Chapter 15 The Rare Elements

Chapter 15 The Rare Elements

CHAPTER 1 5 THE RARE ELEMENTS § 6 7 . Elementary p r o p e r t i e s An e l e m e n t f o f C"(X) w i l l be c a l l e d r a r e i f u ( If ~ that...

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CHAPTER 1 5

THE RARE ELEMENTS

§ 6 7 . Elementary p r o p e r t i e s

An e l e m e n t f o f C"(X) w i l l be c a l l e d r a r e i f u ( If ~

that is, i f ku(lf1)

=

0.

1)

i s lean,

A rare element i s of course l e a n .

We w i l l d e n o t e t h e s e t o f r a r e e l e m e n t s by Ra(X).

As i s

t o b e e x p e c t e d , t h e p r o p e r t i e s o f Ra(X) a r e much s t r o n g e r a n d more s a t i s f y i n g t h a n t h o s e o f t h e s e t o f l e a n e l e m e n t s .

To

b e g i n w i t h , w h i l e t h e l a t t e r s e t i s n o t even a l i n e a r s u b s p a c e , we h a v e :

(67.1)

Ra(X) i s a norm c l o s e d R i e s z i d e a l .

S u ppos e f , g E R a( X ) .

Proof.

ku(lf

+

gl) 5 Rullfl <

Then

atuofl)

< %u(lfl) =

hence !Lu(\f

+

lgl)

+

+

+

u(lgl)l

(49.1)

u(lgl)

(49.1)

u(lgl),

gl) 5 ku(lg/)

=

0.

Thus ( f

+

g) E R a ( X ) .

c l e a r t h a t i f f E R a( X ) , t h e n Xf E Ra(X) f o r a l l A ER. 34 8

I t is So Ra(X)

The Rare Elements That f E Ra(X)

is a linear subspace. g E lia(X)

ideal.

349

and ( g l

follows from the definition. Finally, the operation

(49.16)), hence Ra(X)

So

<

If

1

implies

Ra(X) is a Riesz

u ( . ) is norm continuous (iterate

is norm closed. QED

As is easily seen, Ra(X) is the Riesz ideal generated by

the positive lean usc elements, hence by all the lean usc elements (cf. the comment following (65.2)). Some additional characterizations:

(67.2) For f EC"(X), '1

f E Ra(X);

2'

f',f-

3O

Lu(f)

the following are equivalent:

E Ra(X); =

0 = uR(f);

< 0 < uL(f). 4O Lu(f) -

Proof. If 2'

That l o and 2'

are equivalent is contained in (67.1)

holds, then (by iterating (49.10))Lu(f)

uL(f-)

=

0, and similarly for u!,(f).

course implies 40 . (ku(f))'

=

Finally assume ' 4

0 and Lu(f-)

=

(uk(f))-

=

=

ku(f')

Thus 3' holds. holds. 0.

-

'3

Then k u ( f ' )

Thus 2'

of =

holds. QED

' 4

above should be compared with (65.2).

In that case, the

inequalities were only necessary; here they are a l s o sufficient.

Chapter 1 5

350

( 6 7 . 3 ) Ra(X) i s c l o s e d u n d e r t h e o p e r a t i o n s u ( * ) , a ( . ) ,

(therefore) 6( * )

Proof.

and

.

Then i t h a s p r o p e r t y 4'

S u ppos e f E R a( X ) .

we show u ( f ) a n d k ( f ) a l s o h a v e t h i s p r o p e r t y . k(f) < 0 , and u a ( u ( f ) ) > ua(f) > 0.

in (67.2);

au(u(f))

=

Similarly for k(f).

QED We h a v e s e e n t h a t f,g l e a n d o e s n o t i m p l y t h a t f + g , f v g , fAg a r e l e a n .

(67.4)

However,

I f f i s l e a n a n d g i s r a r e , t h e n f + g , f v g , a n d fAg a r e

lean.

(67.5) Corollary.

I f a R i e s z i d e a l I o f C'l(X) i s c o n t a i n e d i n

t h e s e t o f l e a n e l e m e n t s , t h e n I + Ra(X) i s a l s o .

