A Short Course on Banach Space Theory

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A Short Course on Banach Space Theory N. L. Carothers Department of Mathematics and Statistics Bowling Green State University Summer 2000

ii

Contents 1 Classical Banach Spaces

The sequence spaces p and 0 Finite dimensional spaces . . . The p spaces . . . . . . . . . The ( ) spaces . . . . . . . Hilbert space . . . . . . . . . \Neo-classical" spaces . . . . The big questions . . . . . . . Notes and Remarks . . . . . . Exercises . . . . . . . . . . . . . . . `

c

L

C K

2 Preliminaries

Continuous linear operators . Finite dimensional spaces . . . Continuous linear functionals Adjoints . . . . . . . . . . . . Projections . . . . . . . . . . Quotients . . . . . . . . . . . A curious application . . . . . Notes and Remarks . . . . . . Exercises . . . . . . . . . . . . . . .

3 Bases in Banach Spaces Schauder's basis for The Haar system . Notes and Remarks Exercises . . . . . . . . .

[0 1] ..... ..... .....

C

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4 Bases in Banach Spaces II

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A wealth of basic sequences . . . . . . . . . . . . . . . . . . . iii

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1 2 3 5 7 8 8 9 11

13 13 14 16 17 18 19 22 23 24

27 31 33 35 37

39 39

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CONTENTS Disjointly supported sequences in Lp and `p Equivalent bases . . . . . . . . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . .

5

Bases in Banach Spaces III

6

Special Properties of

Block basic sequences . . . . . . . . . Subspaces of `p and c0 . . . . . . . . Complemented subspaces of `p and c0 Notes and Remarks . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . .

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c0 , `1 , and `1 True stories about `1 . . . . . . . . . . . . . . . . . . . . . . . The secret life of `1 . . . . . . . . . . . . . . . . . . . . . . . Confessions of c0 . . . . . . . . . . . . . . . . . . . . . . . . .

40 43 46 48 49

49 52 54 56 58

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Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59 64 68 69 71

Bases and Duality

73

8

Lp

Spaces

81

9

Lp

Spaces II

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Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Basic inequalities . . . . . . . . . . . . . Convex functions and Jensen's inequality A test for disjointness . . . . . . . . . . . Conditional expectation . . . . . . . . . Notes and Remarks . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . The Rademacher functions . . . Khinchine's inequality . . . . . The Kadec-Pelczynski theorem Notes and Remarks . . . . . . . Exercises . . . . . . . . . . . . . . . .

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78 79

81 82 85 87 90 92

93

. 93 . 95 . 99 . 105 . 107

CONTENTS

10 Lp Spaces III

Unconditional convergence Orlicz's theorem . . . . . . Notes and Remarks . . . . Exercises . . . . . . . . . . . . .

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Strict convexity . . . . . . . . . . . Nearest points . . . . . . . . . . . . Smoothness . . . . . . . . . . . . . Uniform convexity . . . . . . . . . Clarkson's inequalities . . . . . . . An elementary proof that Lp = Lq . Notes and Remarks . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . .

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11 Convexity

12 C (K ) Spaces

The Cantor set . . . . . . Completely regular spaces Notes and Remarks . . . . Exercises . . . . . . . . . . . . .

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109 109 111 116 117

119 120 124 125 127 129 132 134 135

137 137 138 147 148

13 Weak Compactness in L1

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14 The Dunford-Pettis Property

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Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . 154 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . 162 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

15 C (K ) Spaces II

The Stone-C ech compacti cation Return to C (K ). . . . . . . . . . Notes and Remarks . . . . . . . . Exercises . . . . . . . . . . . . . . . . .

16 C (K ) Spaces III

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The Stone-C ech compacti cation of a discrete space A few facts about N . . . . . . . . . . . . . . . . . \Topological" measure theory . . . . . . . . . . . . The dual of `1 . . . . . . . . . . . . . . . . . . . .

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175 175 176 178 180

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CONTENTS The Riesz representation theorem for C ( D) . . . . . . . . . . 182 Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . 184 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

A Topology Review

Separation . . . . . . . . . . . . . Locally compact Hausdor spaces Weak topologies . . . . . . . . . . Product spaces . . . . . . . . . . Nets . . . . . . . . . . . . . . . . Notes and Remarks . . . . . . . .

References Index

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187 188 188 190 191 192 194

195 203

Chapter 1 Classical Banach Spaces To begin, recall that a Banach space is a complete normed linear space. That is, a Banach space is a normed vector space ( k
k) which is a complete metric space under the induced metric ( ) = k k. Unless otherwise speci ed, we'll assume that all vector spaces are over R, although from time to time we will have occasion to consider vector spaces over C . What follows is a list of the classical Banach spaces. Roughly translated, this means the spaces known to Banach. Once we have these examples out in the open, we'll have plenty of time to ll in any unexplained terminology. For now, just let the words wash over you. X;

d x; y

x

y

The sequence spaces `p and c0 Arguably the rst in nite dimensional Banach spaces to be studied were the sequence spaces p and 0. To consolidate notation, we rst de ne the vector space of all real sequences = ( n), and then de ne various subspaces of . For each 1  1 we de ne !1=p 1 X k kp = j n jp `

s

c

x

x

s

p
0, show there exists an N such that 1 K . That is, if K is compact, n N x; en en < " for every x then these \tail series" can be made uniformly small over K . k
k

7!

7!







k

k

k k



k k

k
k

=1

k

k

1

=1

(1)

!

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1

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k k

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(1

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=

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12

CHAPTER 1. CLASSICAL BANACH SPACES

Chapter 2 Preliminaries We begin with a brief summary of important facts from functional analysis; some with proofs, some without. Throughout, , , etc., are normed linear spaces over R. If there is no danger of confusion, we will use k
k to denote the norm in any given normed space; if two or more spaces enter into the discussion, we will use k
kX , etc., to further identify the norm in question. X

Y

Continuous linear operators Given a linear map : T

(i) (ii) (iii) (iv)

T T

X

! , recall that the following are equivalent: Y

is continuous at 0 2 . is continuous. X

is uniformly continuous. is Lipschitz; that is, there exists a constant kY  k kX for all , 2 . (v) is bounded ; that is, there exists a constant k kX for all 2 . T

T

Ty

C

x

y

x

y

T

C

C
0, choose y , y 2 M with kx + y kX  kq (x)kX=M + " and kx + y kX  kq (x )kX=M + ". Then, since y + y 2 M , we get kq(x) + q(x )kX=M = kq(x + x )kX=M  k(x + x ) + (y + y )kX = k(x + y) + (x + y )kX  kq(x)kX=M + kq(x )kX=M + 2": 0

0

0

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0

0

0

0

0

0

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0

0

What we've actually done here is to take the quotient topology on X=M induced by q; that is, the smallest topology on X=M making q continuous.

21 o Indeed, check that q(BXo ) = BX=M , where BXo denotes the open unit ball in X . Thus, a set is open in X=M if and only if it is the image under q of an open set in X . In particular, note that q is an open map. We should also address the question of when X=M is complete. If X is complete, and if M is closed (hence complete), then it's not terribly hard to check that X=MPis also complete. Indeed, suppose that (xn) is a sequence in X for which 1 n, choose yn 2 M such n=1 kq (xn )kX=M < 1. For each P 1 kx + y k < 1 and n . Then, that kxP + y k  k q ( x ) k + 2 n n X n X=M n P n X n=1 1 (x + y ) converges in X . Thus, by continuity, 1 q (x + y ) = hence, n n n n n=1 P1n=1 q(xnn=1) converges in X=M . Since we're using the quotient topology, it's easy to check that a linear map S : X=M ! Y is continuous if and only if Sq is continuous. Said in other words, the continuous linear maps on X=M come from continuous linear maps on X which \factor through" X=M . That is, if T : X ! Y satis es ker T  M , then there exists a (unique) linear map S : X=M ! Y which satis es Sq = T and kS k = kT k. In the special case where T maps a Banach space X onto a Banach space Y and M = ker T , it follows that the map S is one-to-one; thus, X=(ker T ) is isomorphic to Y . Finally, it's a natural question to ask whether the quotient space X=M is actually isomorphic to a subspace of X . Now, in the vector space setting, it's easy to see that if N is an algebraic complement of M in X , then X=M can be identi ed with N . Thus, it should come as no surprise that X=M is isomorphic to a subspace of X whenever M is complemented in X . If X is a Banach space with X = M  N and if we write Q : X ! X for the projection with kernel M , then our earlier observations show that X=(ker Q) = X=M is isomorphic to N . Conversely, if the quotient map q : X ! X=M is an isomorphism on some closed subspace N of X , then Q = (q jN ) 1q, considered as a map from X to X , de nes a projection with range N . In the special case of linear functionals, these observations tell us that (X=M ) = M ? (the annihilator of M in X ). That is, the dual of X=M can be identi ed with those functionals in X  which vanish on M . Indeed, if f : X=M ! R is a continuous linear functional, then g = fq : X ! R is a continuous linear functional that vanishes on M and satis es kgk = kf k. Conversely, if g : X ! R is continuous, linear, and satis es ker g  M , then, in fact, ker g = M (since they're both codimension one) and, hence, there exists a (unique) continuous, linear functional f : X=M ! R satisfying g = fq and kgk = kf k. In either case, it's not hard to see that the correspondence f $ g is linear and, hence, an isometry between (X=M ) and M ? . Alternatively,

CHAPTER 2. PRELIMINARIES

22

check that the adjoint of the quotient map : ! is an isometry from   ? ( ) into with range . A similar observation is that  =  ? . In other words, each continuous linear functionals on can be considered as a functional on (thanks to Hahn-Banach); those functionals in  which agree on , that is, functionals which dier by an element of ? , are simply identi ed for the purposes of computing  . Alternatively, the adjoint of the inclusion : ! is a (quotient) map from  onto  with kernel ? (in fact,  is the restriction map, ( ) = jM ). Thus,  =  ? . Lastly, if is a Banach space with =  , then  is isomorphic to   . Speci cally, if : ! is the projection with range and kernel , it's not hard to see that  :  !  is again a projection with range ? and kernel ? ; that is,  = ?  ? . Thus,  =  ? is isomorphic to ? and, likewise,  is isomorphic to ? . q

X=M

X

X

X=M

M

M

X =M

M

X

X

M

M

M

i

X

i

f

f

M

M

M

X

N

P

X

N

i

M

X

N

N

M

N

X

X

P

N

X

X =M

X

M

M

M

X

N

X

M

M

X =M

M

A curious application We round out our discussion of preliminary topics by presenting a curious result, due to Dixmier [33], which highlights many of the ideas from this chapter.

Theorem 2.2 If :

! embeddings, then  :  !

and j : X  ! X  denote the canonical  is a projection with range isometric to X  ji X X and kernel isometric to i(X )? , the annihilator of i(X ) in X . In short, X  is always complemented in X . i

X

X



Proof. Note that i : X  ! X  and, hence, i j : X  ! X  . To show that j i is a projection, it suces to show that ij is the identity on X  , for then we would have (j i)(j i) = j (ij )i = j i; that is, (j i)2 = j i. So, given x 2 X , let's compute the action of ij (x) on a typical x 2 X : (ij (x))(x) = h x; ij (x) i = h i(x); j (x) i = h x; i(x) i = h x; x i = x(x): Thus ij (x) = x and, hence, ij is the identity on X . It's not hard to see that i is onto, thus the range of j i is j (X ), which is plainly isometric to X . Along similar lines, since j is one-to-one, it's easy

23 to see that ker(ji) = ker i. Finally, general principles tell us that ker i = (range i)? = i(X )?, the annihilator of i(X ) in X .

Notes and Remarks There are many excellent books on functional analysis that oer more complete details for the topics in this chapter. See, for example, Bollobas [17], Conway [24], DeVito [28], Dunford and Schwartz [38], Holmes [62], Jameson [68], Jarvinen [70], Megginson [90], Royden [118], Rudin [119], or Yosida [136]. Megginson's book, in particular, has a great deal on adjoints, projections, and quotients.

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CHAPTER 2. PRELIMINARIES

Exercises Given normed spaces X and Y , we write B (X; Y ) to denote the space of bounded linear operators T : X ! Y endowed with the operator norm (2.1). 1. Given a nonzero vector x in a normed space X , show that there exists a norm one functional f 2 X  satisfying jf (x)j = kxk. On the other hand, give an example of a normed space X and a norm one linear functional f 2 X  such that jf (x)j < kxk for every 0 6= x 2 X . 2. Let Y be a subspace of a normed linear space X and let T 2 B (Y; Rn). Prove that T extends to a map T~ 2 B (X; Rn) with kT~k = kT k. [Hint: Hahn-Banach] 3. Prove that every proper subspace M of a normed space X has empty interior. If M is a nite dimensional subspace of an in nite dimensional normed space X , conclude that M is nowhere dense in X . 4. Prove that B (X; Y ) is complete whenever Y is complete. 5. If Y is a dense linear subspace of a normed space X , show that Y  = X , isometrically. 6. Prove Riesz's lemma: Given a closed subspace Y of a normed space X and an " > 0, there is a norm one vector x 2 X such that kx yk > 1 " for all y 2 Y . If X is in nite dimensional, use Riesz's lemma to construct a sequence of norm one vectors (xn) in X satisfying kxn xmk  1=2 for all n 6= m. 7. Given linear functionals f and (gi )niP =1 on a vector space, prove that T n ker f  i=1 ker gi if and only if f = ni=1 aigi for some a1; : : :; an 2 R. 8. Given T 2 B (X; Y ), show that ker T = ?(range T ), the annihilator of range T  in X , and that ker T  = (range T )?, the annihilator of range T in Y . 9. Given T 2 B (X; Y ), show that T  is an extension of T to X  in the following sense: If i : X ! X  and j : Y ! Y  denote the canonical embeddings, prove that T (i(x)) = j (T (x)). In short, T (^x) = Tcx. 10. Let S be a dense linear subspace of a Banach space X , and let T : S ! Y be a continuous linear map, where Y is also a Banach space. Show that T extends uniquely to a continuous linear map T~ : X ! Y , de ned on all of X , and that kT~k = kT k. Moreover, if T is an isometry, show that T~ is again an isometry.

25 11. Let (X k
k) be a Banach space, and suppose that jjj
jjj is another norm on X satisfying jjj x jjj  kxk for every x 2 X . If (X; jjj
jjj) is complete, prove that there is a constant c > 0 such that c kxk  jjj x jjj for every x 2 X. 12. Let M = f(x; 0) : x 2 Rg  R2. Show that there are uncountably many subspaces N of R2 such that R2 = M  N . 13. Let M be a nite dimensional subspace of a normed linear space X . Show that there is a closed subspace N of X with X = M  N . In fact, if M is non-trivial, then there are in nitely many distinct choices for N . [Hint: Given a basis x1; : : : ; x for M , nd f1; : : : ; f 2 X  with f (x ) =  .] 14. Let M and N be closed subspaces of a Banach space X with M \ N = f0g. Prove that M + N is closed in X if and only if there is a constant C < 1 such that kxk  C kx + y k for every x 2 M , y 2 N . 15. Let M be a closed, nite codimensional subspace of a normed space X . Show that there is a closed subspace N of X with X = M  N . 16. Let M and N be closed subspaces of a normed space X , each having the same nite codimension. Show that M and N are isomorphic. 17. Let P : X ! X be a continuous linear projection with range Y , and let Q : X ! X be continuous and linear. If Q satis es P Q = Q and QP = P , show that Q is also a projection with range Y . 18. A bounded linear map U : X ! X is called an involution if U 2 = I . If U is an involution, show that P = 21 (U + I ) is a projection. Conversely, if P : X ! X is a bounded projection, then U = 2P I is an involution. In either case, P and U x the same closed subspace Y = fx : P x = xg = fx : U x = xg. 19. A projection P on a Hilbert space H is said to be an orthogonal projection if range P is the orthogonal complement of ker P ; that is, if and only if H = (ker P )  (range P ). Prove that P is an orthogonal projection if and only if P is self adjoint ; that is, if and only if P  = P . 20. Let M be a closed subspace of a normed space X . If both M and X=M are Banach spaces, then so is X . We might say that completeness is a \three space property" (if it holds in any two of the spaces X , M , or X=M , then it also holds in the third). Is separability, for example, a three space property? 21. Let T : X ! Y be a bounded linear map from a normed space X onto a normed space Y . If M is a closed subspace of ker T , then there is a n

i

j

i;j

n

CHAPTER 2. PRELIMINARIES

26

(unique) bounded linear map e : ! such that = e , where is the quotient map. Moreover, k ek = k k. If and are Banach spaces, and if : ! is a bounded linear map onto all of , then ker is isomorphic to . Let and be Banach spaces, and let 2 ( ). Then, the following are equivalent: (i) ker is isomorphic to range . (ii) range is closed in . (iii) There is a constant 1 such that inf fk k : 2 ker g  k k for all 2 . Let be a closed subspace of a Banach space and let : ! o be the quotient map. Prove that ( Xo ) = X=M , where Xo is the open o unit ball in and X=M is the open unit ball in . We say that 2 ( ) is a quotient map if ( Xo ) = Yo , where Xo denotes the open unit ball in (respectively, ). Prove that is a quotient map if and only if ker is isometric to . Given 2 ( ), prove that is a quotient map if and only if  is an isometry (into). Let be a closed subspace of a Banach space and let : ! denote the quotient map. Prove that  is an isometry from ( ) into  with range ? . Thus ( ) can be identi ed with ? . Let be a closed subspace of a Banach space and let : ! denote the inclusion map. Prove that  is a quotient map from  onto  with kernel ? . Conclude that  can be identi ed with  ?. Let be a closed subspace of a normed space . For any 2 , show that minfk k : 2 ? g = supfj ( )j : 2 k k  1g. (Please note the use of \min" in place of \inf".) T

X=M

T

22.

X

T

X

X=

X

T

B X; Y

T

Y

C
0, let s = mi=1 aixi 2 S with kx sk < ". Then, for n > m, kx Pn xk  kx sk + ks Pn sk + kPn s Pn xk  (1 + kPnk)
"  (K + 1)
": We conclude this rst pass at bases in Banach spaces with two classical examples, both due to J. Schauder [121, 122]. Schauder's basis for C [ 0; 1 ]

Schauder began the formal theory of bases in Banach spaces in 1927 by oering up a basis for C [ 0; 1 ] that now bears his name. Rather than try to give an analytical de nition of the sequence, consider the following pictures: 1

f2 f0 f1 1

1

f3

1/2

f5

f4

1/2

1/4

CHAPTER 3. BASES IN BANACH SPACES

32

If we enumerate the dyadic rationals in the order t0 = 0, t1 = 1, t2 = 1=2, t3 = 1=4, t4 = 3=4, and so on, notice that we have fn (tn) = 1 and fk (tn) = 0 for k > n. This easily implies that the fn are linearly independent. Moreover, it's now easy to see that span(f0; : : : ; f2m ) is the set of all continuous, piecewise linear or \polygonal" functions with \nodes" at the dyadic rationals k=2m , k = 0; 1; : : : ; 2m . Indeed, the set of all such polygonal functions is clearly a vector space of dimension 2m+1 which contains the 2m+1 linearly independent functions fk , k = 0; : : : ; 2m . Thus, the two spaces must coincide. Since the dyadic rationals are dense in [ 0; 1 ], it's not hard to see that the fn have dense linear span. Thus, (fn) is a viable candidate for a basis for C [ 0; 1 ]. P If we set pn = nk=0 ak fk , then pn 1 = 0max p (t ) ; k n n k k

k

j

j

because P pn is a polygonal function with nodes at t0; : : :; tn. And if we set pm = mk=0 ak fk for m > n, then we have pm (tk ) = pn (tk ) for k n, because fj (tk ) = 0 for j > n k. Hence, pn 1 pm 1. This implies that (fn ) is a normalized basis for C [ 0; 1 ] with basis constant K = 1. Consequently, eachP f C [ 0; 1 ] can be (uniquely) written as a uniformly P 1 convergent series f = 1 a f . But notice, please, that k=0 k k k=n+1 ak fk vanP n ishes at each of the nodes t0; : : : ; tn. Thus, Pn f = k=0 ak fk must agree with f at t0; : : : ; tn; that is, Pn f is the interpolating polygonal approximation to f with nodes at t0; : : : ; tn. Clearly, Pnf 1 f 1 . 



k

k

 k

k

2

k

k

 k

k

f

a2

a1

a0

f = a0f0 + a1f1 + a2f2 + It's tempting to imagine that the linearly independent functions tn, n = 0; 1; 2; : : :, might form a basis for C [ 0; 1 ]. After all, the Weierstrass theorem tells us that the linear span of these functions is dense in C [ 0; 1 ]. But a moment's re ection will convince you that not every function in C [ 0; 1 ] has a uniformly convergent power series expansion; your favorite function that is not




33 dierentiable at 0, for example. Nevertheless, as we'll see in the next chapter, C [ 0; 1 ] does admit a basis consisting entirely of polynomials.

The Haar system The Haar system (hn )1 n=0 on [ 0; 1 ] is de ned by h0  1 and, for k = 0; 1; 2; : : : k and i = 0; 1; : : : ; 2 1, by h2k +i (x) = 1 for (2i 2)=2k+1  x < (2i 1)=2k+1 , h2k +i(x) = 1 for (2i 1)=2k+1  x < 2i=2k+1 , and h2k +i(x) = 0 otherwise. A picture might help: 1 h1

h0

h2

h3

1/2

1

1/2

–1

As we'll see, the Haar system is an orthogonal basis for L2[ 0; 1 ]. Each hn, n  1, is mean-zero and, more generally, hn
hm is either 0 or hm for any n < m. In particular, the hn are linearly independent. Note that the R x Schauder system is related to the Haar system by the formula n 1 fn (x) = 2 0 hn 1 (t) dt for n  1 (and f0  1). In fact, it was Schauder who proved that the Haar system forms a monotone basis for Lp[ 0; 1 ], for any 1  p < 1. While we could give an elementary proof, very similar to the one we used for Schauder's basis (see Exercise 3), it might be entertaining to give a slightly fancier proof. The proof we'll give borrows a small amount of terminology from probability. For each k = 0; 1; : : :, let Ak = f [(i 1)=2k+1 ; i=2k+1 ) : i = 1; : : :; 2k+1 g. Claim: The linear span of h0; : : :; h2k+1 1 is the set of all step functions based on the intervals in Ak . That is, spanf h0; : : :; h2k+1 1 g = spanf I : I 2 Ak g: Why? Well, clearly each hj 2 spanf I : I 2 Ak g for j < 2k+1 , and spanf I : I 2 Ak g has dimension 2k+1. Thus the two spaces must coincide. But it should be pointed out here that it's essential that we take 2k+1 functions at a time! The claim won't be true if we consider an arbitrary batch h0; : : : ; hm.

CHAPTER 3. BASES IN BANACH SPACES

34

This allows us to use the much simpler functions I in place of the Haar functions in certain arguments. For example, it's now very easy to see that the hn have dense linear span in Lp[ 0; 1 ] for any 1 p < . Also, please note that the set I : I Ak is again orthogonal in L2[ 0; 1 ] (although not the full set of I for all dyadic intervals I ). In particular, if Pk is the orthogonal projection onto span h0; : : :; h2k+1 1 , and if Qk is the orthogonal projection onto span I : I Ak , then we must have Pk f = Qk f . The savings here is that Qk f is very easy to compute. Indeed, X f; m(I ) 1=2I m(I ) 1=2I P k f = Qk f = 

2

f

1

g

f

2

f

I 2Ak

=

h

g

i




X

g

1 Z f m(I ) I

I 2Ak




I :

Thus, Pk f is the conditional expectation of f given k , the -algebra generated R by A . That is, P f is the unique  -measurable function satisfying k k k Af = R P f for all A  . k A k Now let's see why Pk is a contraction on every Lp. First consider f L1[ 0; 1 ]: Z 1 Z X 1 PK f = f m(I ) m(I ) 0 I I 2Ak 2

2

j

j




Z

X 

Now, for 1 < p
0, choose a sequence of positive numbers ("n) with 1n=1(1 + "n)  1 + ". We next construct a basic sequence, inductively, by repeated application of Mazur's lemma. To begin, choose any norm one x1 2 X . Now, choose a norm one vector x2 so that kyk  (1 + "1) ky + x2k for all y 2 [ x1 ] and all scalars . Next, choose a vector x3 of norm one so that kyk  (1 + "2) ky + x3k for all y 2 [ x1; x2 ] and all scalars . Choose x4 so that. . . . Well, you get the picture. The sequence (xn) so constructed is a basic sequence with basis constant Q 1 at most K = n=1(1 + "n)  1 + ".

Disjointly supported sequences in Lp and `p In preparation for later, let's consider an easy source for basic sequences: Disjointly supported sequences in Lp, `p, or c0. In case it's not clear, the support of a function f 2 Lp() is the set ff 6= 0g. Two functions f , g 2 Lp() are thus disjointly supported if ff 6= 0g \ fg 6= 0g = ?; that is, if f
g = 0. In `p or c0 this reads: x = (xn ) and y = (yn ) are disjointly supported if xn yn = 0 for all n.

41

Lemma 4.3 Let (fn ) be a sequence of disjointly supported nonzero vectors in Lp(), 1  p < 1. Then (fn ) is a basic sequence in Lp(). Moreover, [ fn ] is isometric to `p and is complemented in Lp () by a norm one projection.

