3.2 Bases of C[0, 1] and L p [0, 1]

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1 70 CHAPTER 3. BASES IN BANACH SPACES 3.2 Bases of C[0, ] and L p [0, ] In the previous section we introduced the unit vector bases of `p and c 0. Less obvious is it to find bases of function spaces like C[0, ] and L p [0, ] Example (The Spline Basis of C[0, ]) Let (t n ) [0, ] be a dense sequence in [0, ], and assume that t = 0, t 2 =. It follows that (3.3) mesh(t,t 2,...t n )! 0, if n!where mesh(t,t 2,...t n ) = max t i t j : t j is neighbor of t i.,2,...n For f 2 C[0, ] we let P (f) to be the constant function taking the value f(0), and for n 2weletP n (f) be the piecewise linear function which interpolates the f at the point t,t 2,...t n. More precisely, let 0 = s <s 2 <...s n = be the increasing reordering of {t,t 2,...t n },thendefinep n (f) by P n (f) :[0, ]! K, with P n (f)(s) = s j s s j s j f(s j )+ s s j s j s j f(s j ), for s 2 [s j,s j ]. We note that P n : C[0, ]! C[0, ] is a linear projection and that kp n k =, and that (a), (b), (c) of Proposition 3..3 are satisfied. Indeed, the image of P n (C[0, ]] is generated by the functions f, f 2 (s) =s, for s 2 [0, ], and for n 2, f n (s) is the functions with the property f(t n ) =, f(t j ) = 0, j 2{, 2...}\{t n }, and is linear between any t j and the next bigger t i.thus dim(p n (C[0, ])) = n. Property (b) is clear, and property (c) follows from the fact that elements of C[0, ] are uniformly continuous, and condition (3.3). Also note that for n> it follows that f n 2 P n (C[0, ]) \N(P n ) \{0} and thus it follows from Proposition 3..3 that (f n ) is a monotone basis of C[0, ]. Now we define a basis of L p [0, ], the Haar basis of L p [0, ]. Let T = {(n, j) :n 2 N 0,j =, 2...,2 n }[{0}. We partially order the elements of T as follows (n,j ) < (n 2,j 2 ) () [(j 2 )2 n 2,j 2 2 n 2 ] ( [(j )2 n,j 2 n ]

2 3.2. BASES OF C[0, ] AND L P [0, ] 7 () (j )2 n apple (j 2 )2 n 2 <j 2 2 n 2 apple j 2 n, and n <n 2 whenever (n,j ), (n 2,j 2 ) 2 T and 0 < (n, j), whenever (n, j) 2 T \{0} Let apple p< be fixed. We define the Haar basis (h t ) t2t and the in L p normalized Haar basis (h (p) t ) t2t as follows. h 0 = h (p) 0 on [0, ] and for n 2 N 0 and j =, n we put h (n,j) = [(j )2 n,(j 2 )2 n ) [(j 2 )2 n,j2 n ). and we let (n,j) =supp(h (n,j) )= (j )2 n,j2 n, + (n,j) = (j )2 n, (j 2 )2 n (n,j) = (j 2 )2 n,j2 n. We let h () (n,j) = h (n,j). And for apple p< h (p) (n,j) = h (n,j) kh (n,j) k p =2 n/p [(j )2 n,(j 2 )2 n [(j 2 )2 n ),j2 n ). Note that kh t k p = for all t 2 T and that supp(h t ) supp(h s ) if and only if s apple t. Theorem If one orders (h (p) t ) t2t linearly in any order compatible with the order on T then (h (p) t ) is a monotone basis of L p [0, ] for all apple p<. Remark. a linear order compatible with the order on T is for example the lexicographical order h 0,h (0,),h (,),h (,2),h (2,),h (2,2),... Important observation: if (h t : t 2 T ) is linearly ordered into h 0,h,..., which is compatible with the partial order of T, then the following is true:

