Let $H$ be a Hilbert space, and $K: H \to H$ a compact opeartor. If $(u_n)$ is a bounded sequence in $H$, then $(Ku_n)$ has a convergent subsequence by definition of compact operator.
Meanwhile, by Banach-Alaoglu we see that $(u_n)$ has a subsequence which converges weakly, say to $u \in H$. Then obviouly $(Ku_n)$ has a subsequence converges strongly to $(Ku)$.
My question is: Is it possible to find a $u \in H$ that, no subsequences of $(u_n)$ converge weakly to $u$, but there exists a subsequence of $(Ku_n)$ converges strongly to $Ku$?
I came across this question when I read some proofs related to compact operators which rely on the existence of weakly convergent subsequences of $(u_n)$ to find strongly convergent subsequences of $(Ku_n)$. I don't know whether this approach is essential.
Edit: As geetha290krm sayed in the comment, you can consider the case where $K$ is zero operator. This question is solved.