yn

_
0

E

$(u) = 0

so

,

i.e., u E N

J , .

i s Archimedean and i f

THEOREM 90.6. If L

following conditions are equivalent: (i) $ I J, , (ii) C c N $

J,

PROOF. (i)

(iii)

'

(ii). If $ I J,

=$

.

with

$

C

E L z and

$

IC

4

(iv)

J I '

*

(ii)

(iii). C c N J

I C

J , .

(iii) c (iv).

L

and

,

$

c Ic $

J

, which

implies CJ,c (N$)dd

$

c4

is equivalent to

,

is Archimedean, we have

(Nj)dd

N

=

E Ln , the N

C c $

E Lz , then C c N

last theorem.

cQ

J,

J,.

4

9 '

by the

is equivalent to

.

c ( N ~ ) Since ~ ~

C c N

Hence

N

$

E :L

J,

J,'

(iv) * (i). The linear functional oo = 101 A I J , ! satisfies +o E L; (since J, E Lx ), so N($O) is a band. Furthermore $o vanishes on N 4 Since C C N this implies that $o vanishes on C @ N and on N so

C

J,'

$

0

N

$

$

is the whole space L But then

$,

J,'

is contained in N ( $ O ) =

0

,

, it

. Since the band

follows now that the band

i.e., 191

lJ,l = 0

A

THEOREM 90.7. (i) If 0 < $ E L"

(ii) If 0

5 $

E L"

c($AJ,) =

and

c nc $

0 5 $

,

=

Bd

and

(AnB)dd = Add n Bdd so

so

0

I J,

$

S J,

N($O)

.

$

C

$

$ '

@

N

$

is equal to L

E L" , then

E L" , then

J,'

(i) Note first that N($VJ,) Theorem 90.4(iii),

and

and

PROOF. We recall that for ideals A d A

generated by

.

= N(@+J,)

C(@vJ,) = C($+J,)

and =

N

$

B

nN J

in L

,

we have

(A+B)d =

by the proof of

. Furthermore,

.

Ch. 12,5901


181

so

x

(ii) Write $ =

$]+x

and

$ =

= $

J,

A

JII+x

for brevity. Let

$

=

1

$-x

q1

and

=

J,-x

. Then

with all functionals involved positive. It follows

that N Furthermore,

+'

=

n NX and N J,

N($])

1 $,

with

0

5 $

1

=

N(JIl) ll NX

E L"

and

.

0i

E Lc

,

so

c($,)

c N($~)

by Theorem 9 0 . 5 . Since

we find

so

EXAMPLE 90.8. We have proved that if

0 < $ E L"

then C($A$) = C($) n C($) and C c N if @ J , necessarily hold if it is only given that $ of

L"

. By

way of example, let

L

$

I JI

and

be the space

JI

and

0 < $

E Li ,

. This does not

are positive elements

C(CO.11)

of all real

182

Ch. 1 2 , § 9 0 1

CARRIER OF A LINEAR FUNCTIONAL

continuous functions on the interval C 0 , l l , let f f L

and let $(f)

I01 f dx

=

for all

, where (rn:n=1,2, ...) is the set

2-"f(rn)

=

$(f)

. Then $ and J, are strictly positive, = C = L . It is easily seen that if J,,(f) = . This holds for all n = 1,2 ,... , s o $ I J,

of all rational numbers in C 0 , l l so

N

$

=

J

N

I

= 2-"f(tn)

and

= {O}

, then

$

C

I J,n

k

are disjoint, neither

J

JIn

J, = supk .InE1

(because

$

,

) . The example shows that although

C$ c N J ,

or

C cN

J

I

nor

$

C($AJI)

C$ fl C

=

It will be indicated in Exercise 90.14 and 90.15 that if

L

JI

and

$

holds.

J,

has the

Egoroff property, then the situation is improving, because in this case C($A$)

C

=

C

$

fl C J

N

generated by If A

$ A JI = 0

holds for all positive $,I) and

,

, but

is perhaps not included in N

JI

J,'

,

is an ideal in an Archimedean Riesz space L

an order dense ideal. In particular therefore, if

+

(Ng)dd

is the band generated by the ideal NQ

some condit

L

.

THEOREM 90.9.

(i) I f

L

projection property, then L (ii) I f

property and

+ ldd

(N

8,

c

+

*

i s an integraZ on

$

L = N eC

N

0

i s a band in

L L

, then

. Hence,

projection property, then L

is

L has the

L has the projection $

i s a normal integral on

L

i n t h i s case we have

=

(N )dd $

C

4

8

= (N+)d

(N+)d

=

For the present proof we may also assume that $ is Archimedean, so have

=

all w n

N

$

@

sup(t$(w):O 0 Then there

,...,an

e x i s t compZex numbers

a1

that

.

,...,v

and elements

...,n

(i) v +...+ v = u and l a I i 1 f o r k = 1, I n k (ii) Z: l ak I vk 5 u , (iii) If-Cn a v I 5 EU and Ilfl-Cy laklvkl 5 E 1 k k

L+ such

in

v1

,

.

U

PROOF. Follows from the preceding lemma by observing that the principal ideal generated by

, where

is Riesz isomorphic with a space C(X)

u

is Hausdorff and compact, such that the image of

X

is the function

u

identically one. (Tcr:a{ a } ) i s a system of reaZ order bounded operators

LEMMA 9 2 . 4 . I f

from the Riesz space L that

i n t o the Dedekind complete Riesz space

exists i n

To = sup T

Lb(L,M)

and i f

u E L+

M

such

, then

n n

where

(al,...,an) runs through aZZ non-empty f i n i t e subsets of

...,

{ a } and,

runs through n-tupZes of f o r each subset of t h i s kind, ( u ~ , u ) n p o s i t i v e elements s a t i s f y i n g u l + ...+ u = u n PROOF. If

{a)

,

..

consists of two elements a ,

and

a2

, so

that

To = sup(Ta ,T

)

case that

has a finite number of elements follows by induction. The

!a}

a2

the formula for TOu was proved in Theorem 83.6. The

general case is now derived by observing that if T

0

system of all TB is directed upwards, then T u

M.

0

=

=

sup TB

sup T u B

, where

the

in the spare

Ch. 12,9921


In what follows we assume again that L

LEMMA 92.5. (i) If T E Lb(L,M)

i s r e a l and

IT1 (lhl) (ii) If T E Lb(L+iL,M+iM)

f E L , then

.

5

and

(ii) Let

T

=

, so

T +iT2 with

TI and

I

T2

has the decomposition Tf = T 1f+iT2f , lTfl

=

h E L+iL, then ITfI

. Since

f,g E L

PROOF. (i) Let h = f+ig with the decomposition Th = Tf+iTg

Tf

is an Archimedean and uni-

is a Dedekind complete Riesz space.

formly complete Riesz space and M

lThl

205

T

THEOREM 92.6. Let

L and

M

be an order bounded operator from

PROOF. We have

IT1

is real,

ITI(lfl)

cose+T2 sine

.

be a s defined above and l e t L+iL i n t o M+iM

. Let

, where

T(t3)

sup(T(8):0 0 , and hence To t sup(T,B) = T+ Since 0 is a kernel operator, it follows from the last theorem

To 2 T+ 2 t3 and To that 'T is also a kernel operator. Let TI(x,y) be the kernel of T+ account of To 2 T+ the corresponding kernels satisfy T+(x,y) 2 T,(x.y)

. On

BAND OF ABSOLUTE KERNEL OPERATORS

222

Ch. 13,9943

almost everywhere. On the other hand it follows from T+ t T that

TI(x,y) t T(x,y)

as well as

TI(x,y)

2

0

T+ t 0

and

hold almost everywhere,

so

' T (x,y> 2 T+(x,y) almost everywhere. The final conclusion is that TI(x,y)= I + This shows that T+ has = T (x,y) almost everywhere, i.e., T+ = To T+(x,y) as kernel.