3 72 CHAPTER 3. BASES IN BANACH SPACES If n 2 N and and if h = a j h j, is any linear combination of the first n elements,thenh is constant on the support of h n. Moreover h can be written as a step function h = NX b j [sj,s j ), with 0 = s 0 <s <...s N, so that Z sj s j h n (t)dt =0. As we will see later, if <p<, any linear ordering of (h t : t 2 T )is a basis of L p [0, ], but not necessarily a monotone one. Proof of Theorem First note that the indicator functions on all dyadic intervals are in span(h t : t 2 T ). Indeed: [0,/2) =(h 0 + h (0,) )/2, (/2,] = (h 0 h (0,) )/2, [0,/4) =/2( [0,/2) h (,) ), etc. Since the indicator functions on all dyadic intervals are dense in L p [0, ] it follows that span(h t : t2t ). Let (h n ) be a linear ordering of (h (p) t ) t2t which is compatible with the ordering of T. Let n 2 N and (a i ) n a scalar sequence. We need to show that a i h i apple a i h i. As noted above, on the set A =supp(h n ) the function f = P n a ih i is constant, say f(x) =a, for x 2 A, therefore we can write A (f + a n h n )= A +(a + a n )+ A (a a n ), where A + is the first half of interval A and A convexity of [0, ) 3 r 7! r p, we deduce that the second half. From the a + an p + a a n p a p, 2

4 3.2. BASES OF C[0, ] AND L P [0, ] 73 and thus Z Z Z f + a n h n p dx = f p dx + a + a n p A + + a a n p A dx ZA c A = f p dx + 2 m(a) a + a n p + a a n p Z Ac Z f p dx + m(a) a p = A c f p dx which implies our claim. Proposition Since for apple p<, and <qapple,with p + q easy to see that for s, t 2 T it is (3.4) hh (p) s,h (q) t i = (s, t), we deduce that (h (q) t ) t2t are the coordinate functionals of (h (p) t ) t2t. Exercises. Decide whether or not the monomial, x, x n are a Schauder basis of C[0, ]. 2. Show that the Haar basis in L [0, ] can be reordered in such a way that it is not a a Schauder basis anymore.

5 74 CHAPTER 3. BASES IN BANACH SPACES 3.3 Shrinking, Boundedly Complete Bases Proposition Let (e n ) be a Schauder basis of a Banach space X, and let (e n) be the coordinate functionals and (P n ) the canonical projections for (e n ). Then a) Pn(x )= hx,e j ie j = h (e j ),x ie j,forn2nand x 2X. b) x = (X,X) lim n! P n(x ),forx 2X. c) (e n) is a Schauder basis of span(e n : n2n) whose coordinate functionals are (e n ). Proof. (a) For n2n, x 2 X and x = P he j,xie j 2X it follows that and thus hp n(x ),xi = hx,p n (x)i = (b) For x 2 X and x 2 X D x, P n(x )= E D E he j,xie j = hx,e j ie j,x hx,e j ie j. hx,xi = lim n! hx,p n xi = lim n! hp n(x ),xi. (c) It follows for m apple n and (a i ) n K, that mx a i e i = sup x2b X = sup x2b X apple mx a i he i,xi a i he i,p m (x)i a i e i kp m kapplesup kp j k j2n a i e i It follows therefore from Proposition 3..9 that (e n) is a basic sequence, thus, a basis of span(e n), Since h (e j ),e i i = he i,e ji = i,j, it follows that ( (e n )) are the coordinate functionals for (e n).

