Frames and Riesz bases for Banach spaces, and Banach spaces of ...

FRAMES AND RIESZ BASES FOR BANACH SPACES 175of the dual space B ∗ s of B s ; cf. [7, Lemma 3.1]. Let (e n ) be the sequence of thecanonical unit vectors and suppose that (e n ) is a Schauder basis for B s . LetY s = {(x ∗ s(e n ))|x ∗ s ∈ B ∗ s}and ‖(x ∗ s(e n ))‖ Ys = ‖x ∗ s‖. Then we see that (Y s , ‖ · ‖ Ys ) is a normed space andthe coordinate functionals for Y s are continuous. Consider the map j s : Y s → Bs∗defined by j s [(x ∗ s(e n ))] = x ∗ s. Then j s is a surjective linear isometry and so Y s isa Banach sequence space. For example, if B s = l p (1 < p < ∞) (resp. l 1 ), thenY s = l p ∗ (resp. l ∞ ), and if B s = c 0 , then Y s = l 1 .Now for every x ∗ s ∈ Bs ∗ and (α n ) ∈ B s( ∑ )x ∗ s((α n )) = x ∗ s α n e n = ∑nnα n x ∗ se n = ∑ nα n (j −1s x ∗ s) n ,where (js−1 x ∗ d ) n is the n-th element of js−1 x ∗ d . Let (f n) be the sequence of thecanonical unit vectors in Y s . Fix k ∈ N. Then for every (α n ) ∈ B sj s f k ((α n )) = ∑ nα n (js−1 j s f k ) n = ∑ nα n (f k ) n = α k .This shows that (j s f n ) is the sequence of the coordinate functionals for B s . If(f n ) is a Schauder basis for Y s , then for every x ∗ s ∈ B ∗ s( ∑ )x ∗ s = j s [(x ∗ s(e n ))] = j s x ∗ s(e n )f n = ∑ nnx ∗ s(e n )j s f n .Throughout this paper we use the objects Y s , j s , (e n ), (f n ), the analysis andsynthesis operators in the introduction. Recall that an operator S from Y ∗ to X ∗is weak ∗ to weak ∗ continuous if and only if there exists an operator T from X toY such that S is the adjoint operator T ∗ of T ; cf. see [17, Theorem 3.1.11]. Wenow haveTheorem 2.1. Suppose that (e n ) is a Schauder basis for B s and let (x n ) be asequence in X. Then the following are equivalent.(a) The analysis operator F (xn) : X ∗ → Y s is well defined.(b) The synthesis operator R (xn) : B s → X is well defined.(c) The analysis operator F (xn) : X ∗ → Y s is well defined and the operator j s F (xn) :X ∗ → Bs ∗ is weak ∗ to weak ∗ continuous.Hence (x n ) ∈ Ys w (X) if and only if (x n ) ∈ B s R(X).Proof. (c)=⇒(a) is trivial.