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
107. Riea& of
L•
in
n
349
and order deniity of
t~
L-
385
n
about Riesz seainores 109. Embedding of L into the order biduol
394
110. Perfect R\eoz &paces
409
1 1 I. Perfect Banach l attices
4 13
11 2. 8taon.cb funct i on spaces
1115
I I 3. Tho Patou property
42 1
I tt.. Ecbedding in the Sanach bid ual; reflexivity
429
115. Adjoint operators
437
116. Di•joint stquences and order concinuout nonDt
44 1
11 7. Copiu
446
108.
~rc
404
455
CIW"l'ER 17. AliSUW:f Lp- SPACES
t18.
Rie•~
apaces with p- additive
456
no~
465
119. Seo.i-H-ap..aces
IZO. J.pretentation by a space of 12:1. Repretentation by a space of
~eaaurabl e (ontin~u•
(unctions functions
471
490 495
CIIAPTER I 8 . COHPACT OPERATORS
12:2. C.ompo.c: t tcu . pr ecompoact set a: and alaao.tt order boundl!d
499
seta 12 3. The bnnd of
AM-co~pact
opera~ors
124. 1'heortmt of Dodd&-i:'rem.l in andl A lipra n tis-Bur k int~ha.w
504 514
'[ABLE 01'
125.
Coc:~pac tness
CONIE~'TS
• AM- compactness and semi -compac tness
xi
524
126. Semi- compac t operators and or de r continuity of the operator
534
norm 127 . St!mi- compa c t operator s and d isj oin t sequences
5 44
128. Semi-compact operators and o r der project ions
550
1Z9. Scl'lli -compact operators and indi ces
556
C1~P ~ER
19 . ORLICl SPACES AND I RREDUCIBLE OPERATORS
571
t30 . Youn.g' s inequality
575
131 . Youog classes
579
132. The
3Ssoeia t~ s p~ ce
of
~n O rl ic~
space
586
133 . The Ban ach d ual of an Odicz space
593
134. The spectrum of an operator in Banach s pace
60 1
13$. The Krein- Rutma n theorem and the spe c tral radius
60$
136. Irreducible operators
613
137 . Spectra l p roper t i es o f
c ~pac t
i rreducib le o perators
13 8. The peripheral spectrum of irreducible pos it i ve op e rato rs CHAl'TER 20. OR·r nOM0Rl'1HSMS AND f - ALCEBRAS
139. &l ementary prope rc iea o f orthomorphi sms
140 . The s pace
Orth(L)
626 63~
645 648
651
141 . E~amp l es of orthomorphisms 142. Pcoperties of f- algobras
663 667
143. The stabi..li zer of a Rics2 spac e
676
1~ 4 .
681
Orthormorphisos in a normed Riesz s pace
145 . Theor ems of Radon- Nikodym type
683
146 . The range of an orthomorphism
692
6 1BLIOGRAPI!Y
70 3
SUBJECT INDEX
717
CHAPTER 1 1
PRIME IDEAL EXTENSION
In the present chapter we consider extensions and generalizations of certain results for prime ideals proved in Volume 1 . Readers interested primarily in the theory of linear mappings between Riesz spaces, in particular when it concerns normed Riesz spaces, may immediately proceed to Chapter 12. (X,e)
Throughout the whole chapter with smallest element
6
,
will denote a distributive lattice X
and L will denote a Riesz space. In section 76
we first introduce the notion of a dual ideal (also called a filter) in (X,O)
. This
x,y E F y E F
implies x
. The
at least of
A
y E F
dual ideal F x
and
y
e ,
X , not containing
is a non-empty subset of
x E F
and such that
is a member of
F
and y
x v y E F
is called prime if
2
x
such that implies
implies that one
. It follows easily that
prime dual ideal if and only if its set theoretic complement w
=
F
is a
X-F
is
a proper prime ideal. The definition in a Riesz space is similar but not quite the same. The non-empty subset F the zeroelement is not in F furthermore u,v E F+ and
ideal F
sup(u,v) E F+
X
=
L-w
implies af E X
is called a dual ideal if
if and only if
w
0
i
If1 E F ;
u 5 v
and
u E F+
is called prime if it follows from
that one at least of
Here we have the theorem that only if
f E F
and
L
implies inf(u,v) E F+ and
. The dual
implies v E F+ u,v E L+
of
u
and v
is a proper prime ideal in L
is in F+
is a prime dual ideal with the extra property that for all real a # 0
. The main
prime ideal separation theorem, stating that if
F is a lower sublattice in exists a prime ideal w
(X,e)
such that
such that
Ic
w
f E X
theorem in section 76 is the I
is a proper ideal and
I rl F
and w r l F
theorem i s due to G . Birkhoff (1933) and M.H.
.
if and
is empty, then there
is empty. This
Stone (1937),
and it was
proved again independently by several other mathematicians (J. Hashimoto, 1952; S . J .
Bernau, 1972). A similar theorem holds in a Riesz space. The
prime ideal extension theorem (cf. Theorem 51.8 for for L ) is an easy corollary.
(X,e)
and Theorem 52.2
CHAPTER I 1
2
In section 7 7 we consider the set A (X,e)
. For any
with
A
,
x E X
let
of all prime dual ideals in
(X:AEA,xEA)
=
. Thesets
, together
,
itself, form a base for the closed sets of a topology in A
the
hull-kernel topology. It will be proved that this is a T -topology such that
A
0
{AIX
and all
are compact. Denoting by
prime ideals, the mapping
X
correspondence between A
and
+
w = X-A
fi
fi
the set of all proper
establishes a one-one
. The hull-kernel topology
R ; the resulting topology in
transferred, therefore, to
in A
can be
is called the
8
dual hull-kernel topology. Some properties of this topology will be discussed. We also introduce the Boolean topology in fi ; this is the weakest topology stronger than the hull-kernel topology and the dual hull{wIx
kernel topology. The collection of all
{wjY , for x,y € X
and all
,
is a subbase for the open sets of the Boolean topology. (X,e)
In section 78 we first prove that every proper prime ideal in
X
has a unique extension to a prime ideal in the Boolean ring generated by (the Boolean ring generated by X
X
is the smallest Boolean ring containing
as a sublattice). The sublattice
(X',8)
(X,8)
of the distributive lattice
is said to have the monotone prime ideal extension property whenever
for every ideal satisfying I'
I' c w '
.
I' in X'
nX
and every proper prime ideal
u)
there exists a prime extension w '
cw
It was proved in Theorem 51.8 that X
extension property with respect to X'
if X
of
in X such that
w
has the unique prime ideal is an ideal in X'
now make this result more precise by showing that X
. We
shall
has the monotone and
unique prime ideal extension property if and only if X
is an ideal in X ' .
A certain special kind of sublattice, called a p-sublattice, has the
monotone prime ideal extension property. By way of example, every subring of a Boolean ring is a p-sublattice. A second example is obtained by considering the lattices X(L) Riesz spaces L L'
.
and
L'
In this situation X(L)
the ideal generated by
and X(L')
L
in L '
. Then L'
ideal P
.
.
L
IL be
Now, let
has the unique prime ideal
u' E L'
,v E
X(L) = X ( I L ) L
.
satisfying
there exists a number o, > 0 and an element u E L+ such that In this case, the unique extension of a given proper prime 2 u
0 < U' 2 v
IL
is a Riesz subspace of
if and only if
Explicitly, this means that for any pair U'
L
is a p-sublattice of X(L')
extension property with respect to
au 2
of all principal ideals in the
respectively, where
.
in L
to a prime ideal in
In the special case that L
IL
is the ideal generated by
is Archimedean and L'
P
in
is the Dedekind
ch. 1 1 1
completion of X(L)
3
PRIME IDEAL EXTENSION
and
L we have
X(L')
L'
=
IL ,
so
X(L')
. The
X(1L)
=
are equal is now exactly the condition introduced by
J.J. Masterson for different purposes (cf. Exercise 3 7 . 1 6 ) . that if
L'
condition that
is the Dedekind completion of
,
L
We shall prove
then the unique prime ideal
extension property is strictly weaker than Dedekind completeness of to the situation that (X,e)
(X,e)
L
(X',8) , we show that
is a sublattice of
has the unique prime ideal extension property with respect to
if and only if the Boolean ring generated by ring generated by
X
(X',f3)
is an ideal in the Boolean
. The results on unique prime
X'
and
. Finally, returning
strictly stronger than the projection property for L
extension are from a
paper by W.A.J. Luxemburg (1973) According to the definition in section 2 1 the ideal A Archimedean Riesz space L
is order dense whenever Add
extended to an arbitrary non-empty subset D dense whenever L
Ddd
=
L
.
of the Archimedean space
{u'Idd n L
. This can be
L
L ; the set D
of
is order
In section 79 we introduce, for a Riesz subspace L'
,
the notion of strong order denseness.
This means, by definition, that for every L'
=
in the
u' > 0
in L'
the intersection
does not consist only of the zeroelement. For an ideal L
this is the same asorderdenseness, but if
L
in
,
is not an ideal in L'
then strong order denseness is a stronger condition than order denseness.
It will be proved that for every
0 < v
5
u'
L
u' > 0 in L'
is strongly order dense in L' there exists an element v
if and only if
in L
such that
holds, and also if and only if for every u ' > 0 in L'
intersection AU, f l L
the
does not consists only of the zeroelement, where u'
denotes the ideal generated by
in L'
. This last result is due to
A. Bigard (1972). In section 80 we assume, for simplicity, that L' = IL (i.e., L Riesz subspace of
L'
such that the ideal generated by
itself). For any ideal I in L in L' L
.
in L'
is a is
is prime in L'
P'
in L
the ideal M'
is strongly order dense in L'
property, then P'
is prime in L'
the particular case that L'
and
L
L'
.A
=
IL
, L'
is
has the projection
for every prime ideal P
is the Dedekind completion of
L
in L
. For
, this result
(due in this case to J.J. Masterson) was already indicated in Exercise 37. 17.
I
in
if and only if for any minimal prime
is a minimal prime ideal in L'
sufficient condition for this is indicated, as follows. If Archimedean, L
L'
I' the ideal generated by
It is not difficult now to prove that for any prime ideal P
the ideal
ideal M
we denote by
L
4
CHAPTER 1 1
Ch. 1 1 1
In section 81 the results in section 80 are rounded off by the L' = IL
following main theorem. If
L
is strongly order dense in L'
L ), then P '
completion of
L
L
if and only if
L'
,
L'
has the projection property and
(all this holds if
is prime in L'
L'
is the Dedekind
for every prime ideal P
in
has the projection property. For the special case that
is the Dedekind completion of
L
, this
result is due to K.K. Kutty
and 3 . Quinn (1972).
In section 82 we return to the lattice
(X,e)
and we first prove
another prime extension theorem. Every proper prime ideal in a sublattice
Y X
of the form Y
.
The lattice
in X
=
(x:x>x ) 0
(X,e)
can be extended uniquely to a prime ideal in
is called normal whenever every proper prime ideal
contains a unique minimal prime ideal. It will be shown that X is d x A y = 0 implies X = {x) v {yjd , and also if
normal if and only if and only if X
holds for any pair of different minimal prime p2 d ideals. Similar conditions (i.e., L = {u} + {vId if u A v = 0 and
L
= p 1 + p2 )
(X,e,e)
= p
1 v
hold for normality of a Riesz space L
with smallest element
8
and largest element
.
The lattice e
is called
extremally disconnected whenever every proper prime ideal in X
is
contained in a unique maximal ideal. We shall prove that a topological space is normal or extremally disconnected (in the usual topological sense)
if and only if the lattice of its closed subsets is normal o r extremally disconnected in the sense as defined here. The lattice
(X,e)
is called completely extremally disconnected
whenever every principal ideal in X for a Riesz space L proper ideal in X
.A
(or
is extremally disconnected. Similarly
necessary and sufficient condition is that any
L ) containing a prime ideal is prime itself.
Another one is that the family of all proper prime ideals containing a given proper prime ideal is linearly ordered by inclusion. Every Riesz space is completely extremally disconnected, but not every lattice (X,e)
is called completely normal whenever every sublattice Y
Y = (x:x?x ) 0
. Finally,
(X,e)
of the form
is normal. A necessary and sufficient condition is that the
family of a l l prime ideals contained in a given proper prime ideal is linearly ordered by inclusion. Another condition is that X = w 1 V w 2 holds for any pair of incomparable proper prime ideals. Complete normality in a Riesz space is defined by these last conditions. Finally, if
(X,e)
i s normal and completely extremally disconnected, then X is completely norma.1. Similarly for a Riesz space. Hence, any normal Riesz space is
Ch. 111
5
PRIME IDEAL EXTENSION
completely normal. It will be shown by examples that all situations considered in the theory can really occur. Some of the results in this section are due to A.A. Monteiro (1954) and some (independently) to W.H. Cornish (1972). It is not without some importance, finally, to observe that the analogy with commutative rings goes still farther than indicated in section 52 (52.6,
52.14 and 52.15) and section 78 (78.16 and 78.17).
I n particular,
regular semi-prime rings are of interest in this respect (the commutative there exists an element r E R
is regular if for every x E R
ring R such that
x
2
x r
=
and R
R has no non-zero
is called semi-prime if
nilpotent elements). The following theorems hold. A. R
is regular if and only if R
ideal I in R
is semi-prime and every principal
is a direct summand (i.e., R = I
B. R is regular if and only if R in R
J
@
for some ideal J ) .
is semi-prime and every prime ideal
is a maximal ideal.
C. If R
is semi-prime and has a unit element, then the following
conditions are equivalent. (i)
Every proper prime ideal in R
is a maximal ideal.
(ii)
Every proper prime ideal in R
is a minimal prime ideal.
(iii) The base sets
{PIr of the hull-kernel topology in the set P
of all proper prime ideals are open and closed. (iv)
The hull-kernel topology in P
(v)
The hull-kernel topology in
(vi)
R
P
is Hausdorff. is a T -topology. 1
is regular.
D. If R
is semi-prime and has a unit element, then the following
conditions are equivalent. (i)
Every proper prime ideal in R
contains a unique minimal prime
ideal. (ii)
Every maximal ideal in R
(iii) R
=
prime ideals. (iv)
M 1 + M2
If r,s E R
(v)
If
and
and M2
rs = 0 , then R
=
of different minimal
+
N(r)
r (i.e., N(r)
=
(x:rx=O) ) .
r,s E R , then N(rs)
=
N(r) + N(s)
is the annihilator of Compare C
contains a unique minimal prime ideal.
for any pair MI
, where
N(r)
.
and D with Theorems 37.6 and 82.5-82.7.
details we refer to C.B. Huiismans ("21, 1974).
N(s)
For proofs and further
[Ch. 11,576
PRIME IDEAL SEPARATION
6
76. Prime i d e a l s e p a r a t i o n A s i n Chapter I ,
element subset
8
. We
of
S
be a d i s t r i b u t i v e l a t t i c e w i t h s m a l l e s t
not containing
X
implies t h a t
x,y E S
X
let
s h a l l i n d i c a t e t h i s by x
d e f i n e d i n s e c t i o n 5 , any
i s c a l l e d a lover s u b l a t t i c e of
8
. According t o t h i s d e f i n i t i o n of X . The lower s u b l a t t i c e S
y E S
A
s e t i s a l s o a lower s u b l a t t i c e
muxima2 whenever
. As
(X,e)
X
if
t h e empty is called
i s n o t p r o p e r l y included i n any o t h e r lower s u b l a t t i c e .
S
D
The non-emuty s u b s e t
of
...
...,
.
x,, x E D that x A A x # 8 n 1 lower s u b l a t t i c e has t h e f i n i t e i n t e r s e c t i o n
property i f i t follows from Obviously, any non-empty
i s s a i d t o have t h e f i n i t e i n t e r s e c t i o n
X
property. There i s another fundamental n o t i o n which we s h a l l d e f i n e now. The subset
F
of
i s c a l l e d a dual i d e a l o r a f i l t e r i f
X
F
s a t i s f i e s the
following c o n d i t i o n s : (i)
F
i s non-empty,
(ii)
8
i s not i n
( i i i ) x,y E F (iv)
x E F
,
F
implies and
x
S
x y
y E F
A
implies
,
.
y E F
I n o t h e r words, a dual i d e a l i s anon-empty property ( i v ) . In algebraic discussions
lower s u b l a t t i c e w i t h t h e e x t r a F
i s usually called a dual ideal,
b u t i n s e t theory one more o f t e n speaks about a f i l t e r . Note t h a t t h e i n t e r s e c t i o n of a family of dual i d e a l s i s e i t h e r empty o r it i s again a dual ideal.
F
The dual i d e a l
i s c a l l e d maximal whenever i t i s n o t p r o p e r l y i s a maximal dual i d e a l i f
included i n any o t h e r d u a l i d e a l . Obviously, F and only i f
F
i s a maximal lower s u b l a t t i c e .
The d u a l i d e a l
i s c a l l e d a prime dual i d e a l i f
F
t h a t one a t l e a s t of
x
and
y
. Proper
F
i s a member of
x v y E F
implies
prime i d e a l s and
prime d u a l i d e a l s a r e r e l a t e d , a s shown by t h e following theorem. THEOREM 76.1.
The proper i d e a l
s e t t h e o r e t i c complement
F = X-I
minima2 prime idea2 i f and only i f
I in
X
i s prime i f and only i f i t s
i s a prime dual i d e a l . Hence, I F = X-I
is a
i s a maximal dual i d e a l (ef.
Theorem 5.4 ( i ) ) . PROOF. We sketch one h a l f of t h e proof. Let w i t h complement
F
. Evidently,
F
I
i s n o t empty and
be a proper prime i d e a l
8
i s not i n
F. I f
Ch. 11.§761
,
x,y E F (because
then
x
that
x
x,y E I
and
x E I (because
would imply
y E I
A
x
7
would imply 5
y
implies
x E I
or
,
y E F
y E I
since
i s an i d e a l ) . This shows a l r e a d y
I
i s a dual i d e a l . Finally, x
F
since
, since
y E F
A
i s prime). Also, x E F
I
would imply
y E I
F
PRIME IDEAL EXTENSION
y E F
V
x v y E I
implies
(because
or
x E F
,
y E F
i s an i d e a l ) . Hence,
I
i s prime. If
g e n e r a t e d by
i E I
f o r some
X
i s an i d e a l i n
I
(I,yo)
and
I
. Note
n o t c o n t a i n i n g t h e element yo
that
c o n s i s t s of a l l
(I,yo)
is x -maximal
I
contained i n
(I,yo)
f o r any
yo
not i n
( i t ) t h a t any x -maximal i d e a l i s prime. 0 Dually, i f F i s a dual i d e a l i n X denote t h e s e t of a l l
(F,yO)
x
satisfying
i s e i t h e r t h e whole s e t
x
, and
so
A
yo
A
i v y
S
s a y now
0
i s maximal
I
xo
,
yo
f o r some
yo = 8
x
is
was proved i n Theorem 5.1
not c o n t a i n i n g f
2
X (if f
xo
the ideal
. To
X
means t h a t
I
. It
I
,
yo
satisfying
can be t h e whole s e t
f o r a given xo n o t i n 0 w i t h r e s p e c t t o t h e p r o p e r t y of not c o n t a i n i n g that
x
f o r some
let
f E F
(F,yO)
. Then
f E F )
or
i t i s a g a i n a d u a l i d e a l which i s t h e n a p p r o p r i a t e l y c a l l e d t h e d u a l i d e a l and
yo
. The d u a l
i d e a l not containing
xo
.
F 'is now c a l l e d x -maximal f o r a 0 n o t i n F whenever xo i s c o n t a i n e d i n (F,yO) f o r any y 0 I n o t h e r words, F i s not p r o p e r l y included i n any o t h e r d u a l
generated by given
xo
F
not i n
F
.
ideal
A e-maximal d u a l i d e a l i s t h e same a s a maximal
dual i d e a l . We g e n e r a l i z e t h e d e f i n i t i o n s of an x -maximal i d e a l and an x -maximal 0 0 I and t h e non-empty lower s u b l a t t i c e F such
dual i d e a l . Given t h e i d e a l that I
I
n
F
i s empty, we s h a l l s a y t h a t
I
i s an F-maximal idea2 whenever
i s n o t p r o p e r l y included i n any o t h e r i d e a l
empty. Note t h a t i f
such t h a t
i s t h e d u a l i d e a l g e n e r a t e d by
F'
F ' = (y:y2x f o r some x E F) ), t h e n i s F'-maximal.
J
I
J
n
F
is
F (i.e.,
I
is t h e F-maximal i f and only i f
I n o t h e r words, we may j u s t a s w e l l assume t h a t
F
is a dual
i d e a l , which makes t h e d e f i n i t i o n more symmetric. This becomes even more e v i d e n t i f we compare t h e d e f i n i t i o n of an F-maximal
i d e a l w i t h t h e now
f o l l o w i n g d e f i n i t i o n of an I-maximal d u a l i d e a l . Given a g a i n t h e i d e a l and t h e dual i d e a l
F
such t h a t
I-moxima2 dual i d e a l whenever ideal
G
such t h a t
I
n
G
F
I
n
F
i s empty, we s h a l l say t h a t F
1 is an
i s not p r o p e r l y i n c l u d e d i n any o t h e r d u a l
i s empty.
PRIME IDEAL SEPARATION
8
Let
THEOREM 76.2 (i)
X
any F-maximal idea2 i n
be a non-empty lower s u b l a t t i c e i n
F
i s prime.
.
(ii) Let 1 be a proper i d e a l i n X i n X is prime. (iii) Let
idea2 and x
.
y E I
A
For
I
, there
.
X
Then
Then any I-maximal dual i d e a l
.If
X
I be a proper i d e a l i n
is not i n F
x
Ch. 11,5761
i s an I-maximal dual
F
y E F
e x i s t s an element
such t h a t
{ e l t h i s i s the weZZ-known f a c t t h a t if F i s a x i s not i n F , there e x i s t s y E F such
=
maximal lower s u b l a t t i c e and that
x
A
y
PROOF.
.
e
=
(i) Let P
xo (cf. Theorem 5.1 (ti)).
, not in
z
P
. The proof
be an F-maximal ideal in X
prime is almost exactly as in the case that F If
P
but such that y
that P
is
consists of only one point
is not prime, there exist elements y
E P
z
A
. The
P
ideal generated by
contains at least one point of the dual ideal F'
.
F
generated by
and
and
,
so
y
.
p, v y E F ' for some p1 E P Similarly, p2 v z E F ' for some p 2 E P Setting p3 = p1 v p 2 , we have p E P with p v y and p3 v z in F' 3 Hence
On the other hand
. Hence
P
P n F'
p3
and
p3 v (yhz)
is in P
(F,y)
at least one point of 2
A
z E I for some
f3
A
y
f 3
and
. Hence, f3
It follows that (iii) If
f3
(yvz)
A
x
F
I ,
so
fq E F in
z
A
On the other hand F
so
P
,
z
. Contradiction, since
is prime.
not in F
I
and
.
f
I
A
.
v
f3
for some
.
,
so
f3
is not prime, y v z E F
f E F
f, E F
= f l A f2
are in F
z
is in I n F
If F
but such that y
A
y E I for some
Setting
. Hence
y
f
v (yhz) is also in
p3
as well as in F'
of all elements majorizing
and
,
be an I-maximal dual ideal in X
there exist elements y
f
are in P
z
A
is empty. It follows that
(ii) Let F set
y
. Similarly,
, we have f3
A
(yvz)
Contradiction, since
. The
contains E F
with
is also in i s empty.
I I7 F
is prime. is not in the I-maximal dual ideal F , the set
contains at least one element of
I , i.e., x
A
.
y E I
f o r Some
(F,X)
y E F
.
Ch. 11,5761
PRIME IDEAL EXTENSION
9
The prime ideal separation theoremwhich follows now is an immediate consequence. The theorem occurs in the famous paper by M.H. Stone ([11,1937, Theorem 6) on distributive lattices. The result stated in the theorem is essentially due to G. Birkhoff ([l1,1933,Theorem 2 1 . 1 ) .
This result has
been rediscovered, in one form or another, by several other mathematicians (for example, J. Hashimoto [11,1952; S . J . Bernau C21,1972). We first recall
I is given, then (by section 5) the prime ideal M
that if the ideal
is said to be I-minimal if any prime ideal M' satisfies M'
=
M
THEOREM 76.3 (Prime ideal separation). I f
and
F
i s a lower sublattice i n
e x i s t s a prime ideal
w
in
X such that
I
(l
F
and
Ic w
also
i s a proper idea2 in X
I
such that
X
I c M' c M
satisfying
.
I
2
i s empty, then there w
n
F
i s empty.
Actually, there e x i s t s an F-maximal prime ideal satisfying these conditions, m d there also e x i s t s an I-minimal prime ideal satisfying these conditions. PROOF. Consider the set of all ideals J
such that
J
3
I and J n F
is empty. It follows from Z o m ' s lemma that this set has a maximal element w
. By
the preceding theorem w
ideal containing
I
.
is prime, i.e., w
Consider the set of all dual ideals G
is an F-maximal prime
such that G
3
F
and
I
n
G
is empty. It follows from Zorn's lemma that this set has a maximal element
A
. By
the preceding theorem X
ideal containing F such that
is prime, i.e., A
. It follows that
w =
X-A
is an I-maximal dual
is an I-minimal prime ideal
n F is empty.
w
Stone's original formulation of the theorem is as follows. Given the proper ideal
I and the dual ideal F
exists a prime ideal w w a I
and
such that
w
such that
I n F is empty, there X = X-w satisfy
and its complement
A 2 F .
The prime ideal extension theorem, already proved in Theorem 5 1 . 8 , may be regarded as an immediate corollary of the prime ideal separation theorem. THEOREM 76.4 (Prime ideal extension). I f
d i s t r i b u t i v e Lattice (X','d) there e x i s t s a prime ideal
w'
and i f w in X'
i s a sub2attice of the i s a proper prime ideal i n X , such that w ' n x = w (X,t?)
.
Ch. 11,5761
PRIME IDEAL SEPARATION
10
F = X-w
PROOF. The s e t generated i n
I
x' 5 p
by
X'
f o r some
i s a lower s u b l a t t i c e . Furthermore, t h e i d e a l
w (i.e.,
t h e s e t of a l l
p E w ) satisfies
x ' E X'
,
I 7 I X = w
satisfying
I ll F
so
i s empty. Hence,
by t h e prime i d e a l s e p a r a t i o n theorem, t h e r e e x i s t s a prime i d e a l such t h a t
X' and
w'
ll (X-w)
w'
Note t h a t
w'
= w'
w ' ll F
and
I
3
i s empty. It follows from
that
ll F = pi
ll X
w'
. This
= w
can be chosen F-maximal o r I-minimal
=J
0
element
of
S
i f desired.
. The
lower s u b l a t t i c e
. As
L
not c o n t a i n i n g t h e zero-
L+
L
i s c a l l e d a lower s u b l a t t i c e i n
inf(u,v) E S
13w
concludes t h e proof.
We b r i e f l y d i s c u s s t h e corresponding n o t i o n s i n a Riesz space d e f i n e d i n s e c t i o n 3 3 , any s u b s e t
in
w'
w'
if
u,v € S
implies
i s maximal i f it i s n o t p r o p e r l y
S
included i n any o t h e r lower s u b l a t t i c e . We a l s o i n t r o d u c e t h e n o t i o n of a
F
dual i d e a l . The s u b s e t
L
of
i s c a l l e d a dual i d e a l o r a f i l t e r i f
F
s a t i s f i e s t h e following c o n d i t i o n s : (i)
F
i s non-empty,
(ii)
0
is not i n
( i i i ) u,v E F+
0 < u
(iv)
S
F
implies
v
and
f E F
and u
Obviously t h e p o s i t i v e p a r t
i n f ( u , v ) E F+
E F+ implies
, v E F+
,
If1 E F
i f and only i f
.
of a d u a l i d e a l i s a non-empty
F+ = F ll L+
lower s u b l a t t i c e w i t h t h e e x t r a p r o p e r t y ( i v ) . The d u a l i d e a l
F
is
maximal whenever it i s n o t p r o p e r l y included i n any o t h e r dual i d e a l . Obviously, F
i s a maximal d u a l i d e a l i f and only i f
lower s u b l a t t i c e . The dual i d e a l follows from is i n
F+
.
u,v E L+
and
i s a maximal
F+
i s c a l l e d a prime dual idea2 i f i t
F
u v v E F+
t h a t one a t l e a s t of
u
and
v
I t can be proved s i m i l a r l y a s i n Theorem 76.1 t h a t t h e complement
X
= L-w
of a proper prime i d e a l
property t h a t if
X
then
F
f E X
implies
i s a prime dual i d e a l w i t h t h e e x t r a
w
af E
i s a prime dual i d e a l such t h a t w = L-X
f E A
Hence, any maximal dual i d e a l
a J: 5
.
Given t h e i d e a l
171 F
implies
. Conversely,
af E
X
for
o
# 0 ,
i s a proper prime i d e a l . It was proved i n Theorem 3 3 . 7 t h a t
i s a maximal d u a l i d e a l i f and only i f
any
a # 0
f o r any r e a l
X
I
F
M = L-F
i s prime and
and t h e non-empty
i s empty, w e say t h a t
I
i s a minimal prime ideaL
f E F
implies
lower s u b l a t t i c e
F
i s a n F-mmimaZ i d e a l whenever
p r o p e r l y included i n any o t h e r i d e a l
J
such t h a t
J
n
F
af E F
for
such t h a t
I
is n o t
i s empty- This
i s t h e same d e f i n i t i o n as i n t h e l a t t i c e case. The proof t h a t any F-maximal
Ch. 11,1761
PRIME IDEAL EXTENSION
11
i d e a l i s prime i s a l s o t h e same a s i n t h e l a t t i c e case. Given t h e i d e a l we say t h a t
F
and t h e dual i d e a l
I
such t h a t
F
i s an I-maximal dual i d e a l whenever
included i n any o t h e r dual i d e a l
such t h a t
G
n F i s empty,
I
i s not properly
F
i s empty. A maximal
I fl G
dual i d e a l i s t h e same a s a 10)-maximal dual i d e a l , where
{O} i s t h e
i d e a l c o n s i s t i n g of t h e zeroelement only. The next theorem, s t a t i n g t h a t dual i d e a l i s prime, i s now analogous to p a r t s ( i i ) and ( i i i )
any I-maximal
of Theorem 76.2. THEOREM 76.5. Let
dual i d e a l and 0
S
v E F
is prime, and
F
. ALSO,
a # 0
i f
u
i s not i n
PROOF. It w i l l be s u f f i c i e n t t o show t h a t the complement
I c P
a prime i d e a l . It i s evident t h a t and
0
5 u S
v
with
it is clear that i f
that
inf(u,w) E I
u+v E P
. By what
such t h a t
v E P u
2
. Now,
u
i s not i n let
u,v
2
,
f
E
F
and
, there
, we
have
w3 E F+
since
are in
I
w,
and
inf(u+v,w ) s inf(u,w ) + inf(v,w3) E 1 3 3 so
u+v
i s not i n
inf(u+v,w ) 3 consequence
F ( s i n c e otherwise
F
i n common). It follows t h a t
P
and
I
u+v E P
i s an i d e a l , and e v i d e n t l y
P
S
. We
and
. Hence,
inf(u,w ) 3
with
F
0
exists
u,v E P
0 with
inf(v,w2)
f E F
P = L-F
i f and only i f
P
. Also,
E P
was j u s t observed, t h e r e e x i s t
inf(u,wl)
w3 = inf(w ,w ) 1 2 B u t then
0
implies
for
, there e x i s t s an element
F
.
I-maximal
af E F
has the property t h a t
F
0
2
inf(u,v) E I
such t h a t
. Then m y
be a proper i d e a l i n L
I
is
,
If1 E P
is Imaximal, such
w E F
show t h a t
w2
in
F+
writing inf(v,w ) 3
in
I.
,
would have t h e element
. As
an immediate
i s then prime.
The prime i d e a l s e p a r a t i o n theorem and t h e prime i d e a l extension theorem follow now, e x a c t l y a s i n t h e l a t t i c e case. The analogy between i d e a l s ( i n p a r t i c u l a r p r i m e i d e a l s ) i n a d i s t r i b u tive lattice
(X,e)
and i n a Riesz space
t h e following remarks. Given a Riesz space
a l l principal ideals i n
L
. Let
4
L
i s made even more evident by
L
, let
X(L)
be t h e mapping from
be t h e s e t of L
onto
X(L)
f E L
assigning to each
. Note
$ ( f ) = Af
satisfying A g
=
the principal ideal Af
that the inverse image $-'(Af) Af
. For
$(inf(u,v))
=
$(sup(u,v))
=
where AU+Av
is the algebraic sum of
It follows easily that $ onto Q(x
g E L
AU
and
A
V
and A
V
=
A
0
(note that AU+Av
is the
) . Evidently, X(L)
is
as smallest element.
= {O}
maps, in a one-one manner, ideals of
L
onto
, maximal ideals onto maximal ideals, x0 -maximal ideals
X(L) 0
f, i.e.,
Au n Av ' $(u+v) = AU + A V '
smallest principal ideal containing AU
ideals of
generated by consists of all
u,v E L+ we have
therefore a distributive lattice with $ ( O )
)-maximal ideals and prime ideals onto prime ideals. Also, rhe
of a lower sublattice F
image $(F) X(L)
Ch. 11,5761
PRIME IDEAL SEPARATION
12
and the inverse image Q-I(G)
is a lower sublattice in L prime dual ideal in X(L)
in L
is a lower sublattice in
of a lower sublattice G
in X(L)
. Finally, note that the inverse image of
is a prime dual ideal in L
a
containing any
non-zero multiple of each of its elements.
The analogy is not complete, however. In theorem 52.4 it was proved that prime ideal extension for Riesz spaces is monotone in the following sense. If
Q,
and
are proper prime ideals in the Riesz subspace K
Q,
satisfying Q, c Q ,
the Riesz space L
extension P, , then Q,
and if Q ,
has a prime extension P2
of
has a given prime ideal
.
satisfying P, c P,
A similar result does not always hold in the lattice case. Indeed, let w1
and
be proper prime ideals in the distributive lattice
w,
such that w I
is properly included in w2
. Assume, furthermore, that
is embedded as a sublattice in the Boolean ring element i3 w;
in
. Then
B , but
w1
and
w ' c w;
1
w2
(X,e)
X
B with the same smallest
can be extended to prime ideals w '
1
and
is impossible because prime ideals in a Boolean
ring are maximal ideals and
so
one maximal ideal cannot be properly included
in another one. That a distributive lattice is embedded in a Boolean ring (even i n a Boolean algebra) is exactly the situation that occurs in the Stone representation theorem, where a given distributive lattive embedded in the Boolean algebra of all subsets of the set i2 prime ideals in X
. For a
(X,e)
of all proper
simple explicit example we refer to Exercise
76.14 at the end of the present section.
is
Ch. 11,5761
13
PRIME IDEAL EXTENSION
EXERCISE 76.6.
The subset
of the commutative ring
S
a rnuZtipZicative s e t if the zeroelement is not in
. Show
xy E S
that if
S
S-maximal ideal in R and
R
is called
x,y E S
and
implies
is a non-empty multiplicative set, then any
is prime. Next, show that if
exists a prime ideal P
in R
I
is an ideal in R
I n S
a non-empty multiplicative set such that
S
S
is empty, there
I c P and P n S
such that
is empty.
This prime ideal separation result is known as Iirull's lemma. EXERCISE 76.7.
The proper ideal
I in the distributive lattice (X,e) (Ia:a€{a)) of ideals
is called completeZy p r i m e whenever for any family it follows from I = n I
that
I
=
I
ideal I is completely prime if and only if EXERCISE 76.8.
Show that an
space as in earlier chapters. The set
. The set
is not in P
is, therefore, the set of all
{PIu
in the Riesz space L {PIu
such that the given element
is assumed to have the (relative)
hull-kernel topology. We recall that, by Corollary 3 5 . 4 , the kernel d Now, let satisfies k({P} ) = (AU) = {uId k({P)U) = n (P:PE{P}u) {PIu
a subset of
.
{PIu
N
w >
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< u 5 v
.
that are u-maximal as well as k(Q
'
U,V
) = {uId
HINT: We have to show that for any w > 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
<
w
5 u
are empty. Assume,
. Observe now
that
(i.e., the set of all u-maximal ideals not containing v ) is the
S
U,V same as
Qu,w
(i.e., the set of all u-maximal ideals not containing w ) .
EXERCISE 76.12.
Let
u > 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. < u and 1. is still in F+ . If u is not in F+ , then inf(u,v) = 0 for some v E F+ . But u 5 v since u is a lower bound of F+ . Hence u = inf(u,v) = 0 , contradicting u > 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
PRIME IDEAL EXTENSION
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
is allowed
, and for a and b finite it will make
.
(x:aSxsb)
no
L has the Riesz
Show that
is not a Riesz space. This is Problem 5882
L
in the Am. Math. Monthly (1972), due to E.S. Langford. HINT: Given that u1+u2 ;nvl;2 existence of C.W.. = v 1 1~
w.. (i,j=1,2) 1J
(j=1,2)
j
define w.. 1J
.
13
.
such that
Set u 1 + u 2
w.. (x)
by
=
, we
in L+
= u.
1
(i=1,2)
and
in Exercise 15.14,
if w(x) # 0 and w. .(x)
u. (x)v.(x)w-'(x) 1
C.w J ij
. Exactly as
v1+v2 = w
=
have to prove the
J
1J
=
0
if w(x) = 0 It remains to prove that w is everywhere differentiable. ij Hence, let If w(xo) # 0 , then w ij is differentiable at xo a < x < b and w(xo) = 0 , s o ui(xo) = v.(x ) = 0 and u!(x ) = 0 and
.
0
v!(x ) J o w(xo+h)
=
0 for
# 0
i,j
, we 0 2
wij(x +h) O
0
h as
h i 0
at
x
.
h
J
> 0
o
,
1
0
and assuming that
have
U. (X +h) 1
. Then, for
= 1,2
-
h
-
w.. (x ) 1J
0
-
ui (X0 +h)
=
u. (x +h) v. (xo+h) 1
ui (xO )
h
Similarly for h < 0
0
J
5
hw (xo+h)
+
u!(x ) 1 0
=
. This shows that
w
0
ij
is differentiable
0 '
EXERCISE 76.17.
Now, let L be the ordered vector space of all real
continuously differentiable functions on is still continuous. Show that
L
(a,b)
,
i.e., the first derivative
has the Riesz decomposition property. It
Ch. 11,9761
PRIME IDEAL SEPARATION
18
can be proved that the same holds for the space of twice continuously differentiable functions on
(a,b)
and even for the space of three times
differentiable functions, but not for the space of three times continuously differentiable functions (A. Nagel and W. Rudin [ I ] ,
1974).
HINT: Determine the derivative w! in a point x where w(x) lj by means of the ordinary differentiation formula, and show that if w(xo)
=
0 then w!.(x) 13
EXERCISE 76.18.
tends to zero as x
tends to
xo
# 0
*
We mention another formulation of the Riesz de-
composition property which makes it more evident why the property is also called the Riesz interpolation property (cf. Exercise 15.12). ordered vector space L
Show that the
has the Riesz decomposition property if and only
if L has the property that for any systems fl,...,f and g l ,...,gm n satisfying g. 5 f for i = I , m and j = I , ,n there exists an 1 j element f E L such that g. 5 f 5 fj for all i and j
...
...,
.
0
HINT: By induction it is sufficient to indicate the proof for m
n = 2
=
gi
5
v2
=
. Assume
that L has the decomposition property, and let
.
f for i,j = 1,2 Set u . = f.-g. (i=1,2) , v I = fl-g2 and j 1 1 1 f2-gl Then u1+u2 = vl+v2 , so there exist wij (i,j=l,2) such that
.
W I I + W I 2 = u1
,
w,I
w21 + w22
,
WI2 +
=
u2
(see the diagram). It follows easily that
+
w21
=
v1
,
w22 = v2
f o = g1
+ w I 2 is the desired
element. The proof in the converse direction will also be evident.
Ch. 11,1771
PRIME IDEAL EXTENSION
19
77. The hull-kernel topology and the dual hull-kernel topology
Let R
be the set of all proper prime ideals in either the dis(X,e)
tributive lattice
. For reasons of
or in the Riesz space L
notationalconvenience we shall discuss the lattice case; the discussion for
R will be denoted by
a Riesz space is quite similar. The elements of the elements of w
X by
x,y,...
. We
recall that
not containing x ; union and intersection of and
{w}
XVY the kernel
{w}xhy
of all w
{w}x
and
respectively. For any non-empty subset S
. For any
satisfying w
2
D
subset D
. The
of
,
are
Y of
R
in S , and the kernel of
is the intersection of all w
k(S)
the empty set is X
w
is the set of all
X , the hull h(D)
,x
family of all
E X
is the set
,
is a base
for the hull-kernel topology (briefly hk-topology) in R ; the closure of any given subset in R
S
of
R
. The hull-kernel
is the set h(k(S))
is a T -topology, the base sets
topology
of which are open and compact
0
(cf. Theorem 7.3 for the lattice case and Theorem 36.3 for the case of a Riesz space). Let A
be the set of all prime dual ideals A
a Riesz space L , let A
in X
. In the case of
be the set of all prime dual ideals A
with the extra property that
f E A
implies af E h
in L
for all real
a # 0
.
This extra condition is included here since we desire to have the situation that the set theoretic complement of
A
is a proper prime ideal w
lattice case this holds without any extra condition, by Theorem 76.1. return to the lattice case. For every x E X { h I x = (A:hEA,xEh)
Note immediately that
Furthermore, if and only if
c {A}’
x = y
{A]@
,
. In the We
let
. is empty, and
holds if and only if x
5
. It follows that the mapping
the Boolean algebra of all subsets of
A
y ,
so
x + {XIx
=
from X
into
is a lattice isomorphism.
We define hulls and kernels similarly as for prime ideals. Explicitely, if
S
is a subset of
A
,
the kernel k ( S )
is defined by
HULL-KERNEL TOPOLOGY
20
k(S)
=n (X:XES)
for
non-empty,
S
and t h e k e r n e l of t h e empty s e t i s s u b s e t of
.
X
Fcr any s u b e t
h(k(S))
3
subset
D
and
of
.
X
Theorem 6 . 2 ) t h a t
h(D)
i s d e f i n e d by
of
S
immediately t h a t A
implies
We i n t r o d u c e a topology i n
together with
the hull
X
,
and
k(h(D))
h(k(4)) = 3
D
6 ,
f o r every
I t can be proved now e x a c t l y as f o r prime i d e a l s ( c f . S = h(D)
.
k(h(D)) = D
is, therefore, a
. Note
h(X) = $
f o r every s u b s e t
S
k(S)
.
h(D) = (h:A3D) Hence, h ( $ ) = A
X.Note t h a t
of
D
Ch. 11,1773
A
h(k(S)) = S
and
D = k(S)
implies
by choosing a l l s e t s of t h e form
{XIx
i t s e l f as a subbase f o r t h e c l o s e d s e t s i n t h e topology.
A
A c t u a l l y t h i s i s a b a s e f o r t h e topology, s i n c e any f i n i t e union of s e t s of t h e form the hull
i s a g a i n of t h e same form. For any s u b s e t
{XjX
is either
h(D)
h(D) =
so
n
(if
A
A
s e t i s t h e h u l l of some
o r a n i n t e r s e c t i o n of s e t s D c X
.
THEOREM 77.1. For any s u b s e t For this reason, fhk-topology)
in
A
h(k(S))
1)
, the
c l o s u r e of
is h(k(S)).
S
h(k(S))
. But
h ( k ( S ) ) ; it i s s u f f i c i e n t ,
i s i n c l u d e d i n any c l o s e d s e t
. Then
i . e . , h(D)
A
every closed
.
of t h i s kind i s of t h e form S
of
i s included i n t h e closed set
PROOF. S
3
S
{XIX , s o
t h e topology is called t h e h u l l - k e r n e l topology
t h e r e f o r e , t o prove t h a t h(D)
,
X
i s c l o s e d in t h e j u s t i n t r o d u c e d topology. Conversely, every
h(D)
S,
of
D
i s empty) o r e l s e
,
({hlx:x€D)
closed set i s e i t h e r
Any
D
h(D) = S1
h(D)
f o r some s u b s e t
,
S,
so
3
h(k(S))
.
D
of
S
1
X
I>
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
PRIME IDEAL EXTENSION
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
PRIME IDEAL EXTENSION
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
HULL-KERNEL TOPOLOGY
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
PRIME IDEAL EXTENSION
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
HULL-KERNEL TOPOLOGY
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
PRIME IDEAL EXTENSION
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
HULL-KERNEL TOPOLOGY
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
PRIME IDEAL EXTENSION
{ 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
PRIME IDEAL EXTENSION
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
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
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.
PRIME IDEAL EXTENSION
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
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
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
PRIME IDEAL EXTENSION
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 UNIQUE PRIME IDEAL EXTENSION PROPERTY
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
of
(B,B)
is a Boolean algebra. THEOREM 78.8.
of a BooZean ring
(B,B)
(i) Every subring
(B',9)
is
a p-subZattice.
is a Riesz subspace of t h e R i e s z space L' , then X(L)
(ii) If L a p-sublattice of
.
X(L')
B will be denoted by
PROOF. (i) The elements of B' by
elements of from 8
to b
... . Now,
a',b',
let b
=
x
...
ac
so
that it follows from b
exists and we have ac
=
b
interval in B
,
in B'
b A ac = (avb') A ac =
then ac E w ' u,v E L+
(ii) Let
where Au,Av u,v
and
B
(sac)v
b' E w '
. Hence, B
u ' E (L')'
and
(b'Aac)
5
sup(v,u')
. Then
a sup(v,u')
, u
= a
and generates v
But then u2
5
=
u2
5
v + (u2-v)
-1
u
b' A ac I b'
=
5
V
v+u'
=
A
2
L'
by
sup(v,u')
2
generate the
a > 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
PRIME IDEAL EXTENSION
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
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
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
This will be used in the proof of the following theorem. THEOREM 78.10. Let
that
X(L) = X(IL)
more
L'
this, L
L
be a Riesz subspace of the Riesz space
, as i n p a r t
L . If, i n addition t o is an i d e a l i n L' ( i . e . , L = IL ).
has the p r o j e c t i o n property, then so has i s uniformly complete, then
L
I f , instead of t h i s l a s t condition, I L
the Dedekind completion of PROOF. Assume that in L
.
L
i s uniformly complete, then
I in L'
in L'
A
will be denoted by
will be denoted by
a projection band in L' ; given u 2 0 in L u
. As
in A'
(
(1)
. The band
let u'
inf(z,u)
_<
u
. Hence,
(
u' = sup v:v€A,o
,
is
is a projection band in L
= X(IL)
such that
,
z
a
1.
this is now equivalent to showingtbat u' hypothesis that X(L)
A , say
satisfying
, and
L . If u' = 0 , . On account of the
there exists a number a
aw < u' < w
by Theorem 24.5
is a member of
this is obviously the case. Hence, assume that u' > 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'
Every v'
IL i s
L' has the projection property. Let A be a band
. The ideal generated by
band generated by of
such
L'
(ii) of t h e preceding theorem. I f f u r t h e r
>
0 and an
. It follows then from formula
(I)
41
PRIME IDEAL EXTENSION
Ch. 11,§781
inf(v,aw):vEA,O5viu aw E A , and hence w E A
so
Then v E A , 0 set
5
v
u
5
. On
(v:vEA,O
. Now,
implies v
let u 1
,
5 u1
the other hand
=
so
I This is the desired result, i.e., L
, and
u1
=
complete by Theorem 4 2 . 6 .
=
In order to show now that
. There exist
=
(
sup v:vEL,Oiv5u'
=
This supremum exists in L = u'
. If not, we
element in L
is
L , consider an such that
u' > 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
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
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
PRIME IDEAL EXTENSION
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' < u , then the projection property for L'
a > 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'
is a Riesz subspace of
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
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
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
PRIME IDEAL EXTENSION
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
is a Riesz subspace of
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
PRIME IDEAL EXTENSION
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
(v)
u' > 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 < p E {u'Idd
fl
. Then
L
n inf(u',p)
.
inf(w,p)
,
zdd r l L # { O }
, which
that w E zd L
>
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
STRONGLY ORDER DENSE RIESZ SUBSPACES
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
is not the supremum of MU,
satisfying u; < u'
MU, element v1 E L
, and
so
,
there exists an upper bound
1 Of (by hypothesis again) there exists an
satisfying 0 < v 1 5 u'-u' , so by induction v+nv
for every v E Mu'
u'
. But E Mul
then v+vl E Mu,
holds
for every v E Mul
PRIME IDEAL EXTENSION
Ch. 11,1791
n = l,2,
and
... . It follows that
is impossible since L'
(iii)
0 < nv
* (iv). Given
u' > 0
in L'
(iv)
*
(i).
Note that A
L
for n
#
fl L
{u'ydd
is strongly order dense in L'
{O}
nL #
, i.e.,
u' = sup MU,
follows immediately from (iii) u'
0< v
for every for every
{O}
... . This
= l,Z,
. This shows that
u' t 0
holds for every
c
U'
Hence, since by hypothesis A follows immediately that
, it
such that
.
n L # 101
u'
is Archimedean. Hence, we have
that there exists an element v E L A,,
5
1
49
in L'
.
in L' , it
u' t 0
u' > 0
,
in L'
.
i.e.,
It remains to show that (v) is equivalent to the other conditions. In B ' + B'
one direction this is easy. Assume that (v) is satisfied, so
, the
is aone-onemapping. Given u ' > 0 in L'
{ u ' } ~n~ L # {0}
{u'}dd # {0) , so follows that
L
band
{u'}dd
satisfies
since the mapping is one-one. It
is strongly order dense.
Conversely, assume that
L
nL
is strongly order dense in L'
. We shall
prove not only that (v) holds, but also that all the additional properties hold that are mentioned at the end of the theorem. Note first that for any subset D
of
L possessing a supremum do in L , the element do is
also the supremum of an upper bound
db
in L'
D
of
D
satisfying 0 < v < do-db
in
. Indeed, if otherwise, there would exist L' such that db < do . Take v E L
(such an element v
order dense in L' ) . Then d+v s do sup(d:dcD)
= do
in L
implies
exists since L
sup(d+v:dED)
=
d +v 0
the thus established fact that suprema of subsets of the same in L the set B' bands in L implies Bi
and L'
=
so
B'
+
and
n
L
(B;)d
B'
in L'
is one-one, we prove that
, then
one of these, say g'
E (B;)d
,
so
onto
B; # B; B; h'
, is
simultaneously. A l s o , it follows from h' > 0
is contained in B;
contained in B; I7 L
(if existing) are
which is not in the other one. Then
and the strong order denseness of {h'ldd
L
B' I7 L maps bands in L'
inf(1 f'I ,/g'l) > 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
STRONGLY ORDER DENSE RIESZ SUBSPACES
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 < u 5 u' It follows that 0 < u 5 u" , and so L is
such that 0
.
.
Ch. 11,9791
51
PRIME IDEAL EXTENSION
.
strongly order dense in L"
(ii) We have to show that for any
n
u' > 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#
is strongly order dense in L'
.
L# = (L')#
L
L'
of
In particular, show that if it is given that L# = ( L ' ) #
is strongly order dense in 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'
.
