By A.M. Fink

ISBN-10: 3540067299

ISBN-13: 9783540067290

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9. There exist a Banach space X with a semi-normalized basis (xn ) and a non-tauberian operator T : X −→ 2 with property (N ) such that (T xn ) is the unit vector basis of 2 . xi ) any Proof. For every n ∈ N and every sequence (xi ) ∈ c0 , we denote by (ˆ decreasing rearrangement of (|xi |), and we consider the semi-norm (xi ) n := ˆ2 + · · · + xˆn x ˆ1 + x . 1 + 12 + · · · + n1 Let X1 be the largest linear subspace of c0 normed by the expression (xi ) L := sup (xi ) n, n∈N (xi ) ∈ c0 and let X be the subspace formed by all the elements (xi ) of X1 for which (xi ) n −→ 0.

Thus the sequence (gn ) of coeﬃcient functionals associated with (xn ) is bounded, and consequently, every gn admits a Hahn-Banach extension fn ∈ X ∗ so that (fn ) is bounded. Since T xn − y −→ 0, passing to a subsequence if necessary, we can n assume that T xn − y · fn ≤ 2−n for each n. Therefore, the formula ∞ fi , x (y − T xi ) Kx := i=1 20 Chapter 2. Tauberian operators. Basic properties deﬁnes a compact operator K ∈ L(X, Y ). Clearly (T + K)xi = y for every i. Thus N (T + K) contains the closed subspace generated by the sequence (xi − x1 )∞ i=2 which is not reﬂexive, hence N (T + K) is not reﬂexive.

Note that if both (xn ) and (yn ) are semi-normalized, then (αn ) is bounded. Obviously, a shrinking basis (xn ) is weakly null. The following lemma is standard, but its proof is included for the sake of completeness. 1. A semi-normalized basic sequence (xn ) is shrinking if and only if all its semi-normalized block basic sequences are weakly null. 26 Chapter 2. Tauberian operators. Basic properties ∞ Proof. Let (xn ) be a semi-normalized basic sequence and write Xn := span{xi }i=n for every n ∈ N.

### Almost Periodic Differential Equations by A.M. Fink

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