I t follows

t h a t i f a R i e s z i d e a l i s maximal w i t h r e s p e c t t o b e i n g c o n t a i n e d i n t h e set of l e a n elements

,

t h e n i t c o n t a i n s Ra(X ).

The Rare Elements Again, i f two e l e m e n t s o f

C'l(X)

351

d i f f e r o n l y be a r a r e e l e -

ment, we c a n s a y more t h a n we c a n i f t h e y d i f f e r by a l e a n element.

( 6 7 . 6 ) Given f , g E C " ( X ) , i f ( f - g ) + i s r a r e , t h e n ( u ( f ) - u ( g ) ) +

and ( P ( f ) - Q ( g ) ) + a r e r a r e .

Proof.

By ( 4 9 . 2 ) , ( u ( f ) - u ( g ) ) + , ( R ( f ) And by ( 4 9 . 1 0 ) ,

(u(f - g))'.

(u(f - g))'

=

R(g))+

-

2

u ( ( f - g)'),

which

i s r a r e by h y p o t h e s i s .

( 6 7 . 7 ) Given f , g E C 1 ' ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t :

1'

(u(f)

u(g)j+ is rare;

2'

(u(f) - u(g))+ i s lean;

3'

Ru(f) 5 Q u ( g ) ;

4O

uRu(f) 5 u R u ( g ) .

-

And t h e f o l l o w i n g a r e e q u i v a l e n t : 11.

(a(f)

21.

( a ( f ) - . ~ ( g ) ) +i s l e a n ;

3'.

u Q ( f ) 5 uR(g);

4'.

QuR(f) < RuR(g).

Proof. 4'.

T h a t 2'

-

~ ( g ) ) +i s r a r e ;

We o f c o u r s e have t h a t 1' i m p l i e s 3'

i m p l i e s 2'

f o l l o w s from ( 6 5 . 7 ) .

and '3

implies

F i n a l l y , suppose

352 4'

Chapter 15

holds; we show ( u ( f )

( u ( f ) - u(g))+

=

-

u(g))'

is rare.

[ u ( f ) - uRu(f) + u R u ( f ) - uRu(g)

+

uRu(g) - u ( g ) ] +

< ( u ( f ) - u R u ( f ) ) + + ( u R u ( f ) - uEu(g))+

+

(u!Lu(g) - u(g))+.

The second term in this last line vanishes by supposition, and uRu(g) 5 u ( g ) , so the last term vanishes.

We thus have:

This last is lean by (65.1), hence, being a usc element, is rare. The equivalence o f l', 2 ' , 3', 4 ' i s proved similarly. QED

Combining (67.6) and (67.7):

(67.8) I f ( f - g ) + is rare, then the eight properties in (67.7) hold.

I f we interchange f and g in the above three propositions,

then combine the statements thus obtained with the original ones, we have

(67.9)

I f f - g is rare, then the following s i x properties

-

The Rare Elements

353

o f which t h e f i r s t t h r e e a r e e q u i v a l e n t and t h e l a s t t h r e e a r e

equivalent - a l l hold: (i)

u ( f ) - u(g) is rare;

(ii)

!,u(f)

iu(g);

=

(iii) uau(f)

utu(g);

=

(iv)

e(f) - a(g) is rare;

(v)

uk(f) = uL(g);

(vi)

kua(f) = k u k ( g ) .

(67.10)

I f r i s r a r e , t h e n f o r e v e r y g E Cl'(X),

C o r o l l a r y 1.

ku(g

2

r ) = ku(g),

uL(g t r ) = uk ( g ) .

I f r i s r a r e , then f o r every usc elementu,

(67.11) Corollary 2.

ku(u

?

r) = k(u),

and f o r e v e r y ksc e l e m e n t L, uk(k

2

r)

=

u(L).

In p a r t i c u l a r , f o r every g E C ( X ) , ku(g ? r ) = g = u k ( g

We c a n a l s o s t r e n g t h e n ( 6 5 . 1 1 )

(67.12)

Theorem.

r).

(Note t h a t u k u ( f ) > uL(f)).

