I work on a bounded domain. Suppose $u_n \to u$ in $H^1$ (weakly, strongly if necessary) and suppose that $\lVert -\Delta u_n \rVert_{L^\infty}$ is bounded uniformly in $n$.
So we know that (for a subsequence at least) $$-\Delta u_n \rightharpoonup^* v \quad \text{in $L^\infty$}$$ for some $v$. Is it possible to identify $v=-\Delta u \in L^2$?
I believe it is, since the weak star convergence implies that $$\int (-\Delta u_n)\varphi \to \int v\varphi$$ for all $\varphi$ in $L^2$. But if $\varphi \in H^1$, then the LHS equals $\int \nabla u_n \nabla \varphi \to \int \nabla u \nabla \varphi$ and by definition of weak derivative we get the result. am I right about it?
The Laplacian is a bounded operator from $H^1$ to $H^{-1}$. So, the convergence (weak or strong) $u_n\to u$ in $H^1$ implies the convergence (weak or strong) $\Delta u_n\to\Delta u$ in $H^{-1}$.
On the other hand, you also have $\Delta u_{n_k}\to v\in L^\infty$ (weakly).
Since a sequence cannot converge to two different distributions (i.e., the distributional limit is unique), it follows that $\Delta u = v$. So, $\Delta u\in L^\infty \subset L^2$.
Note that the final conclusion $\Delta u\in L^2 \implies h\in H^2$ requires some smoothness assumptions on the boundary of $\Omega$; being a bounded domain is not enough.