Let $\Omega$ be a Lipschitz domain in $\Bbb R^n$, is the space $$\mathcal{H}=\{v\in H^1(\Omega):\Delta v\in H^1(\Omega)^*\}$$ a Hilbert space when endowed with the norm $\|\cdot\|_\mathcal{H} = \|u\|_{H^1(\Omega)}+\|\Delta u\|_{H^1(\Omega)^*}$? To be a Banach space it just needs to be a closed subspace of $H^1(\Omega)$. Clearly the space is closed under addition, and if $u_k$ is cauchy in $\mathcal{H}$ then $\Delta u_k$ is cauchy in $H^1(\Omega)^*$ which is a Banach space, and so there is a $u\in H^1(\Omega)$ and a $w\in H^1(\Omega)^*$ such that $$u_k\to u\quad\mbox{in}\quad H^1(\Omega),$$ $$\Delta u_k\to w\quad\mbox{in}\quad H^1(\Omega)^*,$$ is this enough to deduce that $w=\Delta u$?
We can see that for $v\in C^\infty_c(\Omega)$
$\begin{align}\langle\Delta u|v\rangle&=\int_\Omega u\Delta v\\ &=\lim_{k\to\infty}\int_\Omega u_k\Delta v\\ &=\lim_{k\to\infty}\langle\Delta u_k|v\rangle\\ &=\langle w|v\rangle\end{align}$
which I believe implies that $w=\Delta u$ in the $H^1(\Omega)^*$ sense by the definition of the weak derivative.