Appendix A: Bases in Banach Spaces

Chapter 4

Appendix A: Bases in Banach spaces

4.1 Schauder bases

In this section we recall some of the notions and results presented in the course on Functional Analysis in Fall 2012 [Schl]. 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) linear combination of elements in B.This 2 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. 2 Hamel bases are also not very useful for Banach spaces, since (see Exercise 1), the coordinate functionals might not be continuous. Definition 4.1.1. [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,thereisa 2 unique sequence of scalars (an) K so that ⇢ 1 x = anen. n=1 X Examples 4.1.2. 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 of ` n p  1 0 n p and c0,respectively.

Remarks. Assume that X is a Banach space and (en) a basis of X.Then

33 34 CHAPTER 4. APPENDIX A: 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 = N j1=1 anen. So we can identify X with a space of sequences in K .

PropositionP 4.1.3. Let (en) be the Schauder basis of a Banach space X.For n N and x X deﬁne e⇤ (x) K to be the unique element in K,sothat 2 2 n 2 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 following n ! j  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 Pn, n N,arecalledtheCanonical Projections for (en) and (e⇤ ) the Coordinate 2 n Functionals for (en) or biorthogonals for (en).

Theorem 4.1.4. 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 || k 1 2N 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 n + 1 1 : X R0 , aiei aiei =sup aiei , ||| · ||| ! 7! n j=1 j=1 2N j=1 X X X is an equivalent norm under which (ei) is a monotone basis. 4.1. SCHAUDER BASES 35

Deﬁnition 4.1.5. [Basic Sequences] Let X be a Banach space. A sequence (x ) X 0 is called a 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

1 (4.2) 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 anxn. 2N b) If span(xj : j N) is complementedP in X, then so is span(yj : j N). 2 2 36 CHAPTER 4. APPENDIX A: BASES IN BANACH SPACES

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 (y ) of (y ),thaty n⇤ n n⇤ 2 span(x⇤ : j N),forn N. j 2 2 Now we recall the notion of unconditional basis. First the following Proposi- tion.

Proposition 4.1.8. For a sequence (xn) in Banach space X the following state- ments are equivalent.

a) For any reordering (also called permutation) of N (i.e. : N N is ! bijective) the series n x(n) converges. 2N b) For any ">0 thereP is an 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 xnj converges. 2N d) For sequence (" ) 1 the series P 1 " x converges. j ⇢{± } j=1 j nj In the case that above conditions hold weP say that the series xn converges unconditionally. P Deﬁnition 4.1.9. A basis (ej) for a Banach space X is called unconditional,if for every x X the expansion x = e ,x e converges unconditionally, where 2 h j⇤ i j (e ) are coordinate functionals of (ej). j⇤ P Asequence(x ) X is called an unconditional basic sequence if (x ) is an n ⇢ n unconditional basis of span(xj : j N). 2 Proposition 4.1.10. 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,sothatforallﬁniteB N,allscalars(aj)j B K, ⇢ 2 ⇢ and A B ⇢ (4.3) a x C a x . j j  j j j A j B X2 X2 c) There is a constant C0, so that for all ﬁnite sets B N,allscalars ⇢ (aj)j B K,andall("j)j B 1 ,ifK = R,and("j)j B z 2 ⇢ 2 ⇢{± } 2 ⇢{ 2 C : z =1 ,ifK = C, | | } n (4.4) " a x C0 a x . j j j  j j j B j=1 X2 X 4.2. MARKUSHEVICH BASES 37

In this case we call the smallest constant C = Cs which satisﬁes (4.3) for all n n, A 1, 2 ...,n and all scalars (aj) K the supression-unconditional ⇢{ } j=1 ⇢ constant of (xn) and we call the smallest constant C0 = Cu so that (4.4) holds n n for all n, ("j)j=1 1 ,or("j)j=1 z C : z =1,andallscalars n ⇢{± } ⇢{ 2 | | } (aj) K the unconditional constant of (xn). j=1 ⇢ Moreover, it follows that

(4.5) C C 2C . s  u  s

Proposition 4.1.11. If (xn) be an unconditional basic sequence. Then

1 1 (4.6) C =sup a b x : x = a x B and b 1 . u i i i i i 2 X | i| n j=1 i=1 o X X Remark. While for Schauder bases it is in general important how we order them, the ordering is not relevant for unconditional bases. We can therefore index unconditional bases by any countable set.

