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f o r the ele-
MORE ABOUT RIESZ SEMINORNS
402
ment
$
which t h u s corresponds t o
E (L-)'
This formula has a c o u n t e r p a r t .
exists since
x
$' Z $
. On
(note t h a t i t
x x
t h e o t h e r hand i t follows from
satisfying
J,
p
imp 1i e s
$1
0'
, we
get
<
that
i s Dedekind complete). Since p = sup(p : p S p )
L-
so
f o r any
51)
Show t h a t
Denote t h e supremum i n t h e l a s t formula by
HINT:
,
. Hence
.
+ < p > = inf(Ji:p$Zp)
p@r Z p
p E S1
Ch. 16,51081
0
= sup i(:p 5 p
$
5
( x
,
> p
Ji-
EXERCISE 108.8. Show t h a t
and t h e r e f o r e
x
px
5 $
p
, which
. s 1 onto
i s a mapping from
p + $
) . Hence ; t h e seminorms =
(L-)+
w i t h t h e following p r o p e r t i e s
+
f o r every
(i)
$
(ii)
$
( i i i ) $<suppT> = sup +p @inf ( p l
(iv) p,
in
I p2
,.. . ,pn)>
>
i f and only i f
S1
$i = sup($:p$Spi)
+
5
Similarly f o r
x 1 2 /
+
for
J, Z $ < p T >
$2 t $
hence
. For
+pT>
that
p
J,
J, t $<supp >
.
i
P$
Finally, f o r (iv), w r i t e
sI
'
and
we have
$2
respective-
, so
,P$5P2)
5
.
. i s an upper bound of t h e
$<sup p >
any o t h e r upper bound t pT
by
$
= 1,2,
= $
For ( i i i ) , i t i s e v i d e n t t h a t s e t of a l l
and
$
,
al,a2 t 0
exists i n
. .,$
$
(
\
for real
>
2 2 sup p,
if
$2 = sup $+jl:P$
sup x : p 5 p +p
,
inf ( $ < p l > , .
=
HINT: For ( i i ) , denote l y . Then
1
4 E (L-)+ > + a $
for a l l
p* = i n f ( P I , .
T
it f o l l o w s from
J,
,
..,P
so )
pJi 2 sup p
and
, and
Ch.
16,11081
EMBEDDING I N BIDUALS
We have t o show t h a t
Q
=
i s s u f f i c i e n t t o show t h a t
0'
403
holds. Since
$
. Since
$ ' 5 $
is evident, it
Q'
we g e t
Every n - t u p l e
f o r which t h i s h o l d s s a t i s f i e s
EXERCISE 108.9. T
E
{T}
,
( i ) Show t h a t i f
and it i s g i v e n t h a t
p,
I. p
p,po
E S,
and
p,
and
p,
for a l l
E S1
,
then
t h e i n t e r s e c t i o n of
S1
I po
for a l l
T
P I P o .
( i i ) Show t h a t f o r any s u b s e t t h e d i s j o i n t complement
of
Dd
HINT: For ( i ) , n o t e t h a t
D
of
S1
i s a l a t t i c e band i n
D
I. $ < p >
$
. Hence, i n L" , a l l $
and
$
L"
, so
i.e.,
p
T
(since
$ < p T > f $
EXERCISE 108.10.
$
I
$<po>
if X
and
= sup($,$)
Show a l s o t h a t i f
$(f) = f(y) in
*
I $
for a l l 0 d i s j o i n t complement o f
it contains
I po
.
$
(i) L e t L be t h e R i e s z s p a c e of a l l r e a l f u n c t i o n s
on a p o i n t s e t c o n s i s t i n g of two p o i n t s $(f) = f(x)
,
s1
and
L-
,
, so
x and
$
and
y
. For
b e l o n g t o (L")'
$
X(f) = f ( x ) + f ( y )
then
p = sup(pg,pJI)
,
any
f o r every
f E L
,
let
. Show t h a t f 6 L .
then
1 J f o r every mapping
f E L $ + pb
. Derive
that
p
<
p
X
and
p
#
px
. This
d o e s n o t p r e s e r v e a r b i t r a r y suprema.
show t h a t t h e
Ch. 16,51091
EMBEDDING I N THE ORDER BIDUAL
404
( i i ) Let
be an a r b i t r a r y Riesz space and l e t
L
and
S
same as i n t h e e a r l i e r e x e r c i s e s . Show t h a t t h e s e t equivalence c l a s s e s
belong t o t h e same equivalence c l a s s whenever and
p1
$ < p l > = $
belong t o t h e same equivalence c l a s s , then
p2
p1
and
. Show t h a t
sup(pl,p2)
p2
if
and
belong a l s o t o t h a t same c l a s s . Show t h a t every equivalence
inf(p1,p2)
c l a s s contains e x a c t l y one a d d i t i v e Riesz seminorm. and
be t h e
i s decomposed i n t o
S1 p + $
by t h e mapping
S1
,
0,
$
then t h e equivalence c l a s s of
Explicitly, i f contains
po
s1
$0 i s t h e l a r g e s t member of
no o t h e r a d d i t i v e Riesz seminorm. Furthermore, p
49 equivalence
t h a t p a r t i c u l a r equivalence c l a s s . Hence, t h e
Po
and
p
c l a s s of t h e zero
seminorm c o n s i s t s of t h e zero seminorm only. Show t h a t f o r and
E
pl,p2
, the
S,
through t h e equivalence c l a s s e s of by p a r t ( i ) ) t h a t
p;
equivalence c l a s s of If
0 s QT f $
J, = sup $T p o = sup p
pl + p
(L-)
in
, but
exists in
$T
S1
pl,p2
and
p1
E
,
S1
let
p;
p
and
and
3
run
p i
r e s p e c t i v e l y . Show (again
p2
2,
.
,
then
p
t h e system and
po
4JT
in
p4J
+
.
S1
I f , however,
i s not d i r e c t e d upwards, then
satisfies
s
po
may b e d i f f e r e n t ( c f . again p a r t ( i ) ) , show t h a t
$
+ inf(p1,p2)
does not n e c e s s a r i l y run through the whole
+ pi
(L")'
in
p 3 = sup(p,,p2)
belong t o t h e same equivalence c l a s s . Note t h a t
p 4 = p 1 + p2
may be d i f f e r e n t ( c f . p a r t ( i ) ) . For
p4
p
seminorms
p$ po
. Although and
p+
and 0 belong t o p
t h e same equivalence c l a s s .
109. Embedding of Let by of
B
. We
B
. In
f o r every
prove t h a t
every p o s i t i v e
. Hence,
if
i t follows from
$
f
$
f"
in
B
of t h e Riesz space
. Evidently,
B
f"
on
in
, so
L
-
and B
B-
. For
B#
t h e proof,
f"($) = $ ( f ) 2 0
for
is now a p o s i t i v e l i n e a r f u n c t i o n a l on
f" (f-)" B
L
i s defined on
i s a r e a l l i n e a r func-
f"
i s p o s i t i v e , then
i s an a r b i t r a r y element of f" = (f+)"
B
i s a member of t h e a l g e b r a i c dual
belongs t o t h e o r d e r dual
f
p o s i t i v e l i n e a r f u n c t i o n a l s on is that
E
o t h e r words, f "
note t h a t i f t h e given B
L"
be given. The realvalued f u n c t i o n
f"($) = $ ( f )
t i o n a l on
i n t o t h e o r d e r bidual
be an i d e a l i n t h e o r d e r dual
B
f E L
let
L
that
, i.e.,
f"
L (not n e c e s s a r i l y p o s i t i v e ) , i s t h e d i f f e r e n c e of two
f " E B"
. The next
f a c t t o observe
for a l l
f,g E L
. On
account of g = 0
formula ( 1 ) o n l y f o r that
(f')"
405
EMBEDDING IN BIDUALS
Ch. 16,81091
(f")+
=
,
(f-g)"
it i s s u f f i c i e n t t o prove
f"-g"
=
t h a t i s t o s a y , i t i s s u f f i c i e n t t o prove
. For
f E L
f o r every
0
t h i s purpose, l e t
$ E B
S
.
Then
by Theonem 83.9. Since
t h e l a s t e x p r e s s i o n i s equal t o $(f+) =
,
(f+)"($)
f o r every p o s i t i v e
in
$
B
so
d i a t e consequence of formula ( 1 )
(f")+
=
-
,
f E L
a Riesz subspace of
0 5 $=
+
0
u: L
+ B"
, defined
i s a Riesz homomorphism. A c t u a l l y , u
of a l l normal i n t e g r a l s on
Bn
let
a s d e s i r e d . I t i s an imme-
(f')"
t h a t a s i m i l a r formula i s t r u e f o r
I t has been proved t h u s t h a t t h e mapping
space
holds
, i.e.,
inf(f,g)
for a l l
$(f+) =
. This
(f")+($) = (f+)"($)
it f o l l o w s t h e r e f o r e t h a t
in
B" B
contained i n
. It
B BZ
, i.e.,
. For
maps
t h e image
into the
u(L)
of
(f")+($,)
+
is
L
f E L
t h e proof, l e t
i s s u f f i c i e n t t o show t h a t
u(f) = f"
by L
aad 0
.
This f o l l o w s by o b s e r v i n g t h a t
Before proceeding, w e s t a t e two more simple f a c t s about t h e Riesz homomor-
.
u
phism
LEMMA 109. I .
o
that
5
, then it foZZows from o , i . e . , o 5 u(uT) 4 u(u) .
( i ) ~f B c L;
uy o ul' i n
:B
( i i ) The Riesz homomorphism B
separates t h e p o i n t s of L
Q
in
PROOF. B ,
(i) Let
B c:L
u
0
u
4 u
in
L
i s a Riesz isomorphism i f and o n l y i f
, i.e., and
5
i f and o n l y i f 5
u
4 u
in
L
OB = 103
. Then,
.
f o r any p o s i t i v e
406
0
which implies
.
u" 4 u"
5
( i i ) Follows by observing t h a t i t follows from If L
Ch. 16,91093
EMBEDDING I N THE ORDER BIDUAL
$(f) = 0
for all
"
Bn
.
o(L)
f = 0
implies that
E B
$
i s a Riesz isomorphism ( i . e . ,
u
and i t s image
space of
f" = 0
.
f = 0
O B = {O}
if
) , we s h a l l i d e n t i f y
I n o t h e r words, we s h a l l regard
and we s h a l l say t h a t
i s embedded i n
L
i f and only i f
a s a Riesz sub-
L
as a Riesz sub-
BX
space. I n t h i s case, the a d d i t i o n a l condition t h a t B i s contained i n :L i s equivalent t o s e v e r a l o t h e r c o n d i t i o n s , as s t a t e d i n Theorem 106.2. One of these conditions, which we s h a l l need f o r t h e next theorem, s t a t e s t h a t i f
E L and 0 < $ E B a r e such t h a t $(u) > 0 , then t h e r e e x i s t s an v E L such t h a t 0 < v 5 u with $(v) > 0 and $(v) = 0 f o r
0 < u
element all
satisfying
J, E B
J,
I
. Furthermore,
$
a s mentioned i n the same
Theorem 106.2, i t follows from Corollary 105.12 t h a t i n t h i s case
. We
o r d e r dense i n
L: l e n t t o t h e condition t h a t THEOREM 109.2.
If
B
is
s h a l l present now two more c o n d i t i o n s , each equivaB
B
i s contained i n
i s an ideal i n L"
LZ
.
*B = {O}
such that
, then the
following conditions are equivalent. (i) (ii) Of
L
.
i s contained i n L"n The ideat D generated by B
.
( i i i ) The embedding of
L
L
:B
in
in
i s a Dedekind completion
:B
preserves arbitrary suprema and
i n f ima. PROOF. ( i )
*
( i i ) . The o r d e r dual
(Corollary 83.5). Hence, s i n c e
of
B-
B
i s an i d e a l i n
D
i s Dedekind complete
i s Dedekind complete. I n v i r t u e of the d e f i n i t i o n of ment of
D
i s majorized
by an element of
i s a Dedekind completion of t h a t f o r every
0
uo S u-
u E L
X > 0
, it
u-
S u
u- = 0
x
, which
. Hence,
t h e r e e x i s t s an element
. For
for all
D
every p o s i t i v e e l e f o r t h e proof t h a t
A > 0
uo E L
such t h a t
be given. There e x i s t s an element
a l l real
would imply
D
follows t h a t
i s s u f f i c i e n t (by Theorem 32.6) t o show
t h e r e f o r e , 0 < u- E D
satisfying
impossible t h a t for a l l
0 < u-ED
. Let,
L
L
, it
B"
A
,
let
, because
u- = 0
(since
UA
=
then BL
(u^-hu)+
u-
5
Xu
. It
is
would hold
i s Archimedean).
D
Ch. 16,01091
,
A
let
be the n u l l i d e a l of
UA
uX > 0
such t h a t
A > 0
Hence, t h e r e e x i s t s a number value of
40 7
EMBEDDING I N B I D U A L S
uX
,
. For
this particular
i.e.,
uA= and l e t of
in
UA
g r a l on B
be the c a r r i e r of
CA
B
,
B
, i.e.,
ui
.
since
A
UA
. As
CA
0
i s the d i s j o i n t complement
From u- E D c B r i t follows t h a t u^ i s a normal i n t e A A s o t h e n u l l i d e a l UA of i s a band i n B Furthermore,
has t h e p r o j e c t i o n property (since
B = U
CA
,
u^ > 0 A
i s a normal i n t e g r a l on
observed already, uX the carrier
of
CA
9 0 . 4 ) , s o t h e r e e x i s t s an element
.
i s Dedekind complete). Hence,
B
satisfies
uX
$o E
0 <
CA
CA
# {O}
such t h a t
. Hence,
B
(by Theorem
u;($~) > 0
.
I t follows then from
that
$,(u)
cluded i n that
. As
2 u*($ ) > 0
A
0
s t a t e d above, the hypothesis t h a t
L i implies now t h e e x i s t e n c e of an element
$,(v)
>
0
and
$(v) = 0
for a l l
p a r t i c u l a r , t h e r e f o r e , $(v) = 0
J,
for a l l
E UA
$-$A
,
so
The elements on
B
, so
($-$A)(v) = 0
u i = (u^-Au)+
.
(u^-Xu)-
$
in
CA
by
$A
and
(u^-Au)-
Since
$A
E
CA
u*($~) t X$A(u)
.
. Then
CA
,
a r e d i s j o i n t normal i n t e g r a l s of
UA
i s contained i n t h e
t h i s implies
(~^-AU)-($~)
which implies t h a t
Hence
. In
o r , i n o t h e r words,
by Theorem 90.5 the c a r r i e r
n u l l i d e a l of
1 $,
be an a r b i t r a r y p o s i t i v e
$
and denote t h e component of
B
J,
such
J,
For t h e f i n a l p a r t of t h e proof, l e t element of
E B satisfying E UA '
i s in-
B
0 < v s u
I n v i r t u e of (Z),
i t follows t h a t
=
0
,
408
This shows t h a t t h e element element
uo = Xv
=)
( i i i ) . By assumption, the i d e a l
a Dedekind completion of
L
. If
the equality (iii)
0 L
0E
5
in
. The
D
D
in
L
holds i n
is
:B
,
L
then t h e
i s an i d e a l i n
,
:B
as well. S i m i l a r l y f o r i n f i m a .
B;
( i ) . In t h i s p a r t of the proof we denote t h e image of any
u" f 0
. This
. Since
D
holds now i n
under t h e embedding i n
implies for
*
f = sup f
generated by
D
f = sup(f :oE{o})
same holds i n the Dedekind completion
f E L
0 < uo I u^
satisfies
h a s , t h e r e f o r e , t h e required p r o p e r t i e s .
uo
(ii)
on
Ch. 16,51091
EMBEDDING I N THE ORDER BIDUAL
B
BI;
, i . e . , ~ " ( 0 )f , we have $(uT)
shows t h a t
again by
0
$ 2 0
f 0
= u:($)
assumption, u in
, i.e., 0
i s contained i n
B
. By
f"
f o r every
C 0
. Hence,
i s a normal i n t e g r a l
.
LI;
B
We n o t e , f i n a l l y , t h a t i s i s not d i f f i c u l t t o give a d i r e c t proof t h a t ( i ) implies ( i i i ) ; c f . Lemma 1 0 9 . 1 ( i ) . The following theorem i s now an immediate consequence, I t deals with the case t h a t
B = L" n
THEOREM 109.3. D
the ideal and
D
.
('L:)
If L i s a R5esz space s a t i s f y i n g L
generated by
in
. Furthermore,
(LZ);
i s order dense i n
t h e embedding of
(Li)I; preserves arbitrary suprema and infima. I t follows t h a t (L;);
ideal i n
, then
= {O}
i s a Dedekind completion o f
(L;);
under the embedding i f and o n l y i f
L
L L
L
in
i s an
i s Dedekind
complete. EXAMPLE 109.4. We mention one example; f u r t h e r examples w i l l follow
i n t h e next s e c t i o n . Let
L
be the space ( c ) of a l l ( r e a l ) convergent
sequences. A l i t t l e more i n d e t a i l , we may say t h a t (re al ) functions
f = (f(l),f(Z),
converges a s
m
n
+
. We w r i t e
uniform norm, then e,(X,)
, where
e x i s t s an element
for a l l
f
E L
on
X = (l,Z,
. If
such t h a t
becomes a Banach l a t t i c e , s o the o r d e r dual XI = { O } U X
. Precisely,
(a(O),a(l),a(Z),...)
. Any
...)
i s t h e space of a l l
$
with
a(0) = 0
in
f(n)
i s equipped with t h e
L
L* a r e i d e n t i c a l . A s well-known,
t h e Banach dual with
L
...)
L = c(X)
L
L*
f o r any
L,({O}UX)
L"
and
may be i d e n t i f i e d $
E L*
there
such t h a t
i s obviously a normal i n t e g r a l on
Ch. 16,51101
. The
409
EMBEDDING I N BIDUALS
E L has the f i r s t n n o r d i n a t e s z e r o and t h e remaining c o o r d i n a t e s one, then f n + 0 , so
L
$,(fn)
+
converse h o l d s a s w e l l (because i f $ E L :
for
0
, which
= L :
from which i t f o l l o w s t h a t L = c(X)
in
L (:):
=
(L;):
f
implies
a(0)
. The
= La(X)
lae_CX) i s t h e space
in
=
sa(X)
L :():
L , (X) ,
itself. (c,)
of a l l sequences
(:): = L-(X) L = ( c ) = co(X) , t h e n L 0 i s now t h e s p a c e L i t s e l f .
converging t o z e r o , i . e . , L
L :
i d e a l g e n e r a t e d by
Note a l r e a d y t h a t i f we s t a r t w i t h t h e space generated by
0 ) . Hence
=
co-
; the ideal
110. P e r f e c t Riesz s p a c e s
I n t h e preceding s e c t i o n we have i n t r o d u c e d t h e Riesz homomorphism
, where
L + (L-)"-n a l l 4 E Ln
0:
u(f)
. Identifying
f E L
i s defined f o r L
w i t h i t s image
Example 109.4, i t can occur t h a t
i s t h e whole space
u(f)($) = $(f)
,
t h e space
for
is
L
a s a Riesz subspace. The Riesz homomor-
embedded i n t h i s manner i n:):L(
0
phism i s a Riesz isomorphism i f and only i f u(L)
by
o(L)
(L;):
L ():
. As
= {O}
seen i n
i s a Riesz isomorphism and t h e image
u
. In
o t h e r words, we may w r i t e
L = (Lz):
i n t h i s c a s e , where t h e s i g n of e q u a l i t y denotes t h a t t h e homomorphism
i s an isomorphism, Any Riesz space s a t i s f y i n g
L = (L-)n n
is called a
p e r f e c t R i e s z space. The name was i n t r o d u c e d o r i g i n a l l y f o r t h e c a s e t h a t i s an i d e a l i n t h e sequence space
L =
(fl,fg,
...)
such t h a t t h e c a r r i e r of
numbers. I n t h i s c a s e :L Clfngnl f
E L
,
then
L
z/fngnl
PROOF. Let f i r s t
L
wards d i r e c t e d system i n
sup $ ( u ) <
5
P($) <
m
such t h a t
p ( $ ) = sup $(u,)
and
-
sup $(uT) <
f o r every
P ( $ ~ + $ ~=) p ( $ , )
such t h a t
+
g E L" n
f o r every
be p e r f e c t and assume t h a t L+
...)
only i f
is p e r f e c t if and only if
L
.
put
g = (gl,g2,
converges f o r a l l
= { o } a n d it folZows from 0 5 u t and E :L t h a t sup uT e x i s t s i n L
. We
i s t h e s e t of a l l n a t u r a l
is perfect.
THEOREM 110.1. The Riesz space
E :L
L
f =
( c f . Theorem 8 6 . 3 ) . I t i s e v i d e n t t h a t
f E L
. If
(L):
of a l l r e a l sequences
i s t h e space of a l l
converges f o r every
i s included i n
L
(s)
P($,)
0
5
for
(L ' ):
0 5 $ E
i s an up-
(u~:T€{T}) m
$ E L-
0 E
f o r every
. It
9, ,
=
0 5 $ €
is evident that N
$2 E L,
. Hence,
410
PERFECT R I E S Z SPACES
extending
p
i n the obvious manner t o the whole of
f u n c t i o n a l , we f i n d t h a t
. Furthermore,
E (LL)-
p
i t follows t h a t element
. Hence,
p E :L)(:
E L such t h a t
u
0 5 $ E L i
.
u = sup u
p
This shows t h a t
holds i n
,
= u
0 5 u 4 and in
L
. The
s u p $(u,)
<
since i.e.,
since
0
in
S $A 4 $
in
1 u^
0 < $ E
every
follows t h a t i.e.,
u1 E L
0 5 $
5 , so
u
f o r every
4 $(uo)
$(u,)
with
L :():
f o r every
u
E L-n
Since a l s o 0 5 $ E L;
L
i s a band i n
. Note
E L*
f E L
{' (L"):}
and
f^($) = 0
= {O}
f^($) = 0
0 5 $ E L"
now t h a t i f
then i n p a r t i c u l a r elements
$
and
under t h e embedding of
$(uT)
. For
$ = 0
as desired.
Next, l e t
0 < $,
4
"
. Then,
easily that shows t h a t
in
L"
and
N
(L ) n
sup u^($,)
i n p a r t i c u l a r , sup $,(u)
p(u) = sup $ ( u ) p = sup $
<
m
LN
L"
.
it
,
is p e r f e c t .
t h i s purpose, we have f^
, then
E (L-):
for a l l
f-
<
, as
,
E (L-):
explained i n the
$(f) = 0
m
f o r any
f o r any u
E L+
for a l l
0
. It
i s a p o s i t i v e l i n e a r f u n c t i o n a l on
holds i n
,
u^ = uo
t h a t a r e images of N
in
L
is
for
4 u^($)
. But then (L;); .
fa($) = 0 f1
. It .
sup $ ( u T ) <
for a l l
f o r a l l those
E L , and s o
U
(L-)Hence, l e t n n In o t h e r words, $(uT) 4
beginning paragraph of s e c t i o n 109. It follows t h a t
E (L ln
N
(Ln)*
t h e order dua2
L
exists
by hypothesis. Then
L
.
shows t h a t
0
.
T
f o r every
T-
PROOF. We f i r s t prove t h a t
t o show t h a t i f
sup u
i s Dedekind complete.
L
implies t h a t exists i n
$ E Ln
5
u ^ ( $ ) = $(u,)
. This
E L for a l l
THEOREM 1 10.2. For any Riesz space
$ = 0
that
i s a band i n
L
. This
= sup u
O
and t h e r e f o r e
and t h a t i t follows from
0 I $ E L l
i s an o r d e r dense i d e a l i n
s u f f i c i e n t , t h e r e f o r e , t o prove t h a t 2 u
,
(Ln)n
l a s t condition implies immediately t h a t
Hence, by Theorem 109.3, L
1. u - ( $ )
holds i n
(LL) = { O }
f o r every
m
f o r every
A. ,-
is perfect.
L 0
i s p e r f e c t , t h e r e e x i s t s an
L
$ ( u ) = sup $ ( u , )
u = sup u
because
L
Assume now, conversely, t h a t
f
as a linear
L :
implies
:L
0
16,51101
Ch.
u^
E
follows L
, which
Hence, t h e p e r f e c t n e s s conditions mentioned i n the l a s t theorem hold
Ch. 16,11101
in
. It
L
EMBEDDING I N BIDUALS
follows t h a t
411
i s perfect.
L"
THEOREM 110.3. I f the Riesz space
L
is p e r f e c t . In p a r t i c u l a r , f o r any Riesz space L t h e space i s a band i n t h e p e r f e c t space L" ) . f e e t ( s i n c e :L A
PROOF. Let
be a band i n the p e r f e c t Riesz space L = A
i s Dedekind complete (by t h e p e r f e c t n e s s ) , s o o f any normal i n t e g r a l on
. To
A
If
j
show now t h a t
'(A;)
and a l s o
I E
E A
If
0 S $
+(if\) = 0
f o r any
0 2 $
. Then
E A"
0 5 $
n
sup $(uT) <
-
0
0 5 u
exists i n
u = sup u
let
f
E A
(A ' ):
in
4
A
in
4
that
satisfy
f E
0
i s an i d e a l i n
and
-
sup $(uT) <
.
Hence, s i n c e n Evidently, u E A (since
.
L
restriction
(A")
?!
A
. Then . Hence .
f o r every
and, once again by the remark above,
L
E L"
0 5 $
L
. The
Ad
.
u
5
f o r every
. Note
E A,". Then, i n view of the remark above, = {O} E L i , and s o I f ( = 0 because ('L:)
= {O}
Furthermore, l e t
@
L
i s per-
L :
i s , t h e r e f o r e , a normal i n t e g r a l on
, since
(A ' ):
f o r any (A ' ):
,
= {O}
$(lfI) = 0
This shows t h a t
A
to
L
L
i s p e r f e c t , then any band i n
L
is p e r f e c t ,
A
i s a band). The
conditions of Theorem 110.2 a r e s a t i s f i e d t h e r e f o r e . It follows t h a t
A
is
perfect. The next theorem goes i n t h e converse d i r e c t i o n . THEOREM 110.4. I f
A
such t h a t
PROOF. Let 0
5 $
If1
=
0
Q E L"
5
u' + u"
with
o
, and
5 $
in
A
u
@
O(L;)
Q ( I ~ =I )o
, so
f o r every
Ad
=
. Furthermore, T
, so If
0 5 u u' + 'u T
t h a t every p o s i t i v e normal i n t e g r a l $ on A ( s e t $ = 0 on Ad ), it follows t h a t $(u') = 0
L
E :A
hence
Now, l e t we have
I ~ EI
by the Dedekind completeness of L , we have d 0 5 u' E A and 0 5 u" E A , s o Q(u') = 0 f o r every
L = A
can be extended t o = 0
. Then
f E '(L;)
L
is p e r f e c t .
. Observing now
f o r every IJ"
A i s a band i n
i s Dedekind complete and
are p e r f e c t , then L
. Since
E :L
L
a d
I
= u'
+ u"
in
L
with
0
4 T
u' =
sup $(u:)
o
on account of =
and 5
<
-
0
. This
sup Q.:u ) <
u' E A
-
'(A;)
shows t h a t
0
5 $I
E A :
. Similarly .
= {O}
f o r every
andT 0 2 u" E Ad
f o r every
= {o}
'(L;)
0 5 Q E L"
. Hence ,
0 5 'u
. 4
s i n c e every posi-
t i v e n o r m a l i n t e g r a l o n A can be extended t o a p o s i t i v e n o r m a l i n t e g r a l on Land
Ch. 16,§1101
PERFECT R I E S Z SPACES
412
s i n c e 0 5 u ' 5 u . It follows, by the p e r f e c t n e s s of T T exists in A S i m i l a r l y , u" = sup u" e x i s t s i n Ad
.
= sup u
holds i n
. The
L
space
A ,that
. Hence,
u ' = sup u'
'
u' + u" =
s a t i s f i e s , therefore, the perfectness
L
conditions of Theorem 110.2. EXAMPLE 110.5. ( i ) I f
L = Lm and
A s an example, l e t
,
= L
A
i s p e r f e c t and
L
i s an i d e a l i n
A
i s not n e c e s s a r i l y p e r f e c t , even i n the case t h a t
A
so
. Then
A = (c,)
i s p e r f e c t . On t h e o t h e r hand, A:
L
=
:L
=
b,
L
,
then
i s order dense.
A
L,
and
and
(A:)
N
N
(Ln)n =
Lm
1- , so
=
i s not p e r f e c t . ( i i ) If
i s an i d e a l i n the p e r f e c t Riesz space
A
i s p e r f e c t i n i t s own r i g h t , then
L = 1-
an example, l e t
and
i s not n e c e s s a r i l y a band i n
A
f i n i t e number of nonzero coordinates. Then A
L
A
. As
t h e i d e a l of a l l sequences with only a
A
( i ) ) and f o r the proof t h a t
such t h a t
L
i s p e r f e c t (as seen i n p a r t
L
i s p e r f e c t , note t h a t
A" = ( s )
,
the space
of a l l r e a l sequences. ( i i i ) Let
be a Riesz space such t h a t
L
and
4 (n=l,2,...)
0 S u
the e x i s t e n c e of
sup u
words, t h e d i r e c t e d s e t
n
sup $ ( u n ) <
. Then 0 S u
L
(L ' ):
on
f
E L i implies
L
.
0 S u 4 A s an example, l e t X n the Riesz space of a l l bounded r e a l
which vanish o u t s i d e a countable s u b s e t of
X
f ) . Then
subset varying with
L
X (this
i s Dedekind complete (even super Dedekind
-
complete) and i t i s not d i f f i c u l t t o v e r i f y t h a t
. Also,
5 $
i s not n e c e s s a r i l y p e r f e c t . I n o t h e r
110.2 cannot be replaced by a sequence functions
0
i n t h e p e r f e c t n e s s condition of Theorem
4
be an uncountable p o i n t s e t and
and such t h a t
= {O)
f o r every
m
L"
n
=
1 , (X) , s o '(L;)
=
N
sup $ ( u ) < f o r every 0 5 $ E Ln = n n = L,(X) , then the sequence (u :n=l,2, ) i s bounded i n t h e uniform norm n (apply t h e Banach-Steinhaus theorem t o the l i n e a r f u n c t i o n a l s un = {O)
if
0 S u
(X) 1:
-
f o r every u
2 u
i?,(X)
) . It follows t h a t
i s t h e space of a l l bounded functions on
completion of that
...
sup u exists. n (Ln)n = "
i s not p e r f e c t , a s follows by observing t h a t
EXERCISE 110.6. <
and
on t h e Banach space
(n=l,2,...)
Nevertheless, L =
4
L
0
Let
'(L;)
=
{Ol
. Show
that
i f and only i f i t follows from 2
$
for a l l
E -L: T
X
an example.
i s a Dedekind
L :():
0 5 u
t h a t t h e r e e x i s t s an element
. Give
.
N
and
4
u
in
sup $(uT) L
such
Ch. 16,51111
HINT: Assume t h a t
with
0 5 u
4
(L:):~
and sup u
=
-
sup $(uT) <
exists i n
L (:)
,
L
for a l l
uT 5 u
N
5
$ E Ln N
L (:)
by Theorem 110.1. Since
0 < u
I.
and
. Choose an a r b i t r a r y in
L
(Lz):
f o r every
0 5 $ E ; L
such t h a t
u 2 v
u^ 5 u
. This
ideal
D
v
in
in
u E L
(L:):
.
-
f o r every
such t h a t
u
Since t h e i d e a l
5 u
D
by hypothesis, t h e r e e x i s t s a n elemenb
u
u^ = sup(v:u->vEL)
satisfying
u- t v E L u- E L (:) N
-
D = (Ln)n
, which
,
u- 5 u
D
have
. Hence completion of L . L
4
(the i a s t
and s i n c e
shows t h a t every p o s i t i v e
generated by
i s a Dedekind
. Hence,
for a l l
0 5 u
i s a Dede-
(L;):
i s a Dedekind completion of
, we
i s order dense i n:):L(
u^ 2 0
and l e t
such t h a t
u E L
sup $(uT) <
implies t h e e x i s t e n c e of an element
generated by
. Then-
N
L
i s a l s o p e r f e c t , so
.
T
N
0 5 $ E {(Ln)n}n = Ln
t h e r e e x i s t s an element
Conversely, assume t h a t 0 5 $ E L" n for a l l T
0
f o r every f o r every
i s p e r f e c t ) . But
L :
kind completion of and s o
i s a Dedekind completion of
L (:):
sup $(uT) <
e q u a l i t y because U-
413
EMBEDDING I N BIDUALS
L
, so
. It
E L
follows t h a t
i s contained i n t h e implies t h a t
1 1 1 . P e r f e c t Banach l a t t i c e s
The property of a Riesz space t o be p e r f e c t i s obviously s i m i l a r t o t h e property of a Banach space t o be r e f l e x i v e . A s we s h a l l s e e ( i n Theorem 1 1 4 . 9 ) , any r e f l e x i v e Banach l a t t i c e i s p e r f e c t . The converse does not hold; the space
La
i s p e r f e c t , b u t not r e f l e x i v e . Another p o i n t t o observe i s
the s i m i l a r i t y between the weak Fatou property f o r d i r e c t e d s e t s ( i n a normed
-
f i e s z space) and t h e p e r f e c t n e s s condition r e q u i r i n g t h a t i t follows from 0
5
u
t
and
sup $(uT) <
f o r every
0 < $ E L i
that
sup uT
exists.
It i s easy t o s e e t h a t i n a p e r f e c t Banach l a t t i c e t h e norm possesses t h e
weak Fatou property f o r d i r e c t e d sets. The converse is a l s o t r u e , b u t t h i s
i s more d i f f i c u l t t o prove. The d e t a i l s a r e contained i n t h e following theorem, which i s e s s e n t i a l l y due t o T. Mori, I. Amemiya and H. Nakano ([11,1955); 107.
t h i s i s t h e theorem r e f e r r e d t o i n t h e l a s t paragraph of s e c t i o n
PERFECT BANACH LATTICES
414
THEOREM 1 1 1 . 1 .
The normed R i i s z space
Banach Zattice i f and onZy i f
('L:)
t h e existence of
sup u
in
PROOF. Assume f i r s t t h a t
4
( w i t h norm
L
and
= {O}
property f o r directed s e t s ( i . e . , 0 < u
sup $(u,)
<
n e s s of
L
m
,
that
m
implies
i s a p e r f e c t Banach l a a t t i c e . I t i s i n -
L
if
0 5 u
0 < $ E L* = :L
n exists i n
sup u
has t h e weak Fatou
p
L i.
. Hence,
L" = L* n n f o r every
i s a perfect
p
sup $ ( u ) <
and
(L ' :)
cluded i n t h e d e f i n i t i o n of p e r f e c t n e s s t h a t Banach, we have
Ch, 16,51113
L
4
= {O}
with
. This
. Since
sup p(u ) <
is
L
, then
m
i m p l i e s , by t h e p e r f e c t -
. The norm
p
has, therefore,
t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s . Convereely, assume t h a t
0
(L):
and
= {O}
has t h e weak Fatou
p
p r o p e r t y f o r d i r e c t e d s e t s . A s shown i n Theorem 107.5, L complete Banach l a t t i c e . I t f o l l o w s t h a t that
N
L
(since
= L*
.
i s now a Dedekind i s Banach) and
L
= (L*)* I t remains t o show L o i s embedded as an i d e a l i n (L")" n n n n t h a t L = (L*)* For t h i s purpose we r e c a l l formula ( 2 ) i n t h e l a s t n n paragraph of s e c t i o n 107 f o r t h e norm p " i n L According t o t h i s formula
.
.
we have
f o r every norm
p
u
. For
E L+ , and by Theorem 107.7 t h e norm
-
t h e proof t h a t
pl'
i s equivalent t o the
i s p e r f e c t , assume now t h a t
L
0
.
5
u
in
I.
0 < $ E L* = L" Each u i s a bounded n n and su~Iu~($)I= supI$(uT)I < l i n e a r f u n c t i o n a l on t h e Banach s p a c e L;
L
and
sup $(u ) <
f o r every
$
f o r every
, so
E L:
(~"(U~):T€{T})
Banach-Steinhaus theorem. But t h e n p"
a r e e q u i v a l e n t , and hence
i s a bounded
i s bounded s i n c e
(P(U~):T€{T))
sup u
COROLLARY 1 1 1.2.
L*
For any rwrmed Riesz space
L*
has
p
is perfect.
(with mrm p ), the
i s perfect.
PROOF. Note f i r s t t h a t i f p e r f e c t , so
L
L
and
p
e x i s t s by t h e h y p o t h e s i s t h a t
t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s . This shows t h a t
Banach dual
L
i s Banach, t h e n
L* = L"
and
is
L"
i s p e r f e c t . What we have t o prove, t h e r e f o r e , i s t h a t
is perfect also i f
L
be i d e n t i f i e d with
(L-)*
i s n o t Banach. One method i s t o show t h a t
, where
L-
-
s e t of numbers by t h e
i s t h e norm completion of
L* L
L*
may
.A
Ch. 16,51121
415
EMBEDDING I N BIDUALS
second method i s t o apply t h e p e r f e c t n e s s theorem which we proved a moment ago. Note f i r s t t h a t every $(f) = 0
that on
0
2
$
a c t s a s a normal i n t e g r a l on
E L
E L*
implies
,
L*
f
i.e.,
=
. Hence,
0
O{(L*),}
and
L*
t h e normal i n t e g r a l s
. For
= {O}
t h e proof
has t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s , assume t h a t
L*
in
.f
such t h a t
L*
. Writing
u E L+
f o r e:ery that
f
$
s e p a r a t e t h e p o i n t s of
L*
that
for a l l
a
sup p * ( $ , )
=
$ ( u ) = sup $,(u)
i s p o s i t i v e and a d d i t i v e on
$
1 1 1 . 1 , L*
$
, we
have
. Then
m
L
in
,
a.p(u)
it i s e v i d e n t
e x i s t s , t h e r e f o r e , a norm
which extends
$ = sup $,
sup $,(u)
u E L+
for
. There
L+
bounded p o s i t i v e l i n e a r f u n c t i o n a l on f u n c t i o n a l a g a i n by
<
L*
. Denoting
$
. Hence,
this
by Theorem
is perfect.
EXERCISE 1 1 1 . 3 . Show t h a t f o r a normed Riesz space w i t h o r d e r continuo u s norm t h e r e s u l t in Theorem 1 1 1 . 1 may be d e r i v e d e a s i l y from Theorem
110.1.
112. Banach f u n c t i o n s p a c e s I n t h e p r e s e n t s e c t i o n we assume t h a t
i s a u - f i n i t e measure
(X,A,y)
space. As on e a r l i e r o c c a s i o n s , t h e Riesz space of a l l ( r e a l ) w m e a s u r a b l e f u n c t i o n s on denoted by
X
( w i t h i d e n t i f i c a t i o n of u-almost
M(X,u)
. Let
L
b e an i d e a l i n
o r d e r s e p a r a b l e , every i n t e g r a l on u-order
proved i n Theorem 86.3 t h a t i f
L
on
X
is
M(X,p)
every
i s o r d e r continuous. I t was
i s an i n t e g r a l on
$
g
. Since
M(X,p)
i s a normal i n t e g r a l , i . e . ,
L
continuous l i n e a r f u n c t i o n a l on
a umeasurable r e a l function
equal functions) w i l l be
L
,
then there e x i s t s
such t h a t
A
f E L
holds f o r a l l
. Without
l o s s of g e n e r a l i t y i t may b e assumed t h a t
,
and t h e n
g
i s U-almost everywhere
uniquely determined. Hence, i f t h e c a r r i e r of
L
is the s e t
v a n i s h e s o u t s i d e t h e c a r r i e r of
L z (=L-> tions all
L
X
g
i t s e l f , then
may be i d e n t i f i e d w i t h t h e space of a l l ( r e a l ) p-measurable funcg
f E L
on
X
, and
having t h e p r o p e r t y t h a t if
g
and
gf
i s u-smmable over
X
for
6 L z correspond i n t h i s manner, t h e n t h e y
determine e a c h o t h e r uniquely. I t may happen t h a t
L :
c o n s i s t s of t h e n u l l
416
BANACH FUNCTION SPACES
Ch. 16,11123
u
f u n c t i o n a l only. In s e c t i o n 85 i t was shown, f o r example, t h a t i f
no atoms and
i s e i t h e r t h e whole space
L
M(X,u)
has
o r one of t h e spaces
.
f o r 0 < p < 1 , then L" = I03 , and t h e r e f o r e L : = TO} We L (X,u) P s h a l l prove now t h a t t h i s does not happen i f L i s equipped with a norm. P r e c i s e l y , we prove t h a t i f
is
L
YL;)
=
, then
i s a norm i n t h e i d e a l
p
and t h e c a r r i e r
L
L* i s X , i . e . , '(L:) n w i l l follow then immediately t h a t t h e c a r r i e r of L : is
of
X
{03
t h e c a r r i e r of
. It
= {O}
x ,
.
i.e.,
F i r s t we make some remarks. A s observed i n Theorem 110.3, t h e space
L;
is perfect. Also, since
space
L,* i s a band i n t h e p e r f e c t space L* , t h e
i s p e r f e c t . Furthermore, note t h a t whereas
LE
( r e a l and measurable) f u n c t i o n s
, the
f E L
g
possibly smaller space
on
:L
where, a s evident from t h e formula, p * g 5 L i
and any
f
with
p(fl) 5 1
and
number
t h e number
a
sgna = 1
gfldu =
X
such t h a t
c o n s i s t s of a l l
L;
I
IgfIdv < g
c o n s i s t s of a l l
E
5-
denotes t h e norm i n
for a l l
m
such t h a t
. For
L*
p(f) 5 1
t h e function
f
(we r e c a l l t h a t f o r any r e a l o r complex
Igf ldu
i s defined by
sgn a
f l = Ifl/sgn
s g n a = a/lal i f
satisfies
g
a # 0
and
a = 0 ). It follows t h a t we may j u s t as well say t h a t
if
c o n s i s t s of a l l
g E L" n
any
L:
such t h a t
sup(1 l g f l d u : p ( f ) < l and t h i s supremum i s then e x a c t l y the norm THEOREM 1 1 2 . 1 .
on
X
.
Let X
ideal with carrier
L
PROOF. Let
X,
of
g
, as
be a normed Riesz space such t h a t M(X,u)
i n t h e space
Then the c a r r i e r of
p*(g)
L,*
is X
, and hence ):L(' X
-
= '(L:)
such t h a t
i s t h e c a r r i e r of
X
L
i s an
of a l l measurable functions
be a measurable subset of
but otherwise a r b i t r a r y . Since
defined above.
L
,
= TO}
p(X,)
> 0
. ,
it follows from
t h e d e f i n i t i o n of t h e n o t i o n of a c a r r i e r ( s e c t i o n 86) t h a t t h e r e e x i s t s of
such t h a t
0 < p(E) <
a subset
E
function
xE
Hence, A
c o n s i s t s of a l l measurable
of
X, E
belongs t o
L
. Let
A
and t h e c h a r a c t e r i s t i c be t h e i d e a l generated by
and (almost everywhere) bounded
xE.
Ch. 16, §1121
EMBEDDING I N BIDUALS
. It
functions vanishing outside
E
M(X,u)
The i d e a l
as well as i n
.
L
i s obvious t h a t
xE
a normed Riesz space. Evidently
417
,
A
, and
E A:
n a l . Let
g(x)
#
A:
, non-negative
E
$
. Since
{O}
(by Theorem 107.2) i t f o l l o w s t h a t
:A
exists a positive linear functional
, is
p
t h e r e f o r e t h e corresponding
linear functional i s not the n u l l functional, so o r d e r dense i n
i s an ideal i n
A
equipped w i t h t h e norm
#
A:
{O)
A* i s n so there
,
which i s n o t t h e n u l l f u n c t i o -
A:
and n o t i d e n t i c a l l y z e r o on
, be
E
the corre-
sponding f u n c t i o n , i . e . ,
for a l l 5
f E A
II$l\.p(f)
f
0 5 f E L
Let now f
. Denoting
for a l l
E A f o r a l l n , so
for a l l
n
.
Setting
t h e norm of
E A
. There
by
$
.
1 1 $ 1 1 , we 0 5 f
e x i s t s a sequence
g(x) = 0
outside
E
, we
have
+
l+(f)l 5 fXE
such t h a t
find that
hence
1
.
gfdu 5 l l $ l l . p ( f )
X
This shows f i r s t of a l l t h a t Hence, w r i t i n g
0
$(f) =
i s u - s u m a b l e over
gf
gfdu
for
i s r e p r e s e n t e d by t h e f u n c t i o n
c a l l y z e r o on t h e s u b s e t every s u b s e t
XI
of
X
E
g
XI
x. A s observed above, 'L* n
X,
f o r every 0
#
$
f
.
E L and
E :L
which, a s s e e n above, i s n o t i d e n t i -
. We
. This
X
E L , we f i n d t h a t
have proved t h e r e f o r e t h a t f o r
there e x i s t s a function
i s n o t i d e n t i c a l l y z e r o on
such t h a t
of
f
g
in
L*
such t h a t
i m p l i e s t h a t t h e c a r r i e r of
c o n s i s t s of a l l measurable f u n c t i o n s
L*
g
on
g
is
X
i s t h e norm i n t h e Banach d u a l
where p *
sometimes c a l l e d t h e a s s o c i a t e to by
.
p'
Since
,
L*
space
space of
L'
=
L
and t h e r e s t r i c t i o n of
P
, we
. For
p
is
f u n c t i o n s p a c e :L
p*
s i m p l i c i t y of n o t a t i o n ,
and t h e c o r r e s p o n d i n g a s s o c i a t e norm
i s a band (and t h e r e f o r e norm c l o s e d ) i n t h e Banach
L;
t h e space
L'
i s a Banach l a t t i c e w i t h r e s p e c t t o
L'
more, as proved, t h e c a r r i e r of
is
L'
X
. In
p'
.Further-
view of t h e d e f i n i t i o n of
have
f E L
for a l l
and
a s s o c i a t e s p a c e of =
. The
L*
i s then c a l l e d t h e a s s o c i a t e norm of
L*
we d e n o t e t h e a s s o c i a t e s p a c e by
p'
Ch. 1 6 , § 1 1 2 1
BANACH FUNCTION SPACES
418
L'
. Note
= (L')-
(L');
n
. Repeating
g E L'
, we
o b t a i n t h e second a s s o c i a t e s p a c e
h
Note t h a t , f o r
X
on
L"
= (L')'
t h a t t h e l a s t e q u a l i t y f o l l o w s from t h e f a c t t h a t
i s Banach. The c a r r i e r of functions
t h e p r o c e d u r e , i . e . , making t h e
is
L"
X
and
L"
=
L'
c o n s i s t s of a l l measurable
such t h a t
,
0 I h E L"
t h i s may b e w r i t t e n as
hgdp:g>O,p'(g)
=
1 I '
Compare t h i s formula a l r e a d y w i t h formula ( 2 ) i n s e c t i o n 107. If = L"
f
E L , then
I lfgldy
by t h e d e f i n t i o n of
Hence, L
is included i n
look a t t h e embedding of i n Theorem 109.2. in that
N
L
B
such t h a t
Writing
L" L
and into
B = L'
OB = { O }
i s contained i n
<
Lz
m
g E L'
for all
. Furthermore,
L"
,
p" 5 p
L"
on
L
.
, and
so
f
E (L'):
=
It i s i n s t r u c t i v e t o
from t h e p o i n t of view of t h e r e s u l t s
f o r a moment, we s e e t h a t
B
is
X
b e c a u s e t h e carrier of
. Furthermore,
B
L'
i s an i d e a l
,
and such
i s Dedekind complete. Hence,
Ch. 16,§1121
419
EMBEDDING I N BIDUALS
by Theorem 109.2, L
i s an order dense i d e a l i n
(L')-
. This
= L"
i s quite
i n agreement with what we have found now i n a d i f f e r e n t manner. Let us ment i o n some examples. I f L' =
L,
and
.em .
L" =
L
i s t h e space
If
L
f i n i t e l y many nonzero terms, then L' = L,:
L,
=
and
.
Lm
L" =
(c,)
with t h e uniform norm, then
i s t h e space of a l l sequences with only N
Ln
i s t h e space of a l l sequences,
The next question t o consider i s under which conditions with
. If
L"
i s a Banach l a t t i c e with r e s p e c t t o
L
coincides
L
p (i-e.,
if
is
L
a Banach f u n c t i o n s p a c e ) , the answer can be derived immediately from Theorem 1 1 1 . 1 . =
.
L (:):
only i f
I n t h i s case
It follows t h a t
N
N
L ' = L* = L and t h e r e f o r e L" = (L ) * = n n n n L and L" contain t h e same f u n c t i o n s i f and
i s p e r f e c t . Hence, by Theorem 1 1 1 . 1 ,
L
under which conditions
L
and
t h e answer t o t h e question
coincide i s contained i n t h e f i r s t p a r t
L"
of t h e following theorem. THEOREM 112.2.
only i f
L
(i) I f
, then
L
the carrier of
L
L"
is
contain the same functions i f and
i s perfect or, equivalently, i f and o n l y i f
L
X
i s a Banach function space such that
and
Fatou property f o r directed s e t s ( i . e . , 0
5
f
has t h e weak
p
sup p ( f T )
i n L with
4
implies t h a t the function sup f T' which e x i s t s i n M ( X , u ) , i s an element of L 1 . I n t h i s ease the norms p and p" i n L = L " are <
m
P
equivalent. ( i i ) If L"
L
i s a normed Riesz space with carrier
contain the same functions and such that
norms i n
L
PROOF.
, then
L
and
X
, so
('L:)
=
L
{O}
has c a r r i e r
given already t h a t p
L
and
are equivalent
p"
('L:)
=
and
, L
L"
the space
,
the space
L
:L
i s a Banach
c o n s i s t of the i f it is
i s p e r f e c t . By Theorem 1 1 1 . 1 ,
L
{O}
X
. Furthermore,
l a t t i c e by hypothesis. Hence,as observed above, L same f u n c t i o n s i f and only i f i f and only i f
such that
i s perfect.
( i ) Note f i r s t t h a t s i n c e
has likewise c a r r i e r
p
X
P
i s a p e r f e c t Banach l a t t i c e
has t h e weak Fatou property f o r d i r e c t e d s e t s . Observe
now t h a t i n view of formula ( 2 ) near t h e end
of s e c t i o n 107 t h e norm
p"
i s e x a c t l y t h e one occurring i n Theorem 107.7. Since the hypotheses of Theorem 107.7 a r e s a t i s f i e d i f
p
has t h e weak Fatou property f o r d i r e c t -
ed s e t s ( t h i s implies Dedekind completeness), i t follows t h a t a re equivalent i n
L = L"
.
p
and
p"
420
BANACH FUNCTION SPACES
( i i ) Since
and
p
p"
Banach ( w i t h r e s p e c t t o
are e q u i v a l e n t i n
Ch. 16,§1121
L = L"
L" i s
and s i n c e
p" ), i t f o l l o w s immediately t h a t
L
i s Banach
(with r e s p e c t t o p ) . Hence, we a r e back i n t h e s i t u a t i o n of p a r t ( i ) .
I n Theorem 107.5 i t w a s proved t h a t i f t h e normed Riesz space t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s , then Banach l a t t i c e and t h e norm
i s weakly Fatou ( i . e . ,
p
has
L
i s a Dedekind complete
L
t h e r e e x i s t s a con-
s t a n t k ( p ) 5 1 such t h a t 0 < u + u i n L i m p l i e s p ( u ) < < k ( p ) . s u p p ( u ) ). Besides t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s t h e r e e x i s t s a s i m i l a r p r o p e r t y f a r monotone sequences. The normed Riesz space L
i s s a i d t o have t h e weak Fatou property f o r monotone sequences i f i t
follows from
0
_<
u
n
1. (n=l,Z,
.
...)
in
and
L
sup p(un) <
m
that
exists i n L It i s e v i d e n t t h a t t h e weak Fatou p r o p e r t y f o r d i sup u n r e c t e d s e t s i m p l i e s t h e weak Fatou p r o p e r t y f o r monotone sequences. Under c e r t a i n a d d i t i o n a l c o n d i t i o n s t h e converse h o l d s . We s h a l l d i s c u s s t h e s e i n t h e n e x t s e c t i o n , b u t we mention a l r e a d y one case because i t can b e a p p l i e d
L
immediately i n t h e p r e s e n t s i t u a t i o n where i n t h e space Riesz s p a c e
M(X,p)
L
. The c a s e we wish
i s an i d e a l w i t h c a r r i e r
X
t o mention i s t h e one where t h e
i s o r d e r s e p a r a b l e and h a s an a t most countable o r d e r b a s i s .
Then t h e weak Fatou p r o p e r t y f o r monotone sequences i m p l i e s t h e same f o r d i r e c t e d sets. The proof w i l l b e g i v e n i n Theorem 113.2. Applying t h i s r e s u l t t o t h e s i t u a t i o n i n t h e p r e s e n t s e o t i o n , we g e t THEOREM 112.3.
( i ) Let
be an ideal with carrier
be a a-finite measure space, l e t
(X,A,p)
i n the space
p
. Then
p
and
( i i ) A s i n p a r t ( i ) , Zet p
be a Banach function space with L"
con-
L i s perfect or, equivalently,
L has t h e weak Fatou property f o r monotone sequences. I n
t h i s case t h e norms and l e t
L
of a l l ( r e a l ) p-measur-
L and i t s second associate space
t a i n the same functions i f and only i f i f and only i f
M(X,p)
L
and, f i n a l l y , l e t
able functions on X respect t o the norm
X
t h e f o l l o w i n g theorem.
in
p"
L
be a Riesz norm i n L
L = L"
are equivalent.
be an ideal with carrier
. If, i n
t a i n the same functions and t h e norms
p
X
i n M(X,p)
t h i s situation, L and
and
p"
L"
con-
are equivalent, then L
i s perfect. Hence, L has nOzJ the weak Fatou property for directed s e t s (as well as for monotone sequences). We f i n a l l y r e c a l l t h a t a normed Riesz space
L
such t h a t
L
i s an
Ch.
16,11131
EMBEDDING I N BIDUALS
i d e a l i n t h e space
42 1
of a l l u-measurable f u n c t i o n s on
M(X,u)
times c a l l e d a nomed Kb’the space ( t h e
i s some-
X
c a s e of sequence spaces was inves-
t i g a t e d by G. KEthe b e f o r e t h e g e n e r a l theory was developed). It i s of int e r e s t t o compare t h e c o n t e n t s of t h e p r e s e n t s e c t i o n w i t h what was observed (without p r o o f s ) i n s e c t i o n 9 about normed KGthe spaces. I n p a r t i c u l a r , n o t e t h a t i f t h e Fatou c o n s t a n t
k(p)
satisfies
k(p) = 1
i n t h e l a s t theorem, and t h u s we g e t t h e r e s u l t of W.A.J. G.G.
, then
= p
p”
Luxemburg and
Lorentz mentioned i n s e c t i o n 9 .
113. The Fatou p r o p e r t y
A s d e f i n e d e a r l i e r , t h e normed Riesz space
that
sup u
case
L
exists i n
L
. It
and
4
i s a Dedekind complete Banach l a t t i c e and t h e norm I. u
in
implies
L
the l a s t inequality (i.e.,
sup p ( u T ) < m
was proved i n Theorem 107.5 t h a t i n t h i s
Fatou, which means t h a t t h e r e e x i s t s a c o n s t a n t 0 5 u
has t h e weak Fatou
L
0 5 u
p r o p e r t y f o r d i r e c t e d s e t s i f i t f o l l o w s from
k(p) 2 1
.
p ( u ) 5 k ( p ) . s u p p(u,)
If
i s weakly
p
such t h a t
k(p) = 1
satisfies
i f p(u) = sup p ( u ) ) , we s h a l l say t h a t
t h e Fatou property for directed s e t s and
p
has
L
i s t h e n c a l l e d a Fatou nom.
There i s a s i m i l a r s i t u a t i o n f o r monotone sequences i n s t e a d of d i r e c t e d s e t s . The space L i s s a i d t o have t h e weak Fatou property for monotone sequences ( a l s o c a l l e d t h e sequential weak Fatou property) i f it f o l l o w s
...
) and sup p(un) < m t h a t from 0 5 u 4 ( n = l , 2 , n We f i r s t prove a theorem p a r a l l e l t o Theorem 107.5.
THEOREM 113.1.
If
L
sup u
exists i n
n
L
.
has t h e sequential weak Fatou property, then L
i s a Dedekind o-complete Banach l a t t i c e and the norm p i s s e q u e n t i a l l y weakly Fatou, i . e . , there e x i s t s a constant k ( p ) 2 1 such t h a t 0 5 u n 4 u in L
implies
P ( U ) 5 k ( p ) . s u p P(u,)
.
0 5 u 4 5 uo i n L , t h e n sup p(un) 5 p ( u o ) < m , so n e x i s t s by h y p o t h e s i s . This shows t h a t L i s Dedekind o-complete
PROOF. I f SUP
u
n Furthermore, L n = 1,2 SO
,...
sup s
n
and
s a t i s f i e s t h e Riesz-Fischer c o n d i t i o n ( i f
a = C p(un) <
m
e x i s t s by h y p o t h e s i s ) .
We prove now t h a t
p
,
then
n This shows t h a t
0
for a l l
p(s ) 5 a
L
5
u
n
E L
for
sn = Z y \ , i s a Banach l a t t i c e .
i s s e q u e n t i a l l y weakly Fatou. I f n o t , t h e r e
e x i s t s f o r e v e r y n a t u r a l number
k
a sequence
0 5 unk Sn
\
such t h a t
422
FATOU PROPERTY > k
P(\)
3
sup, p(unk)
. It is impossible that
, because this would imply unk dicting the inequality for p(uk)
k
constants, we may assume that every 2
k
. For
s Ck-2
P(v,)
sup unk
=
n = 1,2,
...
for every n
\
Ch. 16,51133
for every
sup, p(unk) for all n , and s o
= 0
. Hence, multiplying by
sup, p(unk)
=
k-2
'PI
, let now vn = , s o sup vn exists k , s o p(sup vn) 2
,
and
so
.
0 for some \ = 0 , contra-
=
appropriate > k
p(uk)
for
Then 0 2 v11 4 and unk by hypothesis. But sup v 2 n
p(uk)
k for every k
>
.
This is impossible. If the smallest possible number k(p) k(p)
=
1 , i.e., if 0
2 u
in the last theorem satisfies
implies p ( u ) = sup p(un) , we
in L
4 u
has the Fatou property f o r monotone sequences (sequential
shall say that L
Fatou property) and the norm
is said to be sequentially Fatou. It is
p
evident that the (weak) Fatou property for directed sets implies the (weak) sequential Fatou property. We shall discuss now several conditions under which the converse holds. THEOREM 113.2. If L
i s an order separable nomed Riesz space possess-
i n g an a t most countable order b a s i s and having the (weak) sequential Fatou property, then
has t h e (weak) Fatou property f o r directed s e t s .
L
PROOF. Let
...)
(bn:bn+12b tO;n=1,2,
let the number k 2 1
in the preceding theorem) and
L
such that
sup p(u,) <
exists. For this purpose, we set v n
. Then
is Dedekind o-complete (as proved
is order separable (by hypothesis), the
is super Dedekind complete. Now, let 0 5 u
directed set in L
and
be a constant corresponding to the weak sequential
Fatou property. Note first that since L space L
be an order basis of L
T ,n
m
. We
I.
be an upwards
have to show that sup u
= inf(uT,nbn)
for all
T
and all
exists in L by the Dedekind completeness and by n V,,n is the supremum of a sequence taken the super Dedekind completeness s s
from the set
(v,,~:, variable). It may be assumed that the sequence is
increasing. It follows that
p(sn)
2
k sup p(u,)
for every n
. Since
5 s 4 in L and sup p ( s n ) 2 k sup pfu,) , we may apply the sequential n exists in L Fatou property once more and we find thus that sup s n Furthermore,
0
.
Ch. 16,11131
423
EMBEDDING I N BIDUALS
2 sup s = sup uT , s o p(sup u ) 2 k sup p(uT) n shows t h a t , under t h e mentioned c o n d i t i o n s , t h e
It i s a l s o easy t o see t h a t
The s p e c i a l c a s e
k
1
=
.
Fatou p r o p e r t y f o r d i r e c t e d s e t s f o l l o w s from t h e s e q u e n t i a l F a t o u p r o p e r t y . The weak Fatou p r o p e r t y f o r d i r e c t e d s e t s i s analogous, i n some s e n s e , t o Dedekind completeness. The l a s t may b e expressed by s a y i n g t h a t every o r d e r bounded upwards d i r e c t e d s e t has a supremum and t h e f i r s t by s a y i n g t h a t every norm bounded upwards d i r e c t e d s e t of p o s i t i v e elements has a supremum. S i m i l a r l y , t h e weak s e q u e n t i a l F a t o u p r o p e r t y and Dedekind u-comp l e t e n e s s a r e analogous. Now, i n Theorem 103.6 we have proved an important r e s u l t of H. Nakano f o r a normed Riesz space
L = L
u-completeness t o g e t h e r w i t h u-order c o n t i n u i t y of
P ' p
s t a t i n g t h a t Dedekind i s equivalent t o
Dedekind completeness (even s u p e r Dedekind completeness) t o g e t h e r w i t h o r d e r c o n t i n u i t y of
p
. We
s h a l l prove t h a t t h e r e e x i s t s a p a r a l l e l theorem
f o r t h e Fatou p r o p e r t y . P r e c i s e l y , t h e weak s e q u e n t i a l Fatou p r o p e r t y w i t h u-order c o n t i n u i t y of
p
i s e q u i v a l e n t t o t h e weak Fatou p r o p e r t y f o r
d i r e c t e d s e t s t o g e t h e r w i t h o r d e r c o n t i n u i t y of
p
. In
Nakano's theorem
t h e r e i s a t h i r d e q u i v a l e n t c o n d i t i o n . This i s t h e c o n d i t i o n t h a t every o r d e r bounded i n c r e a s i n g sequence i n
L
must have a norm l i m i t . S i m i l a r l y , i n t h e
c a s e of t h e Fatou p r o p e r t y , t h e t h i r d c o n d i t i o n s t a t e s t h a t every norm bounded i n c r e a s i n g sequence of p o s i t i v e elements i n
L
must have a norm l i m i t .
This p a r a l l e l theorem was e s s e n t i a l l y known t o Nakano (Theoaem 30.20 i n t h e book C41). We begin w i t h a lemma which p a r a l l e l s Lemma 103.3. The f i r s t p a r t i s even i d e n t i c a l .
LENMA 113.3. ( i ) I f
0 5 uT 6
i s a p-Cauchy system and
sequence o f p o s i t i v e numbers, there e x i s t s a sequence t h a t uT + and
(uT ) n
cn+ 0
in
is a
(uT)
such
n
f o r a%%n
. Furthermore,
any upper bound of t h e sequence
.
(uT 1 n
i s an
upper bound of the system ( u T ) ( i i ) I f every increasing p-Cauchy sequence has a norm l i m i t and 0 2 u f i s a p-Cauchy system, then u = sup u e x i s t s and any increasing sequence (u as i n p a r t ( i ) above s a t i s f i e s sup u = u = sup u Tn n Furthermore p(u-uT) $ 0
.
.
FATOU PROPERTY
424
(iii) If
L
has the weak sequential Fatou property and
i s a p-Cauchy system, then u (u,
n
= sup
u
0 5 u
4
e x i s t s and any increasing sequence sup u
as i n part ( i ) above s a t i s f i e s
)
Ch. 16,51133
n
.
u
= sup
PROOF. (i) A s p a r t ( i ) of Lemma 103.3. 0 s u
( i i ) Let sequence
f
b e a p-Cauchy system. For any i n c r e a s i n g
4
(u ) as i n p a r t ( i ) w e have
in
(u, ) n
)
-u
P\'Tn+m
=
Tn+m
TnJ
,u
)-u ,nJ
'n
}
2 En
,
) i s a n i n c r e a s i n g p-Cauchy sequence. Hence, by h y p o t h e s i s , u n Tn converges i n norm t o some u € L But t h e n , by Theorem 100.4, u = sup u ,n By p a r t ( i ) , u i s now a l s o an upper bound of t h e s y s t e m (u,) Hence,
so
(u,
sup u u
'n
4
e
.
that
p(u-u
'n 0 2 u
(iii) Let
i n c r e a s i n g sequence bounded and
L
By p a r t ( i ) , u = u
. It
= u = sup u
=supu
f o l l o w s immediately from
) C 0
ii
+
,
so
p(u-u,)
+
0
p(u-u
.
b e a p-Cauchy system i n
L
Tn
) + 0
h a s t h e weak s e q u e n t i a l Fatou p r o p e r t y , u = sup u
.
i s a l s o an upper bound of t h e s y s t e m
(u,)
and
(u, )
and l e t
as i n p a r t ( i ) . S i n c e t h e sequence
(u,)
.
.
, so
iP
be an
norm
exists. 'n sup u = n
THEOREM 113.4. The following conditions for t h e nomed Riesz space
L
are equivalent. (i)
-The norm p
i s u-order continuous and
L
has the weak sequential
Fatou property. Every norm bounded increasing sequence o f p o s i t i v e elements i n L
(ii)
has a norm limit. ( i i i ) m e norm
p
i s order continuous and L has t h e weak Fatou
property f o r directed s e t s . If
L
s a t i s f i e s one (and therefore each) of these conditions, then
L
i s a super Dedekind compZete and p e r f e c t Banach l a t t i c e . Note t h a t spaces L s a t i s f y i n g these conditions are exactly the KB-spaces mentioned i n s e c t i o n 95.
-.
PROOF. The proof i s s i m i l a r t o t h e proof i n Theorem 103.6. (i)
*
(ii). Let
0 5 u
4
with
weak s e q u e n t i a l Fatou p r o p e r t y , so
sup p ( u ) < The s p a c e L h a s t h e n exists in L Then
u = sup u
.
Ch. 16,§1131
u-u
,
C 0
EMBEDDING I N BIDUALS
t h e norm l i m i t of (ii)
u
n
since
.
,
n
. Then
E L
uo
in
(uT )
0
5
u
C 0
v
=
. We
such t h a t
(uT)
c o n t r a d i c t i n g our ;resent
i s o r d e r continuous, l e t some
c o n t i n u o u s . Hence, u
i s a p-Cauchy system ( i f n o t , t h e r e would e x i s t a number
and a n i n c r e a s i n g sequence for a l l
i s o-order
p
hypothesis).
, so
(vT)
-u
uo 2 u T C 0
may assume t h a t
uo-uT I. uo
(uT
,
p(uo-vT) C 0
i.e.,
For t h e proof t h a t let
+
0 5 u
with
p(uT)
L
+
0
(iii)
*
L
If n u i t y of
p
tained i n
) >
E
p
for (vT)
.
,
h a s t h e weak Fatou p r o p e r t y f o r d i r e c t e d sets,
sup p ( u ) <
u = sup u
> 0
i s a p-Cauchy system.
m
. Then
(uT)
i s a p-Cauchy system, s o
t h e h y p o t h e s e s of p a r t ( i i ) i n Lemma 113.3 a r e now s a t i s f i e d f o r follows t h a t
E
For t h e f:bofT?hat
The h y p o t h e s e s o f p a r t ( i i ) i n Lemma 113.3 a r e now s a t i s f i e d f o r so
is
( i i i ) Note f i r s t t h a t e v e r y norm bounded upwards d i r e c t e d s e t
9
+
0 S u
C 0
p(u-un)
so
425
(uT)
. It
exists.
( i ) Evident.
s a t i s f i e s t h e s e c o n d i t i o n s , i t f o l l o w s from t h e o r d e r c o n t i 0 t h a t L* = L* so = O(L*) = { O } S i n c e L: i s con(L:) n ' L : , we f i n d t h a t '(L;) = { O } . Combining t h i s w i t h t h e
.
condition t h a t
L
1 1 1 . 1 shows t h a t
h a s t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s , Theorem L
i s a p e r f e c t Banach l a t t i c e . F i n a l l y , i t f o l l o w s from
c o n d i t i o n ( i i ) t h a t e v e r y o r d e r bounded i n c r e a s i n g sequence i n
L
has a
norm l i m i t , which i s one of t h e e q u i v a l e n t c o n d i t i o n s i n Nakano's theorem (Theorem 103.6). Hence, a l l c o n d i t i o n s i n Nakano's theorem h o l d . I n particular, L
i s s u p e r Dedekind complete.
A s observed a l r e a d y , t h e e q u i v a l e n t c o n d i t i o n s i n t h e p r e s e n t theorem imply t h e c o n d i t i o n s i n Nakano's theorem (Theorem 103.6). For t h e c o n v e r s e , n o t e t h a t t h e sequence s p a c e
(c,)
of a l l r e a l n u l l sequences ( w i t h t h e
uniform norm) s a t i s f i e s a l l c o n d i t i o n s of Nakano's theorem, but n o t t h o s e of t h e p r e s e n t theorem. Note a l s o t h a t i f , f o r example, L L,(X,p)
with
p
Lebesgue measure i n
L_-norm,
then
p
i s n o t o-order c o n t i n u o u s , b u t
X = C0,ll
s e q u e n t i a l F a t o u p r o p e r t y and h e n c e , s i n c e f i n i t e order basis, L
L
and w i t h L
i s t h e space p
t h e usual
obviously has t h e
i s o r d e r s e p a r a b l e and h a s a
h a s t h e Fatou p r c p e r t y f o r d i r e c t e d s e t s .
It was proved i n Theorem 83.12 t h a t any o r d e r bounded o p e r a t o r mapping
a Banach l a t t i c e i n t o a Dedekind complete normed R i e s z s p a c e i s norm bounded. I n t h e c o n v e r s e d i r e c t i o n , a norm bounded o p e r a t o r i s n o t o r d e r bounded
FATOU PROPERTY
426
Ch. 16,§1131
i n g e n e r a l ( c f . t h e remark i n Exercise 98.7, where i t i s shown among o t h e r things t h a t a c e r t a i n unitary operator i n i s n o t o r d e r bounded). I n f2 c e r t a i n c a s e s , however, norm bounded o p e r a t o r s a r e a l s o o r d e r bounded. A t y p i c a l example occurs i n Dunford's theorem ( C o r o l l a r y 9 8 . 4 ) , s t a t i n g t h a t
if
and
(X,A,u)
Ll(Y,v)
(Y,Z,v) a r e a - f i n i t e measure spaces and t h e number
1 < p 5
satisfies
into
course t h a t
T
m
,
t h e n any norm bounded o p e r a t o r
L (X,u) 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 , which i m p l i e s of P i s o r d e r bounded. Dunford's theorem i s a s p e c i a l c a s e of
a more g e n e r a l theorem (Theorem 98.3), b e as above and l e t
(Y,Z,v)
a s follows. Let
equipped w i t h a Riesz norm with respect t o
.
p
p
(X,A,u)
b e an i d e a l w i t h c a r r i e r
M
of a l l r e a l u-measurable f u n c t i o n s on
M(X,u)
p
which maps
T
such t h a t
M
X
i n t h e space that
is
M
i s a p e r f e c t Banach l a t t i c e
I f , f u r t h e r m o r e , t h e a s s o c i a t e norm
c o n t i n u o u s , t h e n every norm bounded o p e r a t o r
X
. Assume now
and
T
from
p'
Ll(Y,v)
i s o-order into
M
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 . The hypotheses i n t h i s theorem a r e s t r o n g because i t has t o b e shown not only t h a t T
T i s o r d e r bounded b u t a l s o t h a t
can b e expressed by means of a k e r n e l . We s h a i l prove now a theorem
which i n one r e s p e c t i s more g e n e r a l , d e a l i n g w i t h a b s t r a c t Banach lattices (not n e c e s s a r i l y c o n s i s t i n g of measurable f u n c t i o n s ) , b u t which h a s on t h e o t h e r hand a more moderate aim because i t mentions only c o n d i t i o n s under which norm boundedness of an o p e r a t o r i m p l i e s o r d e r boundedness. Before s t a t i n g t h e theorem, we s t i l l n o t e t h a t i n t h e r e s u l t r e f e r r e d t o above
i s supposed t o b e p e r f e c t , t h i s i m p l i e s t h a t
M
(Theorem 98.3), where
M
h a s t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s . This p r o p e r t y of t h e range space
M
w i l l b e of importance i n t h e proof t h a t c e r t a i n norm bounded
o p e r a t o r s a r e o r d e r bounded. Another important h y p o t h e s i s w i l l b e t h a t t h e domain space i s of L -type. We r e c a l l t h a t t h e Banach l a t t i c e L i s c a l l e d I an a b s t r a c t L - s p a c e i f t h e norm p i s a d d i t i v e , i . e . , p(u+v) = p ( u ) + p(v) 1
for a l l positive
u
and
THEOREM 113.5. Let
v
L
an abstract L -space and
P
in
and
L
.
MA
be Banach l a t t i c e s such t h a t
L
P
is
has t h e weak Fatou property f o r directed s e t s MA 1 (hence, MA i s Dedekind complete by Theorem 107.5). Then any norm bounded mapping L
operator
T
k = k(h)
i s a constant corresponding t o the weak Fatou property i n MA T and IT1 s a t i s f y
P
then t h e operator n o m o f
i n t o MA i s order bounded. Furthermore, i f
,
Ch. 16,51131
EMBEDDING I N BIDUALS
PROOF. Let
0 5 u
427
be given. Denote by
E L
Su
t h e s u b s e t of
M
c o n s i s t i n g of t h e elements x = ITu
1
1 +
... +
[Tun[
f o r a l l p o s s i b l e decompositions u = u 1 + t i v e . Let
... + un
u = u1 +
= uf +
n
Z. u
U. = 1
so
u = Cij
so
sup(x,x')
uij 5
uij
J
(i=l,
i ; uf j
. Then y = C i j ITuijI E . This shows t h a t Su i n Mh . Furthermore, i f
y
s i t i v e elements
ul,.
and
x f = :1
posi-
u;,
...,n ; j , ...,k)
for a l l
ij
.. ,un
with
i.
n x = C 1 ITuil
be two of t h e s e decompositions w i t h There e x i s t elements
...
... + un
= Z.
u
1
Su
for a l l
ij
and
x 5 y
j
.
lTu!I J
such t h a t
,
a s well a s
x' 5 y
i s an upwards d i r e c t e d s e t of pon x = E l ITuil i n Su , then
so
Hence, s i n c e
MA has t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s , sup S
.
exists i n
MA
s e t of
, say
Lp
I t follows now e a s i l y t h a t i f If
i s o r d e r bounded i n
so
ITfI 5 sup S
I
5
MA
. This
for a l l
u
. For
D
i s an o r d e r bounded sub-
E D , then the image f E D . Then
f
shows t h a t
T
ITfI 5 I T I ( l f l )
of
D
i s o r d e r bounded.
For t h e r e l a t i o n between t h e o p e r a t o r norms of IT1 and f i r s t that
T(D)
t h e proof, l e t
f o r every
f E Lp
,
so
T
we n o t e
,
Ch. 16,§1131
FATOU PROPERTY
This shows t h a t
IlTll 5 lllTlIl
.
Now, l e t
sponding t o t h e weak Fatou property i n before, i t follows from ( 1 )
k = k(h)
. With
MA
be a constant corret h e same n o t a t i o n s a s
that
Hence, s i n c e
we g e t
A(IT1 ( u ) ) 5 k. l/TII.p(u)
EXERCISE 113.6. Let
L
. For
L
f
E Lp
t h i s implies
be a normed Riesz space.
( i ) Show t h a t i f t h e norm p then
an a r b i t r a r y
is o-order continuous and
i s Dedekind complete, and hence
L
L
is perfect,
s a t i s f i e s a l l conditions of
Nakano’s theorem (Theorem 103.6).
(ii) Show t h a t i f
L
i s t h e sequence space of a l l r e a l sequences with
only f i n i t e l y many nonzero terms (and with t h e uniform norm), then s a t i s f i e s the hypotheses of p a r t ( i ) , but ( i i i ) Show t h a t i f the norm p e r f e c t , then L
L
p
L
i s not Banach.
i s o-order continuous and
( i ) The space
(Li)-
is
L = (LT;”;
L
i s Banach.
i s Dedekind complete and any band i n a
Dedekind complete space i s Dedekind complete, s : o):L( Hence, s i n c e
L
has the weak s e q u e n t i a l Fatou property ( t h a t i s t o say,
s a t i s f i e s a l l conditions of Theorem 113.4) i f and only i f HINT:
L
by hypothesis, t h e space
i s Dedekind complete. L
i s Dedekind complete.
( i i i ) Any space s a t i s f y i n g the conditions of Theorem 113.4 i s Banach. For t h e proof i n t h e converse d i r e c t i o n , use Theorem 1 1 1 . 1 .
Ch.
16,51143
EMBEDDING I N BIDUALS
429
1 1 4 . Embedding i n t h e Banach b i d u a l ; r e f l e x i v i t y Let L
L-
. Hence,
such t h a t
f E L
(L*)-
,
, i.e.,
O(L*) = { O }
L* i s an i d e a l i n t h e
s e p a r a t e s t h e p o i n t s of
L*
, defined
o: L + (L*)-
t h e Riesz homomorphism
f o r every have
b e a normed Riesz space. The Banach d u a l
L
order dual
by
o(f)($) = $(f)
i s a Riesz isomorphism (Lemma 109.1). Furthermore, we L* i s Banach. The Riesz isomorphism
since
= (L*)*
t h e r e f o r e , e x a c t l y t h e f a m i l a r embedding of
o is,
into the bidual
L
proved i n s e c t i o n 109, t h e embedding i s a c t u a l l y i n t o t h e band
. The
= (L*):
that
L
. As
L** (L*):
=
embedding p r e s e r v e s f i n i t e suprema and infima; it f o l l o w s
i s a Riesz subspace of
. The
(L*):
embedding a l s o p r e s e r v e s t h e
norm; it w i l l c a u s e no confusion, t h e r e f o r e , i f we denote t h e norm i n a g a i n by p
.
THEOREM 1 1 4 . 1 .
( i ) Suprema and infima of countable systems i n
L** i f and only if t h e norm
preserved under t h e embedding i n t o
L
L**
are
is
p
o-order continuous. ( i i ) Suprema and infima o f a r b i t r a r y systems i n
I,**
under t h e embedding i n t o
L
are preserved
i f and only i f t h e norm
ous and a l s o i f and only i f t h e i d e a l generated by
.
Dedekind completion of L ( i i i ) If L i s Banach and L**
L
i s order continuL**
in
is a
hasordercontinuous norm, y e n s o has L .
PROOF. ( i ) Assume t h a t suprema and i n f i m a of c o u n t a b l e systems a r e p r e s e r v e d . For t h e proof t h a t
L
. Then,
0
2
by h y p o t h e s i s , u
$ E L*
i.e.,
L*
=
i s o-order continuous, l e t
p
in
J. 0
L**
,
i.e.,
$(un> J. 0
. This shows t h a t every element of L* L* . The last e q u a l i t y i s e q u i v a l e n t t o
u J. 0 i n n f o r every
i s an i n t e g r a l on t h e statement t h a t
,
L p
is
a-order continuous. Conversely, assume now t h a t
.
u 4 u in L Then u + s u i n n n of L** , t h e r e e x i s t s an element It follows t h a t
we have f o r every
$(un)
$(un) 4 $(u) $
4
i s o-order continuous. L e t f i r s t
p
L**
. Hence,by
u" E L**
f o r every
u"($)
t h e Dedekind completeness
such t h a t
0 5 $ E L*
by t h e o-order c o n t i n u i t y of
E L* , which shows t h a t
u" = u
, and
p
so
4 u" i n
u
. On t h e . Hence u
o t h e r hand u"($)
in
4 u
.
L**
L**
= $(u)
.
S i m i l a r l y f o r d e c r e a s i n g sequences. It h a s been proved t h u s t h a t t h e embedd i n g p r e s e r v e s suprema and infima of monotone sequences. I f n e c e s s a r i l y monotone sequence i n =
sup(u,,
...,un)
for
n = 1,2,
L with
...
sup un = u
satisfies
v
4 u
, then
. But
i s a not
(u,)
v
=
then both
430
Ch. 16,§1141
EMBEDDING I N THE BANACH BIDUAL
...,
v
= sup(ul, u ) and v 4 u h o l d i n n e a s i l y t h a t sup u = u h o l d s i n L**
a s w e l l . It follows
L**
.
( i i ) The proof t h a t suprema and infima of a r b i t r a r y systems i n preserved under t h e embedding i f and only i f t h e norm
are
L
i s o r d e r continu-
p
o u s i s s i m i l a r t o t h e proof i n p a r t ( i ) . A second proof i s o b t a i n e d from
Theorem 109.2. Indeed, choosing embedding of
L
i f and only i f
n
in
B = L*
i n Theorem 109.2, we see t h a t t h e
(L*), = (L*)* p r e s e r v e s a r b i t r a r y suprema and infima n i s contained i n L i , i . e . , i f and only i f L* =
L*
. Since
L* = L* i f and only i f p i s o r d e r continuous, n t h e proof i s complete. A t t h e same time Theorem 109.2 shows t h a t t h i s
= L*
:L
= L:
s i t u a t i o n o c c u r s i f and only i f t h e i d e a l generated by i s a Dedekind completion of
= (L*):
L
in
L
.
(L*):
=
( i i i ) Observe t h a t (by Theorem 104.2) i n a Banach l a t t i c e t h e norm i s o r d e r continuous i f and only i f every o r d e r hounded d i s j o i n t sequence converges t o zero i n norm.
In C o r o l l a r y 105.12 i t was proved t h a t i f is an ideal i n
L i
o r d e r dense i n
:L
space
L
into
(L*):
Obviously
A
. Let
(L*);
=
A
L
s e p a r a t e s t h e p o i n t s of
,
then
A
p
. As
shown above, L
b e t h e i d e a l g e n e r a t e d by
s e p a r a t e s t h e p o i n t s of
L* (because
L
L
. It
is
A
i s embedded in
follows t h a t , f o r every
(L*)F,
.
already separates
L* ) . Hence, by C o r o l l a r y 105.12 a s mentioned, A
(L*):
A
apply t h i s t o t h e c a s e t h a t we have a normed Riesz
w i t h o r d e r continuous norm
t h e p o i n t s of dense i n
such t h a t
. We
i s a Riesz space and
L
0 2 u" E (L*):
, we
i s order have
u" = sup (v":v"€A,O
\
A s t r o n g e r r e s u l t i s t r u e , a s shown i n t h e f o l l o w i n g lemma. LEMMA 114.2. If p
and i f
0
2
Ull
u"
L
i s a normed Riesz space with order continuous norm
, then
E (L*):
= sup (u:uEL,O
\
) -
PROOF. A s observed above, t h e r e e x i s t s a d i r e c t e d s e t
ideal
A
generated by
L
in
is t h e Dedekind completion of
(L*): L
, we
such t h a t have
0
v" 4 u"
(vy)
i n the
. Since
A
Ch.
f o r every T
EMBEDDING I N BIDUALS
16,§1141
. The
T
set o f a l l
u"
u <
satisfying
u E L
i s d i r e c t e d upwards and h a s
43 1
f o r a t l e a s t one
V"
as i t s supremum i n
. The
L**
desired
r e s u l t follows.
A s shown i n Theorem 1 1 4 . 1 , t h e i d e a l g e n e r a t e d by
i s i n f a c t t h e i d e a l g e n e r a t e d by of
L
i f and o n l y i f
L
in
i s a Dedekind completion of u € L
by
L
in
L**
by
L
in
L**
case
. In . It
. This
and only i f
(L*)*) i s a Dedekind c o m p l e t i o n
shows t h a t
o t h e r words,
L
u
(L*):
for a l l
.
L**
= sup p ( u ) <
shows t h a t
u"
. Then
m
, defined
L**
. In
T
m
(L*)E
. Hence
(L*):
i s t h e i d e a l generated
i s a Dedekind completion of
sup $ ( u ) 2 a.p*($)
in
(L*),T
L**
, we
S u
m
p
(L*):
.
We have t o show t h a t
u"
directed s e t
that the ideal
p
i s a Dedekind completion of A = (L*): (uT)
in
L
For t h a t p u r p o s e , l e t belongs t o L
such t h a t
A
0
. By
0 2 u
with
E
, which L* , i s
f o r some
0 2 u
4
i s m a j o r i z e d by some
(uT)
f o l l o w s from t h e o r d e r c o n t i n u i t y of in
u" 5 u
i s o r d e r c o n t i n u o u s and
implies t h a t
0 5
. Then
L L
. Hence, s i n c e a l l u (L*): . This i m p l i e s
L ) that
.
for a l l T
in
I.
0 5 @ E L*
f o r every L**
u" E
have
i s a Dedekind completion o f u
sup p ( u T ) <
prove t h a t
in
u" = sup u
0 5 u
f o r every
u"($) = sup $ ( u )
by
Conversely,assume t h a t with
i f
u E L
t h a t there e x i s t s an element
t h i s case
(L*)E
satisfying
b e l o n g t o t h e band
E L
L
i s a Dedekind completion o f
i s o r d e r c o n t i n u o u s a s observed above. Now, l e t
(since
i s t h e n major-
E (L*):
i t s e l f i s now t h e i d e a l g e n e r a t e d
(L*):
sup p ( u T ) <
with
2 u
a member of
L
u"
belongs t o the i d e a l generated
u"
f o l l o w s t h e r e f i o r e from Theorem 1 1 4 . 1 t h a t i n t h i s
PROOF. Assume f i r s t t h a t
u
itself
(L*):
is order continuous and i f , i n addition, i t follows from
p
in
4
such t h a t by L i n
a
(which
i s of some importance. I n view of t h e
L
THEOREM 114.3. The space
p
L**
i s o r d e r c o n t i n u o u s . More p r e c i s e l y , t h e f o l l o w i n g h o l d s .
p
0 5 u
in
i s o r d e r c o n t i n u o u s . The c a s e t h a t
p
d e f i n i t i o n of a Dedekind completion any p o s i t i v e i z e d by some
L
. Hence,
A
in
u E L
.
L It
g e n e r a t e d by
it is sufficient to
0 2 u" E (L*):
be g i v e n .
t h e l a s t lemma t h e r e e x i s t s a 4 u"
in
L**
, so
sup p(u,)
5
432
S
EMBEDDING I N THE BANACH BIDUAL
p(u") <
u
that
. Hence,
m
in
5 u
which i m p l i e s the ideal
L
by h y p o t h e s i s , t h e r e e x i s t s an element for a l l
u" = sup u g e n e r a t e d by
A
. But
T
in
u
S
L
in
COROLLARY 114.4. The space
then
L** L**
L**
. It .
16,§1141
Ch.
u
E L
such
< u holds as well i n
u
follows t h a t
u"
L**
,
belongs t o
L
i s a Dedekind completion of
i f
and only i f the following three conditions are s a t i s f i e d . (i)
(L*)* = L**
(ii)
p
.
i s order continuous.
s u
(iii) 0
in
L
u E L
.
f
majorized by some
L**
I n t h i s case
sup p(u,)
with
<
implies that
m
i t s e l f i s the ideal generated by
L
(uT) L**
in
We s h a l l d e r i v e now 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 s f o r a n i d e a l ( o r even an o r d e r dense i d e a l ) i n
L
is
. t o be
.
L**
THEOREM 114.5. The following conditions are equivalent.
.
(i)
L
i s an i d e a l i n
(ii)
L
i s Dedekind o-complete and
(iii) L
L**
i s o-order continuous.
p
i s super Dedekind complete and
I f these canditions are s a t i s f i e d , then
Furthermore, note t h a t i f
L
L
i s order continuous.
p
(L*):
i s order dense i n
i s Banach, then order continuii;y of
p
. implies
super Dedekind completeness (by Theorem 103.9). >PROOF. ( i ) =+ ( i i ) The s p a c e space
L**
r a t e d by
; t h e space L
in
L**
L
i s an i d e a l i n t h e Dedekind complete
i s t h e r e f o r e Dedekind complete. The i d e a l gene-
L
, which
now a Dedekind completion of
under t h e p r e s e n t h y p o t h e s i s i s L
, and
so
p
itself, is
L
is o r d e r c o n t i n u o u s by
Theorem 114.1 (ii). (ii)
*
( i i i ) T h i s i s Nakano's
(iii)
(i)
Since
114.l(ii) that the ideal completion of
L
. For
L
A
g e n e r a t e d by
every p o s i t i v e
which shows t h a t t h e set since
theorem (Theorem 103.6).
i s o r d e r c o n t i n u o u s we conclude from Theorem
p
u"
(u:uEL,O
L in
in
A
L**
i s a Dedekind
we have, t h e r e f o r e , t h a t
i s o r d e r bounded i n
L
. Hence,
i s Dedekind complete and s i n c e suprema of a r b i t r a r y systems i n
L
Ch. 16,51141
433
EMBEDDING I N BIDUALS
a r e preserved under the embedding (Theorem 1 1 4 . 1 ( i ) ) , i t follows t h a t
.
u" E L in
L**
I n o t h e r words, we have
. This
A = L
.
shows t h a t
i s an i d e a l
L
I f t h e conditions ( i ) , ( i i ) , ( i i i ) a r e s a t i s f i e d , then L i s an i d e a l 0 I t follows, = I01 i n (L*): and, e v i d e n t l y , (L) = I01 , s o o{(L*);l 0 t h a t L i s o r d e r dense i n (L*); t h e r e f o r e , from '(L) = I(L*):l
.
.
THEOREM 114.6. The following conditions are equivalent.
(i)
L
i s a order dense ideaZ i n
(ii)
L
i s Dedekind o-complete and both
L**
.
and the norm
p
p*
L*
in
are a-order continuous. (iii) L
and
L*
are super Dedekind complete and both
and
p
p*
are
order continuous. PROOF.
( i ) w ( i i ) I f i t i s given t h a t
v a l e n t l y , i f i t i s given t h a t continuous), t h e space dense i n
L**
,
L
i s an i d e a l i n
L**
i s Dedekind o-complete and
i s order dense i n
i f and only i f
L**
i s a band i n i.e.,
L
L
. Hence,
(L*):
i s o r d e r dense i n
(L*):
L**
t h i s i s equivalent t o the condition t h a t
equivalent t o the condition t h a t the norm
p*
in
L*
L
(equi-
i s a-order
p
i s order
. Since
(L*):
(L*)* = L** n i s order
,
L* i s Dedekind complete, t h i s again i s equivalent
continuous. But, s i n c e
by Nakano's theorem t o t h e condition t h a t
p*
i s a-order continuous.
( i i ) e 9 ( i i i ) Nakano's theorem. Note t h a t i f
L =
L1 with t h e corresponding
e 1-norm,
then
L
i s Dede-
kind a-complete and the norm i s o-order continuous, s o (by Theorem 114.5)
e**
.
i s an i d e a l i n = L* I f L = (c ) , then L* = L 1 ; i t follows 1 0 i s an t h a t condition ( i i ) of t h e l a s t theorem i s s a t i s f i e d . Hence (c,)
el
o r d e r dense i d e a l i n
(co)** =
1
, as
a l s o immediately v i s i b l e . This
example shows t h a t t h e conditions i n the l a s t theorem a r e not s u f f i c i e n t t o imply t h a t L
i s r e f l e x i v e , even i f
L
i s Banach. Note, f i n a l l y , t h a t i f
L
i s t h e space o f a l l sequences having only a f i n i t e number of nonzero
terms, with
p
t h e uniform norm, then
L* =
; the condition ( i i ) i n t h e
l a s t theorem i s again s a t i s f i e d , t h e r e f o r e . Hence, L dense i d e a l i n
L** =
Lm , but now
L
i s again an o r d e r
i s not Banach.
F i n a l l y , we prove t h a t each of the equivalent conditions i n Theorem 113.4 i s necessary and s u f f i c i e n t f o r
L
t o be a band i n
L**
.
Since any
434
L
Ch. 16,§1141
ENBEDDING I N THE BANACH BIDUAL
s a t i s f y i n g t h e c o n d i t i o n s o f Theorem 113.4 i s a l s o known a s a KB-space,
we prove t h e r e f o r e t h a t
i s a band i n
L
i f and only i f
L**
is a
L
KB-space.
The following conditions f o r t h e normed Riesz space
THEOREM 1 1 4 . 7 .
are e q u i v a l e n t .
L
.
(i)
L
(ii)
Every norm bounded increasing sequence of p o s i t i v e elements i n
i s a band i n
L**
has a norm l i m i t . ( i i i ) The norm
p
i s o-order continuous and
t i a l Fatou property. ( i v ) The norm
p
i s order continuous and
L L
has t h e weak sequen-
has t h e weak Fatou
property f o r d i r e c t e d s e t s . PROOF. The equivalence of ( i i ) , ( i i i ) and ( i v ) was proved i n Theorem
113.4. (i)
*
( i v ) Since
i s a band i n
L
L**
i t follows a l r e a d y from Theorem 114.5 t h a t
,
and hence an i d e a l i n
i s Dedekind complete and
L
i s o r d e r continuous. Furthermore, according t o t h e same theorem, L dense i n t h e band now t h a t
. Hence,
(L*):
.
L = (L*):
since
L
completion of
,
L
sup p(u,)
since
L
m
L = (L*)*
,
then
0
sup u
+ in L . Hence, L . This
uT
S
L
exists i n
h a s t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s .
L
*
is a Dedekind
i s an o r d e r bounded set i n
(u,)
p
i s order
h a s t h e weak
n
i t follows from Theorem 114.3 t h a t i f
i s Dedekind complete, t h e element
shows t h a t (iv)
<
,
i t s e l f i s a band, i t follows
It remains t o prove t h a t
Fatou p r o p e r t y f o r d i r e c t e d s e t s . S i n c e , e v i d e n t l y , (L*): with
L**
( i ) Since h y p o t h e s i s ( i v ) i m p l i e s t h a t
i s o r d e r continuous
p
and s i n c e t h e weak Fatou p r o p e r t y i m p l i e s Dedekind completeness, i t follows (by Theorem 114.5) from t h e Dedekind completeness of c o n t i n u i t y of
that
p
L
i s an i d e a l i n
i n Theorem 114.3 a r e s a t i s f i e d ; t h e space g e n e r a t e d by Hence
L,
i s a band i n
. But,
L**
. This
L = (L*):
Note t h a t
Ll
in
L
L**
shows t h a t
L
and t h e o r d e r
. Furthermore,
(L*):
a s observed, L
L
the conditions
is, therefore, the ideal i t s e l f i s an i d e a l i n
i s a band i n
L**
.
L**
s a t i s f i e s c o n d i t i o n ( i i ) of t h e l a s t theorem. Hence,
Lr*= Lz
. More p r e c i s e l y ,
=
(Lm)E .
W e proceed t o i n v e s t i g a t e now c o n d i t i o n s under which
L
is r e f l e x i v e .
.
Ch. 16,11141
EMBEDDING I N BIDUALS
It i s evident t h a t
is reflexive, i.e.,
L
435
,
L = L**
i f and only i f
a t t h e same time a band a s w e l l a s a n o r d e r dense i d e a l i n dition that p
i s a band i n
L
L**
con-
i s e q u i v a l e n t t o o-order c o n t i n u i t y of
t o g e t h e r w i t h t h e weak s e q u e n t i a l Fatou p r o p e r t y , which i m p l i e s t h e r e -
f o r e Dedekind o-completeness. in
. The
L**
is
L
L**
The c o n d i t i o n t h a t
i s e q u i v a l e n t t o o-order c o n t i n u i t y of
Dedekind o-completeness. r e f l e x i v i t y of
L
L p
i s an o r d e r dense i d e a l and p* t o g e t h e r w i t h
Combining t h e s e c o n d i t i o n s , i t f o l l o w s t h a t
i s e q u i v a l e n t t o o-order c o n t i n u i t y of
and
p
to-
p*
g e t h e r w i t h t h e weak s e q u e n t i a l Fatou p r o p e r t y . Furthermore, an i n s p e c t i o n of t h e c o n d i t i o n s i n t h e l a s t two theorems shows t h a t t h i s a g a i n i s equiv a l e n t t o o r d e r c o n t i n u i t y of
p
and
p*
t o g e t h e r w i t h t h e weak Fatou
p r o p e r t y f o r d i r e c t e d s e t s . This y i e l d s t h e f i r s t p a r t of t h e f o l l o w i n g theorem. THEOREM 114.8. The following conditions f o r the normed Riesz space
L
are equivalent. (i)
L
(ii)
The norms
i s reflexive.
and
p
p*
are o-order continuous and
L
has the
weak sequential Fatou property. ( i i i ) The norms
p
and
p*
are order continuous and
L
has the ueak
Fatou property f o r directed s e t s . Every norm bounded increasing sequence of p o s i t i v e elements i n
(iv) L
has a norm l i m i t and, similarly, every norm bounded increasing sequence
of positive elements i n
L*
has a norm l i m i t .
Every norm bounded increasing sequence of p o s i t i v e elements i n
(v) L
has a norm l i m i t and every order bounded increasing sequence o f p o s i t i v e elements i n L* has a norm l i m i t . PROOF. ( i ) (i)
=s
- (ii)
(iv) L
( i i i ) This was e x p l a i n e d above.
i s r e f l e x i v e and, t h e r e f o r e , L
i s a band i n
L**
. This
i m p l i e s , by Theorem 114.7, t h a t every norm bounded i n c r e a s i n g sequence of p o s i t i v e elements i n
L
h a s a norm l i m i t . Since
s i m i l a r r e s u l t holds i n (iv) (v)
* (v) Evident. * ( i i i ) From t h e
continuous and
L
L*
.
L* i s a l s o r e f l e x i v e , a
f i r s t p a r t of (v) i t follows t h a t
p
is order
h a s t h e weak Fatou p r o p e r t y (Theorem 114.7). From t h e
second p a r t of (v) i t follows by Nakano's theorem (Theorem 103.6) t h a t i s o r d e r continuous.
p*
436
Ch. 16,§1141
EMBEDDING I N THE BANACH BIDUAL
I t w i l l f o l l o w from o u r f i n a l theorem t h a t r e f l e x i v i t y i m p l i e s p e r f e c t -
ness.
L
THEOREM 114.9. I f
Banach, then i.e., L
m
L** = (L)nn
L
i f
and
i s r e f l e x i v e , then L
L
is
,
(LIZ
=
is p e r f e c t .
PROOF. Since
i s Banach, w e have
L
i s an o r d e r dense i d e a l i n o r d e r continuous, s o Since
L**
i s an order dense i d e a l in
. Hence,
,
L**
L* = (L
1;
L* = L"
. Furthermore,
since
i t f o l l o w s from Theorem 114.6 t h a t
. Hence,
we have a l r e a d y t h a t
L
is
p
L* = (L):
.
i s a l s o o r d e r continuous (Theorem 114.6 a g a i n ) , we can apply
p*
t h e same argument t o
and we g e t
L*
L** = (L)nn
.
The r e f l e x i v i t y theorem (Theorem 114.8) i s due t o T. Ogasawara ( [ l ] , 1942-1944,
i n Japanese; a l s o Ch.V,§4, Theorem 1 i n t h e book [23, i n Japan-
e s e ) . The p r e s e n t proof i s d i f f e r e n t from Ogasawara's p r o o f . For t h e c a s e that
i s a Banach f u n c t i o n space, independent p r o o f s a r e t o b e found i n
L
work by H.W.
E l l i s a n d I . H a l p e r i n ([11,1953),
Luxemburg's t h e s i s (111,1955).
W.A.J.
I. H a l p e r i n ([1],1954)
as p r e s e n t e d above, i s e s s e n t i a l l y t h e same a s i n a paper by W.A.J. burg-A.C.
EXERCISE 114.10. L e t
L
Luxem-
Zaanen (C11,Note 13,1964).
f u n c t i o n s on of
and i n
The proof of t h e r e f l e x i v i t y theorem,
L
b e t h e space
w i t h t h e L2-norm.
CO,l]
i s t h e space
. Show t h a t
L,([O,l])
continuous, b u t t h e norms i n
L
-
C([O,ll)
of a l l r e a l continuous
Show t h a t t h e norm completion
, L*
and
L
t h e norm i n
L** = L
-
L
-
i s not order
a r e o r d e r continuous.
Compare t h i s w i t h Theorem 1 1 4 . l ( i i i ) . EXERCISE 114.11. and o n l y i f
L
( i ) Show t h a t t h e Banach l a t t i c e
i s p e r f e c t and
( i i ) Show t h a t
L
p
L
i s a KB-space i f
i s o r d e r continuous.
i s r e f l e x i v e i f and only i f
L
and
L*
a r e KB-
spaces. HINT:
(i) If
L
i s a band i n
i s a KB-space,
i t f o l l o w s e a s i l y from Theorem 114.7
.
(L*)* = (L)"" I n view of Lemma 114.2 t h e band nn n "-itself. Hence L = (L)&". Conversely, generated by L i n (L)zn- i s ( L )nn nn i f L i s a p e r f e c t Banach l a t t i c e and p i s o r d e r c o n t i n u o u s , t h e n p h a s
that
L
t h e weak Fatou p r o p e r t y f o r d i r e c t e d s e t s (Theorem 111.1). Hence, by Theorem
Ch. 16,01153
114.7, L
EMBEDDING I N BIDUALS
437
i s a KB-space.
115. Adjoint o p e r a t o r s We r e c a l l t h e main p o i n t s of what w a s observed i n s e c t i o n 97 about a d j o i n t o p e r a t o r s . A s b e f o r e , any l i n e a r mapping from one v e c t o r space i n t o
L
a n o t h e r v e c t o r space i s c a l l e d an o p e r a t o r . I f
L#
with a l g e b r a i c duals
and
then t h e a l g e b r a i c a d j o i n t d e f i n e d by
(T#$)(f)
t h e mapping
T
specially, L M
,
M
T
then t h e r e s t r i c t i o n
T"
M"
and
a r e v e c t o r spaces
M
i s a n o p e r a t o r from
i s an o p e r a t o r from
T
and a l l
f E L
M#
into
L
into
into
L"
of
to
Tff
L
i s order
T"
i s l i n e a r and p o s i t i v e . More p r e c i s e l y ,
0 : T + T"
Ln(M",L-)
h(L,M)
; we r e c a l l t h a t
N
T"
i s Is-order continuous o r o r d e r continuous, t h e n
into
is
Ln(M ,L )
t h e band of a l l o r d e r continuous and o r d e r bounded o p e r a t o r s from T
,
i s a n o r d e r bounded
M"
i s true; the operator
i s p o s i t i v e and maps t h e Dedekind complete Riesz space
. If
L#
. Note t h a t a T # . I f , more 2 2
$ E M#
i s an o r d e r bounded o p e r a t o r from
. Even more
t h e Dedekind complete Riesz space L"
,
M
(a T +a T ) # = a T# + 1 1 2 2 1 1 a r e Riesz s p a c e s s u c h t h a t M i s Dedekind complete
continuous. The mapping
Q
of
T
for a l l
if
o p e r a t o r from
and
is linear, i.e.,
T#
and
T#
$(Tf)
=
T E L b ( ~ , ~,) i . e . ,
and i f into
+
M#
M"
-
into
has s p e c i a l
p r o p e r t i e s . This was proved i n Theorem 9 7 . 1 , which we r e p e a t h e r e . ( i ) If T € L (L,M) b
i s Is-oPder continuous, then
( i i ) I f T € Lb(L,M)
i s order continuous, then
We s h a l l b e i n t e r e s t e d h e r e i n t h e r e s t r i c t i o n of t h e r e s t r i c t i o n of
to
T#
d i d i n s e c t i o n 9 7 ) , we have t i v e o p e r a t o r mapping ( i i ) above, T ' jince .'I
E
T'
Ln(
(L-):,
(M):;
.
T ' E Ln(Mi,L")
Lb(L,M)
and
Ln(Mi,L")
Ln(Mi,Li)
N
-f
T"
if
T
-- ,
Ln(Mn,L )
P: T
N
into Lc. N
maps M i i n t o Ln to + T'
.
M" ( e q u i v a l e n t l y , n
T'
( a s we
i s now a posi-
. Furthermore,
by remark
i s o r d e r continuous.
i s o r d e r continuous f o r any
LEMMA 115.1. The p o s i t i v e operator Y:
T"
maps M"
1. Denoting t h i s r e s t r i c t i o n by
L b ( ~ , ~ i)n t o
belongs t o
E Ln(Mi,L")
M"
T"
T E Lb(L,M)
, we
Hence, get
PROOF. L e t
T- C 8
,
be a downwards d i r e c t e d s e t i n
TT
+
t h a t i s t o say, T u
0 5 JI E M :
we have now
T h i s shows t h a t
T'
(g:JI)Cu)
f o r any
0
= JI(T,u)
J. 0
5
holds i n t h e space
8
J-
0
We have i n t r o d u c e d t h e r e s t r i c t i o n
. Of
. If
M
. For
,
so
satisfying
every
T'JI J. 0
holds i n
.
L"
.
to
MX
M;
i s not too small. P r e c i s e l y ,
T'
t h e s i t u a t i o n i s of maximal i n t e r e s t i f
Lb(L,M)
u E L
Ln(M:,L")
course, t h i s i s of i n t e r e s t only i f
r a t e s t h e p o i n t s of
16,§1151
, i s order continuous.
Y(T) = T '
defined by
T-
Ch.
ADJOINT OPERATORS
$38
(M ' ):
of t h e a d j o i n t o p e r a t o r
,
{O}
=
i.e.,
i f :M
sepa-
t h i s c o n d i t i o n i s s a t i s f i e d , t h e n (by Theorem
109.2) t h e i d e a l g e n e r a t e d by M i n M (:): under t h e n a t u r a l embedding i s S i n c e M i t s e l f i s Dedekind complete by hypoa Dedekind completion of M
.
again t h a t
T
i s an i d e a l i n
M
t h e s i s , t h i s means t h a t
f E L
a & observed above. We prove t h a t i f under t h e n a t u r a l embedding of
L
n a t u r a l embedding of
M
THEOREM 1 1 5 . 2 .
,
then
T"f"
i s t h e image of
. Then
and
L
'(Mi)
=
{O}
be t h e r e s t r i c t i o n t o M :
t h e p o s i t i v e operator
M
Tf
Y: T
-+
T'
(Mi):
C
M
.
under t h e
. Then
T"f" = (Tf)"
.
i s Dedekind complete and T'
a s i t s image
E
t h e p r o o f , l e t J, E M Z
( J . Synnatschke,[11,1972). Let
M
T 5 L b ( ~ , ~,) l e t T"
. For
l e t us assume
f" E (L"),
it follows t h a t n ' i s t h e image of Tf
o t h e r words, T"f"
operator
T"f" N
(Mn)n
$ E M"
Since t h i s holds f o r every
spaces such t h a t
N
into
N
(L ) n
NOW,
b e l o n g s t o Lfi((L'7:,(Q;)
has N
into
More p r e c i s e l y , we s h a l l prove t h a t
f o r any
.
M (:):
E Lb(L,M) , so t h a t , t h e r e f o r e , T"
. In
be Riesz
. Furthermore, of t h e a d j o i n t
i s a Riesz homomor-
phism. I n other words, we have ( T ~ V T ~ )=' T ; v
for a l l
T1
and
T2
T I
2
.
Lb(~,~)
in
PROOF. It i s s u f f i c i e n t t o p r o v e t h a t i f then Since
T'
1 8
A
T; = 8
5 S 5
Ti
in and
Ln(M;,L-) Ti
. For
T1
A
T2 = 8
t h e p r o o f , we w r i t e
in
Lb(L,M)
,
S = T' A T' 1 2 ' i s o r d e r continuous, t h e o p e r a t o r S i s a l s o
EMBBDDING IN BIDUALS
Ch. 16,51151
order continuous. Hence, S '
belongs to
image S f 0 (L )n "
in
N
Since M
. For any
L
"
,
N
(Mn)n
into M
So 5 T1
It follows that
A
, so
T2 = 6
.