.

To guarantee a smooth proof of the next theorem we include a lemma. be the set of all u-measurable functions f

Let P(X,u)

on

X

assuming

non-negative values in the extended system of real numbers (i.e., 0 5 f(x) 2

for all x E X ). Functions differing only on a p-null set

m

is partially ordered as usual, i.e., f 5 g

are identified. The set P(X,u) means that

f(x) 5 g(x)

holds p-almost everywhere on X

. Then

evidently a lattice. Note that the positive cone of M(X,p) of P(X,u)

(f : T€{ T} )

LEMMA 94.4. I f

f o = sup f

i s a s e t of non-negative functions i n

e z i s t s i n P(X,p)

and

fo i s already the sup-

r e m of an a t most countable subset of the s e t of a l l PROOF. Let

f

=

T,n

for n = l , 2 ,

inf(fT,n)

Dedekind complete, the supremum u n

, and

= supT

f

fT

... . Since

. M(X,u)

exists in M(X,u)

T,n

is super for any

is already the supremum of an at most countable subset of

u

(fT,n:~E{~})

. It follows easily that

THEOREM 94.5. The s e t

into M

is

is a sublattice

-

M(X,u) , then

P(X,u)

i s a band i n

fo = sup u

is the required supremum.

Lk (L,M) of absolute kernel operators from

Lb(L,M)

.

L

.

PROOF. (i) We prove first that L (L,M) is an ideal in Lb(L,M) If k T is an absolute kernel operator with kernel T(x,y) , then IT1 = sup(T,-T) is likewise a kernel operator and ly, if

T € Lb(L,M)

0 C T+ S IT1

T

= T+-T-

and

IT1

0 C T- S IT1

and

IT1

has the kernel

IT(x,y)I

. Converse-

is a kernel operator, then it follows from that T+ and

T-

are kernel operators, so

is a kernel operator. Note that we have used the fundamental ma-

jorization result in Theorem 94.2 several times. For the proof that the absolute kernel operators form an ideal, assume now that

IS1 C IT1

to show that same

S

is true f o r

in Lb(~,~), where

T is a kernel operator. We have

is a kernel operator. Since T

IT1

. But then

IS1

is a kernel operator the

is a kernel operator by Theorem 94.2

Ch. 13,9941

so

223

KERNEL OPERATORS

.

the same holds for S (ii) Let

0 5 T

T in Lb(L,M)

4

, where

all TT are kernel operators.

F o r the proof that the kernel operators form a band, we have to show now

that

T

is a kernel operator. The system of functions

directed upwards in the Riesz space M(XxY,pxv) functions on

X x Y

functions on

X

. Let

x Y

P(XxY,pxv)

(TT(x,y):rE{~})

is

of all (pxv)-measurable

be the set of all (pxv)-measurable

assuming non-negative values in the extended real num-

ber system. According to the last lemma the function T(x,y) = sup TT(x,y) exists in P(XxY,pxv) and there is even a subsequence (Tn(x,y):n=1,2, ...) in the system of all TT(x,y)

such that

T(x,y)

of the subsequence. Since the system of all

is already the supremum

T (x,y)

almost everywhere on

.

kernel Tn(x,y)

0

Now, let have

(Tf)(x)

X x Y

(Tnf)(x)

2

. Since

f E L

5

(Tnf)(x)

=

. Let

I

Tn

x


so

Tf = sup T f holds in the space M

for all n

, we

almost everywhere on X. Furthermore

Tn(x,y)f(y)dv(y)

4n

Y for almost every

is directed upwards,

0 5 T,(x,y) 4 T(x,y) be the operator corresponding to the

we may assume that the subsequence is increasing,

I

T(x,y)f(y)dv(y)

Y

. Hence

X

. Denoting the function on

the right by

g(x)

, we

have

almost everywhere on X

, so

almost every x

T(x,y)

(5)

in M

. Since

g t sup T f

. Combining

=

it follows already that g(x) t TT(x,y)

for all

T

Tf

( 4 ) and (5),

we get

g = Tf

,

i.e.,

is finite for

, we have

224

Ch. 1 3 , 5 9 4 1

BAND OF ABSOLUTE KERNEL OPERATORS

To conclude the proof, we show that the set on which measure zero, i.e., T(x,y)

, where

E

is a subset of

[ T(x,y)dv(y)

('4, (XI)=

show

xE

Y such that

T(x,y)

so

. It is sufficient to

first that in this case (6)

is of

m

is almost everywhere finite valued,

is a member of the Riesz space M(XxY,pxv) finiteness on X x E

T(x,y) =

f L

. Note

0

x E Xo

(TxE)(x)

and it follows from

. For

X

of

(uxv)(P)

that there

> 0

such that u(Xo) > 0 and

these x we get therefore

T(x,y)dv(y)

=

has positive

for all x E X ,

m

exists a u-measurable subset Xo v(Px)

x E

2

E

1

T(x,y)dv(y)

.

=

pX

is finite almost everywhere on X x Y ,

This contradicts (6). Hence T(x,y)

is the kernel of a kernel operator according to our definition

i.e., T(x,y) in section 9 3 .

The final conclusion is, therefore, that the absolute kernel operators

.

form a band in Lb(L,m)

We present some examples. The first example illustrates that one must not conclude too soon that one measure is absolutely .continuous with respect to another measure. To begin we recall that the measurable set E (Y,I,w)

o-finite measure space measurable subset F

of

is well-known that i f Y

E

is called an atom if v(E)

satisfies either v(F)

= 0 or

v(E-F)

=

. It

0

Y does not contain any atoms and E is a subset of 0 < a < v(E) , then E has a

of finite positive measure and if also

.

For the special case that v subset E l satisfying v(E ) = a I sional Lebesgue measure, the proof is almost trival. EXAMPLE 94.6. Let (Y,C,u) without atoms and let A

in the

0 and every

>

be a finite measure space (i.e., v(Y)

be the semiring of all sets A x B

B w-measurable. For A x B E

and

A(AxB)

r

=

v(AnB)

if A x B

=

. Then

urn( I %xBk)

is n-dimen-

r

we define the number

, I is a countably additive measure on

with disjoint terms, then A n B

=

U;(\nB,)

in Y

X(AxB)

r

x

0

n

.

for

fn(x)

lfn(x)I

n = 1,2

+

0

i s '-measurable

,... and

+

0

+

f(x)

, we

Xnfn(x)

0 5 f

If

have

E M(X,u) n (unfn:n=1,2,

such t h a t

for

...)

I t f o l l o w s now from t h e s e p r o p e r t i e s t h a t

. As

X


f o E M(X,u)

0 < f (x) < A-'f0(x) n

p r o p e r t y ( i ) , so i t f o l l o w s from 0

.. and

n = 1,2,.

0

then t h e r e e x i s t s a sequence of p o s i t i v e r e a l

such t h a t

m

for

p ( x ) = sup p E M(X,u)

everywhere f i n i t e , i . e . ,

almost everywhere on numbers

f E M(X,u) n x E X , then

that

fn

n = 1,2,

...

a

by converges

,

then there

i s bounded above i n

M(X,p)

has t h e s t r o n g

Egoroff p r o p e r t y (Theorem 71.6). We p r e s e n t a p r o o f , u s i n g only p u r e l y measure t h e o r e t i c terminology. THEOREM 96.1. k

in

M(X,u)

> 0

we have

e x i s t numbers in f

M(X,u)

n,k (n)

5

(fnksk=l , 2 , .

such t h a t

. It

-1 n h

'n

> 0

in M M

n

k(n)

..)

k

f o l l o w s t h a t f o r each

. Hence,

setting

such t h a t f o r every

. Then

M

n,k(n)

such t h a t

there exists

n

, we

be an i d e a l i n M(X,u) fnk

there e x i s t s a sequence

c

a s above, and l e t

h

n

...)

