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i have a bounded sequence $(\lambda_n)_{n=1}^{\infty}$ in $\mathbb{R}$ and $T \in \mathrm{L}(\ell^{p})$ with $1\leq p \leq \infty$ such that $Tx=(\lambda_{1}x_{1},\lambda_{2}x_{2},\dots)$ , where $x=(x_{1},x_{2},\dots)$ my question is if $\lambda_{n} \to 0$ then $T$ is a compact operator. A classmate gave me a tip to use Cantor diagonal argument. But i don't know how to do that. Any help?

$\mathrm{L}(\ell^{p})$ is the set of continuous linear operators in $\ell^{p}$.

Thanks

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    Let $T_n x = \sum_{K=1}^n \lambda_k x_k$ which is finite rank and hence compact. Then show that $T_n \to T$. – 2017-02-17
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    Sorry, i made a typo. The post has been edited. – 2017-02-17
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    My comment is still valid. – 2017-02-17
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    Nice. Thanks @cooper. – 2017-02-17
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    You could give an idea to show the reciprocal of this result.? – 2017-02-17
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    What do you mean, the converse? – 2017-02-17
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    If $T$ is compact then $\lambda_{n} \to 0$. – 2017-02-17
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    Show that if $\lambda_k \not\to 0$ then $T$ is not compact. – 2017-02-17
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    Ok @copper.hat. Thanks – 2017-02-17
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    Essentially with the above comment you have a subspace of the domain on which $T$ behaves 'like' the identity, or at least not a compact operator. – 2017-02-17

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