$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\normls}[2]{\norm{#1}\_{L^{#2}}}$
$\newcommand{\normhs}[2]{\norm{#1}\_{H^{#2}}}$

I want to proof the following result.

> 
> **Lemma.**
>  Let $\Omega \subset \subset \mathbb R^n$ of class $C^1, \, A = -\Delta : D\_A \to L^2(\Omega)$ with $$D\_A = \big\{ u \in C^2(\bar \Omega) \, : \, u = 0 \text{ on } \partial \Omega \big\} \subset L^2 (\Omega).$$ Then $D\_{\bar A} \subset H^2 \cap H^1\_0 (\Omega)$, with $$D\_{\bar A} = \left \{ u \in L^2; \, \exists \, (u\_k)\_k \subset D\_A \, : \, u\_k \to u, - \Delta u\_k =: f\_k \to f \text{ in } L^2(\Omega) \right\}$$
> 
> 
> 

With the assumption below I can easily check

$$\normhs {u\_k - u\_l} 2 \leq C \, \normls {f\_k - f\_l} 2 \to 0 \, \, (k,l \to \infty)$$

**I'm trying to see how the limit $u$ is still in $H^2 \cap H^1\_0 (\Omega)$?**

Can I just say since $C^\infty\_c(\Omega) \subset H^1\_0(\Omega)$ is dense, there exists $(v^k\_l)\_k \in C^\infty\_c(\Omega) \, :\, v^k\_l \overset{H^1}\to u\_l \, \,(k \to \infty)$ and hence $v^l\_l \overset{H^1} \to u \, \,(l \to \infty)$?

Or is there an easier way to see this and my attempt is just overkill (or wrong)?

$\,$

> 
> **Assumption.**
>  If $v$ weakly solves $-\Delta v = g$ in $\Omega$, $v = 0$ on $\partial \Omega$, then we have
>  $$\normhs {v} 2 \leq C \, \normls {g} 2.$$
> 
> 
>