' gn

*

i.e.,

for

( i i i ) there

0

k = k(n)

un h n

2 h

such t h a t

obtain the desired result.

as

0

and l e t k

(hn:n=1,2,

X

n,k(n)

gn

as

a positive function

n

. By p r o p e r t y

h E M(X,u)

such t h a t hn(x) + 0 almost everywhere on there e x i s t s a number k(n) w i t h 0 5 f PROOF. Make

be a

and such t h a t

converges h -uniformly,

g = n-lh n

n we have

X

0 < f

with

t k(n,E)

and a f u n c t i o n

COROLLARY 9 6 . 2 . Let

where on X

for all

fnk 5 Ehn

...) +0

there e x i s t s a sequence (gn:n=1,2,

By p r o p e r t y ( i i ) t h e r e e x i s t s f o r every

hn E M(X,u)

fnk

n we have

almost euerywhere on

t h e r e e x i s t s a nwnber

n

PROOF.

. Then

gn(x) C 0

such t h a t

f o r every

E

such t h a t f o r every


+ m

(fnk:n,k=1,2,

(Strong E g o r o f f p r o p e r t y . ] Let

double sequence in M(X,p)

+

m

...)

0 5 fnk

5

fo

almost everyi n the ideal

and such t h a t f o r every

' hn

*

= inf(gn,fo)

.

BLlHVAL0V"S THEOREM

236 THEOREM 96.3. Let

Let

be an i d e a l in M(X,u) and l e t 0 5 fnk < fo in inf f = 0 almost everwhere on X k nk for every n Then there e x i s t f i n i t e subsets

M

such t h a t f o r evevy

M

Ch. 13,1963

n we have

En = (fnk:k=1,2,.. . )

.

.

(n=l,Z,. . . I such t h a t f o r every natural number m n n inf(Um E ' ) = o in M . m n

E' c E

PROOF. Let 0 < gnk

2

fo

everywhere on (hn:n=1,2,

nk

...)

X

. By

in

= inf(fnl, f ) for n , k = l,2, Then nk and for every n we have gnk(x) C 0 as k +

'

we get for any natural number m

and M

(X,A,p)

that

and

e,Z,v)

be ideals in M(Y,v)

assumed that Y

almost

such that h (x) C 0 almost everywhere on X and n there exists a number k(n) with 0 < g n,k(n) hn*

n

Writing now

A s before, let

-

the last corollary there exists a sequence

M

such that for every

let L

... .

...,

g

in M

we have

be o-finite measure spaces and

and

is the carrier of both

M(X,u)

respectively. It is

L and L"

. We

recall the

definitions of convergence in measure and star-convergence. If the sequence (fn:n=1,2,

... )

of functions in L

and the measurable subset E

given, it is said that fn converges on

for every number function f

E >

0

converging pointwise to fn

*

+

f

, and

it is said that

if every subsequence of

of

Y

are

t o zero in measure if

E

fn star-converges

(fn:n=l,2, ...)

f almost everywhere on Y

to the

contains a subsequence

. This is denoted by

. The following lemma shows how convergence in measure and

star-

convergence are related. LEMMA 96.4. If 0

5 un

conditions are equivalent.

u

in

L for

n

=

1,2,.

.. , then

t h e following

Ch. 13,5961

KERNEL OPERATORS

*

0 as

237

.

(a)

un

+

(b)

u

converges t o zero i n measure on every subset of

n

n

+ m

measure, (c) For every subset r

as

]udv+O

of

E

Y satisfying

xE

Y

of f i n i t e

E LL we have

n+-.

E PROOF. (a)

* (b). Let

have to prove that v(Enk)

> 6

be the subset of E

of Y of finite

on which

u (y) 2

6 > 0 and some subsequence (nk:k=1,2, ...)

nk

E

. We

n tends to zero. If this fails to hold we have

"(En)

for some

a subsequence of

0 and the subset E

E >

measure be given and let En

if necessary, we may assume that

zero almost everywhere on E

,

so

t

k

v(lim sup E ) = 0 "k lim sup v(E

nk

u

. Selecting

converges to

"k . Hence

) t 6 > 0

,

which is impossible. (b)

* (a). Let E be a subset

sufficient bo prove that almost everywhere on

E

(un:n=1,2,

of

...)

Y

of finite measure. It is contains a subsequence converging

to zero. By means of the well-known diagonal

procedure it is then easy to obtain a subsequence converging almost everywhere on

Y

to zero. We restrict ourselves, therefore, to points

In view of the convergence in measure on k > 0 an index

and we may assume that

v(Ek)

< 2-k

.

there exist for every integer

such that

-k

we have

E

y € E

,

n l < n2
O)

in L

and let

such

S be the

kernel operator of finite rang with kernel

Then

inf(T,S)

the ideal M

= 8

holds in L b ( ~ , ~ ), and

. By Theorem 83.6 this means

so

{inf(T,S)}(u)

= 0 holds in

that

(3) Since M

is order separable there exists an at most countable set

(vk:O

3 -

-

T' v Ti 1

. Assume

,

L

. For

such t h a t a l l

t o prove t h a t

by p a r t ( c ) .

. Then

> 8

on

Tl A T2 = 8' 0 2 v

Ti on

T" a r e order continuous, the 2

For t h e l a s t p a r t , n o t e t h a t

T :

coincides with

t h e r e e x i s t s a d i r e c t e d system

are in

v

.

.

T;

same holds

-

(T"vT") = e 1

2

.

i n e q u a l i t y , so

(T~vT;)' > 0 by what was proved about (T'1vT')2' t T"I v T; . Hence

in

S

a contradiction.

EXERCISE 9 7 . 6 . With the hypotheses and n o t a t i o n s of t h e pPeceding L

e x e r c i s e , aSsume.in a d d i t i o n t h a t an a b s o l u t e k e r n e l operator from p

:):L(

,

t h e a s s o c i a t e norm i n :L by

p"

, we

have

M

by

p'

p"(f) 5 p ( f )

cases e q u a l i t y holds f o r a l l

as well as f o r

L

. Prove

o p e r a t o r norms s a t i s f y i n g

f f L

now t h a t

and

M

into

M

a r e Banach l a t t i c e s and

. Denoting

t h a norm in

T is

L

by

and t h e second a s s o c i a t e norm i n for a l l

f f L

. Assume t h a t

, but

i n some important

t h i s case occurs f o r

T,T' and T" a r e continuous with IITil = IIT'Il = IIT"II

.

L

Ch. 13,8981

25 7

KERNEL OPERATORS

98. Dunford's theorem

I t was mentioned i n s e c t i o n 95 t h a t the o r i g i n a l method of proving Buhvalov's theorem is based on a theorem of N. Dunford (1936), s t a t i n g t h a t any continuous o p e r a t o r from an L1-space i n t o an L -space, f o r 1 < p 5 P i s an a b s o l u t e k e r n e l operator. More p r e c i s e l y , t h e proof of Buhvalov's

m

,

theorem w a s based on t h e g e n e r a l i z a t i o n of Dunford's theorem t o a r b i t r a r y (not n e c e s s a r i l y separable) measure spaces. We s h a l l prove now t h a t i f one follows t h e method presented h e r e , Dunford's theorem may be derived from Buhvalov's theorem. F i r s t we prove a dual theorem. It i s assumed again t h a t (X,A,p)

and

a r e a - f i n i t e measure spaces.

(Y,Z,w)

LEMMA 98.1. Let L be an ideal i n M(Y,v) equipped with a a-order continuous Riesz norm ( i . e . , u J 0 i n L implies p(un) + 0 1 . Then it n * f o Z l o ~ sfrom 0 5 u 5 u E L and un + 0 t h a t p(un) -+ 0 as n + m n

PROOF. I f

p(un)

and a subsequence

does not tend t o zero, t h e r e e x i s t a number

(u

?(

:k=1,2,

.

