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Let $\{e_n\}$ be an orthonormal basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$ on $l^2$.

Find the condition for $A$ to be compact.

My work:

I want to show that in order for $A$ to be compact, we must have $\alpha_n \rightarrow 0$.

Suppose the contrary.

Then there exists some $\epsilon >0 $ such that we can find some sequence $(\alpha_{n_k})_k$ such that $(\alpha_{n_k})_k> \epsilon$.

I want to show that in this case, we can't find a convergent subsequence for $(Ae_{n_k})=(\alpha_{n_k}e_n)_k$.

And I got stuck here. Not sure how to show that there doesn't exist a convergent subsequence.

I feel like I'm missing some facts in real analysis.

Any help is appreciated! Thank you.

Toasted_Brain
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1 Answers1

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If $\alpha_n$ does not tend to $0$ the there exist $\epsilon >0$ and $n_1 <n_2<...$ such that $|\alpha_{n_k}| >\epsilon $ for all $k$. The sequence $(\frac 1 {\alpha_{n_k}} e_{n_k})$ is bounded but $A(\frac 1 {\alpha_{n_k}} e_{n_k})= e_{n_k}$ has no convergent subsequence since $\|e_{n_k}-e_{n_j}\|^{2}=2$ whenever $j \neq k$.

Converse part: Suppose $\alpha_n \to 0$. Define $A_n(x) =\sum\limits_{k=1}^{n}\alpha_kx_ke_k$ where $x=\sum x_ke_k$. The $A_n$ is compact because it has finite rank. Now, $\|(A-A_n)x\|^{2}=\sum\limits_{k=n+1}^{\infty}|\alpha_kx_k|^{2} \leq \sup_{k>n} |\alpha_k|^{2} \|x||^{2}$ proving that $\|A-A_n\| \to 0$. This implies that $A$ is compact.

Kavi Rama Murthy
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  • May I ask why $|(A-A_n)x|^{2}\leq \sup_{k>n} |\alpha_k|^{2} |x||^{2}$ implies$|A-A_n| \to 0$? Thank you! – Toasted_Brain Mar 01 '21 at 06:41
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    @csiqrucalree By definition of norm of an operator we get $|A-A_n|^{2}\leq \sup_{k>n} |\alpha_k|^{2}$. And $\alpha_n \to 0$ implies $\sup_{k>n} |\alpha_k|^{2}$ also tends to $0$. – Kavi Rama Murthy Mar 01 '21 at 06:43