# Do the limits coincide? ($L^2$ and $H^2$)


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 $D_{\bar A} \supset H^2 \cap H^1_0 (\Omega)$, with $$D_{\bar A} = \left \{ u \in L^2; \, \exists \, (u_k)_k \subset D_A \, : \, u_k \to u, - \Delta u_k =: f_k \to f \text{ in } L^2(\Omega) \right\}$$

I want to show that an arbitrary $u \in H^2 \cap H^1_0(\Omega)$ is also in $D_{\bar A}$

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• choose $(f_k)_k \subset C^\infty_c (\Omega)$ with $$f_k \overset{L^2}\to f = -\Delta u \in L^2(\Omega)$$

• choose $u_k \in H^1_0(\Omega)$, such that $u_k$ weakly solves $-\Delta u_k = f_k$, (existence and uniqueness of $u_k$ is guaranteed by Riesz representation thm)

• with the assumption below there holds $$\normhs {u_k - u_l} 2 \leq C \, \normls {f_k - f_l} 2 \to 0 \, \, (k,l \to \infty)$$

• thus $u_k \overset{H^2}\to \bar u$, for some $\bar u \in H^2(\Omega)$

• hence also $u_k \overset{L^2} \to \bar u$

• now I want to conclude that $\bar u = u$ and therefore $u \in D_{\bar A}$

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How do I show that $\bar u = u$?

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

It just follows from $$\|u_k - u\|_{H^2} \le C \, \|f_k - f\|_{L^2}.$$
• Just apply your very last assumption to $g = f_k - f$. This yields $v = u_k - u$ and the above inequality. – gerw Jun 7 '17 at 14:29