For every fEC"(X), (f - uku(f))'

i s r a r e , and

u L u ( f ) i s t h e s m a l l e s t u s c e l e m e n t f o r which t h i s i s t r u e .

Chapter 1 5

354

Proof.

To show t h a t ( f - u L u ( f ) ) + i s r a r e , i t s u f f i c e s t o

show t h a t ( f - k . u ( f ) ) + i s r a r e .

- ku(f))+

0 < (f

k u ( f ) ) + = u ( f ) - k u ( f ) , which i s l e a n ( 6 5 . 1 ) ,

< (u(f)

-

hence ( u ( f ) - n u ( f )

being a usc element) r a r e . Now suppose u i s a u s c e l e m e n t s u c h t h a t ( f - u ) + i s r a r c , we show t h a t u > ui,u(f). k u ( ( f - u)')

=

I t i s enough t o show u z ku(f).

0 , s o Lu(f - u ) < 0 , s o L(u(f)

whence ( a g a i n by ( 4 9 . 2 ) ) k u ( f )

-

u<

-

u )<

0

(49.2),

0.

QED

I f f E Ra(X), t h e n t h e R i e s z i d e a l g e n e r a t e d by f i s c o n t a i n e d i n Ra(X) ( s i n c e Ra(X) i s i t s e l f a R i e s z i d e a l ) . general, the band g e n e r a t e d b y f i s n o t . f E Ra(X) does n o t imply l l ( X ) f E Ra(X).

In

Otherwise s t a t e d ,

A s an example, l e t X be

a r e a l i n t e r v a l , { r n } t h e s e t o f r a t i o n a l p o i n t s of X , and

u t h e u s c e l e m e n t d e f i n e d by ( u , r I 1 ) (u,x)

=

=

l / n (n = 1 , 2 , . - . ) and

I t i s easy t o v e r i f y t h a t u i s r a r e b u t

0 otherwise.

R ( X ) u (which i s t h e c h a r a c t e r i s t i c e l e m e n t o f { r n l ) i s n o t r a r e - i n d e e d u(ll(X)u) = n ( X ) . We do have t h e f o l l o w i n g :

( 6 7 . 1 3 ) For f EC"(X), f

1'

f E Ra(X);

2'

n(x)

Proof.

(f -

(xi 1

3

0 , t h e following a r e equivalent:

+ E Ra(X) f o r a l l

A > 0.

Suppose 1' h o l d s , and c o n s i d e r 1 > 0 .

Let

The Rare Elements

e

=

n(X)

(XI 1

(f-

0 < e < (l/X)€€

+.

Then, by ( 1 7 . 9 ) ,

(ii)

< f.

e -

Thus

Ra(X).

C o n v e r s e l y , s u p p o s e 2' 0 < f < n(X).

Xe < f

355

h o l d s , a n d , f o r s i m p l i c i t y , assume

Consider X > 0 .

( f - h n ( X ) ) + E Ra(X).

t h e l a t t e r i s i n Ra(X) b y a s s u m p t i o n . ( f - hll(X))+ < ll(X);

f < n ( X ) , hence

the d e s i r e d i n e q u a l i t y then follows.

I t f o l l o w s from ( i ) and ( i i ) t h a t f i s i n t h e norm c l o s u r e o f Ra(X).

S i n c e Ra(X) i s norm c l o s e d , f E Ra(X). QED

While t h e band g e n e r a t e d by a n e l e m e n t o f Ra(X) need n o t be c o n t a i n e d i n Ra(X), i t i s contained i n t h e s e t of l e a n elements. T h i s i s c o n t a i n e d i n t h e f o l l o w i n g t h e o r e m , which i s t h e c l a s s i c C a t e g o r y theorem f o r compact s p a c e s .

(67.14)

(Category theorem).

I f { f n } c Ra(X)+ and f = V n f n ,

then f i s l e a n .

Proof.

I t s u f f i c e s t o show t h a t V n u ( f n ) i s l e a n , s o , f o r

Chapter 1 5

356

s i m p l i c i t y , we can t a k e t u n } c Ra(X)+ and f pose f i s n o t l e a n .