Proof. The linear span of (fn ) is unaected if we replace each fn by fn =kfn kp. Thus, we may assume that each fn is norm one. Next, since the fk are disjointly supported we have:

m

X



k =n

p

k

ak f

p

Z m X

=

k =n

p k

m X

ak f d =

k =n

jak

jp

Z

jfk

jp d

=

m X k =n

jak jp;

P1

for any P scalars (ak ). This tells us that n=1 anfn converges in Lp() if and p only if 1 n=1 jan j < 1. (Why?) Thus, the map en 7! fn extends to a linear isometry from `p onto [ fn ]. In particular, it follows that (fn) is a basic sequence in Lp(). Lastly, the existence of a projection is easy. A sequence (gn ), biorthogonal to (fn), is clearly given by gn = (sgn fn ) jfnjp 1. The gn are disjointly supported, norm one functions in Lq (), where q is the conjugate exponent to p. Moreover, gn has the same support as fn , namely An = fgn 6= 0g = ffn 6= 0g. Now consider:

Pf =

1 X n=1

hf; gnifn =

1 X

Z

An

n=1



fgn fn :

Clearly, P is the identity on [ fn ]. To show that P is norm one, we essentially repeat our rst calculation:

kPf kpp

=

1 Z X n=1

An

fgn

p



1 X

Z

n=1 An

jf jp  kf kpp

(the inequality coming from Holder's inequality). It's also true that a disjointly supported sequence in c0 spans an isometric copy of c0. The proof is a simple modi cation of the one we've just given. Rather than repeat the proof in this case, let's settle for pointing out two such modi cations. First, if (yn) is a sequence of disjointly supported norm one P vectors in c0, then 1 n=1 an yn converges in c0 if and only if an ! 0 and, in this case,

1

X

= sup kanynk1 = sup janj: a y

n n

n=1

1

n

n

CHAPTER 4. BASES IN BANACH SPACES II

42

The natural sequence of biorthogonal functionals in this case is simply an appropriately chosen subsequence of (ek ), up to sign. Speci cally, given yn 2 c0, choose kn so that yn attains its norm in its kn -th coordinate; that is, kynk1 = jhyn ; ek ij = hyn ; ek i: The key fact in our last lemma is that k
kpp is additive across sums of disjointly supported functions. But if all we're willing to settle for \isomorphic" in place of \isometric," then all we need is \almost additive" in place of \additive," or \almost disjointly supported" in place of \disjointly supported." This suggests the following generalization: n

n

Lemma 4.4 Let (fn) be a sequence of norm one functions in Lp(). Sup-

there exists a sequence of disjoint, measurable sets (An) such that Rpose that p j f j d < "pn, where "n ! 0 \fast enough." Then (fn ) is a basic sequence Acn n in Lp (). Moreover, [ fn ] is isomorphic to `p and complemented in Lp ().

Proof. From our previous lemma, we know how to deal with the disjointly supported sequence f~n = fnAn . The ideaRhere is that (fn ) is a \small perturbation" of (f~n). By design, kfn f~n kpp = Acn jfnjp < "pn, thus:

X

a f

n n n

anf~n

n p

X



X n

janj kfn f~n kp 

X n

"n janj:

This gives us a hintPas to what \fast enough" should mean: Given 0 < " < 1=2, let's suppose that 1 n=1 "n < ". Then, X n

"njanj 

X n

!

"n
sup janj  "


X

n

n

janjp

!1=p

:

From this, and our previous lemma, it follows that



X

a f

n n n

p



X



anf~n

n

p

+"


X n

janjp

!1=p

 (1 + ")


X n

janjp

!1=p

:

If we can establish a similar lower estimate, we will have shown that [ fn ] is isomorphic to `p . But,

p

X

a f~

n n n p

=

X n

jan

jp

Z An

jfnjp d 

X n

(1 "pn ) janjp;

43 and 1

"pn

 (1



X

a f

n n n

p

"n )p

 (1

")p,



X

~ 

anfn

n

"

p




hence X n

janjp

!1=p

X

 (1 2")


n

janjp

!1=p :

In order to nd a bounded projection onto [ fn ], we now mimic this idea to show that our \best guess" is another \small perturbation" of the projection given in the previous lemma. The map that should work is Pf

=

1  X n=1

kf~nkp p

Z

f (sgn f~n ) jf~n j p

1

 fn :

Rather than give the rest of the details now, we'll save our strength for a more general result that we'll see later.

Equivalent bases Two basic sequences (xn ) and P1 P1(yn ) (in possibly dierent spaces!) are said to be equivalent if i=1 ai xi and i=1 ai yi converge or diverge together. A straightforward application of the Closed Graph theorem allows us to rephrase this condition: (xn) and (yn) are equivalent if there exists a constant 0 < C < 1 such that

1

X

1 C aiyi

i=1

Y



1

X

ax

i=1 i i

X



1

X

C aiyi

i=1

Y

(4.1)

for all scalars (ai). That is, (xn) and (yn ) are equivalent if and only if the basis-to-basis map xi 7! yi extends to an isomorphism between [ xn ] and [ yn ]. Thus we may (and, in practice, will) take (4.1) as the de ning statement. Stated in these terms, the condition takes on new signi cance: If we start with a basic sequence (xn), and if we nd that some sequence (yn ) satis es (4:1), then (yn ) must also be a basic sequence|equivalent to (xn), of course. Indeed, if (xn) has basis constant K , and if (yn) satis es (4:1), then (yn) has basis constant at most C 2K :

n

X

ay

i=1 i i



n

X

ax C

i=1 i i



m

X

ax CK

i=1 i i



m

X

a y : C 2K

i=1 i i

CHAPTER 4. BASES IN BANACH SPACES II

44

As a particular example, a sequence (xn) is equivalent to the usual basis for `p if and only if 1 X

C 1

i=1

jaijp

!1=p



1

X

ax

i=1 i i



C

1 X i=1

X

jaijp

!1=p

(4.2)

for some constant C and all scalars (ai). If (4:1) holds, we sometimes say that (xn ) and (yn ) are C -equivalent. Thus we might say that a disjointly supported sequence of norm one vectors in Lp is 1-equivalent to the usual basis of `p . Or, to paraphrase Lemma 4.4, an \almost disjoint" sequence in Lp is (1 + ")-equivalent to the `p basis. As another example, note that any orthonormal basis in a separable Hilbert space is 1-equivalent to the usual basis of `2. Next, let's generalize the content of Lemma 4.4.

Theorem 4.5 (The Principle of Small Perturbations) Let (xn) be a normalized basic sequence in a Banach space P X with basis constant K , and suppose yn k =  . that (yn) is a sequence in X with 1 n=1 kxn (i) If 2K < 1, then (yn) is a basic sequence equivalent to (xn). (ii) If [ xn ] is complemented by a bounded projection P : X ! X , and if 8K kP k < 1, then [ yn ] is also complemented in X . Proof. We begin with a \micro-lemma." For any sequence of scalars (ai) and any n, rst recall that

janj = kanxnk = kPn x

P

k  2K kxk;

Pn 1 x

where x = n anxn. That is, the coordinate functionals all have norm at most 2K . In particular, we always have



1 1

X 1 sup ja j 

X

janj:

an xn 

n=1 2K n n n=1

Now, on with the proof. . . . To begin, notice that

X

a x

n n n

X n



an yn



X n

janjkxn

yn

k   sup janj  2K n



X

a x :

n n n

45 Thus, (1



X 2K)

an xn

n





X

a y

n n n





X (1 + 2K)

anxn

; n

(4.3)

and hence, (yn) is a basic sequence equivalent to (xn). P P In other words, we've shown that the map T ( n anxn ) = n anyn is an isomorphism between [ xn ] and [ yn ]. Note that (4:3) gives kT k  1+2K < 2 and kT 1k  (1 2K) 1. To prove (ii), we next note that any nontrivial projection P has kP k  1, and hence the condition 8KkP k < 1 implies, at the very least, that 4K < 1. A bitPof arithmetic will convince you that this gives us kxk < 2kyk, where 1 x= n an xn and y = T x (that is, kT k < 2). In particular, it follows from our \micro-lemma" that the coordinate functionals for the yn have norm at P most 4K (that is, janj  4K kyk, where y = n anyn). that T P is an isomorphism on Y = [ yn ]. Indeed, if y = P Next, we showP a y and x = n n n n an xn , then

kT P y

y

k = kT P (y

x)

k =

! 1

X

T P

an (yn xn )

n=1   X

 kT k kP k sup janj
kyn xnk n n  8K kP k kyk < kyk: It follows (see the next lemma) that S = (T P )jY is an invertible map on Y .

Hence, Q = S 1T P is a projection from X onto Y . Now the proof that we've just given supplies a hint as to how we might further improve the result. The \micro-lemma" tells us that we want kxn k  < 1 for all n, where xn is the n-th coordinate functional (which, by Hahn-Banach, we to be an element of X ). Or, better still, we might ask for P1can take  n=1 kxn k kxn yn k < 1. This sum estimates the norm of the map S : X ! X de ned by 1 X Sx = xn (x) (xn yn ): n=1

What is the map S doing here? Well, if we're given x = [ xn ], then the basis-to-basis map should send x into Tx

=

1 X x (x) y n=1

n

n:

P1

 n=1 xn x xn

( )

in

CHAPTER 4. BASES IN BANACH SPACES II

46

Thus, at least on [ xn ], we have S = I T . If S is \small enough," that is, if T is \close enough" to I , then T should be an isomorphism on [ xn ]. That this is true is a useful fact in its own right, and is well worth including.

Lemma 4.6 If a linear map S : X ! X on a Banach space X has kS k < 1, then I S has a bounded inverse, and k(I S ) 1k  (1 kS k) 1. Proof. The geometric series I + S + S 2 + S 3 +


converges in operator norm to a bounded operator U with kU k  (1 kS k) 1. The fact that U = (I S ) 1 follows by simply checking that (I S )Ux = x = U (I S )x for any x 2 X .

Given this setup, see if you can supply a proof for the following new and improved version of Theorem 4.5.

Theorem 4.7 (The Principle of Small Perturbations) Let (xn) be a basic sequence in a Banach space X , with corresponding coordinate functionals (xn). P 1 Suppose that (yn) is a sequence in X with n=1 kxnk kxn ynk =  . (i) If  < 1, then (yn) is a basic sequence equivalent to (xn). (ii) If [ xn ] is the range of a projection P : X ! X , and if  kP k < 1, then [ yn ] is complemented in X . Hint: For (ii), show that the map A : X ! X de ned by

Ax = x

1 X Px + x (Px) y n=1

n

n

satis es kI Ak < 1 and Axn = yn . The projection onto [ yn ] is then given by Q = APA 1.

Notes and Remarks Mazur's construction (Proposition 4.1 and Corollary 4.2) is part of the folklore of Banach space theory and, as far as I know, was never actually published by Mazur. Instead, we know about it through \word of mouth" supplied by the pioneering early papers of Pelczynski. Pelczynski [104] claims that Proposition 4.1 was rst proved by Bessaga in his thesis, but credits Mazur for the idea behind both Bessaga's proof and Pelczynski's (presented here).

47 In a later paper, Pelczynski [105] refers also to a 1959 paper by Bessaga and Pelczynski [15]. The material on disjointly supported sequences in p and p is \old as the hills" and was well known to Banach. The variations oered by the notion of \almost disjointness" and the principle of small perturbations, however, are somewhat more modern and can be traced to the early work of Bessaga and Pelczynski [14]. As a simple application of the principle of small perturbations, it follows that [ 0 1 ] has a basis consisting entirely of polynomials. The same is true of p[ 0 1 ]. As a nal comment regarding equivalent bases, we should point out that there is no such thing as \the" basis for a given space. Granted, there are some spaces in which a \natural" basis suggests itself, p for example, but, in general, a basis, even if one exists, is far from unique. That this is so is shown rather dramatically by the following theorem, due to Pelczynski and Singer [106]. L

C

L

`

;

;

`

Theorem 4.8 If is an in nite dimensional Banach space with a Schauder X

basis, then there are uncountably many mutually nonequivalent normalized bases in X .

48

CHAPTER 4. BASES IN BANACH SPACES II

Exercises Recall that awsequence (xn) in a normed space X converges weakly to x 2 X , written xn ! x, if f (xn ) ! f (x) for every f 2 X  . It's easy to see that w w xn ! x if and only if xn x ! 0. A sequence tending weakly to 0 is said to be weakly null . 1. Let (fn ) be a sequence of disjointly supported functions in Lp, 1 < p < 1. Prove that Pn fn converges in Lp if and only if Pn kfnkpp < 1. 2. Let (xn) be a disjointly supported, norm one sequence in c0. Prove that [ xn ] is isometric to c0 and complemented in c0 by a norm one projection. 3. Let (fn) be disjointly supported, norm one sequence in C [ 0; 1 ]. Prove that [ fn ] is isometric to c0. Is [ fn ] complemented in C [ 0; 1 ]? Will these arguments carry over to disjointly supported sequences in L1[ 0; 1 ]? 4. Let (xn) be a basis for a Banach space let (yn) be a sequence Pn Xan,yand in a Banach space Y . Suppose that converges in Y whenever n Pn anxn converges in X , where (an) is a sequence of scalars. Use the Closed Graph Theorem to prove that formula (4.1) holds for some constant 0 < C < 1. 5. Prove Theorem 4.7. 6. Let T : X ! X be a continuous linear map on a Banach space X . If T is invertible and if S : X ! X is a linear map satisfying kT S k < kT 1k 1, prove that S is also invertible. Thus, the set of invertible maps on X is open in B (X ). 7. In each of the spaces `p, 1  p < 1, or c0, the standard basis (en) is weakly null but not norm null. In fact, the set fen : n  1g is norm closed. Similarly, in any Hilbert space, an orthonormal sequence (xn) is always weakly null while the set f xn : n  1 g is always norm closed. 8. Let (fn) be a disjointly supported sequence of norm one vectors in Lp(), 1 < p < 1. Prove that fn w! 0. Is the same true for p = 1? p = 1? 9. If T : X ! Y is a bounded linear map, and if xn w! 0 in X , prove that T xn w! 0 in Y . 10. Suppose that X and Y are isomorphic Banach spaces and that Z is a complemented subspace of Y . Prove that X contains a complemented subspace W that is isomorphic to Z .

Chapter 5 Bases in Banach Spaces III Recall Banach's basis problem : Does every separable Banach space have a basis? Although the question was ultimately answered in the negative, several positive results were uncovered along the way. For example, we might ask instead: Does every separable Banach space embed in a space with a basis? Or, does every Banach space contain a subspace with a basis; that is, does every Banach space contain a basic sequence? The answers to both of these amended problems are: Yes, and both were known to Banach. The rst follows from the amazing fact, due to Banach and Mazur [8], that every separable Banach space embeds isometrically into [ 0 1 ]. One of our goals in this short course is to give a proof of this universal property of [ 0 1 ]. The second question, the existence of basic seqeunces, was settled by Mazur, as we saw in the last chapter (Corollary 4.2). In this chapter we give a second solution, due to Bessaga and Pelczynski [14], which features a useful selection principle. Our goal here is to mimic the simple case of disjointly supported sequences in . The only real dierence is the interpretation of the word \disjoint." In a space with a basis ( ), we could interpret \disjointly supported" to mean \having nonzero coecients, relative to the basis ( ), occurring over disjoint P P 1 1 and = =1 subsets of N." That is, we could say that = =1 are disjointly supported, relative to the basis ( ), if = 0 for all . C

;

C

;

`p

xn

xn

x

n

xn

an xn

y

an bn

n

bn xn

n

Block basic sequences Let ( ) be a basic sequence in a Banach space . Given P increasing sequences of positive integers 1 1 2 2 , let = =k k be any nonzero xn

X

p

< q

< p

< q

<




49

yk

q i

p

bi xi

CHAPTER 5. BASES IN BANACH SPACES III

50

vector in the span of xpk ; : : : ; xqk . We say that (yk ) is a block basic sequence with respect to (xn). It's easy to see that (yk ) is, indeed, a basic sequence with the same basis constant as (xn):

n

X

ay

k=1 k k

=

n q k

X X

a b x

k=1 i=p k i i k



m q

k

X X

K ak bi xi

k=1 i=p

=

k

m

X

a y : K

k=1 k k

If (xn) is xed, we'll simply call (yk ) a block basic sequence, or even just a

block basis . By way of a simple example, note that any subsequence (xn ) of k

a basis (xn ) is a block basis. We next show how we to \extract" basic subsequences. The method we'll use is a standard bit of trickery known as a \gliding hump" argument. You may nd it helpful to draw some pictures to go along with the proof.

Lemma 5.1 Let (xn ) be a basis for a Banach space X , and let (xn) be the

associated coecient functionals. Suppose that (zn) is a nonzero sequence in  X such that limn!1 xi (zn ) = 0 for each i. Then, given "k > 0, there is a subsequence (zn ) of (zn) and a block basic sequence (yk ), relative to (xn), such that kzn yk k  "k for every k. Proof. Let n1 = 1. Since zn = P1i=1 xi (zn ) xi converges in X , we can nd k

k

1

1

some q1 > 1 such that

1

X

xi (zn1 ) xi  "1 :

i=q +1 1 Pq1  y 1 = i= p1 xi (zn1 ) xi yields kzn1

Setting p1 = 1 and y1 k  "1 . The idea now is to nd a vector zn2 which is \almost disjoint" from zn1 . For this we use that fact that limn!1 xi (zn) = 0 for each i. In particular, by applying this fact to only nitely many i we can nd an n2 > n1 such that

q 1

X

x(z ) x  " =2:

2

i=1 i n2 i

(The span of nitely many xi is just Rn in disguise!) Let choose q2 > p2 such that

1

X

xi (zn1 ) xi  "2=2:

i=q +1

P Setting y2 = qi=2 p2 xi (zn1 ) xi

2

then yields kzn2

y2

k  "2 .

p2

=

q1

+ 1 and

51 We continue, nding zn3 \almost disjoint" from zn1 and zn2 . And so on. The last proof contains a minor mistake! Did you spot it? The y in the ointment is that we have no way of knowing whether the yk are nonzero! This is easy to x, though: We should insist that the znk are bounded away from zero. The principle of small perturbations tells us what to do next: Lemma 5.2 Using the same notation as in Lemma 5.1, suppose that, in addition, lim infn!1 kznk > 0. Then, (zn ) has a subsequence that is basic, and that is equivalent to some block basic sequence of (xn). Proof. By passing to a subsequence if necessary, we may suppose that kzn k  " > 0 for all n. Now, by taking "k ! 0 \fast enough" in Lemma 5.1, the principle of small perturbations (Theorem 4.5) will apply. By modifying our gliding hump argument just slightly, we arrive at: Corollary 5.3 Let X be a Banach space with a basis (xn), and let E be an in nite dimensional subspace of X . Then, E contains a basic sequence equivalent to some block basis of (xn). Proof. It suces to show that E contains a sequence of norm one vectors (zn) such that limn!1 xi (zn ) = 0 for each i. But, in fact, we'll prove something more: Claim: For each m = 1; 2; : : :, there exists a norm one vector zm 2 E such that xi (zm) = 0Pfor all i = 1; : : : ; m; that is, E contains a norm one vector of the form zm = 1 n=m+1 an xn . T How can this be? Well mi=1 ker xi is a subspace of X of codimension at most m and so must intersect every in nite dimensional subspace of X nontrivially. Alternatively, consider the linear map z 7! (x1(z); : : :; xm(z)) from E into Rm. Since E is in nite dimensional, this map must have nontrivial kernel; that is, there must be some norm one z 2 E which is mapped to (0; : : : ; 0). Once we have shown that every separable Banach space embeds isometrically into C [ 0; 1 ], an application of Corollary 5.3 will yield that every Banach space contains a basic sequence. Indeed, it would then follow that every separable Banach space contains a basic sequence (equivalent to a block basis of the Schauder basis for C [ 0; 1 ]). Since every Banach space obviously contains a closed separable subspace, this does the trick. A similar approach can be used to show what we might call the BessagaPelczynski selection principle :

CHAPTER 5. BASES IN BANACH SPACES III

52

Let wX be a Banach space, and suppose that (zn) is a sequence such that zn ! 0 but kznk 6! 0. Then, (zn) has a basic subsequence.

Corollary 5.4

in X

Subspaces of `p and

c0

Temporarily, X will denote one of the spaces `p, 1  p < 1, or c0. We'll use (en) to denote the standard basis in X , and (en) to denote the associated sequence of coordinate functionals in X  . (Note that (en) is really just (en) again, but considered as a sequence in the dual space.) Let's summarize what we know about the (closed, in nite dimensional) subspaces of X . Let (yn ) be any disjointly supported, nonzero sequence in Then, [ yn ] is isometric to X and is complemented in X by a projection of norm one.

Proposition 5.5

X.

This is immediate from Lemma 4.3. Any seminormalized block basis (yn) of (en) is equivalent to (en). Moreover, [ yn ] is isometric to X and is complemented in X by a projection of norm one. Corollary 5.6

A seminormalized sequence (yn) satis es inf n kynk > 0 and supn kynk < 1. This assumption is needed to check the equivalence with (en). Finally, Corollary 5.7 Every in nite dimensional subspace of X contains a further subspace that is isomorphic to X and complemented in X .

As a brief reminder, the proof of this last fact consists of rst showing that every in nite dimensional subspace contains an \almost disjoint" sequence of norm one vectors. Such a sequence is a small perturbation of a block basic sequence, and so is \almost isometric" to X and is the range of a projection of norm \almost one." In fact, given " > 0, we can nd a subspace that is (1 + ")-isomorphic to X and (1 + ")-complemented in X . The fact that every subspace of X contains another copy of X is just what we need to prove that each member of the family `p, 1  p < 1, and c0 is isomorphically distinct. In fact, no space from this class is isomorphic to a subspace of another member of the class. To see this, let's rst consider a special case:

53

Theorem 5.8 Let 1  1, and let : r ! p be a bounded linear map. Then k nkp ! 0. In particular, is not an isomorphism. The same is true of any map : 0 ! p . p < r
=(1 + ") for all k. Now, for every choice of scalars (ak), we have Pn0 ( k ak yk ) = 0 and hence 1 X k k

k=1

ay



1

X 1 n0



k=1



k k

ay

1

CHAPTER 6. SPECIAL PROPERTIES OF C0, `1, AND `1

64

= n01



P

P

1 X ja j ky k k

k 1

k=1

n0 (1 + ") 1

 (1 + ")

2

1

1 X ja j 

1 X ja j:

k

k=1 k

k=1

Of course, jjj k ak yk jjj  k jak j follows from the triangle inequality. Consequently, (yk ) is (1 + ")2-equivalent to (en); that is, [ yk ] is (1 + ")2-isomorphic to `1.

The secret life of `1 We begin by recalling a useful corollary of the Hahn-Banach theorem: For every vector x in a normed space X we have kxk = supf jf (x)j : f 2 X ; kf k = 1 g: Indeed, for a given x, we can always nd a norm one functional f in X  such that jf (x)j = kxk. Of interest here is the fact that there are enough functionals in the unit sphere of X  to recover the norm in X . The question for the day is whether any smaller subset of X  will work. Our rst result oers a simple example of just such a reduction:

Lemma 6.7 If X is a separable normed space, then we can nd a sequence (fn) in X  such that kxk = sup jfn (x)j (6.1) n

for every x in X . In particular, a countable family in X  separates points in X. Proof. Given (xn) dense in X , choose norm one functionals (fn) in X  such that fn(xn) = kxnk. It's not hard to see that (6.1) then holds. Our interest in Lemma 6.7 is that the conclusion holds for some nonseparable spaces, too. For example, it easily holds for the dual of any separable space (if (yn) is dense in the unit sphere of Y , where X = Y , check that (^yn) works). In particular, `1 has this property. Indeed, if en(x) = xn has its usual meaning, then the sequence (en) ts the bill: kxk1 = supn jen(x)j.

65 If some collection of functionals kxkX

 X

satis es

= sup jf (x)j; f2

we say that is a norming set for X , or that norms X . In this language, Lemma 6.7 says that separable spaces have countable norming sets. It's easy to see that any space with this property is isometric to a subspace of `1 . Indeed, if kxk = supn jfn(x)j, then de ne T : X ! `1 by T x = (fn(x))1n=1. Taken together, these observations produce an easy corollary.

Corollary 6.8 A normed space X has a countable norming set if and only if it is isometric to a subspace of `1 .

There are a couple of reasons for having taken this detour. For one, it's now pretty clear that every normed space X embeds isometrically into `1 ( ) for some , where depends on the size of a norming set in X . For another, these observations lend some insight into the nature of complemented subspaces of `1 . Here's why: T ker(e  T ). If T : `1 ! `1 is continuous and linear, then ker T = 1 n n=1 In particular, any complemented subspace of `1 is a countable intersection of kernels of continuous linear functionals. Our next task is to show that c0 fails to be so writeable; that is, c0 is not complemented in `1 . This result is due to Phillips [109] and Sobczyk [127]. The proof we'll give is due to Whitley [131].

Theorem 6.9 c0 is not complemented in `1 .

T

Proof. We'll show that c0 can't be written as 1 n=1 ker fn for any sequence of functionals (fn) on `1. (We won't need to know very much about (`1) in order to do this.) What this amounts to, indirectly, is showing that `1 =c0 has no countable norming set, hence can't be a subspace of `1 . The proof sidesteps consideration of the quotient space, though. We proceed in three steps:

1 There is an uncountable family (N) of in nite subsets of N such that N \ N is nite for  6= . 2 For each , the function x = N is in `1 n c0 (since N is in nite). If f 2 (`1 ) with c0  ker f , then f  : f (x) 6= 0 g is countable.

CHAPTER 6. SPECIAL PROPERTIES OF C0, `1, AND `1

66 3 IfT c0

T1 ker f , then some x is in T1 ker f , too. Thus, c = n  n 0 n=1 n=1



6

ker fn.

1 n=1

First let's see how 3 follows from 1 and 2 (besides, it's easiest): From 2, we would have that the set : fn (x ) = 0 for some n is countable. Thus, some  isn't in the set. That is, there is an  for which fn(x) = 0 for all n, and so 3 follows. 1 is due to Sierpinski [125]. Here's a clever proof (that Whitley attributes to one Arthur Kruse): Let (rn) be a xed enumeration of the rationals. For each irrational , choose and x a subsequence (rn ) converging to , and now de ne N = nk : k = 1; 2; : : : . Each N is in nite, there are uncountably many N, and N N has to be nite if  = ! Very clever, no? Finally we prove 2. Suppose that f (`1 ) vanishes on c0. For each n, let An =  : f (x) 1=n . We'll show that An is nite. Suppose that 1; : : :; k are distinct elements in An. If we de ne f

6

g

k

f

g

\

6

2

f

j

j 

g

k X

x =

i=1

sgn (f (x )) x ; i




i

then f (x) k=n. But since f vanishes on c0, we're allowed to change x in nitely many coordinates without eecting the value of f (x). In particular, if S we de ne Mi = N j6=i N , then Mi diers from N by a nite set, and so setting 

i

n

i

j

y =

k X i=1

sgn (f (x )) M i




i

gives f (y) = f (x). But now that we've made the underlying sets disjoint, we also have y 1 1 and hence k k



k



nf (x) = nf (y)



nf : k

k

Thus, An is nite. The functionals on `1 which vanish on c0 are precisely the elements of (`1 =c0). In fact, our proof of 2 actually computes the norm of x in the quotient space `1 =c0. Corollary 6.10

c0 is not isomorphic to a dual space.

