Hypercyclic operators are subspace hypercyclic
A bounded operator \(T\) on a separable Banach space \(X\) is called subspace hypercyclic for a subspace \(M\) of \(X\) if there is a vector \(x \in X\) such that the intersection of its orbit and \(M\) is dense in \(M\). The aim of this paper is to solve a question of \textit{B. F. Madore} and \textit{R. A. Martínez-Avendaño} [J. Math. Anal. Appl. 373, No. 2, 502--511 (2011; Zbl 1210.47023)] and prove that if \(T\) is a hypercyclic operator on \(X\), then there is a non-trivial closed subspace \(M\) of \(X\) such that \(T\) is hypercyclic for \(M\). To do this, they show that, if \(A\) is a dense subset of a Banach space \(X\), then there is a non-trivial closed subspace \(M\) such that \(A \cap M\) is dense in \(M\).
scientific article; zbMATH DE number 6522745