Chapter 3
Bases in Banach Spaces
Like every vector space a Banach space X admits an algebraic or Hamel basis,i.e.asubsetB X, so that every x X is in a unique way the (finite) ⇢ 2 linear combination of elements in B. This definition does not take into account that we can take infinite sums in Banach spaces and that we might want to represent elements x X as converging series (with possibly infinite 2 non zero elements). Hamel bases are also not very useful for Banach spaces, since (see Exercise 1 the coordinate functionals might not be continuous.
3.1 Schauder Bases
Definition 3.1.1. (Schauder bases of Banach Spaces) Let X be an infinite dimensional Banach space. A sequence (e ) X is n ⇢ called Schauder basis of X, or simply a basis of X, if for every x X,there 2 is a unique sequence of scalars (an) K so that ⇢
1 x = anen. n=1 X Examples 3.1.2. For n N let 2 N en =(0,...0 , 1, 0,...) K 2 n 1times Then (e ) is a basis of ` ,1 | p<{z } and c . We call (e ) the unit vector n p 1 0 n basis of `p and c0,respectively.
Remarks. Assume that X is a Banach space and (en) a basis of X.
65 66 CHAPTER 3. BASES IN BANACH SPACES
a) (en) is linear independent.
b) span(en : n N)isdenseinX, in particular X is separable. 2 c) Every element x is uniquely determined by the sequence (an) so that x = j1=1 anen. So we can identify X with a space of sequences in N, for which a e converges in X. K P n n Proposition 3.1.3.PLet X be a normed linear space and assume that (en) ⇢ X has the property that each x X can be uniquely represented as a series 2 1 x = anen, with (an) K ⇢ n=1 X (we could call (en) Schauder basis of X but we want to reserve this term only if X is a Banach space). For n N and x X define e⇤ (x) K to be the unique element in K, 2 2 n 2 so that 1 x = en⇤ (x)en. n=1 X Then e⇤ : X K is linear. n ! For n N let 2 n P : X span(e : j n),x e⇤ (x)e . n ! j 7! n n Xj=1 Then P : X X are linear projections onto span(e : j n) and the n ! j following properties hold:
a) dim(Pn(X)) = n,
b) Pn Pm = Pm Pn = Pmin(m,n),form, n N, 2 c) limn Pn(x)=x, for every x X. !1 2 Conversely if (Pn : n N) is a sequence of linear projections satisfying 2 (a), (b), and (c), and moreover are bounded, and if e P (X) 0 and 1 2 1 \{ } en Pn(X) (Pn 1),withen =0,ifn>1, then each x X can be 2 \N 6 2 uniquely represented as a series
1 x = anen, with (an) K, ⇢ n=1 X so in particular (en) is a Schauder basis of X in case X is a Banach space. 3.1. SCHAUDER BASES 67
Proof. The linearity of en⇤ follows from the unique representation of every x X as x = 1 e (x)e , which implies that for x and y in X and 2 j=1 n⇤ n ↵, K, 2 P n n ↵x + y =lim↵ e⇤(x)ej + e⇤(y)ej n j j !1 Xj=1 Xj=1 n 1 =lim (↵e⇤(x)+ e⇤(y))ej = (↵e⇤(x)+ e⇤(y))ej, n j j j j !1 Xj=1 Xj=1 and, on the other hand
1 ↵x + y = ej⇤(↵x + y)ej, Xj=1 thus, by uniqueness, e⇤(↵x + y)=↵e⇤(x)+ e⇤(y), for all j N.The j j j 2 conditions (a), (b) and (c) are clear. Conversely, assume that (Pn) is a sequence of bounded and linear projec- tions satisfying (a), (b), and (c). By (b) Pn 1(X)=Pn Pn 1(X) Pn(X), ⇢ for n N (put P0 = 0)and, thus, by (a), the codimension of Pn 1(X)inside 2 Pn(X) is 1. So if e1 P1(X) 0 and en Pn(X) (Pn 1), if n>1, 2 \{ } 2 \N then for x X,by(b) 2 Pn 1(Pn(x) Pn 1(x)) = Pn 1(x) Pn 1(x)=0, and thus Pn(x) Pn 1(x) (Pn 1) and 2N Pn(x) Pn 1(x)=Pn(Pn(x) Pn 1(x)) Pn(X), 2 and therefore Pn(x) Pn 1(x) Pn(X) (Pn 1). Thus, we can write 2 \N Pn(x) Pn 1(x)=anen, for n N, and it follows from (c) that (letting 2 P0 = 0) n n 1 x =limPn(x)= lim Pj(x) Pj 1(x)= lim ajej = ajej. n n n !1 !1 !1 Xj=1 Xj=1 Xj=1
In order to show uniqueness of this representation of x assume x = j1=1 bjej. From the continuity of Pm Pm 1, for m N it follows that 2 P n amem =(Pm Pm 1)(x)= lim(Pm Pm 1) bjej = bmem, n !1 ⇣ Xj=1 ⌘ and thus am = bm. 68 CHAPTER 3. BASES IN BANACH SPACES
Definition 3.1.4. (Canonical Projections and Coordinate functionals) Let X be a normed linear space and assume that (ei) satisfies the assump- tions of Proposition 3.1.3. The linear functionals (en⇤ ) as defined in Proposi- tion 3.1.3 are called the Coordinate Functionals for (en) and the projections Pn, n N, are called the Canonical Projections for (en). 2 Proposition 3.1.5. Suppose X is a normed linear space and assume that (e ) X has the property that each x X can be uniquely represented as a n ⇢ 2 series 1 x = anen, with (an) K. ⇢ n=1 X If the canonical projections are bounded, and, moreover, supn N Pn < 2 k k 1 (i.e. uniformly the Pn are bounded), then (ei) is a Schauder basis of its completion X˜. ˜ ˜ ˜ Proof. Let Pn : X X, n N, be the (by Proposition 1.1.5 and Exercise ! 2 1 in Section 1.2 uniquely existing) extensions of Pn.SincePn has finite dimensional range it follows that P˜ (X˜)=P (X) = span(e : j n)is n n j finite dimensional and, thus, closed. (P˜n) satisfies therefore (a) of Proposi- tion 3.1.3. Since the Pn are continuous, and satisfy the equalities in (b) of Proposition 3.1.3 on a dense subset of X˜, (b) is satisfied on all of X˜. Finally, (c) of Proposition 3.1.3 is satisfied on a dense subset of X˜, and we deduce ˜ forx ˜ X,˜x =limk xk,withxk X, for k N, that 2 !1 2 2
x˜ P˜n(˜x) x˜ xk +sup Pj x˜ xk + xk Pn(xk) k kk k j N k kk k k k 2 and, since (Pn) is uniformly bounded, we can find for given ">0, k large enough so that the first two summands do not exceed ", and then we choose n N large enough so that the third summand is smaller than ". It follows 2 therefore that also (c) is satisfied on all of X˜. Thus, our claim follows from the second part of Proposition 3.1.3 applied to X˜.
Our goal is now to show the converse of Proposition 3.1.3, and prove that if (en) is a Schauder basis, then the canonical projections are uniformly bounded, and thus that the coordinate functionals are bounded.
