Orbits of semigroups of truncated convolution operators
A family \({\mathcal F} = \{ F_a : a \in A \}\) of continuous maps from a topological space \(X\) to a topological space \(Y\) is called \textit{universal} if there exists \(x \in X\) such that the orbit \({\mathcal O}({\mathcal F}, x) =\{ F_a x : a \in A \}\) is dense in \(Y\). The element \(x\) is called an \textit{universal element} for the family \({\mathcal F}\), the set of all such elements is denoted by \({\mathcal U}({\mathcal F})\). This concept is applied in the particular case where \(X = Y\) is a topological vector space, the maps belong to the set \(L(X)\) of continuous linear operators from \(X\) into itself and the family \(\mathcal F\) is a commutative subsemigroup of \(L(X)\), in which case the word \textit{universal} is replaced by \textit{hypercyclic}. Finally, \(\mathcal F\) is \textit{supercyclic} if the semigroup \({\mathcal F}_p = \{zT : T \in {\mathcal F},\;z\text{ a scalar}\}\) is hypercyclic; hypercyclic vectors for \({\mathcal F}_p\) are called \textit{supercyclic vectors} for \({\mathcal F}\). A~finite set \({\mathbf T} = \{ T_1, T_2, \dots, T_n\}\) of commuting operators in \(L(X)\) is hypercyclic (supercyclic) if the semigroup generated by \textbf{T} enjoys the corresponding property.NEWLINENEWLINEIt is known that, if \(1 \leq p < \infty\), the Volterra operator NEWLINE\[NEWLINE Vf(x) = \int_0^x f(t)\,dt NEWLINE\]NEWLINE in \(L^p(0, 1)\) is not supercyclic [\textit{E. A. Gallardo-Gutiérrez} and \textit{A. Montes-Rodríguez}, Integral Equations Oper. Theory 50, No.~2, 211--216 (2004; Zbl 1080.47011)]. The author defines this paper as a ``quest for supercyclic or even hypercyclic operators as close as possible to the Volterra operator'', ``as close as possible'' understood as ``commuting with''. The operators called into play are truncated convolution operators in \(L^p(0, 1)\) (roughly speaking, restrictions to \([0, 1]\) of convolutions with Borel measures) and the main result is that the semigroup generated by finitely many truncated convolution operators is not supercyclic. There are also results going in the opposite direction that involve the notion of \textit{irregular} vector, weaker than that of hypercyclic vector.
scientific article; zbMATH DE number 6031021