PRIME IDEAL EXTENSION
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
Ch. 11,9801
* (ii). It is sufficient to prove that if
PROOF. (i)
given in L' there exists an element in L Hence, let
0 < v' < u'
0 < v' < u' 5 uo
0
=
. Then (u -vl) 0
+ v'
< (u -v') + u '
0
by hypothesis there exist u1,u2 E L+
u
5
u -v' 0
u2
so
-
u2 5 u '
v'
u -u 1 = u 2 < u ' ,
is an element in L
(ii) If u'
. Then
and
Archimedean and
u'
It follows that
u' E L
If in L' d'.
=
=
u'
.
, we
and
iu' < u'
u'
is an atom, so
.
u'
for
i
1,2
=
sup(d;,d;)
.
i
= 1,2
sup(d;,d;)
Hence
(iii)
5
by hypothesis
. But
u'
L'
is
is now a numerical multiple of
0 and
=
, we have 0
= u'
5
sup(v;,v;) d! < u '
+ sup(-v',-v') 1
By (ii) there exist elements u1,u2 E L+ 2
. .
z
is not an atom, there exist strictly positive elements v;,v;
satisfying inf(v;,v;)
u'-vi
E L such that f u r < z
z
u1+u2 with
have to show that u ' E L
is an atom, we observe that it follows from
that there exists an element
.
such that
such that uo
IL
=
uo E L
is u'
and
,
lying between v'
(iii). Given 0 < u ' E L'
0 < v' < u '
that lies between v'
. First, choose
in L'
so
1
55
PRIME IDEAL EXTENSION
=
u'
* (i).
2
=
_i
u'
. Writing
and
u ' - inf(v' v') = u ' 1' 2
such that
d! < u . < u' 1
1
sup(ul,u2) i u ' , but on the other hand
. It follows that
If L
and L '
decomposition property in L
=
sup(ul,u2)
=
u'
, and
. for
sup(ul,u2) 2 so u ' E L
.
coincide, then condition (i) is the Riesz
L'
.
In the preceding sections we have met the notions of unique prime ideal extension and unique band extension. We shall now define unique ideal extension. Given again that that L
ever, for every ideal L'
L
is a Riesz subspace of
L' , we shall say
has the unique idea2 estension property with respect to L'
such that
J n L
=
the ideal generated by
I in
.
when-
L , there is a uniquely determined ideal J
I In this case we must have J = I' , i.e., J is I in L' It is also evident that every ideal J
.
in
56
in
EXTENSION THEOREMS
i s now of t h e form
L'
f o r some
I'
in
I
The following theorem, f o r t h e case t h a t
, namely
L
for
.
I = J Il L
i s t h e Dedekind comple-
L'
L , i s due t o J . J . Masterson ( [ l l , c f . E x e r c i s e 37.16).
t i o n of
THEOREM 80.4.
Let
L'
.
a > 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
PRIME IDEAL EXTENSION
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'
f o r every prime i d e a l
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
has the projection property.
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
PRIME IDEAL EXTENSION
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 ,
=
.
f o r every prime i d e a l
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
has the projection property.
(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
PRIME IDEAL EXTENSION
,
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
then
A
has a unique prime
in
w
Y
has a unique prime
i s given by
w'
w ' = (x:fix,xvatw
,
Y
i s given by t h e formula
f o r some aEY) =
f o r some bEw
We s h a l l prove t h a t t h e same holds i f
X
does not have a l a r g e s t element.
Of course, the proof i s very s i m i l a r t o t h e proof of Theorem 51.8 ( i i ) i n dualized form, but f o r t h e sake of completeness we include t h e proof. LEMMA 82.1.
sublattice dual i d e a l
Let
1'
in X
and l e t X be a prime dual ideal in the can be extended uniquely t o a pz-ime Explicitly,
xo f (X,6)
Y = (x:xtx ) 0
.
. Then
h
x:x€X,xvaEA f o r a l l aEY PROOF. It is e a s i l y v e r i f i e d t h a t
x:x€X,xva€A f o r a l l aEY
Ch. 11,1821
PRIME IDEAL EXTENSION
is a dual ideal in X
. FOK the proof
(xva) v (yva) E X
for all a E Y
so
X
,
is prime in Y
x v xo E A
X'
that
. But
,
is prime, let x v y E A '
X'
that
in particular f o r
for all a E Y ,
then x v a E X
A'
is prime. It is easy to see that
a = xo
n Y
. This
x E X'
so
= A
X'
so
. Since shows
is an
X . Assume that u ' is another prime extension of X . For the proof that X' is included in p' , take a E
extension of
, so
u' n Y
Y-X
=
X
a is not in u ' ) . Every (since u '
is prime and
the proof that
u'
satisfies x v a E X c u ' ,
x E A'
is not in p ' ) . This shows that
a
is included in
,
1'
let x E !.I'
hold, there would exist an element a E Y
X
. But
y
2
x E u'
and y
Hence x € X' , and s o prime extension of
2
p' c
aE Y
X'
, so
. If
such that y y E u' l Y
. It follows that
u'
=
X
= =
,
, say
or y v xo E X
x v xo E X
it follows that
65
so
. For
X' c u '
x E X' x v a
(so
x E !.I'
would not is not in
. Contradiction.
X' , i.e., the
1. is unique.
In the theory of Riesz spaces we have met several times the conditZon that every proper prime ideal contains a unique minimal prime ideal. Indeed, it was shown in Lemma 37.10 that in a Riesz space the condition is equivalent to the condition that disjointness of of the closures
rw),
and
{w];
{wju
, where
and
{w)v
as usual
implies disjointness denotes the set of
all proper prime ideals not containing the element u E L+ closures are in the hull-kernel topology of the set
,
and where the
of all proper prime
ideals. Also, it follows from Theorem 37.11 that the condition is necessary for the Riesz space to have the projection property. It will be shown in the present section that for lattices the corresponding condition is related to normal topological spaces, and the dual condition is similarly related to extremally disconnected topological spaces. DEFINITION 8 2 . 2 . The distKhutive l a t t i c e
whenever every proper p r i m e ideal in
x
(X,e)
i s called noma2
contains a unique minima2 prime
ideal, i . e . , whenever every pKme dual idea2 i s contained i n a unique
maximal dual ideal. Given the proper prime ideal w
in
(X,8)
, we
shall denote by
the intersection of all prime ideals contained in w , so
k(w)
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
66
is an ideal in X. Note that k(w)
Hence, k(w) 8
.
E k(o)
Note also that k(w)
ideals contained in
Ch. 11,9821
is not empty, since
is the intersection of all minimal prime
w ,
LEMMA 8 2 . 3 . Gioen t h e proper prime i d e a l
(X,6) , we have
in
wo
x:xAy=8 for some yEX-w PROOF. For abbreviation, write I
=
f x:uy=8 \
for some yEX-w
. If
w1 = 8
I is an ideal in X
is prime and
I c k(wo)
shows that xo
A
all
satisfying z
h o = X-wo
.
i.e., y
. Note
that
ideal generated by
ho
.
is not in w '
Conversely, assume now that xo A y
2
is a prime ideal for some y E X-wo, Hence,
. This
xo A y = 8 E w 1 , it follows that xo E w '
for all y E X-wo , i.e., xo not in I
y # 0 z
,
implies that y E X-w'
then w ' c wo since w '
'
and if xo E I , i.e., if xo A y
and observe already that contained in wo
1
0)
for some y
,
is not in F
8
and
xo
so
xo E k(wo) , but
. Let
F be the set of
in the prime dual ideal
F
is a dual ideal, the dual
. We prove now that
I and F
are disjoint.
Indeed, if not, we should have xo A y E I for some y E Xo , i.e., (by the definition of I ) x A y A z = 9 for some y E A. and some z E X o But then w
=
y A z E A.
0
,
so
would imply xo E I
xo A w = 0
contradicts the hypothesis that xo
is not in
I
. Hence, I
. This
and
F
.
are
disjoint. It follows by means of the prime ideal separation theorem that there exists a prime ideal w
containing I such that
disjoint. Then w c X-F c X-A
= wo
wo
. Hence, by
the definition of
implies xo E w (since w
so
w
, we
. On the other hand we have
and F
.
are
is a prime ideal contained in have k(wo) c w
xo E F
,
so
xo
,
xo E k(wo) is not in w so
are disjoint). Contradiction. It follows that xo E k(wo)
implies xo E I , i.e., k(wo) c I that k(wo) = I
,
k(wo)
and F
w
. Hence, finally, we obtain the result
The next lemma is an immediate consequence.
Ch. 11,9821
PRIME IDEAL EXTENSION
LEMMA 8 2 . 4 . The proper prime i d e a l
a unique minimal prime i d e a l i f and only if k(w) is a prime i d e a l in X land i n t h i s case k(w) is the unique minimal prime i d e a l auntained i n w 1. k(w)
PROOF. If
.
minimal prime ideal p
Then
u
=
u' ,
u
so
k(w)
k(w)
contains
is obviously the unique minimal contains a unique
w
by the definition of
k(w)
. On
the
contains at least one minimal prime
u
By the uniqueness of
is included in every
, so
k(w) = ~ - i
that
3
w' c w
.
p' c w' c w
so
(x,6)
. Conversely, assume that
other hand, ever prime ideal
u' ,
w
is prime, then k(w)
prime ideal contained in w
ideal
in
67
w' c w
this implies that
, i.e., u c
k(w)
.
It follows
is prime.
We now state and prove the main theorem about normal lattices. We recall d the disjoint complement of x is denoted by {x} .
that for every x E X
THEOREM 8 2 . 5 . The f o l l m i r g conditions f o r the d i s t r i b u t i v e l a t t i c e
are equiva Zent
(X ,6 )
(i)
.
is normal, i . e . , every proper prime i d e a l i n X contains a
X
unique minimal prime ideaZ.
=
If
y = 6 , then
A
x,y E
x
X =
we have
{XI d
v {y}
{xAyld
=
d
.
w
u1
x
.
d 1x1~ v {y~
For any pair o f d i f f e r e n t minimal prime i d e a l s
(v) X
x
For a l l
(iv)
have
is prime f o r every proper prime i d e a l
k(w)
(ii)
(iii) I f
. pl
and
u 2 we
v 1~. 2 '
contains a largest element, then these conditions are equivalent
t o the following condition. (vi) Every maximal i d e a l i n X contains a unique minimal prime i d e a l . PROOF. (i)(ii)
(ii). This was proved in Lemma 8 2 . 4 above.
(iii). Assume that (ii) holds, but (iii) does not hold. Then
there exist
xo,yo E X
proper ideal in X
. It
proper prime ideal w k(w) =
that xo
.
y = 6 and (x0Id v (yo}d is a Od (xo} v {yo}d is contained in some d On account of {x } c w we may conclude from such that xo
A
follows that
0
(x:xhy=tl
is not in k(w)
)
for some yEX-w
. Similarly, {yo} d c w
implies that yo
is not in
Ch. 11,1821
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
68
. On
k(w)
t h e o t h e r hand we have
a t l e a s t of
xo
and
yo
xo
is in
yo = 8
A
and
k(w)
. Contradiction.
k(w)
i s prime, so one
I t follows t h a t ( i i )
implies ( i i i ) .
*
(iii)
{xId
( i v ) . Since
{yJd c {xAyId
V
x , y E X , we d {xhy} , s o
holds f o r a l l
have only t o prove t h a t t h e i n v e r s e i n c l u s i o n h o l d s . Let z E d d (ZAX) A y = 8 By ( i i i ) t h i s i m p l i e s t h a t X = {ZAX} v {y}
.
t h e form and
y
z = p
q = 8
A
Z
with
z A p
that
z
E
=
with
v q
p E {zAxld
. Hence Z A
=
Z
E { x } ~and
d 1x1 v { y l d
(pVq) = (ZAP)
A
Z
z
and
A
V
(ZAq)
,
so
d q 5 q E {y}
. Hence,
any
z
,
q E {yld
i.e.,
E {yId
z A q
,
so
i s of
z
z A x A p =
. This
e
shows
d {xId v {y}
E {xAyId i s a member of
.
This i s t h e d e s i r e d r e s u l t . (iv)
*
(v). E v i d e n t l y ( i v ) implies ( i i i ) , s o i t i s s u f f i c i e n t t o show
ul
t h a t ( i i i ) implies ( v ) . Hence, assume t h a t ( i i i ) h o l d s , and l e t
x
be d i f f e r e n t minimal prime i d e a l s . There e x i s t s an element
i s not i n
that
x
{xJd
i s contained i n
p2
, p2
maximal lower s u b l a t t i c e y
E X-vl
(i.e.,
x
so (since
. Since
x
is in
1 cannot be adjoined t o
p1
X-ul
(i.e.,
x
x A y
= 3'
so
p1 v p2 =
(v)
*
x
€or an element
without d e s t r o y i n g t h e
.
and
p2
. Hence,
w
and
p1
is
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
by ( v ) , we have
F i n a l l y , assume t h a t X
ul
. X = 1-1
impossible. Hence ( i ) h o l d s . of
y = 6
A
( i ) . Assume t h a t (v) h o l d s b u t ( i ) does n o t hold. Then t h e r e
e x i s t s a proper prime i d e a l p1
p2
such
is not i n the
p r o p e r t y t o be a lower s u b l a t t i c e ) . It follows then from x d by ( i i i ) t h a t X = 1x1 v { y I d Also, s i n c e y i s n o t i n d Hence prime, we have {y} c u1
.
and pl
i s prime) t h e d i s j o i n t complement
p2
), we must have
X-p
in
X
1
v p
2
c w
, so w
= X
. This
is
has a l a r g e s t element. E v i d e n t l y , normality
i m p l i e s ( v i ) . Conversely, l e t ( v i ) hold, i . e . ,
every maximal i d e a l
c o n t a i n s a unique minimal prime i d e a l . Now, let
w
prime i d e a l c o n t a i n i n g t h e minimal prime i d e a l s
I J ~
contained i n a t l e a s t one m a x i m a l ideal
are minimal prime i d e a l s contained i n
J
J
,
, SO
be an a r b i t r a r y proper
and
p2
i t follows t h a t
. Since ul
and
w
p2 p l = p 2 , by ( v i ) . Hence, w
is
69
PRIME IDEAL EXTENSION
Ch. 11,1821
contains only one minimal prime ideal, i.e,, (i) holds. The equivalence of normality and the condition that x A y = 9 implies {xId v {yld = X is mentioned already in a paper without detailed proofs by A.A. Monteiro (c11,1954). The theorem as stated above was proved by W.H. Cornish (I 11,1972) In E%ercise 8.9'it was indicated that, under a certain additional hypothesis (incomparable proper prime ideals in ( X , e ) have disjoint neighborhoods in the hull-kernel topology), X is normal if and only if and disjointness of {w}x and {w}y implies disjointness of {w};
.
{wIThe additional hypothesis can be dropped, but the proof needs some Y modification. We show how this can be done.
THEOREM 82.6. The d i s t r i b u t i v e l a t t i c e
disjointness of
and
{w},
Cw3-
(X,e) i s normal i f and only i f
and {GI implies disjointness of the closures Y the closures taken i n the hulZ-kernel topology of the s e t
iwIx
Y' fi of a l l proper prime ideals.
PROOF. We first recall Theorem 7.6(ii), stating that for any x # 9 we if and only if there exists a minimal prime ideal u o E {wlx have wo E {w}; Assume now that X is normal and such that uo is contained in wo x,y E x have the property that {wIx and {wIy are disjoint, but {wIx n {w}By Theorem 7.6(ii), as stated and {LO}- are not. Let wo E {w}: Y Y above, there exist minimal prime ideals u l and uz , contained in wo , such that u1 E { w } ~ and p2 E {wIy , so ul # u2 since {wlx and are disjoint. On the other hand we have ul = uz by the normality of X
.
.
.
Contradiction. Hence, if X
is normal, then disjointness of
{wJx
and
implies disjointness of their closures. Conversely, assume the
cwIy
disjointness condition satisfied, but X not normal. Hence, there exists a containing different minimal prime ideals y I and Then w 0 3 y2 E {wIx , be an element in p1 but not in p2 E {wJx , once again by Theorem 7.6(ii). On account of x E 1.1, there
proper prime ideal wo
u2 SO
. Let wo
x
.
.
Hence { w I X exists an element y E X-p1 satisfying x A y = 9 = {wIXAy = {w}~ is empty, which implies by hypothesis that {a};
tl
{w), {w};
empty. But w0 3 ul E {wIY implies wo E { w I i , s o wo E Cwl; n {wly Contradiction. Hence, the disjointness condition implies normality.
=
.
is
70
WORMAL AND EXTREMALLY DISCONNECTED LATTICES
ch- 1
' §a2] 9
I n Lemma 3 7 . 1 0 ( i i ) t h e corresponding theorem i s proved f o r Riesz spaces. The proof i s n o t a p p l i c a b l e t o t h e l a t t i c e c a s e , s i n c e i t i s used t h a t i n a Riesz space we can s u b t r a c t . On t h e o t h e r hand, t h e p r e s e n t proof f o r t h e l a t t i c e c a s e i s a p p l i c a b l e i n t h e Riesz space s i t u a t i o n a s w e l l . A s may b e i n f e r r e d from t h e l a s t remark, analogous r e s u l t s a s f o r l a t t i c e s hold f o r Riesz s p a c e s . The Riesz space e v e r every proper prime i d e a l i n For any proper prime i d e a l
k(w)
w
L
i s c a l l e d normal when-
L
c o n t a i n s a unique minimal prime i d e a l .
i n t h e a r b i t r a r y Riesz s p a c e
L
the ideal
i s d e f i n e d t o b e t h e i n t e r s e c t i o n of all prime i d e a l s contained
i n w ; t h e proof t h a t
k(w)
(f :fEL,inf ( I f I ,g)=O f o r some gEL+-o)
=
i s e x a c t l y a s f o r t h e l a t t i c e case. Also, w i d e a l i f and o n l y i f
k(w)
c o n t a i n s a unique minimal prime
i s prime. The main theorem on normal Riesz s p a c e s
i s t h e n a s follows.
The f o l l d n q c o d i f i o n s for the Riesz space L are
THEOREM 82.7.
equivalent. (i) L i s normal, i.e., every proper prime idea2 i n L contains a unique minima2 prime ideat. ( i i ) k(w) i s prime for every proper prime idea2 w (iii)
= 0
If i n f ( u , v )
a2gebraic swn of the bands (iv)
, then
d L = {u} + {vId
d {v} ) .
{ u } ~and
.
( i . e . , L i s the
~f u , v E L+ , then { i n f ( u , v ) T d = { u l d + {v}d For any p a i r of d i f f e r e n t minima2 prime ideats
.
(v) u1 arid p 2 ue haue L = p 1 + p2 If L contains a strong u n i t , then these conditions are equiuaZent t o the foZ2owing condition.
.
( v i ) Every mo2ima2 idea2 i n
contains a unique minima2 prime i d e a t .
L
PROOF. We r e s t r i c t o u r s e l v e s t o t h e proof t h a t ( i i i ) i m p l i e s ( i v ) . d S i n c e [ u I d + {v} c { i n f ( u , v ) l d h o l d s f o r a l l u , v E L+ , w e have only t o prove t h a t t h e i n v e r s e i n c l u s i o n h o l d s . Let v = 0
(whu)
A
form
w = p+q
w h u h p = O
. By
with and
( i i i ) t h i s implies t h a t 0
5
pE
[WAU)~
and
vhq=O.Hence
0
5
L =
w
E
{WAU)
d
inf(u,v)
+ {vjd,
.
, so so w i s of t h e
0 s q E { v } ~ It f o l l o w s t h a t
w with w that w
=
w
A
p E {uId
A
71
PRIME IDEAL EXTENSION
Ch. 11,9821
w = inf(w,p+q) s inf(w,p) + inf(w,q) and w
A
q
q
S
E {vld ,
so
w
[ u } ~+ [vld
{uId + [vId
. Hence, any
w
q E {vId
A
is majorized by an element in [uld + {vld
element of
=
A
p + w
A
q
. This shows
, i.e., w itself is an
w E {inf(u,v)}d
is a member of
. This is the desired result.
We return to the lattice situation, and before introducing now the dual property of normality we make one more remark about the particular case that
(X,e)
normality that x
has alargestelement e A
y
= 8
implies X =
equivalently by saying that x p E {xld and some q E [yId
A
.A
y =
8
. In this case the condition for
{XI d v
[yId may be expressed
implies e
=
p v q
for some
similar remark holds for the case of a
Riesz space with a strong unit. DEFINITION 82.8. The distributive l a t t i c e
(X,B,e) i s called
extremally disconnected whenever every proper prime ideal i n X i s contained i n a unique maximal ideal, i . e . , whenever every prime dual ideal i n X contains a unique minimal prime dual ideal. It is obvious from the definition that
(X,e,e) is extremally
disconnected if and only if X is normal with respect to the dual ordering. The following theorem follows, therefore, immediately from the main theorem for normal lattices. THEOREM 82.9. The folZ&ing are equivalent.
conditions f o r the distributive l a t t i c e
(X,e,e)
X i s extremally disconnected, i . e . , (i) X i s contained i n a unique maximal ideal. (ii) For m y prime dual ideal
prime dual ideals contained i n
A
, the
intersection
k(A)
of a l l
1 i s prime.
(iii) If x v y = e , there e x i s t elements and x v x = y l v y = e . 1
every proper prime ideal i n
(iv) Every minimal prime ideal i n X maximal ideal.
xl,y,
such that
x1 " y 1 = e
i s contained i n a unique
PROOF. Note that (i), (ii), (iii) correspond with (i), (ii), (iii) the main theorem for normal lattices, and (iv) correspond with (vi).
in
72
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
Ch. 11,§821
C e r t a i n p r o p e r t i e s of a t o p o l o g i c a l space a r e r e f l e c t e d i n t h e p r o p e r t i e s of t h e l a t t i c e of i t s c l o s e d s u b s e t s . We r e c a l l t h a t t h e topolog i c a l space
i s c a l l e d normal whenever f o r every p a i r
E
c l o s e d s u b s e t s t h e r e e x i s t d i s j o i n t open s e t s and
of d i s j o i n t
F1 = O1 t h e s p a c e i s a l s o r e q u i r e d t o b e Hausdorff, b u t we
. Sometimes
F2 c O2
Fl,F2
01,02 such t h a t
s h a l l n o t r e q u i r e t h i s h e r e . A l s o , t h e space
is c a l l e d e x t r e m a l l y d i s -
E
connected whenever t h e c l o s u r e of every open s u b s e t i s open. It i s e a s i l y seen t h a t t h i s i s e q u i v a l e n t t o t h e c o n d i t i o n t h a t f o r every p a i r of d i s j o i n t open s u b s e t s t h e r e e x i s t d i s j o i n t c l o s e d s e t s that
O 1 c F1
and
and t h e open s e t
-
closure
so
of
O1
-
. Indeed,
O2 c F2
i s given, t a k e
O1
O I 'O2 such
i f the l a s t condition is s a t i s f i e d t o be t h e complement of t h e
O2
O1 ; i t f o l l o w s t h a t
i s c l o s e d a s w e l l a s open, and
O2
i s open a s w e l l a s c l o s e d .
O1
( i ) The topological space
THEOREM 82.10.
the l a t t i c e
i f the l a t t i c e
E
O1 fl 0 =
2
b e normal, i . e . ,
E
,
F1 fl F2 = @
0
i s extremally disconnected i f and only
o f i t s closed subsets i s extremally disconnected.
X
PROOF. ( i ) Let satisfying
and
F1 c O1
(i.e.,
F1 ). S i m i l a r l y
F;
F1 VF2 01,02 such t h a t
as well as
. Now,
F2 c O2 and
O1
taking f o r
,
F; fl F2 = @
i.e.,
F; U F' = E
2
. Also,
F; E
. Hence,
Fi
and
r e s p e c t i v e l y , we have
O2
i s a member of t h e d i s j o i n t complement
a r e d i s j o i n t , w e have t h e space
given t h e c l o s e d s e t s
t h e r e e x i s t open s e t s
t h e s e t t h e o r e t i c complements of
F; fl F1 = @
i s normal i f and only i f
E
of i t s closed subsets i s normal.
X
( i i ) The topological space
F;
Fl,F2
since
{ F I I d of O1
and
O2
t h e normality condition f o r
can be expressed e q u i v a l e n t l y by s a y i n g t h a t f o r any p a i r
E
FI,F2
i n the l a t t i c e
Fi ,Fh
in
X
X such t h a t
satisfying
F
1
fl F
, F;E
Fi E [F,}d
X = [F,}
equivalent t o the condition t h a t
=
0
t h e r e e x i s t elements F; U F; = E
and
d
v {F21d
, which
. This
is
i s exsctly
c o n d i t i o n ( i i i ) i n t h e main theorem f o r n o r m a l i t y of a l a t t i c e . Hence, t h e topological space
i s normal i f and only i f t h e l a t t i c e
E
( i i ) The l a t t i c e
X
t h e preceding theorem i f and only i f f o r any p a i r F
1
U F
2
= E
t h e r e e x i s t elemehts
= F; U F2 = E
and
complements of
F,
F; f l F; =
and
i s normal.
X
i s e x t r e m a l l y d i s c o n n e c t e d by c o n d i t i o n ( i i i ) i n
FP
0
F;,F;
. It
in
X
FI,F2 such t h a t
in
X
satisfying
F; U F1 =
i s e a s i l y s e e n by t a k i n g s e t t h e o r e t i c
t h a t t h i s is e q u i v a l e n t t o t h e c o n d i t i o n t h a t
f o r any p a i r of d i s j o i n t open sets
01,02
in
E
there exist disjoint
PRIME IDEAL EXTENSION
Ch. 11,5821
closed sets F;,F;
such that
O1 c F ;
and
73
O2 c F ;
. By
one of the
remarks above, the last condition is equivalent to the condition that E extremally disconnected. Hence, the topological space E connected if and only if the lattice X
is
is extremally dis-
is extremally disconnected.
Before introducing more general concepts, we present the following simple theorem. THEOREM 82.11. (i) The d i s t r i b u t i v e l a t t i c e
only i f every i d e a l in X
i s normal i f and
(X,e)
i s normal, and a l s o i f and only i f every
i s normal.
principal i d e a l i n X
(ii) The d i s t r i b u t i v e l a t t i c e
(X,e,e)
and only i f every s u b l a t t i c e of the form
i s extrernally disconnected i f is eztremally d i s -
(x:xo<x<e)
connected. PROOF. (i) Let x,y E I with x
I
A
y
I be an ideal in the normal lattice (X,O) , and let d 0 Then X = {XI V {yId since X is normal, so
.
=
.
v
=
This shows that
I is normal. It is evident that if every ideal is normal,
then every principal ideal is normal. Now assume that every principal ideal is normal. We show that
X
We have to prove that every xl
A
xo
=
y1
A
yo
principal ideal (ii) (X,e,e)
is normal. Let xo,yo E X
with
E X is of the form
x
z
. This is seen to hold by . 0 0
= €I
z =
x
A
y
0 0 v y 1 with
= 8
.
considering the normal
(x:@<x<x vy vz)
is extremally disconnected if and only if
X
is normal
with respect to the dual ordering, i.e., if and only if every sublattice (x:xolx<e)
i s a normal principal ideal with respect to the dual ordering.
This is equivalent to the condition that every
(x:x <x<e) 0-
-
is extremally
disconnected with respect to the original ordering. In Chapter 5 (cf. Theorem 33.3(ii)) L
it was proved that in a Riesz space
every ideal including a given prime ideal is prime, and the family of all
(Prime) ideals including a given prime ideal is linearly ordered by inclusion. These properties are immediately transferred to the distributive lattice X ( L )
of all principal ideals in L
. It is not
every distributive lattice X with smallest element 8
true, however, that has these properties.
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
74
({e}
is a prime
is not prime; furthermore, {€I}
is contained
The simple lattice in Exercise 7 6 . 1 4 is a counter example ideal, but the ideal
such that neither w1 = w2 nor w2 = 1 It is of interest, therefore, to investigate what distributive
in prime ideals w 1 holds).
and
lattices share with (X,0)
if w
{e,xo}
Ch. 11,5821
in X
w2
X(L)
the above mentioned properties. Note already thaf
has these properties, then it follows that for each prime ideal and each x
satisfying w 1
2
.
w
not in w
there exists a unique x-maximal ideal w 1 (X,e) has a largest element e ,
In particular, if
then each proper prime ideal is contained in a unique maximal ideal, i.e., X
is extremally disconnected. The converse is not true; an extremally
disconnected lattice
(X,e,e)
does not necessarily have the abovementioned
properties, as can be shown by a simple example (cf. Exercise 82.20).
We
first prove a simple lemma. LEMMA 82.12. For any
i d e a l generated by of
(X,e)
A
maximal i d e a l in
A
.
PROOF. (i) Let J
in
w
(since x
is not in J
includes w n Ax
=
n Ax
, and
J
uniqueness of the extension from Ax maximal). It follows that Ax c w ' w'
is x-maximal. Hence, w (ii) Let w
in Ax w ll Ax
,
,
), s o if
so
is not in the is not x-maximal,
properly including w
. But
x E w'
,
n Ax
(since J is x which is impossible because w'
=
A
is x-maximal.
be x-maximal in X
. If the ideal
there exists a maximal ideal J
. By
n Ax is a
J is prime and
x
w
w
this inclusion is proper (by the
to X ), s o so
. Then
. Evidently
to X = w
there exists an x-maximal (prime) ideal w ' then w ' fl Ax
holds now.
, the intersection
X
be a maximal ideal in Ax
has a unique prime extension w
ideal w
denote t h e principal
i n Ax , t h e (unique) prime extension
J
(ti) For any x-maximal ideal
J
Ax
. The following
i s x-maximal.
to X
J
, let
= (y:e
For any maximal i d e a l
(i) w
in
x
, i.e.,
x
in Ax
part (i) the prime extension w '
of
w
n Ax is not maximal
properly including
J
is x-maximal, and
w'
properly includes w (this follows from the explicit formula for the extension in Theorem 51.8(ii);
the formula is cited at the very beginning of
the present section). This is impossible, since w w fl
Ax
is maximal in
Ax
-
is x-maximal. Hence,
Ch. 11,5821
75
PRIME IDEAL EXTENSION
COROLLARY 82.13.
The following uniqueness conditions are equivalent.
(i). Given any prime ideal
i n X and any element
w
there e x i s t s a unique x-maximal prime ideal (ii) For any
x E X and for any proper prime ideal
not i n
x
such that
w1
w1
P
3
w
w
.
,
i n the
principal ideal Ax there e x i s t s a unique mmimal ideal J i n Ax such that J 3 P I n other words, every principal ideal A, i s extremally
.
disconnected.
Ax
.
PROOF. (i)
Given x E X
(ii).
=s
In the lattice A
(with 8'
, let P
be a proper prime ideal in
as smallest and x
as largest element)
there exists at least one x-maximal ideal (i.e., maximal ideal) containing
P
. Let
J1
J 2 be ideals of this kind and let
and
. By
unique prime extensions to X
and w2
w1
the lemma above w 1
and
w2
be their are
x-maximal, so (since they both contain the unique prime extension of it follows from (i) that w1 (ii)
, and hence J 1
= w2
* (i). Similarly.
= J2
.
P )
We now generalize the notion of an extremally disconnected lattice. The condition that the lattice must have a largest element can be dropped. DEFINITION 82.14. The d i s t r i b u t i v e l a t t i c e
(X,e)
i s said t o be
completely extremally disconnected whenever every p2.incipal ideal i n X i s extremally disconnected. In view of Theorem 82.9 the lattice
(X,e)
x,y E X there exist elements v y = x v y I = x v y (note that,
disconnected if and only if for every pair X,,yI E X
such that x
A
and
yI = 8
is completely extremally
x
1 I Compared to the formulas in Theorem 82.9, we have interchanged x1 and y 1 1.
It follows that x 1 < x XI = XI A
=
Similarly y 1
(XVY,)
5
Y
A
y
5
= (X
(x1 Ax) v 0 = x 1
x
AX)
1
. Indeed, x 1 5 x I v V
(X
, and
so
y = x v y 1 implies
Ay ) =
1
1
XI
< x .
.
The folZoLling conditions f o r the d i s t r i b u t i v e l a t t i c e are equivalent.
THEOREM 82.15. (X,e)
and y I
NORMAL AND EXTRFNALLY DISCONNECTED LATTICES
76
Ch. 11,5823
i s completely extremally disconnected.
(i)
X
(ii)
Any proper ideal i n X
containing a prime ideal i s prime i t s e l f .
(iii) The f a m i l y of a l l proper prime ideals containing a given proper
prime ideal closure
i s linearly ordered by inclusion. I n other words, the
w
o f the s e t
iw)-
i n fi i s a linearly ordered s e t .
{w}
(iv) The f a m i l y of a l l prime dual ideals contained i n a given prime dual ideal i s linearly ordered by inclusion. If
(v)
and
w1
hull-kernel topology of
fi
.
and
w,
have d i s j o i n t neighborhoods i n the
w2
For any proper prime ideal
(vi)
and f o r any
w
e x i s t s a unique x-maximal ideal containing PROOF. (i) prime ideal w
Hence, let x A y E I =
x v y1 = x v y
. Since
X
so
x E I
(ii)
such that x 1 A y 1 = 0
x ) and x 1 v y
S
*
(iii). Let w 1
and
=
x v y
, we
.
have
be proper prime ideals containing the
w2
i s included in the other.
is included in the ideal w 1 n w 2 , s o
hypothesis (ii).
w1,w2 such that x 1 is in w l
. Then
but not in w l
is in w2
prime) we have
x1 E w 1 fl w2
definition of x I and containing w
w1
Also, since neither of
there exist x 1 , x 2 E X
x2
x
1
A x2
w2
is prime by
is included in the other,
E wI n
or x2 E w 1 f l w2
n
but not in w2 w2
,
so
and x2
(since w 1
n
. This contradicts the
19
is
. Hence the family of all proper prime ideals
(iii)
*
(iv).
These are equivalent by duality (cf. Theorem 76.1).
(v). Let w 1
and
w2
be proper prime ideals, neither of them
included in the other. We have to prove the existence of elements x in w ) and
w2
is linearly ordered by inclusion.
(iii)
1
=
. This is the desired result.
prime ideal w , and assume that neither of w Then w
and x 1 v y
x 1 A y 1 = 0 it follows that x 1 E w or y 1 E w , x 1 E w c I and x A y E I , so x 1 V (xhy) E I Hence,
x1 v x = x (since x 1
observing that
containing the
is completely extremally disconnected by
. From
. Then
w
there
w
to show that x A y E I implies x E I or y E I.
hypothesis, there exist xI,yI E X say x 1
not i n
x
.
w
I be a proper ideai in X
(ii). Let
. We have
, neither of them
are proper prime ideals i n X
w2
included i n the other, then
y (not in w ) such that 2
{ w ) ~
and
(not
are disjoint. Assume
Ch. 11,5821
PRIME IDEAL EXTENSION
that such x not in w I
and
y
do not exist,
and all y not in w2
{wIx n { w }
non-empty for all x Y Then, by Theorem 77.5, the family of
so
.
77
{wIX n {wIy has a non-empty intersection (note that in the
all these
present application of Theorem 77.5 the family T is the union of all x not in w l the intersection of all these
n ( ~ w ~ ~ not : x in
w
and all y not in w2 ). Let
{wIX ll { w }
1
Y
. Now, since
and
0
be in
wo
)
is the set of all prime ideals contained in w 1 it follows that w c w 1
is empty and the set S
,
wo c w2
so
, and and
w1
similarly for w2
of the family of all proper prime ideals containing wo
w2
’
are both members
. By hypothesis
(iii)
this family is linearly ordered by inclusion. This contradicts the assumption that neither of w2
ideal w
,
w
and w2
w1
are proper prime ideals containing the prime
then either w 1 c w2
imply that w1 2
and
(vi). We first go back from (v) to (iii), i.e., asauming (v) we
prove that if
and
w2
w1
and
w4
(vi)
*
Indeed, if not, (v) would
w
w
(cf. Theorem 7.6 (i)),
. Similarly, every neighborhood of
contains w
or
be a prime ideal, let x be not in w
w cw3
4
,
so
implies (iii). We prove now
be x-maximal ideals containing w
w3 c w4
property that w3 = w4
.
is in the closure of
that (iii) implies (vi). Let we have either
w2 c w 1
. Contradiction. Hence, (v)
contains w
and let w3
or
have disjoint neighborhoods. On the other hand
implies that w1
every neighborhood of w2
.
have disjoint neighborhoods in the hull-kernel topology of (v)
w1
is included in the other. Hence, w 1
w1,w2
,
. By hypothesis (iii)
and so it follows from the x-maximal
. Hence, the x-maximal ideal containing
w
is unique.
(i). This is exactly the statement that part (i) implies part (ij
in Corollary 82.13. In Exercise 8.9 the above theorem was partly proved. In fact, it was Proved that (v) implies (iii) and (vi). We mention some examples. Any Boolean ring X disconnected. Indeed, every proper prime ideal in X
is completely extremally is maximal, and so
condition (ii) is trivally satisfied (i.e., any proper ideal containing a Prime ideal is prime itself). Furthermore, as already noted earlier, the lattice X(L)
of all principal ideals in a Riesz space L
is completely
extremally disconnected (any proper ideal containing a prime ideal is prime
78
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
itself by Theorem 33.3(ii)). interval [O,l]
Ch. 11,1821
The lattice of all closed subsets of the unit
(with its ordinary topology) is not extremally dis-
connected, simply because
CO,l]
is not an extremally disconnected topo-
logical space. For an example of an extremally disconnected lattice that is not completely extremally disconnected we refer to Exercise 82.20. The dual notion of a completely extremally disconnected lattice is a completely normal lattice. We present the definition. (X,e)
DEFINITION 82.16. The d i s t r i b u t i v e l a t t i c e
i s said t o be
completely normal whenever every sublattice of the form
(x:xo<x) i s
norma I . By Theorem 82.11 the sublattice (x:x <x) 0-
every principal ideal in sublattice of the form saying that
(X,e)
is normal if and only if
(x:x <x) is normal, i.e., if and only if every 0(x:xo<x<x ) is normal. This is equivalent to 1
is completely normal if and only if every sublattice
(x:x <x<x ) is extremally disconnected with respect to the dual ordering. 0- - 1 This observation is important for the proof of the following main theorem about completely normal lattices. THEOREM 82.17. The following conditions f o r the d i s t r i b u t i v e l a t t i c e
(X,6> 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
.
PRIME IDEAL EXTENSION
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
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
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.
PRIME IDEAL EXTENSION
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
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
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
PRIME IDEAL EXTENSION
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
.
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
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
PRIME IDEAL EXTENSION
{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
,
PRIME IDEAL EXTENSION
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
. Similar formulas hold formulas; for T t 8 =
,
8
L
f E L
0
sup Sf:B<S
/TI = T++T-
.
arbitrary, we have
,
and similar formulas hold for Tfin the last formula.
for T- and for and
and
T(lf1)
. The supremum is attained
90
Ch. 121
CHAPTER 12
defined on a Riesz
Under certain conditions positive operators T subspace K
of
can be extended to positive operators defined on
L
K
for example if the ideal generated by
is majorized by a sublinear operator p (p(f)=p(lfl))
as well as monotone
is the whole space L such that p
(O
M ) . As an application we prove that if Dedekind complete space M , then M
L
in L
,
L
o r if
T
is absolute
implies p(f)
5
p(g)
in
is a Riesz subspace of the
contains the Dedekind completion L-
L (A.S. Veksler, 1969).
of
In many cases (for example, if subset of
L
L
is Archimedean) any order closed
is relatively uniformly closed (ru-closed).
This holds in
particular for ideals. The converse is not true in general; it is rather special if a Riesz space L has the property that every ru-closed ideal in L
is order closed. The space L
Riesz homomorphism from L
has this property if and only if every
into an arbitrary Archimedean Riesz space is a
a-homomorphism (D.H. Fremlin,1977). The relation with norm closed ideals in normed Riesz spaces will be discussed in section 105.
In section 8 4 we introduce order continuous and a-order continuous operators as subsets of continuous if u
J.
Lb
. The operator
T E
Lb is called order
inf(TuTl = 0 holds for any downwards directed system
0 in L and T
is called o-order continuous (sequentially order
continuous) if this property holds for sequences (i.e., u infITu I = 0 ). It will be proved that T only if order convergence of Tfn
to
Tf
in M
fn
to
f
J.
0 implies
is a-order continuous if and
in L
implies order convergence of
. This explains the name. The set of all order continuous
operators will be denoted by
L
(abbreviation for
Ln(L,M)
) and the set
Lc . Evidently Ln is a subset of Lc and Ln are bands in . The disjoint complement of Lc with respect to Lb is denoted by Ls ; the elements of Ls are called singular operators. The disjoint complement of Ln with respect to Lb is denoted by Lsn ; the elements of Lsn are called normal of all 0-order continuous operators by
Lc
. It will be proved
that
singular operators. We have
L If L
b
= L e l c s
= L
n
.
is Archimedean and if order convergence and relatively uniform con-
vergence for sequences in L
-.
for example, if and
e L s n
0 < p <
are equivalent, then
Lc
=
Lb
. This holds,
L ( X , A , u ) , where ( X , A , p ) is o-finite measure space P It does n o t hold for p = m (unless in trivial exceptional
L
=
Ch. 121
ORDER BOUNDED OPERATORS
cases). If L
is order separable, then
L
91
Ln , i.e., every o-order
=
continuous operator is order continuous. In particular, order separability of
L implies that every o-order continuous linear functional on L
order continuous. This holds, for example, if L M(r)(X,A,p)
is
is any ideal in the space
of all real u-measurable functions on X , where again
(X,A,p) is a o-finite measure space. In section 85 we investigate the order dual L" happen that L
is of infinite dimension, but L"
functional. If
(X,A,p)
more closely. It can
contains only the null
L
is a o-finite measure space without atoms and
or L (X,A,p) for 0 < p < 1 , then L" consists P of the null functional only. It can also happen that every linear is either M(r)(X,A,p) functional on
L
L"
is a member of
. This occurs if and only if
Archimedean and every principal ideal in L
(equivalently, by Theorem 61.4, if and only if and every quotient space of
L
L
is
is of finite dimension
L is uniformly complete
is Archimedean). It was proved some years
ago that another equivalent condition is that every ideal in L
is a
projection band (C.B. Huijsmans,1976). In the general case, when it is usually not known whether one of these extreme cases occurs, there is a condition to see if a linear functional on L belongs to L" on L
condition says that the linear functional Q
and only if Q(f ) tends to zero for any sequence n that converges relatively uniformly to zero. The existence.of non zero elements in L-
(fn:n=1,2,
. It is evident that if
the null functional, then p(f)
lQl(lfl)
if
in L
@
E L"
is not
is a non zero Riesz seminorm
. Conversely, if there exists a non zero Riesz seminorm in
L , then contains a non zero element. The proof is based on the Hahn-Banach
in L L"
...)
is related to the existence
of non zero Riesz seminorms in L =
or not. The
satisfies Q E L"
extension theorem. A s a particular case of a general extension theorem for operators in section 8 3 , we get the result that any positive linear functional on a Riesz subspace of L seminorm in L
which is majorized by a Riesz
can be extended to a positive linear functional on
L
. As
an application it will be shown that there exist non zero singular linear
.
functionals on the sequence space Lm If L is a normed Riesz space (i.e., L norm), then the Banach dual L*
of
L
is equipped with a Riesz
is an ideal in the order dual L"
h y normed Riesz space which is norm complete is called a Banach lattice. In a Banach lattice L we have L* = L"
.
.
Ch. 1 2 1
CHAPTER 12
92
Any a-order continuous (order continuous) linear functional on
. The
be called an integral (normal integral) on L
set L:
. In section 86 we
L
will
of all
integrals is a band in L ; the same is true for the set :L integrals on
L
of all normal
investigate integrals on L
, where L
is an ideal in the space M(I) (X,A,p) such that, once more, (X,A,p) is a a-finite measure space. Note that L is order separable, so a-order continuity is the same as order continuity. In particular, every integral on
L
is a normal integral. The subset E
of
X
is called an L-zero
f E L vanishes p-almost everywhere on
set if every
largest L-zero set Em
E
. There exists a
(uniquely modulo sets of measure zero); the set
theoretic difference X-E, is called the carrier of the ideal L integral $
on L
the word. Precisely, for any integral $ real function t
for all
f E L
on
X
on
L
there exists a u-measurable
such that
. The function
-
is unique on the carrier of
t
L (X,A,p) for some p P well-known classical result.
L
case that
. Any
can be written as an integral in the usual sense of
I
satisfying
=
p <
5
L
, we
. In
the
get a
Section 87 is devoted mainly to examples. We first introduce some N
further terminology. The disjoint complements of respect to
Li
L" ) are denoted by
Lc
L i (with
and
and Lin ; the members of
L"
and Lin
are called singular linear functionals and normal singular linear functionals respectively. The point set X
is said to have a measurable cardinal if
there exists a a-additive (non-negative) measure X all subsets of
X
0 < X(X)
such that
or countable subsets E
of
X
.
< =
and
on the collection of
A(E) = 0 for all finite
In all other cases
X
is said to have
a non-measurable cardinal. Not much is known about the existence of sets with a measurable cardinal. There is a conjecture that every set has a nonmeasurable cardinal. It has been proved that if the continuum hypothesis holds, then the set of all real numbers has a non-measurable cardinal. In the first example of section 87 we let L be the space of all real functions on a non-empty point set X
. Then
L" = L z
.
is always true;
has a non-measurable cardinal, then L" = L" = L" In the second c n example we specialize X to be the set of all real numbers and we prove if X
that now
L" = :L
= Lz
, without
again a non-empty point set and
using the continuum hypothesis. If X L
is the space of all bounded real
is
Ch. 121
93
ORDER BOUNDED OPERATORS
functions on X , then L" = L" holds if and only if X has a nonc n measurable cardinal. Furthermore, if L is the space of all real continuous functions on the closed interval If
L
(x:OSx
is super Dedekind complete, then L
.
L-
.
is order separable, and
L" = L" If L is Dedekind o-complete, then L" = L" does not c n c n necessarily hold. It is not known if Dedekind completeness of L implies so
.
If it is true, however, that every set has a non-measurable L" = L" c n cardinal, then L" = L i holds for every Dedekind complete space L (W.A.J. Luxemburg,l967). In section 88 we assume L,M,Lb and its subspaces to be the same as before (in particular, M is still assumed to be Dedekind complete). For any subset A
of
L
the (L,M)-annihilator
A '
is defined by
TELb:Tf=O for all fEA and for any subset
B
of
Lb
the inverse (L,M)-annihilator
'B
is
defined by fEL:Tf=O for all TEB is real number space, then
If M
Lb
=
L"
,
and
so
for A c L
and
B c L"
we get the familiar definitions for annihilator and inverse annihilator. is an ideal in L , then A'
If A
is a band in
Lb
. If
B
is an ideal
Lb , then OB is an ideal in L . We also introduce the ideal La , consisting of all f E L such that If/ 2 un J 0 implies Tun .I. for every positive T in Lb . Obviously, La is the largest ideal in L in
in L
0
on which every order bounded operator is o-order continuous. Similarly, we define Lan as the largest ideal in L
on which every order bounded
and operator is order continuous. It will be proved that La = '(Ls) a 0 = O(Lsn) In the converse direction, (L ) = Ls holds if La is an 0 super order dense in L and (L ) = Lsn holds if Lan is order dense in L
.
Lan
.
In section 89 it is assumed that W and of
W' W
is a real or complex vector space
a linear subspace of the algebraic dual of W the annihilator A'
of
A
is defined by
. For any subset
A
CHAPTER 12
94
Ao
Ch. 1 2 1
($EW':+(f)=O f o r a l l fEA)
=
and f o r any s u b s e t
of
B
W'
the inverse annihilator
O B
of
is
B
d e f i n e d by 'B
(fEW:$(f)=O
=
It i s obvious t h a t O(W')
= {O}
,
.
= I01 ; we assume t h a t
W'
i.e.,
for a l l ~ E B )
s e p a r a t e s t h e p o i n t s of
W'
i s so large that
W'
. This
W
implies t h a t
may b e regarded a s a l i n e a r subspace of t h e a l g e b r a i c d u a l of and
W
a r e i n d u a l i t y , a s i t i s sometimes c a l l e d ) . The weak topology
W'
a(W,W')
is by d e f i n i t i o n t h e weakest topology i n
a r e continuous f u n c t i o n s on
W
. It
such t h a t a l l
W
i s proved t h a t
W
c l o s u r e of any l i n e a r subspace a(W',W) topology i n
Banach d u a l topology
W*
, we
a(w*,tl)
W'
is
A
. If
'(Ao)
in
w*
. Similar
a(W,W')
r e s u l t s hold f o r
i s normed and we choose f o r
W
g e t t h e weak topology
.
We r e t u r n t o t h e s i t u a t i o n t h a t
@ E W'
w i t h t h e topology
i s a Hausdorff t o p o l o g i c a l v e c t o r space and t h a t t h e
a(W,W') the
W
(so
W'
a(W,W*) i n
W
4 E L
i s a Riesz space. For any
L
the
W'
and t h e weak s t a r
the ideal
i s c a l l e d t h e n u l l i d e a l ( o r a b s o l u t e k e r n e l ) of complement
4
and t h e d i s j o i n t
i s c a l l e d t h e c a r r i e r of
C($) =
4 ( n o t t o be confused
w i t h t h e c a r r i e r of an i d e a l of measurable f u n c t i o n s ) . I n s e c t i o n 90 it
i s proved t h a t f o r p o s i t i v e
6
and
The formula
C($A@) = C ( 4 ) flC($)
a t l e a s t of
4
and
(I
J,
in
L" we have
does n o t h o l d g e n e r a l l y . It h o l d s i f one
i s a member of
p r o p e r t y . As observed a l r e a d y , N($)
Lcf
or also i f
i s an i d e a l . I f
is a band; t h e converse does n o t always hold. I f or
L
h a s t h e p r o j e c t i o n p r o p e r t y and i f
if and o n l y i f
N($)
i s a band i n
L
. In
4 E Lz
L
,
t h i s case
L
h a s t h e Egoroff
E L i
, then
N(+)
i s Dedekind a-complete t h e n we have
$
L = N($)
C($)
@
E L :
.
Ch.
12,9831
ORDER BOUNDED OPERATORS
I n s e c t i o n 91 it is proved f i r s t t h a t i f
95
i s Archimedean and
L
uniformly complete, t h e n i t i s p o s s i b l e to d e f i n e i n t h e c o m p l e x i f i c a t i o n
L+iL of
an a b s o l u t e v a l u e
L
I f+igl Equivalently, i f v a l u e of
h
=
If+igi
by
).
sup[f
cose+g sine:O<e<2a
h = f+ig
,
1h.l = supe Re(he-ie)
then
c o i n c i d e s w i t h t h e e x i s t i n g one i f / h l + h 2 / = Ihl I + I h 2 /
t r i a n g l e i n e q u a l i t y and
. The
absolute
i s real, s a t i s f i e s the
h if
Ihl
I I Ih21. Freuden-
t h a l ' s s p e c t r a l theorem can b e extended t o complex Riesz spaces. I f a s u b s e t of
L+iL
Dd =
, the
(hEL+iL: l h l l l f /
i s a band ( i . e . ,
f o r a l l fED
t h e r e a l p a r t i s a band i n
p r o p e r t i e s a s i n t h e r e a l case. F i n a l l y , i f ideal i n
L
, the
h
quotient
h € L + iL
and f o r every
is
D
d i s j o i n t complement
(L+iL)/(A + i A )
r
r
L )
, and
has s i m i l a r
Dd
i s a uniformly c l o s e d Ar admits an a b s o l u t e v a l u e ,
t h e a b s o l u t e v a l u e of t h e equivalence c l a s s of
Ihl
is e x a c t l y t h e e q u i v a l e n c e c l a s s of
.
