# Inverting the Laplacian

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.

## 1 Answer

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 $$-\Delta \frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{1}{|x-y|}f(y)dy=f.$$

• I already see this calculation as given. My question is about the technical aspect behind the inverse laplacian, specifically whether operators defined by functional calculus and defined by convolution are the same especially in terms of their natural domains. The calculation of the inv. FT just shows that both operators coincide on the schwartz functions, but their natural domains are larger. – MrMatzetoni Jan 2 at 21:42
• @MrMatzetoni : Take another look at my latest operator-based revision. – DisintegratingByParts Jan 4 at 2:04