...)

of

(un:n=1,2,

...)

E

> 0

such t h a t

The subsequence contains another subsequence converp(un ) 2 E f o r a l l k k ging t o zero almost everywhere on Y The l a s t subsequence i s , t h e r e f o r e , order convergent t o zero i n

.

L

. Hence,

by t h e a-order

c o n t i n u i t y of

t h e subsequence converges t o zero i n norm, c o n t r a d i c t i n g all

. Hence,

k

p(un)

p(u

tends t o zero.

) 2

?(

E

p

,

for

L

THEOREM 98.2. I f

i s an ideal i n M(Y,v) equipped with a a-order continuous Riesz norm, then every norm continuous operator from L i n t o i s an absolute kernel operator. I n partieuZar, i f 1 5 p < then every continuous operator from L (Y,v) i n t o L_(x,u) i s an

L,(X,v)

&soZute

PROOF. We s h a l l denote the norm i n

L

For

f

E L w e have

almost every (Tv:OSvSu)

ness

Of

into

L-(X,U)

. We

L

by

,

so

x E X. It follows t h a t f o r each

L,(X,u)

¶

t h e function

p

. Let

prove f i r s t t h a t

IITf\lm 5 I [ T / l . p ( f )

i s bounded above i n

,

P

kerne Z operator.

operator from

m

L_(X,u)

T T

be a continuous

i s order bounded.

[ ( T f ) ( x ) l 5 llT[l.d(f)

u 2 0

. Hence,

in L

for

the set

by t h e Dedekind Complete-

258

Ch. 13,1981

DUNFORD'S THEOREM

f o r every

e x i s t s i n L,(X,u) prove t h a t

u

E L+

. As

i n Theorem 83.3, we can e a s i l y

i s an a d d i t i v e mapping from

T

into

L+

(by Lemma 83.1) can be extended t o a p o s i t i v e operator

. It

L,(X,p)

follows t h a t

T = TI

-

(TI-T)

a d i f f e r e n c e of p o s i t i v e o p e r a t o r s , s o and TI we have

, which

(L-(X,u))+ from

T,

into

L

i s a decomposition of

as

T

i s order bounded. The o p e r a t o r s

T

T -T a r e norm continuous, s i n c e i t i s easy t o s e e t h a t f o r u E L+ 1 1 (Tv)(x) I S \IT11 . p ( u ) f o r a l l v s a t i s f y i n g 0 2 v I u , s o

It i s s u f f i c i e n t , t h e r e f o r e , to prove t h a t every p o s i t i v e continuous opera-

t o r from

L

into

L_(X,p)

i s a k e r n e l operator. Hence, l e t

0 2 un I u E L

and continuous. I f

and

un

*

+

,

0

then

be p o s i t i v e

T

p(un)

+

0

by t h e

l a s t lemma, s o i t follows from

X


that

by Buhvalov's theorem, T

(Tun)(x)

+

0

x E X

f o r almost

. Hence,

i s a k e r n e l operator.

By combining the l a s t theorem with Theorem 9 7 . 4 , we s h a l l now derive a theorem (dual t o the l a s t theorem) which includes Dunford's theorem as a s p e c i a l case. THEOREM 9 8 . 3 . Let

c a r r i e r o f both M

M

and

be an i d e a l i n M ( X , u )

M

MX

and such t h a t

(Mi):

such t h a t =

M

be a Banach l a t t i c e w i t h respect t o t h e Riesz nom

a s s o d a t e norm p*

from

LI(Y,v)

let

such t h a t the

p

( i . e . , the r e s t K c t i o n t o MX o f the adjoint norm

p'

i n fl = M"

is the

X

. Furthermore,

is a-order continuous. Then every norm continuous operator into M

is an absolute kernel operator.

PROOF. Note f i r s t t h a t

M*

=

M"

since

M

makes sense t o say t h a t t h e a s s o c i a t e norm

p'

of t h e a d j o i n t norm

T

.

p*

in

fl

=

M"

. Let

i s a Banach l a t t i c e , s o i t

i s t h e r e s t r i c t i o n t o :M

be a continuous operator from

L (Y,v) i n t o M The Banach a d j o i n t o p e r a t o r T* i s a continuous o p e r a t o r 1 is a from M* i n t o L_(P,v) , s o the r e s t r i c t i o n T ' of T* t o :M

continuous o p e r a t o r from :M

into

.

L _ ( Y , ~ ) The norm

p'

in

o-order continuous by hypothesis, s o by the preceding theorem

M : T'

is i s an

Ch. 13,5981

KERNEL OPERATORS

259

a b s o l u t e k e r n e l o p e r a t o r . But t h e n , by Theorem 97.4, T

i t s e l f i s an

absolute kernel operator. COROLLARY 98.4.

Ll(Y,v)

operator from

M = L (X,u) P mentioned i n t h e l a s t theorem.

r a t e d by

. For t h e s p e c i a l . Furthermore, as

t

M(Y,v) L

in

M(X,u)

0 2 t ( x ) E M(X,u)

EXERCISE 98.4. Let

M

every continuous

m

case t h a t

s a t i s f i e s a l l conditions

M

and l e t

be t h e i d e a l gene-

i s i d e n t i c a l l y one, w e g e t

t

i n Theorem 98.2,

equipped w i t h a u-continuous norm

into

5

L ( x , ~ ) is an absoZute kerneZ operator. P

into

PROOF. The i d e a l

M = L_(X,u)

1 < p

(Dunford's theorem.) For

p

let

. Let

L

be an i d e a l i n

T

b e an o p e r a t o r from

such t h a t


f o r every

X

. Show

f EL

that

i s a kernel

T

operator. H I N T : The proof i s s i m i l a r t o t h e proof i n Theorem 98.2.

98.2 t h e f u n c t i o n

i s equal t o t h e c o n s t a n t

t(x)

EXERCISE 98.5. Let in

M(X,u)

every

be an i d e a l i n

M(Y,v)

f E L

,

then

T

(i.e.,

t

into

M

$O(f) =

$o

JY f t d v

b e an i d e a l

L = Ll(Y,vI)

v1

f o r t h e measure

v-measurable

subsets T

E

f E L ). Let

Y

0

of

Y

, defined

. The

by

S

n $,(If\)

for

Ll-norm of

is a kernel operator, i . e . ,

Yo = ( y : t ( y ) > O ) .

= Y (because, o t h e r w i s e ,

T ). Note now t h a t

vl(E) =

i s a norm continuous o p e r a t o r from

by Theorem 98.3, T

O p(Tf)

the

. Show

E L"

i s r e p r e s e n t e d by a non-negative f o r every

i s of no i n t e r e s t f o r t h e o p e r a t o r

Y-Yo

0 5 $

such t h a t

Without l o s s of g e n e r a l i t y we may assume t h a t

hypothesis

M

is a kernel operator.

HINT: The l i n e a r f u n c t i o n a l

the s e t

and l e t

included. Furthermore, l e t

p

is an o p e r a t o r from L

T

function

.

llTll

s a t i s f y i n g a l l t h e c o n d i t i o n s mentioned i n Theorem 98.3,

c o n d i t i o n s on t h e norm that i f

L

I n Theorem

f E L

1,

is

Ll(Y,vl)

tdv

for a l l

$,(If

1)

into

. By M . Hence,

260

s

DUNFORD'

f o r every

f

E L

. In

Ch. 13,1981

THEOREM

t h e l a s t formula we have used t h a t t h e o-algebra of

a l l v -measurable s e t s and t h e o-algebra of a l l v-measurable 1

identical (i.e., measure

sets are

i f we apply t h e Carathgodory extension procedure t o t h e

on t h e o-algebra of a l l v-measurable s e t s , then we do not get

v1

any more vl-measurable s e t s than we had a l r e a d y ) . EXERCISE 98.6.

