Strong convergence due to Compact Operator

Question

Given a sequence $u_{n}$ such that: $u_{n} \rightharpoonup 0$ in $L^{2}(\mathbb R^{n})$ & $A$ is a compact operator. The problem is to show that : $Au_{n} \rightarrow 0$ in $L^{2}(\mathbb R^{n})$ .

What I am thinking is: since {$u_{n}$} is weakly convergent so it is bounded.

Since: $A$ is compact, & {$u_{n}$} is bounded $\implies$ the sequence {$Au_{n}$} has a convergent subsequence; say: {$Au_{n_{k}}$} . Now, does that help in someway??? Please help me to solve the problem.

Answer 1

Hint. Your idea is fine (and you are almost done). Just make use of the following fact:

If $(x_n)$ is a sequence in some metric space $X$ and there is an $x \in X$ such that for every subsequence $(x_{n_k})$, some sub-sub-sequence $(x_{n_{k_j}})$ converges to $x$, then $x_n \to x$ in $X$.

Now apply this to $x_n = Au_n$, $x= 0$ and use your idea to find the converging sub-sub-sequences.