Bases in Banach spaces

Chapter 3 Bases in Banach spaces Like every vector space a Banach space X admits an algebraic or Hamel basis, i.e. a subset B X, so that every x X is in a unique way the (finite) 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. 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 n ) X is called Schauder basis of X, or simply a basis of X, if for every x X, there is a unique sequence of scalars (a n ) K so that Examples 3.1.2. For n N let x = a n e n. n=1 e n = ( 0,... 0, 1, 0,...) K N }{{} n 1 times Then (e n ) is a basis of l p,1 p< and c 0. We call (e n ) the unit vector of l p and c 0, respectively. Remarks. Assume that X is a Banach space and (e n ) a basis of X. 51

52 CHAPTER 3. BASES IN BANACH SPACES a) (e n ) is linear independent. b) span(e n : n N) is dense in X, in particular X is separable. c) Every element x is uniquely determined by the sequence (a n ) so that x = a ne n. So we can identify X with a space of sequences in K N. Proposition 3.1.3. Let X be a normed linear space and assume that (e n ) X has the property that each x X can be uniquely represented as a series x = a n e n, with (a n ) K n=1 (we could call (e n ) Schauder basis of X but we want to reserve this term only for Banach spaces). For n N and x X define e n(x) K to be the unique element in K, so that x = e n(x)e n. Then e n : X K is linear. For n N let n=1 P n : X span(e j : j n),x e n(x)e n. Then P n : X X are linear projections onto span(e j : j n) and the following properties hold: a) dim(p n (X)) = n, b) P n P m = P m P n = P min(m.n), for m, n N, c) lim n P n (x) =x, for every x X. Conversely if (P n : n N) is a sequence of linear projections satisfying (a), (b), and (c), and moreover are bounded, and if e 1 P 1 (X) \{0} and e n P n (X) N (P n 1 ), with e n 0, if n>1, then each x X can be uniquely represented as a series x = a n e n, with (a n ) K, n=1 so in particular (e n ) is a Schauder basis of X in case X is a Banach space.

3.1. SCHAUDER BASES 53 Proof. The linearity of e n follows from the unique representation of every x X as x = e n(x)e n, which implies that for x and y in X and α, β K, αx + βy = lim α e n j(x)e j + β = lim n and, on the other hand αx + βy = e j(y)e j (αe j(x)+βe j(y))e j = e j(αx + βy)e j, (αe j(x)+βe j(y))e j, thus, by uniqueness, e j (αx + βy) =αe j (x) +βe j (y), for all j N. The conditions (a), (b) and (c) are clear. Conversely assume that (P n ) is a sequence of bounded and linear projections satisfying (a), (b), and (c), and if e 1 P 1 (X) \{0} and e n P n (X) N (P n 1 ), if n>1, then for x X, by (b) P n 1 (P n (x) P n 1 (x)) = P n 1 (x) P n 1 (x) = 0, and P n (x) P n 1 (x) =P n (P n (x) P n 1 (x)) P n (X), and therefore P n (x) P n 1 (x) P n (X) N (P n 1 ). Thus we can write P n (x) P n 1 (x) =a n e n, for n N, and it follows from (c) that (letting P 0 = 0) x = lim P n(x) = lim n n P j (x) P j 1 (x) = lim n a j e j = a j e j. In order to show uniqueness of this representation of x assume x = b je j. From the continuity of P m P m 1, for m N it follows that ( ) a m e m =(P m P m 1 )(x) = lim (P m P m 1 ) b j e j = b m e m, n and thus a m = b m.

54 CHAPTER 3. BASES IN BANACH SPACES Definition 3.1.4. [Canonical Projections and Coordinate functionals] Let X be a normed space and assume that (e i ) satisfies the assumptions of Proposition 3.1.3. The linear functionals (e n) as defined in Proposition 3.1.3 are called the Coordinate Functionals for (e n ) and the projections P n, n N, are called the Canonical Projections for (e n ). Proposition 3.1.5. If X is a normed linear space and (e i ) assume that (e n ) X has the property that each x X can be uniquely represented as a series x = a n e n, with (a n ) K. n=1 If the canonical projections are bounded, and, moreover, sup n N P n <, then (e i ) is a Schauder basis of its completion X. Proof. Let P n : X X, n N, be the (by Proposition 1.1.4 and Exercise 1 in Section 1.2 uniquely existing) extensions of P n. Since P n has finite range it follows that P n ( X) =P n (X) = span(e j : j n) is finite dimensional and, thus, closed. ( P n ) satisfies therefore (a) of Proposition 3.1.3. Since the P n 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 for x X, x = lim k x k, with x k X, for k N, that x P n ( x) x x k + sup P j x x k + x k P n (x k ) j N and, since (P n ) is uniformly bounded, we can find for given ε> 0, k large enough so that the first two summands do not exceed ε, and then we choose n N large enough so that the third summand is smaller than ε. It follows 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. We will now show that if (e n ) is a basis for a Banach space X the coordinate functionals, and thus the canonical projections are bounded, and moreover the canonical projections are uniformly bounded. Theorem 3.1.6. Let X be a Banach space with a basis (e n ) and let (e n) be the corresponding coordinate functionals and (P n ) the canonical projections. Then P n is bounded for every n N and b = sup P n L(X,X) <, n N