=

vnun.

Now s u p -

Then t h e r e e x i s t s g EC(X) s u c h t h a t 0 < g < f.

Again f o r s i m p l i c i t y , we c a n assume IIgll > 1 . S e t vn

=

unAg ( n

=

Then t h e v n t s a r e u s c e l e -

1,2,..-).

m e n t s , Cvnl c Ra(X)+, and g

=

Vnvn.

h n } c C(X)+ s a t i s f y i n g :

. . .) ; for all n

We remark f i r s t t h a t , s i n c e IIgll > 1, ( g - l ( X ) )

Hence,

0.

s i n c e ( g - n ( X ) ) E C(X), no l e a n e l e m e n t c a n dominate i t . We o b t a i n t h e h n ' s i n -

We p r o c e e d t o e s t a b l i s h t h e Lemma. ductively.

By t h e above r e m a r k , v 1 i- ( g - IL(X)).

Now v1 i s

t h e infimum o f a l l t h e e l e m e n t s o f C ( X ) between i t and g ; hence t h e r e e x i s t s hlEC(X)

such t h a t

v1 hl

5

h l 5 g9 (g -

n(x) 1 -

That i s , hl s a t i s f i e s t h e Lemma. Assume h l , - . - , h n have b e e n c h o s e n t o s a t i s f y t h e Lemma.

To e s t a b l i s h t h i s , we show t h e s t r o n g e r i n e q u a l i t y : Vn+l

(g

(g - l ( X )

- n(X>) - hn)

Suppose hn

+

Vn+l

>

(g - n ( X ) ) .

hn

+

Write t h i s

t h e l e f t s i d e i s i n C(X) and t h e 5 v ~ + ~Here .

r i g h t s i d e i s l e a n , hence ( g - n ( X )

- hn)

5 0 , o r hn 2 Cg

- 1(X)).

T h i s c o n t r a d i c t s t h e i n d u c t i o n h y p o t h e s i s t h a t hn s a t i s f i e s ( i i i )

357

The R a r e E l e m e n t s i n t h e Lemma.

I t f o l l o w s f r om ( * ) ( a s i n t h e c h o i c e o f h l ) t h a t t h e r e exists

EC(X) such t h a t

hnvvn+1 5 h n + l 5 g , hn+l b

s

-

n(x)),

w hic h e s t a b l i s h e s t h e L e m m a . But t h e n h n + g , s o

by D i n i ' s t h e o r e m , limnllg - hnll

=

0.

A ga in t h i s c o n t r a d i c t s ( i i i ) i n t h e L e m m a .

QED We h a v e s e e n ( 6 5 . 1 ) t h a t u ( f ) l e a n ( a l t h o u g h tE(f)

,

-

f and f

i n general, is not).

-

t ( f ) are always

If o n e o f t h e s e i s

r a r e , we c a n s a y much m o r e :

(67.15) For f E C " ( X ) ,

t h e following are e q u i v a l e n t :

1'

u(f)

2'

f - t ( f ) i s rare;

3'

6(f) is rare;

4'

6(f) is lean.

P roof. -

-

f is rare;

If one of u ( f )

-

f and f - k ( f ) i s rare, then,

s i n c e t h e o t h e r i s l e a n , i t f o l l o w s from ( 6 7 . 4 ) t h a t 6 ( f ) i s l e a n , hence (being a usc element) rare. imply

4O,

w h i c h i m p l i e s 3'.

3'

Thus 1' a n d 2'

o f c o u r s e i m p l i e s '1

each

a n d 2'.

QED A t r i v i a l c o r o l l a r y i s t h a t f o r a u s c e l e m e n t u a n d a n Lsc

Chapter 1 5

358

element R ,

6 ( u ) and 6 ( k ) a r e r a r e .

For e v e r y f E S(X) , 6 ( f ) E Ra(X).

( 6 7 . 1 6 ) Theorem.

Proof.

We have a s t r o n g e r r e s u l t :

Suppose f i r s t t h a t f E s ( X ) , t h a t i s , f

where u1 and u 2 a r e u s c e l e m e n t s . 6(ul)

Then, by ( S O . Z ) ,

6(u2) E Ra(X) ( c f . t h e above r e m a r k ) .

