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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 $\mathbb R^n$.

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    $\begingroup$ Are you familiar with the Friedrichs extension theorem? $\endgroup$ – Nate Eldredge Jul 11 at 21:14
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    $\begingroup$ I am not, but this seems like it would apply, thank you for the reference! $\endgroup$ – Nathanael Schilling Jul 11 at 21:25
  • $\begingroup$ You want to state the question a little differently for other domains $\Omega$, because the Laplacian is typically not essentially self-adjoint on $C^\infty_c(\Omega)$ in that case. $\endgroup$ – Nate Eldredge Jul 12 at 3:56
  • $\begingroup$ I was thinking of the (geodesically) complete Riemannian manifold case, where we do have essential self-adjointness. $\endgroup$ – Nathanael Schilling Jul 12 at 7:20
  • $\begingroup$ If you want simple proofs, I don't think it's wise to ban the Fourier transform. After taking FTs, your operator becomes a multiplication operator, and for these the questions you ask are extremely easy to analyze. $\endgroup$ – Christian Remling Jul 12 at 23:52
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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 $u \in H^1$.

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The Laplace operator is essentially self-adjoint: define $$ \mathcal D(∆)=\{u\in L^2, ∆u\in L^2\}=H^2. $$ Then for $u,v\in \mathcal D(∆)$, $\langle ∆ u,v\rangle=\langle u,∆v\rangle$: 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 $$ \langle ∆ u,v\rangle=\lim_k\langle ∆ u,v_k\rangle=\lim_k\lim_l\langle ∆ u_l,v_k\rangle=\lim_k\lim_l\langle u_l,∆v_k\rangle=\lim_k\langle u,∆v_k\rangle=\langle u,∆v\rangle, \text{qed}. $$ Moreover you define for the symmetric operator $∆$, $$ \mathcal D(∆^*)=\{v\in L^2,\exists C, \forall u\in \mathcal D(∆), \vert\langle ∆u,v\rangle\vert\le C\Vert u\Vert_{L^2}\}. $$ 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 $v\in \mathcal D(∆^*)$. Let $u$ be a smooth compactly supported function: then we have $$ \vert\langle ∆u,v\rangle\vert\le C\Vert u\Vert_{L^2}, \quad \text{i.e.} \quad \vert\langle u,∆v\rangle\vert\le C\Vert u\Vert_{L^2}, $$ proving that $∆ v\in L^2$ and since $v\in L^2$, the ellipticity of the Laplace operator implies that $v\in H^2=\mathcal D(∆)$. The above argument works for any symmetric elliptic operator.

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  • $\begingroup$ I know the Laplacian is essentially self-adjoint, I was maybe a bit unclear but my question was why its domain of definition is a subset of $H^1$, not why the operator is essentially self-adjoint. The definition of $H^2$ you give isn't the standard one (existence of all weak partial derivatives of order 1 and 2 in $L^2$), I would like to know why $\mathcal D(\Delta) \subset H^1$ with the standard definition. $\endgroup$ – Nathanael Schilling Jul 11 at 20:26
  • $\begingroup$ Maybe you might ask yourself what the relation of $H^2$ and $H^1$ is. $\endgroup$ – Fabian Wirth Jul 11 at 20:31
  • $\begingroup$ I think I misunderstood the post above in that I thought it was a definition for $H^2$, whereas the definition was for $\mathcal D(\Delta)$ in terms of $H^2$. Nevertheless, my question still stands as the point I was asking about is contained in the reference to the ellipticity of the Laplace operator; which is nontrivial on general domains as far as I remember. This is why I would like a simple proof of the weaker statement that the domain is contained in $H^1$. $\endgroup$ – Nathanael Schilling Jul 11 at 20:49
  • $\begingroup$ In this situation, essential self-adjointness is exactly that the domain of the unique self-adjoint extension is $H^2$. (More generally, for semi-bounded operators, we can define analogous $H^2$, Friedrichs' extensions, and so on.) $\endgroup$ – paul garrett Jul 12 at 2:07
  • $\begingroup$ @WillieWong Yes, thanks, I did correct that. $\endgroup$ – Bazin Jul 12 at 19:17

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