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

178 K. CHO, J.M. KIM, H.J. LEE(a) The analysis operator F (x ∗ n ) : X → B s is well defined.(b) The synthesis operator R (x ∗ n ) : Y s → X ∗ is well defined and the operatorR (x ∗ n )js−1 : Bs ∗ → X ∗ is weak ∗ to weak ∗ continuous.3. B s -Frames and Riesz basesIn this section we use the results in Section 2 to establish some relationshipsbetween frames and Riesz bases, necessary and sufficient conditions for frames tobe Riesz bases.Recall that an operator T is surjective if and only if T ∗ is an isomorphism,and T ∗ is surjective if and only if T is an isomorphism; cf. [17, Theorem 3.1.22].Then we haveTheorem 3.1. 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 an isomorphism.(b) The synthesis operator R (xn) : B s → X is surjective.(c) The analysis operator F (xn) : X ∗ → Y s is an isomorphism and the operatorj s F (xn) : X ∗ → B ∗ s is weak ∗ to weak ∗ continuous.Proof. (c)=⇒(a) is trivial.(a)=⇒(b) By Theorem 2.1 R (xn) is well defined. Then R ∗ (x n) = j sF (xn) in theproof of Theorem 2.1(b)=⇒(c). Hence by the assumption (a) R (xn) is surjective.(b)=⇒(c) Since R ∗ (x n) = j sF (xn), by the assumption (b) F (xn) is an isomorphism.□From Theorem 3.1, if (x n ) is a B s -Riesz basis for X, then (x n ) is a Y s -framefor X ∗ . For example, if (x n ) is an l p -Riesz basis for X (1 ≤ p < ∞), then (x n ) isan l p ∗-frame for X ∗ , and if (x n ) is a c 0 -Riesz basis for X, then (x n ) is an l 1 -framefor X ∗ . But a Y s -frame for X ∗ does not imply a B s -Riesz basis for X in general.Indeed, consider the sequence (x n ) = (e 1 , 0, e 2 , 0, · · ·, 0, e n , 0, · · ·) in c 0 . Then forevery (α k ) ∈ l 1 ‖(α k )‖ 1 = ‖((α k )x n )‖ 1 . Thus (x n ) is an l 1 -frame for l 1 , but (x n )fails the condition (ii) of a c 0 -Riesz basic sequence for c 0 .In the proof of Theorem 2.4 the synthesis operator R (xn) : Y s → X ∗∗ is theoperator F ∗ (x n) j s, hence we have the following.Theorem 3.2. 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 (xn) : X ∗ → B s is an isomorphism.(b) The synthesis operator R (xn) : Y s → X ∗∗ is surjective, and the operatorR (xn)j −1s: Bs ∗ → X ∗∗ is weak ∗ to weak continuous.(c) The analysis operator F (xn) : X ∗ → B s is weak ∗ to weak continuous and anisomorphism.Remark 3.3. In Theorem 3.2, if (a) holds, then X is reflexive because R (xn)(Y s ) ⊂X in the proof of Theorem 2.4 . Consequently, under the assumption in Theorem3.2, a nonreflexive Banach space X cannot contain a B s -frame for X ∗ .We now consider sequences in dual spaces. The following result extends [11,Theorem 2.4].