Weakly supercyclic operators
An operator \(T\) on a Banach space \({\mathcal B}\) is said to be supercyclic if there is \(x\in{\mathcal B}\) such that \(\{\lambda T^n x: \lambda \in \mathbb C \text{ and } n=0,1,\ldots \}\) is dense in \(\mathcal B\). If norm density is replaced by density with respect to the weak topology, then one obtains the concept of weakly supercyclic operator. If there is no need of scalar multiples to have the required density, one obtains the special concepts of (norm) hypercyclicity and weakly hypercyclicity. The weak versions are, of course, not mere generalizations. Indeed, they are really interesting notions, since a theorem of Mazur, acording to which the norm closure and the weak closure of a convex set coincide, implies that weak supercyclic operators are cyclic. Thus the class of operators that are weakly supercyclic but not supercyclic is a new class of cyclic operators. On the other hand, although some results from supercyclicity can be generalized easily to weak supercyclicity, this is not always the case and, as is well-known, weak topologies are much more difficult to handle than norm topologies. For instance, \textit{F.~Bayart} and \textit{E.~Matheron} [``Hyponormal operators, weighted shifts, and weak forms of supercyclicity'', Proc. Edinb. Math. Soc. (to appear)] have recently proved that some unitary operators are weakly supercyclic in Hilbert space, which is impossible for norm supercyclicity [\textit{H. M. Hilden} and \textit{L. J. Wallen}, Indiana Univ. Math. J. 23, 557--565 (1974; Zbl 0274.47004)]. Indeed, Bayart's and Matheron's result answers the question posed in the paper under review: Is there a weakly supercyclic hyponormal operator? In the paper under review, there are several very nice results. First of all, the author gives a proof that if \(T: \mathcal B\to \mathcal B\) is a bounded linear operator, then the vector \(x\oplus \mu\) in \(\mathcal B\oplus \mathbb C\) is weakly supercyclic for \(T \oplus \alpha I_\mathbb C\), \(\alpha\not=0\), if and only if \(\mu \not=0\) and \(x\) is a weakly hypercyclic vector for \(\alpha^{-1}T\). Thus \(T\oplus \alpha I_{\mathbb C}\) is weakly supercyclic if and only if \(\alpha^{-1}T\) is weakly hypercyclic. The result and its proof are the same as for supercyclic vectors, see Theorem 5.1 in the paper by \textit{M. González, F. León Saavedra} and the reviewer [Proc. Lond. Math. Soc., III. Ser. 81, No. 1, 169--189 (2000; Zbl 1028.47007)]. The weak version of the result has also been stated by \textit{H. N. Salas} and the reviewer [Bull. Lond. Math. Soc. 35, No. 6, 721--737 (2003; Zbl 1054.47009)]. Another interesting result in the paper under review is that if \(T: \ell^p \to \ell^p\), \(1< p < \infty\), is a unilateral weighted backward shift whose weight sequence is strictly increasing, then the set of weakly supercyclic vectors for \(T\) that fail to be norm supercyclic is, in fact, norm dense in \(\ell^p\). It was known that every unilateral weighted backward shift is supercyclic. Thus it can be concluded that there are many weakly supercyclic vectors that are not supercyclic vectors. The author ends with a further nice result that extends the corresponding one for hypercyclic operators due to \textit{C. Kitai} [Invariant Closed Sets for Linear Operators, Thesis, Univ. Toronto (1982)], namely, a hyponormal operator cannot be weakly hypercyclic. \textit{P. S. Bourdon} [Mich. Math. J. 44, 345--353 (1997; Zbl 0896.47020)] proved that no hyponormal operator can be supercyclic. In Bayart and Matheron's article mentioned above it is proved that every weakly supercyclic hyponormal operator must be a multiple of a unitary operator.
scientific article; zbMATH DE number 2081788