# Does distributional convergence imply weak convergence

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.

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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 $H^1$.