4.2 Markushevich bases

Not every separable Banach space has a Schauder basis [En]. But it has at least a bounded and norming Markushevich basis according to a result of Ovsepian and Pelczy´nski [OP]. We want to present this result in this section,

Deﬁnition 4.2.1. A countable family (en,en⇤ )n X X⇤ is called 2N ⇢ ⇥ biorthogonal, if e⇤ (em)=(m,n), for all m, n N, • n 2 fundamental, or complete, if span(en : i N)isdenseinX, • 2 total, if for any x X with e⇤ (x) = 0, for all n N, it follows that x = 0, • 2 n 2 norming, if for some constant c>0 it follows that •

sup x⇤(x) c x , for all x X. x span(e :n ) B | | k k 2 ⇤2 n⇤ 2N \ X⇤

and in that case we also say that (en,en⇤ )n is c-norming, 2N

shrinking, if span(e⇤ : n N)=X⇤, and • n 2

bounded, or uniformly minimal if C =supn en en⇤ < , and we say in • 2N k k·k k 1 that case that (en,en⇤ )n is C-bounded and call C the bound of (en,en⇤ )n . 2N 2N

A biorthogonal, fundamental and total sequence (en,en⇤ )n is called an Marku- 2N shevich basis or simply M-Basis. 38 CHAPTER 4. APPENDIX A: BASES IN BANACH SPACES

Remark. Assume (en,en⇤ ) is an M-bais. It follows from the totality that span(en⇤ : n N)isw⇤-dense in X⇤ Thus in every reflexive space M-bases are shrinking, 2 and shrinking M bases are 1-norming. Our goal is to prove following Theorem 4.2.2. [OP] Every separable Banach space X admits a bounded, norming M-basis which can be chosen to be shrinking if X⇤ is (norm) separable. More- over, the bound of that M-basis can be chosen arbitrarily close to 4(1 + p2)2. Remark. Pelczyński [Pe] improved later the above result and showed that for all separable Banach spaces and all ">0 there exists a bounded M-basis, whose bound does not exceed 1 + ". It is an open question whether or not every separable Bach space has a 1- bounded M-basis. But it is not hard to show that a space X with a bounded and norming M-basis can be renormed so that this basis becomes 1-bounded and 1 norming. Remark. It might be nice to know that every separable Banach space has a bounded and norming Markushevish basis (e ,e ) . Nevertheless, given z X, i i⇤ 2 we do not have any (set aside a good one) procedure to approximate z by finite linear combinations of the ei, we only know that such an approximation exists. This is precisely the di↵erence to Schauder bases, for which we know that the canonical projections converge point wise. Lemma 4.2.3. [LT, Lemma 1.a.6] Assume that X is an infinite dimensional space and that F X and G X are finite dimensional subspaces of X and ⇢ ⇤ ⇢ ⇤ X , respectively. Let ">0. Then there is an x X, x =1and an x X ⇤ 2 k k ⇤ 2 ⇤ so that x 2+", x (x)=1, z (x)=0,forallz G ,andx (z)=0for all k ⇤k ⇤ ⇤ ⇤ 2 ⇤ ⇤ z F . 2 Proof. Let (y )m S be finite and 1/(1 + ") norming the space F .Pick i⇤ i=1 ⇢ X⇤ x ? y⇤ : j =1, 2,...m G⇤ 2 { j }[ = z X : y⇤(z)=0,j=1, 2 . . . m, and z⇤(z)=0,z⇤ G⇤ , with x =1. 2 j 2 k k It follows for all R and all y F that 2 2 y y + x max y⇤(y + x) = max y⇤(y) k k . || k j j 1+" Then define u⇤ : span(F x ) R,y+ x . [{ } ! 7! We claim that u⇤ 2+".Indeed,lety F , y = 0, and R.Then k k 2 6 2 x + y u⇤ = | | x + y ! x + y k k k k 4.2. MARKUSHEVICH BASES 39