So = 6
. Hence,
Sof E M
it follows that
, i.e.,
T' 1
Ti
A
is
So
5 So 2 Tk
=
.
T2
T1 and
PROOF. For every
for all
and
(5AT2 ) "
=
T"
1
A
and M
L
2
T E L (L,M)
the corresponding T '
b
belongs to
such that S 1
will imply S'
A
A
S;
=
8
if
(L")"
1 This is certainly the case; the subset of
of elements
0 E L"
f E L
S
2
= 8
,
then S;
E L , which implies 0
for all
n
n
M
S;
=
6
that T','
T" = 8' 2 S1 A S 2 = 8 A
A
Sq
, beT;
. By
=
the
in
N
f"
,
L"
then $(f)
(because = 0
for all
= 0 ).
Once more, let T E Lb(~,~) and assume that T Then T ' E Ln(M",L")
A
and
separates the points o f L". " , (L consisting of all images
already separates the points of
satisfies f"($) = 0
also that L and
S1
sufficient, therefore, to prove that if
theorem whibh has just been proved, the hypothesis that
if
have
Lb(~,~)
Lb(Mi,L")
f"
, we
T"
cause it will follow then in particular from T' A T; = 6 1 = 8 , and hence it will follow from T, A T2 = 8 that T',' Lb(Mi,LM)
E M" n'
$J
.
in
. It will be
are members of
S$ = 0
shows that
8.
(T vT ) " = T',' v T;' 1 2
Ln(Mi,L-)
. This
Under t h e same hypotheses about
COROLLARY 115.3.
f
the
E L
for k = 1 , 2 = Sof = 0 S'f" But then
and satisfying 8
and all $ E MX
f E L
S = 8
fur a22
+
f E L , which implies
for every
so
0
L
, where f" is the image of f
= S'f"
it
f E L we have
is an ideal in
an operator mapping
for all
S f
is defined by
. This makes """ ( M , ) , ; for any f E L
Ln((L")",(M")")
So:
possible to define a positive operator
439
and, therefore, ):T" ):M(,:)E:L((nL
is order continuous.
. Assuming now
are perfect, the space L may be identified with
A D J O I N T OPERATORS
440
and t h e space
;):L( any
=
,
T"f"
identify
f " F: L (:)
t h e image
(Tf)"
i t follows t h a t i f we i d e n t i f y
T f and
. Hence,
M (:):
may b e i d e n t i f i e d w i t h
M
w i t h image
f E L
(Tf)"
Ch. 16,§1151
f
, but
T
of Theorem 115.2 (and C o r o l l a r y 115.3) i t occurs f o r
i s a r b i t r a r y , then
T E Lb(L,M)
and :M
N
L
respectively).
f"
Tf
satisfies
,
t h e n we must
T"f" as w e l l . I n o t h e r words, we i d e n t i f y
I n g e n e r a l , t h i s s i t u a t i o n does n o t occur f o r
the spaces
of
and
T ' E Ln(M:,L")
since f o r
and
T
T"
.
under t h e c o n d i t i o n s
. Indeed,
T'
if
i s o r d e r continuous and
a r e p e r f e c t (by Theorem 110.2 and Theorem 110.3
It follows t h a t
and
T'
may b e i d e n t i f i e d . This
T"'
i m p l i e s immediately t h a t (T1vT2)"' = TY' v T;" for a l l
T1
and
in
T2
and
Lb(L,M)
115.2 a r e s a t i s f i e d ( i . e . , M
(T,hT2)"' = T',''
, provided
A
Ti'
t h e c o n d i t i o n s f o r Theorem
Dedekind complete and
'(Mi)
) . Similar-
= {O]
l y f o r a l l higher adjoints. If
M
i s Dedekind complete b u t
Ln(Mz,L-)
T1
and
T2
applied t o
with
L"
a r e a r b i t r a r y members of
T' v T '
and
(T'vT')' 1 2
=
1
2
i s n o t s a t i s f i e d , then
= {O}
T E L (L,M) t h a t t h e corresponding T' belongs b Dedekind complete and '((L"):) = {O} Hence, i f
i t remains t r u e f o r any to
(M ' ):
T' 1
T','
THEOREM 115.4. Let
A
v T"
2
L
T' 2
, showing
and
and M
,
Lb(L,M)
(T'hT')' 1 2
.
t h e n Theorem 115.2 may be
that = T','
A
TY
.
be Banach l a t t i c e s ( w i t h norms
p!
and
X r e s p e c t i v e l y ) such t h a t T E L b ( ~ , ~,) then
T'
X i s order continuous. A s before, i f is t h e r e s t r i c t i o n t o Mi; of t h e order a d j o i n t
and T* i s t h e Banach a d j o i n t of T (note t h a t T i s norm bounded since any p o s i t i v e operator on a Banach l a t t i c e i s norm bounded). Then
T"
T*
=
T'
and, for a17. T1
PROOF. Note t h a t T E Lb(L,M)
. Note
and
in
Lb(L,M)
, we have
.
and M" = M* = M* Hence T* = T ' f o r every n '(MT;' = O(M*) = ( 0 ) , and M i s Dedekind
L" = L*
now t h a t
T2
complete (Theorem 103.9),so Theorem 115.2 may b e a p p l i e d t o
Ti
=
T;
.
Ti
=
TT
and
Ch. 16,11161
EMBEDDING I N BZDUALS
44 1
The r e a d e r is a d v i s e d t o compare t h e r e s u l t s i n t h e p r e s e n t s e c t i o n w i t h t h o s e i n E x e r c i s e 97.5, where we have t h e s p e c i a l c a s e t h a t
ind
L
are
M
spaces of measurable f u n c t i o n s .
L
EXERCISE 115.5. Once more, l e t
i s Dedekind complete and
M
O(M;)
under t h e n a t u r a l embedding of
=
=
f"
f"
t h e space
. It
M
,
L
b e Riesz spaces such t h a t image of an element i s denoted by
(L"):
.
f"
L"
seems somewhat s u r p r i s i n g a t f i r s t , t h e r e f o r e , t h a t i n
. Ex-
f " = f " i m p l i e s Tfl = Tf2 f o r every T E'Lb(L,M) 1 2 p l a i n t h i s . Also e x p l a i n t h a t f " = f " does not n e c e s s a r i l y imply
if
If
has n o t h i n g t o do w i t h
s p i t e of t h i s
Tfl = Tf2
f E L
then it i s very well possible t h a t
. This phenomenon
f l # f2
holds f o r points
. The
into
L
does n o t s e p a r a t e t h e p o i n t s of
and M
{O}
('M;)
{ O } does n o t hold.
=
v = f -f s a t i s f i e s v" = f',' - f; = 0 , t h e n v E '(L") . 1 2 T'JI E L" f o r every J, E M" , t h i s i m p l i e s t h a t (T'J,)(v) = 0 , s o
HINT: I f
Since
J,(Tv) = 0 Tfl = Tf2 L"
=
M"
. It . If
f o l l o w s t h e n from (M "' ): M (' ):
= {O}
and w i t h
= {O}
T
=
{O}
Tv = 0
that
i s not r e q u i r e d , we can choose
,
i.e.,
L = M
with
the i d e n t i t y operator.
EXERCISE 115.6. I n Lemma 115.1 i t w a s shown t h a t t h e p o s i t i v e o p e r a t o r 'Y
, defined
by
J . Synnatschke
Y(T) = T ' , i s o r d e r continuous. It w a s observed by ( [ I 1 ) t h a t t h e p o s i t i v e o p e r a t o r Q , d e f i n e d by Q(T)
i s n o t even a-order =
fl,
n for
continuous i n g e n e r a l . Show t h i s by choosing
and d e f i n i n g t h e o p e r a t o r s
f = (f, ,f7,.
...)
(n=I,Z,
T",
L = M =
by
...,fn,O,O ,...
. .) E
. Then
i d e n t i t y o p e r a t o r . Note now t h a t n ( E x e r c i s e 103.16)
T
=
.
Tn 4 I
in
Lb(Lm,Lm)
TZ$ = 0 f o r a n y
, where
4 E (La):
I
i s the
and f o r
116. D i s j o i n t sequences and o r d e r continuous norms I n t h e p r e s e n t s e c t i o n we assume t h a t
The Banach d u a l t h e space
L*
L* (norm
p*)
L
i s a Banach l a t t i c e (norm P >
and t h e o r d e r d u a l
L-
a r e now i d e n t i c a l ;
i s t h e r e f o r e a Dedekind complete Banach l a t t i c e . As proved
DISJOINT SEQUENCES AND ORDER CONTINUOUS NORMS
442
Ch. 16,11163
i n Theorem 104.2 ( a s an immediate consequence of Meyer-Nieberg's t h e norm
in
p
i s o r d e r continuous i f and only i f every o r d e r bounded
L
d i s j o i n t sequence i n thus t h a t t h e norm
converges t o z e r o i n norm. Applied t o
L
in
p*
o r d e r bounded d i s j o i n t sequence i n
Fremlin ([11,1979)
that order
a l s o e q u i v a l e n t t o a c e r t a i n p r o p e r t y of d i s j o i n t sequences i n l y s t a t e d , t h e norm
in
p*
get
converges t o z e r o i n norm. It was
L*
Dodds and D.H.
i s n o t only r e l a t e d t o d i s j o i n t sequences i n
p*
, we
L*
i s o r d e r continuous i f and only i f every
L*
proved now r e c e n t l y by P.G. c o n t i n u i t y of
lemma),
but is
L*
. Precise-
L
i s o r d e r continuous i f and only i f every
L*
norm bounded d i s j o i n t sequence i n
L
converges weakly t o z e r o . We p r e s e n t
t h e proof. THEOREM 116.1. Let
1 . Then
L* (norm p*
be a Banach l a t t i c e (norm
L
1 w i t h Banach dual
p
i s order continuous i f and only i f every norm
p*
bounded d i s j o i n t sequence i n
converges weakly t o zero.
L
PROOF. ( i ) Assume f i r s t t h a t
p*
i s o r d e r continuous. For t h e proof
t h a t every norm bounded d i s j o i n t sequence in
L
converges weakly t o z e r o
i t i s s u f f i c i e n t t o c o n s i d e r p o s i t i v e d i s j o i n t sequences of norm a t most
one. Hence, l e t and
p(un)
prove t h a t
1
5
0
u E L ( n = l , 2 , ...) w i t h a l l un mutually d i s j o i n t n for all n Furthermore, l e t 0 5 E L* We have t o
+(u )
S
+
.
0
as
n t h e Banach b i d u a l t h e space
i m p l i e s t h a t t h e sequence
n
+
L
m
. Under
becomes a Riesz subspace of
(un:n=1,2,
...)
carriers
C(u )
. It
L*
a r e mutually d i s j o i n t bands i n
+
ed d i s j o i n t sequence i n
in
C(un) and
L*
L*
(Qn:n=1,2,
u (+) = un(+ )
n
. Since
n
. Let ...)
+n b e t h e comi s an o r d e r bound-
f o r every
n
, i.e.,
i s o r d e r continuous and t h e
f o r every
sequence
i s o r d e r bounded and d i s j o i n t , we have
p*
. This
(L*):
follows t h a t t h e
$(un) = $n(un) (9,)
n
. Then
into
L
i s a d i s j o i n t sequence of posi-
t i v e o r d e r continuous l i n e a r f u n c t i o n a l s on ponent of t h e given
.
+
t h e n a t u r a l embedding of
p*(+,)
+
0
(by
Theorem 104.2). Hence
( i i ) Assume now t h a t e v e r y norm bounded d i s j o i n t sequence i n v e r g e s weakly t o z e r o . For t h e proof t h a t
p*
L
con-
i s o r d e r continuous, i t i s
s u f f i c i e n t t o show (by Thbocem 104.2 a g a i n ) t h a t
p*(+,)
+ 0
f o r every d i s -
Ch. 16,81161
EMBEDDING I N BIDUALS
j Q i n t sequence
($n)
in
satisfying
L*
Assuming t h i s t o b e f a l s e , l e t p*($,)
>
0
E >
f o r some
p o s i t i v e element m
$n(u)
C,
5 $(u) <
E
such t h a t
E L
mn
f o r every
for a l l
5 $ E L*
$
n.
b e such a sequence s a t i s f y i n g n and a l l n For every n t h e r e e x i s t s a
($ )
> 0
u
s
0
443
.
p(un) 5 1
, we
u E L+
and
$n(un) >
. Since
E
may assume (by passing t o a
subsequence, i f n e c e s s a r y ) t h a t
For
n = 2,3,
...
,
let
0 5 v 5 u f o r every n and (vn:n=1,2, ...) i s a d i s j o i n t sequenn n e x i s t s s i n c e L i s Banach. The d i s j o i n t n e s s follows ce. Note t h a t Z2-ky,
Then
from v n 5 ( ~ ~ urn)+ - 2 ~ and
vm
5
2-"(2"
=
E
'+
um-un,
for
m < n
.
Furthermore,
n 2 E
for
n
-
-
'
n
m
n+l
2-k P*($)
s u f i f i c i e n t l y l a r g e . Hence
-
&€
$(vn) > 1s
-
2-n
for
' 1E
P*($)
n
s u f f i c i e n t l y large.
.
But (v ) i s a norm bounded d i s j o i n t sequence i n L T h e r e f o r e , $(vn) n by h y p o t h e s i s . C o n t r a d i c t i o n . It f o l l o w s t h a t p* i s o r d e r continuous.
A s a c o r o l l a r y we f i n d t h a t t h e norm
in
p**
case
(0,)
E L+
~ " ( 4 ~ + )0
f o r every
converges a l s o weak star t e z e r o , i . e . ,
. Hence,
if
d i s j o i n t sequence i n
p**
L*
i s o r d e r continuous and
,
then
$,(u)
+
0
v e r s e does n o t h o l d , By way of example, l e t L**
Then
=
em . For Qn(u)
+
n = l,Z, 0
...
f o r every
,
let
u E L+
On
(0,)
in
0 s u" € L** $,(u) ($n)
f o r every L = (c ) 0
+
0
p**
con-
L*
. In
this
f o r every
i s a norm bounded u E L+
, so
. The
L* =
be t h e n-th u n i t v e c t o r i n
, but
0
i s o r d e r continu-
L**
ous i f and only i f everynorm bounded d i s j o i n t sequence verges weakly t o z e r o , i . e . ,
+
e,
conand L* =
( t h e Lm-norm) i s n o t o r d e r
Ll
.
444
D I S J O I N T SEQUENCES AND ORDER CONTINUOUS NORMS
16,81161
Ch.
continuous. It i s n a t u r a l , t h e r e f o r e , t o ask what c o n c l u s i o n ( i f any) can be drawn from t h e h y p o t h e s i s t h a t every norm bounded d i s j o i n t sequence i n L*
converges weak s t a r t o zero. The n a t u r a l c o n j e c t u r e i s norm c o n t i n u i t y
of
p
. We
prove t h e c o n j e c t u r e t o b e c o r r e c t . The proof i s s i m i l a r t o ,
b u t s i m p l e r t h a n t h e proof of t h e l a s t theorem. THEOREM 116.2. A s before, l e t L* (norm
Banach dual
L
p* ). Then
be a Banach l a t t i c e (norm L*
every norm bounded d i s j o i n t seqttence i n PROOF. ( i ) Assume f i r s t t h a t Theorem 103.9, t h e space band i n of
0
$
5
n for a l l
as
n
C($n)
-. n
i s o r d e r continuous. Note t h a t , by
p
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
...) w i t h a l l . Furthermore, l e t 0 Since a l l in
u
. Then
and
L
$,(u)
o r d e r continuous and t h e sequence p(un)
+
0
. We have
u E L
5 1
p*($,)
t o prove t h a t
$ (u)
n
-+
0
a r e o r d e r continuous and d i s j o i n t , t h e c a r r i e r s
$n C($n)
d i s j o i n t sequence i n
i s o r d e r continuous. Let
L
mutually d i s j o i n t and
$* I
a r e mutually d i s j o i n t bands i n
given element
we have
converges weak s t a r t o zero.
i s now Dedekind complete (and, t h e r e f o r e , every
L
E L* (n=l,2,
+
with
i s a p r o j e c t i o n band). Also, on account of t h e o r d e r c o n t i n u i t y
L
, every
p
p )
i s order continuous i f and only i f
p
. Let
L
(un:n=l,2, =
gn(un)
be t h e component of t h e
u n
...)
i s a n o r d e r bounded
f o r every
n
. Since
is
p
i s o r d e r bounded and d i s j o i n t
(un)
(by Theorem 104.2). Hence
( i i ) Assume, c o n v e r s e l y , t h a t every norm bounded d i s j o i n t sequence in
converges weak s t a r t o zero. For t h e proof t h a t
L*
t i n u o u s , i t i f s u f f i c i e n t (by Theorem 104.2) t o show t h a t every d i s j o i n t sequence
n
. Assuming
p(un) >
E
(un)
in
t h i s t o be f a l s e , l e t
> 0
for a l l
n
. The
0
n
sequence
. By
L*
n
n
) 5 1
n and, t h e r e f o r e ,
hypothesis
2 $ (u ) = un(+,)
p*($
> E
(4), for a l l
u
p(u,)
L*
, which
for
for a l l
and (0,)
L*
u ~ ( $ ~>) E
implies that the
. For
every
. It may b e
n
, let
assumed
i s a norm bounded d i s j o i n t sequence
converges weak s t a r t o z e r o . But
n
+ 0
5 u E L
i s then a d i s j o i n t sequence
(u,)
a r e mutually d i s j o i n t bands i n
satisfy
$n E C(un)
that in
C(u )
$n E L*
I
I
i s o r d e r con-
(un) be such a sequence s a t i s f y i n g
of o r d e r continuous l i n e a r f u n c t i o n a l s on carriers
0
satisfying
L
p
. Contradiction.
$,(u)
2
Ch. 16,11161
EMBEDDING I N BIDUALS
445
I n t h e s e p r o o f s we have e s t a b l i s h e d s e v e r a l r e l a e i o n s between d i s j o i n t sequences i n
L* .It i s worth w h i l e t o l i s t t h e s e s e p a r a t e l y a s a
and
L
theorem f o r l a t e r r e f e r e n c e . Note how much h e a v i e r i t i s t o o b t a i n d i s j o i n t sequences i n
than i n
L
L*
.
THEOREM 116.3. Once more, l e t L* i n o m
Banach dual (i)
If (un)
that
i s a d i s j o i n t sequence i n L+
=
(and hence also i n
i s given, then the components $n of $ i n form an order bounded d i s j o i n t sequence i n (L*)+ such
C(u )
$,(u,) (ii)
with
p )
0 s $ E L*
(L**)+ ) and i f
the carriers
be a Banach l a t t i c e (norm
L
J.
p*
f o r all n and
$(un)
Assume
$,(urn) = 0 L
i n addition that
for m # n
.
i s DeJekind complete. Then, i f
i s a d i s j o i n t sequence of order continuous elements i n (L*)+ and i s given, the components un of u i n the carriers ~ ( 4 ~ ) form an order bounded d i s j o i n t sequence i n L+ such that $ n ( ~ n )= $n(u) ($n)
i f
0 s u E L
for a l l
n
> 0
$,(urn) = 0 f o r
and
(iii) I f
f o r all n
m
.
# n
i s a d i s j o i n t sequence i n L+
(un)
satisfying
, then there e x i s t s a d i s j o i n t sequence
($n)
p(u ) >
E
n of positive
>
elements i n the u n i t b a l l of $ (u ) = 0
*n zn
2.
for
m
.
# n
p(un) = 1
with
L* such that On(un) > E f o r a l l n and In particular, i f (u,) i s a d i s j o i n t sequence for all n , and i f E > 0 i s given, there e x i s t s
of p o s i t i v e elements i n the u n i t b a l l of L* for a12 n and $,(urn) = 0 for m # n
(0,)
a d i s j o i n t sequence such that
I--E
< $,(u
(iv)
If
($n)
n
)
.
1
5
i s an order bounded d i s j o i n t sequence i n
satis-
(L*)+
fying p*($,) t E > 0 f o r a l l n , and i f 0 < 6 < E , there e x i s t s natural numbers n , ,n2 ,... and a d i s j o i n t sequence (un,,un2 ) of p o s i t i v e elements i n t h e u n i t b a l l of L such t h a t $ ( ) > 6 f o r k = 1,2,
,...
with
p*(On) = 1
j o i n t sequence that I-E < $
f o r a12
(x) )
% %
PROOF.
(ii)
i s an order bounded d i s j o i n t sequence i n n
, and
i f
E
> 0
5
1
for k = I,Z,,..
(L*)+
i s given, there e z i s t s a dis-
of positive elements in the u n i t b a l l of
(U
... .
"k unk
(9,)
I n particular, i f
.
L
such
( i ) This was proved i n p a r t ( i ) of t h e proof of Theorem 116.1.
On account of t h e a d d i t i o n a l c o n d i t i o n on
L
and s i n c e a l l
a r e o r d e r c o n t i n u o u s , t h e proof of p a r t ( i ) can b e taken o v e r ( a s was a c t u a l l y done i n t h e proof of Theorem
116.2).
( i i i ) This was proved i n p a r t ( i i ) of t h e proof of Theorem 116.2. (iv)
This was proved i n p a r t ( i i ) of t h e proof of Theorem 116.1.
+tl
446
COPIES
EXERCISE 116.4. Show t h a t i f
Ch. 16,11171
i s a normed Riesz space b u t n o t a
L
Banach l a t t i c e , t h e n i t may occur t h a t every norm bounded d i s j o i n t sequence converges weak s t a r t o z e r o , b u t t h e norm i n
in L*
L
is not o r d e r
continuous. HINT: Let
satisfying
b e t h e space
L
1 5 p <
EXERCISE 116.5. remains t r u e i f
C([O,ll) w i t h L -norm f o r some P ( c f . a l s o Exercise 114.10).
m
p
Show t h a t t h e s t a t e m e n t i n p a r t ( i i ) of Theorem 116.3
i s only Dedekind o-complete.
L
H I N T : Use Theorem 90.4.
117. Copies Let
L
and
P
LX b e normed Riesz spaces (norms p
b e a Riesz subspace of t h e norms
and
p
X
L
in
been c a r r i e d over from K
and
LA
a copy of space
. If
respect t o
LA
and A ) and l e t
to
LX
A
has
by means of t h e Riesz isomorphism between
K
the case t h a t
K
a r e Riesz isomorphic and
a r e e q u i v a l e n t . Note h e r e t h a t t h e norm
K
L
i s Banach w i t h r e s p e c t t o
K
and
K
t h i s s i t u a t i o n o c c u r s , we s h a l l s a y t h a t
. In
LA
P
such t h a t
and
P
LX
contains P a r e Banach l a t t i c e s , t h e L
(by t h e isomorphism) a s w e l l a s w i t h
A
p (because t h e norms a r e e q u i v a l e n t ) . We s h a l l d i s c u s s s e v e r a l
c o n d i t i o n s under which a Banach l a t t i c e o r i t s d u a l c o n t a i n s ( o r f a i l s t o c o n t a i n ) a copy of THEOREM 1 1 7 . 1 . p
copy of
in L
P
(c,)
or
.
Lw
L contains a copy of (c,) such P i s order bounded i n L if and only i f the
The Bamch l a t t i c e
that the u n i t b a l l of norm
, Ll
(c,)
(c,)
P
i s not order continuous. I n particular, L* contains a i f and onZy if the rwm
p*
i n L* P
P
i s not order continuous.
PROOF. A s proved i n Theorem 104.2, t h e norm p i n L is order P continuous i f and only i f e v e r y o r d e r bounded d i s j o i n t sequence i n L tends t o z e r o i n norm. I n t h i s case i t i s e v i d e n t , t h e r e f o r e , t h a t cannot c o n t a i n a copy of bounded i n
L
type
p
then
P
. In
(c,)
such t h a t t h e u n i t b a l l of
o t h e r words, i f
L
P
c o n t a i n s a copy
(c,) of
(c,)
L
P P
i s order of t h i s
i s n o t o r d e r continuous. For t h e proof i n t h e converse
d i r e c t i o n , we assume t h a t
p
i s n o t o r d e r continuous. Then (once more by
Ch. 16,51171
447
EMBEDDING In BIDUALS
Theorem 104.2) t h e r e e x i s t s an o r d e r bounded d i s j o i n t sequence L
P
5
such t h a t u
E L
P
p(un) = 1
and
(u,)
in
0 5 u 5 n are arbitrary real
does not tend t o zero. We may assume t h a t
p(un)
for a l l
.
n
If
a l ,...,a
n
numbers, then
I t follows t h a t i f
,
( a l , a2,...) E (c,)
then
...+anun :n=1,2,... )
( s =aIuI+
n
)
i s a p-Cauchy sequence i n
L
P i n Banach space theory . ( i f a l l
m
Z 1 aiui
; we w r i t e the l i m i t as
a r e non-negative,
a.
, as
usual
t h i s i s t h e r e f o r e the
supremum of t h e p a r t i a l sums). Note t h a t
...
It i s e v i d e n t now t h a t t h e mapping of
(c,)
in
L
P (c,)
u n i t b a l l of
such t h a t , on account of
i s order bounded i n
F i n a l l y , f o r the r e s u l t about of L*
(c,)
. To
P u E L* P
,
(al,a2, )
then the u n i t b a l l of
L* P (c,)
L
m
+ C1
0 5 u
.
P
, note
n
for a l l
5 u
that i f
generates a copy
aiui
n
, the
contains a copy
L*
P
i s automatically o r d e r bounded i n
s e e t h i s , i t i s s u f f i c i e n t t o prove t h e e x i s t e n c e of an element such t h a t
C
W
I
a.u. 1
1
I
u
holds f o r a l l
(i=1,2 ,...)
0 5 a. 5 1
.
s = :1 u. , the element n 1 1 L* The ( c )-norm of every s i s equal t o one, so P 0 n 5 A f o r some constant A z 0 and a l l n But then, s i n c e
Hence, i t i s s u f f i c i e n t t o show t h a t , w r i t i n g sup s we have
L* P
.
exists in p(sn)
.
i s p e r f e c t (Corollary 111.2),
sup p ( s ) 5 A n
that
sup s
n
exists in
The next theorem d e a l s w i t h result that Copy of
La
i t follows from L*
P
n
I.
and
together. It contains t h e P L contains a copy of L l i f and only i f L* contains a P P The theorem i s due t o B. Kihn ( [ 1 1 , 1 9 7 9 ) . The proof a s presenL
P
and
.
0 5 s
L*
.
t e d h e r e (due t o A.R. except f o r ( i i i )
*
Schep) i s d i f f e r e n t from the proof i n Kihn’s paper,
(iv).
448
Ch. 1 6 , 8 1 1 7 1
COPIES
THEOREM 1 1 7 . 2 . The following conditions f o r t h e Banach l a t t i c e
are equivalent. (i)
L*
contains a copy of
(ii)
L*
contains a copy of
(c ) 0
( i i i ) The nomi i n (iv)
=5
P
.
L* i s not order continuous. contains a copy of b,
L
.
PROOF. We s h a l l prove ( i )
(i)
.
b,
L = L
( t i ) For
n = 1,2,
*
...
(ii)
,
=$
let
(iii) e
*
(iv)
*
*
(iii)
be the element of
(i). (c,)
with
the n-th coordinate equal t o one and a l l o t h e r coordinates zero. Let be
$n under t h e isomorphism. Then n i s a d i s j o i n t sequence of p o s i t i v e elements i n L* and on account
t h e element of
($n)
corresponding t o
L*
e
of t h e norm equivalence t h e r e e x i s t p o s i t i v e c o n s t a n t s
and
A
such
B
that
for a l l NOW,
n = 1,2 all L*
s
.
a = (al,a2
,...)
,... , we have . Hence, s i n c e
E
L:
particular, A 5 p*($
. men,
writing
n
) 5 B
= al$l
s
+
for a l l
... +
n
an$n
. for
0 5 s I. and p * ( s ) 5 B sup(la.1) = B . l l a l l m f o r n n i s p e r f e c t , .the element s = sup s exists in n Each p o s i t i v e element a i n b, generates thus a p o s i t i v e element
n
= s(a)
in
additive, so into
. In
...) E (c,)
(a,,.,,
let
L*
L*
L*
.
S
extends uniquely t o a p o s i t i v e l i n e a r mapping from
. Note
It i s e a s i l y seen t h a t the mapping
t h a t , f o r every
a = ( a l , a 2 , . ..) Eb,
S: a
, we
-t
is
s(a)
b,
have
(where we use t h a t 0 < s I. s implies p * ( s n ) I. p * ( s ) ) . Furthermore, S n preserves d i s j o i n t n e s s , s o t h a t , t h e r e f o r e , S i s a Riesz homomorphism. Combining these f a c t s , S and t h e L,-norm
contains a copy of (ii)
*
i s seen t o be a Riesz isomorphism such t h a t
a r e equivalent i n the range of
( i i i ) If
bm
.
L* contains a copy of
S
. This
,
there e x i s t s i n
shows t h a t
p*
L* L*
an
o r d e r bounded d i s j o i n t sequence of elements of norm one. Hence, by Theorem 104.2 again, the norm
p*
i s not o r d e r continuous.
(iii) * ( i v ) Since the norm
p*
i s not o r d e r continuous, t h e r e e x i s t s
(by Theorem 116.1) a norm bounded d i s j o i n t sequence (u : n = l , 2 ,
n
...)
of
Ch. 16,11171
EMBEDDING I N BIDUALS
p o s i t i v e elements i n 0 2 $ E L*
which does not converge weakly t o zero. Hence,
L
we may assume t h a t
p(un)
and some
the l i n e a r mapping
L,
(the notation
we have
+
T
C
T(a) = E l aiui
. Now
n
for a l l
a
=
define
(a1,"*,...)€
m
Banach, i t follows t h a t such t h a t
B > 0
i s norm bounded, i . e . ,
T
p{T(a)} 2 B l l a l l l
Hence, p that
and t h e
t h e r e e x i s t s a constant
. For
a € L,
. Then
an e s t i m a t e i n
L 1-norm a r e equivalent i n the range of L, .
. It
T
follows
contains a copy of
L
*
(iv)
el
( i i i ) There e x i s t s i n
a norm bounded d i s j o i n t sequence
of p o s i t i v e elements and t h e r e e x i s t s a p o s i t i v e l i n e a r f u n c t i o n a l
(en)
E
=
km such t h a t
contains a copy
L
for a l l
a E kl
t h e converse d i r e c t i o n , l e t
K
+(en) = 1
,
of
for a l l
a p o s i t i v e l i n e a r functional and
lJ,(f)l 2 Ap(f)
on
$
i s a Riesz subspace of
L
, so
$
on
for a l l
(and hence i n
K
L ) a
of p o s i t i v e elements and t h e r e e x i s t s K
f o r some constant
bounded l i n e a r f u n c t i o n a l $(un) = 1
n .Hence, s i n c e by hypothesis
there e x i s t s i n
norm bounded d i s j o i n t sequence (u,)
K
for a l l
E
m
by
L
and a l s o t h a t f o r some
n
$(un) >
a.u i s i n t h e sense of norm convergence). Then 1 i i i s a Riesz isomorphism and, t h e r e f o r e , T i s p o s i t i v e . Since L i s
E k,
)I
for a l l
1
S
> 0
E
T:
449
J,
L
such t h a t A > 0
J,(un) = I
and a l l
for a l l
f E K
n
. The copy
can be extended t o a p o s i t i v e norm (Theorem 85.5). It follows from
n t h a t t h e norm bounded sequence
(un)
converge weakly t o zero. Hence, by Theorem 116.1, t h e norm
in
L
does not
p*
i s not
order continuous. (iii)
*
( i ) By hypothesis, t h e norm
ous. Hence, by Theorem I 1 7.1,
p*
in
L*
contains a copy of
L*
i s not o r d e r continu(c,)
.
In t h e f i r s t theorem of t h e p r e s e n t s e c t i o n we have proved t h a t the norm p
of
in (c,)
L
f a i l s t o be o r d e r continuous i f and only i f such t h a t t h e u n i t b a l l of
(c,)
L
contains a copy
i s o r d e r bounded i n
L
second theorem we have seen t h a t i f a s i m i l a r s i t u a t i o n occurs i n p*
not o r d e r continuous), then
L*
contains n o t only a copy of
. In
the
L* ( i . e . ,
(c,)
, but
COPIES
450
one df i.e.,
La
a s w e l l . I t may be asked, t h e r e f o r e , i f t h e same h o l d s i n
i f t h e hypothesis t h a t
La
c o n t a i n s a Copy of
L
Ch. 16,§1171
,
f a i l s t o be o r d e r continuous i m p l i e s t h a t
p
. The
L
answer i s n e g a t i v e . The supremum norm i n
t h e Banach l a t t i c e ( c ) of a l l ( r e a l ) convergent sequences i s n o t o r d e r continuous. Therefore, ( c ) c o n t a i n s a copy of (c ) w i t h o r d e r bounded 0 I f L i s a Dedekind ( c ) does n o t c o n t a i n a copy of La
.
unit b a l l , but
a-complete Banach l a t t i c e , t h e s i t u a t i o n becomes d i f f e r e n t , a s shown by t h e f o l l o w i n g theorem. THEOREM 117.3.
L
The Dedekind a-complete Banach l a t t i c e
Lm if and only if t h e norm
a copy o f
tinkous. Note t h a t f o r
contains
in L fails t o be order con-
p
t h i s simply means t h a t conditions ( i i ) and
L*
( i i i ) in t h e preceding theorem are equivalent. PROOF. I t i s e v i d e n t t h a t i f
L
La ,
c o n t a i n s a copy of
then
L
c o n t a i n s an o r d e r bounded d i s j o i n t sequence of elements of norm one. Hence, by Theorem 104.2, t h e norm
p
i s n o t o r d e r continuous. For t h e converse,
assume t h a t i n t h e Dedekind a-complete space continuous. We have t o prove t h a t
L
L
t h e norm
c o n t a i n s a copy of
p
La
i s not order
. As
i n the
proof of Theorem 1 1 7 . 1 , t h e r e e x i s t s an o r d e r bounded d i s j o i n t sequence (u,)
in
Cn u.
< u
1
1
f o r each
0 < u
such t h a t
L
for a l l
n
and
L
. The
...) E
(al,a2,
t o some element of
E 1- and t h e mapping phism from
La
f i n i t e sum
lalIul +
into
we f i n d t h a t
Furthermore, s i n c e
L:
S:
L
< u E L and
p(un) = 1
i s Dedekind o-complete,
t h e p a r t i a l sums same h o l d s
( a l , a 2 ...) ,
C y aiui
for a l l
m
L . Since p ( C a.u.) = 1 1 ... + l a n / u n s a t i s f i e s
p(C
. Since
converge i n o r d e r
then f o r an a r b i t r a r y Z 1 aiui
n
i t i s evident t h a t
...) E
(al,a2,
d e f i n e s a Riesz isomorlailui)
and s i n c e any
EMBEDDING I N BIDUALS
Ch. 16,51171
45 1
we g e t
It f o l l o w s t h a t
mapping
S
. This
and t h e La-norm aye e q u i v a l e n t i n l t h e range of t h e
p
shows t h a t
.
c o n t a i n s a copy of
L
We have s e e n i n t h e f i r s t theorem of t h e p r e s e n t s e c t i o n t h a t i n a Banach l a t t i c e
L
o r d e r c o n t i n u i t y of t h e norm
sufficient for
L
n o t t o c o n t a i n a copy of
o r d e r bounded i n
(c,)
. There
L
L
n o t to c o n t a i n a copy of
c a n t i n u i t y of
p
alone i s not s u f f i c i e n t .
THEOREM 117.4. The Banach L a t t i c e (c,)
L
i f and only i f
L
i s a band i n p
w i t h t h e u n i t b a l l of
i s a l s o a simple n e c e s s a r y and s u f f i c i e n t
condition f o r
114.71, i f and only i f
i s n e c e s s a r y and
p
(c,)
L**
(c,)
a t a l l . Obviously, o r d e r
does not contain a copy of
, t h a t i s t o say (by Theorem L
i s order continuous and
has t h e weak
sequential Fatou property. PROOF. Assume f i r s t t h a t
L
does n o t c o n t a i n a copy of
as observed above, i t follows from Theorem 1 1 7 . 1 t h a t tinuous. Therefore, since
i s s u p e r Dedekind complete.
L
Hence, by Theorem 114.5, we s e e a l r e a d y t h a t L
i s o r d e r dense i n
It remains t o prove t h a t
i s an i d e a l i n
L
.
(L*);
m
such t h a t
i s n o t p-Cauchy.
(u,)
such
L**
h a s t h e weak s e q u e n t i a l Fatou p r o p e r t y .
L
Assume, f o r t h i s purpose, t h a t we have a sequence sup p(un) <
i s o r d e r con-
p
i s now a Banach l a t t i c e w i t h o r d e r continuous
L
norm, we d e r i v e from Theorem 103.9 t h a t that
. Then,
(c,)
+
0 5 u i n L with n It i s e v i d e n t t h a t
w" = sup u e x i s t s i n L** , simply by d e f i n i n g t h a t ~ " ( 0 )= sup g(un) n is perfect). f o r every 0 2 L$ E L* ( a l t e r n a t i v e l y , by observing t h a t L** Since W"
0 5 u E L c (L*);E n
for
E (L*)E , which shows t h a t
t o t h e non-p-Cauchy
sequence
i t may b e assumed t h a t Setting
fn = u p**(L;
~ -+ un~
n = l,Z,
0 5 u (un)
n
.
~(U,+~-U,) >
, we
have
... and
4 w"
i s a band, we g e t
(L*)E
holds i n
(L*)E
. We
return
P a s s i n g t o a subsequence i f n e c e s s a r y , E
f o r some
0 5 f 2
n
w"
f k ) = p**\un+l -u 1) 2 p**(w")
E
> 0
and a l l
n
as well a s and
p**(fn) >
6
.
COPIES
452
for a l l
. Hence,
n
t h e sequence
(f ) n
Ch. 1 6 , § 1 1 7 3
lemma (Lemma 104.1) t o
we may apply Meyer-Nieberg's t o o b t a i n a d i s j o i n t sequence
0 5 v" 5 w"
(vi)
.
in
such
(L*):
p**(v") > E f o r a l l n For each n t h e r e n = 1 and v"($ ) > E e x i s t s a p o s i t i v e $n E L* s a t i s f y i n g p*'($,) n n Furthermore, s i n c e L i s o r d e r dense i n (L*): , t h e r e e x i s t s f o r each that
n
an element
@,(zn) > Since
.
E
and
such t h a t 0 i z 5 v" and z ~ ( $ >~ )E , i . e . , n n n On account of p * ( $ ) = 1 i t follows t h a t p(zn) > E n
z E L
i s a d i s j o i n t sequence, so i s
(v;)
j o i n t sequence i n If
...,an
al ,
L
as well as i n
= alzl + n sequence i n
.
. Hence,
(2,)
, majorized
(L*):
i s a dis-
(2,)
by
w"
in
(L*):
.
a r e a r b i t r a r y r e a l numbers, t h e n
As i n Theorem 117.1 i t follows t h a t i f s
.
... + an zn f o r L . Writing Z
before that
Hence, t h e sequence
(2,)
n = 1,2, a.z.
1 1
...
(a,,a2,
,
...)
then
and
E (c,)
(s,)
i s a p-Cauchy
f o r t h e l i m i t , i t f o l l o w s s i m i l a r l y as
g e n e r a t e s a copy of
(c,)
L
in
,
contradicting
our h y p o t h e s i s . We have proved t h u s t h a t n o t only every o r d e r bounded i n c r e a s i n g Sequence i n quence i n
b u t a l s o every norm bounded i n c r e a s i n g se-
L'
i s p-Cauchy and hence i s norm convergent. Since t h e norm
L+
l i m i t i s a l s o t h e supremum, t h i s shows t h a t
L
h a s t h e weak s e q u e n t i a l
Fatou p r o p e r t y .
I n t h e converse d i r e c t i o n , l e t
L
have o r d e r continuous norm a s w e l l
a s t h e weak s e q u e n t i a l Fatou p r o p e r t y and assume t h a t , n e v e r t h e l e s s , L c o n t a i n s a copy of
-
(c,)
(un)
a d i s j o i n t sequence 2 p(un) 5 B <
all n
, and
. The
l a s t assumption i m p l i e s t h e e x i s t e n c e of
of p o s i t i v e elements i n
L
such t h a t
f o r appropriate s t r i c t l y positive constants
.
p s-(u
as
n + m
, since
0 < A 9 and
On account of t h e weak p ( u , + ...+u ) < B f o r a l l n n s = s u p ( u l +...+ u ) e x i s t s , so n- I
also
Fatou p r o p e r t y t h e element
{
A,B
I
+...+ unp
c 0
i s o r d e r continuous. On account of
Ch. 16,11171
EMBEDDING I N BIDUALS
0 i un 5 s
it follows t h a t Hence, L
-
453
(uI+...+u n- 1
p(u ) + 0
, contradicting
cannot c e n t a i n a copy of
(c,)
p(un)
.
5
A > 0
for a l l
.
n
The l a s t theorem, a s a l s o t h e p r e c e d i n g one and t h e r e f l e x i v i t y theorem which f o l l o w s , i s e s s e n t i a l l y due t o G. Ya Lozanovskii ([21,[31,[41, 1967-1969).
Independent p r o o f s occur i n papers by H.P. Lotz ([21,1974) and
P. Meyer-Nieberg
(C 11,1973).
THEOREM 1 1 7 . 5 . The f o l l o v i n g conditions f o r the Banach i a t t i c e
L
are
equivalent. (i)
L
i s reflexive.
(ii)
L
does not contain a copy of
(c,)
o r a copy of
b,
( i i i ) L* does not contain a copy o f
(c,)
o r a copy of
L,
PROOF. ( i ) L
, then
(c,)
L
i s r e f l e x i v e , then
does n o t c o n t a i n a copy of
*
(c,)
o r a copy
( i ) From t h e h y p o t h e s i s t h a t
L
.
1‘
does n o t c o n t a i n a copy of p
i s o r d e r continuous and
h a s t h e weak s e q u e n t i a l p r o p e r t y . It remains t o prove (Theorem 114.8)
that
p*
i s o r d e r continuous. I f n o t , t h e space
(iii)
L
c o n t a i n s a copy of
117.2), c o n t r a r y t o h y p o t h e s i s . Hence, p *
L, (Theorem
*
(c,)
i s o r d e r continuous.
( i i ) We have t o show t h a t i f ( i i ) does n o t h o l d , t h e n ( i i i )
does n o t hold. I n o t h e r words, w e have t o show t h a t i f of
(c,)
i s r e f l e x i v e . Hence, L*
L*
o r a copy of
it f o l l o w s (by t h e l a s t theorem) t h a t
(c,)
L
c o n t a i n s a copy of
are not reflexive. (iii) If
=$
(ii)
L
t h i s copy i s r e f l e x i v e . This i s i m p o s s i b l e , however, s i n c e
and (i)
( i i ) Any norm c l o s e d l i n e a r subspace of t h e r e f l e x i v e
i s r e f l e x i v e . Hence, i f
space of
. .
as w e l l a s a copy of
a copy of
Ll
a copy of
(c,)
, it
Ll , t h e n s o does
L*
L
c o n t a i n s a copy
. Since
f o l l o w s immediately from Theorem 117.2 t h a t
. Since
L
c o n t a i n s a copy of
once a g a i n by Theorem 117.2, t h e space
L*
(c,)
, so
does
c o n t a i n s a copy of
L
contains
L*
contains
. Hence,
L** 1‘
EXERCISE 117.6. Various r e s u l t s i n t h i s c h a p t e r ( n o t r e s t r i c t e d t o t h e l a s t s e c t i o n ) can b e complexified. In t h i s e x e r c i s e we assume t h a t
L
Archimedean and uniformly complete Riesz space. The c o m p l e x i f i c a t i o n of
is an L
COPIES
454
i s denoted by
L"
. For
on
B+iB
f" + ig"
. Similarly
L+iL
any
h
f+ig with
=
as i n s e c t i o n 109, l e t
f,g
E
L
be d e f i n e d by
h"($) = $(h)
with
g"
f"
and
Ch.
, let
b e an i d e a l i n
B
the complexvalued f u n c t i o n
f o r every
belonging t o
16,81173
B-
. It
,
. Then
E B+iB
@
h"
h" =
was proved i n s e c t i o n
(f vf ) " = f',' v f " 1 2 2 (f hf )" = f',' A f " I n p a r t i c u l a r , t h e r e f o r e , we have I f [ " = If"I 1 2 2 For t h e proof t h a t l h l " = Ih"I h o l d s f o r any h = f + i g for a l l f E L 109 t h a t i f
fl
and
and
f2
.
.
a r e elements of
L
then
,
n o t e the following f a c t s . (i) If
f < g
in
L
If cos8+g s i n e l 5 Ihl of t h e s e t o f a l l ( i i ) If f:
+
f"
fn
in
n
-+
for a l l
f
. Hence,
s
E C0,21r)
9
in
+ IgI
B"
. By
h(e1)
.
that
Ihl"
in
L
f o r some
so
s"
5
s"
p < Ihl"
n as well as to
for a l l
,...,h ( 8n ) n
. Show t h a t
. Therefore
Ihl"
t h e supremum of t h e s e t of a l l Note t h a t i f
L
m i l i a r embedding of t h e r e f o r e , with
u
2
p = Ihl"
s"
n
. Assume now
L+iL i n t o
.
.
such t h a t
, and
,
i.e.,
,
then
is
p
,
p 5 Ihl"
it follows t h a t
If" cos8+g" sin81 (L+iL)**
that
converges el'-uniformly
Ihl"
Ihl" = Ih"/
i s a Banach l a t t i c e and we a r e d e a l i n g
If"+ig"I
Ih"l
E L+ , then
E x e r c i s e 91.12 t h e r e e x i s t s a sequence
converges e " - u n i f o m l y t o Ihl" n an upper bound of t h e s e t of a l l If" cos8+g" sin81 -+ m
i s an upper bound
o t h e r words, Ihl"
h ( 8 . ) = If cos8.+g s i n 5 . I f o r a l l j , then 3 J 3 converges e-uniformly t o Ihl a s
such t h a t i f of
. In
h o l d s u-uniformly
e = If1
[0,21r)
t h e supremum
f " < g " . I n p a r t i c u l a r , i t f o l l o w s from
then
If" cos8+g" sin81
h o l d s u"-uniformly
( i i i ) Let
(ej>
,
to
.
p is
with the fa-
I f + i g l correspbnds
It f o l l o w s , f o r example, t h a t t h e r e s u l t s i n
s e c t i o n 1 1 4 and 117 about t h e Banach b i d u a l and about e o p i e s can b e extended immediately t o t h e complex case.
CHAPTER 17 ABSTRACT Lp - SPACES
If
i s a Riesz space equipped w i t h a Riesz norm
L
1 5 p <
number s a t i s f y i n g
pp(u+v) = pp(u) + pp(v) Similarly, p u
and
v
t h e norm
p m
p =
, then
for a l l
t h e norm and
u
is s a i d t o be --additive
in
that
m
L
in
i s p-additive
in
p
is a real
satisfying
L
. If
f o r some
i s included h e r e ) , t h e n
and
inf(u,v)= 0 .
p(u+v) = max{p(u),p(v)}
inf(u,v) = 0
satisfying
L
v
if
p
i s s a i d t o be p-additive i f
p
p
for a l l
i s a Banach l a t t i c e and
L
satisfying
1 2 p 5
m
(note
i s c a l l e d an a b s t r a c t L -space. For P P = 1 , we t h e r e f o r e g e t an a b s t r a c t L -space, which i s a l s o c a l l e d an ab1 s t r a c t L-space o r an AL-space. For p = , an a b s t r a c t L_-space i s a l s o L
c a l l e d an a b s t r a c t M-space o r an AM-space. Among t h e main examples a r e spaces C(X) X
, where
Lp(X,u)
, i.e.,
. One
i s a measure space, and spaces of type
(X,A,p)
spaces of continuous f u n c t i o n s on a compact t o p o l o g i c a l space
of our p r i n c i p a l purposes i n t h e p r e s e n t c h a p t e r i s t o r e p r e s e n t an
L (X,u) P
a b s t r a c t L -space as a space of t y p e P space of type C(X)
.
o r a s a Riesz subspace of a
I n s e c t i o n 118 i t w i l l be proved f i r s t t h a t i f t h e norm i n
p-additive dual
p
satisfying
i s q-additive f o r
L*
1 5 p<m
f o r some
,
then
L
p
-1
1
S
+ q- 1
= 1
This i m p l i e s t h a t i f u
and
space
L
(norm
P(f)
1
implies
space of type
v
L in
i s an AL-space,
L+
. The
. Any
or
C(X)
Lm(X,u)
holds f o r a l l then
L*
i s order
L
and
u
v
p(u+v) = p ( u ) + p(v)
p o s i t i v e element
e
in
L+.
holds
i n t h e normed Riesz p(e) = 1
and
s t r o n g norm u n i t i s a s t r o n g u n i t . I n a
, with
uniform norm, t h e f u n c t i o n i d e n t i -
c a l l y one i s a s t r o n g norm u n i t . I t w i l l be proved t h a t i f space, t h e n t h e Banach d u a l
is
(T. Ando). Furthermore, i f
p ) i s c a l l e d a s t r o n g norm u n i t i f
If1 5 e
L
t h e norm i n t h e Banach
i s super Dedekind complete and t h e norm i n
continuous. Also, pp(u+v) 2 pp(u) + pp(v) for a l l
, then
pSm
L
i s an AL-
i s an AM-space p o s s e s s i n g a s t r o n g norm
unit.
455
Ch. 1 7 , 5 1 1 8 1
CHAPTER 17
456
A s observed a l r e a d y , t h e Banach d u a l of a normed Riesz space w i t h
norm i s an AL-space.
-additive
p r o p e r t y t h a t t h e band l i n e a r f u n c t i o n a l s on
(norm
in
Lg
although
=
1
i t s e l f i s not
L*
It was proved by E. d e Jonge t h a t t h e s e a r e t h e
d e f i n e d ( i n s e c t i o n 119) a s b e i n g normed Riesz spaces
p ) with t h e property t h a t f o r a l l
= p(u2)
possessing t h e
L
( c o n s i s t i n g of a l l norm bounded s i n g u l a r
L*
L ) i s an AL-space,
n e c e s s a r i l y an AL-space. semi-M-spaces,
There e x i s t Riesz spaces
and f o r any sequence
(v,)
.
l i m p(vn) 5 1
u1,u2
with
p
1 5 p <
satisfying
-
with
L+
sup(ul,u2) 2 v
+
L
p(ul) = 0
we have
i s e i t h e r an a b s t r a c t L P o r a Dedekind complete Riesz space
I n s e c t i o n 120 i t i s proved now t h a t i f space f o r some
in
L
p o s s e s s i n g a s t r i c t l y p o s i t i v e o r d e r continuous l i n e a r f u n c t i o n a l , t h e n
i s e i t h e r Riesz isomorphic (and norm isomorphic) t o a space t o an i d e a l i n a space
, where
L1(X,A,u)
compact Hausdorff space
. The
X
b l u s t (1940). F i n a l l y , i f
theorem. I f AL-space.
r e s u l t f o r L -spaces goes back t o F. BohnenP i s an AM-space p o s s e s s i n g a s t r o n g norm u n i t ,
L
therefore
. This
C(X)
such t h a t
X
L
L*
i s an
i s an AM-space w i t h a s t r o n g norm u n i t , and
L**
i s isomorphic t o some
L**
i s Riesz
f o l l o w s from Yosida's r e p r e s e n t a t i o n
i s an a r b i t r a r y AM-space, t h e n t h e Banach d u a l
L
It follows t h a t
L
or
P i s a measure i n t h e l o c a l l y
p
t h e n t h e r e e x i s t s a compact t o p o l o g i c a l space isomorphic t o t h e space
L (X,A,u)
Riesz subspace of a space of type
. Thus L i s isomorphic t o a . The i n v e s t i g a t i o n of AM-spaces goes
C(X)
C(X)
back t o S . Kakutani (1941).
118. Riesz s p a c e s w i t h p - a d d i t i v e norm Let
L
a r e a l number s a t i s f y i n g if
-.
be a Riesz space equipped w i t h a Riesz norm
1 s p <
pP(u+v> = pp(u) + pp(v)
inf(u,v) = 0
and
p
holds f o r a l l
u
and
The norm
holds f o r a l l
v
in
L
satisfying
,
p
if
v
in
L
satisfying
p(u+v) = max{p(u),p(v)}
inf(u,v) = 0 nad
be
p
. If
L
is a
i s p - a d d i t i v e f o r some
i s c a l l e d an a b s t r a c t L -space. For P p = 1 , we t h u s g e t a n a b s t r a c t L 1 -space which i s a l s o c a l l e d an abstract p
satisfying
L-space o r an &-space.
m
For
then
p
i s s a i d t o b e p-additive
p
and
i s s a i d t o be --additive
Banach l a t t i c e w i t h r e s p e c t t o t h e norm
1 5 p s
u
and l e t
p
L
p =
m
an abstract M-space o r an AM-space. p - a d d i t i v e norm a r e t h e space
, an
a b s t r a c t Lm-space i s also c a l l e d
Among t h e main examples of s p a c e s w i t h
Lp(X,p)
, where
(X,A,u)
i s a measure space
Ch. 17,§1181
The s p a c e space
ABSTRACT L -SPACES P
of a l l ( r e a l ) c o n t i n u o u s f u n c t i o n s on a compact t o p o l o g i c a l
C(X)
w i t h t h e uniform norm i s an AM-space.
X
457
Another example of a n AM-
s p a c e i s o b t a i n e d by t a k i n g t h e second commutant
C”(D)
of a set
of
m u t a l l y commuting H e r m i t i a n o p e r a t o r s i n a H i l b e r t s p a c e ( c f . s e c t i o n 57). Abstract L -spaces,
P
1 _< p <
for
m
, were
i n t r o d u c e d by F. Bohnenblust ( [ I ] ,
1940); AM-spaces were i n v e s t i g a t e d i n a p a p e r by S. Kakutani ( [ l I , l 9 4 1 ) , w i t h an immediate s e q u e l b y Bohnenblust and Kakutani j o i n t l y ( [ l ] , l 9 4 l ) .
One
of t h e main purposes i n t h e s e p a p e r s i s t o r e p r e s e n t a n a b s t r a c t L -space a s P a s p a c e of t y p e L (X,u) on an a p p r o p r i a t e measure s p a c e o r a s a R i e s z subP s p a c e of a s p a c e of t y p e C(X) on an a p p r o p r i a t e t o p o l o g i c a l s p a c e X
.
As well-known, o-finite,
p
(i.e.,
q =
t h e Banach d u a l of t h e s p a c e if
m
p
=
LI(X,p)
now g e n e r a l l y t h a t i f
,
1 5 p <
for
, unless
L
L*
p
satisfying
h a s q - a d d i t i v e norm f o r
p
such t h a t
q-additive for
p-l
+ q-l
m
= 1
L
s p e c t i v e l y . Assume f i r s t t h a t > 0
+ q-l
= 1
Lacey and
. Then .
and 1
L* 5
t h e norm i n t h e Banach dual
w i l l be denoted by
p <
m
b e g i v e n . There e x i s t s an element
. Let
p
inf(+,$) = 0
u t 0
in
L
L*
is
and
p*
re-
in
L*
and l e t
such t h a t
p(u) = I
and
Since V
.
be a normed R i e s z s p a c e v i t h p - a d d i t i v e norm f o r
L
1 5 p 2
PROOF. The norms i n
E
p-l
Bernau ([1],1974).
THEOREM 118.1. L e t some
is
i s 1 - a d d i t i v e . We s h a l l prove
(L,(X,p))*
h a s p - a d d i t i v e norm f o r some
t h e n t h e Banach d u a l
= 1
i n very t r i v i a l c a s e s , i t i s not
The p r o o f , due t o T. Ando, i s from a n e x p o s i t o r y p a p e r by H.E. S.J.
and
m
L (X,p) w i t h q determined by p-l + q-l 4 1 ). A l s o , a l t h o u g h t h e Banach d u a l of Lm(X,p)
d i f f i c u l t t o show t h a t t h e norm i n
m
,
i s t h e space
properly l a r g e r than
1 5 p 5
Lp(X,p)
0 = { i n f ( $ , $ ) } ( u ) = inf($(v)+$(u-v):O
E L with
0 S v 5 u
and
$(v)
+ $(u-v)
<
E
,
t h e r e e x i s t s a n element
. Hence
For
p = 1
and, s i n c e
, this
i s majorized by
{(u-v)-inf(u-v,v)}
bracket i s equal t o
p{u-2
Since, conversely, p*($+Q) p
Ch. 17,§1181
RIESZ SPACES WITH p-ADDITIVE NORM
458
i s 1 - a d d i t i v e , then
A
{v-inf(u-v,v)l
2
p*
=
5 p(u) = 1
inf(u-v,v)}
max{p*($),p*($)}
i s --additive.
,
0
,
,
t h e l a s t square
s o (1) becomes
it i s e v i d e n t now t h a t i f
Then, by HBlder’s i n e q u a l i t y , t h e l a s t e x p r e s s i o n i n ( 1 )
pJ (u-v)
*
lP ‘I
- i n f (u-v ,v ) ‘1+ppJv- i n f (u-v
J ‘ 1
S i m i l a r l y a s b e f o r e , i t follows from A
{v-inf(u-v,v)}
=
0
1 < p <
Assume now t h a t
,v)
‘1PP (1 +
{ (u-v)-inf (u-v,v)
3E
1
m
.
is majorized by
.
A
t h a t t h e l a s t square b r a c k e t i s e q u a l t o
so ( 1 ) becomes
T h i s holds f o r every
E
> 0
, and
hence
{p*($+$)lq 5 { p * ( $ > j q
Far t h e i n e q u a l i t y i n t h e converse d i r e c t i o n , l e t such t h a t
p(u)
=
p(v) = 1
and
0 5 u
+
,v
{P*($)lq E L
be
Ch. 17,11181
Put
P
w = { $ ( U ) } ~ - ' U and
Since
459
ABSTRACT L -SPACES
inf(+,$) = 0
,
(w -w A Z 0 0 0
( z -w
.
z = {$(v)}q-'v
w
0
there e x i s t
0
Then
5 w
0 5 zo 5 z
and
such t h a t
Hence
Since
A
0
0
AZ
0
= 0
This h o l d s , t h e r e f o r e , a l s o f o r
i t follows t h a t that i f
1 < p <
The c a s e
+
{p*(+)Iq m
and
p = =
p
, we
0
E =
have
. Observing
5 {p*(++$)}'
{p*($)Iq
i s p-additive,
i s l e f t . Since
p*($+$)
i t i s s u f f i c i e n t t o prove i n t h i s c a s e t h a t for
inf(+,$) = 0
. Again,
let
E >
0
then
. There
now t h a t
. This
5 p*(+)
exist
-1
=
q
-1
,
concludes t h e proof
i s q-additive.
p*
p*(+)
1-p
+
p*($)
+ p*($>
5
always h o l d s , p*($+$)
0 5 u,v E L
holds
such t h a t
R I E S Z SPACES WITH p-ADDITIVE
460
p(u) = p(v) = 1
NORM
Ch.17,§1181
and
inf($,JI) = 0
Furthermore, s i n c e
,
0 2 w 2 u
there e x i s t
and
0
Iz 2
v
such t h a t
Hence
This holds f o r
as w e l l , so
= 0
E
p*($) + p*(JI)
.
2 p*($+$)
-
For 1 2 p < , a b s t r a c t L -spaces have s p e c i a l p r o p e r t i e s . A s Riesz P spaces these spaces a r e super Dedekind complete and t h e norm i s o r d e r continuous. THEOREM 118.2. I f 1 Ip <
m
, then
L
L
i s an abstract L -space f o r some
P
p
i s super Dedekind complete and the norm i n
satisfying L
is
order continuous. PROOF. Let
(un:n=1,2,
p o s i t i v e elements i n m
+ n . Then
sm =
norm by
p
, we
for a l l
m
, so
L
~ ' pun
,
...)
say
be an order bounded sequence of d i s j o i n t
0
5
u
satisfies
n
Iuo
o
for a l l
< sm < uo
n
and
for a l l
m
u Iu for m n Denoting t h e
.
have
p(un)
+
0
as
n
+
-.
Since
L
i s a Banach l a t t i c e with
ABSTRACT L -SPACES
Ch. 17,11181
46 1
P
respect t o
p
by assumption, i t f o l l o w s from Theorem 104.2 t h a t every
o r d e r bounded i n c r e a s i n g sequence i n
i s convergent i n norm. By Theorem
L
103.6, however, t h i s i s e q u i v a l e n t t o s u p e r Dedekind completeness of
L
t o g e t h e r w i t h o r d e r c o n t i n u i t y of t h e norm.
-
I t f o l l o w s from t h e l a s t theorem t h a t i f
f o r some L
, then
p
satisfying
1 5 p <
f o r any p o s i t i v e
u
and
i s an a b s t r a c t L -space P i s a given p o s i t i v e element i n
e
L
i n t h e band g e n e r a t e d by 0 5 s
(by F r e u d e n t h a l ' s s p e c t r a l theorem) a sequence s
a f i n i t e sum
n elements
pi
C a.p.
with the c o e f f i c i e n t s
1 1
mutually d i s j o i n t components of
t h e o r d e r c o n t i n u i t y of
p
, so
t y of
in
4 u
, each
L
nonnegative and t h e
. Then
. Note
there exists
p(u-sn)
+
0
by
t h a t Freudenthal's
i s Dedekind complete. This proper-
L
w i l l b e used i n t h e proof of t h e n e x t theorem.
L
THEOREM 118.3. 1 5 p <
a.
e
4 p(u)
p(sn)
s p e c t r a l theorem may b e a p p l i e d s i n c e
e
m
If
L
p
i s an a b s t r a c t L -space for some P
, then
holds f o r a l l pDsitive elements
u
and
v
in L
. In
such t h a t
other words, t h e
i s p - s u p e r a d d i t i v e . If L i s an abstract L-space (AL-space), holds f o r a l l p o s i t i v e u and v i n L
norm
p
then
p(u+v) = p ( u ) + p(v)
PROOF. Note f i r s t t h a t numbers
.
(a+b)'
2
f o r a r b i t r a r y non-negative
a p + bp
a , b ; hence pp(aw+bw) 2 pp(aw) + pp(bw) w E L+
f o r any
. This
i s obviously a v e r y simple case of t h e theorem t o b e
proved; we s h a l l reduce t h e g e n e r a l proof t o t h i s case. For t h i s purpose,
0
let
5
u
,v E
L
b e g i v e n and s e t
e = u+v
p o s i t i v e elements i n t h e i d e a l g e n e r a t e d by
. Then ,
e
u
and
v
are
and t h e r e f o r e (by
F r e u d e n t h a l ' s s p e c t r a l theorem, a s observed above) u
and
v
may b e
approximated from below by f i n i t e l i n e a r combinations of d i s j o i n t components Of
e
w i t h non-negative c o e f f i c i e n t s . Given
t = 1b.p:
J J
E
> 0
,
let
be l i n e a r combinations of t h i s k i n d such t h a t
s = 1a.p.
1 1
and
pp(s) > pp(u)
-
E
RIESZ SPACES WITH
462
and
-
p P ( t ) > pp(v)
coefficients i n
and
s
d i s j o i n t ) components Zipij
pi
=
. We
E
for all
:3 i,j
s
aipij
= Ci,j
may assume t h a t
Ch. 1 7 , § 1 1 8 1
Cpi = CpJ = e
(some
may t h e n be z e r o ) . I n t r o d u c i n g t h e (mutually
t
p..
p-ADDITIVE NORM
=
pi
A
, we
pj
may w r i t e
CjPij
and
,
t = C.
1,
,
Pi
.
bjpij
j
=
I n view of t h e remark a t t h e beginning of t h e proof we f i n d now t h a t P P (u+v) 2 p P ( s + t ) = pp { c i , j (ai+bj)pij} = C.
1,
Since
E
If
u,v
pP{(ai+bjjpij} 1
i s an AL-space,
. Since
+ c.
pp(aipij)
2
i s a r b i t r a r y , i t follows t h a t
> 0 L
E L+
j
2(bjPij)
1,j
.
pp(u+v) 2 pp(u) + pp(v)
we g e t thus t h a t
p(u+v) t p(u) + p(v)
for a l l
t h e r e v e r s e i n e q u a l i t y h o l d s always, i t follows t h a t f o r all
p(u+v) = p ( u ) + p(v)
.
u , v E L+
As
We now t u r n our a t t e n t i o n t o a b s t r a c t Lm-spaces (AM-spaces). examples we mentioned a l r e a d y t h e space of type spaces of type
=
C(X)
with
compact and
X
and a l s o t h e
Lm(X,u)
t h e uniform norm. Some
p
f u r t h e r examples follow. (i) and
i s a given p o i n t i n
xo (ii)
X
.
.
xo,xb
given points i n
( i i i ) For any i n f i n i t e p o i n t set Riesz space of a l l r e a l (x:lf(x)I>E)
f
on
x
# xn
and
l i m f(xn) = 0
supremum norm i s an AM-space.
X
, we
, where Xo
and
X
define
as
f
E co(X)
{xn} n
+
-. of
again
E
is
X
i s a given r e a l
co(X)
such t h a t , f o r e v e r y
X
i s f i n i t e . Note t h a t
e x i s t s an a t most c o u n t a b l e s u b s e t all
i s compact and Hausdorff
X
C(Xlxo,x~,Xo) = {f:f€C(X),f(xO)=AO f ( x b ) }
compact Hausdorff w i t h number
, where
C(X(xo) = (f:fEC(X),f(xo)=O)
t o be t h e
,
> 0
the s e t
h o l d s i f and only i f t h e r e X
such t h a t
The space
Note t h a t t h e norm i n
f(x) = 0
with the
c (X)
0 co(X)
for
i s o r d e r con-
tinuous. The p o s i t i v e element
e
i n t h e n o m e d Riesz space
L = L
P
is called
Ch.
ABSTRACT L -SPACES P
17,§1183
a strong norm u n i t i n
if
L
p(e) = 1
and
p(f)
It f o l l o w s immediately t h a t a s t r o n g norm u n i t
A
o t h e r words, t h e i d e a l
g e n e r a t e d by
463
1
5
5
.
e
is a strung unit; i n
e
satisfies
e
If1
implies A
= L
. Also,
any
normed Riesz space can have a t most one s t r o n g norm u n i t . I n a space of type
or
Lw(X,u)
t h e f u n c t i o n i d e n t i c a h l y one i s a s t r o n g norm
C(X)
-
u n i t . The spaces mentioned above i n t h e examples ( i ) , ( i i ) and ( i i i ) , w i t h t h e uniform norm, do n o t p o s s e s s a s t r o n g norm u n i t . Also f o r s p a c e s of type
P
I n t h e l a s t theorem we have seen t h a t i n an AL-space satisfies
p(u+v) = p(u) + p(v)
t h e norm
L
and
L
i s an AL-
space, then
is an AM-space w i t h t h e p r o p e r t y t h a t p*($v$) =
L*
i n f ($,$)
=
0
for a l l positive
. Furthermore
THEOREM 118.4. I f
and
$
we show t h a t
$
in
L*
,
and n o t only i f
L* p o s s e s s e s a s t r o n g norm u n i t .
i s an AL-space, then L*
L
,
u
s h a l l prove now t h a t i f
inf(u,v) = 0
v
p
f o r a l l p o s i t i v e elements
. We
and n o t only i f
= max(p*($),p*($))
,
do n o t p o s s e s s a s t r o n g norm u n i t ( u n l e s s i n very
L (X,p)
t r i v a l cases).
1 5 p <
i s an A??-space with
the property that
and
p
strong norm unit
$o , defined by
PROOF. For any
$
f E L
. Furthermore
L*
in
for a22 p o s i t i v e
the space L* has a f o r a22 f E Lo
$O(f) = p(f+) - p(f-)
we d e f i n e
$O(f) = p(f+)
-
p(f-)
. It
follows
from
+
( f + d + + fand t h e a d d i t i v i t y of p{ff+g)+} So
p
-
g- = ( f + g ) - + f
+
+ g
that p{(f+g)-}
$ O ( f + g ) = $ O ( f ) + $o(g)
$ O ( a f ) = a$ ( f ) 0
+
. From -
-
p(f-)
+ p(g+)
-
p(g-)
,
the d e f i n i t i o n i t i s evident t h a t
a 2 0
for a l l real
$ J O ( - f ) = p{(-f)+}
= P(f+)
. Furthermore
p{(-f)-}
=
-{P(f+)-P(f-)}
=
.
.
RIESZ SPACES WITH p-ADDITIVE NORM
464
Hence
$JO(af) = a$ (f)
$o is a positive linear functional on L
already that
$o
. Thus we have proved . For the norm of
for all real numbers a
0
Ch. 17,11181
we f;nd that
Also, if j , E L *
for all u
2
. Hence
0
$o
far that
p*(j,>
and
is a
5
,
1
,
1
then p * ( i J I I )
SO
. It follows from what we strong norm unit in L* . 191 5 0 ,
It remains to prove that p*($v$) For this purpose, let
$,j,
2
=
have proved
max{p*($),p*($)}
$,I) t 0
for $
0 be given. If one at least of
so
.
and j ,
is
the null functional, the proof i s trivial. Hence, we may assume that p*($>
#
0
p*(j,)
and
Similarly $
5
# 0
.
. Since
~ * ( j , ) . $ ~ Hence
p*C$/p*($)}
5 1
, we
get
$ 5 P*($).$~
, which
$ v j , 5 max{p*(+),p*($)}.$o
.
implies
that
The inequality in the converse direction is trivial. The question may be put forward if for any normed Riesz space L
(not
necessarily a Banach lattice) with p-additive norm (for some p satisfying 1 2 p < m ) it remains true that the norm is p-superadditive, i.e.,
.
Similarly, we ask if in a space for all u,v E L+ pp(u+v) 2 pp(u) + pp(v) with --additive norm the equality ~ ( u w )= max{p(u),p(v)} holds for all u,v E L+
. The answer is affirmative.
THEOREM 118.5.
norm
p
for some
holds f o r a l l
u,v E L+ L
L is a normed Riesz space uith p-additive
p satisfying
u,v E L+
holds for all (ii) If
(i) If
. For .
p
1 = 1
p <
5
m
, the
, then
equality
pP(u+v)
t pP(u)
is a normed Riesz space with -additive norm
~(uvv) = max{p(u),p(v)l
holds for a12
u,v E L+
.
+ pp(v)
p(u+v) = P ( U ) + ~ ( v ) p
, then
465
ABSTRACT L -SPACES P
Ch. 1 7 , 9 1 1 9 1
PROOF. (i) The Banach bidual L** is norm complete with p-additive norm (Theorem 118.1)
more, L
and, therefore, L** is an abstract L -space. FurtherP is embedded in L** with preservation of the norm. The result
follows then from Theorem 118.3. (ii) The Banach dual L* norm in L**
in L**
is an AL-space. Hence, by Theorem 118.4,
has the desired property. Once again, since L
the
is embedded
as a Riesz subspace with preservation of the norm, the norm in
L has the desired property.
11 9. Semi- M-spaces
It was shown in the preceding section that if L space with -additive norm (in particular, if L
is a normed Riesz
is an AM-space),
then the
Banach dual L* is an AL-space. In the present section we shall deal with a particular kind of normed Riesz spaces L possessing the property that the band
in L*
L*
functionals on
(consisting of all norm bounded singular linear
L ) is an AL-space, although L*
itself is not necessarily
an AL-space. We recall for this purpose that the order dual L" arbitrary Riesz space L where
is the direct
sum of
the bands L :
of an
and
LL
,
L z consists of all integrals (sequentially order continuous linear
functionals on
L ) and L"
Hence, for any
$ E L"
$ = $c + $,
with
oc
,
is the disjoint complement of
L :
.
in L"
there exists a uniquely determined decomposition
E L"
and
0, E
L :
. The ebements of
LL are called
the singular linear functionals. We shall now need a certain property of these singular linear functionals. This property was actually mentioned already in Exercise 90.12. THEOREM 119.1.
i n t e g r a l component
Hence, $
For any given $c o f
$
i s singular ( i . e . , $
e x i s t s a sequence
0 < u
+
u
$ 2 0
in
L"
and
u 2 0
i n L , the
satisfies
= $s
in L
I i f and only i f f o r any such t h a t
0
$(un) <
PROOF. Denote the right hand side of (1) by $ (u)
E > E
0
there
for all
. Evidently, we
n have
.
SEMI-M-SPACES
466
. We
0 5 $ (u) 5 $(u)
t o see t h a t
inequality, take
> 0
E
l i m $(wn) < $,(u+v) v
=w-u n n n for a l l n
+
forall
$,(u) =
.
O s u + u , O < v 4 v n
s l i m $ ( un ) + l i i n $(vn)
+ $r(v)
$r 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 L
m
$,(u)
= C1
sequence
$r(vn)
. Now,
/
let
, choose
(vnk:k=1,2,
< $r(vn) + ~
and
and
u + v n n
= W
n
=
. Thus
0,
i s a d d i t i v e on
L+
,
can be extended i n t h e obvious manner a s a
. For
t h e proof t h a t
$r
i s countably
we n o t e f i r s t t h a t , by d e f i n i t i o n ,
L+
u E L+
f o r every
i s easy
the inverse
,
E
$r(u) + $ r ( v ) 5 $r(u+v)
from which i t follows t h a t a d d i t i v e on
. It
L+
. For
a sequence
l i m $(wn) < $r(u+v) +
and t h e r e f o r e
u,v E L+
0 s w 4 u+v s a t i s f y i n g n Now, by Theorem 1 5 . 5 ( i ) , i f un = inf(u,wn)
and
E
i s a d d i t i v e on
for a l l
+ $,(v)
n,then
. Hence
4,
show f i r s t t h a t
$ r ( ~ + v ) 5 $,(u)
Ch. 17,51193
...)
u = C
a number
wn
I
v
n
E
in > 0
L+
. For
L+ such t h a t
in
.2Now~ l e t
m
t h e proof t h a t
every
C k v nk
Cp=I Ct=l v j k
=
. For
for a l l
.
,
n
vn
=
there exists a
and
Ck $(vnk) <
. Then
n
0 5 wn 4 u
and $(wn) < C . 4 ( v . ) + E f o r every n It follows t h a t $,(u) J r J The i n v e r s e i n e q u a l i t y i s e v i d e n t . It has been shown t h u s t h a t 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 by
$
. Equivalently,
follows t h a t j o r i z e d by
($c)r
that
0,
($ ) = c r
p
. The
. Hence
space
...)
dent t h a t i f L = 1
L
u1,u2 E L+
(vn:n=1,2,
with p
0,
5 $,
p(ulvu2)
a d d i t i v e on
L
L+
such t h a t
. It
O S $ ~5 $
i s t h e l a r g e s t p o s i t i v e i n t e g r a l ma-
i s majorized by
. It
$
,
$
. The
t h e d e f i n i t i o n of
$
same d e f i n i t i o n i m p l i e s a l s o
follows t h a t
$c = $r
.
i s a normed Riesz space; t h e norm w i l l be denoted
L
i s c a l l e d a semi-M-space i f it has t h e p r o p e r t y t h a t
with
and f o r any sequence
p ( u l ) = p(u2) = 1
u1 v u2 5 vn J. 0
i s @=-addv ite
i s a semi-M-space, that
$c
$c
i s majorized by
Assume now t h a t for a l l
(because
$ ) . Since
implies t h a t
by
i s an i n t e g r a l on
$
$ r < $c
, countably
L
sC.6 ( v . ) . 4 r J 1s a $r and majorized
l i m p(vn) 5 1
we have
(in particular, i f
L
. Also,
i f t h e norm
p
i s evi-
i s a n AM-space),
because i n t h i s c a s e it follows from = 1
. It
then
p(u ) = p(u ) =
1 2 is o r d e r continuous, t h e n
L
46 7
ABSTRACT L -SPACES P
Ch. 1 7 , 8 1 1 9 1
is a semi-M-space, because in this case it follows from vn J. 0 that = 0
lim p(vn)
. There are
less obvious examples of semi-M-spaces; some of
these (the Orlicz spaces) will be presented in section 131. The semi-Mspaces are characterized by the following property. THEOREM 119.2. The normed Riesz space L
if the band Lg i n the Banach dual
in L
and
will be denoted by
L*
,
+ P*($~)
t p*($,) E
p(u ) 1
>
u
=
$2
p
and
respectively. We have to
p*
=
p(u ) 2
=
1
E
. Then
$ = $, + $,
. Writing
for all n
and
u
n
) < 1 + ~ for all
Hence
p*(o1+O2)
that P * ( $ , + $ ~ )
2 2
in L+
2
u E L+
v
2
and w
0
n
(l+c)
2 no
-1
2
)2
such
. Hence, such that
=
=
such that
n we get
for all
{P*($~)+P*($,)-~E}
.
$ E L:
u-w for all n , we have u1 u2 n in view of the hypothesis that L is a
n
. For these values of
P*($~) + p*($,)
5
in L
4 u
= u > v J. 0 , s o lim p(v ) 5 1 n n semi-M-space. Therefore, there exists a natural number no p(v
1
and
by the last theorem, there exists a sequence 0
$(wn) <
then p * ( $ +$
the inverse inequality being evident. For the proof,
and
v u2
,
are positive elements of L z
0 be given. There exists elements u1
let that
Set u
and
is an AL-space.
is a semi-M-space. As before, the norms
PROOF. Assume first that L prove that if
L*
is a semi-M-space if and o n l y
E
> 0
, which
implies
holds. It has been shown thus that P *
is 1-additive on Lz Since L : is norm complete (being a band in the Banach space L* ), it follows that LE is an AL-space. For the proof in the converse direction we assume that L a semi-M-space and we shall prove that u , , u 2 E L+ u1 v u2
2
and a sequence vn 4 0
with
(v,)
lim p(v
Lz in L+ such that
) = a > 1
fails to be
is not an AL-space. There exist
. By
p ( u I ) = p(u2)
=
1
and
the Hahn-Banach theorem there
Ch. 17,81191
SEMI- M-SPACES
468
e x i s t s f o r every
n
p(vn) = JIn(vn)
. Then
The u n i t b a l l
B = ($:p*($)~l)
of accumulation
$,
(0),
gy. The sequence
$o
JI, E L*
an element =
lJInl
p*($,)
of
i s compact i n t h e weak s t a r topolo-
L*
B
, so
t h i s sequence has a p o i n t
every weak s t a r neigborhood of
i n f i n i t e l y many p o i n t s of t h e sequence). I n p a r t i c u l a r , i f E
> 0
a r e a r b i t r a r y , t h e neighborhood
i n f i n i t e l y many $,(u)
$o
2
E B
0
,
.
so
0,
. Since
P*($~) 5 1
.
and l e t
> 0
t 0
$,(u)
This h o l d s f o r a l l
u t 0
{$:
n
E
I $ ( U ) - $ ~ ( U ) I <E)
for a l l
, so
This i m p l i e s t h a t
theref ore
Now, f i x
and
= 1
satisfies
i s contained i n (i.e.,
such t h a t
4,
n 2
0
, it
$o c o n t a i n s u E L+
and
contains
follows t h a t
. Furthermore
0 5 c$ O(vn ) 5 p(vn)
we have
, and
be given. I n t r o d u c i n g t h e weak s t a r open neigh-
borhood
.
w e have
$ E U f o r i n f i n i t e l y many v a l u e s of m Since m t n i m p l i e s m $m(vn) 2 $m(vm) = p(v ) , i t f o l l o w s t h a t p(vm) 5 c$o(vn) + E f o r t h o s e m m t n f o r which $ E U Hence m
.
l i m p(vm) c $o(vn) +
(3)
E
. .
T h e r e f o r e , i n view of (2) and ( 3 ) , lim $ (v ) = l i m p ( v ) = a Observe now O n n t h a t $o = $c + $ s w i t h 0 5 $= E LE and 0 5 $ E L* Furthermore, l i m $ (v ) = 0 on account of v J- 0 , s o c n u 1 v u2 t v f o r a l l n , i t follows t h a t n
Let now of
Qs
osl
s
s
.
l i m $s(vn) = a > 1
. Since
be 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 t h e r e s t r i c t i o n
t o the principal ideal
A
P
generated b y t h e element
p = (u 1-u 2 ) +
,
69'1
d S33VdS- 'I L3WILSBV
[611
§'f I ' q 3
I t i s easy t o s e e t h a t
if
f o r a f i x e d constant
A
complete, i . e . ,
in
I.
L
sup f
and
p(fn) S A
for a l l
E L ) . I t follows t h a t
d i v i d e the i n t e r v a l
n
and
i s norm
L
is
L
i n t o t e n equal i n t e r v a l s
[;,11
[:,;I
i n t o one hundred
'
ui =
u = u1 v u2
:j
3 ?xEi
u(x)
=
23
and
for a l l
,
n
vn = uxFn
,
so
('E. 1 , j :j
E2 = U.
1,j
h > 0
f o r any
. Then
(i=1,2)
satisfies
Fn = CO,n-ll
and
1 , and
p(E ) = u(E2) = 1
.
...,
,j \ E i ,
E l = Ui
p(u.) S I
for a l l v1 = u
i t i s evident t h a t
x
even
1
J *
we have
for
i
. For
4 0
u t v
and
l i m p(v )
and
= 1,2
n = 1,2
.
let
Since
p(vn) =
does not hold. There-
1
S
,
i s not a semi-M-space.
fore, L
EXERCISE 119.4. Let
L
be the space introduced i n the l a s t exampl?.
i s not a semi-M-space,
L
Since
n
, and s o on. Generally, d i v i d e E 2 , 1 , * - * ,E 2,100 i n t o lon equal i n t e r v a l s E (j=l, lon) Let n,j
~2-~,2-"+'1
$
then
possessing t h e s e q u e n t i a l weak
(from l e f t t o r i g h t ) , then divide
E l . l ~ * * * 1.10 ~ E equal i n t e r v a l s
Now, l e t
0 5 f
,
L
i s a Banach function space. For the proof t h a t
L
not a semi-M-space,
Then
i s a norm i n
p
Fatou property ( i . e . ,
=
Ch. 17,§1191
SEMI- M-SPACES
470
i t follows t h a t
Lz
contains o t h e r func-
t i o n a l s besides the n u l l f u n c t i o n a l . But then, s i n c e the inverse a n n i h i l a t o r '(Lf) La
Lt
of
satisfies
f E L
such t h a t
If[ t u
( i ) Show t h a t i f in
'(L;)
i s properly contained i n
,
C0,ll
then
f
( i i ) Show t h a t tends t o zero a s HINT:
1; .
over
[O,ll 4 0
0 < h
S
h
implies satisfies
La
i s t h e s e t of a l l
p(un) 4 0
.
f(x) = 0
on some i n t e r v a l
.
vanish on
JA
C0,6]
and observe t h a t f E L
. Furthermore,
. Set
by Theorem 102.8, i t i s evident t h a t recall that
c o n s i s t s of a l l
La
f
undu
( i i ) Let f i r s t If1 t u
. We
h
-1
f E L
f o r which
h
and l e t
If[ t u
4 0
[0,61 If Idu
h C 0
( i ) Let
p(un) < - 6-I
C 0
f E L
E La
= La
L
f = f
1
satisfy let
+ f2
E
h > 0
, where
h lo IfIdp
-I
. f
undp 4 0
1
since
+
0
as
Then
h-'
1;
= f
on
CO,hcl
n
f
. Prove
h J. 0
If Idu < and
that
i s sumable
E
and l e t for fl = 0
Ch. 17,51201
elsewhere. S i m i l a r l y , decompose 5
p(fl) <
<
E
for
n
u
for a l l
s u f f i c i e n t l y large.
Conversely, l e t
u
=
that
and
f E La
I f l x En
u
as
n
p(un2) J 0
E
n
functions
47 1
ABSTRACT L -SPACES P
.
) < nl by p a r t ( i ) , we f i n d t h a t p ( u n ) <
Then, f o r
I
satisfy
fl t u
n
= unl + un2 , Since
h J. 0 and E n n c 0 , so p(un)
=
p(u
C0.h;
. It
c o
,
the
follows
120. Representation by a space of measurable f u n c t i o n s We r e c a l l some r e s u l t s from Chapter 1 concerning Boolean a l g e b r a s . Let
is a distributive lattice
be a Boolean a l g e b r a , t h a t i s t o say, A
A
with n u l l element ( s m a l l e s t element) and u n i t element ( l a r g e s t element) such
a E A
t h a t every element
has a complement
e
u n i t element a r e denoted by ment
a'
of
ideal
P
a
A
implies
a l v a2 E I
at
and
i s denoted by
a2
P
i s not contained i n
and
belongs t o
.
For any
a'
A
the s e t s
{PIa
a
. This
a3
A
P
5
I
of
a3 E I
. The
a F: A
topology i n
. The
i s an i d e a l i f
A
implies al
a
4
E I
a2 E P
. The
t h a t one
s e t of a l l proper prime i d e a l s
t h e s e t of a l l {PIa
P
A
. The
. The base
P E P
s e t of a l l
such t h a t {PIa
is
s e t s a r e open and
P which i s both open and closed i s one
i s based on t h e formulas
a s well a s on t h e f a c t t h a t i f such t h a t
element and t h e
a v a' = e
and
= 63
i s denoted by
P
closed. C o w e r s e l y , any subset of
a3 F: A
. The n u l l
i s c a l l e d order complete i f every non-empty subset of
A
a base f o r t h e hull-kernel
Of
a'
r e s p e c t i v e l y ; the (unique) comple-
i s c a l l e d a prime i d e a l i f i t follows from
a t l e a s t of in
e
has a supremum and an infimum. The subset
al,a2 E I
a
i s determined by
a
Boolean a l g e b r a A
and
a2
=
8
{P} c {P}a , then t h e r e e x i s t s an element a2 1 a v a = a l , and t h e r e f o r e t h e s e t
and
{P} and {PI i s {PIa3 . The space P i s a1 a? compact and Hausdorff i n i t s hull-kernel topology. Tne space P i s a l s o t h e o r e t i c d i f f e r e n c e of
t o t a l l y disconnected ( i - e . , t h e topology has a base c o n s i s t i n g of s e t s t h a t are open a s w e l l a s c l o s e d ) . I f
A
i s o r d e r complete, t h e hull-kernel
472
REPRESENTATION BY MEASURABLE FUNCTIONS
P is extremally disconnected (i.e. the closure of any open
topology of set i s open). =
0
-
. We
sup a
To see this, let 0
be open,
prove that the closure 0
-
.
so
0
of
0 = UT {PI Let aT is exactly {PI ?O
of
0
-
{PIa
0
is contained in 0
A
-
is contained in the complement of
is of the form Uy {PIa
,
(sup aT) = 0
with
i.e?, a
A
a .a
a
A
V
T
= 0
{PIa
0
.
0
Since
-
The complement of 0
for all
= 0
T
.
Hence
(we use here that i n a Boolean alge-
bra the infinite distributive law holds, cf. Exercise 4.13).
It follows
.
{PI is contained in the complement of {PI aV a0 An important example of a Boolean algebra A is an algebra (or field)
from a
A
. a
a =
.
{PI , it is sufficient to aO or, equivalently, that thecomplement
is evidently contained in the closed set
show that
a
Ch. 17,11201
that
= 8
of subsets of a fixed set X
, partially
element of A
. According
any Boolean algebra A
ordered by inclusion. In this case and the set X
the empty set is the null element of A
itself is the unit
to Stone's representation theorem (Theorem 6.6)
can be represented as an algebra of subsets of some
P is the set of all proper prime ideals in A , then A is lattice isomorphic with the algebra of subsets of P of the
fixed set. Precisely, if form
1
{PIa = (PEP: a not contained in P
j -
The isomorphism is given by A
a
c-f
{P}a
.
of
Note that the null element 8
corresponds with the empty set and the unit element of
A
corresponds
{PI,
E
of aZ1 subsets
A
and l e t
A
{PIa of
P
be isomorphically represented by t h e algebra
. Furthermore,
i n t o t h e non-negative numbers such that v(a va 1 1 2
Then t h e mapping
=
v(a ) + v(a 1
v
a,
A
l e t v be a mapping of a2 = 8 implies
A
.
2
v may be regarded j u s t as well as a mapping from
t h e non-negative nwnbers such t h a t The f u n c t i o n
n
P be the s e t of a l l proper p r i m e i d e a l s i n the
LEMMA 120.1. Let Boolean algebra
1
v
E
into
i s a f i n i t e t y a d d i t i v e measure on E
i s even a o-additive measure on
E
.
.
Ch. 1 7 , § 1 2 0 1
PROOF. By taking
+
v(a2)
element
,
a3 E A
v
i n the formula
a l = a2 = 8'
v(0) = 0
we get
= v(a2) + v(a3)
that i f
473
ABSTRACT L -SPACES P
such t h a t
, which
. Furthermore, a,
a2 = 0
A
v(a,)
implies
1
a
v(a ) 1
v a
. It
2
,
t h e r e e x i s t s an
=
al
a2 5 a l
if and
5
v(a va ) = v ( a , ) +
i s now regarded as a non-negative f u n c t i o n on
a f i n i t e l y a d d i t i v e measure on
. For
E
. Hence
v(a,) =
follows immediately
t h e proof t h a t
v
,
E
then
is
v
i s a-additive,
. Then, s i n c e {PIa i s compact and a l l {PI {PIa = U y [P) an an a r e open, t h e r e e x i s t s a number no such t h a t {PIa = C y { P I , i.e., an Hence, i f a l l an a r e d i s j o i n t , then a = '8 f o r a l l n > no
assume t h a t
.
v({PIa) = E p o v({P} If
v
i s a non-negative
'
an)
= C;
v(IPIa
i s c a l l e d a s t a t e on
v
E
,will
v
,
v*(V)
and the subset
for a l l
W c
P
v
be the
of
V
P
the e x t e r i o r
i s defined by
V
of
, where
P i s c a l l e d v-measurable i f
Vc
denotes the s e t t h e o r e t i c d i f f e r e n c e
f o r these s e t s t h e e x t e r i o r measure v
and
A,P,E
be extended now by means of the f a m i l i a r Carathsodory
r e c a l l the following f a c t s . A l l s e t s i n the algebra measure
having
defined f o r a l l s e t s i n the
extension procedure. We r e c a l l t h a t f o r any subset measure
A
i s not i d e n t i c a l l y zero, then
A ( c f . Exercise 4.18). Let
same as i n t h e l a s t lemma. The measure algebra
).
f u n c t i o n on t h e Boolean a l g e b r a
the p r o p e r t i e s described i n the lemma and
v
n
. Every
v*
E
P-V
. We
a r e measurable and
i s the same a s t h e o r i g i n a l
s e t of e x t e r i o r measure zero i s measurable. The measurable
s e t s form a a-algebra, and t h e r e f o r e f i n i t e o r countable unions and i n t e r s e c t i o n s of measurable s e t s a r e measurable. I n p a r t i c u l a r , any s e t
v
, with Vn E E f o r a l l n , i s measurable. S e t s of t h i s type a r e I n sometimes c a l l e d a-sets. I t follows t h a t any s e t W = fl" W , with W a I n a-set f o r a l l n , i s a l s o measurable. S e t s of t h i s type a r e sometimes c a l l e d 0
= Urn V
6- s e t s .
I t i s an important f a c t f o r many a p p l i c a t i o n s t h a t every measurable
s e t i s t h e s e t t h e o r e t i c d i f f e r e n c e of a a - s e t and a s e t of e x t e r i o r 6
Ch. 17, 1203
REPRESENTATION BY MEASURABLE FUNCTIONS
474
measure z e r o . It i s customary t o w r i t e set
v(V)
i n s t e a d of
i s measurable and t o c a l l t h i s t h e measure of
V
e x t e r i o r measure of
V
.
v*(V)
i f the
i n s t e a d of t h e
V
I n p a r t i c u l a r , s e t s of e x t e r i o r measure z e r o a r e
c a l l e d s e t s of measure z e r o . The measurable s e t s v-almost e q u a l i f t h e s e t t h e o r e t i c d i f f e r e n c e s measure z e r o . I n t h i s c a s e w e s h a l l w r i t e
V
-
and
V
V-W
. The
W
a r e s a i d t o be
W
and
a r e of
W-V
relation t o be
v-almost e q u a l i s a n e q u i v a l e n c e r e l a t i o n . Under c e r t a i n a d d i t i o n a l c o n d i t i o n s on t h e Boolean a l g e b r a function
v
on
P
s e t s of
i t t u r n s o u t t h a t t h e c o l l e c t i o n of
A
i s e s s e n t i a l l y n o t much l a r g e r than t h e c o l l e c t i o n
of t h e form
. More
{PIa
t h a t t h e Boolean a l g e b r a
. These
E
v
tisfies
a
a
to
J 8
,
by
v ( b T ) J. 0
b
,
. Hence,
v(aT) J 0
then
a = sup a
i f now
i s o r d e r con-
(aT:T€{T})
t i n u o u s i n t h e s e n s e t h a t i f t h e downwards d i r e c t e d s e t ed, then
a
A
bT = 8
and
a
v
and
A,P,E
P
a
= a ) , we have
v(aT) = v(a) - v(bT)
and t h e r e f o r e
lemma, showing t h a t e v e r y open s u b s e t of LEMMA 120.2. Let
v bT
+
. We
v(a)
i n A sa-
i s upwards d i r e c t -
(aT)
e x i s t s and, d e n o t i n g t h e complement of
(i.e.,
of sets
E
a d d i t i o n a l c o n d i t i o n s are
i s o r d e r complete and t h a t
A
sub-
s e t i s t h e n v-almost
p r e c i s e l y , any v-measurable
e q u a l t o a s e t from t h e c o l l e c t i o n
and t h e
A
v-measurable
relative bT J 8
, so
f i r s t prove a
i s v-measurable.
be as before, and assume furthermore
A
is order complete and v i s order continuous in t h e sense a s exThen v*(O) = v(O-) plained above. Now, l e t 0 be an open subset of P that
0
where
-
denotes t h e closure of
exists a set since
V
V
c 0 c
€ E
such t h a t
VE c 0
v
0-
0- w i t h
"(o-)
-
. Furthermore,
0
"(VE) <
and
.
f o r any
A
i s o r d e r complete, t h e h u l l - k e r n e l the closure
as w e l l a s c l o s e d . I t f o l l o w s t h a t E A
. In
o t h e r words, 0
makes s e n s e t o w r i t e
u(O-)
t h e topology. The open set
-
0
-
0
-
. The
0
there
i s measurable.
topology i n
of any open s e t
i s of t h e form
i s one of t h e s e t s from
0
0
. Hence,
,
E
extremally disconnected, i.e., a.
E
measurable and
i t follows from a f a m i l i a r theorem i n measure theory t h a t PROOF. S i n c e
>
E
v(VE) > v*(O) -
and
E
sets i n the a l g e b r a
i s t h e r e f o r e o f t h e form
{P}
. Ta0 his
E
0
P
is
i s open
f o r some shows t h a t i t
form a b a s e f o r 0 = ti
{PI
. Note
ABSTRACT L -SPACES P
17,11201
Ch.
t h a t any f i n i t e union of s e t s
{PIa
475
i s again of the form
adding a l l these f i n i t e unions t o t6e s e t of a l l
0
s i t u a t i o n t h a t we may assume t h a t
0
d i r e c t e d and sup v({P} find t h a t
-
.
)
a,
{PI
=
for
aO Since e v i d e n t l y
v(o-)
= V*(O)
a
UT {PI
=
. aT It
= sup a
v(O-)
Z
v*(O)
{PI aT
with
, we
a r e i n the
(aT:T'E{?))
follows t h a t
t v({P}
a,
. Hence,
{PIa
upwards
v(O-)
=
for a l l
)
T
, we
0.
= sup v {PI
a7
A l l r e s u l t s t o be proved follow now immediately.
P
Since
i s compact i f and only i f i t i s
P
i s compact, any s u b s e t of
closed.The u-algebra generated by the compact s e t s ( i . e . ,
the smallest
u-algebra containing a l l compact s e t s ) i s t h e r e f o r e the same a s the ualgebra generated by the open s e t s . Borel sets i n
P
.
The s e t s i n t h i s a-algebra a r e t h e
Since every open s e t is v-measurable by the l a s t lemma,
i t follows t h a t t h e Borel s e t s form a a-subalgebra of t h e a-algebra of a l l v-measurable s e t s . As promised, we prove now t h a t every v-measurable s e t
i s v-almost equal t o a s e t i n the a l g e b r a open-closed
E
,
i.e.,
v-almost equal t o an
set.
THEOREM 120.3. Let
A
be an order co?rplete Boolean algebra, P
the s e t
o f a l l proper prime i d e a l s i n A , E the alqebra of a l l subsets o f P of the form {PIa ( e q u i v a l e n t l y , the algebra of a l l open-closed s e t s i n t h e hull-kernel topology i n P ) and v an order continuous f u n c t i o n on A Then ( a s already proved) P . The measure v has the property t h a t every v-measurable s e t i s v-almost equal t o a s e t i n the algebra E . Hence, i f as usual we i d e n t i f y v-almost equal s e t s , we may say t h a t E i s a c t u a l l y the c o l l e c t i o n of a l l v-measurable s e t s . s a t i s f y i n g the conditions mentioned i n Lemma 120.1. v
i s a measure i n
PROOF. L e t
W =
n"I
(with
W
V
be an a r b i t r a r y v-measurable s e t . There e x i s t s a a o-set f o r every
Wn
n r e t i c d i f f e r e n c e of
W
and a s e t of measure zero. Hence
v-almost equal; i n n o t a t i o n (Wn:n=l,2,
fig,
Wk
...)
, which
n ) such t h a t
V -W
. It
V
and
W
are
may be assumed t h a t the sequence
i s descending, because otherwise we replace i s a l s o a o-set.
a -set 6
i s t h e s e t theo-
V
It follows t h a t
by
Wn
v(Wn) J. v(W)
.
WA =
Every
n'
Ch. 1 7 , 9 1 2 0 1
REPRESENTATION BY MEASURABLE FUNCTIONS
416
i s t h e union of (countably many) open-closed s e t s , so every
v(Wn) = v(Wn)
I n view of t h e preceding lemma we have then t h a t
The sequence of open-closed s e t s
-
Wn
i s descending. The s e t
Wi
i s open.
,
S =
i s t h e r e f o r e c l o s e d , and
n"I
Wn
.
i v(S)
V(W;1)
and so
-
.
Hence v(W) = v(S) and W = fl" W c n" W- = S , s o W S Now, a s we I n I n have s e e n , any open s e t and i t s c l o s u r e have t h e same measure. Taking complements, i t follows t h a t any c l o s e d s e t
and i t s i n t e r i o r
F
int F
have t h e same measure. Furthermore, s i n c e t h e c l o s u r e of any open s e t i s open-closed,
t h e i n t e r i o r of any c l o s e d s e t i s a l s o open-closed.
this t o the s e t S open-closed. that
V
-
, we
get
v(S)
=
, so
v ( i n t S)
S
-
int S
Combining now a l l r e s u l t s , we conclude from
int S
with
int S
V
an open-closed s e t .
Applying
- - with
W
int S
S
int S
The measure theory as developed above w i l l b e a p p l i e d i n t h e f o l l o w i n g s i t u a t i o n . Let a l l bands i n
22.8). K2
a r e bands, t h e n
g e n e r a t e d by of
{O}
K
K
i s t h e band
K
1
K1
A
K2 = K1
. In particular,
U K2
i f and only i f
K1
and
n K1
K2
A
K
P
p r o p e r prime i d e a l s i n sets
K
. The
i s e q u a l t o t h e n u l l element
. Assuming
now t h a t
v
L
. Accor-
i s l a t t i c e isomorphic w i t h t h e
, where P
i s t h e s e t of a l l
isomorphism i s d e f i n e d by
K
++
{ P I K a r e e x a c t l y t h e open-closed s e t s i n t h e h u l l - k e r n e l
P
K1
i s t h e band
a r e d i s j o i n t bands i n
{ P I K of
Boolean a l g e b r a of a l l s u b s e t s
K2
itself. If
L
K 1 v K2
and
K2
d i n g t o t h e Stone r e p r e s e n t a t i o n theorem
of
of
g e n e r a t e d by t h e n u l l
{O}
i s t h e space
K
and t h e u n i t element of
L
K
i s an o r d e r complete Boolean a l g e b r a ( c f . Theorems 2 2 . 7 and
The n u l l element of
element of and
b e a Dedekind complete Riesz space. The c o l l e c t i o n
L L
{P}K ; t h e topology
i s a non-negative r e a l f u n c t i o n on
K
such
that
0
v K1vK2
for disjoint
KI,K2
= v(K,)
and
+ v(K2)
v(KT) i 0
for
KT
+
{O}
,
i t f o l l o w s from t h e
Ch. 17,81201
ABSTRACT L -SPACES
r e s u l t s proved s o f a r t h a t v
p
the s e t
477
P
may be regarded a s a 0-additve measure i n
such t h a t modulo s e t s of v-measure zero t h e c o l l e c t i o n of a l l
procedure
v
of a f u n c t i o n
p
f o r some e
. By
in
i s simply defined t o be equal t o
v({P} ) K
of t h i s kind i s obtained i f
satisfying
1
5
Theorem 118.2 the space e
K
in
by
order p r o j e c t i o n on t h e band where
i s the norm i n
p
v!K
d i s j o i n t and
L
) J. 0
v(K)
i s an a b s t r a c t L -space
L
f i r s t that
L
P has a weak u n i t
i s Dedekind complete and the norm
L
i s o r d e r continuous. Every band
L
the component of
K2
. Assume
p <
. In this . An example
{PIK
v-measurable s e t s i s e x a c t l y the c o l l e c t i o n of a l l s e t s
,we
K
i s a p r o j e c t i o n band; denoting
k = Q e , whete QK i s t h e K For every band K , l e t V(K) = pP(QKe),
k
K . . Then
have
v(K VK ) = v(K ) + V(K2) f o r K1 and 1 2 1 KT 4 10) ; the f i r s t immediately from t h e
for
d e f i n i t i o n of an a b s t r a c t L -space and the l a s t because KT C { O } P e 4 0 and p i s order continuous. Transplanting t h e r e f o r e v
QKT
p
implies t o the
P by d e f i n i n g v({P}K ) = v(K) f o r every K , we o b t a i n a o-additive measure v i n P such t h a t , modulo s e t s of measure zero,
subsets
{PIK of
every v-measurable s e t i s one of t h e s e t s
{PIK
. Since
x{PIK i s t h e c h a r a c t e r i s t i c f u n c t i o n of
where
correspondence between t h e components functions
QKe of
preserves the L -norm.
x{P},
P
{PIK e
, the
one-one
and t h e c h a r a c t e r i s t i c
I t i s now a n a t u r a l question t o ask
whether t h e r e e x i s t s a norm preserving one-one corresponding between t h e elements of
L
and those of t h e space
Lp(P,v)
, an
extension of the one
j u s t e s t a b l i s h e d . Such an extension does indeed e x i s t and t h e r e a r e s e v e r a l methods t o prove t h i s . For t h e method we s h a l l use i t i s necessary t o e s t a b l i s h f i r s t some f a c t s about t h e subspace of combinations of components of t h e weak u n i t
L
formed by the f i n i t e l i n e a r
e ; we s h a l l c a l l these e-step
elements. Once again, l e t weak u n i t
K
e
. All
bands
w i l l be denoted by
ion on
K
sup(kl,k2)
. Note and
combination :1
L
k
that i f inf(k aiki
be a Dedekind complete Riesz space possessing a K
a r e p r o j e c t i o n bands; t h e component of
. Hence kl
e
in
k = Q e , where QK i s the band p r o j e c t K and k2 a r e components of e , then
.
k ) are again components of e Any l i n e a r 1’ 2 of components of e (with r e a l c o e f f i c i e n t s ) i s c a l l e d
Ch. 1 7 , 5 1 2 0 1
REPRESENTATION BY MEASURABLE FUNCTIONS
478
an e - s t e p element. I f a l l C y ki
=
vn k = e ) and i f a l l a . a r e m u t u a l l y d i f f e r e n t , t h e n 1 i i s s a i d t o be w r i t t e n i n standard form. This i m p l i e s , t h e r e f o r e ,
e (i.e.,
C p aiki
a r e nonzero and m u t u a l l y d i s j o i n t such t h a t
ki
t h a t one of t h e c o e f f i c i e n t s may b e z e r o . I f i t i s only g i v e n t h a t a l l
E y ki
a r e m u t u a l l y d i s j o i n t and
standard form. I f
s = Cn o k .
and
s t 0
s
a.k. t 0 1 1
satisfies
,
a. t 0
so
1
,
'i
and
and
= s+
C'
= s
-"C
-
L
a. < 0
such t h a t
.
f o r a t l e a s t one of
and
s+
s
-
s
i
i.e.,
s = Ca. ki
(al-
. Then
s = 'C
'C
-
and
-"C
.
(-"E)
It
s = C
then t h e r e i s no uniqueness
assume, t h e r e f o r e , t h a t n a.k. a r e non-negative 1
1
s
1
m
s
is
i f this
6.k' i n s t a n d a r d form. I ~j i s t r i v i a l , we may assume t h a t a. > 0 f o r a t
0
=
,
s
i s t h e s t a n d a r d form. Assume now t h a t a l s o l e a s t one v a l u e of
if
,
i s w r i t t e n i n standard form, t h e n
s
. We may
a l l coefficients i n
Since the case t h a t
...,n
i s unique.
s
PROOF. I f t h e r e i s no uniqueness f o r positive, i.e.,
I,
c o n s i s t s of a l l t e r m s w i t h
'C
LEMMA 120.4. I f an e-step element t h i s way of w r i t i n g
=
...,n . Furthermore,
, where
are p o s i t i v e and d i s j o i n t elements of follows t h a t
i
for
i
c o n s i s t s of a l l term w i t h
"1
ki i s w r i t t e n i n almost
Cn a . k 1 i i
then
QK s t 0
i = l,
for
,
e
i n s t a n d a r d form o r almost s t a n d a r d form
1
then
m o s t s t a n d a r d ) , l e t s = '1 + "1
a. 2 0
=
. Without
= C
l o s s of g e n e r a l i t y , l e t t h i s be f o r
i = 1
Then
Not a l l kl
A
kl
k; > 0
k'
A
j
a l k l > 0 ) ; w i t h o u t l o s s of g e n e r a l i t y , l e t
vanish (since
. Then
QK; Q K l s = " I \ (k 1 ~ k1 )" = B I \(k 1hk"1) which shows t h a t only one
kl kl
A
Bj
k! > 0 J
. Since
a, =
B1
. It 1
all
,
= (vk!)
J
A
, we
kl
=
Bj
i.e.,
follows t h a t a l l o t h e r
Ck! = vk' = e J j k
. Since B j - al
satisfying
a r e mutually d i f f e r e n t , t h e r e i s k;
k!
J f i n d now t h a t
k'
1
h
k
1 '
'
i s t he only
k' satisfying j a r e d i s j o i n t t o the given
Ch. 17,51201
SO
Blk;
k' 1
2
ABSTRACT L -SPACES P
. But
kl
now, s i n c e
i n s t e a d of f o r
B1
. We
alkl
=
a1 > 0
, we
g e t thus t h a t
It f o l l o w s t h a t e v e r y term o c c u r r i n g i n
479
can r e p e a t t h e argument f o r k
Ca.k.
1
1 1
. Hence
Z k;
with
a. > 0
kl = k;
.
occurs also
CB.k' ; c o n v e r s e l y , e v e r y t e r m i n ZB.k! with B . > 0 o c c u r s i n ~j J J J Ca.k. But t h e n t h e t e r m w i t h c o e f f i c i e n t z e r o ( i f o c c u r r i n g ) i s a l s o t h e in
.