I n s e c t i o n 9 2 , f i n a l l y , we i n v e s t i g a t e t h e s i t u a t i o n t h a t t h e r e i s M
, which
L+iL
into
a n o t h e r Riesz space operator and
T2
T
from
i s of t h e form
r e a l o r d e r bounded o p e r a t o r s
M ) . This i m p l i e s t h a t
Lb(L+iL,M+iM)
We prove t h a t f o r any complex
IT1
i s Dedekind complete, Every o r d e r bounded M+iM
(i.e.,
TI
and
T = Tl+iT2 T2
map
with
TI
into
L
is t h e c o m p l e x i f i c a t i o n of
Lb(L,M)
.
i n t h e space t h e a b s o l u t e v a l u e
T
satisfies IT/ (u) = sup(lTh1 :h€L+iL, IhlSu)
,
f o r every
u E L+ ( W . J .
Dedekind a-complete,
de Schipper; H.H.
S c h a e f e r ; 1973-1974).
If
L
is
t h e proof can be s i m p l i f i e d (Exercise 92.7).
83. Order bounded o p e r a t o r s We r e c a l l t h a t t h e mapping r e a l v e c t o r space if
W
T
of t h e r e a l v e c t o r space
V
into the
is c a l l e d a linear operator ( o r zinear transformation)
96
Ch. 12,1831
ORDER BOUNDED OPERATORS
0
T af+bg = a Tf + b Tg blds for all f
and
g
in V
and all real numbers
a,b
. We
shall
briefly say operator instead of linear operator. Defining addition of operators (from V
into W ) and multiplication of an operator (from V
into W ) by a real constant by (TI+T2)f = Tlf+T2f ; (aT)f fE V
for all L(V,W)
and all real numbers
of all operators from
V
a Tf
=
,
a
it is evident that the set
into W
is a raal vector space. In the
case that W
is the space of real numbers, we get in this manner the
vector space
v#
of all real linear functionals on
called the (real) algebraic dual of V Let L
and PI
Definition 1 1 . 1 ) ; usual, by
L+
and M+ f
5 g
v
. The space
v#
is
be real ordered vector spaces (as defined in
the positive cones of L
and M will be denoted, as
respectively. The operator T for all u E L+
called positive if Tu E M" and only if
.
in L
implies Tf
5
Tg
from L
. Evidently, T
in M
into M
is
is positive if
. Note that any finite
linear combination with non-negative coefficients of positive operators is a positive operator. In the following we assume about L (i) L
generating (i.e-, every with
and M
that
is an ordered vector space such that the positive cone L+ f E L has at least one decomposition f
=
is
u-v
1,
u,v E L+
(ii) L has the dominated decomposition property (i.e., if u,v,w E L+ satisfy 0 < u u, s v
, u2
(iii) M
< w
5
v+w
, then
there exist u 1 and
u in L+ 2
such that
and u,+u2 = u ) ,
is a Dedekind complete Riesz space.
Observe that by Corollary 15.6 every Riesz space satisfies the conditions (i), (ii). For the equivalence of the dominated decomposition property and the Riesz interpolation property we refer to Exercise 15.12 and Exercise 7 6 . 1 8 . Some authors call any ordered vector space with a generating cone and possessing the dominated decomposition property already a Riesz space (cf. for example L. Fuchs,Cll),
but we shall adhere to the conventions established in the earlier chapters that a Riesz space is the
12,5831
Ch.
ORDER BOUNDED OPERATORS
97
t h e same a s a v e c t o r l a t t i c e . For examples of spaces s a t i s f y i n g ( i ) and ( i i ) w i t h o u t b e i n g a Riesz space we r e f e r t o E x e r c i s e 15.14 (where t h e s p a c e of a l l with
p
and
f ( x ) = p(x)/q(x)
r e a l polynomials and
q
t o E x e r c i s e 76.16 f u n c t i o n s on (Ti;)
on t h e closed r e a l i n t e r v a l
L
(where
q(x) > 0
for a l l
L
is
Ca,bl
x E Ca,bl ) and
i s t h e space of a l l r e a l d i f f e r e n t i a b l e
Ca,bl ) . Returning t o t h e c a s e t h a t c o n d i t i o n s (i),
(ii),
h o l d , t h e r e a d e r i s advised t o keep i n mind t h e important s p e c i a l
case t h a t
i s a Rhesz space and
L
M
i s t h e space of a l l r e a l numbers.
F i r s t of a l l we s h a l l prove a u s e f u l e x t e n s i o n lemma. LEMMA 83.1. Let T(U) t 0
be an additive mapping from
T
u E L+
for a22
T(U+V)= T ( U ) +
and
that
T
i s an extension o f
T
,i.e.,
PROOF. We prove f i r s t t h a t a t 0
and every r e a l a d d i t i v i t y of and
r'
T
. If
such t h a t
r u 5 au 5 r ' u
0
. For a
5
i n t o M+
a
, i.e.,
i n t o M such
.
u E L+
holds f o r every
u E L+
t h i s follows e a s i l y from t h e
i s p o s i t i v e and i r r a t i o n a l , choose r a t i o n a l
r < a < r'
.
u,v E L+
Tu = T ( U ) holds for a22
-r(au) = a-r(u)
rational
from L
T
Then there e x i s t s a unique p o s i t i v e operator
L+
f o r a22
T(V)
. Given
u E L+
,
r
i t follows from
that
so
r f a
Letting
r'
and
+
n o t e t h a t we u s e h e r e t h a t
a ,we have
M
rT(u)
+
aT(u)
and
i s Archimedean ( s i n c e
M
r ' T ( u ) J. a-r(u) ;
i s Dedekind
complete by h y p o t h e s i s ) . It follows then t h a t
SO
-r(au) = aT(u)
.
Now observe t h a t i f
.
u y v , u I , v I E L+
sacisfy
T(U)-T(V) = T ( u ~ ) - T ( v ~ ) For t h e p r o o f , n o t e t h a t T ( U ) + T ( V) = 1
T(U+V
1
) = T ( u ~ + v )=
T(U
1
)+T(v)
, from
u-v = ul-vl
,
u+vl = u I + v
then
, so
which t h e d e s i r e d
98
Ch. 12,8831
ORDER BOUNDED OPERATORS
c o n c l u s i o n f o l l o w s , Hence, Since every f = u-v
position
with
, then
Tf = T(u)-T(v)
decomposition of Tu =
. Note
f
f,g E L
f
i s d e f i n e d now on
f = u-v
and
g = u -v
I
T(au) = aT(u)
that
T ( a f ) = aTf
there exist
, for
T
, i.e.,
L
u,v,uI,vI
T(f+g) = Tf+Tg
L+
in
and
L
such t h a t
I Y S 0
all
=
.
Tf + Tg
and a l l
u E L+
holds f o r a l l
bene proved t h a t
and n o t on t h e p a r t i c u l a r
i s a d d i t i v e on
T
T(u)+T(u,)-T(v)-T(v,) From
T
.
u E L+
. Indeed,
it f o l l o w s t h a t i f we d e f i n e
already that
W e s h a l l prove now t h a t for a l l
h a s a t l e a s t one decom-
depends only on
holds f o r every
T(U)
,
u,v E Lc Tf
f E L
f E L
, it
a 2 0
and a l l r e a l
i s a p o s i t i v e o p e r a t o r from
L
follows e a s i l y
. It
a into
M
has thus which extends
T .
F i n a l l y , f o r t h e proof t h a t o p e r a t o r which extends such t h a t
f = u-v
. Given
T
, and
i s unique, assume t h a t
T
f E L
there exist
i s another
TI
and
u
v
in
L+
hence
T f = T u-T v = T ( u ) - T ( v ) = Tu-Tv = Tf 1 I 1 This shows t h a t
TI = T
, so
We a g a i n assume t h a t p o s i t i v e cone and such t h a t Given of
L
fo
and
go
in
L
i s unique.
T
L
i s an ordered v e c t o r space w i t h a g e n e r a t i n g
L
h a s t h e dominated decomposition p r o p e r t y .
satisfying
o r d e r i n t e r v a l of t h e form
C-u,ul
f o r some
symmetric. I n t h e p a r t i c u l a r c a s e t h a t f
f o 5 go
, the
subset
i s c a l l e d an order i n t e r n a l , and i t i s denoted by
E C-U,~]
i f and only i f
The s u b s e t
D
of
L
If1
<
u
.
L
u E L+
(f:fobf5go)
Cfo,gol
is s a i d t o be
i s a Riesz space we have
i s s a i d t o be bounded (order bounded)
i n c l u d e d i n some o r d e r I n t e r v a l . I n t h i s c a s e o r d e r i n t e r v a l . For t h e p r o o f , assume t h a t
. Any
D
if
D
is
i s included i n a s y m e t r i c
D c Cf , f ' l 0 0
. Since
L+
is
Ch. 12,5831
ORDER BOUNDED OPERATORS
ut-vt for appropriate u1,v' E L+ , so f;, u'+vl . Similarly, -f u-v with u,v E L+ , -f u+v ,
generating, we have f' < 0 -
99
and hence
=
f > -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
ORDER BOUNDED OPERATORS
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
of the elements in the set so
u E L+
is order bounded. For any
i.e.,
Given u 1 and u2
T(U)
(Tv:OSv
, we
in L+
T(U1)+~(u2)
=
. We
E M+
of which
prove that
is the supremum,
T(U)
T
is additive on
):o~vl
sup
In the converse direction, if v E L+ v
.
have
satisfies 0
S
S T(U
in L+ such that 0 < v 1 S u 1 , 0 5 v 2 (by thedominated decomposition property), and s o
By taking the supremum on the left for all v
+u ) 1 2
v 5 u1+u2
exist v 1 and v Vl+V2 =
L+
satisfying 0
,
. there
5
u2
and
S
v
u +u 1 2 ’
S
it follows that
It has been shown thus that
T
is an additive mapping from L+
into M+
so by the extension lemma (Lema
83.1) there exists a positive operator
from L
T
into M
which extends
. Write
T2
=
T1-T
. By
definition of
, T
Ch. 12,5831
ORDER BOUNDED OPERATORS
101
, i.e.,
T u = T ( U ) 2 Tu f o r every u E L+ , s o T u t 0 1 2 p o s i t i v e . Hence T = T1-T2 w i t h T I and T2 p o s i t i v e .
we have
is
T2
A s mentioned b e f o r e , we d e n o t e t h e v e c t o r space of a l l o p e r a t o r s from into
L
by
M
o p e r a t o r s by T
T2
2
1
L(L,M)
whenever
and t h e l i n e a r subspace of a l l o r d e r bounded
. For
Lb(L,M)
TI
and
in
T2
Lb(L,M)
we s h a l l w r i t e
i s p o s i t i v e . E v i d e n t l y , t h i s makes
T -T 2 1
ordered v e c t o r space such t h a t t h e p o s i t i v e cone a l l p o s i t i v e o p e r a t o r s from
L
into
. We
M
Lb(L,M)
an
i s t h e s e t of
Li(L,M)
s h a l l prove now t h a t
Lb(L,M)
i s a c t u a l l y a Dedekind complete Riesz space. THEOF34 83.4. The ordered vector space
i s a Dedekind com-
Lb(L,M)
plete R i e s z space. PROOF. The n u l l o p e r a t o r from be an a r b i t r a r y member of
T
into
L
L b (L,M)
w i l l be denoted by
M
. Let
8
TI
and l e t t h e p o s i t i v e o p e r a t o r
be
d e f i n e d a s i n p a r t ( i i ) of t h e preceding theoaem, s o T u = sup/ Tv:O
f o r every
. Then
u E L+
T u t 0 and T u 2 Tu f o r every u E L+ , s o 1 1 and 8 Let T ' b e a n o t h e r upper bound of T
i s an upper bound of
TI
and
8'
. Then
.
T
T'u t T'v t Tv
(
T'u t sup Tv:O
T,
It f o l l o w s t h a t
T1 = sup(T,B) = T+ i s t h e supremum of
0 5 v
for
1
6 u
,
.
= T u 1
i s t h e l e a s t upper bound of
. Hence, T
for
and
T
and
8
,
i.e.,
E L b ( ~ , M ) , t h e mapping
T,S
in t h e
S
so
space
T+(S-T)+
.
L b ( ~ . ~ i) s a
L b ( ~ , ~ )Thus
Riesz space. For t h e proof t h a t
8
2
T
Lb(L,i) element u2
in
4 5 To
. To
in
.
t h i s end, l e t
~ ( u ) exists i n L+
L b ( ~ , ~ i)s Dedekind complete, assume t h a t
L b ( ~ , ~ )We have t o prove t h a t
t h e systems
M+
T(u)
=
sup,, T,,u
since
{T,,ul)
and
M
f o r every
i s Dedekind {Tou2}
sup T
exists i n
uaE L+
. The
complete. F o r
a r e e q u i d i r e c t e d , so
u1
and
ORDER BOUNDED OPERATORS
102
Ch. 12,5831
.c(uI+u2) = sup( Tu ul +T0 u 2
sup T u +sup T u 0 1
by Theorem 15.8(iii).
= T(u~)+T(u~)
This shows that
T
is an additive mapping from L+
into M+ , and so (by the extension lemma) there exists a positive operator T
from L
u E L+
,
of all Tu
.
into M which extends
the operator T
.
T Since Tu = sup T u for every u u is obviously the desired supremum of the system
COROLLARY 83.5. If L i s an ordered vector space w i t h a generating
p o s i t i v e cone and w i t h the dominated decomposition property, then the order dual
L"
L
of
i s a Dedekind complete Riesz space w i t h the s e t of a l l L as p o s i t i v e cone. I n p a r t i c u l a r , the
p o s i t i v e Zinear f u n c t i o n a l s on order dual
L"
L
o f any Riesz space
i s a Dedekind complete Riesz space.
L (L,M) is a Dedekind complete Riesz space is due to b L.V. Kantorovitch ([1],1936) and to H. Freudenthal (C17,1936,section 13). The corollary about L" is due to F. Riesz (Cll,1928;C21,1937). Together The theorem that
with Freudenthal's spectral theorem this was one of the starting points of Riesz space theory. THEOREM 83.6. (i) Let
any (1)
T E Lb(L,M)
and any
Lb(L,M)
be t h e same as before. For
we have
0 0
T+u = sup Tv:OSvSu
-
and
L,M u E L+
(21
-T u = inf Tv:OSv
(3)
lTul 5 ITI(u)
, ,
.
(ii) In p a r t i c u l a r , l e t and any f E L we have
L
be a Riesz space. For any
T E Lb(L,M)
Ch. 12,5831
103
ORDER BOUNDED OPERATORS
and f o r any
we have
u E L+
(7)
in the proof of TI According to that proof, T 1 is an extension of the mapping
PROOF. (i) T+ Theorem 83.3. T
from Lf
for u E L+
is the operator which was called
into M"
, and
T
is defined by
. Hence, the formula
(1)
follows. For the proof of ( Z ) ,
note
that in view of (1) we have
For ( 3 ) , note that Tu
S T+u 5
ITl(u>
and
For the proof of ( 4 ) , write T3 = sup(T 1 ,T2) hence
,
, SO
5
T-u
S
ITI(U)
so
T3
=
(T1-T2)++T2 , and
-Tu
which is the desired result. The proof of (5) is similar. (ii) L and
If1
is now a Riesz space, so for any
f E L
the elements f+,f-
are well defined, and we have
which proves (6). Finally, for the proof of ( 3 ) , note first that
Ch. 12,9831
ORDER BOUNDED OPERATORS
104
Observe now that
0
i
. Hence
(v-wl 5 u
so
v i u
,0
i
w
2
u implies v-w
i
u and w-v
S
u ,
/TI(u) = sup(T(v-w):O
: 1 f 1 iu)
.
Thus formula (7) follows. We present another application of the extension lemma. THEOREM 83.7.
(i) Let
complete, and l e t
positive operator from
Then
T
L
that
8
I
'T . It
T to
TIA (i.e.,
Of
TA
< T
dA f E A
T(A of
M
into M
be Riesz spaces with L
. Furthermore,
. For every
u E L+
let
Dedekind
M
T be a
we s e t
, so there e x i s t s a L i n t o M which extends T . Note and TAf = Tf for a l l f E A , whereas T f = 0 for A f o l Z m s that TA i s an extension of the r e s t r i c t i o n The eztension TA i s the minimal positive extension T I i s another positive extension of T1A , then
i s an additive mapping from L+ i n t o M+
unique positive operator all
L and
A be an ideal i n
A i f
TA from
.
TI
In particular, i f i s a p o s i t i v e Zinear functionaZ on L , then there e x i s t s a positive Zinear functionaZ $ on L such that $ A ( f ) = $(f) for aZZ
f E A and
$,(f)
= 0
f o r aZZ
f /Ad
. The functionaZ
minimaz p o s i t i v e Linear extension of the r e s t r i c t i o n
.
$IA (ii) With the same notations as above, i f the r e s t r i c t i o n
$A i s the
TIA i s
Ch. 12,§831
105
ORDER BOUNDED OPERATORS
sequentiaZZy order continuous on A (i.e., vn E A f o r n = 1 , 2 , ... and v C 0 implies Tv C 0 i n M 1, then the minimal extension TA i s n n sequentially order continuous on L Similarly, i f T I A i s order continuous on A ( i . e . , Tv C 0 holds i n M for any dotlnwards directed system vT C 0 i n A 1, then TA i s order continuous on L
.
.
~ ( u ), a s defined above, e x i s t s f o r every
PROOF. ( i ) Note f i r s t t h a t u E L+ For
by t h e Dedekind completeness of
uI
and
u
in
2
L+
M
. Hence
maps
T
i n t o 'M
L+
.
we have
T(U ) + T(U2) = I
{
sup T(V 1+V 2) :V 1 'v2 EA,0
v E A
satisfies
2 T ( U 1+U
0
4
0
such t h a t
5
v1 5 u1
Tv = Tv +Tv2
1
,0
S
2
v 2 5 u2
, and
L+
T(U
into
opera t o r TIA
any
.
t . By
. Let v E A
TA
L
into
M
prove t h a t
0
T
v
satisfying
u
5
v 2 u1+u2
T
0 2 v S u
n sup TAun e x i s t s i n
, we
have
, i.e.,
,
which extends
. For
TIA
T u 2 T v = Tv
1
1
any
u E L+
, so
= TAu
i s t h e minimal p o s i t i v e l i n e a r e x t e n s i o n of
TA
TIA S
0
i s an a d d i t i v e mapping from
which extends
TJu 2 sup(Tv:v€A,OSv
( i i ) Let
in
v ] +v 2 = v , s o
be another p o s i t i v e e x t e n s i o n of
satisfying
This shows t h a t
v2
t h e e x t e n s i o n lemma (Leuma 83.1) t h e r e e x i s t s a p o s i t i v e
from
TI
there
.
+u2) 5 T ( u ~ + ) T ( u ~ ) Hence
M
v,
T ( U ~+) T ( u ~ )
By t a k i n g t h e supremum on t h e l e f t f o r a l l we g e t
,
and
v 4 uI+u2
e x i s t (by t h e dominated decomposition p r o p e r t y ) elements A
2) '
be s e q u e n t i a l l y o r d e r continuous on f u
sup TAun
M
5
in
L
implies
and s a t i s f i e s
T u 4 TAu An
. It
A
T/A.
. We have
to
i s evident t h a t
T u A
For t h e proof of t h e converse i n e q u a l i t y , l e t
v E A
satisfy
0 < v
5 u
and
106
ORDER BOUNDED OPERATORS
and l e t
Ch. 12,9831
... . Then
v = inf(v,un) f o r n = l , Z , n Tv 4 Tv Since u 2 v implies T u n n n An
.
sup T u An
2
sup Tv
n
= Tv
2
vn 4 v
TAvn = Tvn
,
in
A
,
so
i t follows t h a t
,
and s o
T u = sup TAun , i . e . , T u 4 TAu A An implies o r d e r c o n t i n u i t y of TA
Hence of
. The
TIA
proof t h a t o r d e r c o n t i n u i t y
is similar.
We s p e c i a l i z e t h e l a s t theorem t o t h e c a s e t h a t
A
is a principal
ideal. THEOREM
83.8.
( i ) Let
.
L
M
and
be Riesz spaces with M Dedekind
complete, and l e t v E L+ Furthermore, l e t For every u E L+ we s e t from L i n t o M
.
inf(u,nv)
Then
T~
1
:n=1,2
T
be a positive operator
1.
,...
i s an additive mapping from L+ i n t o M+ , so there e x i s t s a
.
unique positive operator Tv from L i n t o M which extends T~ Note T and Tvf = Tf f o r aZZ f i n the principal ideal A, that 0 S Tv
E (AJd . The operator Tv i s the rmkirnaZ positive linear extension of the r e s t r i c t i o n TIA, . In particular, i f + i s a positive linear functionaZ on L , then there e x i s t s a positive Zinear functional $, on L such t h a t +,(f) = $ ( f ) f o r aZZ f E A, and +,(f) = 0 f o r aZZ f E (Av)d . The functional bV i s the minimal positive Zinear extension of the r e s t r i c t i o n of t o A, ( i i ) If TIAV i s order continuous (or sequentiazzy order continuous) on A, , then Tv i s order continuous (or sequentialZy order continuous) on L . generated by
+
v
, whereas
Tvf = 0
.
PROOF. Given
u E L+
, we
set
for a l l
f
ORDER BOUNDED OPERATORS
Ch. 12,1831
107
and
.
sup W' = T ~ ( U ) It i s e v i d e n t t h a t
Note t h a t
For t h e converse i n e q u a l i t y , n o t e t h a t i f w
that
and s o
I
u
w
I
and
w
nv
I
. It
inf(u,nv)
SUP
T (U) V
W'
Tw E W
f o r some n a t u r a l number follows t h a t
w'
= SUP
w
sup W
5
C
W
sup W'
,so
5
sup W
i s given, we may assume n
(depending on
. Hence
sup W'
w ),
.
A l l d e s i r e d c o n c l u s i o n s f o l l o w t h e r e f o r e from t h e p r e c e d i n g theorem. Once more, l e t and l e t
T E &(L,M)
h o l d s f o r every
L
and
. Then
M
b e Riesz spaces w i t h
M
Dedekind complete
according t o t h e formulas ( 1 ) and ( 7 ) i n Theorem
u E L+
8 3 . 6 . We s h a l l prove now a p a i r of d u a l formulas. For t h e f i r s t of t h e s e formulas and f o r some of t h e f a c t s about t h e o p e r a t o r tional
T
V
(and t h e func-
Qv ) i n t h e l a s t theorem, c f . I. Namioka (C11,Lema 7.7 and
Theorem 7.8). THEOREM 83.9. Let
complete and Zet f E L
T
L
and
M
be Riesz spaces with
be a positive operator from
M
Dedekind
L into M
These suprema are attained, i.e., these suprema are maxima. I n p a r t i c u b , i f 4 i s a positive linear functional on f E L
. For any
wehave
, then
L
and if
.
108
Ch. 12,§831
ORDER BOUNDED OPERATORS
sup(Sf:e<S
Tf+
5
For the converse inequality, let the operator Tf+ be defined as in the preceding theorem i.e.,
1
. Then
for u E L+
8 5
Tf+
5
T
,so
Tf+ is an admissable S
. The operator
yields the desired maximum on account of
Tf+
T f = Tf+fc
f+
-
Tf+f-
=
Tf+ - 0 = Tf+
.
This concludes the proof of the first formula. For
we have
IS1 5 T
lSfl
5
ISl(lfl)
5
T(lfI)
For the converse inequality, consider the operator S
SO
S
is admissable. The thus defined S
, so
=
T +-Tff
. Then
yields the desired maximum on
account of Sf = T f f+
EXAMPLE 83.10. case that L and m
-
T -f f
=
Tf+ + Tf- = T(lf1)
.
As a very simple but illuminating example we consider the and M
are Archimedean Riesz spaces of finite dimension n
respectively (with n
assume then that
L
2 1
and m
2 1
). By Theorem 2 6 . 1 1 we may
is n-dimensional real number space Xin
usual coordinatewise ordering, and similarly M = Xim
with the
. The vectors having
Ch. 12,5831
ORDER BOUNDED OPERATORS
109
one coordinate equal t o one and a l l o t h e r coordinates zero a r e b a s i s v e c t o r s in
Rn
R m ) . Any o p e r a t o r
(and s i m i l a r l y i n
from
T
En
lRm i s
into
represented (with r e s p e c t t o these bases) by a matrix
...,m
( t . . ) : i = 1, 1J where each
....n
; j = I,
,
i s a r e a l number. It i s evident t h a t a l l numbers
tij
+
tij =
[
maxi, i j '
0)
and
tij
+
-
=
max(-tij,O)
.
t. = t -t f o r a l l i and j Hence, t h e ij ij ij o p e r a t o r s T+ a n d T- , having the m a t r i c e s ( t z j ) and (tT.1 r e s p e c t i 1J vely, a r e p o s i t i v e operators (from Rn i n t o Rm') such t h a t T = T+-Ta r e non-negative,
and
.
?Rn
This shows t h a t every o p e r a t o r from
into
i s a l s o evidenet t h a t i n the Riesz space
L (Rn,IRm) b T
i s t h e Jordan decomposition of
T = T+-T-
IT1 = T++T-
negative p a r t s . Hence
i s o r d e r bounded. It
lRm
t h e decomposition
i n t o i t s p o s i t i v e and
(1
has t h e matrix
t.. I )
.
1J This matrix r e s u l t w i l l be considerably generalized i n Chapter 13 t o
, where
L
and
M
a r e Riesz spaces of
measurable f u n c t i o n s on p o i n t s e t s
Y
and
X
r e s p e c t i v e l y and where t h e
o p e r a t o r s from operator
for a l l
T
f
L
into
i s of t h e form
E L
. The
r o l e of t h e matrix the operator IT(x,y)
M
function (tij)
T(x,y)
. One
/TI = sup(T,-T)
, the
k e r n e l of
T
, takes
over t h e
of t h e r e s u l t s which we s h a l l prove i s t h a t i n t h e space
Lb(L,M)
has t h e k e r n e l
I (Theorem 9 4 . 3 ) .
We b r i e f l y consider the case t h a t
L
and
M
a r e normed Riesz spaces.
A more d e t a i l e d d i s c u s s i o n of normed Riesz spaces w i l l follows i n Chapter
14. We r e c a l l t h a t a normed Riesz space i s a Riesz space provided with a Riesz norm. The norm i n t h e Riesz space in
L
implies
llfll
_<
L
i s a Riesz norm i f L
L
into
Ig1
i s a Banach l a t t i c e and
i s a Dedekind complete normed Riesz space, then any o r d e r bounded
o p e r a t o r from
2
IlgIl. Any norm complete normed Riesz space i s c a l l e d
a Bmzach l a t t i c e . It w i l l be proved now t h a t i f
M
1f 1
M
i s norm bounded. F i r s t a simple lemma.
ORDER BOUNDED OPERATORS
110
LEMMA 83.11.
space
L
fn converges i n norm t o
If
fn z g f o r a l l
and i f
n
, then
Ch. 12,§831 f i n the normed Riesz
f t g
.
- -
.
PROOF. For the proof it may be assumed that g = 0 Since If -f I < If-fn 1 , the sequence (fn:n=1,2, ) converges in norm to f But f- = 0 for all n , s o f- = 0 In other words, f 2 0 n
.
...
.
.
THEOREM 83.12. Any order bounded operator
S
T from the Banach l a t t i c e
L
i n t o the Dedekind complete nomned Riesz space M i s norm bounded. In particular, any order bounded linear functional on the Banach l a t t i c e L i s norm bounded, i . e . ,
. Note
the or&r dual
L"
'
i s contained i n the Banach dual
i s not Dedekind complete, then any p o s i t i v e operator from the Banach l a t t i c e L i n t o M i s norm bounded. L*
that i f the nomed Riesz space
M
PROOF. We may assume that T is positive. If T
is not norm bounded,
(un:n=1,2, ...) in L such that Ilunl( = n-2 k for k = 1 , 2 ,... , and /ITunll t n for all n Writing sk = C u 1 n the sequence ( s :k=l,Z, ) is a Cauchy sequence in L , s o s k k 'Onverges in norm to an element s Since sk 2 un for k t n , it follows from the there exist positive elements
... .
last lemma that IITslI t IITunll bounded.
t
.
.
u for all n , so Ts E Tu But then n n n for all n , which is impossible. Hence, T
s
2
is norm
It will be proved later (cf. Exercise 83.29 and Theorem 85.6) any normed Riesz space L the Banach dual L* dual L"
. Hence
L* = L"
if L
that for
is contained in the order
is a Banach lattice.
We present a generalization of the familiar Hahn-Banach extension theorem. THEOREM 83.13.
vector space
V
Let
p be a sublinear mapping from the arbitrary real i n t o the Dedekind complete Riesz space M This means
.
for a12 f,g E v and p(af) = ap(f) real nwRbers a 2 0 . Furthemore, l e t t be an operator from subspace VI of V i n t o M such that t(f> 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
ORDER BOUNDED OPERATORS
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
ORDER BOUNDED OPERATORS
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 < u
which extends
0
observe first that T in L+
.
are of course incomparable. Now choose u
< u 5 u-
.
implies Tu- > 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
ORDER BOUNDED OPERATORS
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
ORDER BOUNDED OPERATORS
by Namioka’s formula in Theorem 83.9, and
=
0
sup Sf:BSS
byformula (1) in Theorem 83.6. The desired equality follows. Hence, y is a Riesz homomorphism. As a final remark we note that it may occur that
is the n u l l
f being the zeroelement, i.e., f # 0 , but
operator without all
f-
. This happens, for example, if
T E Lb(L,M)
Tf = 0 for
Lb (L,M) consists only of
the null operator. Examples will be presented in section 85.
We recall that the positive operator T from L
into M
be sequentially order continuous (a-order continuous) if in M (cf. Theorem 83.7(ii),
I. Tf
implies Tf
is said to
fn I. f
in L
where this notion did
already occur). According to Definition 18.10, any sequentially order continuous Riesz homomorphism is called a Riesz o-homomorphism. Similarly, T
is said to be order continuous if, for any upwards directed system fT
in L
, we
TfT I. Tf
have
. Any
+
f
order continuous Riesz homomorphism is
called a normal Riesz homomorphism by Definition 18.12. The next theorem follows easily from Tucker’s lemma. THEOREM 83.20. The folZowing conditions f o r the Riesz space
L
are
equivalent. (i)
Any p o s i t i v e operator from L
i n t o an Archimedean Riesz space
is a-order continuous. (ii) Any Riesz homomorphism from L i n t o an Archimedean Riesz space is a Riesz a-homomorphism, i . e . , the homomorphism i s a-order continuous. PROOF. (i) (ii) y
*
*
(ii).
Evident.
(i). Assume first that M
is a Dedekind complete space. Let
into L (L (L,M),M) , as defined in the b b is Dedekind complete, the space Lb(Lb(L,M),M) is
be the Riesz homomorphism from L
last lemma. Since M
Dedekind complete as well, and s o this space is Archimedean. Hence, by hypothesis, the homomorphism Y implies :f
=
y(f
) J. 0
for any positive T
in
in
is a a-homomorphism,
Lb(L,M)
so
fn
J.
0 in L
. This implies that )T(:f .L 0 . In other words, Tfn .L 0 . It has been
Lb(Lb(~,~),~)
Ch. 1 2 , § 8 3 1
1 I7
ORDER BOUNDED OPERATORS into M
proved thus that every positive operator from L
is o-order
continuous. Assume now that M complete. Let M^
is Archimedean, but not necessarily Dedekind
be the Dedekind completion of M
T
from L
into M
M^
,
is o-order continuous from L
so
T
above. But then T from L
into M
.
. Any
positive operator
can be regarded as a positive operator from L into M A
into
by what was proved
is also o-order continuous when regarded as an operator
Tne proof of the next theorem is similar. THEOREM 83.21.
equivaZent
.
The folloving conditions f o r the Riesz space
i n t o an Archimedean Riesz space
L
Any p o s i t i v e operator from
(i)
L are
is order continuous. (ii) Any Riesz homomorphism from
L i n t o an Archimedean Riesz space
i s a normal homomorphism, i . e . , the homomorphism i s order continuous. The last two theorems were proved by C.T. Tucker ([11,1977)
different manner, by D.H. Fremlin (121,1977).
and, in a
The conditions in Theorem
83.20 may seem somewhat artificial because of the occurrence of an arbitrary
Archimedean space besides the given space L
. It has been proved, however,
by D.H. Fremlin (in the paper mentioned just before) that these conditions are equivalent to the condition that every relatively uniformly closed ideal in L
is order closed. Before presenting
the subset D
of L
is called order closed or relatively uniformly closed
(ru-closed) respectively if fn
+
implies f E D
f(ru)
the proof, we recall that
...
f E D for n = 1 , 2 , and f n + f or n (cf. section 16 for these definitions which
give rise to the order topology and the relatively uniform topology in L ). in L , a weaker condition at first sight implies already
For an ideal A that A
is closed, as follows.
THEOREM 83.22.
(i) implies
For an ideal
A
in
L the foZZoving hoZds.
A i s order closed i f and only i f u E A
. In other
0 5 u 4 in
n
A
and
un + u
words, A i s order closed i f and only i f
A is a
o-ideal. (ii) A implies
i s ru-closed i f and only i f
u E A
.
0 5 un 4
in A
and
un +tu(ru)
118
ORDER BOUNDED OPERATORS
Ch. 12,1832
i s a normed Riesz space, then the idea2 A i s norm d o s e d i f and onZy i f 0 5 u 4 i n A and u + u in norm impZies (iii) I f
L
n
n
u E A
PROOF. The p r o o f s a r e almost i d e n t i c a l . We p r e s e n t t h e proof f o r order convergence. It i s e v i d e n t t h a t i f with
u
+
n
so
fn
fn
are i n
g
E A and
n
f
+
imalies
u
u E A
. For
A
implies
f:
+
f+
A
g
=
i n f ( f ,fn)
. Set
+
0 5 f -g
"
=
n un = s u p ( g l
Finally, l e t
i s closed, then
with a l l
+ +
+
If -gnl
,... ,gn)
=
0 5 u
t h e converse, n o t e t h a t f+
in
for a l l
+ +
A n
.
n = 1,2
in
I.
so
u
n f- E A
fn
in
. It
f+
+
, and A
-u
5 f+
-
If+-++(
gn 5
n
f E A
implies
. It
f E A
+ +
0 5 u
A
with
.
4 in
n
A ,
'
. Similarly
f+ E A
h a s been proved t h u s t h a t
, -so
n
5 If - f n \
,... . Then
f o l l o w s from our h y p o t h e s i s t h a t
so
If-fnl,
5
0 5 gn 5 f +
+ , f n+) I
and
0 5 f+
A
+I
i f i t i s given t h a t a l l Then
linf(f ,f )-inf(f for
"+
If -f
fn
i s closed.
-+
f
with a l l
P a r t ( i i ) was proved a l r e a d y i n Theorem 60.2. The p r e s e n t proof i s more d i r e c t ; on t h e o t h e r hand i t was proved a d d i t i o n a l l y i n Theorem 60.2 t h a t t h e conditions i n ( i i ) a r e equivalent t o t h e condition t h a t
L/A
is
Archimedean. S i n c e i n an Archimedean space r e l a t i v e l y uniform convergence of
fn to
f
i m p l i e s o r d e r convergence of
i n an Archimedean space i s ru-closed. Archimedean space i s ru-closed.
fn
to
f
, any
In p a r t i c u l a r , every
order closed s e t o - i d e a l i n an
Even more s p e c i a l l y , e v e r y ' b a n d i n an
Archimedean space i s ru-closed.
A s observed, any o r d e r c l o s e d set i n an Archimedean space i s ru-closed. Hence, i n an Archimedean s p a c e any o r d e r c l o s e d i d e a l i s ru-closed. converse i s f a l s e ; i n t h e s p a c e
on t h e i n t e r v a l f(0) = 0
C0,ll
t h e i d e a l of a l l
The
of a l l r e a l continuous f u n c t i o n s f E C (C0,ll)
satisfying
i s ru-closed b u t n o t o r d e r c l o s e d . It i s r a t h e r s p e c i a l , t h e r e -
f o r e , i f a Riesz space L
C(C0,ll)
L
h a s t h e p r o p e r t y t h a t every ru-closed i d e a l i n
i s o r d e r closed.
THEOREM 82.23. The foZZowing conditions f o r the Riesz space L are equivatent. (i) Any Riesz homomorphism from L i n t o an Archimedean Riesz space i s
I19
ORDER BOUNDED OPERATORS
Ch. 12,1833
a a-homomorphism. (ii) Any ru-closed i d e a l in PROOF. (i)
L
. Then
(ii). Let (i) hold and let A be a ru-closed ideal in is Archimedean by Theorem 60.2,
L/A
homomorphism
is order closed.
L
from L
TI
onto L/A
the canonical Riesz
so
is a a-homomorphism. The kernel of
TI
is now a o-ideal in L by Theorem 18.11 (note that this is true because TI
maps
onto L/A and not only into L/A ) . The kernel of
L
is a a-ideal, i.e., A
Hence A (it)
*
(i).
Let any ru-closed ideal in L be a 0-ideal. Assume now
that there exists a Riesz homomorphism space M C
u
such that
Since M E
>
B
=
0
,
from L
TI
such that v
is Archimedean, the inequality v s o there is a number
{wid
E
5 E
inverse image of B. Since M
~
Udoes ~
2 TIU n for all n . not hold for all
. Let
such that w = (v-ETIU) + > 0
> 0
1
be the disjoint complement of w
= TI-'(B)
and let A
B
is Archimedean, the band -1
and so by Theorem 6 3 . 3 the inverse image A The set A
into some Archimedean
is not a o-homomorphism. Then there is a sequence
TI
and an element v > 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
<
w = (v-Enul)+ for all n
,
so
w
=
is disjoint from
Denoting elements of L/A by i.e., [ a u -u I+ = C O I I n homomorphism L + L/A
,
so
~ { ( E u -u
Cfl,Cgl,
[Eu,]
...
run]
2
TI{(un-Eul)'} )
I n
, we
+1
, i. . e . ,
have
for all n
[(EU
.
This implies w = 0
, so 2
, contradicting
nul
C
u1 E A
, which
But
B = {wl
.
1
=
101 ,
canonical
is a a-ideal,
we have a o-homomorphism. It follows therefore from u But then C E U ~ I = L O 1 Cunl c L O 1 Hence w E B on account of w < v
I+
. The
is onto L/A and its kernel A
.
( a u -u ) + E A -u ) I n
SO
0 that
implies n u l E B d , s o w E {wl d
.
.
the fact established above that w > 0
holds. It follows that condition (i) of the theorem is satisfied.
120
ORDER BOUNDED OPERATORS
12,9831
Ch.
The proof of t h e next theorem i s similar. THEOREM 83.24. The f o l l d n g conditions f o r the Riesz space
are
L
equiualent. (i)
L i n t o an Archimedean Riesz space
Any Riesz homomorphism from
is a normu2 homomorphism. ( i i ) Any xu-closed i d e a l i n EXERCISE 83.25. L e t
L
i s a band.
be a p a r t i a l l y ordered v e c t o r space with a
L
generating p o s i t i v e cone and with t h e dominated decomposition property and let
M
implies
Tnf
Tn + T(ru)
+
M
in
Tf
implies
Tnf
+
Assume now t h a t both complete. Let Tfn + T f ( r u ) Tfn
in
M
.
M
in
HINT: Let
let
, but
M
Tf
-f
and
for
M
. Show t h a t
.
n
+
in
T
fn + f(ru)
in fn
T
Lb(L,M)
show t h a t
a r e Riesz spaces with
t h e r e a r e examples t h a t
Dedekind
M
L
implies
f
in
+
does not
L
be t h e Riesz space (c) of a l l convergent ( r e a l ) sequences,
L
be the r e a l numbers and f o r
Tf = l i m f
. Similarly, every f E L .
f E L
Tf(ru) L
T E Lb(L,M)
Tn
L b ( ~ , ~ )Show t h a t
f o r every
and
Tn (n=l,2,
.
be members of t h e Riesz space
imply
...)
be a Dedekind complete Riesz space. Let
...)
f = (fl,f2,
There e x i s t s a sequence i n
L
E (c)
let
converging in o r d e r t o zero f o r
which t h e sequence of images does not converge i n o r d e r t o zero. EXERCISE 83.26. Let be a Riesz seminorm on
f E L
f o r every
pA(f) 5 p(f)
Show t h a t p/A
be an i d e a l i n the Riesz space
. We
. Show t h a t
f o r every
pA
.
A
L
i s a Riesz seminorm on
pA
f E L
, pA
= p
I n a d d i t i o n , show t h a t i f
space
0
0
5
v
L
2
v
4 v
in
. Similarly,
4 v E A
and l e t
p
on
A
and
pA
L
such t h a t
vanishes on
Ad
.
i s t h e minimal Riesz seminorm extension of t h e r e s t r i c t i o n p
has t h e Fatou property on
d e f i n i t i o n of t h e Fatou property, p ( v , ) system
L
set
A ) , then if
implies
p
pA
4 p(v)
A (i.e.,
has t h e same property on t h e whole
has t h e sequential Fatou property on
p(vn) 4 p(v)
by
f o r any upwards d i r e c t e d
, then
pA
A
, i.e.,
has t h e same property on
Ch. 12,1831
ORDER BOUNDED OPERATORS
121
L . H I N T : S i m i l a r l y a s i n Theorem 83.7. EXERCISE 83.27. L
and l e t
Let
v
be a given p o s i t i v e element i n t h e R i e s z space
be a Riesz seminorm on
p
. We
L
set
1 f f o r every pv 5 p
pv = 0
of t h e
. Show
f E L
on
L
on
(Av)d
with
that
pv = p
. Show t h a t restriction plAv .
EXERCISE 83.28. Let
be a Riesz seminorm on on
A (i.e.,
if
v
i s a Riesz seminorm on
pv
on t h e i d e a l
such t h a t
E A for n
p
1,2,
=
Show t h a t t h e minimal extension continuous on
generated by
A
V
such t h a t v
and
i s t h e minimal Riesz seminorm extension
pv
be an i d e a l i n the Riesz space
A L
L
...
of
pA
L
and l e t
p
i s s e q u e n t i a l l y o r d e r continuous v J. 0 , then p(vn) J. 0 ) . n is not necessarily order
and
p
L ; t h i s can occur even i f
A
HINT: Let, f o r example, L = 1- with
i s a principal ideal.
.
p t!he uniform norm on L Let 1 1 1 A be t h e p r i n c i p a l i d e a l generated by (1,) Then p i s 2'3 s e q u e n t i a l l y order continous on A Show t h a t p A = p s o pA i s n o t
.
s e q u e n t i a l l y o r d e r continuous on EXERCISE 83.29 ( i ) . L e t
(ii) Let f = (fl,f2,
...)
L
uniform norm on HINT:
so
4
be a normed Riesz space. Show t h a t t h e
be t h e Riesz subspace of fk # 0
with L
( i ) For
. Show 6E
f = (fl,f2
EXERCISE 83.30. subspace of t h e space
L*
and
,... )
The space
L"
.
c o n s i s t i n g of a l l
f o r only f i n i t e l y many
that
L*
i s o r d e r bounded. ( i t ) For
L
,
.
i s contained i n t h e o r d e r dual
L*
Banach dual
L
-,...,--,... .
, let
L"
p
.
be t h e
we have
$ ( f ) = Ckfk
C = C(C0,ll)
>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
ORDER BOUNDED OPERATORS
, 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
ORDER BOUNDED OPERATORS
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 .
o5
inf
T+U, 5 T+U
0
- sup ~v:o
+
+
Tu-Tu=O, so
inf T+u
=
implies T- E
0
. This concludes the proof for
Ln follows by observing that T-
T+
.
The proof that
= (-T)+
.
T E
Ln
It follows from the definition of order continuity that the sum of two positive order continuous operators is order continuous. Hence, if and
T- E
Ln , then IT1
0 5 T+ i IT/ and
continuity of
8 i
= T+
T-
5
T+ and T-
Finally, we prove that
.
+ T- E Ln
IT1
that order continuity of
IT1 E
T+ E
. Conversely, it follows from
Ln implies T E Ln
IT1
. Let
Ln
implies order
IT1 E
Ln and
Ch. 12,5841
u
0
J.
ORDER BOUNDED OPERATORS
. From
lTuTl
+
IT[ (uT)
2
125
0 it follows that
inflTu 1
=
0
. Hence
TELn. It is now evident that Ln
Lc are linear subspaces of Lb * Indeed, if for example TI and T2 are members of Ln and if u J. 0 , and
then
so
. It can also be
Ln
T +T E 1 2
the linear subspaces Ln THEOREM 84.2. L
PROOF. We present the proof for
Lb
ideal in all a
. For the proof
in the index set
T u 4 Tu
T E
implies inf Tu
0 in M
=
T( (u-u ) < T(u-u ) Y T Tu
T
Keep
a
fact ideals in
and
Ln
. As
Ln is a band, assume that Ta E Ln for 0 5 T 4 T in Lb . This means that
+
a (cf. the proof of Theorem 8 3 . 4 ) . We
and
T
,
T E L a
is positive and
u
a
+
+
0 in L
0 in L
. Then
is directed downwards, the set inf Tu
is Dedekind complete. Furthermore, since u
, we
n
u
a T
have
inf Tu for all
u 2 u
so
(Tu~:T€{T}) is also directed downwards, and :o 8 5
is an
5 T u +Tu-Tu.
fixed. Since T
because M
’
observed already, Ln
. We may assume that
for all a
‘b
.
Ln , i.e., we have to show that
have to show that
(1)
that
{a}
for every u E L
in M+
in
Lc are bands i n Lb
and
II
seen immediately form the last lemma that
Lc are
and
. But .
5
Ta u T J. 0 Tu
-
. Hence, from ( I ) ,
exists in M+
+
0 and
T u
Ta u 4a Tu , i.e., Tu-Tau +a 0
. It follows that
inf Tu = 0
Before proceeding with the further investigation of the bands
Ln , we show now that any order bounded operator T if and only if
Lc and
is o-order continuous
T maps order convergent sequences into order convergent
126
ORDER CONTINUOUS OPERATORS
T E
sequences. We show first that if in M
. It follows from
fn
+
Lc and fn
0 that
Ch. 12,1843
+
0
in L
,
then Tf n
Ifn/ C pn C 0 for appropriate
Pn
so
Lc implies IT1 E Lc ) . This shows that Tfn Lb have the property that fn + 0 implies Tfn
(since T E
+ 0
let T E
+ 0
I n
0 5 ITf
5
C 0
q
implies infITf I n
for appropriate qn = 0
,
T E
so
Lc
.
in M
. This
0
+
’
. Conversely, ,
so
shows that
fn
-+
0
Lc and Ln in Lb will be denoted by l S Lsn respectively. Elements of Ls will be called singular operators (from L into M ) and elements of Lsn will be called n o m a l singular operators (from L into M ). Since Lb is Dedekind complete, the space Lb is the-direct sum of the bands Lc and Ls . Similarly, Lb is the direct sum of the bands 1, and Lsn . Hence The disjoint complements of and
1 = L b
On account of
c
Ln
c
e l s
=n L se l n
.
Lc we have L s n 3 Ls , and it follows that each T E Lb
has a unique decomposition T = T + Tc,sn + Ts with Tn E ln n Tc,sn ‘c,sn = Lc n Lsn is a band in hence it is a band in Lb Furthermore Note that
Evidently, we have
.
L
c,sn
=
Lc
, Ts
E
Lc
‘sn
Ls and *
as well as in
{ e l if and only if L
c
=
Lsn , and
L n ’ i.e., if and only
if every o-order continuous operator is order continuous. For the particular case that L
Lc and Ln are bands in 1,
is Dedekind complete, the theorem that
is due to A . G . Pinsker (cf. the book by
Kantorovitch, Vulikh and Pinsker C17,1950).
The general case, as presented
Ch. 12,0841
ORDER BOUNDED OPERATORS
127
here, is due to R. Cristescu (cf. the book [11,1959). If M
is the
space
of real numbers the theorem expresses the fact that the order continuous linear functionals and the o-order continuous linear functionals form bands in the order dual L"
. This was
already proved by F. Riesz on the last page
of his fundamental paper ([21,1940).
Riesz's proof canndt immediately be
copied in the more general case because in Riesz's proof the infimum of a certain set in real number space is approximated to within a prescribed
E
,
and in a more general Riesz space there is no such approximation available. It is of some interest to mention here a paper by H. Gordon and E.R. Lorch (C11,1957). The situation in this paper is very similar to the one we have here, but not quite the same. The authors call a positive linear functional Q
on the space C(C0,ll)
an integral whenever it
follows from f C 0 pointwise on C 0 , l l that $(fn) + 0 (in accordance n with the general definition of a Daniel1 integral). It is shown then that the so defined integrals form a band C"(C0,ll)
is decomposed into B
B
in the order dual
C"(C0,ll)
and
and its disjoint complement. This
definition of an integral is not the same as our definition of a o-order continuous linear functional, because for functions in C([O,ll)
it is
true that f C 0 pointwise on C 0 , l l implies order convergence of n to the zeroelement, but not conversely (cf. Exercise 18.14). Under certain conditions L into M
operator from L
=
fn
Lb , that is, every order bounded
is o-order continuous. We mention one sufficient
condition of this kind. The condition is on the space L
only.
THEOREM 8 4 . 3 . If L is Archimedean and. i f order convepgence and L m e e q u i v d e n t , then
reZutiveZy uniform convergence f o r sequences i n L = L c b' PROOF. Let
Tun
+
8
< T E Lb and let un C 0 in L
0 holds in M
. We have to
show that
. Since order convergence and relatively uniform
convergence in L are equivalent, it follows from u
an element uo E L+ with real numbers all n
. Hence
Tu < n
Tu n O
E
+
0 in M
E
n
+
C 0 that there exists n 0 such that un 5 for
, which
implies Tun
+
0
.
We recall that by Theorem 16.3 order convergence and relatively uniform convergence in the Archimedean Riesz space L
are equivalent if and only if
128
Ch. 12,0841
ORDER CONTINUOUS OPERATORS
L
o r d e r convergence i n
is s t a b l e , i.e.,
f o r any sequence
t h e r e e x i s t s a sequence of r e a l numbers
0 S An 4
m
and
Anfn
0
+
a - f i n i t e measure space
. It was
(Xn:n=l,2, ...)
proved i n Theorem
fn + 0
such t h a t
order convergence i s s t a b l e i n t h e Riesz
(X,A,u>
X
. In
(s)
M ( ~(X,A,u) )
L
71.4 t h a t f o r any
t h e r e f o r e , o r d e r convergence i s s t a b l e i n the sequence space
space
in
of a l l r e a l u-measurable functions on
particular, of a l l
r e a l sequences (cf. a l s o Exercise 16.16). Order convergence i s s t a b l e i n some i d e a l s of
, but
M(')(X,A,u)
not i n a l l i d e a l s . I n Theorem 71.8 and
Exercise 71.9 i t was proved t h a t f o r
0 < p <
m
t h e space
Lp(X,A,u)
has
t h e diagonal property, which implies by Exercise 16.14 t h a t order convergence is s t a b l e i n t h e space. Hence, i f and
M
L
i s one of these spaces
i s an a r b i t r a r y Dedekind complete space, then
The spaces
L = Lm(X,A,u)
L = lm behave d i f f e r e n t l y . We s h a l l prove
and
i n the next s e c t i o n , a s a consequence of Theorem 85.5, t h a t i f and u n l e s s
X
.
