Chapter 14 The Lean Elements

CHAPTER 1 4 THE LEAN ELEMENTS

565. Elementary p r o p e r t i e s

An e l e m e n t f o f C t ' ( X ) w i l l be c a l l e d

lean

i f a ( If

1)

=

0.

If1 i s t h e n a l s o l e a n a n d , i n f a c t , s o i s e v e r y e l e m e n t of t h e

R i e s z i d e a l g e n e r a t e d by f .

Thus a n a l t e r n a t e d e f i n i t i o n i s

t h a t f i s l e a n i f t h e R i e s z i d e a l which i t g e n e r a t e s i n t e r s e c t s C(X) only i n 0.

There i s a n abundance o f l e a n e l e m e n t s .

We show t h a t € o r

u ( f ) - f and f - a ( € ) a r e l e a n .

every f EC"(X),

l e a n e l e m e n t i s o f t h i s form.

Note

Moreover, e v e r y

t h a t the following proposi-

t i o n c o r r e s p o n d s t o s t a n d a r d t o p o l o g i c a l ones on s e t s w i t h empty i n t e r i o r [ 31,581

.

( 6 5 . 1 ) For f € C l ' ( X ) ,

f > 0 , the following a r e equivalent:

1'

f is lean;

2'

f

=

u ( g ) - g f o r some ~ E c ~ * ( x ) ;

3'

f

=

h - R(h) f o r some h € C " ( X ) ;

4O

f < 6(f).

336

The Lean E l e m e n t s Proof. ___ u(f)

S u p p o s e 1' h o l d s .

p(f)

=

S(f).

u(f) < s(f)

=

u(f)

-

a(f)

=

0.

so f

=

f - k(f).

i n 3'.

Thus 4' -

T h a t 3'

Finally, i f f

t ( u ( g ) - g) 0

plies 1

5

Then k ( f ) = 0 , s o f < u(f) =

holds.

S u p p o s e 4'

e(f) < u ( f ) , so u ( f )

Thus 1' h o l d s .

=

1'

i m p l i e s 3':

i m p l i e s 2'

u(g)

-

337

holds.

Then

u(f) - Elf), so

=

i n effect, k(f)

=

0,

f o l l o w s by s e t t i n g g

=

-11

g , t h e n by ( 4 9 . 2 ) ,

u ( g ) - u ( g ) = 0 , whence k ( f )

=

0 < k(f) =

Thus 2'

0.

im-

. QED

I f f i s l e a n , t h e n s o are f + and f - .

The c o n v e r s e i s f a l s e .

For o n e e x a m p le, l e t X be a r e a l i n t e r v a l and g a n d h t h e c h a r a c t e r i s t i c elements o f t h e sets o f r a t i o n a l p o i n t s and i r rational points respectively h are lean.

( r em em be r, g , h E I J ( X )

Suppose k(g) > 0 ; t h e n e(g),

(555)).

g and

i s a lowersemicon-

t i n u o u s f u n c t i o n o n X w hi ch i s > 0 a n d v a n i s h e s o n t h e i r r a t i o n a l points, an impossibility. Then f +

=

g and f -

=

Similarly for k(h).

11, b o t h l e a n , w h i l e If

I

=

Set f

=

g - h.

R ( X ) , which i s n o t

lean. F o r a n e x a m p l e removed f r o m f u n c t i o n i m a g e r y , l e t X b e d e n s e - i n - i t s e l f and f f - = l.(X)d, f

=

l(X)u

- n(X)d.

Then f +

=

n(X),

and

b o t h l e a n ( c f . t h e comment f o l l o w i n g ( 4 0 . 5 ) ) , w h i l e

n(x).

=

Remark.

N ot e t h a t i n t h e s e two e x a m p l e s , 6 ( f )

T h u s, i n c o n t r a s t w i t h ( 6 5 . 1 ) ,

(65.2)

=

2.n(X).

6 ( f ) is, i n general, not lean.

If f i s l e a n , t h e n k ( f ) < 0 < u(f).

C h a p t e r 14

338

Proof.

f f and f - a r e l e a n , h e n c e , by ( 4 9 . 1 1 ) ,

u(f+) - a(f-)

=

u(f+) > 0 , and L ( f )

=

a(f')

u(f)

=

- u(f-) = -u(f-)

.(

0.

