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Please help me to proof of problem :

Show ‎that ‎the ‎operator ‎‎$‎T:\ell^{‎2‎} \rightarrow ‎\ell^{‎2‎}‎$ ‎‎defined ‎by ‎‎$‎T(\{x_{n}\})=\{2^{-n}x_{n}\}$ ‎is ‎compact‎.

Tanks for your hint.

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2 Answers 2

up vote 2 down vote accepted

$\ell^2$ is Hilbert, thus $T$ is compact iff the image of any weakly convergent sequence under $T$ converges. You can show that $T$ is compact as follows:

Let $x^k\stackrel{\omega}{\rightarrow}x$, since $\ell^2$ is complete, it is enough to show that $\{Tx^k\}_k$ is Cauchy. Notice that $||Tx^k-Tx^l||_{\ell^2}^2=\sum_{n\geq1}2^{-2n}|x_n^k-x_n^l|^2$. We have convergence for every term, since the projection on a term of the sequence is a linear operator and we have weak convergence. For every $\epsilon >0$ we can choose $n_0,\ k_0\in\mathbb{N}$ such that for every $n\leq n_0$ and $k,\ l>k_0$ we have $|x_n^k-x_n^l|<\epsilon$ and $\sum_{n>n_0}2^{-2n}<\frac{\epsilon}{||x^k -x^l||_{\ell^2}^2}$. Then: $$||Tx^k-Tx^l||_{\ell^2}^2=\sum_{n\geq1}2^{-2n}|x_n^k-x_n^l|^2 = \sum_{n=1}^{n_0}2^{-2n}|x_n^k-x_n^l|^2 + \sum_{n>n_0}2^{-2n}|x_n^k-x_n^l|^2<$$ $$<\sum_{n=1}^{n_0}2^{-2n}\epsilon^2 + \sum_{n>n_0}2^{-2n}||x_n^k-x_n^l||_{\ell^2}^2<n_0\epsilon^2 + \epsilon$$ And this concludes the proof.

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Could you please explain why $||Tx^k-Tx^l||_{\ell^2}^2=\sum_{n\geq1}2^{-2n}|x_n^k-x_n^l|^2$ –  El Angel Exterminador Feb 5 '13 at 22:10
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@Panu: By definition of $T$ and of the $\ell^2$ norm. –  Daniel Robert-Nicoud Feb 6 '13 at 6:52

Define $T_k:\ell^2\rightarrow\ell^2$ by $$T_k(\{x_n\})=\Big(\frac{x_1}{2},\frac{x_2}{2^2},...,\frac{x_k}{2^k},0,0,...\Big)$$

What can you say about $T_k$?

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Sine $\ell^{2}$ is Hilbert and $\{Tx_{n}\}$ contains a convergent subsequence of $\{T_{k}x_{n}\}$, consequently T is compact ? –  Ali Qurbani Feb 5 '13 at 19:01
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Think more about the behavior of $T_k$ as operators. Can you show that $T_k$ converges to $T$ in a desirable sense as $k\to\infty$? –  Keaton Feb 5 '13 at 19:03

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