Closure of the Laplacian in $L^2(\mathbb R^n)$

Question

Consider the Laplacian $\Delta$ as an operator on $L^2(\mathbb R^n)$, densely defined on the subspace $C^\infty_0(\mathbb R^n)$.

My questions are: Is the domain of the closure of the Laplacian, in the sense described here: http://planetmath.org/?method=l2h&from=objects&name=ClosedOperator&op=getobj

equal exactly to: $$\{u \in L^2(\mathbb R^n) | \Delta u \in L^2(\mathbb R^n)\}$$ (where $\Delta$ here means in the distributional sense)?

My 2nd question is: does any of the above spaces (which I hope are equal) in turn exactly equal the Sobolev space $W^{2,2}(\mathbb R^n)$, or is $W^{2,2}$ actually a strictly smaller space?

My 3rd question is: does any of the above spaces equal the Friedrichs extension? (See http://en.wikipedia.org/wiki/Friedrichs_extension).

Sorry for all the questions! I am unbounded in my confusion.

Answer 1

I don't know much about Friedrichs extension, so I will only comment on the first two.

I will sketch how to prove that your space with the $2$ replaced by a $p$ is equal to that Sobolev space. For $p = 2$ you can just use Plancherel together with our friend the Fourier transform (try it!).

Using the Riesz transform one can show (See Stein's Singular integrals and differentiability properties of functions) that

Theorem. Suppose $f \in C^2$ and suppose that $f$ has compact support. Then we have $$\left \|\frac{\partial^2 f}{\partial x_j \partial x_k} \right \|_p \leq A_p \|\Delta f\|_p \text{ for $1 < p < \infty$}$$

Using limits and so on we can show that this holds for $W^{2, p}(\mathbf{R}^d)$ and with some PDE tricks also for domains.

From this we get

Corollary. For $1 < p < \infty$ we have, $$W^{2, p}(\mathbf{R}^d) = \{f \in L^p(\mathbf{R}^d): \Delta f \in L^p\}.$$

This is quite easy. Just introduce the norm $|\!|\!|f|\!|\!| = \|f\|_p + \|\Delta f\|_p$ and show that this is equivalent to the Sobolev norm. To show this we can use that in $\mathbf R$ we have that $\|f'\|_p \lesssim \|f\|_p + \|f''\|_p$. This is just by integration by parts. A similar formula holds for $\mathbf R^d$.

Answer 2

To answer your third question, this extension is indeed the Friedrichs extension, because the closure of $\Delta$ is self-adjoint; that is, $\Delta$ is essentially self-adjoint.

The Friedrichs extension is a certain distinguished self-adjoint extension of a symmetric operator which is in some sense the "smallest". However, it is not hard to show that if the closure of an operator is self-adjoint, then it is the unique self-adjoint extension. So the Friedrichs extension must equal the closure of $\Delta$.