Xrfgtjtk

I've had a hard time looking for literature on this, so here's my question:

We take a look at the Laplacian $-Delta$ as an unbounded operator on $mathrm{L}^2(mathbb{R}^3)$. We know that $-Delta$ is unitary equivalent to the multiplication operator with $|p|^2$ in Fourier space, so $-Delta=mathcal{F}^{-1} |p|^2 mathcal{F}$.

So one could now use functional calculus and define an inverse Laplacian by setting $(-Delta)^{-1}=mathcal{F}^{-1} (1/|p|^2 )mathcal{F}$, which will also be unbounded of course.

It is also known from theory of PDEs that one can invert the Laplacian on Schwartz functions using the Green's function $1/4pi |x|$, which is derived as a distributional Fourier transform of $1/|p|^2$. So define $G$ on $mathrm{L}^2(mathbb{R}^3)$ by convolution with $1/4pi |x|$:
$$(Gphi)(x)=int frac{phi(y)}{4pi |x-y|} dy$$
$G$ is also unbounded and coincides at least for the Schwartz functions with the above defined inverse Laplacian, $Gphi=(-Delta)^{-1}phi$ for all $phi inmathcal{S}(mathbb{R}^3)$.

Now my question is: Are both operators the same? Does $G=(-Delta)^{-1}$ hold, i. e. is $D(G)=D((-Delta)^{-1})$ and $Gphi=(-Delta)^{-1}phi$ for all $phi in D(G)=D((-Delta)^{-1})$?

My guess is that $mathcal{S}(mathbb{R}^3)$ is a core of $(-Delta)^{-1}$, and as $(-Delta)^{-1}|_{mathcal{S}(mathbb{R}^3)}=G|_{mathcal{S}(mathbb{R}^3)}$ equality should follow by closing the restrictions.

Any comments, hints on how to proceed or references are welcome! Thank you.

The operator $(-Delta +epsilon I)^{-1} : L^2(mathbb{R}^3)rightarrow W^{2,2}(mathbb{R}^3)$ is a bicontinuous bijection that is equivalently defined by the Fourier transform $mathscr{F}$ as
$$
(-Delta +epsilon I)^{-1}f = mathscr{F}^{-1}frac{1}{|xi|^2+epsilon}(mathscr{F}f)(xi),;;; fin L^2(mathbb{R}^3).
$$
Suppose $fin L^2$. If $g(xi)=|xi|^{-2}(mathscr{F}f)(xi)$ is also in $L^2$, then
$$
g=L^2mbox{-}lim_{epsilondownarrow 0}(-Delta+epsilon I)^{-1}f=mathscr{F}^{-1}frac{1}{|xi|^2}mathscr{F}f = frac{1}{4pi}int_{mathbb{R}^3}frac{1}{|x-y|}f(y)dy.
$$
It is also true that, for such an $f$, the right side of the following converges to $f$ in $L^2$ as $epsilondownarrow 0$:
$$
-Delta(-Delta+epsilon I)^{-1}f=f-epsilon(-Delta+epsilon I)^{-1}f.
$$
And $(-Delta+epsilon I)^{-1}f$ converges to $mathscr{F}^{-1}(|xi|^{-1}mathscr{F}f)$. Because $-Delta : W^{2,2}subset L^2rightarrow L^2$ is selfadjoint, it follows that $mathscr{F}^{-1}(|xi|^{-2}mathscr{F}f)inmathcal{D}(-Delta)$ and
$$
-Deltaleft[mathscr{F}^{-1}frac{1}{|xi|^2}mathscr{F}fright] = f,
$$
which is equal to
$$ -Delta frac{1}{4pi}int_{mathbb{R}^3}frac{1}{|x-y|}f(y)dy=f.
$$