Theorem 3.1.6. Let X be a Banach space with a basis (en) and let (en⇤ ) be the corresponding coordinate functionals and (Pn) the canonical projections. Then Pn is bounded for every n N and 2
b =sup Pn L(X,X) < , n N || k 1 2 3.1. SCHAUDER BASES 69 and thus e X and n⇤ 2 ⇤ Pn Pn 1 2b e⇤ = k k . k nkX⇤ e e k nk k nk We call b the basis constant of (ej).Ifb =1we say that (ei) is a monotone basis. Furthermore there is an equivalent renorming of (X, ) for which ||| · ||| k·k (e ) is a monotone basis for (X, ). n ||| · ||| Proof. For x X we define 2 x =sup Pn(x) , ||| ||| n N k k 2 since x =limn Pn(x) , it follows that x x < for x X. k k !1 k k k k||| ||| 1 2 It is clear that is a norm on X. Note that for n N ||| · ||| 2 Pn =sup Pn(x) ||| ||| x X, x 1 ||| ||| 2 ||| ||| =supsup Pm Pn(x) x X, x 1 m N k k 2 ||| ||| 2 =supsup Pmin(m,n)(x) 1. x X, x 1 m N k k 2 ||| ||| 2
Thus the projections Pn are uniformly bounded on (X, ). Let X˜ ˜ ||| · ||| be the completion of X with respect to , Pn, for n N,the(unique) ||| · ||| 2 extension of Pn to an operator on X˜. We note that the P˜n also satisfy the conditions (a), (b) and (c) of Proposition 3.1.3. Indeed (a) and (b) are purely algebraic properties which are satisfied by the first part of Proposition 3.1.3. Moreover for x X then 2 (3.1) x Pn(x) =sup Pm(x) Pmin(m,n)(x) ||| ||| m N k k 2 =sup Pm(x) Pn(x) 0ifn , m n k k! !1 which verifies condition (c). Thus, it follows therefore from the second part of Proposition 3.1.3, the above proven fact that Pn 1, for n N, and ||| ||| 2 Proposition 3.1.5, that (e ) is a Schauder basis of the completion of (X, ) n ||| · ||| which we denote by (X,˜ ). ||| · ||| We will now show that actually X˜ = X, and thus that, (X, )is ||| · ||| already complete. Then it would follow from Corollary 1.3.6 of the Closed Graph Theorem that is an equivalent norm, and thus that ||| · ||| C =sup sup Pn(x) =sup x < . n N x BX k k x BX ||| ||| 1 2 2 2 70 CHAPTER 3. BASES IN BANACH SPACES
So, letx ˜ X˜ and write it (uniquely) asx ˜ = 1 a e ,wherethis 2 j=1 j j convergence happens in .Since , and since X is complete the ||| · ||| k·k|||·||| P series 1 a e also converges with respect to in X to say x X. j=1 j j k·k 2 Important Side Note: This means that the sequence of partial sums n P a e ) converges in (X, )tox, which means that (a )isthe unique j=1 j j k·k n sequence in K, for which x = 1 ajej. In particular this means that P j=1 n P ˜ Pn(x)= ajej = Pn(˜x), for all n N. 2 Xj=1 But now (3.1) yields that P (x) also converges in to x. n ||| · ||| This means (since (Pn(x) cannot converge to two di↵erent elements) that x =˜x, which finishes our proof.
After reading the proof of Theorem 3.1.6 one might ask whether the last part couldn’t be generalized and whether the following could be true: If k·k and are two norms on the same linear space X, so that , and so ||| · ||| k·k|||·||| that ( ,X) is complete, does it then follow that (X, ) is also complete k·k ||| · ||| (and thus and are equivalent norms). The answer is negative, as k·k ||| · ||| the following example shows. Example 3.1.7. Let X =` with its usual norm and let (b : ) 2 k·k2 2 ⇢ S`2 be a Hamel basis of (by Exercise 4 in Section 1.3, is necessarily uncountable). For x ` define x , 2 2 ||| ||| x = x , ||| ||| | | X2 where x = x b is the unique representation of x as a finite linear 2 combination of elements of (b : ). Since b 2, for , it follows for P 2 k k 2 x = x b `2 from the triangle inequality that 2 2 P x = x = x b 2 x b = x 2. ||| ||| | | k k 2 k k X2 X2 X2 Finally both norms and , cannot be equivalent. Indeed, for arbitrary k·k ||| · ||| ">0, there is an uncountable set , so that b b <", , , 0 ⇢ k 0 k2 0 2 0 ( is uncountable but S is in the -norm separable). For any two `2 k·k2 di↵erent elements , it follows that 0 2 0 b b <"<2= b b . k 0 k ||| 0 ||| Since ">0 was arbitrary this proves that and cannot be equivalent. k·k ||| · ||| 3.1. SCHAUDER BASES 71
Definition 3.1.8. (Basic Sequences) Let X be a Banach space. A sequence (x ) X 0 is called basic sequence n ⇢ \{ } if it is a basis for span(xn : n N). 2 If (ej) and (fj) are two basic sequences (in possibly two di↵erent Banach spaces X and Y ). We say that (ej) and (fj) are isomorphically equivalent if the map
n n T : span(ej : j N) span(fj : j N), ajej ajfj, 2 ! 2 7! Xj=1 Xj=1 extends to an isomorphism between the Banach spaces between span(ej : j N) 2 and span(fj : j N). 2 Note that this is equivalent with saying that there are constants 0 Proof. “ ” Follows from Theorem 3.1.6, since K := supn N Pn < and n ) m n 2 k k 1 Pm aixi = aixm,ifm n and (ai) K. i=1 i=1 i=1 ⇢ “ ” Assume that there is a constant K 1 so that for all m m n n Pm =sup ajxj : n N, (aj)j=1 K, ajxj 1 K. k k ( 2 ⇢ ) j=1 j=1 X X Thus, our claim follows from Proposition 3.1.5. Also note that the proof of “ ” implies that the smallest constant so ) that 3.2 is at most as big as the basis constant, and the proof of “ ”yielded ( that it is at least as large as the basis constant. Remark. It was for a long time an open problem whether or not every separable Banach space admits a Schauder basis. 1973 this was solved by Enflo [En] in the negative. He constructed the first separable Banach space which does not admit a Schauder basis. Every separable Hilbert space has a basis (for example an orthogonal basis). Thus, every subspace of a Hilbert space has also a basis. It was shown [Jo] that only Banach space which in some sense are “very close” to a Hilbert space, have the property that each of their subspaces have bases. Exercises 1. Assume that (e ) is a basis of a Banach space X.Then(ej )iswell j ej defined and also a Schauder basis of X. k k 2. Show that a Banach space X has basis enjoys the Bounded Approxi- mation Property (Exercise 4 in Section 2.5). 3. Let (e : ) be a Hamel basis of an infinite dimensional Banach 2 space X.Show that some of the coordinate functionals associated with that basis are not continuous. Hint: pick a sequence ( ) of pairwise di↵erent elements of and n ⇢ consider 1 n e n x = 2 . e n=1 n X k k 3.1. SCHAUDER BASES 73 4. For n N define in c0 2 n sn = ej =(1, 1,...1, 0,...). Xj=1 n times Prove that (sn) is a basis for c0, but| that{z one} can reorder (sn) so that it is not a basis of c0. (sn) is called the summing basis of c0. Hint: Play around with alternating series of (sn). 