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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?

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    @matgaio I guess weak operator topology.2012-05-09

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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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    sorry, I had wot limit and wot limit point scrambled in my head, shouldn't read maths while decaffeinated2012-05-09