Prove that a given operator is compact Let $H$ be an Hilbert space with scalar product $(\cdot,\cdot)$ and $T : H \to H$ a linear operator defined as follow:
$$
Tx 
=
\sum _{n=1} ^\infty
(x,a_n)b_n
$$
where $(a_n)_{n\in\mathbb{N}},(b_n)_{n\in\mathbb{N}}$ are two sequences in $H$ such that $\sum_{n=1}^\infty|a_n||b_n|<+\infty $.
I have to prove that $T$ is compact, i.e. if $x_n \to x_0$ weakly for $n\to+\infty$, then $\|Tx_0-Tx_n\|\to 0$ for $n\to+\infty$.
My idea was to use somehow Banach-Steinhaus theorem to prove that $(x_0-x_n,a_k)\xrightarrow[n\to +\infty]{}0$ uniformly in $k$ and then deduce the strong convergence. However, the sequence $(b_n)_{n\in\mathbb{N}}$ can be unbounded, so I fear that this approach do not work.
Does anyone know how to solve this exercise?

$\newcommand{scal}[2]{\left({#1};{#2}\right)}\newcommand{abs}[1]{\left\lvert {#1}\right\rvert}\newcommand{nrm}[1]{\left\lVert {#1}\right\rVert}$Call $T_nx=\sum_{k=1}^n \scal x{a_k}b_k$ and consider any $x$ such that $\abs x\le 1$. Then $$\abs{Tx-T_nx}=\abs{\sum_{k=n+1}^\infty \scal x{a_k}b_k}\le \sum_{k=n+1}^\infty \abs{\scal x{a_k}}\abs{b_k}\le \abs x \sum^\infty_{k=n+1}\abs{a_k}\abs{b_k}\le \sum^\infty_{k=n+1}\abs{a_k}\abs{b_k}$$
Therefore, $$\nrm{T-T_n}\le \sum_{k=n+1}^\infty \abs{a_k}\abs{b_k}\stackrel{n\to \infty}\longrightarrow 0$$
$T$ is the strong limit of a sequence of finite-rank operators, hence compact.
