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?

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 $\Delta$.