67 Proof. Suppose, to the contrary, that there is an isomorphism T : c0 ! X  from c0 onto the dual of a Banach space X . Then T  : `1 ! X  is again an isomorphism and, further, T jc0 = T . Now, from Dixmier's theorem 2.2, there exists a projection P : X  ! X  mapping X  onto X . But then T 1P T  : `1 ! `1 is a projection from `1 onto c0, contradicting Theorem 6.9.

As it happens, a closed in nite dimensional subspace of 1 is complemented precisely when it's isomorphic to 1 . We can prove half of this claim by showing that 1 is \injective." This means that 1 has a certain \HahnBanach" property. This, too, is due to Phillips [109]. `

`

`

`

Theorem 6.11 Let be any subspace of a normed space , and suppose that Y

X

: 1 is linear and continuous. Then can be extended to a continuous linear map : = . 1 with T

Y ! `

T

S

X ! `

kS k

kT k

Proof. Clearly,

= ( n( ) )1n=1 = ( n ( ) )1n=1  . If we let   be a Hahn-Banach extension of  , where n = n n n 1  then = ( n( ))n=1 does the trick: Ty

y

e

Sx

x

e

T 2 Y

x

y

2 X

y

x

= sup

1

kS xk



n





= Thus,

y

Ty

. That

kS k  kT k

(

)

jxn x j



sup kx k n

n



sup ky k n

kS k  kT k


kxk


kxk

n



kT k kxk:

is obvious.

As an immediate corollary, notice that if 1 is a closed subspace of some Banach space , then 1 is necessarily the range of a norm one projection on . Indeed, the identity : 1 1 (considered as a map into ) extends to a norm one map : (also considered as a map into ). Clearly, 1 is a projection. In short, 1 is norm one complemented in any superspace. This fact is actually equivalent to the \extension property" of the previous theorem. It's not hard to see that any space 1 ( ) has this same \Hahn-Banach extension property." `

X

`

X

I

P

P

`

! `

X

X ! `

X

`

`

CHAPTER 6. SPECIAL PROPERTIES OF 0, 1, AND

68

C

`

`1

Confessions of c0 Although 0 is obviously not complemented in every superspace, it does share 1 's \extension property" to a certain extent. In particular, 0 is \separably injective." This observation is due to Sobczyk [127]; the short proof of the following result is due to Veech [130]. (If you don't know the Banach-Alaoglu theorem, don't panic: We'll review all of the necessary details in a later chapter. Time permitting, we'll even present a second proof of Sobczyk's theorem that sidesteps certain of these issues.) c

`

c

Theorem 6.12 Let be a subspace of a separable normed linear space . If : ! 0 is linear and continuous, then extends to a continuous linear map : ! 0 with k k  2 k k. Y

T

Y

S

X

c

X

T

c

S

T

Proof. As before, let yn = en  T 2 Y  and let xn 2 X  be a Hahn-Banach extension of yn with kxnk = kyn k  kT k. We would like to de ne S x = (xn(x))1 n=1 , as before, but we need to know that S x 2 c0 . That is, we want to replace (xn) by a sequence of functionals which tend pointwise to 0 on X , but we don't want to tamper with their values on Y . What to do? Since X is separable, we know that B = kT k
BX  is both compact and metrizable in the weak topology. Let d be a metric on B which gives this topology. Now let K = B \ Y ? and notice that any weak limit point of (xn) must lie in K , since xn(y) = yn (y) ! 0 for any y 2 Y . That is, d(xn ; K ) ! 0. This means that we can \perturb" each xn by an element of K without eecting its value on Y . Speci cally, we choose a sequence (zn ) in K such that d(xn; zn ) ! 0 and we de ne S x = ( xn(x) zn (x) )1n=1 . Then S x 2 c0, S y = (xn(y)) = T y for y 2 Y , and kS k  2kT k.

As an immediate corollary we get that c0 is complemented by a projection of norm at most 2 in any separable space that contains it. Our nal result brings us full circle. This is the promised (and deep) result of Bessaga and Pelczynski [14] mentioned earlier.

Theorem 6.13 Let be a Banach space. If

contains a subspace isomorphic to c0 , then Y contains a complemented subspace isomorphic to `1 . Thus,  Y contains a subspace isomorphic to all of `1 . In particular, no separable dual space can contain an isomorphic copy of c0 and, of course, c0 itself is not isomorphic to any dual space. Y

Y



69 Proof. In order to prove the rst assertion, we'll construct a map from Y onto `1 .

The second assertion will then follow easily. To begin, let T : c0 ! Y  be an isomorphism into. Then T  : Y  ! `1 is onto, by Theorem 2.1. But we want a map on Y , so let's consider S = T j . Now h e ; Sy i = h y; T e i for all y 2 Y and all n, and so we must have Y

n

n







= h y; T e1 i; h y; T e2 i; : : : =

Sy



T e1(y ); T e2(y ); : : : ;

where (e ) is the usual basis for c0. Next we \pull back" a copy of (e ), the basis for `1: Since T  is onto, there exists a constant K and a sequence (y) in Y  such that kyk  K and T (y ) = e for all n. We would prefer a sequence in Y with this property, for then we'd be nished. We'll settle for the next best thing: Since K
B is weak dense in K
B  , we can nd a sequence (y ) in Y with ky k  K such that n

n

n

n

n

n

Y

n

Y

X1 n

jT e (y ) 1j < 1=n; and n

n

jT e (y )j < 1=n: i

i

=1

n

n

(We've used the functionals T e1; : : : ; T e to specify a certain weak neighborhood of y for each n. Of course, y(T e ) = T y(e ) = e (e ) =  .) Thus, Sy has a large n-th coordinate and very small entries in its rst n 1 coordinates. What this means is that (Sy ) has a subsequence equivalent to the usual basis in `1 (i.e., an \almost disjoint" subsequence) whose span is complemented in `1 by a projection P . In particular, after passing to a subsequence, we can nd a constant M so that n

k

n

n

n

k

n

k

n;k

n

n

X

a

n

n

yn



 K X ja

n

n

X j  KM

a n

n

Syn





1

X  KM kS k

a

n

n

yn



:

That is, S is invertible on [ y ]  `1 and hence Q = S 1P S is a projection from Y onto [ y ]. This completes the proof of the rst assertion. In other words, we now have Y  `1  X for some X  Y . It follows that Y   `1  X , which nishes the proof. n

n

Notes and Remarks The proof of Theorem 6.1 is closely akin to Banach's proof of the Open Mapping theorem. It's quite clear that Banach understood completeness completely! Schur's Theorem 6.2 is an older gem which also appears in Banach's

70

CHAPTER 6. SPECIAL PROPERTIES OF C0, `1, AND `1

book [6]. It provides a classic example of a gliding hump argument. Our proofs of Theorem 6.4 and its corollary are largely borrowed from Diestel [31], who attributes these results to Lindenstrauss. Theorem 6.6 is sometimes called \James's non-distortion theorem." In brief, it says that `1 and c0 are not \distortable" (a notion that we won't pursue further here). Embeddings into `1 spaces, along the lines of Corollary 6.8, are semi-classical and can be traced back to Frechet; see [45]. Theorem 6.11 and its companion Theorem 6.12 were among the rst results in a frenzy of research during the late 40s and early 50s concerning injective spaces and Hahn-Banach-like extension properties. We will have more to say about such topics in a later chapter. For more on the spaces c0, `1, and `1 , see Day [27], Diestel [31], Jameson [68], and Lindenstrauss and Tzafriri [84].

71

Exercises 1. Find a weakly null normalized sequence in L1 and conclude that L1 and `1 are not isomorphic. 2. Complete the proof of Lemma 6.7. 3. Show that the conclusion of Lemma 6.7 also holds in case X = Y  where Y is separable. 4. If X is separable, show that X  embeds isometrically in `1 . 5. Let X and Y be Banach spaces and let A be a bounded, linear map from X onto Y . Then, for every bounded linear map T : `1 ! Y , there exists a \lifting" T~ : `1 ! X such that AT~ = T . [Hint: Use the Open Mapping theorem to choose a bounded sequence (xn) in X such that Axn = Ten. ~ n = xn then does the trick.] The operator de ned by Te 6. If H is any Hilbert space, prove that every bounded linear operator from H into `1 is compact. 7. Prove Theorem 6.6 for c0. 8. Prove Theorem 6.11 for maps into `1 ( ). That is, prove that `1 ( ) is injective. 9. Prove that the following are equivalent for a normed space X . (i) If X is contained isometrically in a normed space Y , then X is complemented by a norm one projection on Y . (ii) If E is a subspace of a normed space F , then every bounded linear map T : E ! X extends to a bounded linear map S : F ! X with kSk = kT k. [Hint: We've already seen that (ii) implies (i). To prove that (i) implies (ii), rst embed X isometrically in some `1 ( ) space and extend T (as a map into `1 ( )) using Theorem 6.11.]

72

CHAPTER 6. SPECIAL PROPERTIES OF C0, `1, AND `1

Chapter 7 Bases and Duality If (xn) is a basis for X , you may have wondered whether the sequence of coordinate functionals (xn) forms a basis for X . If we consider the pair (en) and (en), then it's easy to see that the answer is sometimes: Yes (in case X = c0 and X  = `1, for example), and sometimes: No (in case X = `1 and X  = `1 , for instance). As it happens, though, (xn) is always a basic sequence; that is, it's always a basis for [ xn ]. Let's see why: P Given a1; : : :; an and " > 0, choose x = 1i=1 cixi with kxk = 1 such that n X i=1

*

aici = x;

n X i=1

+

aixi >

Now if K is the basis constant of (xi), then

K = K kxk  Thus, for any m > n, we have

m

X

 K

aixi

i=1

  =

>

n

X

a x

i

i=1 i

n

X

c x :

i=1 i i



m

n

X

X

c x a x



i=1 i i i=1 i i * n + m X X  i=1

n X

ci x i ;

aixi

aici

i=1n

X

a x

i=1 i i

73

i=1

":

":

CHAPTER 7. BASES AND DUALITY

74

That is, since " was arbitrary, (xn) is a basic sequence with the same basis constant K . It's also easy to see that if Pn is the canonical projection onto [ x1; : : : ; xn ], then Pn is the canonical projection onto [ x1; : : :; xn ]. Indeed, given m, k > n, we have *

k X j =1

bj xj ; Pn

m X i=1

!+

aixi

*

Pn

= *

= = =

n X

k X j =1

!

bj xj ;

bj xj ;

m X

m X i=1

+

aixi

+

aixi

j =1 i=1 n X biai i=1 * + n k X X  bj xj ; aixi : j =1 i=1

That is, Pn ( mi=1 aixi ) = ni=1 aixi . What's more, since Pn x ! x for any x, it's easy to check that Pnx w! x for any x, and hence span(xn) is weakdense in X . If X is re exive, then this already implies that (xn) is a basis for X  since the weak (= weak) closure of a subspace coincides with its norm closure. In any case, (xn) is a basis for X  if and only if [ xn ] = X . Our next result supplies a test for this condition. Theorem 7.1 Let (xn) be a basis for X with coordinate functionals (xn). Then, (xn ) is a basis for X  if and only if limn!1 kxkn = 0, for each x 2 X  , where kxkn is the norm of x restricted to the \tail space" spanf xi : i > n g. Proof . The forward implication is easy. If (xn) is a basis for X  and if x = P  i ai xi , then

P

P

kxkn =

1

X



i=n+1

aixi

! 0:

Now suppose that limn!1 kxkn =P0 for each x 2 X . Given x 2 X , n hx ; x i x vanishes on the span of rst notice that the functional x i i=1 i P1 x1; : : : ; xn. Thus, given x = i=1 cixi, we have *

x; x

n X i=1

+  i

hxi; xi x

=

* 1 X

i=n+1

+ 

cixi; x

 kxkn

1

X



i=n+1



i

cix :

75 But where

1

X

ci x i

i=n+1

n

X kxk +

cixi

i=1



 (1 + ) k k K

x ;

is the basis constant for ( ). Thus,

K

xi



x



hxi; xi xi

 i=1

n X

and hence (  ) is a basis for xn

X



(1 + ) k k ! 0 K

x

n

;

.

We say that a basis ( ) is shrinking if lim !1 k k = 0 for each  2 , as in our last result. That is, ( ) is shrinking if (  ) is a basis for  . The natural basis ( ) is shrinking in 0 and , 1 1, but not in 1 .  If has a shrinking basis, then can be represented in terms of the basis too. xn

x

n

xn

en

x

n

c

`p

X

X

X

xn

< p
2 is essentially identical.

Theorem 8.3

Let

f, g

2 Lp() 1  p < 1 ,

,

p

6= 2

.

Then,

disjointly supported if and only if

k f + g kpp + k f

g kpp




= 2 kf kpp + kgkpp :

f

and

g

are

CHAPTER 8.

86

L

P

SPACES

Proof. The function h = jf + gj

+ jf g j is of constant sign and integrates to give p

Z

h d

2 (jf j + jg j )

p

p

= kf + gk + kf g k p

p

p

p

p




2 kf k + kgk : p

p

p

p

Thus, f and g are disjointly supported if and only if fg = 0 if and only if R h = 0 if and only if h d = 0. If T : L () ! L ( ) is a linear isometry, 1  p < 1, p 6= 2, then T maps disjoint functions to disjoint functions. Corollary 8.4

p

p

Proof. If T is an isometry, then kT f k = kf k , kT gk = kg k , and kT f  = kf  gk for any f and g in L (). Consequently, f and g are disjoint if and only if T f and T g are disjoint. This last result tells us something about the subspace structure of L . We already know that L contains an isometric copy of ` , spanned by any sequence of disjointly supported, nonzero functions, and now we know that this is the only way to get an isometric copy of ` inside L . L contains other natural sublattices. For example, if we identify the space L [ 0; 1=2 ] with the collection of all functions f 2 L which are supported in [ 0; 1=2 ], for example, then L [ 0; 1=2 ] is a complemented sublattice of L which is isometric to all of L . The projection is obvious: P f = f
[ 0 1 2 ] ; and the isometry is nearly obvious: If we set (T f )(t) = 21 f (2t) for 0  t  1=2 and f 2 L , then T f 2 L [ 0; 1=2 ] and

T gk

p

p

p

p

p

p

p

p

p

p

p

p

p

p

p

p

p

p

; =

=p

p

p

kT f k =

Z

1=2

p p

jT f (t)j

p

0

= 2

dt

Z

1=2

jf (2t)j

p

0

dt

= kf k : p p

More generally, L [ a; b ] is isometric to L , and from this it's a short leap to the conclusion that L [ 0; 1) and L (R) are isometric to L . Indeed, if we partition [ 0; 1 ] into countably many disjoint, nontrivial, intervals (I ), then L = (L (I1)  L (I2) 


) = (L [ 0; 1 ]  L [ 1; 2 ] 


) = L [ 0; 1): Of course, L is also isometric to a sublattice of L [ 0; 1). p

p

p

p

p

n

p

p

p

p

p

p

p

p

p

p

87

Conditional expectation Less obvious examples of Lp-subspaces of Lp are generated by sub--algebras of the Borel -algebra B. Given a sub--algebra B0 of B, we de ne the space Lp(B0) = Lp([ 0; 1 ]; B0; m) to be the collection of all B0-measurable functions in Lp. Thus, Lp(B0) is a closed subspace (and even a sublattice) of Lp. In fact, Lp(B0) is complemented in Lp by a norm one projection. What's more, any subspace of Lp which is the range of a norm one projection has to be of this form. We'll prove the rst claim in some detail; the second claim isn't terribly hard to prove, but we'll leave the details for another time. By way of a simple example, suppose that we partition [ 0; 1 ] into nitely many disjoint, measurable sets A1; : : :; An, with m(Ak ) > 0 for each k, and let B0 be the algebra generated by the sets AkP . The B0-measurable functions are precisely the simple functions of the form nk=1 ak A , and the space Lp(B0) is then isometric to `np . It might also help matters if we wrote out the projection onto Lp(B0). In this case it's given by k

Z  n  X 1 Pf = f A : m ( A ) k A k=1 k

k

Recall that kPf kp  kf kp (review the computation we used to show that the Haar system is aR basis forR Lp). Notice that Pf is constant on each Ak , of course, and that A Pf = A f for every B0-measurable set A. Indeed,

Z

Aj

Pf =

! 1 Z f Z  = Z f: A m(Aj ) A A A j

j

j

j

This simple example is not far from the general situation: In some sense, f is \averaged" over all B0-measurable sets to arrive at its B0-measurable counterpart. In order to build a norm one projection from Lp onto Lp(B0), we need some way to map B-measurable functions into B0-measurable functions without increasing their norms. The secret here is what is known as a conditional expectation operator. Given an integrable, B-measurable function f , we de ne the conditional expectation of f given B0 to be the a.e. unique integrable, B0-measurable function g with the property that Z Z f = g for every A 2 B0; (8.4) A

A

CHAPTER 8. LP SPACES

88

and we write g = E (f j B0). The existence and uniquenessR of g follow from the Radon-Nikodym theorem. That is, if we de ne (A) = A f for A 2 B0, and if we think of m as being de ned only on B0, then  is absolutely continuous with respect to m and g is the Radon-Nikodym derivative d=dm. It's often simpler to think of (8.4) as the de ning statement, though; after R all, g is completely determined, as an element of L1, by the values A g as A runs through B0. For example, it's easy to see how (8.4) forces E (f j B0) to be both linear and positive. Consequently, we must also have jE (f j B0)j  E ( jf j j B0 ), and now it's clear that conditional expectation is a contraction on both L1 and L1. Indeed, if f 2 L1, then:

kE (f j B0)k1 =

Z 1

0

jE (f j B0)j 

Z 1

0

E(

jf j j B0) =

Z 1

0

jf j = kf k1:

And if f 2 L1 , then:

jE (f j B0)j  E ( jf j j B0)  E ( kf k1 j B0) = kf k1 ; since constant functions are B0-measurable. Thus, kE (f j B0)k1  kf k1 . As it happens, a positive linear map which is a simultaneous contraction on L1 and L1 is also a contraction on every Lp, 1 < p < 1. That this is so follows from an application of an ingenious interpolation scheme due to Marcinkiewicz in 1939 [88].

Theorem 8.5 Suppose that T : L1 ! L1 is a linear map satisfying: (i) Tf  0 whenever f  0 (i.e., T is positive), (ii) kTf k1  kf k1 for all f 2 L1, and (iii) Tf 2 L1 whenever f 2 L1 and kTf k1  kf k1 . Then, Tf 2 Lp whenever f 2 Lp , for any 1 < p < 1, and kTf kp  kf kp . Proof. Since jTf j  T jf j, it's enough to consider the case where f  0. So, let 1 < p < 1 and let 0  f 2 Lp. Now, for each xed y > 0, we can write f = fy + Ry , where fy = f ^ y and Ry = f fy = (f y)+, and where y is also used to denote the constant function y
1 2 L1.

89 Since fy  y we have that T (fy )  T (y)  y, from (iii), and so Tf

f

= T (fy ) + T (Ry )  y + T (Ry ):

y

But T (Ry )  0 because Ry  0, hence (T f

y )+

fy

 T (Ry ) = T ((f

y )+ ):

Next, integration gives:

Z

(T f

y)

+

Z



T ((f y )

+

) = kT ((f

y)

+

)k  k(f

y)

+

1

Z

k = (f y )

+

1

:

The rest of the proof consists in \upgrading" this inequality to an estimate involving the Lp norms of T f and f . We de ne auxiliary functions y and y by y (x) =



1; if f (x) y > 0; 0; otherwise;

and

y (x) =

Then, (f y) = (f y) y and (T f our last estimate says that

y )+

+

Z

0

1

(T f

y )(x) y (x) dx



Z

1

0

for every y > 0. Consequently,

Z1

y

p 2

Z

0

0

1

(T f (x)





1; if (T f )(x) 0; otherwise:

= (T f (f

y ) y .

y > 0;

In this notation

y )(x) y (x) dx



y ) y (x) dx dy

Z1

y

p 2

Z

0

1

0



(f (x)

y ) y (x) dx dy:

Since all the functions involved are nonnegative, Fubini's theorem yields

Z1 0

y

p 2

Z

1 0

(f (x)



y ) y (x) dx dy

= =

Z Z1 1

yp

Z Zfx 0

2

0

1

( )

yp

(f (x) 2

(f (x)

Z  1 1 = p 1 p 0

0

0

1

y ) y (x) dy dx y ) dy dx

f (x)p dx;

CHAPTER 8.

90 and, similarly,

Z1 0

yp 2

Z 1 0



(T f (x) y) y (x) dx

Thus, T f 2 Lp and kT f kp  kf kp.

 1 dy = p

1

1

p

Z 1 0

LP

SPACES

T f (x)p dx:

Corollary 8.6 Conditional expectation is a simultaneous contraction on every Lp .

Some authors prove Corollary 8.6 by rst proving a version of Jensen's inequality valid for conditional expectations; in particular, it can be shown that jE (f j B0)jp  E ( jf jp j B0). One proof of this fact rst reduces to the case of nonnegative simple functions, where the inequality is more or less immediate, and then appeals to a \monotone convergence" theorem for conditional expectations (recall that conditional expectation is monotone). In any event, integration then yields kE (f j B0)kp  kf kp. There is an another approach to conditional expectation that warrants at least a brief discussion. In this approach, one begins by de ning conditional expectation for L2 functions. Speci cally, we de ne E (
j B0) to be the orthogonal projection from L2 onto L2(B0). Since L2 is dense in L1, we can extend this de nition to L1 functions by taking monotone limits: Since each 0  f 2 L1 is the limit in L1-norm of an increasing sequence ('n) of bounded (hence L2) simple functions, we can take E (f j B0) = limn!1 E ('n j B0). The savings in using Marcinkiewicz interpolation is clear: We have been spared the tedium of the typical \limit of simple functions" calculation. Instead we cut to the heart of the matter: Each f 2 Lp can be written as the sum of an L1 function and an L1 function; namely, f = fy + Ry . Marcinkiewicz then estimates the L1 norm of E (fy j B0) and the L1 norm of E (Ry j B0). Finally, we should at least state the converse, due to Douglas [34] (for p = 1) and Ando [4] (for 1 < p < 1).

Theorem 8.7 Let 1  p < 1, and let P : Lp ! Lp be a positive, linear, contractive projection with P (1) = 1. Then there exists a sub--algebra B0 such that P f = E (f j B0 ) for every f 2 Lp .

Notes and Remarks An excellent source for inequalities both big and small, including a discussion of convex functions and Jensen's inequality (Theorem 8.1), is the classic book

91 Inequalities, by Hardy, Littlewood, and Polya [58]. The results on disjointly supported functions in Lp and their preservation under isometries were (essentially) known to Banach, and form the basis for a more general result, due to Lamperti [81]. Loosely stated, here is Lamperti's result for isometries on Lp.

Theorem 8.8 Let 1

, p 6= 2, and let T be a linear isometry on Lp[ 0; 1 ]. Then, there is a Borel measurable map ' from [ 0; 1 ] onto (almost all of ) [ 0; 1 ] and a norm one h 2 Lp such that 

p
0. Finally, note that if f 2 M (p; "), then so is f for any scalar ; in particular, 0 2 M (p; "). The elements of a given M (p; ") have to be relatively \ at." Notice, for example, that every element of M (p; 1) is constant (a.e.). Indeed, mf jf j  kf kp g  1 =) jf j = kf kp a.e. Along similar lines, notice that M (p; ") doesn't contain the \spike" f =  =p ; for any 0 <  < ". In this case, kf kp = 1 while mf f  " g = . The following Lemma puts this observation to good use. 1

1

[0

2

1

2

0

]

Lemma 9.4 For a subset A of L , the following are equivalent: (i) A  M (p; ") for some " > 0. (ii) For each 0 < q < p, there exists a constant C < 1 such that kf k  kf k  C kf k for all f 2 A. (iii) For some 0 < q < p, there exists a constant C < 1 such that kf k  kf k  C kf k for all f 2 A. p

p

p

q

q

q

q

q

q

Proof. (i) =) (ii): If f 2 A  M (p; "), then

Z

q

q

jf j = q

Z

jf j + q

fjf j"kf kp g

Z f jf j 0, and hence that A  M (p; ) for all 0 <  <  . q

q p

1+

(1+ )

0

0

0

101 Let S = f jf j  "kf kp g. Then, of course, m(S ) < " because f 2= M (p; "). Next, write 1=q = 1=p + 1=r for some r > 0. Thus, 1 = q=p + q=r and so p=q and r=q are conjugate indices. Now let's estimate:

kf k = q q

Z

jf j +

SZ

q

Z

q=pS

c

jf jq

 jf jp
m(S )q=r + "q kf kqp S q=r  (" + "q )kf kqp: Thus, from (iii), kf kqp  Cqq kf kqq  Cqq ("q=r + "q )kf kqp. Since f 6= 0, this means

that " must be bounded away from 0; that is, there exists some 0 such that "  0 > 0. If an entire subspace X of Lp is contained in some M (p; "), then the Lp and Lq topologies evidently coincide on X for every 0 < q < p. In fact, it's even true that the Lp and L0 topologies agree on X , where L0 carries the topology of convergence in measure. Most important for our purposes is this: If 2 < p < 1, and if X is a closed subspace of Lp which is contained in some M (p; "), then the Lp and L2 topologies coincide on X . That is, X must be isomorphic to a Hilbert space. How so? Well, if kf k2  kf kp  C kf k2 for all f 2 X , then the inclusion map from X into L2 is an isomorphism. Since every closed subspace of L2 is isometric to a Hilbert space, X must be isomorphic to a Hilbert space. Now if f 2 Lp, kf kp = 1, and f 2= M (p; "), then we can write f = g + h, where g is supported on a set of measure less than ", and where khkp < ". Indeed, g = f
f jf j" g and h = f g = f
f jf j 1 " while m(supp g) < ". Next, suppose that a subspace X of Lp fails to be entirely contained in M (p; ") for any " > 0. Then, in particular, the set SX of all norm one vectors in X isn't contained in any M (p; "). Thus, given a sequence of positive numbers "n ! 0, we can nd a sequence of norm one vectors fn 2 SX such that fn 2= M (p; "n ). Each fn can be written fn = gn + hn , where gn is supported on a set of measure less than "n, and where khnkp  "n. That is, (fn) is a small perturbation of the sequence of spikes (gn). The claim here is that if "n ! 0 fast enough, then the seqence (gn ) is almost disjoint. If so, then (gn), and hence also (fn ), will be equivalent to the usual basis in `p. This claim is worth isolating as a separate result.