1 1
same i n b o t h e x p r e s s i o n s .
s
COROLLARY 120.5. I f
= C y aiki
i s the unique standard form f o r
Is1
i n standard form, then
.
C y lailk;
THEOREM 120.6. Every e-step element can be w r i t t e n uniquely i n stand-
ard form. PROOF. L e t f i r s t k i j = ki
A
k'
for a l l
j
s = Ca k
i i
. Then
i,j
and
s'
= CB k '
ki = C.k 1 ij
j j and
i n s t a n d a r d form. L e t k! = Cikij I
,
so
and t a k i n g t o g e t h e r kij = 0 terms w i t h e q u a l c o e f f i c i e n t s , w e o b t a i n t h e s t a n d a r d form f o r s + s ' By
i n almost s t a n d a r d form. Omitting terms w i t h induction i t follows t h a t i f then s o can
s1
+
... +
s
s i s t s of one t e r m , i . e . ,
,
e-k
we g e t
s
n s = ak
n e c e s s a r i l y a R i e s z space. L e t that
in T
L
,
into
L
w e have a r e a l v e c t o r s p a c e
i i
T
M (in notation
It f o l l o w s immediately t h a t
we had b e f o r e i s t h a t . T
T
M
, not
yet
b e a mapping from t h e s e t of components q
=
~ ( k ) f o r every
i s a d d i t i v e i n the sense t h a t i f
t h e mapping
con-
can be w r i t t e n i n s t a n d a r d form.
a k.
kl
A
k = QKe )
k2 = 0 ( i . e . , k l
= k l + k 2 ), t h e n
v
s
t h e n by adding a t e r m e q u a l t o z e r o t i m e s
Cn
1
Assume now t h a t b e s i d e s e
can be w r i t t e n i n s t a n d a r d form,
now t h a t i f t h e e - s t e p element
i n s t a n d a r d form. Hence, i n view of t h e r e s u l t a l r e a d y
proved, e v e r y e - s t e p element
of
...,s n
sl,
. Note
.
~ ( 0 =) 0
. The main
V
such k
2
=
d i f f e r e n c e w i t h t h e mapping
i s n o t n e c e s s a r i l y r e a l v a l u e d . W e wish t o extend
t o t h e s p a c e of e - s t e p elements by d e f i n i n g
Ch. 17,11201
REPRESENTATION BY MEASURABLE FUNCTIONS
480
n s = 1 a.k 1 i i
f o r any e - s t e p element
This d e f i n i t i o n makes s e n s e o n l y i f s (i.e,,
, not C
i t should n o t depend on how
n e c e s s a r i l y i n s t a n d a r d form.
a . ~ ( k . ) i s uniquely determined by
1
1
1
i s w r i t t e n as a l i n e a r combination
s
of components). We prove f i r s t t h a t t h i s i s indeed t h e c a s e .
If s
LEMMA 120.7.
aiki
= Cy
PROOF. Assume f i r s t t h a t
+
S I
=
C
1 1
A
k! 1
,
, then
and
s'
=
we have
ZB.k! J J
, both
i n stand-
c ~ (ai+Bj)kij , ~
i n almost s t a n d a r d form. S i n c e t h e a d d i t i v i t y of
s = Ca.k.
kij = k . 1
a r d form. W r i t i n g a g a i n
Bjk;
= C y
k. = C. k 1
that
T
J
a i ~ ( k . )= C . a. ?(kij) 1,j 1
ij
for all
i
, it
f o l l o w s from
.
Similarly
Hence
+
C a. T(k.) 1
1
and u s i n g a g a i n t h a t
T
C B . ~ ( k ! ) = Zi
J
J
i s a d d i t i v e , i t i s e a s y t o see t h a t t h e l a s t
, where C y k" i s t h e s t a n d a r d form P P + s B y i n d u c t i o n t h i s r e s u l t i s extended t o t h e sum s 1 + for n ' e a c h of s l , s i n s t a n d a r d form. n n L e t now s = E l aiki , n o t n e c e s s a r i l y i n s t a n d a r d form. Write each expression is equal t o s+s'.
C yp r ( k i )
...
...,
term s e p a r a t e l y i n s t a n d a r d form by adding a t e r m w i t h c o e f f i c i e n t zero. This g i v e s us sI +
... +
s
n
s
1
,...,sn . Applying
, we
get
the j u s t obtained r e s u l t t o
Ch. 17,51201
where C
6 k' , j i
48 1
P
y k"
C
Hence
ABSTRACT L -SPACES
i s the standard form of P P then we g e t s i m i l a r l y
C oi .r(ki)
=
1
can a l s o be w r i t t e n as
s
.
Z 5 . r(k!) 1
. If
s
3
We s p e c i a l i z e the s i t u a t i o n by assuming now t h a t complete Riesz space possessing a weak u n i t
mapping o f t h e Boolean a l g e b r a of a l l components algebra of a l l components of plained above, because
kl
A
. Then
d
k2 = 0
be an isomorphic
T
k
i s a Dedekind
M
. Let
d
e
of
onto the Boolean
i s a d d i t i v e i n the sense ex-
T
implies
r(kl)
A
r(k2) = 0
by the
isomorphism. Hence
I t follows now from the l a s t lemma t h a t T
can be extended uniquely t o t h e
space of a l l e-step elements by d e f i n i n g T
fZ
a.k
i,
=
.
Z a. ~ ( k . ) 1
1
Every d-step element i s the image of some e-step element, i . e . , of
i s now t h e space of a l l d-step elements.
T
The s i t u a t i o n becomes even more s p e c i a l i f t i v e and u-additive) i n a p o i n t s e t i s the space
M(X,v)
X
i s a measure (non-nega-
v
such t h a t
v(X)
for a l l
x E X
i s now a weak u n i t i n
the c h a r a c t e r i s t i c functions s e t of
X
i s f i n i t e and
of a l l r e a l v-measurable f u n c t i o n s on
t i f i c a t i o n of v-almost equal f u n c t i o n s ) . The f u n c t i o n = 1
the range
. The mapping
x,(x)
M D
X (with iden-
satisfying
and the components of
d(x) = d
algebra of a l l components of s e t s . I n t h i s case, t h e r e f o r e ,
-i(z
e
onto the Boolean algebra of allv-measurable T
may be extended t o t h e space of a l l e-step
.
a . k . ) = Z ai ~ ( k . ) The extended 1 1
T
is
l i n e a r and maps the space of e-step elements i n a one-one manner onto the space of v-step functions.
are
an a r b i t r a r y v-measurable sub-
i s now an isomorphic mapping from t h e Boolean
T
elements by d e f i n i n g t h a t
for
d
M
REPRESENTATION BY MEASURABLE FUNCTIONS
482
Assume now, once more, t h a t satisfying
we s e t
1 5 p <
component of
e
a l g e b r a of a l l
p
. Again,
m
v(K) = pP(QKe)
i s called
k = Q e K
let
, where
i n t h e band
L
p
.
K
i s an a b s t r a c t L -space f o r some p P e b e a weak u n i t i n L . A s b e f o r e ,
i s t h e norm i n L and Q e i s t h e K The isomorphic mapping of t h e Boolean
o n t o t h e Boolean a l g e b r a of t h e s u b s e t s
T
. As
i s ( e x c e p t f o r a s e t of v-measure
P
of
{P}K
and
T
i s a measure on t h e a l g e b r a of a l l {P}K in
Ch. 17,51201
proved, e v e r y v-measurable s e t
z e r o ) one of t h e
{PIK
. Hence,
i s e s s e n t i a l l y a mapping o n t o t h e c o l l e c t i o n of a l l v-measurable s e t s .
A s above, w e e x t e n d
T
t o t h e s p a c e o f a l l e - s t e p e l e m e n t s by d e f i n i n g t h a t
. done t h i s , we compute T(Z a . k . ) = C a . ~ ( k . ) Having 1 1
s =
Z a.k.
1 1
. For
1
1
t h i s p u r p o s e , we may assume t h a t
a r d form. Furthermore, f o r b r e v i t y , w r i t e Then, by t h e a d d i t i v i t y of
M
Once more f o r b r e v i t y , w r i t e
M(f)
s
pp(s)
for
i s w r i t t e n i n stand-
= pp(f)
f o r any
f € L
.
f o r d i s j o i n t elements,
xi
f o r t h e c h a r a c t e r i s t i c f u n c t i o n of
{PIKi
{PI a r e m u t u a l l y d i s j o i n t . The l a s t e x p r e s s i o n Ki i n !the formula above now becomes
and n o t e t h a t t h e s e t s
The one-one l i n e a r mapping now e x t e n d
T
T p r e s e r v e s , t h e r e f o r e , t h e L -norm. We s h a l l P once more, which g i v e s r i s e t o t h e f o l l o w i n g theorem.
THEOREM 120.8. Any abstract L -space f o r a v a h e o f P
1 5 p <
type
m
p
such t h a t
and possessing a weak u n i t i s Riesz isomorphic w i t h a space of
L (X,v) P
, where
v(X)
i s f i n i t e . The isomorphism i s mm preserving.
More i n detaiZ, t h e weak u n i t i n
L
corresponds (under t h e isomorphism)
w i t h t h e f u n c t i o n identieaZZy one on X and the s e t
X
is a compact
Ch. 1 7 , 9 1 2 0
ABSTRACT L -SPACES P
483
Hausdorff t o p o l o g i c a l s p a c e which is e x t r e m a l l y d i s c o n n e c t e d and h a s t h e p r o p e r t y t h a t e v e r y v-measurable s e t is v-almost e q u a l t o an open-closed s e t in t h i s topoZogy. PROOF. The space L ( X , v ) w i l l be t h e space L (P,v) P P a r e as i n t r o d u c e d above. The one-one l i n e a r mapping T
v
i n t r o d u c e d , maps t h e space of a l l e - s t e p elements i n v-step
functions i n
. Note
Lp(P,v)
a r e p o s i t i v e mappings. We e x t e n d f
pose, l e t p(f-s
f o r any
) J. 0 > 0
E
o n t o t h e s p a c e of
L
and i t s i n v e r s e
T
t o t h e whole space
be an a r b i t r a r y p o s i t i v e element of
wards d i r e c t e d system Then
T
t h a t both
, where P and , a l s o already
.
L
. There
L
t h e r e e x i s t s a n e - s t e p element
.
e x i s t s a n up-
I n o t h e r words, t h e e - s t e p elements a r e l y i n g norm dense i n of c o u r s e , t h e v-step f u n c t i o n s a r e l y i n g norm dense i n obvious now how t o extend v i n g . Indeed, i f ments such t h a t in
Lp(P,v)
f
such t h a t
T
p(f-sn)
+
.
0
Then
( s :n=1,2,
n (Tsn:n=1,2,
n o t on t h e approaching sequence. Now d e f i n e Then
T
,
p(f-s
t h u s d e f i n e d on
L T
.
, is
Lp(P,v)
...)
...)
T
-1
T
(sn:n=1,2,
...)
Tf
=
I
Ifl-lsnl
f
,
I
,
f
and
t o be t h i s l i m i t function.
obviously l i n e a r , norm p r e s e r v i n g and a n
b e a p o s i t i v e element of
b e e - s t e p elements such t h a t
I
It i s
i s a Cauchy sequence
T
T i s t h e whole s p a c e L (P,v) P i s a Rie.sz isomorphism, i t remains t o prove t h a t T
If-lsnl we have
.
Extending s i m i l a r l y t h e o r i g i n a l
are p o s i t i v e mappings. L e t
5
1)
If-snl
p(f-sn)
+
0
L
. Since
-1
,
. For
it the
and
and l e t
I
.
L (P,v) P A l l T ( ~ S I ) are p o s i t i v e and t h e p o s i t i v e cone i s norm c l o s e d , s o Tf i s n -1 p o s i t i v e . This shows t h a t T i s p o s i t i v e . The proof f o r T is similar.
p(f-lsnl)
+
0
Tf-T(/s
t e n d s t o z e r o i n norm i n
The s t a t e m e n t s i n t h e theorem about t h e topology i n
X = P
were proved
e a r l i e r i n t h i s section. The theorem which w e have t h u s proved shows t h a t an a b s t r a c t L -space
(I
E.
b e e - s t e p ele-
becomes e v i d e n t t h a t t h e range of proof t h a t
) <
. Similarly,
L
t h e r e f o r e a l i m i t . The l i m i t depends only on
e x t e n s i o n of t h e o r i g i n a l
4 f.
s
5
remains l i n e a r and norm p r e s e r -
T
E L is a r b i t r a r y , l e t
, having
0
It f o l l o w s t h a t
satisfying
s
-1
For t h i s pur-
of e - s t e p elements such t h a t
(sa:a€{a))
by t h e o r d e r c o n t i n u i t y of t h e norm p
T
P
can b e r e p r e s e n t e d (Riesz i s o m o r p h i c a l l y and i s o m e t r i c a l l y ) as a
REPRESENTATION BY MEASURABLE FUNCTIONS
484
Ch. 17,11201
concrete L -space of measurable functions. The question a r i s e s what can be P i n f e r r e d i n the more general s i t u a t i o n t h a t L i s a Banach l a t t i c e with 'order continuous norm and having a weak u n i t . We s h a l l prove t h a t a l s o i n t h i s case
L
i s Riesz isomorphic t o a space of measurable functions. For
t h i s purpose we show f i r s t t h a t under t h e mentioned conditions
L
possesses
a s t r i c t l y positive l i n e a r functional. THEOREM 120.9. Every normed Riesz space
L
with order continuous norm
and having a weak u n i t possesses a norm continuous and s t r i c t l y positive linear functional. Note that the functional i s also order continuous (since
p
the norm i s order continuous). PROOF. In Theorem 103.12 i t was proved t h a t i f
space with order continuous norm and
A
L
t h e r e e x i s t s a norm continuous l i n e a r f u n c t i o n a l on p o s i t i v e on
A
. In
i s a normed Riesz
is a principal ideal i n A
the p r e s e n t s i t u a t i o n , l e t
L
L
, then
which i s s t r i c t l y
be the p r i n c i p a l i d e a l
generated by the weak u n i t . There e x i s t s t h e r e f o r e , a norm continuous linear f u n c t i o n a l $
$
on
such t h a t
L
i s a l s o s t r i c t l y p o s i t i v e on
L
$
i s s t r i c t l y p o s i t i v e on
, because
A
. Then
i s the band generated by
L
A .
For l a t e r a p p l i c a t i o n s i t i s of i n t e r e s t t o consider t h e s i t u a t i o n t h a t L
i s a Dedekind complete Riesz space with a weak u n i t and possessing a
s t r i c t l y p o s i t i v e o r d e r continuous l i n e a r f u n c t i o n a l . Note t h a t i f
L
is
a Banach l a t t i c e with o r d e r continuous norm and having a weak u n i t , then L
s a t i s f i e s these conditions ( t h e Dedekind completeness follows from
Theorem 103.9 and the e x i s t e n c e of a s t r i c t l y p o s i t i v e o r d e r continuous l i n e a r f u n c t i o n a l follows from the l a s t theorem). THEOREM 120.10. Let
unit tional
e
L
be a Dedekind complete Riesz space with a weak
and possessing a s t r i c t l y positive order continuous Zinear func$
.
Then
L
i s Riesz isomorphic with an ideal
L l ( X , v ) of ( r e a l ) v - s m a b l e functions, where
v ( X ) = + ( e ) 1 . The weak u n i t i n
L
the function i d e n t i c a l l y one on
X
V(X)
M
i n a space
is f i n i t e (actually,
corresponds (under the isomorphism) with
. It
folluws that every bounded v-measur-
.
able function on X i s contained i n M This r e s u l t holds i n particular i f L i s a Banach l a t t i c e with order continuous nomi and having a weak u n i t .
Ch. 17,11201
ABSTRACT L -SPACES P
PROOF. Define
in
L
p(f) = + ( I f / )
f o r every
. Then
f E L
algebra
of a l l bands i n
K
.
K
and, f o r every band
L
K
The Boolean a l g e b r a
,
K
let
i s a d d i t i v e and o r d e r continuous on
+
from t h e o r d e r c o n t i n u i t y of a l g e b r a of a l l i.e.,
be t h e
Q,
i s o r d e r complete ( s i n c e
i s Dedekind complete). Likewise a s b e f o r e , t h e f u n c t i o n
,
.
p(u+v) = p(u) + p(v) f o r a l l u , v E L+ P b e t h e s e t of a l l p r o p e r prime i d e a l s i n t h e Boolean
band p r o j e c t i o n o n t o L
i s a norm
p
having t h e p r o p e r t y t h a t
As b e f o r e , l e t
T
485
v(K) = p(QKe)
K ( t h e o r d e r c o n t i n u i t y of
follows
v
) . The isomorphic mapping from t h e Boolean
{PIK i s denoted by
QKe o n t o t h e Boolean a l g e b r a of a l l
T(Q,~) = {PIK f o r a l l
K
. Transferring
the function
to the
v
{PIK by d e f i n i n g
c o l l e c t i o n of a l l
=
,
p(QKe) = $(Q,e)
we o b t a i n once more ( a s i n t h e L Lease) a o - a d d i t i v e measure v such t h a t P a f t e r e x t e n s i o n by means of t h e Carathsodory procedure, any w-measurable s u b s e t of P
i s v-almost
isomorphism
T
T ( S ) = Z;
e q u a l t o one of t h e s e t s
t o t h e s e t of a l l e-step elements
1
1
$(lsl)
= P(S)
I
=
s
. Extending
= Zn
k l ,...,k
f o r a r b i t r a r y components
a. T ( k . )
{PIK
1
n
ci.k i i e
of
the
by d e f i n i n g
, we
get
.
IT(S)ldW
.
For any S i m i l a r l y as i n t h e L -case, we now extend T t o t h e whole of L P of e-step e l e f E L t h e r e e x i s t s an i n c r e a s i n g sequence (s,)
positive
ments such t h a t t h e norm
p
sequence i n
.
.
0 5 s
I. f Then p(f-sn) n It follows t h a t t h e sequence
LI(P,w)
quence. Note t h a t
. We
-rf
define
,
T
e x t e n s i o n of t h e o r i g i n a l
T
by t h e o r d e r c o n t i n u i t y of
( T S ~ ) of images i s a Cauchy
t o b e t h e l i m i t of t h i s Cauchy se-
Tf
depends only on
sequence. Note a l s o t h a t
J. 0
f
and n o t on t h e approximating
thus d e f i n e d on
, is
L
with the property t h a t
obviously a l i n e a r T
-1
preserves t h e
L -norm (and, t h e r e f o r e , t h e i n v e r s e mapping T e x i s t s ) . The proof t h a t 1 -1 T and T a r e p o s i t i v e i s a s i n t h e L -case. Hence, T i s a Riesz isomorP It remains t o prove t h a t M i s phism from L onto i t s image M = r(L)
an i d e a l i n
L1(P,u)
.
.
Denote t h e f u n c t i o n s i n
(with i d e n t i f i c a t i o n of w-almost 5
v
N
with
v" = T ( V ) f o r some
of s t e p f u n c t i o n s such t h a t
LI(P,v)
by
fN,g , u N
e q u a l f u n c t i o n s ) . Assume now t h a t v E L+
0 d s-
+
u"
. There
e x i s t s a sequence
and f o r e v e r y
N
s
n
N
,...
0 d u
N
5
N
(sn)
t h e r e e x i s t s an
REPRESENTATION BY MEASURABLE FUNCTIONS
486
e - s t e p element
2 0
s
such t h a t
0 5 s
such t h a t
0 <
Hence, s i n c e
N
0
follows t h a t
0 < u
5
.
Ll(P,w)
v
N
with
v
s
N
5
by h y p o t h e s i s , w e have N
i s an i d e a l i n
. It
L
4 u”
S”
T h i s shows t h a t i f M
in
4 u
n
.
,
= s Then 0 < s 4 5 v i n L n n L t h e r e e x i s t s a n alzment u E L+
TS
s o by t h e Dedekinnd completeness of
Ch. 17, 51201
N
u
E M , then
N
in
4 TU
,
= TU
E M
tiN
LI(P,v) N
i.e.,
u
.
E M
. Therefore,
.
We e x t e n d Theorem 120.8 ( a b s t r a c t L -space f o r 1 S p < m ) and Theorem P 120.10 (Dedekind complete s p a c e w i t h s t r i c t l y p o s i t i v e o r d e r c o n t i n u o u s l i n e a r f u n c t i o n a l ) t o t h e c a s e t h a t t h e s p a c e does n o t have a weak u n i t . I n e i t h e r of t h e s e c a s e s , l e t by
Ka
. The
. The
L
p o s i t i v e elements i n system
every
t h e element
a
b e a maximal d i s j o i n t system of s t r i c t l y
band g e n e r a t e d i n
L
K
i s a weak u n i t i n
e
measure s p a c e s w i l l b e denoted by
. Hence,
.
Ka
T (K )
a
spaces
(Pa,Aa,va)
t o b e open whenever
P
C
. The
thus obtained
V
n
Pa
i s open i n X
Pa
. The
.
f o r every
Pa
X
. It
of
X
Pa
follows t h a t
of
V
X
Pa
satisfying
f o r every
v
(since every
from
V E A
A
ha
V
n Pa A
E Aa
s e e t h i s , obC
f o r every
a
,
so there
w i l l be
of a l l measurable s e t s i s a
i s a a-a3gebra i n
i n t o t h e extended r e a l numbers by
( t h i s implies t h a t
. To
Pa
v(V) =
m
P
) . We d e f i n e a
w(V)
=
X a va(VnPa)
i f t h e r e a r e more than
a o u n t a b l y many s t r i c t l y p o s i t i v e t e r m s ) . I t w i l l b e shown f i r s t t h a t a non-negative for
V,W E A
and a - a d d i t i v e measure. E v i d e n t l y , v ( $ ) = 0 with
V c W
. For
them-
i s a compact sub-
i s an open c o v e r i n g of
c a l l e d a measurable s e t . The c o l l e c t i o n X
Pa
these
i s a l o c a l l y compact s p a c e . Any compact sub-
X
s e r v e t h a t t h e c o l l e c t i o n of a l l
mapping
Pa
i s c o n t a i n e d i n a f i n i t e union of s e t s
Any s u b s e t
i s easy
i s Hausdorff
X
i s open a s w e l l a s
i n i t s own topology i s compact, Pa
e x i s t s a f i n i t e subcovering.
o-algebra i n
Every
, it
a
topology i n
i s H a u s d o r f f , and f o r p o i n t s i n d i f f e r e n t
c l o s e d , and s i n c e s e t of
Hence, we may a p p l y
Ka
.
s e l v e s a r e d i s j o i n t open neighborhoods).
set
. For
{O}
and t h e c o r r e s p o n d i n g R i e s z
T (K )
t o s e e t h a t t h i s d e f i n e s a topology i n (each
K =
i s e i t h e r t h e s p a c e L (P , A a a p a a%) Or i s an i d e a l i n the space L (P A ) Thinking of t h e t o p o l o g i c a l 1 a ’ a’va Pa a s d i s j o i n t s e t s , we p u t X = !ja Pa D e f i n i n g a s u b s e t V of Ta
isomorphism by
X
w i l l b e denoted
e
Ka , then
i s disjoint to a l l
e i t h e r Theorem 120.8 o r Theorem 120.10 t o e v e r y
a
by
i s a d i s j o i n t system of b a n d s , maximal i n t h e
{Ka}
s e n s e t h a t i f t h e band
{e }
t h e proof t h a t
v
and
i s a-additive,
v
is
v(V ) 5 v(W) let
Ch. 1 7 , 5 1 2 0 1
ABSTRACT L -SPACES P
E A
V =U V and a l l Vn m u t u a l l y d i s j o i n t . Then I n w i t h d i s j o i n t t e r m s , and hence
V,V =
E
U
487
with
(V n P ) 1 n a
We now a p p l y CarathGodory's e x t e n s i o n p r o c e d u r e t o t h e measure does n o t produce any v - m e a s u r a b l e s e t s which a r e n o t a l r e a d y
A
. For
t h e proof, n o t e f i r s t t h a t every s u b s e t
Now, l e t
be a s u b s e t of
V
W c Pa
Then, f o r any
, we
W
Pa
of
V
n Pa
=
. This
v
members of satisfies
which i s v - m e a s u r a b l e ( a f t e r t h e e x t e n s i o n ) .
X have
V*(W) = v *(w) = v*(wnv) + v*
T h i s shows t h a t
V
t h a t t h e a-algebra s e t s of
X
.
n Pa
i s v measurable, i.e.,
A
V ll Pa E
Aa
. It
follows
i s p r e c i s e l y t h e o - a l g e b r a of a l l v - m e a s u r a b l e sub-
As u s u a l , t h e Riesz s p a c e of a l l e q u i v a l e n c e c l a s s e s of r e a l v-almost everywhere f i n i t e v a l u e d v - m e a s u r a b l e f u n c t i o n s on
. Note
M(X,v)
that
v need n o t be o - f i n i t e .
p r o j e c t i o n on t h e band
. We
Ka
In
L
d e f i n e a mapping
X
,
w i l l b e denoted by let
be t h e o r d e r
Pa
from
T
L
into
M(X,v)
by
'1
(Tf)(x) = (Ta(Paf) (x)
I
Since a l l
Ta
and a l l
homomorphism. I f p f = 0 that
f o r every
f = 0
. Thus
Pa
Tf = 0 a T
In the case that
,
whenever
x E P
a r e Riesz homomorphisms, T then
, because
T (P f) = 0 CYa
each
f o r every
( w i t h norm p
( i n view of Theorem 120.8) t h a t i f
i s l i k e w i s e a Riesz a
, and
hence
Tcc i s a one-one mapping. It f o l l o w s
i s a one-one mapping, i . e . , L
.
T
i s a R i e s z isomorphism.
is a n a b s t r a c t L -space, we know P f E L i s g i v e n , t h e n e v e r y Ta(Paf) i s
Ch. 17,§1201
REPRESENTATION BY MEASURABLE FUNCTIONS
488
L (P , v )
an elment of Since
Paf
# 0
P
and the L -norm of Ta(Paf) i s the same as p(Paf). a P holds f o r a t most countably many values of a , the element
i s the o r d e r l i m i t (and hence the norm l i m i t ) of a sequence
f
. It
a f i n i t e sum of elements of the form
s
n Tf
Paf and the L -norm of Tf P P The proof t h a t , conversely, every element g E Lp(X,v) of
f
belongs t o
g = Tf
f o r some
f E L
holds i f
g
each
p(f)
.
i s of the form
.
) f o r some a P a c t L i s a Dedekind complete space possessing a s t r i c t l y
positive l i n e a r functional
,
$
v
the measure
and from t h e weak u n i t
$
,
i s s i m i l a r , s t a r t i n g from the observation t h a t
E L (P , v
In the case t h a t from
i s equal t o
L (X,v)
g = Tf
(sn)
follows t h a t the image
e
i n Theorem 120.10. Since f o r any
in f
Pa i s
i n each
derived
i n a c e r t a i n manner described
Ka
E L each component
P f
belongs t o
.
L (P v ) , i t follows as above t h a t each f E L belongs t o L,(X,v) In 1 a' a g e n e r a l , however, t h e image T(L) of L under the isomorphism T w i l l
not be equal t o the whole space in
. It
LI(X,v)
f
and
g
E T(L)
g
L (X,v). We prove t h a t 1
satisfy
. Observing
t h a t t h e product
0
2
that
fxp
ua E Ka
i n v e r s e image of ness of
L
,
0
. Obviously
f (i.e.,Tu
Tu = Tu
. It
in
, so ,
LI(X,v)
Tu = g
see t h a t
5
u
uo f o r any
2
gxp a
f o r some
, where
uo i s the
a
u
exists i n
a
L
. Since
u = u
+ (u-u,)
+ T(u-ua) = gxp + T(u-ua) a
. This
(Tu)x
&-
= gXpa
concludes t h e proof t h a t
holds f o r
i s an i d e a l
.
M(X,v)
and hence an i d e a l i n
. This T(L)
We f i n a l l y make some remarks about compact s e t s and Borel s e t s i n
B
By d e f i n i t i o n , the a-ring
of a l l Borel s e t s i n
containing a l l compact s u b s e t s of of a l l subsets V X
. It
Hence
8c
of
of
X
such t h a t
i s e a s i l y seen t h a t C
. We
and
Pa
E T(Ka) because
gxpa = T g
follows t h a t
0
with d i s j o i n t terms, which implies t h a t
a
, we
T(Ka)
f ). Hence, i n view of t h e Dedekind complete-
=
0 u = sup
the element
holds v -almost everywhere on
f
2 g 2
belongs t o
with d i s j o i n t terms, we have
every
i s an i d e a l
E T(L) v-almost everywhere on X , then
g(x) 5 f ( x )
i s an i d e a l in" LI(Pa,va)
T(Ka)
non-negat i v e
T(L)
i s s u f f i c i e n t t o prove t h a t i f the v-measurable f u n c t i o n s
prove t h a t
a l l v-measurable s e t s . Let
C C
V E C
X
. Let
X
X
.
i s the s m a l l e s t a-ring
us consider now t h e c o l l e c t i o n
V Cl C E
B
C
f o r every compact subset C
i s a u-algebra containing a l l compact s e t s .
i s a sub-o-algebra
. It
of t h e u-algebra
A of
follows from t h e d e f i n i t i o n t h a t
Ch. 17,81201
V
n Pa
E B
489
ABSTRACT L -SPACES P
for every
. This shows that
a
V n Pa
is a Borel subset of
P (because the relative topology in Pa is the original topology in Pa But then V
n
Let C that C
P, E ha
for all a , so V E A
.
. Then there exist al,.. .,a such (k=l,...,n ) . Since the characteristic
be a compact subset of X
C c P ak P belong to T(L) and T(L) ak it follows that the characteristic function of Ck
= U;
with
functions of the sets
,
M(X,v)
>.
is an ideal in C
belongs to T(L).
We summarize the obtained results in the following theorem. THEOREM 120.11. Let
satisfying
1
5
p <
m
L
be e i t h e r an abstract L -space f o r some P
p
or a Dedekind complete Riesz space possessing a
s t r i c t l y p o s i t i v e order continuous l i n e a r functional. Then L is e i t h e r R i e s z isomorphic t o a space L ( X , A , v ) or t o an i d e a l i n a space Ll(X,A,v), P where X i s a l o c a l l y compact Hausdorff space, v is a ( n o t necessarily u - f i n i t e ) measure and t h e u-algebra A o f a l l v-measurable subsets of X contains t h e o-algebra of a l l s e t s V w i t h t h e property t h a t v n C i s a
Borel s e t f o r every compact subset X
of any compact subset of
.
isomorphism T
1
S
X
. The c h a r a c t e r i s t i c
belongs t o t h e image
T(L)
of
L
function
under t h e
EXERCISE 120.12. Any abstract L -space L (for some p satisfying P S - ) is Banach, and therefore L is uniformly complete. It follows
p
that the complexification L+iL
1)
exists. The norm
. Show that
p(f)
= p(lf
for
1 < p
-
p(f)
for f E L+iL
is
the abstract L -space P is Riesz isomorphic (and norm isomorphic) to a complex space Lp(X,A,u)
defined by t
of
C
Similarly, any Dedekind complete Riesz space L therefore L+iL
exists. Show that if L
<
.
is uniformly complete and
is Dedekind complete and possesses
a strictly positive order continuous linear functional, then L+iL isomorphic to an ideal in a space of type Ll(X,A,lI) HINT: Let L
be isomorphic to the real space L
Note now that in L+iL f = f +if
1
(f,
f
.
and
the ordinary absolute value f2
is
of the desired type. (f:+fi)'
of a function
real) is the same as the function
sup\fl cos8+f2 sin't3:0<8<2a
J'
where the supremum is taken with respect to the order in L
.
Ch. 17,11211
REPRESENTATION BY CONTINUOUS FUNCTIONS
490
1 2 1 . R e p r e s e n t a t i o n by a space of continuous f u n c t i o n s I n t h i s f i n a l s e c t i o n we b r i e f l y d i s c u s s AM-spaces. an AM-space p o s s e s s i n g a s t r o n g norm u n i t L
by
p
, we
have
p(e)
,
1
=
p(f) 5 1
It f o l l o w s immediately t h a t
Xo > 0
number
, i.e.,
p(f)
C(X)
for
X
5
d e n o t i n g t h e norm i n
p(f) 5 1
i f and o n l y i f
If
I
S
, which
X1
p(f) = 0
and t h i s holds a l s o i f space
be
L
implies
Ifl
S
0 <
A,<
. If
e
If\
5 e
the
i s given, then
I f , i n t h i s c a s e , we should have p(Xil f ) 5 1
. Hence,
e
and f u r t h e r m o r e
First, let
.
X 1e
f o r some
lo
then
i s a g a i n s t our h y p o t h e s i s . Hence
I n the p a r t i c u l a r case t h a t
is a
L
compact ( w i t h t h e uniform norm) o r a space
( w i t h t h e c o r r e s p o n d i n g L_-norm),
,
L_(X,p) X
t h e n t h e f u n c t i o n i d e n t i c a l l y one on
i s a s t r o n g norm u n i t . The formula ( 1 ) becomes now
i n the case t h a t
L = C(X)
,
and
A : / f ( x ) l < h f o r p-almost a l l i n the case t h a t
L = L_(X,u)
. L
We r e t u r n t o t h e case t h a t
e
. More
generally, let f i r s t
L
i s a n AM-space w i t h a s t r o n g norm u n i t be a normed Riesz s p a c e w i t h - a d d i t i v e
norm and p o s s e s s s i n g a s t r o n g norm u n i t L
, i.e.,
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
. Then e
e
is a strong unit i n
i s t h e space
L
i t s e l f . This
i s immediately v i s i b l e from formula ( 1 ) . According t o t h e Yosida represent a t i o n theorem (Theorem 4 5 . 3 ) t h e r e e x i s t s now a compact Hausdorff space C(J)
J
. The
such t h a t subspace
h o l d s i f and o n l y i f h o l d s i f and o n l y i f
L "L
L
i s R i e s z isomorphic t o a Riesz subspace i s uniformly dense i n i s e-uniformly
C(J)
,
and s o
-L
-L = C(J)
complete. I n o t h e r words, -L
L i s n o r m complete, i . e . ,
i f and o n l y i f
of
L
=
C(J)
is a n AM-
Ch. 17,11211
49 1
ABSTRACT L -SPACES P
J
space. I n c i d e n t a l l y , we r e c a l l t h a t the p o i n t s of
J
in
L
and the value satisfying
ci
of the f u n c t i o n J E J
f E L ) a t the p o i n t
t o t h e element number
"f(J) cue-f E J
. Under
. The
E J
J
norm
in
p
L
(corresponding
i s the uniquely determined
t h e isomorphism the s t r o n g u n i t
J , i . e . , -e(J)
corresponds with t h e f u n c t i o n i d e n t i c a l l y one on all
a r e the maximal i d e a l s
-f E -L
i s transferred t o
e for
= 1
; according t o ( 1 )
-L
we have
This shows t h a t
i s t h e uniform norm i n
"p
space with a s t r o n g norm u n i t , then phic t o
L
.
C(J)
Now, l e t
L
. Hence,
-L
L*
be an a r b i t r a r y AM-space.
i s an AL-space.
L
i s an
AM-
i s Riesz isomorphic and norm isomorThen the norm i n
t i v e , s o by Theorem 118.1 the norm i n the Banach dual i.e.,
if
L*
L
i s m-addi-
i s I-additive,
B u t then, by Theorem 118.4, L**
i s an AM-space
possessing a s t r o n g norm u n i t . A s seen above, t h e r e e x i s t s a compact Hausdorff space
J
embedded i n
L**
follows t h a t
L
such t h a t
i s isomorphic t o
L**
C(J)
. Furthermore,
is
L
as a Riesz subspace (with p r e s e r v a t i o n of t h e norm). It
i s isomorphic t o a R i e s z subspace of
C(J)
. Combining
a l l r e s u l t s , we have proved t h e following theorem. THEOREM 1 2 1 . 1 . Any AM-space possessing a strong norm u n i t i s Riesz C(X)
isomorphic and norm isomorphic t o a space Hausdorff space space of some
X
. Any
C(X)
arbitrary AM-space i s isomorphic t o a Riesz subX
with
EXERCISE 1 2 1 . 2 . Let
V
compact and Hausdorff. be a ( r e a l or complex) vector space. The k e r n e l
( n u l l space) of any l i n e a r f u n c t i o n a l $I
i s t h e n u l l f u n c t i o n a l , then
such t h a t Let J, =
J,
$(e) = 1
. It
for an appropriate compact
$
on
N($) = V
follows t h a t
f
-
. If
.
EXERCISE 121.3.
Show t h a t and only i f
$
Let
$
and
f o r every
N($) c N($)
e
. Show
E
V
f
E V
0 5 J,
5 $
implies
L $ =
.
that
be a l i n e a r f u n c t i o n a l on the Riesz space
i s a Riesz homomorphism (from $ 2 0
E N($)
. If
N($)
not, there e x i s t s
$(f)e
be another l i n e a r f u n c t i o n a l s a t i s f y i n g
$(e).$
i s denoted by
V
L
i n t o t h e r e a l numbers) i f
a 4 f o r some r e a l number
.
492
a
REPRESENTATION BY CONTINUOUS FUNCTIONS
satisfying
0
HINT: I f
$
.
L"
0 5 (I
now t h a t
that
4
w E L+
a
. Let
implies
J,(u) = $ ( u ) > 0
. The
0
i s sufficient
. We
may
be defined
J,
f E K
$(w) = 0
J, = a$
we have
a = 1
for a l l
0 5 w I u
f o r some r e a l number
, so
J, = $
. Hence
a
.
.
$(v) =
b e a convex s u b s e t of t h e r e a l v e c t o r s p a c e
K
i s c a l l e d an extrernal p o in t of
f = ag + (I-a)h
. In
and
d e s i r e d r e s u l t follows.
EXERCISE 121.4. L e t
. The
. It
...
. By h y p o t h e s i s
$(v) = 0
f = g = h
N
L
$(u+v) = max{$(u),$(v)}
$(u) = $(u) > 0
with
In particular
K
, which
N($) c N(J,)
the positive linear functional
On a c c o u n t of
The p o i n t
. Hence
by
0 5 J, 5 $
= $(v) =
p o s i t i v e . Assume
( i n view of t h e p r e c e d i n g e x e r c i s e ) .
$(wAnu):n=I,Z, Then
0 is
i t f o l l o w s from
b e a p o s i t i v e atom i n
inf(u,v) = 0
$(u) > 0
assume t h a t
. Then
l$(f)l 5 $ ( l f l ) = 0
f o r some number
Conversely, l e t
for
$ i s t h e n a p o s t i v e atom i n
I n o t h e r words,
$(f) = 0
and
$
, so
(I = a$
t o prove t h a t
.
i s a R i e s z homomorphism, t h e n 5
$(lfl) = 0
implies
a 5 1
5
Ch. 17,§1211
g,h E K
for
and a number
a
i f i t f o l l o w s from
K
f
that
can l i e wholly i n
i s denoted by
K
0 < a < 1
satisfying
o t h e r words, no l i n e segment through
s e t of a l l e x t r e m a l p o i n t s of
V.
. We
Ext K
mention
some examples.
V
( a ) Let
be two-dimensional
p(f) = I f l [ + If2\ V
,
i.e.,
K = (f:p(f)
(O,l),(-l,O) (0,O)
.
(b) L e t in f
V
V
f = (f
.
(O,-l)
K+ = ( f : f > O , p ( f ) < l ) and
for
.
then
Let
K+
r e a l number s p a c e w i t h L -norm, i . e . , 1 L e t K b e t h e c l o s e d u n i t b a l l of
.
Ext K
c o n s i s t s of t h e p o i n t s
be t h e p o s i t i v e p a r t o f
Ext K+
that
p(f) = 1 K
and
p(g) = p ( h ) = 1
f o r any
.
K
c o n s i s t s of t h e p o i n t s
be a r e a l H i l b e r t s p a c e and l e t
. Show f i r s t ,
.
Show t h a t
i s an e x t r e m a l p o i n t of
0 < a < 1
,f ) 1 2 Show t h a t
K
(l,O),
i.e.,
(l,O),(O,l)
be t h e closed u n i t b a l l
f E Ext K
f = ag + (I-a)h
,
. Next,
with
g,h
F i n a l l y , assume now t h a t
show t h a t i f
E K
and
p ( g ) = p(h) =
Ch. 1 7 , 5 1 2 1 1
= 1
,
g # h
Show t h a t
49 3
ABSTRACT L -SPACES P and
f = ag + (I-a)h
f o r some
satisfying
a
.
0 < a < 1
p ( f ) < 1 .Combining t h e s e r e s u l t s , show t h a t e v e r y f E K
i s a n e x t r e m a l p o i n t of
P(f) = 1
satisfying
H I N T : For t h e l a s t p a r t of ( b ) , l e t
.
K
p(g) = p(h) = 1
and
g
.
# h
By t h e p a r a l l e l o g r a m law i n H i l b e r t s p a c e we have
Since
, it
> 0
p(g-h)
any p o i n t
f
follows t h a t
121.5. L e t
EXERCISE
closed u n i t b a l l i n
g
then
and
p(f) < 1
# 0 of
f
.
(b) Show t h a t any e x t r e m a l p o i n t
f
of
or
K
.
K
satisfies
K+
f t 0
satisfies
K
be t h e
K
b e t h e p o s i t i v e p a r t of
K"
for
.
h
(norm p ) , l e t
b e an AL-space
L
and l e t
L
( a ) Show t h a t any e x t r e m a l p o i n t p(f) = 1
. But
p($g+ih) < 1
on t h e l i n e s e g m e n t p r o p e r l y between
or
f 5 0 . ( c ) Show t h a t any p o s i t i v e e x t r e m a l p o i n t of of
and any nonzero e x t r e m a l p o i n t of
K+
i s t h e union of
Hence, Ext K
f+ > 0
H I N T : For ( b ) , assume =
, so
p(If1) = 1
and
0 < p(f+)
, we
h = -f-/p(f-)
,
g # h
Ext K+
have
0 < p(f-)
contradicting the hypothesis t h a t
Assume now t h a t g,h
E K
f
0 < a < 1
.
Note t h a t
f
which would imply
p ( f l ) > p(f) = 1
p(g+) 5 1 h = h+
and
. Now
p(h) 5 1
use t h a t
EXERCISE 121.6. e x e r c i s e . Show t h a t and
p(f) = 1 0 5 a S 1
one i n
L
. It .
0
5
with
K
f
that
f
L,K
and
and l e t
K+
. Show now
p(f) = 1
f
K'
. Hence K+ .
+
)
E K and
1
K+.
f = ag
+ (I-a)h
that
g = g+,
= cig+ +
(I-a)h
,
g = g+ and, s i m i l a r l y ,
be t h e same as i n t h e p r e c e d i n g
i s a nonzero e x t r e m a l p o i n t of implies
=
+
On t h e o t h e r hand, i t f o l l o w s from
p(fl) 5 1
i s extremal i n
(a) L e t
g 5 f
.
g,h
p(f-)
g = f /p(f
i s an e x t r e m a l p o i n t i n
would b e p r o p e r l y smaller t h a n
because o t h e r w i s e
.
i s e x t r e m a l . For ( c ) , i t i s
f
# 0 i s a n e x t r e m a l p o i n t of
and
K
minus t h e p o i n t z e r o .
. Then p ( f + ) + < 1 . Writing
f = p(f+).g + p(f-).h
obvious t h a t any p o s i t i v e e x t r e m a l p o i n t of with
Ext (-K+)
f- > 0
and and
< 1
i s a n e x t r e m a l p o i n t of
K+
and
i s an extremal p o i n t
K
g = af
K+
f o r some number
follows t h a t t h e e x t r e m a l p o i n t s of
K
i f and only i f
a
satisfying
are t h e atoms of norm
Ch. 17,51211
REPRESENTATION BY CONTINUOUS FUNCTIONS
494
( b ) Show t h a t if t h e measure
in
li
X
h a s no atoms, t h e n t h e c l o s e d
L (X,li) does n o t have any e x t r e m a l p o i n t s . 1 ( c ) Show t h a t t h e e x t r e m a l p o i n t s of t h e p o s i t i v e p a r t of t h e c l o s e d
u n i t b a l l of
u n i t b a l l of n a t e of
u
(d) L:t
( n = l , Z , ...) , where t h e n-th c o o r d i n i s one and a l l o t h e r c o o r d i n a t e s a r e z e r o .
b, L
a r e a l l elements
u
b e an AM-space and l e t
c l o s e d u n i t b a l l of
L*
. Show t h a t
B+
t h e s e t of a l l e x t r e m a l p o i n t s of
0 S $ E L*
c o n s i s t s of t h e n u l l f u n c t i o n a l and a l l $
i s a R i e s z homomorphism (from
p(g)
Writing
and
p(f-g)
g1 = g / p ( g )
f = p(g).g,
Conversely, l e t
h l = (f-g)/p(f-g)
. Hence p(f) = 1
some
B
satisfying
0 2 B
some
a
satisfying
0 < a < 1
i s a p o s i t i v e m u l t i p l e of
0
with f
ag S f
g,h
,
0 < g
i
f
.
,
have
that
E K+
and
p(g) + p(f-g) = 1
.
0 i g 2 f
. Assume now S
, we
f = g l = g/p(g) and l e t
1
5
Furthermore, on account of g
K+
a r e s t r i c t l y p o s i t i v e and
and
+ p(f-g).hl
B+
of norm one such t h a t
i n t o t h e r e a l numbers).
i s e x t r e m a l p o i n t of
f > 0
HINT: I n ( a ) , i f then
L
b e t h e p o s i t i v e p a r t of t h e
imply
g = Bf
for
f = ag + (I-a)h
for
. Then
.
p(g) = p(h) = 1
i t f o l l o w s from t h e h y p o t h e s i s t h a t
. Similarly
for
h
. Hence
g = h = f
.
CHAPTER 18
COMPACT OPERATORS
The p r e s e n t c h a p t e r d e a l s mainly w i t h c o n d i t i o n s , n e c e s s a r y and ( o r ) s u f f i c i e n t f o r an o r d e r bounded o p e r a t o r from one Banach l a t t i c e i n t o a n o t h e r t o b e compact. A s well-known,
the operator
a n o t h e r i s s a i d t o be compact i f
T
T
from one Banach s p a c e i n t o
maps norm bounded s e t s i n t o precompact
t o t a l l y bounded) sets. I n Banach l a t t i c e s i t makes s e n s e t o i n t r o d u c e
(i.e.,
a l s o o p e r a t o r s which map o r d e r bounded s e t s i n t o precompact s e t s . I n doing s o , i t becomes c l e a r v e r y soon t h a t t h e r e i s s t i l l a n o t h e r c l a s s of s e t s p l a y i n g a p r i n c i p a l r o l e . These a r e t h e almost o r d e r bounded s e t s . The subset
L ( w i t h norm
of t h e Banach l a t t i c e
S
almost o r d e r bounded i f f o r any [p,q ] in
L
such t h a t
S
p(f) 5
E
)
. In
> 0
and u n i t b a l l
p
B ) i s called
t h e r e e x i s t s an o r d e r i n t e r v a l
i s i n c l u d e d i n t h e a l g e b r a i c s u m [ p , @ ] + EB
(where, a s u s u a l , t h e n o t a t i o n satisfying
E
EB
means t h a t
EB
c o n s i s t s of a l l
f
E L
s e c t i o n 122 i t i s observed t h a t precompact s e t s
a s w e l l a s o r d e r bounded s e t s a r e almost o r d e r bounded and almost o r d e r bounded s e t s a r e norm bounded, b u t i n g e n e r a l no i m p l i c a t i o n i n t h e converse d i r e c t i o n h o l d s . Of c o u r s e , i f f o r example L u n i t ( a s p a c e of type
i s an AM-space w i t h s t r o n g norm
Lm ), t h e n o r d e r bounded s e t s , almost o r d e r bounded
s e t s and norm bounded s e t s c o i n c i d e . I n Theorem 83.3 i t was proved t h a t i f space
L
T
i s an o p e r a t o r from t h e Riesz
i n t o t h e Dedekind complete Riesz space
bounded i f and only i f
T = T
,
- T2
with
T1
M and
,
then
T
is order
T2 p o s i t i v e , and i n
t h i s case the s e t L (L,M) of a l l o r d e r bounded o p e r a t o r s (from L i n t o b M ) i s a Dedekind complete Riesz s p a c e . I f M f a i l s t o b e Dedekind comp l e t e , t h e d e f i n i t i o n of a n o r d e r bounded o p e r a t o r from unchanged and
Lb(L,M)
L i n t o M remains
i s now a v e c t o r s p a c e , b u t n o t n e c e s s a r i l y a Riesz
space. I t i s p o s s i b l e i n t h i s c a s e t h a t t h e v e c t o r s p a c e of a l l with
T1
and
T2
T
p o s i t i v e ( c a l l e d t h e s p a c e of J o r d a n o p e r a t o r s ) i s
s m a l l e r t h a n t h e space of a l l o r d e r bounded o p e r a t o r s . Note t h a t i f M
a r e Banach
T
=
L
l a t t i c e s , t h e n every p o s i t i v e o p e r a t o r (and hence e v e r y r. a c
1
-
T2
and
496
Ch., 183
CHAPTER 18
J o r d a n o p e r a t o r ) i s norm bounded ( s i n c e that i f
i s Banach). Furthermore, r e c a l l
L
i s a Banach l a t t i c e w i t h o r d e r continuous norm, t h e n
M
is
M
Dedekind complete (Theorem 103.9) and hence t h e space of J o r d a n o p e r a t o r s (from
L
M ) c o i n c i d e s now w i t h t h e Riesz space
into
. After
Lb(L,M)
t h e s e p r e l i m i n a r i e s we can now g i v e a d e f i n i t i o n t h a t f i t s t h e f a c t s . The Jordan o p e r a t o r
T
(from t h e Banach l a t t i c e
M ) i s c a l l e d AM-compact i f
compact s u b s e t s of compact. Also i f Lm
>,
M
i n t o t h e Banach l a t t i c e
L
maps o r d e r bounded s u b s e t s of
T
. Evidently,
i n t o pre-
L
every compact J o r d a n o p e r a t o r i s AM-
i s an AM-space w i t h s t r o n g norm u n i t ( a space of t y p e
L
t h e n t h e Jordan o p e r a t o r
i s compact i f and o n l y i f
T
i s AM-
T
compact. This e x p l a i n s t h e terminology. AM-compact o p e r a t o r s a r e r e l a t e d Fremlin i n t h e i r fun-
t o t h e AMAL-compact o p e r a t o r s of P.G. Dodds and D.H. damental paper ([1],1979).
I n s e c t i o n 123 i t will be proved f i r s t t h a t t h e
Jordan o p e r a t o r
i s AM-compact i f and only i f
T
i n t o precompact s u b s e t s of
M
T: L + M
o r d e r bounded s u b s e t s of
L
maps almost
. This
shows
a l r e a d y t h a t i f one wishes t o i n v e s t i g a t e compact Jordan o p e r a t o r s , i t w i l l be n e c e s s a r y t o i n t r o d u c e a s w e l l o p e r a t o r s t h a t map norm bounded s e t s i n t o almost o r d e r bounded s e t s (because t h e n , i f
i s AM-compact,
t h e composition
has t h i s p r o p e r t y and
T1
about AM-compact o p e r a t o r s i s t h e theorem t h a t i f t h e space
M
has o r d e r
continuous norm, t h e n t h e s e t of a l l AM-compact o p e r a t o r s (from M ) i s a band i n t h e Riesz space
T2
w i l l b e compact). The main r e s u l t
T2T,
Lb(L,M)
. The
into
L
r e s u l t i s due t o Dodds-
Fremlin. The proof i n t h e i r paper i s a s i m p l i f i e d v e r s i o n of t h e i r o l d e r p r o o f ; t h e s i m p l i f i c a t i o n i s due t o A.R.
Schep. S t i l l assuming t h a t
o r d e r continuous norm, an immediate consequence i s t h a t i f and only i f
T
has
M
i s AM-compact
i s so and a l s o any p o s i t i v e o p e r a t o r majorized by an
IT1
AM-compact o p e r a t o r i s AM-compact i t s e l f . Another consequence i s t h a t i n some c a s e s t h e s e t of a l l o r d e r bounded compact o p e r a t o r s ( f r o m
is a band i n
Lb(L,M)
. This
w i t h s t r o n g norm u n i t and happens a l s o i f
L*
M
happens, f o r example, i f
L
L
into
M)
is a n AM-space
has o r d e r continuous norm and, d u a l l y , i t
has o r d e r continuous norm and
M
i s an AL-space.
The
proof of t h e d u a l theorem i s based on Synnatschke's theorem (Theorem 115.4) stating that
IT/
*
= /T*l
for a l l
a r e Banach f u n c t i o n s p a c e s such
T E Lb(L,M)
that
every a b s o l u t e k e r n e l o p e r a t o r (from
M
L
. Finally,
if
L and
M
has o r d e r continuous norm, then into
M ) i s AM-compact.
In s e c t i o n 124 we i n t r o d u c e o p e r a t o r s napping norm bounded s e t s i n t o almost o r d e r bounded s e t s . These a r e sometimes c a l l e d L-weakly
compact, but
49 7
COMPACT OPERATORS
Ch. 181
we shall call them semi-compact. Evidently, any compact operator is semicompact. Also, if L,M,K and
T2: M
+
tions that
K
are Banach lattices with T,: L
+
M
semi-compact
AM-compact, then it follows immediately from the definiT: L + L
is compact. In particular, if
T2Tl: L + K
compact as well as AM-compact, then T2
is semi-
is compact. It is easy to see that
any positive operator majorized by a semi-compact operator is semi-compact itself. It follows now that if L
is a Banach lattice with order continu-
T are positive operators from L into itself such that T is compact and 0 2 S 2 T , then S inherits both AM-compactness ous norm and
and
S
and semi-compactness from T
so
that, therefore, S2
,
ty the same holds if, instead of L
is compact. By duali-
the dual space L*
nuous norm. The problem arises what happens if neither
has order continor L* has order
L
continuous norm. In this case we have the deeper result that now it follows from 0
5 S 5
T with
T
what happens if both L
compact that and
L*
S3
is compact. Finally, we may ask
have order continuous norm. More generally,
we shall discuss the situation (in section 125) that L lattices such that M
and L*
order bounded operator mapping and only if 0 5 S 5 T
T
and
and M
are Banach
have order continuous norm and T L
into M
. In this case
T
is an
is compact if
is AM-compact as well as semi-compact. It follows that if T
is compact, then
S
is compact. These last results are
among the main Dodds-Fremlin results; the theorems about
S2
and
S 3 (which
are best possible) are due to C.D. Aliprantis and 0. Burkinshaw ([21,1980). The proofs which we present in sections I 2 4 and 125 are somewhat simpler and more elementary than the original proofs (simplifications due to A.R. SchepX It will be observed that even if all norms involved are order continuous,
the set of all semi-compact
operators is not always an ideal. We present an
already classical example in L, but
1
TI
is not. This is caused by the fact that although
compact, 1 TI L*
and
then T Pagter)
M
.
(due to U. Krengel) that T
is compact,
I TI
is AM-
fails to be semi-compact, Finally, we mention that if both
have order continuous norm and
is AM-compact if and only if
T*
T: L
+
M
is order bounded,
is so (a result due to B. de
In section 126 we shall deal with conditions for semi-compactness of an operator. Let L
and
M
be Banach lattices such that, to begin with, M
has only the principal projection property. For any decreasing sequence (B,) Of
of principal bands in M , let order projections. Then I l B
= {O}
(P )
be the corresponding sequence
if and only if
Pn C 0
. We
shall
498
Ch, 181
CHAPTER 18
prove t h a t i f
i s a p o s i t i v e o p e r a t o r from
T
L
into
M
IIPnT/I J 0
and
f o r e v e r y sequence P C 0 as e x p l a i n e d , t h e n T i s semi-compact. I n n I I S I I J O f o r e v e r y sequence (Sn) s a t i s f y i n g particular, therefore, i f
C 0
T 2 S
, then
i s semi-compact. As an a p p l i c a t i o n w e g e t a s i m p l e
T
proof of t h e c l a s s i c a l theorem t h a t k e r n e l o p e r a t o r s w i t h a Hille-Tamarkin k e r n e l a r e compact. I f w e r e q u i r e more from /IS
11
M
,
i s a p o s i t i v e o p e r a t o r from
T
o r d e r continuousnorm and T
and
L
the condition t h a t
i 0 i s n o t only s u f f i c i e n t , but a l s o necessary. I f both
i s semi-compact i f and o n l y i f
S t i l l assuming t h a t
M
and
L*
/IS
/I
M and L* have into
L
J 0 f o r e v e r y sequence
,
M
then
T 2 S
J 0 .
have o r d e r c o n t i n u o u s norm, w e d i s c u s s i n
s e c t i o n 127 s e v e r a l c o n d i t i o n s (due t o Dodds-Fremlin a g a i n ) , n e c e s s a r y s u f f i c i e n t f o r semi-compactness
s a r i l y p o s i t i v e t h e r e f o r e ) . L e t u s mention h e r e o n l y t h a t n + m
f o r a l l d i s j o i n t norm bounded sequences
elements i n
M*
L and
l/QnTII J 0
+
f o r e v e r y sequence
0
L
-f
($n)
as
of p o s i t i v e
M
i s semi-compact i f and only i f
T
(Q,)
and a l s o i f and o n l y i f
order projections i n
and
0
$n(T~n)
r e s p e c t i v e l y . A s an a p p l i c a t i o n w e prove i n s e c t i o n
128 t h a t t h e o r d e r bounded o p e r a t o r Q
(un)
and
( n o t neces-
i s semi-compact
T
i s semi-compact and a l s o i f and o n l y i f
T*
i f and o n l y i f
T
of an o r d e r bounded o p e r a t o r
of p o s i t i v e o p e r a t o r s i n
/IPnTI/
satisfying
+
0
Pn C 0
M
f o r e v e r y sequence
.
satisfying (P,)
of
I f we assume i n a d d i t i o n t h a t
h a s t h e p r i n c i p a l p r o j e c t i o n p r o p e r t y , t h e r e i s a n even b e t t e r theorem.
I n t h i s case quences
P
T
C 0
i s semi-compact i f and only i f and
J 0
Q
i o n s on p r i n c i p a l bands i n
i s e v i d e n t because
lIPnTI/
IIPnTQnII
of o r d e r p r o j e c t i o n s i n L
+
M
-t
0
f o r a l l se-
and o r d e r p r o j e c t -
r e s p e c t i v e l y . The n e c e s s i t y of t h i s c o n d i t i o n s
0
i s already necessary; i t i s the sufficiency
which i s i m p o r t a n t h e r e . The l a s t theorems a r e due t o P. van E l d i k and
J.J. G r o b l e r (1979). A s an a p p l i c a t i o n , l e t s p a c e s such t h a t
L*
and
M
L
and
M
b e Banach f u n c t i o n
have o r d e r c o n t i n u o u s norm
and l e t
T(x,y)
b e t h e k e r n e l o f t h e o p e r a t o r T from L i n t o M . Then T i s compact i f and only i f
i t i s t r u e f o r a l l sequences
E
J I$
in
X
and
F
C $I
in
Y
t h a t t h e norm of t h e o p e r a t o r w i t h k e r n e l
tends t o zero as p a p e r of W.A.J. L -spaces, P
n
+ m
. This
i s e s s e n t i a l l y t h e main r e s u l t i n a 1963
Luxemburg and A.C.
Zaanen ( f o r O r l i c z s p a c e s , i n c l u d i n g
t h i s goes back t o a p a p e r by T. Ando, 1962). These p a p e r s were
Ch. 18,51221
499
COMPACT OPERATORS
one of the motivations for the later developments in Banach lattices. In section 129 the notions of upper and lower index for a Banach lattice are introduced. Let
1 i
p
5
. The Banach
m
said to have the d -decomposition property if P order bounded disjoint sequence (un) in L+
lattice L (norm
p ) is
dP for every
(p(un))
E
and
is said to have the
L
d -composition property if supIp(~;1 anun)}
for every sequence im P The (an) E dp and every norm bounded disjoint sequence (u,) in L+
.
numbers inf(p:L has the R -decomposition property), P sup(p:L has the d -composition property) P are called the upper index and lower index of L respectively. If o(L) < u(L)
=
s(L)
=
then L has order continuous norm. If
=
decomposition property if and only if L* It follows that
1 If L
Is(,*,\- 1 'I I
=
is called the lower and
1
.
1
1
is of infinite dimension, then s(L)
why
1
+
9
o(L)
s(L)
5
o(L)
9
m
If L
and M
, then
implies that L*
and
norm. Hence, the order bounded operator T: L if
T
is AM-compact. In particular, if L
spaces satisfying o(M) < s(L) into M
, then
1
5
=
is an
a(L) o(L)
= p =
m.
r < p i
m
)
.
M -f
M
and M
-f
M
is semi-compact.
have order continuous is compact if and only are Banach function
any absolute kernel operator from L
is compact (for example, if L
space with
=
are Banach lattices of infinite dimension such that any order bounded operator T: L
o(M) < s(L)
Note that
. This explains
the upper index. If L
abstract L -space (l
has order con-
has the d P has the d -composition property. 4
tinuous norm. Furthermore, for p-' + q-'
IU(J1
, then L* , the space L
> 1
s(L)
m,
is an L -space and M
P
is an L -
122. Compact sets, precompact sets and almost order bounded sets As well-known, the linear operator T the Banach space W set in
V
from the Banach space V
into
is called compact if the image of every norm bounded
is a precompact set in W
. Evidently
it is sufficient that the
500
PRECOMPACT AND ALMOST ORDER BOUNDED
Ch. 18,81221
image of t h e u n i t b a l l i s precompact. We r e c a l l t h a t a s u b s e t o f
is
W
s a i d t o b e precompact whenever i t s c l o s u r e ( w i t h r e s p e c t t o t h e norm topology i n
W ) i s compact. For a b e t t e r u n d e r s t a n d i n g we a l s o r e c a l l some pro-
p e r t i e s of compact s e t s and precompact s e t s i n a Banach s p a c e , o r even more g e n e r a l l y i n a complete m e t r i c s p a c e . Hence, l e t s p a c e ; t h e d i s t a n c e between t h e p o i n t s and t h e open b a l l let
(y:d(x,y)
b e a s u b s e t of
S
.
E
x
b e a complete metric
E
and
y
in
w i l l b e denoted by
The s e t
w i l l be
E
d(x,y)
. Furthermore,
B(x;r)
may have one o r s e v e r a l of t h e
S
following properties: S
i s compact, i . e . , e v e r y open cover of
S
i s sequentiazly compact, i . e . ,
quence converging t o a p o i n t of
e v e r y sequence i n
t h a t any f i n i t e
every i n f i n i t e subset
S (with the understanding
automatically has t h i s property).
S
i s totaZZy bounded, i . e . ,
S
f o r any
> 0
E
t h e r e e x i s t a f i n i t e sub-
(x,
,...,xn)
S
i s precompact, i . e . ,
S
i s sequentiaZZy precompact, i . e . , e v e r y sequence i n
set
of
such t h a t
S
i s contained i n
S
-
the closure
S
of
-
Uy=l
B(x.;E)
.
i s compact.
S
v e r g i n g subsequence ( t h e l i m i t need n o t b e a p o i n t of S
h a s a subse-
S
.
S
h a s a t l e a s t one p o i n t of accumulation i n
S
h a s a f i n i t e subcover.
h a s t h e BoZzano-Weierstrass property, i . e . ,
S of
S
S
h a s a con-
S ). Equivalently,
i s s e q u e n t i a l l y compact. h a s t h e conditional BoZzano-Weierstrass property, i . e . , e v e r y i n -
S
f i n i t e subset of
S
not necessarily i n
h a s a t l e a s t one p o i n t of accumulation i n S
,
therefore). Equivalently, S
-- -
s t r a s s property.
These p r c p e r t i e s f o r t h e s u b s e t
-
of
S
sequentiaZly compact
compact
E
-
E (this point
h a s t h e Bolzano-Weier-
are r e l a t e d , a s f o l l o w s
BoZzano-Weierstrass prop
19
cZosed and totalZy bounded.
Hence, i f i t i s n o t knownwhether t h e s e t S i s c l o s e d , t h e f o l l o w i n g c o n d i t i o n s for
are equivalent:
S
precompact
s e q u e n t i a l l y precompact
0
conditiona2 BoZzano-Weierstrass property
-
totaZZy bounded.
The p r o o f s b e l o n g t o t h e e l e m e n t a r y t h e o r y of m e t r i c s p a c e s ; we r e s t r i c t o u r s e l v e s h e r e t o t h e proof t h a t if
S
corrpact. Then point
S
i s s e q u e n t i a l l y comapct i f and o n l y
i s c l o s e d and t o t a l l y bounded. Assume f i r s t t h a t x
1
E S
S
. If
i s o b v i o u s l y c l o s e d . Now, l e t S
is contained i n
t a l l y bounded i s complete. I f n o t , l e t
E
> 0
is sequentially
S
b e given. Choose a
B ( X ~ ; E ), t h e proof t h a t x2
b e a p o i n t of
S
S
outside
is to-
Ch. 1 8 , 5 1 2 2 1
.
B(x~;E)
50 1
COMPACT OPERATORS
If
i s c o n t a i n e d i n t h e union of
S
B(xl;€)
and
B ( x ~ ; E ), t h e
proof i s complete. I f n o t , w e c o n t i n u e i n t h i s way, The p r o c e d u r e ends a f t e r
a f i n i t e number of s t e p s , b e c a u s e o t h e r w i s e t h e sequence b e a sequence i n
converse d i r e c t i o n , w e assume t h a t s = (x1,x2,
. .)
...)
h a s a Cauchy subsequence
S
b e a n a r b i t r a r y sequence i n
.
S
Since
is totally
S
4,
bounded, t h e r e e x i s t s a f i n i t e number o f open b a l l s , each of r a d u s that
such
i s c o n t a i n e d i n t h e union of t h e s e b a l l s . I t f o l l o w s t h a t t h e
S
s1 = (xll,xI2,.,.)
h a s a subsequence
sequence. s
t h e s e b a l l s . S i m i l a r l y , s 1 h a s a subsequence
. We
31
i n a b a l l of r a d i u s (xl
would
i s c l o s e d and t o t a l l y bounded. It i s
S
s u f f i c i e n t now t o prove t h a t e v e r y sequence in Let
(xl ,x2,.
w i t h o u t c o n v e r g e n t subsequence. For bhe p r o o f i n t h e
S
.
c o n t a i n e d i n one of
s 2 = ( X ~ ~ , X ~ ~ , .c o . n. t)a i n e d
c o n t i n u e i n t h i s way. The d i a g o n a l sequence
now a Cauchy sequence.
, X ~ ~ , X ~. ) ~ , i. s
F o r t h e d e f i n i t i o n of a n a l m o s t o r d e r bounded set w e assume t h a t L i s a Banach l a t t i c e (norm p ) . The s u b s e t
bounded s e t i f f o r any g i v e n =
Cp,ql = (g:p
f = f where
+ f2
1
in fl E A
with
and
i s t h e u n i t b a l l of
B
a symmetric o r d e r i n t e r v a l
t o see t h a t t h e s u b s e t f o r any f
E
S
> 0
E
. Indeed,
t h e r e exists -u 5 g < u
u
there exists if
5
E
. In
o t h e r words, S c Cp,ql
such chat
S c [-u,ul
such t h a t
E L+
. Compare
E
. The
+ EB ,
+ EB
. It
i s easy
L+ i s a l m o s t o r d e r bounded i f and o n l y i f u
and n o t e now t h a t
p(h) <
E
A =
i s of t h e form
In t h i s case, t h e r e e x i s t s 05 c o u r s e
s u c h that e v e r y
p(h) <
f = f A u + (f-u)' p{(f-u)+}
i
f E S
p{(f-u)+} <
i s a l m o s t o r d e r bounded and
S
E L+
and
.
[-u,uJ of
S
p(f2) L
i s c a l l e d an almost order
there e x i s t s an order i n t e r v a l
> 0
E
such t h a t every
L
L
of
S
f € S
> 0
E
f o r every
E
i s given, then
i f of the form
f = g+h
with
t h i s decomposition w i t h ( f - u ) + 5 (f-g)+ = h+ 5 / h i
, so
c o n c l u s i o n i n t h e converse d i r e c t i o n i s e v i d e n t .
I t i s a l s o e v i d e n t t h a t t h e a l g e b r a i c sum of a f i n i t e number of almost o r d e r bounded sets i s a l m o s t o r d e r bounded.
THEOREM 122.1. ( i ) The subset
S
o f the Baruch l a t t i c e
almost order bounded i f and o n l y if t h e s e t
IS1 = (1fl:fES)
L
(norm
p)
is
i s aZmost order
bounded. ( i i ) The subset
S
of the Banach Zattice
bounded i f and o n l y i f f o r any p{(lfl-u)+}
E
f o r every
f E
E
s
> 0
.
L
there e x i s t s
p) i s almost order E Li such t h a t
(norm u
502
PRECOMPACT AND ALMOST ORDER BOUNDED
( i ) It i s e v i d e n t t h a t i f
PROOF.
is
IS1
E > 0
. Assume now
there e x i s t s
If1 = g+h w i t h
that
u
fi
0 < f. < u
IS1 i s almost o r d e r bounded, i . e . ,
-u i g 5 u 0
and
i
for
=
1,3
p(h) < ;E
f + i u + Ihl
5
(i=l,2,3,4)
E L+
i s almost o r d e r bounded, t h e n so
S
E L+ such t h a t every
theorem i t follows from exist
Ch. 18,51221
such t h a t
and
0
If1
. By
E
t h e dominated decomposition
0 < f - i u + Ihl
and
f l + f2 = f + , f 3 + f
f . 5 Ihl
f o r any g i v e n
i s of t h e form
IS1
f = ( f l - f 3 ) + (f -f ) w i t h -u 5 f l - f 3 5 u and 2 4 S i s almost o r d e r bounded.
4
-
. Hence
i = 2,4
for
that there
= f
,
p ( f 2 - f 4 ) 5 2p(h) <
E
.
This shows t h a t
i s almost o r d e r bounded i f and only
(ii) S
only i f f o r any f o r every
f
E >
E S
.
0
there exists
E
u
i s so, i . e . ,
IS1
such t h a t
L+
THEOREM 122.2. For subsets of the Banach l a t t i c e
i f and
p I ( l f 1 -u)+l < E
(non
L
p)
the
following inp l i e a t i o n s hold. t o t a l l y bounded
alsmost order bounded * norm bounded.
order bounded
st
For examples showing t h a t no implication i n the converse d i r e c t i o n holds, we r e f e r t o the e x e r c i s e s . PROOF. Assume f i r s t t h a t
S
b e given. There e x i s t
gl,
E >
0
in
Ijy=l
B(gi;E)
such t h a t and
. Hence,
f = g. + f 2
q = sup(gl
f o r any with
,...,gn) , we
f i n d , t h e r e f o r e , t h a t every fl
E
Cp,q]
and
p(f2) < E
i s a t o t a l l y bounded s u b s e t of
... gn 9
f E S
p(f2) < have f E S
. This
in E
such t h a t
S
L
. Let
i s contained
S
t h e r e i s a t l e a s t one of t h e
. Setting
for a l l
p 2 gi 5 q
i s of t h e form shows t h a t
gi
p = i n f ( g l , . . . , g ,)
i
f = f
= 1
,...,n . W e
+ f2
with
1 i s almost o r d e r bounded.
S
I t i s obvious t h a t o r d e r bounded s e t s a r e almost o r d e r bounded and
almost o r d e r bounded s e t s a r e norm bounded. EXERCISE 122.3. We f i r s t p r e s e n t a norm bounded s e t i n a Banach l a t t i c e wh?ch f a i l s t o b e almost o r d e r bounded. For
n
=
1,2,
...
, let
e
be the n c o o r d i n a t e one and
L, ( i . e . , en h a s t h e n-th a l l o t h e r c o o r d i n a t e s z e r o ) . Show t h a t t h e s e t of a l l e i s norm bounded i n n L2 , b u t n o t almost o r d e r bounded. I t follows t h a t t h e u n i t b a l l i n g2 i s n-th u n i t v e c t o r i n t h e space
n o t almost o r d e r bounded.
Ch.
COMPACT OPERATORS
18,91221
EXERCISE 122.4. W e r e c a l l t h a t f o r set
the function
X
a r e a l function
is d e f i n e d by
sgn f
503
sgn f
=
on t h e p o i n t
f
f/lfI
f(x) # 0
where
and z e r o elsewhere. The f u n c t i o n
...) . L e t
i s c a l l e d t h e n-th Rademacher function (n=0,1,2, measure i n function
X
. Neglecting
Ir -r I
t a k e s t h e v a l u e 2 on a s e t of measure
m n z e r o on t h e remaining s e t of measure
...)
S = (rl,r2,
b e Lebesgue
p
a set of measure z e r o , n o t e t h a t f o r
. It
T
II
m > n
f o l l o w s t h a t t h e sequence
does n o t p o s s e s s a p o i n t w i s e (almost everywhere on
converging subsequence. Show t h a t L (X,u) P
f o r any
p
satisfying
1 5 p 2
-.
L2(X,u)
H I N T : From what i s observed above, i t f o l l o w s immediately t h a t
n o t p o s s e s s a subsequence which converges i n norm. T h e r e f o r e , S s e q u e n t i a l l y precompact, i . e . , EXERCISE 122.5. For
e2 . L e t
(~~:n=1,2, )
and
=
S =
E~
S
or,
does
i s not
i s n o t t o t a l l y bounded.
S
n = 1,2,
...
,
let
be t h e n-th u n i t v e c t o r i n en be a sequence of p o s i t i v e numbers such t h a t E ~ 0S
... . Furthermore, l e t un = En e n f o r a l l (ul,u2, ...) i s t o t a l l y bounded (and hence almost C
X )
i s o r d e r bounded (and hence almost
S
o r d e r bounded), b u t n o t t o t a l l y bounded i n t h e Banach l a t t i c e for that matter, i n
the
and t h e v a l u e
m
but n o t o r d e r bounded. Extend t h e r e s u l t t o s p a c e s
n
. Show t h a t
-
o r d e r bounded) i n
Lp
for
1 5 p <
L2 and
a l s o t o t h e s p a c e L (X,u) of t h e previous e x e r c i s e by choosing f o r (en) 2 now an a p p r o p r i a t e d i s j o i n t sequence of non-negative f u n c t i o n s i n L2(X,p) each
e
n
,
of u n i t norm.
HINT: Note t h a t e v e r y subsequence of
S
h a s a subsequence converging
i n norm t o t h e n u l l f u n c t i o n . EXERCISE 122.6. L e t c i s e s and l e t
(X,u)
b u t n o t t o t a l l y bounded and Show t h a t t h e a l g e b r a i c sum bounded, a l t h o u g h
Sl + S2
S2 S
1
S 1 + S2
L2(X,p)
such t h a t
S1
i s o r d e r bounded
i s t o t a l l y bounded b u t n o t o r d e r bounded. + S2
i s n e i t h e r o r d e r bounded n o r t o t a l l y
i s almost o r d e r bounded.
HINT: The a l g e b r a i c sum of
s u b s e t of
b e t h e same as i n t h e two p r e c e d i n g exer-
S1,S2 b e s u b s e t s of
S1
and S2 i s n o t o r d e r bounded. I n f a c t , t h e
c o n s i s t i n g of a l l
fl + f2
, where
f l E S1
i s f i x e d and
AM-CONPACT OPERATORS
504
Ch. 18,91233
r u n s through S2 , f a i l s a l r e a d y t o be o r d e r bounded. S i m i l a r l y , S1 + S f2 2 i s n o t t o t a l l y bounded. S i n c e b o t h S l and S2 a r e almost o r d e r bounded,
"the same h o l d s f o r
S1 + S2
.
( i ) Show t h a t i f i n a Banach l a t t i c e e v e r y o r d e r
EXERCISE 122.7.
bounded s e t i s t o t a l l y bounded, t h e n e v e r y almost o r d e r bounded s e t i s tot a l l y bounded.
eP
( i i ) Show t h a t i n t o t a l l y bounded.
e v e r y almost o r d e r bounded set i s
(lsp<-)
EXERCISE 122.8. In a n AM-space w i t h s t r o n g norm u n i t e v e r y norm bounded s e t i s o r d e r bounded. In t h e converse d i r e c t i a n , show t h a t i f
L
i s a Ba-
nach l a t t i c e i n which e v e r y norm bounded s e t i s almost o r d e r bounded, t h e n there exists i n
norm) L
L
a n e q u i v a l e n t norm such t h a t
( w i t h r e s p e c t t o t h i s new
i s a n AM-space w i t h s t r o n g norm u n i t (communication by B.de P a g t e r ) .
HINT: The c l o s e d u n i t b a l l B i m p l i e s t h e e x i s t e n c e of
e E L+
in
L
i s almost o r d e r bounded, which
such t h a t
B c [-e,el + J B c C-e,e] +
[-he,lel
+
t B c C -3~ e3, ~ e+l a B
,
and s o on. G e n e r a l l y
for
... . It . Since L
n = 1,2,
unit i n
L
follows t h a t
B c [-2e,2el
. Hence,
e
i s a strong
i s a Banach l a t t i c e ( w i t h r e s p e c t t o t h e g i v e n norm
1, t h e s p a c e L i s uniformly complete (Theorem 100.4). But t h e n L i s a Banach l a t t i c e w i t h r e s p e c t t o t h e e-uniform norm 11 . / I , i . e . , L i s an AM-space w i t h r e s p e c t t o 11 .I/ . S i n c e B c 2 Ee,e 1 and [ -e,e] c p(e)B , p
t h e two norms are e q u i v a l e n t .
123. The band o f AM-compact o p e r a t o r s I n D e f i n t i o n 83.2 t h e n o t i o n of an o r d e r bounded ( l i n e a r ) o p e r a t o r T (from a R i e s z space The o p e r a t o r
T
L
i n t o a Dedekind complete Riesz s p a c e
i s c a l l e d o r d e r bounded i f
T
M ) was defined.
t r a n s f o r m s o r d e r bounded sub-
Ch. 18,91233
s e t s of
i n t o o r d e r bounded s u b s e t s of
L
505
COMPACT OPERATORS
M
. It
f o l l o w s immediately t h a t
e v e r y p o s i t i v e o p e r a t o r (and hence any d i f f e r e n c e of two p o s i t i v e o p e r a t o r s )
i s o r d e r bounded. S i n c e i.e.,
T
M
i s Dedekind complete, t h e converse h o l d s a s w e l l ,
i s o r d e r bounded i f and o n l y i f
s i t i v e (Theorem 8 3 . 3 ) . The s e t (from L
into
T = TI-T2
with
TI
and
T2
po-
of a l l o r d e r bounded o p e r a t o r s
&(L,M)
M ) i s a Dedekind complete R i e s z space (Theorem 8 3 . 4 ) .
If
M
f a i l s t o b e Dedekind complete, t h e d e f i n i t i o n of a n o r d e r bounded o p e r a t o r from
L
into
M
Lb (L,M)
remains unchanged and
not n e c e s s a r i l y a Riesz space. Also, i f tain that
T
i s of t h e form
with
T = TI-T2
i s now a v e c t o r s p a c e , b u t
i s o r d e r bounded, i t i s n o t c e r -
T
and
TI
p o s s i b l e , t h e r e f o r e , t h a t t h e v e c t o r s p a c e of a l l
T2
positive.Itis with
T = T,-T2
TI
and
p o s i t i v e ( c a l l e d t h e s p a c e of Jordan operators) i s smaller t h a n t h e
T2 space of a l l o r d e r bounded o p e r a t o r s . Note a l r e a d y t h a t i f
L
and
M
are
Banach l a t t i c e s , t h e n e v e r y p o s i t i v e o p e r a t o r (and hence e v e r y J o r d a n operat o r ) from
L
into
emphasize t h a t i f
i s norm bounded ( s i n c e
M
bounded o p e r a t o r s (from
L
into
M
i s Banach). Once more, we
M ) are t h e same, and t h e v e c t o r s p a c e of
a l l J o r d a n o p e r a t o r s i s now t h e R i e s z s p a c e call that i f
L
i s Dedekind complete, t h e n J o r d a n o p e r a t o r s and o r d e r
M
$,(L,M)
. Furthermore,
we re-
i s a Banach l a t t i c e w i t h o r d e r c o n t i n u o u s norm, t h e n
M
is
indeed Dedekind complete (Theorem 103.9). DEFINITION 123.1.
i n t o t h e Banach Zattice bounded subsets o f
L
The Jordan operator M
T
from the Banach l a t t i c e
i s said t o be AM-compact i f
i n t o precompact subsets o f
M
T
. Note
L
transforms order t h a t the d e f i -
n i t i o n applies t o Jordan operators only, s o t h a t , therefore, every AM-compact operator i s nomi bounded. It i s evildent t h a t e v e r y compact J o r d a n o p e r a t o r i s AM-compact.
The
converse does n o t h o l d , s i n c e i n g e n e r a l norm bounded s e t s are n o t almost o r d e r bounded, l e t a l o n e o r d e r bounded. I n some c a s e s , however, norm bounded
sets a r e o r d e r bounded. Obvious examples a r e t h e AM-spaces s t r o n g norm u n i t p(f) 5 1
implies
e
, because If1 5 e
then ( i f
. This
p
i s t h e norm i n
e x p l a i n s t h e terminology. I f
AM-space w i t h a s t r o n g norm u n i t , t h e n t h e J o r d a n o p e r a t o r
M ) i s compact i f and o n l y i f
T
L
possessing a
L ) the inequality L
T (from
i s an L
into
i s AM-compact.
The f i r s t r e s u l t a b o u t AM-compactness which w e s h a l l prove now shows t h a t an AM-compact o p e r a t o r t r a n s f o r m s n o t o n l y o r d e r bounded s e t s i n t o p r e -
AM-COMPACT OPERATORS
506
Ch. 18,51231
compact s e t s , b u t transforms almost o r d e r bounded s e t s i n t o precompact s e t s a s well. THEOREM 123.2. The Jordan operator
I i s AM-compact i f and only i f
M
the Baruch l a t t i c e
order bounded subsets of
be AM-compact.
T
almost o r d e r bounded s u b s e t of
and
L
by
M
L
and
E
in
S1
L
such t h a t
IlTll i s t h e o p e r a t o r norm of
since
T
S c S
1
+ E
and
BL
respecti-
BM
~
i s an
S
E
. There
,Bwhere ~
E
T (as observed e a r l i e r , T
1
T(S)
e x i s t s an o r d e r =
E / { ~ ~ I T \ Iand }
i s norm bounded
i s a Jordan o p e r a t o r ) . Then T ( S ) c T ( s ~ )+ E ~ T ( )B c T ( s ~ )+ L
The s e t
.
i s given, t h e image
> 0
into
L
transforms almost M
I t i s s u f f i c i e n t t o show t h a t i f
can b e covered by f i n i t e l y many b a l l s of r a d i u s interval
T
i n t o precompact subsets of
L
PROOF. Denote t h e u n i t b a l l s i n v e l y and l e t
T (from the Banach l a t t i c e
T(SI)
E ~ I I T I I B ~=
T ( s ~ )+
~
.B
~
i s precompact ( i n view of t h e d e f i n i t i o n of AM-compactness).
Therefore, T ( S I )
can b e covered by f i n t e l y many b a l l s of r a d i u s
f o l l o w s now from
T(S) c T ( S I ) + ~
many b a l l s of r a d i u s
E
.
E
tBh a t~
T(S)
;E
. It
can b e covered by f i n i t e l y
The compact J o r d a n o p e r a t o r s (from t h e Banach l a t t i c e Banach l a t t i c e
E
L
i n t o the
M ) f o m a v e c t o r space because t h e a l g e b r a i c sum of two
precompact s e t s i s precompact. S i m i l a r l y , t h e AM-compact o p e r a t o r s form a v e c t o r space. One of t h e main r e s u l t s t o be proved i n t h e p r e s e n t s e c t i o n
i s t h a t i f t h e norm i n form a band i n
L(L,M)
M
i s o r d e r continuous, t h e n t h e AM-compact o p e r a t o r s
. Note
t h a t t h i s s t a t e m e n t makes s e n s e because
M
i s now Dedekind complete (Theorem 103.9), and t h e r e f o r e t h e space of Jordan o p e r a t o r s i s now i d e n t i c a l w i t h t h e Dedekind complete Riesz space
Lb(L,M)
I n t h e proof we s h a l l need an i n e q u a l i t y which i n a c e r t a i n s e n s e i s somewhat s t r o n g e r t h a n t h e t r i a n g l e i n e q u a l i t y . The i n e q u a l i t y s t a t e s t h a t i f and
g
a r e a r b i t r a r y elements of a Riesz space
The proof follows by o b s e r v i n g t h a t
L
,
then
f
ch. 18,11231
COMPACT OPERATORS
507
where we have used one of t h e Birkhoff i n e q u a l i t i e s and t h e t r i a n g l e ine q u a l i t y . A f u r t h e r r e s u l t we s h a l l need w a s a l r e a d y mentioned i n E x e r c i s e 83.31. We r e p e a t t h i s r e s u l t i n t h e f o l l o w i n g l e m a ; f o r t h e proof we r e f e r t o t h e h i n t s g i v e n i n t h e exercise. LEMMA 123.3. Let
L
be an arbitrary Riesz space and M
compZete Riesz space. Furthermore, Zet from
L
i n t o M and l e t E
u
=
\ Cn(T I 1u .1) A ( T2u i ):n
is downwards directed and i n f THEOREM 123.4. Let
L
E
u
L
and
T2
cnu
variabze,
l i
be p o s i t i v e operators
=u, a12 uiEL+)
= (T AT ) ( u 1 2
M be Banach h t t i c e s ( n o n s
and
respectiveZy) such t h a t t h e norm AM-compact operators from
TI
. Then t h e s e t
u E L+
a Dedekind
A
and
p
A
i n M i s order continuous. Then t h e
i n t o M form
a band i n
.
$,(L,M)
PROOF. As observed a l r e a d y , t h e AM-compact o p e r a t o r s form a l i n e a r subspace of
$,(L,M)
d i s j o i n t operators i n and
T2
the s e t
. We
s h a l l prove now f i r s t t h a t i f
h(L,M)
TI+T2
such t h a t
TI
and
T2
a r e AM-compact. It i s s u f f i c i e n t t o show t h a t f o r any g i v e n (T1v:Osvsu)
i s a precompact s u b s e t of
are
i s AM-compact, t h e n M
. According
T1 u
E L+
t o the last
lemma, i f we w r i t e n :n v a r i a b l e , c l u i = u, a l l uiEL+\
1’
then T2
inf E
= ( I T I I ~ I T 2) ( u ) = 0 ; t h e l a s t e q u a l i t y h o l d s because
a r e d i s j o i n t . Hence, s i n c e t h e norm
a downwards d i r e c t e d s e t , w e have
now a l s o given, t h e r e e x i s t such t h a t
u1,u2
i s o r d e r continuous and
A
A(EU) .L 0
,...,un
TI
in
. It L+
follows t h a t i f with
EU E
> 0
u +u +...+ u = u 1 2 n
and
is is
508
AM-COMPACT OPERATORS
Ch. 18,11231
Using t h e dominated decomposition p r o p e r t y we s e e t h a t f o r e v e r y
0 i v S u
fying and
0 S v . i u. 1
1
there e x i s t
i = 1,2,
f o r every
v1,v2,
. By
( i = 1 , 2 ,...,n)
...,n
such t h a t
v
=
satis-
v +v +...+ v
1 2 n there e x i s t s T1+T2 of t h e o r d e r i n t e r v a l [O,uil
t h e assumption on
a f i n i t e subset
D.
t h e r e i s a member
0 i v' i u.
such t h a t f o r e v e r y
...,v
v
I n p a r t i c u l a r t h e r e e x i s t s f o r every
v.
w
of
w.
an element
1
E
satisfying
D.
D.
1
such t h a t
(3) I t f o l l o w s by means o f t h e i n e q u a l i t y ( 1 )
i
holds f o r every
, so
1,2,. . . , n
=
above t h a t
Note t h a t on account of ( 2 ) and ( 3 ) t h e norm o f t h e r i g h t hand s i d e does n o t exceed
. Writing
E
X'T
v-T w'
\ I
Let
D
I /
+ w2 +
w = w = h(Zn
\ I
T (v -a')' S X'En I i i J \ I
w , + w2 +
be t h e f i n i t e s e t of a l l
through
D.
0 5 v 5 u
(i=1,2,
...,n ) . We
-
TI
compact. Hence,
, we
have t h e r e f o r e
IT 1 ( v 1 . - w1. ) I > <
... + w
for
w.
have proved now t h a t f o r e v e r y
t h e r e e x i s t s a n element
shows t h a t t h e s e t
... + wn
(T,v:Oiv
w E D
E
.
running
v
satisfying
.
such t h a t
X(T1v-T w) < E This 1 i s t o t a l l y bounded, i . e . , t h e s e t i s p r e - , )
i s now a l s o AM-compact.
i s AM-compact. Of c o u r s e , T2
We may conclude from t h e above t h a t t h e AM-compact o p e r a t o r s form a Riesz s u b s p a c e of and
T
= -T
-
$,(L,M)
. Indeed,
and i t f o l l o w s t h a t
if T+
2 For t h e n e x t p a r t of t h e p r o o f , l e t
o p e r a t o r s such t h a t
0 i T
+
T
in
i s AM-compact, we p u t
T and
T-
b e a s e t of AM-compact
(Ta:aE{a})
Lb(L,M)
T I = T+
a r e AM-compact.
. Let
u
E L+
and
E
> 0
be
Ch. 18,51231
COMPACT OPERATORS
given. It f o l l o w s from f o r some
T u
. Hence
uo E {a)
0 5 v < u
holds f o r a l l
+
in
Tu
that
. By
L
h(Tu-Tau)
E
(Tv:OSv
. In o t h e r words,
+
, so
0
assumption t h e s e t
be covered by f i n i t e l y many open b a l l s of r a d i u s the s e t
509
4.
A(Tu-T
a0
u) <
(Tanv:O
, so
;E
can
i n view of ( 4 )
can be covered by f i n i t e l y many open b a l l s of r a d i u s
i s t o t a l l y bounded. This shows t h a t
(Tv:OSvSu)
is
T
AM-compact. For t h e proof t h a t t h e AM-compact o p e r a t o r s form an i d e a l i n i t i s s u f f i c i e n t t o prove now t h a t i f
then
0 5 S S T
i s AM-compact. We may assume t h a t
S
T > 0
and
. It
follows from t h e f i r s t
p a r t of t h e p r e s e n t proof t h a t U i s AM-compact whenever 0 i . e . , e v e r y componentof
,
Lb(L,M)
i s AM-compact,
T
U 5 T and U A(T-U) =O.
T i s AM-compact. Hence, every f i n i t e l i n e a r combination
of components of T i s AM-compact. By F r e u d e n t h a l ' s s p e c t r a l theorem t h e r e e x i s t s asequence i n L,,(L,M).
...)
of t h e s e l i n e a r combinations such t h a t O C S f S h o l d s n I t follows from t h e p r e c e d i n g p a r t of t h e proof t h a t S i s AM-compact.
(Sn:n=1,2,
F i n a l l y , f o r t h e proof t h a t t h e i d e a l of AM-compact o p e r a t o r s i s a band, 0 < T
let
+
T
with a l l
T
AM-compact. A s proved above, T
AM-compact. Hence, t h e AM-compact o p e r a t o r s form a band i n
M
i s o r d e r continuous i s due t o D.H.
.
Lb(~,~)
The r e s u l t t h a t t h e AM-compact o p e r a t o r s form a band i n norm i n
i s now a l s o
L(L,M)
i f the
Fremlin (1976). H i s o r i g i n a l
proof was r a t h e r complicated. A few months l a t e r P. Dodds o b t a i n e d a more d i r e c t p r o o f . The method of proof g i v e n above, due t o A.R.
Schep, i s some-
what s i m p l e r t h a n Dodds' p r o o f , although i t i s hased on t h e same i d e a s . F o r t h e work of Dodds and Fremlin, c o n t a i n i n g t h e i r main r e s u l t t h a t compactness of and
i m p l i e s corrpactness of
T M
S
if
0 S S
S
T
in
$,(L,M)
and both
L*
have o r d e r continuous norm, we r e f e r t o t h e i r paper (C11,1979). We
w i l l come back t o t h i s r e s u l t and r e l a t e d r e s u l t s i n t h e n e x t s e c t i o n s , b u t f i r s t we c o l l e c t some f u r t h e r p r o p e r t i e s of AM-compactness
i n the following
theorem. THEOREM 123.5.
and
M
( i ) If L
i s an AM-space possessing a strong norm unit
i s a Banaeh L a t t i c e w i t h o r d e r continuous norm, then t h e s e t o f aLL
order bomded compact operators (from L
into M
is a band i n
$(L,M)
.
AM-COMPACT OPERATORS
5 10
( i i ) If L
M
and
Ch. 18,91231
are Banach l a t t i c e s and if $ E L*
and
h E M
are given, then the f i n i t e rank operator T: L + M , defined by Tf = $ ( f ) . h for al7, f E L , i s AM-compact. Hence,if M has order continuous norm, the ~~ by the operators of f i n i t e rank i s contained in band ( L * @ M )generated the band of AM-compact operators. PROOF.
( i ) As already observed, any subset of
i s o r d e r bounded i f
L
T E L (L,M) i s compact i f and b The s e t of o r d e r bounded compact o p e r a t o r s
and only i f i t is norm bounded. Hence, any only if
i s AM-compact.
T
c o i n c i d e s , t h e r e f o r e , with the band of AM-compact o n e r a t o r s . ( i i ) Note f i r s t t h a t t h e given operator (because
L* = L-
is a Jordan operator. Since
T
i s Banach), the given l i n e a r f u n c t i o n a l
L
transforms any o r d e r bounded s e t i n
bounded. Hence, $
is o r d e r
$
i n t o an order
L
bounded subset of t h e r e a l numbers. Since any o r d e r bounded s e t of r e a l numbers i s precompact, t h i s shows t h a t the given o p e r a t o r
i s AM-compact.
T
Our next theorem i s a dual of p a r t ( i ) of the l a s t theorem. I t says that i f
L*
has order continuous norm and
o r d e r bounded compact o p e r a t o r s (from
i s an AL-space,
M
into
L
then the
M ) form a band i n
I n the proof we s h a l l need the c l a s s i c a l r p s u l t t h a t an operator a Banach space
i n t o a Banach space
V
Banach a d j o i n t o p e r a t o r
T* (from W+
i s compact i f and only i f i t s
W into
V* ) is compact. Furthermore,
we s h a l l make use of t h e v a r i a n t of Synnatschke's Theorem 115.4: I f that and i f
and
L
a r e Banach l a t t i c e s (norms
M
TI
ticular
and
belong t o
T2
for a l l
IT/* = IT*l
THEOREM 1 2 3 . 6 .
If
continuous norm and
M
E $,(L,M)
0
5
that
T* S 5 T
(T*$)(f) = $(Tf)
0 5 S 5 T
$,(L,M) and
implies
T
L" = L*
, holding
i s p o s i t i v e whenever in
and
h ) such
i s an AL-space),
(T VT ) * = TT v Tt 1 2
.
i n t o M I i s a band in
PROOF. We f i r s t r e c a l l t h a t
f o r e from
then
p
M
. In
par-
L i s a Banach l a t t i c e such that L* has order i s an AL-space, then the s e t of a l l order bounded
compact operators (from L
that
,
$(L,M) T
as s t a t e d i n
theorem
i s o r d e r continuous ( t h i s holds, t h e r e f o r e , i f
h
L(L,M).
from
T
T
M"
and
for all
$(L,M) =
H
f € L
.
. It
follows there-
and a l l
$
E
i s p o s i t i v e . It i s a l s o c l e a r t h a t
0 5 S* S r*
i s compact. Then
in
0 5 S*
= M-
h(M*,L*) . Assume now
5
T*
and
T*
is
,
18,11231
Ch.
COMPACT OPERATORS
511
P
compact. Furthermore, i n view of Theorem 118.4, t h e space
i s an AM-
space w i t h a s t r o n g norm u n i t , so by p a r t ( i ) of t h e l a s t theorem t h e ope-
s*
rator
i s compact. Thus
i t s e l f i s compact. Assume now t h a t
S
i s compact. Then
E L(L,M)
T
i s compact, and t h e r e f o r e , s i n c e ( a g a i n
T*
by p a r t ( i ) of t h e l a s t theorem) t h e o r d e r bounded compact o p e r a t o r s from into
@
form a band, IT*[
L*
i s compact. But then T
T+
and
IT*l
, so
= ITI*
ITI*
i s compact. We have t h u s proved t h a t compactness
IT1
i m p l i e s compactness of
of
i s compact. But
. Conversely,
IT1
if
i s compact, then
IT1
a r e compact by t h e f i r s t p a r t of t h e p r e s e n t p r o o f , and s o
T-
T
i s compact. It h a s t h u s been shown a l r e a d y t h a t t h e o r d e r bounded compact o p e r a t o r s from
L
Finally, l e t Then
0
into 0
form an i d e a l i n
M T
5
, so
f o r a l l a~ E L+
T u 4 Tu
0 E M* (note t h a t every
5
Lb (L,M)
in
4 T
@ E
P
that
. This
u E L+
pact, so
T*
shows t h a t
T:
Ta
.
compact f o r a l l
4 T*
4 (T*@)(u)
in
i s compact. This shows t h a t
Lb(@,L*) T
a
.
for a l l
@(Tau) 4 @(Tu)
i s o r d e r continuous s i n c e
continuous norm). I n o t h e r words, (T:@)(u) and
L(L,M)
with
f o r every
. All
has order
M
E @
@
are com-
T*,
i s compact.
The l a s t theorem h a s a v a r i a n t which can b e a p p l i e d t o t h e t h e o r y of k e r n e l o p e r a t o r s . We r e c a l l t h a t i f N
the order dual on
L
therefore
Li
=
L :
p'
i s a Riesz s p a c e , then t h e band i n
c o n s i s t i n g of a l l o r d e r continuous l i n e a r f u n c t i o n a l s
L
i s denoted by
b e denoted by
L
LL
. If
L
i s a Banach l a t t i c e , t h e n
. The r e s t r i c t i o n of t h e norm p* i n . I n s t e a d of r e q u i r i n g o r d e r c o n t i n u i t y
L-
L* of
l a s t theorem, we s h a l l now only r e q u i r e o r d e r c o n t i n u i t y of
=
to p* p'
L*
and
L:
will
as i n the
. Instead
of proving t h a t t h e o r d e r bounded compact o p e r a t o r s form a band, w e prove now t h a t t h e o r d e r continuous compact Operators form a band. THEOREM 123.7. If
L
i s a Banach l a t t i c e such that t h e mrm
continuous compact operators (from L i n t o M I i s a band i n contained i n t h e band L
P'
in
i s order continuous and M i s an AL-space, then the s e t of a l l order
L :
L,(L,M)
$,(L,M)
,
of a l l order continuous operators (from
i n t o M ). PROOF. A s i n t h e l a s t theorem, we have
L" = L*
shows a l r e a d y t h a t , f o r any o r d e r bounded o p e r a t o r dual
T"
i s t h e same a s t h e Banach d u a l
TC
. Also,
and T: L
since
+
M-= fi M M
,
. cis
the order
i s an AL-space,
512
AM- COMPACT OPERATORS
t h e norm i n
M
T
=
the order dual
t h e r e f o r e , T*
4
i s o r d e r c o n t i n u o u s . Hence, e v e r y
continuous, i . e . , M * = M" tinuous
Ch.
maps
n T-
n maps
into
M*
.
= M*
M"
By Theorem 9 7 . 1 , M,"
. Since
L,*
E M-
into
Li
. In
=
M*
18,11231
i s order
f o r any o r d e r conthe present case,
i s an AM-space p o s s e s s i n g
M*
a s t r o n g norm u n i t and i s a Banach l a t t i c e w i t h o r d e r c o n t i n u o u s norm, L: w e may a p p l y Theorem 1 2 3 . 5 ( i ) t o conclude t h a t t h e s e t of a l l o r d e r bounded
.
compact o p e r a t o r s (from M* i n t o L* ) i s a band i n h(M*,L:) This makes n i t p o s s i b l e now t o show t h a t t h e s e t of a l l o r d e r c o n t i n u o u s compact operat o r s (from
L
M ) i s a band i n
into
,
$,(L,M)
t h e l a s t theorem, r e p l a c i n g everywhere
L*
by
by i m i t a t i n g t h e proof of L:
.
We r e t u r n f o r a moment t o o p e r a t o r s of f i n i t e r a n k , mapping t h e Banach lattice
L
i n t o t h e Banach l a t t i c e ' M
then the operator
T: L + M
,
. If
d e f i n e d by
$
and
E L*
Tf = $ ( f ) . h
h
E M a r e given, f E L , is
for a l l
4 E LIT
J o r d a n and compact (hence norm bounded). I f t h e s t r o n g e r c o n d i t i o n
is s a t i s f i e d (i.e.,
4
o r d e r c o n t i n u o u s ) , then
T
i s o r d e r continuous and
compact. We F e c a l 1 t h a t e v e r y f i n i t e l i n e a r combination ( w i t h r e a l c o e f f i c i e n t s ) of o p e r a t o r s of t h i s t y p e i s c a l l e d a f i n i t e rank o p e r a t o r . The
s e t of a l l f i n i t e t a n k o p e r a t o r s i s denoted by concerns f u n c t i o n a l $ from compact and e v e r y
T E LE
LE ) . Hence, e v e r y 0
M
L* 0 M ( o r T E L* @ M
if it
L* B M
i s J o r d a n and
i s o r d e r continuous and compact. T h i s l e a d s
now t o t h e f o l l o w i n g c o r o l l a r y . COROLLARY 123.8. M
form a band
A
(i) I f the compact Jordan operators from
, then t h e band
This s i t u a t i o n o c c m s i f e i t h e r M
and
(L*@M)dd
L
into
i s contained i n the band A
.
i s an AM-space w i t h a strong norm u n i t
L
has order continuous norm o r
L*
has order continuous norm and
M
is an AL-space. ( i i ) I f t h e order continuous and compact Jordan operators from M
form a band
B
, then the band
!l%is s i t u a t i o n occurs i f e i t h e r and
M
(L2M)dd L
L into
i s contained i n the band B
.
is an AM-space with a strong norm u n i t
has order continuous norm o r
L:
has order continuous norm and
M
i s an AL-space. The s i t u a t i o n i n p a r t ( i i ) of t h e l a s t c o r o l l a r y o c c u r s i n p a r t i c u l a r
if L
and
M
are Banach f u n c t i o n s p a c e s . L e t
M(X,u)
and
M(Y,v) b e t h e
Riesz s p a c e s of a l l r e a l measurable f u n c t i o n on t h e a - f i n i t e measure s p a c e s (X,A,u)
and
(Y,Z,v)
r e s p e c t i v e l y and l e t
L
and
M
be ideals i n
M(Y,v)
18,51231
Ch.
and
COMPACT OPERATORS
r e s p e c t i v e l y . Without l o s s of g e n e r a l i t y w e may assume t h a t
M(X,p)
t h e c a r r i e r of L
and
513
is
L
Y
and t h e c a r r i e r of
M
is
X
.
Assume a l s o t h a t
are Banach l a t t i c e s w i t h r e s p e c t t o t h e norms
M
spectively (i.e.,
p
and
,A
re-
and
M
a r e Banach f u n c t i o n s p a c e s ) . Note t h a t by
Theorem 1 1 2 . 1 t h e s p a c e s
L
and
t h e c a r r i e r of
L
i n our case
LIT
L i = L*
have t h e s a m e c a r r i e r , i . e . ,
i s a l s o equal t o
. Then
Y
Theorem 95.1
may b e a p p l i e d , s t a t i n g t h a t t h e s e t of a l l a b s o l u t e k e r n e l o p e r a t o r s (from L
into
M ) i s e x a c t l y t h e band
(Lz@M)dd
. Hence,
i f t h e c o n d i t i o n s men-
t i o n e d i n p a r t ( t i ) of t h e l a s t c o r o l l a r y a r e satisfied, e v e r y a b s o l u t e k e r n e l o p e r a t o r (from given t h a t
M
into
L
M ) i s compact. Note t h a t i f i t i s only
c o n t i n u o u s norm, t h e n i t f o l l o w s from Theorem
has order
1 2 3 . 5 ( i i ) t h a t e v e r y a b s o l u t e k e r n e l o p e r a t o r (from
L
M ) i s AM-
into
compact.
If
and M are Banach f u n c t i o n spaces such t h a t e i t h e r L i s an Lm-space and M has order continuous norm o r the associate space LE has order continuous norm and M i s an L -space, then 1 every absolute kernel operator ( f r o m L i n t o M I i s compact. This holds THEOREM 123.9.
L
i n p a r t i c u l a r f o r absolute kernel operators from
Lm
into
.
L
P
( Isp< - )
If it i s o n l y known t h a t M has order conP tinuous norm, then every absolute kerne2 operator (from L i n t o M I is
and from
(p>l) into
L
Ll
AM-compact. The Theorems 123.6 and 123.7 ( w i t h t h e a p p l i c a t i o n s mentioned i n Coroll a r y 123.8 and Theorem 123.9) are due t o A.R.
l i z a t i o n o f Theorem 123.9 i s c o n t a i n e d i n C o r o l l a r y EXERCISE 123.10. According t o Theorem 123.8, i f T: L
(p>l)
-+
P pact. Also, i f bounded, t h e n
Ll
or
T
129.8. T: Lm + L
is an absolute kernel operator, then
P
(p<-) T
or
i s com-
T: L (p<-) + Lm o r T : L l + Lp ( p > l ) and T i s norm P T 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 (Theorem 98.2 and C o r o l l a -
r Y 98.4; t h e c a s e either
and
A genera-
Schep ([41,1980).
Ll
-+
L (p>l) P
i s Dunford's theorem). Assume now t h a t
DODDS-FREMLIN AND ALIPMNTIS-BUFXINSHAW THEOREMS
514
where
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 and
TI
Ch. 18,51241
i s norm bounded. Show
Tp
T2TI i s a compact a b s o l u t e k e r n e l o p e r a t o r . S i m i l a r l y , show t h a t i f
that
or
m
I1
Ll
m
Lp ( P > l )
T I i s norm bounded and
where
l2
L1
’
i s an absolute k e r n e l o p e r a t o r , then
T2
i s a compacf a b s o l u t e k e r n e l o p e r a t o r . More g e n e r a l r e s u l t s a r e con-
T2TI
t a i n e d i n E x e r c i s e 129.10.
EXERCISE 123.11. L e t and T2
T2: M
+
be Banach l a t t i c e s and l e t
L,M,K
TI: L
be J o r d a n o p e r a t o r s . Show t h a t i f one a t l e a s t of
K
i s AM-compact, t h e n
T2TI: L
+
K
+
M and
TI
i s AM-compact.
124. Theorems of Dodds-Fremlin and Aliprantis-Burkinshaw
Let
L
and
M
be Banach l a t t i c e s . One of t h e main theorems w e s h a l l
prove i n t h e p r e s e n t s e c t i o n i s t h a t i f
norms and and
T
S
and
T
a r e o p e r a t o r s from
i s compact, t h e n
S
L*
L
into
M
have o r d e r continuous
M
such t h a t
continuous norm ( b u t n o t alwasy imply t h a t L
nor
L*
compact i m p l i e s t h a t due t o C.D.
