Classical operators on weighted Banach spaces of entire functions (Q2855907)

Denote by \({\mathcal H}({\mathbb C})\) the linear space of all entire functions. Let \(v:[0,+\infty) \to [0,+\infty)\) be a non-increasing continuous function which satisfies \(\lim_{r\to \infty} r^m v(r)=0\) for each positive integer \(m\). Then NEWLINE\[NEWLINE H_{v}^{\infty}({\mathbb C})=\{ f\in {\mathcal H}({\mathbb C}): \| f\|_v:=\sup_{z\in {\mathbb C}}v(|z|)|f(z)|<+\infty\}NEWLINE\]NEWLINE is a weighted Banach space of entire functions and NEWLINE\[NEWLINE H_{v}^{0}({\mathbb C})=\{ f\in {\mathcal H}({\mathbb C}):\lim_{|z|\to \infty}v(|z|)|f(z)|=0\}NEWLINE\]NEWLINE is a closed subspace of it. In the paper under review, the operators NEWLINE\[NEWLINE Df(z)=f^\prime(z), \qquad Jf(z)=\int_{0}^{z}f(\zeta)\,d\zeta,\qquad \text{and}\qquad Hf(z)=\tfrac{1}{z}\int_{0}^{z}f(\zeta)\,d\zeta NEWLINE\]NEWLINE are studied on the weighted Banach space of entire functions with weight \(v(r)=e^{-\alpha r}\) (\(\alpha >0\)). The following results are obtained.NEWLINENEWLINE(1) The differentiation operator \(D\) satisfies \(\| D^n\|=n!(\tfrac{e\alpha}{n})^n\) for each positive integer \(n\), which means that \(D\) is power bounded if and only if \(\alpha<1\). The spectrum of \(D\) is the closed disc of radius \(\alpha\). The operator \(D\) is uniformly mean ergodic on \(H_{v}^{\infty}({\mathbb C})\) and \(H_{v}^{0}({\mathbb C})\) if \(\alpha<1\) and not mean ergodic if \(\alpha>1\); for \(\alpha=1\), it is not mean ergodic on \(H_{v}^{\infty}({\mathbb C})\) and not uniformly mean ergodic on \(H_{v}^{0}({\mathbb C})\).NEWLINENEWLINE(2) The integration operator \(J\) is not hypercyclic on \(H_{v}^{0}({\mathbb C})\). For each positive integer \(n\), one has \(\| J^n\|=1/\alpha^n\), which means that it is power bounded if and only if \(\alpha \geq 1\). The spectrum of \(J\) is the closed disc of radius \(1/\alpha\). If \(\alpha>1\), then \(J\) is uniformly mean ergodic on \(H_{v}^{\infty}({\mathbb C})\) and \(H_{v}^{0}({\mathbb C})\) and it is not mean ergodic on these spaces if \(\alpha<1\). When \(\alpha=1\), then \(J\) is not mean ergodic on \(H_{v}^{\infty}({\mathbb C})\) and mean ergodic but not uniformly mean ergodic on \(H_{v}^{0}({\mathbb C})\).NEWLINENEWLINE(3) The Hardy operator \(H\) is compact and has norm \(1\). Its spectrum is the set \(\{ 0, 1, 1/2, 1/3,\dots\}\). It is power bounded and uniformly mean ergodic for any \(\alpha>0\) and is therefore not hypercyclic on \(H_{v}^{0}({\mathbb C})\).