Yukawa kernel of the free resolvent in three dimensions

Spherical coordinates give $\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1={\int }_0^{\mathrm{\infty }}r{e}^{-\kappa r}dr={\kappa }^{-2},\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_2^2=\frac{1}{4\pi }{\int }_0^{\mathrm{\infty }}{e}^{-2\kappa r}dr=\frac{1}{8\pi \kappa }\mathrm{.}$ Thus both norms are finite; the point value at the origin does not affect the $L^p$ class.

For $\rho =\mathrm{\mid }\xi \mathrm{\mid }>0$, [F1] and the elementary integral ${\int }_0^{\mathrm{\infty }}{e}^{-\kappa r}\sin (ar)dr=a\mathrm{/}({\kappa }^2+a^2)$ for $a>0$ give ${\widehat{G}}_{\kappa }(\xi )=\frac{1}{2\pi \rho }{\int }_0^{\mathrm{\infty }}{e}^{-\kappa r}\sin (2\pi \rho r)dr=\frac{1}{{\kappa }^2+4{\pi }^2{\rho }^2}\mathrm{.}$ At $\xi =0$, direct integration gives ${\widehat{G}}_{\kappa }(0)={\int }_0^{\mathrm{\infty }}r{e}^{-\kappa r}dr={\kappa }^{-2}$, the same formula.

Put $M=\mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1>0$ and $A(x)=\int G_{\kappa }(y)\mathrm{\mid }f(x-y){\mathrm{\mid }}^{2}dy$. Tonelli and translation invariance [F2, F3] give $\int A(x)dx=M\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, so $A(x)<\mathrm{\infty }$ for almost every $x$. For each such $x$, $\int G_{\kappa }(y)\mathrm{\mid }f(x-y)\mathrm{\mid }dy\le (M+A(x))\mathrm{/}2<\mathrm{\infty }$ by $2t\le 1+t^2$, so define $b(x)=\int G_{\kappa }(y)f(x-y)dy$. Expanding the nonnegative integral $\int G_{\kappa }(y)\mathrm{\mid }f(x-y)-b(x)\mathrm{/}M{\mathrm{\mid }}^{2}dy$ gives $0\le A(x)-\mathrm{\mid }b(x){\mathrm{\mid }}^2\mathrm{/}M$, hence $\mathrm{\mid }b(x){\mathrm{\mid }}^2\le MA(x)$. Integrating yields $\mathrm{\parallel }{G}_{\kappa }\ast f{\mathrm{\parallel }}_2^2\le M^2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. Therefore convolution is defined almost everywhere and gives a bounded operator on $L^2$.

If $f$ is Schwartz, then $G_{\kappa }\ast f\in L^1$ by Fubini and translation invariance [F3], and lies in $L^2$ by step 2.2. Fubini in the ordinary Fourier integral gives the integral-transform identity $\widehat{G_{\kappa }\ast f}={\widehat{G}}_{\kappa }\widehat{f}$. Since $G_{\kappa }\ast f\in L^1\cap L^2$, [F4] identifies its integral transform with its Plancherel transform almost everywhere; the same interface identifies $\widehat{f}$ with ${\mathcal{F}}_{2}f$ almost everywhere. Thus ${\mathcal{F}}_2(G_{\kappa }\ast f)={\widehat{G}}_{\kappa }{\mathcal{F}}_{2}f$ as $L^2$ classes. Since $\mathrm{\mid }{\widehat{G}}_{\kappa }(\xi )\mathrm{\mid }\le \mathrm{\parallel }{G}_{\kappa }{\mathrm{\parallel }}_1$ by the integral definition of the Fourier transform, multiplication by ${\widehat{G}}_{\kappa }$ is bounded on $L^2$. Schwartz density and boundedness of convolution therefore extend the identity to all $f\in L^2$.

By [A1], $-\mathrm{\Delta }$ is the multiplier $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$. Therefore $(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2{)}^{-1}$ is the bounded inverse multiplier of $-\mathrm{\Delta }+{\kappa }^2$; its product with the latter multiplier is $1$, and its range lies in the operator domain because $4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2)\le 1$. Step 3.1 identifies the same multiplier for convolution, so convolution is exactly $(-\mathrm{\Delta }+{\kappa }^2{)}^{-1}$.