The three-dimensional free-space scattering kernel is square integrable

If $M=0$, then $(D\times D,{\lambda }_3\times {\lambda }_3)$ is null and the conclusion follows. Suppose $M>0$. Off the Borel diagonal, $G_k(x-y)q(y)$ is product-measurable, and setting it to zero on the diagonal preserves measurability. For $x\mathrm{\ne }y$, [F4] and real $k$ give $\mathrm{\mid }{e}^{ik\mathrm{\mid }x-y\mathrm{\mid }}\mathrm{\mid }=1$; choosing the representative in [F5] yields $\mathrm{\mid }a(x,y){\mathrm{\mid }}^2\le \mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}^2(16{\pi }^2\mathrm{\mid }x-y{\mathrm{\mid }}^2{)}^{-1}$. The diagonal is product-null by [F2] and F3, so its assigned value does not alter the class.

In the positive-measure case $d_D>0$. For each fixed $x\in D$, $D\subseteq \bar{B}(x,d_D)$. By F1 with the Borel function $g(y)=\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}{\mathbf{1}}_{0<\mathrm{\mid }x-y\mathrm{\mid }\le d_D}$, polar coordinates give ${\int }_D\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}dy\le {\int }_0^{d_D}{\int }_{S^2}{r}^{-2}{r}^{2}d\sigma (\omega )dr=4\pi d_D$, using $\sigma (S^2)=4\pi$, which follows by applying polar coordinates to the unit ball and using its volume $4\pi \mathrm{/}3$ (A closed three-dimensional ball of radius $r\ge 0$ has volume $4\pi r^3\mathrm{/}3$).

In the positive-measure case, integrating the bound of step 1.1 in $x$ and applying F2 gives ${\int }_D{\int }_D\mathrm{\mid }a(x,y){\mathrm{\mid }}^{2}dydx\le \frac{\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}^2}{16{\pi }^2}M(4\pi d_D)=\frac{\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}^{2}M{d}_D}{4\pi }<\mathrm{\infty }$. Thus $a\in L^2(D\times D)$. Steps 1.1 and 1.2 cover respectively $M=0$ and $M>0$.