3.1. SCHAUDER BASES 55 and thus e n X and e n X = P n P n 1 e n 2b e n. We call b the basis constant of (e j ). If b =1we say that (e i ) is a monotone basis. Furthermore there is an equivalent renorming of (X, ) for which (e n ) is a monotone basis for (X, ). Proof. For x X we define x = sup P n (x), n N since x = lim n P n (x), it follows that x x < for x X. It is clear that is a norm on X. Note that for n N P n = sup P n (x) x X, x 1 = sup x X, x 1 m N = sup x X, x 1 m N sup P m P n (x) sup P min(m,n) (x) 1. Thus the projections P n are uniformly bounded on (X, ). We also note that the P n 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 (3.1) x P n (x) = sup P m (x) P min(m,n) (x) m N = sup P m (x) P n (x) 0 if n, m n which verifies conditionn (c). Thus, it follows therefore from the second part of Proposition 3.1.3, the above proven fact that P n 1, for n N, and Proposition 3.1.5, that (e n ) is a Schauder basis of the completion of (X, ) which we denote by ( X, ). We will now show that actually X = X, and thus that, (X, ) is aleady complete. Then it would follow from Corollary 1.3.6 of the Closed Graph Theorem that is an equivalent norm, an thus that C = sup sup P n (x) = sup x <. x B X x B X n N

56 CHAPTER 3. BASES IN BANACH SPACES So, let x X and write it (uniquely) as x = a je j, since, and since X is complete the series a je j with respect to in X to say x X. But now (3.1) that P n (x) = n a je j also converges in to x. But this means 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 and are two norms on the same linear space X, so that, and so that (,X) is complete, does it then follow that (X, ) is also complete (and thus and are equivalent norms). The answer is negative, as the following example shows. Example 3.1.7. Let X =l 2 with its usual norm 2 and let (b γ :γ Γ) S l2 be a Hamel basis of Γ (by Exercise 2, Γ is necessarily uncountable). For x l 2 define x, x = x γ, γ Γ where x = γ Γ x γb γ is the unique representation of x as a finite linear combination of elements of (b γ : γ Γ). Since b γ 2, for γ Γ, it follows for x = γ Γ x γb γ l 2 from the triangle inequality that x = x γ = x γ b γ 2 2 x γ b γ = x 2. γ Γ γ Γ γ Γ Finally both norms and, cannot be equivalent. Indeed, for arbitrary ε> 0, there is an uncountable set Γ Γ, so that b γ b γ 2 <ε, γ, γ Γ, (Γ is uncountable but S l2 is in the 2 -norm separable). for any two different elements γ, γ Γ it follows that b γ b γ <ε< 2= b γ b γ. Since ε> 0 was arbitrary this proves that and cannot be equivalent. Definition 3.1.8. [Basic Sequences] Let X be a Banach space. A sequence (x n ) X \{0} is called basic sequence if it is a basis for span(x n : n N). If (e j ) and (f j ) are two basic sequences (in possibly two different Banach spaces X and Y ). We say that (e j ) and (f j ) are isomorphically equivalent if the map T : span(e j : j N) span(f j : j N), a j e j a j f j,