5. Let 1 3.2 Bases of C[0, 1] and Lp[0, 1] In the previous section we introduced the unit vector bases of `p and c0. Less obvious is it to find bases of function spaces like C[0, 1] and Lp[0, 1]. Example 3.2.1. (The Spline Basis of C[0, 1]) Let (t ) [0, 1] be a dense sequence in [0, 1], and assume that t = 0, n ⇢ 1 t2 = 1. It follows that mesh(t ,t ,...t ) 0, if n , where(3.3) 1 2 n ! !1 mesh(t1,t2,...tn) = max ti tj : tj is neighbor of ti . i=1,2,...n | | For f C[0, 1] we let P (f) to be the constant function taking the value f(0), 2 1 and for n 2weletP (f) be the piecewise linear function which interpolates n the f at the points t1,t2,...tn. More precisely, let 0 = s1 Pn(f):[0, 1] K, with ! sj s s sj 1 Pn(f)(s)= f(sj 1)+ f(sj), for s [sj 1,sj]. sj sj 1 sj sj 1 2 We note that P : C[0, 1] C[0, 1] is a linear projection and that P = 1, n ! k nk and that (a), (b), (c) of Proposition 3.1.3 are satisfied. Indeed, the image of P (C[0, 1]] is generated by the functions f 1, f (s)=s, for s [0, 1], n 1 ⌘ 2 2 and for n 2, f (s) is the functions with the property f(t ) = 1, f(t ) = 0, n n j j 1, 2,... t , and is linear between any t and the next bigger t .Thus 2{ }\{ n} j i dim(Pn(C[0, 1])) = n. Property (b) is clear, and property (c) follows from the fact that elements of C[0, 1] are uniformly continuous, and condition (3.3). Also note that for n>1 it follows that fn Pn(C[0, 1]) (Pn 1) 0 2 \N \{ } and thus it follows from Proposition 3.1.3 that (fn) is a monotone basis of C[0, 1]. Now we define a basis of Lp[0, 1], the Haar basis of Lp[0, 1]. Let n T = (n, j):n N0,j =1, 2,...,2 0 . { 2 }[{ } We partially order the elements of T as follows n2 n2 n1 n1 (n1,j1) < (n2,j2) [(j2 1)2 ,j22 ] ( [(j1 1)2 ,j12 ] () 3.2. BASES OF C[0, 1] AND LP [0, 1] 75 n1 n2 n2 n1 (j 1)2 (j 1)2 0 < (n, j), whenever (n, j) T 0 2 \{ } Let 1 p< be fixed. We define the Haar basis (ht)t T and the in Lp 1 (p) 2 normalized Haar basis (ht )t T as follows. 2 (p) n h0 = h 1 on [0, 1] and for n N0 and j =1, 2,...,2 we put 0 ⌘ 2 h(n,j) =1[(j 1)2 n,(j 1 )2 n) 1[(j 1 )2 n,j2 n) . 2 2 and we let n n =supp(h )= (j 1)2 ,j2 , (n,j) (n,j) + n 1 n = (j 1)2 , (j ⇥ )2 (n,j) 2 ⇥ 1 n n = (j )2 ,j2 . (n,j) 2 ⇥ ( ) We let h 1 = h . And for 1 p< (n,j) (n,j) 1 (p) h(n,j) n/p h(n,j) = =2 1[(j 1)2 n,(j 1 )2 n 1[(j 1 )2 n),j2 n) . h 2 2 k (n,j)kp Note that h = 1 for all t T and that supp(h ) supp(h ) if and only k tkp 2 t ⇢ s if s t. (p) Theorem 3.2.2. If one orders (ht )t T linearly in any order compatible 2 with the order on T then (h(p)) is a monotone basis of L [0, 1] for all 1 t p p< . 1 Remark. a linear order compatible with the order on T is for example the lexicographical order h0,h(0,1),h(1,1),h(1,2),h(2,1),h(2,2),.... Important observation: if (h : t T ) is linearly ordered into h ,h ,..., t 2 0 1 which is compatible with the partial order of T , then the following is true: 76 CHAPTER 3. BASES IN BANACH SPACES If n N and and if 2 n 1 h = ajhj, Xj=1 is any linear combination of the first n 1elements,thenh is constant on the support of hn. Moreover, h can be written as a step function N h = bj1[sj 1,sj ), Xj=1 with 0 = s0 sj hn(t)dt =0. sj 1 Z As we will see later, if 1 Proof of Theorem 3.2.2. First note that the indicator functions on all dyadic intervals are in span(h : t T ). Indeed: 1 =(h + h )/2, 1 = t 2 [0,1/2) 0 (0,1) (1/2,1] (h h )/2, 1 =1/2(1 h ), etc. 0 (0,1) [0,1/4) [0,1/2) (1,1) Since the indicator functions on all dyadic intervals are dense in Lp[0, 1] it follows that span(ht : t T ). 2 (p) Let (hn) be a linear ordering of (ht )t T which is compatible with the 2 ordering of T . n Let n N and let (ai) be a scalar sequence. We need to show that 2 i=1 n 1 n a h a h . i i i i i=1 i=1 X X n 1 As noted above, on the set A =supp(hn) the function f = i=1 aihi is constant, say f(x)=a, for x A. Therefore we can write 2 P 1A(f + anhn)=1 + (a + an)+1 (a an), A A + where A is the first half of the interval A and A the second half. From the convexity of [0, ) r rp, we deduce that 1 3 7! 1 a + a p + a a p a p, 2 | n| | n| | | ⇥ ⇤ 3.2. BASES OF C[0, 1] AND LP [0, 1] 77 and thus p p p p f + a h dx = f dx + a + a 1 + + a a 1 dx n n n A n A | | c | | | | | | Z ZA ZA p 1 p p = f dx + m(A) a + an + a an c | | 2 | | | | ZA ⇥ ⇤ f pdx + m(A) a p = f pdx c | | | | | | ZA Z which implies our claim. Proposition 3.2.3. Since for 1 p< ,and1 Exercises 1. Decide whether or not the monomials 1, x, xn are a Schauder basis of C[0, 1]. 2.⇤ Show that the Haar basis in L1[0, 1] can be reordered in such a way that it is not a a Schauder basis anymore. 78 CHAPTER 3. BASES IN BANACH SPACES 3.3 Shrinking, and boundedly complete bases Proposition 3.3.1. Let (en) be a Schauder basis of a Banach space X,and let (en⇤ ) be the coordinate functionals and (Pn) the canonical projections for (en). Then n n a) P ⇤(x⇤)= x⇤,ej e⇤ = (ej),x⇤ e⇤,forn N and x⇤ X⇤. n h i j h i j 2 2 Xj=1 Xj=1 b) x⇤ = (X⇤,X) lim P ⇤(x⇤),forx⇤ X⇤. n n !1 2 c) (e⇤ ) is a Schauder basis of span(e⇤ : n N) whose coordinate function- n n 2 als are (en). Proof. (a) For n N, x⇤ X⇤ and x = 1 e⇤,x ej X it follows that 2 2 j=1h j i 2 nP n P ⇤(x⇤),x = x⇤,P (x) = x⇤, e⇤,x e = x⇤,e e⇤,x h n i h n i h j i j h ji j D Xj=1 E D Xj=1 E and thus n P ⇤(x⇤)= x⇤,e e⇤. n h ji j Xj=1 (b) For x X and x X 2 ⇤ 2 ⇤ x⇤,x =limx⇤,Pnx =limP ⇤(x⇤),x . n n n h i !1h i !1h i n (c) It follows for m n and (ai) K, that i=1 ⇢ m m aiei⇤ =sup ai ei⇤,x x B h i i=1 2 X i=1 X Xn =sup ai ei⇤,Pm(x) x B h i 2 X i=1 n X n aiei⇤ Pm sup Pj aiei⇤ k kj N k k· i=1 2 i=1 X X It follows therefore from Proposition 3.1.9 that (en⇤ ) is a basic sequence, thus, a basis of span(e ), Since (e ),e = e ,e = , it follows that ( (e )) n⇤ h j i⇤i h i⇤ ji i,j n are the coordinate functionals for (en⇤ ). 