CHAPTER 9. LP SPACES II

102

Lemma 9.5 Let (fn ) be a sequence of norm one functions in Lp, 1  p < 1. If m(supp fn) ! 0, then some subsequence of (fn) is equivalent to a disjointly

supported sequence in Lp .

Proof. The key observation here is that if f 2 Lp is xed, then the measure R (A) = A jf jp is absolutely continuous with respect to m. In particular, for each " > 0 there is a  > 0 such that (A) < " whenever m(A) < . Let An = supp fn . Then m(An) ! 0. By induction, we can choose a subsequence of (fn ), which we again label (fn), such that

Z Xn p jfi(t)j dt < 4 n p A +1 i S Now let Bn = An n 1i n Ai and let gn = fn
B . ( +1)

n

=1

for all n = 1; 2; : : :. Obviously, the sets (Bn) are pairwise disjoint and, hence, the sequence (gn) is disjointly supported in Lp. Finally, = +1

kfn

gn kpp

=

Z

An nBn

jfn

(t)jp dt



1 Z X

i=n+1 Ai

jfn(t)jp dt
0, in which case X is isomorphic to ` and the Lp and L topologies agree on X , or (ii) X contains a subspace that is isomorphic to `p and complemented in Lp. 2

2

103

Corollary 9.7 For 2 < p < 1 and 1  r < 1, r 6= p, 2, no subspace of Lp is isomorphic to Lr or to `r .

Proof. Since Lr contains an isometric copy of `r , it suces to check that `r is not isomorphic to a subspace of Lp. But for r 6= p, 2, we know that `r is neither isomorphic to `2 nor does it contain a subspace isomorphic to `p. (Recall the discussion surrounding Theorem 5.8.) Thus, neither alternative of Theorem 9.6 is available. As it happens, every copy of `2 in Lp, p > 2, is necessarily complemented. Curiously, this fails, in general, for p < 2.

Corollary 9.8 Let X be a subspace of Lp, 2 < p < 1. If X is isomorphic to `2 ,

then X is complemented in Lp .

Proof. Since X cannot contain an isomorphic copy of `p , we must have X  M (p; ") for some " > 0. Thus, there is a constant C such that kf k2  kf kp  C kf k2 for all f 2 X . As we've seen, this means that X can also be considered as a subspace of L2. As such, we can nd an orthornormal sequence ('n) in L2 such that X is the closed linear span of ('n ), where the closure can be computed in either Lp or L2, since the two topologies agree on X . Now let P be the orthogonal projection onto X ; speci cally, put

X = h 1

Pf

n=1

i

f; 'n 'n :

Just as we saw with the Rademacher functions, the projection bounded as a map on Lp. Indeed, since P f 2 X , we have

P

is also

kP f kp  C kP f k2  C kf k2  C kf kp: Corollary 9.9 Let 2 < p < 1, and let X be an in nite dimensional closed

subspace of Lp. Then either X is isomorphic to `2 and complemented in Lp or X contains a subpsace that is isomorphic to `p and complemented in Lp .

By considering only complemented subspaces, we can use duality to transfer a modi ed version of these results to Lp for 1 < p < 2. The principle at work is very general and well worth isolating:

Lemma 9.10 Let complemented in

X .

Y X,

be a closed subspace of a Banach space X . If Y is then Y  is isomorphic to a complemented subspace of

CHAPTER 9. LP SPACES II

104

Proof. First recall that Y  can be identi ed with a quotient of X . Speci cally, Y  = X =Y ?, isometrically. (Recall that if i : Y ! X is inclusion, then i : X  ! Y  is restriction and has kernel Y ? . Since i is an isometry into, i is a quotient map.) Now if P : X ! X is a projection with range Y , then P  : X  ! X  is a projection with kernel Y ? . As we've seen, P  is indeed a projection. To see that ker P  = Y ?, consider:

P x = 0 () 0 = h x; P x i = h Px; x i for all x 2 X () 0 = h y; x i for all y 2 Y () x 2 Y ? : Thus, the range of P  can be identi ed, isomorphically, with X  =Y ?, which we know to be Y , isometrically. That is, P  is a projection onto an isomorphic copy of Y . We will also need the following observation:

Lemma 9.11 A closed subspace of a re exive space is again re exive. Proof. Let Y be a closed subspace of a re exive Banach space X . Let i : Y ! X denote inclusion, and let jY : Y ! Y  and jX : X ! X  denote the canonical embeddings. Then i : Y  ! X  is an isometry (into) which makes the following diagram commute.

! X  i" " i Y Y  Y j! X

jX

Thus, all four maps are isometries (into) and jX is onto because X is re exive. Next we compare ranges; for this we will need to compute annihilators (but in X  and X  only). Our calculations will be slightly less cumbersome if we occasionally ignore the formal inclusion i. Now the range of i is Y ?? (because the kernel of i is Y ?). Thus, Y is re exive if and only if jY is onto if and only if i(jY (Y )) = jX (i(Y )) = Y ?? . But, for any subset A of X , it's easy to check that jX (? A) = A? (since jX is onto); thus, jX (? (Y ?)) = Y ?? . In other words, Y is re exive if and only if Y = i( Y ) = ? ( Y ? ) = Y .

105 Let X be an in nite dimensional complemented subspace of Lp, 1 < p < 1. Then either X is isomorphic to `2 or X contains a subspace that is isomorphic to `p and complemented in Lp.

Corollary 9.12

Proof. Of course, we already know this result when 2  p < 1. So, suppose that 1 < p < 2 and suppose that X is a complemented subspace of Lp. By Lemma 9.11 we know that X is re exive and by Lemma 9.10 we know that X  is isomorphic to a complemented subspace of Lq , where 1=p + 1=q = 1 and q > 2. Thus, either X  is isomorphic to `2, or else X  contains a complemented subspace isomorphic to `q . Now if X  is isomorphic to `2, then surely X  = X is too. Finally, if X  contains a complemented subspace which is isomorphic to `q , then X  = X contains a complemented subspace isomorphic to `p. Corollary 9.13

Lp is not isomorphic to Lq for any p 6= q.

Now that we know Lp, 1 < p < 1, contains complemented subspaces isomorphic to `p and `2, it's possible to construct other, less apparent, complemented subspaces. For example, it's not hard to see that `p  `2 and Zp = (`2  `2 


)p are isomorphic to complemented subspaces of Lp. For p 6= 2, the spaces Lp, `p, `2, `p `2, and Zp are isomorphically distinct (although that's not easy to see just now). In particular, Lp does contain complemented subspaces other than `p, `2, and Lp. In fact, Lp contains in nitely many isomorphically distinct complemented subspaces.

Notes and Remarks An excellent source for information on the Rademacher functions is the 1932 paper by R. E. A. C. Paley [102]; see also Zygmund [138]. The \fancy" proof of Khinchine's inequality, which immediately follows the classical proof, is a minor modi cation of a proof presented in a course oered by Ben Garling at The Ohio State University in the late 70s; Stephen Dilworth tells me that this clever proof was shown to Garling by Simon Bernau. The constants Ap and Bp arising from our proof(s) of Khinchine's p inequality are not best possible; for example, it's known that A1 = 1= 2, and that B2m = ((2m 1)!!)1=2m, for m = 1; 2; : : :, are best. See Szarek [129] and Haagerup [57]. Most of our applications of Khinchine's inequality to subspaces of Lp and, indeed, nearly all of the resutls from this chapter, are due to Kadec and Pelczynski [74], but much of the \ avor" of our presentation is borrowed from Garling's masterful interpretation. For more on the subspace structure of Lp, see [85] and [73].

106

CHAPTER 9. LP SPACES II

Lemma 9.11 is due to B. J. Pettis in 1938 [107] but is by now standard fare in most textbooks on functional analysis; see Megginson [90] for much more on re exive subspaces.

107

Exercises 1. 2. 3. 4. 5. 6.

If 0 < "1 < "2 < 1, show that M (p; "1) M (p; "2 ). S Prove that ">0 M (p; ") = Lp . Prove Theorem 9.6. Show that c0 is not isomorphic to a subspace of Lp for 1 p < . However, c0 is isometric to a subspace of L1 . For 1 < p < , p = 2, prove that `p `2 and Zp = (`2 `2 )p are isomorphic to complemented subspaces of Lp. Let Y be a closed subspace of a Banach space X and let i : Y X denote inclusion. Show that i : X  Y  is restriction (de ned by i(f ) = f Y ) and that ker i = Y ?. Further, show that i is an isometry (into) with range Y ?? . Let X be re exive and let jX : X X  denote the canonical embedding. Show that jX (?A) = A? for any subset A of X . Is this true without the assumption that X is re exive? Fill in the details in the proof of Corollary 9.12. Speci cally, suppose that X is a complemented subspace of Lp, 1 < p < 2, and that X is not isomorphic to `2. Prove that X contains a complemented subspace isomorphic to `p . Prove Corollary 9.13. 



1

6



8. 9.






!

!

7.



1

j

!

108

CHAPTER 9. LP SPACES II

Chapter 10

Lp Spaces III As pointed out earlier, the spaces Lp and `p exhaust the \isomorphic types" of Lp() spaces. Thus, to better understand the isomorphic structure of Lp() spaces, we might ask, as Banach did: For p 6= r, can `r or Lr be isomorphically embedded into Lp ?

We know quite a bit about this problem. We know that the answer is always \Yes" for r = 2, and, in case 2 < p < 1, the Kadec-Pelczynski theorem (Corollary 9.7) tells us that r = 2 is the only possibility. In this chapter we'll prove: If p and r live on opposite sides of 2, there can be no isomorphism from Lr or `r into Lp .

This leaves open only the cases 1  r < p < 2 and 1  p < r < 2. The rst case can also be eliminated, as we'll see, but not the second.

Unconditional convergence We next introduce the notion of unconditional convergence of series. What follows are several plausible de nitions. P We say that a series n xn in a Banach space X is:

P

(a) unordered convergent if n x n converges for every permutation (oneto-one, onto map)  : N ! N; ( )

109

110

CHAPTER 10. LP SPACES III

P

(b) subseries convergent if k xnk converges for every subsequence (xnk ) of (xn); P (c) \random signs" convergent if n "nxn converges for any choice of signs "n = 1; P (d) bounded multiplier convergent if n anxn converges whenever an 1. 

j

j 

As we'll see, inPa complete space all four notions are equivalent. Henceforth, we will say that n xn is unconditionally convergent if any one of these four conditions holds. P In Rn, all four of these conditions are equivalent toP n xn < ; that is, all are equivalent to the absolute summability of n xn . In an in nite dimensional space, however, this is no longer the case. It is a deep result, due to Dvoretzky and Rogers in 1940 [41], that every in nite dimensional space contains an unconditionally convergent series which fails to be absolutely summable. We will brie y sketch the proof that (a) through (c) are equivalent (and leave condition (d)Pas an exercise), but rst let's look P at an example or two: If we x an element n an en in `p , 1 p < , then n anen is unconditionally convergent (but not absolutely convergent, in general, unless p = 1). Why? P P Because n anen converges in `p if and only if n an p < , which clearly allows us to change the signs of the an. For this reason, we sometimes say that `p has an unconditional basisP. Similarly, if (en) is any orthonormal P sequence in a HilbertPspace H , then n anen converges if and only if n an < if and only if n anen is unconditionally convergent. k



k

1

1

j

j

1

j

j

2

1

Theorem 10.1 Given a series P xn in a Banach space X , the following are equivalent:

(i) (ii) (iii) (iv)

P x n converges for every permutation  of N. P The series xn converges for every subsequence n < n < n < . P The series "nxn converges for every choice of signs "n = 1. Pi2A xi < " for every For every " > 0, there is exists an N such that

The series

( )

1

k

2

3








nite subset A of N satisfying min A > N .

k

k

Proof. That (ii) and (iii) are equivalent is easy. The fact that (iv) implies (i) and (ii) is pretty clear, since (iv) implies that each of the series in (i) and

111 (ii) is Cauchy. The hard work comes in establishing (i), (ii) =) (iv). To this end, suppose that (iv) fails. Then, there exist P an " > 0 and nite subsets An of N satisfying max An < min An and k i2A xikP " for all n. But then, S A = n An de nes a subsequence of N for which i2A xi doesn't converge. Thus, (ii) fails. Lastly, we can nd a permutation  of N such that  maps the interval [ min An; P max An ] onto itself P in such a way that P (An ) = Bn is an interval. Thus, k i2B x i k = k i2A xik. But then, x n doesn't converge, and so (i) also fails. +1

n

1

n

( )

( )

n

P

If xn convergesPunconditionally to x, then condition (iv) implies that every rearrangement x n Plikewises converges to x. The same, of course, is no longer true of the sums "n xn. On the P other hand, condition (iv) tells us that the set of all vectors of the form "nxn is a compact subset P of X . Indeed, from (iv), the map fP: f 1; 1gN ! X de ned by f (("n )) = "nxn is continuous. In particular, if xn is unconditionally convergent, then ( )

n

X

sup sup

"ixi

n " =1 i=1 i

It also follows from (iv) that if (v)

P

P

< 1:

(10.1)

xn is unconditionally convergent, then

anxn converges for every bounded sequence of scalars (an ) P

and the map T : `1 ! X de ned by T ((an)) = anxn is continuous (moreover, the restriction of T to c is even compact). Since (v) clearly implies (iii), this means that condition (v) is another equivalent formulation of unconditional convergence. Since we have no immediate need for this particular condition, we will leave the details as an exercise. 0

Orlicz's theorem In terms of the Rademacher functions (or had you forgotten already?), we can P write (10.1) another way: If n xn is unconditionally convergent, then

n

X

sup sup

ri(t)xi

n 0t1 i=1

< 1:

Armed with this observation we can make short work of an important theorem due to Orlicz in 1930 [101]. We begin with a useful calculation.

CHAPTER 10. L SPACES III

112

P

Proposition 10.2

! X

jf j A

1=2

n

i

p

2

i=1

f ; : : :; f 2 L , 1  p < 1, we have

For any

1

0Z

1

 @

X r (s)f

dsA  B

X jf j !





n

1

p

i

0

1=2

n

i

i=1

p

1=p

p

n

i

p

2

i=1

p





p

0 < A , B < 1 are Khinchine constants.

where

p

p

. This is a simple matter of applying Fubini's theorem. First,

Proof

0Z @

1

1

X

r (s)f

dsA

i

0

1

=

i

i=1

! Z Z X r (s)f (t) dt ds ! Z Z X r (s)f (t) ds dt :

1=p

p

n

n

1

i

0

p

0

1

=

1=p

p

1=p

i

i=1 n

1

i

0

p

0

i

i=1

And now, from Khinchine's inequality,

! Z Z X r (s)f (t) ds dt 1

1

p

n

i

0

0

1=p

i

0Z ! 1 X  A@ jf (t)j dtA



! X

jf j

: = A



p

i=1

1=2

n

i

p

2

i=1

p

The upper estimate is similar. We could paraphrase Proposition 10.2 by writing:

X



r (s)f (t)

n

i

1=2

i

Lp ([ 0;1 ]2 )

i=1

! X



jf (t)j

n

i

2

i

0

i=1

1=p

p=2

n

1

i=1

2



:

Lp

P

The idea now is to either compare the left-hand side to, say, k f k , as we might foran unconditionally convergent series,  P  or to compare the rightP hand side to kf k or to kf k , as we might for disjointly supported sequences. As a rst application of these ideas, we present a classical theorem due to Orlicz [101]. n

i=1

2

n

i=1

i

1=2

n

i=1

p

Theorem 10.3 (Orlicz's P1Theorem) L 1p

X

j n j2 k

T en

a

< p
0, there is a  = (") > 0 such that

kxk  1; kyk  1; kx

y

k  " =)



x + y

2 1

:

(11.1)

Obviously, any uniformly convex space is also strictly convex, but there are strictly convex spaces that aren't uniformly convex. An easy compactness argument will convince you that if X is nite dimensional, then X is strictly convex if and only if X is uniformly convex. Just as with strict convexity, uniform convexity is actually a property of the norm on X , so it might be more appropriate to say that X has a uniformly convex norm whenever (11.1) holds. Similar to the strictly convex case, we could reformulate our de nition in terms of pairs of vectors x, y with kxk = kyk = 1 and kx yk = ". Likewise, we could use some power of the norm, say k
kp for p > 1. The details in this case are rather tedious, and not particularly important to our discussion, so we will omit them. As we saw at the beginning of this chapter, the parallelogram law implies that every Hilbert space is uniformly convex. In fact, we even computed  (") = 1 (1 "2=4)1=2. We will see that uniformly convex spaces share many of the geometric niceties of Hilbert space. Our primary aim is to prove the famous result, due to Clarkson in 1936 [23], that Lp is uniformly convex whenever 1 < p < 1. Using this observation, we will give a geometric proof (essentially devoid of measure theory) that Lp = Lq . In fact, we will show that every uniformly convex Banach space is necessarily re exive. Note that since L1 and L1 fail to be even strictly convex, they can't be uniformly convex. The same is true of c0, `1, `1 , and C [ 0; 1 ]. Also note that uniform convexity is very much an isometric property; the equivalent renorming of R2 we gave earlier that fails to be strictly convex obviously also fails to be uniformly convex. Along similar lines, it's easy to check that the expression jjj x jjj = kxk1 + kxk2 de nes an equivalent strictly convex norm on `1 , but that `1 supplied with this norm can't be uniformly convex since it's not re exive. We begin with a simple but very useful observation.

CHAPTER 11. CONVEXITY

128

Lemma 11.6 X is uniformly convex if and only if, for all pairs of sequences (xn), (yn) with kxnk  1, kynk  1, we have:

xn + yn

2 !1

=) kxn ynk ! 0:

Proof. One the one hand, if X is uniformly convex, and if k(xn + yn)=2k ! 1, then we must have kxn ynk ! 0, for otherwise, kxn ynk  " implies k(xn + yn)=2k  1 . On the other hand, if X is not uniformly convex, then we can nd an " > 0 and sequences (xn) and (yn) in BX such that kxn ynk  " while k(xn + yn)=2k ! 1.

Corollary 11.7 Let X be uniformly convex. If (xn) in X satis es kxnk  1 and k(xn + xm)=2k ! 1 as m, n ! 1, then (xn) is Cauchy! Note that the condition k(xn + xm)=2k ! 1 also forces kxnk ! 1 by the triangle inequality: k(xn + xm)=2k  (kxnk + kxmk)=2  1. Thus, if X is complete, (xn ) converges to some norm one element of X . Our last two results can be used to prove several classical convergence theorems, but we'll settle for just one such result: In the case X = Lp, it's sometimes called the Radon-Riesz theorem.

Theorem 11.8 Let X wbe a uniformly convex Banach space, and suppose that (xn) in X satis es xn ! x and kxnk ! kxk. Then kxn xk ! 0. Proof. If kxk = 0 there's nothing to show, so we may suppose that x 6= 0. Now, since kxnk ! kxk, it's easy to see that we can normalize our sequence: w If we set y = x=kxk and yn = xn=kxnk, then xn ! x implies that yn w! y; also, kyn yk ! 0 will imply that kxn xk ! 0. Next, choose a norm one functional f 2 X  with f (y) = 1. Then,

 

yn + ym y n + ym

f  2

 1: 2 But f (yn ) ! f (y) = 1 as n ! 1, and so we must have k(yn + ym)=2k ! 1 as m, n ! 1. From Corollary 11.7, this means that (yn ) is Cauchy and, hence, (yn) converges in norm to some point in X . But, since norm convergence implies weak convergence, and since weak limits are unique, we must have yn ! y in norm.

129 Using a very similar argument, we can give a short proof of an interesting fact, due to Milman in 1938 [93] and Pettis in 1939 [108]. (If you're not familiar with nets, just pretend they're sequences; you won't be far from the truth.) Theorem 11.9 A uniformly convex Banach space is re exive. Proof. Let X be uniformly convex and let x 2 X  with kxk = 1. We need to show that x = x^ for some x 2 X . Now, since BX is weak  dense in BX  , we can nd a net (x ) in X with kxk  1 such that x w! x. But since kxk  1 = kxk, it follows that kxk ! kxk. A slight variation on Theorem 11.8 shows that (x) is Cauchy in X , and hence must converge to some x 2 X . It follows that x = x^. Clarkson's inequalities

To round out our discussion of uniform convexity, we next prove Clarkson's Theorem : Theorem 11.10 Lp is uniformly convex for 1 < p < 1. As it happens, the proof of Clarkson's theorem is quite easy for 2  p < 1, and not quite so easy for 1 < p < 2. For this reason, dozens of proofs have been given. We have the luxury of selecting bits and pieces from several of these proofs. Now Clarkson proved several inequalities for the Lp norm which mimic the parallelogram law. We will do the same, but without using his original proofs. It should be pointed out that all of the results we'll give in this section hold for every Lp () space (after all, this has more to do with the Lp norm than with measure theory). We rst prove Clarkson's theorem for the case 2  p < 1; this particular proof is due to Boas in 1940 [16]. Lemma 11.11 Given 2  p < 1 and real numbers a and b, we have j a + b jp + j a b jp  2p 1 ( jajp + jbjp ): Proof. We've seen this inequality before (more or less). See if you can ll in the reasons behind the following arithmetic: ( j a + b jp + j a b jp )1=p  ( j a + b j2 + j a b j2 )1=2 = 21=2( jaj2 + jbj2 )1=2  21=2 21=2 1=p( jajp + jbjp )1=p = 21 1=p( jajp + jbjp )1=p:

CHAPTER 11. CONVEXITY

130

Theorem 11.12 For 2  p < 1 and any f , g 2 L , we have
kf + gk + kf gk  2 1 kf k + kgk : p

p p

Corollary 11.13

p p

p

p p

(11.2)

p p

Lp is uniformly convex for 2  p < 1.

Proof. If f , g 2 Lp with kf kp  1, kg kp  1, and kf g kp  ", then

kf + gkpp  2p 1
2



"p = 2p 1

 " p 

2 :

That is, (") = 1 1 "2 p 1=p, and this is known to be exact. Unfortunately, inequality (11.2) reverses for 1 < p  2, so we need a dierent proof in this case; the one we'll give is due to Friedrichs from 1970 [46]. As with Boas's proof, we begin with a pointwise inequality.

Lemma 11.14 If 1 < p  2, q = p=(p 1), and 0  x  1, then (1 + x) + (1 x)  2 (1 + x ) 1: q

q

(11.3)

p q

Proof. Consider the function f (; x) = (1 + 1 q x) (1 + x)q

+ (1 1 q x) (1 x)q 1; for 0    1 and 0  x  1. Then, f (1; x) is the left-hand side of (11.3) and f (xp 1 ; x) is the right-hand side of (11.3) because (p 1)(q 1) = 1. Thus, since 1  xp 1, we want to show that @f=@  0. But 1

 @f = ( q 1) x
(1  q )
(1 + x)q 2 @ and (1  q )  0 since   1, and (1 + x)q 2 q  2.

(1 x)q

2

(1 x)q

2



;

 0 since

For p  2, the inequality in (11.3) holds with the roles of p and q exchanged. Our proof shows that the inequality in (11.3), as written, reverses for p  2. For 1 < p  2, the inequality in (11.3) easily implies that j a + b jq + j a b jq  2 ( jajp + jbjp )q 1 for all real numbers a and b. But this time we can't simply integrate both sides to arrive at an Lp result. Instead, we'll need Friedrichs' clever extension of this inequality to Lp.

131

Theorem 11.15 For 1 < p  2, q = p=(p 1), and any f , g 2 L , we have p

kf + gkqp + kf

g

kqp  2 kf kpp + kgkpp


q

1

:

Proof. First notice that

kf k =

Z

q p

jf j

p

1=(p

1)

=

Z

jf j

q (p 1)

1=(p

1)

= k jf jq kp

1

and 0 < p 1 < 1! Now, since the triangle inequality reverses in Lp , 1

k f + g kqp + k f

g

kqp = k j f + g jq kp + k j f g jq kp  k j f + g jq + j f g jq kp  2 k ( jf jp + jgjp )q kp
= 2 kf kpp + kgkpp q ; 1

1

1

1

1

1

where the last equality follows from the fact that (q 1)(p 1) = 1.

Corollary 11.16

Lp

is uniformly convex for 1 < p  2.

Notice that if f , g 2 Lp, 1 < p  2, with kf kp  1,
kg
kp  1, and
kf gkp  ", then, just as before, we'd get (") = 1 1 " q =q  q " q , where q = p=(p 1)  2. This, however, is too small. And "p is too big. It's known that (")  (p 1)" =8 is (asymptotically) the correct value for Lp, 1 < p  2. (Notice that this is essentially the value of (") for Hilbert space; the point here is that no space can be \more convex" than Hilbert space.) For the sake of completeness, we list all of Clarkson's inequalities (well, half of them anyway). Let 2  p < 1 and q = p=(p 1). Then, for any f , g 2 Lp , we have 1

2

1

2

2




2 kf kpp + kgkpp  k f + g kpp + k f g kpp  2p kf kpp + kgkpp
k f + g kpp + k f g kpp  2 kf kqp + kgkqp p ;
2 kf kpp + kgkpp q  k f + g kqp + k f g kqp: 1




;

1

1

These inequalities all reverse for 1 < p  2. In particular, all reduce to the parallelogram identity when p = 2.

