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Given a bounded sequence $(\lambda_n)$ in $\mathbb С$ define an operator $S$ in $B(\ell_2)$ by $S(x_1) = 0$ and $S(x_n) = \lambda_n x_{n-1}$ , $n > 1$, for $x = (x_n) \in \ell_2$. Find the polar decomposition of $S$, and characterize those sequences $\{\lambda_n\}$ in $\ell_\infty$ for which $S$ is compact.

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What have you tried? –  Davide Giraudo Nov 16 '12 at 17:35
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Write $\lambda_n=|\lambda_n|e^{i\theta_n}$ and $S=UP$, where $(Px)(n)=|\lambda_n| x(n)$, and $U(x)(n)=e^{i\theta_n}x(n-1)$ if $n\geqslant 1$ and $0$ if $n=0$. Then $P$ is non-negative definite, $U$ is unitary.

Show that $S$ is compact if and only if $\lim_{n\to +\infty}\lambda_n=0$. It's done here.

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is that the only thing we have to do i relly missed some thing –  math Nov 16 '12 at 21:36
    
i mean we it is not necessary to check the compactness –  math Nov 16 '12 at 22:41
    
Why? If I read well, it's asked, isn't it? –  Davide Giraudo Nov 16 '12 at 22:47
    
i mean could u prove the compactness –  math Nov 17 '12 at 1:25
    
Well, I'm still waiting for your attempt. –  Davide Giraudo Nov 17 '12 at 10:35
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