Institute of Mathematics

Abstract:  We are concerned here with relating the spectral properties of a bounded linear operator $T$ on a Banach space to the behaviour of the means $(1/{s(n)})\sum_{k=0}^n(\Delta s)(n-k)T^k$, where $s$ is a nondecreasing sequence of positive real numbers, and $\Delta$ denotes the inverse of the automorphism on the vector space of scalar sequences which maps each sequence into the sequence of its partial sums. In a previous paper, we obtained a uniform ergodic theorem for the means above, under the hypotheses $\lim_{n\rightarrow\infty}s(n)=\infty$, $\lim_{n\rightarrow\infty}{s(n+1)}/{s(n)}=1$, and $\Delta^qs\in\ell_1$ for a positive integer $q$: indeed, we proved that if $T^n/s(n)$ converges to zero in the uniform operator topology for such a sequence $s$, then the averages above converge in the same topology if and only if 1 is either in the resolvent set of $T$, or a simple pole of the resolvent function of $T$. In this paper, we prove that if $\liminf_{n\rightarrow\infty}{s(n+1)}/{s(n)}=1$, and the averages above converge in the uniform operator topology, then $1$ is either in the resolvent set of $T$, or a simple pole of the resolvent function of $T$. The converse is not true, even if the sequence $s$ satisfies all the hypotheses of the theorem recalled above, except membership of $\Delta^qs$ in $\ell_1$ for a positive integer $q$. We also prove that if $\dsmash{\lim_{n\rightarrow\infty}}\root nøf{s(n)}=1$, and the function $h_s(z)=\sum_{n=0}^{\infty}s(n)z^n$ has no zeros in the open unit disk, then operator norm boundedness of the averages of the sequence $T^n$induced by $s$ implies that the spectral radius of $T$ is less than or equal to $1$. This result fails if the assumption about $h_s$ is dropped. Indeed, it may happen that the averages converge in the uniform operator topology for a sequence $s$ satisfying $\lim_ {n\rightarrow\infty}s(n)=\infty$, $\lim_{n\rightarrow\infty} {s(n+1)}/{s(n)}=1$, and $\Delta^qs\in l_1$ for a positive integer $q$, and nevertheless the spectral radius of $T$ is strictly larger than $1$.