Let

X = Y

be t h e s e t of a l l i n t e g e r s , equipped with

the counting measure, i - e . , u(m) = v(m) = 1 = M(Y,v)

M(X,u)

sequences

y = (...,n - l , n o , r t l , n 2 X x Y be defined by

T(m,n)

on

f o r every p o i n t

i s , t h e r e f o r e , t h e space

The space

,...)

m

of

X

=

Y.

of a l l

(s)

of r e a l numbers. Let the f u n c t i o n

-1 T(m,n) = n (m+n+i)-l We take

T(m,n)

t o be the k e r n e l of a k e r n e l operator

( a ) Show t h a t I

Cln- 1

nnl

-’ for n

= (~n~’log~n~)for l

=

(a) Show that for x T

and y

does not exist. Hence,

2,3,.

..

and

n = -2,-3,...

/TI be the operator with kernel

(ITly,x) if

=

IT(m,n)l

S n = O forall other n

and

“n

,

= O for all othern.

.

as defined above the inner product

IT1

is regarded as an operator from

does not map

L2

1, into L2 , and

into itself, then T

so

is not

order bounded. This shows that although every order bounded operator from one Banach lattice into another is continuous, the converse is not generally true. It is only under special circumstances, as in Theorems 98.2 and 98.3 in this section, that a continuous operator from one Banach lattice into another is order bounded. HINT: If equal to

(ITly,x) would exist, then

77

times

(IT/y,x) would be

262

GENERALIZED CARLEMAN OPERATORS

against the assumption that

(ITly,x)

Ch. 13,9993

is finite.

99. Generalized Carleman operators If

(X,A,p)

and

(Y,Z,v)

T

T is a

are a-finite measure spaces and

kernel operator with kernel T(x,y)

mapping

L2(Y,v)

into L2(X,p)

,

then

is called a Carlernan operator if the function

Y is finite for almost every function t

,

x E X

,

t E M(X,u)

if

is u-measuable, even in the case that

positive measure. As well-known, T if

i.e

=

that the

on a set of

m

is called a HiZbert-Schmidt operator

is (uxv)-sumable over X

IT(x,y)l‘

t(x)

. Note

X

Y

, i.e.,

if the function t

(as defined above) is p-sumable. Carleman operators were introduced by T. Carleman ([l],l923).

There exist kernel operators, mapping

L (Y,v) into 2 L2(X,~) , even absolute kernel operators, that are not Carleman, and there exist Carleman operators that are not Hilbert-Schmidt (for exampleswerefer to Exercise 99.10).

T

, mapping

L (Y,v) 2

It was proved by J. Weidman ([11,1970) into L2(X,u)

,

that an operator

is a Carleman kernel operator if and

only if it maps every norm null sequence onto an almost everywhere null sequence (i.e., IIfn(12 + 0 in L (Y,v) 2

every

implies

(Tfn)(x)

+

0 for almost

x E X ) . It is of interest to compare this with the condition for T

to be a kernel operator (not necessarily Carleman), where it is required that every dominated norm null sequence in L (Y,v) 2

is mapped onto an al-

most everywhere null sequence (cf. Exercise 96.13). We shall now investigate a more general situation. Once more, let let

(X,A,u)

L be an ideal in M ( Y , v )

and

(Y,X,v)

, equipped

be u-finite measure spaces and with a Riesz norm

p

.

Ch. 13,5991

KERNEL OPERATORS

THEOREM 99.1. Let

T

263 L

be an operator t h a t maps

.

i n t o M(X,u)

The

fofollyowing conditions are now equivalent. (i)

There e x i s t s a non-negative f u n c t i o n

.

t

E M(X,u)

such t h a t every

s a t i s f i e s 1 (Tf) (x) I 5 p ( f ) t (x) almost everywhere on X ( t h e exceptional s e t depending on f 1 . ( i i ) I f f E L f o r n = 1,2, and p ( f n ) 0 , then ( T f n ) ( x ) f E L

n


...

.

If ( i ) and ( i i ) hoZd f o r

T

, then

-f

i s order bounded. A s a consequence,

T

IT/ s a t i s f i e s ( i ) and ( i i ) j u s t as well as

continuous and ( i ) and ( i i ) hold f o r

T

T

, then

. If T

the norm i n L

is order

i s a kerne2 operator (and

i s a kernel operator s a t i s f y i n g ( i ) and ( i i ) ) .

IT1

hence

+ 0

PROOF. It i s e v i d e n t t h a t ( i ) i m p l i e s ( i i ) . For t h e converse, assuming t h a t ( i i ) h o l d s , we have t o prove t h a t set in if

. According

M(X,u)

(un:n=1,2, ...)

and

(hn:n=1,2,

(An Tu,)(x)

...)

i n d i c a t e d . Then

(An Tun)(x)

+

that if

u

.

= Anp(un)

S

f o r almost every

x

T

0 < v S u

x E X

(/Tvl:0SvSu:1

p(un) 5 1

x E X

+

An

,

0

. This

for a l l

n

An C 0 , t h e n

. Hence,

let

un

and

An

as

so by ( i i ) we have

shows t h a t ( i i ) i m p l i e s ( i ) .

i s o r d e r b o m d e d i f ( i ) and ( i i ) h o l d , we have t o prove

i s g i v e n , then

E L+

Hence, l e t

f o r almost every that

such t h a t

L

i s a sequence of numbers such t h a t

p(Anun)

0

To show t h a t M(X,u)

i s a sequence i n

h o l d s f o r almost every

0

-+

i s an o r d e r bounded

S = (Tu:p(u)

denote that

fn

converges to

+

f , fn

+

f(u-unif)

i n order, u-uniformly or

f

relatively uniformly respectively. In a normed Riesz space norm convergence of

fn

to

f will be denoted by

vergence the limit f

fn

fn

+

The subset D

In all these cases of con-

is uniquely determined (for u-uniform convergence and

relatively uniform convergence because follows from

f(norm).

+

f(ru) of

that L

fn

+

f

L

is Archimedean). Furthermore, it

.

is order closed, ru-closed or norm closed if it

...

follows from f f D (n=I,Z, ) and fn + f (in order, ru or in norm) n that f E D For the order topology and the relatively uniform topology

.

this is the defintion of a closed set (cf. section 16); for the norm topology this is not a defintion, but one of the properties of the norm topology. For an ideal A

in L

a weaker condition implies already that A

is closed.

This was proved in Theorem 83.22 and we recall the result. The ideal A closed (in order, ru

or in norm) if and only if

order closed if and only if THEOREM 100.1. If

fn

sup(f ,gn) + sup(f,g) and nfn + f and lfnl -+ If/

A

+

0

5 u I.

in A

and

. In particular, the ideal

(in order, ru or in norm) implies u E A

is

u + u

A

is

i s a o-ideal.

f and

g

inf(fn,gn)

+ +

. The mappings

therefore continuous mappings from

L

( i n order, ru or i n norm), then

g

inf(f,g) f

+

f+

. I n particuZar,

,f

+

f-

and

f

f+ +

;fi

+

f+ ,

are

into itself.

PROOF. Follows immediately from

and similarly for the infimum.

In Lemma 83.11 it was proved that if all n , then

f

2

g

.

fn

+

f (norm) and

f

n

t

g

for

This holds for the other types of convergence as well.

Ch.