Lc(L,M) = L b ( ~ . ~ ) u(X) > 0
c o n s i s t s of only a f i n i t e number of atoms, t h e r e e x i s t non-
t r i v i a l s i n g u l a r l i n e a r f u n c t i o n a l s on e x i s t non-trivial
Lm(X,A,u)
s i n g u l a r l i n e a r f u n c t i o n a l s on
. In particular, . This i s i n
there
L-
accordance with t h e observation made i n s e c t i o n 16 ( t h e p a r t between order convergence i s not s t a b l e .
Theorem 16.2 and Theorem 16.3) t h a t i n
Another f a c t t o bear i n mind i s t h a t i t may occur t h a t c o n s i s t s only of t h e n u l l operator
i s r e a l number space, s o L- =
that
if
L (L,M)
with
of much a d d i t i o n a l i n t e r e s t then.
=
Ls
L n
=
M
s h a l l prove i n t h e next s e c t i o n
u
a - f i n i t e and without atoms or
and again
1
Lc
. We
= L"
L = M("(X,A,u)
L = L (X,A,u) f o r 0 < p P Of course, 1, = {8} implies if
$,(L,M)
This can happen, f o r example, i f
6 .
p a - f i n i t e and without atoms.
= L = {a} s n
, but
t h i s i s not
The theorem which follows now contains s u f f i c i e n t conditions f o r
Lc
=
L
n
t o hold.
i s order separable f i . e . , by Definition 23.1, L possessing a suprermun has an a t most comtabZe subset having the same supremem as D ), then every a-order THEOREM 84.4.
(i) I f
every non-empty subset
continuous operator (from 1 = L c
.
L
D of L
i n t o M ) i s order continuous, i.e.,
n ( i i ) I f f o r every principal ideal
A
in L
there e x i s t s a positive
linear funcbional +A on A such that 4A is strictly positive on A ( i . e . , i f 0 5 v E A and $,(v) = 0 , then v = 0 1, then L i s order
ORDER BOUNDED OPERATORS
12,§841
Ch.
129
separabZe, and so every a-order continuous operator is order continuous. This hold in particular i f there e x i s t s u s t r i c t Z y positive linear functional on L
.
(i)
PROOF.
If
uT i 0
h y p o t h e s i s a sequence Hence
Tun J. 0
implies
i n f Tu
in
L
(un:n=1,2,
,
...)
t h e r e e x i s t s by t h e o r d e r s e p a r a b i l i t y in
f o r any p o s i t i v e o-order =
. This
0
shows t h a t
(vn:n=1,2,
0 < u
...)
in
and l e t
u
n
.
i 0
, which
4
< u
in
L
implies the
( u ~ : T € ~ T }such ) that the
sequence has t h e same upper bounds as t h e s e t of a l l
0 2 u 4's u
T
i s o r d e r continuous.
T
( i i ) I t i s s u f f i c i e n t t o show t h a t e x i s t e n c e of a sequence
( U ~ : T € { T } ) such t h a t continuous o p e r a t o r
u
. Hence,
let
b e a s t r i c t l y 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 t h e
$
. Let
u
p r i n c i p a l i d e a l generated by
a = sup$(u )
.
. There e x i s t elements
such t h a t v f and $(vn) 4 a Let v b e an a r b i t r a r y upper v = u Tn bound o? t h e s e t of al? v We s h a l l prove t h a t u S v holds f o r a l l n T I f n o t , we have w = sup(u ,v) - v > 0 f o r some T~ , and so it
.
.
TO
follows from
that
sup(u
TO
,vn) - v
n
2
w
2 0
for a l l
n
. Hence
O(vn) .f CI i m p l i e s t h a t Q{sup(uT , v ) } > CL f o r some no . But "0 satisfying ( u ~ : T € { T ~ )is d i r e c t e d upwards, so & e r e e x i s t s an element u 'I u 2 sup(u , v ) , and hence $(uT ) > a This c o n t r a d i c t s t h e d e f i n i t i o n T1 To "0 I of a Thus u r; v holds f o r a l l T S i n c e v i s an a r b i t r a r y upper SO
.
.
bound of t h e s e t of a l l
, THEOREM 84.5. If
v
n '
(X,A,p)
.
t h i s completes t h e proof.
is a a-finite measure spaee, if L i s an
ideal i n the R i e s z space (X,A,u) of a l l reaZ u-measurable functions on X and M is an arbitrury Dedekind complete Riesz space, then L c ( ~ , ~=) L*(L,M)
.
PROOF. I n view of t h e preceding theorem it i s s u f f i c i e n t t o prove t h a t
L
i s o r d e r s e p a r a b l e . I n Example 23.3 ( i v ) i t was shown t h a t
M(r)(X,A,~)
i s super Dedekind complete, which i m p l i e s o r d e r s e p a r a b i l i t y . Any i d e a l
L
Ch. 12,5843
ORDER CONTINUOUS OPERATORS
130
in Mr(X,A,p)
is then also order separable.
In particular, for 0 < p Lp(X,h,u)
5
-
every o-order continuous operator on
is order continuous. As a special case, for 0
p
<
Im
every
o-order continuous linear functional on
L is order continuous. In other P is a normal integral. For 0 < p < 1 there
words, every integral on L P are no non-trivial integrals on L (simply because (L )- = P P case, as we shall prove in the next section), but for 1 5 p
in this 5
m
there
exist many non-trivial integrals on L (assuming that u(X) > 0 ) . Indeed, P if 1 5 p 5 m and q is defined by p-l+q-l = 1 (with q = m for p = 1 $g
and
on L P
for p =
q = 1
) , then every
m
g E Lq defines an integral
by means of the formula $g(f) =
fgdu
X holding for all
fE L
P
,
. The order continuity of
bg is evident because
0 in L P Conversely, it follows from Theorem 83.12 that every
it follms from the dominated convergence theorem that
.
fn
C
implies C$ (f ) + 0 g n order bounded linear functional on L (I
is represented by a function go
C$(f)
=
x
fg$dlJ ,
in L 4
by means of the formula
-
.
holding for all f E L Hence, for 1 I p < every order bounded P is an integral represented by a function in L linear functional on L P 9 For p = m not every order bounded linear functional is an integral, but those that are integrals are represented by an L -function, as will follow 1
from a general theorem whcih will be proved in the next section. EXERCISE 84.6.
If L
is Archimedean and of finite dimension, then L
is order separable, s o any a Dedekind complete space M that every operator from L
o-order continuous operator from L
into
is order continuous by Theorem 84.4. Show into M
is order continuous, either by
deriving it from the general results above or immediately from the matrix representation.
Ch. 12,1851
ORDER BOUNDED OPERATORS
131
85. The order dual of a Riesz space
It was proved in Corollary 83.5 that the order dual L"
space L
of a Riesz
is a Dedekind complete Riesz space. It is evident that if L
, i.e., L"
consists only of the null element, then the same holds for L"
consists only of the null functional. It i s perhaps disappointing at first that also if L
is of positive dimension, and even if L
dimension, it can still happen that
is of infinite
consists only of the null
L"
functional. We present two standard examples. EXAMPLE 85.1.
(The space M(r) (X,p) ). Let
p
be a finite or
a-finite countably additive and non-negative measure in the non-empty point set X
such that
p(X)
0 and
>
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 < p < 1 ) . Let u be a countably P additive and non-negakive measure in the non-empty point set X such that p(X)
> 0 and
p
does not possess any atoms. Furthermore, let p
fixed real number satisfying 0 u-measurable functions f
on
p < 1
<
X
. The collection of
such that f I , !
be a
all real
Ipdu i s finite is
denoted by
L (compare Example 11.2(vi)). Functions that are u-almost P equal are identified, as before. It is evident that every f E L is P is a subset of the Riesz finitevalued u-almost everywhere, and hence L P space M(~)(x,~) . We observe first that
is linear, and hence f,g E L * it follows immediately that L P' P It is sufficient for the desired is an ideal in the space M(r)(X,A)
for all L
P inequality to prove that
.
(a+b)'
S
e b + ' .
holds for all
a,b t 0
, and
hence it is sufficient to prove that
This follows by observing that the derivative f'(x) for x > 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
ORDER BOUNDED OPERATORS
.
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
<
m
i s order bounded i f (un:n=O, 1.2,. .) i n L+
f o r every decreasing sequence
0 5 u
n
5 2-"u0
n = 1,2
for
functional on L , then $ E L" sequence
on L
$
(fn:n=l , 2 , .
.. )
,... . Hence,
.
i f
i s a linear
$
i f and only i f
in L
$(fn) + 0 f o r any that converges reZativeZy uniformly t o
zero. PROOF. Let $
for any sequence n = l,2,
be a linear functional on
L
(un:n=0,1,2,...Isatisfying
... . We have
to show that for any
such that
o
u E L+
< u
supl$(un)l
< 2-"u0
n the set
<
-
for
I $ (v) I :05V5U is bounded. Assuming this to be false, there exists an element uo E L+ that
(
sup I$(v)l:O
0)
= +-
.
such
Ch. 12,1851
so
135
ORDER BOUNDED OPERATORS
there exists an element vI satisfying 0 < v1 < ;u,
I$(v,)l
t 31$(uo)l
+ 2
. Hence, w 1 =
? u -v I
such that
satisfies
Note now that it is impossible that
I are both finite, since this would contradict the formula (1). first of these numbers is case denote w I by
u1
+ m ,
. Then
This procedure is repeated, uo element u2
in L
being replaced by
exist
.
u1
. We
thus get an
satisfying
(un:n=0,1,2,
Proceeding by induction, we obtain a sequence
for n = 1,2,.
Now, if the
u 1 , and in the other
denote v 1 by
.. . By hypothesis, however, a
...)
satisfying
sequence of this kind cannot
The existence of nonzero order bounded linear functionals on a Riesz space L
is related to the existence of nonzero Riesz seminorms on
recall that a Riesz seminorm extra property that
If1
C
lgl
p
on
L
is a seminorm on
implies p(f)
5
p(g)
. We
L
L having the
. The definition of
Riesz norm is similar. The familiar norms on spaces Lp(X,p)
for
1
5
a
p c
-
THE ORDER DUAL OF A RIESZ SPACE
I36
Ch. 12,1851
are Riesz norms. More generally, the norms on Orlicz spaces (cf. section 10) and the function norms on normed KEthe spaces (cf. section 9) are Riesz norms. Also, the uniform norm on the space of all real continuous functions on a topological space is a Riesz norm. Orlicz spaces will be discussed more
in detail in sections 131-133. (i) If
THEOREM 8 5 . 4 .
and
is a linear functional on t h e Riesz space
$
L
is a Riesz seminorm on L such t h a t l$(f)l 5 p(f) hoZds f o r a l l $ is order bounded, i . e . , $ E L" . (ii) Conversely, i f E L" is given, then p(f) = 191 (If/) is a
p
f E L
, then
Riesz seminorm on
L such t h a t
l$(f)l
holds f o r a l l
p(f)
5
(iii) There e x i s t s a nonzero p o s i t i v e linear functional
only i f there e x i s t s a nonzero Riesz seminorm on PROOF. (i) For every
I $(f) I :-usf5u)
sup(
and so
$
(ii)
u E L+
i f and
we have
f
SUP\l $(f) I : If l 5 u
=
.
L
.
f E L
on L
)
5
is order bounded by definition (cf. Definition 8 3 . 2 ) . Evident.
(iii) If p ( f ) = $(lfl)
$
is a nonzero positive linear functional on
is a nonzero Riesz seminorm on
nonzero Riesz seminorm on
L
, then
L
p(u ) > 0 0
L
. Conversely, if
for some u
0
,
E L+
then p
is a
, and
$(auo) = a p ( u o ) is then a nonzero positive linear functional on the Riesz subspace of all real multiples of uo such that I$(auo) I 5 p(auo) holds for all real a to all of $
. By the Hahn-Banach extension theorem
L such that
l$(f)l
5
p(f)
is order bounded by part (i). Then
holds for all $ = $+
Before proceeding, we note that if $ is a linear functional on
then
L
p
satisfying
JI
can be extended
f E L
. Hence,
is positive and nonzero since
is a Riesz seminorm on L l $ ( f ) l 5 p(f)
for all
and
f E L
,
Ch. 12,5851
ORDER BOUNDED OPERATORS
.
f E L
holds for all
137
The first inequality follows from formula (6) in
Making use of (7) in the same theorem, we find
Theorem 83.6.
In the proof of part (iii) of the last theorem we could also have used Theorem 83.17 (extension of a positive operator majorized by a sublinear operator
p
operator p
such that
p
is absolute and monotone). Note that a sublinear
from the Riesz space L
into real number space such that p
is also absolute and monotone is exactly a Riesz seminorm on specializing Theorem 83.17 to the case that M
L
. Hence,
is the space of real numbers,
we get the following theorem. THEOREM 85.5. (Extension of a p o s i t i v e linear functional). Let
be
p
a Riesz seminom on L and l e t $ be a p o s i t i v e linear functiona2 on the K of L such that l $ ( f ) l 5 p(f) holds f o r a l l f E K Then there e x i s t s a p o s i t i v e linear functional J, on L such t h a t J1 =
.
Riesz subspace on
and
K As
I $ ( f ) l S $(lfl)
.ern .
linear functionals on k +
f E L
.
an application we shall prove that there exist nonzero singular
is a Riesz subspace of as
for all
2 p(f)
+
. Then
m
The space (c) of all (real) convergent sequences
.ern . For
f
=
(fl,f2,...) E (c)
,
let $(f)
=
lim f
k
is a norm bounded poiitive linear functional on ( c ) . It
$
is easily seen that there exists a sequence u .I. 0 in (c) such that n for all n , so $ is not order continuous on (c). It follows
$(un) = 1
that if
@
then P
is not order continuous. Hence, if
is any positive linear extension of
continuous and
Ps
singular, then Q
Same argument can be used in L,(X,A,u)
P
to the whole of
$
+ Q
= @
.
if u(X)
>
0 and
X does not
is the disjoint union
(Xn:n=l ,2,. .) of positive measure. The role of (c)
can now be taken over by the Riesz subspace of all functions f a constant value k
+
m
fk
on
lrn ,
QC order
c s is not the null functional. The
consist of only a finite number of atoms, i.e., if X of countably many sets
with
%
(for all k ) and for which
that have
lim fk
(as
) exists.
We briefly return to some facts touched upon already in Exercise 83.29. We recall that a
normed Riesz space i s a Riesz space equipped with a Riesz
norm. If the space is norm complete, it is called a Banach lattice.
THEOREM 85.6.
L* of a nomed Riesz space
The Banach dua2
idea2 i n t h e order dual
L"
, and t h e
i s a Banach Zattice, then L* PROOF. The norms on
L
= L"
and
$ E L*
respectively. For any
so
Cb. 12,5851
THE ORDER DUAL OF A RIESZ SPACE
138
n o m on
.
is an
L
is a Riesz norm. I f
L*
L* will be denoted by
p
and
p*
and any u E L+ we have
is order bounded by definition. This shows already that L*
$
subset of
. Hence, if
L"
$
L
E L* , then 101
is a
is well defined, and for any
f E L we have
/$I
This shows that immediately from = p*(l$l)
p*(+)
E L*
l$(f)l
and 5
. Finally, if
6 5 $
. Conversely, it follows
5 p*($)
p*(1$1)
that P * ( $ )
l$l(lfI)
<
IJJ
E L*
,
5 p*(l$l)
.
Eence
it is immediately derived
from
+ E L*
that
and
p*($)
is a Riesz norm on L*
.
5 p*($)
. Hence, L*
For a Banach lattice L we have L* c L"
is an ideal in L"
L" c L*
by the present result. Hence L*
=
L"
and
p*
by Theorem 83.12 and
.
In Theorem 85.3 we have indicated a condition, necessary and sufficient foralinear functional on L to be order bounded. There are a l s o conditions which guarantee that every linear functional on L
is order bounded. The
following theorem mentions one condition of this kind. THEOREM 85.7.
Every Zinear functiona2 on t h e Riesz space
bounded i f and only i f o f f i n i t e dimension.
is order
L is Archimedean and every principal idea2 i n
PROOF. Assumefirsr that
in L
L
L
L
is
is Archimedean and every principal ideal
is of finite dimension. Let $
be a linear functional on L
and let
ORDER BOUNDED OPERATORS
Ch. 12,5851 u
E L+
. We have
to prove that
The principal ideal AU dimension,
A,
so
generated by
0
I
, Bi
ai
u
is Archimedean and of finite
has a disjoint basis
positive numbers y I , ...,yn satisfying
139
If1 5 u
I yi
(e
such that u
we have
f+
= Zy
for i = I,..,,n
,en) of atoms. There exist ' 'i.. yiei , and for any f
= CI
aiei and
. Hence
f
= C
n
1
B.e. 1
1
with
l$J(f)
xnI
ci
i
Then
Conversely, assume that every linear functional on For the proof that L and let
$
bounded,
so
is Archimedean, let
S
, because
L having u
for any
as one
functional $
S
holds for all n
M
u # 0
L
,
is order bounded.
v
for n
. Then
$
$(u) = 0
so
. It follows that
is order
. But then
L is Archimedean.
is of infinite dimension. Then, for any natural number mutually disjoint elements in A,
is Archimedean)
A,
,
n , there
because otherwise
would be of finite dimension by Theorem 26.9
strictly positive elements contained in A,
. There exists a Hamel basis
. We may
(un:n=1,2,
of
...)
assume that 0
L containing all u there exists a linear functional $ on L such that $ ( u ) = 1 n n Since $ is order bounded, we have $+(un) 2 1 for all n
.
...
such that the principal
Hence, there is a countably infinite disjoint system for all n
l,2,
=
of its elements and hence there exists then a linear
satisfying $ ( u ) # 0
exist more then n (since A,
L S
there is a Hamel basis of the vector space
Assume now that there exists an element v E L+ ideal A,
nu
be an arbitrary linear functional on
That implies that I$(nu)l u = 0
0
of I u
n and so
Iv
for all
. Then
140
THE ORDER DUAL OF A RIESZ SPACE
for k
=
l,2,
... . This is
impossible. It follows that
Av
.
v E L+
dimension for every
Ch. 12,5853
is of finite
Combining the last theorem with Theorem 6 1 . 4 , we find that the following conditions on the Riesz space L (i)
in L
Archimedean (i.e., for any ideal A (ii)
are equivalent:
L is uniformly complete and every quotient space of L
the space L/A
L
is
is Archimedean).
is Archimedean and every principal ideal in L
is of finite
dimension. (iii) L
is Riesz isomorphic to the Riesz space oftall real functions
X such that f(x) # 0 for only finitely many points x E X, these points depending on f
f
on some set
(iv)
.
L is super Dedekind complete and every quotient space of L
is
Archimedean. (v)
Every linear functional on
L is order bounded.
Some further remarks can be made. Several authors call a Riesz space having the property that each of its quotient spaces is Archimedean a hyper-
archimedean Riesz space. We shall follow these authors in adopting this name.
It was observed by C.B. Huijsmans ([3],1976)
that there is still another
condition equivalent to the five conditions mentioned above, as follows. (vi)
Every ideal in L
This can beprovedby showing
is a projection band. (vi)
* (ii) * (iii) * (vi)
. The proof
that
(iii) implies (vi) is not difficult. We briefly indicate how to prove that (vi) implies (ii).
It follows immediately from (vi) that L
projection property and
has the
L is certainly Archimedean. As in the proof
so
of the last theorem, assume now that there is an element v E L+ the principal ideal A,
is of infinite dimension. Then again there exists
a countably infinite disjoint system
. Let
elements contained in A, (un:n=1,2, and
0
5 z
...) . Then E Ad
. Then
0 s
w s a
for appropriate
I
A
1
such that
A
(un:n=1,2, ...) of strictly positive
be the ideal generated by the system
is a projection band, so
u
"1
,...,u"k
+...+a
v = w+z with
0
5 W
E
A
k "nk
...,ak . Hence, if the
and posirive numbers a,,
Ch. 12,1863
z
,
u
so
. This implies
contained in A, positivity of
...,nk , then un is disjoint from is disjoint from v . On the other hand u n is
satisfies n # nl,
natural number n and from
141
ORDER BOUNDED OPERATORS
. Hence
u
A,
u
n
= 0
, contradicting the strict
is of finite dimension for every v E L+
w
.
Note, finally, that by Theorem 37.5 (ii) condition (vi) is equivalent to the following condition. (vii) For every ideal A {PIA
=
in L
the set
U({PIf:fEA)
P
is not only open but also closed in the hk-topology of the set
.
proper prime ideals in L
EXERCISE 85.8. Let v E L+
and let p
be a Riesz seminorm on L on L
Show that there exists a positive linear functional $ l$(f)l all
5
$(lfl)
f Iv
.
5
p(f)
for all
f E L
,
of all
$(v)
=
p(v)
and
.
such that
$(f) = 0 for
86. Integrals on ideals of measurable functions
In the present section it will be assumed that
(X,A,p)
is a o-finite
measure space. The Riesz space M(r) (X,A,u) of all real p-measurable M(X,p).
functions will be denoted briefly by
We recall that p-almast
. The
equal functions are identified. Let L be an ideal in M(X,p) p-measurable subset E
of
X
is called an L-zero set if every
vanishes p-almost everywhere on
E
. For
f E L
the investigation of the ideal
L we may just as well remove such an L-zero set from X
. It is possible
to remove all L-zero sets simultaneously since there exists an L-zero set Em
that is maximal, i.e., the set theoretic difference X-Em
-.
does not
contain any L-zero set of positive measure. For the proof we first consider the case that p(X) and let
<
Denote by
r
the collection of all L-zero sets
0.
a = sup p(E):EEr
There exists an increasing sequence
(En:n=1,2
,... )
of sets in
r such that
142
IKTEGRALS ON IDEALS OF FUNCTIONS
u(En)
t a
. The union
Now assume that
u(X) =
...)
(k=I,Z,
k
let Ek
.
m
Em = U 1 E n is an L-zero set such that p ( E m ) = a is maximal and uniquely determined except for a u-null set.
Evidently, Em sets X
. Then there exist mutually disjoint u-measurable = U l Xk . For each k ,
m
Em
set. We prove that
m
of finite measure such that X
\ . Then
be the maximal L-zero set in
set F
Ch. 12,1861
a
is maximal. If not, then X-Em
of positive measure, and
u(Fn5) > 0
so
contains an L-zero
for some k
is an L-zero set of positive measure contained in \-Ek the maximality of
Ek
of
Hence Em
is maximal.
L
by
. The carrier of
X
L' For any subset E
X , let
of
,
E
of the ideal L
is not a member of
is Lebesgue measure and
1 5 p <
m
,
xE
be the characteristic function = 1
i.e., xE(x)
for x E E
then XL
=
X
L
. As
an example, if
for some p
L = L (X,p) P
that there always exists a sequence Xn 4 for all n
X =
. This is what
5
= 0
xL
,
(-m,-)
satisfying
and the characteristic function of
the function identically one, does not belong to L
E L
and .xE(x)
. In general the characteristic function of the carrier
for x E X-E
x
5
F n
contradicts
is uniquely determined modulo null sets. We shall denote the carrier
(indicator function) of
Fi
.
. Then
, which
is called the carrier of the ideal L
The set X-Em L
in Xk
i s an L-zero
Em = U I Ek
X
,
i.e.,
. We shall prove now
such that
p(Xn>
im
and
is sometimes called an exhaustion
Xn theorem, a particular case of a more general exhaustion theorem, which we
shall now state and prove. Before stating the theorem, we introduce a convenient definition. Given the sequence of measurable sets Xn 4 X shall say that the measurable subset E exists a natural number n THEOREM 86.1.
of
such that E c X
(Exhaustion theorem). Let
X
.
is
(Xn)-bounded
, we
if there
(XA:n=l,Z, ...) be a sequence
.
of measurable subsets of X such t h a t Xi 4 X and ~(x;) < m for a l l n Let (P) be some property which any (xA)-bounded s e t does possess or not. I t i s understood here t h a t if one of two u-almost equal s e t s has (P) , then so has the other one. Asswne furthermore that the following hotds. (i)
If E l and
(ii)
If
E2 possess
E possesses
.
(P)
(P)
, then
, then
El
u
E2
possesses
(P)
.
any measurable subset of E
possesses (P) (iii) Any (XA)-bounded s e t o f p o s i t i v e measure has a subset of p o s i t i v e measure possessing (P)
.
ORDER BOUNDED OPERATORS
Ch. 12,5863
143
Then there e x i s t s a sequence ofmeasurabZe s e t s (Xn:n=1,2, ...) Xn + x , x, c :x for alL n and every Xn has property (P) E
PROOF. Let
be
(X:)-bounded
collection of all subsets F
(P)
E possessing
of
, and
p(E) > 0
with
let
. Note
.
such t h a t
r
that
be the p(E)
is finite. Let
0.
a = sup p ( F ) : F E r
There exists a sequence F (i) we may assume that Fn p(F,)
=
. We prove
a
0
E r
that
E r
, and by assumptkn
p(Fn) + a
such that
is ascending. The set Fm = Urn F satisfies ! n a = p(E) If not (i.e., if a < p ( E ) ) , the
.
is of positive measure, and so by (iii) E-F,
set E-F, F
n
p(FnUFo)
= p(Fn)
+ p ( F o ) > a for n a
the definition of
. Hence
has a subset
r
for all n , and n O sufficiently large. This contradicts
of positive measure. It follows that F a = p(E)
,
U F
E
i.e.,
...)
(Xn:n=l,2,
It follows that there exists a sequence of measurable sets such that Xn c X A , Xn
... . It may
n = l,2,
(P)
has property
and
satisfies XI-X,~x'-x, , so n n But XA-Xm 4 X-X, , s o p(X-Xm) = lim p(X'-Xn,) set X, =
< n-l
Xn
(a
= 0
for
is ascending. The
~(xA-x,)
u I xn
(Xn:n=1,2, ...)
p(XA-Xn)
be assumed, by (i) again, that
< n-1
for all n
. Thus the sequence
.
has the desired properties (sets of measure zero are
neglected). THEOREM 86.2. For.any ideal
e x i s t s a sequence
X
1. XL
i n M(X,p)
L
such that
u(Xn)
<
with carrier m
and
x
XL
, there
E L f o r a22 n
% I
x' +
PROOF. Let
XL
and
say that E property
(X::n=1,2,
p(XA> <
be a sequence of measurable sets such that
...)
for all
has the property
(P)
.
n
(P)
. For any
if
xE E
L
(X;)-bounded
set E
, we
. It is evident that the
shall
satisfies the conditions (i) and (ii) of the exhaustion
theorem. Concerning condition (iii) we observe first that for any measurable subset E
of
with p ( E ) > 0 there exists a function f E L
XL
If (x)l > 0 holds on a subset of exists a number
E
> 0
such that
E
such that
of positive measure. Hence, there If(x)l
>
E
holds on a subset F
of
E
144
Ch. 1 2 , 1 8 6 1
INTEGRALS ON IDEALS OF FUNCTIONS
,
of positive measure. Since qF5 If1 Property
(P)
this implies that
xF
E L .
satisfies, therefore, condition (iii) of the exhaustion
theorem. The exhaustion theorem can now be applied. It follows that there exists a sequence
(Xn:n=1,2,
...)
with the desired properties.
A s a corollary we find that if
-
having the same carrier X L , that p(Xn) <
L2
are ideals in M(X,u)
then there exists a sequence X
E L1 n L2
and
Ll and
.
for all n
'Xn
XL
4
such
We shall now prove that an order continuous linear functional on an ideal in M(X,p)
can be expressed by means of an integral with respect
. This
to the measure p
is one of the reasons why it is meaningful to call
an order continuous linear functional an integral. THEOREM 8 6 . 3 . For any i n t e g r a l
holds for a l l
f E L
. We may
assme that
, and then
L
XL of
carrier
L
on t h e i d e a l
$
there e x i s t s a u-measurable r e a l f u n c t i o n
in
M(X,u)
t(x)
on
t(x)
vanishes outside t h e
X
such t h a t
is u-almost everywhere uniquely
t(x)
determined. is a positive integral on
PROOF. It may be assumed that $
the preceding theorem there exists a sequence X (X )-bounded set E n
satisfies
defined for any set E restrict E
xE
E L
n
4
XL
such that every
. Hence, the number
of this kind. We write A(E)
-
to the subsets of one particular Xn
,
=
. By
L
$(xE)
$(xE)
is well
, and
say for n
=
we first
no
. Given
, we have $(x ) f $(X,) such a set E and given that E 4 E as k + k Ek since $ is a positive integral. In other words, we have A(Ek) + A(E) , is a measure on the u-measurable subsets of X
A
which shows that Furthermore, p(E)
=
0
implies A(E)
= 0
. Hence, by
theorem, there is a u-measurable function t(x) A(E)
on
"0
.
the Radon-Nikodym
x"0 such that for all p-measurable subsets E of X The function no , but it is evident that by varying depends on the choice of "0
=
1E
t(x)dp(x)
t(x) the function t(x) that A(E)
=
.
can be extended u-almost uniquely to Uy Xn
t(x)dp(x)
holds for every (Xn)-bounded set E
=
%
such
. Setting
Ch. 12,5861
= 0
t(x)
145
ORDER BOUNDED OPERATORS
outside XL
, we
get
s(x) is now a step function are measurable and (X )-bounded, then Finally, given 0 5 f E L , there exists a sequence
for every (X )-bounded set E. Hence, if s(x)
=
ZT ai xE. (x)
.
t(x)s(x$du(x)
$(s) =
X 05 s f f
, each
n
s
n
$(sn) 4 +(f)
and
such that all Ei
a step function. Then
since
$
is an integral, as well as
by the theorem on integration of increasing sequences. Hence
The extension to an arbitrary real Conversely, if that matter, only on
I
t(x)
f E L
follows immediately.
is a real v-measurable function on
X (or, for
XL ) such that
t(x)f(x)dp(x)
X
exists as a finite number for every on
,
L
for all
f E L
,
then the linear functional $
defined by
f E L
,
is evidently an integral on
In the particular case that L
L
.
is a Banach lattice, we have
by Theorem 85.6. This implies that every integral on
-
bounded linear functional. Let L = L ( X , p ) P 1 5 p 5 and let Q be an integral on L Hence that
Q
for some p
. Then
L
L"
= L*
is now a norm
L
satisfying
is a Banach lattice.
is norm bounded and there exists a measurable function t(x)
such
INTEGRALS AND SINGULAR LINEAR FUNCTIONALS
146
=
$(f)
1
Ch. 12,1871
t(x)f(x)du(x)
X
.
It i s well-known (and not difficult f E L P that @ is norm bounded) that now t E L (X,LJ) for q particular it follows that every integral on L_(X,w) above with t E Ll(X,u) for all
.
to prove by using -1
p
-1 + q
= 1
.
In
is of the form as
87. Integrals and singular linear functionals In the present section we continue with the investigation of the order dual
L"
of a Riesz space L
.
In accordance with the definitions in
section 84 any sequentially order continuous element of integral and any order continuous element of L" integral. The bands in L" integrals are denoted by
s2
functionals. We have L
Lzn
2
Li ,
and
Lx respectively. The elements in the
L z are called singular linear functionals.
.
L" = L" d L" The elements in the c s are called normal singular linear
is Dedekind complete, we have
disjoint complement L"
of :L = :L
Q
Lin
. In view of
it follows easily that each
so
+
as well as in Lin
.
=
L ;
Q
L ; ,
and
Lin
=
L" c L" we get n c has a unique decomposition
E L"
= L z n Lin is a band in :L Note that L" c,sn hence it is a band in L" Furthermore
L ;
is called an
consisting of all integrals and all normal L ;
disjoint complement L i of Since L"
L-
is called a normal
LL,sn
d
N
Ls
,
and
.
L" = {O} if and only if L ; = L ; , i.e., if and only if every c,sn integral is a normal integral. Finally note that by the band decomposition
Evidently we have
for every
$
E L"
. Similarly for
4-
and
By Theorem 84.4 order separability of
1$1
.
L i s a sufficient condition for
Ch. 12,8871
ORDER BOUNDED OPERATORS
147
every integral to be a normal integral. If for every principal ideal A there exists a strictly positive linear functional on order separable and so every integral on L
,
A
then L
is
is normal. This holds in
particular if there exists a strictly positive linear functional on the whole of
L
.
Finally, given 0
5 $
$
be the restriction of
E L" to A
and given the ideal A
. Then, by
uniquely determined minimal positive extension $A
. If
space L 5 $
E Li and
$[A
L
.
is an integral on A
,
let $ ] A
there exists a
of
$ \ A to the whole
is an integral (or normal integral) on
$IA
is an integral (or normal integral) on
0
in L
Theorem 83.7,
A
, then
$A
Hence, if for example
,
then
$A
=
0
.
We uroceed with some examples. Let X
EXAMPLE 87.1.
be an arbitrary non-empty point set and
Riesz space of all real functions on
X (the values
therefore excluded). The order dual L" any xo E X
the functional J,
First of all we prove that L"
-
and
m
L
the
are
has nonzero elements because for
, defined
is a positive linear functional on
+ m
by
$(f)
=
L which is not
f(xo)
for all f E L
,
the null functional.
consists exclusivelv of integrals. i.e.,
. For this purpose, let 0 5 $ E L" and let un J. 0 in L . We have to show that $(u,) .C 0 . Note that u J. 0 is equivalent to monotone n pointwise convergence of un(x) to zero on X . It is no restriction of the generality to assume that u,(x) > 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
ORDER BOUNDED OPERATORS
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)
x
in A
,
(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)
f be a
.
=
X(A)
=
1
.
is constant A-almost everywhere on the atom A fE L
final result is now that any
is A-almost everywhere on X
.
The
equal
to a X-step function, and we may write $(f) = 1; f(xi).A(Cxi}) The functional 0
5 $ E
the corresponding measure observed already that
$
A
+
L" =:L
Zy f(Ai).A(Ai)
.
is a normal integral if and only if
does not have any uncountable atoms. It was
is normal if
h
does not have uncountable atoms.
Conversely, observe that
for any subset A of A , we have
xB
of X 4
, so
if we let B
xA , and
hence
$(xB)
run through the finite subsets 4
g(xA)
if
$
is normal,
150
i.e., A ( B ) 4 A(A) A(B)
Ch. 12,8871
INTEGRALS AND SINGULAR LINEAR FUNCTIONALS
= 0
. Note now
that if
A
for every finite subset B
do not exist uncountable atoms f o r
is an uncountable atom, then
of
A
A
if
, but
A(A)
. Hence, there
0
>
0 is normal.
As an additional remark, we observe that an infinite set X
is said
to have a measurcrble cardinal if there exists a o-additive non-negative measure and
A
A(E)
on the collection of all subsets of =
X
such that 0 < A ( X ) <
.
X
0 for every at most countable subset of
exist such a measure, then X
m
If there does not
is said to have a non-measurable cardinal.
If in the case above X has a non-measurable cardinal and A is again the on the space
measure corresponding to the positive linear functional Q L
of all real functions on
X , then it is obvious that
any uncountable atoms and hence E
Q
A
does not have
is a normal integral. It follows that
.
is now a normal integral, i.e., L" = L" = L" It is not c n known whether there exist sets having a measurable cardinal, and so it is every
L"
not known either whether in the discussion above the measure
A
really
can possess uncountable atoms. Every countably infinite set has a nonmeasurable cardinal. It is known that, under an extra assumption, there exist uncountable sets having a non-measurable cardinal. It was shown, for example, by S . Banach and C. Kuratowki ([11,1929) that, under the continuum hypothesis, the interval
(x:O<xsl)
in the real line has a non-measurable
cardinal. For further information about non-measurable cardinals we refer to S.M. Ulam ([11,1930) and A. Tarski (Cll.1962). EXAMPLE 87.2. This is a specialization of the preceding example. Let X be the set of all real numbers and on
X
. Let
L
the Riesz space of all real functions
0 be a positive linear functional on
L
with corresponding
A , as in the preceding example, and assume now that A has an Every f E L i s a A-step function, so there exists a uncountable atom A
measure
subset Af
.
of A
such that
implies A(Af) = A(A) all x E X )
> 0
. It would
. We
f
is constant on
A
in this case, and
Q
so
and
A(A-A ) = 0
f
apply this to the function f(x) = x
follow that
contradicts the hypothesis that A atom with respect to
Af
, which (for
consists of one point only, which
,
Af and therefore Af
,
is an uncountable
. Hence, there do not exist any uncountable atoms is of the form
Ch. 12,1871
.
f E L
for every
151
ORDER BOUNDED OPERATORS
The same argument works if X
is n-dimensional real
number space l R n . A s observed in the preceding example, it follows from the s o obtained
formula for $(f)
"
is normal, i.e., L
that every integral on L
N
Lc Note that this has been proved now without using that the set of real =
numbers has a non-measurable cardinal. Note also that the space L
=
Ln. N
in this
.
Hence, order separability example is not order separable, and yet L" = L" c n of L is sufficient but not necessary for L" = Li to hold. EXAMPLE 87.3. It is not known whether there exist sets having a measurable cardinal and s o it is not known either whether in Example 87.1
. There are other examples Li . Let L be the Riesz space of all
the space LL can be properly larger than in
which LL is indeed larger than
L;
real (finitevalued) Lebeseue summable functions on denotes that
f(x)
5
g(x)
holds for all x
in
, where f
[O,ll
. Hence. we
C0,ll
5 g
do not
have identification of functions differing only on a set of measure zero. The Lebesgue integral on
L
is a positive integral in the sense of our
present definitions, but not a normal integral. There exist nonzero normal integrals as well; for any defined by
$(f)
=
f(xo)
,
xo
in C0,ll
the linear functional @
,
is a normal integral. Observe that the space L
in this example is Dedekind a-complete. It follows that Dedekind a-comnleteness nf a Riesz space L
.
does not necessarily imply that
L" = L" Of course, super Dedekind completeness of L implies :L = Li c n since super Dedekind completeness imDlies order senarability. For some remarks about the case that L examp1e
.
is Dedekind complete we refer to the next
EXAMPLE 87.4. Let X be a non-empty point set and of a l l real bounded functions on X for any
xo
all f E L
. The space
E X the linear functional
, is
$
X({x))
X >
the Riesz space
. Let
$(f) = f(xo) 0 5 $ E L"
for
. Exactly
of X , E is a o-additive non-negative measure on the collection of
the function A all subsets of
L
L
Li has nonzero elements;
, defined by
a nonzero normal integral on
as in Example 87.1, if we define A(E) =
which
,
0
$(x
)
for any subset E
. There are at most countably many points x E X for , say xI,x*,... . Hence, writing X 1 = X-U.{xil , we have
Ch. 12,5871
INTEGRALS AND SINGULAR LINEAR FUNCTIONALS
152
for every If
.
f E L
X has a non-measurable cardinal, then X I
measurable cardinal, every
f t L
,
and
X(Xl)
so
0
. This
has likewise a non-
implies that $(fXX ) 1
is now of the form
Q(f)
so
=
on
Conversely, assume that every non-negative integral Q form. We prove that in this case X such that 0 < A ( X ) <
X
countable subset E
of
A (i.e., h(f)
corresponding to
m
. Denoting by
X
=
4
x E X
u (x) J. 0 holds for every n
for every h-summable
X
, we
for any at most
the Lebesgue-Stieltjes integral is A-summable,
. Furthermore, since
L
on the collection of all
A
X(E) = 0
and
IX fdh
evident that every bounded function on certainly defined on
is of this
has a non-measurable cardinal. If not,
there exists a o-additive non-negative measure subsets of
L
0 for
=
u
J. 0
f), it is
$
so
is
in L means that
Q(un) C 0 by the properties
have
of a Stieltjes-Lebesgue integral. This shows that
is as well an integral
$
in the sense of our present definitions. By hypothesis, therefore, Q the form ( 1 )
for every
f E L
. But
that
$(f)
follows from formula ( I )
A({x}) =
for every x E X
= 0
0 for every
f E L
is of
, so
, i.e.,
it
$ is
the null functional. On the other hand 4(x,) = A(X) > 0 by hypothesis. Thus we get a contradiction. Hence, if every non-negative integral is of the form ( l ) , then X
has a non-measurable cardinal.
It follows now immediately that X and only if every
Q E L"
where the real numbers where
C.lail <
(ai:i=1,2 ,...) points in X $(f)
m
. As
Z.la.1 < 1
1
C.a. f(x.) 1 1
Q
depend upon
1
m
and (x.:i=1,2 ,...)
converges for every
$
C.a. f(x.1 1 1
, and
1
,
is a sequence of
f E L
, and
. In addition, one sees integral on L . This implies L
has a non-measurable cardinal if and only if every
is normal integral, i.e., if and only if L" = L" c n
It is an open problem whether Lz complete Riesz space L set X
x.
of this form is a normal
now immediately that X integral on L
and the points
is evidently an integral on
= Cai f(x.)
easily that any
ai
=
a kind of converse, note that if the real numbers
satisfy
, then
has a non-measurable cardinal if
can be written in the form Q(f)
.
Lz holds for every Dedekind An affirmative answer implies that every point =
has a non-measurable cardinal. Indeed, if
present example, then L
.
L
is the space in the
is Dedekind complete, and s o if Dedekind complete-
Ch. 12,1871
153
ORDER BOUNDED OPERATORS
Li , then L" = L" holds now for the space L which c n we consider here. As observed above, L" = L" is equivalent to the c n condition that the underlying point set X has a non-measurable cardinal. ness implies :L
=
In Theorem 87.10 we shall prove in the converse direction that if every point set has a non-measurable cardinal, then Lz Dedekind complete space L
.A
=
Li holds for every
situation, similar to the situation in the
Examples 87.1 and 87.4, is discussed by G.W. Mackey ([11,1944). EXAMPLE 87.5. We shall present now an example in which
L" = Li
.
In other words, there are no integrals except the null functknal. Let L of all real continuous functions on the
be the Riesz space C([O,ll)
interval [O,ll with the natural pointwise ordering. Furthermore, let
...)
be the set of all rational numbers in C0,lI
(rn:n=1,2, pair
(m,n)
. For any
of natural numbers, there exists a function umn
is L
such
that
Let vm
=
sup(uml,
...,umn) . Then
vml 5 vm2 5
...
, the pointwise limit
function f (x) of this increasing sequence exists on C 0 , l l and the m function fm satisfies 0 2 f (x) 5 1 for all x and fm(rn) = 1 for m all r Observe nowthatby condition 1;) the set Fmn = (x:v mn (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
<
1 , which
implies $(wm) >
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($)
contained in the n u l l space (or k e r n e l ) of
9
, which
is is by
Ch. 1 2 , § 8 7 1
ORDER BOUNDED OPERATORS
d e f i n i t i o n t h e l i n e a r subspace of a l l i s a normal i n t e g r a l , then
The d i s j o i n t complement
(N($))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 <
JI
5
L" so sn ' i t follows t h a t
is a s t r i c t l y
.
To prove t h e second p a r t of t h e theorem we observe f i r s t t h a t f o r
$
0 S $
E L z we have
$ =
4,
on
, which
N($c,sn)
shows a l r e a d y t h a t
i s a normal i n t e g r a l on t h e q u a s i order dense i d e a l proof t h a t 'that
A
i s normal on
minimal l i n e a r extension
,
&
O S J I S $c , s n
E Lsn
J,
hence
E Lin
C($)
N($)
L
that
N($) = N ( l $ l )
. It
C($)
(ii) If
with
E :L
,
N
E Lc,sn
$
This holds a l s o i f $
L
then
of
$
N($n) # L that
C($)
and
\$Isn =
lesnl , we
i s normal on
, 2
so
L
.
C($,)
C($,)
#
L :
L
. If
@
{O]
On # 0
E Lz
, so ,
N($)
$
is
s o by
C($) = { O )
.
be Archimedean and C(I$l) =
{OI
. We
o t h e r words, we
, then , then
t h e normal t h e band
i s Archimedean, we have
= {N($n))d
{O}
with
L ,,:
C($) =
$n = 0
. Since
C($)
C($)
$
may assume
i s quasi o r d e r dense, so
E L i with
satisfies
i s not t h e whole of =
N($)
SO
N($sn).
i f and only i f
. In
0 S $
{N($,)Jdd
-
E Lc,sn
, i.e., I $ \ E , i.e., I $ ] E
have t o prove t h a t i f
$ 2 $n 2 0
$
L
E Lz,sn
$n
$
,
Lin
E L i the quasi order dense ideal
i s not Archimdean. Now l e t
$
component
c,sn
C($) tB N($) c N(Osn)
C($) =
have t o prove t h a t
N($n)
.
n
i s i d e n t i c a l l y zero on
is s t r i c t l y p o s i t i v e on
$
follows t h a t
t h e l a s t theorem we have
let
$
.
is p o s i t i v e . Then
$
normal on
J,
N(J,c,sn)
( i ? For every
PROOF. ( i ) Since
n
.
i s Archimedean, then
C($) = { O )
and
i s normal, and s o t h e
$c,snlA
i s included i n the quasi order dense ideal
(ii) I f
. Then
A
IA i s normal. But 'c,sn This implies t h a t JI E L i
i s included i n
A
COROLLARY 87.7.
$
i.e.,
of
J,
so
,
A
$
For the
i s normal, assume
$
i s normal on
$
i s t h e n u l l f u n c t i o n a l . It follows t h a t
, and
A
such t h a t
L
.
N($c,sn)
i s t h e l a r g e s t i d e a l on which
N($c,sn)
i s an i d e a l i n
$c,sn = $-$n
J,
ch. 12,1871
INTEGRALS AND SINGULAR LINEAR FUNCTIONALS
156
#
, contrary
{O]
. It
follows then from
t o our hypothesis. Hence
= o . W e s h a l l prove now t h a t , under t h e hypothesis t h a t every s e t has a
non-measurable c a r d i n a l , every i n t e g r a l on a Riesz space with t h e p r o j e c t i o n property i s a normal i n t e g r a l . For t h i s purpose we f i r s t prove a lemma.
LEMMA 87.8. Let
A
. Then there
be a non-empty subset of the complete Boolean
e h s t s a d i s j o i n t subset C = ( c :aEIa)) of B ( i . e . , a 1 # a2 implies c A c = e , where e i s the smallest element a a2 i n the Boolean algebra) suc!~ that sup C = sup A and such that f o r every algebra
B
Ch. 12,1871
157
ORDER BOUNDED OPERATORS
a
a E {a) there e x i s t s an element
E
PROOF. The c o l l e c t i o n of a l l d i s j o i n t s u b s e t s of element i n t h e subset i s majorized by an element of
.
ca S a
A .satisfying
such t h a t each
B
i s non-empty and
A
p a r t i a l l y ordered by i n c l u s i o n . Each chain i n t h e c o l l e c t i o n has a l e a s t upper bound i n t h e c o l l e c t i o n (obtained simply by taking t h e union of t h e s e t s i n t h e chain). Hence, by Zom's lemma, t h e r e e x i s t s a maximal subset C
of t h e Boolean algebra
Evidently and l e t
c'
c' # 8
and
a.
such t h a t
B
that
be t h e r e l a t i v e complement of c
= sup aT
,
a r e equal t o c'
. Writing
c' = 8
A
so
c'
a
A
=
A
a
c
t o the s e t
A
c'
, we
C
, so
# a.
c
c < a
with r e s p e c t t o
, we
a.
a
0 A
= 8
, i.e.,
a
#
8
0 ' , so
have
by Exercise 4 . 1 3 . Not a l l
c'
. Hence
# a.
c
A = (aT:TE{T))
now
sup(c'AaT)
0 8 , because otherwise
t o our hypothesis t h a t Adjoining
has t h e s t a t e d p r o p e r t i e s .
C
. Assume
c = sup C 5 sup A = a.
,
c' = 8
f o r some
c'
A
aT
contrary
T =
T
0 ' '0 get a l a r g e r subset s t i l l possessing
'0 t h e p r o p e r t i e s s t a t e d a t t h e beginning, thus c o n t r a d i c t i n g t h e maximality
of
C
. Hence
; i n o t h e r words, sup C = sup A
c = a.
THEOREM 87.9.
L
Let
such that f o r euery
.
be a Riesz space with the projection property
f E L
the ideal
Af
f
gemrated by
i s a point
s e t with a non-measurable cardinaZ. Then every integral on L PROOF. I f a positive
has an i n t e g r a l which i s not normal, then t h e r e e x i s t s
L N
which i s not t h e n u l l f u n c t i o n a l . By Theorem 87.6
$ E Lc,sn
thenullideal
N($)
i s q u a s i o r d e r dense. Since
i s then o r d e r dense, i.e., L
S
u
4 u
with
i s Archimedean, N($)
$(u) > 0
uT E N ( $ )
L
r e s p e c t i v e l y . Then
$J(U ) =
. Furthermore
0
sup Q,
= P
d e f i n e a non-negative la)
$(P u) = 0
and
P = sup ?
lemma t h e r e e x i s t s a system
index s e t
T
. Note
i s a complete Boolean algebra. Let
L
PT be t h e p r o j e c t i o n s on the bands generated by
such t h a t
for a l l
i s a complete Boolean algebra. Equivalently, t h e s e t
of order p r o j e c t i o n s on bands i n
$(u) > 0
there
h a s t h e p r o j e c t i o n p r o p e r t y , so by Theorem 30.6 t h e set of a l l
p r o j e c t i o n bands i n and
L
0 < u E L with
f o r any
0
e x i s t s a d i r e c t e d system now t h a t
i s normal.
and T
and
and every p
P
u $(Pu) =
by Thoorem 3 1 . 2 ( i i i ) . By t h e preceding
(Oa:a€{a]) of mutually d i s j o i n t p r o j e c t i o n s
measure
by s e t t i n g
for a l l
u
Qa
i s majorized by some
PT
. We
now
on 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 t h e
p(0) = 0
f o r t h e empty s e t
0
and
158
Ch. I2,§88l
LARGEST IDEAL OF ORDER CONTINUITY
f o r every non-empty subset whoie set
{a) we have
Theorem 3 0 . 5 ( i i ) . Hence a t most countable s e t
of
A
sup A
u(Ia1)
. It
t h e measure
u
implies
. Note
sup PT = P
=
$(u) > 0
that for
, so
A
equal t o t h e
sup 0 u = Pu = u
, whereas
u(A) = 0
has thus been proved t h a t
by
f o r any
{a) has a
= 0 u f o r every a , it follows t h a t a a can be t r a n s f e r r e d t o t h e s e t of a l l p (note t h a t
measurable c a r d i n a l . Writing now
a, # a2
fa]
=
p
a1
# pa2
p
), s o
(pa:aE{a])
has a measurable c a r d i n a l .