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 79 Remark. If X is a space with basis (en) one can identify X with a vector space of sequences x =(⇠n) K.If(e⇤ ) are coordinate functionals for (en) ⇢ n we can also identify the subspace span(e⇤ : n N) with a vector space of n 2 sequences x⇤ =(⌘n) K.Thewaysuchasequencex⇤ =(⌘n) X⇤ acts on ⇢ 2 elements in X is via the infinite scalar product: x⇤,x = ⌘ e⇤ , ⇠ e = ⌘ ⇠ . h i n n n n n n n N n N n N D X2 X2 E X2 We want to address two questions for a basis (en) of a Banach space X and its coordinate functionals (ej⇤): 1. Under which conditions does it follow that X⇤ = span(en⇤ )? 2. Under which condition does it follow that the map J : X span(e )⇤, ! n⇤ with J(x)(z⇤)= z⇤,x , for x X and z⇤ span(e ), h i 2 2 n⇤ an isomorphy or even an isometry? We need first the following definition and some observations. Definition 3.3.2. [Block Bases] Assume (xn) is a basic sequence in Banach space X, a block basis of (xn)is asequence(z ) X 0 ,with n ⇢ \{ } kn zn = ajxj, for n N,where0=k0 Proposition 3.3.3. The block basis (zn) of a basic sequence (xn) is also a basic sequence, and the basis constant of (zn) is smaller or equal to the basis constant of (xn). n Proof. Let K be the basis constant of (xn), let m n in N, and (bi) K. i=1 ⇢ Then m m ki bizi = biajxj i=1 i=1 j=ki 1+1 X X X 80 CHAPTER 3. BASES IN BANACH SPACES n ki n K b a x = K b z i j j i i i=1 j=ki 1+1 i=1 X X X Theorem 3.3.4. For a Banach space with a basis (en) and its coordinate functionals (en⇤ ) the following are equivalent. a) X⇤ = span(e⇤ : n N) (and, thus, by Proposition 3.3.1, (e⇤ ) is a basis n 2 n of X⇤ whose canonical projections are Pn⇤). b) For every x X , ⇤ 2 ⇤ lim x⇤ span(ej :j>n) =lim sup x⇤,x =0. n k | k n x span(e :j>n), x 1 |h i| !1 !1 2 j k k c) Every bounded block basis of (en) is weakly convergent to 0. We call the basis (en)shrinkingif these conditions hold . Remark. Recall that by Corollary 2.2.6 the condition (c) is equivalent with c’) Every bounded block basis of (en) has a further convex block which converges to 0 in norm. Proof of Theorem 3.3.4. “(a) (b)” Let x X and, using (a), write it as ) ⇤ 2 ⇤ x⇤ = j1=1 ajej⇤.Then P lim sup x⇤,x =lim sup x⇤, (I Pn)(x) n x span(e :j>n) x 1 |h i| n x span(e :j>n), x 1 |h i| !1 2 j k k !1 2 j k k =lim sup (I Pn⇤)(x⇤),x n x span(e :j>n), x 1 |h i| !1 2 j k k lim (I P ⇤)(x⇤) =0. n n !1 k k “(b) (c)” Let (z ) be a bounded block basis of (x ), say ) n n kn zn = ajxj, for n N,with0=k0 x⇤,zn sup x⇤,z n 0, by condition (b), !1 |h i| z span(ej :j kn 1), z C |h i|! 2 k k 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 81 thus, (zn) is weakly null. “ (a) (c)” Assume there is an x⇤ SX ,withx⇤ span(e⇤ : j N). It ¬ )¬ 2 ⇤ 62 j 2 follows for some 0 <" 1 (3.5) " =limsup x⇤ Pn⇤(x⇤) > 0. n k k !1 By induction we choose z1,z2,...in BX and 0 = k0 Examples 3.3.5. Note that the unit vector bases of ` ,1 T : X⇤⇤ Y, x⇤⇤ ( x⇤⇤,ej⇤ )j N, ! 7! h i 2 is an isomorphism between X⇤⇤ and Y . If (en) is monotone then T is an isometry. 82 CHAPTER 3. BASES IN BANACH SPACES Remark. Note that if aj = 1, for j N,theninc0 2 n sup ajej =1, n N c0 2 j=1 X but the series j N ajej does not converge in c0. Considering X2 as a subspace of X (via the canonical embedding) the P ⇤⇤ image of X under T is the space of sequences 1 Z := (a ) Y : a e converges in X . i 2 j j n Xj=1 o Proof of Proposition 3.3.6. Let K denote the basis constant of (en), (en⇤ )the coordinate functionals, and (Pn) the canonical projections. It is straightfor- ward to check that Y is a vector space and that is a norm on Y . ||| · ||| For x X and x X we have by Proposition 3.3.1 ⇤ 2 ⇤ ⇤⇤ 2 ⇤⇤ n P ⇤(x⇤)= x⇤,e e⇤ and n h ji j Xj=1 n n n P ⇤⇤(x⇤⇤),x⇤ = x⇤⇤, x⇤,e e⇤ = x⇤⇤,e⇤ x⇤,e = x⇤, x⇤⇤,e⇤ e , h n i h ji j h j ih ji h j i j D Xj=1 E Xj=1 D Xj=1 E which implies that n (3.6) T (x⇤⇤) =sup x⇤⇤,ej⇤ ej =sup Pn⇤⇤(x⇤⇤) K x⇤⇤ . ||| ||| n N h i n N k k k k 2 j=1 2 X Thus T is bounded and T K. k k Assume that (a ) Y . We want to find x X , so that T (x )= n 2 ⇤⇤ 2 ⇤⇤ ⇤⇤ (an). Put n x⇤⇤ = ajej, for n N. n 2 Xj=1 (where we identify X with its canonical image in X⇤⇤ and, thus, ej with (e ) X )Since j 2 ⇤⇤ n xn⇤⇤ X = ajej (ai) , for all n N, k k ⇤⇤ X ||| ||| 2 j=1 X 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 83 and since X⇤ is separable (and thus (BX⇤⇤ , (X⇤⇤,X⇤)) is metrizable by Exercise 8 in Chapter 2 ) (xn⇤⇤ has a w⇤-converging subsequence xn⇤⇤j to an element x⇤⇤ with x⇤⇤ lim sup xn⇤⇤ (aj) . k k n k k||| ||| !1 It follows for m N that 2 x⇤⇤,em⇤ =limxn⇤⇤ ,em⇤ = am, h i j h j i !1 and thus it follows that T (x⇤⇤)=(aj), and thus that T is surjective. Finally, since (en⇤ ) is a basis for X⇤ it follows for any x⇤⇤ n T (x⇤⇤) =sup x⇤⇤,ej⇤ ej ||| ||| n N h i 2 Xj=1 n =sup x⇤⇤,ej⇤ x⇤,ej n N,x BX h ih i 2 ⇤2 ⇤ j=1 X =supsup x⇤⇤,Pn⇤(x⇤) x B n ⇤ X⇤ N 2 2 ⌦ ↵ x⇤⇤ (since P ⇤(x⇤) x⇤ if n ), k k n ! !1 which proves that T is an isomorphism, and, that T (x ) x , for ||| ⇤ ||| k ⇤⇤k x X . Together with (3.6) that shows T is an isometry if K = 1. ⇤⇤ 2 ⇤⇤ Lemma 3.3.7. Let X be a Banach space with a basis (en),withbasiscon- stant K and let (e⇤ ) be its coordinate functionals. Let Z = span(e⇤ : n N) n n 2 ⇢ X⇤ and define the operator S : X Z⇤,x (x) i.e. S(x)(z)= z,x ,forz Z and x X. ! 7! |Z h i 2 2 Then S is an isomorphic embedding of X into Z and for all x X. ⇤ 2 1 x S(x) x . K k kk kk k Moreover, the sequence (S(e )) Z are the coordinate functionals of (e ) n ⇢ ⇤ n⇤ (which by Proposition 3.3.1 is a basis of Z). Proof. For x X note that 2 S(x) =sup z,x sup x⇤,x = x , k k z Z, z 1 |h i| x B |h i| k k 2 k kX⇤ ⇤2 X⇤ By Corollary 1.4.6 of the Hahn Banach Theorem. 