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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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Excuse for my question, but what is a WOT-limit? Maybe you could include the definition of this in the question. – matgaio May 9 '12 at 16:47
@matgaio I guess weak operator topology. – Davide Giraudo May 9 '12 at 18:22
up vote 3 down vote accepted

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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Why is $S$ a wot limit point of the powers of $T$? – user16299 May 9 '12 at 19:21
@YemonChoi The odd powers of $T$ equal to $T$, so the powers contain a constant sub-sequence equal to $T=S$. – Theo May 9 '12 at 19:28
sorry, I had wot limit and wot limit point scrambled in my head, shouldn't read maths while decaffeinated – user16299 May 9 '12 at 20:06

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