Weak Laplacian is a closed map

Question

Consider the map $\Delta : H^{2}(\Omega) \rightarrow L^{2}(\Omega)$ the weak laplacian. My professor has commented that this map is a closed map, that is, the graph $G = \{(u, \Delta u) : u \in H^{2}\}$ is closed in $H^{2}(\Omega) \times L^{2}(\Omega)$.

I don't have any Idea how to proceed. If anyone has a tip or a reference which would help me, I really thank.

Answer 1

Let $\{(u_k,v_k)\} \subset G$ converge to some $(u,v)$ in $H^2(\Omega) \times L^2(\Omega)$. To prove $G$ is closed we need to show $(u,v) \in G$.

From the definition of $G$ we know that $v_k = \Delta u_k$. Then \begin{align*} \| v - \Delta u \|_{L^2(\Omega)} &\leqslant \| v - v_k \|_{L^2(\Omega)} + \| \Delta u_k - \Delta u \|_{L^2(\Omega)} \\ &\leqslant \| v - v_k \|_{L^2(\Omega)} + \| u_k - u \|_{H^2(\Omega)}. \end{align*} Sending $k\to \infty$, we conclude that $v=\Delta u$, so $(u,v) \in G$.