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I'd like to show that a compact linear operator $T: X \to Y$ between normed spaces is bounded. Can you tell me if this is right?

If $T$ is compact, then the closure of the image of $B(0,1)$ and hence of $B(0,n)$ is compact. Then $\|T\| = \sup_{\|x\| = 1} \|Tx\| \leq \sup_{x \in B(0,2)} \|Tx\| \leq K$ for some $K \in \mathbb R_{\geq 0}$ since $TB(0,2) \subset \overline{TB(0,2)}$ and $\overline{TB(0,2)}$ is compact hence bounded.

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    @J.Loreaux Thank you for your comments. I was too embarrassed about the mistake I made there and had to delete my previous question.2012-08-17

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