6 3.3. SHRINKING, BOUNDEDLY COMPLETE BASES 75 Remark. If X is a space with basis (e n ) one can identify X with a vector space of sequences x =( n ) K. If(e n) are coordinate functionals for (e n ) we can also identify the subspace span(e n : n2n) with a vector space of sequences x =( n ) K. Thewaysuchasequencex =( n ) 2 X acts on elements in X is via the infinite scalar product: D X E hx,xi = n e n = X n n. n2n n2n n e n, X n2n We want to address two questions for a basis (e n ) of a Banach space X and its coordinate functionals (e j ):. Under which conditions it follows that X = span(e n)? 2. Under which condition it follows that the map J : X! span(e n), with J(x)(z )=hz,xi, for x2x and z 2 span(e n), an isomorphy or even an isometry? We need first the following definition and some observations. Definition [Block Bases] Assume (x n ) is a basic sequence in Banach space X, a block basis of (x n )is asequence(z n ) X \{0}, with z n = k n X j=k n + a j x j, for n 2 N, where0=k 0 <k <k 2 <...and (a j ) K. We call (z n )aconvex block of (x n )ifthea j are non negative and P k n j=kn + a j =. Proposition The block basis (z n ) of a basic sequence (x n ) is also a basic sequence, and the basis constant of (z n ) is smaller or equal to the basis constant of (x n ). Proof. Let K be the basis constant of (x n ), let m apple n in N, and (b i ) n K. Then mx b i z i = mx Xk i j=k i + b i a j x j

7 76 CHAPTER 3. BASES IN BANACH SPACES apple K Xk i j=k i + b i a j x j = K b i z i Theorem For a Banach space with a basis (e n ) and its coordinate functionals (e n) the following are equivalent. a) X = span(e n : n 2 N) (and, thus, by Proposition 3.3., (e n) is a basis of X whose canonical projections are P n). b) For every x 2 X, lim n! kx span(ej :j>n)k = lim n! sup x2span(e j :j>n),kxkapple hx,xi =0. c) Every bounded block basis of (e n ) is weakly convergent to 0. We call the basis (e n )shrinkingif these conditions hold. Remark. Recall that by Corollary the condition (c) is equivalent with c ) Every bounded block basis of (e n ) has a further convex block which converges to 0 in norm. Proof of Theorem (a))(b) Let x 2 X and, using (a), write it as x = P a je j.then lim sup n! x2span(e j :j>n)kxkapple hx,xi = lim sup n! x2span(e j :j>n),kxkapple = lim sup n! x2span(e j :j>n),kxkapple apple lim n! k(i P n)(x )k =0. (b))(c) Let (z n ) be a bounded block basis of (x n ), say z n = k n X j=k n + hx, (I h(i P n )(x)i P n)(x ),xi a j x j, for n 2 N, with0=k 0 <k <k 2 <...and (a j ) K. and x 2 X.Then,lettingC =sup j2n kz j k, hx,z n i apple sup hx,zi! n! 0, by condition (b), z2span(e j :j k n ),kzkapplec

8 3.3. SHRINKING, BOUNDEDLY COMPLETE BASES 77 thus, (z n ) is weakly null. (a)) (c) Assume there is an x 2 S X,withx 62 span(e j : j 2 N). It follows for some 0 <"apple (3.5) " =limsupkx Pn(x )k > 0. n! By P induction we choose z,z 2,...in B X and 0 = k 0 <k <..., so that z n = kn j=kn + a je j, for some choice of (a j ) kn j=k n + and hx,z n i "/2(+K), where K =sup j2n kp j k.indeed,letz 2 B X \ span(e j ), so that hx,z i "/2( + K) and let k =min{k : z 2 span(e j : j apple k). Assuming z,z 2,...z n and k <k 2 <...k n has been chosen. Using (3.5) we can choose m>k n so that kx Pm(x )k >"/2 and then we let z n+ 2 B X \ span(e i : i 2 N) with hx Pm(x ), z n+ i = hx, z n+ P m ( z n+ )i >"/2. Finally choose and z n+ = z n+ P m ( z n+ ) +K 2 B X k n+ =min k : z n+ 2 span(e j : j apple k). It follows that (z n ) is a bounded block basis of (e n ) which is not weakly null. Examples Note that the unit vector bases of `p, <p<, and c 0 are shrinking. But the unit vector basis of ` is not shrinking (consider (,,,,,...) 2 ` = `). Proposition Let (e j ) be a shrinking basis for a Banach space X and (e j ) its coordinate functionals. Put Y = Then Y with the norm is a Banach space and n (a i ) K :sup n (a i ) =sup n2n o a j e j <. a j e j, T : X! Y, x 7! (hx,e ji) j2n, is an isomorphism between X and Y. If (e n ) is monotone then T is an isometry.