I'm trying to combine two ways of looking at the Laplacian $\Delta$ on $\mathbb R^n$ (and on other domains). 


Firstly, it is well known that this operator is essentially self-adjoint on $C\_c^\infty(\mathbb R^n)$. Secondly, I know that for $f,g \in C^\infty\_c(\mathbb R^n)$ it holds that $\langle u, \Delta u \rangle = \langle \nabla u, \nabla v \rangle $, where the inner-product is the $L^2$ one. 


Now the domain $\mathcal D (\Delta) \subset L^2$ on which the Laplacian is self-adjoint must contain $C\_c^\infty (\mathbb R^n)$, but I can't seem to find much more information on this. By the previous identity I feel like $\Delta u \in L^2$ should imply that $u \in H^1(\mathbb R^n)$, but I can't seem to find a way to prove this. I know it is possible to prove that $\mathcal D(\Delta) \subset H^2(\mathbb R^n$) via elliptic regularity etc, but it seems to me like there "should" be an elementary argument for showing $\mathcal D(\Delta) \subset H^1$ (preferably via the non-Fourier-transform characterisation of weak derivatives) but I can't seem to find it anywhere.


Edit: To clarify my question: I am looking for a simple proof of $\Delta u = f \in L^2$ implies $u \in H^1$ where $\Delta u$ is defined as the self-adjoint extension of the Laplacian $\Delta$ on $C^\infty\_c$, preferrably one which also works on domains other than $\mathbb R^n$. 



Domain of definition of Laplace Operator on $L^2$
Maybe this is the argument you're looking for.


Suppose you know that $\Delta$ is essentially self-adjoint on $C^\infty\_c(\mathbb{R}^n)$. This means $C^\infty\_c$ is a core for $\Delta$, so for any $u \in D(\Delta)$, there is a sequence $u\_n \in C^\infty\_c$ such that $u\_n \to u$ and $\Delta u\_n \to \Delta u$ in $L^2$. In particular, we have
$$|\langle \Delta(u\_n - u\_m), u\_n - u\_m \rangle\_{L^2} | \le \|\Delta u\_n - \Delta u\_m\|\_{L^2} \|u\_n - u\_m\|\_{L^2} \to 0$$
using Cauchy-Schwarz. On the other hand, since $u\_n, u\_m \in C^\infty\_c$, we are perfectly justified in integrating by parts to say that $\langle \Delta(u\_n - u\_m), u\_n - u\_m \rangle = \|\nabla (u\_n - u\_m)\|^2\_{L^2}$. So, since $\|u\_n - u\_m\|\_{L^2} \to 0$ and $\|\nabla (u\_n - u\_m)\|\_{L^2} \to 0$, we have shown that $\|u\_n - u\_m\|\_{H^1} \to 0$; in other words, $u\_n$ is Cauchy in $H^1$. Now $H^1$ is a Hilbert space, so $u\_n$ converges in $H^1$ to some $v \in H^1$. But we already know $u\_n \to u$ in $L^2$, which is weaker than $H^1$ convergence, so we must have $u = v$, thus $u \in H^1$.


