The outgoing three-dimensional Helmholtz kernel is a fundamental solution

Writing $r=\mathrm{\mid }x\mathrm{\mid }$, [F7] gives ${\mathrm{\Phi }}_k^{\mathrm{\prime }}(r)=e^{ikr}(ikr-1)\mathrm{/}(4\pi r^2)$; direct substitution yields ${\mathrm{\Phi }}_k^{\mathrm{\prime }\mathrm{\prime }}+2{\mathrm{\Phi }}_k^{\mathrm{\prime }}\mathrm{/}r+k^2{\mathrm{\Phi }}_k=0$ for $r>0$. Also [F7] gives $\mathrm{\mid }{\mathrm{\Phi }}_k(x)\mathrm{\mid }=1\mathrm{/}(4\pi \mathrm{\mid }x\mathrm{\mid })$, whose integral over each centered ball is finite by [F2]. Hence ${\mathrm{\Phi }}_k\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ and [F6] supplies its regular distribution. Point evaluation is continuous by [F5], so ${\delta }_0$ is also a distribution. Thus $(-\mathrm{\Delta }-k^2){\mathrm{\Phi }}_k=0$ away from the origin as a classical equation and the distributional pairing in the next step is defined.

Choose an outer sphere beyond $\mathrm{supp}\phi$ and apply [F3] on the annulus between it and radius $ϵ$. Apply the real identity to the four real/imaginary pairings to obtain the complex bilinear identity. The outer boundary term is zero, and the inner boundary orientation yields ${\int }_{\mathrm{\mid }x\mathrm{\mid }>ϵ}{\mathrm{\Phi }}_k(x)(-\mathrm{\Delta }-k^2)\phi (x)dx={\int }_{\mathrm{\mid }x\mathrm{\mid }=ϵ}({\mathrm{\Phi }}_k{\mathrm{\partial }}_r\phi -\phi {\mathrm{\partial }}_r{\mathrm{\Phi }}_k)dS$. The first term is $O(ϵ)$ since ${\mathrm{\Phi }}_k=O({ϵ}^{-1})$ and the sphere area is $4\pi {ϵ}^2$. From the derivative formula in step 1.1 and the exponential series, $e^{ikϵ}(ikϵ-1)=-1+O({ϵ}^2)$, so ${\mathrm{\partial }}_r{\mathrm{\Phi }}_k=-1\mathrm{/}(4\pi {ϵ}^2)+O(1)$. The second term therefore tends to $\phi (0)$: after writing $x=ϵ\omega$, it equals ${\int }_{S^2}\phi (ϵ\omega )\mathrm{/}(4\pi )d\sigma +O({ϵ}^2)$. The omitted ball contributes $o(1)$ because $\mathrm{\mid }{\mathrm{\Phi }}_k\mathrm{\mid }=O(1\mathrm{/}r)$ is locally integrable by [F2]. By [F6], this limit is precisely $⟨(-\mathrm{\Delta }-k^2){\mathrm{\Phi }}_k,\phi ⟩=⟨{\delta }_0,\phi ⟩$.

Let $A$ be any bounded measurable set and let $C$ be a bounded set containing the support of $g$. Polar coordinates show ${\sup }_{y\in C}{\int }_A\mathrm{\mid }{\mathrm{\Phi }}_k(x-y)\mathrm{\mid }dx<\mathrm{\infty }$, since $A-y$ lies in a fixed-radius ball and ${\int }_0^R{r}^{-1}{r}^{2}dr<\mathrm{\infty }$. Tonelli then gives ${\int }_A{\int }_C\mathrm{\mid }{\mathrm{\Phi }}_k(x-y)\mathrm{\mid }\mathrm{\mid }g(y)\mathrm{\mid }dydx<\mathrm{\infty }$, so the defining convolution is absolutely finite for almost every $x\in A$ and belongs to $L^1(A)$. Taking $A=B(0,n)$ for each integer $n\ge 1$ and using [F8] on the countable union of exceptional null sets proves absolute finiteness for almost every $x\in {\mathbb{R}}^3$. For every bounded measurable $A$, the same estimate gives $V\in L^1(A)$, hence $V\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. With $A=\mathrm{supp}\phi$, the same bound makes $g(y){\mathrm{\Phi }}_k(x-y)(-\mathrm{\Delta }-k^2)\phi (x)$ absolutely integrable, so [F4] permits Fubini. Step 2.1 translated by $y$ gives $⟨(-\mathrm{\Delta }-k^2)V,\phi ⟩=\int g(y)\phi (y)dy$. The preceding argument proves $V\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ and the a.e. existence of its integral representative.