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Let $X,Y$ be Banach spaces. $T\colon X\to Y$ be a bounded linear operator.

How can I prove that $T$ is compact if and only if there is $\lbrace x_n^*\rbrace\subset X^*$ such that $\|x_n^*\|\to 0$ and $\|T(x)\|\leq \operatorname{sup}_n|x_n^*(x)|$ for every $x\in X$?

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Homework? What have you done so far? Which part are you stuck on? –  Robert Israel Jun 19 '11 at 15:54
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So, if it's the "only if" portion of the proposition that's giving you trouble, I take it you're struggling with showing: "if T is compact, then there is...for every $x \in X$"? Have you managed to confirm the converse? Many users are more motivated to help/give hints/answers when you clarify what exactly you're stuck on, and show some of your efforts...so perhaps you can elaborate (edit your question, if possible, or reply to some comments?) –  amWhy Jun 19 '11 at 16:24
    
Are you assuming $X$ is separable? –  André Caldas Jan 25 '13 at 0:22

1 Answer 1

See theorem 1 in A characterization of compact linear mappings. T. Terzioglu

Click there "look inside" to see the proof.

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