3.1. SCHAUDER BASES 57 extends to an isomorphism between the Banach spaces between span(e j : j N) and span(f j : j N). Note that this is equivalent with saying that there are constants 0 <c C so that any n N and any sequence of scalars (λ j ) n it follows that c. λ j e j λ j f j C λ j e j Proposition 3.1.9. Let X be Banach space and (x n : n N) X \{0}. The (x n ) is a basic sequence if and only if there is a constant K 1, so that for all m<nand all scalars (a j ) n K we have (3.2) m. a i x i K a i x i i=1 In that case the basis constant is the smallest of all K 1 so that (3.2) holds. Proof. ( Follows from Theorem 3.1.6, since K := sup n N P n < and P n m i=1 a ) ix i = m i=1 a ix m, if m n and (a i ) n i=1 K. Assume that there is a constant K 1 so that for all m<nand all scalars (a j ) n K we have i=1 m. a i x i K a i x i i=1 We first note that this implies that (x n ) is linear independent. Indeed, if we assume that n a jx j = 0, for some choice of n N and (a j ) n K, and not all of the a j are vanishing, we first observe that at least two of a j s cannot be equal to 0 (since x j 0, for j N), thus if we let m := min{j : a j 0}, it follows that m a jx j 0, but n a jx j = 0, which contradicts our assumption. It follows therefore that (x n ) is a Hamel basis for (the vector space) span(x j : j N), which implies that the projections P n are well defined on span(x j : j N), and satisfy (a), (b), and (c) of Proposition 3.1.3. Moreover, it follows from our assumption that i=1 { m P m = sup a j x j : n N, (aj ) n K, Thus, our claim follows from Proposition 3.1.5. } a j x j 1 K.

58 CHAPTER 3. BASES IN BANACH SPACES 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. Exercises 1. Let (e γ : γ Γ) be a Hamel basis of an infinite dimensional Banach space X.Show that some of the coordinate functionals associated with that basis are not continuous. Hint: pick a sequence (γ n ) Γ of pairwise different elements of Γ and consider x = n=1 2 n e γ n e γn. 2. Show that the Hamel basis of an infinite dimensional Banach space X must be uncountable. 3. For n N define in c 0 s n = e j = (1, 1,... 1, 0,...). }{{} n times Prove that (s n ) is a basis for c 0, but that one can reorder (s n ) so that it is not a basis of c 0. (s n ) is called the summing basis of c 0. Hint: Play around with alternating series of (s n ). 4. Let 1 <p< and assume that (x n ) is a weakly null sequence in l p with inf n N x n > 0. Show that (x n ) has a subsequence which is isomorphically equivalent to the unit vector basis of l p. And then deduce form this: Let T : l p l q with 1 < q < p <, be a bounded linear operator. Show that T is compact., meaning that T (B lp ) is relatively compact in l q.

3.2. BASES OF C[0, 1] AND L P [0, 1] 59 3.2 Bases of C[0, 1] and L p [0, 1] In the previous section we introduced the unit vector bases of l p and c 0. Less obvious is it to find bases of function spaces like C[0, 1] and L p [0, 1] Example 3.2.1. [The Spline Basis of C[0, 1]] Let (t n ) [0, 1] be a dense sequence in [0, 1], and assume that t 1 = 0, t 2 = 1. It follows that (3.3) mesh(t 1,t 2,... t n ) 0, if n where mesh(t 1,t 2,... t n ) = max { t i t j : i, j {1,..., n}, i j }. For f C[0, 1] we let P 1 (f) to be the constant function taking the value f(0), and for n 2 we let P n (f) be the piecewise linear function which interpolates the f at the point t 1,t 2,... t n. More precisely, let 0 = s 1 <s 2 <... s n =1 be the increasing reordering of {t 1,t 2,... t n }, then define P n (f) by P n (f) : [0, 1] K, with P n (f)(s) = s j s f(s j 1 )+ s s j 1 f(s j ), for s [s j 1,s j ]. s j s j 1 s j s j 1 We note that P n : C[0, 1] C[0, 1] is a linear projection and that P n = 1, and that (a), (b), (c) of Proposition 3.1.3 are satisfied. Indeed, the image of P n (C[0, 1]] is generated by the functions f 1 1, f 2 (s) =s, for s [0, 1], and for n 2, f n (s) is the functions with the property f(t n ) = 1, f(t j ) = 0, j {1, 2...} \ {t n }, and is linear between any t j and and the next bigger t i. Thus dim(p n (C[0, 1])) = n. Property (b) is clear, and property (c) follows form the fact that elements of C[0, 1] are uniformly continuous, and condition (3.3). Also note that for n>1 it follows that f n P n (C[0, 1]) N (P n 1 ) \{0} and thus it follows from Proposition 3.1.3 that (f n ) is a monotone basis of C[0, 1]. Now we define a basis of L p [0, 1], the Haar basis of L p [0, 1]. Let T = {(n, j) :n N 0,j =1, 2..., 2 n } { 0}. We partially order the elements of T as follows (n 1,j 1 ) < (n 2,j 2 ) [(j 2 1)2 n 2,j 2 2 n 2 ] [(j 1 1)2 n 1,j 1 2 n 1 ] (j 1 1)2 n 1 (j 2 1)2 n 2 <j 2 2 n 2 j 1 2 n 1, and n 1 <n 2

