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


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*Is the domain of the closure of the Laplacian, in the sense described here:
https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators,
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)?

*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?

*Does any of the above spaces equal the Friedrichs extension?

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 $\mathbf R^d$.