L* S
T
S 5 T
5
S
in
L L
i s i t s e l f compact. I f
L
has order
0 5 S 5 T
T
compact does
has n o t ) , then
i s compact, b u t now
S2
with
i s compact. F i n a l l y , i f
h a s o r d e r continuous norm, t h e n S3
0
I n p a r t i c u l a r we f i n d t h a t i f
have o r d e r continuous norm, t h e n any p o s i t i v e o p e r a t o r
dominated by a compact o p e r a t o r
neither
and
i s a l s o compact. This r e s u l t i s due t o
P.G. Dodds and D. H. Fremlin ([ll,l979). and
L*
0 5 S 5 T
i s compact. These r e s u l t s a b o u t
A l i p r a n t i s and 0. Burkinshaw ([2]1980).
S2
with and
S3
T are
The AM-compact o p e r a t o r s
p l a y a n i m p o r t a n t p a r t , as w e l l a s o p e r a t o r s mapping norm bounded s e t s o n t o almost o r d e r bounded s e t s . A s observed i n Theorem 123.2, AM-compact o p e r a t o r s map almost o r d e r bounded s e t s o n t o precompact s e t s . I t is. obvious now t h a t
Ch. 18,11241
if
L,M,K
5 I5
COMPACT OPERATORS
a r e Banach l a t t i c e s and T,:
L
M
-f
,
T2: M
+
K
a r e operators possessing the properties t h a t onto almost o r d e r bounded s e t s and
maps norm bounded sets
TI
i s AM-compact,
T2
then
T2TI: L
-f
K
i s compact. a r e Banach l a t t i c e s ; t h e norm c l o s e d u n i t
In what f o l l o w s , L,M,N,K b a l l of
L
norm i n
L
for
M,N
w i l l b e denoted by
, the and
set
BL
. As
(fEL:p(f)iE)
before, i f
E
> 0
w i l l be denoted by
and
.
E B ~ Similarly
.
K
is the
p
LEMMA 124.1. Let 0 i S i T
.If
then so does
T S
.
S and T be operators from L i n t o M s a t i s f y i n g maps mrm bounded s e t s onto a h o s t order bounded s e t s ,
PROOF. It i s s u f f i c i e n t t o prove t h a t
maps t h e u n i t b a l l
S
onto
BL
-
.
an almost o r d e r bounded s e t . Let B+ = (fEB :f>O) and BL = (fEBL:f
.
.
f o r e , t o show t h a t
S(Bi)
i s almost o r d e r bounded. By h y p o t h e s i s
i s almost o r d e r bounded, i . e . , such t h a t f o r any A(h)
i E
u E BE
(where
E
we have
> 0
i s given, there e x i s t s w i t h -w i g i w
Tu = g+h
i s t h e norm i n t h e space
h
(Tu-w)+ i (g-w) so
if
(Su-w)+ i (Tu-w)'
i Ihl
+
+
+ h
, and
+
= h
T(Bi)
w E M'
and
M ). I t f o l l o w s t h a t
S l h l ,
therefore
A{(Su-w)+)
i
E
. This
shows
that Su = w
with
0 i w
A
SU
A
S w
Su + (Su-w)+
and
A{(Su-w)+}
c E
S(B+) i s c o n t a i n e d i n t h e a l g e b r a i c sum L almost o r d e r bounded. COROLLARY 124.2. I f the operators
S
. In o t h e r words,
t h e image
r0,wl + E B , ~ i.e.,
and
T
(from L
S(Bt)
into M
is
)
sa-
516
DODDS-FREMLIN AND ALIPRANTIS-BURKINSHAW THEOREMS
0
tisfy
2
S 5 T
T
and
i s compact, then
Ch. 18,11243
maps norm bounded s e t s
S
onto almost order bounded s e t s . PROOF. The image T(B ) i s precompact, and t h e r e f o r e T(BL) i s a l m o s t L o r d e r bounded. But then S(BL) i s a l m o s t o r d e r bounded i n view of what we have j u s t proved. THEOREM 124.3 ( i ) If the operators
S2: M
and
+
, and
K
i f the image
i s AM-compact, then ( i i ) Let
that
0 S S
1
TI
S
mapping M i n t o in
K
S2SI: L
and
S1
.
T,
+
S1: L
satisfy
S2
such t h a t
S2 and
0 5 S
2 TI
0 5 S s T
continuous, then compactness of
S2
T
and L
in
such
T2
be p o s i t i v e operators
T2
be compact. Then
. Furthermore,
T2
5
M
l e t the norm
and the norm i n
implies compactness of
S2S1 i s
L i s order
.
S'
PROOF. ( i ) I t f o l l o w s immediately from t h e h y p o t h e s e s t h a t
i s precompact, i . e . ,
M
K i s compact.
be order continuous and l e t
compact. In p a r t i c u l a r , i f
-f
i s almost order bomded and
be p o s i t i v e operators mapping L i n t o
Similarly, l e t
K
and
S1
S 1 (BL)
S2SI(BL)
i s a compact o p e r a t o r .
S2S1
i s a l m o s t o r d e r bounded. F u r t h e r 1 L i s o r d e r c o n t i n u o u s , t h e AM-compact o p e r a t o r s
( i i ) By C o r o l l a r y 124.2 above, S (B ) more, s i n c e t h e norm i n from
M
into
K
K
form a band (Theorem 123.4). Hence, s i n c e
p a c t (by h y p o t h e s i s , T 2 rator
S2
i s AM-compact.
i s even compact) and s i n c e
i s AM-com-
T2
< T2 , t h e ope-
0 < S
S2SI i s
I n view of p a r t ( i ) i t f o l l o w s t h a t
compact. The same i n c l u s i o n as i n p a r t ( i i ) h o l d s i f i n s t e a d of assuming o r d e r c o n t i n u i t y of t h e norm i n L*
K
.
THEOREM 124.4. Let
n m t h a t the norm i n
L*
S.
we assume o r d e r c o n t i n u i t y of t h e norm i n
(i=1,2)
PROOF. The p o s i t i v e o p e r a t o r s satisfy
0 5 S* < T;
satisfy
0 < S* < T*
2 1
1
T.
and
( i = 1 , 2 ) be a s above and assume
i s order continuous. Then again
. Similarly, . Since T7
S;
S.7
and
and and T;
T;
T*
I
map map
K*
M*
S2SI i s compact. into
into
L*
M*
and they and t h e y
are compact and t h e norm i n
i s o r d e r c o n t i n u o u s , t h e l a s t theorem shows t h a t
SyS;
L*
i s compact, i . e . ,
Ch. 18,51243
i s compact. I t follows t h a t
(S2S1)*
517
COMFACT OPERATORS
i s compact.
S2SI
In t h e n e x t theorem we drop a l l assumptions about o r d e r c o n t i n u i t y of norms. THEOREM 124.5. Let
and Si ( i = 1 , 2 , 3 ) be operators s a t i s f y i n g
Ti
0 5 Si 5 T.
Furthermore, l e t
S3S2SI: L
pact. Then
and
T
K
M
,
(i=1,2,3)
Ti
0
5
be corn
S 5 T
and
i s a compact o p e r a t o r mapping
TI
.
'M
Hence, i f
E >
0
S (B+) i s an alrrost I L i s given, t h e r e e x i x t s
A,
I
(B+) c [-w,wl L
+ SIBM for
be the principal ideal i n
E
1
=E/(II~,II.IIS
g e n e r a t e d by
M
w
II
2 )
. The
ideal
a normed Riesz space w i t h r e s p e c t t o t h e w-uniform norm, d e f i n e d by W
L
i t f o l l o w s from C o r o l l a r y 124.2 above t h a t
s
llfll
in L
such t h a t
E 'M
Let
s3
i s compact.
0 5 S1 5 T I
o r d e r bounded s u b s e t of w
( i = 1 , 2 , 3 ) and Zet a22
i s compact. I n p a r t i c u l a r , i f S3
i s compact, then
PROOF. Since into
-f
s2
s1 L-M-N-K.
and
=
inf(A:lfl
for a l l
f
E
. This
A,
norm i s --additive
A
W
is
on
i s Banach w i t h r e s p e c t t o t h e t h u s i n t r o d u c e d
and we show now t h a t
A, norm. For t h i s purpose, l e t
(fn:n=1,2,
...)
be a Cauchy sequence, i . e . , ,
.
we have If -f 1 5 6w f o r s u f f i c i e n t l y l a r g e m,n n m A(f -f ) 5 6.A(w) , i . e . , ( f n ) i s a Cauchy sequence w i t h n m r e s p e c t t o t h e norm A i n M Since M i s Banach, t h e r e e x i s t s an
f o r any given
6 > 0
It follows t h a t
element
f
.
E M such t h a t
A ) . It f o l l o w s now from n 2 m for
fn fm
t h a t t h e norm l i m i t large
m
, and
hence
llf\lw
. The
-
of
f
Banach d u a l
6w 5 f
A,
A(f) 5 h(w)
,
C-w,wl. i.e.,
fn
f + 6w m satisfies
5
for sufficiently large
i s an AM-space w i t h r e s p e c t t o t h e w-uniA :
Note t h a t t h e c l o s e d u n i t b a l l of order interval
n
f (with r e s p e c t t o t h e norm
f m - 6 w 5 f 5 f + 6w m i s t h e w-uniform l i m i t of t h e sequence (f,)
f
I t h a s been shown t h u s t h a t
form
converges t o
h a s t h e r e f o r e a n o r d e r continuous norm.
A,
(with respect t o
Ilfllw )
This i s a s u b s e t of t h e s e t of a l l
f
is the satisfying
i t i s a n o m b o u n d e d s e t (with r e s p e c t t o A ) . Now con-
518
DODDS-FREMLIN AND ALIPRANTIS-BURKINSHAW THEOREMS
s i d e r the operator
T2
and i t s r e s t r i c t i o n t o
A,
. Since
Ch. 18,91241
i s compact
T2
by h y p o t h e s i s , t h e image of e v e r y norm bounded s e t ( w i t h r e s p e c t t o precompact i n N.
N ; i n p a r t i c u l a r , t h e image of
I n o t h e r words, t h e r e s t r i c t i o n
T2: A,
+
X )
is
i s precompact i n
(f:Ilfl,Fl)
i s a compact o p e r a t o r . W e
N
get th e r e f o r e t h e following s i t u a t i o n : T A w - -T2 - - + N & K
with
T2,T3
and
compact and t h e norm i n
from Theorem 124.4 t h a t
S S ([-w,wl) 3 2
S3S2: A,
+
o r d e r continuous. It follows t h e n
K
of t h e c l o s e d u n i t b a l l
( e q u i v a l e n t l y , t o t a l l y bounded i n
K
i s compact, i . e . , t h e image
[-w,w]
in
>.Applying
i s precompact i n
A,
the operator
s3s2
to
and
. The
right
K
formula (11, we f i n d
where
11 S211
11
and
SdI
are t h e o p e r a t o r norms of
S2
Sj
hand s i d e (and t h e r e f o r e t h e l e f t hand s i d e ) c a n b e covered by f i n i t e l y many b a l l s of r a d i u s
, i.e.,
ZE
S3S2S,:
L
+
K
i s compact.
W e mention some immediate a p p l i c a t i o n s ( a l s o due t o A l i p r a n t i s -
Burkinshaw). THEOREM 124.6.
t h e operator
T: L
(i) +
L
If L i s a Dedekind complete Banach l a t t i c e and has the property t h a t I T ( i s compact, then T3 i s
compact. If it i s known i n a d d i t i o n t h a t t h e norm i n L
or i n
L * is order
i s compact. ( i i ) fie i d e n t i t y operator i n a Banach Z a t t i c e of i n f i n i t e dimension
continuous, then
T~
cannot be dominated by a compact operator. PROOF. ( i ) Each term i n T
3
(T+-T-l3
=
=
... +
(T+l3 +
- 3
(-T )
i s a product of t h r e e f a c t o r s ; e a c h f a c t o r i s e i t h e r 0 s T" s IT1
and
0 s T-
S
IT1
and
IT1
T2
i f t h e norm i n
L
or in
or
T-
. Since
i s compact, i t f o l l o w s from t h e
l a s t theorem t h a t e a c h term i s compact, and t h e r e f o r e larly for
T+
L*
T3
i s compact. Simi-
i s o r d e r continuous.
Ch. 18,51243
519
COMPACT OPERATORS
(ii) If the identity operator I in the Banach lattice L dimension is dominated by a compact operator, then I is impossible since the closed unit ball in L
=
of infinite
I3 is compact. This
is not precompact.
For the discussion of certain conditions under which any positive operator dominated by a compact operator is compact itself, we need the notion of precompactness with respect to a seminorm. For the definition, assume
. The
is a seminorm in the vector space V
that p
subset D of
V
is
said to be p-precompact if every sequence in D has a subsequence, say
,...
, which
fl,fg
is p-Cauchy, i.e., p(fn-fm)
+
0 as m,n
+
=
. We
explici-
tly mention the case that the vector space is a Banach lattice and the seminorm is derived from a linear functional. DEFINITION 124.7. Let
the seminorm p6
L
in
L be a Banach l a t t i c e , l e t
be defined by
p (f)
=
$
6 E L* and l e t
0 5
$(If/) f o r a l l
.
f E L
l?ae subset D of L i s said t o be p precompact i f every sequence i n D 6 has a p -Cauchy subsequence ( i . e . , i f f , , f 2 , . . i s the subsequence, then
.
6
$(lfn-fml)
+
as
0
m,n
+
m
1.
It is obvious that precompactness of D D
for every 0
in L
5
6
E L*
implies p -precompactness of
. Under certain conditions for
6
D
and for the norm
the converse holds.
LEMMA 124.8. Let
lattice
L
only i f
D is
D be an a h o s t order bounded subset of t h e Banach
w i t h order continuous norm p p precompact f o r every 6
PROOF. Assume first that
that D
D
. Then
D
6E
L*
0 5
i s precompact i f and
.
is an order bounded subset of L
is p -precompact for every
6
0
5
6E
L*
. Assume
also that
to be precompact. This implies the existence of a nmber sequence '(fn:n=1,2, Since D
...)
in D
satisfying p(fn-fm)
is order bounded, we have
-$u
5
fn
.
5
hu
E >
> E
such D
fails
0 and a
for all m # n
for some u E L+
and
If -f 1 5 u for all m,n Furthermore, since the norm n m is order continuous, there exists (by Theorem 103.12) a positive linear
all n ; hence functional $I ideal A,
on L
such that
generated by
u
6 is strictly positive on the principal
. The sequence
(fn)
is included in D
is p -precompact, so there exists a subsequence fn ,fn 6 1 2
,...
and D
such that
.
520
MDDS-FREMLIN AND ALIPMNTIS-BURKINSHAW THEOREMS
+ ( l f -f nk Ilk+l' ) < complete (because
'-' p
... . The
k = 1,2,
for
Banach l a t t i c e
i s o r d e r c o n t i n u o u s ) and
If
L
f o r every 5
. But
+(wn) + 0
that
c:
p(w ) C 0 n p ( f -f ) + 0 "k nk+l m # n Hence, D hence
n
+(I
fn;fnk+l
then
wn
-
1)
+
as
k +
+
and
D
0 5 0 E L*
D-
of
i s s t r i c t l y p o s i t i v e on p
. It
AU
1,
follows t h a t
p(fn-f,)
>
E
for
. Let
i s almost o r d e r bounded and
L
D+ = (f+:fED)
and
D- = (f-:fED)
+
a r e almost o r d e r bounded and p -precompact
I f-gl i.e.,
t o t a l l y bounded. For t h i s purpose, l e t
+
for a l l k,
f o l l o w s from
, contradicting the fact that
+ + E L* ( t h e l a s t because I f -g 1 and 1. I t w i l l be s u f f i c i e n t , t h e r e f o r e ,
0 s
5 u
i s precompact.
for a l l
D+
+
0 (since
Assume now t h a t t h e s u b s e t p -precompact
i s Dedekind
L
I
s 2 -n+ 1
i n view of t h e o r d e r c o n t i n u i t y of
.
Note t h a t
+ . It
0 < w
and we have
nk+l
"k
so
exists in
-f
Ch. 18,51243
1 f--g-l
u E L+
i s almost o r d e r bounded, t h e r e e x i s t s
a r e m a j o r i z e d by
> 0
i s precompact,
D+
t o prove t h a t E
be g i v e n . S i n c e
such t h a t every
f+
E
D+ D+
satisfies f+
=
(f'hu)
+
+ ( f -u)
+
with
p((f+-u)+}
<
AE ,
and t h e r e f o r e
Note now t h a t t h e o r d e r bounded s e t
0 5 $ E L* (because
(f+Au:fED)
is p
/ f + ~ u - g + h u l i s m a j o r i z e d by
+- precompact
If+-g+l
f o r every
for a l l
f , g 5 D ) . Hence, a s shown i n t h e f i r s t p a r t of t h e p r o o f , t h e s e t (f+Au:fED)
i s precompact and c a n be c o v e r e d , t h e r e f o r e , by a f i n i t e number
of b a l l s of r a d i u s
1~
. It
f o l l o w s from formula ( 2 ) above t h a t
covered by a f i n i t e number of b a l l s of r a d i u s bounded.
E
.
f o r every
, i.e.,
D+
D+
can b e
is totally
COMPACT OPERATORS
Ch. 18,51241 COROLLARY 124.9.
Let
and
L
52 1
be Bmach Zattices with norms
M
and
p
X be order continuous and l e t T be an order bounded operator from L i n t o M . Then T is AM-compact i f and only if the image of any almost order bounded s e t in L i s p -peeompact i n M f o r every 4 0 5 $ E M* . Equivakntly, T i s AM-compact if and only if the image of any order bounded s e t in L i s p -peeompact in M for every 0 5 $ E M* . 6 A
respectively, l e t
A s an i n t r o d u c t i o n t o t h e n e x t theorem, we r e c a l l some f a c t s about quotient spaces. I f L
and
p
$
Riesz seminorm i n ideal N by
in
6
Cfl,Cgl,
L
...
f ) and d e f i n e Riesz norm i n
completion of
p ( f ) = $ ( / f l ) for a l l
. The s e t of
L
. Denote
(i.e.,
$
all
satisfying
[fl
[fl
L/N
i s t h e e q u i v a l e n c e c l a s s modulo
$
and
[gl
in
L/N+
. Now, p
i s a Riesz space i n which
i s an &-space
p
let
i s an L/N4
containing
$
$
p
(L/N ) $
L
i s a Riesz norm). It i s
4
f o r p o s i t i v e elements i s p r e s e r v e d .
$
w i t h r e s p e c t t o t h e norm
ond
M
be t h e norm
( c f . s e c t i o n 100 f o r t h e
p
. We now s t a t e
$
be Banach 2attices:such that the norms
m d in M are order continuous. Furthermore, l e t i n t o M such that
operators from L
0 5 S 5 T
and
S(Bl)
of
S
T
and
T
be
is compact. Then
is compact. PROOF. W e prove f i r s t t h a t t h e image
f o r every
0 5 $ E M*
l a r l y a s above, l e t norm completion f
N
.
with r e s p e c t t o t h e norm
THEOREM 124.10. Let
T"
is a
p (f) = 0 $
and prove t h e Dodds-Fremlin theorem.
S
p4
t h e elements of t h e q u o t i e n t Riesz s p a c e
e a s y t o s e e t h a t t h e a d d i t i v i t y of
in L*
then
4
(L/N$)
Hence, (L/N$)-
f E L
,
f E L
p a l s o i n L / N + by p $ ( C f l ) = p $ ( f ) Then p is a $ $ L/N+ w i t h t h e p r o p e r t y t h a t p i s I-additive, i . e . ,
for all positive proof t h a t
i s a p o s i t i v e l i n e a r f u n c t i o n a l on t h e Riesz space
$
i s d e f i n e d by
. For
(M/N )$
Tf
. The
of
T
into
p
$
) of t h e q u o t i e n t
M/N$
, defined
by
i s t h e e q u i v a l e n c e c l a s s modulo
[Tfl
operator
L
0 5 $ E M*
T"
i s p$-precompact be given. Simi-
be t h e AL-space which one g e t s by t a k i n g t h e
(with r e s p e c t t o
be t h e o p e r a t o r from
E L , where
t h i s purpose, l e t
BE
M/N$
. Now
TNf = [ T f l N+
let
f o r every
containing
i s compact. Indeed, i t follows from t h e compactness
t h a t t h e image of any norm bounded set i n
L
c o n t a i n s a norm con-
522
WDDS-FREMLIN AND ALIPRANTIS-BURKINSHAW THEOREMS
v e r g i n g sequence
(Tfn:n=1,2
p -convergent i n
M/N+
A
S f = [Sf]
for a l l
in
,
M
. The o p e r a t o r S" . This l e a d s t o
and t h e n
i s compact, where
L
(M/N+)-
into
([Tfnl:n=1,2
,...)
is
i s defined si m i l a r l y , i . e . ,
f E L
a r e p o s i t i v e o p e r a t o r s from T-
,...)
Ch. 18,91241
the s i t u a t i o n t h a t
(M/N+)-
0
such t h a t
i s an AL-space and
2
S" N
S
and
T"
N
and
I T
L* has o r d e r continuous
norm. By Theorem 123.6 t h e s e t of a l l o r d e r bounded compact o p e r a t o r s from L
into
(M/N#)-
t h e operator minology i n
+
i t s e l f ( i n s t e a d of i n
S(B;)
124.8 t h a t
N
and
T
2
i s compact,
T"
S(B+) of B; L Hence, i f we can prove
M / N ) t h e image
i s p+-precompact. This holds f o r every that
0 I S
i s l i k e w i s e compact. I n o t h e r words, r e t u r n i n g t o t h e t e r *
S-
M
N
i s a band. Hence, s i n c e
0 I
+
E fl
.
i s almost o r d e r bounded, i t w i l l f o l l o w by means of Lemma i s a precompact s u b s e t of
S(B:)
t h e n compact. The almost o r d e r boundedness of
1 2 4 . 1 . Indeed, T
maps
BE
compact), and hence s o does
, i.e.,
M
S(B+)
L
the operator
f o l l o w s from Lemma
o n t o an almost o r d e r bounded s e t ( s i n c e S
T
is
by t h e lemma r e f e r r e d t o .
As mentioned a l r e a d y , Theorem 124.5 ( t h e theorem about Aliprantis-Burkinshaw.
is
S
S
3
) i s due
to
The proof p r e s e n t e d h e r e , due t o A . R . Schep, i s
d i f f e r e n t from and s i m p l e r t h a n t h e o r i g i n a l p r o o f . The l a s t theorem (about S
S2
i t s e l f ) i s t h e Dodds-Fremlin and
S3
due t o A.R.
. The
theorem, which was p r i o r t o t h e r e s u l t s about
p r e s e n t p r o o f , making an a p p e a l t o Theorem 123.6, i s a g a i n
Schep.
F i n a l l y , we p r e s e n t s e v e r a l examples showing t h a t t h e theorems proved i n t h i s s e c t i o n a r e b e s t p o s s i b l e . I t i s convenient f o r t h i s purpose t o make use of t h e Rademacher f u n c t i o n s
r (x) = sgn s i n 2"x, 1 on
\
A s u s u a l , we denote
max(rn(x),O)
measure
TI
= [0,2.rrl
for
n
=
1,2
,... .
.
by
r+(x) N e g l e c t i n g a s e t of Lebesgue +n h a s t h e v a l u e one on a s e t of n and t h e v a l u e z e r o on t h e remaining s e t of measure 'TI Similarly,
measure zero, n o t e t h a t t h e f u n c t i o n the function
x
++
r r
(m#n)
mn v a l u e z e r o elsewhere.
r
.
h a s t h e v a l u e one on a s e t of measure
Our f i r s t example w i l l show t h a t t h e Dodds-Fremlin n e c e s s a r i l y h o l d i f one a t l e a s t of t i n u o u s norm. Define
L*
and
M
~ I I
and t h e
theorem does n o t
f a i l s t o have an o r d e r con-
18,11243
by
S l ( a ) = C 1 anrn
m
+
(
S2f =
by
f o r every f r i dx, X
f E L2(X)
f o r every
T a = C 1
m
an.xx
(where
1
1
and
2
dx,
fdx,
...)
E t, and
...1 )
define
xx(x) = 1
fdx,
for a l l
x E X ) and
...) .
X
a r e compact (each i s of rank one). Furthermore,
T2
i
for
fr'
. Similarly,
X Evidently, T I
I
a = (al,a2,
X
(
T2f =
0 5 S. 5 T.
523
COMPACT OPERATORS
Ch.
= 1,2
.
Of 'course, a l l
S.
and
T.
(i=1,2)
a r e norm
bounded, being p o s i t i v e operators between Banach l a t t i c e s . Denoting t h e u n i t vectors i n
Ll
( e l y e 2 ,...)
by
S S (e -e ) = S (r+-r+) 2 1 n m 2 n m S S (e -e ) i s equal t o 2 1 n m
, which
, we
have
S e
=
r:
for a l l
n
S2SI
pact. Note t h a t f o r
.
S1
t h e f a i l u r e i s due t o t h e f a c t t h a t the norm i n
i s not o r d e r continuous, whereas f o r t h a t t h e norm i n
Lm of
S S (e -e ) f o r m # n i s a t l e a s t ;IT It f o l 2 1 n m i s not compact and, t h e r e f o r e , n e i t h e r S 1 nor S2 i s com-
and hence t h e Lm-norm of lows t h a t
. Hence
shows t h a t the n i t h coordinate i n
.em
the f a i l u r e comes from t h e f a c t
S2
.e?;
i s not o r d e r continuous.
I n t h e second example we show t h a t the Aliprantis-Burkinshaw theorem about
S3
i s b e s t p o s s i b l e . Let
a s i n t h e f i r s t example and l e t
Note t h a t n e i t h e r operator
S: L + L
L
nor
L*
and
Si
L
T.
(i=1,2)
be t h e same o p e r a t o r s
be t h e d i r e c t sum
has o r d e r continuous norm. Now d e f i n e the
t o be equal t o
S1
on
Ll
, equal
to
S2
on
L2(X)
and
524
COMPACT, AM-COMPACT AND SEMI-COMPACT
em. S i m i l a t l y ,
z e r o on T2 S
on
L2(X)
let
and z e r o on
T: L
+
em . Then
i s n o t . Furthermore, t h e o p e r a t o r
z e r o on
L2(x)
,
@
be e q u a l t o
L
0 5 S 5 T
L1
@
,
L2(X)
S
i t can be shown t h a t t h e theorem
Ll
S ~ S I on
s 2 i s n o t compact. of c o u r s e , ~3
so
S i m i l a r l y , bymeans of analogous o p e r a t o r s L =
i s compact and
holds, T
i s equal t o
S2
I, , e q u a l t o
on
TI
and
18,51251
Ch.
and
i s compact.
on
T
about
o
=
i s b e s t possi-
S2
ble. t h e f i r s t example i s a
These examples are due t o Aliprantis-Burkinshaw, v a r i a n t upon a n o l d e r example of P. Dodds.
125. Compactness, AN-compactness If
a r e Banach l a t t i c e s and
L,M,K
TI: L
-f
T2: M
+
and semi-compactness Tl,T2
a r e o p e r a t o r s such t h a t
M
maps norm bounded sets i n t o almost o r d e r bounded s e t s and
K
i s AM-compact, t h e n
compact s e t s , i . e . , note t h a t i f
T: L
+
T2TI: L
K
maps norm bounded s e t s i n t o pre-
i s a compact J o r d a n o p e r a t o r , t h e n
M
t h e mentioned p r o p e r t i e s ( i . e . , o r d e r bounded s e t s i n
+
i s a compact o p e r a t o r , I n t h e converse d i r e c t i o n
T2T1
M
and
T
i s a l s o AM-compact).
T
T
maps norm bounded s e t s i n
h a s both o f i n t o almost
L
It i s a n a t u r a l
q u e s t i o n t o a s k i f t h e converse of t h e l a s t s t a t e m e n t h o l d s , i . e . , Jordan operator
T: L + M
i f the
has both p r o p e r t i e s , i s i t t r u e then t h a t
is
T
compact? The proof of t h e theorem t h a t under c e r t a i n norm c o n d i t i o n s ( o r d e r c o n t i n u i t y of t h e norms i n
L*
i z e d by a compact o p e r a t o r
T
and i n
M ) any p o s i t i v e o p e r a t o r
S
major-
i s i t s e l f compact makes i t p l a u s i b l e t h a t t h e r e
must e x i s t some r e s u l t s i n t h i s d i r e c t i o n , because obviously t h e compactness of
S
is related t o the f a c t that
S
i n h e r i t s from
both p r o p e r t i e s w e
T
have j u s t mentioned. I f , however, t h e o r d e r c o n t i n u i t y c o n d i t i o n s f o r t h e norms are only p a r t i a l l y s a t i s f i e d ( e x p l i c i t l y , i f norm, b u t
compact t h a t
S
last section) S t h e spaces
L
M
h a s n o t ) , i t f o l l o w s j u s t a s w e l l from
L*
h a s o r d e r continuous
0 5 S 5 T
with
T
h a s b o t h p r o p e r t i e s , b u t ( a s s h a m i n t h e examples i n t h e
i s now n o t n e c e s s a r i l y compact ( a l t h o u g h and
M
S2
i s compact i f
are i d e n t i c a l ) . This shows t h a t o r d e r c o n t i n u i t y and
of t h e norm i n only one o f t h e s p a c e s
M
c o n j e c t u r e t o b e t r u e . We s h a l l prove
t h a t i f both
L*
i s n o t enough f o r o u r M
and
L*
have o r d e r
continuous norm, then t h e c o n j e c t u r e i s t r u e . It i s n e c e s s a r y t o do some p r e l i m i n a r y work. W e b e g i n by o b s e r v i n g t h a t i n g e n e r a l t h e norm c l o s e d u n i t b a l l i n t h e Banach l a t t i c e
L
i s n o t almost o r d e r bounded, even i n t h e case t h a t
Ch.
18,51251
t h e norm
COMPACT OPERATORS
in
p
L
and t h e norm
in
p*
L*
525
a r e b o t h o r d e r continuous ( a s
shown i n E x e r c i s e 122.3; s e e 122.8 f o r c o n d i t i o n s when norm bounded s e t s a r e almost o r d e r bounded). This i s d i f f e r e n t i f i n s t e a d of a norm we have a c o l l e c t i o n of R i e s z seminorms. L e t Banach l a t t i c e
L
bounded i f f o r any f
E
E >
subset
0
D
of
i s s a i d t o be p-almost order
L
f = f
+ f2
1
p{(lfl-u)+} < E
with
f o r every
L*
1 . Then
B
in
L
L
p
; norm
@ ) there exists
B
E
> 0
E L*
0 5 @
4
elements i n
p(f2) <
E
4
p -almost order bounded f o r e v e w
PROOF. Assume f i r s t t h a t
@
such t h a t e v e r y and
f E D ). A s b e f o r e , i f
be a Banach l a t t i c e (norm
i s o r d e r continuous, but
p*
0 S @
p -almost o r d e r bounded f o r some fixed
u
p*
in
i s order continuous i f and only i f the closed u n i t b a l l
p*
is
E L+ 5
p ( f ) = @ ( I f I ) f o r every
f E L . THEOREM 125.1. Let
u
-u 5 f l
E L* , t h e seminorm p4 i s d e f i n e d by
I$
be a R i e s z seminorm i n t h e
t h e r e e x i s t s a n element
can b e w r i t t e n a s
D
(equivalently, i f
0 5
. The
p
E L*
. It
and a sequence
.
B
f a i l s t o be
follows t h a t ( f o r t h i s
...)
(un:n=1,2
of p o s i t i v e
such t h a t
4(ul)
> E
,
@{(u2-4ul)+} >
,
E
@{(u3-8(ul+u2)
+}
> E,...
.
Generally uir}
(1)
for
> E
Note now t h a t f o r any n a t u r a l number
n = 2,3,
... .
t h e norm l i m i t
n
Cm
n+ 1
sequence of p a r t i a l sums ( Z nk+ l
exists i n
L
2-iui:k=n+l,n+2,.
, because
L
n = 2,3,
... . Then,
of t h e
..
i s a Banach l a t t i c e ( t h e norm l i m i t i s a l s o t h e
supremum of t h e p a r t i a l sums). P u t
for
2-iui
v
1
i n view of ( 1 )
=
0
, w1
= u1
above, $(wn) >
and
E
f o r every
n
.
526
COMPACT. AM-COMPACT AND SEMI-COMPACT
Ch. 18,51251
Furthermore,
... . Hence, $(wn-vn) . This i m p l i e s t h a t $(vn)
for
n = 2,3,
<
1~
for a l l
n
ate
n
> ie
for a l l
n t no
0 t o observe is t h a t t h e sequence
. Indeed,
L
for vn
5
m < n
we have
f\un-2"
urn)'
and
v
d i s j o i n t and
1~
$(vn) >
for a l l
Conversely, assume now t h a t
. Let
val
CO,$l
(cpn:n=1,2,...)
L*
in
. For
s u f f i c i e n t t o show t h a t By h y p o t h e s i s
${(I
that
u
and
c o n t r a d i c t s Theorem 116.1,
i s p a l m o s t o r d e r bounded f o r every $
b e a d i s j o i n t sequence i n t h e o r d e r i n t e r -
t h e proof t h a t p*($,)
.
is
B
+ 0
i s o r d e r continuous it i s
p*
n +
as
m
(Theorem 104.2). L e t
E
> 0.
i s p -almost o r d e r bounded, s o t h e r e e x i s t s u E L+ such $ 1~ f o r a l l f E B S i n c e 2'; gn(u) 5 g ( u ) f o r t h i s
B
and a l l
a n a t u r a l number n t no
i s norm bounded and
i m p l i e s t h a t e v e r y norm bounded
p*
0 5 Q1 E L* B
.
fl -u)+] <
particular
next thing
converges weakly t o z e r o . It f o l l o w s t h a t
L
p -almost o r d e r bounded for e v e r y
0 5 $ E L*
(vn:nmo)
. This
n t no
a c c o r d i n g t o which o r d e r c o n t i n u i t y of d i s j o i n t sequence i n
. The
i s a d i s j o i n t sequence i n
2-n(2num-un)t
5
m
i s o r d e r continuous, t h e sequence
Hence, p *
$
...)
(vn:n=1,2,
exceeding an appropri-
no
f E B
k
, we
have
such t h a t
Q1,(u)
$,(u)
<
-t
1~
0
as
for all
n
-+
m
, so
n t no
there e x i s t s
. Then,
if
a r e given, we have
There e x i s t s a d u a l theorem, a s f o l l o w s .
THEOREM 125.2. The norm
p
f o r every =
I @I(u)
0 5 u
L is order continuis p -almost order bounded
i n the Banach l a t t i c e
ous i f and only if t h e closed u n i t b a l l i n
L*
E L , where t h e seminorm pu i n L* is defined by
f o r every
cp E L*
.
~ ~ ( =$ 1
18,11251
Ch.
COMPACT OPERATORS
527
PROOF. The proof i s q u i t e s i m i l a r t o t h e proof of t h e l a s t theorem, u s i n g now t h a t o r d e r c o n t i n u i t y of j o i n t sequence in
i m p l i e s t h a t every norm bounded d i s -
p
converges weak s t a r t o z e r o (Theorem 116.2).
L*
COROLLARY 125.3. Let
be a p o s i t i v e operator, mapping the Banach
T
, and
l a t t i c e L i n t o the Banach l a t t i c e M order continuous. Then the image T(B) i s p -almost order bounded f o r every
l e t the n o m
i n L* be
p*
of the elosed u n i t b a l l 0
5 $
E M*
B
.
in L
~
PROOF. L e t $ ( f ) = $(Tf)
be g i v e n and l e t
0 5 $ t: M*
for a l l
. Since
f E L
(by Theorem 125.1) f o r any given ${(lfl-u)+} <
for a l l
E
Note now t h a t
(Tg)'
for a l l
. It
f E B
,
f E B
Tg'
5
i.e.,
f o r every
follows t h a t
an element
> 0
E
0 s I$ t: L*
b e d e f i n e d by
i s o r d e r continuous, t h e r e e x i s t s
p*
u
${T(lfl-u)+} < g E L
. For
such t h a t
E L+ E
for a l l
g = If1
-
f E B
.
t h i s gives
u
T(B) i s p -almost o r d e r bounded. Note t h a t $
t h e l a s t i n e q u a l i t y i n (2) i s somewhat s t r o n g e r t h a n t h e s t a t e m e n t t h a t
i s p. -almost o r d e r bounded.
T(B)
J,
THEOREM 125.4. Let
i n L*
and
M
be Banach l a t t i c e s such t h a t t h e n o m s
are order continuous and l e t t h e operator T
order bounded. Then bounded s e t i n
and M
L
L
i s p precompact i n M f o r every $
PROOF. I f t h e image of any norm bounded s u b s e t of f o r every s e t of
L
$
Conversely, l e t 0 5 J, E M*
T(B)
be
0 5 JI € M* L
.
i s p -precompact
9
b e AM-compact. I t w i l l b e s u f f i c i e n t t o prove t h a t
T
of t h e u n i t b a l l
s e c t i o n 122, T(B)
...,hn
M
in
i s AM-compact.
T
. Similarly as
bounded, i . e . , h,,
-f
JI E M* , t h e n c e r t a i n l y t h e image of any o r d e r bounded subi s p -precompact f o r e v e r y 0 s $ € M* . T h i s i m p l i e s immediately,
0 5
by C o r o l l a r y 124.9, t h a t t h e image
T: L
i s AM-compact i f and o n l y i f t h e image of any n o m
i s p -precompact
f o r any M
J,
B
in
L
i s p -precompact f o r e v e r y $
observed f o r a d i s t a n c e i n a m e t r i c space i n
E >
such t h a t
0
i f and only i f T(B) i s p - t o t a l l y J, t h e r e e x i s t s a f i n i t e s e t of elements
COMPACT, AM-COMPACT AND SEMI-COMPACT
528
Note f i r s t t h a t t h e space
i s Dedekind complete ( s i n c e
M
continuous norm); i t f o l l o w s t h a t t h e o p e r a t o r
0 5 $ E M*
let
and
E
> 0
b e given. Write
o r d e r continuous norm, t h e s e t f = g+h
t h e form
compact, t h e image of a f i n i t e s e t
with
g E
g ,,. ..,gn
be a r b i t r a r y . Then
f E B For
gi
gi
u E L
[-U,U]
++
such t h a t
+(I h l ) <
and
in
[-u,u]
,
f o r which
. Since
every
f E B
. Since
JE
such t h a t f o r every $(\Tg-Tgi
1)
L*
has
T
i s of
i s AM-
g E [-u,u]
AE .
<
g E C-u,ul
f = g+h w i t h
T(B)
(Tf:fEB,$(l
Tf-Tgi]
Now, l e t
+ ( l h l ) < 1~
and
.
J,
and
T: L + M
set
D
of
)<E
i s p - t o t a l l y bounded
It has been shown t h u s t h a t i f
L
L*
and
M
i s AM-compact, t h e n t h e image i s p -precompact f o r every J,
t o compare t h i s r e s u l t w i t h Lemma 124.8,
0
have o r d e r continuous norm
T(D) 5 $
of any norm bounded sub-
€ M*
. It
which i s almost o r d e r bounded a s w e l l a s p -precompact $
t o conclude t h a t compact f o r every
T(D)
T(D)
i s precompact, because a l t h o u g h
0 5 $ E M*
and
T(D)
i s n o t almost o r d e r bounded u n l e s s
c o n d i t i o n . I n o t h e r words, even i f
L*
i s compact. I n mentioned above
i s p -pre-
T(D)
J,
i s a l s o (of c o u r s e ) norm bounded, T and
s a t i s f i e s an a d d i t i o n a l M
have o r d e r continuous norm,
i t i s n o t t r u e i n g e n e r a l t h a t an AM-compact o p e r a t o r bounded s e t s i n t o precompact s e t s , i . e . ,
i s of importance
according t o which any s u b s e t of
g e n e r a l we cannot apply t h i s l a s t r e s u l t t o t h e s e t
T(D)
e x i s t s . Now,
i s covered by t h e f i n i t e union Ui,ln
M
!TI*$
=
has o r d e r
M
M
as above, t h i s g i v e s
Hence, T(B)
i.e.,
4
-t
i s t o t a l l y bounded, which i m p l i e s t h e e x i s t e n c e
T[-u,uI
t h e r e i s one t h e s e , say
ITI: L
i s p -almost o r d e r bounded (Theorem
B
Therefore, there e x i s t s
125.1).
Ch. 18,11251
T: L
-f
M
maps norm
i t i s n o t t r u e t h a t an AM-compact
o p e r a t o r i s always compact. Obviously, t h e e x t r a h y p o t h e s i s we need ( b e s i d e s
Ch. 18,81251
T ) i s that i f the subset
AM-compactness of t h e image
529
COMPACT OPERATORS
of
D
i s norm bounded, t h e n
L
i s almost o r d e r bounded, because t h e n
T(D)
0
o r d e r bounded a s w e l l a s p -precompact f o r e v e r y jl
124.8 can be a p p l i e d t o conclude t h a t
5
i s almost
T(D)
Ji E M* , s o L e m a
i s compact. Order bounded o p e r a t o r s
T
from one Banach l a t t i c e i n t o a n o t h e r , mapping norm bounded s e t s i n t o almost o r d e r bounded s e t s a r e sometimes c a l l e d L-weakly compact operators ( f o l l o w i n g P . Meyer-Nieberg,
[21,1974), but we s h a l l p r e f e r t o call these operators
The r e s u l t mentioned above w i l l now be f o r m u l a t e d as a theorem.
semi-compact.
The theorem i s due a g a i n t o Dodds-Fremlin THEOREM 125.5.
(Z: I ] ) .
If L and M are Banach Lattices such t h a t L*
compact i f and only i f
is
T
and
T: L
M have order continuous norm, then the order bounded operator
+
M
AM-compact and semi-compact.
is
To conclude t h i s s e c t i o n , we d i s c u s s some c o n d i t i o n s under which AM-
compactness of
T
i m p l i e s AM-compactness of
and c o n v e r s e l y . The theorem
T*
which f o l l o w s i s due t o B. de P a g t e r . THEOREM 125.6. Let
norms
and
p*
(i)
If
L
and
M
L*
and
M* ) .
be Banach l a t t i c e s (norms
T*
and
M
is
i s AM-compact,
T
then
T: L
A
are order continuous and t h e operator
T
i s AM-compact i f and only i f
T = T
follows t h a t
p*
( i ) Note f i r s t t h a t
- T2
with
> 0
TI
lTfl 5 S ( l f 1 )
W e have t o prove t h a t E
p*
i s Jordan (since
T
and
T
for a l l
T*[O,$]
b e given. S i n c e
2
(in
f E B , i. . e . ,
M ) of t h e s e t
f E L
. Now,
let
T*
T: L
0 5 $
${S((h/)} <
31 "
. Since
T
, it
+ T2
1
E fl b e g i v e n . L*
. Let
B
in
g E [-u,ul
i s AM-compact,
i s compact, Denote t h i s c l o s u r e by
L
also
is
u E L+
Therefore, there e x i s t s with
M
i s AM-compact), S = T
i s a t o t a l l y bounded s u b s e t of
f = g+h
+
i s AM-compact.
i s o r d e r c o n t i n u o u s , t h e u n i t ball
i s of t h e form
T[-u,u]
T
p o s i t i v e . Writing
p S * y a l m o s t o r d e r bounded (Theroem 125.1). such t h a t e v e r y 1 ( S * $ ) ( l h l ) < 3'
AM-
M has the property t h a t
+
and
PROOF.
,
i s AM-compact.
i s order bounded, then
hence
A
i s order continuous ( s o t h a t , therefore, M i s Dedekind
A
( i i i ) If
+
p
i s AM-compact.
compLete) and the order bounded operator T*
T: L
i s order continuous and the operator
p*
compact, then (ii) I f
in
A*
and the closure X
. Hence,
5 30
X
COMPACT, AM-COMPACT AND SEMI-COMPACT
Ch. 18,81251
i s a compact m e t r i c space (with r e s p e c t t o t h e d i s t a n c e f u n c t i o n for a l l points
d ( p I , p 2 ) = A(p -p ) Tg
1
f u n c t i o n on C(X)
2
g E [-u,ul
for
. Therefore,
X
pI,p2
in
LO,$]
X
a r e uniformly bounded and equicontinuous on Theorem t h e s e t
[O,$l
$ E
Now, l e t
f E B
f = g+h
and
that
Q E [O,$l
5
E
,
(i2iIn)
T*Qi
,
and
of
i.e.,
for a l l
$i
. Hence
1 5~
be a s above. We
. This
f E B
implies
i s covered by t h e union of t h e b a l l s
T*[O,$]
and r a d i u s
such t h a t
.
g E [-u,ul
E
. This
C(X) LO,$]
, satisfying
$i
${S(lhl)} <
<
CO,$l
in
Q
s o t h a t i n view of the
b e a r b i t r a r y , and l e t
I (T*$)(f)-(T*Qi)(f) I
p*(T*Q-T*Qi)
with centers
X
(Ql,...,Qn)
for a l l
g E [--u,ul
with
It follows t h a t
The f u n c t i o n s
t h e r e i s one of t h e s e , say
lQ(Tg)-Qi(Tg)l< E;1
have
.
i s t o t a l l y bounded i n
[O,$l
i m p l i e s t h e e x i s t e n c e of a f i n i t e s u b s e t f o r each
i s a r e a l continuous
Q E @
i s a s u b s e t of t h e s p a c e
= ($:O<$I$)
of a l l r e a l continuous f u n c t i o n s on
Ascoli-Arzela
X ) and t h e p o i n t s
form a dense s u b s e t . Every
E
. This
shows t h a t
i s to-
T*[O,$I
t a l l y bounded. ( i i ) The proof i s s i m i l a r , Note f i r s t t h a t Dedekind complete). Furthermore, s i n c e
A
theorem (Theorem 115.4) may be a p p l i e d , i . e . , we have t o prove t h a t given. Since
A
TIO,u]
i s o r d e r continuous, t h e u n i t b a l l
$ E B*
I $ " I ( I T I U<) Since
1
3E ,
i s of t h e form
i.e.,
the
function
on
$ = $ ' + $"
X
X ( t h e v a l u e of
i s a s u b s e t of t h e space
f
. Every
. The
X
i m p l i e s t h e e x i s t e n c e of
v
each
. This
v E C0,ul
in
B*
with
0 Iu E L ,
also is
M*
E
> 0
be
'ITlu-
E @ such Q' E [-$,$I and $
1
L* ) of t h e s e t
x
, so
t h e r e i s one of t h e s e , say
E X
functions
,,...,v
C0,ul
v.
T*[-$,$l
.
is
i s a r e a l continuous
f E L
a t the point C(X)
formly bounded and equicontinuous on C(X)
. Given
. Let
M
Therefore, t h e r e e x i s t s
closure (in
compact. Denote t h i s c l o s u r e by C0,uI
is
M
U T * I ( I Q I ~ I ) I=( {~I )T I * ( I $ ~ ~ I ) } ( ~=) I $ " I ( I T I <~ )?E
i s AM-compact,
T*
ITI* = \T*l
i s t o t a l l y bounded i n
almost o r d e r bounded (Theorem 125.2). t h a t every
e x i s t s (since
IT1
i s o r d e r continuous, Synnatschke's
in
,
is v
X(f) ). T h e r e f o r e , in
[O,ul
a r e uni-
i s t o t a l l y bounded i n C0,ul
satisfying
such t h a t f o r
Ch. 18,11251
Now, l e t
g
have
COMPACT OPERATORS
Q E B* =
g ' + 9''
It f o l l o w s t h a t
X(Tv-Tv.) centers
5 E
Tv.
and
v E [O,u]
be a r b i t r a r y , and l e t
g ' E [-$,+I
with
Ig(Tv)-g(Tvi)l<
, i.e.,
TC0,ul
(I
531
and
b e a s above. We
{~T*~(~g"~)}(u)
4 E B*
for a l l
E
v.
. This
implies t h a t
i s covered by t h e union of t h e b a l l s w i t h
and r a d i u s
E
. This
shows t h a t
TC0,ul
i s totally
bounded. ( i i i ) Follows immediately from p a r t s ( i ) and ( i i ) . It i s of some i n t e r e s t t o observe a l r e a d y t h a t i t w i l l b e proved i n
Theorem 127.4 t h a t i f o r d e r bounded, t h e n if
T*
so
T
p*
X
and
a r e o r d e r continuous and
i s semi-compact i f and only i f
T
i s AM-compact a s w e l l a s semi-compact,
T*
i s compact by Theorem 125.5, which i m p l i e s t h a t
T*
T*
EXERCISE 125.7. Let a s w e l l a s semi-compact (i)
S
and
Show t h a t
S
Show t h a t
S2
is T
,
i s compact. Note
i s o r d e r continuous. For t h e same reason,
@*
0 < S < T
M
i t s e l f , s i n c e we do
p a t t ( i i ) of t h e l a s t theorem does n o t f o l l o w immediately from
i n t o i t s e l f such t h a t
+
t h e n t h e same holds f o r
t h a t Theorem 125.5 cannot immediately be a p p l i e d t o n o t know whether t h e norm i n
T: L
i s s o . Hence
T
part (i).
be o p e r a t o r s from t h e Banach l a t t i c e
. Assume
furthermore t h a t
(so t h a t , t h e r e f o r e , T 2 i s compact i f b o t h
L*
T
L
i s AM-compact
i s compact). and
L
have o r d e r c o n t i -
nuous norm. (ii)
i s compact i f
L
h a s o r d e r continuous norm.
( i i i ) Show t h a t , w i t h o u t c o n t i n u i t y assumptions, S4 t h i s purpose, n o t e f i r s t t h a t
o r d e r bounded (where unit ball such t h a t
B
of
B+
S
i s semi-compact.
i s compact. For
Hence, S(B+)
i s almost
i s t h e set of a l l p o s i t i v e elements i n t h e c l o s e d
L ). I t f o l l o w s t h a t f o r
E
> 0
given t h e r e e x i s t s
w E L+
532
COMPACT, AM-COMPACT AND SEMI-COMPACT
S(B+) c C-w,wl
+ clB
with
E]
&/l\SII
=
3
Ch. 18,51251
.
4 + For t h e proof t h a t S (B ) i s t o t a l l y bounded, i t i s s u f f i c i e n t , t h e r e f o r e , 3 4 + i s precompact (because S (B ) i s i n c l u d e d i n t o show t h a t S C-w,wl 3 S [-w,wl
A,
+ EB ). The p r i n c i p a l i d e a l
g e n e r a t e d by
w
i s an AM-space
as i t s c l o s e d u n i t b a l l ( s e e t h e p r o o f of Theorem 1 2 4 . 5 ) .
[-w,w]
with
Denote t h e r e s t r i c t i o n s of
and
S
to
T
by
A,
T1
and
S1
respectively.
I t i s s u f f i c i e n t t o prove t h a t S S S I : A,
+
L
+
L
-+
L
3
i s compact, b e c a u s e t h e n t h e image
C-W,~]
unit ball
in
by showing t h a t T2
A
S;(S*)2r
2
S C-w,wl
= S S
A,
i s AM-compact. S i n c e
i s an AM-space,
TT: L*
-t
A*
W
of t h e c l o s e d 2 S S 1 can b e proved
i s compact. For t h i s p u r p o s e , n o t e t h a t
+ A:
i s compact from which i t f o l l o w s t h a t ST
C-w,wl
i s precompact. Compactness of
L*
(T*)’
semi-compact), which i m p l i e s semi-compactness of prove t h a t
1
the operator
TI
i s compact (and hence (S*)’
i s AM-compact
TI
.
I t remains t o
(by h y p o t h e s i s ) and
i s compact. It f o l l o w s t h a t
is compact, and hence AM-compact. But t h e n
i s a l s o AM-
ST
compact (because t h e AM-compact o p e r a t o r s from
L* i n t o t h e AL-space A: form a band). T h i s proof i s due t o P . van E l d i k ( r e p l a c i n g a s i m p l e r proof
that
S5
i s compact).
EXERCISE 125.8. L e t
i t was proved t h a t i f o p e r a t o r s (from
M
L into
L
and
M
b e Banach l a t t i c e s . I n Theorem 123.4
h a s o r d e r c o n t i n u o u s norm, then t h e AM-compact M ) form a band i n t h e s p a c e of o r d e r bounded
o p e r a t o r s . I n Theorem 125.5 i t w a s proved t h a t i f c o n t i n u o u s norm, t h e n and semi-compact.
T: L + M
T
L* have o r d e r
i s compact i f and o n l y i f
second proof of t h e r e s u l t ( i n Theorem 124.10) and
and
i s &compact
T
Show t h a t by combining t h e s e r e s u l t s w e can o b t a i n a
o r d e r c o n t i n u o u s norm and
0 5 S < T
M
S,T
i s compact, t h e n
EXERCISE 125.9. For
that i f
a r e o p e r a t o r s from
n = 0,1,2,
S
...
L
M
and
into
have
L*
M
such t h a t
i s compact.
,
let
Ln
b e a Zn-dimensional
real
v e c t o r s p a c e w i t h t h e u s u a l c o o r d i n a t e w i s e o r d e r i n g and t h e norm d e r i v e d from t h e u s u a l i n n e r p r o d u c t . For e v e r y i t s e l f i s g i v e n by t h e m a t r i x
An
n
, where
,
an o p e r a t o r mapping
Ln
into
Ch. 18,81251
533
COMPACT OPERATORS
Since the matrix B = 2-n/2A is orthogonal, Bn determines a norm presern n ving operator in Ln It follows that Cn = 2-"An determines an operator
.
I(Tnf 1 1
in Ln preserving orthogonality and such that
Tn
for every
f E Ln
. Since all entries in the matrix
of
=
2-"12 11 f 11
ITn/ are equal to
2-n , we have lTnlenk where
enk (k=1,2,...,2n)
Z&
= 2-n/2
,
. Note that
are the basis vectors of Ln
enk satisfies I/ p ( 1
f = Z a enk E Ln
if
2" Ek=l enk
2-
=
IT Ip,
and
= 1
p,
=
p
=
. Furthermore:
is arbitrary, then
.
ITn If = 2-n/2( ikak)Pn
Note now that, by the familiar Schwarz inequality,
It follows that
I/ I Tnl f 1)
with
, show
ITnlpn = p,
11 ITn/11
= 1
.
The space
=
2-n(Cak)2
b2 can be written as a direct
Define the operator T:
.
a T For n O n the operator norm of = Cm
result that T /TI
m
= Co 8
ITn[
Ekat
=
L2 + b,
,...
0,1,2
T?Sn
8
Show that
derive from this result that
let
Tn on each Ln , i.e.,
Sn = T 8 T, 0
tends to zero as n ITlp, = p,
IT1
ITn/ satisfies
sum
to be equal to
,
. Combining this result
I/ f It2
... .
+
... 8
(where
p,
.
Show that T n and derive from this
8
IT\ exists in L,
is compact. Show now that
.
=
that the operator norm of
Cm8L = L 8 L @L O n 0 1 2
T
5
and satisfies
is as defined above) and
is not compact. It follows that the set
L2 i s not a band in the Space all order bounded operators in L2 (although L2 and L; have order
of all order bounded compact operators in of
continuous norm). This example is due to U. Krengel ([21). that
IT/
is a positive kernel operator in
L, , so
Finally, note
that by Theorem 123.8
534
SEMI-COMPACTNESS AND ORDER
the operator
CONTINUITY
(TI is AM-compact. It follows that
Ch. 18,11261
IT/ must fail to be
semi-compact. This can be seen in a direct way by observing that =
, where
for all n
pn
(pn:n=0,1,2,
...)
IT\pn= is norm bounded, but not almost
order bounded. EXERCISE 125.10. In Theorem 125.6 it is shown that if L Banach lattices such that L* T: L
+
M
and M
are
has order continuous norm and the operator is AM-compact. If the norm in L*
is AM-compact, then T*
is
not order continuous, this does not necessarily hold. A s an example, let T: L ,
+
where
L [0,27~1 be defined by 1
r
bounded. Since every order bounded subset of 122.7), it is clear that in L1
is positive and norm
is the n-th Rademacher function. Then T
by
, we
en
have
T
is AM-compact. Denoting the n-th unit vector
Ten
=
+
for all n
r n
Idx This shows that
Ll
is precompact (Exercise
=
71
.
Hence
for m # n
.
T is not compact. Show now that T*: Lm
AM-compact (because AM-compactness of
is not
+
T would trivially imply compactness).
126. Semi-compact,operators and order continuity of the operator norm
In this section L and M
A
are again Banach lattices (norms
and
respectively). A s defined in the previous section, the order bounded ope-
rator T: L unit ball
+
B
M
o f the closed
is called semi-compact if the image T(B)
in L
there exists 0
2
is almost order bounded in M
v C: M
(1)
, i.e.,
for all
c E
exist such that
T
f E B
is semi-compact, but
in Lemma 124.1, if 0 IT1
exists and
S S 2
IT1
if for any
E
> 0
such that
.
As we have seen in Exercise 125.9, it is possible that T
if
p
T
and T
and
IT1
both
\TI is not. Furthermore, as shown
is semi-compact, then so is S
is semi-compact, then so is
T
.
. Hence,
Ch. 18,11261
5 35
COMPACT OPERATORS
We shall firstindicate a sufficient condition for a positive operator to be semi-compact, holding under rather weak hypotheses for the spaces L and M
.
Under much stronger assumptions the condition is also necessary, as will be shown later. To begin with, assume only that M has the principal projection property (besides the general assumption that L Banach lattices). Let (Pn:n=1,2,
...)
principal bands (B :n=1,2,...)
in M
nnBn
=
COI ). Let
order projection on
R=
pn
-
P,+]
d Bn+l
Bn
and M
are
be any sequence of order projections on such that P
for n = 1,2,
.
+
0 (equivalently,
... . Therefore, Qn
is the
L rmd M be Banach l a t t i c e s such that M has the p r h ~ c i p a lprojection property and l e t T: L + M be a p o s i t i v e operator. For any descending sequence (Bn:n=1,2, ) o f principal bands i n M satisfying n B = {O} , l e t Pn and Qn be defined as explained above. n n (i) I f , for every choice of the descending sequence (B,) with n Bn = { O } , the operator norm /lQnTII of QnT tends t o zero as n + m , then T i s semi-conpact. (ii) I f , f o r eve22 choice of the descending sequence (B,) with ll Bn = { O } , the operator norm IIPnTII of PnT tends t o zero as n + m , then T i s semi-compact. (iii) I f , f o r every sequence (Sn:n=1,2, ) of operators from L THEOREM 126.1. Let
...
i n t o M satisfying i s semi-compact.
T 2 Sn J. 0
, the norm
PROOF. It is sufficient to prove (i). (i) holds, but T
...)
2 E
and
T
Assume that the hypothesis in
is not semi-compact. Then there exists a number
and a sequence (u :n=l,2, n L such that h(Tu 1
...
llSnll tends t o zero, then
E
of positive elements in the unit ball
X
-2" Zf'
Tq)f}
z
E
for n 2 2
> 0
B
.
Let w
n
= (Tun-2"
'-:X
Tt-L
m
2-kTtj)f
n+1
Note that r - 01 n - 'n+l 2-kTq exists in M an order limit) because M is Banach and
for n = 1,2,..
..
as a norm limit (and hence as
T is norm bounded. Since rn
of
536
tends t o zero i n norm a s n
n
-+
s u f f i c i e n t l y l a r g e , say f o r
sequence
(wn:n=1,2,
wm
-
...)
2-"(
5
-
,
Ch. 18,11261
AND ORDER CONTINUITY
SEMI-COMF'ACTNESS
i t follows from (2) t h a t
. Note
n 2 no
A(wn) t
for
;E
a l s o t h a t t h e norm bounded
i s d i s j o i n t . Indeed, f o r
m < n
we have
2nTum-Tuny = 2-n(Tun-2nTum)-
..
n = I,.?.,. , l e t Bn be the band i n M generated by -k e = C 2 wk Then f l Bn = {O} and, with t h e n o t a t i o n s agreed upon, n n Qn By hypothesis i s t h e order p r o j e c t i o n on t h e band generated by w
Now, f o r
IIQnTII
.
-+
as
0
n
119nTI1 n t no
for
-.
-+
.
On the o t h e r hand,
X(Q Tu ) t X(Qnwn) = X(wn) t n n
2
. Contradiction.
Hence, T
:E
i s semi-compact.
The theorem j u s t proved has a p p l i c a t i o n s t o k e r n e l o p e r a t o r s from one Banach function space i n t o another. Let spaces
M(Y,v)
and
M
(Y,C,v)
and
r e s p e c t i v e l y . Note t h a t
(X,h,u)
a r e Dedekind complete. We may assume t h a t t h e c a r r i e r of
t h e c a r r i e r of
M
is
X
. Assume
with r e s p e c t t o t h e norms
M
be i d e a l s i n t h e Riesz
of a l l r e a l measurable f u n c t i o n s on t h e o-fi-
M(X,p)
n i t e measure spaces
M
and
L
and
p
also that
X
L
and
M
L
is
L
to
Y
has t h e same c a r r i e r as
. Furthermore,
o p e r a t o r s (from
L
and hence a l s o operator (from
into
(LFM)dd L
, i.e.,
by Theorem 95.1,
a r e Banach l a t t i c e s
M
, is
M
and L* = L"
n
L* i s a l s o equal n t h e space of a l l absolute k e r n e l
t h e c a r r i e r of
M ) i s e x a c t l y t h e band
123.5(ii) i t was shown t h a t i f I n o t h e r words, i f
L
and
r e s p e c t i v e l y . I n o t h e r words, L
a r e Banach f u n c t i o n spaces. By Theorem 112.1 t h e a s s o c i a t e space
of
and
L Y
. I n Theorem
(L:BM)dd
has order continuous norm, then
dd (L*@M)
contained i n t h e band of AM-compact o p e r a t o r s .
has order continuous norm, then every absolute k e r n e l
into
M ) i s AM-compact. Our next purpose i s t o f i n d
reasonable conditions under which every absolute k e r n e l operator i s semicompact. Assume f i r s t t h a t theorem, T n + Pn f 0
P
in
M
T
i s p o s i t i v e . Then, according t o t h e l a s t
i s semi-compact i f t h e o p e r a t o r norm
f o r every sequence
. Since
M
n
(P,)
IlP TI1 tends t o zero as
of order pEojections i n
i s an i d e a l i n t h e space
M(X,p)
,
M
satisfying
any order p r o j e c t i o n
can be w r i t t e n a s a m u l t i p l i c a t i o n by a c h a r a c t e r i s t i c f u n c t i o n
Ch. 1 8 , 5 1 2 6 1
xE
f o r some p-measurable
= xE(x)f(x)
subset
E
almost everywhere i n
that the s e t
of X
has the kernel
xE(x)T(x,y)
projections i n
M
X
, and
operator
that is t o say, (Pf)(x) =
f o r every
. To
f E M
M(X,p) ; t h e s e t T
j.
i s now simply
,
T(x,y)
any sequence E
see t h i s , note
E
has t h e k e r n e l
. Furthermore,
P
0
j.
0
then
PT
of o r d e r
of measurable s e t s
conversely. Therefore, i t i s s u f f i c i e n t f o r the p o s i t i v e k e r n e l T(x,y) Z 0
with kernel
T
x
f o r e v e r y sequence
E
0
j.
in
X
t o be semi-compact (x) T(x,y)
EIl Of c o u r s e , i f
.
i s o r d e r bounded b u t n o t p o s i t i v e , and t i s f i e s t h e mentioned c o n d i t i o n , t h e n T
,
c o r r e s p o n d s w i t h a sequence
norm of t h e o p e r a t o r w i t h k e r n e l
so i s
X
i s an i d e a l i n
(Pf:fEM)
t h e c a r r i e r o f t h a t i d e a l . Hence, i f
in
537
COMPACT OPERATORS
.
Assume now t h a t b o t h
L*
and
M
T
compact, t h e n
T
L
T(x,y) ) sa-
and hence
have o r d e r c o n t i n u o u s norm. Then,
T: L
i s AM-compact a s w e l l a s semi-compact.
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 from
+ m
IT(x,y)/ )
i s semi-compact,
IT1
n
(with k e r n e l
(with k e r n e l
IT1
a c c o r d i n g t o Theorem 125.5, any o r d e r bounded o p e r a t o r i f and o n l y i f
t h a t the operator
t e n d s t o z e r o as
into
M
+
M
i s compact
I f , therefore, T
such t h a t
T
i s semi-
i s a l r e a d y compact. As observed above, i t i s s u f f i c i e n t
f o r semi-compactness
of
T
t h a t , f o r e v e r y sequence
En
j.
0 in
X
,
the o p e r a t o r norm of t h e o p e r a t o r w i t h k e r n e l
tends t o zero a s satisfying
n
1 < p <
. As
+ m m
a n example, l e t L = L (Y,v) f o r some p P M = L (X,p) f o r some q s a t i s f y i n g 1 5 q < q, -1 p-I + (p ) = 1 Note t h a t L* and M have
and
.
L* = L , ( Y , v ) for P o r d e r c o n t i n u o u s norm. Assume now t h a t t h e ( u x v ) - m e a s u r a b l e f u n c t i o n
Then
T(x,y)
satisfies
It i s e a s i l y s e e n t h a t
T(x,y)
i s t h e k e r n e l of a n o r d e r bounded k e r n e l
operator T: L ( Y , v ) P and
IT(x,y)l
-t
Lq(X,v)
is t h e k e r n e l of
, IT1
. Furthermore,
m.
t h e o p e r a t o r norm of
5 38
SEMI-COMPACTNESS AND ORDER C O N T I N U I T Y
i s majorized by
IT1
, because
I!ITlllpq
f o r any
f
E
Ch. 18,51267
t h e norm
L
we have
1 5
where
p (f) S
P
1
(ITlf)(x) PPl{TX(Y
Tx(y) = T(x,y)
for
x
f i x e d . Hence
which i m p l i e s
Therefore, t h e corresponding o p e r a t o r norm tends t o z e r o a s w e l l . I n view of t h e above remarks, t h i s shows t h a t
and
IT(
T
a r e compact.
We formulate t h e obvious g e n e r a l i z a t i o n t o Banach f u n c t i o n s p a c e s , which w i l l a l s o cover t h e case
p =
m
i n t h e example above. The proof f o r
t h i s more g e n e r a l case i s n o t a l t o g e t h e r t r i v i a l s i n c e i t i s based on a n o t
s o obvious m e a s u r a b i l i t y r e s u l t , namely, t h e Luxemburg-Gribanov r e s u l t A s mentioned above, we may assume t h a t
i n Theorem 99.2 and C o r o l l a r y 99.3. both Lt
L
and t h e
associate space
i s t h e n sometimes denoted by
r e s t r i c t i o n of t h e norm L = L_(Y,v)
,
then
in
p*
I
if
X
p
and
have c a r r i e r
L
and X
and we r e c a l l t h a t
Y
and t h e norm i n
L'
L* ) i s then denoted by
(i.e., p'
the
. Note
that
(Theorem 118.1; l a s t
i s o r d e r continuous (Theorem 118.2). The
p*
a s s o c i a t e space i s now t h e space THEOREM 126.2. Let
L'
i s an a b s t r a c t L-space
L*
p a r t of t h e p r o o f ) , and so
abooe) with noms
L:
LI(Y,v) M
.
be Banach function spaces (as defined
respectivezy such t h a t the norm
p*
i n L*
i n M are order continuous (hence, the norm p ' i n L' i s likewise order continuous). Fwthermore, Let the (pw)-measurable real function T(x,y) on X x Y s a t i s f y and the n o m
Ch. 18,51261
Tx(y) = T(x,y)
where x
5 39
C W A C T OPERATORS
for f i x e d
x
, i.e.,
f i x e d . Then the absolute kernel operator
T(x,y) T: L
as a function of
* N with kernel
y
for
T(x,y)
is compact. PROOF. By t h e r e s u l t of Luxemburg i n Corollary 99.3 t h e f u n c t i o n
i s a p-measurable non-negative
t ( x ) = p'iTx(y)} that
IIITlllpk
A(t) <
, i.e.,
f o r any
f E L
and t h e r e f o r e of
IT1
f u n c t i o n on
X ; i t follows
i s a w e l l defined number. By our hypothesis, w e have the function with
t
X(ITIf)
5
i s a menher of the space
M
. Furthermore,
w e have
p(f) 5 1
X(t) =
!llTlllph
. It
follows t h a t the o p e r a t o r norm
satisfies
I f t h e sequence
(En)
t xEn(x)t(x) i 0
, xhi& ,
of s u b s e t s of
implies
X
x(x
satisfies t) 4 0
0 ,
En J.
bscausr
h
then
t(x) 2
is order c o n t i -
mobs. In o t h e r word6
i.e.,
IT1
i s semi-compact by what was observed before. I t follows t h a t
i s compact and, t h e r e f o r e , T The number
IIITll/pA i s c a l l e d t h e double norm of the kennel operator
T (more p r e c i s e l y , t h e ph-double
norm). Operators with
lllTlllpA f i n i t e a r e
-
and
c a l l e d operators of f i n i t e double norm. For k e r n e l o p e r a t o r s with
lllTlllpq <
because t h e theorem t h a t E. H i l l e and J . D .
lllTlllpq <
-
1 < p 2
m
1 5 q <
m
,
a r e c a l l e d HilZe-Tamrkin operdtors,
Tamarkin (C11,1934).
implies compactness of
T
i s due t o
Their proof i s q u i t e d i f f e r e n t from
the proof as presented here. Note t h a t f o r
i s equal t o
IT1
i s compact.
p = q = 2
the number
lllTlllpq
SEMI-COMPACTNESS AND ORDER CONTINUITY
540
Ch. 18,51261
XXY
t h e corresponding o p e r a t o r s a r e t h e Hilbert-Schmidt o p e r a t o r s . I n t h e b e g i n n i n g of t h e p r e s e n t s e c t i o n i t w a s shown t h a t i f
/ISnlI
t h e p r i n c i p a l p r o j e c t i o n p r o p e r t y and i f
J- 0
M
has
f o r any sequence
+
(Sn)
L into M satisfying T 2 S 0 , then T i s s e m i n compact. We s h a l l prove now t h a t t h e converse h o l d s under t h e much s t r o n g e r
of o p e r a t o r s from
condition t h a t both
L*
THEOREM 126.3. Let
A
and L
M
have o r d e r continuous norm. M
and
be Banach l a t t i c e s (with norms
r e s p e c t i v e t y l such t h a t t h e norm p* T
ordercontinuous. Furthermore, l e t M
. Then
T
of operators s a t i s f y i n g
and the n o m
(1 n
T
/IS
S
2
n
C 0
PROOF. I t i s only n e c e s s a r y t o prove t h a t i f
t h e mentioned c o n d i t i o n h o l d s . Hence, l e t The image
i.e.,
T(B)
of t h e u n i t b a l l
there exists
{
0 5 vo E M
A (Tu-v )
o+>
<
E
Furthermore, s i n c e t h e norm closed u n i t b a l l i n
@
is p
in
VO
Now, t a k e any a r b i t r a r y
0 5 u E B
that
A*($)
< 1
. It
L
into
holds f o r any se-
. T
i s semi-compact,
b e semi-compact and
then
M
E @
.
i s o r d e r c o n t i n u o u s , t h e norm
f o l l o w s from
such t h a t
and any a r b i t r a r y
0 5 $ E
fi
.
T2 S C 0 n M ,
-almost o r d e r bounded (Theorem 125.2), i . e . ,
0 S @
0
0
and
i s almost o r d e r bounded i n
0 < u E B
for all
A
T L
C
p
i n M are
such t h a t
t h e r e e x i s t s a n element
ing
in
B
A
be a p o s i t i v e operator from
i s semi-compact i f and only i f
(Sn)
quence
L*
in
satisfy-
Ch. 18,91261
54 1
COMPACT OPERATORS
(3)
By h y p o t h e s i s
Sn C 0
$o(Snw) C 0
hence
,
0 5 w 5 L
has o r d e r continuous norm, s o
,
i.e.,
p*(S;$*)
(3) t h a t t h e r e e x i s t s a n a t u r a l number
such t h a t
0
5
IJJ E
$(Snu) < 3~
M*
holds f o r
holds f o r a l l
satisfying
A*($)
11 Snll
0
as
C 0
S;+,
+
0
no n
2
holds i n
C 0
. It
, and
L
b0
is
no
, all
1
. But
L*
L*
follows t h e r e f o r e from
(not depending on
. The l a s t f a c t 5 u 5 B . But t h e n n + . 5
and a l l
n Z no
I n o t h e r words,
in
h a s o r d e r continuous norm). This shows t h a t
M
f o r any
C 0
n
$o t h e same a s above ( n o t e h e r e t h a t
o r d e r continuous s i n c e (S;$o)(w)
w Z 0
f o r any a r b i t r a r y
S w C 0
so
for
0
5
5
A(Snu) 5 3~
for
3~
$ )
and a l l
u E B
implies t h a t
1 1 Sn 11
or
u
n 2 n
0 '
m
The r e a d e r i s warned a g a i n s t t r y i n g t o avoid t h e o r d e r c o n t i n u i t y of p*
$o
by a r g u i n g t h a t
C 0
S
S u C 0
implies
and hence
$J~(S,U)C 0
i s o r d e r continuous i n view of t h e o r d e r c o n t i n u i t y of
t h a t t h e r e e x i s t s a n a t u r a l number
n 1. n,
. We
nl
such t h a t
cannot be s u r e , however, t h a t
nl
$,(Snu)
1.
A
<
E
(since
I t follows
for a l l
does n o t depend on
u
.
The r e s u l t i n t h e l a s t theorem o c c u r s i n t h e Dodds-Fremlin paper el1 (combination of Theorem 5.1 and 5.2 of t h a t paper; Theorem 5 . 2 i s based on t h e r a t h e r complieated Theorem 2.5). The d i r e c t proof p r e s e n t e d h e r e goes back t o Fremlin (1975). If
L
and
M
and ( a s b e f o r e )
a r e Banach l a t t i c e s such t h a t
bounded o p e r a t o r s from
M
i s Dedekind complete
i s t h e Dedekind complete Riesz s p a c e of a l l o r d e r
Lb(L,M)
L
into
M
,
t h e n ( a s f o l l o w s from Theozem 83.12)
T E L (L,M) i s norm bounded. To each T E L (L,M) we a s s i g n now t h e b b o p e r a t o r norm of IT1 and we denote t h i s number by / / T l l r , i . e . ,
every
The number IlTll
= IITll,
i s c a l l e d t h e r-norm of
IlTll
Riesz norm i n
Lb(L,M) if
T
. Furthermore,
IlTll
T 5
. Obviously,
ikTllr f o r every
t h e r-norm i s a T
and
i s p o s i t i v e . According t o Theorem 102.8 t h e s u b s e t of
SEMI-COMPACTNESS AND ORDER CONTINUITY
542
Lb(L,M)
of all T
,
in Lb(L,M)
11s I I J . 0 is an ideal
IT1 2 Sn C 0 implies
such that
n
the largest ideal on which the r-norm is
Our last theorem shows that if the norms in L* o u s , this ideal is exactly the set of all
corresponding
IT1
T
18,81261
Ch.
a-order continuous.
and M
are order continu-
L (L,M)
in
for which the b is semi-compact. Note that this set may be properly
smaller than the set of all semi-compact T because, as we have seen, there exist semi-compact T
IT1
for which
is not semi-campact.
Although it is somewhat outside of the main subject in the present discussion, we include a proof that
Lb(L,M)
i s a Banach lattice with
respect to the r-norm. THEOREM 126.4. Let
such t h a t
M
L and
M
be Banach l a t t i e e s [norms p
i s Dedekind complete. Then the Riesz space
X I
and
Lb(L,M)
is a
Lb(L,M)
with re-
Banach l a t t i c e w i t h respect t o the r-norm. PROOF. Let
(Tn:n=1,2,
...)
be a Cauchy seqeunce in
spect to the r-norm. By passing to a subsequence if necessary we may assume that 11 Tn+l-TnIIr 5 2-(n+1)
for all n
. The sequence
(T,)
is Cauchy with
respect to the ordinary operator norm as well and, therefore, Tn to a norm bounded operator T (from L
converges
into M ) with respect to the
ordinary operator norm. It follows that for any
-
, the
f E L
.
element
T f is the norm limit of Zq (Tn+,-Tn)f Hence, I Tf-T f 1 is the P n=p P norm limit of IZ:,p(Tn+l-Tn)f I .Now, let u E L+ be given. Then the sum
Tf
is increasing as q for k > q > p
increases and
we have
Hence, the norm limit s
of
s
Q
s
4
converges in norm as q
is equal to
. All
sup s
q
X(s ) 5 2-’.p(u) , which implies h ( s ) 2 2-’.p(u) since s 9 limit of s Finally, let f E L satisfy If/ 5 u Then
. 4
11; (Tn+I-Tn)fl
.
5
Ef, ITn+,-Tn 1(u) = s9
5
s
.
s
9
+
- 7
because
satisfy
is the norm
Ch.
18,51261
COMPACT OPERATORS
It f o l l o w s t h a t t h e norm l i m i t ( a s
s
i s a l s o majorized by
, i.e.,
q
-+ m
ITf-T f l P
5
543
) of t h e e x p r e s s i o n on t h e l e f t
. Hence
s
Taking norms, we g e t
{
X IT-Tpl(u)}
.
X(s) 5 2-’.p(u)
5
I i s a norm bounded o p e r a t o r w i t h o p e r a t o r norm a t P It f o l l o w s t h a t T i s t h e l i m i t most e q u a l t o 2-’ , i . e . , I[T-T 11 5 2-’ P r bf (T : p = l , 2 , ) w i t h r e s p e c t t o t h e r-norm. P This shows t h a t
IT-T
.
...
A t t h i s p o i n t i t i s perhaps of some i n t e r e s t t o make t h e f o l l o w i n g
remarks about semi-compact o p e r a t o r s and AM-compact o p e r a t o r s (from
L
into
M ). Semi-compactness has t h e p l e a s a n t p r o p e r t y t h a t any p o s i t i v e o p e r a t o r
majorized by a semi-compact o p e r a t o r i s i t s e l f semi-compact
(Lemma 124.1).
b u t a t t h e s a m e time i t i s u n p l e a s a n t t h a t (even i f a l l norms involved a r e o r d e r continuous) t h e semi-compact o p e r a t o r s may f a i l t o form an i d e a l , i.e.,
i t may happen t h a t
125.9). M
T
i s semi-compact and
IT1
i s not (Exercise
On t h e o t h e r hand, AM-compactness h a s t h e p l e a s a n t p r o p e r t y t h a t i f
has o r d e r continuous norm, t h e n t h e AM-compact o p e r a t o r s form a band.
I f , however, t h e norm i n
M
i s n o t o r d e r continuous, i t may occur t h a t a
p o s i t i v e o p e r a t o r majorized by a n AM-compact o p e r a t o r is n o t AM-compact (because o t h e r w i s e i t would always f o l l o w from that
S2
EXERCISE 126.5. L e t
L
p r o j e c t i o n p r o p e r t y and l e t with t h e property t h a t
T
L
compact
T
b e a p o s i t i v e AM-compact o p e r a t o r i n
IIPnTII f O
p a c t f o r any p o s i t i v e o p e r a t o r if
with
b e a Banach l a t t i c e p o s s e s s i n g t h e p r i n c i p a l f o r every sequence
p r o j e c t i o n s on p r i n c i p a l bands. Show t h a t S
in
S
in
L
satisfying
0 < S
T2
S2 5
T
.
HINT: Theorem 126.1 and E x e r c i s e 125.7.
P,
4 0
i s compact and
satisfying
L
has o r d e r continuous norm, t h e n
operator
0 5 S 5 T
i s compact, which i s n o t t h e case a s shown i n examples i n s e c h i o n 124).
0 2 S 2 T
L
of o r d e r S4
i s com-
. Show t h a t
i s compact f o r any p o s i t i v e
SEMI-COMPACT OPERATORS AND D I S J O I N T SEQUENCES
544
. 18,11273
Ch
127. Semi compact operators and d i s j o i n t sequences A s i n t h e p r e v i o u s s e c t i o n , w e assume t h a t
ces (norms
p
and
r e s p e c t i v e l y , norms
X
L and M a r e Banach l a t t i -
p*
and
i n the Banach
X*
d u a l s ) . I n Theorem 3 of t h e previous s e c t i o n i t w a s proved t h a t i f the norms and
p*
M
,
then
a r e order continuous and
X T
T
i s a p o s i t i v e operator from
i s semi-compact i f and only i f
.
llSnllS 0
L
into
holds f o r every se-
quence (Sn) satisfying T 2 S S 0 I n the p r e s e n t s e c t i o n we s h a l l deal n with s e v e r a l o t h e r c o n d i t i o n s , necessary and s u f f i c i e n t f o r T t o be semicompact. In these conditions
i s not n e c e s s a r i l y p o s i t i v e . A s seen
T
e a r l i e r ( i n formula ( 1 ) of the previous s e c t i o n ) , i t i s an immediate consequence of t h e d e f i n i t i o n of semi-compactness t h a t the o r d e r bounded operator T
(from
L
-.
into
M ) i s semi-compact i f and only i f f o r any
E
> 0
there
v = v
for a l l
f
exists a positive i n the u n i t b a l l I T I(w)
for a l l
B
of
in L
f o r an appropriate
.
M
such t h a t
I n some cases t h e element
w E L+
i s Dedekind complete (so t h a t every
. For
0 5 f E B
and l e t f i r s t
Also, since
so
v
i s of the form
, so
f E B. This l a s t condition i s c e r t a i n l y s a t i s f i e d i f t h e space
e x i s t s an element u E L+
for a l l
X{(ITfl-v)+} < E
0 5 f E B
lTfl
-
T
has a corresponding
M
[ T I ) and t h e r e
such t h a t
t h e proof t h a t (2) implies ( l ) ,
. It
follows now from
ITI(u) 5 ITfI - IT(fAu)l
f = f
, we
have
A
l e t ( 2 ) be s a t i s f i e d u
+ (f-u)+
that
Ch.
18,11271
If
f
E
and s o all
COMPACT OPERATORS
i s a r b i t r a r y , then
B
hICITf [ - I T 1 (2u)l+}
f E B
for a l l
545
0
. If,
f o r any
f E B
2
, we
THEOREM 1 2 7 . 1 . If
<
.
E
This s h m s t h a t ( I ) holds f o r
> 0 , there exists
E
s h a l l say t h a t M
T
and
i s strongZy semi-compact. T
has order continuous norm and
compact order bounded operator from L
w = 2u
such t h a t (2) holds
u E L+
into
, then
M
T*
i s a semii s strongZy
semi-compact. PROOF. Note f i r s t t h a t
M
i s Dedekind complete and, t h e r e f o r e , IT1
i s w e l l d e f i n e d . S i m i l a r l y , IT*l
i s w e l l d e f i n e d . I t makes s e n s e , t h e r e f o r e ,
t o speak of s t r o n g semi-compactness. Now, l e t M
v
for all
i n thellunit b a l l
B
of
continuous norm, t h e u n i t b a l l i n particular
for all arbitrary
Hence
0
, i.e.,
v
S $
f
E
E B
M*
> 0
b e g i v e n . There e x i s t s
such t h a t
a positive
f
in
E
L
. Furthermore, 0
S $
5 1 (Theorem
A*($)
and a n a r b i t r a r y
0
M
has order
i s pv-almost o r d e r bounded f o r t h i s
M*
t h e r e e x i s t s an element
satisfying
since
9
$
E M*
E M*
such t h a t
125.2). Now t a k e an
satisfying
A*($)
S
1 , Then
SEMI-COMPACT OPERATORS AND DISJOINT SEQUENCES
546
f E B
Since this hold for all
, the norm
of
This shows that the operator T*: M* + L*
T*($-$€)+
18,51271
Ch.
satisfies
is strongly semi-compact.
There exists a dual theorem, as follows.
If M i s Dedekind complete and
THEOREM 127.2.
norm, and i f
T
then
T
IT1
M
such t h a t
being Dedekind complete, the
has order continuous norm, the unit ball in
is pe-alrnost order bounded for every positive $ COROLLARY 127.3.
Let
L*
and
M
T be an order bounded operator from
T and
into M
exists. The proof is now similar to the proof of the last
theorem, using that since L*
L
has order continuous
i s strongly semi-compact.
PROOF. Note again that in view of
operator
L
i s an order bounded cperator from
T* i s semi-compact,
L*
in L*
.
have order continuous norm and l e t L
into M
.If
now one a t l e a s t of
T* i s semi-compact, then both T and T* are strongly semi-
compact. The main theorem which follows now is essentially Theorem 5.2 in the Dodds-Fremlin paper c11. The proof is different. THEOREM 127.4.
Let
L* and
M
have order continuous norm and l e t
be an order bounded operator from L
into M
. The foZZuwing
T
conditions
are now equivalent.
i s semi-compact.
(i)
T
(ii)
T* i s semi-compact.
(iii) A(Tun)
+
0 f o r every d i s j o i n t norm bounded sequence
p o s i t i v e elements i n
L
.
(iv)
p*(T*$,) + 0 f o r every d i s j o i n t nom bounded sequence p o s i t i v e elements i n M* .
(v>
$n(Tun)
+
0
f o r a l l d i s j o i n t norm bounded seguences
(Qn) of p o s i t i v e ezements i n
If t h e s e conditions hold, then
T and
(u,)
($,) (u,)
and M* r e s p e c t i v e l y . T* are strongly semi-compact. L
of
of and
Ch.
18,§1271
547
COMPACT OPERATORS
PROOF. The equivalence of (i) and (ii) was proved in the last corollary. It was also proved that if (i) or (ii) holds, then T
and
T*
are strongly
semi-compact. (i)
*
(iii) Let
> 0 and let
E
be a disjoint norm bounded
(u,)
. We may assume that 0 5 un E B for all n , where B is in L . Since T is strongly semi-compact, there exists an
sequence in L+ the unit ball
element v E L+
such that
Furthermore, since M 'ITlv that
has order continuous norm, the unit ball in M*
-almost order bounded, i.e., there exists an element 0 5 J,
(I)-$~)+(ITI (v)) <
E
for all 0 ' J ,
E M*
satisfying A*($)
E M* 5
1
is
.
such
Hence
and, furthermore, if J, E M*
for all n
I*($)
1
5
is positive and satisfies
, then
(4)
In view of the disjointness and order boundedness of the sequence (unrtv:n=l,2,.
..)
-
'n= 1
the partial sums of the series
( I TI *oE)(UnAv)
are majorized by (I TI *$I,) (v) and, therefore, (IT1 *$(, U, IAV) tends to zero It follows from ( 4 ) that there exists a natural number as n + m such no that J,(IT(u~Av)~) < 2~ f o r all n t no and all 0 5 I) E M* satisfying
.
A*($)
?,(Tun)
5 1 5
. Then
3~
X{T(U,AV)}
for n
Z
n
0 '
s 2~
for n
2
no
In other words, X(Tun)
and hence, in view of ( 3 ) , +
0 as n
+
.
SEMI-COMPACT OPERATORS AND DISJOINT SEQUENCES
548
(ii) (iii)
* *
Ch.
18,11271
(iv) Similarly. (v)
and a l s o ( i v )
*
(v) a r e e v i d e n t , s o t h a t w e have a l r e a d y
proved t h a t t h e f o l l o w i n g i m p l i c a t i o n s h o l d .
(i)
Q
(iii) ( i v ) e (v)
e
(ii)
*
I t w i l l b e proved now t h a t (v) i m p l i e s ( i i i ) and ( i i i ) i m p l i e s ( i ) .
(v)
*
Gun)
( i i i ) Let
the u n i t b a l l
of
B
. We
L
be a d i s j o i n t sequence of p o s i t i v e elements i n have t o prove t h a t Tun
norm. For t h i s purpose we b e g i n by o b s e r v i n g t h a t t o zero, i . e . , (\TI*$)(un)
+
${ITl(un)}
. This
0
+
0
(by Theonem 116.1) u -f
.
0
converges weakly
E P o r , equivalently,
$
f o l l o w s by o b s e r v i n g t h a t t h e norm i n
continuous and t h e sequence (lTl*$)(un)
f o r every
converges t o z e r o i n IT1 (un)
in
(un)
L
i s order
L*
i s norm bounded and d i s j o i n t , s o
converges weakly t o z e r o . This i m p l i e s
It f o l l o w s t h a t
Assume now t h a t
Tu
/Tun/
converges weakly t o z e r o as w e l l .
does n o t tend t o z e r o i n norm. P a s s i n g t o a sub-
sequence i f n e c e s s a r y , w e may assume t h a n t h a t t h e r e e x i s t s a number such t h a t
X(Tun) > 4~
of elements i n
fl
n
. Then
A*($
) 5 1
for a l l
such t h a t
t h e r e i s a l s o a sequence and
IOn(Tun)I > 4~
E
> 0
($n)
for a l l
n
.
It f o l l o w s from
t h a t one a t l e a s t of
$ :
Iqn(Tun)l > 2~
that
. Note
t \JIn(Tun)l > 2~
a subsequence n
. Since
for all
(uln)
of
4:
and
( c a l l t h i s one
A*($,) n
. Since
(un)
Qn ) s a t i s f i e s ) 5 1 and JIn(ITunl) t n tends t o z e r o , t h e r e i s
= A*($
5
JII(ITunl)
such t h a t
$ l ( I T u l n l ) < ~ / ( n . 2 ~ + ' )f o r a l l
$ J ~ ( I T u , I ) t e n d s t o z e r o , t h e sequence
(u
In
)
h a s a subsequence
( u ~ ~such ) that $ J ~ ( I T u ~ < ~~ /\ ()n . 2 " + ' ) f o r a l l n . Of c o u r s e , n+ 1 J I I ( I T u 2 n l ) < ~ / ( n . 2 ) h o l d s then as w e l l . Continuing i n t h i s manner, t h e d i a g o n a l sequence n 2 m n 2 m
,
(unn)
satisfies
, m ( ~ ~ u n n<~ E/(n.2"+') )
Hence, w e may j u s t as w e l l assume t h a t
h o l d s from t h e b e g i n n i n g on. Now, l e t
for all n+ 1
J, (ITunl) < ~ / ( n . 2
m
)
for
Ch. 18,11271
COMPACT OPERATORS
The sequence
i s a d i s j o i n t and norm bounded sequence of p o s i t i v e
(Yn)
P
elements i n
549
. Since
we have (writing t h i s as 'Yn(Tun)
-
Yn = $n
-
N
$ ,
)
,
N
$ n ( T ~ n ) = $,(Tun)
where
n
for a l l
$n(Tun)\ > 2~
Since n 2 n
exceeding some
0
no
for a l l
. Hence
n
,
i t follows t h a t
This c o n t r a d i c t s hypothesis (v). Hence, Tu
norm, i e . , (i i )
( i ) Assume t h a t ( i i i ) holds and
t i v e elements i n t h e u n i t b a l l of
for
for
L
6
> 0
T
f a i l s t o be s t r o n g l y semi-
and a sequence
(u,)
of posi-
such t h a t
. Writing
n = 1,2,
for a l l n
E
converges t o zero i n
( i i i ) holds.
*
compact. Then t h e r e e x i s t s a number
for a l l n
iYn(Tun)l >
...
,
the sequence
. Furthermore,
(vn)
i s d i s j o i n t and we have
i t follows from
0 5 v
n
5
u
n
SEMI-COMPACT OPERATORS AND ORDER PROJECTIONS
550
18,§ 1281
Ch.
that
for
n
s u f f i c i e n t l y l a r g e . This c o n t r a d i c t s h y p o t h e s i s ( i i i ) . Hence, T
i s s t r o n g l y semi-compact. EXERCISE 127.5.
implies ( I ) , T
In t h e b e g i n n i n g of t h e s e c t i o n i t was shown how (2)
i . e . , how s t r o n g semi-compactness
i m p l i e s semi-compactness.
of t h e o r d e r bounded o p e r a t o r
Show t h a t t h e proof becomes s i m p l e r i f
is
T
p o s i t i v e . S i m i l a r l y , show t h a t t h e p r o o f s i n Theorem 127.4 become s i m p l e r if
is positive.
T
128. Semi-compact o p e r a t o r s and o r d e r p r o j e c t i o n s A s i n t h e p r e v i o u s s e c t i o n s we assume t h a t
l a t t i c e s (norms
and
p
X
e assume a l s o t h a t t h e norms duals). W T
be a p o s i t i v e o p e r a t o r from
compact i f and only i f
.
IISnll
+
X
and
p*
a r e Banach
M
and
i n t h e Banach
A*
a r e o r d e r continuous. Let
L
into
0
f o r every sequence
. Conversely,
,
M
projections i n
(Sn)
of o p e r a t o r s
n
IIP TI/ .L 0 f o r every sequence P C 0 of o r d e r n n t h e n T i s semi-compact (Theorem 126.1). There e x i s t s if
a v a r i a n t upon t h e l a s t r e s u l t , a s f o l l o w s . Let quence of o p e r a t o r s i n semi-compact.
Then
compactness of
-.
Q T
1
M
such t h a t
0 2 QnT 2 Q T I
Q
and
J- 0
QIT
(Qn:n=1,2, ...)
and l e t
T
be a se-
b e p o s i t i v e and
i s semi-compact
( t h e semi-
follows by observing t h a t a p o s i t i v e o p e r a t o r maps
almost o r d e r bounded s e t s i n t o almost o r d e r bounded s e t s ) . Hence as
n
+
i s semi-
M. A s observed e a r l i e r , T
compact, then M
and
p*
T 2 S C 0 This i m p l i e s i n p a r t i c u l a r t h a t i f T i s semin IIPnTII C O f o r e v e r y sequence P J- 0 of o r d e r p r o j e c t i o n s
satisfying in
L
r e s p e c t i v e l y , norms
operators i n
Conversely, i f M
,
IIQ TI1
n
+
then i n p a r t i c u l a r
of o r d e r p r o j e c t i o n s i n
M
, and
0
f o r every sequence
IIP TI1 n therefore T
+
0
Q
n
of
f o r e v e r y sequence
i s semi-compact.
r e a s o n a b l e t o ask whether s i m i l a r r e s u l t s h o l d i f
IIQnTII C 0
C 0
T
Pn
i s o r d e r bounded b u t
n o t n e c e s s a r i l y p o s i t i v e . Our f i r s t purpose i s now t o prove t h a t i f T: L
+
M
+
It i s
i s o r d e r bounded (and not n e c e s s a r i l y p o s s i t i v e ) , then
T
is
0
Ch. 18,§1281
55 1
COMPACT OPERATORS
semi-compact i f and only i f operators i n
M
.
0
d i s j o i n t l i n e a r f u n c t i o n a l s on
+
f o r every sequence
and t h e i r c a r r i e r s . Let
M
be a d i s j o i n t sequence of p o s i t i v e elements i n
. Then,
Jln
be t h e c a r r i e r of
all
so
.
m < n
nn
.
Bn
.
{O}
=
On t h e o t h e r hand fl Bn = { O }
have
Pn t
that
P n
-
E n Bn B1
0
. It
B1
note t h a t
THEOREM 128.1. Let
L*
Bn+l
@
Cn = Bn ‘n
+
M
satisfying
I1 PnTII
+
into M
L
Pn
ions i n M s a t i s f y i n g
+
Qn
.
0
+
0
.
PROOF. We s h a l l prove t h a t ( i i )
. The following
=$
pn
, we shows
T
(iii)
(i)
*
(i)
conditions
of positive of order project-
( i i ) . Since ( i i )
*
*
(iii)
(ii).
( i ) I n view of Theorem 127.4 i t i s s u f f i c i e n t t o show t h a t
(iii) +
,...)
(Pn:n=l ,2,.. . )
i s e v i d e n t , w e can r e s t r i c t ourselves t o ( i i i ) $,(Tun)
. . Hence
*
(Qn:n=1,2
f o r every sequence
0
m ‘
, which
n
Bn
and
and M have order continuous norm and l e t
f o r every sequence
0
by
,
m
u = 0
Bn
for a l l
are now equivalent. ( i ) T i s semi-compact. ( i i ) IIQnTII
for a l l
follows t h a t
Cn
B J. and we n u I Cm f o r
implies
u I Cm
denoting the o r d e r p r o j e c t i o n on
be an order bounded operator from
(iii)
let
generated by t h e s e t of a l l
i s t h e order p r o j e c t i o n on
Pn+l
E B
u
2
satisfies
i s an element of
u
. Therefore, 0 . Finally,
operators i n
,
n
. Then
m 2 n
for
Cm
Note f i r s t t h a t
0 S u
Hence, any
i s d i s j o i n t from t h e band
u
(J,n:n=1,2,...)
each
since a l l
be t h e band generated by the s e t of a l l prove t h a t
. For
@
J, a r e o r d e r continuous, n i s a d i s j o i n t sequence of bands i n M For each n , l e t
...)
(Cn:n=1,2,
+
Q 0 of n Before p r e s e n t i n g the p r o o f , we make some remarks about
IIQnTII
0
for a l l disjoint
p o s i t i v e elements i n
L
and
norm bounded sequences
M*
(un)
r e s p e c t i v e l y . Hence, l e t
and
(J,,)
of
(un)
and
(Jln)
u and a l l J,n a r e of norm a t n most one. As above, l e t C be the c a r r i e r of J, let B be t h e band n n ’ generated by the s e t of a l l C (m2n) and l e t Pn be the o r d e r p r o j e c t i o n m on Bn Then P 0 and hence I(PnTI/ + 0 as n + by hypothesis. I t
be a s i n d i c a t e d . W e may assume t h a t a l l
.
-
+
-.
follows t h a t t h e o r d e r p r o j e c t i o n llRnTII
$,(Tun)
+
0
=
as
n
+
I$n (RnTu n )
Since
for a l l
$,(g) n
R = P - P n + l on Cn n n = JIn(Rng) f o r every
, which
implies
satisfies g
E
M
, we
have
as
n
+
. It
m
(i)
follows t h a t
i s semi-compact.
T
( i i ) We assume now t h a t
i s semi-compact
T
semi-compact by Theorem 127.4) and t h a t operators i n
satisfying
M
z e r o we may assume t h a t
Ch. 18,51281
OPERATORS AND ORDER PROJECTIONS
SEMI-COMPACT
552
+
Q
11QJ
.
0
i s a sequence of p o s i t i v e
(Qn)
For t h e proof t h a t
(hence
5 1
(and hence s t r o n g l y
llQnll
llQ,II
5
IIQ TI1
n
tends t o
for a l l
5 1
n ).
Before s t a r t i n g on t h e p r i n c i p a l p a r t of t h e p r o o f , not;? thi-t F o r a l l pcsitive
u
and
v
in
L
we have
IQ,T(MV)/ 5 Q , ( I T ( M V ) I ) 5 \ l T \ ( u )
(1)
as well as
(2)
1
IQnT(uhv)
5
Denote t h e u n i t b a l l i n s t r o n g l y semi-compact, A{T(u-v)+} <
L
A'Q Tf' \ n
1 <-
by
and l e t
B
E
> 0
0 5 u E B
. Observing
ilQnI1.h(Tf) 5 A(Tf)
b e given. Since
0 5 v E L
t h e r e e x i s t s an element
for a l l
E
I)
Q,(IT(UAV)
is
T
such t h a t
now t h a t
for a l l
f E L
and a l l
n
,
we g e t h(QnT(u-v)+)
(3) Since
M
< E
for a l l
n
and a l l
M* i s
h a s o r d e r continuous norm, t h e u n i t b a l l i n
o r d e r bounded f o r every
0 5 w E M
. This
pw-almost
holds i n p a r t i c u l a r f o r
,
i.e.,
t h e r e e x i s t s an element
0 < J,
E fl
satisfying
w = Q , IT1 (v)
.
0 5 u E B
0 5 J,
E
P
such t h a t
(4) for a l l
A*($)
5 1
.
A f t e r a l l t h e s e p r e l i m i n a r i e s , we choose an a r b i t r a r y an a r b i t r a r y and
0 5 J, E
J, = (J,-J,c)+
+ J,
A
fl
satisfying
aE . From
A*($)
5 1
. Note
that
0 5 u E B
and
u = (U-V)++UAV
Ch.
553
COMPACT OPERATORS
18,11283
Q
n
Q ~ ( u - v ) ++ Q T(~AV)
TU =
n
i t follows t h a t
where
Writing
To prove now t h a t Qn C 0
. This M
continuous s i n c e p*
.
0 5 w E L
every in
L*
p*(SE$,)
QnlTlw C 0
that
0 5 w E L
every
f o r b r e v i t y , we have
Q n I T / = Sn
+
0
n +
as
for a l l all
tC/
<
E
for a l l
, all
S*$
C
IIQnTII
5
1 2 ~f o r
f E B
n 2 no
0 5 u E B
L
0
holds i n
,
n 2 no
@
, i.e.,
S w C 0
n
(note t h a t L*
$E
(S,*$,)(w) C 0
. Then
and a l l
. This
0 5 Q EM*
shows t h a t
for
is order for
and, s i n c e the norm
p*(S*$ ) C 0 n E no (not depending on Q
i$(QnTf)i < 1 2 ~ f o r a l l
i n the u n i t b a l l of and a l l
in
QE(Snw) C 0
n E i s o r d e r continuous, i t follows t h a t
n t no
$J
w
t h a t i t follows from
has o r d e r continuous norm), i . e . ,
B u t then
It follows e a s i l y t h a t n 2 no
, note
f o r every p o s i t i v e implies t h a t
by ( 5 ) , t h e r e e x i s t s a n a t u r a l number such t h a t
m
. Hence, o r on
satisfying
n 2 no
, all
f
u )
A*($) 5 1 E B and
A(Q Tf) 5 1 2 ~ f o r a l l n
-.
s o t h a t , f i n a l l y , we a r r i v e a t the r e s u l t t h a t
. In
o t h e r words, IIQnTII
-t
0
as
n
+
In t h e f i r s t p a r t of t h e proof of the l a s t theorem we had t o show t h a t
.
SEMI-COMPACT OPERATORS AND ORDER PROJECTIONS
554
$,(Tun) (un)
tends t o zero a s and
($n)
n
-f
-
f o r a l l d i s j o i n t norm bounded sequences
of p o s i t i v e elements i n
d i s j o i n t n e s s of
18,91281
Ch.
and
L
@
r e s p e c t i v e l y . The
h a s n o t been used i n t h e p r o o f . T h i s may l e a d one
(u,)
t o s u s p e c t t h a t t h e r e e x i s t s a more p r e c i s e theorem p r o v i d e d i t i s p o s s i b l e t o l e t t h e d i s j o i n t n e s s o f t h e sequence ions i n
(un)
correspond t o o r d e r p r o j e c t -
on d i s j o i n t p r i n c i p a l bands. I n t h i s d i r e c t i o n we s h a l l prove
L
and
M
, it
h a s t h e p r i n c i p a l p r o j e c t i o n p r o p e r t y and
T
i s an
now t h a t i f , b e s i d e s o r d e r c o n t i n u i t y of t h e norms a l s o assumed t h a t
L
o r d e r bounded o p e r a t o r from only i f
IIPnTQnII
L
into
tends t o zero a s
order projections i n
M L
LEMMA 128.2. Let
L
. For
and
is
i s semi-compact i f and
T
Pn C 0
f o r e v e r y sequence
n +
Qn J. 0
of
of o r d e r p r o j e c t i o n s
t h i s purpose we need a s i m p l e lemma.
...
(u : n = l , 2 , ) be a d i s j o i n t norm bounded n L Then there e x i s t s a sequence
.
sequence of p o s i t i v e elements i n Qn J. 0
L*
be a Banach l a t t i c e possessing the principal
projection property and l e t
..)
then
and e v e r y sequence
on p r i n c i p a l bands i n
(Qn:n=l,2,.
,
M
in
of order projections on principal bands i n L such that
Qnun = un
PROOF. For
for a l l
n = l,2,
...
n
, let
. B
a
b e t h e p r o j e c t i o n band g e n e r a t e d
n
vn = 2-kx ( n o t e t h a t v e x i s t s s i n c e L i s Banach) and l e t ‘k=n n Q be t h e o r d e r p r o j e c t i o n on Bn Then Q u = u S i n c e un I Bm f o r n n n n satisfies w I u f o r a l l n , which a l l m > n , any 0 2 w E il B m m n a -k = v 1 , i . e . , w 1 B1 On t h e o t h e r hand, w i s implies w I E l 2
by
.
contained i n
B1
. It
.
follows t h a t
THEOREM 128.3. Let
L
T
and
Qn C 0 L
nBn =
. Hence,
L*
and
M
IIPnTQnII
-+
L
Qn C 0
has the
have order continuous norm. 0
into M
fop a l l sequences
. Then
Pn C 0
of order projections i n M and order projections on principal respectively.
PROOF. Denote o r d e r p r o j e c t i o n s i n p r i n c i p a l bands i n IIPnTII + 0
, i.e.,
T be an order bounded operator from L
i s semi-compact i f and only i f
bands in
w = 0
and M be Banaeh l a t t i c e s such that
principal projection property and Furthermore, l e t
.
L
Q
by
. Let
f o r e v e r y sequence
theorem. Hence, i f
Qn
+
0
,
P .I 0 n
then
M
first
by T
in
M
P
and o r d e r p r o j e c t i o n s on
be semi-compact.
, as
Then
shown i n t h e p r e v i o u s
.
n
as
+
. Note
m
t h a t we use o n l y t h a t
Conversely, assume t h a t Qn C 0
in
555
COMPACT OPERATORS
Ch. 18,11283
M
and
IIPnTQnII
lIQnlI +
0
i t i s s u f f i c i e n t t o show t h a t di s j o i n t norm bounded sequences
and
(u,)
Pn C 0
(Gn)
and
i s semi-compact
T
tends t o zero a s
bn(Tun)
.
f o r a l l sequences
r e s p e c t i v e l y . For t h e proof t h a t
L
n
for a l l
i 1
n
+ m
for a l l
of p o s i t i v e e l e m e n t s i n
un and a l l $n a r e of norm a s t most one. By t h e l a s t lema t h e r e e x i s t s a sequence Qn J 0 i n L
and
r e s p e c t i v e l y . W e may assume t h a t a l l
M*
L
such t h a t Q u = u n n n t h e r e e x i s t s a sequence = $,(Rng)
$,(g)
for a l l
P J 0 n for a l l g E M
n in
,
. As M
i n t h e proof o f t h e l a s t theorem such t h a t
i n particular
R = P - Pn+l s a t i s f i e s n n $,(Tun) = JIn(RnTun)
.
Hence
n + -
as
It follows t h a t
L*
If
and
T: L
operator T
.
-f
M
i s semi-compact.
T
have o r d e r c o n t i n u o u s norm and i f t h e o r d e r bounded
i s AM-compact, a l l c o n d i t i o n s f o r semi-compactness
M
we have met s o f a r become c o n d i t i o n s f o r compactness of L
that
M
and
such t h a t
M
have o r d e r c o n t i n u o u s norm and l e t
a b s o l u t e k e r n e l o p e r a t o r from T
L
into
M
. As
of t h e type t h a t
llPnTll
+
0
. This h o l d s
T
IIPnTQnII
or
+
0
X
E
of
X
f o r every
such t h a t g E M
there e x i s t s a subset almost everywhere i n kernel
T(x,y)
,
then
F Y
Y
of
T
is, therefore,
. In
s e c t i o n 126 i t was alP
in
M
there exists
xE ( x ) g ( x ) h o l d s almost everywhere f o r any o r d e r p r o j e c t i o n Q i n L
such t h a t
f o r every PTQ
be an
(Pg)(x) =
. Similarly, of
T
i n p a r t i c u l a r f o r conditions
ready o b s e r v e d t h a t f o r any g i v e n o r d e r p r o j e c t i o n
a subset
Assume now
observed i n Theorem 1 2 3 . 9 ,
i s AM-compact. Every c o n d i t i o n f o r semi-compactness
a c o n d i t i o n f o r compactness of
in
.
a r e Banach f u n c t i o n s p a c e s ( n o t a t i o n s a s i n s e c t i o n 126)
and
L*
T
of
f E L
has t h e k e r n e l
( Q f ) ( y ) = XF(y)f(y)
. It
follows t h a t i f
holds T
has the
556
SEMI-COMPACT OPERATOFS AND I N D I C E S
Ch. 1 8 , 9 1 2 9 3
From t h e thoerems proved e a r l i e r i n t h i s s e c t i o n w e can d e r i v e , t h e r e f o r e , t h e f o l l o w i n g theorem f o r a b s o l u t e k e r n e l o p e r a t o r s ( n o t n e c e s s a r i l y p o s i t i v e kernel operators). THEOREM 128.4. Let M
and
L
M
and
have order continuous norm and l e t
absolute kernel operator
T from
T
i s compact
and
Fn
+
i f and only
xE
into
L
only it i s true for every sequence of the operator with kernel
L*
be Banach fwzction spaces such t h a t
En
+ 0
T(x,y) M
be t h e kernel of the
. Then
T
i s compact if and
of subsets of
(x)T(x,y) tends t o zero as
i t is"t r u e f o r aZZ sequences
-.
t h a t t h e norm
X
n
+
0 in
En J
Also, X
0 in Y t h a t the norm o f the operator w i t h kernel
tends t o zero as
n
+ m
.
theorem i s e s s e n t i a l l y t h e main r e s u l t i n a paper on compactness
Thelast
of k e r n e l o p e r a t o r s by W.A.J.