QED We n o t e t h e u s e f u l c o r o l l a r y t h a t i f a u s c e l e m e n t u i s l e a n , > 0 , and i f an Esc e l e m e n t R i s l e a n , t h e n R < 0. then u -

(Note

that (59.5) i s a corollary of t h i s . ) Note a l s o t h a t t h e c o n c l u s i o n o f ( 6 5 . 2 ) i s n o t a s u f f i c i e n t c o n d i t i o n f o r f t o be l e a n , a s t h e examples p r e c e d i n g ( 6 5 . 2 ) show. These examples a l s o show t h a t t h e s e t o f l e a n e l e m e n t s i s not c l o s e d under a d d i t i o n o r t h e l a t t i c e o p e r a t i o n s i s n e i t h e r a l i n e a r subspace nor a s u b l a t t i c e of

V,A.

C'l(X).

So i t

I t con-

t a i n s w i t h e a c h e l e m e n t t h e R i e s z i d e a l g e n e r a t e d by t h a t c l e ment.

I t i s t h u s a union of Riesz i d e a l s , a c t u a l l y t h e union

o f a l l t h e R i e s z i d e a l s i n t e r s e c t i n g C(X) o n l y i n 0 .

I t is

t h u s a s o l i d s e t , and c o u l d be d e f i n e d a s t h e l a r g e s t s o l i d s e t i n t e r s e c t i n g C(X) o n l y i n 0 . An o b v i o u s , b u t u s e f u l , consequence o f t h e s o l i d n e s s i s t h a t i f f i s l e a n , t h e n f o r e v e r y band I o f Cl'(X), f I i s a l s o lean. Another p r o p e r t y :

( 6 5 . 3 ) The s e t o f l e a n e l e m e n t s i s norm c l o s e d .

T h i s f o l l o w s from t h e e a s i l y v e r i f i e d f a c t t h a t i f a s o l i d

s e t i n t e r s e c t s C(X) o n l y i n 0 , t h e n s o does i t s norm c l o s u r e .

The Lean Elements

339

The examples preceding (65.2) actually show that the sum of two lean elements f,g need not be lean even if f A g

=

How-

0.

ever, under "stronger" disjointness of f and g, their sum

%

lean.

> 0 are lean and u ( f ) A g (65.4) If f,g -

Proof.

By (49.1), k(f

+

g)

-

u(f)

(using Exercise 4 in Chapter 1) R(f u(f)Af

R(f

+

+

g)

u(f)Ag =

=

f.

Thus 0

<

k(f

+

0, then f

=

+

t(g)

u(f).

=

g) < u(f)A(f

+ g) < L(f)

g i s lean.

+

Hence g)

+

<

0, whence

=

0.

QED

(65.5) Corollary 1.

Suppose f,g

0 are lean.

< R and fAR Rsc element R such that g -

Proof.

By (49.7), u(f)AR

=

=

If there is an

0, then f

0 , so u(f)Ag

=

+

g is lean.

0 and (65.4)

applies.

Let I be a u-band or an R-band.

(65.6) Corollary 2 . f EC"(X),

i f fI and f

Proof.

I

For every

are lean, then f is lean.

For concreteness, let I be a u-band, and we can

Chapter 1 4

340 a s su me I f 1
II

II

USC,

a n d If

=

Ifl

1

and If1

1

=

I

If

I

dl,

s o , by t h e

a r e l e a n . We show u ( I f 1 ) A l f d I I i t w i l l f o l l o w from ( 6 5 . 4 ) t h a t I f 1 i s l e a n . I f I I .Y(X),,

which i s

If

If

so, a fortiori, u(lflI)Alfl

I

=

0;

= 0.

QED

I n t u i t i v e l y , a l e a n e l e m e n t s h o u l d be " n e g l i g i b l e " .

We

n e x t e x p l o r e t h e c o n s e q u e n c e s o f two e l e m e n t s o f C1l(X) d i f f e r i n g by a l e a n e l e m e n t .

Proof. is lean.

0 5 ( L ( f ) - u ( g ) ) + 5 ( f - g)',

hence (L(f) - u ( g ) ) +

S i n c e i t i s a n ~ s ecl e m e n t , i t f o l l o w s ( a ( f ) - u ( g ) ) + = 0.

QED A f o r t i o r i , i f f - g is lean, then L(f) < u(g).

This gives

us :

( 6 5 . 8 ) C o r o l l a r y 1.

