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{\Large Department of Mathematics, BGU}

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{\Huge Probability and ergodic theory (PET)}\\[0.2\baselineskip]

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\textbf{On} \emph{Tuesday, December 22, 2015}
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\textbf{At} \emph{10:50 -- 12:00}
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\textbf{In} \emph{Math -101}

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{\large\scshape Guy Cohen 
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  (BGU)
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will talk about
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{\Large\bfseries REMARKS ON RATES OF CONVERGENCE OF POWERS OF CONTRACTIONS\par}
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\textsc{Abstract:}
We prove that if the numerical range of a Hilbert space contraction
\$T\$ is in a certain closed convex set of the unit disk which touches the unit
circle only at 1, then \$\textbar{}T\^{}n(I-T)\textbar{} =\textbackslash{}mathcal O(1/n\^{}\{\textbackslash{}beta\})\$ with
\$\textbackslash{}beta \textbackslash{}in {[}\textbackslash{}frac\{1\}\{2\}, 1)\$.
For normal contractions the condition is also necessary.
Another sufficient condition for \$\textbackslash{}beta=\textbackslash{}frac\{1\}\{2\}\$, necessary for \$T\$ normal,
is that the numerical range of \$T\$ be in a disk
\$\{z: \textbar{}z-\textbackslash{}delta\textbar{} \textbackslash{}le 1-\textbackslash{}delta\}\$ for some \$\textbackslash{}delta \textbackslash{}in (0,1)\$.
As a consequence of results of Seifert, we obtain that a power-bounded
\$T\$ on a Hilbert space satisfies \$\textbar{}T\^{}n(I-T)\textbar{} = \textbackslash{}mathcal O(1/n\^{}\{\textbackslash{}beta\})\$
with \$\textbackslash{}beta \textbackslash{}in (0,1{]}\$ if and only if
\$\textbackslash{}sup\_\{1\textless{}\textbar{}\textbackslash{}lambda\textbar{} \textless{}2\} \textbar{}\textbackslash{}lambda -1\textbar{}\^{}\{1/\textbackslash{}beta\} \textbar{}R(\textbackslash{}lambda,T)\textbar{}\textless{} \textbackslash{}infty\$.
When \$T\$ is a contraction  on \$L\_2\$ satisfying the numerical range
condition, it is shown that \$T\^{}nf /n\^{}\{1-\textbackslash{}beta\}\$ converges to 0 a.e.
with a maximal inequality, for every \$f \textbackslash{}in L\_2\$. An example shows
that in general a positive contraction \$T\$ on \$L\_2\$ may have an
\$f \textbackslash{}ge 0\$ with \$\textbackslash{}limsup T\^{}nf/\textbackslash{}log n \textbackslash{}sqrt\{n\} =\textbackslash{}infty\$ a.e.








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