The volume integral operator is well defined and bounded on L2

On the product space, Cauchy–Schwarz gives ${\int }_{D\times D}\mathrm{\mid }a(x,y)u(y)\mathrm{\mid }dxdy\le \mathrm{\parallel }a{\mathrm{\parallel }}_2(\mathrm{\mid }D\mathrm{\mid }{\int }_D\mathrm{\mid }u(y){\mathrm{\mid }}^{2}dy{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Hence the section integral defining $T_{a}u$ exists absolutely for almost every $x$ and is measurable by F3.

If $u=u^{\mathrm{\prime }}$ almost everywhere, then $a(x,y)(u-u^{\mathrm{\prime }})(y)=0$ for product-almost every $(x,y)$ by Tonelli on $D\times \{u\mathrm{\ne }{u}^{\mathrm{\prime }}\}$; F3 makes their section integrals equal for almost every $x$. Changing $a$ on a product-null set likewise changes only a.e.-zero sections. Thus the output class depends only on the kernel and input classes. Integral linearity on $L^1$ proves complex linearity.

For almost every $x$, Cauchy–Schwarz in $y$ yields $\mathrm{\mid }{T}_{a}u(x){\mathrm{\mid }}^2\le ({\int }_D\mathrm{\mid }a(x,y){\mathrm{\mid }}^{2}dy)\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. Integrating this inequality in $x$ with Tonelli gives $\mathrm{\parallel }{T}_{a}u{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }a{\mathrm{\parallel }}_2^2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. Step 2.1 and this estimate prove the general assertion.

Apply step 3.1 to $a(x,y)=G_k(x-y)q(y)$ and then use The three-dimensional free-space scattering kernel is square integrable for its explicit square norm. If $\mathrm{\mid }D\mathrm{\mid }=0$, both domain and codomain are the zero $L^2$ space.