60 CHAPTER 3. BASES IN BANACH SPACES and whenever (n 1,j 1 ), (n 2,j 2 ) T 0 < (n, j), whenever (n, j) T \{0} Let 1 p< be fixed. We define the Haar basis (h t ) t T and the in L p normalized Haar basis (h (p) t ) t T as follows. h 0 = h (p) 0 1 on [0, 1] and for n N 0 and j =1, 2... 2 n we put and we let h (n,j) =1 [(j 1)2 n,(j 1 2 )2 n ) 1 [(j 1 2 )2 n ),j2 n )). (n,j) = supp(h (n,j) )= [ (j 1)2 n,j2 n), + (n,j) = [ (j 1)2 n, (j 1 2 )2 n) (n,j) = [ (j 1 2 )2 n,j2 n). We let h ( ) (n,j) = h (n,j). And for 1 p< h (p) (n,j) = h (n,j) =2 n/p( 1 h (n,j) [(j 1)2 n,(j 1 p 2 )2 n ) 1 ) [(j 1 2 )2 n ),j2 n ). Note that h t p = 1 for all t T and that supp(h t ) supp(h s ) if and only if s t. Theorem 3.2.2. If one orders (h (p) t ) t T linearly in any order compatible with the order on T then (h (p) t ) is a monotone basis of L p [0, 1] for all 1 p<. Remark. a linear order compatible with the order on T is for example the lexicographical order h 0,h (0,1),h (1,1),h (1,2),h (2,1),h (2,2),.... Important observation: if (h t : t T ) is linearly ordered into h 0,h 1,..., which is compatible with the partial order of T, then the following is true:

3.2. BASES OF C[0, 1] AND L P [0, 1] 61 If n N and and if n 1 h = a j h j, is any linear combination of the first n 1 elements, then h is constant on the support of h n 1. Moreover h can be written as a step function h = N b j 1 [sj 1,s j ), with 0 = s 0 <s 1 <... s N, so that sj s j 1 h n (t)dt =0. As we will see later, if 1 < p <, any linear ordering of (h t : t T ) is a basis of L p [0, 1], but not necessarily a monotone one. Proof of Theorem 3.2.2. First note that the indicator functions on all dyadic intervals are in span(h t : t T ). Indeed: 1 [0,1/2) =(h 0 + h (0,1) )/2, 1 (1/2,1] = (h 0 h (0,1) )/2, 1 [0,1/4) =1/2(1 [0,1/2) h (1,1) ), etc. Since the indicator functions on all dyadic intervals are dense in L p [0, 1] it follows that span(h t : t T ). Let (h n ) be a linear ordering of (h (p) t ) t T which is compatible with the ordering of T. Let n N and (a i ) n i=1 a scalar sequence. We need to show that n 1. a i h i a i h i i=1 As noted above, on the set A = supp(h n ) the function f = n 1 i=1 a ih i is constant, say f(x) =a, for x A, therefore we can write i=1 1 A (f + a n h n ) = 1 A +(a + a n )+1 A (a a n ), where A + is the first half of interval A and A the second half. From the convexity of [0, ) r r p, we deduce that 1[ a + an p + a a n p ] a p, 2

62 CHAPTER 3. BASES IN BANACH SPACES and thus f + a n h n p dx = f p dx + a + a n p 1 A + + a a n p 1 A dx A c A = f p dx + 1 A c 2 m(a)[ a + a n p + a a n p] f p dx + m(a) a p = A c f p dx which implies our claim. Proposition 3.2.3. Since for 1 p<, and 1 <q, with 1 p + 1 q easy to see that for s, t T it is (3.4) h (p) s,h (q) t = δ(s, t), we deduce that (h (q) t ) t T are the coordinate functionals of (h (p) t ) t T. Exercise 1. Decide whether or not the monomial 1, x, x 2,... are a Schauder basis of C[0, 1]. 2. Show that the Haar basis in L 1 [0, 1] can be reordered in such a way that it is not a a Schauder basis anymore.