Luxemburg and A.C.
Zaanen (C11,1963). For
O r l i c z s p a c e s ( i n c l u d i n g L -spaces) t h e themrem goes back t o T. Ando (C31, P 1962). The methods of proof i n t h e s e papers were s t i l l r e s t r i c t e d t o t h e s p e c i a l case of f u n c t i o n s p a c e s , b u t t h e y provided a m o t i v a t i o n f o r t h e l a t e r developments i n Banach l a t t i c e s . A s w e have s e e n , t h e r e c e n t c o n t r i b u t i o n s of Dodds-Fremlin and of Aliprantis-Burkinshaw are of fundamental imp o r t a n c e i n t h i s r e s p e c t . Theorems 128.1 and 128.2 (about lIPnTQnl/
-+
IIPnT(I
+
0
and
0 ) are due t o P. van E l d i k and J.J. G r o b l e r (C11,1979).
129. Semi-compact o p e r a t o r s and i n d i c e s Let
L
b e a Banach l a t t i c e (norm p ) . As w e know, p
i f and only i f i.e.,
(c,)
p(un)
+
0
f o r e v e r y o r d e r bounded d i s j o i n t sequence i n
i f and only i f t h e sequence
. This
t o t h e space definition.
P
f o r some
p 2 1
DEFINITION 129.1. Let 1 5 p the L -decomposition property if P
(p(un):n=1,2,
condition is certainly s a t i s f i e d i f
a.
i s o r d e r continuous
. This
5
-.
...)
L+ ,
belongs t o t h e space
( p ( u ):n=l,2,
n
...)
belongs
i s the motivation f o r t he following
The Banach l a t t i c e
L (norm p ) has
COMPACT OPERATORS
Ch. 18,11291
f\p(un)
557
..
:n=l,2,.
(un:n=l ,2,.
holds f o r every order bounded d i s j o i n t sequence
. .)
in
L+
We l i s t some f a c t s t h a t a r e immediately e v i d e n t from t h e d e f i n i t i o n Every Banach l a t t i c e h a s t h e Lm-decomposition p r o p e r t y . I f t h e Banach l a t t i c e L
L -decomposition p r o p e r t y , then
has t h e
P
. Furthermore,
r t p
p r o p e r t y f o r every p r o p e r t y f o r some
p <
m
,
then
L
L
L -decomposition L has t h e L -decomposition has t h e
if P has o r d e r continuous norm. The number
p:L has the d -decomposition p r o p e r t y ) P i s c a l l e d t h e upper index o f the Banach l a t t i c e If
u(L) <
m
,
then
L
L
. We
1 S o(L) I
have
m
.
h a s o r d e r continuous norm. We pkoceed w i t h a dual
definition. DEFINITION 129.2. Let
1 2 p S
m
.
(norm
L
!The Banach l a t t i c e
p
has the L -composition property i f P
for every sequence (an:n=l ,2,. . .) E L and every norm bounded d i s j o i n t P in L+ . sequence (un:n=I,2, . . . I The following f a c t s a r e immediately e v i d e n t . Every Banach l a t t i c e h a s the .f? -composition p r o p e r t y . I f t h e Banach l a t t i c e 1
h a s t h e L,-composition
property, then
L
1 5 r I p
number
. The
L
f
s ( L ) = sup p:L h a s t h e 1-composition p r o p e r t y \ P
i s c a l l e d t h e lower index o f the Banach l a t t i c e ( e q u i v a l e n t l y , o(L) <
m
. We
r
P
satisfying
1
have
1 S s(L) I m
L
L has t h e !k -composiP p > 1 ( e q u i v a l e n t l y , s(L) > 1 ), then t h e Banach
follows i m m d i a t e l y from t h e d e f i n i t i o n s . Dually, i f t i o n p r o p e r t y f o r some dual
L*
.
has t h e 1 -decomposition p r o p e r t y f o r some P ) , then L h a s o r d e r continuous norm. This
We have s e e n above t h a t i f p <
L
L -composition
h a s the
p r o p e r t y f o r every
h a s o r d e r continuous norm. The proof i s n o t q u i t e t r i v i a l .
558
SEMI-COMPACT OPERATORS AND INDICES
THEOREM 129.3. If the Banach l a t t i c e
valently, i f
a. -composition
has the
P
property
, then the dual space L* has order continuous norm. Equi, then L* has order continuous norm.
p > 1
f o r some
L
ch. 18,11291
s(L) > 1
PROOF. Assume t h a t t h e norm
in
p*
i s n o t o r d e r continuous. Then
L*
t h e r e e x i s t s a norm bounded d i s j o i n t sequence i n
L
n o t converging weakly
t o z e r o (Theorem 116.1). It f o l l o w s t h a t t h e r e e x i s t s a number
E
a d i s j o i n t sequence
such t h a t
$(un) >
E
sequence
for a l l
in
(u,) n
0 5 4
and some
(an:n=1,2, ...) E
. Since
l;
y > 0
t h e r e e x i s t s a number
with
Lf
E L*
. Now
has the
L
such t h a t for a l l
2 y <
for a l l
p(un) 5 1
n
and
> 0
choose an a r b i t r a r y
eP-composition
property,
m
It f o l l o w s t h a t
xy
an +(un) =
T h i s shows t h a t By a well-known
Hence, p*
an
+nl1 5
v.p*(+)
for all
m
. ep .
m
a $(u ) converges f o r every sequence (a,) E I n n theorem t h i s i m p l i e s t h a t t h e sequence ($(un) : n = l , 2 , . C
b e l o n g s t o t h e space follows t h a t
+(xy
$(un)
. Note
Lq(p-'+q-l=l) +
0
as
n
+ m
,
that
q <
contradicting
m
since
$(un) >
p > 1 E
..)
. It
for a l l
n
.
i s o r d e r continuous.
The L -decomposition p r o p e r t y and t h e -composition p r o p e r t y , t o g e t h e r P P w i t h t h e d e f i n t i o n of t h e upper and lower index were i n t r o d u c e d f o r Banach f u n c t i o n spaces by J.J. Grobler l a t t i c e s by P.G.
(c
Dodds (C21,1977).
11,1975) and extended t o a b s t r a c t Banach I n d i c e s f o r Dedekind complete Banach
l a t t i c e s have been e a r l i e r d e f i n e d by T.Shimogaki (C11,1965) and i t can be proved t h a t t h e Grobler-Dodds i n d i c e s c o i n c i d e w i t h those of Shimogaki. Some
L -decomposition and -composition P P and upper and lower p-estimates w i l l b e made l a t e r i n t h i s s e c t i o n .
f u r t h e r remarks about t h e r e l a t i o n between
F o r a p p l i c a t i o n s of t h e n o t i o n s i n t r o d u c e d above t o
o p e r a t o r s from one
Banach l a t t i c e i n t o a n o t h e r i t i s of importance t o know t h a t f o r every Banach l a t t i c e
L
s(L)
u(L)
of i n f i n i t e dimension. For t h e proof we f i r s t
need a n o t h e r r e s u l t , a s follows.
18,51291
Ch.
COMPACT OPERATORS
THEOREM 129.4. L e t ' L
satisfy
p-l
+ q-l
-1
=
m
5
m
as uswrll. Then L has the L P has t h e L -cornpasition property.
0
L*
decomposition property i f and only i f
,q
1 5 p
be a Banach l a t t i c e and l e t
(with
= 1
559
9
PROOF. Since e v e r y Banach l a t t i c e h a s t h e Lm-decomposition p r o p e r t y and t h e .t? -composition p r o p e r t y , we may r e s t r i c t o u r s e l v e s t o t h e case that
p <
some
p c
1
m
m
. Assume f i r s t t h a t L h a s . Let (an:n=1,2 ,...) E L i
,
L -decomposition p r o p e r t y f o r
L*
satisfying
P and l e t
j o i n t sequence of p o s i t i v e elements i n
n
the
,...)
':$,:n=1,2
p*($
be a d i s -
We have t o show t h a t t h e r e exists a f i n i t e p o s i t i v e c o n s t a n t
that
an $n)
p*(Zy
space
L
for a l l
5 C
m
. Since
i s Dedekind complete and a l l
$
for a l l
) 5 1
C
such
i s o r d e r continuous, t h e
p
a r e o r d e r continuous l i n e a r
. I t follows t h a t t h e c a r r i e r s of t h e $, a r e mutually d i s j o i n t bands i n L . Let 0 5 u E L be given. Denote t h e component of u i n t h e c a r r i e r of $n by un . By assumption ( p ( u ):n=1,2 ,... ) E Lp and, n s i n c e ( p * ( a n $,):n=1,2, ...) E lq, we have f u n c t i o n a l s on
L
11
Hence
m
f o r every
. It
p o s i t i v e number
follows t h a t f o r every yf
f E L
But then, i n view of t h e Banach-Steinhaus p o s i t i v e constant
there exists a f i n i t e
such t h a t
such t h a t
a
theorem, t h e r e e x i s t s a f i n i t e
.
4 )
5 C for a l l m I n n For t h e proof i n t h e converse d i r e c t i o n we assume t h a t L*
C
p*(Cm
L -composition p r o p e r t y f o r some q 4
sequence i n
L
such t h a t
that
(p(un):n=l,2,
that
1;
an p ( u n )
...)
E
0 i u
Lp
. For n
> I
5 u
. Let
L*
satisfying
(un)
in
p*(4 ) 5 1
n
...)
. We
has t h e be a d i s j o i n t
have t o prove
t h i s purpose i t i s s u f f i c i e n t t o show
converges f o r every sequence
t o t h e Eiven sequence
sequence i n
n
E L for a l l n
(an) be such a sequence. We may assume t h a t ponding
(u : n = l , 2 ,
L
,
(an:n=1,2,
...)
E
Li
.
. Let
a # 0 f o r a l l n Corresn l e t (4,) be a p o s i t i v e
for all
n
and
560
SEMI-COMPACT OPERATORS AND INDICES
Every element of
Dedekind complete space are disjoint i n
of
u n
that
. Then
p*($,)
L*
L* . Hence, t h e c a r r i e r s of t h e d i s j o i n t elements . Let Jin b e t h e component of On i n t h e c a r r i e r
1
for a l l
Jln(un) = $,(Un)
n
cm a 1 n
p(un)
By t h e assumption on an Jln) 5 y
5
for a l l
...
, we
c m1 an qn(Un)
m
lower index of L*
n
have
c mI
5 1 +
<
an J,n (u) 5
.
y
L
we have
i s the upper index of L and
PROOF. I t is easy t o s e e (from t h e l a s t theoaem) t h a t
. We
only i f
s(L*)
For any
p > a(L)
t h e apace
for
+
,
p-l
therefore
=
q-l =
q 5 s(L*) p- 1 = 1
%is h o l d s f o r a l l
-
such t h a t
m
COROLLARY 129.5. For every Banaeh l a t t i c e
where (as defined above) o ( L )
such
. Hence
1 + y.p(u)
11 an P(U,)
2-n a-l n for a l l
there e x i s t s a f i n i t e constant
L*
m
-
m = l,2, 1 +
L*
and
2 p(u n )
It follows t h a t , f o r every
p*(Cy
i s a d i s j o i n t p o s i t i v e sequence i n
($n:n=l,2,...) 5
18,11291
a c t s a s an o r d e r continuous l i n e a r f u n c t i o n a l on t h e
L
u
Ch.
may assume, t h e r e f o r e , t h a t
o(L) <
s(L*) i s the
a(L) = and
m
i f and s(L*) > I .
h a s t h e k? -decomposition p r o p e r t y . Hence, P t h e space L* h a s t h e -composition p r o p e r t y , and 4 It follows t h a t L
e
.
q- 1 5 1
p > o(L)
-
{s(L*)l
, so
1-I
{a(L)}-'
5 1
- {s(L*)}-'
, i.e.,
a. 18,91291
COMPACT OPERATORS
For any
q < s(L*)
for
+
p
-1
q
-1
t h e space
, the
= 1
p t o(L)
therefore
L*
has the
L
has the
space
. It
follows t h a t
THEOREM 129.6. FGP any Eanach l a t t i c e
(i)
eq-composition P
L
t h e following holds. a(L) = 1
L
i s of f i n i t e dimension, then
(ii) I f
L
is of i n f i n i t e dimension, then
L
space
.
1 5 p 5
be of i n f i n i t e dimension and assume t h a t
L
r
Then t h e r e e x i s t numbers
Therefore, L
has the
s
property. Let
-1
+ r
5
m
m
.
i s of f i n i t e dimension e a c h d i s j o i n t sequence i n
L
h a s t h e k?. -decomposition p r o p e r t y a s w e l l as t h e P
(ii) Let
s(L) =
and
< s(L) < o(L)
1
h a s o n l y f i n i t e l y many terms. It f o l l o w s t h a t f o r e v e r y property
p r o p e r t y . Hence,
-decomposition p r o p e r t y , and
If
PROOF. ( i ) I f
56 I
3
arid
p
1
.
L
the
eP-composition s ( L ) z o(L)
.
such t h a t
-composition p r o p e r t y and t h e =
m
and l e t
(an:"= 1,2,.
. .)
e -decomposition b e a sequence belong-
Then t h e r e e x i s t s a sequence .l? ,+ b u t n o t t o b+ P (En:n=1,2, ...) E l: such t h a t Cm a = m
ing t o
sequence
I n n
L*
Since
($n:n=1,2,...)
s e r v e d a l r e a d y , t h e space t h e es-composition constant
y1
t h e norm
p
pmperty for
.
i s of i n f i n i t e dimension, t h e r e e x i s t s a d i s j o i n t p o s i t i v e
L
L
L*
such t h a t
has th e
p * ( Z y E n $n) 2 y 1
(En)
L
E
e:
for all
i s o r d e r continuous ( s i n c e
r > 1 ). I t f o l l o w s t h a t
= 1
p*($,)
for all
n
. As
L -decomposition p r o p e r t y , so L*
p r o p e r t y . Hence, s i n E e
such t h a t in
in
L
, m
obhas
there e x i s t s a f i n i t e
. Next,
observe t h a t
h a s t h e b -decomposition
i s Dedekind complete and a l l
are o r d e r c o n t i n u o u s . T h e r e f o r e , t h e c a r r i e r s of t h e $ n
$
are m u t u a l l y d i s -
n
SEMI-COMPACT OPERATORS AND INDICES
562
j o i n t bands i n and
On
4
$n(un) >
. Then
p(vn)
S
. For
L
and l e t
(v :n=1,2,
n and
1
n v
...)
$n(vn) >
such t h a t
4
for a l l
,
m
a v ) p(Cm 1 n n
e x p l a i n s why
o(L)
n
(a,) E
,
Cm a
I
n
s ( L ) 5 o(L)
and
.ep'
for a l l
un E L+
6
n
=
. Hence,
satisfy
p(un)
1
S
since
L
has the
there e x i s t s a f i n i t e constant m
.
I t follows t h a t
. Hence
m
holds (for a l l
s ( L ) 5 o(L)
L
.
of i n f i n i t e dimension)
a r e c a l l e d upper and lower index r e s p e c t i v e -
s(L)
[s(L),o(L)I
l y . The i n t e r v a l
let
f o r every S y2
contradicting
The f a c t t h a t
,
be t h e component of u i n t h e c a r r i e r of n 9 s a d i s j o i n t sequence i n L+ s a t i s f y i n g
k? -composition p r o p e r t y and P y2
.
1,2,..
=
Ch. 18,51291
i s then c a l l e d t h e i n d e x interval of
L
.
In some i n t e r e s t i n g examples t h e index i n t e r v a l c o n s i s t s of one p o i n t , i . e . , s ( L ) = o(L)
. This happens
s i o n f o r some
p
if
satisfying
i s an a b s t r a c t L -space of i n f i n i t e dimenP m I n t h i s case, i f (un:n=1,2, )
L
i s a p o s i t i v e d i s j o i n t sequence i n
. It
.
1 2 p <
Lf
, we
...
have
h a s t h e L -decomposition p r o p e r t y a s w e l l P as t h e L -composition p r o p e r t y . T h e r e f o r e , s ( L ) = o ( L) = p I f L i s an P i s a p o s i t i v e d i s j o i n t sequence AM-space of i n f i n i t e dimension and (u,) for a l l
in
L+
for a l l 1 5 p 5
m
, we
m m
follwws t h a t
L
follows t h a t
L
.
have
. It
. Hence
s(L) =
,
has the
and s o
k?
-composition p r o p e r t y f o r a l l P a(L) = m as well.
A t t h i s p o i n t we i n s e r t some remarks about t h e r e l r i t i o n of
L -decompo-
P s i t i o n and 1 -composition p r o p e r t i e s w i t h some o t h e r n o t i o n s . For conveP n i e n c e , l e t us denote ( f o r 1 I p 2 m ) t h e norm of a n element
...
.
) E Lp by 1 1 ~ 1 1 We b e g i n by r e c a l l i n g t h e a -composition (x:k=1,2, P P and L -decomposition p r o p e r t i e s . The Banach l a t t i c e L h a s t h e L -decompoP P s i t i o n p r o p e r t y i f ( p ( u ):n=l,Z, ...) E Lp f o r e v e r y d i s j o i n t o r d e r bounded
a. 18,51293
COMPACT OPERATORS
....)
(un:n=1,2
sequence
E L+
and
563
L has the L -composition property if
P .) E L+ satissup, p ( u I + . .+u ) < -a for any disjoint sequence (un:n=l ,2,. n ) E There exist also two properties, called fying ( p ( u n):n=1,2, the strong k? -decomposition property and the strong -composition property, P P as follows. The Banach lattice L has the strong a. -decomposition property P if there exist a finite constant K1 such that if ul, u n is any finite disjoint system in L+ , then
.
...
.
ep .
...,
and L
has the strong e-composition property if there exists a finite P u is any finite disjoint system in L+ such that if u l , n
...,
constant K2
,
then
= K = 1 if L 1 2 is an abstract L -space. It is evident that the strong 1 -decomposition P P property implies the t -decomposition property, and similarly for the P -composition property. It has been proved (essentially by P.G. Dodds [21) P that actually the -decomposition property and the strong l? -decomposition P P property are equivalent, and similarly for the 1 -composition property. Also, P it is not only true that L has the -decomposition property if and only p -1 if L* has the 1 -composition property (p +q-l=l) , but also L has the 4 -composition property if and only if L* has the 1 -decomposition properP q ty. It follows that not only {o(L)I-’ + Is(L*)}-’ = 1 , but also
Note that there is equality in both inequalities with K
e
e
{s(L)I-’ [I])
+ {o(L*)) = 1
. Some authors
(T. Figiel-W.B. Johnson C11, B. Maurey
use a somewhat different terminology. Instead of saying that L
e
has
the (strong) 1 -decomposition property or the (strong) -composition P P property it is said now that L satisfies a Lower p-estimate or an upper p-estimate respectively. For further details (as well as for the relations between these notions and the notions of type and cotype of a Banach lattice) we refer to the book
c11 by
J. Lindenstrauss and L. Tzafriri.
There is a relation between semi-compactness of an order bounded operator from one Banach lattice L ces of L
and M
index, then T
. Roughly
into another Banach lattice M
and the indi-
speaking, if the operator T decreases the
is semi-compact. More precisely, if the index interval of M
is disjoint from and below the index interval of
L
on the real line. then
SEMI-COMPACT OPERATORS AND INDICES
564
any order bounded operator T: L
+
M
18.91291
Ch.
is semi-compact, as we shall prove
now. THEOREM 129.7. Let
such t h a t
o(M)
< s(L)
L and
. Then any
be Banach l a t t i c e s of i n f i n i t e dimension
M
I T\
compact (note t h a t i n this case
PROOF. Denote the norms in L and the norms in M <
o(M) that
that
m
X
and M*
T: L
order bounded operator
L*
and
X
by
+
i s semi-
M
is serrL-compact as w e l l ) . by
X*
and
and
p
p*
respectively
respectively. It follows from
is order continuous (hence, M
I T / exists) and it follows from s ( L )
is Dedekind complete, s o
1
that p *
is order conti-
nuous. The criteria in Theorem 127.4 for semi-compactness of the order
bounded operator T: L--
M
not semi-compact. Then
I TI
can be applied, therefore. Assume that T also is not semi-compact,
Theorem 127.4, there exists a number (un)
sequences n
Furthermore, since M*
in L
E
> 0
and M*
are of norm at most one and
and all J, o(M)
(J,n)
and
o(M)
e s(L)
, there
. Therefore, M
is
that, in view of
so
and there are positive disjoint respectively such that all
J,n(lTlun) >
for all n
E
exist numbers r
and
p
.
u
such that
has the 1-decomposition property and -1 has the L -composition property for r + s -1 = 1 Also, L has the < r < p < s(L)
.
L -composition property. Note now that on account of r < p there exist P and (B,) E L i such that ,Ym a B = m . Hence, sequences (a,) E i?; I n n returning to (J,,) and (u,) , and using now that M* has the L -composition property and L has the L -composition property, it follows that there exist finite constants y,
for all m
. But
for all m
=
1,2,
are semi-compact.
P and y 2
such that
then
...
, which
contradicts Cm a I
n
B
n
=
-
. Hence, T
and
IT1
Ch.
18,51291
565
COMPACT OPERATORS
COROLLARY 129.8. A s above, l e t
L
and
o(M) < s(L)
f i n i t e dimension such t h a t operator from L i n t o M
. Then
T
be Banach l a t t i c e s of i n -
M
be an order bounded
T
and l e t
i s compact i f and onZy i f L
AM-compact (hence, the compact order bounded operators from
T
is
into M
form a band i n Lb(L,M) 1 . In p a r t i c u l a r , i f L and M are Banach f u n c t i o n spaces ( a s i n section 126) s a t i s f y i n g o(M) < s ( L ) , then any absoZute ( i . e . , order bounded) kernel operator from that i f into
1 s r i: p
Lr(X,u)
L
into
M
i s compact. I t f o l l o w s
, then any absoZute kernel operator from
~0
i s compact.
PROOF. The f i r s t statement follows by observing t h a t i f have o r d e r continuous norm, t h e n any o r d e r bounded o p e r a t o r compact i f and only i f
L (Y,v) P
and
M
L*
is
T: L + M
i s AM-compact and semi-compact (Theorem 125.5).
T
For t h e s t a t e m e n t about a b s o l u t e k e r n e l o p e r a t o r s , n o t e t h a t every a b s o l u t e k e r n e l o p e r a t o r i s AM-compact and n o t e a l s o t h a t i f p
1
satisfying
5 m
,
then
f o r some
L = L (Y,v) P
.
s(L) = o(L) = p
The l a s t theorem and i t s c o r o l l a r y were proved f o r a b s o l u t e k e r n e l o p e r a t o r s between O r l i c z spaces by T. Ando ([31,1962).
This was extended t o
Banach f u n c t i o n spaces by J . J . Grobler (111,1975). There e x i s t r e l a t e d theorems f o r norm bounded (not n e c e s s a r i l y o r d e r bounded) k e r n e l o p e r a t o r s beL -type ( c f . , €or example, t h e book [ I ] by M.A. KrasnoP Zabreiko, E . I . P u s t y l n i k and P.E. S o b o l e v s k i i ) . There a r e some
tween spaces of s e l s k i i , P.P.
r e s u l t s about compactness of norm bounded o p e r a t o r s i n t h e Dodds-Fremlin paper C11; f o r some recer.t r e s u l t s about norm bounded k e r n e l o p e r a t o r s and i n d i c e s we r e f e r t o a paper by P.G.
A s an a p p l i c a t i o n , l e t M
i s an i d e a l i n
measure space ( i . e . , with
1s q < p s
included i n
M
Jf = f
Banach l a t t i c e
m
.
m
) and l e t
(and, of c o u r s e , MI
J
f o r every M1
M2
be Banach l a t t i c e s such t h a t let
M1 = Lp(X,u)
and
i s no* an i d e a l i n
be t h e i d e n t i t y o p e r a t o r from
f E MI
. Since
J
i n t o t h e Banach l a t t i c e
M
M2 ).
s u b s e t of
MI
M2
, but
Returning t o
into
M1
M2
,
i s a p o s i t i v e o p e r a t o r from t h e M2
,
the operator
J
i s norm
i s a norm MI a l s o every compact ( o r almost o r d e r bounded)
bounded. Note t h a t n o t o n l y every norm bounded s u b s e t of bounded s u b s e t of
MI
be a f i n i t e
(X,A,p)
= L (X,p) 2 q It follows t h e n from H6lder's i n e q u a l i t y t h a t M1 i s
p(X) <
2 the general case, l e t i.e.,
and
1
. To mention an example,
M2
Scheu (C11.1982).
Dodds and A.R.
i s compact ( o r almost o r d e r bounded) i n
M2
-
566
Ch. 18,11291
SEMI-COMPACT OPERATORS AND INDICES THEOREM 129.9. A s above, l e t
M1 and M2 be Banach l a t t i c e s such is an i d e a l i n M2 and l e t J be t h e i d e n t i t y operator from M, i n t o M2 .Assume furthermore t h a t u(M2) < s ( M 1 ) . Then .J is semicompact. Hence, i f M3 i s another Banach l a t t i c e and T: M -+ M1 i s a 3 Jordan operator (which implies t h a t T i s norm bounded), then JT: M3 + M2 i s semi-compact. Also, i f T: M3 -+ M1 i s AM-compact (which i m p l i e s , by our d e f i n i t i o n , t h a t T is a Jordan operator), then JT: M M2 i s als o 3 AM-compact. Hence, i f M'; has order continuous nomi and T: M + M1 i s M1
that
-+
AM-compact, then
JT: M
3
-+
M2
3
is compact.
PROOF. For the last assertion, note that M; continuous norm and
JT: M
3 Hence, by Theorem 125.5, JT
+
M2
have order
is compact.
Again by way of example, let
, M2
and M2
is AM-compact as well as semi-compact.
(X,A,u)
be a finite measure space and
and M3 = Lr(X,u) with 1 5 p,q,r 5 m , 9 s(M ) = o(M1) = p and s(M ) = a(M2) = q 1 2 Therefore, M1 is an ideal in and the condition u(M ) < s(M ) is saM2 2 1 tisfied. Furthermore, M'; and M2 have order continuous norm. As in the let M
r > 1
= L (X,p) 1 P and q < p
=
. Then
last theorem, let J
L (X,u)
.
into M2
be the identity operator from M,
follows that if T: M3
-f
M,
. It
is an absolute kernel operator, then
JT: M - + M 2 is also an absolute kernel operator (in fact, JT has the same 3 kernel as T ) . Hence, since M2 has order continuous norm, JT is AMcompact (Theorem 123.9). By the last theorem, JT JT
is semi-compact and hence
is compact. In other words (formally less precisely), the kernel opera-
tor T
,
as an operator from M j
semi-compactness of
J (with its consequences for kernel operators) is due
to J.J. Grobler ([11,1975). space and
T
into M2 , is compact. The theorem about
In the case that
(X,A,p)
is a finite measure
is an absolute kernel operator on the Hilbert space L2(X,ti)
the operator T
is compact from L2(X,u)
into L l ( X , p )
. This particular
,
case, called (2,1)-compactness, is discussed in the book (C11,§13) on integral operators (i.e., kernel operators) by P.R. Halmos and V.S. Sunder. EXERCISE 129.10. This is an extension of Exercise 123.10. All spaces L
P
(l
mentioned are with respect to the u-finite measure space
. Assume that all
L are of infinite dimension. According to P Corollary 129.8, if T: L + L (q
Ch. 18,81291
i s norm bounded, t h e n
T
ah&
567
COMPACT OPERATORS
98.2 and C o r o l l a r y 98.4;
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 (Theorem
T
the case
Ll
+
Assume now t h a t e i t h e r
(p>l)
L
P
i s Dunford's theorem).
or
where
TI
that
T,T.
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 and
T2
i s norm bounddd. Show
i s a compact a b s o l u t e k e r n e l o p e r a t o r . S i m i l a r l y , show t h a t i f
OK
T
Ll
where
-
i s norm bounded and
T1
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 , then
T2
i s a compact a b s o l u t e k e r n e l o p e r a t o r .
T2TI
EXERCISE 129.11. Assume t h a t t h e Banach l a t t i c e
w i t h o r d e r continu-
M
ous norm i s embedded as an i d e a l i n t h e Archimedean Riesz space
more, assume t h a t t h e sequence M
and
that
fn f
(fn:n=1,2,
converges i n o r d e r i n t h e s p a c e
E M and
h(f-fn)
H I N T : There e x i s t s
-+
u
0
.
such t h a t
E 'M
thermore, t h e r e is a sequence For
w e have
m 2 n X(qn)
such t h a t hence i n
+
,
so
X(g-fn)
-+
0
. Since
X(fm-fn)
in
E
f l E C-u,ul
E
. Further-
i s o r d e r bounded i n
M
t o a n element
f
E E
f E [-u,u] for all n n such t h a t If-fnl 5 pn
0
. Hence
q
+
. Show
. Fur0 .
and If -f I S 2u , s o Ifm-fnI 5 m n E M f o r a l l n , i t f o l l o w s from qn
n
+
f
as
0
"k
m,n
+ m
. Then
there e x i s t s
converges i n o r d e r t o
g
E ) f o r an a p p r o p r i a t e subsequence. I t f o l l o w s t h a t
EXERCISE 129.12. with
(p,)
E
in
lfm-fnj 5 2pn
2 i n f ( p n , u ) = qn J. 0 that
...)
and
I n t h e Banach l a t t i c e h(f2)
< 6
. Let
M (norm
A),
h = (fhu)V(-U)
let
. Show
in
g
+
0
E M
M (and
g = f
.
f = f1 + f2 that
568
SEMI-COMPACT OPERATORS AND INDICES
If-hl
If-f
5
1
I
. Hence,
HINT: Note t h a t
and
f
A
f = f
h E [-u,ul
with
u + (f-u)+
A
u = (fAu)V(-u) + (-U-fAU)+
f = h + r
Hence
f = h+r
and
X(r) < 6
.
with
with
.
0 5 r 5 If I 2
with
18,§1291
Ch.
EXERCISE 129.13. T h i s e x t e n d s E x e r c i s e 129.11. Again, l e t t h e Banach lattice
( o r d e r c o n t i n u o u s norm
M
R i e s z space
E
f E M
and
HINT: L e t
X(f-fn) E
+
b e g i v e n . It i s s u f f i c i e n t t o prove t h a t t h e r e e x i s t s
> 0
no
such t h a t
that all
f
are i n c l u d e d i n
X(fm-fn) <
E
C-U,~]
for
m,n
.
no
2
+ ~ E B (where
Choose
u E M'
such
i s the unit b a l l i n
B
.
g = f A u and h = ( f AU)V(-U) f o r a l l n Similarly, let n n n u and h = (fAu)v(-u) By t h e p r e v i o u s e x e r c i s e , f n = hn + rn
M ). Let
g = f
.
.
0
a number
n
...
t h a t t h e sequence ( f :n=l,2, ) i n M i s almost n and f n converges i n o r d e r i n E t o f E E Show
M
o r d e r bounded i n that
X ) b e an i d e a l i n t h e Archimedean
. Assume
.
A
.
with
h E C-u,ul and h ( r n ) < :E Furthermore, by one of t h e B i r k h o f f n f o r a l l n , s o t h e o r d e r bounded L n e q u a l i t i e s , Ih-h I 5 /g-g I 5 I f - f n / n n It f o l l o w s (by E x e r c i s e sequence (h,) converges i n o r d e r t o E t o h 129.11) t h a t h(hm-hn) <
X(h-hn)
BE
for
+
0
,
m,n 2 no
.
s o t h e r e e x i s t s a number
h ( f -f ) I A(hn-hn) m n EXERCISE 129.14. L e t
M
L
no
. Then
for
+ X(rm) + h ( r n ) < E
and
M
such t h a t
m,n 2 n
0 '
be Banach l a t t i c e s such t h a t
L*
and
have o r d e r c o n t i n u o u s norm. Furthermore, as i n t h e p r e c e d i n g e x e r c i s e ,
let
M
b e an i d e a l i n t h e Archimedean R i e s z s p a c e
o r d e r bounded opeEator
T: L
+
M
p e r t y t h a t , f o r any norm bounded sequence h a s a subsequence converging i n o r d e r i n T: L
-f
M
i s compact.
. Assume
E
now t h a t t h e
i s semi-compact and h a s t h e f u r t h e r pro(fn) E
in
L
,
t h e sequence
t o an element of
(Tf,)
E . Show t h a t
Ch. i 8 , § 1 2 9 1
569
COMPACT OPERATORS
HINT: U s e the preceding e x e r c i s e . EXERCISE 129.15. I f (fn:n=1,2, on
X
,
...)
then
and fn
i s a a - f i n i t e measure space and
(X,A,p)
f
a r e ( r e a l and f i n i t e valued) p-measurable functions
i s s a i d t o converge i n measure t o
holds f o r every p o s i t i v e number converges i n measure to
f
a
. It
(f )
n
pointwise almost everywhere on
X
on
X
if
i s a well-known theorem t h a t
on every subset of
only i f every subsequence of
f
fn of f i n i t e measure i f and
X
has a subsequence converging t o
. The
set
D
of
f
p-measurable functions
i s s a i d t o be compact in measure i f every sequence i n
contains a sub-
D
sequence which converges i n measure on every subset of
of f i n i t e measure
X
( d i f f e r e n t subsequences may converge t o d i f f e r e n t l i m i t f u n c t i o n s ) . In o t h e r words, D
i s compact i n measure i f and only i f every sequence i n
subsequence converging pointwise almost everywhere on Now, l e t
L
and
M
X
.
D
has a
be Banach function spaces (which implies t h a t
L
i s an i d e a l i n t h e space of a l l ( r e a l and finitevahued) v-measurable functions corresponding t o a a - f i n i t e measure space M
(Y,I,v)
and, s i m i l a r l y ,
i s an i d e a l i n t h e space of p-measurable functions on a a - f i n i t e
Assume t h a t t h e norms i n order bounded o p e r a t o r b a l l of
L
L*
T: L
i n t o a s u b s e t of
and +
M
M
M
(X,A,p).
a r e o r d e r continuous. Show t h a t i f the
i s sed-compact and maps the closed u n i t which i s compact i n measure, then
T
is
compact.
HINT: Use the preceding e x e r c i s e . The space a l l ( f i n i t e v a l u e d ) p-measurable functions on
X
.
E
i s now the space of
CHAPTER 19
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
The f i r s t p a r t of t h e p r e s e n t c h a p t e r d e a l s w i t h O r l i c z spaces. The t h e o r y of O r l i c z spaces h a s grown o u t of s e v e r a l a t t e m p t s t o g e n e r a l i z e t h e t h e o r y of L -spaces (llp<m) P f i r s t that 1 < p < m The p r o p e r t i e s of
. Let
L
(X,h,u)
and =
p-'
b e a o - f i n i t e measure space and assume
+ q-'
Lp(X,h,p)
1 (equivalently, (p-l)(q-l)
=
and i t s Banach d u a l
P obviously r e l a t e d t o t h e p r o p e r t i e s of t h e f u n c t i o n s u 2 0
,v
inverse
2 0
. Observe
u = vq-'
that the derivative
e x a c t l y t h e d e r i v a t i v e of
t h e m o t i v a t i o n f o r W.H.
let
u = $(v)
$(u)
v = $(u)
for
u,v t 0
[ $(t)dt
and
Y(v) =
u
vq
p-luP
has a s i t s
o b s e r v a t i o n was v = up-'
such t h a t tends t o
are
for
+
a
$(O) = 0 m.
For
,
$
v t 0,
,
1
$(t)dt
0
t h e Young c l a s s
Yo
c o n s i s t s , by d e f i n i t i o n , of a l l
( r e a l o r complex) u-measurable f u n c t i o n s
f
uv 5 @ ( u ) + Y(v)
t y ) , which i m p l i e s t h a t i f
,
i.e.,
fg
f G YG
X
on
i s f i n i t e . The Young c l a s s
easy t o see t h a t L1
of
. This
and
V
0
belongs t o
q
be t h e i n v e r s e f u n c t i o n . W r i t i n g now
@(u) =
I @ ( l f ( x )l ) d p
u 2 0
tends t o i n f i n i t y as
U
for a l l
q-lvq
up
1 ).
=
Lq(X,h,u)
=
Young (1912) t o i n t r o d u c e i n s t e a d of
s t r i c t l y increasing function
i s continuous and
v = up-'
L
Yy holds f o r a l l
and
such t h a t
M$(f) =
i s d e f i n e d s i m i l a r l y . It i s
g E Yy
i s summable. F o r $ ( u )
u,v t 0 (Young's i n e q u a l i -
, =
t h e n t h e product up-'
we g e t
.
fg
$ ( v ) = vq-';
t h e Young c l a s s e s a r e now t h e s p a c e s L and L A t f i r s t sight the P 4 g e n e r a l i z a t i o n does n o t seem v e r y s a t i s f a c t o r y because i n g e n e r a l t h e Young c l a s s e s f a i l t o b e v e c t o r s p a c e s and even i f t h e s e c l a s s e s a r e v e c t o r s p a c e s
i t i s n o t so e v i d e n t how t o i n t r o d u c e n o m s . Norms were i n t r o d u c e d by
w.
O r l i c z (1932 f o r a p a r t i c u l a r c a s e , 1936 f o r a more g e n e r a l s i t u a t i o n ) .
I n t h e d i s c u s s i o n which f o l l o w s we s h a l l n o t r e q u i r e i n c r e a s i n g and continuous b u t only r e q u i r e t h a t
571
$
0
t o be s t r i c t l y
i s non-decreasing and
CHAPTER 19
512
4
,
191
I$ i s t h e n t h e l e f t continuous gene-
continuous from t h e l e f t . The f u n c t i o n ralized inverse of
Ch.
This h a s t h e e f f e c t t h a t t h e s p a c e s
L1
and
La
are now a l s o i n c l u d e d i n t h e t h e o r y . I t i s proved i n s e c t i o n 131 t h a t i f s a t i s f i e s a A -condition
@
2
O(2u)
5
M@(u) f o r a l l
(i.e.,
there e x i s t s a constant
u 2 0 ), t h e n
i t i s observed t h a t t h e set
number
k > 0 (depending on
p(f) = po(f)
The norm
for
f
Lo
d e f i n e d analogously. The complementary s p a c e s t h e s e norms, a r e Banach l a t t i c e s . E v i d e n t l y @
= L
@
if
L
0 .
A s shown i n s e c t i o n 1 1 2 , t h e space
4
o r d e r bounded) l i n e a r f u n c t i o n a l s s p a c e of a l l p-measurable f u n c t i o n s
i s summable f o r e v e r y
f
C
. Under
Lo
i s c a l l e d t h e a s s o c i a t e space of Lt
in
The norm
.
p i
.
The
LQ
Y0
w i t h norm
Ly
and
L@ which
Ly
i s contained i n
is
P,,,
, equipped L
with and
.
MO(f)
norm c l o s e d u n i t b a l l of
i s a band i n
f o r which
s a t i s f i e s a A - c o n d i t i o n . It w i l l b e proved t h a t p ( f ) 5 1 2 s 1 Hence, t h e s e t B0 = ( f : M @ ( f ) < l ) i s e x a c t l y t h e
0
i f and o n l y i f
p$
f
kf E Y0
i s g i v e n by
E Lo
we t h u s o b t a i n i s c a l l e d a n O r l i c z space. The s p a c e
Y
of a l l
f ) such t h a t
i s c a l l e d t h e Luxemburg norm and t h e normed s p a c e
p
such t h a t
i s convex. This makes i t
B@ = ( f : M o ( f ) < l )
p o s s i b l e t o i n t r o d u c e a norm i n t h e v e c t o r space t h e r e e x i s t s a number
M > 0
i s a v e c t o r space. Furthermore,
Yo
L$ and t h e norm For any
in
Li
w i l l be proved t h a t i c a l l y speaking
g
Li
C
Li
on
on
g
X
having t h e p r o p e r t y t h a t
(Lo); The s p a c e L; i s t h e r e s t r i c t i o n of t h e norm
in
t h e norm
L' 0 pi(g)
L;
.
i s given by
i s c a l l e d t h e O r l i c z norm i n
Li
.
I n s e c t i o n 132 it
c o n t a i n t h e same f u n c t i o n s , i . e . ,
Li and
L,,,
and
a r e t h e same. I n g e n e r a l t h e Luxemburg norm
L,,,
fg
t h i s i d e n t i f i c a t i o n t h e space and i s denoted by
L0 0;
of a l l o r d e r c o n t i n u o u s (and (L,),*1 La may be i d e n t i f i e d w i t h t h e
algebra-
are n o t t h e same, b u t t h e s e norms a r e L,,, = L i with e q u i v a l e n t ( p r e c i s e l y , p y 5 p i 5 2pY ) . S i m i l a r l y , we have Lo = L;
and t h e O r l i c z norm i n e q u i v a l e n t norms.
A s noted a l r e a d y , t h e a s s o c i a t e s p a c e
band
(Lo):
Li
c a n be i d e n t i f i e d w i t h t h e
of a l l o r d e r c o n t i n u o u s l i n e a r f u n c t i o n a l s on
Lo
.
In section
ORLICZ SRACES AND IRREDUCIBLE OPERATORS
Ch. I91
133 we prove that if
573
satisfies a A -condition, then L z = Li (note that 2 is a necessary and sufficient condition
4
order continuity of the norm
po
for L ’o = Li to hold). Under some extra conditions there is a converse result. As an example we mention that if u ( X ) = m and L i = LG , then satisfies a A -condition. There are also examples, however, that Lk is 2 properly contained in L* (such as when $ ( u ) = eU - 1 ) . In this case the 0
o
disjoint complement We prove that
(Lo):
(Lo):
of
in L$
contains non-null elements.
is an AL-space (T. Ando, 1960). In other words, every
(Lo):
Orlicz space is a semi-M-space. In the second part of the chapter we deal with irreducible operators. The order bounded operator T
in the Riesz space L
ducible (band-irreducible) if
T
and
{O}
then T
L
is said to be irre-
leaves no band in L
itself. It is easy to see that if L
is irreducible if and only if /TI is
invariant except
is Dedekind complete, One of the main theorems
so.
in section 136 is the Ando-Krieger theorem (T. Ando, 1957; H . J . 1969), stating that if
L
L
such that L i separates the points of operator in L
Krieger,
is a Dedekind complete complex Banach lattice and
T
is a positive irreducible
( L 3 I d d in
belonging to the band
, then T has
Lb(L,L)
a strictly positive spectral radius. It is important for the proof to note that if
To
and
such that To
T
...)
(n=1,2,
are positive operators in a Banach lattice
corresponding spectral radii satisfy that L”
0
is compact and order continuous and
separates the points of
r
f
ro
T
S
. In a Banach
I. To
, then the
lattice L
such
L we can distinguish between several
degrees of irreducibility in the class of positive irreducible operators T in L of all
.
There exists a subclass consisting belonging to the band (L-@L)dd n T with the stronger property that for every non-null u in L+
the image Tu of all
.
is a weak unit in L
The subclass has
T such that T is a weak unit in
(L>L)dd
a subclass consisting
. In general these sub-
classes are proper as shown by examples for positive kernel operators. The Positive kernel operator T with kernel T(x,y), space
(X,A,u)
,
operating on the measure
is irreducible if and only if
for every subset S
of
X
satisfying p(S) > 0 and
T
has the stronger property that Tu
L
if and only if the subset of X
x
X
p(X-S)
> 0
i s a weak unit for every
on which T(x,u)
u
and >
0 in
vanishes contains
574
a. I91
CHAPTER 19
no r e c t a n g l e
A
of p o s i t i v e measure (modulo s e t s of measure z e r o ) .
B
X
i s a weak u n i t i n t h e band
Finally, T
T(x,y) > 0
'and only i f
(L?Lldd
of k e r n e l o p e r a t o r s i f
h o l d s almost everywhere on
X
x
.
X
I n t h e s e c t i o n s 137 and 138 i t w i l l be a s s m e d a g a i n t h a t Dedekind complete Banach l a t t i c e such t h a t and t h e p o s i t i v e o p e r a t o r
belongs t o
T
s e p a r a t e s t h e p o i n t s of
Lz
. The main
(L2L)dd
compact and i r r e d u c i b l e , t h e n t h e s p e c t r a l r a d i u s theorem, i s an eigenvalue of
, which
r(T)
L
result in
s e c t i o n 137 i s t h e g e n e r a l i z e d J e n t z s c h theorem s t a t i n g t h a t i f by t h e Ando-Krieger
is a
L
is also
T
is positive
of m u l t i p l i c i t y one
T
and t h e r e e x i s t s a corresponding eigenelement which i s a p o s i t i v e weak u n i t in
L. I f
image r(T)
Tu
T
has t h e s t r o n g e r p r o p e r t y t h a t f o r any
i s a weak u n i t , then any e i g e n v a l u e
satisfies
1x1
< r(T)
. For
A
L
the
d i f f e r e n t from
T
k e r n e l o p e r a t o r s t h i s r e s u l t goes back
t o R . J e n t z s c h (1912; continuous k e r n e l on an i n t e r v a l If
in
u > 0
of
L = E n f o r some n a t u r a l number
, the
n
kernel
[a,b;a,b]
T(x,y)
in
R2 ) .
becomes a m a t r i x
( t j k ) ; t h e theorem i s t h e n 0. P e r r o n ' s famous theorem on e i g e n v a l u e s of m a t r i c e s w i t h p o s i t i v e e n t r i e s (1907). I f
i s i r r e d u c i b l e b u t does n o t
T
have t h e s t r o n g e r p r o p e r t y t h e r e may be many e i g e n v a l u e s of v a l u e e q u a l t o t h e number
. These
r(T)
of a b s o l u t e
T
form t h e p e r i p h e r a l spectrum of
T
.
I n s e c t i o n 138 we prove t h a t i f t h e p e r i p h e r a l spectrum c o n s i s t s of
,
A, , . . . , A k =
0
then t h e s e numbers a r e t h e r o o t s of t h e e q u a t i o n
. Furthermore,
t h e spectrum of
complex p l a n e by t h e a n g l e
2nk-'
T
. For
Ak
-
{r(T)lk =
i s i n v a r i a n t under r o t a t i o n of t h e t h e m a t r i x c a s e t h i s i s G. Frobenius'
theorem (1 908-1 91 2 ) . The r e a d e r w i l l observe t h a t o n l y a t one p l a c e i n o u r d i s c u s s i o n of i r r e d u c i b l e o p e r a t o r s we s h a l l use a r e p r e s e n t a t i o n theorem. This w i l l be i n Theorem 136.8, where we have a Dedekind complete Riesz space
L
that
T E (L>)dd
and
'(L;)
= {O}
and a p o s i t i v e o p e r a t o r
i s a weak u n i t i n
Tu
i n t h i s case
T2
L
f o r any
i s a weak u n i t i n
u > 0
(L3)dd
isomorphic w i t h an i d e a l i n a space of type image
T
T
in
L
in
L
such t h a t
. For
t h e proof t h a t
we s h a l l u s e t h a t L,(X,v)
. In
such
L
i s Riesz
t h e isomorphic
becomes a k e r n e l o p e r a t o r and i t f o l l o w s t h e n from Theorem 136.4
t h a t t h e k e r n e l of
T2
the desired result for
i s s t r i c t l y p o s i t i v e almost everywhere, which g i v e s T2
. To make
t h e t h e o r y of i r r e d u c i b l e o p e r a t o r s ( a t
l e a s t t h e p a r t we have d i s c u s s e d ) independent from any r e p r e s e n t a t i o n theor e m , i t would be d e s i r a b l e t o f i n d a d i f f e r e n t proof f o r Theorem 136.8. There i s t h e obvious q u e s t i o n whether t h i s d e s i r a b i l i t y makes much sense s i n c e
i n t h e i n t r o d u c t i o n of complex Riesz spaces and complex o p e r a t o r s ( s e c t i o n s
Ch. 19,11301
575
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
91 and 92) we have a l s o make use of r e p r e s e n t a t i o n s . For Dedekind complete spaces (even f o r Dedekind a-complete s p a c e s ) , however, t h i s c a n b e complet e l y avoided by u s i n g E x e r c i s e 91.12 i n s t e a d of Lemma 91.1 and E x e r c i s e 92.7 i n s t e a d of Lemma 92.3. For t h e f u r t h e r r e s u l t s about t h e p e r i p h e r a l spectrum of ( n o t necessar i l y compact) o p e r a t o r s we r e f e r t o t h e book [ 1
3 by
H.H.
Schaefer.
130. Young’s i n e q u a l i t y Let q
be a a - f i n i t e measure s p a c e and l e t t h e numbers p and -1 1 < p < m and p + q-’ = 1 I t w i l l be obvious
(X,A,u)
.
be given such t h a t
t h a t t h e p r o p e r t i e s of t h e Banach f u n c t i o n space
and i t s Banach P d u a l L (X,A,v) a r e c l o s e l y r e l a t e d t o t h e p r o p e r t i e s of t h e f u n c t i o n s up q -1 Since p + q-1 = 1 i s e q u i v a l e n t t o ( p - l ) ( q - l ) = and uq f o r u t 0 L (X,A,u)
.
=
,
1
(for
i t i s of some importance t o n o t e h e r e t h a t t h e f u n c t i o n u = vq-l
u t 0 ) has
motivated W.H.
v t 0 ) a s i t s i n v e r s e . These o b s e r v a t i o n s
Young (C11,1912) t o i n t r o d u c e t h e f o l l o w i n g g e n e r a l i z a t i o n . u Z 0
v = $(u)
for
u = C(v)
b e t h e i n v e r s e f u n c t i o n . W r i t i n g now
such t h a t
I
$(O)
,
$(t)dt
Y(v) =
u,v t 0
I
0
and
+(u)
+ m
as
u
-t
Let
$(t)dt
0
0 for a l l
=
V
U
O(u) =
-.
we t a k e a s t r i c t l y i n c r e a s i n g continuous f u n c t i o n
v = up-’
I n s t e a d of
(for
v = up-’
, we
d e f i n e t h e Young c l a s s
Y
0
= Y (X,h,u) 0
t o be t h e s e t
of a l l ( r e a l o r complex, a s t h e c a s e may b e ) u-measurable f u n c t i o n s on
f
f o r which t h e i n t e g r a l over
+
5
X
. The Young
class
@(u) + Y(v) f
holds f o r a l l and
i s p-summable, we get
g
i.e.,
*(v) = vq-’
.
YY
i s f i n i t e , t h e i n t e g r a l b e i n g taken
i s d e f i n e d s i m i l a r l y . Sketching t h e graph of
i t i s e a s y t o read o f f from t h e drawing t h a t
i n t h e uv-plane,
functions
0 ( l f ( x ) I)du
satisfy fg
E L1
, and
X
uv 5
u , v t 0 (Young’s i n e q u a l i t y ) . Hence, i f t h e f
. In
E Y0 and
g E Yyr
,
t h e n t h e product
t h e p a r t i c u l a r example t h a t
t h e Young c l a s s e s
Y0
and
fg
$(u) = up-’
YY a r e now t h e spaces
and L The g e n e r a l i z a t i o n does n o t seem v e r y s a t i s f a c t o r y a t f i r s t P q s i g h t , however,because i n g e n e r a l t h e Young c l a s s e s f a i l t o be v e c t o r s p a c e s ,
L
and even i n t h e c a s e t h a t t h e s e f u n c t i o n c l a s s e s are v e c t o r spaces it was n o t c l e a r u n t i l a f t e r 1930 how t o i n t r o d u c e a norm i n t h e s e spaces. This
576
YOUNG'S INEQUALITY
was done in 1932-1936 by W. Orlicz ( C l l , [ 2 1 ) .
ch
.
19,11301
The norm introduced by Orlicz
is a Riesz norm. The thus obtained normed Riesz spaces (Banach lattices, actually) are called Orlicz spaces. A little before 1950 a further generalization was obtained by no longer requiring $ to be continuous and strictly increasing but only requiring that $ should be continuous from the left and non-decreasing. We shall begin by looking at the proof of Young's inequality in this more genral situation. For this, we need the notion of the ordinate
set of a non-negative function. Let v
f(u)
=
Ord(f)
If
of
f
f
be a non-negative
. The ordinate set
Lebesgue measurable function, defined for all u t 0
is the subset of the uv-plane defined by
is Lebesgue summable, the integral
dimensional Lebesgue measure of
Ord(f)
I fdu
. This
is equal to the two-
is evident if
f
is a
Lebesgue summable step function and the general case follows by observing that an arbitrary summable
f
2
0 can be approached from below by an in-
creasing sequence of step functions. It is evident how to reformulate this result if the integration is not extended over
. Applying
to some interval C0,u 1 0
CO,uol and letting
0
J-
E
,
c
, but
[O,-)
f(u) +
the result to
0 f(u)du
it follows that
is restricted in the interval
E
is not only equal f , but also equal
to the two-dimensional measure of the ordinate set of
to the two-dimensional measure of the somewhat larger set consisting of all satisfying 0
(u,v)
u s uo
5
holds then also for u
=
0
.
m
measure zero. Recall that the graph of that v
=
Let v $(O)
.
f(u)
= $(u)
,u
for all
u > 0
0
u = +(v)
is summable, this
is of two-dimensional
is the set of all
(u,v)
$
$(a-)
< v 5 $(a+)
u < a
, then $(c)
$(O)
=
0 and
as
u
is discontinuous at
, and = +a,
a
such
a non-decreasing real function such that
is continuous from the left (i.e., $ ( u )
$
$(u) = 1
for all
u > 0
we denote the generalized left continuous inverse of
means that if
limit L
, be
f
f
f
=
$(u-)
does not vanish identically. A s an example we
$
and
mention the case that
J,
t
0 and assume that
=
. Since
, 0 s v s f(u)
Hence, the graph of
if
$(u) = c
u = a
for
. Furthermore, J , ( O )
then $(v)
is non-decreasing and
$
=
-
,
then
a < u =
5
b
$(v)
0 and if
for v > L
.
a
=
but
. By $ . This
for
$(u) < c
$(u)
for
has a finite
The thus defined function
is the generalized inverse of
.
In the
Ch. 19,B 1301
$(O) = 0
example t h a t
0 5 v 5 1
577
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
and
$(u) = 1
and
$(v) = =
for
u > 0
for
.
v > 1
$(v) = 0
we have
for
I t i s e v i d e n t from t h e s e assumptions
t h a t t h e p o s i t i v e q u a d r a n t i n t h e uv-plane i s t h e d i s j o i n t union of t h e ordinate sets
all
E,
and
E2
of
JI
and
$
r e s p e c t i v e l y and t h e s e t
l y i n g on a t l e a s t one of t h e graphs of
(u,v)
and
$
J,
. The
E3
of
set
E3
i s of two-dimensional measure z e r o . For
and
$
J,
as defined, let V
U
Q(u)
,
$(t)dt
=
Y(v) =
for a l l
u,v 2 0
.
(
$(t)dt
0
0 The f u n c t i o n s
@
and
are c a l l e d complementary Young
Y
fmctions. THEOREM 130.1.
(Young's i n e q u a l i t y . ) If
0 and
ape complementary
Y
Young f u n c t i o n s , then uv 5 @ ( u ) + Y(v)
holds f o r a l l equalities
v = $(u)
PROOF. L e t by
A
u,v t 0
,
uo t 0
, with
e q u a l i t y if and only i f one a t l e a s t of the
u = $(v) i s s a t i s f i e d .
, vo
t 0
be given. Denote t h e i n t e r v a l
and t h e two-dimensional measure of a measurable s e t
E
by
~O,uo;O,vol m(E)
.
Using t h e same n o t a t i o n s as above, we have
0 + m\AllE2/ 0 .
uovo = m(A) = m\AnEl,
Since
A
n
,
t h e number
El
i s a s u b s e t of t h e o r d i n a t e s e t o f
$
u
for
IA0
r u n n i n g through
m(AflE ) i s a t most e q u a l to $ ( u ) d u = O(uo) , w i t h 1 0 e q u a l i t y i f and o n l y i f vo t + ( u o ) S Y(vo) , w i t h m(AnE2) Similarly [O,uo]
e q u a l i t y i f and o n l y i f
uo 2 J,(vo)
w i t h e q u a l i t y i f and o n l y i f
. . Hence
and
vo t $ ( u o )
l e n t t o t h e c o n d i t i o n t h a t one a t l e a s t of h o l d s , because
vo > $(uo)
implies
uo
5
uo t J,(vo)
vo = +(u,) JI(vo)
.
. This
and
i s equiva-
uo = +(vo)
YOUNG'S INEQUALITY
578
COROLLARY 130.2. For a l l
where
u,v t 0
my be replaced by
sup
-
max
ch. 19,§1301
we have
$(v) <
if
m
.
Before proceeding, we make two remarks. Note f i r s t t h a t to infinity as
u
+
i f and only i f
and t h i s a g a i n i s e q u i v a l e n t t o that for
l i m +(u) = v >
. For
1
note t h a t
L i s f i n i t e as
, we
u +
0 5 u 1 < u2
and
have define'd
b r e v i t y we s h a l l say i n t h i s c a s e t h a t u3 =-aul + (I-a)u2
if
. In
v > 0
for a l l
m
i s a convex f u n c t i o n , i . e . ,
@
tends t o i n f i n i t y a s
Q(u)
$(v) <
Q(u)/u
u1,u2
f o r some a
u
tends +
,
m
t h e case
$ ( v ) = Y(v) = Y
m
jumps. Secondly,
a r e given such t h a t
satisfying
0 5 a 5 I
,
then
I n o t h e r words, t h e a r c between can be r e w r i t t e n as
equality ( I )
I
u
1
and
u2
i s below t h e chord. The i n -
u3
uI
,
$ ( t ) d t 2 (I-a) f 2 Q ( t ) d t
r3
uI
t h a t i s t o say, (2)
a
Q(t
u1
u3
Observing t h a t i n (2) t h e l e f t hand s i d e i s l e s s t h a n o r e q u a l t o a$(u3)(u3-u,)
and t h e r i g h t hand s i d e i s g r e a t e r than o r e q u a l t o
(I-~)$(u~)(U~-U~), t h e
= ( l - a ) ( u -u ) Y
2 3 is finite.
i n e q u a l i t y (2)
. Similarly,
follows now, because
the function
Y
a ( u -u ) = 3 1 i s convex i n t h e i n t e r v a l where
Ch. 19,81311
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
579
131. Young c l a s s e s I n t h i s s e c t i o n we assume t h a t
(X,A,u)
preceding s e c t i o n . Let p-measurable
,
$(lfl)
function and
0(lfl)
t h a t even i f
If/
on
f
,
X
"(If])
and
I$,$,@
Y
a r e d e f i n e d as i n t h e
be a u - f i n i t e measure space. Given t h e r e a l o r complex, t h e f u n c t i o n s
a r e non-negative
assumes only f i n i t e v a l u e s , t h e f u n c t i o n
+
assume t h e v a l u e
m
.
DEFINITION 131.1. The Young class
Y f
the s e t of a l l u-measurable functions
i s f i n i t e . The Young class
Yy
Y'(If1)
,
. Note
X
may
i s , by d e f i n i t i o n ,
= YO(X,A,u)
0
$(If])
and p-measurable on
on X f o r which the nwnber
i s defined similarly. Thhse s e t s of functions
are called complementary Young classes. We mention some simple examples. I f
1 5 p <
m
and
,
0 ( u ) = p-luP
Y0 c o n s i s t s of t h e same f u n c t i o n s a s L . For p > 1 we have P , where p - 1 + q-l = 1 , s o t h a t t h e complementary Young c l a s s Y(v) = q-lv' then
YY c o n s i s t s of t h e same f u n c t i o n s a s Y(v) = 0
tisfies that
Yy
0 < v < 1
for
c o n s i s t s of a l l
L
and
. For
q
Y(v)
p-measurable
s e t of
La
X . In t h i s case, therefore, . One might b e l i e v e f o r a moment
X = [O,ll
Y0 , b u t
The Young c l a s s and
0 < a
5
I
,
then
yQ af
convexity of t h e f u n c t i o n fying
MO(f)
If(x)/
5
1
a f + (I-a)g E Bo
2f
. In
i s convex, i . e . ,
s u f f i c i e n t condition f o r
Yo
almost sub-
0(u) = e' f(x) =
- u - 1
1
l o g x-'
is n o t .
+ (I-a)g E Y0
. Similarly
1
t h a t n o n - l i n e a r i t y of a Young
i s a convex s e t of f u n c t i o n s , i . e . , 0
2
sa-
Y
follows
YY i s a p r o p e r non-linear
w i t h Lebesgue measure, t h e n t h e f u n c t i o n
i s contained i n
. It
v > 1
I jumps. This i s n o t so. I f
c l a s s can occur only i f and
the function
for
satisfying
f
everywhere on
p = 1
=
. This
particular the s e t if
if
f , g E Y0
f o l l o w s immediately from t h e
f , g E B0
.
and
of a l l
B0 0
5
a
5
1
f
satis-
, then
W e s h a l l now i n d i c a t e a and By Yy t o be a v e c t o r space.
for
DEFINITION 131.2. The Young function
condition i f there eccists a constant
M > 0
0
is said t o s a t i s f y a A2such that
Q(2u) <M@(u) f o r
YOUNG CLASSES
580
.If
u z 0
all
, then
u t uo
for all
uo > 0
there e x i s t s
THEOREM 131.3. I f
then
s a t i s f i e s a A -condition, then
0
PROOF. Let
,
2f E YQ
implies
Af+jg E Yo (convexity of Y f C YQ
Q if
2f E Y0
implies
.
yY
af Lc Y0
for any constant a
implies f+g E Yo
f,g E Y0
prove that
for
u 2 uo , we write X
and
If(x)l
>,
. It follows from f+g E Y0
and hence
p(X) <
, where the value e
a,B 2
=
In particular Mo(af) modular My
0 a+B
5 aM
0
kMO(f)
(i) MO(kf)
MO(kf)
5 1
almost everywhere on (iii) I f = 0
(iv)
f,g E Y0
that
For the proof that
2
If(x) I
5
f
for
I a1
,
then
= 1
(f)
if
if k
= 1
05 a 2
=
x
Mo(af)
X
1
. Equivalently,
5 1
k z 0
for
u > 0
for all X
.
.
b$,
k t 0
for
My
.
kO(u)
i f and only i f
i f and only i f MQ(f)
a > 0
f i x e d , i s a convex (and continuous) f u n c t i o n
t
for the
listed in the next lemma.
, then
i s f i n i t e f o r some
implies
Q
0(ku)
. Similar properties hold
. Similarly
. Similarly
almost everywhere on If
the non-negative number
and the convexity of
0 holds f o r a11
holds f o r a l l
0(u) > 0
u 0
is also admitted, is called a modular. It is
0 almost everywhere on
(ii)
f(x)
m
0
It remains to
and the A -condition holds only
m
. Some further simple results are
LEMMA 131.4. =
t
+
M (f)
0 and
and hence Mo(kf)
f(x>
f E Y
> u 0 '
evident that M (xf) that if
.
.
as the union of the sets on which
The mapping assigning to each measurable MQ(f)
,
u
2
2
and hence
.
i s linear.
YQ
2
satisfy a A -condition. It is then obvious that
0
u
s a t i s f i e s a A -condition f o r large
0
i s linear. Similarly f o r
Yo
holds only
.
Y
i s o f f i n i t e measure add
X
0 ( 2 u ) < MQ(u)
such t h a t
i s said t o s a t i s f y a A2-condition f o r large
Q
A s i m i l a r d e f i n i t i o n holds f o r
If
ch. 19,51311
of
=
, then k for
0
f(x)
0
=
i f and only if
MO(kf)
, for
0 < k < a
.
f
Ch. 19,11311
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
PROOF. (i) Let MQ(kf)
=
0 for all
581
k t 0 and assume that
on a subset of positive measure. Then there exists a number that
If(x)l
Ikf(x)l
holds on a set E
E
vo
2
,
on E
O(vO).u(E)
2
0(lkf(x)l)
so
> 0
@(vo)
2
, contradicting our
almost everywhere. (ii) Let Mo(kf)
0
all k
t
M@(kf)
> 1
M (kf)
=
@
. If not,
2
1
0 for all
for all k
implies f
=
0
k t 0
@(u) > 0
, which
for all
>
0 such
0
2
on
,
. We
.
0
=
0
= 0
for k t 1
(f)
0
O > O
It follows that
prove that M (kf)
2 kM
for
that
contradicting our hypothesis. Hence
implies f(x) u >
E
. Then
hypothesis. Hence, f(x)
it follows from M@(kf)
for sufficiently large k
(iii) Let
6
of positive measure. Now, let v
O(vo) > 0 and choose k > 0 such that kE t vo
satisfy M@(kf)
2
> 0
If(x)l
0
=
0 almost everywhere.
. The proof
that M@lf)
=
0
almost everywhere is analogous to (and even simpler than)
the proof in part (i). (iv) Evident. Let case
Lo
be the set of all u-measurable
k > 0
number
(depending on
@(lkf(x)I)
is summable, which shows that
everywhere on X for any constant kg
>
and
Mo(k2g)
is finite. In this
If(x)l
is finite almost
0
Thus it has been provedthat Lo J- 0 as
k J- 0
crf E LQ 0 and
are finite. This implies that
.
,k ) It follows that f+g E La. 1 2 is a vector space. Note that for f E L
is finite, where k
we have M@(kf)
for which there exists a
. It is imediately evident that f E La implies a . Furthermore, if f,g E Lo , there exist k l >
0 such that Mo(klf)
MQ(lkOf+lkOg)
f
f ) such that Mo(kf)
.
=
min(k
Since
If1 < lgl
with
g E Lo
@
implies
, the space L@ is a Riesz space. Precisely, L@ is an ideal in the (complex) Riesz space of all almost everywhere finitevalued u-measurable
f E Lo
functions on X
. The Riesz
space Ly
i s defined similarly. The spaces
La and L,,, are called compZernentary O r l i c z spaces. We introduce norms in La
and
Ly
.
THEOREM 131.5. For
f E Lo
,
by
Then
p
is a R i e s z norm in
-
Let t h e n m b e r
p(f)
=
pO(f)
be d e f i n e d
YOUNG CLASSES
582
PROOF. Note first that p(f) 2 0
evident that p(f) (note that p ( f )
for all
0 implies f
=
= p([f[)
Furthermore p(f)
is finite for every p(f)
f and
and
=
=
0 by part
p(af)
=
5
. It is
1;)
f
0
=
of the last lemma).
\al.p(f)
for every constant a
, then
k-' If I
+
.
If \/p(f),
it follows from M (k-'f) 5 1 for these values of k that MQCf/p(f)l Q 1 . This can be used in the proof that lgl 5 If I implies p(g) 5 p(f)
Indeed, it follows from
so
f F: L0
0 if and only if
The next point to observe is that if k f p(f) > 0 so
19,51311
Ch.
p(g)
5
p(f)
f and
Writing
p(f+g)
so
g
p(f)
5
p(f)
+ p(g)
and
ay
p(g)
. For the proof
p
of the triangle
we may now restrict ourselves to the case
= By
(a+B=l)
separate results, we have shown that p The norm
p(g)
by definition, i.e., p(f+g)
y
.
that
are non-negative. We may assume that
=
5
If1
5
by the definition of
inequality p(f+g) that
(gl
5
, we
5
p(f) + p(g)
= y >
0
.
have
. Collecting all
p(f) + p(g)
is a Riesz norm in
*
in LQ we have defined here i s not exactly the same as the
original norm introduced by Orlicz in 1932. The method followed by Orlicz is different from the method as explained here. Our method, due to W.A.J. Luxemburg ([11,1955) a
is based on the observation that B0
convex set, and the norm
norm
p
p
=
(f:M (f)
is now the Minkowski functional of
is sometimes called the Lurcembtirg norm in Lm
. As we
BQ
is
. The
shall see, the
original Orlicz norm and the Luxemburg norm are equivalent. It should be observed that in the abstract theory of modulared vector spaces, as developed by H. Nakano ([5 1,1950), the two equivalent norms a l s o emerge. The definitions of Nakano and Luxemburg are different, however. We recall that (by definition) the normed Riesz space L (norm the sequential weak Fatou property if it follows from 0 sup ~(u,)
<
the space L
m
that
u
=
sup u
n
exists in L
. According
5
u 4 n
p)
has
and
to Theorem 113.1
i s then a Dedekind o-complete Banach lattice and there exists
19,51311
Ch.
ORLICZ 4PACES AND IRREDUCIBLE OPERATORS
5 u
583
sup u
in L with p(u) 5 n 5 k sup p(un) . We shall use this result to show that is a Banach L0 latticewithrespect to the Luxemburg norm p and for the constant k we such that 0
a constant k
2
1
may take k
1
.
=
u
=
, equipped w i t h t h e Luxemburg norm
Lg,
THEOREM 131.6. T ~ space E
4
n
p
,
i s a Banach l a t t i c e having t h e sequential Fatou property, i . e . , if 0 5 u 4 and p ( u n ) 5 ci < m for a l l n , then t h e pointwise supremum n u (x) = sup un(x) is contained i n Lo and p(un) .f p ( u o ) . Similarly 0 f o r Ly . PROOF. We begin by proving that L0 has the sequential Fatou property. in L@ with p ( u ) 5 a < m for For this purpose, assume that 0 5 u I-
. We may
all n
assume that
the sequence un(x) ; n Since p(un)
ci
1,2,
=
for all n
5 ci
> 0
n
. Let
... . Then
, we
u (x) u
0
0- 1
n
be the pointwise limit of
is non-negative and measurable
have MO(a un)
5
1
for all n (by the
Hence, according to the theorem on integration of -1 monotone sequences, M ( a u ) 5 1 This shows that uo E LQ and 0 0 . As p(uo) 5 a Letting a J. sup p(un) , we see that p(u0) = sup p ( u n is a Banach already observed, it follows now from Theorem 113.1 that L0 lattice. definition of
p ).
.
.
By way of example we look once more at the case that Q(u)
=
+(O)
$(v)
L0
p-luP
. For
p-’ + q-l = 1
by
=
+(u)
0 and
=
for v > 1
= m =
p > 1
. Hence
L1 and
Ly
=
Lm
1
we have
Y(v)
=
q-lvq
and L = L L y q P for u > 0 Then $(v) Lo
=
.
, where
. For
=
p
1
5
q
p
and
< m
is determined 1
=
0 for 0
we have 5
v
5
1
and
. Hence Y(v) = +(v) for all v . It follows that . An easy computation shows that for L@ = LP (~21)
is the familiar L -norm 11 fll for p > 1 , p-’ + q” = 1 ( Iflpdu)’/p . Hence = (l/q)l/qllf 1 1 pY 4 For L0 = L1 and L = Lm we get p Y = 11 film.
.
subset E
is
we have
p = p0 =
( l / ~ ) ” ~llfllp, where
Y We proceed with some additional remarks. First, note that for any of X
of finite measure the characteristic function of E
Contained in L0 and Ly such that
0
Then 1Y(lfl) Ly Ly
. We
indicate the proof for Ly
. Choose
v
0
Y
0
Y(vo) < m and let f(x) = v on E and zero elsewhere. 0 is summable, i.e., f E Yy c Ly I n view of the linearity of
5
.
it is evident now that the characteristic function of E
. It follows immediately that
belongs to
any measurable function, bounded and
ch. 19,§1311
YOUNG CLASSES
584
v a n i s h i n g o u t s i d e a s e t of f i n i t e measure, belongs t o
u
more, s i n c e t h e measure (En:n=1,2,
...)
. The
En I. X
in
of s u b s e t s of
i s b-finite,
X
, each
X
(x
corresponding sequence
and
Lo
t h e r e e x i s t s a sequence
...)
:n=1,2,
Lo
f u n c t i o n s i s then a c o u n t a b l e o r d e r b a s i s i n t h e same time t h a t i f we r e g a r d
. Further-
Ly
of f i n i t e measure, such t h a t
En
En
Lo and
of c h a r a c t e r i s t i c
and
Ly
. This
shows a t
a s i d e a l s i n t h e space
Ly
M(X,u)
of a l l u-measurable (and almost everywhere f i n i t e v a l u e d ) f u n c t i o n s on then
X
i s t h e c a r r i e r of
Lo
and
. Observing
Ly
a r e super Dedekind complete ( s i n c e
M(X,u)
now t h a t
,
X
Lo and
LY
i s super Dedekind complete),
we can apply Theorem 113.2 t o conclude t h a t
and
LQ
Ly p o s s e s s t h e Fatou
property f o r d i r e c t ed s e t s . THEOREM 131.7. The spaces
La
Ly
and
have t h e p o i n t s e t X
c a r r i e r . Every measurrrbZe bounded function on of f i n i t e measure, beZongs t o Lo
Ly
and
.
Furthermore, La
the Fatou property f o r directed s e t s , i .e . , if 0 sup p ( u T ) <
ZarZy f o r
m
L~
, then
.
uo = sup u
exists i n
5 uT+
Lo
PROOF. The super Dedekind completeness of
and
i n Lo p(uT)
Ly
and
+
with p(uo)
implies t h a t
La
have
. Simiis
LQ
has a c o u n t a b l e o r d e r
o r d e r s e p a r a b l e . Furthermore, a s observed above, La b a s i s . F i n a l l y , by t h e preceding theorem, La
as
X
, vanishing outside a s e t
has t h e s e q u e n t i a l Fatou
p r o p e r t y . A l l c o n d i t i o n s f o r a p p l y i n g Theorem 1 1 3 . 2 a r e s a t i s f i e d , t h e r e f o r e . We conclude t h a t for
L
Lo
has t h e Fatou p r o p e r t y f o r d i r e c t e d s e t s . S i m i l a r l y
Y *
I n our n e x t theorem we compare norm convergence of w i t h moduZar convergence of
fn
to
f
,
i.e.,
fn
to
f
MQ(f-fn) + 0
as
n +
a p r e l i m i n a r y remark we r e c a l l t h a t any sequence L
In
converging i n norm t o
LQ t h i s means t h a t
sequence
(p,)
in
LQ
almost everywhere t o
f
(gk(x)-f(x)l
1 < p(fj 5 M (f) 5 6
m
5 1
. As
i n a Banach l a t t i c e
5 pk(x) .C 0
t h e subsequence
( g =f ) k "k f o r an a p p r o p r i a t e
(g,)
.
converges p o i n t w i s e
.
THEOREM 131.8. ( i ) I n
More p r e c i s e l y , p ( f )
m
Lo
has an o r d e r convergent subsequence
. Hence,
f
(fn)
in
we have p ( f ) 5 1 if and only i f MQ(f) 5 1 . implies Mo(f) 5 p ( f ) i 1 rmd p ( f ) > 1 impZies La
. Sikzarzy
for
Lp
.
Ch. 1 9 , § 1 3 1 1
(ii)
M (f) = 1
, and similarly
p(f) = 1
impZies
@
converse does not always hold. However, if M (k f ) < then
p(f)
n
+
p(f-fn) k
f o r every
m
n
as
Ly
. It
, it is
n
expression is i n f i n i t e f o r srnalz (iv)
If
k
MQ(f-fn)
as n
0
+
MQ(f) 5 k = p ( f )
1 < k < p(f)
that, for
, we
.
+ m
.
i.e.,
(ii)
. Letting
k
+
implies p ( f ) = 1
I n t h e example where t h e space
L,+,
is
Y
Y(v) = 0
La
find that
. Let
MQ(kf)
2
p(f) = 1
but
py(f) = 1
p(kf) > 1
. For
0
k
+
by dominated convergence), w e g e t 5 p(f) = 1
by p a r t ( i ) . Hence
MQ(f) = 1
Y(v) =
n
w e have
+
m
).
=
.
.
. Si-'
m
for
,
v > 1
can o n l y assume t h e v a l u e s M (f) = 0
Y
M (k f ) 0
w e have
. Fix
as
no
+
and
Y(f)
MQ(f) 2 1
0
p(f-fn)
implies
p(f) = 1
implies
0
i n t h i s c a s e . Re-
i s f i n i t e f o r some
p(kf) = k > 1
1 (and o b s e r v i n g t h a t now
which i m p l i e s t h a t f o r some
( i i i ) Let
m
converse does n o t always hold.
implies
1 < k < ko
. Letting
. This
MO(k-lf) =
M@(f) 2 p ( f )
M (f) = 1
. The
t u r n i n g t o t h e g e n e r a l c a s e , assume now t h a t ko > 1
+
*
p(f) > 1
> 1 (possibly
0 5 v 2 1
for
w i t h La-norm,
z e r o and i n f i n i t y , s o t h a t
LY
n
as
, , we
p(f)
0
+
,
now t h a t
0
I t f o l l o w s from p a r t ( i ) t h a t
r n i l a r l y , M,+,(f) = 1
MQ[k(f-fn)l
p(f-f,)
SimiZarly f o r
M (k-If)
MQ(f) t MQ(k-lf) > 1
M@(f) > k
.
Then
. Assume
have
Therefore, k-'
0
.
k-lMQ(f) 2 MQ(k-lf) = M g [ f / p ( f ) ] 5 1 which i m p l i e s
+
s t i Z Z very w e l l p o s s i b l e t h a t t h e same
n
0 < k = p(f) 2 1
PROOF. ( i ) L e t
0
2
s a t i s f i e s a A2-condition, then
@
if and onZy if
LY
should be kept i n mind hbre t h a t if
becomes small f o r Zarge
,
ko > 1
i f and onZy if Molk(f-fn)l
m
f o r every
MQ(kf) 5 l i m i n f M (kf ) Q n
SimiZarZy f o r
+
f o r some
m
0
SimilarZy for
. The
Ly
for
, and in t h i s case
0
2
0
+
0
.
M@(f) = 1
implies
= 1
( i i i ) We have as
5 85
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
. On
k > 0
. Then 2 1
so
Mg(kf)
t h e o t h e r hand
pCk(f-fn)l
,
+
pCk(f-fn)l for
Mo(f)
MQ(f) 2
n t no
+
0
,
. Hence,
ch. 19,§1321
THE ASSOCIATE SPACE OF AN ORLICZ SPACE
586
for
n t n
0 ’
I t follows t h a t
assume t h a t
Mo[k(f-fn)l
MOCk(f-fn)l
+
-1
for
I 1
+
+
,
I n t h e converse d i r e c t i o n ,
f o r every
MoC~-I(f-fn)l i 1
n t no
Assuming now t h a t
n
- n
as
as
Our h y p o t h e s i s i m p l i e s t h a t p [ ~ (f-fn)l
0
+
0
p(f-f
. Hence
) + 0
p(f-fn)
k
for i
. Choose
0
2
,
n 2 no(€) for
E
n t n k > 0
h o l d s and f i x i n g
0 ‘
, we
(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 l i m M (kf ) O n f i n i t e number (because t h e r e remains n o t h i n g t o prove i f
.
> 0
E
i.e., may assume
exists as a
l i m i n f M (kf ) = = ) . By once more p a s s i n g t o a subsequence we may assume O n For j = 1,2, , t h a t f n converges pointwise almost everywhere t o f
.
...
let
MO(kq.) 2 l i m M (kf ) f o r e v e r y j , s o t h a t on account of k q . I. kl f J O n J almost everywhere we g e t Mo(kq.) .f MO(kf) , and t h e r e f o r e MO(kf) S J The S l i m Mo(kf ) Note t h a t we have used only t h e l e f t c o n t i n u i t y of O n same proof h o l d s , t h e r e f o r e , f o r Y , even i f Y jumps. Then
.
.
(iv) M > 0
Let
and a l l
Mo(f-fn)
-+
0
s a t i s f y a A - c o n d i t i o n , i . e . , O(2u) 5 M@(u) f o r some 2 u > 0 I n view of p a r t ( i i i ) i t i s s u f f i c i e n t t o prove t h a t 0
.
implies
MoCk(f-fn)]
o b s e r v i n g t h a t f o r any given on
k
,
I
such t h a t
+
k > 1
O(ku) i C(k).O(u)
0
f o r every
k > 1
.
t h e r e e x i s t s a number for a l l
u t 0
.
This follows by C(k)
, depending
132. The a s s o c i a t e space of an O r l i c z space Let Ly
Lo
and
be complementary O r l i c z s p a c e s . The norms i n
Ly
w i l l b e denoted by
and
po
pyr
r e s p e c t i v e l y . Although
0
and
of course i n t e r c h a n g e a b l e , we s h a l l assume (as b e f o r e ) t h a t i f one of Y
jumps, i t w i l l b e
and t h e Banach d u a l
Y
:L
.
Since
LO
L i
and
and are
Y
and
0
i s a Banach l a t t i c e , t h e o r d e r dual
c o i n c i d e . Hence, t h e bands
o r d e r continuous elements i n
LO
Lz
coincide
(Lo):
and
. According
(Lo):
of
L;
t o t h e general
r e s u l t s about Banach f u n c t i o n spaces i n s e c t i o n 1 1 2 we know t h a t f o r every
ch. 19,11321
0E
t h e r e e x i s t s an (almost everywhere uniquely determined)
(LQ)-n = (LO):
measurable function
4
Identifying space
and
M(X,p)
g
on
g
,
such t h a t
X
the space
of a l l p-measurable
. As
functions on
X
space of
and t h e n o t a t i o n
L;
Li
L8
i s a band i n
. Therefore,
if
(La):
i s t h e r e f o r e an i d e a l i n the
(and almost everywhere f i n i t e v a l u e d )
observed i n s e c t i o n 112, bhis i d e a l i s the a s s o c i a t e
$I
and
g
in
ph
correspond, p ' ( g ) 0
Our f i r s t observation w i l l be t h a t every
For t h e p r o o f , l e t
,
g
be given. Then
E Ly
I Ifgldp
5
2py(g)
Taking t h e supremum over a l l such L
Y
g
i s contained i n L'
0
It follows then from
p'(g)
Q
space in
pt
i s given by the same formui s contained i n
E Ly
5 1
M,+,[g/py(g)]
f o r every f
, we
get
f
i 1
+ %(g)
, where
satisfying
pk(g)
. Hence,
L;
.
if
5
2py(g)
M8(f) < 1
. This
.
shows
(not n e c e s s a r i l y w i t h the same norm). We
derive s t i l l another i n e q u a l i t y , as follows. Let
that
. The
then
I t follows t h a t
that
(La):
i s t h e r e s t r i c t i o n of
Lk
(I).
in
pg(@)
i s used i n s t e a d of
Lk
L t ; the norm
l a a s we had f o r
M,+,(f) 5 1
587
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
g
E L'
0
and l e t
M8(f) i 1
t h e r i g h t hand s i d e may be f i n i t e . The l a s t
i n e q u a l i t y i s known a s Arnehya's inequaZity. We c o l l e c t the r e s u l t s obtained so f a r i n a lemma.
is contained i n Lk . For any g E 1, I 5 2pp(g) and for any g E L; ue have p k ( g ) 5 I + $(g) . sii s contained i n LG . For any f E L0 we have p;(f) 5 2p0(f)
LEMMA 132.1. The space
have p k ( g ) milarly, Lo
Ly
.
5 88
THE ASSOCIATE SPACE OF AN ORLICZ SPACE
f E L;I
and f o r any
we have
p;I(f)
1 + MQ(f)
5
Our n e x t purpose i s t o show t h a t
Ch. 19,§1321
Ly
i s n o t only contained i n
Li
,
b u t t h e s e s p a c e s a c t u a l l y c o n t a i n t h e same f u n c t i o n s . The same h o l d s f o r
L;'
and
LQ
for
. On
account of t h e p o s s i b i l i t y t h a t
L;'
i s somewhat s i m p l e r t h a n f o r
and
LQ
Y
and
Ly
s e n t t h e s i m p l e r p r o o f . S i n c e w e know a l r e a d y t h a t
i t i s s u f f i c i e n t t o show t h a t such t h a t
M (k-lf)
Q'
L;'
Lo LQ
i s contained i n
0 # f F: L(,
c i e n t t o show t h a t i f
may jump t h e proof
. We
Li
f i r s t pre-
i s contained i n
, i.e.,
i s g i v e n , t h e r e e x i s t s a number
i s f i n i t e . We s h a l l prove t h a t
$
9
it is suffi-
k > 0
satisfies
k = p;'(f)
t h i s condition. THEOREM 1 3 2 . 2 . The space
La
valent
.
L;'
and t h e associate space
contain the same functions. The corresponding nomns
pQ
and
of Ly p;'
are equi-
Precisely,
holds f o r every
f E L
Q '
PROOF. It i s s u f f i c i e n t t o prove t h a t
pQ(f) 5 p i ( f )
f o r every
E L;' . This i s e q u i v a l e n t t o p r o v i n g t h a t M Q [ f / p 4 ( f ) l < 1 h o l d s f o r # f E L; . For t h e p r o o f , l e t f E L;l and g E Yy b e given. Note t h a t
f 0
M (g) Y
i s f i n i t e , t h e r e f o r e . W e have
I If
Ifgldu
5
p;(f)
if
1 < M (g) = a , t h e n M (a-'g) Y Y a-'. Ifgldu 5 p;'(f) Hence
.
I
Ifgldp
5
pi(f).My(g)
My(g) 5
a
if
-1
5
1
.
My(g) = 1
, which
My(g) > 1
.
It follows t h a t i n both cases
I n o t h e r words, w r i t i n g
k = p'(f) Y
f o r b r e v i t y , we have
implies
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
.Ch. 19,51321
589
B e f o r e p r o c e e d i n g w i t h t h e p r o o f , r e c a l l t h a t , by d e f i n i t i o n , Q ( u ) = =
1:
qi(t)dt
and t h a t Young's i n e q u a l i t y
equality i f
v = $(u)
. Assume now
uv I O(u) + Y(v)
becomes a n i s bounded
f C L$
f i r s t t h a t t h e given
and v a n i s h i n g o u t s i d e a s e t of f i n i t e m e a s u r e ' ( w i t h o u t b e i n g z e r o almost everywhere)
. Then
t h e bounded f u n c t i o n s
summable. For convenience, w r i t e N
a
= maxwy(g),1
J
. I n view
1f I)
P, (k-l
g = $(k-l If
1)
and
1 f 1) I a r e
Y [ $ (k-l
and, a s above, w r i t e
of (2) above, we have
Note t h e e q u a l i t y i n Young's i n e q u a l i t y . It f o l l o w s immediately t h a t f o r N
a
= My(g)
. Hence
s 1
we have
Mo[f/p$(f)l
L e t now
u
Mm(k-l If 1) 2
0
N
and f o r
a
1
=
we have
MP,(k-' If
1)
5
i n both cases,
, without
0 # f E L$
is o-finite,
1
=
f u r t h e r r e s t r i c t i o n s . S i n c e t h e measure
t h e r e e x i s t s a sequence
0 5 fn 4
I
If
s u c h t h a t each
i s bounded and v a n i s h i n g o u t s i d e a s e t of f i n i t e measure ( r r i t h o u t b e i n g
fn
.
M [f / p ' ( f ) I I 1 f o r a l l n On account @ n Y n p ' ( f ) 5 p l ; ( f ) , h o l d i n g f o r a l l n , i t f o l l o w s t h a t M [f / p ' ( f ) l 5 1 Y n O n Y for a l l n But t h e n , s i n c e 0 I f .f I f ( and P, i s c o n t i n u o u s , we g e t
z e r o almost everywhere). Hence of
.
MP,rf/pl;(f)l I 1
.
This i s t h e d e s i r e d r e s u l t .
We s h a l l prove now t h a t
Ly
t a i n t h e same f u n c t i o n s and t h a t If
Y
and t h e a s s o c i a t e s p a c e
does n o t jump, t h e proof o f f e r s no
p r e c a u t i o n s a r e n e c e s s a r y . Note t h a t i f f i n i t l i m i t as @ ( u )I Lu
u + m
for a l l
,
i.e.,
u 2 0
.
l i m $(u) =
l i m $ ( u ) = .l?
LEMMA 1 3 2 . 3 . I f
PROOF. Assume t h a t Then
g
by
of
L0
con-
<
m
u +
as
is f i n i t e a s
u +
. More
on
E
m
m
. In
t h i s case
f
and i f
preciseZy, I f
If(x)l > . l ? p i ( f ) on a s e t g ( x ) = [Lu(E)]-l
@ ( u ) tends t o a
jumps, t h e n
Y
L
then f i s bounded abost everywhere on X s L almost everywhere on x .
measure. D e f i n e
Li
I 2 p ( f ) f o r every f E L y . Y new a s p e c t s . I f Y jumps, some
p y ( f ) I p;(f)
and
E
E
l/p;(f)
b' , 2
of f i n i t e p o s i t i v e g(x) = 0
elsewhere.
590
THE ASSOCIATE SPACE OF AN ORLICZ SPACE
Ch. 19,11321
and
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 of THEOREM 132.4. The space
Ly
.
pi(f)
and t h e associate space
t a i n t h e same functions. The corresponding norms
and
py
L ' of L a c o n pi
0
are equiva-
Zent. Precisely,
hoZds f o r every
.
f E L~
PROOF. We may assume t h a t
'Y
jumps, i . e . ,
f i n i t e . It i s s u f f i c i e n t t o show t h a t
E Li
.
For t h i s purpose, l e t
proof w i t h
and
B
that
M,$f/pk(f)] and
k
=
p'(f) 0
5
1
holds f o r
-+
m
) is
0 # f
E
b e given. Then, a s i n t h e
f o r b r e v i t y . Assume now t h a t
v a n i s h e s o u t s i d e a set
f
g f Ya
l? ( a s u
interchanged,
Y
where we have w r i t t e n i s such t h a t
f E Lk
l i m +(u) =
0
# f
E Li
of f i n i t e p o s i t i v e measure. Note -1 i s bounded by t h e lemma above; more p r e c i s e l y , k I fl 5 l? almost
f
everywhere. This does n o t imply t h a t possible that functions
Y(6k-'l
6
satisfying fl )
. Therefore,
the s e t
E
writing
g = $(6k-'l
fl )
and
i s bounded, s i n c e i t i s
$(k-'l f ( )
tends t o i n f i n i t y as
$(v)
choose a number
E
0 < 6 < 1
6$(6k-'l fl )I
v
-+
l?
. For
t h a t reason, we
and we observe t h a t now t h e a r e bounded and v a n i s h i n g o u t s i d e
these functions a r e s u m a b l e over
X
.
Hence,
f o r b r e v i t y , we have
(3) where once more t h e r e i s e q u a l i t y i n Young's i n e q u a l i t y . Since t h e l a s t term on t h e r i g h t i s
Mlp(g)
,
i t f.ollows t h a t
ch. 19,91321
implying t h a t
at
v =
maxCMo(g),ll = 1
6 4 1
Letting
L
59 I
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
. Then,
i s continuous from t h e l e f t ( a l s o
and o b s e r v i n g t h a t Y
), we f i n a l l y g e t
[@I
MI f / p ’ ( f )
5
1
My(k-’f)
from ( 3 ) ,
< 1
,
i.e.,
. 0 # f E Li
The e x t e n s i o n t o a n a r b i t r a r y
i s as b e f o r e .
I t has been shown t h u s t h a t t h e O r l i c z space
Lo
v a l e n t norms. The f i r s t one i s t h e Luxemburg norm
carries two equi-
and t h e second one i s
or, equivalently,
It i s t h e second norm, as i n ( 4 1 , which o c c u r s i n t h e o r i g i n a l d e f i n i t i o n of W. O r l i c z (C11,1932). I n h i s f i r s t paper (1932) O r l i c z r e s t r i c t e d himself 0 s a t i s f i e s a A - c o n d i t i o n (and t h e n La and t h e Young 2 c o i n c i d e ) ; i n a l a t e r paper (121,1936) t h e A - c o n d i t i o n w a s Yo 2 Although t h e dropped. We s h a l l c a l l t h e norm p;I t h e OrZicz nomi i n Lo t o t h e case t h a t
class
notation
.
p;’
might s u g g e s t o t h e r w i s e , n o t e t h a t t h e O r l i c z norm in
can be d e f i n e d by t h e formula ( 4 ) w i t h o u t knowing t h a t Riesz s p a c e w i t h r e s p e c t t o t h e Luxemburg norm modular
%(g)
=
Y(lgl)du
i n t h e Young c l a s s
Ly
Lo
i s a normed
p y ; a l l one needs i s t h e
Yy
.
The c a s e t h a t
Y
may
jump w a s n o t y e t d i s c u s s e d by O r l i c z . I n c l u s i o n of t h i s c a s e i n t h e g e n e r a l t h e o r y i s due t o A.C. p i
hold f o r t h e norms
Zaanen (C11,1949). S i m i l a r remarks a s f o r p,,,
and
p i
in
Ly
. Summarizing
P@
and
t h e s i t u a t i o n now,
592
THE ASSOCIATE SPACE OF AN O R L I C Z SPACE
we s e e t h a t i f we have t h e O r l i c z s p a c e pQ
, then
i t s a s s o c i a t e space i s
a s s o c i a t e s p a c e of
equipped w i t h t h e Luxemburg norm
Lo
w i t h t h e O r l i c z norm. S i m i l a r l y , t h e
LI
w i t h Luxemburg norm
Ly
19,§1321
Ch.
is
PI
LQ w i t h O r l i c z norm.
The n a t u r a l q u e s t i o n a r i s e s t o f i n d o u t what are t h e a s s o c i a t e s p a c e s of Lo
and
i f t h e s e are equipped w i t h t h e i r O r l i c z norms, I n o t h e r words,
LI
L@ o r
s t a r t i n g from
Lg w i t h e i t h e r Luxemburg norm o r O r l i c z norm, w e a s k
f o r t h e second a s s o c i a t e s p a c e . THEOREM 132.5.
i s equipped w i t h the Euxemburg norm, then i t s w i t h the Orlicz norm (as we have already s e e n ) . I f
If
LQ
associate space i s
L,,, LQ is equipped w i t h the OrZicz norm, then i t s associate space is
I with t h e Luxemburg norm. Similar f a c t s hold w i t h Q and Y interchanged. Hence, t h e second associate space of Lo w i t h e i t h e r Luzernburg or Orlicz norm i s Lo
again
.
w i t h the same norm. Simi2arZ-y for LI
PROOF. For b r e v i t y , d e n o t e t h e Luxemburg norm p @ i n i t s a s s o c i a t e norm
of
p’
by
p”
oi
. Note
in
by
L k = Ly
that
L
by
Lo
p
,
denote
and d e n o t e t h e a s s o c i a t e norm
p’
has the Fatou property f o r d i r e c t e d s e t s
LQ
(Theorem 131.7). Hence, a p p l y i n g Theorem 112.3, w e see t h a t c o n t a i n s t h e same f u n c t i o n s a s
and t h e norms
Lo
l e n t . Even more i s t r u e . The e q u i v a l e n c e of
and
p
and
p
Lb = ( J , @ ) “
a r e equiva-
p“
f o l l o w s from t h e
p”
weak F a t o u p r o p e r t y f o r d i r e c t e d s e t s as e x p l a i n e d i n Theorem 107.7, where i t i s proved t h a t
p“ 5 p 5 k L ( p ) . p ”
with
k(p)
t h e Fatou c o n s t a n t occurk ( p ) = 1 (because
r i n g i n t h e weak F a t o u p r o p e r t y . I n t h e p r e s e n t c a s e h a s t h e Fatou p r o p e r t y
LQ p” = p
. We
and n o t o n l y t h e weak F a t o u p r o p e r t y ) . Hence
have proved t h u s t h a t t h e second a s s o c i a t e s p a c e of
Luxemburg norm i s a g a i n f i r s t a s s o c i a t e s p a c e of
LQ Ly
L
Q
with the
w i t h t h e Luxemburg norm. E q u i v a l e n t l y , t h e w i t h t h e O r l i c z norm i s
Lo
w i t h t h e Luxemburg
norm. Taking a s s o c i a t e spaces once more, i t f o l l o w s immediately t h a t t h e second a s s o c i a t e s p a c e of
LI w i t h
t h e O r l i c z norm i s a g a i n
O r l i c z norm. O f c o u r s e , t h e same r e s u l t s h o l d w i t h The l a s t theorem i s due t o W.A.J. EXERCISE 132.6. L e t Q ( u ) = p-luP
. Then
1
c
p <
I ( v ) = q-lvq
m
and
Y
with the
interchanged.
Luxemburg (C1],1955). and
. Hence
p
L
-1 Q
+ q = L
P
-1
= 1
and
. Furthermore, l e t . In section 9
LI = L
pQ i n L i s g i v e n by P / ( f l ( p i s t h e f a m i l i a r L -norm lflpdp) I/P P
131 we have s e e n a l r e a d y t h a t t h e Luxemburg norm p Q ( f ) = ( l / p ) ” p l ~ f ~ l p, where
@
LI
(i
.
Ch. 19, § 1331
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
S i m i l a r l y , w e have pk(f) in
in
L
q
i s g i v e n by
L P
. According
pi(f)
.
\If11
pY(f) = (I/q)'Iq
q ' / q ~ l f l l p. Similarly, pk(f)
=
. This
5 2p0
p 0 5 p;'
pl/Pllflh
=
,
i.e.,
.
( l / p ) l / p 5 q l / q 5 2 ( l / p ) 1/P
p = q = 2
Show t h a t t h e O r l i c z norm
t o Theorem 132.2 w e must have
Prove d i r e c t l y t h a t
593
ql/q 5 2 ( l / ~ ) ' / ~ h o l d s , w i t h e q u a l i t y i f and o n l y i f
shows t h a t t h e c o n s t a n t 2 i n Theorem 132.2 cannot b e lower-
e d . Note t h a t f o r
p
1
J-
and f o r
+
p
both
m
and
(I/p)"'
ql/q
tend
t o t h e l i m i t one. HINT: W r i t i n g
I/p = x
and
t o be proved i s e q u i v a l e n t t o xlogx + ( I - x ) l o g ( l - x )
2
I/q = y (hence, x+y = 1 ) , t h e i n e q u a l i t y
xxyy
log;
for
2
4 , i.e.,
.
0 < x < 1
equivalent t o
133. The Banach d u a l of an O r l i c z s p a c e As b e f o r e , l e t
Lo and
LY b e complementary O r l i c z s p a c e s w i t h r e s p e c t
t o t h e o - f i n i t e measure s p a c e Y
jumps, i t w i l l b e
Y
0(u) = 0
It i s p o s s i b l e t h a t
for small
5
uo
and
c r e a s i n g on
0(u) > 0
. This
[uo,-)
for
uo
2
0
. In
, but
0(u)
since
0
-f
0
i m p l i e s t h a t f o r any g i v e n
v
as
and
0
u
-f
m
.
i s not identi-
0(u) = 0
such t h a t t h i s case
a u n i q u e l y determined number
u > uo
u
-
assume t h a t i f one of
0 i s c o n t i n u o u s and
c a l l y z e r o , t h e r e e x i s t s a number
0 5 u
. We
(X,A,u)
. Hence,
for
i s s t r i c t l y in-
I
> 0
.
there exists
s u c h t h a t O(u ) = v 1 We s h a l l use 1 1 t h i s t o show f i r s t t h a t t h e c h a r a c t e r i s t i c f u n c t i o n xE of any g i v e n
p-measurable
set
tends t o zero as
we w r i t e
E
LEMMA 133.1. Let 0 < u(E) <
SoZution of
m
. !7%en
> 0
of f i n i t e measure b e l o n g s t o
Po
and t h e norm
p(xE)
E
xE E
1-
be a measurable subset of Lo
0 ( k - l ) = Cp(E)I-'
and
. It
p(xE) = ko
, where
follows t h a t i f En
tends t o zero if and only i f
tends t o zero.
p(En)
X
such t h a t ko
i s t h e (unique)
(En:n=1,2,
...)
i s of f i n i t e measure, then
sequence of subsets such t h a t every
PROOF. Note t h a t
L0
t e n d s t o z e r o (as l o n g a s t h e r e can b e no c o n f u s i o n ,
u(E)
instead of
p
u
is a p(xE ) n
594
THE BANACH DUAL OF AN ORLICZ SPACE
k 5 ko
i f and only i f p(xE) = ko
.
, where
ko
satisfies
In Theorem 131.3 i t was proved t h a t i f then t h e Young c l a s s
s a t i s f i e s a %-condition,
0
i s l i n e a r . I n o t h e r words, L
Yd
a A -condition only f o r large 2
Q
=
. Furthermore,
YQ
condition, we have Note t h a t i f
<
-
then
p(X)
-
and
+
0
'3
in
and
satisfies
0
i s again l i n e a r , i . e . ,
Yrp
satisfies a A
0
i f and only i f
Lo
i n t h i s case.
= Yrp
'3
<
according t o Theorem 131.8, i f
p(f-fn)
p(X)
,
u
. Hence
Q(k0') = [u(E)]-'
I n t h e same theorem i t was proved a l s o t h a t i f L
Ch. 19,11331
M (f-fn) d
.
u
,
s a t i s f i e s a A -condition f o r l a r g e
then a s i m i l a r r e s u l t holds f o r any sequence
.
2
f
C 0
n
in
Lrp
2
+ 0
-
. Precisely,
M (f ) c 0 For the proof i t i s s u f f i c i e n t t o O n show t h a t M (f ) C 0 implies M (kf ) C 0 f o r every k > 0 , because rpn d n Mm(kfn) -t 0 , holding f o r every k > 0 , i s equivalent t o p(fn) -t 0 , with
p(fn) C 0
i f and only if
k > 0
o r without a A -condition (Theorem 1 3 1 . 8 ( i i i ) ) . Hence, l e t Then
M (kf )
m
1
follows t h a t Lo
.
2
i s f i n i t e (because
MQ(kfn) C 0
Yo
i s l i n e a r ) and
kf
(dominated convergence).
1
be given.
+
t kfn
. It
0
The preceding remarks w i l l enable us t o determine the subspace
This subspace
i s a norm closed i d e a l i n t h e Banach l a t t i c e
La
m
L
m
a set
Ef
f E Lo
of f i n i t e measure ( d i f f e r e n t
THEOREM 133.2. We have
L
a Q
= r
m .
We
L Let Lb be the norm 0 . '3 t h a t a r e bounded and vanishing o u t s i d e
introduce (temporarily) another subspace of closure of t h e s e t of a l l
of
La
We r e c a l l (Theorem 102.8) t h a t
Lb p
f
may have d i f f e r e n t
Ef ) .
.
Lb i s inclused i n La , i t i s s u f f i c i e n t t o observe rp 0 t h a t vanishes outside a s e t of f i n i t e measure i s inclu-
PROOF. To see t h a t
t h a t any bounded ded i n
Li
. For
f
the proof we need only show t h a t
i s of f i n i t e p o s i t i v e measure. Hence, l e t write
Fn = (xEE:u ( x ) > E )
for
n = l,2,
xE
2
xE
un C 0
... . Then
belongs t o
.
Fn C
Choose
0
and
La E
m
> 0
if and
E
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
Ch. 19,11331
Since
p(x
Fn for
2€p(xE)
by t h e p r e c e d i n g lemma, t h e r i g h t hand s i d e i s l e s s t h a n
) C 0
n
595
s u f f i c i e n t l y l a r g e . Hence
p(un) C 0
, i . e . , xE E
.
Li
For t h e proof i n t h e converse d i r e c t i o n i t i s enough t o show t h a t b L e t 0 I f E L: b e given. There e x i s t s a i m p l i e s f E Lo
.
0 I f E Lz sequence
0 I f
4 f
such t h a t every
a s e t of f i n i t e measure, i . e . , follows from
f 2 f-f
C 0
f
and
fn
b E Lo
f
i s bounded and v a n i s h i n g o u t s i d e
for
all
n
. Furthermore, i t C 0 . This shows
t h a t p(f-fq) n @ b Since L: f i s t h e norm l i m i t of a sequence i n LO b d e f i n i t i o n ) , i t follows t h a t f E LQ
E La
.
.
THEOREM 1 3 3 . 3 . I f @
O
therefore, the norm in
i s norm c l o s e d (by
s a t i s f i e s a A2-condition or i f
s a t i s f i e s a A2-condition f o r large Lo
, then
u
L:
that
= LO
i s a-order continuous. Since
u(X) <
. In
m
and
these cases,
Lo
i s super
Dedekind complete (as an ideal i n the super Dedekind complete space of a l l u-measurable functions on X I , t h i s is equivalent t o saying that the norm in
i s order continuous.
L~
PROOF. In both cases we know t h a t
,
= LO
Y
and a l s o t h a t i f
.
fn C 0
We have t o prove t h a t p ( f ) C 0 i f and only i f Mo(fn) C 0 n f i n Lo belongs t o For t h i s purpose, l e t L; 0 5 f E Lo . There e x i s t s a sequence 0 5 f .f f such t h a t every f n i s b bounded and v a n i s h e s o u t s i d e a s e t of f i n i t e measure, i . e . , f E Lo = Lz n From f Z f-f C 0 and Mo(f) < m i t follows t h a t M (f-f ) t 0 (dominan O n t e d convergence). But t h e n p(f-fn) C 0 , and t h e r e f o r e f E La ( s i n c e L i in
Lo
then
.
every p o s i t i v e
.
0
i s norm c l o s e d ) .
v
I n t h e case t h a t t h e measure
in
i s f r e e of atoms, t h e converse
X
of t h e l a s t theorem h o l d s . An e q u i v a l e n t way of s a y i n g t h a t a t o m i s t h e s t a t e m e n t t h a t f o r any wmeasurable s e t
6
and f o r any number subset
of
B
THEOREM 0 < @(u) <
(i)
If
A
such t h a t
satisfying
0 < 6 < p(A)
p(B) = 6
3 3 . 4 . Let the measure
1-1
for all u > 0
. Then the
v(X)
L t = LO
=
m
and
.
in
X
A
u
i s f r e e of
of p o s i t i v e measure
t h e r e e x i s t s a u-measurable
be free of atoms and l e t
follo3;ing holds.
, then
O
s a t i s f i e s a A2-condition.
596
THE BANACH DUAL O F AN O R L I C Z SPACE
u(X) <
(ii) If
f o r Large
.
u
and
m
, then
L i = Lo
PROOF. ( i ) Assume t h a t
ch. 19,11331
s a t i s f i e s a A -condition
Q
2
does n o t s a t i s f y a A - c o n d i t i o n . Then, f o r 2 every n = l , 2 , , . . , t h e r e e x i s t s a number un > 0 such t h a t @(2un) > > n@(un) Now choose numbers a > 0 (n=1,2, ) such t h a t Ca = 1 and n Cna = and t h e n choose d i s j o i n t s u b s e t s En of X s u c h t h a t n O(un) p(En) = an f o r e v e r y n . Define t h e r e a l f u n c t i o n f on X by
.
