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I'm having a hard time finding a solution for the following problem:

Prove that the operator $ T \in \mathcal{L}(\ell_2) $ defined with the formula $$ T((x_1, x_2, \dots, )) = (0, x_1, x_2/2, x_3/3, \dots) $$

is compact.

That's the trouble I'm having with functional analysis - how do I ever know how to approach a problem like this? I'm supposed to show that $T$ turns bounded sets into relatively compact sets (it's enough that it does so for a unit ball). However I can't seem to find a way to solve that

Jytug
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  • You can use ideas from this question http://math.stackexchange.com/questions/320751/trying-to-prove-that-operator-is-compact. – Cortizol Jan 15 '15 at 20:52

2 Answers2

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Let $T_n(x) = \sum_{k=1}^n {1 \over k} x_k e_{k+1}$. $T_n$ has finite rank, so it is compact.

We have $\|T-T_n\| \le {1 \over n+1}$, hence $T_n \to T$ and so $T$ is compact.

copper.hat
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  • Since I was not sure about your last statement, let me refer to one of your different answers: http://math.stackexchange.com/questions/416046/how-to-show-that-the-norm-limit-of-compact-operators-is-compact – Quickbeam2k1 Jan 15 '15 at 21:25
  • @Quickbeam2k1: Thanks! – copper.hat Jan 15 '15 at 21:28
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Hint: You might show the image of the unit ball is totally bounded. Given $\epsilon > 0$, take $N$ large enough, and cover the projection of the image of the unit ball onto the first $N$ coordinates by small balls ...

Robert Israel
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