Let $X$ be a separable reflexive Banach space and let $T$ be a power-bounded operator on $X$ ($\sup_n \|T^n\|<\infty$.) Let $S$ be a WOT-limit point of $(T^n)$. Suppose for some $n$ we have $T^n=S$. Does it follow from this that $T^n$ is an idempotent?
Power bounded operators
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functional-analysis
banach-spaces
operator-theory
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0@matgaio I guess weak operator topology. – 2012-05-09
1 Answers
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Here's a counterexample, if I understand the question correctly.
- $X=\mathbb C^2$
- $T(x,y)=(y,x)$
- $S=T$
- $n=1$.
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0sorry, I had wot limit and wot limit point scrambled in my head, shouldn't read maths while decaffeinated – 2012-05-09