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It is well known that the power of a weakly compact operator is compact. Is the spectrum of a weakly compact operator is the same as a compact operator?

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we know for a compact operator $K$, every nonzero $ λ \in \sigma(K)$ is an eigenvalue of $K$. Can I have the same property for a weakly compact operator? –  salma Jan 10 '13 at 19:27
    
I think he is asking if the spectrum of a weakly compact operator behave like the spectrum of a compact operator. –  Tomás Jan 10 '13 at 19:27
    
yes, I mean that Thomas says –  salma Jan 10 '13 at 19:52
    
Looking at the definition online, it looks like an operator is "weakly compact" when it maps the unit ball into a pre-weakly compact set. Now, if you take operators on a Hilbert space, then the weak and weak-$*$ topologies agree, so that the unit ball is weakly compact. This would imply that every bounded operator is weakly compact. Am I missing something? –  Martin Argerami Jan 13 '13 at 2:01
    
@MartinArgerami you should post this as answer –  no identity Nov 21 '13 at 17:18

1 Answer 1

If you consider operators on a Hilbert space, then the weak and weak-∗ topologies agree, so that the unit ball is weakly compact. This implies that every bounded operator is weakly compact. In particular, any compact subset of $\mathbb C$ can be the spectrum of a weakly compact operator.

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