-
f(x) = u
0
...
for
n
x E E
M@(f) =
(n=l,2,
n
I
...)
and
f(x) = 0
@ ( f ) d u = CQ(un) u ( E ) n
Zan = 1
=
It follows t h a t
p @ ( f ) = 1 (Theorem 131.8). For
function
b e d e f i n e d on
rn(x)
X
by
outside
rn(x) =
. Then
.
n = l,2,
\
U y En
...
on e v e r y
,
let the
E
(ktn) k which i m p l i e s
and
r (x) = 0 e l s e w h e r e . Then r (x) J. 0 f o r e v e r y x , n n p Q ( r n ) .C 0 ( s i n c e L: = La by h y p o t h e s i s ) . I t f o l l o w s t h a t t h e r e e x i s t s no
a number for
n t no
f o r every
such t h a t
p (2r ) 5 1
O
. But
.
n
C o n t r a d i c t i o n . T h i s shows t h a t
( i i ) It i s g i v e n now t h a t
u(X)
n 2 no
for all
n
a <
=
.
m
.
. Hence
M (2r ) S 1 Q n
s a t i s f i e s a A -condition. 2 Assume t h a t 0 does n o t @
T h i s i m p l i e s t h a t , f o r e v e r y number s a t i s f y a A -condition f o r large u 2 u > 0 and e v e r y n = 1 . 2 , . , t h e r e e x i s t s a number u t u such t h a t n n Q(2un) > n Q ( u n ) Again, l e t Ca = 1 and Cna = . For n = l , 2 , , n determine u such t h a t Q(ui).(a/Z ) = an and t h e n d e t e r m i n e u 2 u n n n a s i n d i c a t e d above. Now, l e t s a t i s f y Q(un).Sn = an (hence f3 5 a / 2 ) .
..
N
.
-
N
F i n a l l y , choose t h e d i s j o i n t s u b s e t s p(En)
=
fin
f o r every
. Note
n
Finally, define the functions
as above t h a t
p (r )
@
n
+
0
E
(n=l,2,
N
...
N
...)
of
X
t h a t t h i s can b e done s i n c e f
, but
rn ( n = l , 2 ,
and
M (2r ) =
a
n
m
such t h a t C 5
S a = p(X). n ) as above. I t f o l l o w s
...
for all
We add some remarks. A s n o t e d b e f o r e , t h e s p a c e
n Lo
.
i s s u p e r Dedekind
complete (and, t h e r e f o r e , Lo
i s o r d e r s e p a r a b l e ) . This i m p l i e s t h a t
coincides with the space
,
:L
c o n s i s t i n g of a l l
t y t h a t any downwards d i r e c t e d s y s t e m
L: f E LQ h a v i n g t h e proper-
{uT} s a t i s f y i n g
If/ t u
+
0
also
Ch. 19,
§
1331
. Furthermore,
p (u ) t 0
satisfies
O
T
one may ask f o r e x p l i c i t examples. Of
constant, 1 < p
course, O(u) = Cup (C > 0
<m
La = LO holds. I f O(u) = e U - u - 1
which
0
Li
the space
597
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
) i s a standard example f o r
(and
u
i s f r e e of atoms),
conhains many, b u t n o t a l l , functions of
LO
. In
fact,
L:
i s the norm c l o s u r e of t h e s e t of a l l measurable functions t h a t a r e bounded and vanishing o u t s i d e a s e t of f i n i t e measure (Theorem 133.2). However, since
, we
does n o t s a t i s f y a A -condition f o r l a r g e u 2 A f u r t h e r problem a r i s e s i f i s a proper subset of LO O
.
see t h a t
Lt
jumps. A p a r t i a l
Y
answer i s contained i n t h e following theorem.
u is f r e e of atoms and if
THEOREM 133.5. If L;
.
= {O}
PROOF. Assume t h a t
i m p s , then
'4
contains a f u n c t i o n t h a t i s not zero almost
everywhere. Then (since
L; L;
function of some s e t
of f i n i t e p o s i t i v e measure. The f u n c t i o n
i.e.,
E
i s an i d e a l ) L; v >0
t h e r e e x i s t s a number
,
f = 2v0xE
Let
let
0
(En:n=1,2,
p o s i t i v e measure such t h a t f t u
0
j.
and
f E L;
n %(un) < 1 ) f o r a l l = Y(2vo) u(En)
=
E
, we
Note t h a t i f
0
t
have
such t h a t
v > v
be a sequence of s u b s e t s of and l e t
py(un)
+
un = 2voxE 0
, so
p
5
0 of
E
for a l l
($ ) Y n
jumps,
Y
for a l l
m
. Since
n
1 (and hence
no . On the o t h e r hand M $ l n ) = . Contradiction. Hence L: = { O } .
n
has atoms, t h e l a s t theorem does not hold any longer
p
( i f , f o r example, Ly,
i s t h e sequence space
,
then
L;
i s the spaee
of a l l sequences converging t o zero).
(co)
We r e p e a t t h a t the o r d e r s e p a r a b i l i t y of =
Y(v) =
n exceeding some for a l l
m
... )
contains the c h a r a c t e r i s t i c
(LO):
,
i.e.,
t h e band ( i n
L0
implies t h a t
(LO): =
L* ) of a l l o r d e r continuous l i n e a r f u n c t i o n a l s 0
i s the same as t h e band of a l l o-order continuous l i n e a r f u n c t i o n a l s . S i m i larly for
if
LY
. The
o r d e r s e p a r a b i l i t y a l s o implies t h a t
Li = L r
If1 t u t 0 implies p,(un) t 0 n f o r any downwards d i r e c t e d s e t If1 2 u C 0
f E LO has the property t h a t
then a l s o
p,(u,)
larly for
Ly
i f and only i f space
Lk
, we
t 0
. Recall
now t h a t i n Theorem 102.7 i t was proved t h a t
L* = (LO): O
see now t h a t
words, the norm i n l i n e a r f u n c t i o n a l on
,
. Since L
0
= Li
L F = Li
and
,
. SimiLO = L r
i s the associate
(La):
i f and only if
i.e.,
L* = 0
L i . In o t h e r
i s o r d e r continuous i f and only i f every continuous
LO L O i s o r d e r continuous. A s before ( f o r example i n
598
THE BANACH DUAL OF AN ORLICZ SPACE
s e c t i o n 102) we denote t h e d i s j o i n t complement of L$
by
on
L0
.
(Lo):
and we c a l l
Similarly f o r
(Lo):
L* = L‘ d (Lo): 0 @
and
L*y = LI;
A s we have s e e n , t h e r e a r e cases L* = L‘
. Then
in
L& = (L@): = (LJz
t h e space of 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
. Hence
LY
19,51333
Ch.
.
(L ) * Y s
(B
(A2-condition)
i n which
L
any continuous s i n g u l a r l i n e a r f u n c t i o n a l on
f u n c t i o n a l . There a r e a l s o c a s e s , however, i n which
=
0
Lo
(Lo):
,
Li
i.e.,
is the n u l l
does not c o n s i s t
of t h e null f u n c t i o n a l o n l y . This happens i n p a r t i c u l a r i f t h e measure i s f r e e of atoms and
@
. As
only f o r large
u
prove now t h a t
(Lo):
elements of = P*($~) +
(Lo):
.
does n o t s a t i s f y a A - c o n d i t i o n , 2
i s an AL-space,
and
space i f and only i f
v u
> v
p
i.e.,
if
i s t h e norm i n
Lt
.I 0
,
t h e band
then
u1
and
in
u2
L+
THEOREM 133.6. “he OrZicz space
an AL-space.
L0
S i m i l a r l y for
LY
. We
0
or
shall
+$ ) = 1 2 L i s a normed L*
i f and only i f
with
p(uI)
l i m p(vn)
t h e number
u z
$2 a r e p o s i t i v e p*($
i.e.,
= p(u2)
i s an AL-
is a semi-M-space.
has
L
and
= 1
l i m p(vn)
satisfies
1 2 - n prove, t h e r e f o r e , t h e f o l l o w i n g theorem.
(LQ)E i s
and
,
i n t h e Banach d u a l
Lz
i s a semi-M-space,
L
the property that for u
- u - 1
P * ( $ ~ ) According t o Theorem 119.2, given t h a t
Riesz space w i t h norm
for
p*
for a l l
0 ( u ) = e’
an example we mention
S
1
. We
EquivaZentZy,
.
PROOF. As b e f o r e , we denote t h e Luxemburg norm and t h e O r l i c z norm i n Lo
by
p0
f o r every f E Lo
and
r e s p e c t i v e l y . By Theorem 132.2 we have
p i
f E L0
. Hence
and by L e m a 132.1 we have
Assume now t h a t t h e non-negative such t h a t
MO(uI) in u
L0
5 1
.
functions
p O ( f ) i p{(f)
p i ( f ) 5 1 + M@(f) f o r every
u1
and
in
u2
L0
a r e given
p (u ) = p (u ) = 1 Hence, i n view of Theorem 131.8, 0 1 0 2 and M (u ) S 1 Let (vn:n=1,2, ) b e a sequence of f u n c t i o n s
.
such t h a t
i s equal t o
0 2 u , v u2 Z vn J. 0
u1
complementary s u b s e t
.
...
Denote
on a c e r t a i n s u b s e t X-A
. Hence
A
u 1 v u2 of
X
by
u
. The
and e q u a l t o
u2
function on t h e
ch. 19,81331
C 0
Since
u
hence
Q[v ( x ) ] C 0
2
559
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
v
(i.e.,
J. 0
v
f o r almost e v e r y
x ) , i t f o l l o w s now t h a t
,
X
p o i n t w i s e almost everywhere on
and
MQ(vn) C 0
(dominated convergence), and t h e r e f o r e l i m p (v )
a
n
< 1 + l i m M (v ) Q n
= 1
.
LQ i s semi-#.
This shows t h a t
The f i r s t t o o b s e r v e t h a t f o r a n O r l i c z s p a c e an AL-space was T. Ando
t h e space
(Lo):
is
([21, 1960). A s observed i n s e c t i o n 119, t h e more
g e n e r a l p r o p e r t y t h a t f o r a normed R i e s z s p a c e an AL-space
Lo
i f and o n l y i f
L
i s a semi-M-space
L
t h e band
L:
in
is
L*
i s due t o E. de Jonge
([21,1977). EXERCISE 133.7. I f
Q(u)
and
Y(v)
a r e complementary Young f u n c t i o n s ,
i t may be asked whether always one o f t h e s e s a t i s f i e s a A - c o n d i t i o n , 2 least for large u o r v Show t h a t t h e answer i s n e g a t i v e .
.
...
H I N T : Let 1 < M < M2 such t h a t l i m M 1 @ ( u ) c o n s i s t of l i n e s e g m e n t s c o n n e c t i n g t h e p o i n t s
...,, where
u3 = M2u2
and
-.
v3 = 2v2:u4 = 2u3
, Y(v 3 )
EXERCISE 133.8.
> iM2Y(v2)
( i ) Let
a p-measurable s u b s e t of
and
v 4 = M3v3
, and
v2 = M v * 1 1 ’ and s o on. Show t h a t and
s o on.
b e Lehesgue measure i n
p
. Assume,
lRn
,
L e t t h e graph of
(0,0),(u,,v,),(u,,v2),
a r e a r b i t r a r i l y p o s i t i v e ; u2 = 2 u l
u 1, v l
@(u,) > i M l @ ( u l )
=
at
furthermore, t h a t
.
and l e t
IE? @
X
be
i s a Young
Show t h a t f o r any f u n c t i o n s a t i s f y i n g a A - c o n d i t i o n . Let f E LQ(X,p) 2 g i v e n E > 0 t h e r e e x i s t s a c o n t i n u o u s f u n c t i o n g p o s s e s s i n g a compact c a r r i e r such t h a t
pQ(f-g) <
pQ(f(x+h)-f(x)) HINT:
+
0
as
E
.
X = IRn,
( i t ) Assume now t h a t h
+
0
i.e.,
f E Lo (lRn,p)
.
( i ) There e x i s t s a s t e p f u n c t i o n on i n t e r v a l s
. Show t h a t t (i.e.,
t
f i n i t e l i n e a r combination of c h a r a c t e r i s t i c f u n c t i o n s of i n t e r v a l s i n
is a lRn )
600
.
p m ( f - t ) < 4s
such t h a t
a continuous
g
It i s n o t d i f f i c u l t t o s e e now t h a t t h e r e e x i s t s
(with a compact c a r r i e r ) such t h a t > 0
,
such t h a t
E
f o r every
h
. Finally,
EXERCISE 133.9.
<
--E
p@(g(x+h)-g(x)) <
f
E La
F(t) =
( i ) Let
3
for
and
(En,p)
/
g
E Ly (lRn,p)
I
f(t-x)g(x)dx =
s u f f i c i e n t l y small,
0
.
lRn
and l e t
and
@
s a t i s f i e s a A -condition. 2 Show t h a t
f(x)g(t-x)dx
lRn
i s a. c o n t i n u o u s f u n c t i o n on ( i i ) Assume t h a t
h
b e Lebesgue measure i n
En
as
1
-E
b e complementary Young f u n c t i o n s such t h a t
Assume t h a t
.
$E
w i t h compact c a r r i e r
1;
i s uniformly c o n t i n u o u s .
g
g
.
po(f (x+h)-g(x+h)
since
p@(t-g) <
there e x i s t s a continuous 1 Then p@(f(x)-g(x)) < JE
( i i ) Given
Y
Ch. 19,91331
THE BANACH DUAL OF AN ORLICZ SPACE
Y
P.
a l s o s a t i s f i e s a A -condition. 2
Show t h a t
F(t)
+
0
t + m .
HINT: (i) Note t h a t
( i i ) Assume f i r s t t h a t
where we have
to
+ m
+ m
, we
and
At t
or
+ m - m .
Bt
n = 1
t e n d t o z e r o as
. For
t
+
t > 0
m
w e have
. Similarly
if
.
t < 0
i f and o n l y i f one a t l e a s t of t h e c o o r d i n a t e s of I f , f o r example, t h e f i r s t c o o r d i n a t e
w r i t e ( a s above)
tl
of
t
If t
n > 1, tends
tends t o
Ch. 19,91341
OBLICZ SPACES AND IRREDUCIBLE OPERATORS
60 1
134. The s p e c t r u m of an o p e r a t o r i n Banach space I n t h e s e p r e l i m i n a r y remarks, l e t
b e a Banach space ( n o t necessa-
V
r i l y a Banach l a t t i c e ) ; we d e n o t e t h e norm of an element
As
usual, the series
s
i f t h e p a r t i a l sums
series
1;
1; s
fn =
(all
f
fk
1;
.
/If 1 1
by
V ) i s s a i d t o converge w i t h sum
E
IIs-snI/ + 0
satisfy
i s s a i d t o converge a b s o l u t e l y i f
fn
f E V as
n
. The
-t
) / f n J I converges. It
1;
i s e v i d e n t t h a t a b s o l u t e convergence i m p l i e s convergence. The s e r i e s Cz=o
anzn
(all
a
E V
and
complex) i s c a l l e d a power s e r i e s w i t h
z
V. E x a c t l y a s f o r a power series w i t h complex c o e f f i c i e n t s
coefficients in
t h e r e e x i s t s a number
R (05RSm)
,
t h e r a d i u s of convergence of t h e power
s e r i e s , such t h a t t h e s e r i e s converges (even a b s o l u t e l y ) f o r diverges for
IzI > R
.
and
IzI < R
The r a d i u s of convergence depends of c o u r s e on t h e
coefficients ; precisely ,
0-I
with the usual conventions about
and
m
. The
-1
t h e same as i n t h e complex c a s e . Assume now t h a t c o n t i n u o u s ) o p e r a t o r mapping
.
b e denoted by
llTll
and t h e r e f o r e
/ I T n I / l / n 5 IlTll
i s a bounded ( i . e . ,
i n t o i t s e l f . ThB o p e r a t o r norm of
V
Note a l r e a d y t h a t
. It
Denote t h e i d e n t i t y o p e r a t o r i n
A # 0
T
proof i s p r a c t i c a l l y
V
llTnll 5 l l T i l n
for
T
n = 1,2,
will
...
,
follows t h a t
by
I
. We
a s k f o r which complex
the series
-A
-1
I - A
-2
T - h
-3 2 T
-...
converges. Note t h a t t h i s i s a power s e r i e s i n t h e Banach s p a c e
A
-1
with coefficients i n
of a l l bounded o p e r a t o r s mappin?,
B(V)
V
into itself.
The formula f o r t h e r a d i u s of convergence shows t h a t t h e series converges for
/ A / > l i m ~ u p / / T ~ (and / / ~ h/ e~n c e s u r e l y f o r
values of f o r e , maps
,
A V
d e n o t e t h e sum of t h e s e r i e s by
/ A / > //TI/ ) . For these
S
. The
operahor
i n t o i t s e l f . A s i m p l e c a l c u l a t i o n show t h a t
,(,-AT>
=
(,-A,>,
= I
.
S
,
there-
602
SPECTRUM OF AN OPERATOR / A / > l i m s ~ p ( / T ~ l I, l t/h ~ e operator
Hence, f o r verse
(T-AI)-'
in
,
B(V)
The s e t of a l l complex
i s c a l l e d t h e resolvent s e t of
all
. From what
o(T)
1x1
satisfying
A
number, a l l numbers
T-XI
h a s a bounded i n v e r s e i n
all
An
1x1
f o r which
t h a t i n t h i s case
A
. The
p(T)
f o r which
A
B(V)
complementary does n o t
T-XI
) i s c a l l e d t h e spectrwn of
B(V)
T
and de-
we have observed a l r e a d y , i t i s e v i d e n t t h a t
> IITi/
belong t o
IAn/
satisfying
An
h a s a bounded i n -
T-A1
and denoted by
T
s e t ( t h a t i s t o s a y , t h e s e t of a l l complex p o s s e s s a bounded i n v e r s e i n n o t e d by
19,51341
g i v e n by
f o r which
A
Ch.
> IITnlll/n
. Hence,
p(T)
if
belong t o
. It
p(Tn)
is a natural
n
> l/Tn/l belong t o
p(Tn)
,
i.e.,
f o l l o w s t h e n from
E p(T) , b e c a u s e
We u s e h e r e t h a t t h e f a c t o r s on t h e r i g h t i n (2) commute. T h i s f o l l o w s from the general observation t h a t i f and
S1
-1
on b o t h s i d e s by some and
n , then An
SIS2 = S2SI
-1 S l Sq =
h a s a bounded' i n v e r s e , t h e n S1 A
for
Sl
s~s;'
) . We have proved t h u s t h a t i f
. Applying
E p(T)
i n s t e a d of t o
T
and
A
,
and
in
S2
(multiply / A / > /IT
s2sI for
11
t h e same r e a s o n i n g as above t o
i t follows a l s o t h a t f o r
B(V)
Tn
/ A 1 > IITn/Il/n
we have -A-nI
- A
-2n n T A
M u l t i p l y i n g b o t h s i d e s by holds f o r a l l satisfying
1x1
A
Tn-
satisfying > infnllTn/II'"
+ ATn-'
+
... +
An-'I
A 1 > / I T n / l * / n . Hence,
. Comparing t h i s
, we
( I ) holds f o r a l l
to
infnIITnII
set
p(T) o f
(I)
that
T
.All
. As
A
such t h a t
A
w i t h t h e formula f o r t h e n I/n l i m supIITnll 5 i n f IIT I / .
r a d i u s of convergence, we conclude t h a t n I/n r ( T ) = limllT / I e x i s t s and t h e number
T h i s shows t h a t
see t h a t ( I )
I A / > r(T)
r(T)
i s equal
belong t o the resolvent
a f i n a l remnrk, h o t e t h a t i t f o l l o w s immediately from
Ch.
19,11341
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
f E V
for a l l
t h e norm i n
V
, where
t h e s e r i e s on t h e r i g h t converges w i t h r e s p e c t t o
.
A s s e e n above, t h e s e t set
.
p(T)
(A:IAlSr(T))
i s a s u b s e t of
r(T)
. We
b r i e f l y r e c a l l t h e proof t h a t
i s an open s u b s e t of t h e complex p l a n e . F o r convenience, we i n d i c a t e
p(T)
(T-AI)-'
RA
by
. The
LEMMA 134.1. If m
RA = C (A-Ao) 0
f o l l o w i n g r e s u l t h o l d s now.
A.
E p(T)
and
hoZ& i n
n R n+l AO
PROOF. The convergence i n
Writing
S = C(A-AO)n
R n+l
AO
hence (T-AoI)[I-(A-Ao)RAo
i.e.,
[(T-A = I
S(T-AI)
S = R
a(T)
g e n e r a l , t h e r e w i l l a l s o be p o i n t s of t h e r e s o l v e n t
s e t i n s i d e t h e c i r c l e of r a d i u s
and
i s a s u b s e t of t h e r e s o l v e n t
(A:lAl>r(T))
I n o t h e r words, t h e spectrum
. In
603
0
I)-(A-A
. It
B(V)
, we
I
IA-AoI B(V)
< I I R A ~ -I 1/
. Hence
p(T)
, then A E
p(T)
is open.
of t h e s e r i e s f o l l o w s from
have
S = I
,
.
)I]S = I , and t h e r e f o r e (T-11)s = I Similarly, 0 f o l l o w s t h a t S i s t h e i n v e r s e of T - A 1 , i . e . ,
A' COROLLARY 134.2.
...) .
(An:n=1,2, IIRA 11 0
+ m
If A.
E a ( T ) and if there e x i s t s a sequence o f compZex numbers in p(T) such t h a t An + A. , then
-
PROOF. I f does n o t h o l d , we may assume ( p a s s i n g t o a \ I R A 11 0 subsequence i f n e c e s s a r y ) t h a t IIRA 1 1 5 M f o r some M > 0 and a l l n n -f
604
SPECTRUM OF AN OPERATOR
Then
. It
< [IRA
for
< M-I
Ihn-hoI
n
Ch. 19,91341
s u f f i c i e n t l y l a r g e , which implies
follows from t h e l e m a above t h a t
ho
Iho-XnI
.
E p(T)
<
Contra-
dictio:.
has
R
(T-XI)-'
=
X
. It
/ A / < r(T)
1x1
as i t s sum f o r
> r(T)
m
series
-IoA
-(n+l) n T
and diverges f o r
can be proved now ( s i m i l a r l y as f o r a s e r i e s with complex r(T) > 0
coefficients) that for spectrum
. The
r(T) = l i m l l T n / ( ''n
We r e t u r n t o the number
o(T)
t h e r e m u s t be a t l e a s t one p o i n t of t h e
on the boundary of the c i r c l e of r a d i u s
. The
r(T)
s m a l l e s t closed c i r c u l a r d i s c around the o r i g i n which contains the spectrum o(T)
has t h e r e f o r e
the spectral radius of t h a t t h e spectrum of of
c o n s i s t of
T
as i t s r a d i u s . This e x p l a i n s why
r(T)
X
. We
T
=
f i n a l l y r e c a l l t h a t i t can a l s o be proved
i s never empty. Hence, i f
T
0
is called
r(T)
, the
r(T) = 0
spectrum
only. To conclude t h i s preliminary s e c t i o n we
p r e s e n t a simple lemma, a p a r t i c u l a r case of a much more general theorem (the s p e c t r a l mapping theorem).
o(T2)
LEMMA 1 3 4 . 3 . ( i ) The spectrwn
which
.
E o(T) ( i i ) Let ( A o \ A
consists of a21
to
2
and Let f o r which
T - XI
( i ) I f both
PROOF.
T2 -
> r(T)
A(AO-h)-I
of
T2
S = T(A I-T)-I 0 h E o(T)
and
.
T + XI
p(T)
,
then
belongs t o
X2
,
A
2
p(T )
T - XI
then
has
. Conversely, A
1
=
X2
for
. The spectrwn of
S
have bounded i n v e r s e s , then
X I has a bounded inverse. I n o t h e r words, i f
bounded i n v e r s e
consists o f a l l
X
and
-X
if
T2
- X
2
belong I
has a
(T+AI)A as inverse. Note t h a t
(T+XI)A = A(T+AI) , s o t h a t indeed A (T-XI) = 1 2 2 = (T-XI)AI = I Hence, X E p(T ) i f and only i f both X and - A belong 2 2 I n o t h e r words, h E o(T ) i f and only i f one a t l e a s t of X t o p(T) (T+AI)A i s bounded and
.
.
and
-X
belongs t o
( i i ) Assume i.e.,
let
o(T)
X # Xo
S 1 = S-X"I
commute; t h e r e f o r e , S ,
X 0 I - T and
(XOI-T)-I
-
.
and
write
X w = A ( A -A)-' 0
. Let
X- E p ( S )
have a bounded i n v e r s e . Note f i r s t t h a t and
T
commute. It follows t h a t
s1
9
S -1
s1
,
and
,
a l l mutually commute. The following a s s e r t i o n s
a r e now e q u i v a l e n t .
A- E p ( S ) (S-A-I)-' e x i s t s c9 [(S-X-I)(XoI-T)]-l CT-X(XO-A)-l(XOI-T)I-' e x i s t s (9 C(XO-X)T-X(AOI-T)I-1
exists, i.e., exists, i.e.,
T
Ch. 19,§1351
ORLICZ SPACES AND IRREDUCIBLE OPERATORS exists
[XO(T-XI)]-’
X E p(T) we have
I)
X # Xo
Hence, for
605
. A F: p(T)
X(XO-A)
if and only if
-1
p(S).
is of the form
It remains to investigate whether every complex number -1 X(Ao-X) for some complex X For A- # -1 this is true (take
.
X
= A
0
X-(A-+l)-’ A-
number
) . This leaves
=
-1
1-
-
X
=
. As
-1
immediately visible, the
belongs to the resolvent set of
therefore, that the spectrum of
.
which X E a(T)
S
S
. The final result is,
consists of all points
A(XO-A)
-1
for
135. The Krein-Rutman theorem and the spectral radius Let L L
be a complex Banach lattice. For any bounded operator T
, a(T) , r(T)
p(T)
the notations
and
RX
=
(T-XI)-’
in
have the same
meaning as in the preceding section. T be a p o s i t i v e operator i n
LEMMA 135.1. Let
L
(therefore, T i s
bounded). Then t h e following holds. (i)
If
1x1
r
>
(ii) The n m b e r PROOF. Writing +
, then llRhfli 5 IIRIXI(IfI) 11 holds f o r every 1 1 ~ 1 15 1/RlXl/I -
=
r(T)
r
=
. It follows t h a t
f E L
RXf in norm as
Sk(X)
k
+
It follows that
]lSk(X)(f)ll 5
/IRXf11
5
] / R II (AI f 1 ) RX
operator norms of
-(h-lI+X-’T+.
=
f 5
for every
m
we obtain
r(T) bezongs to
and
r
. Let
An
+
Xo
.
. .+X-(k+’)Tk) L . Also
, we
.
]~Sk(lA~)(/f~)]~Letting k
I] .
R I A l satisfy
such that
A.
(An[
/IRhII 5 I / R l h /I l
have
.
f E L , the
in the spectrum o(T) >
r
for all n
.
Sk(h)(f)
tend to infinity,
Since this holds for every
(ii) There is at least one point [AO[ =
o(T)
and
satisfying
(An[ 4 r
. All
A n belong, therefore, to the resolvent set p(T) I n view of Corollary 134.2 we have /IRA 11 + m Then also / / R I X 11 + m (since I/RX I/ 5 IIRIX I1 ) . Since n it is impossible (by what was proved in Lemma 134.17 that r lXnl 4 r
.
belongs to
p(T)
.
Hence, r belongs to
o(T)
.
606
THEOREM 135.2. (M.G. Krein-M.A.
s i t i v e and compact operator i n e x i s t s a p o s i t i v e element
Rutman, [11,1948). Let
L
r
given t h a t
An
. Then
r
j.
llRX
-.
tence of a sequence
113 frill +
and
.
n
I( +
11%
n
...)
in
Note now t h a t
vn/l +
= -
n u t 0 n
and
11%
,
= Ifn/
nk
...)
(k=1,2,
. Let
m
,
, we
from
Tu
+
have
u
a s well a s t o
Since
r > 0
for a l l
lISn(If n l ) /I
I/v 11 5
for a l l
1
.
n
for a l l
T
n
.
Furthermore,
i s compact, t h e r e e x i s t a subsequence
converges i n norm t o some element
ru
i s p o s i t i v e . Since
+ u.
, we
Then
rTu
+
find that
rTu ru
+ Tu
. Hence,
Tu = r u
by hypothesis, we have
. On
k > 1
.
-
Tu
ru
u (I L +
0
. All
Tun
and
t h e o t h e r hand i t follows
since
rTu
converges t o
Tu
. Furthermore,
~~u~~ > 0
. This
We r e f e r t o Exercise 135.10 f o r t h e case t h a t not compact f o r some
5
and
1
which converges i n norm. Passing t o t h e subsequence, we
n that also
ru
II%nf, I I
hence v t 0
5
is a p o s i t i v e
now
for a l l
are positive; therefore, u + u
IlfJl
= -(T-X~I.)-~
vnII n
Ilunl; = 1
Tun
may assume t h a t Tu
T
p o s i t i v e . For t h e proof,
such t h a t
L
-pA
we have
which tends t o zero i n norm. Since Tu
.
by Corollary 134.2. This implies t h e exis-
m
R~ vn
Then
there
Tu = r u
i t follows already from t h e , l e m a above t h a t t h e
(f::n=1,2,
Writing v
and a l s o
n
. Then
and
i s compact) the spectrum of
T
operator (sge’formula (1) i n s e c t i o n 134 for a l l
be a po-
only of eigenvalues. Hence, s i n c e i t i s
t h e r e e x i s t s a corresponding eigenelement t h a t i s let
T
i s an eigenvalue. The s p e c i a l property t o be proved here i s t h a t
r
number
,
0
= r(T)>
,
0
=
= r(T) > 0
u # 0
suck t h a t
PROOF. Note f i r s t t h a t (because
X
r
suck t h a t
L
in
u
c o n s i s t s , besides perhaps
n
a. 19,91351
KREIN-RUTMAN THEOREM AND SPECTRAL R A D I U S
T
shows t h a t i t s e l f but
u # 0 Tk
. is
Ch. 19,51351
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
607
The proof of t h e Krein-Rutman theorem a s p r e s e n t e d h e r e i s e s s e n t i a l l y B o n s a l l ([11,1958).
due t o F.F.
The remaining p a r t of t h e s e c t i o n w i l l b e
is a positive operator i n
To
devoted t o t h e r e s u l t t h a t i f
o r d e r c o n t i n u o u s and compact and
0
such t h a t
,+
T
,
To
then
which i s
L
i s a n upwards d i r e c t e d system
(T,:TE{T})
.
I t i s e a s y t o see t h a t k k r ( T ) I r ( T ) f o r a l l T , e i t h e r by o b s e r v i n g t h a t /ITT1l 5 IITOll f o r a l l 0 k = l,Z, o r by means of t h e s e r i e s e x p a n s i o n of RA f o r I X i > r ( T o ) 5
r(T,)
1- r ( T o )
.
...
By a s i m i l a r argument we s h a l l s e e t h a t
i s upwards d i r e c t e d . There
r(TT)
w i l l b e s e v e r a l lemmas, and f o r b r e v i t y we w r i t e for
R(TT)
.
ro
for
r(To)
r
and
F i r s t , however, w e r e c a l l D i n i ’ s theorem ( i n t h e form a s w e
s h a l l need i t ) . L e t X b e a compact s u b s e t of a m e t r i c s p a c e ( o r e v e n , more gene-
let (f
rally,ofatopologicalspace)and of non-negative
T h e n t h e convergence i s u n i f o r m , i . e . , w r i t i n g ( I f T ( (j. 0
be a downwards d i r e c t e d s y s t e m
:T€{T?)
f (x)
c o n t i n u o u s f u n c t i o n s on X such t h a t
. We
indicate
t h e proof. L e t
llf,ll > 0
E
+0
we have
= max(fT(x):xEX),
x E X
b e given. F o r each
i n t h e system (f,) s u c h t h a t f - r ( x ) (x) < E (X) Hence, by c o n t i n u i t y , t h e r e i s a n open neighborhood Ux of x such t h a t there i s a function
f
T
(x)
(y) <
(y) <
E
T
y E U
for a l l
E
from t h e system f
f
(f,)
for a l l
. Cover
such t h a t y E X
TO
. Hence
fT
X
by
jfTll <
E
LEMMA 135.3. A s already mentioned, l e t L
in
To
such t h a t
,...,U
U
2 inf(f
T
xn
x1
and t a k e
.
)
( x l ) * * * y f T (xn) for all f 2 f
To
‘0
f
Then
.
‘0
.
be a p o s i t i v e operator
i s order continuous and compact and l e t
be an upwards d i r e c t e d system o f operators i n
L satisfying
(TT:~€{~)) 0
2
TT
+
To
Then
PROOF. L e t V
i 0
, and
0 I u E L
therefore
compactness of
To
T v 0
be given. Writing T
i 0 (since
To
i m p l i e s t h a t t h e system
v
= (T -T )u O T
, we
have
i s o r d e r c o n t i n u o u s ) . The (TovT)
i s Cauchy (because
E > 0 and a d e c r e a s i n g sequence v i n (v,) Tn v -T v I/ > E f o r a l l n , and hence 1lTovT -T Ov T /I > ‘IT0 T n 0 T n + l > E for m = l,Z, i n view of v I v ; on t h e o t h g r ‘n+m ‘n+l t h e sequence (T v ) must have a converging subsequence). Choose now a 0 Tn sequence (En:n=l,Z, ) of numbers such t h a t E i 0 . S i n c e (TOvT) i s
otherwise there e x i s t s such t h a t
...
...
Cauchy, t h e r e e x i s t s a sequence
(vT ) n
in
.
f o r each x E X
(v,)
such t h a t
vT i n
and
.
608
KREIN-RUTMAN THEOREM AND SPECTRAL RADIUS
E v i d e n t l y , (vT )
i s a Cauchy sequence. L e t
sequence. Fromn(l) w e s e e t h a t
w
shows t h a t
w = inf Hence
for a l l
5 E
v
T
5 v
(TovT)
(by Theorem 100.8). A s observed above, w e have
, i.e.,
IITOvT\I C 0
.
19351351
b e t h e norm l i m i t of t h e
w
i s t h e norm l i m i t of t h e d i r e c t e d s y s t e m
T v
O T w = 0 T
IITov T -w\l
-
T
?
. This
Therefore,
In o t h e r words, IITO(TO-TT)ull C 0
LEMMA 135.4. Under t h e same conditions for
.
To
and
TT
.
TOvT C 0
.
we have
~ / T ~ ( T ~ - T , ) cT 0~ / / PROOF. L e t
S
be the subset
b e i t s image under of
L+
. We
. The
To
( u : u X l , ~ ~ u ~ ~ of 5l) L
norm c l o s u r e of
d e n o t e t h i s compact s u b s e t by f T ( x ) = (IT 'T
o\
0
-T 'x(I
TJ
X
T (S) 0
. For
and l e t
TO(S)
i s a compact s u b s e t every
x
E X we have
C O
by t h e l a s t lemma. On account o f D i n i ' s theorem t h e convergence i s uniform,
i.e.,
Since
T u 0
for
u E S
belongs t o
X
,
t h i s implies t h a t
T h e r e f o r e , we have a l s o
f
f
s u p \ j ( T T -T O\ o
)T TJ
EXAMPLE 135.5. L e t
Let
e(x)
all
x
hn I. e
and
f / / : f E L , ~ ~ f ~ I S l 0' C,
/
L = L-([O,Il,p)
n
hn(x) = 1
, but
Ile-hnII Tof =
A!
for = 1
...)
, where
p
i s Lebesgue measure.
be d e f i n e d on C0,Il by e ( x ) = I f o r -1 0 2 x 5 I-n and h n ( x ) = 0 e l s e w h e r e . Then
(h (x):n:n=l,Z,
and
d e f i n e d by
o
for all
To(x,y)f(y)dy
n . . The p o s i t i v e o p e r a t o r
, where
the kernel
TO(x,y)
To:
L
-f
L
is
i s g i v e n by
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
ch. 19,51353
609
. Hence
TO(x,y) = e ( x ) e ( y )
I
1
T f = e(x) 0 Note t h a t then
f(y)dy
.
0
i s compact and o r d e r c o n t i n u o u s ( i f
To
u (x) C 0
u
+
1 uT(x)dx 4 0
a l m o s t everywhere; hence
0
,
S i m i l a r l y , t h e p o s i t i v e o p e r a t o r Tn: L + L i s d e f i n e d by 1 Hence Tnf = T n ( x , y ) f ( y ) d y , where Tn(x,y) = h n ( x ) e ( y )
I,
in
L = Lm
i.e., T u
O'T
,
C 0 ).
.
1 Tnf
= hn(x)
f(y)dy
.
0 f t 0
For
we have
T f f Tof n
(i.e., 1
I. I
e(x)-hn(x)
O n
0
T f To ) and n f(y)dy ;
hence
It follows t h a t f 2 0
l e t again a
. Then
T f = 0
IIT -T
O n
/I
ae
and
2 T f = ae(x). 0
T f
I
e(y)dy = ae(x) = T f 0 0 1
T T f = ah (x).
n O
hence (T -T
n
'
J
11 O n 0
To
t h e l a s t lermna,
,
e ( y ) d y = a h n ( x ) = Tnf
,
0
.
) T - f = (T -T ) f = a(e-hn) But t h e n 11 (T -T )T f 11 = a , i . e . , O n O n 0 = 1 for all n This shows t h a t i n t h e l a s t lemma t h e f a c t o r
01-10
ll(T -T )T
. To s e e t h e r e l a t i o n w i t h L . For b r e v i t y , denote = ah . It follows t h a t
= 1
b e given i n
i n f r o n t of
.
(TO-TT)TO cannot be o m i t t e d . I n accordance w i t h t h e lemma
we f i n d i n o u r example t h a t T (T -T T ' o o n!
of
= T
0
o
nf
)dy = n - 1
= a e ( x ) "(e-h J
/T -T 'f
O\
n
=
aT (e-h ) = O n
ae(x)
,
KREIN-RUTMAN THEOREM AND SPECTRAL RADIUS
610
hence
IIT ( T -T ) T
O O n O
f l l = n-la
and
IIT ( T -T
)T
O O n O
PROOF. M u l t i p l y i n g b o t h s i d e s of ( 2 ) by TT -
11
=
n
-1
19,51351
Ch.
+
0
n
as
+
-.
on t h e l e f t and by
To - XI
XI on t h e r i g h t , t h e l e f t hand s i d e becomes
and t h e r i g h t hand s i d e becomes
Hence, t h e r e i s e q u a l i t y i n ( 2 ) . LEMMA 135.7. Once more, l e t
0 5 TT
f
with
To
continuous. Furthermore, l e t t h e p o s i t i v e number vent s e t
p(TO)
of
and l e t
To
are uniformly bounded, i . e . ,
> sup r T
X
supT / I R A ( T T ) I I <
To
belong t o t h e resol-
A
. Then the .
-
A > sup r
m
0
=
A -(n+l)T:
; therefore
2
) w e have
A
(T ) f T
. It
follows
( I I R X ( T T ) I l ) o f numbers i s upwards d i r e c t e d . Assume now t h a t
t h a t the system
SUP~IIR~(T~)I/
= Zo
nurnbers I I R A ( T T ) I I
-:
PROOF. Note f i r s t t h a t (on account of -RA(TT) = (XI-TT)-I
conpact and order
m
. Then,
and, by L e m a 135.4,
trivially,
Ch. 19,51351
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
611
Hence, by a d d i t i o n and using t h e i d e n t i t y ( 2 ) i n t h e l a s t lemma,
R (T ) ] I $ 0 O A 0
/ / R (T )II-'.ilT
Since, t r i v i a l l y ,
A
T
, we
see now t h a t
(3) As observed a l r e a d y , -R S
-To
RA(TT)
A
(T )
. Therefore,
i s a p o s i t i v e o p e r a t o r ; hence
T
-T R (T ) 5 T i
from ( 3 ) ,
T
(4) Note now t h a t T R ( T ) = (T -AI)R (T ) T
i
A
A
T
+ AR (T A
T
) = I
+ AR (T A
T
)
. Therefore,
The l e f t hand s i d e tends t o zero i n norm by ( 4 ) , the same holds f o r the f i r s t term on the r i g h t . Hence, the r i g h t hand s i d e tends i n norm t o
This would imply t h a t It follows t h a t
THEOREM 135.8. Let
Banach l a t t i c e 0 S TT 4 To
A = 0
.
Contradiction, s i n c e
sup//RA(TT)II <
L
To
and
such t h a t
. Then
m
To
(T,:T€{T})
sup, r
.
by hypothesis.
be p o s i t i v e operators i n the
i s compact and order continuous and
the s p e c t r a l r a d i i s a t i s f y
PROOF. Assume t h a t
A > 0
.
A
< rO
. Since
r
T
To
4 r
0 '
i s compact, t h e number
ro (which i s s t r i c t l y p o s i t i v e b y our assumption) i s an i s o l a t e d p o i n t i n (note t h a t ro E o(T ) by t h e Krein-Rutman theorem). 0 0 Therefore, t h e r e e x i s t s a number A > 0 i n the r e s o l v e n t s e t of To such t h e spectrum of
that
sup, r T < A < ro
a f i n i t e number
m 2 0
T
and every
M
. Then,
such t h a t T
,
according t o t h e l a s t lemma, t h e r e e x i s t s
sup,
IIR (T A
T
)I1
5 M
. Hence,
f o r every i n t e g e r
THEOREM AND SPECTRAL RADIUS
5
/IR~(T~)//
5
M
B e f o r e p r o c e e d i n g , we o b s e r v e now t h a t from
... .
ch. 19,11351
. 0
5
T
4 To
it follows t h a t
.
2,3, We i n d i c a t e t h e proof f o r n = 2 Obviously, 2 2 f o r u 2 0 g i v e n , T u i s a n upper bound of t h e system (TTu) Let v 0 be any o t h e r upper bound. Given T t h e element v i s an upper bound of 0 ' (TToTTu:T >T ) ; hence, v t T T u . S i n c e t h i s h o l d s now f o r a l l T~ , we 2 g e t v 2 ;iuT? T h i s shows ChatToT;u = sup T u The proof f o r n 2 3 i s by TI: 4 T t
n
for
=
O + u E L
i n d u c t i o n . It f o l l o w s t h a t , f o r
-
r
and
.
.
m
f i x e d , we have
Then
where ( s i m i l a r l y a s i n Lemma 135.3) t h e compactness of
To t h e system of e l e m e n t s on t h e l e f t i s a Cauchy system i n L
norm t o i t s supremum ( i . e . ,
implies t h a t
,
converging i n
c o n v e r g i n g i n norm t o t h e element on t h e r i g h t ) .
Every element i n t h e system has a norm a t most e q u a l t o
11 TO//.M.I/ u I/
by ( 5 ) ; hence, by t h e norm convergence,
This holds f o r every
T h i s h o l d s f o r any
... .
for
m
for
m = 2,3,.
=
1,2,
..
,
m
u
E L+
. We
.
Therefore,
may w r i t e , t h e r e f o r e ,
It f o l l o w s t h a t
i.e.,
m
1 1 To / I
2
(2M.11 Toll) Am.
This implies t h a t
r
0
=
Ch. 19,11361
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
613
m 1Jm ro = limm - t m I/ToI/ 2 X , which contradicts our assumption that The final conclusion is, therefore, that supT rT = rO , i.e., r
X
+
T
r
<
r
0 '
0 '
The last theorem (about the spectral radii) is due to H.J. Krieger ( c1 1,1969) ; the proof as prese3ted here is due to A.R. Schep ( [4],1980).
EXERCISE 135.9. Let To and T (n=1,2, ...) be the positive operators n from Example 135.5. Show that the corrresponding spectral radii satisfy -1 ro = 1 and r = 1-n
.
EXERCISE 135.10. Show that the Krein-Rutman theorem continues to hold if
T
is a positive operator in L
such that
Tk
with spectral radius
is compact for some k
>
r
.
1
=
r(T) > 0
and
HINT: As when T itself is compact, we get Tun - ru + 0 , and hence - rTk-1 u + 0 We may assume that Tku converges. Since r > 0 ,
.
k T I+, Tk-t
u
converges as well. Continuing, we see that Tu
converges.
136. Irreducible operators We immediately present a definition. DEFINITION 136.1. The order bounded operator
i s said t o be i r r e d u c i b l e i f {O}
and
L
If L
B
= {O}
for any
T E Lb(L,L)
ant if and only if T+
or B
and
invariant if and only if v
T-
B (since B
B
L
i n v a r i a n t except Tf E B f o r e v e q
L 1.
do so. Equivalently, T
form a
exist B
in
invari-
leaves the band
B
/TI does so. The proof is easy. Assume first that 2
is a band).
ant. The same holds then for Tband
L
such t h a t
Lb(~,~) . Hence, T ,T and IT/ . In this case, T leaves the band
belong to B ; hence, all Tv
belongs to
=
L
+ -
T leaves B invariant. For any 0 All
in
B
is Dedekind complete, the order bounded operators in L
Dedekind complete Riesz space Lb(L,L)
leaves no band i n
i t s e l f ( i . e . , any band
satisfies either
f E B
T
T i n t h e Riesz space
and
u E B
we have
TCu
=
.
sup(Tv:O2v2u)
belong to 3 , and therefore T+u It follows that
IT1
T+
leaves B
. Conversely, if
invariant, it follows immediately from 0
2
T+
2
IT1
invari-
IT1 leaves the and
0
T-
5
614
S
Ch. 19,51361
IRREDUCIBLE OPERATORS
IT1
that
T+
and
T-
leave
THEOREM 136.2. Let
L
,
L
p o s i t i v e operator i n
B
be a (complex) Banach Lattice and l e t T be a not the null operator, such t h a t T i s order
continuous and i r r e d u c i b l e . Let t r a l radius of T
invariant.
, and Let
> r(T)
X
S = T(xI-T)-‘
, where
. Then
r(T) S
denotes t h e spec-
i s a p o s i t i v e and
irreducible operator i n L having t h e property t h a t f o r any t h e image
Sfo
PROOF. On account of and t h e r e f o r e
X
S = T(XI-T)-’
> r(T) =
0 < fo E L
.
is a weak u n i t in L
we can w r i t e
ZT X-nTn
, where
(XI-T)-’
Tn-’,
1;
=
t h e s e r i e s converges i n
norm. A l l terms a r e p o s i t i v e ; t h e p a r t i a l sums form t h e r e f o r e a n i n c r e a s i n g sequence. The norm l i m i t i s t h e n a l s o t h e o r d e r l i m i t , i . e . , p o s i t i v e o p e r a t o r , n o t t h e n u l l o p e r a t o r . L e t now We prove f i r s t t h a t =
0
,
If0Idd
then
Tfo = 0
g
.
g e n e r a t e d by
=
.’ Indeed,
Sfo # 0
T h i s would imply f
0
i s a n i n v a r i a n t band under The f i r s t i s i m p o s s i b l e , s i n c e {fOIdd = L
. As
f E {fOIdd= L
. Then
T
,
i.e.,
T
(AI-T)-’Tf
for a l l
f
i n t h e band
-1
X sup, (h-2T2+X-3T3+.
{fo}dd = {O}
or
{gjdd = L
,
Idd
{f
= L.
{fo}ddo. Hence Tf = 0
T
{gjdd
i.e.,
. Note
g = Sfo > 0
implies
+X -2 T 3+. . .+X -(n-I) n \
2
for a l l
now t h a t
=
Jfo
..
i s i n v a r i a n t under
T
. Since
i s a weak u n i t i n
g = Sfo
.
= XSfO = Xg
‘X-1T+X-2T2+...+X-nTn)fo I t f o l l o w s t h a t t h e band
=
i s o r d e r c o n t i n u o u s ) . Thus,
i t s e l f i s contained i n
fo
0
would be t h e n u l l o p e r a t o r , which c o n t r a d i c t s
Tg = TSfO = supn(X
must have
T
be given.
, i.e.,
a l r e a d y o b s e r v e d , we should have t h e n t h a t
o u r h y p o t h e s i s . Hence, 0 < f o E L
=
Tf = 0
(we u s e h e r e t h a t
is a
S
0 < fo E L
Sfo = 0
if
i s t h e sup-
S
remum of t h e sequence of p a r t i a l sums. T h i s shows a l r e a d y t h a t
We s h a l l now c o n s i d e r i n p a r t i c u l a r t h e c a s e t h a t
L L
g > 0
.
, we
i s an i d e a l i n
t h e Dedekind complete R i e s z s p a c e o f a l l ( a l m o s t everywhere f i n i t e v a l u e d )
, where
u - m e a s u r a b l e complex f u n c t i o n s on t h e measure s p a c e
(X,h,u)
assume ( a s b e f o r e ) t h a t t h e measure
A s u s u a l , u-almost
p
i s o-finite.
e q u a l f u n c t i o n s a r e i d e n t i f i e d . Note t h a t
L
we
i s Dedekind complete. Without
l o s s of g e n e r a l i t y we may a l s o assume t h a t t h e c a r r i e r of
L
is
X
. Hence,
Ch.
19,§1361
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
t h e r e e x i s t s a sequence
X
such t h a t
I. X
c h a r a c t e r i s t i c f u n c t i o n of
belongs t o
X
t h a t f o r any measurable s u b s e t such t h a t
u(Dn) <
belongs t o Let
L T
n
T(x,y) 2 0
<
for a l l
m
f o r every
L
n
. It
and t h e
n
follows
t h e r e e x i s t s a sequence
X
Dn .f D
and t h e c h a r a c t e r i s t i c f u n c t i o n of
n
.
be a p o s i t i v e k e r n e l o p e r a t o r i n ( u x u ) m e a s u r a b l e on
f i n e d and
g = Tf
f o r every
of
D
for a l l
m
u(Xn)
615
XXX
,will
h o l d s a l m o s t everywhere on
L ; t h e k e r n e l of
be denoted by
, and
XXX
T(x,y)
f o r any
T
Dn
, de-
. Then
f E L
t h e image
i s g i v e n by
t h e i n t e g r a t i o n i s performed o v e r
and t h e n o t a t i o n
X
t h a t t h e i n t e g r a t i o n i s with respect t o
du(y)
.
y
THEOREM 136.3. The p o s i t i v e kernel operator
T
in
L
indicates
, as defined
above, i s i r r e d u c i b l e i f and only i f
1 {1
x-s
T ( x , y ) d u ( y ) ‘Id u b ) > 0
J
s
holds f o r every subset
S
of
satisfying
X
and
u(S) > 0
I t i s p o s s i b l e t h a t t h e repeated i n t e g r a l asswnes t h e value PROOF. Assume f i r s t t h a t
e x i s t s a subset
It follows t h a t i f
I
of
S
E
which i m p l i e s t h a t f F: L
on
X-S
S
> 0
and
such t h a t
f o r almost e v e r y
X-S
u(X-S)
xE
> 0
such t h a t
E L , then
shows t h a t t h e band
x E X-S
. Since
any non-
can be approximated from below by l i n e a r
s e e now ( s i n c e
almost everywhere on
. This
u(S)
,
(TxE)(x) = 0
xE , we
.
+ m .
i s i r r e d u c i b l e . Assume a l s o t h a t t h e r e
i s a s u b s e t of
v a n i s h i n g on
combinations of such (Tf)(x) = 0
T
satisfying
(TxE)(x)du(x) = 0
x-s negative
X
~(x-S) > 0
X-S
T
f o r any
i s o r d e r continuous) t h a t f E L
satisfying
f(x)
=
0
IRREDUCIBLE OPERATORS
616
i s i n v a r i a n t under since
hypothesis, B # { O }
and
B
. Contradiction,
# L
i s i r r e d u c i b l e . Hence, t h e i n t e g r a l i n ( 1 ) is p o s i t i v e i f b o t h
T
and
u(S)
. By
T
Ch. 19,81363
are positive.
p(X-S)
Conversely, assume t h a t t h e i n t e g r a l i n ( 1 ) i s p o s i t i v e f o r a l l satisfying
p(S) > 0
a n t band f o r
and
. Let
T
. Assume
> 0
p(X-S)
b e t h e c a r r i e r of
X-S
also that B
S
i s an i n v a r i -
B
, i.e.,
f:fEL,f(x)=O f o r a l m o s t e v e r y xES 1
J '
and l e t to
S
I. X-S
for all
B
n
f o r almost every
such t h a t t h e c h a r a c t e r i s t i c f u n c t i o n of
. Then x E S
1 T(x,y)dp(y)
that
extended over
X-S
=
.
belongs t o
TX Sn
, where
0
. This
a l m o s t everywhere on
time
T
shows t h a t
T
T(x,y)du(y) = 0
sn
. It
follows
i f the integration i s
S
or
p(X-S)
L (where
=
0
, i.e.,
B = L
or
is irreducible. T
is a positive kernel
i s t h e same as i n t h e l a s t theorem), and t h i s
L
s a t i s f i e s t h e c o n d i t i o n t h a t f o r any
(Tf)(x) > 0
belongs
This implies t h a t
I n t h e n e x t theorem we assume a g a i n t h a t operator i n
, i.e.,
the integration i s over
Hence, by o u r assumption, u(S) = 0 B = CO}
B
Sn
f o r almost every
x
. Note
0 < f E L
we have
t h a t t h i s i s a much s t r o n g e r
p r o p e r t y than i r r e d u c i b i l i t y . Observe a l s o how t h i s i s r e l a t e d t o Theorem 136.2, where we have m e t an o p e r a t o r p o s s e s s i n g t h i s s p e c i a l p r o p e r t y . THEOREM 136.4. Let
L
be as i n the Last theorem and l e t
p o s i t i v e kernel operator i n L such t h a t on
X
f o r any
in
L
such t h a t the kernel
T2(x,y) > 0
f > 0
in
L
. Then
T2(x,y)
almast everywhere on
(Tf)(x) > 0
T be a
almost everywhere
T2
i s a p o s i t i v e kernel operator
of
T2
XxX
.
i s stri,etly p o s i t i v e , i . e . ,
PROOF. It i s e a s y t o v e r i f y (by means of F u b i n i ' s theorem) t h a t
is a kernel operator with kernel
T2
is
Ch. 1 9 , 9 1 3 6 1
I
T2(x,y) =
i s t h e k e r n e l of
T(x,y)
t h a t f o r almost e v e r y
y E X
.
a b l e f u n c t i o n of
z
i s a p-measurable
f u n c t i o n of
every
T
almost everywhere on
T2(x,y) > 0
y E Xo
,
T(x,z)T(z,y)dv(z)
X where
617
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
Let
we have t o prove o n l y t h a t
. From
F u b i n i ' s theorem i t f o l l o w s
g ( z ) = T ( z , y ) i s a p-measurY be t h e s e t of a l l y E X such t h a t g y ( z )
the function
Xo
the function
. Hence, XxX
. We
z
gy(z)
s h a l l prove f i r s t t h a t f o r almost
i s n o t i d e n t i c a l l y z e r o almost everyof XI xO w e have g y ( z ) = 0
where. Suppose, on t h e c o n t r a r y , t h a t t h e r e e x i s t s a s u b s e t p(X1) > 0
such t h a t
f o r almost e v e r y almost e v e r y
z
, i.e.,
for all
y E XI
. Then
E X
z
y E XI
and such t h a t f o r a l l
E X
w e have
T(z,y) = 0
for
Hence, by F u b i n i ' s theorem,
T(z,y)dp(y) = 0
which i m p l i e s t h a t
Note now t h a t , on account o f p(E) > 0
such t h a t
and
(TxE)(z)
xE
p(Xl)
E L
> 0
. Then
h o l d s f o r almost e v e r y
,
z E X
there e x i s t s a subset
E
of
. XI
T(z,y)dv(y) = 0
=
E almost everywhere on
X
. On
t h e o t h e r hand, s i n c e
f u n c t i o n , i t f o l l o w s from o u r h y p o t h e s i s t h a t z
. This
function
gy(z)
every
n
gn
,
(TxE)(z) > 0
i s n o t i d e n t i c a l l y z e r o . F i x such a p o i n t
f o r almost e v e r y functions
i s not the n u l l
c o n t r a d i c t i o n shows t h a t f o r almost e v e r y
0 5 gn E L (n=I,Z,
e x i s t s a sequence
every
xE
z
. Without
...)
such t h a t
f o r ahpost
y E Xo y E
0 5 gn(z)
the
% . There
+
g,(z)
l o s s of g e n e r a l i t y we may assume t h a t t h e
are n o t i d e n t i c a l l y z e r o (almost everywhere). Then, f o r
618
I R R E D U C I B L E OPERATORS
f o r almost every
. Hence,
x
also
J
T(x,z)T(z,y)du(z) =
T2(x,y) = X
f o r almost every
i.e.,
. This
x E X
y E X
last r e s u l t holds f o r almost every
. Thus, x E X
f o r almost every
T (x,y) > 0 2
find, therefore, that
T(x,z)gy(z)du(z) > 0
X
f o r almost every
T2(x,y) > 0
19,51361
Ch.
f o r almost every
. Once
y E X
y E Xo
,
we have
more by F u b i n i ' s theorem we
h o l d s a l m o s t everywhere on
.
XXX
I n o r d e r t o a p p l y t h e r e s u l t s proved so f a r f o r k e r n e l o p e r a t o r s t o m o f e g e n e r a l s i t u a t i o n s , w e r e c a l l t h e n o t i o n of a n a d j o i n t o p e r a t o r as d e f i n e d i n s e c t i o n 97. L e t and
and
L
bounded o p e r a t o r from
L
the algebraic adjoint
T#
The o p e r a t o r
b e R i e s z s p a c e s such t h a t
L
i s Archimedean
T
E L(L,M) , i . e . , i f
T
i s an o r d e r
,
then t h e r e s t r i c t i o n
T"
M
i s a n o r d e r bounded o p e r a t o r from T
,
o r d e r continuous, then T"
T-
to
maps
M i by
i s order continuous, then
Mi into
. Hence,
T'
L i
. As
if
T
belongs t o
T'
to
M"
M"
into
s e p a r a t e s t h e p o i n t s of
M
THEOREM 136.5. Let
L
Archimedean, M T
M
and
N
N
Ln(Mn,Ln)
.
$,(L,M)
(M ' ):
= {O}
in
u > 0
TI$ i s a weak u n i t i n PROOF. L e t
. Then
$(Tu) = 0
L
, the image
LL f o r every
0 < u E L $(g) = 0
and
0
S $
for all
Tu
space that i f
M
.
Tu
.
(T'$)(u) > 0
E
L
. This
and
L
0 < $
E :M
g
E
is
M
into
.
i n t h e i d e a l g e n e r a t e d by $(g) = 0
for a l l
By h y p o t h e s i s , t h e band g e n e r a t e d by
Hence $ ( g ) = 0 f o r a l l
0 < u
L
if
M
,
.
Then
i.e.,
. It
$ = 0
a r e given, then
shows t h a t t h e l i n e a r f u n c t i o n a l
Tu
. Since
in the
i s the
f o l l o w s now
$(Tu) > 0
J1
Tu g
let
such
b e g i v e n and assume t h a t
E M"
g
is
M :
i.e.,
i s a weak u n i t i n M in M i
$ > 0
i s o r d e r continuous, i t i s t r u e even t h a t
band g e n e r a t e d by
,
. Furthermore,
0
and
It was observed
be Riesz spaces such t h a t
i s Dedekind complete and
.
b e f o r e , we denote t h e belongs t o
(ML) = { O } be a p o s i t i v e and order continuous operator mapping
t h a t , f o r any
$
.
L"
i t s e l f is a l s o
a l r e a d y i n s e c t i o n 115 t h a t t h i s s i t u a t i o n i s of i n t e r e s t o n l y i f n o t t o o small. I n p a r t i c u l a r , t h i s i s t r u e i f
of
I t w a s proved i n
i s o r d e r c o n t i n u o u s . I f , moreover, T
T"
r e s t r i c t i o n of
into
i s c a l l e d t h e o r d e r a d j o i n t of
T"
s e c t i o n 97 t h a t
T
M
i s Dedekind complete. I f
M
= TI$
, i .e., is strictly
19,
Ch.
§
1361
p o s i t i v e on i.e.,
. Therefore,
L
the c a r r i e r
C
we have
$ ,$ E L i 1 2
619
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
of
$
the n u l l i d e a l satisfies
J,
I Q2
C
N
of
$
$
i f and o n l y i f
satisfies
$
. Recall
= L
$1 $2 I n the p r e s e n t case, t h e r e f o r e , i t follows from $ 1 I J, f o r some
that
,
I C$ = L
C
i.e.,
C
$1
i s a weak u n i t i n
words, $1 J, = TI$
. This .
= {O}
L;
$
{o} ,
=
(Theorem 9 0 . 6 ) .
I C
C
N
how t h a t f o r
Q1 = 0
implies t h a t
. In
E :L
other
We add s e v e r a l remarks. Note f i r s t t h a t under the conditions of t h e l a s t theorem
i s s t r i c t l y p o s i t i v e on
TI$
$ =
which shows t h a t t h e f u n c t i o n a l of
. Hence
L
v E M
('L:)
the operator
T = J,
. The band
f E L
for a l l
,
= {O}
t o r s (from
E Mi
L
ad :
into and
M
LEMMA 136.6. Let
B
T = $
D
v
. For J,
0 5 T I. 5 T n
,
f o r every
n = 1,2,
Tf = J,(f).v
v. E M ) i s denoted
and a l l
of measurable f u n t t i o n s and
then
T'
= v
@
$
, i.e.,
and
vo
T = J,
TI$ = $ ( v ) . $
be p o s i t i v e weak u n i t s i n i s a p o s i t i v e weak u n i t in
B
v
for a l l
L i and dd ( L ~ M )
.
M
.
.
t h i s purpose i t i s s u f f i c i e n t t o prove t h a t every
E L i
and
v E M ) i s contained i n
and
0
(vnnvo)
T
n
0 2 u
E B
E L
for
for a l l
. It
n
. It
... . Since
vo
B
. Let
.. .
n = I ,2,.
follows from
Tnu
i s a weak u n i t i n
. Fix M
T
5 T
n T u I. Tu
remains t o prove t h a t
be an a r b i t r a r y upper bound of the s e t of a l l
for a l l
i s defined by
u n i t s w e prove the following lemma.
$,
f
Then
N
$i E Ln
consist
= $ J o B V
Tn = \y.nQ0)
T u 5 Tu
M
generated by the s e t of a l l f i n i t e
L n (L,M) generated by To Evidently ( L - B ? ~ ) ~ ~We have t o prove, t h e r e f o r e , t h a t (L3Mldd
B
(with
into
and
E L;
J,
be the band i n
B
i s contained i n
i s contained i n
we r e c a l l t h a t f o r
M ) . F i n a l l y , i t i s easy t o see t h a t i f v E M
respectively. Then To
,
0 < $ E:M
i s t h e band of a l l absolute k e r n e l opera-
. Concerning weak
PROOF. Let
L
n (all
the band
L
$ E L i
with $
. If
(L?)dd
from
L (L,M)
CJ,. B vi
l i n e a r combinations by 0 L ():
v
B
in
f o r any
on i t s own already s e p a r a t e s t h e p o i n t s
J,
. Furthermore,
= {O}
L
nl
, we
. Then
have
.
that Let
w
620
v
A
IRREDUCIBLE OPERATORS
. Therefore w 1,2, ... . S i n c e $o i s
nvo
nl =
+
Therefore
v
w
a weak u n i t i n
. This
$ ( U ) . V = Tu
2
We now tunn t o t h e case belonging t o
(L>L)dd
. This
{(lbAnl$JO) (u)h
2
Li
shows t h a t
L = M
. In
19,51361
Ch.
holds f o r a l l
, we
have
.
T u 1. Tu
+
($hn$JO)(u)
$(u).
t h i s c a s e t h e product of o p e r a t o r s
. An
belongs i t s e l f t o ( L 2 L ) d d
even somewhat more
g e n e r a l r e s u l t can b e s t a t e d as follows. LEMMA 136.7. Let
L
be Dedekind complete, l e t L
bounded and order continuous operator i n T
~ ET ( ~ ~
for n
. I)n p~a r t i~c u t a r , ... . 3
= 2,3,
~
PROOF. Any o r d e r bounded o p e r a t o r i n ence of p o s i t i v e o p e r a t o r s ( s i n c e
,
L
3
. Then
T2 E (L>L)dd
and l e t
T E (
i f
b e an order
TI
, then ) ~
T~~
E ( L ~ L ) ~ ~
can b e expressed as a d i f f e r -
i s Dedekind complete). W e may assume
L
L e t 0 I S I En$ 0 e. l i 1 ' 0 2 $. E L" and 0 5 e . E L f o r i = 1 n Then 0 2 T S I i n 1 Zy+i," T e. This shows t h a t 'TIS E ( L 2 L ) d d , Furthermore, s i n c e 0 5 T Ad1 2 , t h e r e e x i s t s an upwards d i r e c t e d system 0 2 S T f T2 , each E (LnOL) therefore
where
that
and
T2
are positive.
,..., .
.
of t h e type as T,
TI
. As
S
0
above. Then
proved, e v e r y
5
T S I
T l S T belongs t o
+
T
T T
ST
by t h e o r d e r c o n t i n u i t y of
. Therefore,
(L$L;dd
Our n e x t lemma c o n t a i n s a key r e s u l t
E
TIT2
belongs
which w i l l b e used i n t h e proof
of o u r p r i n c i p a l theorem d e a l i n g w i t h s p e c t r a l r a d i i . Unfortunately, w e s h a l l need h e r e a r e p r e s e n t a t i o n theorem (Theorem 120.10) i n which an abs t r a c t Riesz space i s r e p r e s e n t e d ( i s o m o r p h i c a l l y ) by a space of measurable f u n c t i o n s . There seems t o b e no d i r e c t proof a v a i l a b l e .
THEOREM 136.8. Let
in
L
in L
belonging t o t h e image
weak mit i n
(
be a Dedekind complete Riesz space ( r e a l o r
L
'(L;)
complex) such t h a t
= {O}
. Furthermre,
Tu
i s a weak u n i t i n
~
3
.
)
~
arbitrarily i n
L
,
T
be a p o s i t i v e operator
L
. Then
the operator
T2
u > 0
is a
~
PROOF. For t h e proof w e may assume t h a t u > 0
let
and having the property t h a t f o r any
(L>L)dd
t h e element
e
= Tu
L
i s a r e a l space. Choosing i s a weak u n i t i n
L
.
Ch. 19,51361
62 1
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
N
L (on account of '(LI;' = { O } t h i s can be done i n n many ways), t h e element $ = T'J; i s an o r d e r continuous s t r i c t l y p o s i t i v e
Choosing
Q > 0
in
l i n e a r f u n c t i o n a l on L
L ( a s shown i n t h e proof of Theorem 136.5). Hence,
i s a Dedekind complete Riesz space p o s s e s s i n g a weak u n i t
o r d e r continuous 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
$
p r o p e r t i e s Theorem 120.10 may b e a p p l i e d , showing t h a t p h i c w i t h an i d e a l where
i n a space
M
t h e isomorphism t h e element X
. Hence,
in the ideal fy
L
I n view of t h e s e
i s Riesz isomor-
L
of r e a l w - s m a b l e f u n c t i o n s ,
1
and
M
M
.
.
corresponds w i t h t h e f u n c t i o n i d e n t i c a l l y
e
every bounded v l n e a s u r a b l e f u n c t i o n on
0 5 T E (L'@L)dd
. Furthermore,
t i v e kernel operator i n
L = M
f o r any
X
, having
u > 0
in
by h y p o t h e s i s ,
. We
L = M
the special property t h a t its kernel
s t r i c t l y p o s i t i v e almost everywhere on
XXX
.
S
,
I T2
i.e.,
2 i n f ( T ,S) = 0
if
,
S = 0
. This
(Tu)(x) > 0
may now a p p l y
shows t h a t
S
is
S ( x , y ) ), t h e n t h e
S(x,y) = 0
i s a weak u n i t i n
T2
is
T2(x,y)
then the function
must be z e r o almost everywhere. I t f o l l o w s t h a t where, i . e . ,
i s a posi-
T
R e c a l l now t h a t i f
another p o s i t i v e kernel operator i n L = M (with kernel 2 k e r n e l of i n f ( T , S ) is t h e f u n c t i o n
Hence, i f
that
is likewise a positive kernel operator
T2
Theorem 136.4 t o conclude t h a t L = M
is c o n t a i n e d
X
I n view of t h e Riesz isomorphism w e may and s h a l l i d e n t i -
I t follows t h e n from
almost everywhere on in
and an
i s a f i n i t e o - a d d i t i v e meausre ( p r e c i s e l y , w(X) = $ ( e l ) . Under
w
one on
L (X,v)
.
e
S1(x,y)
almost every(L>)dd
.
We r e t u r n t o i r r e d u c i b l e o p e r a t o r s i n a Banach l a t t i c e . THEOREM 136.9.
p o s i t i v e operator in L such t h a t c i b l e , the spectral radius
if L L
r(T)
'(L-)
T
is i r r e d u c i b l e , then
PROOF. Choose a number
X
0
=
.
T E (L2L)da
of
i s a Banach f u n c t i o n space and
such t h a t
L
(Ando-Krieger spectral radius theorem). Let
Dedekind complete Banach l a t t i c e such t h a t
> r(T)
T
satisfies T
{Ol
and l e t
Then, if T r(T) > 0
.
be a T
be a
i s irreduI n particular,
i s a p o s i t i v e kernel operator i n
r(T)
;0
and l e t
.
622
IRREDUCIBLE OPERATORS
T
The o p e r a t o r
Ch. 19,51361
i s p o s i t i v e , o r d e r continuous and i r r e d u c i b l e by hypothe-
s i s . Hence, by Theorem 136.2, t h e p o s i t i v e o p e r a t o r
u > 0
t h a t f o r any
t h e image
s i n c e by Lemma 136.7 a l l p a r t i a l sums since to
(L-@L)dd r(T)
r(S) = 0
L
. Furthermore,
belong t o
(L>)dd
i s t h e supremum of t h e s e p a r t i a l sums, t h e o p e r a t o r
S
A(,l0-X~-' if
T
Zn
has t h e p r o p e r t y
S
i s a weak u n i t i n
Su
a s w e l l . The spectrum of
X
f o r which
.
and
r(S)
S
c o n s i s t s of a l l numbers
belongs t o t h e spectrum of
T (Lemma 134.3). Hence,
a r e t h e s p e c t r a l r a d i i , we have
r(T) = 0
i f and o n l y
. Since
r(S) > 0
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 prove t h a t
h a s t h e s p e c i a l p r o p e r t y of t r a n s f o r m i n g every p o s i t i v e non-null i n t o a weak u n i t , we may apply t h e preceding theorem t o that the positive operator
i s a weak u n i t i n
S2
(once more by Lemma 134.3) t h e s p e c t r a l r a d i u s of
2
r ( S ) = [r(S)12
. Hence,
S
. We
(L"@L)dd
0 2 $ E Ln
For t h i s purpose, choose
and
0 5 e E L
S
element f i n d then
. Furthermore,
S2n i s g i v e n by
t h e proof i s reduced t o showing t h a t N
and
belongs
S
such t h a t
2 r(S ) > 0
.
$(el = 1
( t h i s can be done i n many ways, a s seen i n t h e proof of t h e p r e c e d i n g theorem). Denote t h e o p e r a t o r n = 1,2,
... w e
=
$
@
e
by
To
. For
any
f E L and f o r
have
@(f)T:-2e
For t h e s p e c t r a l r a d i u s
=
... =
r(T
0
@(f)e
we f i n d , t h e r e f o r e ,
-.
i s c o n t a i n e d i n t h e band (L>L)dd and S2 i s a weak u n i t 2 Hence, by i n t h e same band, we have 0 2 i n f ( n S ,To) To a s n + 2 Theorem 135.8, t h e s p e c t r a l r a d i i s a t i s f y rCinf(nS ,To)] .f r(To) . From
Now, s i n c e
To
+
r ( T ) = 1 i t f o l l o w s t h e n t h a t t h e r e e x i s t s a n a t u r a l number 0 2 t h a t rCinf(noS ,To)] > 0 . Hence
no
such
19,11361
Ch.
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
2 2 r ( n S ) = nor(S )
From
0 L
If
cause
i s Dedekind complete be-
L
i s now an i d e a l i n a Riesz space c o n s i s t i n g of a l l ( r e a l o r
L
complex) measurable f u n c t i o n s on some p o i n t s e t r a l i t y i t may be assumed t h a t N
= LE
g
on
thiscase L functions
f E L
every
.
N
The space
L ' = Ln
,
i.e.,
(L ' :)
= {O?
. Any
of
L'
such t h a t t h e product
X
. Without
X
i s t h e c a r r i e r of
X
i s t h e a s s o c i a t e space
L
L
. Furthermore
,
c o n s i s t i n g of a l l
has t h e same c a r r i e r
as
X
Banach f u n c t i o n space
belongs t o tor in
. Thus,
(L2L)dd
,
L
then
if
.
r(T) > 0
for
X L
itself,
s a t i s f i e s , therefore,
L
i s t h e same a s s a y i n g t h a t
L
in
s e p a r a t e s t h e p o i n t s of
L' = L z
t h e c o n d i t i o n s of t h e p r e s e n t theorem. F i n a l l y , s a y i n g t h a t t i v e kernel operator i n
l o s s of gene-
i s summable over
fg
and i t f o l l o w s immediately from t h i s t h a t L
.
2 r(S ) > 0
i t follows f i n a l l y t h a t
i s a Banach f u n c t i o n space, t h e n
623
T
T
i s a posi-
i s p o s i t i v e and
i s a p o s i t i v e i r r e d u c i b l e k e r n e l opera-
T
I t i s of some i n t e r e s t t o n o t e t h a t i f
i s a positive irreducible
T
k e r n e l o p e r a t o r i n a Banach f u n c t i o n space and we want t o prove t h a t r(T) > 0
,
t h e n i t i s n o t n e c e s s a r y t o make an a p p e a l t o Theorem 136.8,
i n which we have used a r e p r e s e n t a t i o n theorem (Theorem 120.10) t o reduce t h e a b s t r a c t s i t u a t i o n t o t h e s i m p l e r s i t u a t i o n t h a t we have t o d e a l w i t h k e r n e l o p e r a t o r s . I n s t e a d we can immediately use Theorem 136.4 t o conclude -1 2 t h a t i f h > r ( T ) , t h e n CT(XOI-T) 1 has an almost everywhere s t r i c t l y 0 positive kernel. The Ando-Krieger
s p e c t r a l r a d i u s theorem was proved f o r compact opera-
t o r s by T. Ando ([11,1957);
H.J.
t h e compactness c o n d i t i o n was removed by
K r i e g e r (El 1,1969). EXERCISE 136.10. Let
be a u - f i n i t e measure space and l e t
(X,A,p)
L
be an i d e a l i n t h e space of a l l (almost everywhere f i n i t e v a l u e d ) U-measura b l e f u n c t i o n on T
by
X.
It may b e assumed t h a t
be a p o s i t i v e k e r n e l o p e r a t o r i n T(x,y)
. Assume
that
elsewhere on
rally, l e t
u(l),
assume t h a t elsewhere on
...,u(n)
T(x,y) = 0 X
X
X
T
T(x,y) = 0
X
x
X
. Show t h a t
on
Xi
X.
. Show a g a i n 1
x
X
u(i) that T
for
X.
X.
(i=l,
(i=l,
...,n)
...,n)
and
i s i r r e d u c i b l e . More gene-
T
be a permutation of
on
x
. Let
L
w i l l b e denoted
i s t h e d i s j o i n t union of s e t s
X
of p o s i t i v e measure such t h a t T(x,y) > 0
i s t h e c a r r i e r of
X
L ; t h e k e r n e l of
i
1,
...,n
=
1,
where
...,n
is irreducible.
and
n
2
3
and
T(x,y) > 0
IRREDUCIBLE OPERATORS
624
EXERCISE 1 3 6 . 1 1 . L e t once more, l e t We d e n o t e by
T
L
be t h e same a s i n t h e p r e c e d i n g e x e r c i s e and,
be a p o s i t i v e k e r n e l o p e r a t o r i n t h e s e t of a l l p o i n t s
A
19,81361
Ch.
(x,y)
L
with kernel
f o r which
T(x,y).
T(x,y) = 0
. Show
t h a t t h e following conditions a r e equivalent. (i)
A
does n o t c o n t a i n a r e c t a n g l e of p o s i t i v e measure, i . e . ,
does n o t have a s u b s e t
A x B
, where
B
and
A
a r e s u b s e t s of
A
X
of
p o s i t i v e measure ( t h i s has t o be u n d e r s t o o d modulo s e t s of measure z e r o ) . u > 0
( i i ) For any
in
serve that the operator
T
L
t h e image
Tu
i s a weak u n i t i n
L
. Ob-
i n t h e p r e c e d i n g e x e r c i s e does n o t s a t i s f y ( i )
and ( i i ) . H I N T : For ( i i ) and in
B
*
( i ) , assume t h a t (ii) h o l d s and assume a l s o t h a t
a r e s u b s e t s of
X
of p o s i t i v e measure s u c h t h a t
z e r o (xqhich does n o t a f f e c t
A
. The
set
B
contains a subset
. Since
x E A
For ( i ) L
Tu
x E A
i s contained i n
Show now t h a t
xs
6 L.
(fxs)(x) = 0
u > 0
( i ) h o l d s and t h a t f o r some
in
i s of
A = (x:(Tu)(x)=O)
i s a l s o e f p o s i t i v e measure. For
B = (y:u(y)>O)
we have
T(x,y) = 0
T h i s shows t h a t we have
AxB
of p o s i t i v e measure s u c h t h a t
f a i l s t o be a weak u n i t . Then
p o s i t i v e measure. The s e t any
A of measure
i s of p o s i t i v e measure, we g e t a c o n t r a d i c t i o n .
A
( i i ) , assume t h a t
t h e image
S
i s a weak u n i t by h y p o t h e s i s .
T h e r e f o r e , TxS for a l l
T ) , we may assume t h a t
i s contained
AxB
e x c e p t f o r a s e t of measure z e r o . Making a change i n
A
A
T(x,y) = 0
f o r almost every
almost everywhere on
AxB
. Contradiction.
EXERCISE 1 3 6 . 1 2 . The s u b s e t
caZZy dense i f
y E B
f o r almost e v e r y
u(DnU)
i s Lebesgue measure i n
D
. Thus,
y € B
. It
of t h e r e a l l i n e
p o s i t i v e measure, we s h a l l c a l l
A AXB
and
B
T(x,y) = 0
R i s c a l l e d metri-
i s p o s i t i v e f o r e v e r y open i n t e r v a l R ) . If
x E A
f o r any
follows t h a t
a r e s u b s e t s of
U
(where
1-1
R of f i n i t e
a measurable r e c t a n g l e i n
R2
. For
b r e v i t y , we s h a l l s a y t h a t t h e s u b s e t , E of R2 has p r o p e r t y (*) i f u 2 { E n ( A x B ) } > 0 f o r e v e r y m e a s u r a b l e r e c t a n g l e AxB (where 1-r’ i s Lebesgue 2 measure i n R ) . I f D i s any s u b s e t o f R , t h e s e t E = {(x,y):x-yED)} c o n s i s t s of a l l s t r a i g h t l i n e s of s l o p e one i n in
D
. Show t h a t
this set
E
X2
i n t e r s e c t i n g t h e x-axis
has p r o p e r t y (*) i f and o n l y i f
D
i s metri-
Ch.
19, 51361
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
c a l l y dense. I t f o l l o w s t h a t i f to
IR , t h e n t h e s e t
with respect
D
c o n t a i n s no measurable r e c t a n g l e (nor u 2 -almost e q u a l t o a measurable
that is
TK2
i s m e t r i c a l l y dense.
D
r e c t a n g l e ) i f and o n l y i f HINT: I f
i s t h e complement of
DC
{(x,y):x-yEDC}
does i t c o n t a i n a s u b s e t of
625
i s any measurable r e c t a n g l e , then
AXB
D
f(u) =
where
,f
XA(u+y)XB(y)dy = u{An(u+B)}
. The
function
f
i s non-nega-
t i v e and c o n t i n u o u s . The c o n t i n u i t y f o l l o w s from
For
D = IR we g e t
= R2 ; hence
E
t h a t the continuous function Therefore !.i2{En(AxB)}
I
=
JIR
2 f ( u ) d u = 1.1 (AxB) > 0
. It
follows
i s s t r i c t l y p o s i t i v e on some i n t e r v a l .
f
f(u)du > 0
D
if
D
i s m e t r i c a l l y d e n s e . Conversely, i f t h e r e i s a n open i n t e r v a l
IR such t h a t
u(DnU)
=
0
,
t h e n t h e s e t of a l l
c o n t a i n s a measurable r e c t a n g l e i n t e r s e c t i g n g
(x,y)
such t h a t
{(x,y):x-yED}
measure z e r o . T h i s proof i s due t o P. Erd6s and J . C .
in
U
x-y E U
i n a s e t of
Oxtoby ([11,1955),
but
t h e p r e s e n t r e s u l t goes back t o and i s r e l a t e d w i t h t h e much o l d e r r e s u l t that i f
A
and
are s u b s e t s of
B
difference set (i.e.,
all
x-y
W of p o s i t i v e measure, t h e n t h e
with
and
x E A
y E B ) c o n t a i n s an
interval. EXERCISE 136.13. The r e s u l t i n t h e p r e c e d i n g e x e r c i s e i s of i n t e r e s t o n l y i f t h e complement
of
DC
D
i s of p o s i t i v e measure. We i n d i c a t e a
method t o f i n d a m e t r i c a l l y dense s e t interval
F
D
with
u(DC) > 0
. Given
a closed
of p o s i t i v e measure, w e s h a l l s a y t h a t t h e C-procedure
procedure) i s applied t o
F
whenever an open i n t e r v a l
0 < p ( U ) < y(F) and w i t h t h e same c e n t r e as r a l Cantor s e t i n
[0,ll
F
,
U
(Cantor
, satisfying
i s removed from
F
.A
i s made by f i r s t a p p l y i n g t h e C-procedure t o
gene-
626
Ch. 19,11371
COMPACT IRREDUCIBLE OPERATORS
[ O , 11 then t o t h e two remaining closed i n t e r v a l s (such t h a t the two removed
open s u b i n t e r v a l s have the same measure), next t o the four s t i l l remaining closed i n t e r v a l s , and s o on. The s e t which remains a f t e r a countable number of s t e p s is a Cantor s e t . Now, l e t (A
:n=l,2,
n n = O,l,Z,
...)
...
, make
t h e (n+l)-th s t e p equal t o mentary Cantor s e t
Show t h a t i f
u(S) [O,ll
in
S'
.
An
=
a
. Hence,
satisfies
i s m e t r i c a l l y dense and
(S+n)
D = U n=-m
and l e t the sequence
2nAn = a
. For
the length of each of the removed open i n t e r v a l s a t
removed open i n t e r v a l s , then m
0 < a < 1
of p o s i t i v e numbers be such t h a t :Z
i s the union of a l l
S
the measure of the comple-
p(S')
I-a > 0
=
. Show t h a t
i n t e r s e c t s each i n t e r v a l of
DC
u n i t length i n a s e t of p o s i t i v e measure. EXERCISE 136.14. Use t h e r e s u l t s i n the l a s t two e x e r c i s e s t o d e f i n e
T
a positive kernel operator
such t h a t the s e t
A
on which the k e r n e l
vanishes i s of p o s i t i v e measure but does not contain a measurable r e c t a n g l e .
137. S p e c t r a l p r o p e r t i e s of compact i r r e d u c i b l e operators I n t h e p r e s e n t s e c t i o n t h e assumptions concerning t h e (complex) Banach lattice
w i l l be the same a s i n the l a s t s e c t i o n , t h a t i s t o say, L
L
Dedekind complete and operator i n
L
('L:)
{O}
=
belonging t o
(L?Idd
we s h a l l sometimes assume t h a t i s a weak u n i t i n unit i n
. Furthermore,
w i l l be assumed t h a t
as before, T
is
w i l l be a positive
and, depending upon t h e s i t u a t i o n ,
i s i r r e d u c i b l e and sometimes t h a t
T
f o r every
L
(L2L)dd
. Also
u > 0
in
,
L
o r even t h a t
T
Tu
i s a weak
i n most of the theorems t o be proved i t
i s compact. This w i l l imply t h a t the non-zero
T
p o i n t s i n the spectrum of
a r e eigenvalues, each of these of f i n i t e
T
m u l t i p l i c i t y ( i . e . , each eigenspace corresponding t o a non-zero eigenvalue
i s of f i n i t e dimension). The a d j o i n t operator therefore, the reset r i ct i o n
T'
of
T*
to
s h a l l prove a s one of t h e main r e s u l t s t h a t i f i r r e d u c i b l e and
belongs t o
T
i s an eigenvalue of
T
(L2L)dd
,
Tu
eigenvalue
i s a weak u n i t i n h
of
T
i s likewise compact. We
i s p o s i t i v e , compact and
T
then t h e s p e c t r a l r a d i u s
r(T)
with an eigenspace of dimension one; each non-null
element i n t h e eigenspace i s a weak u n i t i n that
i s then a l s o compact and,
T* Lr
L
satisfies
f o r each
L , I f i n a d d i t i o n i t i s given
u > 0
( A / < r(T)
. If
in L
L
,
then every o t h e r
i s a Banach function
627
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
ch. 1 9 , 9 1 3 7 1
space, these r e s u l t s become statements about p o s i t i v e k e r n e l o p e r a t o r s , g e n e r a l i z a t i o n s of a famous theorem, due t o R. Jentzsch (1912), continuous k e r n e l s
T(x,y)
about
t h a t a r e s t r i c t l y p o s i t i v e on a square i n
R2.
The theorem of Jentzsch i s i t s e l f a g e n e r a l i z a t i o n of a c l a s s i c a l r e s u l t of 0. Perron (1907) about matrices with p o s i t i v e e n t r i e s . I n the f i r s t theorem we do not y e t assume THEOREM 137. I . Let
'(L;)
such t h a t L
L
(L2L)dd
belonging t o
.
T'
w i l l be demoted by
be a Dedekind complete aomplex Banach l a t t i c e
and l e t
= {O}
. As
be a p o s i t i v e i r r e d u c i b l e operator i n
T
we did before, the r e s t r i c t i o n of
O((L-)-) = 103 and T ' i s a p o s i t i v e i r r e d u c i b l e n n - -dd Hence, i n view of the Andobelonging t o ((L,lnQLn)
operator i n L;
.
r(T)
Krieger theorem, both the s p e c t r a l r a d i i positive.
L-
i s a norm closed band i n
) and every
= L*
f
continuous l i n e a r f u n c t i o n a l on
L-
$T(u) C 0
L ). The space
u t 0
f o r every
f o r e , a s a subspace of
i t follows t h a t
we have
0
5
T
n
for
n
i = 1,
...,n ,
in
(LN)-
E 2
(because
. Since
.
L
0
2
T =
Zn14.1
Q
ei
T;
4 T'
.
in
LL
implies
may be regarded, there-
0 2 $. E LL
n Zlei
.
Z$.
1
Q
e . ) . Hence, i f 1
0 5 T 0 5 $
f T
E L;
(T:@)(u) = @ ( T T ~4 ) @(Tu) = (T'$)(u) i.e.,
C 0
L
L;
t r i v i a l l y s e p a r a t e s the p o i n t s of
(with
there e x i s t s an upwards d i r e c t e d system a r b i t r a r y , then
a c t s as an o r d e r
0 $i ) , because then 0 5 T I = n 0 2 T ' S Zlei B $i Finally, i f
then
by a f i n i t e sum
$T
E L
E (L3L)(Id h o l d s , we have
This i s evident i f 5 Cl$i B ei
are s t r i c t l y
((L ) ) = { O } As observed s e v e r a l times a l r e a d y , n n as soon as T i s o r d e r continuous. I n the p r e s e n t
T I : L i + L-
case, where
0
r(T')
and
i s a Dedekind complete Banach l a t t i c e (since
Li
PROOF. The space
,
L i
T* t o
i s now a l s o a Dedekind complete
L;
The space
Banach l a t t i c e such t h a t
Li
t o be compact.
T
It follows t h a t , f o r any
0 5 T
0 5 e. E L
and
. Hence,
if
0 5 T 5
0 5 T E (L2L)dd (every and
TT
,
then
majorized
0 5 u E L
are
, E ( L - c ~ L ) ,~ ~the correspond-
628
COMPACT IRREDUCIBLE OPERATORS
T'
ding
. To
0 5 T ' E (LBL;))'
satisfies
we choose a number
(hence, A.
X o > IlTll
Theorem 136.2,
the positive operator
property that
Su
i s a weak u n i t i n
$ > 0
n
in
and s i n c e
+ -
> IIT*II
2
I-T)-l
0 f o r any
L
is irreducible,
T'
IIT'II). A s shown i n m -k k = CIAo T has the
u > 0
in
L
. Then,
by
S'$ i s a weak u n i t i n L; n -k k L" We d e t e r m i n e S ' S i n c e CIAO T 4 S f o r k kn (T ) ' = f o r a l l k , we f i n d t h a t S ' = Zy X i k ( T f ) .
h a s now t h e p r o p e r t y t h a t
Theorem 136.5, S ' f o r any
show t h a t
S = T(X
19,51371
Ch.
.
.
From t h i s f o r m u l a i t i s c l e a r t h a t any band i s a l s o i n v a r i a n t under
S'
. Since
i n v a r i a n t , t h e same h o l d s f o r
, i.e.,
T'
B-
in
L;
i n v a r i a n t under
{O}
leaves only
S'
and
T'
itself
L ;
is irreducible.
T'
LEMMA 137.2. Once more, assume t h a t L i s a complex Banach l a t t i c e 0 such t h a t (Lz) = I 0 1 Let To be a p o s i t i v e operator i n L such that
.
To
i s a weak u n i t i n
and
lTOf\ = To(/fl)
. Furthermore,
(L>)dd
. Then
PROOF. For r e a l
f = e
ia
f E L
let
for some r e a l number
If/
CY
# 0
f
satisfy
.
t h e proof i s e a s y , because i t f o l l o w s immediately
f
from = lTOfI = T ( l f l ) = T f + + T f -
/Tof+-Tof-l that and
Tof+ I T f -
0
f-
one of
0
. Both
and
f-
Tof-
a r e weak u n i t s , u n l e s s one of
must i n d e e d v a n i s h . Hence f
Tof+ I Tof-
Assume f i r s t t h a t aTo - T I
T2
lTOf/ = To(lfl)
1
5
aTo
. Then
, we
find that
arbitrary positive operator i n (L2L)dd
,
f o r some number
(L2L)dd
. Since
To
t h e r e e x i s t s a n upwards d i r e c t e d s y s t e m
m a j o r i z e d by a p o s i t i v e
m u l t i p l e of
To
. It
T
a > 0
lTlfI = T l ( l f l )
or in
.
. Let
holds
.
(L>)dd
. For
convenience,
now
T
be an
i s a weak u n i t i n 0 5 TT 4 T
,
f+
holds,
f = -If1
ITfI = T ( l f / )
b u t a l s o f o r any p o s i t i v e
To
0 5 T by
f = If1
we show f i r s t t h a t t h e e q u a l i t y
not only f o r t h e given
Since
and
v a n i s h e s . T h e r e f o r e , i n t h e p r e s e n t c a s e where f+
For n o n - r e a l
denote
Tof+
0
0
each
TT
i s not d i f f i c u l t t o see t h a t
ch. 1 9 , 9 1 3 7 1
The l e f t hand s i d e does n o t depend on
T
. Hence
.
ITfI = T ( l f 1 )
We apply t h i s r e s u l t t o a n a p p r o p r i a t e l y chosen
.
T
. Let
0 < $ E LL
$ = TI$ Then $ i s a s t r i c t l y p o s i t i v e l i n e a r 0 L (Theorem 136.5). Now, l e t T = $ @ u , where 0 < u E L
be a r b i t r a r y and w r i t e f u n c t i o n a l on
629
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
i s a r b i t r a r y . From
ITfI = T ( l f 1 )
i t follows t h a t
l$(f)I = $ ( I f [ )
. Note
t h a t t h i s number i s d i f f e r e n t from z e r o s i n c e If1 > 0 and $ i s s t r i c t l y ia w i t h r > 0 and 0 S a < p o s i t i v e . T h e r e f o r e , we can w r i t e $ ( f ) = r e -ia < ZII The element f l = e f s a t i s f i e s I f l / = If1 and $ ( f l ) = r
.
.
Therefore,
T h i s shows t h a t hl
It f o l l o w s t h a t
or, writing
$(lfll-gl) = 0
and
f l = g1 + i h
1
(1))
=
0
,
I f l / = g1
o b s e r v e t h a t t h e l a s t p a r t of t h e proof
h o l d s a s w e l l f o r any o t h e r p o s i t i v e
holds f o r every p o s i t i v e f o r every
. Since
$(hl) = 0
is s t r i c t l y p o s i t i v e , we f i n d already t h a t
$
h
$(lfll-fl) = 0
(with
g1
and
I f l I - g1 2 0
and
real),
in
L i
.
$
in
Lx
L i
. Hence,
$(hl) = 0
$(hl) = 0
holds
s e p a r a t e s t h e p o i n t s of
i h l = g1 = I f l l
. The
i s , therefore, that
From h e r e on we add t h e h y p o t h e s i s t h a t
show now t h a t
( u n t i l we come t o formula
This i m p l i e s t h a t
. S i n c e , by h y p o t h e s i s , h = 0 . Thus f l = g1 + 1
$ E L i
i t follows t h a t
$
. To
T
i s compact.
L
,
final result
630
COMPACT IRREDUCXBLE OPERATORS
ch. 19,51371
THEOREM 137.3 (Generalized Jentzsch theorem). Let
complete Banach l a t t i c e such t h a t
'(L;)
L
and l e t
= {O}
be a Dedekind be a p o s i t i v e
T
operator i n L such t h a t T i s compact and i r r e d u c i b l e and T belongs T i s an eigenvalue of t o ( L 2 L ) d d . Then t h e spectral radius r ( T ) of
of m u l t i p l i c i t y one, i.e., the corresponding eigenspace is of dimension one. Furthermore, each non-null element i n the eigenspace is a weak u n i t T
.If
in L L
T
i s irreducible in the stronger sense t h a t f o r any Tu
t h e image
i s a weak u n i t in L r(T)
d i f f e r e n t from
satisfies
1x1
, then any eigenvazue X
PROOF. A s proved i n Theorem 137.1, t h e s p e c t r a l r a d i i r(T')
of
T
and
T'
Krein-Rutman Tu
T
i s p o s i t i v e and compact w i t h
theorem may be a p p l i e d t o conclude t h a t
T
and
Tu = r ( T ) u
f
Since
t h e image
in
.
u
g e n e r a t e d by
u
i n t h e i d e a l g e n e r a t e d by T
{ u } ~i s~ s e e n t o be i n v a r i a n t under
ducible, t h i s implies t h a t
{uIdd = L
,
i s an eigen-
. Evidently,
L
I t follows t h a t f o r
T
f
. Since
, assume
u
f = g+ih
r e a l ) , then
(g
and
h
that
Tf
=
r(T)f
Tg = r ( T ) g
Tf = r ( T ) f
i t follows t h a t
r(T)/ f
Th = r ( T ) h
1
= ITf
f
I
applied a l s o t o
T'
. This
'(6)
. Since
in
$
i s 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
r(T')
Hence
2
satisfying
r(T)
, we
. If
Hence,
in
Tf = r ( T ) f and i t
i n t h i s in-
theorem may b e
TI$ = r ( T ' ) $
= {O}
L
,
t h e weak u n i t
. Observing
that
get
r ( T ' ) = r(T)
s i t i v e . Therefore,
.
is a
l e a d s t o t h e e x i s t e n c e of a p o s i t i v e weak u n i t
$
Lx
.
r(T)
T(I f l )
2
w i l l be o u r f i r s t o b j e c t i v e t o prove t h a t t h e r e i s e q u a l i t y e q u a l i t y . For t h a t purpose, o b s e r v e t h a t t h e Krein-Rutman
L
f # 0
f o r some
and
we may assume w i t h o u t l o s s of g e n e r a l i t y t h a t t h e element i s r e a l . From
is irre-
T
To show t h a t any o t h e r e i g e n e l e m e n t b e l o n g i n g t o t h e e i g e n v a l u e c o n s t a n t m u l t i p l e of
{ujdd.
i n t h e band
i s a weak u n i t i n
u
i.e.,
, the
i s contained i n
Tf
i s o r d e r c o n t i n u o u s , t h e same h o l d s f o r any
. Thus
and
r(T) > 0
r(T)
u > 0
h o l d s f o r some
i s c o n t a i n e d i n t h e band
any
T
r e s p e c t i v e l y a r e s t r i c t l y p o s i t i v e . Hence, s i n c e
we know t h e r e f o r e t h a t v a l u e of
r(T)
in
0
of
.
< r(T)
u
and T(I f
${T(Ifl)-r(T)If
I)
= r(T)I f
1
. It
11
=
0
. But
$
i s s t r i c t l y po-
h a s been proved t h u s t h a t t h e
Ch.
19,11371
eigenspace in
,
Lo
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
Lo
of
.
L
satisfying
if
fl
and
, which
L
has t h e property t h a t i f
is
f
i s a Riesz
Lo
has t h e p r o p e r t y t h a t any v > 0
implies t h a t i f
inf(vl,v2) = 0
,
a r e given i n
Lo
f2
r(T)
T h i s i m p l i e s immediately t h a t
I n a d d i t i o n , Lo
i s a weak u n i t i n Lo
.
1f I
t h e n so i s
subspace of
belonging t o
T
63 1
v1
v1 = 0
then
m
and i f
and
or
in LO a r e elements of
v2
v2
=
0
( o r b o t h ) . Hence,
i n f ( f ,f ) 1 2 Lo and i n f ( f m , f -m) = f l - m and f 2 - m belong t o 1 2 or f 5 f l f l - m = 0 or f 2 - m = 0 , i.e., f 5 f 2 t h u s s e e n t o be simply o r d e r e d . I n p a r t i c u l a r , g i v e n f E
, t h e elements
f t 0
in
=
.
0
Therefore,
. The
. It
f 5 0
or
remains t o show t h a t i f
Lo
0 < v 5 u
.
space
,
is
Lo
we have
,
Lo
then
v = a u f o r some a. s a t i s f y i n g 0 < a. 5 1 I t i s well-known how t o do 0 0 , i t i s impossit h i s (compare w i t h Theorem 2 6 . 4 ) . S i n c e au J. 0 a s a ble that
satisfying
a
implies t h a t v
2 au
for a l l
v 5 au
For any
, we
a f a.
Lo
c l u d e s t h e proof t h a t in
. Given
(L3L)dd
element
f # 0
and t h e r e f o r e t h e proof t h a t here with
If
an e i g e n v a l u e
in
L
-
I
.
such t h a t
X
T
Tf = Xf
.
r ( T ) I f I = T( If
1) f
, so
I
r ( T ) If
i t f o l l o w s t h a t 1 Tf 1 = T( If ia If I f o r some r e a l a
.
1)
5 T( If
f = e
Tf = Xf
i s a weak u n i t i n
Hence
unit i n f o r which that
(L:d,)dd
I
Then
2
s o l u t e value
=
con-
i s e q u a l to
ITf
.
I
=
I A I.
I Tf I
f
as well as t o
If I ,
is real T( If
1) ,
lemma t h i s i m p l i e s t h a t
. u > 0
in
Then, a s shown i n Theorem 136.8, T
Furthermore,
If1 = r ( T )
t h i s i s so b e c a u s e we d e a l
. By t h e p r e c e d i n g . T h e r e f o r e , we have
= r(T)
, we
= r(T)
e x a c t l y a s t h e f i r s t p a r t of
1)
X
1X/
i t does n o t m a t t e r whether
t h e spectrum of
b e l o n g s t o t h e spectrum of
A
r(T )
.
L
.
i s a weak u n i t
T
satisfying
Assume now t h a t i t i s g i v e n o n l y t h a t f o r any Tu
a u 0
i s an e i g e n v a l u e , t h e r e e x i s t s an
and n o t w i t h
o r complex. S i n c e now
as well as
of
X
Since
. It follows . Note t h a t
r(T)I f
5
i s of dimension one.
= r(T)
X
v
, which 0 , i.e.,
. Then (v-au) = a u . Hence v = a u . T h i s 0 0
I n t h e proof of t h e second p a r t we assume f i r s t t h a t have t o show t h a t
Then
0 < (v-au)+ E Lo
L
v 2
find
.
= inf(a:~au)
a.
we have
i s a weak u n i t i n
(v-au)'
. Letting
. Let
> 0
CY
0 < a < a.
+
T
T2
, which
2
L
.
is
t h e image a weak
c o n s i s t s of a l l
X2
implies i n p a r t i c u l a r
[ r ( T ) ] ' . A s shown above, t h e o n l y e i g e n v a l u e of T2 of ab2 r ( T ) i s t h e number r ( T 2 ) i t s e l f . I t f o l l o w s t h a t , b e s i d e s
632
Ch.
COMPACT IRREDUCIBLE OPERATORS
r(T)
itself, the only possible eigenvalue of
is the number
. We
-r(T)
T
of absolute value
prove now that -r(T)
Recall first that every eigenelement of
19,51373
r(T)
is not an eigenvalue of T
T2 belonging to the eigenvalue
.
is a constant multiple of a fixed positive weak unit u in L Assume now that Tf = -r(T)f for some f # 0 . Then T2f = [r(T)] 2f = 2 = r(T )f . Therefore, f is a constant multiple of the weak unit u , say r(T2)
f
= au
. It follows then from
Tf
=
-r(T)f
.
impossible since Tu > 0
and
any eigenvalue X
different from r(T)
of
T
< 0
that Tu
-r(T)u
assume that X
is the carrier of L
to
(L?LIdd
. As
(X,A,u)
is a Banach function
mentioned before, we may
. The condition that
T
are now the kernel operators in L
satisfying T(x,y)
2
is
.
satisfies I X 1 < r(T)
is automatically satisfied. The positive operators T T(x,y)
, which
-r(T)u
This concludes the proof that
We shall now briefly discuss the case that L space on the o-finite measure space
=
) ; L ( '
=
in L
{O}
belonging
possessing a kernel
0 almost everywhere on XxX
.
The operator
of this type is irreducible if and only if
satisfying u(S) > 0
> 0
for any subset
S
136.3)
satisfies the much stronger condition that T
and
unit in XXX
T
of
X
(L2L)dd if and only if T(x,y) > 0
. There
and
almost everywhere on subset of XxX
X
u(x)
in L
T(x,y)
> 0
(Tu)(x)
such that u(x)
. This condition i s
on which
(Theorem is a weak
holds almost everywhere on
is also the intermediate condition that
almost everywhere for any
u(X-S)
holds
does not vanish
satisfied if and only if the
vanishes does not contain a measurable
rectangle of positive measure (Exercise 136.11). THEORM 137.4.
( J e n t z s c h ' s theorem f o r kernel operators.) Let
a Banach f u n c t i o n space on and l e t
(X,A,u)
such t h a t
T be a p o s i t i v e kerneZ operator i n
b l e and compact. Then the spectral radius r(T)
i s an eigenvalue of
T
.
X L
r(T)
i s t h e c a r r i e r of
such t h a t of
L
be L
is irreduciT i s p o s i t i v e and T
There e x i s t s a corresponding eigenfunction
uo(x) such t h a t uo(x) > 0 almost everywhere on x , and any other eigenf u n c t i o n belonging t o the eigenvalue r(T) is a constant m u l t i p l e of u,(x) . I f T has the stronger property that for any non-negative u(x) in
.
Ch. 19,11371
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
633
(not i d e n t i c a l l y zero almost everywhere) the image (Tu)(x) i s p o s i t i v e almost everywhere, then any eigenvalue X of T d i f f e r e n t from r ( T ) saL
.
I h l < r(T)
tisfies
I n t h e o r i g i n a l theorem o f R. J e n t z s c h ([11,1912) an i n t e r v a l
[a,bl
,
t h e measure
XxX = [ a , b ] x [ a , b l
p o s i t i v e and c o n t i n u o u s on a compact s u b s e t of
,
IR'
T(x,y) t c
such t h a t
. Since
. Similarly,
XxX
therefore
uo(x)
2
and
d
c
d
on
. The
.
e x i s t e n c e of
I n t h i s c a s e a l r e a d y t h e r e a r e no such numbers
a proof i n t h e in
L2
, but
L2
c a s e , s e e t h e book ([21,1953) by A.C.
L
i s t h e space
c
and
W.
n
. For
Schmeidler.
We s p e c i a l i z e s t i l l f u r t h e r by assuming t h a t t h e p o i n t s e t of
d
Zaanen. A proof
o n l y f o r k e r n e l s t h a t a r e c o n t i n u o u s i n t h e mean, i s c o n t a i n e d
(I: 11,1955) by
i n t h e book
that
w i t h t h e s e p r o p e r t i e s i s used i n t h e o r i g i n a l p r o o f .
Many y e a r s l a t e r t h e theorem was extended t o t h e c a s e t h a t
L2(X,A,u)
c > 0 uo(x)
, so
X = Ca,bl
d > 0
is
i s then
XxX
the eigenfunction
f o r some number
X
p o i n t s , each p o i n t of u n i t measure. The s p a c e
complex f u n c t i o n s on
X
X
consists
c o n s i s t s of a l l
L
w i t h pointwise o r d e r i n g ( f o r the r e a l functions i n
i s t h e complex n-dimensional
L ) . I n o t h e r words, L
space
En = IRn + iIRn
w i t h t h e n a t u r a l c o o r d i n a t e w i s e o r d e r i n g f o r t h e p o i n t s i n t h e subspace ( s i n c e a l l norms i n
C2
e l = (l,O,
basis vectors
be a s u b s e t of
,...,
a r e m u t u a l l y e q u i v a l e n t anyhow). We i n t r o d u c e t h e
...,0 )
(l,2,
,
...,0 )
e2 = ( O , l , O ,
...,n) . The
e
more, bands and i d e a l s a r e t h e same ( i . e . , L = En
T
in
(el
)
w i l l b e denoted by
l e a v e s t h e band
A( 1,2)
combinations of
el
for
j = 3,4,
non-negative. (l,Z,
...,n)
and
...,n . The
and s o on. L e t
En
. Note
En
that this i s a
i s of t h i s form. F u r t h e r -
e v e r y band i s an i d e a l ) . For any
i t s m a t r i x elements w i t h r e s p e c t t o t h e b a s i s
operator
,...,e
,
s e t of a l l l i n e a r combinations
e. i s an i d e a l A ( j l , . . . , j k ) in j1 Jk p r i n c i p a l i d e a l . Note a l s o t h a t e v e r y i d e a l i n of
Rn.
i s n o t of g r e a t i m p o r t a n c e f o r our p u r p o s e s
L = Cn
I n t r o d u c i n g a norm i n
jl,...,jk
is
X
T(x,y)
t h i s i m p l i e s t h e e x i s t e n c e of a number
on
mentioned above i s now p o s i t i v e and c o n t i n u o u s on numbers
the point s e t
i s Lebesgue measure and
LI
...,n ) .
t (j,k=l, jk i n v a r i a n t , t h e n Tel
e2
. This
operator
and
means t h a t a l l
I f , f o r example, T
Te2
must be l i n e a r
t
and
jl
t
i s p o s i t i v e whenever a l l
T
To d e s c r i b e i r r e d u c i b i l i t y of a p o s i t i v e o p e r a t o r i n t o d i s j o i n t subsets
Sl
and
S2
. It
j2 t
vanish are jk T , divide
i s n e c e s s a r y and s u f f i -
634
ch. 19,91371
COMPACT IRREDUCIBLE OPERATORS
cient for
t o be irreducible that
T
f o r e v e r y c h o i c e of t h e l a s t theorem t h a t
S1
and
a s i n d i c a t e d . The s t r o n g e r c o n d i t i o n i n
S2
T(x,y) > 0
now t h e c o n d i t i o n t h a t
> 0
t
h o l d s a l m o s t everywhere on
holds f o r a l l
j,k
. The
XXX
becomes
intermediate
jk c o n d i t i o n mentioned i n t h e t h e same theorem c o i n c i d e s i n t h e p r e s e n t
s i t u a t i o n with the s t r o n g condition t h a t a l l any o p e r a t o r
T
the s p e c t r a l radius
r(T)
r u n n i n g through t h e e i g e n v a l u e s o f
A
. Applying
T
p r e s e n t s i t u a t i o n , we see t h a t i f (I) u
t. 2 0 Jk r ( T ) > 0 and
above i s s a t i s f i e d , t h e n
p o s s e s s i n g an e i g e n v e c t o r
u = (ul,
satisfy t > 0 jk jk i s t h e maximal v a l u e of t
...,u
.
