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Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?

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Let $T_n$ such that $T_n(e_k)=\begin{cases}\lambda_ke_k&\mbox{ if }k\leq n\\\ 0&\mbox{ if }k>n \end{cases}$. Then $T_n$ is finite ranked hence compact and for $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq n+1}|\lambda_k|^2\cdot|v_k|^2\\\leq \sup_{k\geq n+1}|\lambda_k|^2\lVert v\rVert_{\ell^2}^2 $$ so $\lVert T-T_n\rVert\leq \sup_{k\geq n+1}|\lambda_k|$ and we conclude that $T_n\to T$ in norm. A norm-limit of compact operators is compact so $T$ is compact.

Conversely, if $T$ is compact, then you can extract from $\{Te_n\}$ a converging subsequence so you can extract from $\{\lambda_ne_n\}$ a converging subsequence. Since $\lVert \lambda_{n+1}e_{n+1}-\lambda_ne_n\rVert^2=|\lambda_{n+1}|^2+|\lambda_n|^2\to 0$, we should have $\lambda_n\to 0$.

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