14,11001

NORMED RIESZ SPACES (i) If fn

THEOREM 100.2.

for all

n

, then

.

f t g

+

. Hence,

283

or i n nomn) and

f ( i n order, ru i f

fn

+

f and

f

for a l l

fn t 0

n

g

t

,

n

This shows t h a t the p o s i t i v e cone L+ i s closed. (ii) I f D i s a subset of L and fn + f (in order, ru o r i n n o m ) such t h a t fn I D f o r a l l n , t h e n f I D . Hence, every disjoint complement i s closed or, equivalently (since L i s Archimedean), every band is cLosed. then

f

2

0

PROOF. (i) It may be assumed that

...

g

0

=

f-

the sequence (fi:n=l,Z, ) converges to so f = 0 In other words, f 2 0

.

.

It was already observed above that fn so every order closed subset of be a normed Riesz space

given

E

i E.~(u)

> 0

,

we have

for all

n t

If-f I n(E)

and

fn

for all

5 EU

It follows that every norm closed subset of

fn

+

f(norm),

+

. This shows that

Having observed thus that

fn

But

If -f

I

S

If-fnl ,

for all n ,

f- = 0

implies fn

+

f , and

is stronger than the order topology. Now,

L P

relatively uniform topology in L

.

f(ru)

+

- -

Since

is ru-closed. In other words, the

L

relatively uniform topology in L let L

.

L

in L

f(u-unif)

,

so

. For

any

~(f-f,) 2

n t

n(E)

fn

converges to

f

in norm.

is ru-closed, and so the

is stronger than the norm topology. +

f(ru)

implies fn

+

f

as well as

it may be asked if there is any relation between norm conver-

gence and order convergence, We shall prove first, by means of an example, that norm convergence does not necessarily imply order convergence, and hence does not imply ru-convergence. EXAMPLE 100.3.

x

=

[O,ll

Let u

and let L

be Lebesgue measure in the closed interval

be the space L I ( X , p )

f (x) i 0 as n

and only if

...

+

. Note

that

fn

holds for p-almost every

+

0 holds if

x E X

.

Now, 1

let (En:n=1,2, ) be the sequence of closed intervals [O,fl,[~,lI,[O,jl, 1 2 1 [j,31,[2,ll,[O,-],. . , and let f be the characteristic function of En 3 4 for all n Then fn converges to zero in norm but not in order.

.

.

For monotone sequences the situation is different because now norm convergence implies order convergence. Details are contained in the first part of the next theorem. In the second part it is proved that a u-uniform Cauchy sequence has a u-uniform limit if and only if the sequence converges in norm.

NORElED RIESZ SPACES

284

THEOREM 100.4. (i) I f

.

Ch. 1 4 , § 1 0 0 ]

fn4 i n the normed Riesz space

and fn P S i m i l a r l y f o r decreasing sequences. I n other L

(norm), then

f

words, i f

i s the norm l i m i t of a monotone sequence, then

f

=

sup f

the order l i m i t of t h e sequence.

f

f

+

i s also

.

,...

(ii) I f f o r some u E L+ the sequence (fn:n=1,2 ) in L is a P P u-uniform Cauchy sequence, then the sequence i s a l s o Cauchy i n norm. In t h i s

ease, t h e sequence has a u-uniform l i m i t i f and only i f t h e sequence has a norm l i m i t , and these l i m i t s are then the same. Any Banach l a t t i c e i s theref o r e uniformly complete. (iii) Similarly, i f

Cauchy, then

fn

v-uniformzy,

(f ) n

i s u-uniformly Cauchy as w e l l as v-uniformly

converges u-uniformly i f and only i f

f (norm). Since f 2 f for all n 2 m, n m by Theorem 100.2. This holds for all m , s o f is an

PROOF. (i) Let we have

f f

and

fn

f s'f m upper bound of the sequence. Let for all

fn converges

and the l i m i t s are then the same.

n

,

so

f 2 g

+

g

be another upper bound. Then f

by Theorem 100.2. It follows that

.

upper bound of the sequence, i.e., f = sup f

f

2

g

is the least

(ii) It is evident that any u-uniform Cauchy sequence is also Cauchy in norm. It is also evident that if i n this case the sequence has a u-uni-

form limit

f

, then f is also the norm limit. It remains to prove, there-

fore, that if the u-uniform Cauchy sequence then f If -f I m n

is also the u-uniform limit of 4 EU

for m,n

2 n(E)

fn

.

(f ) has the norm limit f n For any E > 0 we have

and also (n fixed, m

-z m

,

) the sequence

(Ifm-fnl :m=n,n+I,...) converges in norm to If-f fn

I < E U for

n 2

If-fnl n(E)

. Hence, by

Theorem 100.2, we find that

. This shows that

f

is the u-uniform limit of

*

(iii) Similarly. It was shown above (Example 100.3) that norm convergence does not

always imply order convergence. In the converse direction, we have similarly that order convergence does not necessarily imply norm convergence, even for monotone sequences. To mention an example, let the usual supremum norm) and let u

n

(for n

L

P

= 1,2,

be the space Lm (with

...

) be the elemen0 with

Ch. 14,11001

NORMED RIESZ SPACES

the first n u

+

0

285

coordinates equal to zero and all other coordinates one. Then

, but the sequence does not converge in norm. The sequence is not

p(un-um) = 1 for n # m . For an example with an order convergent monotone sequence such that the sequence is Cauchy in

even Cauchy in norm because

norm but not convergent in norm we refer to the exercises at the end of this section. The examples presented show that norm convergence and order convergence are to a great extent independent. Fortunately, as we prove now, if a sequence is convergent in norm as well as in order, then the limits are the same. THEOREM 100.5. If fn L~

, then

f

g

=

.

PROOF. Writing un exists a sequence pn u

=

+

p(v-un) 0

5

+

0

.

f (norm) and

, we

lfn-gl

0 such that

converges in norm to

we have

+

/f-gl

.

fn

have

0

5

+

u

un

i n t h e normed R i e s z space

g

0 by hypothesis, so there

+

pn

5

Hence, writing

If-gl = v

It has to be proved that v

inf(v,u ) s inf(v,pn)

5

for all n =

0

. Note

. Furthermore,

for brevity, that

v ,

so (by one of the Birkhoff inequalities)

o

5

v - inf(v,pn)

5 v

- inf(v,u )

2

Iv-u

I ,

which implies

The decreasing sequence

(inf(v,pn):n=1,2, ...) converges, therefore, in

order to zero and in norm to v

. It follows from Theorem 100.4(i)

that

v = o .

-

It was shown in Example 100.3 that a norm convergent sequence need not be order convergent. Thinking of L -spaces (for I 5 p 5 ) , where each norm P convergent sequence has a pointwise converging subsequence, it is a natural conjecture that in a Banach lattice each norm convergent sequence has a sub-

NORMED RIESZ SPACES

286

Ch. 14,§lOOl

sequence converging in order. We prove that the conjecture is right. THEOREM 100.6. Every norm convei,ent

sequence i n a Banach l a t t i c e has

a subsequence which converges i n order. As a consequence, every order closed s e t i n a Banach l a t t i c e i s nomi closed. PROOF. Let x.

11

-t

f (norm) in the Banach lattice L

(gk=fnk:;=l,2, ...) from

sequence for

f

.

(fn:n=1,2,

... )

. Choose

P

such that

It is evident now that the partial sums

s

= 1:;

the sub-

p(f-gk) f-gkl

: 2-k

of the

seriz, -,f-gkj form a Cauchy sequence in norm, s o the series converges in norm. Let 5 s-s

k- 1 order.

s = sup s s o that n ’ Clf-g 1 cannot cause any confusion. Since If-g I 5 k k k + , it is evident now that gk converges to f in

be the sum. By Theorem 100.4(i) we have

s

the notation s

=

C 0 as

-

Let us assume now that proof that D all n

D

is norm closed, assume that

. We have

to show that

fn

+

f (norm)

P

with

For the f

E D

for

. As proved above, there exists a sub(fn:n=1,2, ...) such that gk converges to f f E D

(g :k=I,Z,...) of k in order. Since g E D for all k k that f E D

sequence

.

is an order closed set in L

.