For
1x1
for
t h e l a s t theorem t o t h e
for a l l
j,k
and c o n d i t i o n
i t s e l f i s an e i g e n v a l u e
r(T)
having a l l i t s coordinates
)
s t r i c t l y p o s i t i v e . Each o t h e r e i g e n v e c t o r b e l o n g i n g t o t h e e i g e n v a l u e
j
i s a c o n s t a n t m u l t i p l e of
r(T)
. It
u
i s given i n a d d i t i o n t h a t a l l
are s t r i c t l y p o s i t i v e , t h e n e v e r y o t h e r e i g e n v a l u e Ihl
. In
< r(T)
that a l l
t
of
X
T
satisfies
t. Jk
t h e o r i g i n a l theorem of 0. P e r r o n (C11,1907) i t was assumed
are s t r i c t l y p o s i t i v e .
jk
i s compact i n
T
We make o u r f i n a l remark about t h e h y p o t h e s i s t h a t
o u r main theorem. This h y p o t h e s i s i s n o t r e a l l y n e c e s s a r y ; i t i s s u f f i c i e n t if
i s compact f o r some p o s i t i v e i n t e g e r
Tk
. To
k
see t h i s , o b s e r v e
t h a t t h e Krein-Rutman theorem c o n t i n u e s t o h o l d i f i n s t e a d of
i s compact f o r some
Tk
EXERCISE 137.5.
j
for all A2 = 0
and
t
jk
=
X I = n-1 9
n-l
C2
then
T
I
for
and
,
j # k
X2 =
... =
-1
)
,
Te2 = e 3 , , . . , T e
then
r ( T ) = n-l
are t h e r o o t s of
Xn
n
-
absolute value equal t o
En
1 = 0
T and
has T
,
e2-e3,
the p o s i t i v e operator
,
then
. Hence .
r(T) = 1
T
itself
has
X
respectively.
1
t. = 1 Jk = 2 and
= 0 for j = k jk has eigenvalues
t
w i t h e i g e n v e c t o r Zn el-e2
T
has eigenvalues
t h e m a t r i x of
An = - 1
= el
the matrix of
el-e2
.
( i i i ) Show t h a t if i n
e2
En
and
and
) and i n d e p e n d e n t e i g e n v e c t o r s
eigenvalue =
r(T) = 2 el+e2
Show t h a t i f i n
T
.
( i ) Show t h a t i f i n
with eigenvectors
(ii) and
,
k
k > 1
j=l
e
j
(for
...,en-l -e n T
X
1
=
(for the
i s given by
Te 1 =
i s i r r e d u c i b l e and t h e e i g e n v a l u e s r(T) = 1
and e v e r y e i g e n v a l u e h a s
T
( i v ) Let the p o s i t i v e o p e r a t o r Te2 = e 3
and
for
a > a.
are
A l = $4
a l l eigenvalues of A2,A3
EXERCISE 137.6. the Banach space T*
of
some
,
g E V
p(f) = 0
then
-
A = a-ib
For r e a l g E V
,
,
. It
T
, i.e.,
if
i.e.,
that
the eigenvalues one eigenvalue and
f o r some
of )I
IA21
= IA31
-
-
f = Tg
and T*
< 1.
and
a
and
and
f = Tg
-
Ag
,
Tf = 0
b
r e a l ) and
a n n i h i l a t e s the range
A1
for
annihilates
E V*
A = a + i b (with T*
T*$ = A$
.
T*$ = 0
if
the range
$ = T*$
follows t h a t i f
$(f) = 0
a = a.
3
. Show
i s a norm bounded o p e r a t o r i n
T
. Similarly,
t h i s means t h a t i f
then
Tel = e2
142
=
XI > 1
Show t h a t 6 = 1 5 '
,
T
then t h e n u l l space of
A
a.
i s the a d j o i n t o p e r a t o r , then the n u l l space
T*
$(f) = 0
the
A
( i ) Recall t h a t i f
a n n i h i l a t e s the range of
the n u l l space of
be given by
. Let
0 < a < a.
and f o r
we have
and
V
C3
a > 0
are real, for
T
a r e non-real.
450a = 91
Show t h a t f o r
3
A 2 = A 3 = -1J4
and
in
, where
Te3 = e l + 3ae2
i s r e a l and
Al
6 35
O R L I C Z SPACES AND IRREDUCIBLE OPERATORS
19,11381
Ch.
of
T-XI.
f o r some
( i i ) L e t the hypotheses of the generalized Jentzsch theorem be s a t i s f i e d . As proved, t h e s p e c t r a l radius T
of m u l t i p l i c i t y one, i d e . ,
and each non-null
r = r(T)
T-rI
such t h a t every element i n t h e n u l l uo i s a constant m u l t i p l e of uo Show now t h a t every element
i n the n u l l space of of
T-rI
i s an eigenvalue of
T
element i n the eigenspace i s a weak u n i t . I n o t h e r words,
there e x i s t s a p o s i t i v e weak u n i t space of
of
the corresponding eigenspace i s one-dimensional
(T-rI)2
.
i s already contained in t h e n u l l space
r = r(T)
i t s e l f ( i n more t e c h n i c a l terms, the eigenvalue
i s not
only of geometric m u l t i p l i c i t y one, b u t a l s o of a l g e b r a i c m u l t i p l i c i t y one). HINT: Observe t h a t the r e s t r i c t i o n on
L
r
T'
of
T*
to
LL
= LE
has spec-
and t h e r e e x i s t s 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 2 such t h a t @ E LL and T I $ = r $ Hence, i f (T-rI) f = 0 , then
t r a l radius g = (T-rI)f
satisfies
$(g) = 0
,
.
i.e.,
g = 0
.
138. The p e r i p h e r a l spectrum of i r r e d u c i b l e p o s i t i v e o p e r a t o r s
if^
I n Theorem 137.3 ( t h e generalized Jentzsch theorem) i t was proved t h a t T
i s a p o s i t i v e o p e r a t o r i n the space
reducible and belongs t o
i s an eigenvalue
(L2L)dd
,
L
such t h a t
T
then t h e s p e c t r a l radius
, l o of m u l t i p l i c i t y one. I f , furthermore, T
i s compact, irr(T)
of
T
i s irreducible
636
PERIPHERAL SPECTRUM
i n the s t r o n g e r sense t h a t
T
u n i t , then every eigenvalue T
Ch. 19,51381
u > 0
transforms e v e r y h
#
of
ho
T
satifies
in Ihl
i n t o a weak
L
= r(T)
< A.
.
If
i s i r r e d u c i b l e b u t does n o t s a t i s f y t h e l a t t e r s t r o n g e r c o n d i t i o n , t h e r e
may be many e i g e n v a l u e s o f a b s o l u t e v a l u e E x e r c i s e 1 3 7 . 5 ( i i i ) i n which
,
r(T) = 1
and a l l r o o t s of t h e e q u a t i o n
An - I
all
satisfying
i n t h e spectrum o f
X
pe ripheral spectrum of
T
.
T
, as
r(T)
shown f o r example i n
i s a g i v e n n a t u r a l number
n
a r e e i g e n v a l u e s of
1x1
T
.
The s e t of
i s cal l ed the
= r(T)
I t i s n o t by a c c i d e n t t h a t t h e p o i n t s of t h e
p e r i p h e r a l spectrum a r e e v e n l y d i s t r i b u t e d on t h e circumference of t h e c i r c l e of r a d i u s
.
r(T)
For t h e f i n i t e dimensional c a s e ( t h e m a t r i x c a s e )
i t i s a fundamental theorem, due t o G . Frobenius (C11,1908;C21,1912), if
T
i s p o s i t i v e and i r r e d u c i b l e (and
t h e n t h e r e e x i s t s a n a t u r a l number of
T
c o n s i s t s e x a c t l y of a l l
k
r(T) = 1
, say
that
f o r convenience)
such t h a t t h e p e r i p h e r a l spectrum
k-th
r o o t s of u n i t y . The theorem h a s been
extended t o p o s i t i v e i r r e d u c i b l e k e r n e l o p e r a t o r s i n Banach f u n c t i o n s p a c e s and, g e n e r a l l y , bo o p e r a t o r s
T
s a t i s f y i n g t h e hypotheses mentioned i n t h e
g e n e r a l i z e d J e n t z s c h theorem. There e x i s t even e x t e n s i o n s t o c e r t a i n i r r e d u k c i b l e p o s i t i v e o p e r a t o r s T such t h a t T and a l l i t s powers T ; k = 2,3, that
...
,
f a i l t o be compact. W e r e s t r i c t o u r s e l v e s h e r e t o t h e c a s e
T ( o r a power of
T ) i s compact. I n the proof t h e r e i s a c e r t a i n
f e a t u r e which i n t h e c a s e t h a t than i n t h e g e n e r a l c a s e t h a t this, recall that i f d e f i n e d by satisfies
sgn z = e
z = reia
ia
.
i s a Banach f u n c t i o n space i s much easier
L
i s an a b s t r a c t
L
# 0
Banach l a t t i c e . To e x p l a i n
i s a complex number, t h e n
is
sgn z
-
Note t h a t t h e c o n j u g a t e complex number sgn z -ia sgn z = s g n z = e Now, l e t L b e a Banach f u n c t i o n space
(on t h e measure s p a c e f u n c t i o n on
X
.
(X,A,p)
such t h a t
) and l e t
q(x) # 0
q(x)
for a l l
be a g i v e n measurable x
. The
operator
i s d e f i n e d by (Tqf)(x) = (sgn q (x )). f(x )
for a l l
x E
x
T
q
in
L
.
-1 has an in v ers e o p erato r T ( m u l t i p l i c a t i o n by sgn q ( x ) ) . q -1 Furthermore, 17 f l = I T f l = If1 f o r e v e r y f E L I f q i t s e l f i s con4 q t a i n e d i n L , t h e n T ( l q l ) = q and T-lq = I q / . The case t h a t L i s 4 q o f f i n i t e dimension ( i . e . , L = Cn f o r some n a t u r a l number n ) i s of Obviously
7
q
.
course i n c l u d e d . I t i s n o t inunediately e v i d e n t how t o d e f i n e a b s t r a c t Riesz space
L
,
T i n an q even though w e s h a l l assume from t h e b e g i n n i n g
that
637
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
Ch. 19,91381
i s Dedekind complete. One way t o s o l v e t h e problem i s t o r e p r e s e n t
L
L h a s a f u n c t i o n space (as w e w e r e o b l i g e d t o do i n Theorem 136.8). T h i s , however, o b s c u r e s t h e r e a l i s s u e . Although i t makes t h e d i s c u s s i o n somewhat l e n g t h i e r , we s h a l l a d h e r e t o a d i s c u s s i o n i n s i d e
i t s e l f . The f i r s t
L
t h i n g t o do i s
t o d e f i n e a n o p e r a t o r i n a n a r b i t r a r y (Dedekind complete)
R i e s z space
having similar p r o p e r t i e s as the operator
above
L
.
Let
e
pl
Denoting by
B
,. . . ,pk
j
z
# 0 ) . The o p e r a t o r s
j
d e f i n e d as f o l l o w s . F o r any
f = C
are g i v e n by
n-f
Note t h e f o l l o w i n g p r o p e r t i e s of
-
(i)
TI
(ii)
For
n (Isl)
k f l j
and
f = Zk (sgn z . ) f 1 ~j
TI
, we
p
e
such t h a t
have
B
,..., .
complex numbers, a l l and
mentioned
Q
...
:Z p j Q
Bk
= e
TI
-
n
(wihh
nif and
IT
in
(sgn z . ) f
TI
I
.
j
L
are
n f
f . E B. ) the elements J J
1;
=
-
and
.
L
=
j I k can b e w r i t t e n ( u n i q u e l y ) a s f = Z f with k l j k Now, l e t s = Z z j p j b e g i v e n ( a l l z
f E L
j = I
for
~j
b e d i s j o i n t components of
t h e band g e n e r a t e d by
T h e r e f o r e , each f. E B
q
b e a p o s i t i v e weak u n i t i n t h e Dedekind complete Riesz s p a c e
and l e t
L
n
j *
.
n = n IT = I (where I i s t h e i d e n t i t y o p e r a t o r i n L ) . s s s s f = i s ( w e have I s / = ( C z . p . 1 = C / z . p . \ = Z l z . 1 ~ Hence
-
and
= s
J J use h e r e t h a t i f
. We
Is1
IT s =
g
3 3 I g2
in
~j L
,
.
then
1 + I g2 I (Theorem 9 1 . 4 ) . 1 ( i i i ) For any f E L we have
/g,+g2/ = l g
Similarly
-
IT f I
=
( i v ) For any
If1
.
u t 0
we have
I ITs I (u Hence
ITI I (v)
= I
. Similarly,
-
I
= I
.
I t f o l l o w s immediately from t h e d e f i n i t i o n s t h a t i f
,
then
nsf I g
If
I 181: )
n
a r e orthomorphisms i n
-
In
next chapter.
L
. Similarly ,
for
IT-
. The
operators
f I g (i.e., n
and
a type of o p e r a t o r s t o b e d i s c u s s e d i n t h e
PERIPHERAL SPECTRUM
638
The c o n s t r u c t i o n of
. Instead
s
q"
r e a l ) t o b e given such t h a t
w e now assume a f i x e d element
s
a r b i t r a r y ) . Note t h a t both val q"
C-lql, l q l ]
. We
q'
and
q = q ' + i q " (4'
i s a weak u n i t i n
(91
q"
as u n i t , i . e . ,
Iql
a r e contained i n the o r d e r i n t e r -
b e lower sums f o r
q'
. These
of t h e i n t e r v a l [ - 1 , 1 ] k S'I = E l yjpj with a l l components of Iql s = s'
+ is"
for
and
q'
and
a l l approximating lower and upper
sums are f i n i t e l i n e a r combinations of components of
s"
and
L (but o t h e r w i s e
now apply F r e u d e n t h a l ' s s p e c t r a l theorem t o
with respect to
19,51381
i s n o t d i f f i c u l t because of t h e s p e c i a l n a t u r e
IT
of
of
Ch.
Iql
. Let
s'
and
corresponding t o t h e same p a r t i t i o n
q"
sums may be w r i t t e n as
s' =
k
x j p j and x y i n [-1,11 and a l l p . m u t u a l l y d i s j o i n t J j' j W r i t i n g z = x j + i y j , t h e approximating sum j k Let O < E l < t q = q ' + i q " becomes s = E l z j p j
.
'1
.
b e given. We may assume t h a t t h e approximation i s a l r e a d y so c l o s e t h a t
0
S
q'
-
s'
S
0
~ ~ 1 4and 1
S
q
i s a g a i n t h e band g e n e r a t e d by
(where B q. E B ~j j q = C q j = C(q!+iq'!) Hence J J
ponents
.
q" - s" S ~ ~ 1 9 1 Decomposing
.
i n t o comp. ) , w e have J
for a l l
This shows t h a t of
[-l,l]
12.1
, each
2 1
-
2~
J 1 s e p a r a t e term
1
t
for a l l
zjpj
j
. Refining
j
the p a r t i t i o n
i s r e p l a c e d by a f i n i t e sum
z ' p ' ( t h e elements p: m u t u a l l y d i s j o i n t components o f IqI w i t h m m m Em p: = p . ) such t h a t e a c h z ' s a t i s f i e s Iz'-z.I S Z E ~ (because t h e new J m m J approximating lower sums f o r q ' and q" m a j o r i z e t h e o r i g i n a l ones b u t C
cannot exceed
q'
and
q" ) . Now, choose
1z'-z.1 S Z E ~ f o r a l l m m J have lsgn zA-sgn z.1 < E J
, it
E
> 0
follows t h a t f o r
for a l l
m
. Hence
. Because
I z . I > 1 and J s u f f i c i e n t l y s m a l l we
i n t h e o r i g i n a l approximating sum, This h o l d s n o t only f o r one t e r m zjpj b u t f o r a l l terms s i m u l t a n e o u s l y . S i m i l a r l y , i f f E L i s a r b i t r a r y and
Ch. 19,11381
6 39
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
i s t h e decomposition of f i n t o d i s j o i n t terms c o r r e s p o n d i n g t o j i s decomposed i n t o CmfA , and w e g e t t h e o r i g i n a l p a r t i t i o n , then e a c h f j
f = Xf
Denoting t h e o r i g i n a l approximating sum by
w e have
j
for all
) ( s g n z.)f.-X (sgn zA)fAl < Elfjl J J ~
.
and i t s r e f i n e m e n t by
s
.
-,
s
(sgn z . ) f and IT f = X.{X (sgn zA)f;) Hence ~j SJ m (sn:n=1,2, ...) i s a ITisf-rs-f I < E X 1 f . 1 = E / f 1 T n i s shows t h a t i f J sequence of approximating lower sums, each s corresponding t o a refinen+ 1 t h e n t h e sequence (a f : n = 1 , 2 , ) is ment of t h e p a r t i t i o n of s n ' sn I f [ - u n i f o r m l y convergent. W e denote t h e l i m i t b y TI f Thus, f o r any E > 0 q there exists n such t h a t TI f-71 f l 5 Elf1 f o r a l l n 2 n Note t h a t E q 'n n does n o t depend on f S i m i l a r l y , (Tsnf : n = l , 2 , . .) converges t f I-uniIT
f = Z
j
.
...
.
.
formly. L e t u s d e n o t e t h e limit by
-
and
. For
TI
q (i)
convenience, w e w r i t e
Since
lnnfl = I f /
hand s i d e converges f
every
E L
(ii)
Hence Hence
.
f o r every
If I-uniformly t o IT-fI = If1 q u 2 0 w e have
-
.
IT I
for
IT
Similarly
For any
. We
n-f 9
.
f
lnqfl
q (iv)
let
E >
for
n
f I g
. Similarly
We prove t h a t 0
b e g i v e n . Then
for IT
q
.
l i s t some p r o p e r t i e s of
.
IT
n ' and a l l
, we
n
have
IT
q
and since t h e l e f t
ITI f l q
= If1
for
.
= I S i m i l a r l y /IT I = I q q ( i i i ) I f f I g ( i . e . If1 I lgl ) , then IT
.
IT
-
.
4 (lql) = q
and
nnf I g 71-q = l q ( q
( a s a l r e a d y proved).
. For
t h i s purpose,
s u f f i c i e n t l y l a r g e . Hence
for large
n
. Since
also
Isn-ql
S
Elql
for large
n
, it
follows t h a t
PERIPHERAL SPECTRUM
640
(v)
f o r every
for
-
. Similarly,
a (141) = q q
Given f E L
and a l l
. This
(vi)
For t h e proof t h a t a--a n-\f q q nnI
n
for a l l
-
q q
n
, we
= I .
(vii)
f o r every
= (a -a
q
-
n
)a-f q
+
-a ) f q n
n t n
, we
I
5
clf I
,
u t 0
.
write
- -
71
n
( a -a ) f = t + t q n 1 2 '
I (a a--a a-)fI 5 2E / f I f o r l a r g e n q q n n find that a a f = f , i.e., a a = I q q q q in
T
u
z o
we have
L
. Similarly
i s given i n s o does
rrf
< u
-
and
L
. Indeed,
,
then
g
If1 = la gl = Igl 5 u
T h i i shows t h a t
f
a
.
IT-TI = I T /
ITal
.
= IT1
. Since ana-fn . Similarly,
a For any o r d e r boundq This f o l l o w s from
. Note
f i r s t that i f
u t 0
r u n s through a l l elements s a t i s f y i n g If1 5 u If1
5
u
i s of t h e form
. It
/Tal = IT1
l n f l = If1 < u
implies
g = nf
. Similarly
From ( v i i ) and ( v i i i ) we g e t
(x)
For
k = 1,2,
for
... i t
1Trr-l
f = a-g
= (TI
with
.
1aTa-l = In-Tal
f o l l o w s from
~a
-
-
,
and c o n v e r s e l y ,
follows t h a t
(ix)
= f
.
i n s t e a d of
ITTI = IT1
( v i i i ) It i s a l s o t r u e t h a t
lgl
for
S EI
a a = a a = I q q q q
F o r convenience, w r i t e
ed operator
if
f o r any
sufficiently large,
I t follows t h a t
'7171
la -a I q -n
shows t h a t
I (a
such t h a t
n
. Hence,
n t n
n Z n
(. For
.
a q = 191 q > 0 , there exists
E
Ch. 19,11381
=
= a v = I
IT1
.
that
then
Ch.
( ~ r T n - )= ~ ITT~II- and Hence, f o r
'
k = l,2,
(IT-TIT= ) ~x-Tkz
.
complex,
h
k
k
-?)
-
\ITTIT
for
64 I
ORLICZ SPACES AND IRREDUCIBLE OPERATORS
19,11381
= {IT(T-AI)~T-'
I
... . This
complete), then
k - n(T-hl)
shows t h a t i f
(T-hI)k
L
i s a Banach l a t t i c e (Dedekind
.
has a bounded i n v e r s e i f and only (ITTIT--XI)~
has a bounded i n v e r s e . I n p a r t i c u l a r , these o p e r a t o r s have the same spectrum
i s compact, they have the same eigenvalues
T
and i f
# 0
,
not only with
the same geometric m u l t i p l i c i t i e s , but a l s o with the same a l g e b r a i c multiplicities. THEOREM 138.1. A s
in the generalized Jentzsch theorem, l e t L be a = I01 and l e t T be a Dedekind complete Banach l a t t i c e such t h a t (L'):. p o s i t i v e operator i n L such t h a t T is compact and i r r e d u c i b l e and T belongs t o r
= r(T)
that
IS1
. Then
(L$)dd
i s p o s i t i u e . Let 5
T
and l e t
corresponding
IT
be an order bounded operator i n
S
be an eigenualue of
A
there e x i s t s an element
(as we know already) the s p e c t r a l radius
i n L such t h a t
q
satisfies
4
PROOF. The number
h
S = hr
-1
-
n TIT q
q
.
S
Iql
i s an eigenvalue of
sponding eigenelement, i . e . ,
Sq = hq
. It
S
such t h a t
L
such
/A1 = r
. Then
is a weak u n i t and t h e
. Let
q # 0
be a corre-
follows t h a t
A s shown i n the proof of the J e n t z s c h theorem, t h e r e e x i s t s a s t r i c t l y posit i v e l i n e a r functional r e s t r i c t i o n of
This shows t h a t
T*
to
C$
on
L"
). Hence
n
T(lq1) = r l q l
T(lql) =
1s
L
such t h a t
. Therefore,
(Iql) = rlql
T'q = r$
(where
T'
i s the
from
.
Again from the Jentzsch theorem, i t follows t h a t
Jql
i s a weak u n i t i n L .
PERIPHERAL SPECTRUM
642
T - IS1
Since
(T-ISl)(g)
both
T
and
IS1
(T-lSl)(g) = 0 Since
=
0
g
f o r every
i n t h e i d e a l g e n e r a t e d by
are o r d e r c o n t i n u o u s ( s i n c e
g
f o r every
i n t h e band
i s a weak u n i t , w e have
141
i s the n u l l operator, i . e . ,
9
(141) = q
Writing tor that
U
.
= rlq/
and
=
u1
.
g e n e r a t e d by
. It
(lql)dd = L
. But
E ( L 2 L ) d d ) ; hence,
IS1 5 T
(Iql)dd
find
Iql
.
Iq/
T - /SI
follows t h a t
IS1 = T L e t IT be the o p e r a t o r i n L 4 i n t r o d u c e d and d i s c u s s e d above. R e c a l l t h a t - -1 n-o, = Iql Now, w r i t e U = Xr IT S n Then q q q
corresponding t o TI
, we
(T-lSl)(lql) = 0
i s a p o s i t i v e o p e r a t o r and
that
Ch. 19,51381
+
q
, as
iu
(U1
.
.
and
U2
r e a l ) , i t follows t h a t
U (/q1)
=
I U i = In Sn 1 = = T and U 5 I U 1 / 5 IUI , t h e operaq q 1 i s p o s i t i v e and s a t i s f i e s ( T - U l ) ( / q l ) = 0 As above i t f o l l o w s
Since
T-UI
.
i s t h e n u l l o p e r a t o r . Hence
T-UI
, i.e., the real part
U1 = T = I U /
.
of U = U. + i U 2 s a t i s f i e s U 1 = IU/ This i m p l i e s I s i m p l e remarks about t h i s c o n c l u s i o n we r e f e r t o E x e r c i s e - -1 have U = U = T , i . e . , T = Ar IT SIT It follows t h a t 1 q q Observing now t h a t hr-] = r1-I , w e f i n d t h a t S = Xr-*n U1
.
U2
0 ( f o r some
=
138.3). Thus we
-
-
.
n Tn = Xr-IS q- q Tn q q
.
THEOREM 138.2 (Generalized Frobinius theorem). Once more, Zet
L
be
a Dedekind complete Banach Zattice such t h a t '(LC;' = I01 and Zet T be a p o s i t i v e operator i n L such t h a t T i s compact and i r r e d u c i b l e and T
belongs t o
. The spectral, radius
I f the peripheral spectrum of A A
of
, then these nunbers are exactly the
k' -.- r' t k= 0 . Furthermore,
the spectrum
d i f f e r e n t eigenvalues
k
roots of the equation
k
T
of
o(T)
is denoted by r = r ( T ) .
T
c o n s i s t s of t h e
T
i s i n v a r i a n t under
2rk-I
the r o t a t i o n of the ocmplex p Z m by t h e angZe
(multiplieities in-
cZuded). This implies i n particuZar t h a t a12 eigenvalues
XI,.
.., A k
i n the
peripheral spectrum are of m u l + i @ l i e i t y one. PROOF. P-assing. from T t o r-'T
i f n e c e s s a r y , w e may assume t h a t t h e
spectral radius r = r ( T ) s a t i s f i e s r = 1 p e r i p h e r a l s p e c t r u m of A = a
,X
=
T
. Applying
a
6 be points i n the
and
5 , we f i n d t h a t t h e r e e x i s t e l e m e n t s qa
the corresponding operators sati s f y
. Let
t h e l a s t theorem t o
n
q
( c a l l these
n
and
and
n
5
S = T
q5
and such t h a t
for brevity)
ch. 19,51381
643
ORLICZ SPACES AND IRREDUCIBLE OPERATORS T = a n Tn
-
and
C X C I
-
T = Bn Tn
B
B ’
--
. Hence
T = f3agTnB
From t h e l a s t e q u a l i t y i t f o l l o w s t h a t
-
wich i m p l i e s
Applying ( x ) from t h e l i s t of p r o p e r t i e s a t t h e b e g i n n i n g t o we f i n d t h a t t h e n u l l s p a c e s of Since t h e n u l l space of
T-I
and
T
-
-
0.6
Ba’
and
TI
T = aBa n Tn
n
IT
B ’
C I ~have I t h e same dimension.
i s of dimension one (by t h e J e n t z s c h theorem),
T-I
.
t h e same hold6 t h e r e f o r e f o r t h e n u l l s p a c e of T - aEI I n o t h e r words, -1 a; = aB i s a n e i g e n v a l u e of T of m u l t i p l i c i t y one. Obviously, ai b e l o n g s t h e n t o t h e p e r i p h e r a l spectrum. I t h a s been shown t h u s t h a t i f CI
and
b e l o n g t o t h e p e r i p h e r a l spectrum, t h e n s o does
B
. It
-1
af3
f o l l o w s t h a t t h e p e r i p h e r a l s p e c t r u m i s a (commutative) group w i t h r e s p e c t t o m u l t i p l i c a t i o n a s group o p e r a t i o n . S i n c e t h e group c o n s i s t s of t h e elements
A ,...,Ak
equation
Ai
-
For
CI
= e
1 =
, these 0 , i.e.,
we may assume t h a t
i t f o l l o w s from
o(T) = a (uaaTn:)
T =
= au fnaTn-
a/
CIT
An = e
Tn-
a
, i.e.,
...,k) .
, T
i s invari-
i n v a r i a n t under r o t a t i o n of t h e
2n/k ( a l l m u l t i p l i c i t i e s i n c l u d e d ) . F i n a l l y , n o t e t h a t
p l a n e by an a n g l e t h e non-positive
a
(n=l,
that
o
= aa(T)
2ninlk
a g a i n by p r o p e r t y (x) from t h e l i s t . Hence, t h e s p e c t r u m of a n t under m u l t i p l i c a t i o n by
k
k d i f f e r e n t r o o t s of t h e
numbers must b e t h e
e i g e n v a l u e s i n t h e p e r i p h e r a l spectrum cannot have a
positive eigenelemnt.
A s o b s e r v e d e a r l i e r , t h e same r e s u l t s h o l d i f pact, but
Tn
i s compact f o r some n a t u r a l number
T n
i t s e l f i s n o t com-
. The p r o o f s
of
Theorems 138.1 and 138.2, a s p r e s e f i t e d h e r e , a r e m o d i f i c a t i o n s of t h e proof f o r t h e m a t r i x c a s e by H. Wielandt (C11,1950). EXERCISE 138.3. L e t s p a c e and l e t and l e t
L+iL
f = g + ih
L
b e an Archimedean and uniformly complete R i e s z
be i t s c o m p l e x i f i c a t i o n . Furthermore, satisfy
H I N T : I t f o l l o w s from
If1 = l g l
. Show
that
h = 0
let
.
g,h E L
PERIPHERAL SPECTRUM
644
that
l h j s i n e 5 (I-cosO)lg(
lhlcosi8 5 Ig/siniO
,
for all (hl
i.e.,
5
Ch. 19,51383
satisfying
'8
Igltg!e.
Let
0
0 < 0
5
+0.
. Hence
in
EXERCISE 138.4. With t h e n o t a t i o n s of E x e r c i s e 1 3 7 . 5 ( i i i ) , l e t the positive operator
T
in
Cn
b e g i v e n by
T e l = e2
, Te2
= e3,
...,T en
I n E x e r c i s e 1 3 7 . 5 ( i i i ) i t was asked t o p r o v e t h a t t h e e i g e n v a l u e s
X ],...,X
of
Xk = ak n f o r belonging t o +
... +
h;-'el
T
are t h e r o o t s of
k = I,
Xk
.
...,n
, where
An
-
1 = 0
a= e 2ni'n
c o n s i s t s of a l l m u l t i p l e s of
. We may assume . Show t h a t t h e e
n
+
Xk en-1
-
that eigenspace 2 + k n-2
+ A e
CHAPTER 20
ORTHOMORPHISMS AND f -ALGEBRAS
The o p e r a t o r
i n t h e Archimedean Riesz space
IT
s e r v i n g i f t h e image in
L
.
of
n(B)
satisfies
B
i s s a i d t o b e band pre-
L
s(B) c B
f o r e v e r y band
I n t h e f i r s t theorem i n s e c t i o n 139 we prove t h a t
v i n g i f and o n l y i f
f Ig
in
implies
L
. Any
Tf I g
B
i s band p r e s e r -
T
operator
in
IT
L
which i s band p r e s e r v i n g and o r d e r bounded i s c a l l e d a n orthomorphism. The
set
Orth(L)
of a l l orthomorphisms i n
v e c t o r space ordering i n
i s i n h e r i t e d by
Lb(L)
i s Dedekind complete, t h a n
. Every
Riesz space
Lb(L)
sense t h a t
uo 2 u
J. 0
t u r e of t h e s p a c e
Orth(L)
TI
(nu)
and a l l
t
,
IT
-
u E L+
u = (nu)-
. In
s e c t i o n 140 t h e s t r u c -
i s i n v e s t i g a t e d . It i s shown t h a t ( a l t h o u g h
Orth(L)
i s an Archime-
( T I , V I T ~ ) (=U )( a , u ) v (a u> f o r a l l TI ? , n 2 i n 2 f o r n1 A IT^ I n p a r t i c u l a r IT'U =
. Similarly
lai(u) = lnul
and
.
.
The method of p r o o f , due t o
Luxemburg, does n o t make use of any r e p r e s e n t a t i o n by f u n c t i o n s (as
W.A.J.
i n e a r l i e r methods). Denoting n u l l s p a c e and range of an orthomorphism by
L
i s o r d e r continuous i n t h e
inflru I = 0
implies
dean Riesz s p a c e such t h a t Orth(L)
L. The p a r t i a l
I t i s e a s y t o see t h a t i f
i s a band i n t h e Dedekind complete
orthomorphism
Orth(L)
.
Orth(L)
i s n o t a Riesz s p a c e i n g e n e r a l ) t h e s p a c e
Lb(L)
=
i s a l i n e a r subspace of t h e
L
of a l l o r d e r bounded o p e r a t o r s i n
Lb(L)
Na
and
Furthermore
r e s p e c t i v e l y , w e have
Ra
N
T
= I
Rd
T
I
and
IT^
I
IT^
in
N
= Nlal
O:th(L)
and
i s a band i n
Na
implies
IT
R,I1l
RTI
L.
2 in
It i s i m p o r t a n t f o r t h e a p p l i c a t i o n s t o observe t h a t i f t h e orthomorphisms
n1
and
IT^
c o i n c i d e on a n o r d e r dense s u b s e t of
on t h e whole s p a c e The space
L
.
Orth(L)
L
( l a t t i c e ordered algebra) i f t h e r e e x i s t s i n
A
A
Riesz a l g e b r a
A
an a s s o c i a t i v e m u l t i p l i c a uv t 0
i s c a l l e d commutative i f
645
To
i s c a l l e d a Riesz a l g e b r a
t i o n w i t h t h e u s u a l a l g e b r a p r o p e r t i e s and such t h a t
. The
then they coincide
i s n o t o n l y a Riesz s p a c e b u t a l s o an f - a l g e b r a .
e x p l a i n t h i s , we mention t h a t t h e Riesz s p a c e
u,v E A+
,
a
for all
fg = fg
for all
CHAPTER 20
646
. The
f,g E A
t i o n a l property t h a t
u
w E A+
C(X)
. The
c a l space
A
Riesz a l g e b r a algebra
implies
(uw)
v = (wu)
A
Archimedean f - a l g e b r a i s commutative. The s p a c e f-algebra with the i d e n t i t y operator i n formly complete, t h e n s o i s then so has
. In
Orth(L)
orthomorphisms. I f n ( f ) = pf
i s of type
or
0
of a s e t
for a l l
We prove t h a t every
i s a n Archimedean
Orth(L)
a s u n i t element. I f
L
i s uni-
L
has the p r o j e c t i o n property,
L
s e c t i o n 141 we s h a l l d i s c u s s some examples of
,
L
t h e n t h e r e e x i s t s an element f
. This
E L
p
E
such
L
r e s u l t covers t h e c a s e t h a t
L
i s t h e s p a c e of a l l r e a l u-measurable f u n c t i o n s on
L
a s e t c a r r y i n g a measure
c"(0)
. If
Orth(L)
holds f o r a l l
C(X)
v = 0
i s an Archimedean f - a l g e b r a w i t h u n i t element and
L
n i s an orthomorphism i n that
A
h a s t h e addi-
of a l l r e a l continuous f u n c t i o n s on a topologi-
i s a s t a n d a r d example of an f - a l g e b r a .
X
A
i s c a l l e d an f - a l g e b r a i f
v = 0
A
ch. 201
a s w e l l a s t h e case t h a t
L
i s t h e bicommutant
of mutually commuting Hermitian o p e r a t o r s i n a H i l b e r t
space. In s e c t i o n 142 we prove s e v e r a l simple p r o p e r t i e s o t t - a l g e b r a s . I t 2 2 i s t r u e i n any f - a l g e b r a A t h a t f 2 0 f o r e v e r y f E A and (UAV) = 2 (uvv)2 = u v v2 f o r a l l u , v E A+ I f A has a = u2 A v2 as w e l l a s u n i t element -1
u v
, we
,
then
u-'
E A'
e
have
.
e
2
0
and
and f o r any u
-1
-1
< e
-1
u
. If
E A+ u,v
-1
p o s s e s s i n g an i n v e r s e
E
A'
and b o t h
-1
-1
= u A v and (UAV) = u (uvv) 2 i s semiprime ( i . e . , f = 0 i f and only i f f = 0 ) , then
have an i n v e r s e , t h e n
If
u
A
A
u
and
.
v v
-1
f
and
a r e d i s j o i n t i f and only i f f g = 0 and 0 2 u 2 v h o l d s i f and only 2 In an Archimedean f - a l g e b r a we have f g = ( f v g ) ( f h g ) f o r a l l i f u 2 v2
g
f,g E A
. If
.
A
i s Archimedean and has a u n i t element,
t h e n A i s semiprime,
s o t h a t i n t h i s case a l l t h e above mentioned p r o p e r t i e s h o l d . In p a r t i c u l a r ,
a l l these properties
hold i n
. It
Orth(L)
i s an f - a l g e b r a w i t h u n i t element
e
w i l l a l s o b e proved t h a t i f u E A+
and
..
, i s i n c r e a s i n g and converges sequence u = i n f ( u , n e ) ; n = 1 , 2 , . n 2 u -uniformly t o u . Hence, i f n i s a p o s i t i v e orthomorphism i n L n
n
=
then
inf(n,nI) nn
for
n = l,2,
...
(where
converges n2-uniformly t o
T
I
and
i s the identity operator i n
. This
n
in
L = Lq(X,p) ; 0 i q < q t h e r e e x i s t s a f u n c t i o n p E L_(X,p)
L
9 f E L , we have ( n f ) ( x ) = p ( x ) . f ( x ) q We prove a l s o i n t h i s s e c t i o n t h a t f o r any
such t h a t , f o r any g i v e n everywhere on
X
.
t h e a d j o i n t o p e r a t o r n-
belongs t o
duce i d e a l p r e s e r v i n g o p e r a t o r s
71
Orth(L-)
L ),
-.
r e s u l t makes i t p o s s i b l e t o
determine a l l orthomorphisms i n a space o f type For any orthomorphism
A
i s given, then t h e
. In
almost TI
E Orth(L)
s e c t i o n 143 we i n t r o -
i n t h e (Archimedean) Riesz space
L
a. 201
647
ORTHOMORPHISMS AND f-ALGEBRAS
(i.e.,
ITU
f o r every
E A
U
g e n e r a t e d by
u ). Any
u
, where
E L+
i s the p r i n c i p a l i d e a l
AU
having t h i s property i s c a l l e d a contractor
IT
L . E v e r y c o n t r a c t o r i s o r d e r bounded. Hence, e v e r y c o n t r a c t o r i s an
in
orthomorphism ( b u t n o t c o n v e r s e l y , i n g e n e n a l ) . The s e t
. The
i s an ideal i n
L
contractors i n L
ideal
Z(L)
generated i n
i s i n general s t i l l smaller than c e n t r e of in
,
L
. If
L
then
and i s c a l l e d t h e s t a b i l i z e r of
OrthfiL)
by t h e i d e n t i t y o p e r a t o r
S (L )
. The
ideal
Banach l a t t i c e , t h e n e v e r y orthomorphism i n Orth(L) = Z(L)
L
.
i s a n orthomonphism
IT
.
E Z(L)
IT
I
i s c a l l e d the
Z(L)
i s normed ( s e c t i o n 1 4 4 ) and
L
of a l l
Orth(L)
i s norm bounded i f and o n l y i f
IT
S(L)
If
L
is a
i s norm bounded, i . e . ,
I n s e c t i o n 145 we d i s c u s s some of t h e r e l a t i o n s between orthomorphisms and Radon-Nikodym t y p e p r o p e r t i e s . I f
L
i s Dedekind complete, i f
L" and J, i s c o n t a i n e d i n t h e i d e a l g e n e r a t e d by $ , t h e n t h e r e n e x i s t s an orthomorphism IT i n L s u c h t h a t J, = $IT Furthermore, J', =
0 5 $
E
.
= $IT+ and
that
J,-
[$I c"(D)
= $ I n [ ) . This i s applied t o t h e case
= $IT- (and hence
i s the bicomutant
L
of a non-empty
muting H e r m i t i a n o p e r a t o r s i n t h e H i l b e r t s p a c e r e p r e s e n t e d by t h e element all
A E L
t e d by
,
.
H
i n the sense t h a t
then every p o s i t i v e l i n e a r f u n c t i o n a l
i s of t h e form
$
x E H
D of m u t u a l l y comI f now 0 5 $ E L" is
set
$(A) = (Ay,y)
$(A) = (Ax,x) J,
for
i n t h e band genera-
f o r an a p p r o p r i a t e
y € H
.
Furthermore we prove t h a t i t f o l l o w s from t h i s r e s u l t t h a t f o r e v e r y p o s i tive
in
$
L :
holds f o r a l l
t h e r e e x i s t s a n element A
E
L
t h a t every p o s i t i v e
. In
in
$
x E H
such t h a t
$(A) =
AX,^)
t h e terminology of von Neumann a l g e b r a s t h i s means
L i
i s a normal s t a t e . T h i s i s R. P a l l u de l a
B a r r i s r e ' s theorem (1954); o u r method of proof i s due t o P.G. Dodds (1974).
IT
i n t h e Archimedean R i e s z s p a c e
a Riesz subspace. I f
IT
an i d e a l i n g e n e r a l . I f
i s p o s i t i v e , then L
. I n general
L
(A. B i g a r d , 1972) and r e c e n t l y (C.B.
Huijsmans and B.
s t r o n g e r r e s u l t was proved t h a t
i s an i d e a l i f
te
RIT i s n o t even
RIT i s a Riesz s u b s p a c e , b u t n o t
i s Dedekind complete, then RIT
of an orthomor-
Rx
I n t h e f i n a l s e c t i o n 146 w e i n v e s t i g a t e t h e r a n g e phism
RIT i s an i d e a l de P a g t e r , 1980) t h e
i s uniformly comple-
L
and normal ( r e c a l l t h a t L i s s a i d t o be normal i f u A v = 0 i m p l i e s d d + { v } ) . We s h a l l o b s e r v e t h a t t h e combination of uniform comple-
L = {u)
t e n e s s and n o r m a l i t y i s e q u i v a l e n t t o t h e o - i n t e r p o l a t i o n p r o p e r t y ( i . e . ,
f 1. 5g
n n for a l l
C
i m p l i e s t h e e x i s t e n c e of a n e l e m e n t n )
. In
h
satisfying
f
5 h 5 g
t h e e x e r c i s e s we i n d i c a t e a proof t h a t e v e r y p o s i t i v e
element i n a u n i f o r m l y complete Archimedean f - a l g e b r a
A
n
w i t h u n i t element
648
ELEMENTARY PROPERTIES
h a s a p o s i t i v e square r o o t . More g e n e r a l t y , i f h a s no u n i t e l e m e n t ) , every product f
2
+ g
2
, has
(f,gEA)
20,51391
Ch.
i s semiprime (even i f
A
,
uv (u,vEA+)
A
as w e l l a s every sum
a p o s i t i v e square r o o t .
139. Elementary p r o p e r t i e s of orthomorphisms Let
b e an Archimedean Riesz space. The o p e r a t o r
L
n: L
+
L ) i s c a l l e d band preserving i f t h e image
n(B)
C
B
f o r every band
B
in
THEOREM 139.1. For an operator
equiva Zen t
.
(i)
n
(ii)
n(D ) c D
n(B)
.
L
L
in
T
of
(iii) u
A
d
f o r every non-empty subset
v = 0
(iv)
f Ig
(v)
nf E
i n L implies L
in
rf I g
implies
f E L
f o r every
. .
I nf I
If these conditions are s a t i s f i e d , t h e n
D
.
nu 1 v
L (i.e.,
satisfies
B
the f o z l m i n g conditions are
i s band preserving. d
in
n
of
L
.
I a( I f I ) I ) hoZds f o r every
=
f E L . PROOF. ( i ) (ii)
=+
( i i ) This i s e v i d e n t s i n c e
( i i i ) If
=+
u
A
v = 0
,
Dd
u E {vId
then
i s a band.
. Hence
nu E {v)
d
, i. . e . ,
n u l v . (iii)
= 0
. Then (iv)
fore
*
(v)
and
(v) Let
=$
nf I g
f I g
( i v ) From
af+ I g
. This
( i ) Let
g
, and
hence
be any element i n
shows t h a t B
i t follows t h a t
nf- I g
nf €
L
be any band i n
{fId
.
/gl = 0
and
rrf = (nf+-nf-)
f+
1g
A
. Then
f E B
and l e t
, and
f I g
. Then
f-A l g l =
.
there-
nf E { f l d d C
C B . a r e s a t i s f i e d . and l e t
Assume now t h a t c o n d i t i o n s ( i ) - ( v ) given. I t f o l l o w s from
f + I f-
lnf++nf-l = laff-nf-l
But t h e n
nf+ I f -
that
. i.e.,
Any band p r e s e r v i n g o p e r a t o r i n
, and
hence
InfI = I n ( l f l ) l
L
.
f E L
nf+ I nf-
be
.
which i s a t t h e same time o r d e r
bounded i s c a l l e d an orthornorphisrn. From t h e above theorem i t i s e v i d e n t that i f
'IT
i s a positive operator i n
L
, then
T
i s an orthomorphism i f
ch. 20.11391
ORTHOMOWHISMS AND f-ALGEBRAS
and only i f
u
A
v = 0
in
implies
L
s e t of a l l orthomorphisms i n
by
L
nu
A
. With
L
.
Orth(L)
i s a l i n e a r subspace of t h e v e c t o r space operators i n
v = 0
649
. We
s h a l l denote t h e
I t i s obvious t h a t
Orth(L)
of a l l o r d e r bounded
$(L )
r e s p e c t t o t h e p a r t i a l o r d e r i n g i n h e r i t e d from
Orth(L)
i f TI u 5 a u f o r a l l u E L+ , the v e c t o r space n2 1 2 i s an o r d e r e d v e c t o r space. Note t h a t i f 0 5 TI 5 n2 i n $(L)
and
i s an orthomorphism, t h e n s o i s
,
Lb(L)
i.e.,
a2
TI
1
i s Dedekind complete, t h e space
%(L)
I n t h i s case t h e o r d e r bounded o p e r a t o r nTI+ and
and only i f =
il-
1x1
t h e s p e c i a l case t h a t
in
71
L
. In
L
i s an orthomorphism i f
a r e so. This follows by observing t h a t u E L+
sup(nw:O~ww
i f and only i f
. In
TI^
i s a Dedekind complete Riesz space.
o t h e r words, n
n+u =
i s an orthomorphism
i s so. I t f o l l o w s , t h e r e f o r e , immediately t h a t Orth(L)
i s an i d e a l (and even a band) i n t i o n 142) t h a t i n f a c t
. It
$(L)
w i l l be proved l a t e r ( i n sec-
i s t h e band g e n e r a t e d i n
Orth(L)
$(L)
by t h e
i d e n t i t y operator. EXAMPLE 139.2. We p r e s e n t a band p r e s e r v i n g o p e r a t o r which f a i l s t o b e o r d e r bounded. Let
C0,I)
b e t h e v e c t o r space of a l l r e a l f u n c t i o n s
L
f o r which t h e r e e x i s t s a p a r t i t i o n
depending on
fore, a t the points ing, L
x.
v a t i v e of
f
rr
( a t each p o i n t of
af I g
, i.e.,
S
fn 5 e
sequence
for a l l (nfn:n=1,2,
to
may be d i s c o n t i n u o u s , t h e r e respect t o the pointwiseorderrrf
t o be t h e r i g h t d e r i -
n (where
...)
and
'II
i s li-
i n t o i t s e l f . Furthermore, f I g
L
i s band p r e s e r v i n g . However,
n
= 1 (n
x
C X ~ - ~ , X i~s ) li-
C 0 , l ) ) . C l e a r l y , nf E L
i s an o p e r a t o r from
bounded, because t h e r e e x i s t s a sequence
0
f
i s an Archimedean Riesz s p a c e . We d e f i n e
near, so implies
. The f u n c t i o n f ( i = l , ...,n-I) . With
i = I,..,,n
n e a r f o r each
... <
0 = x0 < x 1 <
f ) such t h a t t h e r e s t r i c t i o n of
on
f
e(x) = 1
TI
in
(fn:n=1,2, ...) for a l l
x E [O,l)
i s not o r d e r bounded i n
L
.
in
i s not order L
such t h a t
) , whereas t h e
Orthomorphisms a r e o r d e r continuous. To s e e t h i s , we f i r s t prove a lemma.
LEMMA 139.3. If
uo 2 uT C 0 + dd
nT {(UT-EUO)
}
=
{o}
in
.
L
and
@ <
E
< 1
, the%
L
650
o s
PROOF. Let
Since
(ELI
,
i.e.,
=
0
O
-u
v
v E {(u
T
-Eu
o
)+ldd
for a l l
) + 4 &uO on account of
T
A
0 S
u
=
0
1,
0
-EU
\ T
.
, it
C 0
,
i.e.,
follows t h a t
v
A
EU
,
( u -&u0)+ = 0
v
i.e.,
. Hence
.
v E { ( U ~ - E U ~ ) + } ~ Onthe o t h e r
.
v = 0
THEOREM 139.4. Every orthomorphism a i n the Archimedean Riesz space is order continuous i n the sense t h a t uo t u T + 0 in L implies
inflau
I
.
= 0
PROOF. S i n c e
la11 5 p
number
E
such t h a t
0 S f 5
E U ~
prove t h a t
. Now
Therefore,
for a l l
0 <
E
f < 1
suppose t h a t
. It
w = 0
5
i s o r d e r bounded, t h e r e e x i s t s an e l e m e n t
TI
such t h a t
. Then
Ep
I.rrfI 5
0 5 w 5 lau u
for a l l
EP
1
for a l l
T
f
. We
p
E
L+
a real
satisfying have t o
A E U ~that
= ( u T - ~ u o ) ++ u
.
I ja(u - E U ~ ) + ~ f o r a l l
(w-cp)'
. Choose
0 5 f 5 uo
satisfying
f o l l o w s from
(n(uT-Eug)+/ +
phism, we have
T
2
. Since
R
i s a n orthomor-
a ( u -mO) E { ( U ~ - E U ~ ) + ] ,~ ~which i m p l i e s nowl t h a t
0 5 (w-Ep)+ E
"T{(UT-EUO)t}dd .
I n view of t h e l a s t lemma t h e i n t e r s e c t i o n on t h e r i g h t i s e q u a l t o It f o l l o w s t h a t satisfying
(w-~p)+ = 0
0 <
E
< 1
,
i.e.,
. Hence,
0 5 w 5
since
L
. This
EP
{O}
holds f o r every
i s Archimedean, we f i n d t h a t
w = o . Every p o s i t i v e orthomorphism i n
if
=
)+
v E {(AT-~uo)+ldd
E
0
O/'U~'UO'
we f i n d t h a t
A
u
T
Observing now t h a t
hand we have
L
ch. 20,51391
ELEMENTARY PROPERTIES
TI
is a p o s i t i v e orthomorphism i n
i s a R i e s z homomorphism. Indeed,
L L
and
u
A
v = 0
holds i n
L
,
.
Ch.
ORTHOMORPHISMS AND f-ALGEBRAS
20,81401
then nu
v
A
= 0
, and
therefore nu a # 0
If, for example, the number is defined by
(rrf)(x)
f(x+a)
=
A TV =
. The converse does not hold.
0
is fixed, if L
L ( R ) and
=
, then
f E L
for all
651
1
morphism, but not an orthomorphism. The identity operator in L tive orthomorphism. Since any orthomorphism n n
lows immediately that if P
a P = Pa ). Conversely, if
L
order bounded operator n
in L
B
f E L , i.e., operator n
that
which commutes with all order projections u E L+
and v
140. The space
. Hence
{"Idd
nf E {fIdd
for every
I is the identity operator). Indeed, if n+I
in L+
Therefore, since 0 s = 0
ITUE
be given
. It follows then from
{uIdd
is an orthomorphism if and o n l y if IT+I is a Riesz
thomorphism, then so is
inf(au,v)
then
is band preserving. We finally observe that the positive
IT
in L
homomorphism (where Choose u
,
in L
has the principal projection property, any
be the order projection on = u
is a posi-
is band preserving, it fol-
on principal bands is an orthomorphism. For the proof, let
ITPU= Pnu and Pu
L
-P
commutes with any order projection in L (i.e.,
is the order projection on the projection band
and let P
L
IT:
is a Riesz homo-
I I
ITU 5
.
. Conversely, let
such that
(n+I)u
inf(u,v)
n+I = 0
be a Riesz homomorphism
. Then
0 S v 5 (n+I)v
and
is an or-
T
, we find that
Orth(L)
As before, let
Orth(L)
be the ordered vector space of all orthomor-
. We
phisms in the Archimedean Riesz space L
shall prove that Orth(L)
is
itself a l s o an Archimedean Riesz space. We begin with an extension lemma. LEMMA 140.1. Let
L be an Archimedean Riesz space and
M
a Dedekind
complete Riesz space. The Dedekind completion of L w i l l be denoted by L-
. Furthermore,
l e t t h e operator
continuous i n t h e sense t h a t M
. Then
T, T+ and
T-
u
T: L
t uT
+
+
0
M be order bounded and order
in L implies
infITu I
have unique order continuous extensions
(T+)- and (T-)- t o L^ ( i . e . , these extensions map t h a t (T->+ = ( T + > - and (T-1- = (T-)-
.
LA i n t o
=
0
TI
,
M ) such
in
65 2
THE SPACE
PROOF. Since
Orth(L)
i s Dedekind complete, the space L,,(L,M)
M
kind complete Riesz space, and t h e r e f o r e and
T-
rators
. On
sup(-T,O)
s
T+ and
ch. 20,§1401
T = T
+
-
- T
account of what was prmved i n Lemma 84.1 t h e ope-
,
L
any
. It
(T+)^fa
Denote the common value by
L^ i n t o
T+
,
i s easy t o see t h a t
M which extends
E L s a t i s f y i n g f0A
fi
E L+
tain
, which
(T-)-
(T+)-f;
implies
which extends
T-
.
> g 2 fma f o r a t l e a s t one {go} ) w i t h go 0
4 0
. By
T- = (T+)^
continuous extension of
T
, where
( T - ) + < (T+)^
that
to
. Hence,
LA
.
(T+)^
T
+
. This . Hence
(T+)-
S i m i l a r l y a s we g o t
.
set
, we
ob-
the order c o n t i n u i t y these extensions
a r e unique. Writing now T^ = (T+)^ - (T-)-
operator
f a ). f - J 0 i n L0 - T Now consider the
i s downwards d i r e c t e d (denote the s e t by T+go .1 0
is a
(T+)-
. The
T+
i s Also o r d e r continuous. For the proof, assume 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 g
i s the
L-
E LA s a t i s f i e s
f-
Hence, i n view of the o r d e r c o n t i n u i t y of
p o s i t i v e o p e r a t o r from
Dede-
T+ = sup(T,O)
a r e o r d e r continuous. Furthermore, s i n c e
T-
Dedekind completion of
s e t of a11
is a
with
-
(T-)-
. We
(T+)- and
,
the operator
prove t h a t (T-)^
i s an order
T-
(Ta)+ = (T+)-
. From
a r e p o s i t i v e , i t follows
i t i s s u f f i c i e n t t o prove now t h a t
(T+)^u 5
< (T^)+u f o r every u E (LA)+ . By o r d e r c o n t i n u i t y we may assume t h a t E L+ , s o i t remains t o prove t h a t T+u S (Ta)+u holds f o r u E L+ ,
u
i.e.,
\ )*
This i s e v i d e n t l y t r u e . S i m i l a r l y we prove t h a t W e s h a l l apply the lemma i n the case t h a t
morphism from 'TI+
L
and
TI
into
-
TI
in La
map
. The
L
operator
TI
M = LA and
and can, t h e r e f o r e , be w r i t t e n as L
into
LA
. By
T
. i s an ortho-
can then be regarded as an operator TI
=
'TI
+
-
-
'TI
, where
means of the extension lemma the remark-
a b l e r e s u l t w i l l be proved t h a t , a c t u a l l y , s+ itself.
(TA)- = (T-)-
and
-
TI
map
L
into
L
Ch. 20,51401
653
ORTHOMORPHISMS AND f-ALGEBRAS
L E W ! 140. 2. Any ortkomorpkism
i n the Archimedean Riesz space L ITi n t h e Dedekind comple-
IT
can be uniquely extended t o an orthomorphism tion
of
.
L
PROOF. Let
LA
,
be a n orthomorphism i n
IT
L
. As
an o p e r a t o r from
L into
i s o r d e r bounded and o r d e r c o n t i n u o u s . Hence, by t h e p r e c e d i n g
IT
lemma, n
as well as
a-,(n+)-
and
and
'TI
(IT-)- t o
For t h e proof t h a t
L^
IT
-
have unique o r d e r c o n t i n u o u s e x t e n s i o n s
such t h a t
(a+)^ E Orth(L^)
.
and (nA)- = (IT-)-
= (a+)-
u^
we assume t h a t
v- = 0
A
in
L^
and we show t h a t
Since
u* = sup(wEL:O<w
,
i t f o l l o w s from t h e o r d e r c o n t i n u i t y of
(IT+)^and t h e formula f o r t h e p o s i t i v e p a r t of a n o p e r a t o r t h a t = sup
(Ir+)^U^
For a l l
w,z E L
, and
= 0
v-
, we
satisfying
therefore
{(n+)-u-)
A
z = 0
w
A z =
. Since
0 < w
0
. This
and
[
i s a p o s i t i v e orthomorphism i n
L-
.
1 )
0 < z w< v-
*
w
we have
z =
A
i m p l i e s , by t h e l a s t f o r m u l a , t h a t
. As
z
E L
satisfying
0 < z < v^
a l r e a d y mentioned above, t h i s
(IT+)-i s a p o s i t i v e orthomorphism i n
shows t h a t
. Similarly,
L-
The f i n a l c o n c l u s i o n i s t h a t
(IT-)^i s a n orthomorphism i n
= (a+);-
sup Trw:wEL,O3$su-
t h i s holds f o r a l l
{(IT+)^u-) A v^ = 0
find that
f
f IT+w:wEL,O~ww
L^
t h a t extends
~r
.
(n-)IT-
=
We have o b t a i n e d now, by e x t e n s i o n , a n orthomorphism i n a Dedekind complete s p a c e . The main p r o p e r t i e s of a n orthomorphism of t h i s k i n d w i l l now be i n v e s t i g a t e d f i r s t .
LEMMA 140.3. Let
Riesz space
L
. Then
IT
be an orthomorphism i n the Dedekind complete
(a+)u = (nu)+
and
( n - ) u = (nu)-
for e v e q
U E L + . PROOF. S i n c e
that
(n+)u
2 ITU
(a+)u = sup(Trw:~sw
,
and hence
(a+)u t ( n u ) +
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
0 2 w s u
. First
assume t h a t
w = p
my
u
. To prove
< (nu)+
E L+
it is clear
equality, i t i s
for a l l
i s a component of
,
w u
satisfying
,
i.e.,
654
p
Orth(L)
THE SPACE
(u-p)
A
. Then
0
=
and t h e r e f o r e
np I n(u-p)
0
=
, which
. Next,
np 5 ( r p ) + 5 (nu)'
Ch.
implies t h a t
assume t h a t
s = C T aipi
w i t h real c o e f f i c i e n t s
(i=l,.,.,n)
and mutually d i s j o i n t components
npl,
a r e now a l s o m u t u a l l y d i s j o i n t , we have
...,npn
Finally, let
w
theorem t h e r e e x i s t s a sequence
0 5 s
such t h a t E
"
4 0
+
n nsn
. Then
5 (nu)'
5 (as )
4 w
=
(nu)
-
. This
for a22
Orth(L)
E Orth(L)
. Since
u
Freudenthal's
spectral
of sums as d e s c r i b e d above
i.e., 0 5 w
-
s
S E u n n so that, since
find that
(n+)u = (nu)'
of
. Finally,
TIW
=
ns
. This
5 (au)'
(n-)u
with n
5
is
(-a)+u = (-m)'=
L
be an Archirnddean Riesz space. Then the ordered
i s a c t u a l l y an Archimedean Riesz space such t h a t ,
n l , n 2 E Orth(L)
and every
(n+)u = (nu)'
I n particular TI
, we
n
0 5 ai 5 1
concludes t h e p r o o f .
THEOREM 140.4. Let
vector space
. By
l n l (u)-uniformly,
holds f o r a l l
t h e d e s i r e d r e s u l t . Hence
. .)
(sn:n=I ,2,.
holds
+ my
p l , ...,pn
0 S w 5 u
h o l d s u-uniformly,
i s of t h e form
w
such t h a t
al,.. . , a n
b e given such t h a t
20,§1401
and every
PROOF. L e t n
u
,
E L+
(n-)u = ( n u ) -
.
be an orthomorphism i n
extended t o a n orthomorphism
, we have
u E L+
n^
L
and
Inlu =
. By
Inul
for every
Lemma 140.2, II
i n t h e Dedekind completion
LA
can be such
t h a t (Lemma 140.3)
f o r every
u € L"
It follows t h a t
. This (n^)'f
shows t h a t and
(n^)-f
(n^)+u
and
belong t o
(n^)-u L
belong t o
f o r every
f E L
L
.
.
ch. 20, 51401
655
ORTHOMOWHISMS AND f-ALGEBRAS
Hence, t h e r e s t r i c t i o n s
and
T~
of
n
and
(a^)+
(nA)-
to
map
L
L
i n t o i t s e l f . C l e a r l y , these r e s t r i c t i o n s a r e p o s i t i v e orthomorphisms i n and the given orthomorphism an upper bound of
li
3
i s a r b i t r a r y , then
u 2 (nu)+ = (n^)+u =
71
.
u
1
bound of the s e t c o n s i s t i n g of
n2
Now, l e t therefore
(a
=
-71
1
2
(nl-a2)
) + + a2
(n
u
f o r every
v71
2
Orth(L) L
2
,
n v 0
=
71'
i.e.,
-
a
exists in
Orth(L)
Orth(L)
-71
1
2
)+U
71 u 2 0 ; 3 i s t h e l e a s t upper
2
I+ u E L
al
exists in
712=
+ n u = (n u) v (a2u) 2 1
U)+
e x i s t s i n Orth(L) and (a A n )u = 1 2 may conclude, t h e r e f o r e , t h a t
n2
A
2
.
- n2 E Orth(L) , and
al
u-71
1
v 0
Orth(L)
It follows t h a t
= (a
= 1~
. We
i s a Riesz space. Evidently, the space i s Archimedean (because
i s Archimedean). Ra
We proceed with some p r o p e r t i e s of the range (kernel) N
of
an orthomorphism
.
n
THEOREM 140.5. Once more, l e t
n E Orth(L) (i)
has the following properties.
= N
. Furthermore,
1711
an i d e a l i n If
, where
f
L
E
. Hence,
Na = N
1711
='(a+)lfl 2 I s l ( l f l ) that
Rn
Nn c Na+
n
Nn-
=
of
a
i s a band i n
i s the range o f
( i ) I t follows from
PROOF. 71
and n u l l space
be an Archimedean R i e s z space. Any
L
The n u l l space ( k e r n e l ) Nn
d ( i i ) Nn = Rn
N
i s another
n3
and
+ a u
Similarly, a
.
.
L
is
n1
a s well as
3
2 = n be given. Then
f o r every
= ( a l u ) A (a u )
a u 2 nu
E Orth(L)
) u = (n
.
E L+
. Obviously,
n1 and the n u l l o p e r a t o r , i . e . ,
a
v 0
= (-a)
exists in
1
n2
This shows t h a t
In o t h e r words, with ohe usual n o t a t i o n , S i m i l a r l y we g e t
-
a = al
as w e l l a s of the n u l l o p e r a t o r . I f
'TI
u E L+
upper bound and hence
satisfies
n
L
and
.
n
/ a f I = I d f l l = I a l ( I f 1 ) = I/nlfl
i t i s obvious from these e q u a l i t i e s t h a t since
,
then
0
and
n
i s o r d e r continuous, N n
laI(If1) = 0 In-fI
. Conversely,
=
,
(a-)lfl
S
= n
is
In+fl =
lal(1fl) = 0 71
Nn
i s a band.
and t h e r e f o r e
i t follows from
that
+
-
. This
-
71
that
shows
.
656
THE SPACE
~
Nn+
N
-
(i:)
C
N
.
f i r s t that
Not:
. Indeed,
= Rd
:R
Orth(L)
I TI
Ch. 20,91401
we have
g IR
0
.
If 0 =
E Rf , t h e n u
u
5
,
0
i.e.,
nu = 0
, which
t h a t now
u E Nn
A
, so
nu = 0
g i v e n . We have t o show t h a t
u A nv = 0
f o r every
let
v E L+
d
0 5 u
. It
lnul
= Rn
immediately i m p l i e s t h a t
. Conversely,
I
g
N , = N ~ ~ I
f o r a l l ~ E L + - ~ I1n111 f o r a l l ~ E L + - ~ I R Since d d In1 and R n = R l n l , we may suppose t h a t n 2 0 i n t h e proof t h a t N
.
nu =
ITU A
E
Nn
be
follows
from
that
Since
A
TIV
u
A
n - l v .L 0
It f o l l o w s now from COROLLARY 140.6. nf = 0
,
0
n(nuAv) = n(nu)
=
that
u
A
( i ) The orthomorphism
nv = 0
.
n
L i s injective (i.e.,
in
f = 0 ) i f and only i f the band generated by
only f o r
Rd:
i f and o n l y i f
i t s e l f , i.e.,
t h i s implies t h a t
( i i ) Let
n1,n2
n1 = n 2 ' (iii) I f
TI
= L
.
Rn i s
L
E Orth(L) and l e t D be a subset of L such t h a t Ddd = L . If it i s given nou t h a t n , f = IT f f o r a l l f E D , then 2 n 1 = n2 . In p a r t i c u l a r , i f e is a weak order u n i t i n L and n 1e = n 2e , then n2
=
o
2
implies n =
(iv) PROOF.
o .
n l I n2
If
f E L
f o r some
i n Orth(L)
, then
, then
R
i t follows t h a t (iii) If
, and
IT = IT
L = D 2 n f = 0
ddIC-,:a
therefore
,
71
then nf
=
0
. In p a r t i c u l a r ,
nf = 0
1 R
"1
( i ) T h i s f o l l o w s immediately from
( i i ) Writing
nf I nf
f = 0
'2
d N n = Rn
. .
. S i n c e NT L . Hence
, we have D - Nn , i . e . ,
c Nn
nf
f o l l o w s now from
.
E
Nn
. It
N,=
i s a band, =
n2
;1
Nn = Rn
that
Ch. 20,§1401
ORTHOMOE'HISMS AND f-ALGEBRAS
TI^
( i v ) Let
This shows t h a t
I r2 i n
R
and l e t
f,g E L
be a r b i t r a r y . Then
.
IR
'TT1
EXAMPLE 140.7.
Orth(L)
657
Tz
A s w e have s e e n , t h e k e r n e l of any orthomorphism
i s a band. I n c o n t r a s t t o t h i s , t h e range
L
t h e Archimedean R i e s z s p a c e
i s i n g e n e r a l n o t even a R i e s z subspace. As a n example, l e t
RTI C(C0,Il)
,
let
p E L
be d e f i n e d by
l e t t h e orthomorphism f E L
and a l l
in
T
orthomorphism
, however,
TI
L
. Then
x E C0,ll
in
TI
p ( x ) = x-1
b e d e f i n e d by p E R,
t h e range
for a l l
L =
x E C0,ll and
(nf)(x) = p(x)f(x)
but
IpI 6! R , .
for a l l
For any p o s i t i v e
i s a R i e s z subspace of
RTI
L (since
more g e n e r a l l y , t h e range o f any Riesz homomorphism i s a Riesz subspace RTI L = C ( C 0 , l l ) and l e t
E L be d e f i n e d by
q
Now d e f i n e t h e p o s i t i v e orthomorphism for a l l
f E L
i s d e f i n e d by 0 in L
5
u L
5
of
i s i n g e n e r a l n o t a n o r d e r i d e a l . F o r example, l e t a g a i n
L ), b u t
q
. It
,q E
and a l l
x
is a
u
L
by
(rf )(x)
x E C0,ll =
.
q(x)f(x)
E 1 0 , I l . I f t h e c o n t i n u o u s f u n c t i o n u on [ O , l l
u(x) = Ix s i n x - ' ( RTI b u t
in
TI
for a l l
q(x) = x
RT
for
. This
0 < x 5 I
u(0) = 0
and
,
then
i s not an o r d e r i d e a l
shows t h a t
RTI n a t u r a l q u e s t i o n t o ask under what a d d i t i o n a l c o n d i t i o n s on
t h e range of e v e r y orthomorphism i n
L
i s a n o r d e r i d e a l . T h i s problem
A
i s ca%led a Riesz algebra ( l a t -
w i l l b e d i s c u s s e d i n s e c t i o n 146. DEFINITION 140.8.
The Riesz space
t i c e ordered algebra) if there e x i s t s i n A an associative m u l t i p l i c a t i o n w i t h t h e u s m % algebra properties and such t h a t The Riesz algebra
A
uv 2 0
for a l l
u , v E A+
for aL% f , g E A
i s called c o m u t a t i v e if f g = gf
.
The Riesz algebra A is calZed an f-algebra i f A has the additional property t h a t u A v = 0 implies (uw) A v = (wu) A v = 0 f o r a l l w E A+ Examples o f Riesz a l g e b k a s a r e t h e a l g e b r a o p e r a t o r s i n a Dedekind complete Riesz s p a c e
L
h(L)
i s an f-algebra.
X
CfX)
. Obviously,
of
C(X)
I t w i l l b e proved soon (in Theorem 140.10) t h a t an Archi-
medean f - a l g e b r a i s n e c e s s a r i l y commutative. S i n c e
.
of a l l o r d e r bounded
and t h e a l g e b r a
a l l r e a l c o n t i n u o u s f u n c t i o n s on a t o p o l o g i c a l s p a c e
.
L,, (L)
i s Archimedean
658
Orth(L)
THE SPACE
ch. 2 0 , § 1 4 0 1
but not commutative (unless i n a very t r i v i a l c a s e ) , t h i s i s an example of a Riesz algebra which i s not an f-algebra. Let
there e x i s t s a n a t u r a l m u l t i p l i c a t i v e s t r u c t u r e i n h e r i t e d from
,
L,,(L)
be an Archimedean Riesz space. In the Archimedean Riesz space
L
Orth(L)
i.e.,
(a a )f = r 1 ( a 2 f ) 1 2
THEOREM 140.9. Let
for a l l
n l , a 2 E Orth(L)
and a l l
f E L
.
be an Archimedean Riesz space. With respect t o
L
t h e above defined m u l t i p l i c a t i o n i n Orth(L) , the space
Orth(L)
i s an
Archimddean f-algebra w i t h the i d e n t i t y operator as u n i t element. PROOF. I t i s e a s i l y v e r i f i e d t h a t
i s a Riesz a l g e b r a with
Orth(L)
the i d e n t i t y operator as u n i t elememt. For the proof t h a t f-algebra, that
1~
l e t the p o s i t i v e orthomorphisms
. We
= 0
A T
TT
1
u
This implies t h a t
(a a ) u
1 u E L+
f o r every
= 0
,
a u = 0 2
A
A
a u = 0
2
. Hence
f o r every
a a 1
A
a
a 1 a 2 = a2a1
.
for a l l
,
u E L+
Orth(L)
. This
n1,a2 E Orth(L)
and
i.e.,
(a
.
2 = 0
We s h a l l prove now t h a t m u l t i p l i c a t i o n i n i.e.,
be given such
a2 0
a a A a2 = 0 1 u E L+ we have a u A a u = (a AT ) u = 0 , 1 2 1 2 i . e . , (aa A T ) u = 0 . For the second, note t h a t 1 2 ntrl A n 2 =
1 2 The f i r s t i s easy; f o r any hence
and
a,al
have t o prove t h a t
i s an
Orth(L)
ITAT
1
2
)u =
i s commutative,
w i l l follow from the
more general r e s u l t t h a t every Archimedean f-algebra i s commutative.
THEOREM 140.10. Any Archimeclean f-aLgebra
A
i s comutative.
PROOF. It i s s u f f i c i e n t t o show t h a t p o s i t i v e elements i n Note f i r s t t h a t an f-algebra, E {vjd
. On
Hence
uv = 0
u
A
v = 0
in
i t follows from
A
u
A
uv = 0
implies v = 0
that
t h e o t h e r hand, i t follows from
. Similarly,
uv
. Indeed, A
v I {vjd
of course, vu = 0
.
v = 0
A
since
,
commute. A
uv E d
i.e.,
uv I { v l
that
is
. .
uw = wu
f o r every
u E A+
For t h i s purpose, we d e f i n e the o p e r a t o r s
aL
and
T
in
71
and
aL
and
n
a r e p o s i t i v e ortho-
L e t now
T
f = fw
w E A+ for a l l
be given. We prove t h a t f E A
. Evidently,
A
by
e
f = wf
20,81401
Ch.
ORTHOMORPHISMS AND f-ALGEBRAS
wv = vw = 0
L
E {wid
of
v E A+
0
for a l l
. Also
and
w
any
n w =
{wjd
1-
.
1~
Then
5
E {wid
v
. Let
w = w2
r
,
Dd = { O }
. T h i s i m p l i e s , by A . I n o t h e r words,
c o i n c i d e on
the algebra
0
c u l a r , t h e band
then
TI
Z(L)
n
E Z(L) in
number
n
n E Z(L)
, hence
w = 0
A
follows t h a t
i.e.,
holds f o r a l l
u
A
v = 0
e
,
then
implies
u
A
is
Idd g e n e r a t e d by t h e u n i t element
I
I
in
. Hence,
Z(L)
In1 5 n I
belongs t o
such t h a t
Z(L)
Orth(L)
for a l l
r
consisting
A
nL
uv = 0
nr
ne and
=+nr
.
u,w E A
e = 0
{ejdd
On
. Hence,
if
u = 0
,
implies
itself. In parti-
A
in
is
Orth(L)
is called thecentre
i f i t i s given t h a t
n
f o r some n a t u r a l number
i s i t s e l f an Archimedean f - a l g e b r a . L
L
. Furthermore,
Ddd = A
f o l l o w s t h a t t h e band
i f and o n l y i f
n f = n f
b e t h e s u b s e t of
D
i t s e l f . The i d e a l g e n e r a t e d by
that tor
. It
and i t i s denoted by
L
of
h a s a u n i t element
A
{eld =
. It
wu = uw
A s observed i n t h e l a s t p r o o f ,
Orth(L)
v
we have
Corollary 140 . 6 ( i i ) , t h a t
D
t h e whole of
i.e.,
{wid
in
by t h e remarks a t t h e b e g i n n i n g of t h e p r o o f . I n o t h e r words,
n v = nrv = 0
f
. For
A
morphisms i n
659
E Orth(L), n
. Note
It i s e v i d e n t t h a t an opera-
i f and o n l y i f t h e r e e x i s t s a n a t u r a l
l n u l < nu
holds f o r every
leaves a l l i d e a l s i n
u E L+
. Note
t h a t any
invariant.
L
I n 1950 H. Nakano (C21) d e f i n e d i n a Dedekind o-complete Riesz s p a c e L
t h e n o t i o n of a d i h t a t o r ,
a n o t i o n which i s s i m i l a r t o t h e n o t i o n of
a n orthomorphism. P r e c i s e l y , a d i l a t a t o r i n L
L
is a positive operator i n
commuting w i t h a l l o r d e r p r o j e c t i o n s . I n 1969 A. B i g a r d and K. Keimel
([I])
d e f i n e d an orthomorphism i n a n Archimedean h e s z s p a c e a s t h e d i f f e r -
e n c e of two p o s i t i v e orthomorphisms, where a p o s i t i v e orthomorphism was d e f i n e d t o b e a p o s i t i v e o p e r a t o r which l e a v e s a l l bands i n v a r i a n t . A few y e a r l a t e r , i n 1971, P.F.
Conrad and J . E .
Diem ( [ I ] )
i n t r o d u c e d t h e same
n o t i o n and they c a l l e d t h e s e o p e r a t o r s polar p r e s e r v i n g endornorphisrns. I n b o t h p a p e r s i t i s proved t h a t t h e s e t of a l l orthomorphisms i s a n Archimedean f - a l g e b r a w i t h u n i t e l e m e n t . The p r o o f s a r e b a s e d on r e p r e s e n t a t i o n theory; t h e given Riesz space
L
of r e a l f u n c t i o n s on a p o i n t s e t w i s e m u l t i p l i c a t i o n by a f u n c t i o n
i s isomorphically represented as a space X
and an orthomorphism i s t h e n a p o i n t -
.
~ ( x ) The fundamental Theorem 140.4 and
i t s consequences can b e found i n t h e s e p a p e r s . M. M e i j e r was t h e f i r s t t o c o n s i d e r o r d e r bounded band p r e s e r v i n g o p e r a t o r s i n an Archimedean R i e s z s p a c e . He c a l l e d t h e s e o p e r a t o r s s t a b i 2 i s a t e u r . s . ( a s d e f i n e d by Bigard-Keimel-Conrad-Diem)
Obviously any orthomorphism
i s a s t a b i l i s a t e u r . M e i j e r proved
660
THE SPACE
OrthfL)
Ch.
20,§1401
(C11,1976) t h a t , i n f a c t , t h e converse h o l d s a s w e l l , and s o t h e n o t i o n s of orthomorphism and s t a b i l i s a t e u r are t h e same. I n t h e p r e s e n t d i s c u s s i o n w e have adopted Meijer's d e f i n i t i o n of an orthomorphism. The method of p r o o f , l e a d i n g up t o Theorem 140.4, i s almost t h e same as t h e proof g i v e n by W . A . J .
Luxemburg (C61 ,1979,Theorem 4 . 1 0 ) .
Note t h a t no use i s made of any r e p r e s e n t a t i o n theorem. h e of t h e main f e a t u r e s of t h e method i s t o prove f i r s t t h a t a n orthomorphism i s o r d e r continuous (Theorem 139.4), which makes i t p o s s i b l e t o e x t e n d an orthomorphism t o t h e Dedekind completion (Lemma 140.2) i n which F r e u d e n t h a l ' s s p e c t r a l theorem can b e a p p l i e d (Lemma 140.3). Another proof of Theorem 140.4 a v o i d i n g r e p r e s e n t a t i o n t h e o r y can b e found i n a p a p e r by S . J .
Bernau
(131,1979). The example of a non-order bounded o p e r a t o r l e a v i n g a l l bands i n v a r i a n t i s due t o M. M e i j e r ([2],1979,Example
2.6).
The d e f i n i t i o n of a n f - a l g e b r a a s p r e s e n t e d i n D e f i n i t i o n 140.8 above f i r s t appeared i n a p a p e r by G. Birkhoff and R.S. P i e r c e (C11,1956,section 8 ) . These a u t h o r s ( i n Theorem 12 of t h e i r p a p e r ) prove t h a t a Riesz a l g e b r a
A
i s an f - a l g e b r a i f and o n l y i f
A
i s isomorphic t o a s u b d i r e c t union of
t o t a l l y o r d e r e d a l g e b r a s . Using t h i s r e s u l t (which depends on Zorn's a c e r t a i n number of p r o p e r t i e s of f - a l g e b r a s
lema),
can e a s i l y b e proved by
checking t h e s e p r o p e r t i e s o n l y f o r t o t a l l y o r d e r e d a l g e b r a s . S e v e r a l o t h e r a u t h o r s d e f i n e an f - a l g e b r a immgdiately a s a R i e s z a l g e b r a which i s isomorp h i c t o a s u b d i r e c t union of t o t a l l y o r d e r e d a l g e b r a s ( a s , f o r example, i n t h e book [ 1 1 by A. B i g a r d , K. K e i m e l and S . W o l f e n s t e i n ) . We adopt t h e o r i g i n a l Birkhof f - P i e r c e
d e f i n i t i o n , w i t h o u t u s i n g anywhere i n t h e p r o o f s t h e
r e p r e s e n t a t i o n as a s u b d i r e c t union of t o t a l l y o r d e r e d a l g e b r a s . The comut a t i v i t y of an Archimedean f - a l g e b r a was f i r s t proved ( u s i n g t h e terminology of s p e c t r a l f u n c t i o n s ) by I. Amemiya ([1],1953,Theorem and R.S. P i e r c e show i n t h e i r p a p e r t h a t
nlfg-gfI
5
f2
19.2). G . Birkhoff
+ g2 ( n = l , 2 , ...)
h o l d s i n any f - a l g e b r a by o b s e r v i n g t h a t t h i s i n e q u a l i t y h o l d s i n any t o t a l l y o r d e r e d a l g e b r a . It f o l l o w s immediately from t h i s i n e q u a l i t y t h a t f g = gf
i n any Archimedean f - a l g e b r a .
S.J. Bernau p r e s e n t s an e l e m e n t a r y
( b u t n o t s o simple) proof of t h e mentioned i n e q u a l i t y (C11,1965),Lemma 3 ) . The commutativity proof as given i n Theorem 140.10 i s due t o A.C.
Zaanen
(C31,1975 ,Theorem 2 ) .
IT^
EXERCISE 140.11. Show t h a t i f i n t h e Archimedean Riesz s p a c e
L
,
and
r2
are p o s i t i v e orthomorphisms
then the n u l l spaces (kernels) s a t i s f y
Ch. 20,11401
66 1
ORTHOMORPHISMS AND f-ALGEBRAS
N
ITlVIT2 =
1
= N
+a 2
= N
al
H I N T : Since a l l e x p r e s s i o n involved a r e bands, we may r e s t r i c t our-
, it
(Nal+NT2>ddc N N
f i r s t formula i s e a s y . For t h e proof t h a t
i s s u f f i c i e n t t o show t h a t
“IT2 i s a band. Now, N
1~141~
since
. The
L+
s e l v e s t o elements of
”
+ N
N
C
N
*l a2 ‘7182 ’ i s e v i d e n t , and on account of
c N
.
For t h e proof t h a t N c *la2 ITIT2 a a u = 0 , i . e . , IT u E N = 1 2 2 a1 Tl”T2 a1n2 = Rd , from which i t f o l l o w s t h a t a 2 u A a 1u = 0 , i . e . , (a 1A a 2 ) u = 0 a1 c (N +N )dd = (Nd nNd 1‘ . I n o t h e r I t remains t o prove t h a t NTlAIT2 71 IT1 a1 “2 words, we have t o prove t h a t u A v = 0 whenever 0 u E NT and 1 2 d d 0 5 v f N T ll Na2 For t h i s purpose, l e t w ,w E L+ . Then a
c
IT
1 2
t h e same h o l d s f o r
=
,
N
let
.
1
c N
a1
. Then
u E N
0
N
‘
(
.
{
i n f u,T w a w < i n f u , a (w vw ),a (w vw ) 1 1 ’ 2 2 ) 1 1 2 2 1 2
because
u E N = (RITIAn ) d a 1 AT2 2 inf(u,a!wl,v) = U slnce
implies = N
ITl
.
0
On t h e o t h e r hand
EXERCISE 140.12.
. Hence v
u
S
A
, which inf(u,.ir w ) E Rd = N 1 1 IT2 a2 This shows t h a t U A V t Kd = a1 IT’ d v 2 v E N Hence u A v = 0 . al
.
Nd
.
Show t h a t i f t h e Archimedean Riesz space
formly complete, t h e n
,
Orth(L)
and hence
,
Z(L)
i s uni-
L
i s l i k e w i s e uniformly
complete. HINT: Let
0
C
(nn)
L
Cauchy sequence i n prove t h a t to
IT
IT^
be a .rr-uniform Cauchy sequence i n
. Then,
a E Orth(L)
if
,
(Tnf)
Orth(L)
f o r some
i s a rlfl-uniform
having t h e r e f o r e a l i m i t . C a l l t h e l i m i t
F Orth(L)
. Finally,
show t h a t
IT
n
a f and 0 converges a-uniformly
0 ‘
L
EXERCISE 140.13. Let B
f E L , t h e sequence
i n Orth(L)
,
The d i s j o i n t complement (i)
Show t h a t i f
then t h e c a r r i e r s (ii)
be an Archimedean Riesz s p a c e . For any band
t h e nu22 band
N:
of
N;
B1,B2
and
Show now t h a t i f
of
NB NB
B1
i s d e f i n e d by
is called
a r e bands i n of
N;
B
and
B1
and B2
NB = fl(N,:*EB)
t h e c a r r i e r of
B
.
Orth(L)
B2
.
such t h a t B 1 I B2 d d s a t i s f y N 1 I N2 .
a r e d i s j o i n t complements i n
,
662
TIIE SPACE
,
Orth(L)
then
the carriers ( i i i ) Let
,
n C Orth(L)
and
N1
and
N;’
B then
and
B1
and
Show t h a t i f IT
. Hence
‘1 IT] d T h e r e f o r e N 1 c ll N
add
6 B1 R 713
IT2 =
c
r2
n
N
.
N2
IT1
F
B2 we have
=
N1
.
n
, i.e.,
N2
follows t h a t (iii)
IT
B2 u =
E B
. Hence 0 .
that
satisfies phism
. If
N2 = Nd
u E (N+N2)+
,
B1
and
t h e i r null bands be
N,
for a l l (v)
Let
i o n s on
and
N,
+ nP 2
B
= Orth(L)
1
0
B
f
2
N2
RIT
L
such t h a t
N,
t h e n s o has Orth(L).
,
i.e.,
R
c Kd IT2
=
C
=2 N
IT2
N 2 = { O } . Assume t h a t
.
.
0 2
7if = 0
IT
9 I
2T
. Since
IT
and
2
Let
with
S IT
for a l l
f 5 N
IT* E
B
IT
,
it
B1
may b e assumed of
PI2
2 . It follows t h a t r2
=
0
. Contradiction.
N2
. Let
P
1 IT
and
P2
, we
Orth(L)
E Orth(L)
2
u =
can b e w r i t t e n a s
,
ITP E B2
Similarly
2
IT
and l e t
I T P ~ a r e orthomorphisms. S i n c e ,
IT u = 0
have
be t h e o r d e r p r o j e c t -
. Hence
ITP2 f = 0
be an orthomorphism i n t h e Archimedean R i e s z
. Show t h a t
i s idempotent, i.e., IT’ = IT IT
i s t h e o r d e r p r o j e c t i o n on
HINT: Denote t h e i d e n t i t y o p e r a t o r i n
I-IT i s a l s o an idempotent
. It
t h e n u l l band
be d i s j o i n t complements i n
xP2 E B1
i s a p r o j e c t i o n band and
S IT
I T ~ = ~ U 0
rlTu =
by o r d e r c o n t i n u i t y of
and
IT
0
with
f E N (by d e f i n i t i o n ) . Conver-
. Therefore,
r e s p e c t i v e l y . Any
IT
+
B2 does n o t h o l d , t h e r e e x i s t s an orthomor-
E B
B2
.
EXERCISE 140.14. space
so
, where ITP, E N1 , we have
= aP1 for a l l
IT
IT
. Hence,
u E L+
0
n
1T
for a l l
B2 = Bd
0 < IT
such t h a t
IT*
for a l l =
.rrf = 0
satisfies
TI
IT
+
i s p o s i t i v e . Let
IT
1’ 2 i f and o n l y if
RITl I RIT
( I T ~ ~ + I T ~ ~u) U
implies
s e l y , assume t h a t
.
. Show
u = IT u = 0 f o r a l l IT E B , and r 2 E B Since 1 2 and t h e i d e n t i t y o p e r a t o r I belongs t o Orth(L) ,
(Bl@B2)ad = Orth(L)
< IT^^ E
N
IT
t h e r e e x i s t s an upwards d i r e c t e d system 0
Show t h a t if
f 6 N N
I t f o l l o w s t h a t 2N: N1
.
N
i f and o n l y i f t h e c a r r i e r s
has t h e p r o j c e t i o n property,
I t i s d u f ? i c i e n t t o show t h a t
(ii)
u E N
L
.
for a l l
Orth(L)
.
N
=
and
nf = 0
w i t h n u l l bands
a r e d i s j o i n t complements i n
B2
H I N T : For
S
i f and o n l y i f
be bands i n Orth(L)
. Equivalently,
L L
w i t h n u l l band
N2
(v)
c Rd
Orth(L)
a r e d i s j o i n t complements i n L , i . e . , d a r e d i s j o i n t complements i n L
N2
and
N;‘
E B
IT
( i v ) Let B I , B 2 that
a r e d i s j o i n t complements i n
be a band i n
20,51401
Ch.
a r e d i s j o i n t complements i n
N2
d N2
Orth(L)
L
orthomorphism and
by N,,
I =
RIT
t h e range *
. Show f i r s t
%-, .
that
S i m i l a r l y , We
ORTHOMORPHISNS AND f-ALGEBRAS
20,§1411
Ch.
have
R
=
IT
NITI7 Rn
N
I-IT
,
= {O}
' and therefore RIT is a band. the algebraic
so
sum. Finally, it follows from
EXERCISE 140.15. Let
f
sum of
It is easy to see that
the bands
(I-.rr)f+nf
=
663
NIT and NIT
that
RIT is a direct
RIT= L
.
n2 be positive orthomorphisms in the dd Archimedean Riesz space L such that N t N Show that n2 E IT1 172 HINT: Every positive orthomorphism T C {nlId satisfies IT 1 T = 0 (Exercise 140.11), so T
IT,}^
E
sl and
CB
R
t
. It follows that '
N 1'
.
IT^}
, which implies
cN
n
IT,}^'
IT,)^ cIT,?IT^} d , i.e., .
T =
0
.
, and hence
.
c {n1}dd
141. Examples of orthomorphisms We first prove a theorem which covers several special cases. THEOREM 41 . I . Let e ( i . e . , fe = ef
p E L
=
f E L
f o r all
pf
.
i f the corresponding element
to
i f and only i f
Z(L)
1. I f
p E L
ment to L
L
by
p
=
ae
f E L
for every
p belongs t o t h e i d e a l i n
L
.
Since
the set that
p+
, i.e.,
L (indeed, if 0
S u
{el
rrf
pf
=
with
{eldd
for every
=
e
L , we have
f E L
.
p
are posi-
is an orthomurphism in
p1
IT
IT =
'II
1
and
Cef
E Eeld , then u
as observed in the proof of Theorem 140.10). set
generated by
L
and multiplication by
i s an orthomorphism in L
=
belongs
is an f-algebra). Hence, multiplica-
Ire = IT e = p , the orthomorphisms 1 {el , consisting of e only. The set
{eldd
IT
.
. Since L i s an algebra, the product pf belongs . Therefore, IT1 f = pf defines an operator IT^ in -
. Evidently, multiplication by p
define'd
in L be given. We define the ele-
IT
tive orthomorphisms in L (since L tion by
L
in
IT
is p o s i t i v e i f and only
IT
is p o s i t i v e . Furthermore,
PROOF. Let the orthomorphism
f E L.
holds f o r every
.rrf = pf
The orthornorphism p
L ,
i s an orthomorphisrn i n
IT
such t h a t
p F: L gives r i s e t o an orthomorphism
Conversely, every nf
f E L
for a l l
there e x i s t s an element by
be an Arehimedean f-aZgebra w i t h u n i t element
L
IT^
coincide on
has the property A
e
=
Hence, since
0
, so ue
IT =
IT^
by Corollary 140.6(ii),
=
0
on the i.e.,
664
EXAMPLES OF ORTHOMORPHISMS
EXAMPLE 142.2. Let t h e Riesz s p a c e
non-empty p o i n t s e t
L
consist
of r e a l f u n c t i o n s on a
( w i t h t h e v e c t o r s p a c e s t r u c t u r e and t h e o r d e r s t r u c -
X
t u r e i n t h e u s u a l p o i n t w i s e manner). Furthermore, l e t
L
b e an a l g e b r a w i t h e
r e s p e c t t o p o i n t w i s e m u l t i p l i c a t i o n and l e t t h e u n i t f u n c t i o n ing
e(x) = 1
for all
, belone
x € X
f - a l g e b r a w i t h u n i t element 'TI
to
. Hence,
e
. Then
L
if
that
(af)(x) = p(x)f(x)
a t o p o l o g i c a l space f u n c t i o n s on
X
h o l d s on
i s the space
L
. The
f o r every
X
Cb(X)
orthomorphism
corresponding function
i s an Archimedean
L
. This
f E L
L
,
p E L
then such
holds i n
of a l l bounded r e a l continuous
belongs t o
'TI
Z(L)
i f and o n l y i f t h e
i s bounded.
p
EXAMPLE 141.3. L e t
satisfy-
of a l l r e a l c o n t i n u o u s f u n c t i o n s on
C(X)
o r the space
X
,
i s an operator i n
71
i s a n orthomorphism i f and only i f t h e r e e x i s t s a f u n c t i o n
particular i f
20,81411
Ch.
L
be t h e s p a c e
f u n c t i o n s on t h e p o i n t s e t
X
, where
of a l l r e a l u-measurable
M(X,u)
i s a measure i n
1-1
X
(and where
u-almost e q u a l f u n c t i o n s a r e i d e n t i f i e d ) . The orthomorphisms i n p o i n t w i s e m u l t i p l i c a t i o n s . The same h o l d s i f of a l l bounded f u n c t i o n s i n
.
M(X,u)
If
L = M(X,p)
a r e t h e m u l t i p l i c a t i o n s by f u n c t i o n s from
L
i s t h e subspace
L
L=(X,u)
.
,
are the
L,(X,u)
t h e e l e m e n t s of
Z(L)
EXAMPLE 141.4. The g e n e r a l Theorem 141.1 a l s o covers t h e c a s e t h a t
i s one of t h e sequence s p a c e s ( s )
, .em
t h e s p a c e of a l l r e a l sequences and
or
(c)
. We
...)
with
f ) . The same h o l d s i f
depending on f = (fl,f2,
f = (fl,f2,
...)
such t h a t
fn
fn L
(s)
L
is
i s t h e s p a c e of a l l convergent
(c)
i s t h e s p a c e of a l l
L
r e a l sequences. The theorem can a l s o b e a p p l i e d i f r e a l sequences
recall that
constant f o r
n 2 no
n
(with
0
i s t h e s p a c e of a l l r e a l sequences
assumes o n l y f i n i t e l y many d i f f e r e n t v a l u e s
(these values dependingon f ) . EXAMPLE 141.5. I f once more t h e r e a l number orthomorphism
a
L
i s an Archimedean Riesz s p a c e and
i s g i v e n , t h e n m u l t i p l i c a t i o n by
(i.e.,
if
.rrf = a f
for all
f € L
,
a then
i s obviously an 'TI
i s a n orthomor-
trivia2 orthomorphism. It may b e asked i f t h e r e
phism). T h i s is c a l l e d a
e x i s t n o n - t r i v i a l Riesz s p a c e s p o s s e s s i n g o n l y t r i v i a l o r t h o m o r p h i s m . The answer i s a f f i r m a t i v e . We p r e s e n t a n example. L e t val i n f
7R and l e t
on [ a , b ]
L
such t h a t
[a,bl
be a closed i n t e r -
b e t h e R i e s z s p a c e of a l l r e a l c o n t i n u o u s f u n c t i o n s f
i s piecewise l i n e a r ( i . e . ,
t h e graph o f
f
s i s t s of a f i n i t e number of l i n e segments). I t i s e a s y t o v e r i f y t h a t
conL
is
20,11411
Ch.
ORTHOMORPHISMS AND f-ALGEBRAS
uniformly dense i n of numbers L
J. 0
E
0
such t h a t
2
. Therefore,
C([a,b])
f E C([a,bl)
I
Erie , where ,
L
(ITS,)
g
sequence
(s,)
. Indeed,
d e f i n e t h e element every
f in
n-f
. It
E C([a,bl) C([a,b])
= pf
0 I s -s I E e for m n n This shows t h a t t h e sequence
i t f o l l o w s from
.
i t i s e a s y t o see t h a t in
C(Ca,bl)
f
by
n-f
i s e v i d e n t now t h a t
such t h a t
f o r every
particular
extends
n1
The l i m i t
E C([a,bl)
nf = pf f o r e v e r y
f
.
This can b e done f o r
i s a p o s i t i v e orthomor-
. Writing
E L
. From
r e a l continuous f u n c t i o n s
n
f
on
,
have
E L i t follows t h a t
p = ne f E L
pf
. This
i s p o s s i b l e only
[a,bl
i s t h e Riesz s p a c e of a l l
L
such t h a t
n
On t h e o t h e r hand, i f
i s t h e space o f a l l r e a l continuous f u n c t i o n s
such t h a t
L
h a s only t r i v i a l orthomorphisms.
i s p i e c e w i s e a polynomial (without r e s t r i c t i o n s on t h e degree)
f
orthomorphisms i n
are t h e m u l t i p l i c a t i o n s by e l e m e n t s of
L
EXAMPLE 141.6. Given t h e complex H i l b e r t space
,
H
L
let
ff
s i t i v e cone of (Ax,x) t 0
for a l l
x
E H
.
A s shown i n Chapter 8, ff
u n l e s s i n the t r i v i a l case that t h e H i l b e r t s p a c e
D
be a non-empty
mutually ( i . e . , Let
r e m 55.2,
s u b s et of
AB = BA
for all
ff
c"(D)
D
D
. As
commute
i s Hermitian). shown i n Theo-
and
B
j o i n t ) i f and only i f It follows t h a t i f
i n C"@)
AB
A,B,C
= 8'
c"(D)
i t f o l l o w s immediately t h a t
, we
have
, where
0
are members of
H
po-
i s n o t a R i e s z space
E D ; n o t e t h a t now AB
commutative r i n g . w i t h t h e i d e n t i t y o p e r a t o r i n A
. The
i s one-dimensional.
such t h a t a l l members of
A,B
H
satisfying
c"(D)
i s a Dedekind complete Riesz space. The members o f
commute m u t u a l l y (Leme 55.1); more, f o r
H
be t h e second commutant (bicommutant) of
C"(D)
A
c o n s i s t s of a l l Hermitian o p e r a t o r s
.
be t h e or-
ff
dered v e c t o r space of a l l norm bounded H e m i t i a n o p e r a t o r s i n
Let
f
i s an Archimedean f - a l g e b r a w i t h u n i t e l e m e n t , and t h e r e f o r e t h e
L
then
then a g a i n
i s piecewise a poly-
f
nomial of degree a t most
L
p
must a l s o
i s trivial.
i s f i x e d and
n
, we
p = ne
i n view o f o u r g e n e r a l Theorem 141.1. I n
i s constant. It follows t h a t I f t h e n a t u r a l number
. We
g = sup(ns:sEL,s
IT^ IT
b e continuous and piecewise l i n e a r f o r e v e r y p
.
[a,bl
i s a continuous and p i e c e w i s e l i n e a r f u n c t i o n . The p r o d u c t if
in IT
i s t h e n continuous and does n o t depend on t h e approximating
function
phim
(s,)
i s t h e u n i t f u n c t i o n . Hence, i f
e
t h a t 0 I I T S - I T S I E ne f o r m 2 n m n n of continuous f u n c t i o n s converges uniformly on
m t n
and t h e sequence
a r e g i v e n , t h e r e e x i s t s a n i n c r e a s i n g sequence f-s
i s a p o s i t i v e orthomorphism i n
n-f
if
665
is a
a s u n i t element. F u r t h e r -
A I B (i.e.,
A
and
B
are d i s -
i s t h e n u l l o p e r a t o r (Lemma 55.3).
c"(D)
such t h a t
A I B
,
then
EXAMPLES OF ORTHOMORPHISMS
666
AC
IB
. Hence,
i s a Dedekind complete f - a l g e b r a w i t h u n i t e l e m e n t .
c"(D)
Therefore, the operator
c"(D) .
of
.
c"(D)
such t h a t
n(A) = BA
holds f o r a l l
i s p o s i t i v e i f and o n l y i f
IT
B
i s i n the
i s R i e s z isomorphic t o
E v i d e n t l y , Orth{C"(D))
i t s e l f . I n f a c t , w e have Z{c"(D)}
= Orth{C"(D)}
EXERCISE 1 4 1 . 7 . Let i n g . Every o p e r a t o r
in
IT
non-negative. L
= C"(D)
.
i s given by a m a t r i x
L
(pij:i,j=l,
...n)
Hence, e v e r y o p e r a t o r i n
a Dedekind complete R i e s z s p a c e . I f
IT
has the matrix
IT
.
(lp.,l) Show t h a t 1J i f and o n l y i f t h e m a t r i x ( p i i ) of p.
ii
=
0
i
for
p. are 1j i s o r d e r bounded. Furthermore,
L
1;
t h e main d i a g o n a l v a n i s h ) , then
( p . . ) , then IT 1J dd i s an orthomorphism, i . e . , IT E I
IT
E Id
IT
o p e r a t o r . Hence,
p,l 0
shows t h e decomposition of
IT
is
L
has the matrix
I,...,n
=
in
i s p o s i t i v e i f and o n l y i f a l l
IT
i s Dedekind complete, t h e v e c t o r s p a c e of a l l o p e r a t o r s i n
Show t h a t i f
,
with the usual coordinatewise order-
L = En (1122)
t h e u s u a l manner. The o p e r a t o r since
i s an orthomorphism i f and o n l y i f
C"(D) C"(D)
The orthomorphism
p o s i t i v e cone of c"(D!
in
IT
B
t h e r e e x i s t s a member A E
ch. 20,81411
;22
satisfies (i.e.,
. Here
p..
11
=
0
,
.
i # j
for
a l l m a t r i x c o e f f i c i e n t s on
... 0 ... 0
I
denotes the i d e n t i t y
PI2 0
\ +
... Pnn:
P n 1 P,2
i n t o components i n
Idd and
]:I:
... ... -
0
I
-
d
0 respective-
ly. H I N T : For t h e proof t h a t =
0
for
i # j
that
pql = 1
E L+
satisfies
, use and
ITU =
p
IT
i s an orthomorphism i f and o n l y i f
p
Theorem 1 4 1 . 1 . For t h e second p a r t , w e may assume
ij
(O,U1,O
=
0
for all other
p
ij
,...,0 ) .
Furthermore av+w:v~O,wrO,v+w=u
. Then
u = (u1,u2
ij
,...,u
=
) E
Ch. 20,11421
Choose
667
ORTHOMORPHISMS AND f-ALGEBRAS
v ' = (ul,O
,...,0 )
, w'
U-v'
=
and
,...,un)
v" = ( 0 , u 2
, w"
=
u-vgt
Then i n f f TV'+W',TV"+W"
\
Hence
(TAI)(u)
=
. It
0
follows t h a t
I
B A
=
0
.
d
, i.e, T E
I
142. P r o p e r t i e s o f f - a l g e b r a s I n t h e p r e s e n t s e c t i o n w e p r o v e some s i m p l e p r o p e r t i e s of f - a l g e b r a s a n d from t h e s e w e s h a l l d e r i v e some f u r t h e r p r o p e r t i e s of orthomorphisms. THEOREM 142. I . Any f-algebra
M u l t i p l i c a t i o n by a p o s i t i v e element i s a Riesz homomorphism,
(i)
i.e.,
has the f o l l m i n g p r o p e r t i e s .
A
f o r any
u €'A+
U(fAg) = ( u f )
(ii)
.
f v g
Similarly for
(ug)
A
E
+ +
- -
+ f g
( i i i ) If f , g , h E A and
in
A
and
(vii) u PROOF.
2
A V
A
, then
5 (uv) A (vu)
2
(gu)
f+u = (fu)+
. Fz&thermore,
hf I g
and
.
.
f+g- + f-g+
(fg)-=
.
. Hence,
fh I g
is a two-sided ring i d e a l (besides being a
and
( i ) I t f o l l o w s from
. In p a r t i c u l a r ,
fg = 0
f f + = f + f = (f
v2 = ( w v )
A
and
f,g E A
, then
f E A . U2
(uf)+
=
for aZZ
f I g
any d i s j o i n t compZement i n A
f 2 t 0 and 2
(fAg)u = ( f u )
I n partirmlar, u f +
(fg)+ = f g
f I g
we have
A
and
lfgl = Ifl.Igl
We have
If
f,g
and a l l
and
+ 2 U
2
t
o
for every
V V2
A
E
2 (UV) V (VU)
u2 v v 2 = (uvv)2 {f-(fAg)}
f
+ -
f f
for a l l
{g-(fhg)} = 0
A
=
- +
f f
=
.
fOY'
u,v that
all
E A+
.
0
668
and t h e r e f o r e
, we
- (fAg)
u(fAg) = ( u f )
. Observing now
(up)
A
that
u(fvg) = (uf) v (us)
g e t immediately t h a t
f v g = f+g
-
. Multiplication
on t h e r i g h t i s t r e a t e d s i m i l a r l y .
u
by
a.20,51421
f-ALGEBRAS
(ii)
Let
f,g
f-algebra t h a t
E A b e g i v e n . I t f o l l o w s from t h e d e f i n i t i o n of an
,
f+g+ I f+g-
,
f + g + I f-g+
f-g-
I f+g-
and
f-g-
.
I f-g+
Hence
u1 = Note now t h a t
(f+g++f -g- ) I
[
/ f +g -+f-g+) [ / = ' 2 '
. The
f g = u1-u2
elements
and d i s j o i n t and t h e i r d i f f e r e n c e i s and
.
( f g ) - = u2
u1,u2
fg
are t h e r e f o r e p o s i t i v e
. Then,
as well-known
(fg)'
f I g
(iii) If
lgl = 0
A
larly, fh I g (iv)
.
If
f l g
and
i s a r b i t r a r y , then
h
,
i.e.,
/hfI
,
then
fg I g
f
E A w e have
lgl = 0
A
For any
f 2 = /f+-f-)
2
. In
If1
+ -
f f
= f-f+
- (f+)2 + ( f - p
) -
2 0
=
0
.
(vi) A
It f o l l o w s from
{(v-u)+v} = 0
formula
f+
A
,
i.e.,
(u-v)+
{u(u-v)]+
fg I fg
A A
.
implies
. Simi-
, i.e.,
. Hence
.
Furthermore, f f + = ( f + - f - ) f + = ( f + ) 2 - f - f + = ( f + ) ' = (f+)2
lgl = 0
A
o t h e r words, hf I g
by ( i i i ) , and hence
f g = O . (v)
u1
This i m p l i e s t h a t
Ifgl = ( f g ) + + ( f g ) - = u , + u 2 = (f++f-)(g++g-) = I f l . I g l
(lhl.lfl)
=
(v-u)+ = 0
. Similarly,
that
{(v-u)v)+ = 0
{~(u-v)'}
f+f =
A
.
I n view of t h e g e n e r a l
9
vu
g+ = (fAg)+ w e t h u s g e t
Hence
This shows t h a t 5
(uv)
A
(vu)
.
u2
A
v
2
uv
The proof f o r
. Similarly u2 v v 2
u2
A
v2
,
and hence
i s s i m i l a r by o b s e r v i n g t1.at
u2
2
A V S
ch. 20,§1421 (u-v)+
ORTHOMORPHISMS AND f-ALGEBRAS
o
(v-u)+ =
A
669
implies
( v i i ) We have
u2
since
v2 I (uv)
A
. The
(vu)
A
proof f o r
THEOREM 1 4 2 . 2 . I n any f-algebra
(UVV)~ i s similar. e
w i t h u n i t element
A
t h e following
properties hold.
u
-I
E
(i)
e 2 0
(ii)
If
u
and
A
u
(uAv)-'
-1
IS
E A+
u,v
.
of
Idd
u =
exists i n
I01 , i . e . , exist,
v-'
A , then {uIdd = A
(ti)
-1
v
A
(i) e = e2 t 0
PROOF.