Since t h i s i s a subset of t h e p r i n c i p a l i d e a l t r i v i a l l y be extended t o t h e set
A,
. It
AU
, the
measure
follows t h a t
A,
u
can
has a measurable
c a r d i n a l , contrary t o our hypothesis. Hence t h e assumption t h a t
L
has a
non-normal i n t e g r a l leads t o a c o n t r a d i c t i o n , and s o every i n t e g r a l on
L
i s normal. COROLLARY 87.10. Asswne that every point s e t has a non-measurable cardinal. Then every integral on a Riesz space with the projection
property i s a normal integral. In particuZar, every integral
071
a Dedekind
complete Riesz space i s normal. This theorem i s due t o W.A.J. EXERCISE 87.11.
Let
X
Luxemburg ( [ 4 1 , 1 9 6 7 ) .
be an a r b i t r a r y t o p o l o g i c a l space and
t h e Riesz space of a l l real continuous functions on u
n
for
E L
$(un)
-+
o
n = 1,2, f o r every
...
and
4 E L"
.
u (x) n
J.
0
X
. Show t h a t
holds pointwise on
X
,
L
if then
HINT: Similar t o t h e proof of Example 87.1.
88. The l a r g e s t i d e a l on which every order bounded o p e r a t o r i s order continuous
Let
L
be an a r b i t r a r y Riesz space and
M
a Dedekind complete Riesz
space. We use t h e same n o t a t i o n s as i n the preceding s e c t i o n s . I n p a r t i c u l a r
Lb
i s an abbreviation f o r
k(L,M)
. Similarly f o r
W e r e c a l l t h a t t h e null o p e r a t o r i s denoted by
on
L
, however,
Lc,Ls,Ln and Lsn 8 ; f o r the n u l l functional
.
we u s u a l l y w r i t e an ordinary zeno. I n a l l t h a t follows t h e
Ch. 12,1881
case t h a t
is t h e space of real numbers i s of s p e c i a l importance.
M
A
DEFINITION 88.1. For any non-empty subset
A0 = (T:TELb,Tf=O for
all fEA,1
of Lb
B
For any non-empty subset B
L
the (L,M)-anni-
.
the inverse (L,M)-annihiZator
0 B
of
i s defined by OB = (f:fEL,Tf=O f o r a l l E B )
. Note
L
A c L
that i f B c L"
and
Lb
i s a l i n e a r subspace of
Evidently, Ao of
of
of A i s defined by
Ao
hilator
159
ORDER BOUNDED OPERATORS
. and
i s a l i n e a r subspace
OB
Lb = L - ,
i s r e a l number space, then
M
so for
we have now
t h e usual d e f i n i t i o n s f o r a n n i h i l a t o r and i n v e r s e a n n i h i l a t o r i n t h i s s p e c i a l case ( f o r a f u r t h e r d i s c u s s i o n of t h i s case we r e f e r t o t h e next
A c A cL 1 2 implies o B 1 3 OB2
s e c t i o n ) . Note t h a t BI c B
Lb
of fl
2
c
L
b
, then
Tf = 0
# f2 in L
for a l l
implies t h a t
3
OLb
, it
is evident t h a t i f
does t h e l a r g e r s e t LEMMA 88.2.
.
A
0 1
3
A
0 2
3
Lo =
{el
and
I f OB = {O} f o r some subset B E B implies f = 0 I n o t h e r words, T f l # Tf2 f o r a t l e a s t one T E B . This i s
o f t e n expressed by saying t h a t OB
implies
.
T B
separates t h e p o i n t s of B
s e p a r a t e s t h e p o i n t s of
L
. Since
L
, thenso
Lb'
If there i s a subset of
that separates the points Lb of L , then L i s Archimedean. I n particular, i f there i s a subset of L" that separates the points of L , then L is Archimedean.
have
0
5
nv
0 5 nTv 5 Tu
for all
T E
Lb . Then OLb = I01 . . For any 8 < T E Lb we 'n 1,2, ... , so Tv 0 . It follows t h a t T v = 0 implies v = 0 on account of OLb = {O] . Hence L
OB = 10)
PROOF. Let
Assume t h a t
for
& , which
B
f o r some subset
< u for n =
= 1,2,
...
in
of
L
=
LARGEST IDEAL OF ORDER CONTINUITY
160
ch. 12,8883
i s Archimedean. If
and
OB = (0)
f E L
satisfies
. I n p a r t i c u l a r , i f u E L+ . The f o l l o w i n g lemma shows
f = 0 u = 0
Tf = 0
satisfies
for a l l
Tu = 0
T
E B , then
for a l l
T E B
,
then
t h a t under c e r t a i n c o n d i t i o n s t h e i n v e r s e
holds. LEMMA 88.3. If T
u E L+
E B and
Lb
i s an idea2 i n
B
PROOF. It i s s u f f i c i e n t t o show t h a t f o r t h i s it i s enough t o prove t h a t
E B
0 5 T S
E B , so
. Hence,
let
Sf = 0
0 5 T
E B
. For
{B)
A
IT1
ITI(u) = 0
for
A
. It
E A'
lSul
5
. Hence,
,T E
0 5 u
Tu = 0
for
e
8 5 T A
Ao
T(lfl) = 0
B
and
for we have
IS1 5 T
Lb f
have '
T
. We have
0 2 u E A
in
let
0
Su
Lb
0
IS1
, so
with
Tu = sup T,u
, i.e.,
{B] =
.
B
. If
show f i r s t t h a t
T E Ao
E A and
T E Ao
implies
If1
5
u
. Then
I t follows t h a t
and =
0
then OB i s a band i n L
n'
0 S u
i s a band
Ao
is an idea2 i n L
, then
. We
L
, then
OB
S
IT1
t r i v i a l l y . Since
IS1 E Ao
, we
is an ideal i n L e t now
for
, then
b e an i d e a l i n
S E
IS1 (u) = 0
0 B
.
i s s u f f i c i e n t t o show t h a t
0 S u E A
j u s t proved, and s o
Ao
0
=
Lb
E A , s o Tf = 0 by h y p o t h e s i s .
Assume now t h a t
for a l l
If1 E
satisfying
S
i s an idea2 contained i n L
( i ) Let
PROOF.
implies
Tu = 0
implies
implies
any
i s the band generated by the idea2 B
B
i s an idea2 i n L
i s an idea2 i n
B
If
( i i i ) If
f
( i ) If
.
(ii)
0
f E
0
.
by h y p o t h e s i s . It follows by Theorem 83.9 t h a t
THEOREM 88.4.
Lb
OB = { O ]
B
f E
T ( l f 1 ) = sup(lSf1:ISIST)
in
such t h a t
, then
u = 0
impZies
T E Ad
S E Ao
TT E Ao
. Then 0
/TI E Ao a s s u E A implies
. This
f o r a11
shows a l r e a d y t h a t T
,
so
T u = 0
( c f . t h e proof of Theorem 83.4),
. This
shows t h a t
Ao
i s a band.
so
.
ORDER BOUNDED OPERATORS
Ch. 12,1881
B be an ideal in Lb
(ii) Let
f E
already that and
lgl
.
If
S
0
In the last lemma it was proved
implies If1 E B 0 Then If1 E B , and s o B
. Assume now that lgl E
follows now from 6 5 T E B that lTg\ 5 T(1gl) 0 so g E B This shows that OB is an ideal.
.
{B}
Let
0
B
.
For this purpose, let
Now, let
6
S
0 2 u
g E L , f E
0
B
trivially. Since it
=
be the band generated by the ideal
c O{B}
B '
161
0
0
, we
B
. We
E B , so
Tu
=
Tg = 0
have
,
have to prove that 0
for all
.
T E B
. There exists an upwards directed set {TT} in B . Hence T0u = sup T u = 0 . This shows that u E OIB} . HenceT OB c O I B I . (iii) Let B an ideal contained in Ln . To show that the ideal OB 0 is actually a band, let 0 5 u I. u with u E B for all T . Since all T
in B
T
in B
To E {B}
6 5 T
such that
.
To
4
are now order continuous, we have Hence
Tu
=
DEFINITION 88.5.
s e t of a l l
f E L
0
, so
u
E OB
.
La (more precisely
By
such that
+
If1 t u
0 = Tu
La(L,M)
implies
0
Tu
4
)
for any positive
we shall denote the
Tun
+
0 f o r every
TELb.
Similarly, we shall denote by the s e t of a l l
f E L
such t h a t
/fI t uT
L
Lan
+
0
(as an abbreviation f o r
I
Lan(L,M)
1 i s any dowmards directed s e t i n , then inflTu I = 0 f o r any T E L,, .
such that i f
Iu
Note that La
=
(,EL:
If Itu +O
implies Tu J.0 for every 8STELb
Lan
=
(fEL:If12uT+0
implies TuT+O for every 8STELb
THEOREM 88.6.
La and
n
1.
Lan are ideals i n
L
. For
any
T E Lb
r e s t r i c t i o n TILa i s a-order continuous (by d e f i n i t i o n ) . Hence largest ideal i n L on which every Lan i s the largest ideal i n PROOF. We prove that La
T E Lb
the i s the
i s o-order continuous. Similarly
on which every
L
La
T E Lb
i s order continuous.
is an ideal. The proof for Lan
is similar.
It follows immediately from the definition of La that f E La holds if and only if If( E La holds. Also lgl 5 If1 and f E La imply g E La It remains to prove that fl,f2 E La implies f +f E La Hence, let fl,f2 E La
and
If,+f21
t un J. 0
. We have
1 2 to show that
.
Tun
J.
0 for
.
162
LARGEST IDEAL OF ORDER CONTINUITY
. It follows from our assumptions that
8 5 T E Lb
Ch. 12,1881
I f l /+
I f 7I
t u
n 4 0
,
so
v
Ifll
=
-
+ If21
un 4
By Theorem 15.5 there exist sequences 0 such that vn l+vn2 w
nl
=
vn
= /flI
for all n
-
v
5
. Hence
v i 0 , wn2 nl
If 1
4
nl
1
If2/ -
=
.
.
If21
IflI +
0
and
5
v n2
+ If2/
v 40 n2
and w +wn2 = un for all n In view of fl,f2 E La we have Twnl nl and TWn2 J 0 for every T t 9 in Lb By addition Tu 4 0 for 0 5
T E
.
+
O
. This is the desired result.
Lb
COROLLARY 88.7. La
if and only if Ln
=
=
.
Lb
Lc
L if and only if
=
Lb
. Similarly, Lan = L
The main theorem in the present section follows now. THEOREM 88.8. La
and
OLS
=
Lan
=
OL
.
sn
. We prove first that La . For this purpose it is sufficient to show that for 9 5 T E L . Given f E La and 8 5 T E L s '
PROOF. We present the proof for La contained in OL implies Tf
=
0
restriction TILa
is o-order continuous on the ideal La
minimal positive linear extension TI of L
and it satisfies 8
Ls
is an ideal), so
(since f E La and
T
S
TI E TI
=
1
5
T
. But then La )
T, E Ls
.
= 9
2
so
Tu
9 5
T E
4 0
u
. Then all
u
=
n 0 holds for any 9
,
i.e., f E La
and s o the
(since T E Ls
. Hence
For the proof of the converse direction. assume that If1
the
T / La is o-order continuous on
Lc I7 Ls , i.e., T 1
T on
,
is f E La
are members of
Tf
=
1
f E OLS
OLS (since O L S
and
T f = 0 and
is an ideal),
5 T E L and all n . Assume now that Lb . Then T has a unique decomposition T = Tc+T s with Tc E Lc and Ts E Ls , s o T u J. 0 and T u = 0 for all n . It follows that s n c n
n
Tun J. 0
.
It has thus been proved that O L S c La
It follows immediately from the definitions that A c '(Ao)
non-empty subset A
of
L
and
B c (OB)'
.
for any
for any non-empty subset B
of
Ch. 12,1881
163
ORDER BOUNDED OPERATORS
. In particular, O L
.
c (L~~)'. L~ implies ts c (L~)' Similarly, Lsn It is, therefore, a problem of some interest to determine when one or both
tb
of the equalities
=
(La)'
=
Ls
(Lan)O
or
Lsn will hold. There is one
=
particular case for which the answer is easy to give. A s observed above, we have La
=
L
Lc
=
Lb , i.e., Ls
{el
=
L
if and only if
case the equality
(La)'
=
0
L
=
a 0 (L )
THEOREM 88.9. The equality
=
{ e l , and in this
holds. =
Ls
holds i f and onZy i f
.
d i f f e r e n t members of Lc h a w d i f f e r e n t r e s t r i c t i o n s t o La Similarly, (Lan) 0 = Lsn holds i f and only i f d i f f e r e n t members of Ln have d i f f e r e n t restrictions t o
.
L~~
PROOF. Note first that the statement about La may a l s o be formulated a 0 holds if and only if it follows from 8 # T E Lc (L ) = L
by saying that
L a ) ' . In view of L 0 L = { 9 ) it is evident that if tbat T is not in ( c s a 0 (L ) = Ls and 8 # T E Lc , then T is not in (La)' Conversely, let it
.
Lc the operator T is not in (La)' , and is properly larger than Ls , i.e., there exists an
be given that for 8 # T E (La)'
assume not that operator T E (La)'
, because E Lc and
IT/ Tc
such that T
(La)'
Ls
and
. Then
Ts E Ls
is not in Ls
Tc # 9
since
Hence, by our hypothesis, Tc is not in follows from 0 2 T shows that
(La)'
:s
IT1 E (La)'
S
Lb
are ideals in
IT1
(La)'
that Tc E
not properly larger than
. The same holds then for . Let IT1 = Tc+T with is not a member of
Ls
. On the other hand, it (La)' . This contradiction a 0 Ls , so (L ) = Ls .
There exists a close connection between order denseness or super order denseness of (Lan)O
=
Lsn
La
or
dense if the band generated by
A
is L
there exists an upwards directed set The ideal A
THEOREM 88.10.
Lc
*
such that
(i) I f
Ls
or is order
in A
such that 0
u f L+
if for any
0s u
n
I. u
.
u E L+
La i s super order dense i n
L
5
u
+
u
.
there exists
, then
(La)'
=
Ls.
La i s super order dense i n L and if L separates the a 0 L , then (L ) = Ls and t h e p d i n t s of L are aZready separated
(ii) I f
by
in A
=
in L
itself, i.e., for any
{uT}
is super order dense in L
a sequence {un}
points of
(La)'
Lan and the conditions that
should hold. We recall that the ideal A
164
LARGEST IDEAL OF ORDER CONTINUITY a 0 (L ) = Ls
(iii) ConverseZy, if
then La i s order dense i n L
.
PROOF. (i) Assume that La show that (La)'
=
Ch. 12,0881
Lc separates the points
and
is super order dense in L
. We have
to
Ls holds, which by the preceding theorem is equivalent
to showing that if a a-order continuous operator T vanishes on
. Hence, let
T = 0
of L,
Lc and Tf
T E
=
0 for all f E La
La
. For every
, then
0 c v E La we have
and since
If1
5
v E La
satifying
If1
5
v
implies f E La , we have
f
Tf = 0 for all
. Hence ITI(v) = 0 . We shall prove now that ITI(u> = 0 . Given u E L+ , there exists by hypothesis a sequence (un:n=1,2,...) in La such that 0 5 u + u . By the result just n established we have IT1 (u,) = 0 for all n (since u E La) . Furthermore n for every u E L+
we have
ITI(un) 4 ITI(u)
IT1 (u) = 0 (ii)
since
. This holds for every
Assume that La
IT1
is a-order continuous. Hence
u E L+
,
IT1
so
0
=
is super order dense in L
have to show only that OLC
,
= {O}
sufficient to show that if u E L+
,
and hence T = 8
and
. We
OLb = {Ol
.
i.e., in view of Lemma 8 8 . 3 , it is and
Tu
=
0 for all T E L
then
c'
. Hence, let us assume that u E L+ satisfies Tu = 0 for all T E Lc . By hypothesis there exists a sequence 0 5 u + u with all u n n La . Since Tu = 0 for every T satisfying 8 5 T E L , we have Tu = n for 8 5 T E L and all n . Also Tu = 0 for every T satisfying u = 0
8 5
T E L
an:
in 0
all n (since u E La = L ). By addition it follows that n satisfying 8 5 T E Lb and for all n But
.
Tun = 0 for every T
"L
= { O } by hypothesis, so b it follows that u = 0
u
(iii) Assume that
=
.
(La)'
=
0 for all n
Ls and
0
Lc
. On account of
= {O}
. Then
L
0 2 un .f u
is Archimedean,
order denseness of La is the same as (La)dd = L , i.e., the same as ad ad We have to show, therefore, that (L ) = { O } holds. Assume (L ) = { O )
so
(La)d # { O }
. . Then
(La)d
collows then from O L such that
Tu > 0
contains a non-zero positive element u
= {O}
that there exists an operator
. If necessary, we may replace
linear extension of
TI (La)d
,
Lc
T by the minimal positive
s o we may assume that
which implies that T vanishes certainly on La
. It
0 C T E
T vanished on
. This
shows that
(Laldd,
Ch. 12,1881
T E (La)'
165
ORDER BOUNDED OPERATORS
. Hence
Ls
=
T E Lc
(La)d # { O }
It follows that
Il
so
T
, which contradicts Tu
= B
is impossible, so
(La)d
(ii) Lan is order dense i n (Lan)'
=
L and
= {O}
L
.
an 0 (L )
, tlzen
separates the points of
Lsn and Ln
.
> 0.
= Lsn Lb separates the points of L
(i) If Lan i s order dense i n
THEOREM 88.11.
adn only i f
,
Ls
L
.
i f
PROOF. Similarly as the proof of the preceding theorem.
(i) Lc = Ln implies La = Lan . Ls holds, then the eonverse of the above statement holds as well. I n other words, i f (La)' = Ls holds, then La = Lan i f and only i f Lc = Ln . I n t h i s case ( i . e . , i f a l l these conditions are an 0 s a t i s f i e d ) , the condition (L ) = Lsn i s also s a t i s f i e d . COROLLARY 88.12. (ii) I f
(La)'
=
PROOF. (i) lC
L~ implies L~
=
Ls
Note first that
. We have
=
Ls
c
Lsn ,
(La)'
(ii) Let
so it
.
0
L~ = O L sn = L~~ to show that La = Lan implies =
follows from La = Lan
and
(La)'
Lc =
=
Ls
L n.
that (Lan)O = (Lalo
=
Ls c Lsn
an 0 On the other hand the inclusion Lsn c (L ) is always true. Hence an 0 (L ) = Lsn , which immediately implies
Ls and so
L
c
=
L
=
n
a 0
(L )
an0
= (L
=
Lsn ,
(by taking disjoint complements).
For examples that the conditions in Theorems 88.10 and 88.11 and Corollary 88.12 are best possible in some respects, we refer to Exercises 88.14-88.16
at the end of this section.
THEOREM 88.13.
Let
L~(D) the s e t of a l l f o r every
that
T E D
If1 2 u
I ~ zI
denote by
Lan(D)
. Similarly,
J. 0
Lb and denote by c o implies TU n + o
D be a non-empty subset of f E L such t h a t
implies
u
the s e t of a l l T E D
infITu I = 0 f o r every
.
f
E L such
166
The s e t s
(i)
and
La(D)
continuous respectively (ii) Denoting by
, we have
D i n Lb
.
and
BD
{Dl
La(D) = La(BD)
.
are ideals i n L , the largest
Lan(D)
T E D i s u-order continuous o r order
ideals i n L on which every
=
Ch. 12,9881
LARGEST IDEAL OF ORDER CONTINUITY
=
the ideal and the band generated by and
L~({D})
L~"(D) = Lan(BD)
=
L~~(IDI)
Lb , then
(iii) If B i s an ideal i n
Lan(B) = Lan(BnLsn) =
=
0
(BllLsn)
PROOF. We restrict ourselves to indicating the proof that La(B ) D holds. Let f E La(BD) We have to prove that f E La({Dl) La({D})
.
For this purpose, assume that If1 t u J. 0 n we have
,
Taun Gn
so
T_ 4 T with all T
9 S
0 for all
Tu = (T-Ta )u
+
a
. For
in BD
.
.
. Let
any fixed a
(T-Ta) (If I ) + Taun
Ta un
=
,
.
inf Tu < (T-T )(If/) This holds for every a Since , we get inf Tu = 0 This shows that f E La({D}) = n as desired. so
.
infa(T-Ta);lfl)
EXERCISE 88.14. of
La
It was shown in Theorem 88.10 that super order denseness OLb
together with
implies OLc = 103
= {O}
shown in Theorem 88.11 that order denseness of
. Show that if
implies OLn = { O }
,
OLb = { O }
Lan
. Similarly, it was
together with
OLb
= {O}
is omitted, then these con-
clusions do not necessarily follow. HINT: If
Lb
the case that M dual L"
,
=
0
=
Lan
=
L
,
so
a an L = L
is super order
.
= L = {e} does not deparate the points of L In n is the space of real numbers the space Lb is the order
so
=
{Ol
L
b
= {9}
,
is now the same as L"
= {9}
. For examples
for
holds we refer to Examples 85.1 and 85.2.
EXERCISE 88.15. OL
La
, but Lc
dense in L
which L"
; then
=
then La
It was shown in Theorem 88.10 that if is order dense in L
a 0 (L )
. Similarly, it was
=
Ls
shown in
and
167
ORDER BOUNDED OPERATORS
Ch. 12,9881
= Lsn and OLn = 1 0 ) , then Lan is order Theorem 88.11 that if (Lan)' n 101 it is only dense in L . Show that if instead of OL = { O } or o ~ =
assumed that
OLb
,
= {O}
functions on
C0,ll
do not necessarily
then the same c:nclusion
follow. For an example, choose for L
the space of all real continuous
and use the fact that L"
consists only of singular
linear functionals (as proved in Example 87.5). HINT: L" =
' ( L "
L"
=
s"
=
) =
{Ol
=
LLn
L
Li
=
. But
separates the points of
Lyn
, i.e.,
La = Lan La
=
Lan
L
,
'(L")
so
. It follows that
= {O}
=) ; L ( '
(La)'
is not order dense in L
=
.
=
(Lan)O =
EXERCISE 88.16. It was shown in Corollary 88.12 that L c = L n implies an a an = L I n the converse direction, it was shown that L = L together a 0 an 0 with (L ) = Ls implies Lc = L and (L ) = Lsn Show now that if
.
a L
.
(Lan)O = Lsn
holds, it may occur that
Lc # Ln
HINT: Let L be the space of a l l real bounded functions f
on the
La = Lan together with and
(La)'
interval
# Ls
.
(x:O<xil)
for every (i.e., Q
f E L
,
f
such that
most countable set. Let M
is an integral on
then Q
, but
is in LL
has a constant value
be real number space, s o
L
, but
af outside an at
Lb
= L"
. If $(f)
=
, and s o L; # Lyn recall that on the sequence space
not in L" ) . Hence Lc # Ln N
n
For the determination of
La
, we
af
not a normal integral
N
.
Lrn there exists a positive linear functional that is not an integral (this was proved in section 85). Applied to the present situation it means that if f E L
and
satisfies f(x ) n
t E >
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
s e p a r a t e s t h e p o i n t s of
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
s e p a r a t e s t h e p o i n t s of
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
ORDER BOUNDED OPERATORS
6f-u
n
, so
4 Bf
in
Bf
is
Bn
_<
. We may . Since
n and i t f o l l o w s t h a t , t h e component
and
f
2
un
,
pn 4 6f by
C 0.
the
the
Bn
(because
, so
.
assume t h a t
. Denoting pn
exhausting
La = La
f E La
Tfin 4 0
n
f
E La
).
t h e component of
(unlnn
of
u
n
, so
is
O < ( u ) S 6 f . n nn The component Tu
(un)in
,
< T(6f) + T f i n
6 > 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
ORDER BOUNDED OPERATORS
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
ANNIHILATORS AND WEAK TOPOLOGIES
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
ORDER BOUNDED OPERATORS
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
ANNIHILATORS AND WEAK TOPOLOGIES
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
separates the points
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
ORDER BOUNDED OPERATORS
. 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
separates the points
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
ORDER BOUNDED OPERATORS
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}
<
.
m
L+
in (a)
0 2 u
Then, f o r any
5
> 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
_<
.
.
exist sequences {v,} 0 S v 5 z n n
such that
let
, we
S
...,um )
n,m is a X-Cauchy
increases. For brevity, write
. Then
inf(x l,...,x )
=
u
5
yn
n
z
n n,m For every" n
we get un
twn3
= sup(un,
set u
en)
S u , so
increases as n
We may assume that m z
,
.
n n,m m sequence. Hence, there exists a natural number m
n,m x =nu
{vnl
,
be given. For every n
> 0
E
,
n
f o r aZZ
n
C 0
p(zn)
for all
< E
= v +w
n
and
C
n
for m > n . Then u
n
and
e x i s t sequences
such that
0 5 v
x = u
...
n = 1,2,
for
and
{wn}
and
0 5 wn 5 yn
in L+
for all n
such that u
n
= v +w
. The conditions
n
n
with
(a), (b), (c)
are now satisfied. COROLLARY 90.2. ~(2,)
j.
(i) I f
0 i n concZusion
(ii) If
A(f)
I Cp(f)
p
is a Riesz norm i n L
, then
(a) of the l e m a implies t h a t f o r some
concZusions of t F z lemma hold i f
un
C > 0
zn
f E L
and a l l
E L+ f o r all n and
PROOF. (i) Evident. (ii) The lemma becomes trivial. Since ZA(un) <
-
zn C
.
and
j.
0
,
then the
CP(U,)
, there is
<
a natural
CARRIER OF A LINEAR FUNCTIONAL
I78
number N v
n zn
=
N
0
, wn
=
z
=
sup (u,,
n
=
such that X(un) u
=
n
...,UN) .
and for n
Let
COROLLARY 90.3.
(c)
u
$(f)
n
5
already i f
...,N
. For
,
n > N
take wn = 0 , v
=
n
take
u
and
n
0
...
n
c
u E L+ for a l l n
PROOF. A(f) = $(lfl)
,
n
v +w f o r a l l n n
~p(f) f o r some
+
p(zn)
5
=
I,
N
_<
(b)
0
=
2
be a Riesz seminomi i n L and l e t
P 0
v 5 z + and n n $(wn) < E for a l l
(a)
If
o
>
E
for all n
E
E L".
5 $
u 2 u E L f o r n = l,Z, and Cp(un) < m n , t h e r e e x i s t sequences {vn} , {wn} , {znl i n L+
Furthermore, l e t again Then, for any such t h a t
<
Ch. 12,5901
0
.
,
. and a l l
> 0
n and
f E L , the conclusions hold
Cp(un) <
m
satisfies the conditions of the lemma.
Let L be a Riesz space As observed already in section 87, for any $ E L"
the set = (f:fEL,l$l Ifl)=O)
N($)
,
is an ideal in L
called the n u l l i d e a l or absolute kernel of is called the carrier of
disjoint complement C ( $ ) = {N($)Id
$
$I
. The
. This
notion should be distinguished from the notion of the carrier of an ideal of measurable functions, as defined in section 86. The null ideal of should also be distinguished from the n u l l space of of all f E L
satisfying $(f) = 0
null space of
$
of
C($)
N($)
and
.
0
i s the band i n L" 0
( t h e r o l e s of (ii)
C($)
=
$
E L"
generated by
and
W
C(l$l)
I t follows then t h a t
(iii) F o r
and
.
$
and
W'
We have $
and
o~
N($)
$
which is the set is contained in the
If convenient, we shall write N 4
THEOREM 90.4 (i) For any
of
. Evidently, N($)
$
=
C
$
N(l$l)
instead
= ' 8 ,
where
i s the inverse a n n i h i l a t o r
are now taken over by
L
and L" 1.
and p ( f ) = I $ l ( l f l ) i s a Riesz norm on C($) $ i s Archimedean. $ p o s i t i v e t h e nu12 i d e a l of sup($,$) i s the
C($)
.
ORDER BOUNDED OPERATORS
Ch. 12,1901
i n t e r s e c t i o n of null ideal
and
N($)
L
For (v) (vi) For C
PROOF.
E L"
c (N )dd J
the
$ E L"
.
then N($)
Archimedean and
$,$
$
I t follows t h a t f o r any
i s the i n t e r s e c t i o n o f N ( $ + ) and N($-) is a o-ideal. I f $ E L i , then N ( $ )
N ( $ ) = N(I$/)
(iv) If 4 E L,: i s a band.
.
N($)
179
,
0 # $ E L:
- -
we have
# I03
C($)
.
we have
C IC $
J
,
C c (N$)
dd
J,
.
(i), (ii) and (iv) are evident from the definitions. In (ii)
we use that every normed Riesz space is Archimedean. This follows immediately nu S v
by observing that
0
all n
is the norm),
(where p
(iii) For
(v) For
5
J,
and
$
for n so
...
l,Z,
=
implies p(nu)
, i.e.,
p(u) = 0
t
=
L ,
E L i and L Archimedean the null ideal N 8 C is L itself. Hence, if $ 0 $ = 0 It follows that 0 # $ E L i implies d c ( N ~ )implies ~ ~ C$ I ( N ~ =) (N$) ~ ~ = ~c $
c Ic +
J
,
.
so
c
(vi)
$
C IC $
*
{ll;}
and
Hence
C$(un)
Since p(f) 90.2(i). {Vn}
,
-
$ = 0
(inf($,$)}(u)
such that u = u +u'
in L+
< =
$(If!)
and
0
n
5
u n
n 5 u
and J, and
Cb # I01
, we
have
.
. Conversely,
are positive. Let $(un)+$(u;)
for all n
is a norm on C
, {zn} in
0
= {O}
$
L+
0 5 u
E
that there exist sequences {un}
, we
As a result we find that for any
{w,}
4
C
J,
PROOF. We may assume that
It follows from
J,
is a band and
N
JI E L,: then C c N w i t h $ E L-and $ J , C c (N ) dd by p a r t (vi) of t h e l a s t theorem). $ 4
THEOREM 90.5. If $ 1 (and hence
.
( ~ $ 1=~ ( N ~ dd )
implies C$ c
for
positive we have
the band generated by N
.
u = 0
p(v)
5
such that 0
5
5
. Now we
2-n
C $ *
and
for all n
.
apply Corollary 90.3.
can combine this with Corollary E
v
n
> 0 5 z
there exist sequences n
J. 0
, v(wn)
<
E
and
I80
Ch. 12.1901
CARRIER OF A LINEAR FUNCTIONAL
. Then
for all n u =v+w n n n which implies that
u:
= u-u
=
n
u-vn-wn ,
.
u-v
so
n
ul+w n n'
=
-
u Z u-v Z u-z +u Hence, since J, is u-order continuous by n , so $(u) 5 E This holds hypothesis, we have J,(u-vn) + $ ( u ) as n +
with
for every
,
> 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
ORDER BOUNDED OPERATORS
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<%u,wEC in Cg
I
. Hence, if
L
,
so
(N9)dd
case the proof is trivial. Let us assume now that
a
Ad
$ '
PROOF. (i) By definition we have
Let
0
C+
0
i s Dedekind o-complete o r i f =
L i s Dedekind a-complete or i f
i f and only i f
0
. Under
then A
E L"
is an order dense ideal. The same is true for is equal to
implies that
it is included in the o-ideal
L
@
C
$
has the in this
is Dedekind o-complete
is positive. The space L
is order dense. Hence, given 0 < u E L
C
p
4)
, we
. There exists a sequence 0 < wn 4 < u with . Since L is Dedekind o-complete,
such that +(wn) I. a
Ch. 12,1901
183
ORDER BOUNDED OPERATORS
the element wo
=
sup wn
exists in Cb
. Then
wo
u
, and we shall prove
and
sup(wo,w) > w 0 ’
S
that
If n o t , there exists w E C so
it follows from
0 5 w 5 u
such that
0
SUP(W0,W) - sup(wn,w)
5
wO-wn
that
which implies (since 0
is strictly positive on
~(suP(wn,w~-wn)
> 0
O(sup(wo,w)-wo)
2
C
0
) that
.
But the expression on the left tends to zero (since $(wn) 4 a ,
so
that
,w)) tends likewise to a ). Thus we get a contran diction. Hence formula (1) holds. Observe now that for 0 5 v E N and
necessarily Qi(sup(w 0 5 w
E C0 the condition v+w
conditions v
S
u
and w 5 u
u
0
is equivalent to the separate
(since v+w
sup(v,w)
=
in view of the
condition that v I w ) . Hence, writing
we find for any fixed v1 E N
and so
u
0
with
0
S
v1
5
u
that
is an upper bound of A . Evidently, every upper bound of
greater than or equal to u , s o
u = sup A
. Then
A
is
CARRIER OF A LINEAR FUNCTIONAL
184
so
wo E C
(ii) If
$
L = N
0
Hence $
u = (u-w ) + wo
It follows that L = (Nm)Bd
$ *
N
. Hence
u-wo E (N$)dd
$
@
u-wo E (N$)dd
with
C
0 ‘
dd is a band, s o N = (Ng) 4 qJw by part (i). Conversely, assume that 0 5 $ E L and
C
$
N $
=
(B
by part (i).
C 4
L
$(vn)
J. a
0 < vo
5
N
=
= a
$
@
C
it may be assumed now that u
$
. There
. Let us
is a sequance v
such that
.
inf(vo,uo) < vo =
’0
6
is strictly positive on
- inf(v
v
T
in L
. In
for all
T
4 0
E C
$
-
u ) 2 vo
n’ 0
. Let
such that vn C and
vo Then there exists also an element
4 (vo-inf(v o,uo)) since $
u
Tn assume that there exists an element
for all n
vn
(u,:T€{T})
= u
.
We have to prove that
$ ( u T ) J. 0 holds for any downwards directed system
infT $ ( u , )
and
is a normal integral, then N
is a band. Then L
view of
Ch. 12,1901
such that
, so
1
C
$ ’
It follows from
inf(vo,uO)
that
.
for all n Now we have $(v ) C a so $(inf(vn,uo)) < $(vn) - B n there exists a natural number no such that < a-48
$(inf(vno,uo)) The system of all u u T!
This contradicts inf, $ ( u , ) = a
a
=
0
,
i.e.,
EXERCISE 90.10. and
$*(f)
=
f(0)
functionals on L $,
and
+2
so
.
is directed downwards, so there exists an element
in the system satisfying u
Hence
,
$(u,)
Let
C 0
5
inf(v no,~O)
. Hence
. It follows that .
‘1
L = C(E0,ll)
for all f E L such that N ( $ , )
,
. Show that
then
and
is a band and
$ ( u T ) < a-16 1
vn 4 0
,
so
$(vn)
ji’2
.
4
0
if
$l(f)
$2
are positive linear
N($2)
=
.
f(x)dx
i s not. Note that
are singular linear functionals by Example 87.5.
Ch. lZ,§9Ol
ORDER BOUNDED OPERATORS
185
EXERCISE 90.11. Let L be the RBesz space of all real functions f on [0,11 any
E
[O,ll
>
for which there exists a real number f(m)
0
, we
have
. Show that if
If(x)-f(m)l $l(f)
=
for all f E L
f(m)
such that, given
for at most finitely many
> E
, then 0,
x
in
is a non-
normal positive integral on L with
, but
an order dense ideal in L
not a band. Show also that if
(an:n=1,2,...) is a sequence of positive numbers such that Xan < $ (f) = .Eanf(n-l) is a positive normal integral on L such that 2
.
m
, then
is a positive non-normal integral such
N($*) C N($l) Hence, $ = that N($) = N ( I $ ~ ) is a band.
EXERCISE 90.12. For the proof of Exercise 90.14 we need the following 0 5 $ E L- and
result, a particular case of much more general result. Let 0
5
u
E L
.
Show that the integral component $
of
$
satisfies
n HINT: Denote the right hand side by Show first that Or(u+v)
5
or
is additive on
+ gr(v)
$,(u)
. For
L+
$ (u)
. Evidently
. It is evident that
0
5
$,(u)
5
$(u).
the inverse inequality, apply the Riesz
decomposition theorem to a sequence 0 < w 4 u+v satisfying n lim $(wn) < $,(u+v) + E . It follows that $r can be extended as a ppsitive linear functional on L
.
Show now that $r is countably additive on L+
by observing first that, by definition,
.
.
Let u = Cv For the proof that $,(u) = Z$r(vn) , let E > 0 For every n n , there exists C vnk = vn with Ck $(vnk) < $r(vn) + c/Zn Then w = Cn j=l CA=l B vjk satisfies 0 s w t u and $(wn) < C Ipr(v.) + E for all n J n Hence $,(u) is < Z $,(v.) The inverse inequality is evident. Thus $r J It-follows that $r 5 OC an integral satisfying 0 5 $ 5 $ Since It follows also Q c 5 $ , the definition of $r implies that ($=)= 5 $r
.
from the definition that
.
.
($c)r
.
=
$c
. Hence
$c
2
$r
.
.
.
186
CARRIER OF A LINEAR FUNCTIONAL
EXERCISE 90.13.
Show t h a t i f
Ch. 12,1903
has t h e Egoroff p r o p e r t y , t h e n i n
L
t h e p r e c e d i n g e x e r c i s e t h e infimum i s a t t a i n e d . HINT: For every n a t u r a l number
that
s Q c ( u ) + n-'
$(un,)
for a l l
0 5 v
p r o p e r t y t h e r e i s a sequence
. Then
m,n
. Hence
j = j (m,n)
appropriate
Let
Show t h a t i f
0 5 Q E L"
(ii)
Show t h a t i f
0
Q
ll C
JI'
= 0
$,(u)
0 5 u i.e.,
n
4 u
u
to the
.
. By t h e d e f i n i t i o n of u
such t h a t
v
u
such
t h e Egoroff
s u
m .) 5 Qc(u) + n'j "3 l i m $(vm) Qc(u)
5 Q(u
, so
m
and
,
$ E L"
implies
0 5 u E L
f o r an for all
.
Q
then
$
, but
I CJI
b e given. S i n c e
C
.
, then
0 s JI E L"
and C
i s singular, then C(Q) = C($c)
0
.
= {O]
C(QhJI) =
n o t conversely. i s s i n g u l a r , we have
fj
Furthermore, by t h e Egoroff p r o p e r t y , t h e r e e x i s t s a sequence
0 5 Q(un) I. QC(u)
such t h a t
E N for all n n Q o - i d e a l g e n e r a t e d by
(ii)
m
0 s Q E L"
Hence, Q I JI
HINT: ( i ) Let
+
+
unk
have t h e Egoroff p r o p e r t y .
L
(i)
( i i i ) Show t h a t i f = C
t h e r e i s a sequence
k
Q(vm)
Q(vm) 5 Q C ( d f o r a l l
EXERCISE 90.14.
n
Since
0 2 $ = $;+QS
0
( i i i ) Note t h a t i f
. This N
Ib'
, so
$(un) = 0
shows t h a t every It f o l l o w s t h a t
= Qc v
Q = Qc+$,
Qs
and
and
0
C
u E L+
belongs
=
= {O}.
0
(Nib) d
, we
C(Qs) = { O }
JI
JI +JI
=
n
for a l l
E
S
,
have
, then
so
EXERCISE 90.15. implies
Q I$ (i) that
C
Q
It w a s proved i n Theorem 90.4 t h a t
CQ c (N$)dd
Show t h a t i f
L L
and
Q I JI
implies
and conversely.
has t h e Egoroff p r o p e r t y , t h e n
i s c o n t a i n e d i n t h e o-ideal generated by
( i i ) Show t h a t i f
Q,$ E L"
N
JI'
has t h e Egoroff p r o p e r t y and
N(I
$1
hlJI1)
i s order
dense in L HINT:
,
then C
= 0
that J,(u,!
$
+
IN 1
is contained in the band
(i) We may assume that $
then from $ I $ J,,(u)
187
ORDER BOUNDED OPERATORS
Ch. 12,§911
that C$ c N(Jlc)
and
, 80
N
$ '
are positive. It follows
J,
any
generated by
0
. But then, as above, there exists a
<
u E C
satisfies
$
sequence 0
5
u -t u
such
n
= O for all n , and so u is a m e m b e r o f t h e o - i d e a l g a n e r a t e d b y N
. There exists a directed system 0 < u 4 u for all T . Note that uT E C for every T such that u E N(l$lhl$I) $ and also that $ I $ holds in the ideal N ( l $ l h l J I I ) . Hence, by part (i), every u belongs to the o-ideal generated by N . Then u E IN 1 for $ T $ all T , 5 0 u E {Nql . (ii) Let 0
5
u E C+
$
EXERCISE 90.16. Let the Archimedean space L have the Egoroff property and let $,$ E L". dd I~(l$lhl$l)1 = L
.
Prove that C I C
$ 4 J
if and only if
91 Complex Riesz spaces All Riesz spaces investigated in the preceding sections are reel vector spaces. We shall now extend some parts of the theory to complex Riesz spaces. Given the Riesz space L L
x
of all pairs
L
(f,g)
with
, we
f and
g
equip the Cartesian product in L with a vector space
structure in the usual manner by defining that
(a+iB)(f,g)
=
(af-Bg,Bf+ag)
for a,B
real numbers,
and the so defined complex vector space is denoted by
L+iL
with
f nad
(f,O) E L+iL
g in L
, the
, we
space L
write
. Instead of
. Also, if h E L + iL and h f+ig € Re h . Identifying f E L and
(f,g) we appropriately write f+ig
=
=
is embedded in L+iL
as a real linear sub-
space. The so defined complexification of a given (real) Riesz space L
is
only of interest if it is possible to define the notion of the absolute value
Ihl
of an element h E L + iL
in such a manner that for h E L
new absolute value coincides with the existing one in complex numbers we would like to define
L
. Just as for
the
I88
Ch. 12,1911
COMPLEX RIESZ SPACES
Ihl ,Note th
=
(
sup Re(he-ie):0<8<2n
).
if this supremum exists in L
, then
Ihl
occur in pairs, for 8 and
elements Re(he-ie)
)
=
2
, since t ie , and such that
0
B+n
-Re(he-').
Hence, if the supremum exists, then
For the special case that =
, so
h cos 0
h
is an element of
, we
L
-i€l have Refhe ) =
the supremum exists and is equal to the already existing
absolute value
Ihl
=
sup(h,-h)
.
To derive a simple condition sufficient for the existence of the desired supremum, we first consider the complexification of the Riesz space C(X) space X
of all real continuous functions on a compact Hausdorff
. Taking the complex function
the ordinary abslute value
Ihl
/ h /= (f2+g2)1/2
h
=
f+ig with
f,g
in
C(X)
,
satisfies
E C(X) ,
but also Ihl
=
supg Re(he-ie)
=
supe (f cose+g sine)
,
where the supremum must be taken with respect to the ordering in the Riesz space C(X) function
.That the last supremum exists and that it is equal to the is seen by observing that, for every fixed x E X
(f2+g2)l12
the number Re(h(x)e-ie)
is less than or equal to
with equality holding for exactly one value of h(x) that
# 0 , and equality for all 0 if h(x) Ihl
is on X
f cose+g sine
, and
,
=
0
0
(depending on x ) if
. It follows, therefore
the pointwise supremum of the functions so
Ihl
is also the supremum of this set of functions
Ch. 12,5911
ORDER BOUNDED OPERATORS
with respect to the ordering in
189
.
C(X)
We observe furthermore that, bytheYosida representation theorem in section 4 5 , if the Riesz space L
is Archimedean and has a strong unit,
then there exists a compact Hausdorff space X isomorphic to a linear subspace of complete, then L
. If
C(X)
such that L in addition L
is Riesz is uniformly
, as proved
is isomorphic to the whole space C(X)
in Theorem 45.4. This implies the following result.
If
LEMMA 91.1.
L
is Archimedean and uniformly complete, then
f cos8+g sin8:0<8<2n
exists i n
L
for a l l
PROOF. Let
f
f and
and
g
g in
L
I
in L be given. For every
we
8(/0<8<2n)
have
,
If cos8+g sinel < If1 + lgl so all elements
in L by
f cos8+g sin8 are members of the ideal A
generated
. According to what was observed above, the ideal
If1 + lgl
is Riesz isomorphic to a space C(X)
with
f and
Hence, denoting the images of
g
X
compact and Hausdorff.
in C(X)
by
f^
and
g-
respectively, the supremum
exists in C(X)
, and
sup8(f
)
cos8+g sine
exists in the ideal A THEOREM 91.2.
so by the isomorphism
, and
hence exists in L
If L i s Archimedean and uniformly complete, then ):0i852n)
e x i s t s for every
.
h
=
f+ig E L+iL
, and
=
sup(f
we have
cos8+g sin8:0<8<2*
A
I90
Ch. 12,§911
COMPLEX RIESZ SPACES
Furthermore, Ihl
only for h
0
=
= 0
, ;Ah1 = Ihl.Ihl f o r
a complex
A
nwnber, and the t r i a n g l e i n e q u a l i t i e s
hold. Since any Dedekind o-complete space i s Archimedean and uniformly complete (as proved i n s e c t i o n 3 9 ) , these r e s u l t s hold i n p a r t i c u l a r i f
L
i s Dedekind o-complete. PROOF. The existence of already by taking 0
and
Ih( follows from the last lemma. A s observed together, we get
8+s
= sup(1f
IRe(he-ie)l:05052n) Taking
8 = 0
and
8
= I *, we
find that
If1
cos8+g sinel : 0 5 8 < 2 ~ lhl
2
and
lgl
5
. The
lhl
second form of the triangle inequality follows by observing that IhlI - Ih21 2 Ihl-h21 and
-
.
Ih2-hlI = Ih1-h2 I In Exercise 91.12 it will be indicated how to prove the existence of
Ihl
lh21
lh,l
5
without using that any principal ideal in L
space C(X)
is isomprphic to a
.
The dominated deoomposition property (also called the Riesz decomposition property) carries over to the complex case. THEOREM 91 . 3 . (Dominated decomposition property. 1 I f
Archimedean and uniformly complete, and i f
in that
L+ s a t i s f y h
=
Ihl
IhlI 5 f l
hl+h2 w i t h
PROOF. The ideal A to the space C(X) space X
, and
md
Ih21
generated in L
5
by
h l and
f2
.
f-+ig^
h2 i n
fl,f2 L+iL such
fl+f2 is Riesz isomorphic
of all real continuous functions on a compact Hausdorff
function identically one on X =
f+ig E L+iL and
=
the isomocphism is such that the image of
corresponding to and h-
h
f I+f2 , t h e n there e x i s t
5
is
L
...
f,g,
. We denote the functions in
in A
. Note that
fl+f2 is the
by
f-,g-,
...
,
so
1h-l = Ihl- 2 (f 1+f 2 ) -
C(X)
(fl+f2)= 1
. We
= 1
on X
define h;
ORDER BOUNDED OPERATORS
Ch. 12,8911
and h;
in
and
are the images of elements h l and
so
h2
and 5
by
h;
h-f;
=
and
h;
=
h2
. Furthermore, since the images of
hl+h2 = h
Ih;l Ih,I
C(X)+iC(X)
I91
Ih21
f l and
,
it follows from
Ih21
5
Ih;l
.
f2
5
f;
and
.
h-fThe functions 2 h; in A+iA , and
IhlI and lh21
Ih2j
f;
5
are
that
In view of these theorems it is reasonable to assume now, in the further investigation of a complex Riesz space L+iL
,
that L
is
Archimedean and uniformly complete. h l and
THEOREM 91.4. (i) I f
h2 i n
L+iL s a t i s f y
IhlI J.
then Ih,-h21 = Ihl+h21 = IhlI + Ih21 h = f+ig with
(ii) I f
f and
=
Ilhll-lh21I
sup(1h
=
g positive, then
f cosO+g sinO:O
.
(iv)
I f
lgll 5 1g21
h = f+ig and
h'
If I + ilgl with
=
f +igl and h2 1 , then IhlI 5 Ih21 hl
=
.
PROOF. (i) For arbitrary 2 inf(f
Hence, if h l and
f )
1' 2
h2
0 = 2 inf(lh
=
f2
in L we have
fl+f2 + Ifl-f21
in L+iL
1
2
f l and
g in
with Ifl[ 5 If2/
f +ig2
=
f and
I,Ih
2
I)
satisfy =
Ih
1
I
+ Ih2b
- I IhlI-lh21I
.
L
, then
and
COMPLEX R I E S Z SPACES
192
Ch. 12,1911
Combining these results we get the desired equalities. (ii)
Since f and g
are positive, we can restrict ourselves to
o s e s f ~ i . (iii) Let and
0
5 8 5
$a be fixed. We consider also the numbers
. For the given
-(Ti-e)
8
-0,v-8
in L
we take the element f cose+g sine
and we take also the corresponding elements f o r the other three values. We compute the supremum pe -8
, we
of these four elements. Taking first
8
and
get
Similarly, the supremum of the two remaining elements is
Hence
by part (i ), it follows that (iv) 1by
lg2
Ihl = Ih'I
Let h l = f +igl and
. Then
f,I
= f +ig,
f +ig2 are disjoint
2
1
Note as a corollary that h
.
h2 = f +ig2 with
and h2 = 1 (i.e., lh,l I /h21 ) if and only if both f l and g l k Also, if h = Z both f2 and g2 with all n=l 'nhn joint, then
.
1
5
If2]
and
2
are disjoint from h
mutually dis-
193
ORDER BOUNDED OPERATORS
Ch. 1 2 , § 9 1 1
Freudenthal's spectral theorem can also be extended to the complex case, as follows. THEOREM 9 1 . 5 . Let h
=
L be Dedekind o-complete and l e t
( w i t h f and
f+ig
g in
L
)
satisfy
Ihl
5
e
e E L+
. Furthermore,
be a p o s i t i v e number. Then there e x i s t d i s j o i n t components e and complex numbers
with
Iy
n
I
5
1
for
n
PROOF. Since L
yl,
=
I,
...,yk
and let
...,ek
E
of
el,
such t h a t
...,k
is Dedekind u-complete, we can apply Freudenthal's
spectral theorem for the real case. Hence there exist disjoint components ei -1
,...,e'P S
a
m
5 1
,..., aP
e and real numbers a l
of
for m
= 1,
...,p
such that
satisfying Cp e' = e and 1 m
It may be assumed here that for a 2 0 we have a e t 5 f + and for m m m a < 0 we have la e'I 5 fHence l a e'] 5 If1 for m = 1 , ...,p m m m m m Similarly, there exist disjoint components ey, e" of e and real
.
numbers j = I , ...,q
,,...,
satisfying Cq e" 1 j such that fiq
=
.
e and
..., 4 -1
5
.
B.
J
5 1
for
Note that some of the am for j = 1 , . . . , q and l!'ejBl 5 lgl and j ' may be zero. Let e = inf(e' e ' ! ) for all m and j involved. Then all mj m' J = e and e are mutually disjoint, C . e mj m~ mj
Hence, writing
ymj =
am+iB , we have j
COMPLEX RIESZ SPACES
194 Ih-C y e I mj mj mj
IBjemjI
.
e'I 5 If/ and IB.e'!l 5 lgl that lamemjl 5 I f 1 J J m m l g / for all m and j Hence, by part (iv) of the
It follows from
and
5 Ee
Ch. 12,§911
5
la
.
preceding theorem, we find for all m
and
j
that
.lemj = Iymjemj I = lamem j+iB.e jmj'' mJ
(y
If+igl = Ihl
e
5
= C
mjemj
'
which implies
(
Iy .lemj = inf I y .lemj,ly
mJ
.
mJ
. lemj
mJ
This shows that lymjl 5 1 for all m and j .lemj 2 e mJ mj # 0 , we get satisfying e # 0 Renumbering the elements e mj mj so
Iy
.