+

holds f o r s(X).

=

u

1 - u2

0 < 6(f) <

Thus t h e theorem

That i t h o l d s f o r i t s norm c l o s u r e S ( X ) f o l l o w s

from t h e f a c t s t h a t t h e o p e r a t i o n 6(.)

i s norm c o n t i n u o u s ( 5 0 . 1 0 )

and R a ( X ) i s norm c l o s e d . QED

( 6 7 . 1 7 ) C o r o l l a r y 1, 1'

f is rare;

2'

f i s lean;

3'

u(f) is lean;

4'

k(f)

Proof. u(f) = f Thus '3

+

For f E S ( X ) , t h e f o l l o w i n g a r e e q u i v a l e n t :

is lean.

'1

o f c o u r s e i m p l i e s 2'.

Suppose 2'

holds.

Then

( u ( f ) - f ) , hence i s l e a n by ( 6 7 . 1 6 ) above and ( 6 7 . 4 ) .

holds.

hence i s r a r e .

Suppose 3'

holds.

> 0 by ( 6 5 . 2 ) , Then u ( f ) -

Since S ( f ) i s a l s o r a r e ( 6 7 . 1 6 ) ,

L(f)

=

u(f) - 6(f) is rare.

k(f)

5

0 by ( 6 5 . 2 ) ,

F i n a l l y , suppose 4'

hence - k ( f )

therefore rare, so k(f) is rare.

i t follows

holds.

Then

i s a p o s i t i v e l e a n usc element,

Since f - k ( f ) i s a l s o r a r e ,

The R a r e E l e m e n t s by ( 6 7 . 1 6 ) , i t f o l l o w s f

=

L(f)

+

(f

-

359

!,(f)) i s r a r e . QED

For a g e n e r a l e l e m e n t f o f C t l ( X ) ,

a u ( f ) a n d u a ( f ) h a v e no

particular order relation, that is, a l l possible order relations occur.

However:

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

Proof.

For f E S ( X ) ,

QU(f)

-

uf,(f).

u ( f ) - t ( f ) is l e a n , by (67.16),

so a ( u ( f ) - k(f)) = O .

Applying (49.2) g i v e s us a u ( f ) - uL(f) < 0.

QED

We c l o s e t h i s 5 w i t h t h e p r o m i s e d p r o o f o f ( * ) i n ( 6 6 . 2 ) . 0 < u - gvu, hence u i s l e a n , and t h e r e f o r e (being a p o s i t i v e U

S

~e l e m e n t )

rare.

I t f o l l o w s from t h e C a t e g o r y theorem (67.14)

t h a t I i s contained i n the s e t of lean elements.

Thus g i s l e a n .

QED

568. A " l o c a l i z a t i o n " t h e o r e m

The p r o p e r t i e s e s t a b l i s h e d i n 566 f o r l e a n e l e m e n t s h o l d , w i t h one e x c e p t i o n , f o r r a r e elements.

Note t h a t w e h a v e b e e n

f o r c e d t o change t h e o r d e r of development.

360

Chapter 1 5

(68.1) Theorem. f o r fEC"(X), fAR,

L e t {La}

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

rare f o r a l l

c1

i m p l i e s t h a t Va(fARa)

Then

i s rare.

A g a i n , we p r o v e t h e f o l l o w i n g e q u i v a l e n t t h e o r e m i n s t e a d .

L e t {La} b e a c o l l e c t i o n o f Rsc e l e m e n t s , a n d g,

= VIJP,il.

I f fAk,x i s r a r e f o r a l l u , t h e n fnil i s r a r e .

P roof.

Again, f o r s i m p l i c i t y , w e c a n assume t h a t f < 8 . ~

We show i ? u ( f ) < 0 < uR(f)

(cf. (67.2)).

Ru(f) < 0: W e f i r s t p r o v e t h a t i t s u f f i c e s t o show P u ( f ) A Rc 0.. Assume Ru(f)AR < 0. au(f)Au(R) < 0. comes R u ( f ) < 0.

