How to show that the Laplacian is not closed? Let us define the Laplace operator $A:D(A)\subset L^2(\Omega)\to L^2(\Omega)$ by setting $Au:=-\Delta u$ and $D(A):=H^2(\Omega)$ (without any boundary condition), where $\Omega$ is a bounded domain with smooth boundary. From Weyl's lemma, we can infer that this operator is closable. However, the proposition 4.1 in the paper, titiled "Boundary value problems for elliptic partial differential operators on bounded domains", shows that $A$ is not closed. The approach there is not direct and has to use the concept of "quasi-boundary triple".
I just wonder if there is a simple, direct method to show that $A$ is not closed.
 A: In the multidimensional case the domain of the maximal operator associated to
$-\Delta$ in $L^2(\Omega)$ is given by 
$$
D_{\max}:=\{f\in L^2(\Omega):\Delta f \in L^2(\Omega)\},
$$
and here $\Delta f$ is understood in the sense of distributions. This space is larger than $H^2(\Omega)$ as e.g. all harmonic functions in $L^2(\Omega)$ belong to $D_{max}$, but these
harmonic functions are not necessarily in $H^2(\Omega)$ (since their weak derivatives may not be square integrable).
Now equip $D_{max}$ with the graph norm $\Vert f\Vert:=\Vert f\Vert_{L^2}
+\Vert\Delta f\Vert_{L^2}$. 
It is known that the space $C^\infty(\overline\Omega)$ is dense in $D_{max}$
and hence also $H^2(\Omega)$ is dense in $D_{max}$ with respect to the graph norm, but $H^2(\Omega)\not=D_{max}$ by the argument above. The maximal operator associated to $-\Delta$ coincides with the adjoint of the minimal operator defined on $H^2_0(\Omega)$ and hence the maximal operator is closed.
It follows that the closure of the Laplacian defined on $H^2(\Omega)$ is the maximal operator and that the Laplacian defined on $H^2(\Omega)$ is not closed.
In the one-dimensional case the situation is different: If $\Omega=(a,b)$, then the maximal domain $D_{max}$ coincides with $H^2((a,b))$, and hence
in this case the Laplacian $-\Delta=-\frac{d^2}{dx^2}$ defined on $H^2((a,b))$ is closed.
