Does $u'_k \to v$ weakly imply $u_k \to u$ weakly with $u' = v$?

Question

Let's assume $u'_k \to v$ weakly where $u_k \in L^2(\Omega)$ and $\Omega \subseteq \mathbb{R}^n$. Here $'$ denotes the weak derivative. Then, for an arbitrary $\phi \in C^\infty_c(\Omega)$, $\langle u_k, \phi' \rangle = - \langle u'_k, \phi \rangle \to - \langle v, \phi \rangle$. If we can deduce that $u_k$ converges weakly, then we obtain that $u_k' \to v$ weakly implies $u_k \to u$ weakly with $u' = v$. However, I'm not sure that we can ensure the weak convergence of $u_k$. Is it actually true? I tried to construct a counter example but failed.

Answer 1

The example suggested by PhoemueX, namely, $u_k \equiv (-1)^k$, shows that we may not have the weak convergence of $\left(u_k\right)_{k\geqslant 1}$ (but a subsequence converges to a $u$ whose derivative is $0$).

It is also possible that no subsequence of $\left( u_k\right)_{k\geqslant 1}$ converges weakly, for example if $u_k(x)=k$ for all $x\in\Omega$.