Ih-C y e I I n n
k
5 Ee
...,ek
for appropriate disjoint components e l , yI,...,yk
and comp ex numbers
bf absolute va ue at most one. Since
follows from the triangle inequality that
II
k
I c 1Ynen
k hl -CnI ynI en
The subset A
of
5 Ee
.
L iL is called an i d e a l if A is a complex linear IhlI 5 Ihl with h E A and h l E L+iL implies
subspace of L+iL and if hl E A
. This is the same definition, therefore, as
for an ideal in a real
Riesz space. The set of all real elements in the ideal A will be denoted by
Ar
,
i.e., Ar
=
(f:f€AllL)
.
Evidently, h = f+ig € A with f are members of
A
r'
and of course
and Ihl
g
in L
implies that f
is also a member of A,
.
and
g
Ch. 12,5911
195
ORDER BOUNDED OPERATORS
If A i s an ideal i n L+iL, then Ar is an idea2 i n Ar+iA . Conversely, i f B i s an ideal i n L , then B+iB
THEOREM 91.6.
L
, and
A
=
i s an ideal in
L+iL, and
PROOF. Let A A
A n L
=
g E A =
. It is easy to see that . Note now that for any g E L we have . It follows that for real f and g we
ig E A
if and only if
if and only if
.
A +iA r r Conversely, let
1
h E A
f +ig, with
=
1
Ihl I If1 + lgl
If A
.
and
and
B
A + B so
A+B
=
f
= f
(A+B)r = Ar+Br f E
and
and
E B
1
. Also,
.
It follows
satisfies
. This
=
g1+g2
.
Let
A
If1
8
5
(that is to
B having the decampo-
. Then
lgl
A IB ,
If1 = I f l /+ If2/ and
lgl
IflI 5 1g,1 so
=
If1
of
L+iL
and
1g,/ + 1g21
is called a band if the real part Ar subset D
.
A n B = {O}
) if and only if
follows by observing that
in L+iL
,
.
is a
the d i s j o i n t complement
is defined similarly as in the real case by =
d (
hEL+iL:lhlIlfl
It is evident that Dd fo = sup(f:fEE) f E E' satisfies
).
for all FED
is an ideal in L+iL
For this purpose, let E that
g
. Furthermore, A I B
A and g E B
. For any non-empty D
Every
Similarly
g be elements of
g
have the decompositions
D
If1 + lgl E B
so
(Ar+B ) + i(A,+B,)
lgl
Dd of
.
=
IfZ[ 5 1g21
The ideal A
,
B
fl E B
so
+ Br + iBr
+ iAr
A
and
band in L
,
the algebraic sum A+B
for all +f
so
and A = B+iB . Then A is . firtermore, let h = f+ig and lhll IIhI . We have to prove that h l E A .
,
In this case, let f sitions
Ar , and
are members of
are ideals in L+iL
is an ideal and
say, If1 I lgl
g
g are members of
f
IflI
that h l E A
and
L+iL
and
The elements IIhli S
f
B be an ideal in L
a complex linear subspace of h
.
= B
be an ideal in L+iL
is an ideal in L
have f+ig E A A
(B+iB),
. We
prove that Dd is a band. d (D )r such d We have to prove that fo E (D Ir
be a set of positive elements in
exists in L f I Igl
.
for all
g E D
,
so
fo
=
.
sup(f:fEE)
196
COMPLEX RIESZ SPACES
satisfies
f o I lgl
for all d fo E (D )r
is real, so If
D
.
is a non-empty subset of
the disjoint complement Dd Di
D
of
. This
g E D
in L
. It is
of
D
L
Ch. 12,1913
, we
in L+iL and the disjoint complement d = (D
.
( i ) If A i s an ideal i n L+iL, then A d dd LL hence (A lr = ((A,) lr = (A~)' AZSO (A )r = (A,) (ii) Every band k i n L+iL s a t i s f i e s A = Add . n (iii) If A1 A are i d e a l s in L+iL , then n = n;=, dd
.
.
PROOF. (i) Let A be an ideal in L+iL d . Conversely, if f E (Ar)d that A c
. If
g E Ar
h
=
gl+ig2 E A
Ihl S lgll + 1g21 , so d A = (Ar)' Replacing
.
with
g1 and
A
by
Ad
dd (A )r (ii) Let =
A
Ar
i.e., the bands Add = A
.
=
,
be a band in L+iL
II )
=
(A,)
in Ar
.
u
(Ar)
d
, and
Aj)dd =
A cA
I f 1 I lgl fzr all
then
g2
,
then
It follows that d (A )r = (Ar)'
in the formula
d l = ((A,) ((A 1),
=
. It follows from
/fI I Ihl , i.e., f E Ad
7
so
d
,...,
.
fo
have to distinguish between
evident that D'
THEOREM 91.7. d
. But
fo E Dd
shows that
, we
get
*
. The real part
since L
is a band in L dd is Archimedean. It follows that (A )r = Ar ,
Add
A
and
(iii) Let A ],...,A ideal in L+iL , so
n
Ar
have the same real parts. This implies
be ideals in L+iL
. Then ng=l Aj
is also an
by part (i). Using Theorem 19.3 (in which the intersection formula was proved for the real case) and part (i) again, the right hand side becomes
The real part being equal, it follows now that
A.)dd (nn J=1 J
=
dd fly=, Aj
.
,
Ch. 1 2 , 8 9 1 1
The ideal A ideal Ar Br
is included i n the band
in L+iL
in L
is included in the band
is the band generated by
generated by
A
Br
.
B
if and only if the
It follows that
B
is the
A (i.e., the smallest band containing A ) if and only
band generated by if
197
ORDER BOUNDED OPERATORS
if and only if
Ar B,
=
. In other words, B (Ar)u . But (Ar)u
is the band dd = (A ) r by
part (i) of the last theorem. Hence, B is the band generated by A if and dd As in the real case, only if Br= (A )r , i.e., if and only if B = Add
.
the ideal A
is called order dense if the band generated by
.
A
is the
This holds if and only if whole space L+iL , i.e, if Add = L+iL i.e., if and only if A, is order dense in L (Ar)LL = L
.
For any ideal A in L , so
A
fB
Ad
The ideal A
in L+iL
L
8
.
(A )
.
is order dense
is called a projection band if
in L+iL
projection bandin
the real ideal A
is order dense in L+iL
I
. In this case
.
A
Ar
is a
is a band, and it follows from
'
is the decomposition h +h 2 d of an arbitrary h E L+iL into h l E A and h2 E A , then [hi , so Ihl = lhll + Ih21 is the decomposition of
Ar
(A )'
B)
=
L
that A
Conversely, if A
(B
Ad
=
L+iL
If
is a band in L+iL
then h l E A
=
with the property that every
h E L + iL admits a decomposition h satisfy ( I ) ,
h
=
and h2 E A
d
h +h I 2
, so
such that A
fOllows immediately from Theorem 30.1 that if A bands in L+iL Particular, if in L+iL if
L
,
then A n B
L
and A+B
,
then A
8
and
We still assume that L
B
are projection
are also projection bands. In B
has the projection property and A
be an ideal in L
A
and
B
are bands
is a band in L+iL is a band in L+iL
,
. Hence, then A
@
Ad =
is Archimedean and uniformly complete. Let A,
. The quotient space
L/Ar
is uniformly complete
(Corollary 5 9 . 4 ) , but perhaps not Archimedean. By Theorem 60.2
Archimedean if and only if
Ih21
is a projection band. It
has the projection property and
satisfying A I B
IhlI and
L/Ar
is
is uniformly closed. Hence, the quotient A, L/Ar is Archimedean and uniformly complete if and only if A, is uniformly satisfies this condition, SO we have closed. We shall assume now that Ar
198
COMPLEX RIESZ SPACES
the situation that both
L
and LIA
Ch. 12,9911
admit complexification with an
absolute value. More generally, let first A
. We
Ar
denote the elements of
it follows that
[h]
(L+iL)/A This shows that
[fl+ iCgl
=
=
be an ideal in L+iL with real part
(L+iL)/A
,
L/A + iL/Ar
(L+iL)/A
by
h
f+ig
=
so
.
is the complexification of
, if
is real, i.e., Chl E L/Ar
the element [hi
... . From
[f],[g],
. For
L/Ar
and only
[g]
=
h = f+ig
,
[Ol
i.e.,
gEAr Assume now, as above, that h
=
1
Cf,]
fl+igl and h = f +ig2 2 2 [f,] and Cg,l = Cg,l
=
\hl-h21E A
is uniformly closed. Note first that Ar satisfy Ch,l = [h,] if and only if
,
and in this case we have h -h E A ,
. It follows then that
1
, so
lhll - Ih2J E A
2
Clh,c1
=
[lh211
We shall prove now that the absolute value of the equivalence class Chl
is exactly the equivalence class of I Clhll
is easy to see that
l[hll
f cose+g sine
that
Ihl
2
, i.e.,
Ihl
, because
C(X)
such that
E
>
0
in the system such that
0
I
po(x$
-
p
80
compactness X
(x) <
E
=
. It
=
, so
Cfl cos8+[gl sine
is compact and Hausdorff
is a system of non-negative functions in
(p,:e€{e))
supe p,(x)
continuous. Given
Clh19
=
exists for each xo E X
0 I p (x ) - pe (x,) 0
X , where po
pointwise on
po(x)
, there
<
0
0
for all x
, and
E
is also
a function p
hence
in a neighborhood of
x
. By
eO
the
can be covered by a finite number of these neighborhoods, SO
there exists a function q, , a finite supremum of functions pe , such It follows that there exists a that 0 I po(x) qE(x) < E on X sequence (\:n=1,2, ) of non-negative functions in C(X) , each qn
-
.
...
a finite supremum of functions pe converges uniformly to C(X)
po
sup np8
=
po
,
such that
. Hence, if
onto the Archimeden space M
relatively uniformly to and hence
.
it follows from
Clhll t Cf cos8+g sinel
To get equality, we return to the situation that X and we assume that
lChll
so
?rpo in M
,
n
0
qn I.
5
then 0
I
rqn
4
and
. This implies that
. Applying this to
and
qn
is a Riesz homomorphism of nq n converges sup aqn = sp0 ,
the situation that
Ch. 12,1911
pe = If cose+g sinel
pe
image of of
199
ORDER BOUNDED OPERATORS
in
and pi
C(X)
, as
, we
derive from
L onto L/Ar
[lhll = s u p r p e
(2)
If^ cosB+g- sinel
=
in Lemma 91.1,
=
is the isomorphic
and taking for
lhl = sup pe
supllfl cose+C:l
the mapping
?i
that
sinel
.
IChll
=
Let L be a uniformly complete Riesz space equipped with a Riesz
. Then
norm p
then np(u)
L
0
is Archimedean (if
p(v)
5
, so
for all n
nu 5 v
5
p(u) = 0
for n = 1,2
, which implies
,...
,
u = 0 ).
Hence L+iL admits an absolute value, as explained. Defining p(h)
=
sup(lfl,Igl) only if
,
for all h E L+iL
p(lh1)
satisfying p(hl)
5
p(h2)
If+igl
5
5
for
If1
+
the function p
IhlI 5 /h21 Igl
is a norm on
. It follows from
that L+iL
,
L+iL
is norm complete if and
is norm complete.
L
EXERCISE 91.8. functions f
Show that if
on [O,l]
for all x
L
in a neighborhood of
then the complexification of same holds if
L
such that f (x)
consists of all real continuous
such that f(x) L
0
a+bx
=
(a and
b
real numbers)
(the neighborhood depending on f ),
does not admit an absolute value. The
consists of all real continuous functions on [O,ll a cos (x-b)
=
in a neighborhood of
HINT: Take (I+x) + i(1-x) check first that
L
0
.
in the first example. In the second example,
is a vector space and secondly that L
is a vector
f ,f E L and fI(G) = f2(G) , then either fl and f2 1 2 coincide in a neighborhood of 0 or f l and f2 have different derivatives lattice (if
at
0 ). Now take cos x+i sin x
EXERCISE 91.9.
.
L
Show that if
is the space C(X)
continuous functions on the topological space X then
Ihl
=
supg (f cose+g sine) exists in L
Ihl (x) = (f2(x)+g2(x))1/2 Let
for all x E
the Riesz space M(r)(X,A,p) and
every x E
.
and
of all real h = f+ig E L+iL
Ihl
,
satisfies
(X,A,p) be a o-finite measure space an let L be an ideal in
Show that if in L
x
and
Ihl
x
of all real p-measurable functions on X
h = f+ig E L+iL
.
satisfies
,
then
lhl(x)
=
.
supe (f cos@+g sine) exists (f2(x)+g 2 (x))”~ for p-almost Ihl
=
200
COMPLEX RIESZ SPACES
EXERCISE 91.10.
Let
Ch. 12,1911
. It
f,g,fo,go be elements of a Riesz space L
is an immediate consequence of the Birkhoff inequalities that
There is a sharper inequality. Show that
and by induction
I
p
HINT: For p V q - p
A
q
,...
sup(f
I
fn)-sup (g
sup(f,g)
=
, write
p
A
and
q
q
=
sup(f
A
q 2 (fAfo) v (gAgo)
Ip-ql
5
p v q-r
EXERCISE 91.11. h
=
f+ig E L+iL
.
,g ) 0
, use
that
Ip-ql
=
as the supremum of four terms and derive
from this that p
=
0
=
r
. Hence
I(fvfo)v(gvgo)} 'I
-
r
=
Let L be Archimedean and uni Drmly complete. For
0
and
S 8 S
277, let h(8) = If cos8+g sin81
...)
and
Show that if 8 (n=l,Z, converges to 8 , then e = If1 + l g l n h(8 ) converges e-uniformly to h(8) Show also that if (On:n=I,2, n dense in [0,2771 and s = sup h(8 ) exists in L , then n s = sup(h(e):ose
.
...)
is
.
EXERCISE 91.12. With the same hypothesis and the same notations as in
...
the preceding exercise, let P (n=l,2, ) be the partition n e2" ) with e k = 2nk/Zn for all k , defined by P = ( e l , n h(Pn) = sup(h(O1) h(8 )) Show that (h(Pn):n=1,2 ) 2" increasing e-uniform Cauchy sequence. Hence, by the preceding
...,
,...,
the e-uniform limt
.
jh/ satisfies
,...
of
[0,2nl
and let is an exercise,
ORDER BOUNDED OPERATORS
Ch. 12,5921
This method to prove the existence of
Ihl
201
without representation theory
is due to B. de Pagter. Use this to proveagain the formula Clhll (:ormula
(2) above) in the quotient space
(L+iL)/A
.
=
Clhl]
HINT: It is evident that h(P n ) is increasing. Let m < n , and let be the corresponding partitions. By Exercise 91.10,
Pm and P n
where the supremum is taken for j the points of
I f3
-8.1
Pn
.
=
l,...,2m
situated in the interval
(Bj-l,8jl
< 21~/2~ It follows that for any given
k J Ih(Pn)-h(Pm)l
c
Ee
if m
Bk
and where
is large enough, so
E
>
,
runs through
so
0 we have
(h(P ):n=1,2, ...) is
e-uniformly Cauchy.
92. Complex order bounded operators If T
is a (linear) operator from the real vector space V
real vector space W
, then
operator from V+iV
into W+iW
into the
T has a unique extension as a (linear) by defining
T(f+ig) = Tf+iTg for all V
f and
into W
L(V+iV,W+iW) h y w+iW
T
g
in L
. The set
L(V,W)
of all (linear) operators from
can be regarded thus as a real linear subspace of the space of all complex (linear) operators from V+iV
in this set is called a real linear operator from
. An arbitrary
T
in L(V+iV,W+iW)
T = Tl+iT2 , where T 1 and
T2
has a unique decomposition
Tf
for f E L
into real and imaginary
Parts; Tf = T f+iT f It follows then that Th = T h+iT h for all 1 2 1 2 , and so T = T +iT2 This shows that the space L(V+iV,W+iW)
h E L+iL
the complexification of
1
.
L(v,w)
.
into
are real. Indeed, the operators TI and
T2 appear in the decomposition of
.
into w+iw V+iV
.
is
Now, let L and M be Riesz spaces such that L is Archimedean and uniformly complete and M is Dedekind complete (hence, L and M have complexifications admitting an absolute value). The operator T = T +iT 1 2 from L+iL into M + i M is order bounded if and only if T1 and T2 are
so,
202
COMPLEX ORDER BOLlNDED OPERATORS
Ch. 12,5921
.
hence L (L+iL,M+iM) is the complexification of Lb(~,~) The operator b T is called positive if T is real and the restriction of T to L is positive; T ever
is said to be order continuous (or o-order continuous) when-
TI and T2 are s o . For any order bounded /TI
value
T = T +iT2 the absolute 1
is, by definition, the operator
f cosO+T2 sinO:W852~1 1 IT( = sup\T1
) -
The supremum exists since Lb(L,M) (i.e., if
T
for any
E L+
u
is real), IT1
. We
is Dedekind complete. In the real case
satisfies
shall prove that the corresponding formula holds if
T is complex. For this purpose we start with some lemmas, one of which is similar to the complex case of Freudenthal's spectral theorem (Theorem 91.5). The difference is that we do not assume Dedekind o-completeness but only the Archimedean property and uniform completeness. In return we get a somewhat weaker result (Lemma 92.3), sufficient for what we wish to prove. LEMMA 92.1. I f
Hausdorff space
,... ,gn
(Ul,
, then
X
...,U
such t h a t
vanishes outside
Uk (k=l,
on
i s an open covering of the compact
there e x i s t non-negative continuous functions
x
g1
)
(gl+...+gn)( x) = I
x
on
...,n) . Such a system of
and each
gk
functions i s called
a p a r t i t i o n of unity subordinate t o the open covering
...,Un) .
(U1,
PROOF. The space X, being Hausdorff and compact, is normal, i.e., disjoint closed sets have disjoint open neighborhoods. Equivalently, if is contained in the open set U
the closed set E open set V
such that E c V c V- c U
In the present situation, En in the open set Un
.
, so
=
X-(UIU...UUn-I)
open set Un-l En-l cVn-l
C
, so -
V-
, then
,...,Un-l,Vn)
there exists an
is the closure of
V
.
is a closed set contained
there exists an open set Vn
E c V c Vi c U Note that (U1 n n n of X Hence, En-l = X-(U,U...UUn-2W
.
, where
such that
is still an open covering
) is a closed set contained in the n there exists an open set Vn-l such that
Vn-l C Un-l
, and
(Ul,...,Un-2,Vn-1,Vn) is still an open
ORDER BOUNDED OPERATORS
Ch. 12,1921
covering of
...,Vn )
. Proceeding in this manner we -
X
of X
(Vl,
writing Fk
-
=
covering of
Vk
such that V
c Vk c Uk
k
, the
for all k
such that F c Uk
X
get an open covering
= I , ...,n . Hence, ...,Fn ) is a closed k . By Urysohn's lemma there
for k
(FI,
system
203
for every
k exists, for each k = I,...,n , a continuous real function fk on X that 0 s fk 5 1 on X , f = 1 on Fk and f = 0 outside It follows that
k
I,
=
=
k f 1 +...+En
...,n . Then all Let
LEMMA 92.2.
X
Hausdorff space
...,an
al,
%
k
. Let
on X
t 1
gk
on
fk/f
=
'
for = 1
...,n) .
(k=l,
k
f be a complex continuous function on the compact
and l e t
. Then
> 0
E
there e x k t eompZex nwnbers
und non-negative continuous functions
g l , ...,gn
Iakl s llfll
gl+...gn = 1 on X and is the s u p r e m norm i n C(X)
(i)
, i.e., and
PROOF. For every x E X, le
Ux
for
= 1
k
X
on
that
llfll
k' X
are continuous and non-negative, gl+...+gn
gk vanishes outside U
X and
on
f
such
,...,n , where
llfll =sup(lf(x)/:xEX)
!If(x) 1-z:
la
k
I
such
gk(x)l
5 E
,
x
on
.
be the open neighborhood of x
defined by
ux
=
(y€x: I f (y)-f (x)
<€)
.
BY the compactness of X there exists an open covering (UXl,".,U 1 of xn Let (gl,...,gn) be a partition of unity subordinate to this covering
X
.
and let ak
=
f(5)
for k
i=
I,...,n
. Then
lakl
5
llfll
for all
k
and
Hence, conditions (i) and (ii) are satisfied. For the proof of (iii), note first that
II f(dI g,(x)-
lak Igk(x)
since gk vanishes outside Uk
I
If (x)g,(x)-a,g,(x)
. Hence
I
5
EPk(X)
,
204
for all
LEMMA 92.3. Let L be an Archimedean and uniformZy complete Riesz s?ace, Let u E L+ , f E L+iL with If1 i u and E > 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
ORDER BOUNDED OPERATORS
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<8<2a)
=
has
so
I
5
(lfl)
is real, Th
real. Since f
supelTlfcose+T2f sine = sup81(TI cosB+T cose ( l f l ) 2
IT1
5
T = Tl+iT2
u E L+
=
. Then
TI cose+T2 sine
I
Hence, by Lemma 92.4 above,
IT[(u) = sup Z. T(8
(1)
J
)u j j '
where the supremum is taken over all finite systems all u
j
are positive and
let p = X.(cosO.)u J ~j TIp+T2q = Re{T(p-iq)}
and
.
(ej,uj)
such that
X.u = u If {(O.,u.)} i s one of these sysiems, ~j J J q = Z.(sinej)uj , s o Ip-iql 5 Z.u = u and
J . This implies
~j
206
COMPLEX ORDER BOUNDED OPERATORS
T p+T q 1 2
But
T(ej)uj
= C.
J
,
Ch. 12,5921
so
IT(f+ig)/:lf+ig/
For the inequality in the converse direction it is now sufficient to holds for all h E L+iL
prove that lThl 5 ITI(lh1)
. This
is the more
complicated part of the proof, in which it is necessary to appeal to the approximation property proved in Lemma 92.3. Note that although in general an element h E L+iL does not have a polar decomposition of the form h
=
alhl
with a
a complex number, Lema 92.3 is a kind of substitute
and as such it will be of use in the proof which follows now. We prove first that
/Thl < 21TI(Ihl)
holds for every h E L+iL
.
Indeed, we have
and so
lThl 5 21TI(Ihl)
. It follows that if
h j holds u-uniformly in L+iL (all h and h in L+iL ), then Th j j holds v-uniformly in M+iM , where v = 21TI (u) This implies that by Lemma 92.5(i),
.
ITh.1 J
p E M
.+
IThl
holds v-uniformly in M ITh.1 5 p
such that
J
. Hence, if
for all j
, then
+
h
+
Th
there exists an element
lThl
5 p
.
Now, to return to theinequality we want to prove, let h E L+iL be given such that
Ihl 5 u
exists a sequence
. By
the approximation result in Lemma92.3 there
(h.:j=1,2,...)
in L+iL such that h converges j is a finite sum Z a v with all ak u-uniformly to h , and each h j k k k complex, all vk positive with Ckvk = u and Cklaklvk 5 u Hence, by J
.
what was observed above, lThj I
converges
IT1 (u)-uniformly to
lThl , so
that it is sufficient to prove now that ITh.1 5 IT1 (u) for la11 j , J because this will imply lThl 5 ITI(u) For any h. = kkvk we have by
.
Lemma 92.5(ii)
that
J
Ch. 12,1921
ORDER BOUNDED OPERATORS
lTh.1 = /Xkak Tv 1 3 k
S
207
0
Xk lak/./Tvk/S IT1 Z k / a k / v k
S
.
/TI(u)
This concludes the proof. A s already defined, T = Tl+iT2 is said to be order continuous (or
a-order continuous) whenever
TI and
T2 are
so.
It follows that T
order continuous (or a-order continuous) if and only if
IT1
before, the band of all real order continuous operators (from L is denoted by
Ln
=
. The set of
Ln(L,M)
operators (from L+iL
is
is s o . A s into M )
all complex order continuous
into M+iM ) is now the band
Ln+iLn
. Similarly, Lc+iLc .
the set of all complex a-order continuous operators is the band
Ln and tc are projection bands in Lb(L,M) , the bands Ln+iLn and Lc+iLc are projection bands in L (L+iL,M+iM) . b
Since
We apply the above results to the case that M numbers, s o
M+iM
is now the order dual L" is of the form $
=
. Any complex linear functional
$l+i$2 , where
L+iL of real linear functionals on on
L+iL
For any
If
if and only if
and
$ = $l+i$2E (L+iL)-
u E 'L
is the space of real
is the space of complex numbers. The space Lb(L,M) and
L
$
on
L+iL
$2 are the extensions to
. Evidently, +
is order bounded
$2 are order bounded on L ; in formula
we have
, then
The sets of all order continuous or o-order continuous linear functionals on
L+iL are the bands
L2iL:
or L-+iL" c c
respectively.
Complexifications of Banach lattices and of operators between Banach lattices were introduced by U. Krengel ([11,1963) and H.P. Lotz ([11,1968).
208
COMPLEX ORDER BOUNDED OPERATORS
The case that L
Ch. 12,1921
is an Archimedean and uniformly complete Riesz space, and
not necessarily a Banach lattice, was treated by W.A.J. Luxemburg A.C. Zaanen ([31,1971) and W.J. de Schipper ([11,1973). Theorem 92.6, stating that and u
ITl(u)
positive, is due to W . J .
=
The important
sup(lThl:hEL+iL,lhjiu)
for T
complex
de Schipper and, independently, to
H.H. Schaefer (cf. the book [11,1974,Ch.IV,Theorem 1.8). EXERCISE 92.7. If L
is Dedekind o-complete (which implies that
uniformly complete and Archimedean) and M
L
is
is Dedekind complete, the proof
of Theorem 92.6 may be simplified. Instead of using Lemma 92.3, the proof of which depends on representing a principal ideal in L
as a space C(X)
of continuous functions on a compact space, we can now simply use Freudenthal's spectral theorem (Theorem 91.5). the proof that
lTh1
5
ITI(lh1)
Show first that if h E L+iL
(i)
...,ek
disjoint components e l , (with 0
5
lynl
Ihl
-
5
1
Clynlen
of
and
~lhl
.
(ii) Show now that if the operator T
M+iM ) is order bounded and h
=
f+ig
,
E
.
are given, there exist
> 0
and complex numbers y I ,
Ihl
for all n ) such that 5
The change occurs only in
holds for any h = f+ig
Ih-Cynenl =
T +iT
then
lTh1
1
5
2
Elhl
...,yk
and
(from L+iL into
I!TI(Ihl)
.
CHAPTER 13
KERNEL OPERATORS
Let M(X,p)
and M(Y,v)
be the Riesz spaces of all real measurable
functions on the o-finite measure spaces (X,A,p) L
and
M be ideals in M(Y,v)
and M(X,p)
shall mainly deal with operators which map the carrier of
L
is the set Y
on X
x
(Y,C,v)
, and
let
respectively. Since we L
into M we may assume that
itself. The operator T: L
called a kernel operator if there exists a real T(x,y)
and
+
M
is
(pxv)-measurable function
such that, for every f E L , the equality
Y
holds for p-almost every x E X The function T(x,y)
(the exceptional set depending on
is called the kernel of the operator T
is a kernel operator with kernel T(x,y) operator Ta mapping L
into M(X,u)
, then
. If
f ). T: L
-+
M
is the kernelof an
IT(x,y)l
. If the kernel operator
T: L + M
,
then
T is called an absolute kernel operator. Evidently, the set Lk(L,M)
of
has the property that the corresponding Ta also maps L
into M
all absolute kernel operators is a linear subspace of the space Lb(L,M) of all order bounded operators (from L into M ). Furthermore, the kernel operator T
is positive if and only if the kernel T(x,y)
negative almost everywhere on and only if
T(x,y)
=
X
x
Y
and
T
= 0
is non-
(the null operator) if
0 almost everywhere.
In section 94 it will be proved that the set Lk(L,M) of all absolute kernel operators is a band in Lb(L,M) The proof as we shall present it
.
is based on a majorization theorem of A.R. Schep (1977), 0 5 S 5 T
and
operator. If S and T(x,y) S(x,y)
and
T
is a kernel operator, then
and
, then T(x,y).
S
stating that if
is likewise a kernel
T are absolute kernel operators with kernel S(x,y)
the kernel of
sup(S,T)
is the pointwise supremum of
In particular, the kernel of
209
IT1
is
IT(x,y)l
.
CHAPTER 13
210
Ch. 131
In section 86 it was proved that the band L" of all integrals on L n (i.e., all order bounded linear functionals that are order continuous) may consisting of all v-measurable
be identified with the ideal in M(Y,v) functions g
. If now the functions
fE L
hi E M
and
fg is v-summable over Y
such that the product
for all
i
,
gi and
...,k)
h.
for every N
satisfy g. E L i n
(i=l,
then
is evidently the kernel of an absolute kernel operator T which maps
L
into M
. Any
T
of this type is called a kernel operator of finite
rank. In section 95 we prove that if the ideal :L
is
large that its
so
carrier is the whole set Y , then L (L,M) is exactly the band generated k by the kernel operators of finite rank. This was proved by A.V. Buhvalov (1974) in a different manner. A sequence
(un:n=1,2,...) of v-measurable functions on Y
to star-converge to zero (notation u
*
+
is said
0 ) if every subsequence contains
a subsequence converging pointwise almost everywhere on
Y
to zero. This
notion occurs in Buhvalov's theorem (1974) which we prove in section 96. The theorem states that i f , once more, the carrier of L" is Y and T n is an operator mapping L into M , then T is a kernel operator (not necessarily an absolute kernel operator) if and only if it follows from
*
...
0 5 u S u E L for n = 1.2, and u -+ 0 that Tu converges pointn n n wise almost everywhere on X to zero. The proof is presented first for
T order bounded (in which case T
is an absolute kernel operator). If T
is not order bounded as an operator from L
into M
but
the mentioned condition (Buhvalov's condition), then T as an operator from L
, so
into M(X,u)
T
satisfies
is order bounded
that the already established
result for the order bounded case can still be applied. If L and if T
and M
are vector spaces with algebraic duals L#
is an operator mapping L
is an operator mapping M# all
fE L
and
into L#
all 41 E M#
into M
, defined
spaces (M Dedekind complete) and
T
of M
T#
to the order dual M"
The operator T"
N
maps M
by
. In the case that
T"
of
, the
into L"
and M#
algebraic adjoint T# (T#$)(f)
I=
$(Tf)
for
L and M are Riesz
is order bounded, the restriction and
is called the order adjoint of T . T"
but even order continuous. If, in addition, T
is not only order bounded, is order continuous, then
Ch. 131 T"
21 I
KERNEL OPERATORS
maps
Mi
,
into :L
so
the restriction T'
order continuous operator mapping M ; and M
be ideals in M(Y,v)
carrier of both L
of
.
into :L
and M(X,p)
T"
to Mi
and assume that Y
is the
Li and X is the carrier of both M
and
Under these hypotheses, if T into M ) with kernel T(x,y)
is an
ow, let once more L
.
and MZ
is an absolute kernel operator (mapping L
,
then T'
is an absolute kernel operator
.
(mapping MZ into L" ) with kernel T(y,x) Conversely, if also n (Mn)n = M and T is an operator (from L into M ) such that the "
N
restriction T' MX
of
T#
, then
into :L
T
to MZ
is an absolute kernel operator mapping
-
is an absolute kernel operator. As applications,
we prove in section 98 that for
1 5 p <
-
every norm continuous operator
from L (Y,v) into Lm(X,p) is an absolute kernel operator and for P 1 < p 2 every norm continuous operator from LI(Y,v) into Lp(X,p)
is
an absolute kernel opezator (Dunford's theorem). Assume now that the ideal L g E M(Y,v)
For any
The ideal L' space of norm of
L p
x
Y
of all
g
satisfying p'(g)
and the restriction of
. If, for example, L =
then L' = L X
in M(Y,v)
9
for p
-1
+q-'
=
I
L
p'
for these values of
every x
-
is defined by
is called the associate
L'
y
t(x)
. In this case the function
on Y
x ), so =
t
-
is called the associate
for some p
finite or infinite number for these x generalized Oarleman kernel if
<
to
p'(g)
satisfying
P . The measurable function
is a v-measurable function of
we write T (y)
is equipped with a norm p .
the (finite or infinite) number
1 5 p 5
T(x,y)
on
for p-almost every x p'(Tx(y))
. We shall say that
, (and
is a well defined T(x,y)
is a
is finite for almost p'(Tx(y)) is measurable on X , i.e.,
(W.A.J. Luxemburg,l958). In section 99 we discuss some results
t E M(X,u)
due to A.R. Schep (1979).
It will be proved that if
then the following conditions for the operator T
p
is order continuous,
from L
into M(X,p)
are equivalent:
every
(i)
T
(ii)
There exists a non-negative function t E M(X,p)
fE L
is a kernel operator with a generalized Carleman kernel. satisfies
(iii) If f E L n for almost every x
.
l(Tf)(x)l
for n = l , Z , .
p(f).t(x)
.. and
such that
for almost every x P(fn)
*
0
,
then
.
(Tfn)(x)
+ 0
KERNEL OPEUTORS
212
we have L ' = L
For L = L2(Y,v) kernels T(x,y) over Y
, so
Ch. 13,1931
in this particular case we get
with the property that the integral of
. These are
is finite for almost every x
introduced by T. Carleman (1923).
IT(x,y)l
2
the Carleman kernels,
Several results are proved, relating
Carleman kernels to orthonormal systems in L2(Y,v)
.
93. Kernel operators
Let
(X,A,p)
and
(Y,Z,p)
be 0-finite measure spaces. The correspon-
ding Riesz spaces of all real measurable functions will be denoted by M(X,p)
and M(Y,v)
respectively. We recall that almost equal functions
are identified. The product measure of product
X x Y
on the semiring
is denoted by
r
r
and
= m.0
v
in the Cartesian
A E A
of all sets A x B with
(with the convention that 0.m on the semiring
p
. The product measure
v
p x
=
and
is first defined
B E E
by
0 ) , and the thus obtained measure
is then extended by means of the Carathzodory exten-
sion procedure. be a realvalued (uxv)-measurable function on X x Y
Let T(x,y) any
f E M(Y,v)
the function T(x,y)f(y)
is (pxv)-measurable, which
implies by Fubini's theorem that for p-almost every x E X T(x,y)f(y)
is v-measurable as a function of
is well defined for these values of
x
f E M(Y,v)
the function
y , It follows that
(the value
permitted) and by Fubini's theorem the function h The set of all
+m
for h(x)
is
is v-measurable on X
for which the corresponding h
Obviously, domy(T) h
T(x,y)
, and
it will be denoted by
is an ideal in M(Y,V)
in ( 1 ) is finite p-almost everywhere on
. For X
f E domy(T)
, and
so
.
in ( 1 ) is
finitevalued p-almost everywhere on X (i.e., for which h E M(X,u) be called the Y-domain of
. For
)
domy(T)
will
.
the function
Ch. 13,5931
KERNEL OPERATORS
213
. The function
is also finite u-almost everywhere on X
g
is u-measurable.
This follows from g(x)
=
1
(T(x,y)f(Y))+
dv(y)
-
Y
1
(T(x,y)f(y))-
,
dv(y)
Y
where the terms on the right are p-measurable by Fubini's theorem. Hence, formula (2) defines an operator T: f
+
g with
doy(T)
as domain and with
. The operator T is called a kernel operator and T(x,y) is called the kemeZ of T . Kernel operators are often called integraZ
range in M(X,u)
operators, but this name has the disadvantage that it is also used for other purposes. If T
is a kernel operator with kernel T(x,y)
are ideals in M ( Y , v )
and M(X,y)
a kernel operator from L T maps
L
for every
into M
f E L
. This
into M
=
I
if L
L
and M
is said to be
is included in doy(T)
and
implies that
. I n this case
(Tlf)(x)
and if
respectively, then T
T
, as
and T2
1
T+(x,y)f(y)dv(y)
,
T-(x,y)f(y)dv(y)
,
defined by
Y
(T2f)(x)
=
1
Y
are also kernel operators from L and
T2 are positive operators such that T
where
Ta
is the kernel operator from L
as kernel. If T and Ta both map
L
into M )
mapping L the set
into M
, we
TI-T2 and T = T +T a 1 2 ' having IT(x,y)I into M
shall say that T
. Then
Ta
and
such that T
is an absolute kernel
T -T are positive operators
, so T = Ta-(T a-T) is order bounded. It follows that of all absolute kernel operators from L into M is
into M
Lk(L.M)
a linear subspace of is a band in
Lb
T(x,y)
.
Lb(L,M)
. It will be
proved in the next section that
. For this purpose we first prove a few simple facts.
THEOREM 93.1. Let
kerneZ
=
. It is evident that
into M(X,u)
is a kernel operator from L
operator (from L
Lk
into M(X,u)
T be a k e r n 1 operator from Then t h e f o l l d n g hoZds.
L into
M with
214
Ch. 13,9933
KERNEL OPERATORS
(i)
is p o s i t i v e i f and only if T(x,y) Z 0 hoZds (uxv)-aZmost everyuhere on x x YL , where YL denotes the carrier o f L . T
( i i ) T = 8 ( t h e nu22 operator) i f and onZy if (pxv)-aZmost everywhere on
.
X x YL
PROOF. ( i ) It i s evident t h a t
T
T(x,y) = 0
is positive i f
everywhere. For t h e converse, assume t h a t
holds
T(x,y) 2 0 almost
i s p o s i t i v e . By Theorem 86.2
T
t h e r e e x i s t s a sequence all
. Since
n
that
u(Xn) <
u
m
Y .f YL such t h a t v(Yn) < and Xy E L f o r n i s o - f i n i t e , t h e r e a l s o e x i s t s a sequence X I. X such n for all n It i s s u f f i c i e n t t o prove t h a t T(x,y) 2 0
.
holds almost everywhere on every that
u(X)
that
jy
add
v(Y)
Xn x Yn
a r e f i n i t e and
IT(x,y)l dv(y)
xy
,
so we may j u s t a s well assume
E L
. From xY
i t follows
E L
i s f i n i t e almost everywhere on
X
. Hence,
i f we
write
Y
for
k
=
1,2,
...
,
then
9 I.
,
X
so it i s s u f f i c i e n t t o prove t h a t
i s non-negative almost everywhere on every
\
i s (uxv)-summable over
x Y
, so
9x
Y
. The f u n c t i o n
f o r any measurable
E C \
and
T(x,y) T(x,y) F c Y
we have
( ~ x ~ ) ( x ) d u ( x2 ) 0
.
E
From t h i s it follows e a s i l y t h a t t h e i n t e g r a l of T(x,y) over any (pxv)measurable subset of which
T(x,y)
q x yi s
non-negative. This implies t h a t t h e set on
i s s t r i c t l y negative i s of measure zero, and so
T(x,y) t 0
holds almost everywhere. ( i i ) Follows by applying t h e r e s u l t i n p a r t ( i ) t o
T
and
-T
.
94. The hand of a b s o l u t e k e r n e l o p e r a t o r s A s i n t h e preceding s e c t i o n , w e assume t h a t
(X,A,u)
and
(Y,C,v)
are
Ch. 13,1941
KERNEL OPERATORS
0-finite measure spaces and L
and M
215
are ideals in
and M(X,P)
M(Y,V)
respectively. For the investigation of operators from L
into M
we may
assume without restriction of the generality that the carrier of the set Y
LEMMA 94.1.
let
L
is
itself.
Let
S
be a p o s i t i v e operator from L i n t o M
E x F be a measurahle subset of
I
x
Y sueh that
xF
and
E L
and
.
<
(SxF)(x)du(x)
X
E
Furthermore, l e t that
A
1' be the semiring of a l l subsets
i s u-measurable and
B i s v-measurable.
A x B of
E x F
such
Then
A
is a non-negative and f i n i t e Z y additive s e t function on A i s a f i n i t e Z y additive measme on r . PROOF. It is evident that also that A x B c A, x B
1
0 < A(AxB) <
. In
other words,
for all A x B E
m
r
and
. The additivity
< A(AIxBI)
implies A(AxB)
r
proof seems simple, but there i s a small complication, caused by the fact n that if fo,fl, f are functions in L such that fo(y) = C 1 fk(y) n = Zy(Sfk)(x) holds only for p-almost for all y E Y , then (Sfo)(x)
...,
every x E X
, and
not necessarily for all x E X
We present the additivity proof. Let A x B and all A
x Bk k disjoint. Then
for all
members of
(x,y) E E
x
F
, so
r
holds for u-almost every x E X depends on xo (1,2,
...,n)
and such that all
if we fix xo
,
.
= U ;
(%xBk)
%x
Bk
with A x B are mutually
then
. The exceptional set of measure zero
. More precisely, denoting by
consisting of those k
D(xo)
for which
the subset of
((xo,y):y€B)
has a non-
216
BAND OF ABSOLUTE KERNEL OPERATORS
4, x
empty intersection
Bk
, the
Ch. 13,5941
exceptional null set depends on D ( x o )
.
For x varying in E , there are only finitely many different D ( x o ) , 0 so there are only finitely many different exceptional null sets. The union
of these null sets is again a null set, so for almost every x E X we have
Hence
In other words, h(AXB) = In h ( q c X B k )
.
The next theorem is of fundamental importance for the further results in this chapter. The proof which we present here is due to A.R. Schep (Theorem 4.2 in C11,1977,cf. also [21,1979). THEOREM 94.2. If 8
5 S 5
T in
T is a kernel operator,
Lb(~,~) and
t h e n S is a kernel operator. I n o t h e r words, any p o s i t i v e operator majorized by a kernel operator is a kernel operator. PROOF. The proof is divided into five steps. (i) Let E such that
xF
and
E L
I
F
be measurable subsets of
X and Y
respectively
and
T(x,y)d(uxv)
=
EXF
I
(TxF)(x)dv(x)
-.
E
, defined
Then the set function h
for any measurable subset P
of E x F
by
is a countably additive measure on the o-algebra of all (pxv)-measurable The measure A is absolutely continuous with respect subsets of E x F to x v , i.e., (vxv)(P) = 0 implies h ( P ) = 0 .
.
A s before, let
r
be the semiring of all sets A
a u-measurable subset of
E
and
B
is a +measurable
B
, where
subset of
F
A
is
. For
A x B E l'
we define
A (AXE)
(2)
1
'
=
J
.
(SXB)(x)du(x)
A
It follows from the last lemma that
r
. Since
0 5 XI
r
on
A
5
with disjoint terms, and n
is a finitely additive measure on
Al
A
and
is countably additive, the measure
r
is actually countably additive on
A1
217
KERKEL OPERATORS
Ch. 13.1941
. Indeed, if
A x B
=
is a natural number, then A x B
xD ) can be written as a finite disjoint union Up(C l i i if E > 0 is given, then
m
Ul(%xBk)
-
UY(\XBk)
. Hence,
r
of sets of
sufficiently large, so in view of
for n
Al(%XBk)
it follows that :1
4
AI(AxB)
as n +
m
.
We now apply the Carathsodory extension procedure to the measures A A1
and
r
on
denoted by
A
; the exterior measures corresponding to
A*
A:
and
respectively. It is evident that of E x F
holds for every subset P
What we have done is to restrict
A
and
0
2
A,
will be
A*(P) 5 A*;P) 1 A*
. Let us take a closer look at
to the sets of
r
.
and then extend
the restriction again by means of the Carathgodory extension procedure. The question arises whether in the first place every (uxv)-measurable set P (for which
A
was originally defined by formula ( 1 ) ) becomes
after the extension and secondly, whether formula ( 1 ) A(P)
replaced by
A*(P)
. The answer is obviously affirmative if
o-set (i.e., a finite or countable union of sets from p
A*-measurable
continues to hold with
r
)
, hence
P
is a
also if
is a finite or countable intersection of u-sets. It follows that the
Same holds if P
is a set of (vxv)-measure zora, because in this case there
exists a set Q
P
3
, also
of (pxv)-measure zero, such that Q
is the
intersection of a decreasing sequence of u-sets. Since any (pxv)-measurable set differs only in a set of measure zero from such a countable intersection, the formula ( 1 ) holds therefore for any (pxv)-measurab%e set P with replaced by
A*(P)
. Hence, A*(P)
A(P)
has the same value as the original A(P)
.
BAND OF ABSOLUTE ERNEL OPERATORS
218
Ch. 13,1941
As usual with X*-measurable sets, we drop the asterisk, so we write
A(P)
as before. The next fact to observe is that every (pxv)-measurable set P h*-measurable. This is evident if P
is also P
is a o-set, hence also if
1
is a finite or countable intersection of a-sets. It remains to prove
that every set of (pxv)-measure zero is A*-measurable. This is easy; if
,
(pxv) (N) = 0 0
< A*1
X*
2
then A (N) = 0
. This
by ( 1 ) ,
1
X:(N) = 0 on account of is a X*-null set, so N is h*-measurable
shows that N
so
1
A,(P)
For any (pxv)-measurable set P we shall write
1
instead of
ht(P)
.
Recapitulating, we have proved that any (pxv)-measurable subset P of E x F
is X*-measurable as well as
h*-measurable and 1
0
2
A,(P) < h(P)
holds for any set P of this kind. Note now that if we combine these inequalities with the fact that A the same for X I
, i.e.,
XI
is (pXv)-absolutely continuous, we get
is (pxv)-absolutely continuous on the
a-algebra of all (pxv)-measurable subsets of
E x F
. It follows then from
the Radon-Nikodym theorem that tihere exists a (pxv)-measurable non-negative function S(x,y) A,@)
on =
E x F
I
such that
S(x,y)d(pxv)
P
for all (pxv)-measurable subsets P A x B E
r , we
of E
x
F
. In particular, for
have
X I (AxB) in formula (2) and observing is an arhitrary u-measurable subset of E , we get
Comparing this with the defintion of that A (3)
(SxB)(x)
=
1
S(X,Y) xB(Y)dv(Y)
Y for almost every x subset B of
F
.
in
E
. Note that
(ii) We assume again that F that
xF
E L
. Furthermore, since
disjoint union of sets X. ( j = l , with kernel T(x,y)
J
is from L
(3)haldsfor every v-measurable
is a v-measurable subset of Y
X is of a-finite measure, X
...)
such is the
of finite measure. The operator T
into M
and
xF
is a member of L
,
so
KERNEL OPERATORS
Ch. 1 3 , § 9 4 1
I
219
T(x,y)dv(y)
F
is finite u-almost everywhere on X E jk
=
1
. For
(x€X. :k-I< T(x,y)dv(y)
j,k = l,2,
... , let
.
F
Then, except for a set of measure zero, X
... .
is the disjoint union of the
sets E (j,k=l,2, ) Note that T(x,y) is (pxv)-summable over E x F jk jk for all j,k Hence, by step (i), there exists a (pxv)-measurable non-
.
negative function S . (x,y) on E Jk jk subset B of F the equality
x
.
F
such that for every v-measurable
holds u-almost everywhere on E Defining now S(x,y) on X x F by jk S(X,Y) = S (x,y) on E x F for j,k = 1,2 , we find for every jk jk u-measurahle subset B of F that (SxB) (x)
=
I
,...
S(x,y)xB(y)dv(y)
Y
holds u-almost everywhere on X
. It follows immediately that if
measurable step function on Y vanishing outside F ,i.e., t(y) finite sum of terms aiXBi(Y) (St)(x)
=
/
t
is a
is a
with all B. measurable subsets of F
, then
S(x,y)t(y)dv(y)
Y holds u-almost everywhere on X
xy
.
(iii) By Theorem 86.2 there exists a sequence Y .f Y such that n € L for all n (we recall that by assumption the carrier of L is
thg set Y
for n = 2 , 3 , itself). Writing D = Y l and D = Y -Y 1 n n n-l ) are disjoint, their union is Y and each
... . Applying the result in step (ii) to each
the sets D (n=1,2, n xD is a member of L
X x Dn
SeFarately, we obtain a (uxv)-measurable non-negative function S(x,y)
x
x Y
such that (St)(x)
=
[
JY
S(x,y)t(y)dv(y)
... , on
220
Ch. 1 3 , 5 9 4 3
BAND OF ABSOLUTE KERNEL OPERATORS
holds almost everywhere on X such that
t
vanishes outside some
(iv) We prove that if
0 < f (y)
4
n
for each measurable step function t on
f(y)
n'
...
f,fl,f2,
*
are functions in L
holds almost everywhere on
. Since
Y
,
then
such that
(Sfn)(x)
holds almost everywhere on
X
holds almost everywhere on
X , it is sufficient to prove that
(Tfn)(x)
almost everywhere on X
1. (Tf)(x)
Y
. For almost every
4
(Sf)(x)
x E X we
have
Y ,
almost everywhere on over Y
.
(v) Given 0
5
f E L
functions tn (n=l,2,.-.) on Y
(Tfn)(x)
so
,
1. (Tf)(x)
there exists a sequence of non-negative step
satisfying 0 < t (y) 4 f(y)
. It may be assumed that for each
outside Yn
, where
Yn
follows by integration
almost everywhere
the step function
n
tn
vanishes
is the same set as in step (iii). Hence, by step
(iii),
holds almost everywhere on X ges almost everywhere on X
for each n to
(Sf)(x)
side converges almost everywhere on X
Y
, by
. Now,
the left hand side conver-
by step (iv), and the right hand to the integral of
S(x,y)f(y)
over
the theorem on integration of monotone sequences. Hence
.
This shows that S is a kernel operator. Note that by for almost every x is uniquely determined modulo (pxv)-null Theorem 93.1 (ii) the kernel S(x,y) functions.
Ch. 13,5941
22 1
KERNEL OPERATORS
A s noted earlier, the absolute kernel operators from
a linear subspace of the Riesz space Lb(L,M)
L
into M
form
. More precisely, since any
absolute kernel operator is order continuous, the absolute kernel operators form a linear subspace in the supremum sup(T,S) space. Hence, if exists in
T
Lb(L,M)
.
Lc(L,M)
. For
L,(L,M)
=
T
and
exists in Lb(~,~) since Lb(L,M)
and
in
S
is a Riesz
are absolute kernel operators, then sup(T,S)
S
It is a reasonable conjecture now that if
have kernels T(x,y)
and
Lb(L,M)
S(x,y)
T
and
S
i s also a
respectively, then sup(T,S)
kernel operator with kernel equal to the pointwise supremum of
T(x,y)
S(x,y)
, the con-
. For. S = t3
almost everywhere. Similarly for inf(T,S)
jecture is, therefore, that T+
=
i s a kernel operator with kernel
sup(T,B)
equal to the pointwise supremum of
T(x,y)
zero, i.e., with kernel the positive part
and
and the function identically T+(x,y)
of
T(x,y)
. The proof
that this i s true was given by W.A.J. Luxemburg-A.C. Zaanen ([31,1971), before the theorem about a positive operator majorized by a kernel operator was known. The details of the Luxemburg-Zaanen proof are not simple. It is a pleasant fact, therefore, that by means of our last theorem we can give now a short and transparent proof of the conjectured result about as mentioned above. Note already that this will imply that if absolute kernel operator with kernel T(x,y) kernel operator with kernel THEOREM 94.3.
.
IT(x,y)\
, then
T
sup(T,S),,
i s an
IT1 = sup(T,-T)
is a
If T and S are absolute kernel operators from L
i n t o M w i t h kernels
T(x,y)
and
S(x,y) r e s p e c t i v e l y , then
sup(T,S)
i s likewise a kernel operator w i t h kernel equal t o t h e pointwise supremum of T(x,y) and S(x,y) almost e v e w h e r e . PROOF. It is sufficient to present the proof for S
=
t3
, since
the
general result will then follow from the formula sup(T,S) The positive part
=
S + sup(T-S,B)
T+(x,y)
of
T(x,y)
zero function, so the operator To satisfies To 2 T
and
. majorizes
T(x,y)
as well as the
corresponding to the kernel T+(x,y)
.
