# Domain of definition of Laplace Operator on $L^2$

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 $$\mathbb R^n$$.

• Are you familiar with the Friedrichs extension theorem? – Nate Eldredge Jul 11 at 21:14
• I am not, but this seems like it would apply, thank you for the reference! – Nathanael Schilling Jul 11 at 21:25
• 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. – Nate Eldredge Jul 12 at 3:56
• I was thinking of the (geodesically) complete Riemannian manifold case, where we do have essential self-adjointness. – Nathanael Schilling Jul 12 at 7:20
• 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. – Christian Remling Jul 12 at 23:52

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 $$u \in H^1$$.
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.
• 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. – Nathanael Schilling Jul 11 at 20:26
• Maybe you might ask yourself what the relation of $H^2$ and $H^1$ is. – Fabian Wirth Jul 11 at 20:31
• 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 $H^1$. – Nathanael Schilling Jul 11 at 20:49
• 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.) – paul garrett Jul 12 at 2:07