If the normed Riesz space L

P

and D

is order closed, it follows

fails to be a Banach lattice, there may

exist norm convergent sequences without order convergent subsequences and that fail to be norm closed. For there may exist order closed sets in Lp an example we refer to the exercises at the end of this section. Recapitualting, we have found that in a Banach lattice L

P

wing implications hold for any subset: (1)

order closed -norm closed

but not if

L

P

the follo-

* ru-closed,

is a normed Riesz space which fails to be norm complete. In

this case, however, the implications in ( 1 ) are still true for ideals, as we shall prove now. It will be proved later (in Example 105.8) that in a Banach lattice every norm convergent sequence has a relatively uniformly convergent subsequence. Hence, a subset of a Banach lattice is norm closed if and only if it is ru-closed.

NORMED

Ch. 14,§1ool

THEOREM 100.7. For any idea2

A

i n a normed Riesz space the fol2owing

ho Zds : order closed

A

(2)

=$

nomi closed

A

287

RIESZ SPACES

-

A

ru-closed.

I n p a r t i c u l a r , therefore, every o-idea2 is nomi c2osed ( t h i s i s an improve-

ment upon Theorem 100.2(ii)). PROOF. It was observed in the beginning of the section that every norm closed set is ru-closed, so we have to prove only that if the ideal A order closed, then A with all un

in A

0

u

5

u 4

and

i s norm closed. Let

A

be order closed. For the

is norm closed it is sufficient to show that if

proof that A

+

and u

+

n Theorem 1 0 0 . 4 , i.e., 0 < un 4 u u E A .

then u E A E A for all n

u(norm),

u (norm) with

.

u

is

0

5 u

. Hence, assume . Then u = sup un

4

by is order closed, this implies

Since A

The inverse implications of those in (2) do not hold. The ideal of all null sequences is norm closed in =

(c,) not order closed. In the

of all (real) continuous functions on

space C(C0,ll) p(f)

em , but

i0 If (x) ldx 1

the ideal of all

f

[O,ll

satisfying f (0)

=

with norm

0 is uniformly

closed, but not norm closed. Under certain additional conditions, which will be discussed in section 105, one or both of the inverse implications hold. A s an application of the lasc theorem we mention the case that

$

is a

positive linear functional on

L such that the null space (kernel) of @ P is a o-ideal. In this case the null space of $I is therefore norm closed,

which implies that

@

is norm continuous. This holds in particular if

@

,

as a mapping from L into the real numbers, is a Riesz o-homomorphism. P Actually, this is the only possible case (i.e., if the kernel op the positive linear functional

@

is a o-ideal, then

@

is a o-homomorphism; cf.

Exercise 100.23). A further simple remark concerns operators from one normed Riesz space

.

L into another, say LA If L is a Banach lattice, then every positive P P operator from L into LA is continuous (the proof in Theorem 83.12 P fails to be norm complete. In this applies), but not necessarily if L P case, i t is sometimes easy to know that if S and T are operators from Lp

into LA

satisfying

€I 5

S 5 T

and

T

is continuous, then

S

is con-

2aa

Ch. 14,81001

NORMED RIESZ SPACES

f E Lo

tinuous. This follows by observing that for any X(Sf)

so

=

X(ISf+-Sf-l)

is continuous and

S

jection band in L B

Evidently

u

for every

operator in L

P

.

IITll

llS// 5

h(Slfl) 2

=

As an example, let

E Li ,

. It follows that

llPBll = 1

B # {O}

if

PB 2 I

8 2

so

PB

.

, where I

T

is a positive operator from L

tor in the ideal generated by

T

in

.

Then

is the identity

is a norm continuous operator in L

It follows now also easily that if, in addition, LA plete and

B be a pro-

with corresponding band projection PB

P

0< P u 2 u

h(Sf++Sf-)

5

we have

LA

into

P

Lb (L p ' Lh )

is Dedekind com-

,

then every opera-

is continuous.

Before discussing some properties of the norm completion of

Lp

, we

observe that the result in Theorem 100.4 about monotone sequences may be extended to directed sets. The upwards directed set to converge i n norm to in the set such that

f

E >

for all

.

p(f,-f)

< E

f

wards directed sets. THEOREM 100.8. If

converges to

(fT :n=l,Z,...)

from

. Also,

for an appropriate

The sequence

(fT ) n

T'

'

t f

To

f

=

sup f

(f':'€{~})

such that

choose a fixed

fT 0

C

{TI ,

+

L

P

and

fT

and a sequence

0

f

'0

Similarly for down-

.

E~

is said

0 an element f

fTS in the normed Riesz space

f in norm, then

PROOF. Choose a sequence of numbers all" fT 2 fT n

(f?:~E{-i})

if there exists for any

1-

. Thenn

and

p(f-f,)

2

SO

i s increasing and converges in norm to

f , so

E~

for

.

Ch. 14,51001

f =

sup f

NORMED RIESZ SPACES

289

by Theorem 100.4. For an arbitrary

n

SUP(fT,f) - f

(

sup fT'fT

=

.f)

fT we have

- SUP(fT

n

)

,f) s SUP(fT>fT n

- fT n

n

,

so

Since

+

E

0

follows that Hence

f

, this implies pIsup(fT,f)-fl f

f

5

.

sup f

=

for all

. We

fT

= 0

,

so

sup(fT,f)

know already that

. It

f

=

.

f = sup f

T

We shall now derive some simple properties of the norm completion of an arbitrary normed Riesz space L

.

P

It will be proved that the norm comple-

tion is a Banach lattice under the natural definitions. For brevity we write instead of Lp ; the norm completion is denoted by L and the extended

L

(as well as the original norm in L ) is denoted by

norm in Lof

L-

are denoted by

... . By

fo,go,

defintion, an element uo

called positive if there exists a sequence in L+ Note that for elements in L

LEMMA 100.9.

The s e t

in L-

sure of L+

.

a uo + a2u2 E (L-)+ 1 1

for uo 1

numbers. We prove now that if There exist sequences

(u,)

in L-

is

converging in norm to u

0

.

.

o f all p o s i t i v e elements i n

(L-)'

containing

PROOF. It is evident that 0

Elements

this definition is in agreement with the

existing notion of positiveness in L

generating cone in L-

.

p

L+

L+

cone

u2

uo

and

(L-)'

in

(L-)+ . .

is a

and also that

al,a2 non-negative

-uo are positive, then uo in L+

(v,)

and

L-

is t h e norm clo-

(L-)'

is contained in 0

and and

. The

=

converging in norm to

0 u

.

0

converges in norm to zero. It follows then -uo respectively, so u +v n n from 0 s u 5 u +v and p(un+v ) 0 that p(un) + 0 , so uo = 0 n n n Having thus proved that (L-)+ isna cone in L- , it remains to show that and

.

-f

(L-)+

is generating. For this purpose, let

(fn) be a sequence in L

and 0 U2

(fi) in L-

converging to

fo

0

f

E L- be given, and let

in norm. The sequences

are Cauchy sequences in L , converging to elements uy

.

The elements uo 1

0 and u2

( f : )

and

are positive by definition. Since

NORMED RIESZ SPACES

290

Ch. 14,§1001

.

0 0 f+-f- = f converges in norm to fo , we have ul-u2 = fo This shows that n n n every element in L- is the difference of elements in the cone (L-)' .

Hence, (L-)+

is generating. L-

The lemma shows that

f0

called equivalent (as usual). converges to

go whenever

5

go-fo

Note now that if

go (all fn

and all

(hn=sup(f n ,gn ):n=l,2,

fn

in L

gn

and

1

changed into an equivalent sequence, that

ho

ho

so

2

bound of so

fo

.

fo and

go

.