84 CHAPTER 3. BASES IN BANACH SPACES On the other hand, again by using that Corollary of the Hahn Banach Theorem, we deduce that x =sup w⇤,x k k w B |h i| ⇤2 X⇤ =sup lim w⇤,Pn(x) w B n |h i| ⇤2 X⇤ !1 =sup lim Pn⇤(w⇤),x w B n |h i| ⇤2 X⇤ !1 sup sup Pn⇤(w⇤),x n N w BX |h i 2 ⇤2 ⇤ sup sup z,x = K S(x) . n N z span(e :j n), z K |h i k k 2 2 j⇤ k k Theorem 3.3.8. Let X be a Banach space with a basis (en), and let (en⇤ ) be its coordinate functionals. Let Z = span(e⇤ : n N) X⇤. Then the n 2 ⇢ following are equivalent a) X is isomorphic to Z⇤, via the map S as defined in Lemma 3.3.7 b) (en⇤ ) is a shrinking basis of Z. c) If (aj) K, with the property that ⇢ n sup ajej < , n N 1 2 j=1 X then j1=1 ajej converges. In that caseP we call (en) boundedly complete. Proof. “(a) (b)” Assuming condition (a) we will verify condition (b) of ) Theorem 3.3.4 for Z and its basis (e ). So let z Z . By (a) we can write n⇤ ⇤ 2 ⇤ z⇤ = S(x) for some x X.Sincex =limn Pn(x), where (Pn) are the 2 !1 canonical projection for (en), we deduce that sup z⇤,w =supS(x),w w span(e :j>n), w 1h i w span(e :j>n), w 1h i 2 j⇤ k k 2 j⇤ k k =supw, x w span(e :j>n), w 1h i 2 j⇤ k k =supw, (I Pn)(x) w span(e :j>n), w 1h i 2 j⇤ k k 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 85 (I Pn)(x) n 0. k k! !1 It follows now from Theorem 3.3.4 that (ej⇤) is a shrinking basis of Z. “(b) (c)” Assume (b) and let (aj) K so that ) ⇢ n n (aj) =sup ajej =sup aj (ej) < . ||| ||| n N n N 1 2 j=1 2 j=1 X X The sequence (x ) X ,withx = n a (e ), is bounded in X n⇤⇤ ⇢ ⇤⇤ n⇤⇤ j=1 j j ⇤⇤ and must therefore have an (X ,X )-converging subnet whose limit we ⇤⇤ ⇤ P denote by x⇤⇤. It follows that aj = x⇤⇤,e⇤ , for all j N. h j i 2 Let z⇤ be the restriction of x⇤⇤ to the space Z (which is a subspace of X⇤). Since by assumption (ej⇤) is a shrinking basis of Z and since by Lemma 3.3.7 (S(ej))j N are the coordinate functionals we can write z⇤ in a unique 2 way as 1 z⇤ = bjS(ej). Xj=1 But this means that aj = x⇤⇤,e⇤ = z⇤,e⇤ = bj, for all j N. and h j i h j i 2 since S is an isomorphism between X and its image it follows that j1=1 ajej converges in norm in X. P “(c) (a)” By Lemma 3.3.7 it is left to show that the operator S is surjec- ) tive. Thus, let z Z .Since(e ) is a basis of Z and (S(e )) Z are ⇤ 2 ⇤ n⇤ n ⇢ ⇤ the coordinate functionals of (en⇤ ), it follows from Proposition 3.3.1 that z⇤ is the w⇤ limit of (zn⇤)where n z⇤ = z⇤,e⇤ S(e ). n h j i j Xj=1 Since w⇤-converging sequences are bounded it follows that n sup z⇤,ej⇤ S(ej) < n N h i 1 2 j=1 X and, thus, by Lemma 3.3.7 n sup z⇤,ej⇤ ej < . n N h i 1 2 j=1 X 86 CHAPTER 3. BASES IN BANACH SPACES By our assumption (c) it follows therefore that x = n z ,e e converges j=1h ⇤ j⇤i j in X, and moreover P n S(x)= lim z⇤,e⇤ S(ej)=z⇤, n j !1 h i Xj=1 which proves our claim. Theorem 3.3.9. Let X be a Banach space with a basis (en). Then X is re- flexive if and only if (ej) is shrinking and boundedly complete, or equivalently if (ej) and (ej⇤) are shrinking. Proof. Let (en⇤ ) be the coordinate functionals of (en) and (Pn) be the canon- ical projections for (en). “ ” Assume that X is reflexive. By Proposition 3.3.1 it follows for every ) x X ⇤ 2 ⇤ x⇤ = w⇤ lim P ⇤(x⇤)=w lim P ⇤(x⇤), n n n n !1 !1 w which implies that x⇤ span(e⇤ : n N) , and thus, by Proposition 2.2.5 2 n 2 k·k k·k x⇤ span(e⇤ : n N) . It follows therefore that x⇤ = span(e⇤ : n N) 2 n 2 n 2 and thus that (ej) is shrinking (by Proposition 3.3.1). Thus X⇤ is a Banach space with a basis (ej⇤) which is also reflexive. We can therefore apply to X⇤ what we just proved for X and deduce that (en⇤ ) is a shrinking basis for X⇤. But, by Theorem 3.3.8 (in this case Z = X⇤) this means that (en) is boundedly complete. “ ” Assume that (e ) is shrinking and boundedly complete, and let x ( n ⇤⇤ 2 X⇤⇤.Then n x⇤⇤ = (X⇤⇤,X⇤) lim x⇤⇤,e⇤ (ej) n j !1 h i Xj=1 By Proposition 3.3.1 and the fact that X⇤ = span(e⇤ : j N) j 2 "has (ej⇤) as a basis, since (ej)isshrinking # n = lim x⇤⇤,e⇤ (ej) (X) n j k·k !1 h i 2 Xj=1 n Since supn N j=1 P ⇤⇤(x⇤⇤),ej⇤ ej < , and 2 k h i k 1 since (e ) is boundedly complete j P which proves our claim. 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 87 The last Theorem in this section describes by how much one can perturb a basis of a Banach space X and still have a basis of X. Theorem 3.3.10. (The small Perturbation Lemma) Let (xn) be a basic sequence in a Banach space X, and let (xn⇤ ) be the coordinate functionals (they are elements of span(xj : j N)⇤)andassume 2 that (yn) is a sequence in X such that 1 (3.7) c = x y x⇤ < 1. k n nk·k nk n=1 X Then a) (yn) is also basic in X and isomorphically equivalent to (xn),more precisely 1 1 1 (1 c) a x a y (1 + c) a x , n n n n n n n=1 n=1 n=1 X X X for all in X converging series x = n N anxn. 2 P b) If span(xj : j N) is complemented in X, then so is span(yj : j N). 2 2 c) If (xn) is a Schauder basis of all of X, then (yn) is also a Schauder basis of X and it follows for the coordinate functionals (yn⇤) of (yn), that y⇤ span(x⇤ : j N),forn N. n 2 j 2 2 Proof. By Corollary 1.4.4 of the Hahn Banach Theorem we extend the func- tionals x⇤ to functionalsx ˜⇤ X⇤,with x˜⇤ = x⇤ , for all n N. n n 2 k nk k nk 2 Consider the operator: 1 T : X X, x x˜⇤ ,x (x y ). ! 7! h n i n n n=1 X Since 1 x y x < 1, T is well defined, linear and bounded and n=1 k n nk·k n⇤ k T c<1. It follows S = Id T is an isomorphism between X and it k kP self. Indeed, for x X we have, S(x) x T x (1 c) x and 2 n k 0 k k k k k·k k k k if y X,definex = 1 T (y)(T = Id)then 2 n=0 1 P 1 1 1 (Id T )(x)= T n(y) T T n(y) = T n(y) T n(y)=y. n=0 n=0 n=0 n=1 X ⇣ X ⌘ X X 88 CHAPTER 3. BASES IN BANACH SPACES Thus Id T is surjective, and, it follows from Corollary 1.3.6 that Id T is an isomorphism between X and itself. (a) We have (I T )(xn)=yn, for n N, this means in particular that (yn) 2 is basic and (xn) and (yn) are isomorphically equivalent. (b) Let P : X span(xn : n N) be a bounded linear projection onto ! 2 span(xn : n N). Then it is easily checked that 2 1 Q : X span(yn : n N),x (Id T ) P (Id T ) (x), ! 2 7! is a linear projection onto span(yn : n N). 2 (c) If X = span(xn : n N), then, since I T is an isomorphism, (yn)= 2 ((I T )(x )) is also a Schauder basis of X. n Moreover define for k and i in N, k n y⇤ = y⇤,xj x⇤ = (xj),y⇤ x⇤ span(x⇤ : j N). (i,k) h i i j h i i j 2 j 2 Xj=1 Xj=1 It follows from Proposition 3.3.1, part (b), that w⇤ limk y⇤ = yi⇤, !1 (i,k) which implies that y (x)= 1 y ,x x ,x , for all x X, and thus for i⇤ j=1h i⇤ jih j⇤ i 2 k i P yi⇤ y(⇤i,k) =sup yi⇤ y(⇤i,k),x k k x B |h i| 2 X 1 =sup yi⇤,xj xj⇤,x x B h ih i 2 X j=k+1 X 1 =sup yi⇤,xj yj xj⇤,x x B h ih i 2 X j=k+1 X 1 y⇤ x y x⇤ 0, if k . k i k k j jk·k j k! !1 j=Xk+1 so it follows that yi⇤ = limk y⇤ span(xj⇤ : j N) for every i N, k·k !1 (k,i) 2 2 2 which finishes the proof of our claim (c). Exercises 3.3. SHRINKING, AND BOUNDEDLY COMPLETE BASES 89 1. Prove that Y with , as defined in Proposition 3.3.6 is a normed ||| · ||| linear space. 