Sums of hypercyclic operators.
A (bounded) operator \(T\) on a complex infinite-dimensional separable Banach space \(X\) is said to be hypercyclic if there is a (hypercyclic) vector \(x \in X\) such that its orbit \(O(T,x):=\{x,Tx,T^2x,\dots\}\) is dense in \(X\). The operator \(T\) is called chaotic if it is hypercyclic and the set of periodic points of \(T\) is dense in \(X\). Motivated by the fact that the existence of a single hypercyclic vector implies the existence of a dense set of such vectors, one might ask if a similar statement holds for operators. However, if we endow the space \(L(X)\) of all operators on \(X\) with the operator norm, there are no hypercyclic operators in the closed unit ball of \(L(X)\). On the other hand, \textit{J. Bès} and \textit{K. C. Chan} [Houston J. Math. 29, 195--206 (2003; Zbl 1037.47005)], extending previous work by K. C. Chan for a Hilbert space \(X\), showed that the set of hypercyclic vectors is dense in the space \(L(X)\) endowed with the strong operator topology. More recently, J. Bès and K. C. Chan, and M. de León independently, proved that the set of sums of hypercyclic operators is dense in \(L(H)\) for the strong operator topology, if \(H\) is an infinite-dimensional separable Hilbert space. These authors asked whether every operator on \(X\) could be written as the sum of two hypercyclic (or even two chaotic) operators. In the interesting article under review, the author answers this question in the following way. Theorem 1: Every operator on an infinite-dimensional separable Hilbert space is the sum of two hypercyclic operators. In fact, she shows in Theorem 2 that it can even be written as the sum of two chaotic operators. Using the work of Gowers and Maurey about hereditarily indecomposable Banach spaces, she proves that there is a separable infinite-dimensional Banach space on which not every operator is the sum of two hypercyclic operators.
hereditarily indecomposable Banach spaces
scientific article; zbMATH DE number 1974808
scientific article