CHAPTER 11. CONVEXITY

132

An elementary proof that Lp = Lq In our discussion of the duality between strict convexity and smoothness, we made the claim that smoothness had something to do with dierentiability of the norm. It should come as no surprise, then, that uniform convexity is dual to an ever stronger dierentiability property of the norm. Rather than formalize the condition, we'll settle for a few simple observations. Most of these results are due to McShane from 1950 [86]. We begin with a General Principle: If a linear functional attains its norm at a point of dierentiability of the norm in , then its value is actually given by an appropriate derivative. This is the content of: X

Lemma 11.17 (McShane's Lemma) Let be a normed linear space and let 2 . Suppose that , 2 satisfy (i) k k = 1 and ( ) = k k, k + kp k kp exists for some  1. (ii) lim t!0 X

T

X

f

g

g

T g

g

tf

X

T

g

p

pt

Then

k + kp k kp . ( ) = k k
lim t!0

T f

g

T

tf

g

pt

It should be pointed out that the -th power in (ii) is purely cosmetic|the case = 1 is enough|but it will make life easier when we apply the Lemma to the p norm. The number limt!0(k + kp k kp) is the directional derivative of k
kpp at in the direction of . Proof. First we use l'H^opital's rule to compute ( ) as a derivative. p p p p ( ( ) + ( )) ( ( )) ( ( + )) ( ( )) = lim lim p

p

L

g

g

tf

g

=pt

f

T f

T g

tf

t!0

T g

T g

tT f

t!0

pt

T g

pt

= lim ( ( ) + ( ))p 1
( ) t!0 = ( ( ))p 1
( ) = k kp 1
( ) T g

tT f

T g T

T f

T f

T f :

Now, since k k k k = ( ) and k k k + k  ( + ), we get T

g

T g

T

g

tf

T g

tf

kp k kp  lim ( ( + ))p ( ( ))p p k + lim k k
t!0+ t!0+ T

g

tf

g

pt

T g

tf

T g

pt

133 = kT kp 1
T (f ) (T (g + tf ))p (T (g))p = tlim !0 pt k g + tf kp kgkp : p
 tlim k T k pt !0 McShane's Lemma gives us a schedule: First we show that each T 2 Lp, 1 < p < 1, actually attains its norm, and then we compute the limit given in (ii). Curiously, the fact that each element of Lp attains its norm is equivalent to the fact that Lp is re exive|which doesn't actually require knowing anything about the dual space Lp ! First, a general fact:

Lemma 11.18 If X is a uniformly convex Banach space, then each T 2 X  attains its norm at a unique g 2 X , kg k = 1. Proof. Choose (gn ) in X , kgn k = 1, such that T (gn ) ! kT k. Then

2  kgn + gmk  kT k 1jT (gn + gm)j ! 2 as m, n ! 1. Thus, (gn ) is Cauchy and hence converges to some g 2 X . Clearly, T (g) = kT k and kgk = 1. Uniform (or even strict) convexity tells us that g must be unique. Next, let's do the Calculus step:

Lemma 11.19 Let f , g 2 Lp, 1 < p < 1. Then Z p kg kp k g + tf k p 1 sgn g: lim = f
j g j t!0 pt Proof. For any a, b 2 R, the function '(t) = j a + bt jp is convex and has ' 0(t) = pj a + bt jp 1
b
sgn(a + bt). The limit now follows from the Dominated (or even Monotone) Convergence theorem:

k g + tf kp kgkp = lim t!0 pt =

Z

Z

j g(x) + tf (x) j lim t!0 pt

p

f
jgjp 1 sgn g:

jg(x)jp dx

CHAPTER 11. CONVEXITY

134

Alternatively, we could use the standard approach: Let F (t) = k g + tf k = p p

Z

j g(x) + tf (x) j dx = p

R

Z

(x; t) dx:

Then, F is dierentiable and F 0(t) = t(x; t) dx provided that is integrable. But for, say, 1  t  1, we have

t

exists and

j t(x; t)j = p f (x) jg(x) + tf (x)jp 1sgn (g(x) + tf (x))
 p 2p 1 jf (x)j jg(x)jp 1 + jf (x)jp ; which is integrable because jgjp 1 2 Lq . Now, given T 2 Lp, take g 2 Lp with kgkp = 1 and T (g) = kT k. Then, for every f 2 Lp, k g + tf kp kg kp T (f ) = lim kT k t!0

Z

pt

= kT k f
jgjp 1 sgn g: That is, T is represented by integration against kT k jgjp 1 sgn g 2 Lq , and, of course, k jgjp 1 kq = kgkpp 1 = 1. This proves that Lp = Lq .

Notes and Remarks Our presentation in this chapter borrows from a great number of sources, for example, the books by Beauzamy [10], Day [27], Diestel [29, 31], Hewitt and Stromberg [61], Holmes [62], Kothe [78], and Ray [112], as well as the several original articles cited in the text.

135

Exercises 1. Let f : I ! R be a continuous function de ned on an interval I . If f satis es f ((x + y)=2)  (f (x) + f (y))=2 for all x, y 2 I , prove that f is convex. 2. A convex function f : I ! R is said to be strictly convex if f (x + (1 )y ) < f (x) + (1 )f (y ) whenever 0 <  < 1 and x 6= y . Show that f (x) = jxjp is strictly convex for 1 < p < 1. 3. Give a direct proof that Lp() is strictly convex. 4. Prove that a normed space X is strictly convex if and only if k
kX is a strictly convex function. 5. Show that every subspace of a strictly convex space is strictly convex. Is the same true for uniformly convex spaces? For smooth spaces? 6. Prove that jjj x jjj = kxk1 + kxk2 de nes an equivalent strictly convex norm on `1. 7. Suppose that T : X ! Y is continuous and one-to-one and that Y is strictly convex. Show that jjj x jjj = kxk + kT xk de nes an equivalent strictly convex norm on X . 8. Give a direct proof that `p, 1 < p < 1, is strictly convex. Generalize your proof to conclude that if Xn is strictly convex for each n, then (X1  X2 


)p is strictly convex for any 1 < p < 1. 9. Let 2  p < 1 and de ne g 2 Lp by g(t) = 1 for 0  t  1 ("=2)p, and g(t) = 1 for 1 ("=2)p < t  1. If we set f  1, show that kf kp = kgkp = 1, kf gkp = ", and k(f + g)=2kp = (1 ("=2)p)1=p. Conclude that (") = 1 (1 ("=2)p)1=p is best possible for Lp. 10. Choose a sequence (pn ) with 1 < pn < 1 such that pn ! 1 or pn ! 1, and de ne Xn = `pnn . Show that (X1  X2 


)2 is strictly convex (and even re exive) but not uniformly convex. 11. If xn w! x in X , show that kxk  lim infn!1 kxnk. Give an example where this inequality is strict. If kxnk  kxk for all n, conclude that we actually have kxk = limn!1 kxnk. Show that the same holds for a  w sequence xn ! x in X . 12. Apply McShane's Lemma to show that each linear map T : Rn ! R is given by inner product against a vector x 2 Rn with kxk2 = kT k. 13. Let x 2 `1 with kxk1  1. Show that x is an exposed point of the closed unit ball of `1 if and only if x = ek for some k.

CHAPTER 11. CONVEXITY

136

14. Let f , (f ) be in L , 0 < p < 1. If f ! f a.e. and kf k ! kf k , show that kf f k ! 0. [Hint: Use the fact that 2 (jf j + jf j ) jf f j is nonnegative and tends pointwise a.e. to 2 +1jf j .] Show that the result also holds if we assume instead that f ! f in measure. 15. (a) Show that a point x in the closed unit ball of `1 is an extreme point if and only if x = ("1; : : : ; " ), where " = 1 for i = 1; : : : ; n. Prove that each point in the unit ball of `1 can be written as a convex combination of extreme points. (b) Show that the set of extreme points of the unit ball of `1 consists of all points of the form (" ), where " = 1 for all i = 1; 2; : : :. Show that the set of extreme points of the unit ball of `1 consists of the points e , k = 1; 2; : : :. For 1 < p < 1, every norm one vector in ` is an extreme point of the unit ball. (c) In sharp contrast to the previous cases, show that the closed unit ball in c0 has no extreme points. n

n

p

n

n

p

p

p

n

p

p

n

n

n

n

k

p

i n

i

p p

p

n

p

Chapter 12

C (K ) Spaces The Cantor set In this chapter we will catalog several important properties of the Cantor set
. Our goal in this endeavor is to uncover the \universal" nature of C (
). For starters, we'll prove that
is the \biggest" compact metric space by showing that every compact metric space K is the continuous image of
. Now the presence of a continuous, onto map ' :
! K tells us something about C (K ). Composition with ', i.e., the map f 7! f  ', de nes a linear isometry (and an algebra isomorphism) from C (K ) into C (
). Since
is itself a compact metric space, this means that C (
) is \biggest" among the spaces C (K ), for K compact metric. To begin, recall that the each element of the Cantor set has a ternary (base P 1 3) decimal expansion ofPthe form n=1 an =3n , where an = 0 or 2, and that P 1 an=2n+1 de nes a continuous map n the Cantor function '( 1 n=1 an =3 ) = n=1 from
onto [ 0; 1 ]. This proves:

Lemma 12.1 The interval [ 0; 1 ] is the continuous image of
. As an immediate corollary, we have that C [ 0; 1 ] is isometric to a closed subspace (and subalgebra) of C (
). Does this sound backwards? If so, have patience! For our purposes, a convenient representation of the Cantor set is as the countable product of two-point discrete spaces. This representation is easy to derive from the ternary decimal representation of
. That is, since the elements of
are sequences of 0s and 2s, we have
= f0; 2P gN1. This representation also yields a natural metric on
: Setting d(x; y) = jan bnj=3n , n=1

137

CHAPTER 12.

138

( ) SPACES

C K

where n, n = 0, 2 are the ternary decimal digits for , 2
, respectively, de nes a metric equivalent to the usual metric on
. (That is,
is homeomorphic to the product space f0 2gN supplied with the metric .) Moreover, this particular metric has the additional property that ( ) = ( ) now implies that = . Whenever we speak of \the" metric on
, this will be the one we have in mind. Many authors take
= f 1 1gN, since this choice makes
a under (coordinatewise) multiplication. Now since N can be partitioned into countably many countably in nite subsets, the product space representation of
yields an immediate improvement on our rst Lemma. a

b

x

y

;

d

d x; y

y

d x; z

z

;

group

Corollary 12.2 The cube [ 0 1 ]N is the continuous image of
. ;

Again, this means that ([ 0 1 ]N) is isometric to a closed subspace of (
). C

;

C

Completely regular spaces Recall that a topological space is said to be completely regular if is Hausdor and if, given a point 2 outside a closed set  , there is a continuous function : ! [ 0 1 ] such that ( ) = 1 while = 0 on . In other words, is completely regular if ( ; [ 0 1 ]) separates points from closed sets in . Since singletons are closed in , it's immediate that ( ; [ 0 1 ]) also separates points in . From Urysohn's lemma, every normal space is completely regular. Thus, metric spaces and compact Hausdor spaces are completely regular. Locally compact Hausdor spaces are completely regular, too. In fact, nearly every topological space encountered in analysis is completely regular. Even topological groups and topological vector spaces. From our point of view, there's no harm in simply assuming that every topological space is completely regular. As it happens, the class of completely regular spaces is precisely the class of spaces which embed in some cube [ 0 1 ]A . To see this, we start with: X

X

x

f

F

X

X

F

;

f x

X

C X

X

C X

;

X

f

;

X

X

;

Lemma 12.3 (Embedding Lemma) Let X be completely regular. Then, is homeomorphic to a subset of the cube [ 0; 1 ]C (X ; [ 0;1 ]) .

X

Proof. For simplicity, let's write C = C (X ; [ 0; 1 ]). Let e : X ! [ 0; 1 ]C be the evaluation map, de ned by e(x)(f ) = f (x) for x 2 X and f 2 C . That is, x

139 is made to correspond to the \tuple" e(x) = (f (x))f 2C . That e is one-to-one is obvious because C separates points in X . That e is continuous is easy, too, since each of its \coordinates" f  e = f , f 2 C , is continuous. Now let U be open in X . We want to show that e(U ) is open in e(X ). Here's where we need complete regularity: Given x 2 U , choose an f 2 C such that f (x) = 1 while f = 0 on U c and let V = f 1(f0gc ). Clearly, V is open in the product [ 0; 1 ]C , and e(x) 2 V because e(x)(f ) = 1 6= 0. Finally, e(y ) 2 V =) f (y ) 6= 0 =) y 2 U and hence e(x) 2 V \ e(X )  e(U ). Our proof of the Embedding Lemma shows that if F  C (X ; [ 0; 1 ]) and if e : X ! [ 0; 1 ]F is de ned by e(x)(f ) = f (x), then e is continuous. If, in addition, F separates points, then e is one-to-one. Finally, if F separates points from closed sets, then e is a homeomorphism (into). The Embedding Lemma also tells us that X carries the weak topology induced by C (X ; [ 0; 1 ]). In terms of nets: x ! x in X

() () () ()

e(x) ! e(x) in [ 0; 1 ]C f (e(x)) ! f (e(x)) for every f 2 C (X ; [ 0; 1 ]) f (x ) ! f (x) for every f 2 C (X ; [ 0; 1 ]) x ! x in the weak topology induced by C (X ; [ 0; 1 ]):

Let's consolidate several of these observations.

Theorem 12.4 For a Hausdor topological space X , the following are equivalent:

(a) (b) (c) (d) (e)

X is completely regular; X embeds in a cube; X embeds in a compact Hausdor space; X has the weak topology induced by C (X ; [ 0; 1 ]); X has the weak topology induced by Cb (X ).

Proof. The Embedding Lemma shows that (a) implies (b) implies (d). Of course, (b) and Tychono's theorem imply (c). Since compact Hausdor spaces are completely regular, and since every subspace of a completely regular space is again completely regular, (c) implies (a).

CHAPTER 12. C (K ) SPACES

140

Next, let's check that (d) implies (a). Suppose that X has the weak topology induced by C (X ; [ 0; 1 ]) and let x 2 X be a point outside the closed set F  X . Then, for some f1 ; : : : ; f 2 C (X ; [ 0; 1 ]) and some " > 0, the basic open set U = N (x; f1; : : : ; f ; ") is contained in F . That is, n

c

n

U = N (x; f1; : : : ; f ; ") = f y 2 X : jf (y ) f (x)j < "; i = 1; : : : ; n g  F : c

n

i

i

Now each of the functions g = jf f (x)j is in C (X ; [ 0; 1 ]), as is the function g = maxfg1 ; : : : ; g g. Moreover, g (x) = 0 while g (y )  " for all y 2 F  U . Hence h = " 1 minf "; g g 2 C (X ; [ 0; 1 ]) satis es h(x) = 0 and h(y) = 1 for all y 2 F  U . Thus, X is completely regular. Finally, the fact that (d) is equivalent to (e) is easy. Since C (X ; [ 0; 1 ])  C (X ), it's clear that each \C (X ; [ 0; 1 ]) open" set is also \C (X ) open." On the other hand, given x 2 X , f 2 C (X ), and " > 0, put M = kf k1 and notice that g = (f + M )=2M 2 C (X ; [ 0; 1 ]) satis es i

i

i

c

n

c

b

b

b

f y 2 X : jf (x) f (y)j < " g = f y 2 X : jg(x) g(y)j < "=2M g: Thus, each \C (X ) open" set is also \C (X ; [ 0; 1 ]) open." If X is completely regular and if some countable family F  C (X ; [ 0; 1 ]) separates points from closed sets in X , then X embeds in the cube [ 0; 1 ]N and, hence, is metrizable. (Compare this with Urysohn's metrization theorem: A normal, second countable space is metrizable.) This result has a converse of sorts, too. If, for example, X is a separable metric space, then some countable family F  C (X ; [ 0; 1 ]) will separate points from closed sets in X . In particular, X will embed in [ 0; 1 ]N. More to the point for us is: b

Lemma 12.5 Every compact metric space is homeomorphic to a closed subset of [ 0; 1 ]N.

Proof. Let K be a compact metric space, and let (x ) be dense in K . We may assume that the metric d on K satis es d(x; y)  1. Given this, de ne : K ! [ 0; 1 ]N; by (x)(n) = d(x; x ). Clearly, is one-to-one and continuous (each coordinate (
)(n) = d(
; x ) is continuous). Since K is compact and [ 0; 1 ]N is Hausdor, is a homeomorphism (into) and the result follows. Lest you be fooled into thinking that only countable products of intervals are separable, here is an easy counterexample: Example. [ 0; 1 ][ 0 1 ] is separable but not metrizable. n

n

n

;

P

141

Proof. Consider the collection D of all functions of the form =1 q  , where q1; : : : ; q are rationals in [ 0; 1 ], and where J1 ; : : : ; J are disjoint closed intervals in [ 0; 1 ] with rational endpoints. Then D is a countable subset of [ 0; 1 ][ 0 1 ]. Now the typical basic open set in [ 0; 1 ][ 0 1 ] is given by: n

n

i

i

Ji

n

;

;

N (f ; x1 ; : : : ; x ; ") = n

f g : jg (x ) i

f (x )j < "; i = 1; : : : ; n g; i

where x1; : : : ; x 2 [ 0; 1 ], " > 0, and f : [ 0; 1 ] ! [ 0; 1 ]. Given a basic open set N (f ; x1; : : : ; x ; "), choose rationals q1 : : : ; q in [ 0; 1 ] with jq f (x )j < " choose disjoint rational intervals J with x 2 J . Then g = Pfor =1eachq  i, and 2 D \ N (f ; x1; : : : ; x ; "). Thus D is dense and so [ 0; 1 ][ 0 1 ] is separable. To see that [ 0; 1 ][ 0 1 ] is not metrizable, note that it's not sequentially compact. Your favorite sequence of functions f : [ 0; 1 ] ! [ 0; 1 ] with no pointwise convergent subsequence will do the trick. The sequence (1 + r )=2, where r is the n-th Rademacher function, comes to mind. n

n

n

i

i

i

i

i

n

i

i

Ji

;

n

;

n

n

n

We're now ready to establish our claim that the Cantor set is the \biggest" compact metric space.

Theorem 12.6 Every compact metric space K is the continuous image of
. Proof. We know that K is homeomorphic to a closed subset of [ 0; 1 ]N and, hence, that K is the continuous image of some closed subset of
. To nish the proof, then, it suces to show that each closed subset of
is the continuous image of
. P So, let F be a closed subset of
and let d(x; y) = 1=1 ja b j=3 be \the" metric on
. Given x 2
, notice that the distance d(x; F ) = inf 2 d(x; y) is attained at a unique point y 2 F (and, in particular, y = x whenever x 2 F ). De ne f (x) = y for this unique y. To check that f is continuous, let x ! x in
. Now, since F is compact, we may assume that y = f (x ) converges to some point z 2
. But then d(x ; y ) ! d(x; z ) and d(x ; y ) = d(x ; F ) ! d(x; F ) = d(x; y ), so we must have y = z ; that is, y ! y = f (x). n

y

n

n

n

F

n

n

n

n

n

n

n

n

n

Corollary 12.7 If K is a compact metric space, then C (K ) is isometric to a closed subspace (even a subalgebra ) of C (
).

CHAPTER 12.

142

( ) SPACES

C K

This completes the rst circle of ideas in this chapter: (
) is \universal" for the class of spaces ( ), compact metric. This may seem odd in view of the small role that
typically plays in a rst course in analysis. Indeed, [ 0 1 ] is usually given much more emphasis. The reader who is uncomfortable with this turn of fate can take heart in the fact that (
) embeds isometrically into [ 0 1 ]. That is, [ 0 1 ] is universal, too. We'll need a bit more machinery before we can give a proof|note that we can't hope to prove this claim by nding a continuous map from [ 0 1 ] onto
! Now each 2 (
) extends to a continuous function on [ 0 1 ]. This already would follow from Tietze's extension theorem, for example, but there may be many such extensions. We want to choose a method for extending that will lead to a linear map from (
) into [ 0 1 ]. And the most natural extension does the job. The complement of
in [ 0 1 ] is the countable union of disjoint open intervals. The endpoints of these open intervals are, of course, elements of
. Simply \connect the dots" in the graph of across the endpoints of each of these open intervals to de ne an extension ~ for . (This is quite like the procedure used to extend the Cantor function to all of [ 0 1 ].) Note that if ~( ) is the \average" of two values of on
. Clearly, then, 2
, then ~ sup0x1 j ( )j = supx2
j ( )j. What's more, it's not hard to see that if we're given , 2 (
), then + = ~ + ~. We'll take this as proof of: C

C K

C

K

;

C

C

;

C

;

;

f

C

;

f

C

C

;

;

f

f

x =

f x

f x f

f x

g

C

^

f

f

;

f

g

f

g

Lemma 12.8 The extension map ( ) = ~ from (
) into [ 0 1 ] is a E f

linear isometry.

f

C

C

;

Theorem 12.9 (
) is isometric to a complemented subspace of [ 0 1 ]. C

C

;

Proof. To complete the proof, we need only note that restriction to
de nes a linear map R : C [ 0; 1 ] ! C (
) with the property that P = E R is the identity on E (C (
)). That is, E (C (
)) is an isometric copy of C (
) and is the range of a bounded projection P : C [ 0; 1 ] ! C [ 0; 1 ].

Corollary 12.10 If is a compact metric space, then ( ) is isometric to K

C K

a closed subspace of C [ 0; 1 ]. In particular, C (K ) is separable.

Our last Corollary should be viewed as a rough analogue of the Weierstrass theorem valid in ( ) for a compact metric space . What's more, the converse is also true: If ( ) is separable, for a given compact Hausdor C K

K

C K

143 space K , then K is metrizable. We'll forego the details just now, but this issue will come up again. The universality of C (
) (or C [ 0; 1 ]) reaches beyond the class of C (K ) spaces; in fact, every separable normed space is isometric to a subspace of C (
). (Even more is true: Every separable metric space is isometric to a subset of C (
).) By our last Corollary, we only need to check that each separable normed linear space embeds in some C (K ), K compact metric. At least part of this claim is easy to check and is valid for any normed linear space. As a consequence of the Hahn-Banach theorem, a normed linear space X is isometric to a subspace of Cb(BX  ), under the sup norm, via point evaluation: x 7! x ^jBX , where x^(x) = x(x). That is, the embedding of X into Cb(BX  ) is nothing other than the canonical embedding of X into X  with the additional restriction that each element of X  is to be considered as a function just on BX  . What remains to be seen is whether BX  can be given a suitable topology that will turn it into a compact metric space (thus making Cb(BX  ) = C (BX  )). For an in nite-dimensional space X , the norm topology on BX  won't do (it's never compact). We'll need something weaker. The topology that ts the bill is the weak topology on X , restricted to BX  . The weak topology is the topology that X  inherits as a subset of the product space RX. That is, each x 2 X  is identi ed with its range (x(x))x2X = (^x(x))x2X , considered as an element of RX. Under this identi cation, x^ is the projection onto the \x-th coordinate" of X . Thus, the weak topology on X  is the smallest topology on X  making every x^ continuous (or, in still other words, the weak topology is the weak topology on X  induced by Xb = X ). A neighborhood base for the weak topology is generated by the sets N (x; x1 ; : : : ; xk ; ")

= f y  2 X  : j(y 

j

x )(xi ) < "

for i = 1; : : : ; k g;

where x 2 X , x1; : : : ; xk 2 X , and " > 0. In terms of nets, x

! x in

w

X

() x^(x) ! x^(x) for each x^ 2 X^ () x(x) ! x(x) for each x 2 X () x ! x in RX:

Of particular merit here is that X is still isometric to a subspace of Cb(BX  ), even when BX  is supplied with the weak topology (since each x^ is weakcontinuous). That this observation has brought us one step closer to our goal is given as:

CHAPTER 12.

144

C (K )

SPACES

Theorem 12.11 (Banach-Alaoglu) If X is a normed linear space, then BX

is compact in the

weak

topology on

X .



Proof. As we've already seen, X  , under the weak topology, is homeomorphic to a subset of RX. If we cut down to BX  , then we can do a bit better. Again its range (x(x))x2X , but now considered we identify each x 2 BX  with Q as an element of the product x2X [ kxk; kxk ]. This identi cation is still a homeomorphism (into), by virtue of the de nition of the weak topology. Q Since x2X [ kxk; kxk ] is compact, we will be done once we show that the image of BX  is closed in the product topology. But what does this really mean? We need to check that the pointwise limit of a net x 2 BX  of linear functions on X is again linear and has norm at most one|which is clear.

Corollary 12.12 Every normed linear space is isometric to a subspace of

C (K ),

for some compact Hausdor space K .

We want to embed a separable normed space X into C (K ) where K is a compact metric space. Our next result shows the way.

Theorem 12.13 If X is a separable normed linear space, then the weak topol-

ogy on BX  is both compact and metrizable.

Proof. That BX  is compact in the weak topology P is immediate. Now let (xn) be dense in SX , and de ne d(x; y) = 1n=1 j(x y)(xn )j=2n for x, y  2 BX  . It's easy to see that d is a metric on BX  . Next, notice that d(x ; y )

 1max j(x nM
0, there exists a  > 0 such that A jf j < " for all f 2 F and for all Borel sets A  [ 0; 1 ] with m(A) < .

