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

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.

• What if $u_k \equiv (-1)^k$? – PhoemueX Nov 9 '18 at 15:08
• @PhoemueX Thanks! It was so elementary. – lionce Nov 9 '18 at 15:19

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 $$x\in\Omega$$.