(h,)

is

is not changed. We shall prove

.

fo and

.

If

there exists a sequence

k' t f for all n n n converging in norm to ko =

ho

. If

in L-

go in LSince h -f t 0 0 on h -f converges in norm to ho-fo , we have h -f t 0 , n n 0 Similarly h 2 go It follows already that ho is an upper

and

that

kn

so

is the least upper bound of

for all n

fo and

...

are replaced by equivalent sequences, then

(9,)

are

then the sequence

is Cauchy in norm, and s o the sequence has a norm limit ho (fn)

-

converges to ) ?

as

is positive.

converging t o the same element in L

Norm Cauchy sequences in L gn

(L-)'

is an ordered vector space with

positive cone. By definition, we have

ko

.

is another upper bound, we have in L

0

k -f

converging in norm to ko

( k : ) Similarly, there exists a sequence

.

0

t 0

) : k (

in L

and such that k" t g for all n Let n n for all n Then kn converges in norm to ko and

sup(kA,k:) k n

2

sup(f .g ) n n

.

=

h n

for all n ,

.

0

,

and such

0

k t ho This shows that ho is the least upper bound of fo and g in L . As a special case,note that if fn converges to fo , then 0 0 sup(fn,-f ) converges to sup(f ,-f ) , i.e., Ifn\ converges to lfol n Having proved thus that L is a Riesz space containing L as a norm dense Riesz subspace, we prove finally that p is a Riesz norm in L so

.

o

+

so the norm

is absolute. To prove that

p

implies p(lf

0

Since p(fo-fn)

There exist sequences 0

0

p(v -u -dn)

+

0

. Then

(u,) un+dn

and

I-ifnl)

(d,)

+

o , we

p

in L+

.

get

is monotone, let 0 such that

converges in norm to

v

0

,

0

p(u -un) so

5

uo -f

0

5

v

0

and

.

NORMED R I E S Z SPACES

Ch. 14,§1001 0

) = lim p ( u +d ) 2 lim p ( u )

p(v

n

n

0

.

p(u )

=

n

29 I

We formulate the final result.

-

THEOREM 100.10. The norm completion L

P

i s a Banach l a t t i c e containing

L

of the nomned Riesz space

(L , p )

as a norm dense Riesz subspace.

P

A s an application we prove the following theurem. THEOREM 100.11. D~

D

of

D

If

i s solid i n

P

. In particular,

s o l i d in L-

of A i n L

and in

P

vely.

i s a solid set i n L and the norm closure

L

(L-,p)

i f

If/ E D 1

fn

+

f(norm).

.

Then

lfnl

. There exists a

u E D1

02 u

sequence

lfnl E D E D1

Iv

(~-,p)

respecti-

, we

f E D1

Given

in D

(~-,p) i s

the norm closures

and i n

.

D1

(fn)

If 1 (norm) and

+

. Next, if

D in

of P '

L~

are i d e a l s i n

There exists a sequence

If1 E D I

solid), so

D

i s an idea2 i n L

A

PROOF. We first present the proof for

that

then the norm cZosure

'-

p

n (since D

for all

is

, we have to show that

of positive elements in D

(v,)

show

such that

such

are members of D and that v + v(norm). Then all u = inf(vn,u) n + u(norm) , so u E D1 T o prove now that D1 is solid, we have to show

.

u

f E Dl

that g E DI

if it is given that

,

lgl E D 1

that

so

If1

/gl and

i

If] E D l

be a sequence of positive elements in D f+ 2

If I

of D 1

add

. Let

in norm to

f-

5

(un)

f+

.

If I

,

. It follows from

g E D1

by what was proved already. Let

the elements


f+

and

f-

u

5

f+

and

.

u

n

i

w

n

converging

for all n

(otherwise, replace u

; this does not change the

convergence to

by inf(un,wn,f+) f+ ) . Similarly, let (v,)

elements in D


be a sequence of positive and such that v 2 f and

V

IW

for all n

n n inf(u ,v ) n n

=

o

.

Since inf(f+,f-)

for all n ,

Un + Vn =

.

so

f=

0

(w,) Since

are likewise members

be a sequence of positive elements in D

We may assume that

If1

, this implies now that

it follows from

sup(un,v ) 2 w

E D

that u +v E D Then Jun-v 1 i u +v E D , s o u -v E D since D is n n n n n n +" solid. The sequence (an-vn) converges in norm to f -f = f , SO it follows

292

Ch. 14,51001

NORMED RIESZ SPACES

at last that

.

f E D1

For the investigation of the closure Dduce the subset D2 0

If 1

of

L- consisting of all

of

D

in

fo E L-

.

we intro-

(L-,P)

such that

for some g E D Evidently, D is contained in D2 and a solid subset of L We prove that D2 i s a subset of D . Let 5

fo E D to

fO 2

lgl

.

.

D2

is

be given. There exists a sequence (fn) in L converging in norm 0 o p .ithermore, since f E D2 , we have If I C: lgl for some g E D.

We may assume that

note that :f E D

/fnl

,

lgl

S

since D

is solid in L

for all n , and so the limit proved therefore that solid set D2

is

D-

for all n (otherwise, replace fn by

fo of

,

D c D2 c D-

.

is contained i n

fn

D-

the norm closure in

so

fn E D

). It follows that

P

. We have

(L-,p)

of the

is solid in L-

Hence, by what was proved above, D-

of an ideal A , the

If one is only interested in the closure A1 proof is simpler. Since A l

is a linear subspace, it follows immediately

from f+ E A 1

that

f- E A l

and

The following remark about

L

f

P

=

f+-f- E A

and

I '

(L-,p)

may be of use for a better

understanding of the difference between convergence in norm and convergence in order. A sequence in L

P

f E Lp

is convergent in norm to an element

if and only if the sequence is convergent in norm to

f

in

(L-,p)

, but

for convergence in order the situation is different. If the sequence con-

-

verges in order in L

P '

versely. If all to

f

in L-

,

fn and then

f

If-f I

are members of 5

is not necessarily a member of order in

, but not con-

then it converges in order in L L

P

and

fn

po J. 0 for appropriate :p

L

P

converges in order

E L-

. Since

0

Pn , the sequence may fail to converge in

LP

In some cases the properties of a Riesz space equipped with a Riesz seminorm can be derived from the properties of a normed Riesz space. We recall Theorem 62.3. If is an ideal in L manner in L / A

by

,

L

is a Riesz space with Riesz seminorm p

then the quotient seminorm X

, defined

and

A

in the usual

.

Ch. 14,§1001

NORMED RIESZ SPACES

[fl E L/A

for every

,

i s a Riesz seminorm in L/A

. If

for n

and

i.e., if it follows from

,

f E A

that

A

then

The space L every sequence

(fn)

in L

f E L

Clp(fn)
0

holds for all

in

fo

T(U)

such

converges to zero

)

T

is an

T

there exists an 2 T(~,E)

. Note

is automatically a net; the partial

(fT:TE{Tj)

is determined by the partial ordering in the system ofall T2

whenever

fT

S

1

is uniquely determined.

HINT: By the definition of f:/pa(f-fo)-al<E

f ‘2

. If

T ,the sets

fT converges to

fo

in T

,

Ch. 14,51001

NORMED RIESZ SPACES

for all a E { a ) , f o E L , real T , and

so

d. 6i

for i

=

= 1,

and positive

E

, form a subbsase for

the finite intersections form a base. Let

be a non-empty base set. Let

Let

a

297

=

Ipa . (f0-f i)-a.1 < 1

E.-d. 1

1

for

is contained in A

E

...,n

min(fil,...,6 ) n

. Note

that

B

for

i

1,...,n

=

, and let f

=

E L

such that

is open in the topology T

a

and

since by the

are continuous. It follows that the

d = pa (f,-f2) > 0

/ f z p (f-fI)

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