Proposition 13.1

Proof. First R suppose that F is uniformly integrable. Given " > 0, choose a such that fjf j>a g jf j < "=2 for all f 2 F and now choose  > 0 such that a < "=2. Then, for f 2 F and A  [ 0; 1 ], we have Z

Z

Z

jf j = jf j + jf j A\fjf j>a g A\f jf ja g  "=2 + a
m(A) < " whenever m(A) < . Thus, F is uniformly absolutely continuous. Now suppose that F is uniformly absolutely R continuous. Then F is norm bounded. Indeed, choose  > 0 such that A jf j < 1 whenever f 2 F and A

151

m(A) < , and partition [ 0; 1 ] into M = 1 + [2=] subintervals, each of length at most =2. It follows that kf k1 < M for each f 2 F. R Given " > 0, choose  > 0 such that A jf j < " whenever f 2 F and m(A) < . Next, from Chebyshev's inequality, notice that for f 2 F we have mf jf j > a g  kfak1 < Ma : R Thus, if a is chosen so that M=a < , then fjf j>a g jf j < ". Thus, F is uniformly integrable. To place our discussion in the proper context, it will help to recall the measure algebra associated with ([ 0; 1 ]; B; m). To begin, it is an easy exercise to check that d(A; B ) = m(A4B ) de nes a pseduometric on B. Thus, if we de ne an equivalence relation on B by declaring A  B if and only if m(A4B ) = 0, and if we write Be to denote the set of equivalence classes under this relation, then d de nes a metric on Be. In fact, d is a complete metric Be. This is easiest to see if we rst make an observation: For A and B in B, we have m(A4B ) = d(A; B ) = k A B k1: Thus, if we agree to equate sets that are a.e. equal, then Be inherits a complete metric from L1. The complete metric space (Be; d ) is called the measure algebra associated with ([ 0; 1 ]; B; m). In this setting, a Borel measure  is absolutely continuous with respect to m if and only if  is continuous at ? in (Be; d ). (Measures behave rather like linear functions on Be. In particular, we only need to worry about continuity at \zero.") And a subset F of L1 is uniformly integrable if and only if the corresponding set of measures F is equicontinuous at ? when considered as a collection of functions on (Be; d ). A special case is worth isolating:

Lemma 13.2 Given f 2 L1, de ne  : Be ! R by (A) = RA f . Then,  is uniformly continuous on (Be; d ). Proof. (Note that  is well-de ned.) This is a simple computation: Z

A

f

Z

B



f =

Z

AnB

f

Z

B nA



f 

Z

A4B

jf j:

Thus, j(A) (B )j can be made small provided that d(A; B ) = m(A4B ) is suciently small (where \small" depends on f but not on A or B ).

CHAPTER 13. WEAK COMPACTNESS IN L1

152 Corollary 13.3

in (Be; d ).



Given f 2 L1 and " > 0, the set A : j

R

Afj"



is closed

Finally we're ready to make some connections with weak compactness (or, in this case, weak convergence) in L1. Let (fn ) be a sequence in L1 such that limn!1 in R for every Borel subset A of [ 0; 1 ]. Then

Proposition 13.4

R

A fn

exists

(a) (fn) is uniformly integrable and R R (b) (fn) converges weakly to some f in L1; in particular, A fn ! A f for every Borel subset A of [ 0; 1 ]. Proof. For " > 0 and N = 1; 2; : : :, de ne

FN =



Z e:

A2B

A

(fm

fn )



 " for all m; n  N :


R S Since A fn is Cauchy for each A 2 Be, we have Be = 1N =1 FN . But each FN is closed and Be is complete. By Baire's theorem, then, some FN0 must have nonempty interior. Thus, there exists some A0 2 Be and some r > 0 such that m(A4A0) < r implies A 2 FN0 . Suppose now that m(B ) < r. Then, for m, n  N , we have Z

and also

B

(fm fn) =

Z

A0 [ B

(fm fn)

Z

A0 nB

(fm fn)

m((A0 [ B )4A0) < r and m((A0 n B )4A0) < r: Hence, B (fm fn )  " + " = 2". Applying the same argument to the sets B \ f fm fn  0 g and B \ f fm fn  0 g R then yields B jfm fn j  4" whenever m(B ) < r and m, n  N0. Now the set F = f fj : R1  j  N0 g is uniformly integrable and so there exists an s < r such that B jfj j < " whenever 1  j  N0 and m(B ) < s. Finally, if m(B ) < s and j > N0, we have R

Z

B

Z

Z

jfj j  jfj fN j + jfN j B B  4" + " = 5": 0

0

153 Thus (fn) is uniformly integrable, which proves (a). R Next, for each A 2 Be, put (A) = limn!1 A fn . Then  is countably additive (this follows from the fact that (fn ) is uniformly integrable) and absolutely continuous with respect to m. Thus, yRm theorem, R by the Radon-Nikod R there exists an f 2 L1 such that (A) = A f . That is, A Rfn ! A fR for every A 2 B. It follows (from linearity of the integral) that fn g ! fg for every simple are dense in L1, we get R R function g . Since the simple functions w fn g ! fg for every g 2 L1 . That is, fn ! f in L1, which proves (b).

Corollary 13.5 L1 is weakly sequentially complete.

R

Proof. Suppose that (fn ) is weakly Cauchy in L1. Then, in particular, A fn converges for every A 2 B. Hence, (fn ) converges weakly to some f 2 L1 by Proposition 13.4. Finally we're ready to characterize the weakly compact subsets of L1.

Theorem 13.6 A subset F of L1 is relatively weakly compact if and only if it is uniformly integrable.

Proof. Suppose that F is not uniformly integrable. Then there exists an R " > 0 such that for each n we can nd an fn 2 F with f jfnj>n g jfnj  ". In particular, (fn) has no uniformly integrable subsequence. It then follows from Proposition 13.4 that (fn) has no weakly convergent subsequence. Thus, F cannot be relatively weakly compact. (This follows from the Eberlein-Smulian theorem: A subset of a normed space is weakly compact if and only if it's weakly sequentially compact.) Now suppose that F is uniformly integrable. We will show that the weak closure of F is contained in L1 (considered as a subset of L 1 ) and, hence, that F is weakly compact. To this end, supposeRthat G is in the weak closure of F . Given " > 0, choose  > 0 such that A jf j < " whenever f 2 F and m(A) < . Since G is a weak cluster point of F , it follows that jG(A )j < " whenever m(A) < . Thus, the measure A 7! G(A) is absolutely continuous with respect toR m. By the Radon-Nikodym theorem, there is a g 2 L1 such that G(A) =R A g and it now follows (from the linearity and continuity of G) that G(h) = hg for every h 2 L1. That is, G = g 2 L1. The same proof applies to any L1() where  is a positive nite measure. Thus, if  is positive and nite, the weakly compact subsets of L1() are precisely the uniformly integrable subsets. This leads us to our nal (but most useful) corollary.

CHAPTER 13. WEAK COMPACTNESS IN L1

154

(Vitali-Hahn-Saks Theorem) Let (n ) be a sequence of signed measures on a  -algebra  such that (A) = limn!1 n (A) exists in R for each A 2 . Then  is a signed measure on . Corollary 13.7

Pn=1 jnj=2nknk (where kk = jj(X ), the total variation Proof. Let  = 1 of  applied to the underlying measure space X ). Then each n is absolutely continuous respect to ; that is, dn = fn d for some fn 2 L1(). Thus, R fn d = with ( A ) ! (A) for each A 2  and so, by Proposition 13.4, there n A exists an f 2 L1() such that Z

A

f d = nlim !1

Z

A

fn d = (A)

for every A 2 . That is, d = f d and it follows that  is a signed measure (i.e.,  is countably additive) on .

Notes and Remarks Our presentation in this chapter borrows heavily from some unpublished notes for a course on Banach space theory given by Stephen Dilworth at the University of South Carolina. The Eberlein-Smulian theorem can be found in any number of books, but see also [105] and [132].

155

Exercises 1. Prove that the sequence (n[ 0 1 ]) is not uniformly integrable in L1. 2. If (f ) converges to f in L1, give a direct proof that (f ) is uniformly integrable. 3. Let (f ) be a sequence in L1. If f ! f a.e., show that the following are equivalent: (a) (f ) is uniformly integrable; (b) kf f k1 ! 0 as n ! 1; (c) kf k1 ! kf k1 as n ! 1. 4. If (f ) converges weakly to f in L1, show that (f ) is uniformly integrable. 5. Prove that d(A; B ) = m(A4B ) de nes a complete pseudometric on B. 6. Given a measure  : B ! R, prove that the following are equivalent: (a)  is absolutely continuous with respect to m; (b)  is continuous at ? in (Be; d ); (c)  is uniformly continuous on (Be; d ). 7. Prove that a subset F of L1 is uniformly integrable if and only if the corresponding set of measures F = f f dm : f 2 F g is equicontinuous at ? in (Be; d ). ; =n

n

n

n

n

n

n

n

n

n

156

CHAPTER 13. WEAK COMPACTNESS IN L1

Chapter 14 The Dunford-Pettis Property A Banach space is said to have the Dunford-Pettis property if, whenever ! 0 in and ! 0 in , we have that ( ) ! 0. Our main result in this chapter will show that the Dunford-Pettis property is intimately related to the behavior of weakly compact operators on . Recall that a bounded linear operator : ! is said to be weakly compact if maps bounded sets in to relatively weakly compact sets in . Thus, is weakly compact if and only if ( ) is weakly compact in . Since weakly compact sets are norm bounded, it follows that a weakly compact operator is bounded. Our rst result provides several equivalent characterizations of weak compactness for operators. X

xn

w

X

fn

w

X

fn xn

X

T

X

Y

T

X

Y

T BX

T

Y

Theorem 14.1 Let :

T X ! Y be bounded and linear. Then, compact if and only if any one of the following hold: (a) T (X )  Y ; (b) T  : Y  ! X  is weak -to-weak continuous; (c) T  is weakly compact.

T

is weakly

Proof. Suppose that T is weakly compact. Then T (B ) is relatively weakly compact in Y . Regarding T (B ) as a subset of Y , we have X

X

(Y  ;weak )

( ) because weakly compact sets in convex, this simpli es to read ( ) T BX

T BX

Y

= ( ) are weak compact in

weak

T BX

= (

T BX

157

(Y ;weak)

) 

Y:

Y



. Since (

T BX

) is

CHAPTER 14. THE DUNFORD-PETTIS PROPERTY

158

But X is weak dense in weak continuous, so B

T



B

(

X  ,

by Goldstine's theorem, and

X  )

B

T



is weak-to-



 ( X ) weak  T B

Y:

Thus, ( )  . Conversely, if ( )  , then ( X  ) is a weakly compact subset of by the Banach-Alaoglu theorem and the weak-to-weak continuity of  (and, again, the observation that the weak topology on , when considered as a subset of , reduces to the weak topology on ). Thus ( X ) is relatively weakly compact in , being a subset of ( X  ). This proves that is weakly compact if and only if (a) holds. Now, from (a), T

X

Y

T

X

Y

T

B

Y

T

Y

Y

Y

Y

T

T

is weakly compact ()

() () ()

()

T T



(

X



  x

)

T

B

Y

2 for each Y



x

2

X



is weak continuous for each  2     ( ) !  ( ) for each  2   whenever  w!  in   (   ) ! (  ) for each  2   whenever  w!  in     w   whenever  w!  in   !  is weak-to-weak continuous.

T

 

T

x

x

x

y

T

x

y

x

T y

x

T y

y

;

y

X

x

X

;

Y

T y

y

() ()

T B

;

y

x

X

;

Y

T y

y

y

Y

T

This proves that is weakly compact if and only if (b) holds. Finally, from (b), if is weakly compact, then  is weak-to-weak continuous. But Y  is weak compact in , so we have that ( Y  ) is weakly compact in . Thus,  is weakly compact. Conversely, if  is weakly compact, then ( X  ) is weakly compact in . Thus, ( X ) is relatively weakly compact in . T

T

T

Y

B

X

T

B

T

T

B

T

Y

T B

Y

Armed with these tools, we can now provide our main result in this chapter.

Theorem 14.2 A Banach space

X has the Dunford-Pettis property if and only if every weakly compact operator from X into a Banach space Y maps weakly compact sets in X to norm compact sets in Y ; i.e., if and only if every weakly compact operator is completely continuous.

159 Proof. Suppose rst that X has the Dunford-Pettis property and let T : X ! Y be weakly compact. To begin, we consider the action of T on a weakly null sequence (xn) in X . For each n, choose a norm one functional gn 2 Y  such that gn (T xn) = kT xnk. That is, T gn (xn) = kT xnk. Now T  is weakly compact, so there is an f 2 X  and a subsequence (gnk ) of (gn) such that w! f . But then  T gnk n

( n )=

g k Tx k



( n ) = ( nk ) + (

n

T g k x k

f x



n

T g k

f

Indeed, ( nk ) ! 0 because ( n) is weakly null and ( because has the Dunford-Pettis property. Thus, f x

x

)( nk ) ! 0 x



n

T g k

f

:

)( nk ) ! 0 x

X

( n ) = k nk k ! 0

n

g k Tx k

Tx

:

That is, ( n) has a norm null subsequence whenever ( n) is weakly null. By linearity, it follows that ( n) has a norm convergent subsequence whenever ( n) is weakly convergent. Consequently, ( ) is norm compact whenever is weakly compact. Now suppose that every weakly compact operator on maps weakly compact sets to norm compact sets. Given n w! 0 in and n w! 0 in , consider the map : ! 0 de ned by = ( n( )). It's easy to see that  :  satis es  1 ! n = n . In particular, given  2  , we have Tx

x

Tx

x

T K

K

X

x

T

T

`

X

X

c

T e

T

  x

X

Tx

f

f

( n) = e



(



X

x

x

x

f

X

n ) = x(fn ) ! 0

T e

since ( n ) is weakly null. That is,   2 0. In other words, is weakly compact. But then maps weakly compact sets in to norm compact sets in 0 and it follows that we must have k nk1 ! 0. (Why?) Consequently, n ( n ) ! 0. f

T

x

c

T

c

f

X

Tx

x

Suppose that X has the Dunford-Pettis property. If is weakly compact, then T 2 : X ! X is compact.

Corollary 14.3 X

T

T

:

X

!

Proof. Since T is weakly compact, T (BX ) is relatively weakly compact. And, since X has the Dunford-Pettis property, T 2(BX ) is then relatively norm compact. Suppose that X has the Dunford-Pettis property. If complemented re exive subspace of X , then Y is nite dimensional.

Corollary 14.4

Y

is a

CHAPTER 14. THE DUNFORD-PETTIS PROPERTY

160

Proof. If P : X ! X is any projection onto Y , then P is weakly compact. Indeed, P (BX ) = BY is weakly compact since Y is re exive. But then P 2 = P is compact. Consequently, BY is norm compact and it follows that Y must be nite dimensional.

Corollary 14.5 In nite dimensional re exive Banach spaces do not have the Dunford-Pettis property. Our task in the remainder of this chapter will be to show that C (K ) and L1 have the Dunford-Pettis property. Given this, it will follow that C (K ) and L1 have no in nite dimensional, re exive, complemented subspaces.

Theorem 14.6 Let K be a compact Hausdor space. Then C (K ) has the Dunford-Pettis property.

Proof. Suppose that fn w! 0 in C (K ) and that n w! 0 in C (K ). Let

 =

1 X

n=1

jn j ; n 2 kn k

where kn k = jnj(K ). Then (n ) is uniformly absolutely continuous with respect to  (by Propositions 13.1 and 13.4). Thus, given " > 0, there exists a  > 0 such that jn j(A) < ", for all n, provided that (A) < . Now fn ! 0 pointwise on K . Thus, by Egorov's theorem, fn ! 0 uniformly on some set K n A where (A) < . Next,

n (fn ) = and

R

K nA fn dn

Z

K

fn dn =

Z

K nA

fn dn +

Z

A

fn dn

! 0 as n ! 1 because fn ! 0 uniformly on K n A. Finally,

Z

A

n

fn d  kfn k1
jn j(A)  "
sup kfk k1 k

because (A) < . Thus, n (fn) ! 0.

Corollary 14.7 If X is a separable in nite dimensional re exive Banach space, then X is isometric to an uncomplemented subspace of C [ 0; 1 ].

161 We next attack the Dunford-Pettis property in L1. To this end, we will need some additional information about weak convergence in L1. Suppose that gn w! 0 in L1 . Then, given " > 0, there exists a Borel set A  [ 0; 1 ] with m(A) > 1 " such that gn ! 0 uniformly on A. Proposition 14.8

Proof. By repeated application of Lusin's theorem, we can nd a Borel subset B of [ 0; 1 ], with m(B ) > 1 "=2, and a sequence of functions g~n , each continuous on B , such that gn = g~n a.e. on B . Moreover, by the Lebesgue density theorem, we may assume that each point x 2 B has density 1; that is, m((x r; x + r) \ B ) = 1 lim r!0 2r for each x 2 B (thus, in particular, B has no isolated points). It follows that n X



ak g~k (x)  ess:sup B

n X



ak g k 

n

X



ak g k





1 for all scalars (ak ) and all x 2 B . Hence, for each x 2 B , the map gn 7! g~n(x) extends to a bounded linear functional on [ gn ]. But then, since (gn ) is weakly null, we must have g~n (x) ! 0 as n ! 1 for each x 2 B . Hence, gn ! 0 a.e. on B . Finally, by Egorov's theorem, there exists a Borel set A  B with m(A) > 1 " such that gn ! 0 uniformly on A. k=1

Corollary 14.9

k=1

k=1

L1 has the Dunford-Pettis property.

Proof. Let fn w! 0 in L1 and let gn w! 0 in L1 . Then (fn ) isR uniformly integrable in L1. Thus, given " > 0, there is a  > 0 such that B jfnj < ", for all n, provided that m(B ) < . By Proposition 14.8, there is a Borel set A with m(A) > 1  such that gn ! 0 uniformly on A. Thus, Z 1

0



fn gn 

Z

jfn gn j + jfn gn j A  "
sup kgn k1 + sup kgn A k1
sup kfnk1; Ac

which tends to 0 as n ! 1. Corollary 14.10

nonre exive.

Z

n

n

n

Every complemented in nite dimensional subspace of L1 is

162

CHAPTER 14. THE DUNFORD-PETTIS PROPERTY

Notes and Remarks Our presentation in this chapter borrows heavily from some unpublished notes for a course on Banach space theory given by Stephen Dilworth at the University of South Carolina. Theorem 14.1 is often called Gantmacher's theorem, after Vera Gantmacher, who proved the theorem for separable spaces. The general case was later settled by Nakamura [97]. Theorem 14.9 (in dierent language) is due to Dunford and Pettis [37]. The Dunford-Pettis property was so-named by Grothendieck [56], who gave us Theorem 14.2, Theorem 14.6, and a wealth of other results. For more on the Dunford-Pettis property (as well as its history), see Diestel and Uhl [32].

163

Exercises 1. If T : X ! Y is bounded and linear, prove that T  : X  ! Y  is weak-to-weak continuous. 2. If K is weakly compact in X , prove that K is weak closed as a subset of X . 3. If K is a subset of X such that the weak closure of K is again contained in X (when considered as a subset of X ), prove that K is relatively weakly compact. 4. If K  X is weak compact as a subset of X , prove that K is weakly compact in X . 5. Prove the `1 has the Dunford-Pettis property. 6. Let A be a Borel subset of [ 0; 1 ] with Lebesgue density 1 and let g : A ! R be continuous. Prove that jg(x)j  ess:supA jgj for every x 2 A. 7. Prove that if X  has the Dunford-Pettis property, then X does too. Thus, c0 has the Dunford-Pettis property. 8. Prove that a Banach space X has the Dunford-Pettis property if and only if every weakly compact operator T : X ! c0 maps weakly compact sets in X to norm compact sets in c0. (In other words, we need only consider Y = c0 in Theorem 14.2.)

164

CHAPTER 14. THE DUNFORD-PETTIS PROPERTY

Chapter 15

C (K ) Spaces II By now, even a skeptical reader should be thoroughly sold on the utility of embeddings into cubes. But the sales pitch is far from over! We next pursue the consequences of a result stated earlier: If X is completely regular, then Cb(X) completely determines the topology on X. In brief, to know Cb(X) is to know X. Just how far can this idea be pushed? If Cb(X) and Cb(Y ) are isomorphic (as Banach spaces, as lattices, or as rings), must X and Y be homeomorphic? Which topological properties of X can be attributed to structural properties of Cb(X) (and conversely)? These questions were the starting place for Marshall Stone's 1937 landmark paper, Applications of the theory of Boolean rings to general topology [128]. It's in this paper that Stone gave his account of the Stone-Weierstrass theorem, the Banach-Stone theorem, and the Stone-C ech compacti cation. (These few are actually tough to nd among the dozens of results in this mammoth 106 page work.) Paraphrasing a passage from his introduction: \We obtain a reasonably complete algebraic insight into the structure of Cb(X) and its correlation with the structure of the underlying topological space." Stone's work proved to be a goldmine|the digging continued for years!|it's in uence on algebra, analysis, and topology alike can be seen in virtually every modern textbook. Independently, but later that same year (1937), Eduard C ech [22] gave another proof of the existence of the compacti cation but, strangely, credits a 1929 paper of Tychono for the result (see Shields [124] for more on this story). To a large extent, we will be faithful to C ech's approach, which leans more toward topology than algebra. 165

166

CHAPTER 15.

C (K )

SPACES II

The Stone-C ech compacti cation Given a Hausdor topological space X , any compact Hausdor space Y which contains a dense subspace homeomorphic to X is called a Hausdor compacti cation for X . What this means, in practice, is that we look for any compact Hausdor space Y which admits a homeomorphic embedding f : X ! Y from X into Y ; the closure of f (X ) in Y then de nes a compacti cation of X . There is a hierarchy of compacti cations, the full details of which aren't necessary just now. It shouldn't come as a surprise that, for locally compact spaces, the one-point compacti cation is the smallest compacti cation in this hierarchy. What we're after is the largest compacti cation.  comGiven a completely regular space X , we de ne X , the Stone-Cech C X ; pacti cation of X , to be the closure of e(X ) in [ 0; 1 ] . From the Embedding Lemma (Lemma 12.3), X is then homeomorphic to a dense subset of the compact Hausdor space X . Note that if X is compact, then e is a homeomorphism from X onto X . Strictly speaking, the compacti cation is de ned to be the pair (e; X ), but we will have little need for such formality. In fact, we often just think of X as already living inside X , and simply ignore the embedding e. X is characterized by the following extension theorem: (

;[ 0 1 ])

Theorem 15.1 (Extension Theorem) Let X be a completely regular space

and let e : X ! X be the canonical embedding.

(a) Every bounded, continuous function f : X ! R extends to a continuous function F : X ! R in the sense that F  e = f . (b) If Y is a compact Hausdor space, then each continuous function f : X ! Y extends to a continuous function F : X ! Y in the sense that F  e = f . If Y is a compacti cation of X , then F is onto. In particular, every Hausdor compacti cation of X is the continuous image of X . Proof. (a): Suppose that f : X ! R is continuous and bounded. By composing with a suitable homeomorphism of R, we may assume that f : X ! [ 0; 1 ]. But then f is one of the \coordinates" in the product space [ 0; 1 ]C X ; . We claim that the coordinate projection f , when restricted to X , is the extension we want. Indeed, f : X ! [ 0; 1 ] is continuous, and f , when restricted to e(X ), is just f since f (e(x)) = e(x)(f ) = f (x). Note that F = f j X is unique since e(X ) is dense in X . (

;[ 0 1 ])

167 (b): Now suppose that Y is a compact Hausdor space and that f : X ! Y is continuous. Let CX = C (X ; [ 0; 1 ]) and CY = C (Y ; [ 0; 1 ]), and let eX : X ! [ 0; 1 ]CX and eY : Y ! [ 0; 1 ]CY be the canonical embeddings (which we may consider as maps into X and Y , respectively). Note that since Y is compact, eY is a homeomorphism from Y onto Y . In order to extend f to X , we rst \lift" f to a mapping  from [ 0; 1 ]CX to [ 0; 1 ]CY . To better understand this lifting, recall that f (x) 2 Y corresponds to the \tuple" (g(f (x))g2CY in [ 0; 1 ]CY under the evaluation map eY . Also note that g  f 2 CX whenever g 2 CY . [ 0; 1 ]CX - [ 0; 1 ]CY id

6

X

6

id

j X -

Y

6@ 6 @@ e 1 eY eX Y F @ R@ ? -Y X f

We now de ne  : [ 0; 1 ]CX ! [ 0; 1 ]CY by specifying the coordinates of its images: g((p)) = gf (p) for each g 2 CY and each p 2 [ 0; 1 ]CX . The map  is continuous (since its coordinates are) and   eX = eY  f : g ((eX (x))) = gf (eX (x)) = g (f (x)) = g (eY (f (x))): Next, since (eX (x)) = eY (f (x)) 2 eY (Y ), we have that ( X )  eY (Y ) = eY (Y ). Consequently, F = eY 1  j X  eX extends f . Again, uniqueness of F follows from the fact that e(X ) is dense in X . Finally, if f : X ! Y is a homeomorphism from X onto a dense subspace of Y , then F : X ! Y maps X onto a dense, compact subset of Y . That is, F is onto. The conclusion of part (b) of the Extension Theorem is that the StoneCech compacti cation is the largest Hausdor compacti cation of X (in a categorical sense). The Extension Theorem tells us something new about the space Cb(X ); note that the map F 7! F  e de nes a linear isometry from C ( X ) onto

168

CHAPTER 15.

C (K )

SPACES II

Cb (X ).

That is, each f 2 Cb(X ) is of the form F  e for some F 2 C ( X ). In particular, we now know that Cb(X ) is a C (K ) space for some (rather speci c!) compact Hausdor space K .