y 2 kx+ky 2(1 + ")if 2 y , k k  | | k k 8  > 2 y < 2 y k k y =2 if > 2 y . k kk k | | k k > Letting now x⇤ be a Hahn Banach: extension of u⇤ onto all of X, our claim is proved.

Lemma 4.2.4. ([Ma], see also [HMVZ, Lemma 1.21]) Let X be an inﬁnite dimensional Banach space. If (z ) X and (z ) X are n ⇢ n⇤ ⇢ ⇤ sequences so that span(zn : n N and (z⇤ : n N) are both inﬁnite dimensional 2 n 2 and so that

(M1) (zn) separates points of span(z⇤ : n N), n 2

(M2) (z⇤) separates points of span(zn : n N). n 2 Let N N be co-inﬁnite, and ">0. ⇢ Then we can choose a biorthogonal system

(xn,x⇤ ) span(zn : n N) span(z⇤ : n N) n ⇢ 2 ⇥ n 2 with

(4.7) span(zn :n N) span(xn :n N) and span(z⇤ :n N) span(x⇤ :n N) 2 ⇢ 2 n 2 ⇢ n 2 (4.8) sup xn⇤ xn⇤ < 2+". n N k k·k k 2

Remark. Note that (xn,x⇤ ) is an M-basis if span(zn : n N)isdenseinX, n 2 and if (z ) separates points of X. It is norming if B span(z ) is norming X. n⇤ X⇤ \ n⇤

Proof. Choose s1 =minn N : zn =0. x1 = zs1 / zs1 .If1 N choose { 2 6 } k k 2 x⇤ SX with x⇤(x1) = 1. Otherwise choose x⇤ = z⇤ with t1 =min m N : 1 2 ⇤ 1 1 t1 { 2 z (x ) =0 (which exists by (M1)). m⇤ 1 6 } Write N N = k1,k2,... and proceed by induction to choose x1,x2,...xn \ { } and x1⇤,...xn⇤ as follows. Assume that x1,x2,...xn and x1⇤,...xn⇤ been been chosen. If n +1 N,thenletF = span(x : i n) and G = span(x : i n) and 2 i  ⇤ i⇤  choose xn+1 and xn⇤+1 by Lemma 4.2.3. If n +1=k2j 1 N N let 2 \

s2j 1 =min s : zs span(xi : i n) . 62  Then deﬁne n

xn+1 = zs2j 1 xi⇤(zs2j 1 )xi, Xi=1 40 CHAPTER 4. APPENDIX A: BASES IN BANACH SPACES

which implies that xi⇤(xn+1) = 0 for i =1, 2,...n. Next choose (using (M2))

t2j 1 =min t : zt⇤(xn+1) =0 { 6 } and let n zt⇤2j 1 i=1 zt⇤2j 1 (xi)xi⇤ xn⇤+1 = , zt⇤2j 1 (xn+1) P which yields that xn⇤+1(xi) = 0, for i =1, 2 ...n, and xn⇤+1(xn+1) = 0. If n +1=k2j N N then we ﬁrst choose 2 \

t =min s : z⇤ span(x⇤ : i n) . 2j s 62 i  Let n

xn⇤+1 = zs⇤2j zs2j 1 (xi)xi⇤, Xi=1 and hence xn⇤+1(xi) = 0, for i =1, 2 ...n, and then let (using (M1)

