Spectrum of a weakly hypercyclic operator meets the unit circle
Functional Analysis
2007-05-23 v1
Abstract
It is shown that every component of the spectrum of a weakly hypercyclic operator meets the unit circle. The proof is based on the lemma that a sequence of vectors in a Banach space whose norms grow at geometrical rate doesn't have zero in its weak closure.
Cite
@article{arxiv.math/0208193,
title = {Spectrum of a weakly hypercyclic operator meets the unit circle},
author = {S. J. Dilworth and Vladimir G. Troitsky},
journal= {arXiv preprint arXiv:math/0208193},
year = {2007}
}
Comments
3 pages, to appear in Proceedings of the Conference "Trends in Banach Spaces and Operator Theory", Memphis, 2001