Corollary 15.2 Let X be completely regular. (i) Cb(X ) is isometrically isomorphic to C ( X ). (ii) If Y is completely regular, then each continuous function f : X ! Y lifts to a continuous function F : X ! Y satisfying F  eX = eY  f . (iii) If Y is any Hausdor compacti cation of X enjoying the property of X described in part (a) of the Extension Theorem, then Y is homeomorphic to X . Proof. Only (iii) requires a proof. Let Y be a compact Hausdor space, and suppose that f : X ! Y is a homeomorphism from X onto a dense subspace of Y . Suppose further that each h 2 Cb (X ) extends to a continuous function g 2 C (Y ) with g  f = h. Then, in particular, each h 2 C (X ; [ 0; 1 ]) is of the form g  f for some g 2 C (Y ; [ 0; 1 ]). This implies that the \lifting"  of f , constructed in the proof of part (b) of the Extension Theorem, is one-to-one. Thus, the extension F : X ! Y of f is both one-to-one and onto, hence a homeomorphism.

For later reference, we next present two simple methods for computing the Stone-C ech compacti cation.

Lemma 15.3 Let X be completely regular. (a) Let T  X  X . If each bounded, continuous, real valued function on T extends continuously to X , then T = cl X T (the closure of T in X ). (b) If X  T  X , then T = X . Proof. (a): By design, each bounded, continuous, real valued function on T extends all the way to X , hence also to cl X T . Since cl X T is a compacti cation of T , it must, then, be T by Corollary 15.2 (iii). (b): First observe that T is dense in X and so X is a compacti cation of T . Next, each bounded, continuous, real valued function on T extends continuously to X since its restriction to X does. The fact that X is dense

169 in T takes care of any uniqueness problems caused by extending the restriction (or restricting the extension. . . ). Thus, by Corollary 15.2 (iii), X = T . We're way overdue for a few concrete examples.

Examples

1. (0; 1) 6= [ 0; 1 ]. Why? Because sin(1=x) has no continuous extension to [ 0; 1 ]! But sin(1=x) does, of course, extend continuously to (0; 1), whatever that is! As we'll see in a moment, (0; 1) is much larger than [ 0; 1 ]. 2. If D is any discrete space, then `1 (D) = Cb(D) = C ( D), isometrically. In particular, `1 = C ( N). Since `1 isn't separable, we now have a proof that N isn't metrizable. In fact, as we'll see, N is in no way sequentially compact. 3. card( N) = 2 . Here's a clever proof: Recall that Y = [ 0; 1 ][ 0;1 ] is separable. Hence, there is a continuous map from N onto a dense subset of Y . This map extends to a continuous map from N onto all of Y . Consequently, 2 = card(Y )  card( N), while from the construction of N it follows that card( N)  card [ 0; 1 ][ 0;1 ] = 2 . c

c

N

c

4. card( (0; 1)) = card( R) = card( N). The rst equality is obvious, since (0; 1) is homeomorphic to R. Next, as above, R is the continuous image of N, because R is separable. Thus, card( R)  card( N). To nish the proof, we'll nd a copy of N living inside R. Here's how: Each bounded (continuous) real-valued function on N extends to a bounded continuous function on all of R. (No deep theorems needed here! Just \connect the dots.") Thus, by Lemma 15.3 (a), cl RN = N. Hence, card( R)  card( N). 5. Banach limits (invariant means). We will use the fact that `1 = C ( N) to extend the notion of \limit" to include all bounded sequences. First, given x 2 `1, let's agree to write x~ for its unique extension to an element of C ( N). Now, given any ( xed) point t 2 N n N, we de ne: Lim x = x~(t). This generalized limit, often called a Banach limit, satis es: Lim x = lim x; if x actually converges; lim inf x  Lim x  lim sup x; Lim(ax + by) = a Lim x + b Lim y; Lim(x y) = (Lim x) (Lim y):

CHAPTER 15. C (K ) SPACES II

170

Just for fun, let's check the rst claim. The key here is that t is in the closure of fn : n  mg for each m. Thus, if L = lim x exists, and if " > 0, then xn 2 [ L "; L + " ] for all n  m for some m. Hence, x~(t) 2 [ L "; L + " ], too. That is, x~(t) = L. What we've actually found is a (particularly convenient) Hahn-Banach extension of the functional \lim" on the subspace c of `1 . What makes this example interesting, as we'll see later, is that no point t 2 N n N can be the limit of a sequence in N. 6. From our discussion of completely regular spaces in the last chapter, we can give an alternate de nition of the Stone-C ech compacti cation. Each completely regular space T lurks within Cb(T ) under the guise of the point masses: P = f t : t 2 T g. It follows that we can de ne T to be the weak closure of P in Cb(T ). Why? Well, since P = weak-cl P is a compacti cation of T , we only need
to show that each element f 2 Cb (T ) ^ extends to an element f 2 C P , and this is easier than it might sound: Just de ne f^(p) to be p(f )! In other words, the canonical embedding of Cb(T ) into Cb (T ) supplies an embedding of Cb(T ) into C ( T ). 7. Finally, here's a curious proof of (a special case Q of) Tychono's theorem based on the Extension Theorem: Let X = 2A X, where each X is a (nonempty) compact Hausdor space. Then, X is completely regular. Hence, the projection maps  : X ! X have continuous extensions ~ : X ! X. But this means that the map p 7! (~(p))2A, from X onto X , is continuous! Thus, X is compact. Return to

C (K ).

N is a most curious space, and will play a major role in the next chapter. More generally, as our examples might suggest, the Stone-C ech compacti cation of a discrete space is a potentially useful tool. This is further highlighted by the following observation:

Lemma 15.4 Every compact Hausdor space K is the continuous image of

D for some discrete space D. Consequently, C (K ) is isometric to a subspace of C ( D) = `1 (D). Proof. Let D0 be any dense set in K , and let D be D0 with the discrete topology. Then, the formal identity from D into K extends to a continuous

171 map ' from D onto K ! The composition map f 7! f  ' de nes a linear isometry from C (K ) into C ( D). As an immediate Corollary, we get a result that we've (essentially) seen before. Corollary 15.5 If K is a compact metric space, then C (K ) embeds isometrically into `1 . Hence, every separable normed linear space embeds isometrically into `1 .

In the next chapter we will compute the dual of C (K ) by, instead, computing the dual of `1 (D). That is, we will prove the Riesz representation theorem for `1 spaces, and then transfer our work to the C (K ) spaces. This being the case, we might be wise to quickly summarize a few features of the \fancy" versions of the Riesz representation theorem. First, the spaces C (K ) and its relatives CC (X ), Cb(T ), and so on, are probably best viewed as vector lattices . Under the usual pointwise ordering of functions, C (K ) is an ordered vector space: f; g 2 C (K ); f  g =) f + h  g + h for all h 2 C (K ); f 2 C (K ); a 2 R; f  0; a  0 =) af  0:

and

In addition, C (K ) is a lattice: f; g 2 C (K ) =) f

_ g = maxff; gg and f ^ g = minff; gg are in C (K ) This last observation allows us to de ne positive and negative parts: f + = f _0 and f = (f ^ 0). Thus, each f 2 C (K ) can be written as f = f + f . Moreover, jf j = f + + f .

Now C (K ) enjoys the additional property that its usual norm is compatible with the order structure in the sense that

jf j  jgj =) kf k1  kgk1 : Since C (K ) is also complete under this norm, we say that C (K ) is a Banach lattice . As it happens, the dual of a Banach lattice can be given an order structure, too: We de ne an order on C (K ) by de ning S  T to mean that S (f )  T (f ) for all f  0. In particular, a linear functional T on C (K ) is positive if

CHAPTER 15.

172 T (f )  0 whenever f

C (K )

SPACES II

 0. It's easy to see that every positive linear functional

is bounded; indeed, if T is positive, then

jT (f )j  T (jf j)  T (kf k1
1)  kf k1
T (1); where 1 denotes the constant 1 function. Hence, kT k = T (1). What's a little harder to see is that the dual space will again be a Banach lattice under this order. We rst need to check that each bounded linear functional can be written as the dierence of positive functionals. Given T 2 C (K ), we de ne T + (f )

= supf T (g) : 0  g  f g for f  0:

It's tedious, but not dicult, to check that T + is additive on positive elements and that T +(af ) = aT +(f ) for a  0. Now for arbitrary f we de ne T + (f ) = T + (f + ) T +(f ). It follows that T + is positive, linear, and satis es T + (f )  T (f ) for every f  0. Thus, T = T + T is likewise positive and linear. That is, we've written T as the dierence of positive linear functionals. Consequently, a linear functional on C (K ) is bounded if and only if it can be written as the dierence of positive linear functionals. Finally, let's compute the norm of T in terms of T + and T . Clearly,

kT k  kT +k + kT k =

T + (1) + T

(1):

On the other hand, given 0  f  1, we have j2f 1j  1 and hence kT k  T (2f 1) = 2T (f ) T (1). By taking the supremum over all 0  f  1 we the get kT k  2T +(1) T (1) = T +(1) + T (1): Hence, kT k = T +(1) + T (1). If we de ne jT j = T + + T , as one would expect, then we have kT k = k jT j k = jT j(1). It follows from this de nition of jT j that C (K ) is itself a Banach lattice. In terms of the Riesz representation theorem, all of this tells us that we only need to represent the positive linear functionals on C (K ). As you no doubt already know, each positive linear functional on C (K ) will turn out to be integration against a positive measure. The generic linear functional will then be given by integration against a signed measure. In terms of measures,  = +  is the Jordan decomposition of , while jj = R+ +  is the total variation of . Not surprisingly, we de ne kk = jj(K ) = K 1 d jj = k jj k.

173

Notes and Remarks

The Stone-C ech compacti cation is discussed in any number of books; see, for example, Folland [44, Chapter 4], Gillman and Jerison [49], Wilansky [133], or Willard [134]. For more on Banach limits and their relationship to N, see Nakamura and Kakutani [98]. Banach lattices are treated in a number of books; see, for example, Aliprantis and Burkinshaw [3], Lacey [80], MeyerNieberg [92], or Schaefer [120].

174

CHAPTER 15.

C (K )

SPACES II

Exercises 1. Prove Corollary 15.2 (ii). 2. Let X be completely regular. Show that X is locally compact if and only if X is open in X . 3. Complete the proof of the claims made in Example 5 concerning the Banach limit Lim x on `1 .

Chapter 16

C (K ) Spaces III In this chapter we present Garling's proof [48] of the Riesz representation theorem for the dual of C (K ), K compact Hausdor. This theorem goes by a variety of names: The Riesz-Markov theorem, the Riesz-Kakutani theorem, and others. The version that we'll prove states:

Theorem 16.1 Let K be a compact Hausdor space, and let T be a positive linear functional on C (K ).R Then there exists a unique positive Baire measure  on K such that T (f ) = K f d for every f 2 C (K ). As we pointed out in the last chapter, our approach will be to rst prove the theorem for `1 spaces. To this end, we will need to know a bit more about the Stone-C ech compacti cation of a discrete space and a bit more measure theory. First the topology.

The Stone-C ech compacti cation of a discrete space A topological space is said to be extremally disconnected , or Stonean , if the closure of every open set is again open. Obviously, discrete spaces are extremally disconnected. Less mundane examples can be manufactured from this starting point:

Lemma 16.2 If D is a discrete space, then D is extremally disconnected. Proof. Let U be open in D, and let A = U \ D. Then A is dense in U , since U is open, and so cl DA = cl D U . Now we just check that cl DA is

175

176

CHAPTER 16. C (K ) SPACES III

also open. The characteristic function A : D ! f0; 1g (a continuous function on D!) extends continuously to some f : D ! f0; 1g. Thus, by continuity, cl D A = f 1(f1g) is open. By modifying this proof, it's not hard to show that a completely regular space X is extremally disconnected if and only if X is extremally disconnected. Since we won't need anything quite this general, we'll forego the details. Notice that if A and B are disjoint (open) sets in a discrete space D, then cl DA and cl DB are disjoint in D. Indeed, just as in the proof of Lemma 16.2, the function A extends continuously to a function f : D ! f0; 1g which satis es cl D A = f 1 (f1g) and cl DB  f 1(f0g). In particular, any set of the form cl DA, where A  D, is clopen ; that is, simultaneously open and closed. In fact, every clopen subset of D is of this same form.

Lemma 16.3 Let D be a discrete space. Then the clopen subsets of D are of the form cl D A, where A is open in D. Further, the clopen sets form a base for the topology of D. Proof. If C is a clopen subset of D, then, just as in Lemma 16.2,

C = cl DC = cl D (C \ D): Now let U be an open set in D, and let x 2 U . Since D is regular, we can nd a neighborhood V of x such that x 2 V  cl D V  U . Since cl D V is clopen, this nishes the proof.

A few facts about N We can now shed a bit more light on N. Note, for example, that N is open in N. Indeed, given n 2 N, the set cl Nfng is open in N. But fng is compact, hence fng = cl Nfng. That is, fng is open in N, too. In particular, each n 2 N is an isolated point in N. It follows that a sequence in N converges in N if and only if it is eventually constant ; that is, if and only if it already converges in N. Suppose, to the contrary, that (xn) is a sequence in N which is not eventually constant, and suppose that (xn) converges to a point t 2 N n N. Then the range of (xn) must be in nite, for otherwise (xn) would have a subsequence converging in N. Thus, by induction, we can choose a subsequence (xnk ) of distinct integers;

177

xn = xn if i = j . But now the sets A = xn2 and B = xn2 1 are disjoint in N, while t is in the closure of each in N; a contradiction. In particular, we've shown that no point t N N can be the limit of a sequence in N. As a consequence, N isn't sequentially compact (and thus isn't metrizable). i

6

6

j

f

2

k

g

f

k

g

n

A similar argument shows that, for any t N N, the compact set t isn't a G in N. Indeed, if t were a G , then we could nd a sequence of clopen T 1 sets of the form Bn = cl NAn, where An N, such that t = n=1 Bn . The sets (An) have the nite intersection property, so we can choose a sequence of distinct points xn TA1 An. Putting A = xn : n N , we would then 1 have cl NA N n=1 Bn = t . But since A is an in nite subset of N, it follows that cl NA is homeomorphic to N and, in particular, has cardinality 2 , a contradiction. What we've shown, of course, is that every nonempty G subset of N N has cardinality 2 . This observation will be of interest in our discussion of measures on N. 2

n

f g

f g



2

n

f g

\


\



f

2

g

f g

c

c

n

If A is an in nite subset of N, then cl NA is homeomorphic to N. Since N can be partitioned into in nitely many disjoint, in nite subsets (An), it follows that N contains in nitely many pairwise disjoint clopen S sets Bn = cl NAn, each homeomorphic to N. Note, however, that T = 1n=1 Bn is not all of N since a compact space can't be written as a disjoint union of in nitely many disjoint open sets. Since N T N, we do have that T is dense in N; moreover, T = N. 



Using this observation, we can build a copy of N inside the closed set N N. To see this, let tn Bn N = cl NAn N and set D = tn : n 1 . Obviously, D is a discrete subspace of N and, as such, is homeomorphic to N. Thus, D is homeomorphic to N. Now, given a bounded real-valued function f on D, we can easily extend f to a bounded continuous function on T by setting f (x) = f (tn ) for every x Bn. Since D T N = T , it follows that D = cl ND. But since D is a subset of the closed set N N, so is D. In short, we've just found a copy of N in N N. n

2

n

n

2

f





g



n

n

As a very clever argument demonstrates, there are, in fact, c disjoint copies of N living inside N N. Indeed, recall that we can nd c subsets (E)2A of N such that each E is in nite, and any two E have, at most, a nite intersection. For each , the set F = cl NE N is then homeomorphic to N N, and so contains a copy of N. Finally, notice that the F are pairwise disjoint since, for each  = , the set cl N(E E ) diers from cl NE in only a nite subset of N. n

n

n

6

n

CHAPTER 16. C (K ) SPACES III

178

\Topological" measure theory Now for some measure theory. Our job, remember, is to compute the dual of C ( D), where D is discrete. We know that there are enough clopen sets in D to completely determine its topology and, so, enough clopen sets to completely determine C ( D). It should come as no surprise, then, that there are also enough clopen sets to determine C ( D). The clopen sets in D form an algebra of sets which we will denote by A; the -algebra generated by A will be denoted by . Two more -algebras will enter the picture: B, the Borel  -algebra on D, and B0, the Baire  -algebra on D. The Baire  -algebra is the smallest -algebra B0 such that each f 2 C ( D) is B0-measurable. It's not hard to see that B0 and  are sub--algebras of B. The next lemma shows that we also have B0  :

Lemma 16.4 Each f 2 C ( D) is -measurable. Moreover, the simple functions based on clopen sets in A are uniformly dense in C ( D). Proof. Let f 2 C ( D) and let  2 R. Then:

ff  g =

\f f >  1

n=1

1=n g 



\ ff >  \f f   1

n=1

1

n=1

1=n g 1=n g (by continuity)

= f f   g: Thus we have equality throughout. It then follows from Lemma 16.2 that the set f f   g is the countable intersection of clopen sets and, as such, is in . Hence, f is -measurable. The second assertion follows from the fact that the nitely-many-valued functions are dense in `1(D). Since each f 2 C ( D) is -measurable, we must have B0  . On the other hand, since each clopen subset of D can be realized as a \zero set" for some f 2 C ( D), we also have A  B0, and hence   B0. Thus, the  -algebra of Baire sets on D coincides with the  -algebra generated by the clopen sets in D. Please note that our proof also shows that f f   g is a compact G in D. The Baire -algebra on any \reasonable" space turns out to be the -algebra generated by the compact G sets. In contrast, note that the Borel -algebra

179 on any compact Hausdor space could be de ned as the -algebra generated by the compact sets. We brie y describe a few such cases below. For a locally compact space X , the Baire -algebra B0 is de ned to be the smallest -algebra on X such that each element of CC (X ) is measurable, where CC (X ) is the space of continuous real-valued functions on X with compact support.

Lemma 16.5 Let X be a locally compact Hausdor space. (a) If f 2 C (X ) is nonnegative, then f f   g is a compact G for every  > 0.

C



(b) If K is a compact G in X , then there is an f 2 CC (X ) with 0  f  1 such that K = f 1 (f1g). (c) The Baire -algebra in X is the -algebra generated by the compact G sets in X . Proof. (a): For  > 0, the set f f   g is a closed subset of the support of f , T 1 hence is compact. And, as before, f f   g = n=1f f >  1=n g is also a G . T (b): Suppose that K = 1n=1 Un, where Un is open. Apply Urysohn's lemma to nd an fn 2 CC (X ; [ 0; 1 ]) with fn = 1 on K and fn = 0 o Un. P 1 Then, f = n=1 2 n fn is in CC (X ; [ 0; 1 ]) and f = 1 precisely on K . (c): Let G be the -algebra generated by the compact G sets in X . From (a), each f 2 CC (X ) is G -measurable, hence B0  G . From (b), each compact G is a Baire set, and so G  B0 .

If D is discrete, then the Baire sets in D are typically a proper sub-algebra of the Borel sets in D. If D is in nite, then a cardinality argument, similar to the one we used for N, would show that, for t 2 D n D, the compact set ftg is not a G in D. Our next Lemma explains why we never seemed to need the Baire sets before: On R, or on a compact metric space, the Baire sets coincide with the Borel sets.

Lemma 16.6 Let X be a second countable, locally compact Hausdor space.

Then:

(i) Every open set in X is a countable union of compact sets.

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SPACES III

(ii) Every compact set in X is a G . (iii) The Baire and Borel -algebras on X coincide. Proof. Since X is locally compact, it has a base of compact neighborhoods. Since X is second countable, we can nd a base consisting of only countably many such compact neighborhoods. Thus, (i) follows. And (ii) clearly follows from (i) by taking complements. Finally, (iii) follows from (i), (ii), and Lemma 16.5 (c).

For good measure, here's another example:

Example

On an uncountable discrete space D, the Baire sets are a proper sub-algebra of the Borel sets. Proof. Since D is discrete, the only compact subsets of D are nite. It follows that the Baire -algebra on D is the -algebra generated by the singletons:

B0 = f E : E or E c is countable g: Since we can write D as the union of two disjoint subsets, each having the same cardinality as D, we can obviously nd an (open) subset of D which is not a Baire set.

The dual of `1 We are now well-prepared to compute the dual of `1(D). As you might imagine, the continuous linear functionals on `1 (D) should look like integration against some measure on D. As a rst step in this direction, we introduce the space ba (2D ), the collection of all nitely additive signed measures of nite variation on 2D , supplied with the norm kk = jj(D), where jj is the total variation of . The name \ba" stands for \bounded (variation and nitely) additive." Recall that the total variation of  is de ned by

jj(E ) = sup

(X n

i=1

j(Ei )j : E1; : : : ; En disjoint, E 

which applies equally well to nitely additive measures.

[ ) n

i=1

Ei

;

181 But what isR meant by integration against such measures? Well, if  2 ba (2D ), then D f d is well de ned and linear for simple ( nitely-manyP valued) functions f . Now, given a simple function f , write f = ni=1 aiE , where E1; : : : ; En are disjoint and partition D. Then: i

Z

D

f d



=

n

X

i=1

jai(Ei )j  kf k1

n

X

i=1

j(Ei)j  kf k1 kk:

R

Thus, D f d de nes a bounded linear functional on the subspace of simple functions in `1 (D). Since the simple functions are dense in `1 (D), this means R that D f d extends unambiguously to allR f 2 `1 (D). This unique linear extension to `1 (D) is what we mean by D f d. With this understanding, our work is half done!

Theorem 16.7

`1 (D) = ba (2D ), isometrically.

Proof. As we've justR seen, each  2 ba (2D ) de nes a functional x 2 `1 (D) by setting x(f ) = D f d, for f simple, and extending to all of `1 (D). And, as the calculation above shows, kxk  kk. Next, the \hard" direction: Let x 2 `1(D) , and de ne (E ) = x(E ), for E  D. Clearly,  is nitely additive, we just need to check that  is of bounded variation. Given disjoint subsets E1; : : : ; En of D, we have n

X

i=1

n

j(Ei)j =

X

=

X

i=1 n i=1

= x

jx(E )j i

"i x (E ); i

n

X

i=1

for some "i = 1

!

"i E

i

 kxk; since

by linearity

;

X



n

i=1

"i E



i

= 1:

1

Thus, kk = jj(D)  kxk. Also, by linearity, we have that x(f ) = D f d for any simple function f . Since both functionals are continuous and agree R  on a dense subspace of `1 (D) we necessarily have x (f ) = D f d for all f 2 `1 (D). R

CHAPTER 16. C (K ) SPACES III

182

Combining the two halfs of our proof, we arrive at the conclusion that the correspondence x $  is a linear isometry between the spaces `1 (D) and ba (2D ). Please note that our proof actually shows something more: The positive linear functionals in `1 (D) correspond to positive measures in ba (2D ).

The Riesz representation theorem for C ( D) While knowing the dual of `1(D) should be reward enough, we could hope for more from our result. It falls just short of the full glory of the Riesz representation theorem for C ( D): Optimistically, we'd like to represent the elements of C ( D) as regular, countably additive measures on D. As you can imagine, we might want to explore the possibility of applying, say, Caratheodory's extension theorem to the elements of ba (2D ). But notice, please, that the natural -algebra associated to integration on C ( D) is the Baire -algebra; in this case . The approach we'll take only supplies a Baire measure. It's a fact, however, that every Baire measure on a compact Hausdor space is regular; moreover, there is a standard technique for extending a Baire measure to a unique regular Borel measure (see [44, Chapter 7], for example). Now while the elements of ba (2D ) are not typically countably additive, they do satisfy a somewhat weaker property. Given a sequence of disjoint sets (Ai) in D, we have: 1 X

i=1

i

(A ) 

1 X

i=1

j(Ai)j  jj

1 [

i=1

!

Ai  kk < 1:

Hence if  is nonnegative, then 1 X

i=1

(Ai)  

1 [

i=1

!

Ai :

If Sis not countably additive, then what could the sum 1i=1 (Ai) miss that  ( 1i=1 Ai) picks up? The answer comes from D. The fact that disjoint sets in D have disjoint closures in D allows us to de ne a \twin" of  on D: We identify each subset A of D with A = cl DA in D and we de ne (A ) = (A). The set function  is a nitely additive measure on A, the algebra of clopen subsets of D. We can use  to explain how  might fall short of being countably additive. P

183 If (Ai) is a sequence of disjoint subsets of D, then, in general, 1 [

1 [

Ai  6 = i=1

i=1

Ai :

Why? Because the union on the left is open while the union on the right is compact ; equality can only occurif all but nitely many of the Ai's are empty ! S  P Thus, in general, (Ai)   Ai ; that is,  fails to account for the S closure of Ai . But this same observation shows us thatS is actually countably additive on A, the algebra of clopen sets. Indeed, if Ai is clopen, then it's actually S a nite union and so must equal Ai. In the terminology of [44],  is a \premeasure" on A. Thus, we can invoke Caratheodory's theorem to extend  to a (regular) countably additive measure on , the -algebra generated by A. The extension  will P still satisfy (A ) = (A) whenever A  D, of course. In particular, if fP= ni=1 aiA is a simple function based on disjoint subsets of D, then f~ = ni=1 aiA is a simple function based on disjoint sets in A, and we have i

i

Z

D

f d =

n X i=1

ai(Ai) =

n X i=1

ai(Ai) =

Z

D

f~ d:

What this means is that we can represent the elements of C ( D) as integration against regular, countably additive measures on . If T : `1 (D) ! C ( D) is the canonical isometry, notice that T maps the characteristic function of a set A in D to the characteristic function of cl DA in D. Thus, T maps simple functions based on sets in 2D to simple functions based on clopen sets in A. Now a functional x 2 C ( D) induces a functionalRx  T on `1 (D). Hence, there is a measure  2 ba (2D ) such that x(Tf ) = D f d for every f 2 `1 (D). If f is a simple function, then Tf = f~ in the notation we used above, and so Z Z  ~  f~ d: x (f ) = x (Tf ) = f d = D

D

Since the simple functions R based on clopen sets are uniformly dense in C ( D),  we must have x (g) = D g d for every g 2 C ( D). That is, we've arrived at the Riesz representation theorem for C ( D). In symbols: C ( D) = `1 (D) = ba (2D ) = rca (); where rca () denotes the space of regular, countably additive measures on .

CHAPTER 16. C (K ) SPACES III

184

Theorem 16.8 Given a continuous linear functional x 2 C ( D), there exists a unique signed measure  on the Baire sets in D such that

x(f )

Z

=

Moreover, kxk = jj( D).