Then u ( k u ( f ) A L ) < 0 , s o by ( 4 9 . 6 ) ,

But f < R , so Ru(f) < u(k), so t h i s last beNow t o p r o v e k u ( f ) A k < 0. ku(f)AR

(the l a s t inequality

Ru(f)A(VaRa) = Vcx(Ru(f)Aila) < Vclku(fARa) f o l l o w i n g from ( 4 9 . 4 )

and ( 4 9 . 6 ) ) .

=

S i n c e LU(fAka) < 0 for

a l l a , Vaku(fARm) 5 0 , a n d we a r e t h r o u g h . uR(f)

0 : Choose cto a r b i t r a r i l y .

Then u k ( f ) > v R ( f A R c l )-> 0 . 0

QED

A g a i n we h a v e t h e c o r o l l a r i e s :

( 6 8 . 2 ) C o r o l l a r y 1.

Let

{Ia} b e a c o l l e c t i o n o f R - b a n d s a n d I

t h e b a n d w i t h I l ( X ) I = Vall(X)I implies t h a t f I i s rare.

a

.

Then f I

a

rare f o r a l l a

The R a t e Elements (68.3) Corollary 2.

For e a c h f E C " ( X ) ,

361

there exists a largest

t - b a n d I f o r which f I i s r a r e .

Again ( 6 8 . 3 ) and ( 6 7 . 1 2 ) r e d u c e t o t h e same t h e o -

Remark.

rem i n t o p o l o g y . ( 6 6 . 1 ) h o l d s w i t h " l e a n " r e p l a c e d by " r a r e " .

We r e c o r d

the f a c t :

( 6 8 . 4 ) I f fAg i s r a r e , t h e n s o a r e f+Ag. f - A g ,

IflAg,

If I A I g l ,

and f - .

The p r o o f i s t h e same. (66.2) a l s o c a r r i e s over t o r a r e n e s s , but only f o r L > 0:

( 6 8 . 5 ) Given a n Rsc e l e m e n t il > 0 , l e t I b e t h e band which i t generates.

P roof. -

Then f o r f € 1 , i f fAR i s r a r e , f i s r a r e .

By ( 6 8 . 4 ) , ) f l A L i s r a r e , s o f o r s i m p l i c i t y , we c a n

assume f > 0.

For e v e r y n E N , 0 < fAnR < n ( f A k ) , s o fAna i s

rare.

=

Since f

Vn(fAnL), i t f o l l o w s from ( 6 8 . 1 ) t h a t f i s r a r e QED

The a b o v e d o e s n o t h o l d € o r a g e n e r a l Lsc e l e m e n t R : be a r e a l i n t e r v a l and { r n } t h e r a t i o n a l p o i n t s o f X .

Let X

L e t f be

36 2

Chapter 1 5

t h e c h a r a c t e r i s t i c element of t h e s e t { r n l , and u t h e u s c e l e ment d e f i n e d by ( u , r n )

wise.

Finally, l e t R

by R a n d fAL

= =

l / n (n

-u.

=

and ( u , x )

=

0 other-

Then f i s i n t h e b a n d g e n e r a t e d But u ( f ) = n ( X ) , s o f i s n o t

R , w hi ch i s r a r e .

=

1,2,...)

rare. A s with l e a n elements ( c f .

( 6 6 . 6 ) ) , t h e f o l l o w i n g theorem

i s analogous t o (68.1), and while they a r e a c t u a l l y d i f f e r e n t , t h e y r e d u c e t o t h e same t h e o r e m i n t o p o l o g y .

( 6 8 . 6 ) Theorem.

u

= Aauu.

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

t h a t ( f - u)'

Proof. Aauu =

u.

Let {uul be a c o l l e c t i o n of usc elements, and (f

-

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

i s rare.

By ( 6 7 . 8 ) , R u ( f ) Thus ( f - u)'

5 ua f o r a l l a, h e n c e Ru(f)

< ( f - Ru(f))'

<

<

u ( f ) - k u ( f ) E Ra(X)

( t h i s l a s t by ( 6 7 . 1 6 ) . QED