s =min s : x⇤ (x ) =0 , 2j { n+1 s 6 } and n zs x⇤(zs ))xi x = 2j i=1 i 2j , n+1 x (z ) Pn⇤+1 s2j which implies that xi⇤(xn+1) = 0, for i =1, 2 ...n and xn⇤+1(xn+1) = 1. We insured by this choice that (xi,x⇤):i N is a biorthogonal sequence i 2 in X X⇤ which also satisﬁes (4.8) and, since for any m N we have span(zi : ⇥ 2 i m) span(xk2j 1 : j m) and span(zi⇤ : i m) span(xk⇤ : j m),  ⇢   ⇢ 2j  (xi,x⇤):i N it also satisﬁes (4.7). i 2 2n For n N we consider on ` the discrete Haar basis 2 2 r 1 h h ,r =0, 1,...,n 1, and s =0, 1,...2 1 , { 0}[{ (r,s) } with

n/2 h0 =2 1,2,3...,2n { } 2s2n r 1+1,2s2n r 1+2 ,...(2s+1 ) 2 n r 1 (2s+1)2n r 1+1,(2s+1)2n r 1+2 ,...(2s+2 ) 2 n r 1 h = { } { } (r,s) (n r)/2 2 if r =0, 1, 2,...n 1 and s =0, 1, 2 ...2r 1. 2n The unit vector basis (ei)i=1 as well as the Haar basis

r 1 h h ,r =0, 1,...,n 1,s=0, 1,...2 1 { 0}[{ (r,s) } 4.2. MARKUSHEVICH BASES 41

2n (n) are orthonormal bases in `2 . Thus the matrix A = A with the property that A(e1)=h0 and A(e2r+s+1)=h(r,s) is a unitary matrix. If we write

A(n) =(a(n) :0 i, j 2n 1) (i,j)   then it follows for k =0, 1, 2 ...2n 1 that a(n) = h (k)=2 n/2 and if r = (k,0) 0 0, 1, 2 ...n 1 and s =0, 1,...2r 1 that (n) a(k,2r+s) = h(r,s)(k) 2 (n r)/2 if k 2s2n r 1 +1, 2s2n r 1 +2,...(2s+1)2n r 1 , 2{ } = 2 (n r)/2 if (2s + 1)2n r 1 +1, (2s+1)2n r 1 +2,...(2s+2)2n r 1 , 8 { } <>0ifk 2s2n r 1 or k>(2s+2)2n r 1.  > and thus :

n/2 n/2 (n 1))/2 1/2 2 2 2 0 2 0 . . . ··· ··· . 2 n/2 . . . 21/2 . 0 ··· ··· 1 . . (n 1))/2 . . B . .2 . 0 . C B ··· ··· C B . . (n 1))/2 . . . C B . . 2 . . . C B ··· ··· C B ...... C B ...... C B . ··· . ··· . C B . n/2 (n 1)/2 . . C B .2 2 0 . . C A = B . ··· . ··· . C B . n/2 (n 1))/2 . . C B . 2 02 . . C B . . . . ··· . ··· . C B ...... C B ...... C B . . . ··· . ··· . C B . . .2. (n 1))/2 . . C B C B . . . ··· . ··· . C B . . . 2 (n 1))/2 . . C B C B . . . . ··· . ··· C B . . . . . 2 1/2 C B ··· ··· C B 2 n/2 2 n/2 0 2 (n 1)/2 0 2 1/2C B C @ ··· ··· A It follows therefore that for all k =0, 1 ...,2n 1wehave 2n 1 n 1 n i (n) (n r)/2 1 p2 (4.9) a = 2 = =1+p2, (k,j) p  p j=1 r=0 i=1 2 2 1 X X X ⇣ ⌘ because, leaving out the ﬁrst column, in each row and for each r 0, 1, 2 ...n 1 (n r)/2 2{ } the value 2 is absolutely taken exactly once. This implies the following: 42 CHAPTER 4. APPENDIX A: BASES IN BANACH SPACES

