Presentation Abstract

Fame 2026 Conference
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Recurrence of multipliers on Banach algebras

Mr Enrique Jordá (Universitat Politècnica de València , Spain) ORCID
Session: Dynamics of linear operators Talk
Abstract

{\bf Definition} [Costakis, Manoussos, Parissis]
Let $X$ be a Banach space and $T\in \mathcal{L}(X)$. We say that $T$ is
\begin{enumerate}
\item \emph{recurrent} if for every open subset $U\subseteq X$ there exists $k\in \mathbb{N}$ such that $\mbox{$U\cap T^{-k}(U)\neq \emptyset$}.$
\item \emph{rigid} if there exists a sequence of positive integers $(k_n)_n$ such that $\mbox{$\lim_n \norm{ T^{k_n}x-x}=0$}$ for every $x\in X$.
\item \emph{uniformly rigid} if there exists a sequence of positive integers $(k_n)_n$ such that $\lim_n \norm{T^{k_n}-I}=0$.
\end{enumerate}



We report on ongoing work with M.J. Beltr\'an and J. Galindo where we study the recurrence of multipliers on some Banach algebras. Our main result can read as follows.

{\bf Theorem} Let $G$ be a locally compact amenable group and let $\phi\in B(G)$ be power bounded. We consider the multipliers $M_\phi:A(G)\to A(G)$, $M_\phi^0:C_0(G)\to C_0(G)$ and $M_\phi^\infty: C_b(G)\to C_b(G)$.
\begin{itemize}
\item[(a)] $M_\phi$ is recurrent if and only if $M_\phi^0$ is recurrent if and only ifor any compact subset $K\subseteq G$ there exists an increasing sequence $(n_k)\subset \N$ such that $\lim_k \sup_{z\in K}|\phi^{n_k}(z)-1|=0$.
\item[(b)] $M_\phi$ is rigid if and only if $M_\phi^0$ is rigid if and only if there is an increasing sequence $(n_k)\subset \N$ such that $\lim_k \sup_{z\in K}|\phi^{n_k}(z)-1|=0$ for any compact subset $K\subseteq G$.
\item[(c)] $M_\phi$ is uniformly rigid if and only if $M_\phi^{\infty}$ is uniformly rigid if and only if $M_\phi^{\infty}$ is recurrent if and only if there is an increasing sequence $(n_k)\subset \N$ such that $\lim_k \sup_{z\in G}|\phi^{n_k}(z)-1|=0$.

\end{itemize}


If $G$ is locally connected we can get rid of power boundedness in (a) above. However, there is $\phi\in B(\mathbb{Z})$ such that $M_\phi^0$ is rigid but $M_\phi$ is not.

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