2. A Banach space X is said to have the Approximation Property if for every compact set K X and every ">0 there is a finite rank ⇢ operator T so that x T (x) <"for all x K. k k 2 Show that the bounded approximation property (Exercise 4 in Section 2.5) implies the approximation property. 3. Show that (ei) is a shrinking basis of a Banach space X,thenthe coordinate functionals (ei⇤) are boundedly complete basis of X⇤. 4. A Banach space is called -space, for some 1 p and some ⇤ L(p, ) 1 1, if for every finite dimensional subspace F of X and every ">0 there is a finite dimensional subspace E of X which contains F and so dim(E) that dBM(E,`p ) < + ". Show that L [0, 1], 1 p< is a -space. p 1 L(p,1) Hint: Firstly, the span of the first n elements of the Haar basis is n isometrically isomorphically to `p (why?), secondly consider the Small Perturbation Lemma. 90 CHAPTER 3. BASES IN BANACH SPACES 3.4 Unconditional Bases As shown in Exercise 2 of Section 3.2 there are basic sequences which are no longer basic sequences if one reorders them. Unconditional bases are defined to be bases which are bases no matter how one reorders them. We will first observe the following result on unconditionally converging series Theorem 3.4.1. For a sequence (xn) in Banach space X the following statements are equivalent. a) For any reordering (also called permutation) of N (i.e. : N N ! is bijective) the series n N x (n) converges. 2 b) For any ">0 there is anP n N so that whenever M N is finite with 2 ⇢ min(M) >n, then n M xn <". 2 k P c) For any subsequence (nj) the series j N xnj converges. 2 d) For sequence (" ) 1 the series P 1 " x converges. j ⇢{± } j=1 j nj In the case that above conditions hold we sayP that the series xn converges unconditionally. P Proof. “(a) (b)” Assume that (b) is false. Then there is an ">0 and for ) every n N there is a finite set M N, n + “(c) (d)” If ("n) is a sequence of 1’s, let N = n N : "n =1 and ) ± { 2 } N = n N : "n = 1 .Since { 2 } n "jxj = xj xj, for n N, 2 j=1 j N +,j n j N ,j n X 2 X 2 X and since j N +,j n xj and j N ,j n xj converge by (c), it follows that n 2 2 "jxj converges. j=1 P P “(d) (b)” Assume that (b) is false. Then there is an ">0 and for every P ) n N there is a finite set M N, n Thus at least one of the series n1=1 xn and n1=1 "nxn cannot converge. “(b) (a)” Assume that : N N is a permutation for which x (j) is ) P! P not convergent. Then we can find an ">0 and 0 = k0 Proposition 3.4.4. For a sequence of non zero elements (xj) in a Banach space X the following are equivalent. a) (xj) is an unconditional basic sequence, b) There is a constant C,sothatforalln N,allA 1, 2,...,n and n 2 ⇢{ } all scalars (aj) K, j=1 ⇢ n (3.8) a x C a x . j j j j j A j=1 X2 X n c) There is a constant C0,sothatforalln N,all("j)j=1 1 and n 2 ⇢{± } all scalars (aj) K, j=1 ⇢ n n (3.9) " a x C0 a x . j j j j j j=1 j=1 X X In that case we call the smallest constant C = Ks which satisfies (3.8) the supression-unconditional constant of (xn) for all n, A 1, 2,...,n and all n ⇢{ } scalars (aj)j=1 K and we call the smallest constant C0 = Ku so that (3.9) ⇢ n n holds for all n, ("j) 1 and all scalars (aj) K the unconditional j=1 ⇢{± } j=1 ⇢ constant of (xn). Moreover, it follows (3.10) K K 2K . s u s Proof. “(a) (b)” Assume that (b) does not hold. We can assume that ) (xn) is a basic sequence with constant b. Then (see Exercise 5) we choose recursively k0 k n 1 ajxj 1 and ajxj for all n N. n2 2 j An j=kn 1+1 X2 X For any l m k j 1 k m i t a x a x + a x + a x s s s s s s s s s=l s=l t=i+1 s=kt 1+1 s=kj 1+1 X X X X X 3.4. UNCONDITIONAL BASES 93 (where the second term is defined to be 0, if i j 1) j 1 2b 1 2b + + (i 1)2 t2 (j 1)2 t=Xi+1 It follows therefore that x = j1=1 ajxj converges, but by Theorem 3.4.1 (b) it is not unconditionally. P “(b) (c)” and (3.10) follows from the following estimates for n N, n() n 2 (aj) K, A 1, 2,...,n and ("j) 1 j=1 ⇢ ⇢{ } j=1 ⇢{± } n n n " a x a x + a x and j j j j j j j j=1 j=1,"j =1 j=1,"j = 1 X X X 1 a x a x + a x + a x a x . j j 2 2 j j j j j j j j 3 j A j A j 1,2,... A j A j 1,2,... A X2 X2 2{ X }\ X2 2{ X }\ 4 5 “(b) ”(a) First, note that (b) implies by Proposition 3.1.9 that (x ) is a ba- ) n sic sequence. Now assume that for some x = 1 ajxj span(xj : j N) j=1 2 2 is converging but not unconditionally converging. It follows from the equiv- P alences in Theorem 3.4.1 that there is some ">0 and of N, M1, M2, M3 etc. so that min M > max M and a x ", for n .On n+1 n j Mn j j N k 2 k 2 the other hand it follows from the convergence of the series 1 ajxj that P j=1 max(Mn) P lim sup ajxj =0, n !1 j=1+max(Mn 1) X and thus a x j Mn j j sup 2 = , max( Mn) n P a x 1 !1 j =1+max(Mn 1) j j which is a contradiction to P condition (b). Proposition 3.4.5. Assume that X is a Banach space over the field C with an unconditional basis (en), then it follows if j1=1 ↵nen is convergent and ( n) C : =1 that 1 n↵nen is also converging and ⇢{ 2 | | } j=1 P P ↵ e 2K ↵ e . n n n u n n n N n N X2 X2 Proof. See Exercise 1. 