The volume integral scattering operator is compact

By [F11] and AC, $X=L^2(D;\mathbb{C})$ is a Banach space. Normalize $D$ to the probability measure $d\nu =dx\mathrm{/}M$. By [F2], the operator $K_k$ is unitarily equivalent to the integral operator on $L^2(D,\nu )$ with kernel $\tilde{a}=Ma$. Its kernel norm equals $\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2(D\times D,dxdy)}$ by [F2]. Applying [F3] to this probability-space kernel gives compactness of the normalized operator and its norm bound; unitary equivalence transfers both to $K_k$.

We also verify the singular-truncation route. By [F12], choose a measurable representative $q_0$ and set it to zero on the measurable null set where $\mathrm{\mid }{q}_0\mathrm{\mid }>\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}$. This does not change the operator, since the kernel changes only on a product-null set, and now $\mathrm{\mid }{q}_0(y)\mathrm{\mid }\le \mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $y$. For $\epsilon >0$, put ${\chi }_{\epsilon }(r)=0$ for $r\le \epsilon$, ${\chi }_{\epsilon }(r)=(r-\epsilon )\mathrm{/}\epsilon$ for $\epsilon <r<2\epsilon$, and ${\chi }_{\epsilon }(r)=1$ for $r\ge 2\epsilon$. Let $K_{\epsilon }$ have kernel ${\chi }_{\epsilon }(\mathrm{\mid }x-y\mathrm{\mid })G_k(x-y)q_0(y)$. The difference kernel is supported where $\mathrm{\mid }x-y\mathrm{\mid }<2\epsilon$ and bounded in modulus by $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(4\pi \mathrm{\mid }x-y\mathrm{\mid })$. By [F9] its row and column integrals, for every fixed row and column, are at most $A=B\le (\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(4\pi ))4\pi {\int }_0^{2\epsilon }rdr=2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}{\epsilon }^2$. For each $x$, weighted Cauchy–Schwarz gives $\mathrm{\mid }{T}_{h}v(x){\mathrm{\mid }}^2\le (\int \mathrm{\mid }h(x,y)\mathrm{\mid }dy)(\int \mathrm{\mid }h(x,y)\mathrm{\mid }\mathrm{\mid }v(y){\mathrm{\mid }}^{2}dy)$; integrating and using Tonelli yields $\mathrm{\parallel }{T}_{h}v{\mathrm{\parallel }}_2^2\le AB\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$. Hence $\mathrm{\parallel }K-K_{\epsilon }\mathrm{\parallel }\le 2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}{\epsilon }^2$.

Fix $\epsilon >0$. Extend the everywhere-bounded representative $q_0$ by zero to ${\mathbb{R}}^3$; it lies in $L^2({\mathbb{R}}^3)$ because $q\in L^{\mathrm{\infty }}(D)$ and $\mathrm{\mid }D\mathrm{\mid }<\mathrm{\infty }$. By [F5], choose $q_j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{C})$ with $\mathrm{\parallel }{q}_j-q_0{\mathrm{\parallel }}_{L^2(D)}\to 0$. The kernel $g_{\epsilon ,j}(x,y)={\chi }_{\epsilon }(\mathrm{\mid }x-y\mathrm{\mid })G_k(x-y)q_j(y)$ extends continuously to the compact set $\bar{D}\times \bar{D}$ by assigning zero on $x=y$; the cutoff vanishes near the diagonal. For $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le 1$, the integral is independent of the representative of $v$, and by [F13] its outputs obey $\mathrm{\mid }{T}_{\epsilon ,j}v(x)\mathrm{\mid }\le \sqrt{\mathrm{\mid }D\mathrm{\mid }}\mathrm{\parallel }{g}_{\epsilon ,j}{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\mid }{T}_{\epsilon ,j}v(x)-T_{\epsilon ,j}v(x^{\mathrm{\prime }})\mathrm{\mid }\le \sqrt{\mathrm{\mid }D\mathrm{\mid }}{\sup }_y\mathrm{\mid }{g}_{\epsilon ,j}(x,y)-g_{\epsilon ,j}(x^{\mathrm{\prime }},y)\mathrm{\mid }$. Uniform continuity of $g_{\epsilon ,j}$ makes these outputs uniformly bounded and equicontinuous. Applying [F6] separately to real and imaginary parts gives compact closures $C_R,C_I$ in $C(\bar{D};\mathbb{R})$. The product $C_R\times C_I$ is compact by finite-product compactness. Under the homeomorphism $w\mapsto (\mathrm{Re}w,\mathrm{Im}w)$, the closure of the complex output family is a closed subset of this compact product, hence compact in $C(\bar{D};\mathbb{C})$. The inclusion is continuous because $\mathrm{\parallel }w{\mathrm{\mid }}_D{\mathrm{\parallel }}_2\le \sqrt{\mathrm{\mid }D\mathrm{\mid }}\mathrm{\parallel }w{\mathrm{\parallel }}_{C(\bar{D})}$; hence $T_{\epsilon ,j}:L^2(D)\to L^2(D)$ is compact.

Since $\mathrm{\mid }{\chi }_{\epsilon }{G}_k\mathrm{\mid }\le (4\pi \epsilon {)}^{-1}$, the square norm of the kernel difference between $K_{\epsilon }$ and $T_{\epsilon ,j}$ is at most $\mathrm{\mid }D\mathrm{\mid }(16{\pi }^2{\epsilon }^2{)}^{-1}\mathrm{\parallel }q-q_j{\mathrm{\parallel }}_{L^2(D)}^2$. [F8] therefore gives $\mathrm{\parallel }{K}_{\epsilon }-T_{\epsilon ,j}\mathrm{\parallel }\to 0$. By [F7], $K_{\epsilon }$ is compact.

As $\epsilon \downarrow 0$, step 1.2 gives $\mathrm{\parallel }K-K_{\epsilon }\mathrm{\parallel }\le 2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}{\epsilon }^2\to 0$. Applying [F7] again shows $K$ is compact. Step 1.1 independently gives the displayed norm estimate and compactness with the finite-measure Hilbert–Schmidt normalization.