Spectrum of a Weakly Hypercyclic Operator Meets the Unit Circle

Suppose that T is a bounded operator on a nonzero Banach space X . Given a vector x ∈ X , we say that x is hypercyclic for T if the orbit OrbTx = {T x}n is dense in X . Similarly, x is said to be weakly hypercyclic if OrbTx is weakly dense in X . A bounded operator is called hypercyclic or weakly hypercyclic if it has a hypercyclic or, respectively, a weakly hypercyclic vector. It is shown in [CS] that a weakly hypercyclic vector need not be hypercyclic, and there exist weakly hypercyclic operators which are not hypercyclic. C. Kitai showed in [K] that every component of the spectrum of a hypercyclic operator intersects the unit circle. K. Chan and R. Sanders asked in [CS] if the spectrum of a weakly hypercyclic operator meets the unit circle. In this note we show that every component of the spectrum of a weakly hypercyclic operator meets the unit circle. Lemma 1. Let X be a Banach space and let c > 1. Suppose that xn ∈ X satisfies ‖xn‖ > c for all n > 1. Then 0 / ∈ {xn}n w . Proof. Let N be the smallest positive integer such that c > 2. We shall prove that there exist F1, . . . , FN ∈ X such that max 16k6N ∣