T > 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
almost everywhere on
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
<
E
for almost every
x
. Assume now that the subset P of X = on P . Writing P = (y:(x,y)EP)
measure and
T(x,y)
the set Px
is a subset of
E
for all
> 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
<
m
)
Y with by
(note that with
KERNEL OPERATORS
Ch. 13,9941
disjoint terms). (vxv)(AxB)
=
The measure
is finite, i.e., A(YxY)
A
0 implies A(AxB)
=
(because (vxv)(AxB)
0
one at least of the numbers u(A)
and
dory's extension procedure to
v
extended measure
A
V X
v(B)
and
and
<
m
=
0 implies that
is zero). We apply CarathEoA
to
. One might
expect that the
will be (vxv)-absolutely continuous, but this is not
the case. Indeed, writing sets AI and A2
225
v(Y)
=
a
for brevity, there exist disjoint sub-
such that their union is Y
Then the union D 1 of
A1
X
A 1 and A
=
0
2
x A2
and
v(A ) = v(A ) = ;a I 2 2 satisfies (vxv)(D ) = ;a
.
1
and
since A(AIxA2) = A(A2XAI)
. Repeating this ...)
obtain a descending sequence (D :n=1,2, (vxv)(Dn)
in Y x Y
such that
A(D ) =na for all n . Then the intersection n (vxv)(D) = 0 and A(D) = a . This shows that A is
2-"a2
=
decomposition procedure, we
and
D = nmD
satisfies I n not (uxv)-absolutely continuous on the o-algebra generated by smallest o-algebra containing
r
It is evident, therefore, that if one has measures, A such that
A(AxB)
= 0
not guarantee that however, 0
5
XI 5 A
XI
r
(i.e., the
).
for A x B E
r
implies AI(AxB)
=
and
0
,
A1
say,
then this does
will be A-absolutely continuous after extension. If, '
holds on
r
(as in the proof of Theorem 9 4 . 2 ) , there will
be no unpleasant surprises; after extension X I
will be A-absolutely
continuous. In our second example we consider the case that are the same a-finite measure space and identity operator
I in L
L
=
M
(X,h,p)
and
(Y,C,v)
is an ideal in M(Y,v)
. The
is a positive operator and it may be asked now
I is a kernel operator. We shall prove that if Y doesnot contain atoms, then I is not a kernel operator. The proof is due to A.R. Schep if
(C 1 3 , 1 9 7 7 ) .
THEOREM 9 4 . 7 . Let L be an idea2 i n M(Y,v) , #here Y does not cont a i n any atoms. Then the i d e n t i t y operator I i n L i s d i s j o i n t t o the band Lk(~,L) of absoZute kernez operators i n I i s not a kernel operator.
$,(L,L)
. It
foZZous that
226
BAND OF ABSOLUTE KERNEL OPERATORS
PROOF. Let
b e a p o s t i v e k e r n e l o p e r a t o r (from
T
. Then
S = inf(T,I)
and l e t
holds i n
8' 5 S 5 T
94.2 t h e r e e x i s t s a (vxv)-measurable f u n c t i o n that
is t h e k e r n e l of
Sfx,y)
p o s i t i v e on a s u b s e t of
Y x
assume t h a t t h e c a r r i e r of Y
xy E
such t h a t
1. Y
one a t l e a s t of t h e s e
n
Yn
and
,
v(Yn) <
c a l l it
For any (vxv)-measurable s u b s e t X(E) =
I
,
that
0
2
itself)
Y x Y
on
S(x,y)
L
s o by Theorem such
is strictly
Y i t s e l f , so t h e r e e x i s t s a sequence
is
L
L
S(x,y)
. Assume now
S
into
L
Lb(L,L)
Y of p o s i t i v e measure. A s b e f o r e , we may for a l l
m
, we
2
have
i s s t r i c t t y p o s i t i v e on a s u b s e t of
S(x,y)
Ch. 13,9941
of
E
. It
n
v(2) <
2 x 2
m
follows t h a t f o r
, xz E
and
L
of p o s i t i v e measure.
we d e f i n e t h e number
2 x 2
X(E) by
.
S(x,y)d(vxv)
E
Hence, f o r
and
A
5
I
B
measurable subsets of 2 , we have
1
(IxB)(x)dv(x) =
A
xBdv = v(AIIB1
-.
,
A
in particular
A(Zx2) 5 3(Z)
a non-negative
countably a d d i t i v e measure on t h e o-algebra of a l l (vxv)-
measurable s u b s e t s of 5
v(AIIB)
for
A
and
<
2 x 2
B
From t h e s e f a c t s we conclude t h a t
such t h a t
, we
have
0 < a 5 B <
m
v x v
. Since A1
e x i s t measurable d i s j o i n t s u b s e t s is
and
2
A2 x A2
i s f i n i t e and
2 . By . Writing
measunable s u b s e t s of
a b s o l u t e l y continuous w i t h r e s p e c t t o
B = v(2)
X(Zx2)
v(A ) = v(A2) = 16
. Then
1 s a t i s f i e s (vxv)(D ) = l B 2 1
and
its definition
a = X(Zx2)
A2
t h e union
of
DI
X
2
such t h a t D =
of
that
S(x,y)
A1 x A1
and
and
.
and X ( D ) = a n s a t i s f i e s (vxv)(D) = 0 and
(Dn:n=1,2,
(vxv)(Dn) = 2-nf32
for a l l
n"ID n
X(D)
d i c t s t h e a b s o l u t e c o n t i n u i t y of
is
X
and
such t h a t t h e i r union
2
Repeating t h e procedure, we o b t a i n a descending sequence intersecti,on
X(AxB) 5
does n o t c o n t a i n atoms, t h e r e
Y
A(DI) = X(A,XAI) + X(A xA ) = X(ZX2) = a 2 2
2
is
X
X with respect t o
in
. Then t h e . T h i s contra-
n
= a
v x v
...)
. The
assumption
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 i s t h e r e f o r e
Ch. 13,5941
KERNEL OPERATORS
S(x,y) = 0
f a l s e . Hence S
I
inf(T,I) = 8
227
Y
almost everywhere on
Y
x
f o r any p o s i t i v e i n t e g r a l o p e r a t o r
. This shows t h a t T . In o t h e r words,
i s d i s j o i n t t o t h e band of a b s o l u t e k e r n e l o p e r a t o r s . As before, l e t
ideal i n
.
M(Y,v)
(Y,C,v) Since
be a a - f i n i t e measure space and l e t
t h e e x i s t e n c e of an a b s o l u t e v a l u e i n t h e c o m p l e x i f i c a t i o n f i e d . Hence, any
f = f +if2 1
s a ti s fying
The elements
fl
and
f2
L
be an
i s super Dedekind complete, t h e c o n d i t i o n s f o r
L
in
L+iL
a r e r e a l f u n c t i o n s on
,so
X
are satis-
L+iL
has an a b s o l u t e value
If
I
in
L+
t h e supremum can be
taken pointwise (almost everywhere), and we g e t
f o r almost every
y
. We
E Y
d e f i n e complex k e r n e l o p e r a t o r s . I f
i s another a - f i n i t e measure space and
T
o r d e r bounded o p e r a t o r
from
L+iL
M
i s an i d e a l i n M+iM
into
M(X,p)
(X,A,v)
, then
the
i s s a i d t o be an a b s o l u t e
k e r n e l o p e r a t o r i f t h e r e e x i s t s a complexvalued (uxv)-measurable f u n c t i o n T(x,y)
on
X x Y
T
Assume t h a t T2
r e a l and l e t
such t h a t
s a t i s f i e s t h e s e c o n d i t i o n s . Let T(x,y) = T1(x,y) + iT2(x,y)
r e a l . It follows from
ITI(x,y)I S ( T ( x , y ) (
jlTl(x,y)f(y)I
dv(y)
T = T + i T 2 with 1 w i t h T I ( x Y y ) and
that
E M for all
f E L
,
Y and i t follows from t h e d e f i n i t i o n of (Tlf)(x) =
j Y
TI
T,(x,y)f(y)dv(y)
that for a l l
f
E L
.
TI
and
T2(x,y)
KERNEL OPERATORS OF FINITE RANK
228
Hence, T I is an absolute kernel operator (from L
. Similarly, T2
TI(x,y)
Conversely, if
1
and
T (x,y) 2
is an absolute kernel operator from L+iL = TI(x,y)
If
into M ) with kernel
is an absolute kernel operator with kernel T2(x,y). T2 are real absolute kernel operators from L
T I and
with kernels T (x,y)
into M
Ch. 13,9951
.
respectively, then T
into M+iM
with kernel
=
Tl+iT2
T(x,y) =
iT2(x,y) T = T +iT2 is an order bounded operator, the absolute value +
IT1
1
satisfies
T I cosB+T sinB:O
T
is a kernel operator, then so are
follows from
/TI 5 IT1[+ IT2/ that
kernel T(x,y) = T I(x,y)+iT2(x,y)
IT1
ITII and
IT2/
. Hence, it
is a kernel operator. Since the
satisfies
(x,y)cosB+T (x,y)sinB:O<€I<2~ 2 almost everywhere, it is evident that If T
IT1
has
IT(x,y)l
as kernel.
T is a non-absolute kernel operator from L+iL into M+iM , then
is an absolute kernel operator from L+iL into M(X,!.I) + iM(X,u)
the above remarks for
IT1 and
EXERCISE 94.8. Let an ideal in M(Y,v) that
IT(x,y)l
(Y,C,v)
,
so
hold in this situation.
be a o-finite measure space and let L be
such that the carrier of L
is the set Y
itself. Show
(Y,Z,v) contains an at most countable number of atoms. More pzecisely,
show that Y = Y atoms and Y2 to Y
1
U Y2
, where
Y 1 is the union of at most countably many
does not contain any atoms. Show that Y 1 is v-almost equal
if and only if the identity operator in L
is a kernel operator.
95. The band generated by the kernel operators of finite rank A s in the preceding section, we assume that
measure space and Y
L
is an ideal in M(Y,v)
itself. The band in the order dual L”
,
is a o-finite
L:
. If
0
is an integral on
there exists (by Theorem 86.3) a uniquely determined v-measurable
function g
on Y
such that
L is
consisting of all integrals (order
continuous linear functionals) is denoted by L
(Y,C,v)
such that the carrier of
Ch. 13,8951
KERNEL OPERATORS
It is evident that satisfy g
,
2
if and only if the corresponding g1 and
$ , 2 $,
. Hence,
g2
sup($,,$,)
.
inf($ ,$ ) 1 2 corresponds with lgl
similarly for
1$1
corresponds with
In particular, if
. This
fore, the band
on
L
L i in L”
-
$
sup(gl,g2)
corresponds with
82
and
g
,
then
shows that
simply because the integral on the left is is a v-measurable function on Y with an integral $
229
I$l(lfl)
. Conversely, if
such that ( 2 ) holds, then g
by means of formula ( I ) . with the ideal in M(Y,w)
We may identify, thereconsisting of all g
L (Y,v) for some p P satisfying 1 5 p 2 , then L” may be identified with the ideal L (Y,u) n q in M(Y,v) , where q is determined by p-l + q-l = 1 satisfying condition (2).
Let
(X,A,u)
in M(X,u) h
As an example, if L
g
corresponds
=
.
be another o-finite measure space and let M
. If the function
g
on
Y
is a member of :L
be an ideal
and the function
is a member of M , then the (pxu)-measurable function T(x,y)
on X
h(x)g(y)
into M
is the kernel of an absolute kernel operator from L f E L
since for every
=
,
we have
Any finite linear combination (with real coefficients) of kernel operators of this elementary kind is called a kernel operator o f f i n i t e rank. For the set of kernel operators of finite rank we use the notation L”
Q
M (tensor
product notation). The set L” B M is evidently a linear subspace of n L b ( ~ , ~ ) The band generated by L” 0 M in L b ( ~ , ~ ) is the second disjoint n Complement (L-BM)dd Since L” 0 M consists of absolute kernel operators, n n the band (L”BM)dd is contained in the band of all absolute kernel operators. n It may be asked whether there are other absolute kernel operators besides
.
.
those in (L2)dd
. We shall prove that under a rather natural condition
there are no other absolute kernel operators. Before mentioning the condition, we note that if kernel T(x,y)
,
T is an absolute kernel operator from L into M with then T(x,y) vanishes (uXu)-almost everywhere outside
230
5x
5
Y , where
xz
such that
E L
is the carrier of M
1,
, then
. Indeed, if
/T(x,y)/ du(y)
x E X-%
,
Letting
Z run through a sequence Yn 4 Y
and
so
IT(x,y)l
over
outside
x
the integral of
(X-$)
.
Y
X
Y
is zero, s o
to
Z i s a subset of
Y
vanishes for almost every
IT(x,y)l
THEOREM 95.1. If the c a r r i e r of
set
Ch. 1 3 , § 9 5 1
KERNEL OPERATORS OF FINITE RANK
over
, we
T(x,y)
(X-%)
x
Z is zeEo.
find that the integral of vanishes almost everywhere
Li i s the s e t
Y
L
lk(~,~) o f a l l absolute kernel operators from
i t s e l f , then the i n t o M i s equal
( L ~ M .) I~n ~other words, i n t h i s ease the s e t o f absolute kernel
operators i s the band generated by the kernel operators of f i n i t e rank. PROOF. It is sufficient to prove that every positive kernel operator is a member of and let V
n
(L>)dd Y
4
the carrier of
. Let
such that
E L for all n , 4 Y such that x Un E L l for all n (this is possible since
U
x
Vn Li is Y ). Then Y
.
= Un
n Vn satisfies Yn
4
Y
and
E L n:L for all n Furthermore, let Xn 4 % such that x E M Xn n' for all n and let be the kernel operator of finite rank with kernel Sn
x
Now, let T be an arbitrary positive kernel operator. For n = 1 , 2 , the operator T = inf(T,Sn)
...
,
is a kernel operator with kernel
n
T(x,y),Sn(x,y)
1.
and S E L" 8 M for all n , we have Tn E (L2M)dd Since 0 < T 5 S n n n n Taking into account now that T(x,p) = 0 almost everywhere
for all n outside
.
5x
Y
, it
alsmost everywhere on space Lb(L,M) T E (L2)dd
.
is evident that T (x,y) 4 T(x,y) holds pointwise n X x Y This implies that T I. T holds in the Riesz
.
. Hence, since
Tn E (L2)dd
We shall prove now that the carrier of
n
for all n
, it
follows that
L';I is the set Y
itself if
L" (when regarded as a subspace of the space of all linear n functionals on L ) separates the points of L ,i.e., if and only if
and only if
Ch. 13,8951
KERNEL OPERATORS
THEOREM 95.2. The c a r r i e r of :L
I,
f E L and
fgdv = 0 for a l l
I,
1
f E L
i t s e l f if and only
i f and only i f it follows from that
g E:L
PROOF. Note first that if
g E:L
Y
is t h e s e t
L , i.e.,
separates t h e p o i n t s of
if :L
23 1
f
, then
=
0
fgdv
JY
.
=
0 holds for all
. It is evident
Y lfgldv = 0 holds for all g E L i lfgldv = 0 implies fgdv = 0 Conversely, let
if and only if
.
1,
1
that fgdv = 0 Y for all g E L" i.e., in the terminology of linear functionals, +(f) = n' for all E:L The result proved in Theorem 83.9 imp ies now that
+
.
o
It follows that :L
separates the points of
g E:L
for all
L
implies f =
if and only if
o
.
LL
Assume now that the carrier of
is Y
f E L
1,
f = 0
points of
.
L
.
lfgldv = 0
-
f lfldv = 0 for all n 'Yn observed above, this implies that Ln separates the
. As
Conversely, assume that if then f = 0
I,
. Then there exists a . Hence, if f E L
for all n sequence Y t. Y such that xy E: L n satisfies Ifgldv = 0 for all g E:L , then and so
and
f E L
and
I,
I fgl dv
If, under this condition, the carrier of
ly smaller than Y
,
= 0
,
for all R E L;,
Li would be proper-
there would exist a subset F1 of Y
of positive
.
measure such that any g E L" vanishes almost everywhere on F 1 The set n Every g E L" F 1 has a subset F of positive measure such that xF E L vanishes almost everywhere on F which implies the carrier
";x of
=
n
, so
1,
.
(XFgldv= 0 for all g E L';I ,
0 by hypothesis. This contradicts v(F) > 0 is Y
.
. Hence,
A special case of the result that the band of absolute kernel operators is generated by the kernel operators of finite rank was proved by
G.Ya. Lozanovskii ([11,1966).
His proof is based on a theorem of N. Dunford
(1936), stating that anynorm continuous operator from Ll([O,ll,v)
Lp([O,ll,p)
, with
1 < p
5
m
and
1-1
into
Lebesgue measure, is an absolute
kernel operator, Dunford's theorem was generalized to more general measure spaces (by G.Ya. Lozanovskii and by A.V. Buhvalov, 1974); on the basis of this generalization A.V. Buhvalov fC13,1974) gave a proof of the general result in the present section. Our proof, as presented here, is quite
KERNEL OPERATORS OF FINITE RANK
232
Ch. 13,8951
different, since it is founded essentially on Schep's majorization theorem (Theorem 94.2). Dunford's theorem (in its generalized form) will appear as a corollary of some further results in section 98. The caae investigated by Lozanovskii deals with a class of normd Riesz spaces, the KB-spaces, to which a good deal of attention has been given in the Soviet literature. The normed Riesz space L (with the norm of noted by f
C
p(f) ) is called a KB-space if
0 implies p(fn)
constant M
+
0 ) and
0
5
p
de-
is o-order continuous (i.e., with p(f )
f 4
f
.
5
M
for some
sup f exists in L It follows immediately n that any KB-space is Dedekind o-complete. In the situation investigated implies that
by Lozanovskii v subspace of
is Lebesgue measure in Y
M(Y,v)
such that
L
=
C0,ll
and
i s a Riesz
L
is a KB-space with the additional
property that
The main result is now that if M
i s another KB-space of the same kind and
T is an order bounded operator from L rator if and only if where L*
T
into M , then T
is a kernel ope-
is contained in the band generated by
denotes the Banach dual of
L
.
L*
@
,
M
For the proof that this is a
special case of our general result, we refer to the exercises at the end of the present section. Another special case was investigated by R.J. Nagel and U. Schlotterbeck ([1],1972).
They consider order bounded operators from a Banach lattice
L
into a Banach lattice M
L
is order continuous (i.e., p ( u )
u
+
0 in L ) and M
under the conditions that (i) the norm
+
is Dedekind complete, (ii) L
and M
interior points, i.e., there exists a positive element u u
the ideal generated by
in
p
0 for any downwards directed system
is norm dense in L
,
have quasi-
in L
such that
and similarly for M
(iii) there exists a strictly positive normal integral on M therefore, a more general situation insofar as L
and M
. This
,
is,
do not necessarily
consist of measurable functions on a measure space. Nagel and Schlotterbeck use a representation of
L
and M
as spaces of continuous extended real-
valued functions to show that an order bounded operator (from L belongs to
(L*@M)dd
into M )
if and only if it can be represented as a kernel ope-
rator on these representation spaces. We say a few words about complexification of the results in this section,
Ch. 13,1951
KERNEL OPERATORS
s t i l l assuming t h a t t h e c a r r i e r of
L (L+iL,M+iM)
is
LL
233
i t s e l f . The s e t
Y
of a l l complex e b s o l u t e k e r n e l o p e r a t o r s (from
k M+iM) i s now t h e c o m p l e x i f i c a t i o n of t h e band t o s e e t h a t t h i s i s t h e band g e n e r a t e d by EXERCISE 95.3. where
As i n Theorem 94.7,
(L-+iLL) n
let
L
into
L+iL
and it i s e a s y
.
(M+iM)
be an i d e a l i n
M(Y,v)
,
does n o t c o n t a i n any atoms. It was proved i n t h e theorem r e f e r r e d
Y
t o t h a t t h e i d e n t i t y o p e r a t o r i s d i s j o i n t t o t h e band kernel operators i n
.
L
L k ( ~ , ~ of ) absolute
I t f o l l o w s t h a t any o p e r a t o r i n t h e band g e n e r a t e d
by t h e i d e n t i t y o p e r a t o r i s d i s j o i n t t o T
,
(L2M)dd
.
L k ( ~ , ~ ) Show t h a t any o p e a r a t o r
i n t h e band generated by t h e i d e n t i t y o p e r a t o r i s a m u l t i p l i c a t i o n , i . e . ,
t h e r e e x i s t s a v-measurable f u n c t i o n almost everywhere f o r every
f E L
t
. This
on
Y
such t h a t
(Tf)(y) = t ( y ) f ( y )
l a s t result is also true i f
Y
has atoms.
EXERCISE 95.4.
i s an o p e r a t o r i n t h e i d e a l generated by t h e
T
H I N T: Assume f i r s t t h a t
i d e n t i t y o p e r a t o r and t h a t
xy E
L
.
L
Show t h a t any KB-space f E L
H I N T : Denote t h e norm of
by
p(f)
i s a Banach l a t t i c e .
. It
i s s u f f i c i e n t t o prove
I f n w i t h terms i n L i s a b s o l u t e l y convergent i n norm k Cp(f ) < m 1, t h e n Cf i s convergent. Writing sk = Z If I , n I n p[=sk) S CY f o r a l l k , s o s = sup sk e x i s t s by one of t h e pro-
that i f the series (i.e.,
CY =
we have
p e r t i e s of m - s p a c e s . converges w i t h sum as w e l l , so
Cfn
L
. It
+
s-sk
0
implies
+
p(~-s,)
follows now e a s i l y t h a t
,
0
If:
and
so
Zfi
Zlfnl converge
converges.
EXERCISE 95.5. and l e t
Then s
Let
(Y,Z,v)
be a f i n i t e mpasure s p a c e , i . e . ,
be a Riesz subspace of
the additional property that (a) Show t h a t i f
fn
Y
almost everywhere on
.
+
M(Y,v)
0
in
L
, then
(b) Show t h a t i f
0 < g
( c ) Show t h a t
L
i s an i d e a l i n M(Y,v)
(d) Show t h a t
Y
is t h e c a r r i e r of b o t h
5
f
in
L
such t h a t
c L c LI(Y,v)
L,(Y,v)
M(Y,v)
fn(y) and
.
L
.
+
0
m
,
i s a KB-space w i t h holds pointwise
f E L
and
v(Y) <
:L
,
then
g E L
.
.
HINT: ( a ) I f f (y) = l i m f (y) i s p o s i t i v e on a s e t o f p o s i t i v e 0 n measure, t h e r e e x i s t s a number E > 0 and a s e t F of p o s i t i v e measure such
234
BUHVALOV' S THEOREM
that f
n
fo(y)
EX^
2
on
2 E
for a l l
. It
F n
and s i n c e
(b) I t follows from
0 5 gn4
Hence
go = sup gn
in
f o l l o w s from
,
L
exists in
in
, we
for a l l
go-gn J. 0
so
L
in
almost everywhere, so
(c) I f
i s measurable and
f
(by p a r t ( b ) ) , s o since
f E L
Y
. It
f o l l o w s from
for
that
xy E
t h a t f o r any
L
, so
f E L
....
1,2,
=
J. 0
(go-gn)(y) as well a s
,
f E L
n
idplies that
f-
are i n
L
then
If1 = s u p ( f , - f )
Y
i s t h e c a r r i e r of
t h e i n t e g r a l of
E L ,
/fl
over
i s f i n i t e , so
xy E
This shows t h a t
N
so
Ln
EXERCISE 95.6. Let
M
L, c I,
. Hence
.
Conversely, i f
L c L1
. This
gn(y) 1. go(y) g = g E L 0 If1 E L , t h e n f + and
i s a Riesz subspace.
L
(d) It follows from
L
.
. Since
E L
get a contradiction. n
L
almost everywhere by p a r t ( a ) . But t h e n gn(y) 4 g f y )
€xF
gn = i n f ( g , n ) E L
p(gn) < p ( f )
and
L
that
that
Lm c L
fn 9 0
L_ c L
Ch. 13,5961
L
i s t h e c a r r i e r of
Y
L"n
b e t h e same a s i n t h e preceding e x e r c i s e and l e t
be a s p a c e of t h e same type. It f o l l o w s a l r e a d y from t h e r e s u l t s i n t h i s
s e c t i o n and from what was proved i n t h e p r e c e d i n g e x e r c i s e t h a t an o p e r a t o r T
from
member of
(L>)dd
HINT:
f o r any
M
into
L
$
i s an a b s o l u t e k e r n e l o p e r a t o r i f and only i f
. Shm that
It f o l l o w s from
E L*
. This
L;
f n J. 0
shows t h a t
i s e q u a l t o t h e Banach d u a l in
L
L* c L i
that
. For N
is norm complete by E x e r c i s e 95.4, so L* = L L" c L" = L* n
. It
follows t h a t
N
Ln = L*
.
p(fn) J. 0
,so
is a
T L*
L.
of
$(fn)
+
0
t h e converse, n o t e t h a t
L
by Theorem 85.6, and hence
96. Buhvalov's theorem I n t h i s s e c t i o n we s h a l l prove a remarkable theorem, due t o A.V. Buhval o v ([1],1974)
d e a l i n g w i t h a n e c e s s a r y and s u f f i c i e n t c o n d i t i o n f o r an
o p e r a t o r t o be a k e r n e l o p e r a t o r . Even f o r p o s i t i v e o p e r a t o r s i n L -spaces 2
t h e c o n d i t i o n was n o t known b e f o r e . F i r s t we l i s t some p r o p e r t i e s of t h e space
M(X,u)
of a l l r e a l +measurable
f u n c t i o n s on t h e s e t
X
, where
( X , A , u ) i s a a - f i n i t e measure space. These p r o p e r t i e s w e r e d i s c u s s e d and
Ch. 13,1961
KERNEL OPERATORS
235
proved i n s e c t i o n 71. ( i ) (Lemna 71.1).
If
h o l d s f o r u-almost every and '-almost
( i i ) (Theorem 7 1 . 4 ) .
Xn
4
If
,
X
consequence, w r i t i n g
E M(X,p)
f
Xnfn(x)
f o ( x ) = sup,
f -uniformly t o z e r o . ( i i i ) (Theorem 71.5).
un
e x i s t numbers
.
M(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
almost everywhere on
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
almost everywhere on X
+ 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 <
for n 2
nk'
... . Hence, if
so
that is, w(lim s u p E ) = 0 . This is the desired result, since u (y) "k k to zero for all y € (E-lim sup Ek) .
tends
BUHVALOV’S THEOREM
238
Ch. 13,9961
.
If the (a) * c). Let E be a subset of Y such that xE E Li integral of u over E does not tend to zero, there exists a number E
>
0 such that
for some subsequence sequence Since 0
(vm:m=1,2, 5
v 2 u E L m
-
. This subsequence contains a
(un :k=l,2,...)
... )
ksuch that
vm(y)
+
*
. This
m
Y
.
and
it follows now from the dominated convergence theorem that
m
sub-
0 almost everywhere on
contradicts ( I ) ,
+
v du m
and so (c) holds.
+
0 as
.
Let Y Y such that xy E Ln for all m It is m sufficient to prove that (un:n=1,2,. ..) $as a subsequence converging to (c)
(a).
N
zero almost everywhere on a fixed Ym
,
since by means of the diagonal
procedure it is then easy to obtain a subsequence converging to zero almost everywhere on
Y
. Fix
m
and, for the purposes of this proof only, let
the integration sign denote integration over Ym n + =
Hence
,
so there exists a subsequence
W
(u
%
. We have
:k=l,2,...)
almost everywhere on
u du
I n
* 0 as
such that
Ym , which implies that
holds for almost every y E Y m
.
The first main result in this section follows now. THEOREM 96.5. (BuhvaZov’s theorem f o r order bounded operators.) For
an order bounded operator
T
from L i n t o M
t h e foZZowing conditions
are equivalent. (a) T is a kernel operator. ( b ) If
uhere on X
.
0
5
PROOF. (a)
un 5 u E L
and
un
0
, then
(b). We may assume that T
(Tun)(x)
*
0
ahfost every-
is positive (because if
KERNEL OPERATORS
Ch. 1 3 , 8 9 6 1
(a)
(b)
holds for T+ and
T- , then it holds for T ). Hence, let T
be a kernel operator with kernel with
u
*
n
X with
+
0 as
p(X-X
n +
) = 0
0
m
. Since
<
, where
holds for all xo E Xo it follows for every
n
from
almost everywhere on
Y
that
holds for all
k
(vm:m=1,2, ...)
2
0 and let
,
05 u
n
5 u
E
there exists a subset
L
xo
of
zero as n
-
the integration i s over Y
Assume now that for some point
to
T(x,y)
05 u E L
such that
T(xo,Y)u(y)dv(y)
does not tend
239
+ m
xo E Xo
. Then there exists a number
+
xo E X
0 ’
the sequence
. By hypothesis, (un :k=l,Z,...)
such that vm(y)
. If
E
> 0 and
has a subsequence
0 kalmost everywhere,
so
by the dominated
convergence it follows that
This contradicts (2). We have proved, therefore, that
E XO , i.e., (Tun)(x) + 0 as n + m for ,almost every x E X 0 * (a). Assume first that T is positive, so let the positive
for all x (b)
.
2.40
operator T
satisfy condition (b). We have
.
8 5 Tq E
kernel operator and
kernel operators form the band
,
Lb(L,M)
It follows from 8
2
T 5
Let Y
T2
2
0
S
I. Y
such that
x
*
+
=
0
{T(u-uxyn))
(XI
+
for a l l
n
u
T
. There exists a
is the carrier of both
L
and
,
Ln
.
so
0
, i.e., {T(ux )}(x) + (Tu)(x) for almost every n' it evident that we need only prove that (Tu)(x) = 0 for all
05 u E L
vanishes outside some fixed Y
. We
satisfying the condition
fix therefore a natural number
and we assume now that we have a non-negative function u
that
Hence
x E X
. This makes
u
.
E L fl :L
, then (u-ux )(y) + 0 almost everywhere on Y n' 0 holds certainly. Hence
holds almost everywhere on X that
T I a positive
and we have to prove that if
u E L
for almost every x E X
TI+T2 with
that T2 also satisfies condition (b).
T E (L2M)d
n' sequence with this property since Y (u-ux ) n'
=
(L-@M)d This decomposition holds since the n(Ln8M)dd in the Dedekind complete space
satisfies condition (b), then T
If
T
so
we may assume now that 8
no
Ch. 13,1961
BUHVALOV'S THEOREM
vanishes outside Yn
. Let
Fo = (x:(Tu)(x)>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
such that
Ch. 13,1967 From
241
KERNEL OPERATORS
(Sv )(x)
(Sv +T(u-vk))(x) almost everywhere on k 0 almost everywhere on X , i.e.,
5
k
inf (Sv )(x) k k
=
(Tu) (x)
.(infk /vkdv)
almost everywhere on
X
, where
it follows that
= 0
the integration is over Y
"0
that for every positive integer n we have infk{(Svk+T(u-vk))(x):
(5)
X
/vkdvcn-'}
alsmost everywhere on the set Fo
=
This implies
0
. It follows
(x:(Tu)(x)>O)
=
.
that for every
n we have infk{(T(u-v almost everywhere on
k ) J ( x ) : Ivkdv
Fo
. Note now
that we may assume that no vk
Fo
It follows that if for n = 1,2,.
=
0 we have
.. we
any given vk = 0
set
,
En = (T(u-vk) : /vkdv
1 vkdv
is
Sv +T(u-v ) = Tu , s o the k k in (5) does not change if we leave the zerofunction out.
identically zero, because for vk infimum on
= 0
can occur in o n l y finitely many
, which
functions in En
would imply vk
=
vanish outside Fo
0
. Note
En
, because
otherwise
that for every n
. It follows
the
from (6) that for each n
is zero almost everywhere on En is a common majorant of all func-
separately the infimum of the functions in
X and the element Tu in the ideal M tions in all
En
together. Hence, Theorem 9 6 . 3 can be applied. By this
...
theorem, there exists finite subsets E' c E (n=1,2, ) such that n n inf(UmE') = 0 for every m Returning to the original notation, this means m n that there exists a subsequence (vkn:n=1,2, ) such that
.
(7)
(8)
(y)dv + 0
infn>m {T(u-v
kn
...
as n
)](x)
=
+ m
,
0 almost everywhere on Fo
for every m
.
242
BUHVALOV'S THEOREM
Ch. 13,5961
*
It follows from (7) by the last lemma that v -+ 0 as n + m , s o kn (Tv )(x)- 0 almost everywhere on X A l s o , it follows from ( 8 ) that, kn for all m ,
.
almost everywhere on where on Fo (Tu)(x) X
. Hence
> 0
. This
Fo
,
and
(Tu)(x)
so
lim sup(Tvk )(x)
=
. Combining these results, we get (Tu)(x) n= 0 . But Fo is, by definition, the set of all x
where on Fo
ii(Fo) = 0
, and
so
(Tu)(x)
=
almost everyalmost everysatisfying
0 almost everywhere on
is the desired result.
It remains to extend the proof to the case that
i s order bounded
T
but not necessarily positive. It is sufficient to prove that if order bounded operator such that Tu
-+
n + Tu
n
5
u
have this property for the sequence
0
2
. Then
, which
v = u-u n implies
satisfies 0 2 v
(un:n=1,2, 5 u
from v
*
-+
u
that
Hence, noting that
(10)
+
*
inf(v,v ) v , so T(inf(v,vn)) n T(inf(v,v ) ) 5 T+vn , we get n -+
0
v
5
-+
so
T+U =
0
sup ~ v : ~ s v i u5 lim inf
Combining (9) and ( l o ) , we obtain T+u T+u
=
T+(u-v )
-+
=
T+V
...)
for all
For an inequality in the converse direction, let
and
*
E L and u
T
is an
0 implies
0 pointwise almost everywhere, then T+ has the same property. Hence,
let T u
02 u
.
lim T+vn , so
0
pointwise almost everywhere. This concludes the proof.
S
satisfying
n
,
so
u
.
It follows
0
5
T+v
n
Tv by hypothesis.
2
KERNEL OPERATORS
Ch. 13,§961
243
As noted already, the first proof i?f the above theorem was given by An independent proof is due to A.R.
A.V. Buhvalov ([1],1974).
Schep ( [ I ] ,
1977). Their proofs are similar. The proof presented here closely follows Buhvalov's theorem possesses points of
Schep's method (cf. also C21,1979).
analogy with a theorem on bilinear forms proved much earlier by H. Nakano ([31,1953). Nakano proved (in Theorem 5.2 of his paper) that a positive bilinear form is contained in the band generated by the finite rank bilinear forms if and only if it satisfies a certain continuity condition analogous to condition (b) in Buhvalov's theorem. The analogy extends itself to the proof; the formulas (3)-(8)
above correspond to analogous formulas in Naka-
no's proof. Our final a i m in this section is the proof that if
T
is an operator
(not necessarily order bounded) which maps the ideal L
into the ideal M
and which satisfies Buhvalov's condition (i.e., 0 2 u
u E L
implies
(Tun)(x)
+
0
almost everywhere), then T
2
and
*
u
+
n is a kernel operator.
0
The first step is to derive from Buhvalov's condition that T ,when regarded as an operator from L
into M ( X , p )
,
is order bounded. For this purpose
. We
we need more information about order bounded sets in M ( X , p ) mention an Egoroff type result for uniform divergence.
first
-
LEMMA 96.6. Let 0 2 un E M(X,u) s a t i s f y un(x) 4 as n + m on a E of X of f i n i t e p o s i t i v e measure. Furthermore, l e t E > 0 be
subset
given. Then there e x i s t s a measurable subset and t h e sequence C > 0
(un:n=1,2,
...)
there e x i s t s a natural number
x E A and
all
A
of
E
diverges uniformly on N
such that
such that p(E-A) < A
, i . e . , f o r any
u (x) 5 C
for all
n 5 N .
PROOF. Apply Egoroff's theorem to the sequence of functions
c ( I + ~)-':n=1,2,
...} .
THEOREM 96.7. Let
D
be a non-empty subset of
order bounded i f and only i f f o r any sequence
hnfn(x)
(fn:n=l ,2,,..)
+
from
M(X,p)
. Then
D
0
almost everywhere on X
I)
and a n y sequence of numbers
xn+o. PROOF. For the purposes of the present proof we shall say that
is
holds
D
E
244
BUHVALOV'S THEOREM
satisfies condition
if
(a)
...)
sequence (f :n=l,Z,
from
that order boundedness of satisfies
,
(a)
so
implies that D
satisfies
.
0
(a)
for any
It is evident
. For
the
(If]:fED) likewise
we may just as well assume that D
consists of non-
the upwards directed set obtained
Dl by taking finite suprema, i.e., each g E D l is of the form
g = sup(fi:i=l,
...,k)
fni E D
for all
pi = X I pi =
Then pi J. 0
and let
. Every
x
almost every
,
A2
so
fi E D
with
be a sequence in D l
with
+
satisfies
. Then the set
(a)
negative functions. We now denote by from D
X
0 almost everywhere on
+
and any sequence A n
D
D
D
converse, assume that
Anfn(x)
Ch. 13,5961
for
An C 0
i
= I,
. We prove
...,k . Let that
(gn:n=l, 2,
Angn(x)
+
. . .)
0 for
is of the form
g
n
i. Define
for
i
for i
= I, =
...,k 1
*
'
k +I,...,kl+k2;... 1
in view of condition
(a)
.
the sequence
...,u k lf 1 , k l t u k l + l f 2 1 ~ ~ ~ +k ~'uk f2,k27'*'
Ulfl1,
1
converges pointwise almost everywhere on X
2
to zero. This implies that the
sequence
...
converges pointwise almost everywhere on X
to zero, i.e., Algl,A2g2,
converges pointwise almost everywhere on
to zero. It has thus been shown
that D 1 satisfies condition that the given set D
f
0
=
sup D
X
we may just as well assume immediately
i s directed upwards. By Lemma 94.4 the function
exists (it is not excluded in this lemma that
on a set of positive measure) and there is a sequence
D such that f
( a ) , so
4
fo
f = sup f 0
. Assume now that
. Since
D
f (x) =
...)
(fn:n=1,2,
,...)
in
is directed upwards, we may assume that holds on a set E
n 0 Then, by the last lemma, there exists a subset A
ofpositive measure and
fo has the value (fn:n=1,2
of
E
of positive measure. such that A
diverges uniformly to
m
on
is A
.
Ch. 13,5961
n 1 < n2 <
Hence, there exist natural numbers holds on A A
for k
245
KERNEL OPERATORS
for all n t nk (k=l,Z,
...
= 1,2,
...
such that
...) . This implies
fn(x) t k
k-'fn, (x) t 1
on
, which is impossible since D satisfies Eondition
. It follows that fo is finite almost everywhere on sup D = fo E M(X,u) . In other words, D is order bounded. (a)
X
,
so
Everything is now available to prove the final main theorem. L
THEOREM 96.8. (Buhvalov's theorem.) A s before, Zet
and M(X,p)
i d e a l s i n M(Y,v) of both L and
Lx
r e s p e c t i v e l y such t h a t
and l e t
i.e., 0 every
u
5
x E X
5
.
u E L and
PROOF. If T(x,y)
,
un
*
-f
0
implies
into M
L
. Then
s a t i s f i e s Buhvazov's condition, (Tun)(x)
f o r almost
0
-f
T is a kernel operator from L into M with kernel
then (as observed already when kernel operators were defined) there
exists a kernel operator Ta from L kernel, and if then T
T
be
i s the c a r r i e r
Y
T be an operator from
T i s a kernel operator i f and only if
and M
=
into M(X,u)
having
IT(x,y)l
as
T is now regarded as an operator from L into M(X,u) ,
T~-(T~-T)with T
and
T -T
positive,
T
so
is order bounded.
The proof that T satisfies Buhvalov's condition can now be taken over from Theorem 96.5. Conversely, let
T be an operator from L
1w's condition. Note first that if into M(X,u) that T
, then
T
T
into M
still satisfies Buhvalov's condition. We shall prove
is now order bounded. For this purpose, let in L and let D = (Tv:O
val (v:Osv
have to show that (hn:O
satisfying Buhva-
is regarded as an operator from L be an order inter-
S
. We
be its image in M(X,u)
. To do this, let C 0 . Then implies X v 0 . n n
D is an order bounded set in M(X,p)
be an arbitrary sequence in D and let A
v X u .L 0 , so A v -+ 0 , which certainly n n n n n Hence X fin = T(Xnvn) + 0 pointwise almost everywhere by Buhvalov's n condition. This shows that D satisfies the condition ( a ) in the last 0 5
theorem, so
D is order bounded in M(X,p)
image of any order interval in L follows that T
. Having shown
thus that the
is an order bounded set in M(X,u)
is an order bounded operator from L
into M(X,u)
, it
. But
then, by Buhvalov's theorme for order bounded operators (Theorem 96.5), T is an absolute kernel operator from L into M
into M(X,u)
by hypothesis, it is evident that T
.
Since T maps
(as a mapping from L
L
into
246
Ch. l 3 , § 9 6 1
BUHVALOV'S THEOREM
M ) is a kernel operator. We add some remarks about the case that the ideals L equipped with Riesz norms, i.e., L
and M
is norm complete (in other words, if L an absolute kernel operator from by Theorem 83.12 T
M
is a Banach lattice) and
into M , then T
L
and
are
are normed Riesz spaces. If T
is norm bounded. It remains to investigate the case that
T
THEOREM 96.9. Let
+
L
f i n norm i n
and
L
r e s p e c t i v e z y such t h a t fn
is
is order bounded, so into M .
is an arbitrary (possibly non-absolute) kernel operator from L
operator mapping
L
M
L and M
be i d e a l s i n M(Y,v)
into M
. Then
and
Tfn + g
L
and
are Banach l a t t i c e s . Let
M(X,y)
be a kernel
T
T i s a cZosed operator ( i . e . , if i n norm, then
g = Tf
1. Hence, by t h e
closed graph theorem, T i s n o m bounded. PROOF. Let
f
+
f
and
Tfn
+
irl norm. By Theorem 100.6 (which will
g
follow in the next chapter, but to the proof of which we refer because it can be read independently) any norm convergent sequence in a Banach lattice has a subsequence converging in order,
(passing to a subsequence if neces-
so
sary) we may assume that there exists a sequence If-f 1 < hn
that
+
0 pointwise almost everywhere on Y
certainly 0 s If-f I
5
hl E L
for all n
seen, the operator Ta with kernel from L on X
into M(X,y)
. Hence
. On
(since M
,
i.e., Tfn
the other hand we have
and
IT(x,y)l
Ta(lf-fnl)
-t
by Buhvalov's condition. Then also
wise almost everywhere on X on X
Tfn
+
+
g
1,2,3 and f o r each
let T and into L2 of
S
*
+
in L
such
. This implies that 0
.
As we have
is a positive kernel operator /T(f-fn)I
Tf
< Ta(lf-fnl)
-t
0 point-
pointwise almost everywhere
in norm in M
by hypothesis, so
is a Banach lattice) there exists a subsequence Tf
EXERCISE 96.10. Let =
If-fnl
...)
0 pointwise almost everywhere
converging pointwise almost everywhere on X
i
(hn:n=1,2,
to
g
. Hence
Tf
...)
(i=l,2, =
g
.
(Xi,Ai,yi) be a-finite measure spaces for i
,
let Li be an ideal in M(Xi,yi)
. Furthermore,
S be order bounded and order continuous operators from L l into L3 respectively. Show that if one at least
and from L2
and
L l into L j
T is a kernel operator, then ST
.
is a kernel operator mapping
Ch.
13,9961
EXERCISE 96.11. L e t
u
gue measure
and l e t
be k-dimensional r e a l number space w i t h Lebes-
X
be an i d e a l i n
L
v a r i a n t under t r a n s l a t i o n s , i . e . , =
f(x-a)
24 7
KERNEL OPERATORS
for a l l
,
x E X
i d e n t i t y operator i n
L
then
if
M(X,u)
Ta(aEX)
Taf E L
such t h a t
f o r every
f E L
i s in-
L
i s d e f i n e d by
(Taf)(x) =
. We
denote t h e
by I. Show t h a t any t r a n s l a t i o n o p e r a t o r
d i s j o i n t from any a b s o l u t e k e r n e l o p e r a t o r
L
T (from
is
Ta
i n t o i t s e l f ) . For
t h i s i s Theorem 94.7.
a = 0
i s a k e r n e l o p e r a t o r by t h e preceding e x e r c i s e , so
HINT: T T-a
i n f ( T T-a,I)
by Theorem 94.7.
= 9
I t f o l l o w s t h a t , f o r every
u E L+
0 = l i n f ( T T-a,I)
=
1
=
,
inf(Tv'+T wl:vt ,w'EL+,v'+w'=u)
.
= {inf(T,Ta)}(u)
EXERCISE 96.12. With t h e same d e f i n i t i o n s and n o t a t i o n s a s i n t h e p r e c e d i n g e x e r c i s e , show t h a t a l l t r a n s l a t i o n o p e r a t o r s i n d i s j o i n t . Hence, i f l a t i o n operator Lk(L,L)
Ba
,
Ta
i s t h e band g e n e r a t e d i n
then each
Ba
Ln(L,L)
a r e mutally
L
by t h e t r a n s -
i s n o t only d i s j o i n t from t h e band Ba
of a b s o l u t e k e r n e l o p e r a t o r s , b u t a l l
a r e a l s o mutually
d i s j o i n t . This r e s u l t was noted by B. d e P a g t e r .
HINT: For s i m p l i c i t y of n o t a t i o n , l e t space. We i n d i c a t e t h e proof t h a t
X
b e one-dimensional number
inf(Ta,I) = 9 f o r
a > 0
.
For
n
integer, l e t
A = U(A :n even) and B = U(An:n odd). Then n j o i n t w i t h union X Note now t h a t , f o r any u E L+
and l e t
.
A
and
B
are dis-
,
. For
v = uxA
x E A
,
so
and
w = uxB we have
{inf(Ta,I)](u)
f u n c t i o n v a n i s h e s on EXERCISE 96.13. ideal
L
in
M(Y,v)
B
Let
.
v+w = u
v a n i s h e s on
(Y,x,u)
A
and
. Prove
(lav+w)(x) = 0
for a l l
similarly that the
be a o - f i n i t e measure space and l e t t h e
be equipped w i t h an o r d e r continuous Riesz norm
P
248
ADJOINT OPERATORS
such t h a t
i s a Banach l a t t i c e with r e s p e c t t o
L
(i)
Ch. 13,§9751
Show t h a t
p(fn)
+
0
implies
.
p
*
, but
fn + 0
not conversely i n
general. (ii) p(fn)
Show t h a t i f i f and only i f
0
+
0 S If
( i i i ) Let an i d e a l i n
fn
n
I
f E L O
5
.
3 0
for
...
n = l,Z,
then
(X,Ayu) be another o - f i n i t e measure space and l e t
. Show
M(X,u)
t h a t t h e operator
T
, mapping
implies
0
0
(Tfn)(x)
+
HINT:
( i ) To show t h a t
be
M
,
into
L
sequence onto an almost everywhere n u l l sequence ( i . e . , 0 p(fn)
M
is
T maps every dominated norm n u l l
a k e r n e l o p e r a t o r i f and o n l y i f and
,
almost everywhere on
i
p(fn) + 0
implies
*
fo E L
5
X ).
, use
fn + 0
I
If
S
again (as i n
t h e proof of Theorem 96.9) than any norm convergent sequence i n a Banach l a t t i c e has a subsequence converging i n order. For t h e converse, l e t Y = [O,ll
on
f n (y) =
, but
y
L = L (Y,v) . Now choose 2 and zero elsewhere. Then f n ( y ) + 0 f o r a l l
with Lebesgue measure
CO,n-'l
IlfnlI2 = 1
0
( i i ) Let
for all
n
If I 5 f o E L
5
and l e t
w
.
and
0
fn
. Assume
that
pff,)
+
0
does not hold. Passing t o a subsequence i f necessary, we may assume t h a t p(fn) t
f o r some
E
a subsequence
f
> 0
E
and a l l
, i.e.,
that
If
I
If
"k %
5
L
"k . Since
k
+
m
I
+ p
5
f
hk
+
20
fn
of
"k
M(Y,v) . Then . In o t h e r words,
0
in
(\:k=1,2,
in
L
there exists
y E Y
. By
converges t o zero i n order i n t h e space
t h e r e e x i s t s a sequence
0
. In view
converging t o zero f o r almost every
nk
Theorem 7 1 . 3 t h i s means t h a t M(Y,v)
n
0 f
2
"k
...)
M(Y,w)
gk = i n f ( h k , f o )
E L
such and
converges t o zero i n order i n
i s order continuous, t h i s implies t h a t
. Contradiction.
in
p(f
"k
)
+
0
as
9 7 . Adjoint operators Let L#
and
L M#
and
M
be a r b i t r a r y ( r e a l ) v e c t o r spaces with a l g d b r a i c duals
and l e t
d e f i n i n g $ ( f ) = $(Tf)
i.e.,
$J
E L#
i f we w r i t e T#
. Any
$ = T#g
be an operator from
T
for a l l
f E L
L
into
g e t a mapping
i s l i n e a r ; i n o t h e r words, T#
R e c a p i t u l a t i n g , we have
. For
any $ E M#
we get a l i n e a r f u n c t i o n a l on
6 E M # thus determines an element
, we
M
T#
from
Mfl
$
E L#
into
i s an operator from M#
L# into
, so
L
that
, ,
. Obviously, L#
.
Ch. 13,8971
KERNEL OPERATORS
(T#@)(f)
@(Tf)
=
fE L
for all
249 and
@
E M#
.
Note also that
for all T1,T2 and all real coefficients al,a2 called the algebraic adjoint of Assume now that L M
and M
.
T
T# is
are Riesz spaces such that, in addition,
is Dedekind complete, The additional hypothesis ensures that any
T E Lb(L,M)
is of the form T = TI-T2 with TI and T2 positive. Given
any T E Lb(L,M)
, the
The operator Td
maps
let Ji
where
=
. For
T"$
Jil
and
Q2 Ji E L"
shows that
restriction of T# into L"
M-
. For
to M*
are obviously positive linear functionals on
. The operator
the general case that T
T"
+
L
is also
is positive ( s o
T"
is positive). Let
satisfying
$T
For the theorem which follows now we recall that :L
n
+ 0 . We
taking an arbitrary and
L"
the bands of o-order continuous and order continuous members of
and Me
. This .
is order continuous. For
. This follows by (TV$T) (u) = $T(Tu) .L 0 .
0 in L"
E L+ and observing :hat
Similarly for :M
.
are "" L .
THEOREM 97.1. Let L m d M be Riesz spaces such that, i n addition, i s Dedekind complete. I f T E l,(L,M) i s o-order continuous, then T" maps Mi n t o L: Similarly, i f " T i s order continuous, then T- maps :M i n t o L" M
n
.
.
and
is positive. Hence, in
is order bounded, the operator T"
be a downwards directed system in M" T"+
N
T
is called the order adjoint of T
is positive if T
this purpose, we may assume that T
($T:~€{~l)
E M"
the proof, let 0 5 @
order bounded. More is true ; we show that T"
have to show that
will be denoted by
f E L we have
any
The proof shows a l s o that T"
u
. The operator
.