D

f d

for all f 2 C ( D):

Although we have not addressed uniqueness in this representation, it follows the usual lines. The fact that we have equality of norms for the representing measure can again be attributed to the fact that the simple functions are dense in C ( D). Now we're ready to apply this result to the problem of representing the elements of C (K ) as integration against Baire measures on K . To begin, let K be a compact Hausdor space. Next, we choose a discrete space D and a continuous, onto map ' : D ! K . Then, as you'll recall, the map f 7! f  ' de nes a linear isometry from C (K ) into C ( D); in other words, each continuous f : K ! R \lifts" to D by way of f ' : D ! R. Thus, each x 2 C (K ) extends to a functional y 2 C ( D) satisfying x(f ) = y(f  '), for all f 2 C (K ), and kyk = kxk. That is, y is a Hahn-Banach extension of the functional g 7! x(g  ' 1) de ned on the image of C (K ) in C ( D). (And it's not hard to check that a positive functional has a positive extension.) From the Riesz representation theorem for C ( D), we can nd a measure , de ned on the Baire sets in D such that

y  (g ) Hence

x(f )

=

y  ( f  ')

=

Z

=

Z D

D

g d

(f  ') d =

for all g 2 C ( D): Z K

f d (  ' 1 )

for all f 2 C (K );

and  (A) = (' 1 (A)) de nes a Baire measure on K . The uniqueness of  follows from the regularity of  and Urysohn's lemma; R it requires checking that K f d = 0 for all f 2 C (K ) forces   0. The details are left as an exercise.

Notes and Remarks Our presentation in this chapter borrows heavily from Garling's paper [48], but see also Diestel [31], Gillman and Jerison [49], Hartig [59], Holmes [62],

185 Kelley [76], and Yosida and Hewitt [137]. An approach to Riesz's theorem that would have pleased Riesz can be found in Dudley's book [35].

186

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SPACES III

Exercises 1. Show that X is extremally disconnected if and only if disjoint open sets in X have disjoint closures. 2. Let X be completely regular. Show that X is extremally disconnected if and only if X is extremally disconnected. 3. If D is an in nite discrete space, prove that D is not sequentially compact and, hence, not metrizable. 4. Let K be a compact Hausdor space and let B and B0 denote the Borel and Baire -algebras on K , respectively. Prove that B0  B.

Appendix A Topology Review We denote a topological space by ( T ), where is a set and T is a topology on . That is, T is a collection of subsets of , called open sets , Ssatisfying (i) ?, 2 T , (ii) , 2 T =) \ 2 T , and (iii) A  T =) A 2 T . The closed sets in are the complements of the open sets; that is, a subset of is closed if c is open. As shorthand, reference to the topology T is often only implicit, as in the phrase: \Let be a topological space. . . ." Every set supports at least two topologies. Indeed, it's easy to check that f? g is a topology on , called the indiscrete topology , and that P ( ), the power set of , is a topology on , called the discrete topology . We say that is a discrete space if is endowed with its discrete topology. Please note that every subset of a discrete space is both open and closed. Once we have the notion of an open set, we can consider continuous functions between topological spaces: A function : ! from a topological space to a topological space is continuous if 1 ( ) is open in whenever is open in . The collection of all continuous functions from into is denoted by ( ; ). In case = R, we shorten ( ; R) to ( ). Various subsets of ( ), such as b( ), C ( ), etc., have the same meaning as in the introductory chapter. We can also consider compact sets: A subset of a topological space is said to be compact if every covering of by open sets admits a nite subcover; is, is compact if, given any collection of open sets U satisfying Sf : 2that U g  , we can always reduce to nitely many sets 1 n2U with 1 [


[ n  . It's easy to see that compact sets are necessarily closed. X;

X

X

X

X

U

V

U

V

X

E

X

E

X

X

;X

X

X

X

X

X

X

f

X

Y

U

X

f

Y

U

X

Y

X

C X Y

C X

Y

C

X

C X

C

C X

X

K

X

K

K

V

V

V

K

V

V ;:::;V

K

187

Y

APPENDIX A. TOPOLOGY REVIEW

188

Separation Recall that a topological space ( T ) is said to be Hausdor if distinct points in can always be separated by disjoint open sets; that is, given 6= 2 , we can nd disjoint sets , 2 T such that 2 and 2 . Please note that in a Hausdor space, each singleton f g is a closed set. More generally, each compact subset of a Hausdor space is closed. Just as with metric spaces, \closed" will mean \closed under limits" (we'll make this precise shortly), while \compact" will mean \limits exist in abundance." We would prefer those existential limits to land back in our compact set, so it's helpful to know that a compact set is closed. For this reason (among others), it's easier to do analysis in a Hausdor space. In fact, it's quite rare for an analyst to encounter (or even consider!) a topological space that fails to be Hausdor. Henceforth, we will assume that ALL topological spaces are Hausdor. Be forewarned, though, that this blanket assumption may also mean that a few of our de nitions are fated to be nonstandard. Metric spaces and compact Hausdor spaces enjoy an even stronger separation property; in either case, disjoint closed sets can always be separated by disjoint open sets. A Hausdor topological space is said to be normal if it has this property: Given disjoint closed sets , in , there are disjoint open sets , in T such that  and  . Normality has two other characterizations, each important in its own right. The rst is given by Urysohn's lemma: In a normal topological space, disjoint closed sets can be completely separated . That is, if and are disjoint closed sets in a normal space , then there is a continuous function 2 ( ; [ 0 1 ]) such that = 0 on while = 1 on . The second is Tietze's extension theorem: If is a closed subset of a normal space , then each continuous function 2 ( ; [ 0 1 ]) extends to a continuous function ~ 2 ( ; [ 0 1 ]) on all of . Urysohn's lemma and Tietze's theorem are each equivalent to normality. The interval [ 0 1 ] can be replaced in either statement by an arbitrary interval [ ]. Further, it follows from Tietze's theorem that if is a closed subset of a normal space , then every 2 ( ) extends to an element of ( ) (simply by composing with a suitable homeomorphism from R into (0 1)). X;

X

x

U

V

x

U

y

y

X

V

x

E

U

V

E

U

F

F

X

V

E

F

X

C X

;

f

f

E

f

F

E

f

C X

;

X

C E

;

f

X

;

a; b

E

X

C X

f

C E

f

;

Locally compact Hausdor spaces If has \enough" compact neighborhoods, then C ( ) will have enough functions to take the place of ( ) in certain situations. In this context, X

C

C X

X

189 \enough" means that X should be locally compact . A locally compact Hausdor space is one in which each point has a compact neighborhood; i.e., given x 2 X , there is an open set U containing x such that U is compact. It's an easy exercise to show that a locally compact Hausdor space has a wealth of compact neighborhoods in the following sense: Given K  U  X , where K is compact and U is open, there is an open set V with compact closure such that K  V  V  U . This observation (along with a bit of hard work!) leads to locally compact versions of both Urysohn's lemma and Tietze's extension theorem. Here's how they now read (please note that CC (X ) is used in place of C (X ) in each case): Let X be a locally compact Hausdor space, and let K be a compact subset of X . (Urysohn) If F is a closed set disjoint from K , then there is an f 2 CC (X ; [ 0; 1 ]) such that f = 0 on K while f = 1 on F . (Tietze) Each element of C (K ) extends to an element of CC (X ). An alternate approach is to consider the one point compacti cation of X . The one point compacti cation X  of a topological space X is de ned to be the space X  = X [ f1g, where 1 is a distinguished point appended to X , and where we de ne a topology on X  by taking the neighborhoods of 1 to be sets of the form f1g [ U , where U c is compact in X . It's easy to see that X  is a compact space that contains X as an open, dense subset (this is what it means to be a compacti cation of X ). It is likewise easy to see that X  is Hausdor precisely when X is locally compact and Hausdor. Thus, if X is a locally compact Hausdor space, then X is a dense, open subset of the compact Hausdor (hence normal) space X . Consequently, we can now take advantage of such niceties as Urysohn's lemma and Tietze's extension theorem in X  and then simply translate these properties to X . In particular, the locally compact versions of Urysohn's lemma and Tietze's theorem, stated above, are direct consequences of considering the \full" versions in X  and then \cutting back" to X . If X is locally compact, then the completion of CC (X ) under the sup norm is the space C0(X ), the functions in C (X ) that \vanish at in nity"; that is, those f 2 C (X ) for which the set f jf j  " g is compact for every " > 0. Clearly, C0(X ) is a Banach space (and a Banach algebra) under the sup norm. The phrase \vanish at in nity" becomes especially meaningful if we consider X  , the one point compacti cation of X . In this setting, the space C0 (X ) is (isometrically) the collection of functions in C (X ) that are zero at 1. It may come as a surprise to learn that the discrete spaces are a very important class of locally compact spaces. In the case of a discrete space D, the various spaces of continuous functions are often given dierent names. For example, since every function f : D ! R is continuous, the space Cb(D) is

APPENDIX A. TOPOLOGY REVIEW

190

simply the collection of all bounded functions on D, and this is often written as ` (D) in analogy with the sequence space ` = ` (N). The space CC (D) is the collection of functions with nite support in ` (D), and the space C0(D) is often written as c0(D), again in keeping with the sequence space notation c0 = C0(N). As a last curiosity, notice that the space c of all convergent sequences is just a renaming of the space C (N ). 1

1

1

1

[ f1g

Weak topologies

A familiar game is to describe the continuous functions on X after we've been handed a topology on X . But the inverse procedure is just as common and perhaps even more useful. In other words, given a collection of functions from a set X to some xed topological space Y , can we construct a topology on X under which each element of will be continuous? If X is given the discrete topology, then every function from X to Y is continuous, while if X is given the trivial, or indiscrete topology, then only constant functions are continuous. We typically want something in between. In fact, we'd like to know if there is a smallest (or weakest ) topology that makes each element of continuous. As we'll see, the answer is \Yes," and follows easily from an important bit of machinery that provides for the construction of topologies having certain predetermined open sets. F

F

F

Lemma A.1 (Subbasis Lemma) Suppose that X is a set and that is a colS

lection of subsets of X . Then, there is a smallest topology T on X containing S . Moreover, S = f?; X g [ S forms a subbase for T . In other words, the sets of the form S1 \


\ Sn , where Si 2 S for i = 1; : : : ; n, are a base for T . 0

0

Proof. Let T1 denote the intersection of all topologies on X containing S (or S ). It's easy to see that T1 is itself a topology on X containing S (as well as S ) and, clearly, T1 is the smallest such topology. Consequently, T1 also contains the collection T2, which we de ne to be the set of all possible unions of sets of the form S1 \


\ Sn, where n  1 and where Si 2 S for i = 1; : : :; n. All that remains is to show that T1 = T2. But since T2 contains S , it suces to show that T2 is a topology on X . In fact, the only detail that we need to check is that T2 is closed under nite intersections (since it's obviously closed under arbitrary unions). Here goes: Let U , V 2 T2 and let x 2 U \ V . Take A1; : : :; An and B1; : : :; Bm in S such that x 2 A1 \


\ An  U and x 2 B1 \


\ Bm  V . Then x 2 A1 \


\ An \ B1 \


\ Bm  U \ V . That is, U \ V 2 T2. 0

0

0

0

191 And how does Lemma A.1 help? Well, given a set of functions F from X into a topological space Y , take the smallest topology on X containing the sets S = f f 1 (U ) : f 2 F ; U is open in Y g: Since each element of S will be open in the new topology, each element of F will be continuous. This topology is usually referred to as the the weak topology induced by F . We don't really need the inverse image of every open set in Y ; we could easily get by with just the inverse images of a collection of basic open sets, or even subbasic open sets. In particular, if Y = R, then the collection of sets

N (x; f1; : : :; fn ; ") = f y 2 X : jfi(x) fi(y)j < "; i = 1; : : :; n g where x; 2 X , f1; : : : ; fn 2 F , and " > 0, is a neighborhood base for the weak

topology generated by F . If X carries the weak topology induced by a collection of functions F from X into Y , then it's easy to describe the continuous functions into X ; note that f : Z ! X is continuous if and only if g  f : Z ! Y is continuous for every g 2 F. Finally, it's worth pointing out that our construction of weak topologies in no way requires a xed range space Y . In particular, given a collection of functions (f)2A, where f maps X into a topological space Y, we can easily apply the subbasis lemma to nd the smallest topology on X under which each f is continuous. In this setting, we consider the topology generated by the collection

S = f f 1 (U) : U is open in Y for  2 A g:

Product spaces The subbasis lemma readily adapts to more elaborate applications. The product (or Tychono) topology provides an excellent example of such an adaptation. First, recall that the (Cartesian) product of a collection of (nonempty) sets (Y)2A is de ned to be the set of allQfunctions f : A ! [2AY satisfying f () 2 Y; the product space is written 2A Y. If we identify an element f of the product space with its range (f ())2A , then we recover the familiar notion that the product space consists of \tuples," where the -th \coordinate" of each tuple is toQbe an element of Y . We also have the familiar coordinate projections  : 2A Y ! Y , de ned by the formula  (f ) = f ( ) (or,

192

APPENDIX A. TOPOLOGY REVIEW

 ((f ())2A) = f ( )). If each Y is the same set Y , we usually write the product space as Y A , the set of all functions from A into Y . each Y is a topological space, we topologize the product space Q In Ycase by giving it the weak topology induced by the 's. That is, we 2A 

de ne the product topology to be the smallest topology under which all of the coordinate projections are continuous. In terms of the subbasis lemma, this means that each of the sets  1(U), where U is open in Y and  2 A, is a subbasic open set in the product. The basic open sets in the product are, of course, nite intersections of these. This particular choice for a topology on the product spaceQresults in several useful consequences. For example, a function ' : X ! 2A Y, from a topological space X into a product space, is continuous if and only if each of its \coordinates"   ' : X ! Y is continuous. We'll see other bene ts of the product topology shortly.

Nets Now this being analysis (or had you forgotten?), we need a valid notion of limit, or convergence, in a general topological space. An easy choice, from our point of view, is to consider nets. The reader who is unfamiliar with nets would be well served by thinking of a net as a \generalized sequence." We start with a directed set D; that is, D is equipped with a binary relation  satisfying (i)    for all  2 D; (ii) if    and    , then    ; and (iii) given any ,  2 D, there is some  2 D with    and    . Several standard examples come to mind. N, with its usual order, is a directed set. The set of all nite subsets of a xed set is directed by inclusion (i.e., A  B if A  B ). The set of all neighborhoods of a xed point in any topological space is directed by reverse inclusion (i.e., A  B if A  B ). And so on. As usual, we also write    to mean   . Now, a net in a set X is any function into X whose domain is a directed set. A sequence, recall, is a function with domain N, and so is also a net. Just as with sequences, though, we typically identify a net with its range. In other words, we would denote a net in X by simply writing (x)2D, where D is a directed set and where each x 2 X . In de ning convergence for nets, we just tailor the terminology we already use for sequences. For example, we say that a net (x)2D is eventually in the set A if, for some  2 D, we have fx :   g  A. And (x)2D is frequently in A if, given any  2 D, there is some  2 D with    such that x 2 A.

193 Finally, a net (x)2D in a topological space X converges to the point x 2 X if (x)2D is eventually in each neighborhood of x. As with sequences, we use the shorthand x ! x in this case. Many topological properties can be characterized in terms of convergent nets. Indeed, sequential characterizations used in metric spaces can typically be directly translated into this new language of nets. Here's an easy example: A set E in a topological space X is closed if and only if each net (x) in E , that converges in X , must actually converge to a point of E . On the one hand, suppose that E is closed and let (x) be a net in E converging to x 2 X . If x 2 E c, an open set, then we would have to have (x) eventually in E c, an impossiblility. On the other hand, suppose that each convergent net from E converges to a point in E . Now let x 2 E c and suppose that for every neighborhood U of x there is some point xU 2 U \ E . If we direct the neighborhoods of x by reverse inclusion, then we've just constructed a net (xU ) in E that converges to a point x not in E , yielding a contradiction. Thus, some neighborhood U of x is completely contained in E c; that is, E c is open. Another example that's easy to check: A function f : X ! Y between topological spaces is continuous if and only if the net (f (x)) converges to f (x) 2 Y whenever the net (x) converges to x 2 X . Suppose rst that f is continuous. Let x ! x in X and let U be a neighborhood of f (x) in Y . Then f 1 (U ) is a neighborhood of x in X and, hence, (x) is eventually in f 1(U ). Consequently, (f (x)) is eventually in U . That is, (f (x)) converges to f (x). Next suppose that f (x) ! f (x) whenever x ! x. Let E be a closed set in Y and let (x) be a net in f 1 (E ) that converges to x 2 X . Then (f (x)) is a net in E that converges to f (x) 2 Y . Hence, f (x) 2 E or x 2 f 1(E ). Thus, f 1 (E ) is closed and so f is continuous. Nets will prove especially useful in arguments involving weak topologies. If X carries the weak topology induced by a family of functions F , it follows that a net (x) converges to x 2 X if and only if (f (x)) converges to f (x) for each f 2 F . (Why?) Now since the product topology is nothing more than a weak topology, our latest observation takes on a very simple guise in a product space. A net Q (f) in a product space 2A Y converges to f if and only if (f) converges \coordinatewise"; that is, if and only if (f()) converges to f () for each  2 A. For this reason, the product topology is sometimes called the topology of pointwise convergence . Beginning to sound like analysis? The only potential hardship with nets is that the notion of a subnet is a bit more complicated than that of a subsequence. But we're in luck: We will have no need for subnets, and so we can blissfully ignore their intricacies.

194

APPENDIX A. TOPOLOGY REVIEW

Notes and Remarks There are many excellent books on topology (or on topology for analysts) that will provide more detail than we have given here. See, for example, Folland [44, Chapter 4], Jameson [68], Kelley [76], Kelley and Namioka [77], Kothe [78], Simmons [126], or Willard [134].

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Index Zp, 107 X , 166, 186 N, 169, 176

Ap, 95, 105, 112 B(X; Y ), 24 Bp, 95, 105, 112, 115 C (K ), 5, 160 C [ 0; 1 ], 48, 79, 121, 126, 127, 160 C [ a; b ], 5 C0(X ), 6, 11 Cb(X ), 6, 11 CC (X ), 6, 11 c, 2 c0, 2, 27, 37, 41, 48, 52, 54, 58, 59, 63,


, 137, 148 `1, 37, 59, 62, 63, 68, 71, 79, 121, 126, 127, 135, 163 `p, 1, 27, 37, 40, 48, 52, 54, 58, 135 `p-sum, 54 `p( ), 4 `np, 2 `1, 2, 11, 59, 64, 65, 68, 71, 79, 117, 120, 121, 126, 127, 169, 171 `1(D), 170 `1( ), 65, 67, 71 supp f , 6 + , 172  , 172 kk, 172 Xb , 17 xb, 16 ba (2D ), 180 rca (), 183

65, 66, 68, 71, 107, 121, 123, 126, 127, 163 c0-sum, 55 f + , 171 f , 171 J , 17 L0, 3 L1, 71, 92, 121, 126, 127, 149, 152, 153, 161 Lp, 3, 40, 81, 92, 93, 107, 109, 117, 123, 126, 127, 129{131 Lp(), 4, 135 L1, 3, 121, 126, 127, 161 M (p; "), 100, 107 M ? , 21 s, 1 T , 17 T +, 172 T , 172 T , 18, 157, 163 X=M , 19 X , 16 X , 16

adjoint, 17 algebra of sets, 178 algebraic complements, 18 algebraic dual, 16 almost disjoint sequence, 42, 44, 50, 52, 61, 69, 102 almost isometry, 63 Baire  -algebra, 178, 186 Banach lattice, 171 Banach limit, 174 203

204

Banach-Alaoglu theorem, 144, 147, 158 Banach-Mazur theorem, 145 Banach-Saks theorem, 115 basic sequence, 27 basis, 27 basis constant, 30, 37 basis problem, 28, 49 Bessaga-Pelczynski selection principle, 51 biorthogonal, 29 biorthogonal sequence, 79 block basic sequence, 50 block basis, 50, 58 bounded linear map, 13 bounded multiplier convergent, 110 boundedly complete basis, 76, 79 Clarkson's inequalities, 131 Clarkson's theorem, 129 clopen set, 176 closed sets, 187 coecient functionals, 28 compact operator, 53, 71, 111 compact set, 187 compacti cation, 189 complemented subspace, 19, 25, 48, 62, 159, 161 complemented subspaces of c0, 54 complemented subspaces of Lp , 41, 87, 98, 103, 105, 107 complemented subspaces of `p , 54 complemented subspaces of `1 , 65 completely continuous operator, 58, 158 completely regular, 138, 148 completely separated, 188 concave function, 84 conditional expectation, 34, 87, 90, 92 convex function, 83, 92, 135 convex sets, 125 coordinate functionals, 28 coordinate projection, 29

INDEX direct sum, 18 directed set, 192 discrete space, 186, 187, 189 discrete topology, 187 disjointly supported functions, 37, 40, 48 disjointly supported sequence, 49, 52, 102 disjointly supported sequences, 40 Dixmier's theorem, 22 dual norm, 16 dual of a quotient, 21 dual of a subspace, 22 dual space, 16, 24, 66, 68, 76, 103 Dunford-Pettis property, 157{161, 163 Eberlein-Smulian theorem, 153 embedding lemma, 138 epigraph, 83 equicontinuous, 155 equivalent bases, 43, 48 exposed point, 123, 124, 126, 135 extension of an operator, 18, 24, 67, 68 Extension Theorem, 166 extremally disconnected, 175, 186 extreme point, 123, 124, 136 nite codimensional subspace, 25 nitely additive measures, 180 Frechet metric, 2, 5 functional, 16 Gantmacher's theorem, 157 generalized Holder inequality, 82, 92 gliding hump argument, 50, 51, 61 Goldstine's theorem, 147, 148, 158 Grothendieck's theorem, 158 Haar system, 33, 37 Hahn-Banach extension property, 67, 70 Hamel basis, 28

INDEX Hausdor compacti cation, 166 Hausdor topological space, 188 Hilbert space, 7, 119, 126, 127, 131 Holder's inequality, 81 independent random variables, 94, 95 indiscrete topology, 187 injective spaces, 67, 70, 71 involution, 25 isometry, 14 isomorphism, 14 isomorphism into, 14 isomorphism theorem, 21, 26 James's non-distortion theorem, 63, 70 Jensen's inequality, 84 Jordan decomposition, 172 Kadec-Pelczynski theorem, 102, 103 Khinchine's inequality, 95, 112 Lamperti's theorem, 86, 91, 92 Liapounov's inequality, 82, 92, 97 locally compact topological space, 174, 189 Marcinkiewicz's theorem, 88 Mazur's lemma, 39 Mazur's theorem, 40 McShane's lemma, 132, 135 measure algebra, 151, 155 Minkowski's inequality, 82, 92 monotone basis, 30 multinomial coecient, 96 nearest points, 124, 125 net, 192 norm dual, 16 normal topological space, 188 normalized basis, 35 norming functional, 17, 24, 64 norming set, 65 nowhere dense, 24

205

one point compacti cation, 7, 189 open sets, 187 operator norm, 13 order interval, 150 Orlicz's theorem, 112 orthogonal projection, 25, 34, 92 orthonormal basis, 35, 44, 94 parallelogram law, 119, 129 Pelczynski's decomposition method, 55, 56, 58 Phillip's lemma, 64, 71 Pitt's theorem, 53, 56, 58 point mass, 146 polygonal functions, 32 positive linear functional, 171 prime spaces, 56 principle of small perturbations, 44, 46, 47, 51, 52, 102 product topology, 192 projection, 18, 25 quotient map, 20, 26, 61 quotient norm, 20 quotient space, 19 quotient topology, 20 Rademacher functions, 93, 111 Radon-Riesz theorem, 128, 136 random signs convergent, 110 re exive space, 17, 77, 104, 107, 127, 129, 147, 149, 160 Riesz representation theorem, 171, 172, 175 Riesz's lemma, 15, 24 Riesz-Kakutani theorem, 175 Riesz-Markov theorem, 175 Schauder basis, 27 Schauder's basis for C [ 0; 1 ], 31, 37 Schroder-Bernstein theorem for Banach spaces, 56, 57 Schur's theorem, 60

206

self adjoint, 25 seminormalized sequence, 52 separably injective spaces, 68 separates points, 138 separates points from closed sets, 138 shrinking basis, 75, 79 small isomorphism, 63 smooth space, 125, 126 smoothness, 125 Sobczyk's theorem, 68 Stone-C ech compacti cation, 166 Stonean topological space, 175 strictly convex function, 135 strictly convex norm, 120 strictly convex space, 120, 122, 124{ 127, 135 strictly singular operator, 54, 58 subbasis lemma, 190 sublattice of Lp , 86, 87 subseries convergent, 110 subspaces of c0, 52 subspaces of Lp , 41, 86, 87, 102, 103, 109, 113, 115 subspaces of `p, 52 subspaces of `1 , 65 support, 6 supporting line, 84, 92 three space property, 25 Tietze's extension theorem, 188 topological space, 187 topology, 187 topology of pointwise convergence, 193 total variation measure, 172 uncomplemented subspace, 61, 99, 160 unconditional basis, 110 unconditionally convergent, 110, 117 uniform convexity, 127 uniformly absolutely continuous, 150 uniformly convex norm, 127 uniformly convex space, 127{131, 133

INDEX uniformly integrable, 149, 152, 153, 155 unordered convergent, 109 Urysohn's lemma, 188 vector lattice, 171 Vitali-Hahn-Saks theorem, 154 Walsh functions, 94 weak convergence, 48, 60, 128, 135, 152, 161 weak topology, 146, 190, 191 weak basis, 79 weak compact sets, 147, 163 weak convergence, 79, 135 weak topology, 146, 148, 163 weakly compact operator, 157{159, 163 weakly compact sets, 147, 149, 152, 153, 163 weakly null sequence, 48 weakly sequentially complete, 153 Weierstrass theorem, 142