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 $1/4\pi |x|$, 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 $1/4\pi |x|$: $$(G\phi)(x)=\int \frac{\phi(y)}{4\pi |x-y|} dy$$ $G$ is also unbounded and coincides at least for the Schwartz functions with the above defined inverse Laplacian, $G\phi=(-\Delta)^{-1}\phi$ for all $\phi \in\mathcal{S}(\mathbb{R}^3)$.
Now my question is: Are both operators the same? Does $G=(-\Delta)^{-1}$ hold, i. e. is $D(G)=D((-\Delta)^{-1})$ and $G\phi=(-\Delta)^{-1}\phi$ 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 $$ (-\Delta +\epsilon I)^{-1}f = \mathscr{F}^{-1}\frac{1}{|\xi|^2+\epsilon}(\mathscr{F}f)(\xi),\;\;\; f\in L^2(\mathbb{R}^3). $$ Suppose $f\in L^2$. If $g(\xi)=|\xi|^{-2}(\mathscr{F}f)(\xi)$ is also in $L^2$, then $$ g=L^2\mbox{-}\lim_{\epsilon\downarrow 0}(-\Delta+\epsilon I)^{-1}f=\mathscr{F}^{-1}\frac{1}{|\xi|^2}\mathscr{F}f = \frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{1}{|x-y|}f(y)dy. $$ It is also true that, for such an $f$, the right side of the following converges to $f$ in $L^2$ as $\epsilon\downarrow 0$: $$ -\Delta(-\Delta+\epsilon I)^{-1}f=f-\epsilon(-\Delta+\epsilon I)^{-1}f. $$ And $(-\Delta+\epsilon I)^{-1}f$ converges to $\mathscr{F}^{-1}(|\xi|^{-1}\mathscr{F}f)$. Because $-\Delta : W^{2,2}\subset L^2\rightarrow L^2$ is selfadjoint, it follows that $\mathscr{F}^{-1}(|\xi|^{-2}\mathscr{F}f)\in\mathcal{D}(-\Delta)$ and $$ -\Delta\left[\mathscr{F}^{-1}\frac{1}{|\xi|^2}\mathscr{F}f\right] = f, $$ which is equal to $$ -\Delta \frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{1}{|x-y|}f(y)dy=f. $$