Corollary 4.2.5. If (x ,x ):i =0, 1,...2n 1 is a biorthogonal sequence of i i⇤ length 2n, in a Banach space X and we let 2n 1 (n) (4.10) ek = a(k,j)xj and Xj=0 2n 1 (n) n (4.11) e⇤ = a x⇤ for k =0, 1,...2 1. k (k,j) j Xj=0 Then

n/2 (4.12) max ek < (1 + p2) max xk +2 x0 , 0 k<2n k k 0 k<2n k k k k   n/2 (4.13) max ek⇤ < (1 + p2) max xk⇤ +2 x0⇤ , 1 k<2n k k 0 k<2n k k k k   n (4.14) (e ,e⇤):j =0, 1,...,2 1 is biorthogonal j j n n (4.15) span(ej :0 j<2 ) = span(xj :0 j<2 ) and  n  n (4.16) span(e⇤ :0 j<2 ) = span(x⇤ :0 j<2 ). j  j  Proof of Theorem 4.2.2 . Let >0 and put M =2+. We start with a fundamental sequence (z ) X and a w -dense sequence (z ) B ((B ,w )is i ⇢ ⇤ i⇤ ⇢ X⇤ X⇤ ⇤ separable if X is norm separable) , which we choose norm dense if X⇤ is norm separable. Then we use Lemma 4.2.4 to choose a norming (reps. shrinking) M- basis (xn,x⇤ ):n N of X which satisfies for N being the odd numbers the n 2 conditions (4.7) and (4.8). Without loss of generality we assume that x = 1, k nk for n N. Now we will define a reordering (˜xn, x˜⇤ ):n N of (xn,x⇤ ):n N 2 n 2 n 2 as follows: ` mj By induction we choose for ` N anumberm` N and define q` = 2 2 2 j=1 and q0 = 0, and choose x˜q` 1 , x˜q⇤` 1 , x˜q` 1+1, x˜q⇤ +1 ,..., x˜q` 1, x˜q⇤ 1 as fol- ` 1 ` P lows Assume that for all 0 r<`, ` 1, mr, and (˜x0, x˜0⇤), (˜x1, x˜1⇤),...(˜xqr 1 1, x˜q⇤r 1 1)  have been chosen. Put

s` =min s :(x2s,x2⇤s) (˜xt, x˜t⇤):t q` 1 1 62 {  } (recall that ((x2s,x⇤ ):s N) are the elements of ((xs,x⇤):s N for which we 2s 2 s 2 do not control the norm) and choose m` N, so that 2 m /2 m /2 2 (1 + p2)M + x 2 ` (1 + p2)M + x⇤ 2 ` < (1 + p2)M + . k 2s` k · k 2s` k ⇥Then let (˜x , x˜ )=(x ⇤ ,x⇥ )while(˜x , x˜ ⇤) ...(˜x , x˜ ) con- q` 1 q⇤` 1 2s` 2⇤s` q` 1+1 q⇤` 1+1 q` 1 q⇤` 1 sist of the elements of (x2t 1,x2⇤t 1):t N ) which are not in the set (˜xt, x˜t⇤): m 2 { t q` 1 1 and have the lowest 2 ` 1indices.  } 4.2. MARKUSHEVICH BASES 43

By that choice we made sure that all elements of (xt,x⇤):t N appear t 2 exactly once in the sequence x˜t, x˜⇤ : t N . t 2 Then we apply Corollary 4.2.5 and deﬁne for k =0, 1, 2,...2m` 1 2m` 1 2m` 1 e = a(m`)x˜ and e = a(m`)x˜ . q` 1+k (k,j) q` 1+j q⇤` 1+k (k,j) q⇤` 1+j Xj=0 Xj=0 m 1 It follows then from (4.12) and (4.13) for k =0, 1, 2,...2 ` that p 2 2 eq` 1+k eq⇤` 1+k (1 + 2) M + . k k·k k Choosing >0 small enough we can ensure that (1+p2)2M 2+<4(1+p2)2+". Since (xn,x⇤ ):n N is a norming M-basis, it follows from (4.14), (4.15) n 2 and (4.16) that (en,e⇤ ):n N is a norming M basis which is shrinking if n 2 (xn,x⇤ ):n N is shrinking. n 2