Suppose $(a_{ij})_{i,j\in \mathbb N}$ satisfy $\sum_{i,j}|a_{ij}|^2<\infty$ and define $A:\ell ^2 \rightarrow \ell ^2$ by $(Ax)_i)=\sum _j a_{ij}x_j$. I need to prove $A$ is compact.
Unfortunately, I really have not idea what to do here. I don't see how playing around with the $\ell^2$ norm will be of any use. On the other hand, using $\|\cdot\|_2\leq \|\cdot\|_1$ only gives me $\|Ax_n-Ax_m\|_2\leq \sum _{ij}|a_{ij}||x_j|$ which doesn't seem to help either. I'm pretty sure I'm not supposed to use this equality anyway, but I don't even have intuition for why the operator should be compact. (My strategy was to take a bounded sequence $x_k$ from the unit ball and show its image has a convergent subsequence.)
You have to understand what $A$ looks like. It acts as an infinite matrix $(a_{ij})_{i,j=1}^{\infty}$ on your elements of $\ell^2$. Since $\ell^2$ is a Hilbert space you know that $A$ is compact if it is a limit of finite-rank operators. You can show that $A$ is the limit of $P_nAP_n$ where $P_n$ is the projection on the span of the first $n$ standard basis vectors.