94 CHAPTER 3. BASES IN BANACH SPACES Proposition 3.4.6. If X is a Banach space with an unconditional ba- sis, then the coordinate functionals (en⇤ ) are also a unconditional basic se- quence, with the same unconditional constant and the same suppression- uncondtional constant. Proof. Let Ku and Ks be the unconditional and suppression unconditional constant of X. Let x⇤ = n N ⌘nen⇤ and ("n) 1 then 2 ⇢{± } P 1 "n⌘nen⇤ =sup "n⌘nen⇤ , ⇠nen X⇤ x= ⇠ e B n N n1=1 n n X n N n=1 X2 2 D X2 X E P =sup "n⌘n⇠n x= ⇠ e B n1=1 n n X n N 2 X2 P =sup ⌘nen⇤ "n⇠nen x= ⇠ e B · n1=1 n n X n N n N 2 X2 X2 P K ⌘ e⇤ . u n n n N X2 Using the Hahn Banach Theorem we can similarly show that if Ku⇤ is the unconditional constant of (en⇤ )then ⇠n"nen Ku⇤ ⇠n"n X X n N n N X2 X2 Thus Ku = Ku⇤. A similar argument works to show that Ks is equal to the suppression unconditional constant of (en⇤ ). The following Theorem about spaces with unconditional basic sequences was shown By James [Ja] Theorem 3.4.7. Let X be a Banach space with an unconditional basis (ej). Then either X contains a copy of c0,oracopyof`1 or X is reflexive. We will need first the following Lemma (See Exercise 2) Lemma 3.4.8. Let X be a Banach space with an unconditional basis (en) and let Ku its constant of unconditionality. Then it follows for any con- verging series n N anen and a bounded sequence of scalars (bn) K,that 2 ⇢ n N anbnen is also converging and 2 P P 1 anbnen K sup bn anen , n N | | n N 2 n=1 X2 X where K = Ku,ifK = R,and K =2Ku,ifK = C. 3.4. UNCONDITIONAL BASES 95 Proof of Theorem 3.4.7. We will prove the following two statements for a space X with unconditional basis (en). Claim 1: If (en) is not boundedly complete then X contains a copy of c0. Claim 2: If (en) is not shrinking then X contains a copy of `1. Together with Theorem 3.3.9, this yields the statement of Theorem 3.4.7. Let Ku be the constant of unconditionality of (en) and let Ku0 = Ku,if K = R, and KU0 =2Ku,ifK = C. Proof of Claim 1: If (en) is not boundedly complete there is, by Theorem 3.3.8, a sequence of scalars (an) such that n 1 sup ajej = C1 < , but ajej does not converge. n N 1 2 j=1 j=1 X X This implies that there is an ">0 and sequences (mj) and (nj)with1 nk m1 n1 k k nk jyj 2Ku max j yj 2KuKs sup j aiej 2KuKsC1 sup j . j k | | j k | | j k | | j=1 j=1 i=1 X X X On the other hand for every j n that 0 n 1 " jyj j0 yj0 max j . Ks k k Ks j n | | j=1 X Letting c = "/Ku and C =2KuKsC1, it follows therefore for any n N and n 2 any sequence of scalars ( j)j=1 that n c ( )n y C ( )n , k j=1kc0 j j k j=1kc0 j=1 X which means that (yj) and the unit vector basis of c0 are isomorphically equivalent. 96 CHAPTER 3. BASES IN BANACH SPACES Proof of Claim 2. (en) is not shrinking then there is by Theorem 3.3.4 a bounded block basis (yn) of (en) which is not weakly null. After passing to a subsequence we can assume that there is a x X , x = 1, so that ⇤ 2 ⇤ k ⇤k " =inf x⇤,yn > 0. n N |h i| 2 We also can assume that yn = 1, for n N (otherwise replace yn by k k 2 y / y and change " accordingly). n k nk We claim that (yn) is isomorphically equivalent to the unit vector basis n of `1. Let n N and (aj) K. By the triangle inequality we have 2 j=1 ⇢ n n a y a , j j | j| j=1 j=1 X X On the other hand we can choose for j =1, 2,...,n " = sign(a x ,y )if j jh ⇤ ji K = R and "j = aj x⇤,yj / aj x⇤,yj ,ifK = C (if aj = 0, simply let " = 1) h i | h i| and deduce from Lemma 3.4.8 n n n n 1 ajyj "jajyj "jaj x⇤,yj " aj , Ku0 h i | | j=1 j=1 j=1 j=1 X X X X which implies that (yn) is isomorphically equivalent to the unit vector basis of `1. Remark. It was for long time an open problem whether or not every infinite dimensional Banach space contains an unconditional basis sequence. If this were so, every infinite dimensional Banach space would contain a copy of c0 or a copy of `1, or has an infinite dimensional reflexive subspace space. In [GM], Gowers and Maurey proved the existence of a Banach space which does not contain any unconditional basic sequences. Later then Gowers [Go] constructed a space which does not contain any copy of c0 or `1, and has no infinite dimensional reflexive subspace. Exercises. 1. Prove Proposition 3.4.5. 2. Prove Lemma 3.4.8 3.4. UNCONDITIONAL BASES 97 3. Prove that a block sequence (of non zero vectors) of an unconditional basic sequence is also a unconditional basic sequence. 4. Show that every separable X Banach space is isomorphic to the quo- tient space of `1. 5. Assume that (xn) is a basic sequence in a Banach space X for which (b) of Proposition 3.4.4 does not hold. Show that there is a sequence of scalars (aj) and a subsequence (kn) of N, so that k n 1 ajxj 1 and ajxj for all n N. n2 2 j An j=kn 1 X2 X 6. Let 1 3.5 James’ Space The following space J was constructed by R. C. James [Ja]. It is a space which is not reflexive and does not contain a subspace isomorphic to c0 or `1. By Theorem 3.4.7 it does not have an unconditional basis. Moreover we will prove that J ⇤⇤/ (J) is one dimensional and that J is isomorphically isometric to J ⇤⇤ (but of course not via the canonical mapping). We will define the space J over the real numbers R. For a sequence (⇠n) R we define the quadratic variation to be ⇢ l 1/2 2 (⇠n) qv =sup ⇠nj ⇠nj 1 : l N and 1 n0 =sup (⇠nj ⇠nj 1 : j =1, 2,...l) `2 : l N and 1 n0 l 1/2 1 2 2 (⇠n) cqv =sup ⇠n0 ⇠nl + ⇠nj ⇠nj 1 : l N and 1 n0 Note that for a bounded sequences (⇠n), (⌘n) R ⇢ l (⇠n + ⌘n) qv =sup (⇠ni + ⌘ni ⇠ni 1 ⌘ni 1 )i=1 2 : l N,n0 (⇠ + ⌘ ) (⇠ ) + (⌘ ) . k n n kcqv k n kcqv k n kcqv and we note that 1 (⇠n) qv (⇠n) cqv p2 (⇠n) qv. p2k k k k k k Thus and are two equivalent semi norms on the vector space k·kqv k·kcqv ˜ J = (⇠n) R : (⇠n) qv < ⇢ k k 1 3.5. JAMES’ SPACE 99 and since (⇠ ) =0 (⇠ ) =0 (⇠ ) is constant k n kqv () k n kcqv () n and are two equivalent norms on the vector space k·kqv k·kcqv J = (⇠n) :lim⇠n = 0 and (⇠n) qv < . R n ⇢ !1 k k 1 Proposition 3.5.1. The space J with the norms and is complete k·kqv k·kcqv and, thus, a Banach space. Proof. The proof is similar to the proof of showing that `p is complete. Let (xk)beasequenceinJ with k N xk qv < and write xk =(⇠(k,j))j N, 2 k k 1 2 for k N. Since for j, k N it follows that 2 2 P ⇠(k,j) =lim ⇠(k,j) ⇠(k,n) xk qv n | | !1 | |k k it follows that ⇠j = ⇠(k,j) k N X2 exists and for x =(⇠ ) it follows that x c (c is complete) and j 2 0 0 l 1/2 2 x qv =sup ⇠nj ⇠nj 1 : l N and 1 n0 1 x 0 for m . k kkqv ! !1 k=Xm+1 Proposition 3.5.2. The unit vector basis (ei) is a monotone basis of J for both norms, and . k·kqv k·kcqv Proof. First we claim that span(ej : j N)=J.Indeed,ifx =(⇠n) J, 2 2 and ">0 we choose l and 1 n0 nl+1 x ⇠ e = (0, 0,...0 ,⇠ ,⇠ ,...) <". k j jk k nl+2 nl+3 k j=1 X (nl+1) times | {z } n In order to show monotonicity, assume m ai if i m ai if i n ⇠i = and ⌘i = (0 otherwise (0 otherwise. For x = 1 ⇠ e and y = 1 ⌘ e we need to show that x y i=1 i i i=1 i i k kqv k kqv and x cqv y cqv. So choose l and n0 Then we can assume that nl >n(otherwise replace l by l + 1 and add nl+1 = n + 1) and we can assume that nl 1 m (otherwise we drop all the nj’s in (m, n] ), and thus l l 2 2 2 x qv = ⇠nj ⇠nj 1 = ⌘nj ⌘nj 1 y qv. k k | | | | k k Xj=1 Xj=1 The argument for the cyclic variation norm is similar. 