-I
. -A s
=
u
-1
v v
-1
.
-1 u E A+ and u e x i s t s , then -I + u = f Then e = uf = u f -uf
.
and
e
= uf
-
{uId = { O }
For t h e proof t h a t
v
that
=
0
,
so
-
u-'
with
uv = 0
v = u-luv = 0
, assume
that
0 I v E
=
0
. It
uf+ = e
{uId
2
I\
. Hence
e = v
2
A
{uld
=
. Then
.
COI
v
E
,
and t h e r e f o r e
e 5 u
A+
e 2 = (vAe)
. Applying
f o r every u2
A
it
by p a r t ( i v ) of the preceding theorem. It follows
For t h e l a s t p a r t of the proof, note t h a t v
E A'
i n p a r t ( i i ) o f the preceding theorem, t h i s implies t h a t
. But e 2 0 , s o e- = 0 , i . e . , uf follows t h a t e = e+ = uf' . S i m i l a r l y e = f+u . From f + u = -1 -1 i s evident now t h a t u = f + , and t h e r e f o r e u t 0 .
e+ = uf+
A
and
by p a r t (v) o f the preceding theorem.
W e prove f i r s t t h a t i f
uf- = 0
-1
(UAV)
and
For n o t a t i o n a l convenience, w r i t e
u
-1
(uvv)
then
.
e x i s t . These i n v e r s e s s a t i s f y -1
A
-1
u-'
and both
(UVV) = u
uf+
u
. Furthermore
5 e
u
A
{eldd = A
In particular (iii)
. E A+ and the inverse
u
2
=
(VAe)(VAe) < ve = v
t h i s t o t h e given element A
u
-1
= u
-1
u E A+ 2 -1 ( u he) 5 u u = e
.
, we
get
670
f -ALGEBRAS
( i i i ) We assume t h a t p a r t ( i i ) that
(u-lv)
A
Ch. 20,81423
u,v E A+
and
-1 -1 u ,v
u) 2 e
and
(vu-I)
-1
(v
e x i s t . I t follows from (uv-I)
A
. This
e
S
implies
that
and s i m i l a r l y
(uvv)(u-'Av-')
t h e i n v e r s e of u-'
that
if
f
=
. Applying
u v v
. It
e
this result to
i s t h e i n v e r s e of
v v-'
i s e v i d e n t now t h a t
u
A
v
.
u
-1
We r e c a l l t h a t an element f i n t h e a l g e b r a A k = 0 f o r some n a t u r a l number k The a l g e b r a
.
u
and
v
A
v
, we
-1
is
find
i s c a l l e d nilpotent A
i s s a i d t o b e semi-
i f t h e only n i l p o t e n t element i n A i s t h e n u l l element. E v i d e n t l y , 2 i s semiprime i f and only i f f = 0 i n A i m p l i e s f = 0
prime A
-1
-1
.
THEOREM 142.3. In a semiprime f - a l g e b r a hold. (i)
f I g
u,v E A'
( i i ) For u2 = v2
fg = 0
i f and only i f
if a n d only i f
we haue u = v
.
A
the f o l l o w i n g properties
.
u2 S v2
if
and only i f
u
PROOF. ( i ) It was proved a l r e a d y i n Theorem 142.1 t h a t fg = 0
. Conversely,
that
(IflAlgI)
i.e.,
f I g
.
2
=
0
assume now t h a t
and hence, s i n c e
fg = 0
A
.
v
2
. Hence,
f I g
implies
I t follows from
i s semiprime,
If1
A
lgl = 0
,
0 S u S v , then u2 = uu S uv i w = v2 . For t h e converse, 2 i v2 , b u t u 2 v does not h o l d . I n t h i s case, t h e r e f o r e , 2 u A v < u Then u = ( m v ) + w f o r some w > o , s o u2 2 (uhv)' + w 2 2 2 2 w i t h w 2 > 0 s i n c e A i s semiprime. Hence u = u A v = ( m v ) ' < u ,
( i i ) If
assume t h a t
.
u
which l e a d s t o a c o n t r a d i c t i o n . I n a l l the above theorems i t i s n o t assumed t h a t (and hence i t i s n o t assumed t h a t
A
A
i s commutative
i s Archimedean). I n a commutative
Ch. 20,81421
ORTHOMORPHISMS AND f-ALGEBRAS
61 1
f - a l g e b r a w e have t h e f o l l o w i n g e x t r a p r o p e r t y . THEOREM 1 4 2 . 4 . I n
f,g E A
f o r a22
a commutative f-aZgebra
.
PROOF. Using t h a t
f + g = ( f v g ) + (fAg)
= (f-fAg)(g-fAg)
find that
.
= 0
The l a s t e q u a l i t y f o l l o w s from
, we
f g = (fvg)(fAg)
we have
A
(f-fAg) I (g-fAg)
S i n c e an Archimedean f - a l g e b r a i s commutative, t h e l a s t theorem h o l d s i n particular i f
i s Archimedean. We prove some a d d i t i o n a l p r o p e r t i e s
A
h o l d i n g i n an Archimedean f - a l g e b r a . THEOREM 1 4 2 . 5 . Let
be an Arehimedean f-algebra and l e t
A
s e t o f all n i l p o t e n t elements i n (i)
f E N
(ii)
N
. !Then the .
A
i f and onZy i f
f2 = 0
i s a band and a r i n g ideal i n A
(iii) f E N
fg = 0
impZies f
u n i t element, then
E
be the
following holds.
.
. Hence, i f A . I n other iJords,
f o r all g E A
i f and onZy i f
N
N
f = 0
has a any
Arehimedean f-algebra possessing a u n i t element i s semiprime (and hence
aZZ properties mentioned i n t h e above theorem hold then i n A 1 . k PROOF. ( i ) I t i s s u f f i c i e n t t o show t h a t f = 0 f o r some k > 2 fk- 1 implies = 0 , and we may assume t h a t f t 0 M u l t i p l y i n g t h e second
.
f a c t o r by
that
nf
, it
f o l l o w s from
(nfk-’ -f k-2 ) +
A
(nf
k- 1
for
) = 0
l-fk-2\+
k- 1
(nfk-l -f k-2 ) + = 0
,
i.e., 0
i s Archimedean, t h i s i m p l i e s t h a t
.. . Since
’
n = 1,2,.
< nf k-l
fk-l
s fk-’
= o .
...) . S i n c e
(n=l,z,
A
612
(ii)
To s e e t h a t
f,g E N
It follows e a s i l y (since
.
E N
Since
f'
i f and o n l y
0
v2
S
E
. Furthermore,
and
v E N
0
shows t h a t
+
uu
2
assume t h a t
N
0 5 v 5 u
if
.
and
.
2
f
g2 = 0 .
=
E
then
i s an o r d e r
N
,
so
i s a l s o a r i n g i d e a l . For t h e p r o o f t h a t
N
u 2 0
in
(gf)2
u
g2f2 = 0
=
+
0 5 u
and
A
uuo ( s i n c e m u l t i p l i c a t i o n by
in
uo
A
,
i s a p o s i t i v e orthomorphism
and orthomorphisms a r e o r d e r c o n t i n u o u s ) . Now, l e t
. From
,
u E N
It follows already t h a t
i s a r b i t r a r y , then
g E A
i s a band, o b s e r v e f i r s t t h a t i f then
f+g E N
i s commutative) t h a t ( f + g ) 4 = 0 Hence f + g 2 2 If1 = If 1 = 0 , we see t h a t f E N
A
and hence
f E N
. This
N
,
u2 = 0
2
implies
i f and o n l y i f
0
=
If( E N
ideal. If gf
ch. 20,91421
f-ALGEBRAS
+
0 5 u
uo
with
u u + T To T '0 T '0 u u i t follows then t h a t u u = 0 This h o l d s f o r a l l T~ . Hence,. 0 TO 2 OTO 2 This concludes since 0 = u u uo , we f i n d t h a t uo = 0 , i . e . , uo E N u
+
E
for all
N
T
N
(iii) Let
f
g
u2
5
=
+
O T
t h e proof t h a t by
u u
i s a band.
E
N
, i.e., f
2
0
=
0
.
for a l l
u
and
2
.
. Define
t h e orthomorphism
in
nf
n g = f g f o r a l l g E A. Then IT f = 0 and IT g = 0 f o r a l l f f f d Hence nf v a n i s h e s on t h e s e t D c o n s i s t i n g of f and
.
E If}
Since i.e.,
Ddd = A
,
i t f o l l o w s from C o r o l l a r y 140.6 t h a t
for all
fg = 0
g E A
.
If)
v a n i s h e s on
rf
A
.
d
,
A
EXAMPLE 142.6. We mention some f u r t h e r examples of f - a l g e b r a s .
(i) space
Let
L
Orth(L)
b e an Archimedean Riesz s p a c e . A s w e have proved, t h e
of a l l orthomorphisms i n
with the i d e n t i t y operator i n
L
i s an Archimedean f - a l g e b r a
a s u n i t e l e m e n t . Hence, by t h e l a s t theo-
L
i s a semiprime f - a l g e b r a . 2 L e t A = R w i t h l e x i c o g r a p h i c a l o r d e r i n g . Then
rem, O r t h ( L ) (ii)
A r c h i m d e a n R i e s z s p a c e . D e f i n i n g t h e p r o d u c t of t h e p o i n t s (x2,y2)
A
in
t o be
(xIx2,xIy2)
,
the space
A
A
i s a non-
( x I 2 y I ) and
becomes a non-commutative
f-algeb ra.
2 ( i i i ) I f , once more, A = IR w i t h l e x i c o g r a p h i c a l o r d e r i n g and m u l t i p l i -
A
cation i n
i s d e f i n e d by
(xI,yI).(x2,y2) = (x1x2,0)
,
then
A
becomes
a commutative non-Archimedean f - a l g e b r a . Many Archimedean f - a l g e b r a s are Riesz i s o m o r p h i c t o a s p a c e of
(iv) type
C(Y)
,
t o a s p a c e c o n s i s t i n g of a l l ( r e a l ) c o n t i n u o u s f u n c t i o n s
i.e.,
on some t o p o l o g i c a l s p a c e
property. Let
L
on t h e p o i n t s e t Then
L
Y
. The
b e the space X
, where
f o l l o w i n g f - a l g e b r a does n o t have t h i s
M(X,p)
of a l l ( r e a l ) p-measurable
functions
i s a o - f i n i t e measure p o s s e s s i n g no atoms.
i s a Dedekind complete f - a l g e b r a and, as shown i n Example 85.1,
the
Ch. 20,§1421
order dual
satisfies
L-
isomorphic t o some
A
If = {O}
,
L"
{eldd = A
. In
,...
only s o t h a t
s
uniformly t o
u
,
0 5 u E A
converges t o
.
If u , v
then
nv(u-nv)+ 5 u2
.
, then
n(u-ne)
+
0 5 u - i n f ( u , n e ) = (u-ne)
PROOF. From
0
S i m i l a r l y , i t follows from
COROLLARY 142.8. Let
0 2
where
I
-
'i~
s
n
+
2
u
2
for
inf(8,nI)
n = l,2,
verges a2-wziformlg t o
.
IT
+
... . Hence, ,... , i . e . ,
5 n
-Iu2
for
n = l,2,
If
A
i f
, A
... .
i t follows t h a t
2
that
.
n
-lr2
for
n = 1,2,
... ,
. Hence, w r i t i n g = ... , t h e sequence i s increasing and con. Note t h a t IT^ belonngs t o the centre Z ( L ) f o r 71
IT
I t follows t h a t any orthomorphism IT i n L can be every n w z i f o n l y appro&mated by a sequence of elements i n the centre COROLLARY 142.9.
A
converges r e l a t i v e l y
d e w t e s t h e i d e n t i t y operator i n L
= inf(.rr,nI)
{eld =
is
be a p o s i t i v e orthomorphism i n the Archimedean
IT
. Then
e
for n = l,2, 2 f o r n = 1,2
(u-nv)(u-nv)+
S
nv(u-nv)+ 5 u(u-nv)
L
5 u
0 5 {(u-nv)+}2 = (u-nv)+(u-nv)
(u-nv)+nv 5 (u-nv)+u 5 u
Riesz space
then
are p o s i t i v e elements i n the f-algebra n(u-nv)+v 5 u2
and
e
has a unit element
,
s = inf(u,ne) n We s h a l l prove now t h a t i t i s n o t
The d e t a i l s f o l l o w .
THEOREM 142.7.
e
i s given and we s e t
i n order,but
u
is Riesz
.
o t h e r words, t h e band g e n e r a t e d by
0 5 s 4 u n
then
L
c o n t a i n s any l i n e a r f u n c t i o n a l
E C(Y)
i s an Archimedean f - a l g e b r a w i t h u n i t element
i.e.,
n = 1,2
i s i m p o s s i b l e now t h a t
{C(Y)}-
$y(f) = f(y) ; f
i t s e l f . It follows t h a t i f for
. It
= {O}
, because
C(Y)
of t h e form
$y (y€Y)
673
ORTHOMORPHISMS AND f-ALGEBRAS
L
1111'-
Z(L)
i s a Dedekind compZete Riesz space, then
.
f-ALGEBRAS
674
Orth(L)
i s the band generated by the i d e n t i t y operator Lb(L)
complete space Orth(L)
that
ch. 20,51421 I i n the Dedekind
o f a l l order bounded operators i n
PROOF. I t was obse.rved a l r e a d y i n s e c t i o n 139 t h a t
.
band i n t h e Dedekind complete space %(L) Orth(L)
. It
L
follows
i s aZso Dedekind complete.
inf(n,nI) f n
and s i n c e
i t follows t h a t
Orth(L)
Since
h o l d s f o r every p o s i t i v e
i s t h e band generated by
i s now a
Orth(L)
i s a member of
I
in
n
.
I
Orth(L),
We s h a l l apply t h e r e s u l t s o b t a i n e d s o f a r t o t h e t h e o r y of o r d e r adj o i n t o p e r a t o r s . For t h i s purpose we r e c a l l t h a t i f Riesz space w i t h o r d e r d u a l order adjoint operator all
and a l l
f E L
of
T-
a d j o i n t of t h e i d e n t i t y o p e r a t o r L"
.
(T"$)(f)
i s p o s i t i v e , then s o i s
I
in
(i)
If
E Z(L)
IT
, it follows (ii) I f
, it follows
PROOF. ( i ) I f r e a l number
N
n IT-
,
follows t h a t
t h e converse, s i n c e by assumption L
that
-
TI
...
, we
-
rn
n
TI-
. Taking
e v i d e n t from
TI"
f TI"
that
n
N
IT
n
L-
=
k(L-)
N
E Z(L ) L
-
L
. For
, note n
that
to
n" E Z(L-1
inf(n,nI)
, which
for immediate-
. Furthermore,
i n t o account t h a t
Orth(L-)
by p a r t
is a
by C o r o l l a r y 142.9, i t i s
i s a member of
r. converse d i r e c t i o n i s a s i n p a r t ( i ) .
.
n E Z(L)
r e l a t i v e l y uniformly i n
band i n t h e Dedekind complete space
S n
r e s t r i c t i o n of
r e l a t i v e l y uniformly i n
Z(L-)
0
s i n c e i t follows from
be given. W r i t i n g
4 n
4 TI-
, i.e.,
. The
E Orth(L).
IT
f o r some non-negative
hI
s e p a r a t e s t h e p o i n t s of
L-
in
.
that
L"
XI"
may conclude now t h a t
have n
belong t o
E Orth(L") 5
order
I-
separates the
IT
. Hence,
0 5 n E Orth(L)
l y implies t h a t (i), a l l
is
, we
E Z(L-)
( i i ) Let n = l,2,
L
N
TI
E Z(L)
L"
0 5
5
for
. The
T-
the fozlowing
TI
.If
0 2 n N
$(Tf)
the
separates the points
that
E Orth(L")
then
L
L"
E Z(L")
i s embedded a s a Riesz subspace i n
t h e Riesz subspace
. If
E Z(L")
conversely from
0 S n E Z(L)
. It
h
n"
E Orth(L) , then
IT
L
p o i n t s of
, then
conversely from
,
L
i s the i d e n t i t y o p e r a t o r
L
THEOREM 142.10. For an Archimedean Riesz space
L
=
T
properties hold.
of
i s an Archimedean
L
i s a linear operator i n
T
i s d e f i n e d by
T
. If
E L"
$
and
L"
Orth(L").
The proof i n t h e
Ch. 20,51421
675
ORTHOMORPHISMS AND f -ALGEBRAS
The l a s t theorem i s due t o A.W.
Wickstead ([11,1977,Proposition 3 . 1 and
Theorem 3 . 3 ) . The p r e s e n t proof f o r p a r t ( i t ) i s d i f f e r e n t from Wickstead's proof. The r e s u l t i n C o r o l l a r y 142.8 t h a t t i v e orthomorphism space of type
IT
holds f o r any posi-
e n a b l e s u s now t o determine a l l orthomorphisms i n a
IT
,
Lq(X,u)
0 < q <
THEOREM 142.11. Let
0 < q <
inf(IT,nI) .f
m
.
be a a - f i n i t e rneasurs i n the p o i n t s e t
LI
X
.
, let
and l e t L be the ( r e a l ) Riesz space Lq(X,u) Then, i f the orthomorphism TI i n L i s given, there ezLsts a real f u n c t i o n p E L ~ ( x , ~ ) m
q f E L q '
such t h a t , f o r any
IT
. Conversely, L . 4 x
holds almost everywhere on t o an orthomorphisrn
the e q u a l i t y
in
PROOF. Assume f i r s t t h a t tion
e
, satisfying
e(x) = 1
w(X)
p E L,
every
gives r i s e thus
i s f i n i t e . I n t h i s case t h e u n i t func-
for a l l
x E X
,
i s a member of
t h a t Theorem 141.1 i s n o t a p p l i c a b l e , because i n g e n e r a l gebra. We proceed, t h e r e f o r e , somewhat d i f f e r e n t l y . Let t h e p o s i t i v e orthomorphism h o l d s on
2 p(x)
n = 1,2 x E X f E L q
...
, we
. Since ,
X
, and
set
inf(n,nI)
=
IT
IT
b e given. Write
the function
the function
p
and
pn(x).f(x)
i t i s evident that the operator
i s an orthomorphism i n
IT
. Note
n '
IT IT
n
and
n =
T I:
IT'
L
d e f i n e d by
4
0
5
for all
f o r every
,
coincide on t h e s e t
{e}
, i.e.,
(anf)(x) = pn(x).f(x)
. Then . For
now t h a t
e = inf(rre,ne) = i n f ( p , n e ) = pn
which shows t h a t
i t follows t h a t
L 4
L
q pn(x) = min(p(x),n)
IT'
. Note
re = p
i s a member of
i s a member of
L 9
i s n o t an a l -
L 9
for a l l
f E L q
.
. Since
{e}dd = L
,
676
STABILIZER OF A RIESZ SPACE
Now, l e t
0 C u E L 0
5
0
be g i v e n . S i n c e
q
p (x).u(x)
4 (au)(x)
5 'TI u 4 'TIU
in
L
q
, we
h o l d s p o i n t w i s e on
4 p(x).u(x)
that
h o l d s a l m o s t everywhere on
It remains t o prove t h a t
.
E L
function
q t h e n so i s
g
For t h i s p u r p o s e , c o n s i d e r a n a r b i t r a r y
.
on
. As
pq.g
u(X) = =
,
then
of f i n i t e p o s i t i v e measure, and pk E L, 'TI
on
. The f u n c t i o n X . Evidently, p
p
E L,
pq
,
i.e.,
i s t h e d i s j o i n t union of s e t s
X
on
E L,
i s r e p r e s e n t e d on
n X
is
g
well-known f o r a measure s p a c e of f i n i t e
o r o - f i n i t e measure, t h i s i s p o s s i b l e o n l y i f If
X. I t f o l l o w s
. Hence
X
X Then l g l l / q E L , and hence q by t h e r e s u l t a l r e a d y e s t a b l i s h e d . I n o t h e r words, i f
( r e a l ) u-sumable p-summable,
p E L,
have
.
Furthermore, 0 5 pn(x).u(x) ('TIu)(x) = p ( x ) . u ( x )
20,91431
Ch.
, equal
. The
to
'i,
p E L,
'i, (k=l,2,...)
.
by a f u n c t i o n
'i, , r e p r e s e n t s
on e a c h
pk
if
e x t e n s i o n t o an a r b i t r a r y orthomorphism
( n o t n e c e s s a r i l y p o s i t i v e ) i s immediate. I t i s a l s o e v i d e n t t h a t , c o n v e r s e l y , every function
p E Lm
g i v e s r i s e t o a n orthomorphism
'TI
in
L
q
.
143. The s t a b i l i z e r of a Riesz s p a c e I n s e c t i o n 139 w e have i n t r o d u c e d band p r e s e r v i n g o p e r a t o r s i n a n Archimedean R i e s z s p a c e
L
. Orthomorphisms
are those o p e r a t o r s i n
L
t h a t are
band p r e s e r v i n g as w e l l as o r d e r bounded. I n Example 139.2 i t was shown t h a t t h e r e e x i s t band p r e s e r v i n g o p e r a t o r s t h a t a r e n o t o r d e r bounded. It i s q u i t e n a t u r a l now t o i n t r o d u c e a l s o i d e a l p r e s e r v i n g o p e r a t o r s . By d e f i n i tion, the operator
tractor i f
'TIU
E A
i d e a l g e n e r a t e d by every
u E L+
t h i s case given
xf
f E L
'TI
i n t h e Azichimedean R i e s z s p a c e f o r any
u
. In
u E L+
, where
o t h e r words,
holds f o r every
w e have
only i f
'TI
. To
nf+ E A f + c A l f I = Af
leaves a l l i d e a l s i n
L
denotes t h e p r i n c i p a l
AU
Xu
f E L
It f o l l o w s immediately t h a t t h e o p e r a t o r
i s c a l l e d a con-
i s a c o n t r a c t o r whenever f o r
'TI
t h e r e e x i s t s a r e a l number
E Af
L
'TI
such t h a t
~ ' T I U5~
h u
. In
s e e t h i s , n o t e t h a t f o r any
and, s i m i l a r l y , vf- E Af in
L
.
i s a c o n t r a c t o r i f and
invariant. Evidently, every contractor
is band p r e s e r v i n g . W e s h a l l prove now f i r s t t h a t c o n t r a c t o r s a r e automati-
Ch. 20,51431
677
ORTHOMORF'HISMS AND f-ALGEBRAS
c a l l y o r d e r bounded. F o r t h e proof w e r e c a l l s e v e r a l f a c t s about maximal i d e a l s i n a n Archimedean Riesz s p a c e If
i s a maximal i d e a l i n
J
L
possessing a strong unit
L
, every
f E L
, we
f = j + A e (jEJ,Xf r e a l ) . S e t t i n g now $ ( f ) = Af f morphism $: L + IR such t h a t $ ( f ) = 0 f o r a l l Conversely, i f $(f) = 0
fying
i s one-one.
$
f o r a l l these ideals i n
L
i s a maximal i d e a l It follows t h a t i f
,
$
$(f-) = 0 > 0
$(f-) =
0
1
and
J
g E L
f+
A
2
and
J
$(g) = 0
has the property t h a t
=
0
$(f+)
that
A
$(f-)
, we
{O}
$(f) 2 0 $(f-)
i s i m p o s s i b l e , however, t h a t
f t 0
satis-
correspondence between
has the property t h a t
o c c u r s i m u l t a n e o u s l y . Hence implies
$
t f
f
i s c o n t a i n e d i n t h e i n t e r s e c t i o n of a l l maximal
g
f-
.
$(el = 1
=
= 0
$(f)
0
2
,
i.e.,
0
,
for a l l
(and hence
$ ( f l ) t $(f2)
. It
$
$(f+) = 0
$(f+) = 0
,
$
for a l l
i.e.,
, f- =
$(f) t 0 $
.
g = 0
find
for a l l
by what was observed above. I t h a s b e e n shown t h u s t h a t
all f
. It
. The
J
t h i s intersection i s equal t o
f E L
f o l l o w s t h e n from or
then
. Since
Assume now t h a t
f E
o b t a i n a R i e s z homo-
i s such a Riesz homomorphism, t h e s e t of a l l
$
.
e E L+
can b e w r i t t e n u n i q u e l y as
for
implies
).
THEOREM 143.1. E v e 3 contractor i n an Arehimedean R i e s z space
is
order bounded. n
PROOF. L e t
L
assume f i r s t t h a t
-A
b e a c o n t r a c t o r i n t h e Archimedean R i e s z s p a c e has a strong u n i t
u 5 n u 5 A u ) and l e t
$(el = 1
.
Similarly
The e l e m e n t
W
W
and t a k i n g Then
e E L+
. Let
u E L+
and
(hence
IR b e a R i e s z homomorphism such t h a t
v = u - $(u)e
satisfies
$(v) = 0
t h a t f o r any
w E L+
, hence
i t f o l l o w s from
that
w = v+
$(nv) = 0
+
. Observing now
$(v-) = 0
-A w 5 nw 5 A w
$: L
L
,
and i.e.,
w = v
-
r e s p e c t i v e l y , we g e t
$(a"+)
-
= $(nv ) =
0
.
678
STABILIZER OF A RIESZ SPACE
Furthermore, i n view of
e
-A
, we
ne 5 A e
5
-xe
have
Hence, from ( I ) ,
Q: L
T h i s h o l d s f o r e v e r y R i e s z homomorphism
+
2Q,§1431
Ch.
.
5 ~ ( ~ 5e A)e
IR s a t i s f y i n g
. On
Q(e) = 1
account of t h e remarks made a b o u t t h e s e orthomorphisrns we may conclude t h a t u 5 nu 5 A u
-A
n
To prove t h a t
i s o r d e r bounded, we show now t h a t t h e image of t h e o r d e r
[-u,u]
interval
.
i s contained i n
[-Aeu,Aeu1
.
Indeed, i f
f E [-u,ul
,
then
+
InfI = lnf -nf where we use t h a t If
f+
A
-
I
+
= Inf +nf
f- = 0
-
I
= In(1fl)l
implies
n f + I nf-
5
Xelfl
[-u,ul
i s contained i n a n o r d e r i n t e r v a l . AU
, possessing
a s a s t r o n g u n i t . By t h e r e s u l t a l r e a d y proved t h e image of contained i n
[-AUu,AUul
.
COROLLARY 143.2. Every contractor The s e t
. Evidently,
S(L)
nl then
and
is an orthomorphism.
of a l l c o n t r a c t o r s i n
L
i s c a l l e d t h e s t a b i l i z e r of
t h e following inclusions hold:
n2
.
i s an o r d e r i d e a l i n
Orth(L) Orth(L)
a r e ot-thomorphisms s u c h t h a t
n 1 E S(L)
u
is
L
i s a l i n e a r subspace of
S(L)
S(L)
[-u,ul
in
Z(L) c S(L) c Orth(L)
that
,
.
For t h e proof we may r e s t r i c t o u r s e l v e s t o t h e i d e a l
The s e t
xeu
does n o t p o s s e s s a s t r o n g o r d e r u n i t , we s t i l l have t o prove
L
t h a t t h e image of any i n t e r v a l
L
5
. For
any
u E L+
. For t h e p r o o f , note . I t remains t o prove
Inl[
5
In2/
t h e following holds:
and
first that i f
n2 E S ( L )
,
Ch.
20,81431
Since
Z(L)
and
Orth(L)
a r e the i d e a l and the band r e s p e c t i v e l y generated
by the i d e n t i t y o p e r a t o r
i s evident t h a t
I
L1
the Riesz space
L2
and the i d e a l
i s o r d e r dense i n
S(L)
EXAMPLE 1 4 3 . 3 . L e t
and
1 Orth(L ) = ( s ) = L1 1 = S(L ) = ( s ) 2
and
and
be t h e Riesz space
Ly
..,$n)
E Ly
E
L1 = ( s )
= L2
N
f o r any
= (nl,n2,
a E S(L1)
IT
E S(LI)
IT
, IT
,...
= (IT~,IT~)
E S(L2)
N
n
... +
.ern
S(LI) =
and
Ip a f
n n n ’
N
IT
every sequence i n
(s)
or
1~
i s the a d j o i n t
t h a t there e x i s t c o n t r a c t o r s
f a i l s t o be a c o n t r a c t o r i n
the s t a b i l i z e r i s i n c o n t r a s t w i t h t h a t of E Orth(L)
E
...) E La = Z(L2) . I t follows t h a t IT- E Z(L2) . Every sequence i n brn = Z(L2) occurs as
. Similarly, . This shows
the a d j o i n t
Orth(L2) =
we have
N
n
. Furthermore
.
( n 4 ) l f ) = + ( n f ) = @ , n l f l+
hence
satisfy
# Orth(L1) ,
f = ( f l , f 2 ,...)
IT
of a l l r e a l sequences
(s)
and
. Hence
Z(L2) # S(L2) = Orth(L2)
=
Ly
Z(L1) = S(LI) =
Z(L2) =
Z(L1) =
if
.
L ” = L2 and 1 (under the usual i d e n t i f i c a t i o n s ) . I t i s not d i f f i c u l t t o see t h a t
Ly = L
4
i s between these two, i t
S(L) Orth(L)
of a l l sequences with only f i n i t e l y many non-zero
coordinates. Note t h a t the o r d e r duals
For
6 79
ORTHOMORPHISMS AND f-ALGEBRAS
E Z(L) , then
N
n
IT
L y = L1
in
. This
Orth(L) and
E Orth(L-)
L2
IT-
f o r some of some
such t h a t
behaviour of
Z(L)
or
N
n
c S(L2)
, because
E Z(L-)
r e s p e c t i v e l y (by Theorem 142.10). This example is due t o A.W.
Wickstead
(C 11,1977,section 4). If
i s an orthomorphism i n
n
elements
f
E
L
f o r which
nf E Af
L
b u t not a c o n t r a c t o r , then t h e r e a r e
holds and t h e r e a r e elements
f
f o r which
rrf E Af
does not hold. We s h a l l prove t h a t the s e t of a l l
f o r which
n f E Af
holds i s an o r d e r dense i d e a l i n
THEOREM 143.4. I f
space
L
, then
TI
L
E L
.
i s an orthomophisn i n t h e Arehimedean R i e s z
J = (fEL:nfEA ) f
is an order dense i d e a l i n
L
.
f
E L
6 80
STABILIZER OF A R I E S Z SPACE
PROOF. Assume f i r s t t h a t that i f
and
g E J
t h e orthomorphism ( g l E N,
1
=
. In
0
,
then
, we
A1
, and
If/ 5 lg/
get
therefore
. This
shows t h a t
For t h i s p u r p o s e , n o t e t h a t
f,g E J
t h e f i r s t p a r t of t h e proof t h a t To prove t h a t
0 < u E L
any
Hence, l e t that fies
0 < u E L
, i.e.,
n
5 nv
, and
= nI-n
hence
Finally, l e t s e t of a l l +a1
any
. As
f
we have
implies
Ifl,lg/
E
.
J
.
It f o l l o w s
, we
.
is
J
f+g E J
d e r i v e from
v E J L
0 < v
such t h a t
S
.
u
i s Archimedean, i t i s i m p o s s i b l e
... . T h e r e f o r e (nu-au)+ > 0 ( n I - r ) + u > o . Then v = n - I v l
for
v1 =
satis-
.irf E Af
, J1
+
f
IT]IT; =
IT^)^ . I t
follows t h a t
E J
-
I
f+
A
f- = 0
f E J
If1 E J 1
. Since
i f add o n l y i f
i s an o r d e r dense i d e a l i n
The l a s t theorem i s due t o A. B i g a r d ([1],1972),
L
f
.
i s an i d e a l ,
JI
.
lr
). Hence
= /lTf++ITf-l = l I T ( l f l ) I = l a l ( I f l )
i f and o n l y i f
J
i s the
J 1 be the corresponding s e t f o r
(on account of
t h i s i s equivalent t o saying t h a t
. Hence,
. Let
i s an o r d e r d e r s e i d e a l . Note now t h a t f
a f + I nf-
ITfl = IITf -ITf
words, J = J 1
t h e proof t h a t
b e an a r b i t r a r y orthomorphism. A s b e f o r e , J
satisfying
T h i s shows t h a t
implies t h a t
n = 1,2,
satisfies
.
v E J IT
proved a l r e a d y
f E L
. By n,(Ifl)
and
where we use t h a t ITV
, so
i.e.,
If+gl S If1 + l g /
f+g E J
be g i v e n . S i n c e
holds f o r a l l
0 < v 5 u
,
that
. Introducing
lg/
i s o r d e r dense i t i s s u f f i c i e n t t o show t h a t f o r
J
some n a t u r a l number
i t follows
J
{pidd c N T 1
f,g E J
implies
t h e r e e x i s t s a n element
nu5 nu
. For
f E J
. Since
If1 + lgl E J
E
a,(Igl) = 0
, which
s u p { n ( l f I ) , A / f / } = AlfI
l n f l = a ( l f 1 ) 5 hlrfI
g
1rgl i
=
If1 E {g}dd c N n l
a n i d e a l , i t i s s u f f i c i e n t t o show now t h a t then e a s i l y t h a t
From
n(lg1)
i s a band, t h i s i m p l i e s t h a t
NT1
o t h e r words
-
.
f E 3
such t h a t
h
= sup(a,XI)
. Since
hypothesis
i s a p o s i t i v e orthomorph sm. We prove
T
/ f I 5 Igl
t h e r e e x i s t s a r e a l number
20,91431
Ch.
E
J1
. In
other
with a d i f f e r e n t proof.
ch. 20,11441
ORTHOMORPHISMS AND f-ALGEBRAS
68 1
144. Orthomorphisms i n a normed Riesz space
I n the present section
i s assumed t o b e a normed Riesz s p a c e (norm
L
p ) . The i d e n t i t y o p e r a t o r i n
a norm bounded o p e r a t o r n
LEMMA 144.1. Let
in
T
i s denoted by
L L
I ; t h e o p e r a t o r norm of
i s denoted by
.
IlTll
. Then
L
be a p o s i t i v e orthomorphism i n
norm bounded i f and onZy if there e x i s t s a p o s i t i v e r e a l n d e r
that =
.
XI
5
'TI
inf(A:n
.
PROOF. Let
x
any
>
o
(n-nI)+u
=
n
.
n 2 XI
Then
such
A
II'TII! =
satisfies
(n-nI)+ > B
t h e r e e x i s t s an element
. It
> 0
is
be norm bounded and assume t h a t t h e r e does n o t e x i s t
'TI
satisfying
and f o r each
Ilnll
In t h i s case the operator nomi
n
u n
follows t h a t
E
n = I,Z,
for a l l satisfying
L+
...
v = n
.
'TIV t nv S i n c e v > 0 f o r a l l n , t h i s shows t h a t 'TI i s n o t n norm bounded. C o n t r a d i c t i o n . Hence n s X I f o r some p o s i t i v e r e a l number
hence
A . Conversely, assume t h a t f
E L , we have
follows t h a t
n 2 XI
InfI = n ( l f l ) 2 AlfI
f o r some
,
and hence
i s norm bounded w i t h norm
n
. Then,
A 2 0
f o r every
.
p ( s f ) 2 Ap(f)
Ilnll s a t i s f y i n g
It
.
1/n11 5 A
This i m p l i e s immediately t h a t
Denoting t h e r i g h t hand s i d e i n t h e l a s t formula by t o prove t h a t (1) E
>
llnll t Ilnll_
. If
llnlIm = 0
, it
~
~
, nwe
( ( ~ 1=l ( I n I ( _ = 0 . Assume, t h e r e f o r e , t h a t \ I T / / _ 0 such t h a t A = 1 1 ~ 1 -1 ~E > 0 . By t h e d e f i n i t i o n of that
, so
(n-XI)'
> 0
v E L+
such t h a t
Since
E
now t h a t
s~ t i l~l have _
follows immediately from > 0
. Choose
11n11_
we have
t h a t e x a c t l y a s above we f i n d t h a t t h e r e e x i s t s an element nv t Xv > 0
. This
shows t h a t
11x11
2
i s a r b i t r a r y i n t h e i n t e r v a l between z e r o and
1 1 ~ 1 1t 11'~111_
,
X
=
l l ~ l -l E~
Ilnll_
, we
.
see
682
ORTHOMORPHISMS I N A NORMED R I E S Z SPACE
THEOREM 144.2. Let
. Then
L
be an orthomrphism i n
'TI
Ch. 20,11443
A
bounded i f and only i f there e x i s t s a p o s i t i v e r e a l nwnber
. In
5 XI
1711
tiall
t h i s case the norm
tlall
satisfies
'TI
i s norm
such t h a t
= inf(X:Inl
.
PROOF. Taking i n t o a c c o u n t t h e above lemma, i t i s e v i d e n t l y s u f f i c i e n t
i s norm bounded i f and o n l y i f
t o show t h a t
'TI
t h i s case
and
e d and
71
i s a r b i t r a r y , then
f € L
This shows t h a t 'TI
b e n o r m b o u n d e d . The e l e n e n t s
3
P{l'TIl(f
= P{lT(lfl)3 5
of
L
L c o n s i s t s exactly o f
Z(L)
The operator norm i n
i n the space
L
,
/ I In1 1 1
5
Iird(lfl)
PROOF.
L
II*II.P(f)
and
.rr(lfl)
let
have t h e
/I
. In/
11
5
11d
.
all norm bomded orthomorphisms i n
, because
( i i ) Let
Z(L)
does not depend on t h e norm
i n f a c t the operator norm i n Orth(L) = Z(L)
Z(L)
i s the
.
b e an orthomorphism i n t h e Banach l a t t i c e
'TI 'TI
L
i s a Riesz norm s a t i s f y i n g
i s a Banach l a t t i c e , then
have t o prove t h a t
. Hence
. Conversely,
I-uniform norm. (ii) I f
is norm bound-
i s a normed Riesz space, t h e centre
I n particuZar, the operator norm i n p
11'~111
i s norm bounded and
In
COROLLARY 144.3. ( i ) I f Z(L)
la(lfl)l = I n l ( l f l )
=
lal(f)
1711
T h e r e f o r e , t h e y have t h e same norm, s o
sane absolute value
It follows t h a t
I d
i s norm bounded and
'TI
i s s o and t h a t i n
1711
have t h e same o p e r a t o r norm. I f
1711
i s norm bounded. This i s e a s y . S i n c e
'TI
L
. We
is a differ-
e n c e of p o s i t i v e o p e r a t o r s and e v e r y p o s i t i v e o p e r a t o r i n a B h a c h l a t t i c e
i s norm bounded (Theorem 83.12), Note t h a t a l s o i f normed, t h e c e n t r e
Z(L)
L
the operator
a
i s norm bounded.
i s an Archimedean R i e s z s p a c e , n o t n e c e s s a r i l y c a r r i e s t h e I-uniform norm
ch. 2 0 , 5 1 4 5 1
683
ORTHOMORE'HISMS AND f-ALGEBRAS
In Exercise 1 4 0 . 1 2 it was observed that if L
is Archimedean and uniformly
is uniformly complete. In other words, Z ( L )
complete, then Z(L)
Banach lattice with respect to the norm I l . I l m
. In particular, if
is then a L
is
is a Banach lattice with respect to I I . / l m
a Banach lattice, then Z(L)
,
since every Banach lattice is uniformly complete. EXERCISE 1 4 4 . 4 . Let L
be the sequence space consisting of all (real)
sequences with only finitely many nonzero terms (and with the natural Riesz space structure). Let L
be equipped with the uniform norm. Show that
L has a positive orthomorphism that is not norm bounded.
145. Theorems of Radon-Nikodym type
There exist several theorems related to the now already classical Radon-Nikodym theorem. In section 41 it was shown, for example, how the Radon-Nikodym theorem and Freudenthal's spectral theorem are related. We present another example in which there occur orthomorphisms. LEMMA 145.1. Let
o
5 J, 5
that C$
*
@ E
L"
@no
=
of 0
L
be a Dedekind complete Riesz space and l e t
. m e n t h e r e exists a p o s i t i v e orthomorphisrn . The orthomorphism IT^ i s uniquely determined
*
PROOF. Since L
is Dedekind complete and
$
is order continuous, the
already
@
. The set
@IT 5
J,
longs to
. We
show now that if
0
= S U ~ ( I T , , ~ T belongs ~)
to
0
is not empty, since the null operator be-
IT^
and
. For the proof,
on the band generated by the range of so
@
TI =
satisfying
aqd
.
and the carrier C of $ satisfy N CB C = L Setting @ @ 0 on N , we may now just as well assume that L = Co , i.e., @ strictly positive. Let @ be the set of all positive orthomorphisms IT
null ideal N @
i n L such on t h e c a r r i e r
-
IT^
belong to @ , then IT = 3 let P be the order projection
(.rrl-s2)+
. Note
that
(nl-n2)+
( I T , - I T ~ ) - , being disjoint positive orthomorphisms, have disjoint ranges, P(IT -IT ) f 1 2
= 0
for every
f E L
. For every
u E L+
we have now
THEOREMS OF RADON-NIKODYM
684
(.rrl-IT2)+u = P ( n 1-'TI 2) + U From
a3
P ( n 1-IT 2) + U
=
71
P ( n -n ) u + 1 2
=
3
IT
2
P ( a 1-71 2)-u = p ( * 1-7 2 ) u
it f o l l o w s t h e r e f o r e t h a t
sup(7i,,x2) = (a1-v2)+ + a 2
=
+
Ch. 2 0 , 1 1 4 5 1
TYPE
u = Pa u + (I-P)n u = 1 2
IT
1
,
Pu + a (I-P)u 2
where we u s e t h a t orthomorphisms and o r d e r p r o j e c t i o n s commute ( a s observed a l r e a d y i n s e c t i o n 1 3 9 ) . Then $a u = 3 which shows t h a t
$IT
1
(Pu) +
+n3 <_ J,
$IT
, and
2
{(I-P)ul
hence
IT
J,(Pu) + J,{(I-P)u}
I -<
f 0
3
. Thus,
the s e t
be upwards d i r e c t e d . Our n e x t purpose i s t o show t h a t above by t h e i d e n t i t y o p e r a t o r n o t hold. Then
(71-I)+u > 0
I
. Assume
that
u E L+
f o r some
71
E 0
, i.e.,
,
$(u)
=
i s seen t o
0
i s bounded from
0
, but
IT 5
(au-u)'
> 0
I
does
. Let
be t h e o r d e r p r o j e c t i o n on t h e band g e n e r a t e d by (au-u)+ , s o P1 P (nu-u) = (nu-u)+ > 0 It f o l l o w s t h a t P u < P ITU = nP u , which i m p l i e s 1 1 1 1 is s t r i c t l y p o s i t i v e . But t h e n ( s i n c e that $(Plu) < $(aPlu) , since
.
+
$(Plu) < $n(Plu) i J,(Plu)
$x i J, ) we g e t
R e c a p i t u l a t i n g , w e have proved t h a t {Orth(L>}+ such t h a t
, which
@
i s bounded from above by
0
contradicts
J, 5 $
I
.
Since
we have that
a.
= sup 0
+n 2 J,
i t y from
0 <J,l =
(Jil-s$)+
,
(J,,-E$)+
so
€P2 + r0 $ = $ITo
.
)I
-
IT^
IT
E 0
, and
S $
Orth(L)
that
IT^
. It f o l l o w s by o r d e r c o n t i n u . Assuming t h a t $no < JI ,
s o t h e r e e x i s t s a number
> 0
. Let
i s n o t t h e n u l l o p e r a t o r . For e v e r y
P2
$ ( E P ~ + I T ~ J,)
belongs t o
0
,
> 0
E
such
be t h e o r d e r p r o j e c t i o n on t h e c a r r i e r of
u E L+
, which
we have
.
( $ J ~ - E $ ) - Hence
i s c o n t a i n e d i n t h e n u l l i d e a l of
P u 2
Lb(L) ) ,
5 J,
P2
where we use t h a t
It f o l l o w s t h a t
exists in
for a l l
is
Orth(L)
Dedekind complete ( b e i n g an i d e a l i n t h e Dedekind complete s p a c e t h e supremum
.
i s an upwards d i r e c t e d s e t i n
shows t h a t t h e orthomorphism
contradicting the fact that
a.
=
sup 0
. Hence
ORTHOMORPHISMS AND f-ALGEBRAS
ch. 20,51451
It remains to prove that
IT^
for this purpose, that
IT^
$r1 =
, i.e.,
the element
no
$(IT1-IT 2) = 0 ( I T ~ - ~ T ~also )~
g =
.
is uniquely determined 6n C
IT^
and
0 since
=
$
are orthomorphisms in L C
O IT^-^^) =
4' 0 that
.
and
$(g)
is strictly positive on C
= IT
* Assume* satisfying
$
belonging to C
IT^
(because
f holds for every f E C This shows that 1 2 $ $ = $no is uniquely determined on C f
IT
f
Note now that for any
belongs to
band preserving). It follows from which implies g
685
v0
= $(IT,-IT,)~ =
$ *
4
IT2 are 0,
In other words
in the formula
$ *
THEOREM 145.2. Let
L be a Dedekind complete Riesz space, l e t
0
.
$
5
E
Then there E Lz and l e t I) be contained i n the idea2 generated b y $ e x i s t s an orthomorphism IT i n L such t h a t IJJ = $IT. Furthermore, = $IT+ and
=
also s a t i s f i e s
X$
5
IIJJ I
(and therefore
$- = $IT-
satisfying
/IT/
=
$11~1
s XI
).
F i n a l l y , any r e a l
X
and conversely.
PROOF. From the preceding lemma it is evident that there exist positi-
IT^
ve orthomorphisms
IJJ
=
$fnl-r2)
. Without
on the null ideal
N
n
that
=
IT^3
$
IT^
such that
$+
=
=
. The orthomorphism IT^ -
/TI
the corresponding
X$
5
I)
=
loss of generality we may assume that =
inf(nl,r2)
<
Finally, it follows from the lemma above that
I$/
-
$IT, and
$r2
IT^
=
.
Hence
IT^
=
0
satisfies
IJJ+ and $n3 5 $IT* 5 IJJ . Hence $n3 = 0 . It follows as above 0 , i.e., inf(x1,v2) = 0 . Therefore, writing IT = IT 1-- IT2 1 we + IT and IT^ = IT , which shows that $+ = $IT+ and IJJ = $IT .
$ 1 5~ $IT, ~ = have
and
satisfies
if and only if
EXAMPLE 145.3. Let
/IT 5
(X,A,p)
/IT/
XI
.
s I
. This
be a &finite
be the corresponding space L 1 ( X , p )
$
if and only if
implies immediately that
measure space and let L
of all (real) p-sumable functions
(with the L -norm). We have L* = L" since L is Banach and L* = L* n 1 since the norm is order continuous. As shown in Example 142.11, every orthomorphism
n
p E L-(X,p) $(f)
=
1,
in L
. The
fdu
is of the form
, has
. By
(ITf)(x)
=
p(x).f(x)
strictly positive linear functional 4
in L*
= L"
by
satisfies $
L
, where on
L , given by
itself as its carrier. Therefore, $ is a weak unit
IJJ in the ideal generated r , i.e., if IT corresponds with multiplication by the function p E L z , we have $
our last theorem every positive =
$IT for some positive orthomorphism
THEOREMS OF RADON-NIKODYM TYPE
686
be an arbitrary positive linear functional on
J,
Now, let
band generatad by $ is L" Qn
inf(J,,n@)
=
Let
J,,(f)
=
. Write
on X
ch. 2 0 , 8 1 4 5 1
=
L*
Ji
0 5
itself, we have
is contained in the ideal generated by for a l l
pnfdp p,(x)
f E L
. Since the
4 J, , where 0 for n = 1 , 2 ,
n
0 < p (x) 4
....
almost everywhere
. It follows thatnfor every
lim p (x)
=
. Then
L
f E L+ we
have
This is one of the variants of the classical Radon-Nikodym theorem. ff
EXAMPLE 1 4 5 . 4 . Let
be the (real) vector space of all Hermitian
H is partially order-
operators in the complex Hilbert space H. The space ed by defining that A < B (Bx,x)
S
for all
all Hermitian A tor
in H
6
holds for A
. The positive
x E H
and
H
cone in
satisfying (Ax,x) 2 0
is the null element in
in H
B
(Ax,x) <
whenever
is therefore the set of
for all x E H
. The null opera-
H . Because of its importance for
what follows, we first repeat the statement of Theorem 5 3 . 4 (in Chapter 8 on Hermitian Operators): (i)
If (AT:.rEC.rl) i s an upwards directed subset of
t h e subset i s bounded from above, then every
E
> 0
and every
such t h a t 1 1 Ax-ATxll <
A
=
there e r i s t s an-index
x E H
T(E,X)
such t h a t
H , and for IT}
in
.
holds f o r a l l
E
H
exists i n
sup AT
AT 2 ATrE,x) (ii) Furthermore, i f the Hilbert space H i s separable, there e x i s t s (A :n=l,i', ...I i n n '
an increasing sequence -+
A
Ax Tn
From (i) it follows immediately that
x E H
. It is now also evident
and all AT J.
$x
(Ax,x) by
satisfy A
2
holds for every $,(A)
=
(Ax,x)
linear functional on $x
(AT:~€{~l)such t h a t
holds f o r a l l x E H simultaneously ( i . e . , converges strongly to the operator A 1.
8
that if
, then
x E H
for all
H
is order continuous.
. Since
A
=
(ATx,x) 4 (Ax,x)
AT:^€{^}) inf A
. If, for any
A E-H
, then
A
x +
Tn
the sequence of operators
exists and fixed x E H @x
$x(A ) t 0 for A
for every
is downwards directei (ATx,x> C
, we
define
is obviously a positive
+
8,
the functional
Ch. 20,81451
Assume now, as i n Example 141.6,
D
such t h a t a l l members of
2) i s
that
a non-empty s u b s e t of ff
commute m u t u a l l y . L e t
commutant (bicommutant) of
. Then,
2)
C"(2))
n
u n i t e l e m e n t . Furthermore, t h e o p e r a t o r
C"(D)
in
for all
rr(A) = BA
c"(D)
A E C"(2))
. For
B E C"(D)
b r e v i t y we s h a l l d e n o t e
i s t h e Dedekind complete R i e s z s p a c e
denote t h e c o r r e s p o n d i n g p o s i t i v e l i n e a r f u n c t i o n a l f o r all
0
continuous, i . e . , tional
$
A E L ) simply by
s
. As
E L"
c$
n
c a t i o n by some
B
, we
in
L
$
all
y E H
. In
well-known
elements of
L =
.
B € L+
A € L
for a l l L = C"(D)
B
B' E
H+ B
$(I)
$x(A) =
is order
n
$ = $r
i s a multipli-
A E L
for all L+
,
. (By,y)t 0
i.e.,
for
. The o p e r a t o r B ' i s t h e s t r o n g . Hence, s i n c e any s t r o n g l i m i t of
belongs i t s e l f t o
C"(2))
, where
L
, we
see t h a t
we have used t h a t
I
B'A
= AB'
I
,
B' E L+
when-
s i n c e t h e members of
b e a p o s i t i v e l i n e a r f u n c t i o n a l on
J,
and l e t
. Note
$n
that
particular, for 4
( g i v e n by
and
commute.
the functional
A E L
x E H
i s of t h e form
$
belongs t o
i s c o n t a i n e d i n t h e band g e n e r a t e d by
...
. Fix
R e t u r n i n g t o t h e formula ( l ) , w e g e t
Finally, l e t = l,Z,
$x
o t h e r words, s i n c e
(BAx,x)
=
l i m i t of a sequence of polynomials i n ever
,
L
(and proved f o r example i n s e c t i o n 5 4 ) e v e r y
h a s a unique s q u a r e r o o t
B E ff'
by
have
p o s i t i v e t h e corresponding
. As
c"(2))
proved e a r l i e r i n t h e s e c t i o n , e v e r y func-
$(A) = ${n(A)I = $(BA)
(1)
L"
shown above, $
c o n t a i n e d i n t h e i d e a l g e n e r a t e d by
f o r some orthomorphism
For
. As
$
such t h a t
which we s h a l l c o n s i -
d e r somewhat c l o s e r now. I n p a r t i c u l a r we i n v e s t i g a t e (Ax,x)
as
H
i s a n orthomorphism
C"(2))
i f and o n l y i f t h e r e e x i s t s a n element
because
=
b e t h e second
a s shown i n t h e example r e f e r r e d t o ,
i s a Dedekind complete f - a l g e b r a w i t h t h e i d e n t i t y o p e r a t o r i n
C"(D) in
687
ORTHOMORPHISMS AND f-ALGEBRAS
, i.e.,
Cn = B/ E L+ according to 8 5 C
A = I
4
and
. Write
such t h a t
$, = i n f ( $ , n $ )
b e the operator i n
H
for
J,
n =
corresponding t o
(Z), i . e . , $,(A) = (ACnx,Cnx) f o r a l l 0 < $n(A) 4 $(A) f o r any A E L+ . I n
n (the identity operator i n
(Cnx,Cnx) 4 $(I)
sequence of numbers
$
L
H ), we g e t
o r , i n o t h e r words,
( 1 1 CnxI/ :n=l ,Z,.
. .)
0 5 $n(I)
I/CnxlI2 4 $(I)
.
The
i s , t h e r e f o r e , a Cauchy sequence.
THEOREMS OK RADON-NIKODYM
6 88
n 2 m
For
a.20,11451
TYPE
w e have I l C n x I ~ ~+ IICmx1l2 -
j/cnx-cmx/12= (3)
f c x , c x) - f c x , c m /
\ n
\ m
XI
n /
=
IICnxII 2 + IICmxIl2 - 2[c x , c x) \ n m / '
=
where w e have used t h a t
fc
\ n
and
Cn
x , c x) m /
(c
m
commute. Furthermore,
Cm
x , c x) = m /
({cn-cm}c;x,c~x) m /
2 0
*
Hence, from ( 3 ) , j/cnx-clnxI/ Since
2 5
(IICnxII : n = 1 , 2 , .
(Cnx:n=1,2,
. But
y € H
...)
llcnxli2
. .)
+ /lCmXl12
-
2/ICmxlI2 = Ilcnxl12
i s a Cauchy sequence i n
i s a Cauchy sequence i n
-
.
2 'lcmxll
IR, i t f o l l o w s now t h a t
H . Hence, Cnx
+
y
f o r some
then
A E L. We have proved, t h e r e f o r e , t h e f o l l o w i n g theorem.
for a l l
THEOREM 145.5.
C"(D)
mutant
If t h e Dedekind complete Riesz space
D
of a non-empty s e t
and if
H
t o r s in t h e H i l b e r t space
L
is t h e bicorn-
of mutuaZly commuting Hermitian opera-
s
Li i s represented by the element x E H i n the sense t h a t $(A) = (Ax,x) for a l l A E L , t h e n every p o s i t i v e linear functional $ in t h e band generated by @ is of t h e form $(A) = (Ay,y) for an appropri,ate y E H . 0
@ €
The q u e s t i o n a r i s e s whether e v e r y p o s i t i v e menber of
N
Ln
i s of t h e
i n d i c a t e d form. The answer w i l l b e p o s i t i v e i f t h e r e e x i s t s a n element x
E H
such t h a t t h e c o r r e s p o n d i n g
(Ax,x) > 0 r a t e d by
A > 0
f o r every $
i s t h e space
in
L i
r e c a l l s e v e r a l f a c t s about for all
n
2
L ) , because i n t h i s c a s e t h e band gene-
L = C"(I7)
I
L (i.e.,
i t s e l f . Before d i s c u s s i n g t h i s q u e s t i o n , we
n e r a l . I n t h e f i r s t p l a c e w e have = C(n+2)(0)
@x i s s t r i c t l y p o s i t i v e on
and about H e r m i t i a n o p e r a t o r s i n ge-
C'(D)
=
. Furthermore,
C""(D) , C"(P)
and hence
C(")(D)
is contained i n
C'(Q)
=
20,51451
Ch.
c'(D)
and any p r o d u c t of members of lated for and
A
L
B
689
ORTHOMORE'HISMS AND f-ALGEBRAS
,
t h i s implies t h a t
belonging t o
. Note
L'
L = L"
L
=
L c L'
and
AT3
L'
E L'
i s order separable, i . e . ,
L
for a l l
do n o t commute i n
. As
L'
a
t h a t i t f o l l o w s from ( i i ) a t t h e
b e g i n n i n g of t h e p r e s e n t example t h a t i f t h e H i l b e r t s p a c e then
. Formu-
C'(0)
commute w i t h members of
L
, note
C"(0)
,
t h a t members of
g e n e r a l , b u t of c o u r s e members of f i n a l remark about
i s i t s e l f a member of
any s e t
in
D
i s separable,
H
which i s bounded
L
from above h a s an a t most c o u n t a b l e s u b s e t p o s s e s s i n g t h e same supremum
. We
D
as
H ) l e a v i n g t h e c l o s e d l i n e a r subspace
operator ( i n t h e H i l b e r t space
A
i n v a r i a n t . Then w e l l (since
,
P
by
i s a Hermitian
A
now t u r n o u r a t t e n t i o n t o t h e s i t u a t i o n t h a t l e a v e s t h e o r t h o g o n a l complement of
H1 invariant as
HI
i s H e r m i t i a n ) . Denoting t h e o r t h o g o n a l p r o j e c t i o n on
A
t h i s implies t h a t
AP = PA
.
H1 The proof i s t r i v i a l . F i n a l l y , l e t
PI
and
P2
be t h e o r t h o g o n a l p r o j e c t i o n s on t h e c l o s e d l i n e a r s u b s p a c e s
H1
and
H2
r e s p e c t i v e l y . Then t h e subspaces
(x,,x2) = 0
i.e.,
PIP2 = 8
,
for a l l
x
and i n t h i s c a s e
THEOREM 145.6. Let
,
E HI
and
P2Pl = 8
and
x2 E H2
, if
H2
are orthogonal,
and o n l y i f
holds as well.
C"(D) be order separabZe (as observed, sepa-
L =
H
r a b i l i t y of the Hilbert space
i s a s u f f i c i e n t condition f o r t h i s t o xo E H
there e x i s t s an element
h o l d ) . %en
HI
such t h a t the linear functional
I$o , defined by I$O(A) = (AxO,xO) f o r all A E L , i s s t r i c t Z y p o s i t i v e on L Hence, as shown already, t h i s implies t h a t any p o s i t i v e order continuous
.
linear functional
i s o f the form
on L
$
.
te
$(A) = (Ay,y) f o r an appropria-
y E H In the terminology of t h e theory of von Nemann algebras t h i s means t h a t every p o s i t i v e i n :L is a nonnal s t a t e . PROOF. ( i ) Assume t h a t
(Pa:aE{a})
o r t h o g o n a l p r o j e c t i o n s and s u c h t h a t
i s a s u b s e t of
papB = 8
a #B
A s o b s e r v e d above, t h i s i s e q u i v a l e n t t o assuming t h a t i f t h e o r t h o g o n a l p r o j e c t i o n s on t h e c l o s e d l i n e a r subspaces H P
,
then
2
a # B
8
H
and
, we
. In
5
P P
a B
= 0
that
o t h e r words, t h e s y s t e m I
5 I
in
belongs t r i v i a l l y t o
holds i n
k i n d complete, t h e supremum
P
a
A
P
B
(Pa:aE{a})
(3 5 P
H
a # 6
are orthogonal f o r
d e r i v e from
Furthermore, operator
H
L
f o r every
Po = sup P
L
=
. Since
=
c o n s i s t i n g of
L
for a l l
.
in {a} and
Pa Ha all P
8 holds i n
and
F' H
B B
are of
satisfy L
for
i s a d i s j o i n t system i n
L'.
a (note t h a t the i d e n t i t y
c " ( 0 ) ). Since
exists in
L
. On
L
i s Dede-
a c c o u n t of t h e
690
THEOREMS OF RADON-NIKODYM
of a l l
Pa
Pa
n
,
=
supa Pa
20,91451
Ch.
. .)
(Pn:n=l ,2,.
o r d e r s e p a r a b i l i t y t h e r e e x i s t s a sequence of t h e
TYPE
i n t h e system
Po = sup P . Assuming t h a t there s t i l l e x i s t s one n which i s not one of t h e Pn we have Pc A Pn = 8 f o r a l l
such t h a t
#
B
and hence
.
P
A
.
Po = I3
On the o t h e r hand
P
Po
C
,
since
Po =
(Pa:a€{a}) i s an a t most count-
Contradiction. It follows t h a t
a b l e system. x E H
( i i ) For any of
H
we denote by
H'cxl
(A'x:A'€L')
. Note
spanned by the s e t
x (since
the closed l i n e a r subspace that
x E H'Cxl
f o r every
E H i s an element i n the orthoH ' [ x , l , then H'Cx 1 and H'Cx 1 a r e orthogonal 1 2
I E L ' ) . Note a l s o t h a t i f
gonal complement of
x2
subspaces ( t h i s follows i m e d i a t e l y from t h e r e s u l t t h a t any product of members of H'Cxl
L'
belongs t o
L ' ) . Furthermore, every
. For
t h i s purpose, l e t
(xa:aE{a})
r e s p e c t t o the property t h a t a l l leaves every H'Cxal
that
P
3
Pa
, we
L" = L
E
PUPB = I3
H'[x
by for a
with
B' E
L'
i n v a r i a n t . Hence, denoting t h e orthogonal p r o j e c t i o n have
. Since
# 6 in
PaB'
all
= B'Pa
f o r every
. Note
implies
by p a r t ( i ) , the s e t (xa:aE{a})
(Pa:aE{a}) i s a t most
t h a t t h i s i s an orthogonal system i n
i t s e l f , because otherwise
would s t i l l contain a s e t of type
. This
B ' E L'
a r e mutually orthogonal, w e have
H'[xa]
. Therefore,
{a)
( x , , ~ , ,...)
H
, maximal
H
Zorn's lemma. Every
H"
By t h e maximality the closed l i n e a r subspace together i s
H'[xol =
a r e mutually orthogonal. The
i s a t most countable. This implies t h a t the s e t countable, say
such t h a t
be a s u b s e t of
I
H'Cx
xo € H
e x i s t e n c e of such a maximal s e t follows from on
leaves every
i n v a r i a n t ( t h i s follows from t h e same r e s u l t about products). We
s h a l l prove now t h a t t h e r e e x i s t s an element = H
E L'
B'
H
. Without
H"
loss of g e n e r a l i -
i n H , where n s = Ey % Note now t h a t f o r every A ' E L' and every x the element n n A'x i s contained i n H'Cxo] , because A'x = A'PnxO and A'P, E L' on t y we may assume t h a t
Ellxnl12
.
account of
A' E L'
It follows t h a t ( i i i ) Let some A = 0
0 xo E H
. Then
A E L
(A'xO:A'EL')
. Hence,
and
H'Lx
1
P
E
.
be as
L c L'
.
H
8 < A E L
,
H'Cx,]
for a l l
, we thus get , then Axo # $o
I n o t h e r words, t h e l i n e a r f u n c t i o n a l = (AxO,xO)
Hence
xo = l i m s
c H'[xOI
i n p a r t ( i i ) and assume t h a t
AA'xo = A'AxO = 0
i s dense i n if
n
= H
i s f i n i t e . Let
i s s t r i c t l y p o s i t i v e on
L
.
A'
Ay = 0
0 on
.
H'[xnI
t h e orthogonal complement of
# {O)
H'cxl
spanned by a l l
E L'
.
for a l l
Axo = 0
Since the
for a l l
y
E
H
n
for set
, i.e.,
, and hence (AxO,xO) > 0 L , defined by $O(A)
.
.
ch. 20,§1457
1
ORllHOMOWHISMS A N D f -ALGEBRAS
W e now drop t h e condition t h a t
H
$
be
holds f o r a l l
be a p o s i t i v e order conti-
$
y E H
there e x i s t s an element
A
. In o t h e r uords,
E L
such
every p o s i t i v e
i s a noma1 s t a t e .
L :
in
and l e t
. Then
L
nuous l i n e a r functional on
L = C"(0)
Let
D of mutually c o m u t i n g Hermitian
t h e b i c o m t a n t of t h e non-empty s e t operators i n the Hilbert space $(A) = (Ay,y)
i s o r d e r separable.
L = c"(0)
Pallu de l a BarriBre,C11,19541.
THEOREM 345.7.1R.
that
69 1
PROOF. The n u l l i d e a l
complete. Hence, L = N
Q
i s s t r i c t l y p o s i t i v e on
C
of
N+
C
8
+
i s a band i n
Q
, where
C
the space
$ '
Q
and
L
i s the c a r r i e r of
Q
L
i s Dedekind
. Since
Q
i s o r d e r separable (Theorem
C
8 4 . 4 ) . As observed e a r l i e r , t h e i d e n t i t y o p e r a t o r
in
I
. Let
belongs t o
H
I = P + (I-P ) be the decomposition of I i n t o Po E C 0 0 $ and I-Po E N Then P A ( I - P ) = 8 holds i n L , which implies 0 2 O P (I-P ) = 8 , i . e . , P = Po Since, t h e r e f o r e , Po i s a p r o j e c t i o n as 0 0 0 w e l l as a Hermitian o p e r a t o r , t h e o p e r a t o r i s an orthogonal p r o j e c t i o n Po Then P = I-Po i s likewise an orthogonal p r o j e c t i o n . Furthermore, in H 1 i f we denote t h e ranges of Po and P 1 by Ho and H 1 r e s p e c t i v e l y , L" = L
+':
.
.
then
H
8
HI = H
A = A + A 0 1
Since A
with
. Any A.
E C
(I-P ) = 8 0 0 hence a l s o A. = PoAo A
and s i m i l a r l y
A
arbitrary
4
and
A E L
has a (unique) decomposition
A1 E N
Note already t h a t
$ * A (I-P ) = 8 0 0 because elements of L
, we
= P A P 1 1 1 1
have
. It
follows t h a t
,
i.e.,
A
$(A) = $(Ao) 0 0
0
commute). Hence A.
and
Al
leave
.
.
(and
= A P
A
o
= P
Ho
A P
0 0 0 7
and
A = A on Ho and A = 8 on H , It w i l l 0 0 cause no confusion, t h e r e f o r e , i f w e denote t h e r e s t r i c t i o n AIH = A IH 0 0 0 again by A. The s e t Lo of a l l these r e s t r i c t i o n s ( e s s e n t i a l l y , there-
HI
i n v a r i a n t . Note t h a t
.
f o r e , the s e t
C
8
) i s a ( r e a l ) a l g e b r a of mutually commuting Hermitian
operators i n the H i l b e r t space bicomutant P
0
(since
Li P
0
. Note
E L
conmuting with every
. In
A.
prove t h a t
E Lo
B
Bo = B I H O
. In
Lo
B
leaves
E L'
i s equal t o i t s own commuted with
E L' Ho
S i m i l a r l y , we have
B E L'
o t h e r words, any
(set
(L")o = {(L')o}'
i n v a r i a n t . This
i s a Hermitian o p e r a t o r i n Bo E (L'l0
the converse d i r e c t i o n , i t i s e v i d e n t t h a t any
t h e r e s t r i c t i o n of some find that
. We
) . Therefore, every
implies t h a t t h e r e s t r i c t i o n (Lo)'
Ho
f i r s t t h a t every operator
B = 8
. Hence,
on
H
1
Bo
>.Hence
observing t h a t
HO belongs t o
E XLo)' (L')o L" = L
=
is (Lo)'
, we
.
692
THE RANGE OF AN ORTHOMOWHISM
{(Ll)o}'
(Lo)" = { ( L o ) ' } l =
To r e c a p i t u l a t e , we know now t h a t
(Lye
=
L = (Lo)" 0
20,51461
Ch.
= Lo
.
and
Lo
is order separable.
I n view of t h e p r e c e d i n g theorem t h e given p o s i t i v e o r d e r c o n t i n u o u s l i n e a r functional
( o r r a t h e r t h e r e s t r i c t i o n of
$
$(A ) = (A y , y )
form
0
0
Hence, s i n c e Lo
w e have
=
A = A
y E Ho
and
,
0
) i s now of t h e
and f o r a l l
A.
E L
0 '
o o ' +
for all
+ A1
0 $(Ao) = (AOy,y) = (Ay,y)
Since
L
to
$
y E Ho
),
:A EC
(%lH
$(Ao) = (Aoy,y)
a r b i t r a r y and =
f o r an a p p r o p r i a t e
with
-
A. A.
(Aly,y)
v a n i s h e s on
A1
E C E C
$
$
. It
,
Ho
.
Finally, i f
and
A1 E N
remains
to
4
,
is
A E L
then
prove t h a t
$(A) = (A,,y,y) = O
.
t h i s is clear.
The p r e s e n t method of p r o v i n g P a l l u de l a B a r r i P r e ' s theorem, q u i t e
d i f f e r e n t from t h e o r i g i n a l method, i s due t o P.G. Dodds ([11,1974).
146. The range of an orthomorphism A s o b s e r v e d i n Example 140.7, t h e range i n an Archimedean R i e s z s p a c e ral then
RT R
L
of an orthomorphism
R,
i s a l i n e a r s u b s p a c e of
i s n o t even a Riesz subspace. I f
L
, but
71
i n gene-
i s a p o s i t i v e orthomorphism,
71
i s a Riesz s u b s p a c e , b u t i n g e n e r a l n o t a n i d e a l . We s h a l l prove
now f i r s t t h a t i n a Dedekind complete s p a c e t h e range of any p o s i t i v e o r t h o morphism i s a n i d e a l . The theorem i s due t o A. B i g a r d ( [ 1 1 , 1 9 7 2 , P r o p o s i t i o n I ) , w i t h a somewhat d i f f e r e n t p r o o f .
THEOREM 146.1. I f
complete Riesz space
a(luol)
, we
, then t h e range
L
0 < v
PROOF. L e t =
i s a p o s i t i v e orthomorphism i n t h e Dedekind
5
nuo
f o r some
may assume t h a t
uo t 0
For t h i s p u r p o s e , l e t
u
=
f u:uEL + \
.
R,
i s an i d e a l .
uo E L
.
. We have
Since
0 5 m0 = \ m o ' =
t o prove t h a t
v E R,
.
Ch. 20,51461
ORTHOMORPHISMS AND f-ALGEBRAS
z = inf U
The element we have
exists i n
2 v (since
TIZ
i s Dedekind complete) and
L
i s o r d e r c o n t i n u o u s ) . We prove now t h a t
'II
0 < TIZ-v 2 nz
I f n o t , w e have
(since
L
693
. Note
n
now t h a t
TIZ
=
v.
behaves a s a c o n t r a c t o r
on an o r d e r dense i d e a l (Theorem 143.4). Hence, t h e r e e x i s t s an element w E L+ Let
satisfying
0 < w saz-v
. Then
p = inf(n-'w,z)
w 1 az
, which
and so
a(2-p)
z-p 2 i n f U = z
p > 0
contradicts 2
v
f o r some n a t u r a l number
p 5 0
n.
w 1 z , and hence -1 ap 2 a ( n w) 4 w 2 72-v ,
(because o t h e r w i s e
khows t h a t
. This
nz-v > 0
our assumption about
nw 5 nw
0 < w 5 az ). A l s o
. This
. Hence
and
i s a member of t h e set
z-p
contradicts
i s f a l s e . Hence
p > 0 nz = v
. It
,
,
U
so
follows t h a t
i.e.,
.
v E Rn
To p r e p a r e t h e way f o r a more g e n e r a l theorem, w e r e t u r n t o Archimedean f - a l g e b r a s w i t h u n i t element
,
e
i n particular t o the case t h a t
such an a l g e b r a i s e-uniformly complete. LEMMA 146.2. Let
e
, let
4 n
-1
u E A+
u u mn
and
forall
PROOF. For
u - u m n
and hence
that and
.
un = i n f ( u , n e )
n = 1,2,
for
2"nen 0 < u -u
m n
A
we have (ne-u ) = ( um Am)
i s d i s j o i n t from
u
m
-
un
-
-1 n u u mn
=
u - u = n n
.
o ,
It f o l l o w s t h e n from
-1 0 2 u - u 4 n u u m n mn p 1 r , then p 2 q
. We
THEOREM 146.3. Let
b e an Archimedean f - a l g e b r a w i t h u n i t element
.
A
use h e r e t h a t i f
e Let A be e-uniformly complete. Then, i f -1 t h e inverse u exists in A
.
PROOF. We assume f i r s t t h a t For
... .
m t n .
m 2 n
(um-un)
b e an Archimedean f - a l g e b r a w i t h u n i t element
A
n
= 1,2
(e-a-lu)'
=
,... , l e t s n e . For m t n
=
e
5
u
Z:=o(e-a-lu)k
we have
4
ae
p , q , r E L+
u F A
satisfies
f o r some number
, where
,p
5
q+r
u 2 e
at
a s u s u a l we set
1
.
,
694
THE RANGE OF AN ORTHOMORPHISM
This shows t h a t since that
i s an e-uniform Cauchy sequence. Hence,
s C A
i s e-uniformly complete, t h e r e e x i s t s an element
A
s
...)
(sn:n=l,2,
Ch. 20,§1461
s
r'
(e-a-'u?s
holds u
Then
i t f o l l o w s now t h a t -1
= s-e
e x i s t s (and
e-uniformly.
-1
=
IY
such
s = Zm(e-a-lu)k , and from 0 -1 IY u s = e . T h i s shows t h a t u -1
s ).
A s s u m e now t h a t i t i s o n l y given t h a t u 2 e h o l d s . Put u = -1 = inf(u,ne) f o r n = 1,2, Then u e x i s t s f o r e v e r y n ( i n view
... .
o f what w a s proved above). Furthermore, by t h e p r e c e d i n g lemma, 0 -1
s n
u u mn
holds f o r a l l
m t n
-1 T h i s shows t h a t . (un :n=1,2, t h e r e e x i s t s an element
...)
w F: A+
.
i s an e-uniform Cauchy sequence. Hence, -1 u 4 w h o l d s e-uniformly,
.
e
u n i t element
( i ) Let again
and l e t A
can be w r i t t e n as a product ( i i ) Let
A
A
n- l u 2 .f u
uw u-u
=
vw
and holds and
1. Hence, -i -
- (u-un)un =
w
-
1.
be an Arehimedean f-algebra w i t h u F. A+
be e-uniformly complete. Th5n any u
S
such t h a t
On t h e o t h e r hand, i t f o l l o w s from 0 s u - u 5 n -1 u u 5 m n mn -1 u2 f o r a l l n , and so 0 5 u u 1. u t h a t 0 s u-u 5 n n m -1 uZ-uniformly. Note now t h a t uu converges u-uniformly t o n -1 -1 converges u 2 - u n i f o m l y t o 0 ( s i n c e (u-un)un s (u-u )un -1 2 -1 uu-ln- (u-u ) u converges u -uniformly t o uw But uun n n n -1 = e . It f o l l o w s t h a t uw = e , i . e . , u e x i s t s (and u-' COROLLARY 146.4.
u -u m n
5
Hence
such t h a t
0 5 v s e
w-l
and
be an Arehimedean f-algebra w i t h element ' e
. Then
exists. A
i s e-uniformly complete if and only if A is r e l a t i v e l y uniformly complete. PROOF. ( i ) By Theorem 142.4 w e have V = u h e
and
w=uve,wehave
u = ue = (uhe)(uve)
O < v s e
and
.
w T e , s o
Setting -1 w
exists. ( i i ) If
A
i s r e l a t i v e l y uniformly complete, t h e n (by d e f i n i t i o n ) i t
i s so t h a t f o r e v e r y
u E A+
e v e r y u-uniformly Cauchy sequence h a s a u-
uniform l i m i t . This h o l d s t h e n i n p a r t i c u l a r f o r e-uniformly A
u = e
. Therefore,
A
is
complete. 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
i s e-uniformly
complete. L e t
u E L+
be a r b i t r a r y
(u#O)
and l e t
Ch. 2O,gl46]
ORTHOMORPHISMS AND f-ALGEBRAS
695
(fn:n=1,2, ...) be a u-uniformly Cauchy sequence. Asseen above, we can write
u = vw
, where 0 2 v S e and w -1
(w-lf :n=l,Z,...)
exists. Then
is v-uniformly Cauchy and hence e-uniformly Cauchy (on account of
0 < v s
e ) . Since A is e-uniformly complete, there exists an element g such -1 that w fn + g (e-uniformly). It follows that fn + wg (w-uniformly). We
5
now recall Theorem 100.4(iii)
according to which any sequence, e-uniform-
ly Cauchy as well as u-uniformly Cauchy,and which converges e-uniformly also converges u-uniformly. Hence, in the present case, the u-uniform Cauchy sequence shows that A
has a u-uniform limit. Since
(f )
u
is arbitrary, this
is relatively uniformly complete.
The result in part (ii) has been proved by different methods by E.R. Aron and A.W. Hager ([1],1981). Archimedean Riesz space L L
. The
We now return to orthomorphisms in an of a l l orthomorphisms in
space Orth(L)
is an Archimedean f-algebra having the identity operator
unit element. If
L
is uniformly complete, then
so
I
is Orth(L)
in L
,
as
as indi-
cated in Exercise 140.12. We shall prove now first that in this case the of a positive orthomorphism
range R,
generated by
Rll
*
THEOREM 146.5. Let
be a p o s i t i v e orthomorphism in t h e uniformly
complete Arehimedean Rissz space u E L+ 2
,
there e x i s t sequences
qn for a l l
n
and
vpn
+
v
L
.
p, f
,
Then, if 0
and
f-algebra with unit element Theorem 146.3 that
From
pn
=
I
(,+n-’I)-’
(T+n-’I)-’v
for n
qn
nqn c v
PROOF. As observed above, Orth(L)
Write
is uniformly dense in the ideal
T
+
5
TU
-for some p, <
hold r e l a t i v e l y uniformly. is a uniformly complete Archimedean
. Since
T
-1 -1 + n I 2 n I
exists in Orth(L) =
v
in L+ such t h a t
... . Then
1,2,
, it follows from
for n
0
5
p
I.
=
1,2,...
and
.
--
696
THE RANGE OF AN ORTHOMORPHISM
5
w
5 TU
...
exactly the same manner that =
u-p:
satisfies 0
that
pn
get
qn - pn
5
.
Then vpn I. v holds u-uniformly. Let now w = TU-v and, writing p ’ = (n+n-’I)-’w for n = 1,2, , we find in
it follows then that
0
20,91461
Ch.
qn
5
q
C
I. w
and
holds for all n =
u
-
pi
-
.
pn z u-u
+
Tq
holds u-uniformly. Hence
-
=
v
Since pn + p~
=
TU
w
.
= 0
-
qn It remains to prove -1 (T+n-’I) n u 5 u , we
.
To state the main result in this section, we introduce the following
definition. DEFINITION 1 4 6 . 6 . The Rie sz space
L
i s s a i d t o have t h e o-interpola-
t i o n property i f whenever t h e inc re asing sequence decreasing sequence n
,
(gn:n=1,2,
...)
in
L
then t h e r e e x i s t s an e%ement h E L
n (equivalently, i f
that
fn
5
gm
satisfying
f
(fn:n=1,2,
are such t h a t satisfying
fn
fn 5
h
...)
5
for a%%
g
are a r b i t r a r y sequences i n L
(fn)
and
(9,)
f o r a21 m
and
n , the n t h e r e e x i s t s an element
5
h
We recall that
5
g m
L
for a%l m
and t h e
gn f o r alZ
5
and
sueh
h E L
n ).
is called order complete if every order Cauchy
sequence has an order limit. As indicated in Exercise 10f 8, order completeness is equivalent to the condition that if in L , then there exists an element h E L all n
fn + 9 gn + satisfying
. Evidently, the o-interpolation property
+
and
gn - fn 0 for
14 h 5 g
n
implies order completeness.
It is also evident that order completeness implies uniform completeness (i.e., for every
u C L+
every u-uniform Cauchy sequence has a u-uniform
limit). Hence, in an Archimedean Riesz space the following implications ho Id : Dedekind complete
* Dedekind o-complete * a-interpolation property *
* order complete * uniformly complete As we have seen, the range Rn
.
of any positive orfhomorphism
T
in a
Dedekind complete space is an ideal. We prove now that the o-interpolation property in an Archimedean space L orthomorphism in L
an ideal.
is sufficient to make the range of any
Ch. 20,51463
ORTHOMORPHISMS AND f-ALGEBRAS
L
THEOREM 146.7. I f
697
i s an Archimedean Riesz space possessing the Rn of any orthomorphism T i n
u- interpolation property, then the range
i s an -CdeaZ i n L
L
.
PROOF. L e t f i r s t
be a p o s i t i v e orthomorphism i n
TI
u E L+
c i e n t t o prove t h a t i f
0 S v S nu
and
,
.
L
It i s s u f f i -
. Since
v E R
then
L
i s Archimedean and uniformly complete, t h e r e e x i s t (by t h e l a s t theorem above) sequences and
+
npn
,
v
pn I.
and
Rqn J- v
g
J-
in
L+
such t h a t
a - i n t e r p o l a t i o n p r o p e r t y , t h e r e e x i s t s a n element pn
holds f o r a l l
z S q
2
v F R
rqn - rpn
S
/V-IIZ/
.
71
L e t now /nfI
S
f o r some
= InI(lfl)
R
Once more, s i n c e
,
n+h
-
Ti-h =
. We
?I
h =
-IT
-
1111
T h i s shows t h a t L
satisfy
implies
lkl
I Infl
k+ E R
Hence k F RIT
.
z C: L
and
n-(h+z)
and
L
.
such t h a t
so
v
71
. Finally,
. Then k-
t
E R,IT1
L
0 I g h E L+
, we
have
This i m p l i e s t h a t
-
and
L
r-
and
T+
i.e.,
r+h = ?r+z
and
satisfies
assume t h a t t h e elements
0 I k+ 2 I r f l
Rr
g
,
k
0 5 k- 5 Infl
and
f
in
, which
on account of what w a s proved a l r e a d y . RT
i s an i d e a l .
The r e s u l t i n t h e l a s t theorem i s due t o B. de P a g t e r "11, ideal i n
i.e.,
Inf/ = i s ( l f l ) / = f o r some
are d i s j o i n t , s i n c e
T h i s concludes t h e proof t h a t
r e c a l l t h a t t h e Riesz space ever
,
nz
=
and l e t f i r s t
/ I T / (h)
n+(h-z) = n (h+z) = 0
f o l l o w s t h a t t h e given element
g C R
,
i 0
gn= I d ( h )
/nhl =
n
has t h e
it follows then t h a t
. Since
g € R
f o r some
z E L+
I nq
L
, i.e.,
T'(h-2)
. It
z
ITZ
- 71pn
nq
i s a n i d e a l and
have d i s j o i n t r a n g e s . Hence
-
ITP, i
prove t h a t
nh = I n l ( z )
+ IT-z
IT'Z
. From n . But
i s a n i d e a l , w e have
In1
R
i.e.,
The elements
for a l l
f E L
and
nh F R I T l
n
be a n a r b i t r a r y orthomorphism i n
TI
for a l l
pn I qn
h o l d r e l a t i v e l y uniformly. Now, s i n c e
1981). We
i s c a l l e d normal i f e v e r y p r o p e r prime
c o n t a i n s a unique minimal prime i d e a l o r , e q u i v a l e n t l y , whend i n L i m p l i e s t h a t L i s t h e a l g b e r a i c sum of {u}
inf(u,v) = 0
,
698
THE RANGE OF AN ORTHOMORPHISM
Ch. 20,51461
{ v } ~(Theorem 8 2 . 7 ) . I t can be proved, and we s h a l l i n d i c a t e how t o do
and
t h i s , t h a t an Archimedean Riesz s p a c e i f and o n l y i f
has t h e o - i n t e r p o l a t i o n property
L
i s uniformly complete and normal (C.B. Huijsmans and
L
B. de P a g t e r ([1],1980)).
Hence, i f t h e Archimedean s p a c e
i s uniformly
L
complete and normal, then t h e range of e v e r y orthomorphism i n
i s an
L
i d e a l . A d i r e c t proof of t h i s , w i t h o u t u s i n g t h e o - i n t e r p o l a t i o n p r o p e r t y ,
i s g i v e n by t h e same a u t h o r s i n ([21,1982). EXERCISE
be an Archimedean R i e s z s p a c e .
Show t h a t i f V,W,p
(i)
p = 0
then
L
46.8. Let
E L+
.
( i i ) Show t h a t i f
L
and
p C inf(V,W-whnv)
for
n = l,2,
h a s t h e o - i n t e r p o l a t i o n p r o p e r t y , then
...,
is
L
normal. p S v
H I N T : For ( i ) , n o t e t h a t
n = 1,2,
... . Hence
2p < w
hence
(2v)
A
C
w
-
, which
. For
w
= inf(w,nu+nv) 5
inf(w,nv) t
. By
. Then
v C w
A
i m p l i e s 3p
inf(u,v) = 0
For ( i i ) , assume
+ inf(w,nv)
2p < p + w
as well as
for
w
any
E
n = l,2,
nv C w
A
w
2p
p + w
6
p+w
as well as
we g e t
L+
,
(2v) 5 w
A
... , and
for 2p C 2v,
and s o on.
inf(w,nu) +
0
so
5
inf(w,nu) 4 5
the o-interpolation property there e x i s t s
z E L+
such t h a t
n .
w ~ n u < z < w - w ~ nf v orall
0 2 z
I t f o l l o w s t h e n from i.e.,
.
z E { v } ~ Similarly
A
v
inf(v,w-wmv)
5
0 2 w - z E {uId
. Hence
EXERCISE 146.9. L e t t h e Archimedean s p a c e g t c u in L 1" 2 z E L satisfying I z I 5 -u and f n - 3u 3 t h i s purpose, n o t e t h a t
normal. L e t
-u
c f
4
=
u1 + u2
Then
IzI
with 1
c 3u
2 t -u 1 z + -u 3 3
, we
inf(f,g) = 0
0c u E 1 and ( s i n c e have a l s o
(gn
2 3
L = {ul
Z A
d + {vl
v = 0,
.
be uniformly complete and
.
1-
L
0 < u
+
5
i s normal, we have
.
1
u =
.
L e t z = -(u -u ) 2 + 2 3 2 1 (fn-z--u) C Since 3 I + I + (f --u) But and (f -n 3 n 3u)
2
5 u ) we have
-2--u)
that d
Show t h a t t h e r e e x i s t s a n element 2 5 z C g + -u f o r a l l n For n 3
. Since
and f
L
n
g = E y 2-n(gn+ 31 u/
and
e x i s t uniformly and
.
for a l l
E {g}d
.
.
Ch. 20,11461
ORTHOMORPHISMS AND f-ALGEBRAS 2 + = 0 (f - z - - u ) n 3
a r e d i s j o i n t . Hence
, i.e.
699
f n - j2 U I z .
EXERCISE 146.10. Show t h a t every uniformly complete normal Archimedean Riesz s p a c e -u 5 f
.f i gn 4 I u
such t h a t 2 k u I yk - (3) L
yo = 0
has the
L
Iyk-yk-II I 2 ku f o r gn + (?)
and suppose
k = O,I,Z,..
yo,yI, ...,yk
n = l,2,
... . Then
- ( 52 )k u
.
...
and a l l
, as
n
Yk
in
fn f o l l o w s . Let
w i t h t h e p r o p e r t i e s ( i ) , ( i i ) have a l -
ready been d e f i n e d . For t h e d e f i n i t i o n of
for
.
b e given. Define f o r k = O,I,Z,.. elements 2 k (7) u for k = l,2, and ( i i )
(i)
I
o - i n t e r p o l a t i o n p r o p e r t y . For t h i s purpose, l e t
I
preceding e x e r c i s e , there e x i s t s
f * I. n z
5
E L
, set
yk+l
2 k g* .L 2 (-) u
n
and hence by t h e 1 2 k -(-) u and 3 3
3
such that
121 I
Y ~ =+ yk ~ + z s a t i s f i e s ( i ) and ( i i ) . Condition ( i ) shows t h a t (y :k=O, 1,2,. .) i s u-uniformly Cauchy. Hence, yk converges u-uniformly
Then
.
k
t o some
h E L
. Condition
EXERCISE 146.1 I . L e t b r a w i t h u n i t element
( i i ) shows t h a t A
f
n
I
h 2 g
n
for a l l
.
n
b e an Archimedean uniformly complete
f-alge-
. We
i n d i c a t e how t o prove t h a t every u E A+ has 2 a p o s i t i v e square r o o t w , i . e . , w E A+ and w = u . The a l g e b r a A i s 2 semiprime (Theorem 142.5) and hence w = w2 i f and only i f w2 = w2 f o r
elements i n
L+
e
(Theorem 142.3).
1
1
I f e x i s t i n g , t h e square t o o t i s unique,
t h e r e f o r e . One meLhod of proof i s s i m i l a r t o t h e u s u a l method of proving t h e e x i s t e n c e of t h e s q u a r e r o o t of a p o s i t i v e Hermitian o p e r a t o r ( s e c t i o n
v
n
.
a e 5 u I e f o r some 0 < a < 1 Define uo = 2 u = u + ~(u-s) f o r n = 0, l , 2 , . I n t r o d u c i n g v = e-u and n+l n 2 = e-u , we have 0 5 v 5 (1-a)e and vn+l = i(v+vn) Then vn t 0
54). Assume f i r s t t h a t
.. .
and
.
.
and vo = 0 5 i v = v 1 Assuming t h a t v t v f o r some n n n-l 2 we g e t v2 2 v (Theorem 142.3), so v v = ~(V:-V:-~) t 0 This n n-l n+l n shows t h a t (vn:n=1,2, ) i s i n c r e a s i n g . Furthermore, i f v 5 (I-a)e 2 ( t r u e f o r n = 0 ) , t h e n vn+l = $ ( v + v ) I (I-a)e Assume f o r some n n ~ f o r some n ( t r u e f o r n = 1 ) . Then now t h a t v - v ~ 5- (1-a)"e n for a l l
n
...
.
.
,
700
THE RANGE OF AN ORTHOMOWHISM
Therefore, w = e-z
i s a n e-uniform Cauchy sequence, and hence
(v,)
. Then
z E A
t o some
e-uniformly
satisfies
w
2
Assume now t h a t
.
u
=
un
142.3). Then
m 2 n
shows t h a t
e-uniformly
Cauchy, s o
u
+
wL
as w e l l a s
Finally, l e t = inf(u,ne)
0 2 r
r
u
n
+
n + w
u
,
u'
n
I. w
w E A+
EXERCISE 146.12.
, we
J,
n
have
u E A+
fying
wL
so
n = 1,2,
. Then
( i ) Let
-1 (l+n ) u
=
(Theorem
.
.
u
=
... . S i n c e
u' of u = n n2 h o l d s u -uniform-
(u:)
Cauchy, so
again
A
u 4 u n i s u-uniformly
I
w
2
u
=
.
b e the Archimedean f - a l g e b r a
consisting
t h a t are piece-
(x:O5x
has a u n i t element, there e x i s t s
A
,
let
A
f o r some
be the i d e a l o f a l l
a > 0
and a l l
Archimedean uniformly complete f - a l g e b r a . u E A+
In r = u2 J, n n
h a v i n g no s q u a r e r o o t .
C(C0,ll)
If(x)/ 2 ax
u
.
of a l l r e a l continuous f u n c t i o n s on t h e i n t e r v a l
(ii) In
converges follows t h a t
Therefore, r 5 (n- 1-m- I ) 1, (r ) is m +n Then h o l d s e-uniformly f o r some w E A
w i s e polynomials. Show t h a t , although an e l e m n t
v
be a r b i t r a r y . Then t h e s q u a r e r o o t
u EnA+
exists for all
f o r some
u
. It
e-z
the square r oot of
-
l y (Theorem 142.7), i t f o l l o w s t h a t I
converges t o
. Then
0 2 u 5 e
e x i s t s by t h e above argument. S i n c e
for
Ch. 20,11461
satis-
f E C(C0,ll)
x E C0,ll
. Then
A
i s an
Show t h a t t h e r e e x i s t s a n element
h a v i n g no s q u a r e r o o t .
( i i i ) Let
A
b e an Archimedean uniformly complete f - a l g e b r a w i t h u n i t
element. Show t h a t i t f o l l o w s from
that
p 5 v
type
C(x)
.
,q
A
h a s t h e m u z t i p l i e a t i v e decomposition property, i . e .
u,v,w E A+ 2
w
and
and
pq = u
u S vw
. Apply
H I N T : F o r ( i i i ) , take f o r example
EXERCISE 146.13. I f
A
that there e x i s t
p,q E A+
A
t h i s t o t h e case t h a t
{u+i(v-~)~>'
f
i(v-w)
such
is of
.
i s a n Archimedean uniformly complete and s e m i -
prime f - a l g e b r a w i t h o u t u n i t element, t h e n
A+
can c o n t a i n e l e m e n t s h a v i n g
no s q u a r e r o o t ( a s t h e p r e c e d i n g e x e r c i s e shows). It i s t r u e , however, t h a t
Ch.
20,11461
70 1
ORTHOMOWHISMS AND f-ALGEBRAS
i n t h i s case every product
uv (u
and
v
A+ ) h a s a s q u a r e r o o t . The
in
proof we s h a l l i n d i c a t e does n o t depend o n E x e r c i s e 146.11. Hence, we s h a l l element, then every (i)
Let
t h a t f o r some
u E A+
has a square r o o t .
...,
(f : k = l , , n ) b e a sequence i n t h e Riesz s p a c e k + we have I f k - f k + l l < u ( k = l , n-I) and u E L
,..., f ) S u i n f ( t f l l , ..., lfnl)
sup(fl
as well as
inf(fl
<
. For
u
terms, each one of t h e form
2n
inf(-fl,
where
...,-f
(kl
(a)
(1
)
i s a p e r m u t a t i o n of
show t h a t e a c h term i s m a j o r i z e d by
inf(fl,
u
(1
(kl,
...,k p )
. Cases
...,f
as
sup(fk,-fk)
a s t h e supremum or
)
(b)
0 < v
2
u
j
i s s u f f i c i e n t to
( a ) and (b) a r e t r i v i a l . I n (1,.
and t h e o t h e r one i n
f o r example t h a t t h e s m a l l e s t one, s a y t h e term i n ( c ) i s a t most Let
lfkl
,...,n) . It
c a s e ( c ) t h e r e a r e two c o n s e c u t i v e i n d i c e s i n
(ii)
Show t h a t
I ,. .. ,I f n l )
f
such
or
)
,...,k
contained i n
.
) t -u
the proof, w r i t e each inf
L
...,
,...,f
and use t h e d i s t r i b u t i v e law t o w r i t e of
has a u n i t
A
o b t a i n i n p a r t i c u l a r a second proof f o r t h e r e s u l t t h a t i f
. . ,n)
one of which i s
. Suppose (k l , . . . , k ) . Then P iu < u .
( k p + l , . . .,kn)
, belong
to
f . A (-fj+l) 5 k(fj-fj+l) J i n t h e R i e s z s p a c e L and l e t
n
2
I
b e a na-
t u r a l number. Show t h a t
...,n)
2
n -1 u
.
( i i i ) L e t A be anArchimedean u n i f o r m l y c o m p l e t e a n d semiprime f - a l g e b r a and l e t
u , v E A+. W e i n d i c a t e a proof t h a t
has a p o s i t i v e square
uv = (uvv) (UAV) , we may assume t h a t
r o o t . Since
n a t u r a l number
w
Using t h a t
uv
n
n 2 1 =
1
,
define
w
n
0
S
v S u
. For
any
by
i n f (au+a- ' v : a = ( k / n ) ' , k = l ,
\
( P ~ A - . . A P ~=) p ~2 1A
... APH
for
E A+ P ,...,p n I
n 5
i n inf{(v-kn-lu)':k=l,.
. .,n} .
, we
get
702
THE RANGE OF AN ORTHOMORPHISM
-
Ch.
20,91461
uv 2 0
a s a n infimum of s q u a r e s . It follows now fi-om t h e 1 -1 2 i n e q u a l i t y i n ( i i ) t h a t 0 < w 2 - uv < -n u , s o w2 converges u2-uni4 formly t o uv Also
Note t h a t
w2 n
.
Hence, s i n c e
A
Cauchy. Then
w
0
wL = uv numbers
.
i s u-uniformly
I t follows t h a t n i d e a behind t h e proof i s t h e formula, f o r p o s i t i v e r e a l
. The u
-?
i s semiprime, Iw -w I < In 'u , s o (w,) n m converges u-uniformly t o some w E A+
and
,
v I
1
(uv)' =
(
i n f au+a
-1
v:O
I'
.
of
u
Show t h a t t h e square r o o t ( i v ) Let A be a s above and l e t u,v E A+ 2 2 + v2 e x i s t s . Hence, t h e s q u a r e r o o t of f + g2 e x i s t s f o r a l l
f,g E A
. For
t h e proof, note t h a t
u+v
-
I
(2uv)' 2 0
and
2 + v2 = { u + v + ~ 2 u v ) ~ ~ { u + , ( 2 u v ) ~ }
J
(v)
Let
f,g
b e elements i n t h e Archimedean Riesz space
L
. Show
for
k =
that
)
(
i n f If sine-g coseI:O<e<2~1 = 0
ek
For t h e p r o o f , l e t
...,2 n .
= 0,1,
w
n
w n
=
infl\l
+
0
as
= Irkn-'
Note t h a t
-
for n +
Ihk-\+ll
= 0,1,
k (L
for a l l
f,g E A
. For f
2
5
m
-1
-
g cos
(Ifl+Igl)
...,2n . Show t h a t A
wn 2
ek
for a l l -1
Irn
k
. Let
(lfl+lgl)
and
b e a s i n ( i i i ) and ( i v ) . Show t h a t
the proof, note t h a t
+ g2
=
t h a t t h e a b s o l u t e v a l u e of 2 2 ; l f + i g l = (f + g )
.
hk = f s i n Bk
i s Archimedean).
( v i ) L e t the f-algebra
p a r t ( v ) . Hence
and
.
s u p ( f cose+g f+ig
i n f ( f sine-g c o s 8 ) 2 = 0
. Note
by
t h a t i t f o l l o w s now
i n the complexification
A+iA
satisfies
The r e s u l t s i n t h i s e x e r c i s e a r e taken from a p a p e r by F. Beukers, C.B.
Huijsmans and B. de P a g t e r (C11,1983).
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149 (1963),
150-180 (5128).
Notes on Banach f u n c t i o n spaces 1-13, Notes 1-7 i n 66 ( 1 9 6 3 ) , 135-147,
655-668,
669-681
104-119,
360-376,
Proc. Netherl. Acad. S c i . (A);
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239-250,
251-263,
496-504,
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530-543 (Indag. Math. 26,
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C31
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VULIKH I n t r o d u c t i o n t o t h e t h e o r y of p a r t i a l l y o r d e r e d s p a c e s , Moscow, 1961; E n g l i s h t r a n s l a t i o n , Wolters-Noordhoff,
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Unzerlegbare nicht-negative Matrizen, Math. Z. 5 2 ( 1 9 5 0 ) , 642-648 (§
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P a r t i a l l y ordered topological v e c t o r spaces (Oxford Mathematical Monographs), Oxford University P r e s s , 1973.
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SUBJECT INDEX
a b s o l u t e , a. seminorm 394 a d j o i n t , a. o p e r a t o r 249; a . s p a c e 311 a l g e b r a , f-a.
657; Riesz a . 657
AL, AL-space 456 AM, AM-space 4 5 6 ; AM-compact o p e r a t o r 505
a n n i h i l a t o r , 159, 169; i n v e r s e a . 159,169 a s s o c i a t e , a. norm 418; a. s p a c e 418 atom 148 Banach, B. d u a l 311; B. f u n c t i o n s p a c e 252; B. l a t t i c e 109; complex B. l a t t i c e 294 band, b . p r e s e r v i n g o p e r a t o r 648 Bolzano-Weierstrass, Boolean, B.
B.-W.
p r o p e r t y 500
topology 26
c a r d i n a l , m e a s u r a b l e c . 150 Carleman, C. o p e r a t o r 262; g e n e r a l i z e d C. o p e r a t o r 271 c a r r i e r , c . of a n i d e a l of f u n c t i o n s 142; c . of a l i n e a r f u n c t i o n a l 178 Cauchy, p-C.
sequence 519
c e n t r e 659 c l a s s , Young c . 579 c o n d i t i o n , p-Cauchy
c . 335; A -c. 2
579; Riesz-Fischer
c . 304;
weak R i e s z - F i s c h e r c. 306 c o n t r a c t o r , 676 copy, c. of a s p a c e 446
d u a l , a l g e b r a i c d. 96; Banach d. 311; o r d e r d. 99; d . i d e a l 6 , l O ;
x -maximal d. i d e a l 7; I-maximal d. i d e a l 7,lO; prime d . i d e a 1 6 , l O ; 0
d. h u l l - k e r n e l topology 22
717
718
SUBJECT INDEX
e x t r e m e a l l y disconnected, e . d . d i s t r i b u t i v e l a t t i c e 71; completely e . d . d i s t r i b u t i v e l a t t i c e 75 f , f - a l g e b r a 657
Fatou, F. p r o p e r t y 120; F. p r o p e r t y f o r d i r e c t e d s e t s 421; s e q u e n t i a l F. p r o p e r t y 120; s e q u e n t i a l weak F. p r o p e r t y 421,422; weak F. p r o p e r t y f o r d i r e c t e d s e t s 390; F. seminorm 387; weakly F. seminorm 387 f i l t e r 6,10 Hilbert-Schmidt, Hille-Tamarkin,
H.-S. H.-T.
h u l l - k e r n e l , h.-k.
o p e r a t o r 262 o p e r a t o r 539
topology 19; d u a l h.-k.
topology 22
hyperarchimedean, h . Riesz space 140 i d e a l , d u a l i . 6,lO; maximal d u a l i. 6,lO; x -maximal i. 7; 0 x -maximal d u a l i. 7; F-maximal i . 7,lO; I-maximal d u a l i . 6,lO; 0 prime d u a l i . 6,lO; n u l l i. 154,178 index, upper and lower index of a Banach l a t t i c e 557 i n t e g r a l 123 i r r e d u c i b l e , i . o p e r a t o r 613; i . s e t i n a t o p o l o g i c a l space 28 J o r d a n , J. o p e r a t o r 505 k e r n e l 154; a b s o l u t e k . 154,178; k . o p e r a t o r 213;
k . o p e r a t o r of f i n i t e rank 229 L, L-weakly
compact o p e r a t o r 529
d , d -composition p r o p e r t y 557; d -decomposition p r o p e r t y 556 P P P lower, 1. s u b l a t t i c e 10; maximal 1. s u b l a t t i c e 10
m e t r i c a l l y , m. dense s e t 624 minimal, m. p o s i t i v e e x t e n s i o n of an o p e r a t o r 104 monotone, m. complete space 305; m. seminorm 394 m u l t i p l i c a t i v e , m. decomposition p r o p e r t y 700 norm, p-additive
n. 456; - a d d i t i v e
n. 456; a s s o c i a t i v e n. 418;
Luxemburg n. 582; o r d e r continuous n. 336, a-order continuous n. 336; O r l i c z n. 591 normal, n. d i s t r i b u t i v e l a t t i c e 65; completely n. d i s t r i b u t i v e l a t t i c e 78;
719
SUBJECT I N D E X
n. i n t e g r a l 123; n . R i e s z s p a c e 70 normed, n . R i e s z s p a c e 281 operator,
o f f i n i t e double-norm 539; a l g e b r a i c a d j o i n t
0.
AM-compact Carleman
505; a b s o l u t e k e r n e l
0.
0.
Hille-Tamarkin kernel
0.
539; i r r e d u c i b l e
0.
126; o r d e r a d j o i n t
0.
order continuous
0.
s t r o n g l y semi-compact d u a l 99;
0.
o r d e r bounded, 0 . b .
613; J o r d a n compact 0.
0.
0.
648; 0.
505; k e r n e l
529; l i n e a r
249; o r d e r bounded
105,123; r e g u l a r
0.
9 9 ; semi-compact
545; s i n g u l a r
0.
126
0.
0.
271; H i l b e r t - S c h m i d t
0.
s e t 98; a l m o s t 0.b.
order continuous.
O.C.
norm 336;
o r d e r dense, s t r o n g l y 0.d. 0.
O.C.
529;
s e t 501;
o p e r a t o r 105,123
46
63
o r d e r bounded s e t 525, p-Cauchy s e t 519; p - s u b l a t t i c e
sequence 519;
36
p e r f e c t , p. R i e s z s p a c e 409 p e r i p h e r a l , p. s p e c t r u m 636 p r o p e r t y , Bolzano-Weierstrass
p. 500; Egoroff p. 342;
s t r o n g Egoroff p . 342, F a t o u p .
120; s e q u e n t i a l F a t o u p.
120;
s e q u e n t i a l weak F a t o u p . 421,422; F a t o u p. f o r d i r e c t e d s e t s 421;
L - c o m p o s i t i o n p . 557; P L -decomposition p . 556; monotone prime i d e a l e x t e n s i o n p. 34;
weak F a t o u p. f o r d i r e c t e d s e t s 390;
P m u l t i p l i c a t i v e d e c o m p o s i t i o n p. 700; a - i n t e r p o l a t i o n p. 696;
unique i d e a l e x t e n s i o n p. 55; unique prime i d e a l e x t e n s i o n p . 32 pseudonorm, R i e s z p . 298 r a d i u s , s p e c t r a l r. 604 r e s o l v e n t , r . s e t 602 R i e s z , R. a l g e b r a 657 Riesz-Fischer,
R.-F.
c o n d i t i o n 304; weak R.-F.
c o n d i t i o n 306
95;
0.
orthomorphism 648
p-precompact
213;
105,123;
O r l i c z , 0. s p a c e 581
p , p-almost
0.
0.
i n t e r v a l 98
o p e r a t o r 99; 0.b.
262;
99;
0.
s e t 525
p-almost 0.b.
ordering, dual
213; band p r e s e r v i n g 0.
105,123; s e q u e n t i a l l y o r d e r c o n t i n u o u s
0.
o-order continuous
0.
0.
o f f i n i t e rank 229; L-weakly
normal s i n g u l a r
order,
0.
262; g e n e r a l i z e d Carleman
249;
0.
SUBJECT INDEX
720
o , o - i n t e r p o l a t i o n p r o p e r t y 696 semi-compact,
s .-c.
operator 529; strongly s.-c.
o p e r a t o r 545
seminorm, a b s o l u t e s . 3 9 4 ; Fatou s . 3 8 7 ; weakly Fatou s . 3 8 7 ; monotone s . 394 s i n g u l a r , s . l i n e a r f u n c t i o n a l 146; normal s . l i n e a r f u n c t i o n a l 146 s p a c e , AL-s.
456; a b s t r a c t L - s . 456; a d j o i n t s . 311; P a s s o c i a t e s . 4 1 8 ; hyperarchimedean Riesz s . 1 4 0 ; semi-M-s. 4 6 6 ; 4 5 6 ; AM-s.
O r l i c z s . 581 s p e c t r a l , s. r a d i u s 6 0 4 spectrum 6 0 2 ; p e r i p h e r a l s . 636 s t a b i l i z e r 678 strong,
S.
norm u n i t 463
s t r o n g l y , s . o r d e r dense 46 s u b l a t t i c e , lower s . 10; maximal lower s . 10; p-s. topology, Boolean t . 2 6 ; h u l l - k e r n e l Lebesgue t . 3 5 3 ; a-Lebesgue t o t a l l y , t . bounded s e t 500 u n i t , s t r o n g norm u . 463 Young, Y.
c l a s s 579
36
t . 19; d u a l h u l l - k e r n e l
t . 3 5 3 ; l o c a l l y s o l i d t . 296
t. 22;
