Why is $S\colon\ell^{2}\to\ell^{2}$ defined by $(Sx)_{m}:=\sum_{n\in\mathbb{N}}a_{m,n}x_{n}$ a compact operator?

Question

Suppose that $a_{m,n}\in\mathbb{C}$ are chosen such that $$\sum_{m\in\mathbb{N}}\sum_{n\in\mathbb{N}}|a_{m,n}|^{2}<\infty.$$ How do I prove that $S\colon\ell^{2}\to\ell^{2}$ defined by $$(Sx)_{m}:=\sum_{n\in\mathbb{N}}a_{m,n}x_{n}$$ is a compact operator?

I tried to prove that, if $(x^{i})_{i\in\mathbb{N}}$ is a bounded sequence in $\ell^{2}$, then $(Sx^{i})_{i\in\mathbb{N}}$ has a convergent subsequence. I observed that $$(Sx)_{m}=\langle(a_{m,n})_{n\in\mathbb{N}},(\bar{x}_{n})_{n\in\mathbb{N}}\rangle,$$ so I was hoping that I could use the Cauchy schwarz inequality.

Any suggestions are greatly appreciated.

Answer 1

Define operators $S_k$, $k\in\mathbb N$, by $$ (S_kx)_m=\begin{cases} (Sx)_m,&\ m\leq k\\ \ \\0,&\ \text{otherwise} \end{cases} $$ Then $S_k$ are finite-rank. For $x$ with $\|x\|_2\leq1$, we get via Cauchy-Schwarz \begin{align} \|(S-S_k)x\|_2^2 &=\sum_{m>k}|(Sx)_m|^2=\sum_{m>k}\left|\sum_na_{mn}x_n\right|^2\\ \ \\ &\leq \sum_{m>k} \sum_n|a_{mn}|^2. \end{align} So, by choosing $m$ big enough, you can get $\|S-S_k\|_2$ as small as you want.

Answer 2

Just some further explanation of @MartinArgerami good answer:

The fact that $S_{k}$ being finite rank comes from the observation that \begin{align*} S_{k}x&=\left(\left<(a_{1n}),(x_{n})\right>,...,\left<(a_{kn}),(x_{n})\right>,0,0,...\right)\\ &=\left<(a_{1n}),(x_{n})\right>e_{1}+\cdots+\left<(a_{kn}),(x_{n})\right>e_{k}, \end{align*} where $e_{i}$ are the standard $i$th coordinate sequence.

Furthermore, the double sum $\displaystyle\sum_{m,n}|a_{mn}|^{2}$ being convergent will surely imply that $\displaystyle\sum_{m,n>k}|a_{mn}|^{2}\rightarrow 0$ as $k\rightarrow\infty$, but we have further stronger property:

By letting $b_{m}=\displaystyle\sum_{n\geq 1}|a_{mn}|^{2}$, the double sum is no more than saying that $\displaystyle\sum_{m\geq 1}b_{m}<\infty$, so $\displaystyle\sum_{m>k}b_{m}\rightarrow 0$ as $k\rightarrow\infty$, that is, $\displaystyle\sum_{m>k}\sum_{n\geq 1}|a_{mn}|^{2}\rightarrow 0$ as $k\rightarrow\infty$.