2,(P

f

I f h i s l e a n , t h e n f o r e v e r y gE C"(X ),

h) 5 u ( g > .

(65.9) Corollary 2 .

If h i s l e a n , then f o r every usc element u,

a(u

i h)

<

u,

The Lean Elements

and for every Rsc element u(R

t

341

k,

h) > k.

In particular, for every gEC(X), R(g

?

h)<- g < u(g

(65.10) Corollary 3.

II s

+

*

11).

If h is lean, then for every gEC(X), hll L

I1 gII -

This last follows from the preceding and (49.14).

We will be interested in knowing when a given f EC"(X) differs from a usc element or an ksc element by a lean element. Now for every f, f ially lean.

<

u(f),

so (f - u(f))+

We can expect that if u(f)

0, hence is triv-

=

is replaced by a

smaller usc element u , then (f - u ) + will no longer be lean How much smaller must u be?

We have a simple answer.

(65.11) Theorem, F o r every f EC"(X),

(f

-

uk(f))+

is lean, and

uk(f) is the smallest usc element for which this is true.

Proof.

o

<

(f

-

ua(f))+

<

(f

-

.t(f))+

=

Since

f - a(f).

this last is lean (65.1), it follows (f - uk(f))+

is lean.

Now

suppose u is a usc element such that (f - u ) + is lean; we show u > u&(f).

It is enough to show that u 2 R(f).

k((f

-

u)')

=

0,

34 2

SO

Chapter 1 4 I,(f

-

U)

< 0 (49.11), -

SO

a(f) - u < 0 (49.2).

Thus u > ?(f).

QED Remark.

D u a l l y , u s i n g t h e f a c t t h a t u ( f ) - f i s l e a n , we

have: (ku(f)

-

f)'

i s l e a n , and L(f) i s t h e l a r g e s t fsc element

f o r wh ich t h i s i s t r u e .

A l l t h e s t a t e m e n t s i n P a r t VI h a v e

similar dual s t a t e m e n t s . ( 6 5 . 1 1 ) h a s a c o r r e s p o n d i n g t h e o r e m i n t o p o l o g y : G i ven a s e t A i n a t o p o l o g i c a l s p a c e T , l e t I: t h e i n t e r i o r of A.

=

Int,t h e c l o s u r e o f

Then A \ F h a s empty i n t e r i o r , a n d I: i s t h e

s m a l l e s t c l o s e d s e t f o r w hi ch t h i s i s t r u e . i n t h e n e x t 5 , we exam i ne t h i s f u r t h e r .

For t h e d i s c u s s i o n

Let u s say t h a t A has

empty i n t e r i o r a t a p o i n t t o f T i f t h e r e e x i s t s a n e i g h b o r h o o d

W o f t s u c h t h a t A n W h a s empty i n t e r i o r .

Then t h e s e t F

a b o v e i s p r e c i s e l y t h e s e t o f p o i n t s o f T a t w h i c h A d o e s n__ ot h a v e empty i n t e r i o r . (A

n

Thus A h a s a u n i q u e d e c o m p o s i t i o n A

=

i n t o t h e s e t o f i t s p o i n t s a t which i t does n o t

F ) IJ ( A \ F )

h a v e empty i n t e r i o r a n d a s e t w i t h empty i n t e r i o r .

(65.11)

g i v e s u s a c o r r e s p o n d i n g decomposition o f e a c h f EC"(X) : f = fAuk(f)

+

(f

-

uk(f))+.

For a n e l e m e n t o f & ( I L ( X ) ) ,

d e c o m p o s i t i o n i s i n t o two e l e m e n t s o f E ( l l ( X ) ) r e p l a c e d by

V,

and

+

the

can be

making t h e a n a l o g u e c l e a r e r .

566. A ' l l o c a l i z a t i o n " t h e o r e m

The f o l l o w i n g " l o c a l i z a t i o n " and e a s i l y v e r i f i e d ( c f .

theorem i n topology i s obvious

[31], S8,VII):

I f a s e t A h a s empty

34 3

The Lean E l e m e n t s i n t e r i o r a t e v e r y p o i n t , t h e n A h a s empty i n t e r i o r .

The

( ( 6 6 . 3 ) bel ow ) r e q u i r e s more

correspond-ing theorem f o r C"(X) work t o p r o v e .

( 6 6 . 1 ) L e m m a 1.

I f fAg i s l e a n , t h e n s o a r e f + A g , f - A g ,

lf/Ag,

IflAIgl, and f - .