3.5. JAMES’ SPACE 101 Our next goal is to show that (en) is a shrinking basis of J.Weneed the following lemma Lemma 3.5.3. For any normalized block basis (ui) of ei in J,andm N m 2 and any scalars (ai)i=1 it follows that m (3.11) a u p5 (a )n . i i k i i=1k2 i=1 X Proof. Let (⌘j) R and k0 =0 ki ui = ⌘jej. j=ki 1+1 X Let for i =1, 2, 3 ...m and j = ki 1 +1,ki 1 +2,...ki put ⇠j = ai ⌘j, and · n kn x = aiui = ⇠jej. Xi=1 Xj=1 For given l N and 1 n0 For i =1, 2,...,m define Ai = j 1:ki 1 kl(j) 1 2 2 ⇠nj ⇠nj 1 = am(j)⌘nj al(j)⌘nj 1 | | | | 102 CHAPTER 3. BASES IN BANACH SPACES 2 2 2 2 2 2 2am(j)⌘nj +2al(j)⌘nj 1 2am(j) +2al(j) (for the last inequality note that ⌘ 1since u = 1). For j, j B it | i| k jk 0 2 follows that l(j) = l(j ) and m(j) = m(j ), j = j and thus 6 0 6 0 6 0 l 2 2 2 ⇠nj ⇠nj 1 = ⇠nj ⇠nj 1 + ⇠nj ⇠nj 1 | | | | | | j=1 j A j B X X2 X2 n n a2 +2 a2 +2 a2 5 a2, i l(j) m(j) i i=1 j B j B i=1 X X2 X2 X which finishes the proof of our claim. Corollary 3.5.4. The unit vector basis (en) is shrinking in J. Proof. Let (un) be any block basis of (en), which is w.l.o.g. normalized. Then by Lemma 3.5.3 n 1 uj p5/pn 0if n . n qv ! !1 j=1 X By Corollary 2.2.6 (un) is therefore weakly null. Since (un) was an arbitrary block basis of (en) this yields by Theorem 3.3.8 that (en) is shrinking. Definition 3.5.5. (Skipped Block Bases) Assume X is a Banach space with basis (en). A Skipped Block Basis of (en) is a sequence (un) for which there are 0 = k0 Proposition 3.5.6. Every normalized skipped block sequence of the unit vector basis in J is isomorphically equivalent to the unit vector basis in `2. Moreover the constant of equivalence is p5. Proof. Assume that kn 1 un = ajej, for n N. 2 j=kn 1+1 X 3.5. JAMES’ SPACE 103 with 0 = k0 ln 2 2 un qv = a (n) a (n) =1. k k pj 1 pj 1 j=1 X m (n) Now let m N and (bi) R we can string the p ’s together and deduce: 2 i=1 ⇢ j m m l m 2 n 2 2 2 bnun bi a (n) a (n) = bi . qv pj 1 pj 1 n=1 i=1 j=1 i=1 X X X X On the other hand it follows from Lemma 3.5.3 that m m 2 2 bnun 5 bi . qv n=1 i=1 X X Corollary 3.5.7. J is hereditarily `2, meaning every infinite dimensional subspace of J has a further subspace which is isomorphic to `2. Proof. Let Z be an infinite dimensional subspace of J. By induction we choose for each n N, zn Z, un J and kn N, so that 2 2 2 2 4 n (3.13) z = u = 1 and z u < 2 , k nkqv k nkqv k n nkqv un span(ej : kn 1 Having accomplished that, (un) is a skipped block basis of (en) and by Proposition 3.5.6 isomorphically equivalent to the unit vector basis of `2. Letting (u ) be the coordinate functionals of (u ) it follows that u p5, n⇤ n k n⇤ k for n N, and thus, by the third condition in (3.13), 2 1 4 u⇤ u z p52 < 1, k nkk n nk n=1 X which implies by the Small Perturbation Lemma, Theorem 3.3.10, that (zn) is also isomorphically equivalent to unti vector bais in `2. We choose z1 SZ arbitrarily, and then let u1 span(ej : j N), with 2 4 2 2 u1 qv = 1 and u1 z1 qv < 2 .Thenletk1 N so that u1 span(ej : j< k k k k 2 2 k1). If we assume that z1,z2,...,...zn, u1,u2,...,un, and k1 have been chosen we choose zn+1 Z e1⇤,...ek⇤ (note that this space is 2 \{ n }? infinite dimensional and a subspace of span(ej : j>kn+1)) and then choose u span(e : j>k ), u = 1, with u z < 242 n 1 n+1 2 j n+1 k n+1kqv k n+1 n+1kqv and let kn+1 N so that un+1 span(ej : j n (3.15) J ⇤⇤ = (⇠n) R :sup ⇠iei < ⇢ n N cqv 1 n 2 j=1 o X and for x =(⇠ ) J ⇤⇤ n 2 ⇤⇤ (3.16) x⇤⇤ J⇤⇤ =sup (⇠1,⇠2,...,⇠n, 0, 0,...) cqv k k n N k k 2 l 1/2 2 2 =supmax (⇠k0 ⇠kl ) + (⇠kj 1 ⇠kj ) , l N,k0 (n) (n) (n) (n) ⇠j if j n x =(⇠1,⇠2,...,⇠n, 0, 0,...), thus x =(⇠j ), with ⇠j = . (0else Now we let l and 1 k1 (n) (n) kl 1) and we note that ⇠ = 0, while ⇠ = ⇠k for all j l 1, and thus kl kj j l l 1 1/2 1/2 (n) 2 1 2 2 2 2 2 x cqv = ⇠k0+ (⇠kj 1 ⇠kj ) = ⇠k0+ ⇠kl 1+ (⇠kj 1 ⇠kj ) , k k 2 ⇣ Xj=1 ⌘ ⇣ Xj=1 ⌘ which, after renaming l 1tobel, leads to the second term above “max”. Remark. Note that there is a di↵erence between (⇠ ,⇠ ,...) k 1 2 kcqv and sup (⇠1,⇠2,...,⇠n, 0, 0,...) cqv n N k k 2 and there is only equality if limn ⇠n = 0. !1 It follows that for all x⇤⇤ =(⇠n) J ⇤⇤, that e⇤ (x)=limn ⇠n exists, 2 1 !1 that (1, 1, 1, 1,...) J J, and that 2 ⇤⇤ \ x⇤⇤ e⇤ (x)(1, 1, 1,...) J. 1 2 Theorem 3.5.8. J is not reflexive, does not contain an isomorphic copy of c0 or `1 and the codimension of J in J ⇤⇤ is 1. Proof. We only need to observe that it follows from the above that J ⇤⇤ = (⇠j) R : (⇠j) cqv < ⇢ k k 1 1 = ⇠j)+⇠ (1, 1, 1 ...) R : (⇠j) cqv < limj ⇠j =0⇠ R , p2 1 ⇢ k k 1 !1 1 2 where the second equality follows from the fact that if (⇠n) has finite quadratic variation then limj ⇠j exists. !1 It follows therefore from Theorem 3.4.7 Corollary 3.5.9. J does not have an unconditional basis. Theorem 3.5.10. The operator T : J ⇤⇤ J, x⇤⇤ =(⇠j) (⌘j)=( e⇤ (x⇤⇤),⇠1 e⇤ (x⇤⇤),⇠2 e⇤ (x⇤⇤),...) ! 7! 1 1 1 is an isometry between J ⇤⇤ and J with respect to the cyclic quadratic varia- tion. 106 CHAPTER 3. BASES IN BANACH SPACES Proof. Let x =(⇠ ) J and ⇤⇤ j 2 ⇤⇤ z =(⌘j)=( e⇤ (x⇤⇤),⇠1 e⇤ (x⇤⇤),⇠2 e⇤ (x⇤⇤),.... 1 1 1 By (3.16) p2 x⇤⇤ k k l 1/2 2 2 =supmax (⇠k0 ⇠kl ) + (⇠kj ⇠kj 1 ) , l N,k0 l 1/2 2 2 =sup (⌘k0 ⌘kl ) + (⌘kj ⌘kj 1 ) , l N,k0 = p2 z . k kcqv Since T is surjective this implies the claim. Exercises 1. Define the James Function space as JF = f C[0, 1] : f(0) = 0 and f < , { 2 k kqv 1} where l 1/2 2 f qv =sup f(tj) f(tj 1) . k k t0 2. Show that the unit vector basis in J is also monotone with respect to the cyclic quadratic variation . k·kcqv 3. (The Gliding Hump Argument) Assume that Y is an infinite dimen- sional subspace of a Banach space X with a basis (ei) and ">0. Show that there is an infinite dimensional subspace Z of Y which has a normalized basis (zn), which is (1 + ")-equivalent to a block basis of (ei). 4.* Show that J has a boundedly complete basis.