PROOF. Let T be o-order continuous. We may assume that T
is posi-
250
t i v e . T o show t h a t in
Ch. 13,1971
A D J O I N T OPERATORS
implies
L
T-$ E L z
(T-g)(un)
+
for
0
$
for
f M i , we have t o prove t h a t $ E :M We may assume t h a t
.
u $I
n
+
0
is
p o s i t i v e . Then
since
Tun
+
. The
0
proof f o r o r d e r c o n t i n u i t y i s s i m i l a r .
W e r e t u r n t o t h e s i t u a t i o n a s discussed i n t h e e a r l i e r p a r t s of t h i s chapter. Let and
M
(X,A,u)
(Y,C,v)
and
be i d e a l s i n
M(Y,v)
i s t h e c a r r i e r of both
.
and
and
L
be a - f i n i t e measure spaces and l e t M(X,v)
= L;
L-
L
r e s p e c t i v e l y . We assume t h a t
Y
i s t h e c a r r i e r of both
M
and
X
M" = M" Note t h a t t h e l a s t c o n d i t i o n is obviously s a t i s f i e d i f L c n M a r e of t h e L -type f o r some p such t h a t 1 S p 2 m , because i f P f o r example L = L (Y,v) , then LL may be i d e n t i f i e d w i t h L (Y,v) for P 9 p-'+q-' = I Under t h e s e hypotheses, l e t T be an a b s o l u t e k e r n e l opera,., t o r from L i n t o M Then T i s o r d e r continuous, so T i s a l s o order and
and
.
.
:M
continuous and maps
into
i s , t h e r e f o r e , an o p e r a t o r from i s an a b s o l u t e k e r n e l o p e r a t o r .
L" "N
Mn
THEOREM 97.2. Let L and M be clyl absolute kernel operator Then the r e s t g e t i o n T ' of Tfrom :M i n t o :L such that f o r T
n
a h o s t everywhere on Y PROOF. For
f E L
. The r e s t r i c t i o n i n t o ;,I
g E M i we have
s o by F u b i n i ' s theorem t h e f u n c t i o n h(y) =
I
X
TI
of
T"
to
s h a l l prove t h a t
M : T'
s a t i s f y the mentioned conditions and l e t from L i n t o M uith kernel T(x,y) . t o :M i s an absolute kernel operator any g E M L U e have
.
and
. We
IT(x,y)g(x)ldv(x)
Ch. 1 3 , § 9 7 1
KERNEL OPERATORS
i s a measurable function on
Y
I
25 1
with t h e property t h a t
h(y)lf(y)ldv(y) <
m
Y for all
f E L , so
. Writing
h E :L
T'(x,y)
=
, it
T(y,x)
follows t h a t
L f o r every g E M i , and t h e same holds with T(x,y) n replaced by IT(x,y)l This shows t h a t t h e k e r n e l T'(x,y) d e f i n e s an N
i s a member of
.
absolute kernel operator f E L
and
,
M g E ;
M"
from
S
T'
is exactly the r e s t r i c t i o n
of
into
" , T
to
L';I
Mi
. We have . For t h i s
t o prove t h a t
S
purpose, t a k e
and observe t h a t
L e t us assume now t h a t
L
, besides
s a t i s f y i n g t h e same conditions as
i n t h e l a s t theorem, i s a Banach l a t t i c e with r e s p e c t t o a norm which we p ; the
shall call by
.
p(f)
norm of an element
f E L
w i l l be denoted t h e r e f o r e
We s h a l l denote t h e norm i n t h e Banach dual by
p*
. Since
L
is a Banach l a t t i c e , t h e Banach dual i s t h e same as t h e order dual (by
so
Theorem 85.6), t h e norm
p*(g)
N
L* = L
. It
follows t h a t
i s given by
L" c L* n
. For
any
g E :L
Y
Now, € o r any g(Y)
2
= lfgl
0
and and
g E :L
and
f E L
f j ( y ) = -If(y)l p(fl) = p(f)
. It
,
let
at a l l
f J ( y ) = If(y) I y
where
g(y)
at all i
0
y
. Then
where flg =
follows t h a t we may j u s t a s well w r i t e
,
252
ADJOINT OPERATORS
This shows that p *
, because it is evident from the
is a Riesz norm on :L
last formula that p*(g)
=
.
implies p*(gl) s p*(g2)
Ch. 13,1973
for all g E:L
p*(lgl)
and
0
5
g1
N
2
g2 E Ln
It is perhaps of some interest to note that a Banach lattice L
as
introduced here is often called a Banach f u n c t i o n space. The space L' all v-measurable functions g
is then called the associate space of
norm of
p
. From what was proved
hypotheses the space L' of
p*
to
N
L' = Ln
L
and
p'
is called the associate
above, it is evident that under the present
is the same as :L
. For a
of
satisfying
and
p'
is the restriction
further investigation of Banach function spaces
we refer to section 112. Next, assume that both
L
and M
, besides
satisfying the same con-
ditions as in the last theorem, are Banach lattices. Any norm continuous by Theorem 83.12,
the operator T*
N
and the order adjoint T
the algebraic adjoint T# =
M"
, we
T € Lb(L,M) is the Banach adjoint operator T* exists;
is a continuous mapping from M*
the Banach adjoint T* M*
so
T*
find that
to M* =
T"
and M"
into L*
. Observing that
are the restrictions of
respectively and noting that
. The restriction
T' of T"
N
to Mn
,
as introduced in the last theorem, is therefore the same as the restriction of
N
T* to Mn
. As
a consequence of all these observations, the following
theorem is now a variant of the last theorem.
Let L and M be Banuch l a t t i c e s as introduced above T be an absolute kernel operator from L i n t o M w i t h kernel T(x,y) . Then T i s norm continuous and t h e r e s t r i c t i o n T' o f T* t o :M i s an absolute kernel operator from M" i n t o Ln such t h a t f o r any g E M"n n THEOREM 97.3.
and l e t
N
we have
almost evemhere on Y
.
13,5971
Ch.
KERNEL OPERATORS
253
In Theorem 97.2 and 97.3 we have met w i t h c o n d i t i o n s from which one may conclude t h a t
i s a k e r n e l o p e r a t o r i f it i s known t h a t
T'
is a
T
k e r n e l o p e r a t o r . We prove one r e s u l t i n t h e converse d i r e c t i o n . B e f o r e , doing s o , we observe t h a t i f ( a s b e f o r e ) L t h e c a r r i e r of b o t h N
N
(Ln)n (Lz):
,
so
L
and
f o l l o w s by observing t h a t
functions
f
and every
f E L
then
L (:):
i s a l i n e a r subspace of
L
. That
i s included i n
L
c o n s i s t s of a l l v-measurable
THEOREM 97.4.
in M(Y,v)
and i n
N
ll and
(Ln)n
T
.
(c,)
of a l l sequences con-
. We assume a l s o t h a t
and M i
M
be an operator from L
of the algebraic dual
T'
=
Once more, l e t us a s s m e t h a t L and M are i d e a l s M(X,p) such t h a t Y i s the c a r r i e r o f both L and
i s t h e c a r r i e r o f both
Furthermore, l e t
--
i s t h e sequence space
L
Ln =
striction
, L (:):
such t h a t
obviously s a t i s f i e s t h i s c o n d i t i o n . A s an example, we
verging t o z e r o . Then
X
Y
M(Y,v)
satisfying the condition t h a t
mention t h e c a s e t h a t
and
is
L;
i s a l s o t h e c a r r i e r of
Y
i s an i d e a l i n
M :
operator which maps
T#
. Then
i n t o :L
into M
M (:):
= M.
such t h a t the re-
t o M i i s an absolute kerneZ T
i s an absolute kernel operator.
PROOF. Note t h a t , b e s i d e s l i n e a r i t y , w e have assumed n o t h i n g about t h e s t r e n g t h of t h e hypotheses i s a l l on Denoting t h e r e s t r i c t i o n of Theorem 97.2 t h a t into
(MZ):
. We
to
(TI)-
T'
:L )(:
. We apply Theorem 97.2 by
T"
,
s h a l l prove now t h a t t h e r e s t r i c t i o n of
exactly t h e given operator
T
. For
t h i s purpose, l e t
T"
f E L
to
T"f.gdp
=
( T # g ) ( f ) = g(Tf) =
T ; T'
(L;); L
and
be given. Then
f
to
i t f o l l o w s from
i s a n a b s o l u t e k e r n e l o p e r a t o r which maps
T"
L';I
is g E M" n
= (T"f)(g) = ( T ' g ) ( f ) =
A
J
Tf.gdu
.
X For %ixed f E L
Mi
t h i s holds f o r a l l
i s t h e vhole s e t
X
,
g E Mi
it f o l l o w s t h a t
o p e r a t o r , so t h a t , d e n o t i n g t h e k e r n e l of
. Hence,
T"f = Tf T"
by
s i n c e t h e c a r r i e r of
. But
T(x,y)
T"
, we
is a kernel
have
.
ADJOINT OPERATORS
254
for a l l
. To
f E L
I
Ch. 13,8971
complete t h e proof, we s t i l l have t o show t h a t
lT(x,y)f(y)\dv(y) E M
f o r all
f E L
,
(MX):
=
Y
and t h a t i s where w e use t h e hypothesis t h a t
M
. Since
T"
i s an
a b s o l u t e k e r n e l o p e r a t o r , we have
Y for a l l
f E (Li):
,
and hence c e r t a i n l y f o r a l l
f E L
.
Teh condition i n t h e l a s t theorem t h a t (Mw)- = M holds may seem somen n what a r t i f i c i a l . I n some important cases, however, the condition i s s a t i s -
M = L ( X , p ) f o r a value of
f i e d . For example, i f then
(MX):
that
T
= M
. If
P
p
such t h a t
i s order bounded, then t h e f i n a l conclusion t h a t
o p e r a t o r (from
L
1 5 p
2 m
,
t h e condition i s not s a t i s f i e d , but i n s t e a d i t i s given
into
T
i s a kernel
M ) i s s t i l l v a l i d . The proof i s t h e same u n t i l t h e
conclusion t h a t
Since
T"
i s an absolute k e r n e l o p e r a t o r , t h e operator
IT"I
has
a s kernel. Hence, t h e l e f t hand s i d e of (1) may be r e w r i t t e n a s
IT(x,y)l
IT"I ( f ) I ,
which i s equal t o
I f , in p a r t i c u l a r , we have a r e a l s o members of
L
0 s f E L
and s o
t h a t t h e element i n (2) i s now
, then
all
1T"gI = / T g /
ITI(f)
, which
g
satisfying
f o r these
lgl
2
f
g . I t follows
by our new hypothesis i s a
13,1973
Ch.
member of
and s o
KERNEL OPERATORS
M
. Hence,
I
we f i n d t h a t
returning t o ( I ) ,
IT(x,y)lf(y)dv(y) E M
255
0 S F E L ,
for a l l
i s an a b s o l u t e k e r n e l operator.
T
(Me)= M and t h e r e i s no e x t r a hypothesis, n n i t i s s t i l l of i n t e r e s t sometimes t o conclude a t l e a s t t h a t T , when
I f it i s not given t h a t
regarded a s an o p e r a t o r from
,
N N
into
L
(Mn>,
i s an a b s o l u t e k e r n e l ope-
rator. The extension of t h e r e s u l t s i n t h i s s e c t i o n t o t h e complex case i s not d i f f i c u l t . The only p o i n t which i s perhaps not immediately evident
i s where we have a Banach function space with norm
i s given, then
g E :L
that i f
I f g l dv:p I n t h i s case, it i s a p p r o p r i a t e t o define
z/[zl i f
sgn z =
,
f E L and
let
and we have t o prove
p
# 0 and
z
sgn z =
, since
p(fl) = p(f)
1
if
. It
IflI = ( f l
f o r any complex
sgn z
.
z = 0
for all
f l ( y ) = / f ( y ) I / s g n g(y)
(f)s1
y E Y
Given
by
z
g E Li
. Then
and
f l g = Ifgl
follows a s before t h a t hhe
e q u a l i t y i n (3) holds. EXERCISE 97.5.
As before, let L
(X,A,u)
and
M
be i d e a l s i n
r e s p e c t i v e l y . Furthermore, l e t
Y
be t h e c a r r i e r of both
M
and
X
let
be t h e c a r r i e r of both
L
(a) Prove t h a t dense i d e a l i n
L (:)
. Similarly
, we
fl n
maps
T' = 8
Mi
T'
denote by into
L';[
. Similarly,
into
Lx
, then
.
T4
t h e r e s t r i c t i o n of
. Prove
that
T = 0
S
show t h a t i f
into
S = 0
i f and only i f
TI'
( c ) Show t h a t i f
TI'
M
such t h a t
T,
A
so
M(X,p)
:L
and
L
i s an order
e
S' = 8
T2 = 8
T2
,
into
that
T'
. Derive a s a s p e c i a l
i s from L
T
.
and
. We r e c a l l
:M
( t h e n u l l o p e r a t o r ) i f and only i f
i f and only i f =
to
and
i s an order continuous o p e r a t o r from
case t h a t i f t h e order continuous o p e r a t o r T = 0
, and
M(Y,v)
L
i s an order continuous operator from L
T
(b) As b e f o r e , i f M
M
for
n
and
M(Y,v)
.
M"
i s order dense i n
be a - f i n i t e
(Y,.Z,v)
and
measure spaces and l e t
into
M
,
then
a r e order continuous o p e r a t o r s from then
Ti'
A
T" = 9.. 2
L
Ch. 1 3 ,5 9 7 1
ADJOINT OPERATORS
256
(d) Show t h a t i f
T, and T2 a r e order continuous o p e r a t o r s from L (T VT ) " = T','v T; . S i m i l a r l y f o r TI A T2 Show f i n a l l y 1 2 t h a t (T VT ) ' = T' V Ti S i m i l a r l y f o r TI A T2 1 2 1 For more general r e s u l t s we r e f e r t o s e c t i o n 115.
into
,
M
then
.
HINT: For ( c ) , n o t e t h a t
which implies t h a t
0 5 u E L (:): for
TY
A
(i=1,2)
T;
T','A TY coincides with
. Since
L
T" 2
,
and
T','
so
For ( d ) , w r i t e The l e f t hand s i d e i s
which i s equal t o
8
T3
=
-
(T'vT')
T1 v T2
. We have
1
2
T :
p a r t (b). Note a l s o t h a t
4 u
L
T'
>
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
almost everywhere on
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
almost everywhere on
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
<
, where
m
i s a member of
y E (s)
the symbol
domy(T)
T
.
i f and only i f n = 0
denotes t h a t t h e term with
'C
omitted i n t h e summation. (b) Show t h a t f o r any included i n that
domy(T)
m
Cn,,Innlp
<
m
p
. Note .
satisfying
Lp
that
1
5
p <
m
the space
i s t h e space of a l l
Lp
y E (s)
is such
(c) Shuw t h a t
(d) Show t h a t
T , a s an o p e r a t o r from
L2 i n t o
e2 , i s
a unitary
operator (and t h e r e f o r e a norm continuous o p e r a t o r ) . The o p e r a t o r one of t h e v a r i a n t s of t h e HiZbert transform. HINT:
(b) Use Hglder's i n e q u a l i t y .
(c) zn(m+n+i)-2 = zn(n+:)-2
8 E;(2n+l)-* For
ml f m2
,
set
zn (m
=
T
2
.
a = m2-ml
+n+4
= 4 zn(2n+1)-2 =
. Then
(m2+n+1 )-I
= zn(n+4)-l
(a+n+l)-l =
T
is
is
(d) It follows from part (c) that
L2 consisting
subspace of
y E
of all
preserves inner products on the
T
L2
coordinates. By continuity this restricted tor
e2
on
Tex
nm
Zn T(m,n)S,
=
. This .
,
for all m
in (c) it follows now that
L2
with only finitely many nonzero T
T2
=
so
) = t = T
(77
Tex = T
, where
I
implies that the range of
T
[,,
can be extended to an opera-
such that Tex prserves inner products in
more by continuity it follows that if then
26 I
KERNEL OPERATORS
Ch. 13,598:
x ex
for x
=
and once
(5 ) E
e2
. By means of the formulas
,
I is the identity operator in
is the whole of
L2
. Hence, T
is
unitary
EXERCISE 98.7. Under the same hypotheses as in the preceding exercise, let x
=
(5
) E
En “n
Let
e2
=
and y
(n,)
E L2 be defined by
I
(n’log n>-’ 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
almost everywhere on X
...
.
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)
t o Theorem 96.7 i t i s s u f f i c i e n t t o prove t h a t
,
so
(ITvl:OSv
i s an o r d e r bounded s e t i n
. Then
L
lTvl S p ( u ) . t
holds i n
M(X,u)
i s ccmtained i n t h e o r d e r i n t e r v a l
. This shows . Note
CO,p(u).tl
now t h a t
SO
c o n d i t i o n ( i ) h o l d s f o r IT1
.
Assume, f i n a l l y , t h a t t h e norm
first that if
0 S u
< uo E L -
and
in
p
un
*
+
L
,
0
i s o r d e r continuous. We prove then
p(un)
n may assume (by p a s s i n g t o a subsequence i f n e c e s s a r y ) t h a t some
E >
0
and a l l
n
. I n view
of
u
*
n
+
0
+
0
. If
n o t , we
P(U) 2 E
for
t h e r e e x i s t s a subsequence
264
u
Ch. 1 3 , 1 9 9 1
GENERALZZED CARLEMAN OPERATORS
.
(k=1,2,. . )
nk
. By
y E Y
converging to zero for almost every
Theorem
u converges to zero in order in the space M(Y,v) , "k i.e., there exists a sequence (h :k=l,2,. . . ) in M(Y,v) such that k u 5 hk i 0 in M(Y,v) Then 0 5 gk = inf(h , u ) E L and u 5 gk i 0 k O "k .nk in In other words, u converges to zero in order in L Since p is
7 1 . 3 this means that
.
.
.
-
nL
order continuous, this impTies p(un ) + 0 as k + , which contradicts k p ( u n ) 2 E for all k . Assume now that conditions (i) and (ii) hold for k T. Let 0 5 u I uo E L and u -+ 0 Then p(un) + 0 by what was proved n a moment ago, and hence (Tun)(x) -t 0 almost everywhere on X by condition
*
.
T satisfies Buhvalov's condition,
(ii). This shows that
T
so
is a kernel
operator. A s observed above, /TI satisfies conditions (i) and (ii) just as well as
in condition (i)).
T (and with the same function t
/TI is also a kernel operator. We recall that if
that T(x,y)
,
then
IT1
has the kernel
We shall prove now that if the norm
T has the kernel
.
IT(x,y)l
in L
p
It follows
is order continuous and
the equivalent conditions (i) and (ii) in the last theorem hold for T then T
is not only a kernel operator, but in this case T
,
is a kernel
operator of a special type. For this purpose we recall that if we define, Y , the finite or infinite number
on
for any v-measurable real function g ~ ' ( g ) by
p'(g)
= SUP(
1
1
lfgldv:p(f)
Y then the space L' ate space of te norm of T(x,y)
on
L p
y
satisfying p'(g) to
p'
<
L'
-
is called the associis called the associa-
(cf. section 97). If the real (uxv)-measurable function
X x Y
(we write Tx(y)
tion of
g
and the restriction of
for u-almost every x
of all
,
is given, then x
,
so
T(x,y)
p'(Tx(y))
to indicated that
for some fixed
is a v-measurable function of
y
is well defined for these values of T(x,y)
is regarded here as a func-
x ) . Let u s denote p'(Tx(y))
by
, as
x
question arises whether the function t
a function of
u-measurable. An affirmative answer (for any Riesz norm
p
t(x) on
, not
X
. The ,
is
necessarily
order continuous) is contained in a paper by W.A.J. Luxemburg ([2],1958). Luxemburg's result was generalized by Ya.1. Gribanov (Cl],1970). This more general result will enable u s to prove now that if then conditions (i) and (ii) hold for of
T has the property that
t(x)
=
T
p
is order continuous,
if and only if the kernel T(x,y)
p'(Tx(y))
is finite for almost every
Ch. 13,5991
x
. In other words, taking into account
that
find that (i) and (ii) hold if and only if that
L
=
265
KERNEL OPERATORS
L (Y,v)
and
2
is p-measurable, we shall
t t
E M(X,p)
.
In the special case
is the corresponding L -norm we have
p
2
This shows that conditions (i) and (ii) hold if and only if
T
L' = L
is a Carle-
man kernel operator. We present the proofs, first of all the proof of the Luxemburg-Gribanov result.
(Luxemburg-Gribanov.) Let
THEOREM 99.2.
be o-finite measure spaces, l e t function on
x
x Y
functions on Y
and l e t
. Then,
S
T(x,y)
(Y,C,v)
(X,h,p) and
be a non-negative (pxv)-measurable
be a s e t of non-negative v-measurable
for u-ahost e v e q
x E X
, the f i n i t e or i n f i n i t e
non-negative number d(x)
= sup(
T(x,y)v(y)dv(y):vES Y
i s well defined. Furthermore, there e x i s t s a sequence
(vk:k=l,2,
...)
in
such that d(x)
=
(I
sup
holds for p-almost a l l
Y
x E X
x
p-measurable function on PROOF. Since T(x,y) of
X
T(xO,y) x E Xo
T(x,y)vk(y)dv(y):k=1,2
satisfying p(X-X
0
.
,...
simultaneously. I t foZZows that
is a
is (uxv)-measurable, there exists a subset xO ) = 0 such that for every xo E Xo the function
is non-negative and v-measurable on Y and every v E S
the integral of
. It follows that
T(x,y)v(y)
over
defined, and so 0 5 d(x)
d(x)
= SUP(
T(x,y)v(y)dv(y):vES)
5
-
Y
for every i s well
S
266
Ch. 13,1991
GENERALIZED CARLEMAN OPERATORS
for every x E X
0 '
(i) Let now X n = l,Z,.
n
+
For each x E Xo
n
dn(x)
5
dn+,(x)
. Furthermore
,
Y
for all
Tn(x,y)vn(y)
.
n
4
u(Yn)
and
,
On the other hand
lim dn(x) n
-.
limn + w dn (x)
so
as
T(x,y)v(y)
n
-f
Taking the supremum over all v E S on the left, we get =
over Y
Tn(x,y)vn(y)
exists for each Hence, for any
Y
Y
d(x)
non-
and each x E Xo , let
,
x E Xo,and any v E S
finite for
let
the integral of
so
1,2,..
=
u(Xn)
,
, as a function of y , is
the function Tn(x,y)
is well defined. For each n
Note that
Y with
and each v E S
n
negative and v-measurable on
x E Xo
+
X and Y
.. . For each
dn(x)
5
d(x)
for all n
for all x E Xo
prove that for each n
=
1,2,
...
,
so
d(x)
lim d (x) n n
5
S
limn dn(x)
d(x)
. It follows immediately that
. Hence
if we can
there exists a countable subset
sn
of
s
such that
for p-almost all x E X
p-almost everywhere on
simultaneously, then
X , which will then prove the theorem. For the re-
maining part of the proof we may restrict ourselves, therefore, to the case X and Y
that in
S
are of finite measure, T(x,y)
is bounded and the functions
are uniformly bounded. (ii) We assume now that
u(X)
and
u(Y)
.
are finite and there exists
Ch. 13,5991
0
a constant a > 0 such that x,y
267
KERNEL OPERATORS
and all v E S
. Let first
5
T(x,y)
and
5 CY
0
v(y)
5
for all
5 CY
be a step function on
T(x,y)
X X Y
of
the form
non-negative, all A. p-measurable subsets of X i B; v-measurable subsets of Y Then T(x,y) can be written as P with X I , X disjoint u-measurable subsets T(x,Y) = CiE1 xX2(x)gi(Y) I of X and all g. non-negative v-measurable functions. It is easy to see
with all coefficients c and all
...,
.
that in this case the corresponding function d(x) X
possessing the desired properties (note that
is a step function on is constant on each
d(x)
xi 1. (iii) We assume again that u(X) and
0
5
v(y)
.
for all v E S
a
5
v(Y)
and
are finite, 0
Then T(x,y) E LI(XxY,pxv) ,
so
...
; n = 1,2,
well-known result there exists a sequence Tn(x,y)
T(x,y) < a
5
by a
, each
Tn a step function as in part (ii), such that
XXY
I
I
T(x,y)-Tn(x,y)
d(uxv)
.
0
+
+ 0 By passing to a subsequence it may be assumed that /T(x,y)-Tn(x,y)l holds (uXv)-almost everywhere on X x Y It is also evident that we may
.
assume that
0
5
T (x,y) n
IT(x,Y)-T~(x,~)~ 5
CY
almost everywhere on p(X-X ) 1
=
on
on
a
5
x
X x Y
x Y
,
f o r all
X x Y f o r all
n
n
, and
. Since IT-T,~
so
there exists a subset X I of
0 such that, for every x E X I a 2 IT(x,y)-Tn(x,y)1
-+
o
+
X
pointwise satisfying
, we have
0 v-almost everywhere on
Y
.
Hence, by the dominated convergence theorem,
/
IT(x,y)-Tn(x,y)ldv(y)
+
0
Y for all x E XI
.
Since 0
for x E XI fixed,
5
v(y) s a
for all v E S
,
it follows that,
268
Ch. 13,§991
GENERALIZED CARLEMAN OPERATORS
uniformly with respect to v E S
.
In other words, introducing the uniform
norm in the space of all real bounded functions on
S , we have
[lt-tn/l+O.
Still keeping x E X 1 fixed, we have
d(x) = sup(t(v):vES)
=
#It11
.
It follows from Illtll-llt / I I 5 llt-tnll + 0 that IltnII +IItI/ , i.e., n dn(x) -+ d(x) Since this holds for each x E X I and since all dn are
.
p-measurable on
X by part (ii), we see already that the function d
p-measurable on X S
. For each
n there exists a countable subset
of
such that Tn(x,y)v(y)dv(y):vES
nJ
Y
almost everywhere on X
almost everywhere on
. Hence, writing
X
for all
n
m
S = U 0
S 1 n
,
we have
simultaneously. Arguing now with the
uniform norm in the space of all real bounded functions on on
sn
is
S
, we
get exactly as above that
Y =
limn{sup(
I
lltll = limlltnll
,
So
instead of
i.e.,
Tn(x,y)v(y)dv(y):vES
Y
holds p-almost everywhere on X
. But
lim d (x) = d(x) n n
,
SO
T(x,y)v(y)dv(y):v€So
Y This is the desired result. Luxemburg's original result is contained in the following corollary.
Ch. 13,5991
KERNEL OPERATORS
COROLLARY 99.3. Let
L be an i d e a l i n
equipped w i t h a Riesz norm
=
M(Y,v) such t h a t
. Furthermore, Zet T(x,y) x Y . Then t h e pointwise
p
(uxv)-measurable f u n c t i o n on X t(x)
269
P'(T~(Y)) =
I
SUP(
is
L
be a r e a l supremm
1 T(x,y)f (y) 1 dv (y) :P (f)< 1
Y
i s u-measurable and there e x i s t s a sequence that
p(fn)
5 1
for a l l
t(x)
=
(fn:n=1,2, ...) i n
such
L
and
n
P'(Tx(y))
1
sup(
=
1 d v ( y ) :n=l,Z,. ..
1 T(x,y)fn(y)
Y
holds f o r u-almost a l l
x E X
simultaneously.
We return to the properties (i) and (ii) in Theorem 99.1. THEOREM 99.4. Let
Riesz norm
L
and l e t
p
be an i d e a l i n M(Y,v)
T be an operator mapping
w i t h order continuous
.
i n t o M(X,p)
L
The
following conditions are now equivalent. (i) There e x i s t s a non-negative f u n c t i o n
I (Tf) (x) I
satisfies
f E L
(ii) If
f
n
E L
for
almost everywhere on X
.
5 p(f).
n
=
l,2,
and
p(fn)
(iii) T i s a kernel operator w i t h kernel P'(Tx(y))
E M(X,u)
such t h a t every
almost everywhere on X
t (x)
...
t E M(X,p)
.
+
0 ,
T(x,y)
then
.
(Tfn)(x)
satisfying
PROOF. If the equivalent conditions (i) and (ii) hold, then T kernel operator by Theorem 99.1. Let observed in Theorem 99.1, IT1
. By
is a
. As
T
. Note
that the kernel of
IT1
is
the Luxemburg-Gribanov theorem there exists a sequence
(fn:n=1,2, ...)
in L
such that
for u-almost every x
. For each
(i) that
T
is also a kernel operator satisfying the
Same conditions (i) and (ii) as
1 T(x,y)l
be the kernel of
T(x,y)
+
p(fn)
5 1
n = l,2,
for all n
...
and
, we derive from condition
0
270
GENERALIZED CARLEMAN OPERATORS
Iiolds f o r p-almost
every
Hence, t a k i n g f o r
Xo
x
,
Ch. 13,1991
t h e e x c e p t i o n a l pi-null s e t depending upon
t h e union of t h i s countable number of e x c e p t i o n a l
s e t s , we f i n d t h a t t h e formulas i n t h e l a s t l i n e s h o l d f o r a l l neously o u t s i d e t h e u-null s e t we g e t
p'(Tx(y)) 5 t ( x )
Conversely, i f
T
Xo
. Taking
f o r almost every
n
fn
simulta-
t h e supremum w i t h r e s p e c t t o n ,
, so
x
E M(X,p) . E L b e given.
p'(Tx(y))
s a t i s f i e s condition ( i i i ) , l e t
f
Then
COROLLARY 99.5.
If t h e equivaLent conditions ( i ) , ( i i ) , ( i i i ) hold f o r T , we s h a l l c a l l any 0 5 t E M(X,p) such t h a t , f o r e v e r y f E L , t h e i n e q u a l i t y I (Tf) (x) 1 L p ( f ) . t ( x ) holds almost everywhere on X , a majorizing function. Then p'(Tx(y)) = i n f ( t : t is majorizing)
.
PROOF. It was shown i n t h e proof of t h e l a s t theorem t h a t 5
t(x)
almost everywhere on
X
f o r any m a j o r i z i n g
a s e v i d e n t from t h e l a s t l i n e of t h e p r o o f ,
p'(T
t
(y))
. On
p'(T (y)) 5
t h e o t h e r hand,
is itself a
ma j o r i z i n g f u n c t i o n . I f i n t h e l a s t theorem p'
is also the
1
L,-norm. L
IT(x,y)l
2
L
is the mace
L2(Y,v) w i t h L2-norm p
,
then
Condition ( i i i ) e x p r e s s e s now t h e f a c t t h a t dv(y)
Y
i s f i n i t e f o r almost every
x
. i.e.,
T
i s now a Carlernan o p e r a t o r . A s ob-
s e r v e d a l r e a d y , i t w a s proved by J. Weidman (C11,1970) t h a t i n t h i s c a s e
KERNEL OPERATORS
Ch. 13,9991
27 1
condition (ii) is necessary and sufficient for T
It was proved by V.B. Korotkov ([11,1965)
to be a Carleman operator.
that condition (i) is necessary
and sufficient to be a Carleman operator. The present generalization in Theorem 99.4 and also the theorems which will be proved next are due to A.R. Schep (C21,1979).
We shall call any operator satisfying the conditions
in Theorem 99.4 a generalized Carleman operator. The following theorem is an immediate corollary of what was proved above. L
Let
THEOREM 99.6.
be an ideal i n M(Y,v)
with order continuous
.
Riesz norm P and l e t M be an ideal i n M ( X , u ) with Riesz norm h Furthermore, l e t T be an operator mapping L i n t o M The following
.
conditions are now equivaZent. (i) There e x i s t s a non-negative function
t
E M such that every
.
satisfies 1 (Tf) (x) 1 5 ~(f).t(x) almost everywhere on X is a kernel operator with kernel T(x,y) s a t i s f y i n g
f E L
-.
(ti) T p'(Tx(y))
EM
PROOF.
, so
(i)
X{P'(T~(Y))I
<
* (ii). According to Theorem
with kernel T(x,y)
, mapping
99.4 T
into M ( X , v )
L
is a kernel operator
and such that p'(Tx(y))
is a majorizing function. Since by hypothesis there exists a majorizing function t
such that
t
E M and since p'(T
X
(y))
is the minimal
majorizing function by the last corollary, it follows that (ii)
(i). If (ii) holds, then
everywhere on
X
E
By hypothesis p'(Tx(y)) function belonging to M Any operator
f E L
for any
T
.
M
,
so
. Hence
I (Tf)(x)] p'(T
2
(y))
p'(Tx(y))
p(f).p'(Tx(y))
E M
.
almost
is a majorizing function.
t(x) = p'(Tx(y))
is a majorizing
satisfying condition (ii) in the last theorem is
sometimes called an operator of f i n i t e doubZe norm. Let us take a brief look at the case that
M
=
Lq(X,u)
L = L (Y,v) P
for some q
order continuoas for
I 5 p <
M = L
P'
for some p
-.
satisfying
1 5 p <
satisfying
1 5 q <
m
. Note
m
and
that the L -norm is
P
from L = L into P , where II IIT ( Y ) I / ~ , / I ~ <
The operator T
is now of finite double norm whenever q -1 is the conjugate exponent of p , i.e., p +(p')''
= 1
. These operators
Our last theorem states are known as Hille-Tamarkin operators ([1],1934). that T is a Hille-Tamarkin operator if and o n l y if there exists a nonnegative function g E L ( X , v ) q
such that for any
f E L
P
the ineqUalitY
272
GENERALIZED CARLEMAN OPERATORS
I ( T f ) ( x ) l s Ilf 1 ) . g ( x )
h o l d s almost everywhere on
P
We r e t u r n t o Carleman o p e r a t o r s , i . e . , L2(Y,v)
into
,
M(X,u)
I
/T(x,y)l
.
X
kernel operators
such t h a t t h e k e r n e l
2
Ch. 13,5991
T(x,y)
T
, mapping
satisfies
.
dv(y) E M(X,u)
Y
every
(i)
T
(ii)
There e x i s t s a non-negative f u n c t i o n
x (iv)
,...,e
.
If
n
.
i n t o M(X,u)
i s a Carleman operator.
( i i i ) If
e
L2(Y,v)
conditions are now equivalent.
f E L2(Y,v)
any
be an operator mapping
T
THEOREM 99.7. Let
The following
, the
IlfnlI2
I(Tf)(x)/ 5 Ilf/I2.g(x)
inequality +
0
g E M(X,v)
, then
(Tfn)(x)
(ea:aEIa)) is an orthonorma2 system i n i s a f i n i t e subset of t h e system and
then the s e t
of a l l
S
t h i s case, t h e s e t
S
i n t e g r a t i o n i s over
Y
hoZds f o r almost
almost everywhere on
0
-f
such t h a t , f o r
L2(Y,v)
,i
.
X
f
.
s ( a l , ...,a ) i s order bounded i n M ( X , y ) In n 2 i s bounded above by IT(x,y)I du(y) , where t h e
I
.
PROOF. I t f o l l o w s from Theorem 99.4 t h a t t h e c o n d i t i o n s ( i ) , ( i i ) and ( i i i ) a r e e q u i v a l e n t . The proof t h a t (i) i m p l i e s ( i v ) f o l l o w s by observing t h a t , by B e s s e l ' s i n e q u a l i t y , s(a
5
,,...,an ) ( X I
1
=
11
:Z
2 T(x,y)ea.(y)dv(y)/ 1
Y
/T(x,y)I
2
dv(y)
.
Y
This shows a t t h e same time t h a t i n t h i s c a s e t h e s e t by
I IT(x,y)l
2
dv(y)
.
S
i s bounded above
For t h e converse, assume t h a t ( i v ) h o l d s . We s h a l l prove t h a t t h i s i m p l i e s ( i i i ) . Hence, l e t
f n E L2(Y,v)
There e x i s t s an orthonormal system
for
(el,e2,
n = 1,2,
...)
in
...
with
L (Y,u) 2
Ilfn112
such t h a t
-t
0.
Ch. 13,9991
KERNEL OPERATORS
...
every finite linear combination of
(fl,f2, ) is a finite linear combinaand conversely (Gram-Schmidt). Let S be the set in
... )
tion of
(el,e2,
M(X,y)
273
corresponding to this orthonormal system according to our hypothesis
about condition (iv) and let
be an upper bound of
g2 E M(X,v)
function fn is afinite sum Cc.e.
, so
Tfn = Cc.Tei
, and
. The
S
by Schwarz's in-
equality we have
for almost every
.
x
Evidently we may assume now that
IJf I] .g(x) holds almost everywhere on X n 2 Since \lfn112 -t 0 , it follows that (Tfn)(x) 5
f o r all +
n
5 l(Tfn)(x)l simultaneously.
0 for almost every x ,
so
(iii) holds. THEOREM 99.8. Let
and l e t 2et
T be a Carleman operator a s in the l a s t theorem
be an orthonormal b a s i s i n
(ea:a€{a))
be onee more the s e t of a l l
S
(i)
sup S =
1,
2
IT(x,y)l
dv(y)
s(a1,
.
...,an ) . Then the following
holds.
... , f o r
a E {a} , say
(ii) There are a t most eountabZy many
which
. Furthermore,
L2(Y,v)
a1,a2,
Tea is not zero almost everywhere, and
holds f o r a l m o s t every
x
x E
.
PROOF. (i) Note first that according to the proof of the last theorem
I (T(x,y)l 2 dv(y)
. Furthermore,
S
is an upper bound of
Theorem 99.5, the same integral is the infimum of all any X
,
f E L2
. Now,
let
the inequality f E L2(Y,v)
I (Tf)(x)l
5
according to
8'
such that, for
))f)I2.g(x) holds for almost every
be given. The inner product
c
=
(f,ea)
, i.e., ilf-fkl/ + 0 k Hence, by property (iii) of a Carleman operator in the last theorem f
is the norm limit (as k
(Tfk)(x) we have
+
(Tf)(x)
+
m
) of
f = Ctciei
almost everywhere on
X
. Denoting now
satis-
,... . Then
ea , say for el,e2
fies ca # 0 for at most countably many
sup S
by
.
h2
3
274
GENERALIZED CARLEMAN OPERATORS
almost everywhere on
X
for k x
S
l,2,
=
...
/ (Tf)(x)/
on the left, we get
+ m
. This shows that
1 (Tf)(x)/
for k
the function h
simultaneously. Taking the limit
s llf112.h(x)
1
sup S = h2 2 inf g2 =
for almost every
is one of the functions g
almost everywhere on
Ilfi/ .g(x) 2
Ch. 13,5993
/T(x,y)l
X 2
.
satisfying
It follows that
dv(y)
.
Y Combining this with the earlier observation that the last integral is an S
upper bound of
, we
get the desired result.
(ii) On account of the order separability of
M(X,u)
an at most countable subset possessing the same supremum as
the set S
(An:n=1,2, ...) of finite subsets of
there exists a sequence
has
. Hence,
S
{a) such
that (T(x,y)/
2
dv(y)
=
I (Tea)(x)I2:n=1,2,
sup
...
Y
It follows immediately that
(Tea)(x)
in the complementary set of
UIAn
as
, we
ctl,a2,...
m
= 0
almost everywhere for every
. Numbering the elemenbs
get the frlnal result
for almost every x E X F o r the case that
a
in
a
u>n
. L (Y,v) 2
i s separable (i.e., every orthonormal basis
is at most countable), the equivalence of (i) and (iv) in Theorem 99.7 and the results in Theorem 99.8 are due to M. Schreiber and G. Targonski ( [ l ] ,
1970). EXERCISE 99.9. Let L2(X,p)
T
L (Y,u) into 2 are o-finite measure spaces. Show
be a linear operator mapping
, where (X,A,p) and
(Y,C,v)
that the following conditions are equivalent. (i) T
is
a
Hilbert-Schmidt operator, i.e., T
with kernel T(x,y)
satisfying
is a kernel operator
Ch. 13,1991
KERNEL OPERATORS
1
IT(x,y)l
2
d(uxu) <
XXY
-.
There exists a non-negative function g E L2(X,p)
(ii)
f E L2(Y,v)
every
275
.
most every x € X
, the inequality I (Tf)(x)/
(iii) For every orthonormal system
5
It is easy to see
that (i) implies (iii) Assume that (iii) holds and let
. It follows from
holds for at most countably many
*<
,
Cy\lTekII
,
(ea:a€{a})
(iii) that
say for al,a2
be any
I/Teall > 0 Then
,... .
i.e.,
1 (Tek) (x) 1
Cy
a E {a}
holds for al-
in L (Y,u) we have 2
(ea:a€{a})
HINT: (i) and (ii) are equivalent by Theorem 99.6. orthonormal system in L2(Y,v)
such th t, fo
llfl12.gk)
2
,
dub) <
X and
m
C1
so
(Te )(x)
=
I (Tek)(x)I2
i s finite for almost every
0 almost everywhere far any
... , we
T
. Hence, by
T is Carleman operator. It follows from Theorem 99.8 (ii)
i s Hilbert-Schmidt.
EXERCISE 99.10. Let A
be the interval [ O , l l
measure. Furthermore, let 0 < a < 1
(i) Show that T(x,y)
and
T(x,y)
, equipped with Lebesgue = Ix-yl
-a
on
A x A
1
(ii) For
5
a < I
the function T(x,y)
a kernel operator T mapping
-1
=
Ix-~I-~
$
and
f
is
i s the kernel of
L (A) into itself such that 2 Carleman operator. For the proof, note the following facts. If
.
i s the kernel of a Hilbert-Schmidt operator
(from L (A) into itself) for 0 < a < and TI(x,y) = x T(x,y) 2 the kernel of a Carleman operator which is not Hilbert-Schmidt.
(a)
that
a E { a } which is not one of
by the function E y 1 (Tek)(x)I2
is bounded above in M(X,u) that
. Observing now
find that the set of all
a1,a2,
Theorem 99.7,
x
T
is not a
are (real or complex) Lebesgue measurable functions
276
Ch. 13,1999
GENERALIZED CARLEMAN OPERATORS
on the real line, both of period one and Lebesgue s m a b l e over A , then h(x)
=
i
+(x-y)f(y)dy
A
is Lebesgue sumable over A (b) then hp (c)
It1 2 and
4
llhlll 2 ~ ~ $ ~ ~ .l . ~ ~ f ~ ~ l
and
is sumable for some p satisfying 1 <
If in addition fp
llhll 2 I I $ I I , . I I f I l P be defined for real t
is sumable and Let and
$(t)
$(t+l) = $(t)
for all
for
,
2 $(t)
It1 5 1
t
.
P by
.
Then
$(t)
11$11,
Za(l-a)
-1
is finite
so
f (y)dy
A
This implies that
Ix-~I-~
is the kernel of a kernel operator mapping
L (A) into itself. For p = 2 and P operator.
1
HINT: For (a), note first that $(x-y variables, is measurable. Hence, since $
..
For (b), let
q be defined by
p-'+q-l
Holder's inequality and by part (a),
so
this is not a Carleman
2 a < 1
= 1
,
as
m ,
for
= =
<
a function of two
is periodic,
and let g E L q
. Then, by
Ch. 13,1991
KERNEL OPERATORS
277
EXERCISE 99.11. Show that if in the Luxemburg-Gribanov theorem the measure space
(Y,C,v)
is separable, then there exists a simpler proof than
the one in Theorem 99.2. HINT: Parts (i) and (ii) of the proof remain the same.
may be assumed now that Ll(Y,v) L (Y,v) 1
S
is separable,
...
is also separable. Let vj,v2,
so
In part (iii) it
the subset S
be a sequence in
with respect to the L -norm, and let X 1 be the subset of
X
is a v-measurable function of
y
I
of all x u(X-Xl)
=
such that T(x,y) 0
.
one of the v
1
x E XI , v E S
I f now
and
E >
< E
.
Y
It follows that d(x) = sup(
=
sup(
/
/
T(x,y)v(y)dv(y):vES
T(x,y)vn(y)dv(y):n=l,2,.
)
consisting
,
so
0 are given, there exists
such that T(x,y).lv(y)-vn(y)ldv(y)
of
S , dense in
=
1
.. .
CHAPTER 14
NORMED RIESZ SPACES
A Riesz space
L
then L
(L,p)
L
or
P
.
is called a normed Riesz
If
L
is norm complete,
P
is called a Banach lattice. In section 100 we compare the norm to-
P
pology in L
with the order topology and the uniform topology. We recall
P
of a Riesz space L
that, by definition, the subset D (ru-closed) if it follows from fn E D
p
equipped with a Riesz norm
space and will be denoted by
for all n
f E D
that
fn
.
+
f
is order closed
in order (relatively uniformly) and
In the normed Riesz space L
,
P
if
.
fn
+
f
f 2 g for all n , then f 2 g This n implies that the positive cone L+ is closed in the topologies mentioned. (in order, ru
or in norm) and
Furthermore, fn
-f
P
f(ru)
implies fn
+
f
(in norm) as well as
fn
+
f
(in order). There is, in general, no such implication between norm convergence and order convergence of a sequence. F o r monotone sequences the situation is nicer; if the monotone sequence fn then f then f
converges to =
f
in order (i.e., if
order, then the limits are the same, i.e., if If
L
P
fn4 and
fn
+
f
f
in norm, in norm,
) . Finally, if a sequence converges in norm as well as in
sup f
in order, then
converges to
f
=
g
.
is a Banach lattice, then L
norm convergent sequence in L
P
P
fn
+
f
in norm and
fn
+
g
is uniformly complete and any
has a subsequence that converges in order
Furthermore, every order closed set in a Banach lattice is norm closed and every norm closed set is ru-closed. In general, this does not hold in a normed Riesz space which is n o t a Banach lattice, but i t remains true for ideals. Hence, every order closed ideal in a normed Riesz space is norm closed. It will be proved a l s o that the Riesz space structure in L extended i n a naturpl manner to the norm completion makes
(L-,p)
into a Banach lattice containing L
P
(L-,p)
of
Lp
P
can be
. This
as a norm dense Riesz
subspace. In theexercisesbelonging to this section most of the results are extended to the complexification L
P
+ iL P
and to the case that
L is a
Riesz space with a topology which is not determined by one single norm, but 279
Ch. I41
CHAPTER 14
280
by a system of Riesz seminorms or, even more generally, by a system of Riesz pseudonorms. In section 101 we discuss norm completeness. It is recalled first that (un:n=1,2, . . . )
if L
such that
is a sequence of positive elements in the Riesz space
supn Zyuk
m
Z u
exists, then this supremum is denoted by
I n '
The normed Riesz space L is now said to satisfy the Riesz-Fischer conP for every sequence (un:n=1,2, ...) of podition if Emu exists in L I n
mp
sitive elements for which Z1p(un)
is finite and
L
m
the weak Riesz-Fischer condition if
EIp(un)
<
is said to satisfy
P
implies the existence of
m
.
E L+ such that un s vo for all n It is proved that the O P normed Riesz space L is a Banach lattice if and only if L satisfies
an element v
P
P
the Riesz-Fischer condition (I. Halperin-W.A.J. Luxemburg, 1956) and also
L
that
P
is a Banach lattice if and only if
L
Since any normed Riesz space L P tion LA of L exists. The norm p P
is uniformly complete and
P
satisfies the weak Riesz-Fischer condition.
is Archimedean, the Dedekind complemay be extended to a Riesz norm
P-
on LA by defining
The normed Riesz space
(L^,p^)
is now a Banach lattice if and only if
satisfies the weak Riesz-Fischer condition.
L P
In section 102 we investigate the properties of the Banach dual (Banach
.
The Banach dual of adjoint space) of a normed Riesz space L P denoted by L* , the norm in L* by p* and the order dual of P
It i s proved that
P
L;
is an ideal in L"
, not
L
will be
Lp
by
P
L".
necessarily a band. Further-
more, the space L* , with the order structure inherited from L" , is a P Dedekind complete Riesz space and p* is a Riesz norm in L* . If L saP
L
P
is a Banach lattice, then L* = L" P
ed Riesz space L if
0
P
Everymonotone sequence in the norm-
converging weakly to zero converges to zero in norm, i.e.,
P
2 u J.
.
in L
P
and 4 (u,)
+
0 for every
Similarly for downwards directed sets in L
P
4 E L*
.
In section 87 it was s h m that the order dual L
satisfies
L"
=
P
. In particular, if
tisfies the weak Riesz-Fischer condition, then L* = L"
L" m L;
=:L
m L;~
,
P
L"
,
then p(un)
J.
0
.
of the Riesz space
NORMED RIESZ SPACES
Ch. 14,51001 N
where
Lc
and
28 1
N
are the bands of all integrals and all normal integrals Ln respectively and Ls and Lsn are their disjoint complements. In the case N
N
is a normed Riesz space L , the Banach dual L* P PN N and the intersections of :L , Ln , Ls and Lsn with
that N
L
L
L*
and L*
by Lp*,c 7 LE,n 1 p , s sections are ideals in L”
We introduce
the set
p,sn
f E L
P
such that
E L*
L
P
P
. The
and bands in L* P
L,” of all
If/ 2 u
sets La
on which every
P
I$ E LE
P
are denoted
P
such that
f E Lp such that
. Similarly, Lan P
0 implies I$(u,)
C
and Lan
L*
respectively. This implies thattheseinter-
@(un) C 0 for every positive 4 E Lp* @
is an ideal in
N
If1
2
u
J-
0 implies
is the set of all
0 for every positive
C
, the largest ideals in
are ideals in L
P
is an integral. or a normal integral respective-
ly. It will be proved that La is also the largest ideal in L on which P P the norm p is a-order continuous, i.e., La consists of all f E L such
.
P
P
(fl 2 u C 0 implies p(un) C 0 Similarly, Lan is the largest n P ideal in Lp on which p is order continuous. Furthermore, La is the P inverse annihilator of L* and Lan is the inverse annihilator of L* P,S P p,sn (i.e., La consists of all f E L such that @(f) = 0 for all @ E L* * P P P,S ’ similarly for Lan ) . This implies that La and Lan are norm closed. In
that
P
P
P
the converse direction, it is of interest to know under which conditions
L*
P,S
I$ E L* satisfying + ( f )
consists exactly of all
This holds if and only if weak star topology of
L* P
L*
P
. Super order denseness of P,S
L* P
for all
f E L ~ . P
a(L* L ) , the P’ P
in L is a P consists exactly of all 4 E L* La
sufficient condition. Similarly, L* P,sn satisfying $(f) = 0 for all f E La* if and only if the weak star topology of
o
=
is closed in the topology P
P
L* is closed in p,sn Lan is sufficient. Most
P
. Order denseness of
P
of the results in this section are extended to the complexification L + i L P
P
and the case of a Riesz space with a topology that is determined by a system of Riesz seminorms.
100. Normed Riesz spaces A Riesz space
L equipped with a Riesa norm space and will be denoted by (L,p) or L P
is called a z o n e d
p
.
If
L
P
is norm complete,
NORMED RIESZ SPACES
282
is called a Banach l a t t i c e . Any normed Riesz space is Archimedean,
then L
P
because it follows from 0 < nu 5 v for all
,
n
,
p(u) = 0
so
i.e., u
We shall assume now that L
Ch. 14,§1001
L
for n
.
= 0
l,2,
=
...
that
is an Archimedean Riesz space (also when
is not assumed to carry a Riesz norm) and we shall adhere to the conver-
gence notations introduced earlier. We recall that fn and
p ( u ) 5 n- 1 p(v)
fn
+
f(ru>
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
converging in norm to
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
converging in norm to
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 the