Proof.

5

-IfAg/

fAg < f+Ag < I fl A g

<

IflAlgl 5 /fAgl.

S i n c e t h e s e t o f l e a n e l e m e n t s i s s o l i d , i t f o l l o w s f+A g,

/flAg,

f - = 1 - f - 1 , s o , by w hat w e h a v e j u s t

and IflAlgj a r e l e a n .

shown, t o p r o v e t h a t f - A g i s l e a n , i t s u f f i c e s t o show t h a t (-f-)Ag i s lean.

(-f-)Ag

=

i s l e a n s i n c e fAg i s l e a n .

o

so

<

f-

=

(fA0)Ag

=

(fAg)AO

=

- ( f A g ) - , w hi ch

Finally, f > fAg, s o - f < -(fAg),

(-f)+ < (-(fAg))+

=

(fAg)-;

s i n c e (fAg)- i s l e a n ,

it follows f - i s lean. QED

No te t h a t f A g l e a n d o e s n o t i m p l y t h a t f + i s l e a n : l e t f

=

n(X)

and g

that f is lean. the

=

0.

I t f o l l o w s t h a t fAg l e a n d o e s n o t i m p l y

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

implication does hold.

Lemma (66.2) - 2 . G i ven a n Rsc e l e m e n t k , l e t I b e t h e band w h i c h it generates.

Then f o r f E I , i f f A k i s l e a n , f i s l e a n .

Chapter 1 4

344

Proof. (49.4)

Case I , R > 0 : By ( 6 6 . 1 ) ,

f a ( 1fI)AR = R(lflAR)

=

0.

IfIAR i s l e a n , h e n c e

Since t h e o n l y element o f I

d i s j o i n t from R i s 0 , i t f o l l o w s R ( l f 1 ) = 0 . < 0: Case 11, R -

Thus f i s l e a n .

By t a k i n g n e g a t i v e s , t h i s c a s e c a n be

writ t e n : Given a u s c e l e m e n t u > 0 , l e t I be t h e band which i t

(*)

generates.

Then f o r g € 1 , i f gvu i s l e a n , g i s l e a n .

We w i l l p r o v e t h i s a t t h e end of 567. We p r o c e e d t o p r o v e ( 6 6 . 2 ) f o r t h e c a s e o f a g e n e r a l k s c L e t H be t h e band g e n e r a t e d by R'.

element R .

Then H i s a n

R-band, s o by ( 6 5 . 6 ) , i t i s enough t o show t h a t f H ahd f lean. fHAR+

=

H

are

fHARH = (fA!L)H, which i s l e a n by t h e comment p r e -

ceding (65.3).

S i n c e '.9

i s a n Rsc e l e m e n t , Case I above g i v e s

us t h a t f H i s lean. f dA(-R-) = f dAR = (fAR) d , w h i c h , a g a i n , i s l e a n by H H H H t h e comment p r e c e d i n g ( 6 5 . 3 ) . Now - R - = 9,110, h e n c e i s a n Rsc e l e m e n t , and s i n c e R ed by - k - ,

i s c l e a r l y c o n t a i n e d i n t h e band g e n e r a t H Case I 1 above g i v e s u s t h a t R is lean. H QED

We a r e now i n a p o s i t i o n t o e s t a b l i s h o u r l o c a l i z a t i o n theorem.

( 6 6 . 3 ) Theorem. Then f o r f E C " ( X ) ,

Let

{aa]

be a c o l l e c t i o n o f Rsc e l e m e n t s .

fARalean f o r a l l a i m p l i e s t h a t V a ( f A R a )

is

The Lean E l e m e n t s

34 5

lean.

We p r o v e t h e f o l l o w i n g t h e o r e m , which i s e a s i l y shown t o be e q u i v a l e n t t o t h e above.

L e t { L a } b e a c o l l e c t i o n o f Rsc e l e m e n t s a n d R = v,Lu.

If

FARcx i s l e a n f o r a l l a , t h e n f A k i s l e a n .

Proof.

< k ; s o we For s i m p l i c i t y , we c a n assume t h a t f -

have t o show t h a t f i s l e a n . Case I , L > 0 and f i s i n t h e band g e n e r a t e d by Lemma 2 , i t i s enough t o show t h a t If \ A l l i s l e a n .

