By Robert C. Gunning, Hugo Rossi
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Basic properties / T (E); otherwise, if T x0 = T e and en ≤ 3 for all n, T x0 ∈ 3 · T (BE ). But T x0 ∈ for some e = ∞ c e ∈ E, we would have x − e ∈ N (T ) = N (Q), so 0 i=1 i ni ∞ 0 = Q(x0 − e) = Qx0 − ci i=1 ∞ = 1− ci 1 Qxni + Qx0 ni ∞ Qx0 − i=1 i=1 ci Qxni , ni ∞ i=1 ci and ci = 0 for all i, a and since (Qxni ) is basic, we would get 1 = contradiction. (c)⇒(a) Let us assume that N (T ) is reﬂexive but T is not tauberian. As in the proof of (b)⇒(c), we will ﬁnd a closed subspace E of X so that T (BE ) ⊂ T (E).
A delicate improvement, due to Zippin , states that a Banach space with a basis is reﬂexive if all its bases are shrinking or all its bases are boundedly complete. In this section, we characterize the tauberian operators and the operators with property (N ) in terms of their action on boundedly complete basic sequences and shrinking basic sequences. We notice that most of the results in the previous sections are of a subsequential nature in the sense that, if there is a sequence satisfying a certain property P , then those results assert that it contains a subsequence that possesses a certain property Q.
4. Let T ∈ L(X, Y ) be an operator. The following statements are equivalent: (a) T is tauberian; (b) if (xn ) is a bounded sequence in X such that (T xn ) is weakly convergent, then (xn ) contains a weakly convergent subsequence; (c) if (xn ) is a bounded sequence in X such that (T xn ) is convergent, then (xn ) contains a weakly convergent subsequence. Proof. (a)⇒(b) Let (xn ) be a bounded sequence in X such that (T xn ) is weakly convergent to y ∈ T (BX ). 7, we have y ∈ R(T ). 2 that y = T x.