Abstract
It is well known that if T is a hypercyclic operator, thenp(T � ) = �. We prove in this paper that this is not true for subspace-hypercyclic oper- ators. We show that if T is subspace-hypercyclic, thenp(T � ) may be empty or not. Moreover we show that for every scalarwith |�| > 1, there exists a subspace-hypercyclic operator T such that ||T|| = |�| andp(T � ) 6 �.
Sign-in/Register to access full text options 
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a callMore From: International Journal of Pure and Apllied Mathematics
Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.