:

By

il = V a i l a

v t r ( R w ) + , and by Lemma 1 , I f \ A ( L a ) + i s l e a n f o r e v e r y a. f o l l o w s L ( l f [ ) A ( e w ) += 0 f o r a l l a ( 4 9 . 4 ) , whence L( whence ( a g a i n by ( 4 9 . 4 ) )

=

It

f()AL

=

0,

IfjAR i s l e a n .

Case 11, L < 0 : Choose a n a r b i t r a r y cto.

fARa 0

5 f <- R <- O ,

hence, t a k i n g n e g a t i v e s , 0 < -f < -(fARa ) , which i s l e a n . 0

Now remove t h e r e s t r i c t i o n of p o s i t i v e n e s s o r n e g a t i v e n e s s on R .

L e t I b e t h e band g e n e r a t e d by R + .

By ( 6 5 . 6 ) , we n e e d

are lean. I We r e d u c e t h e c a s e o f f I t o Case I a b o v e . R + = v w ( R a ) + and

o n l y show t h a t f I and f

I -< R I = R + , s o i t s u f f i c e s t o show t h a t f I A ( R a ) + i s l e a n f o r a l l a. (Ra)+ 5 R + , so (Ra)+ = ((La)+),, so fIA(La)+ = f

fIA((ka)+)I

=

( f A ( L a ) + ) , , which i s l e a n by ( 6 6 . 1 ) a n d t h e

comment p r e c e d i n g ( 6 5 . 3 )

.

We r e d u c e t h e c a s e o f f

I

t o Case I 1 a b o v e .

-1

= LAO

=

34 6

Chapter 1 4

Va(LaAO) = V c l ( - ( ~ c l ) -and ) ~ fId 5

=

- f

, so

i t s u f f i c e s to

A(-(Lcl)-) i s l e a n f o r a l l a. f I d = c - f - ) d I I ( f + E I ) , s o we have t o show t h a t ( - f - ) d A ( - ( t 1 - 1 i s l e a n . I

show t h a t f

f i r s t that (-f-)A(-(ila)-) =

(fAo)A(LaAo)

Note

= (fAka)Ao = - ( f A k , t ) - ,

S i n c e - f - 5 ( - f - ) d , we have ( - f - ) A ( - ( k a ) - ) < I ( - f - ) d ~ ( - ( k a ) - ) 0 , h e n c e , by t h e p r e c e d i n g s t a t e m e n t , I ( - f - ) d ~ ( - ( ~ c l ) i-s) l e a n . I QED

which i s l e a n .

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

C o r o l l a r y 1. (66.4) -

Let {I,} be a c o l l e c t i o n o f il-bands and I

be t h e band w i t h ll(X)I

=

V ~ I ( X )( ~ s o I i s a l s o an il-band).

If

c1

fI

i s l e a n f o r a l l a, t h e n f I i s l e a n . c1

(66.5) Corollary 2 .

For e a c h f EC"(X), t h e r e e x i s t s a l a r g e s t

R-band I f o r which f I i s l e a n .

Analogous t o ( 6 6 . 3 ) , we h a v e :

( 6 6 . 6 ) Theorem..

u =

A

c1

ua'

Let Iua} be a c o l l e c t i o n o f u s c e l e m e n t s and

Then f o r f E C " ( X ) ,

t h a t (f - u ) + i s lean.

( f - ua)+ l e a n f o r a l l a i m p l i e s

The Lean E l e m e n t s Remark.

A t f i r s t glance,

347

( 6 6 . 6 ) a n d ( 6 6 . 3 ) seem t o b e t h e

same t h e o r e m , a n d i t i s t r u e t h a t t h e y r e d u c e t o t h e same t h e o -

rem i n t o p o l o g y . Proof o f

-

k(f),

u)')

5

fAu

~ +

t h e y do n o t c o i n c i d e .

By ( 6 5 . 7 ) , 9 ( f a ) 5 ua f o r a l l a.

(66.6).

follows L(f) < ~

E((f

However, i n C " ( X ) ,

=

(f

uu, h e( n c e~ 2 ( f ) 5 f A u . -

whence 2 ( ( f - u ) '

u)+ =

=

Then n ( f )

f , whence t ( f )

+

k((f

It

+ -

u)')

5

0.

QED Remark.

Note t h a t a l s o ( 6 6 . 5 ) a n d ( 6 5 . 1 1 ) r e d u c e t o t h e

same t h e o r e m i n t o p o l o g y .