let $g_k,g\in H^1(\Omega)$ (bounded domain) be given, with $g_k\to g$ in $L^2(\Omega)$. Unfortunately, I don't know whether the $g_k$ are uniformly bounded in $H^1$. I want to show that $g_k\rightharpoonup g$ in $H^1(\Omega)$. But this seems to be intricate if not impossible.
However, with the strong $L^2$ convergence I am able to prove that
$$\partial_j g_k \to \partial_j g\text{ in } \mathcal{D}'(\Omega)$$
Can one deduce from that (or with a different approach) that the convergence is weak in $L^2(\Omega)$?
I fear that this is not possible since one needs to put the derivatives on the test function to prove the distributional convergence.
Distributional convergence is very weak indeed, because it is implied by $L^1$ convergence, and passes to derivatives of all orders. For example, since $n^{-1}\sin n^2x\to 0$ in $L^1(0,\pi)$ (and in $L^2(0,\pi)$ as well), it follows that $n\cos nx\to 0$ in the sense of distributions. But the sequence $n\cos nx$ is not bounded in $L^2$, hence cannot converge weakly. Hence, $n^{-1}\sin n^2x$ does not converge weakly in $H^1$.