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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 $mathbb R^n$.

Maybe this is the argument you're looking for.

Suppose you know that $Delta$ is essentially self-adjoint on $C^infty_c(mathbbR^n)$. 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 $u in H^1$.

The Laplace operator is essentially self-adjoint: define
$$
mathcal D(∆)=uin L^2, ∆uin L^2=H^2.
$$
Then for $u,vin mathcal D(∆)$, $langle ∆ u,vrangle=langle u,∆vrangle$: to prove this, consider $lim u_k=u, lim v_k=v text in $H^2$$ with $u_k, v_k$ smooth compactly supported. Then you have
$$
langle ∆ u,vrangle=lim_klangle ∆ u,v_krangle=lim_klim_llangle ∆ u_l,v_krangle=lim_klim_llangle u_l,∆v_krangle=lim_klangle u,∆v_krangle=langle u,∆vrangle, textqed.
$$
Moreover you define for the symmetric operator $∆$,
$$
mathcal D(∆^*)=vin L^2,exists C, forall uin mathcal D(∆),
vertlangle ∆u,vranglevertle CVert uVert_L^2.
$$
Then as for all symmetric operators we have $mathcal D(∆^*)supset mathcal D(∆)$. To get self-adjointness, you must prove $mathcal D(∆^*)subset mathcal D(∆)$. Let $vin mathcal D(∆^*)$. Let $u$ be a smooth compactly supported function: then we have
$$
vertlangle ∆u,vranglevertle CVert uVert_L^2, quad texti.e. quad
vertlangle u,∆vranglevertle CVert uVert_L^2,
$$
proving that $∆ vin L^2$ and since $vin L^2$, the ellipticity of the Laplace operator implies that $vin H^2=mathcal D(∆)$. The above argument works for any symmetric elliptic operator.