On weak orbits of operators

Let T be a completely nonunitary contraction on a Hilbert space H with r(T ) = 1. Let an > 0, an → 0. Then there exists x ∈ H with |〈Tnx, x〉| ≥ an for all n. We construct a unitary operator without this property. This gives a negative answer to a problem of van Neerven. Let X be a complex Banach space. Then each operator T ∈ B(X) has orbits that are ”large” in the following sense [M1], [B]: Let (an) be a sequence of positive numbers such that an → 0. Then there exists x ∈ X such that ‖Tnx‖ ≥ anr(T) for all n. Moreover, for each ε > 0 it is possible to find x ∈ X with ‖x‖ < supn an + ε. The corresponding question for weak orbits 〈Tnx, x∗〉 was considered by J. van Neerven [N], see also [M3]. (1) Let T ∈ B(X). Let (an) be a sequence of positive numbers such that an → 0. Do there exist x ∈ X and x∗ ∈ X∗ such that |〈Tnx, x∗〉| ≥ anr(T) for all n? There are several interesting cases when the answer is positive. In [N], it was proved for positive operators on Banach lattices. In [M2] and [M4] the statement was shown for Banach space operators satisfying T → 0 in the strong operator topology and r(T ) = 1. In the present paper we consider Hilbert space operators and generalize this for operators satisfying T → 0 in the weak operator topology. As a consequence, we get that (1) is true for any completely non-unitary contraction with r(T ) = 1. Note that for unitary operators questions concerning weak orbits reduce to questions concerning Fourier coefficients of L1 functions. We show that if μ is a Rajchman measure (in particular, an absolutely continuous measure) on the unit circle, then there is a positive function f ∈ L1(μ) such that |f̂(n)| ≥ a|n| for all n (the statement is a folklore in case of the Lebesgue measure, see [K, p. 22 and 26]). However, the previous statement is not true in general. We construct an example of a Kronecker measure ν and a sequence (an) of positive numbers, an → 0 such that there is no function f ∈ L1(ν) with the above property. This also gives a negative answer to question (1) of van Neerven. Let H be a complex Hilbert space and let T ∈ B(H). We say that T → 0 in the weak operator topology if 〈Tnx, y〉 → 0 for all x, y ∈ H. Mathematics Subject Classification (2000): 47A05, 47B15