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

FRAMES AND RIESZ BASES FOR BANACH SPACES 179Theorem 3.4. Suppose that (e n ) is a Schauder basis for B s and (f n ) is a Schauderbasis for Y s and let (x ∗ n) be a sequence in X ∗ . Then the following are equivalent.(a) The analysis operator F (x ∗ n ) : X → B s is an isomorphism.(b) The synthesis operator R (x ∗ n ) : Y s → X ∗ is surjective and the operator R (x ∗ n )js −1 :Bs ∗ → X ∗ is weak ∗ to weak ∗ continuous.Proof. From the same argument as in the proof of Theorem 2.4, we see that thesynthesis operator R (x ∗ n ) : Y s → X ∗ is the operator F(x ∗ ∗ )j s, hence the proof isnestablished.□In view of Corollary 2.6 and Theorem 3.4, if (x ∗ n) ∈ B w∗s (X ∗ ) and (x ∗ n) is a Y s -Riesz basis for X ∗ , then (x ∗ n) is a B s -frame for X, and if B s is reflexive and (x ∗ n)is a Y s -Riesz basis for X ∗ , then (x ∗ n) is a B s -frame for X. Consider the sequence(x ∗ n) = (e 1 , 0, e 2 , 0, · · ·, 0, e n , 0, · · ·) in l 1 . Then (x ∗ n) is a c 0 -frame for c 0 , but (x ∗ n)is not even an l 1 -Riesz basic sequence for l 1 .In Theorem 2.1, we have shown that a sequence (x n ) in X is a B s -Riesz sequencefor X if and only if (x n ) is a Y s -Bessel sequence for X ∗ . But even for a Y s - Rieszbasis in a dual space, it may not be a B s -Bessel sequence.Example 3.5. Let x ∗ 1 = e 1 and for every n ≥ 2 let x ∗ n = (1, 0, · · · , 0, 1, 0, · · · ),where the second 1 is the n-th element. Consider the sequence (x ∗ n) in l 1 . Thenfor every (α n ) ∈ l 1∑n ‖α nx ∗ n‖ 1 = ∑ n 2|α n| < ∞ and so ∑ n α nx ∗ n converges inl 1 ,( ‖(α n )‖ 1 ≤∑)∥ ( ∥∥1 ∞∑ )∥ ∥∥1∥ α k , α 2 , · · · , α n , · · · + ∥ − α k , 0, · · ·and(α n ) =(α 1 −∞∑k=2≤k∥ ∑ nα n x ∗ n∥ +1∞∑|α k | ≤ 2∥ ∑ nk=2k=2α n x ∗ n∥ ,1)α k , 0, · · · + (α 2 , α 2 , 0, · · · ) + · · · + (α n , 0, · · · , α n , 0, · · · ) + · · ·=(α 1 −∞∑α k)x ∗ 1 +k=2∞∑α n x ∗ nwhich shows l 1 = span{x ∗ n}. Hence (x ∗ n) is an l 1 -Riesz basis for l 1 . But x ∗ n(e 1 ) = 1for every n and so (x ∗ n) does not weak ∗ converge to 0 in l 1 . Hence (x ∗ n) is not ac 0 -Bessel sequence for c 0 .We next establish necessary and sufficient conditions for Bessel sequences (resp.frames) to be Riesz basic sequences (resp. Riesz bases).Theorem 3.6. Suppose that (e n ) is a Schauder basis for B s and (f n ) is a Schauderbasis for Y s and let (x n ) ∈ Ys w (X). Then the following are equivalent.(a) (x n ) is a B s -Riesz basic sequence for X.(b) For (β n ) ∈ B s , if ∑ n β nx n = 0, then β n = 0 for all n, and R (xn)(B s ) is closedin X.(c) F (xn)(X ∗ ) = Y s .n=2