Simple question about $H^2 \cap H^1_0 (\Omega)$

Question

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

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 $$\normhs {v} 2 \leq C \, \normls {g} 2.$$

Answer 1

I think you might be missing the point that $H^1_0(\Omega)$ is a Banach space.

So if you have $u_k \overset {H^2} \to u$, this implies that $u_k \overset {H^1} \to u$.

Since $H^1_0(\Omega)$ by definition is closed by $\|\cdot\|_{H^1}$ we have that $u \in H^1_0(\Omega)$.

Also $H^2(\Omega)$ is closed by $\|\cdot\|_{H^2}$ so we have $u \in H^2(\Omega)$, too.

Answer 2

If you want to check $u \in H^2 \cap H_0^1(\Omega)$, the density does not yield anything.

However, you have $u_k \in C^2(\Omega)$ with $u_k = 0$ on $\partial\Omega$. Hence, $u \in H^2\cap H_0^1(\Omega)$. Moreover, $H^2 \cap H_0^1(\Omega)$ is a closed subspace of $H^2(\Omega)$.