The retarded wave kernel is a distribution supported on the cone, solves the point-wave equation, and convolves to the retarded solution

If $\psi =0$, the defining integral is zero; otherwise fix $\psi \in \mathcal{D}$ with nonempty support. The support of $\psi$ is compact in $Q$, hence closed and bounded; choose $R>0$ with $\mathrm{supp}\psi \subseteq B_R(0)\times [-R,R]$ and put $M:=\sup \mathrm{\mid }\psi \mathrm{\mid }<\mathrm{\infty }$, the supremum taken over the nonempty compact support and finite by the extreme value theorem. Since $\mathrm{\mid }\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }\le M$ for every $y$ and $\mathrm{\mid }y{\mathrm{\mid }}^{-1}$ is nonnegative, the triangle inequality and the order rules for nonnegative integrals bound $\mid ⟨K,\psi ⟩\mid \le \frac{M}{4\pi }{\int }_{B_R}\frac{dy}{\mathrm{\mid }y\mathrm{\mid }}$. The polar formula applied to the nonnegative Borel function $y\mapsto \mathrm{\mid }y{\mathrm{\mid }}^{-1}{\mathbf{1}}_{B_R}(y)$ gives ${\int }_{B_R}\mathrm{\mid }y{\mathrm{\mid }}^{-1}dy={\int }_0^R{r}^{-1}\sigma (S^2)r^{2}dr=\sigma (S^2)R^2\mathrm{/}2<\mathrm{\infty }$, using $\sigma (S^2)=4\pi$. Hence the defining integral converges absolutely, and the bound is a fixed-support seminorm, so $K$ is continuous for the LF topology and $K\in {\mathcal{D}}^{\mathrm{\prime }}$.

Fix $h\in C_c^{\mathrm{\infty }}(Q)$ and $(x,t)\in Q$. Since $\mathrm{supp}h$ is compact, there is $\rho >0$ with $\mathrm{supp}h\subseteq {\bar{B}}_{\rho }(0)\times [-\rho ,\rho ]$, and $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ by the extreme value theorem. If $h(y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)\mathrm{\ne }0$, then $y\in {\bar{B}}_{\rho }(0)$ and $\mathrm{\mid }t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c\mathrm{\mid }\le \rho$, so $\mathrm{\mid }x-y\mathrm{\mid }\le c(\mathrm{\mid }t\mathrm{\mid }+\rho )$; therefore the absolute integral is at most $\frac{\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}}{4\pi }{\int }_{B(x,c(\mathrm{\mid }t\mathrm{\mid }+\rho ))}\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy=\frac{\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}}{4\pi }\cdot \frac{\sigma (S^2)c^2(\mathrm{\mid }t\mathrm{\mid }+\rho {)}^2}{2}=\frac{\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}{c}^2(\mathrm{\mid }t\mathrm{\mid }+\rho {)}^2}{2}<\mathrm{\infty }$ by the polar formula and the order rules. Hence $(K\ast h)(x,t)$ is well defined by an absolutely convergent integral for every $(x,t)\in Q$.

Fix $\psi \in \mathcal{D}$ and put $F:=K⊛\psi$, that is, $F(y,s)=\frac{1}{4\pi }\int \psi (y+z,s+\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }z{\mathrm{\mid }}^{-1}dz$. Choose $M>0$ with $\mathrm{supp}\psi \subseteq {\bar{B}}_M(0)\times [-M,M]$ and let $A:=\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. If $\mathrm{\mid }y\mathrm{\mid }\le 2M$, every nonzero contribution has $\mathrm{\mid }z\mathrm{\mid }\le M+\mathrm{\mid }y\mathrm{\mid }\le 3M$, so $\mathrm{\mid }F(y,s)\mathrm{\mid }\le \frac{A}{4\pi }{\int }_{B_{3M}}\mathrm{\mid }z{\mathrm{\mid }}^{-1}dz=\frac{9}{2}{M}^{2}A$; if $\mathrm{\mid }y\mathrm{\mid }>2M$, then $z\in B(-y,M)$ forces $\mathrm{\mid }z\mathrm{\mid }\ge \mathrm{\mid }y\mathrm{\mid }-M>M$, so $\mathrm{\mid }F(y,s)\mathrm{\mid }\le \frac{A}{4\pi }{\int }_{B(-y,M)}\mathrm{\mid }z{\mathrm{\mid }}^{-1}dz\le \frac{A}{4\pi }\cdot \frac{1}{M}\cdot \frac{4\pi }{3}{M}^3=\frac{M^{2}A}{3}$. Hence $F$ is bounded on all of $Q$ by $\frac{9}{2}{M}^{2}A$.

Fix $(y_0,t_0)\in C$ and a neighbourhood $U$ of it. Since $C\subseteq Q$ is closed, take an open ball $W$ with $(y_0,t_0)\in W\subseteq U$. If $y_0\mathrm{\ne }0$: choose $0\le \psi \le 1$ in $\mathcal{D}(Q)$ with $\psi =1$ on a neighbourhood of $(y_0,t_0)$ and $\mathrm{supp}\psi \subseteq W$. Then $\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)=1$ for all $y$ in a neighbourhood $V$ of $y_0$ (the map $y\mapsto (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)$ is continuous at $y_0\mathrm{\ne }0$), and $y\mapsto \psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$ is nonnegative, continuous on $V$, and bounded below by $1\mathrm{/}(2\mathrm{\mid }{y}_0\mathrm{\mid })$ on a smaller ball around $y_0$ of positive measure; hence $⟨K,\psi ⟩>0$ by the criterion that a nonnegative measurable function with positive values on a set of positive measure has positive integral, together with the finiteness of step 1.1 and the positivity of ball measure. If $y_0=0$, then $t_0=0$; choose $0\le \psi \le 1$ in $\mathcal{D}(Q)$ with $\psi =1$ on a neighbourhood of the origin contained in $W$. Then $\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)=1$ for $\mathrm{\mid }y\mathrm{\mid }$ small, and the same argument, using the finiteness of step 1.1 again, gives $⟨K,\psi ⟩\ge \frac{1}{4\pi }{\int }_{0<\mathrm{\mid }y\mathrm{\mid }<\delta }\mathrm{\mid }y{\mathrm{\mid }}^{-1}dy>0$ for some $\delta >0$. In both cases $\mathrm{supp}\psi \subseteq U$ and $⟨K,\psi ⟩\mathrm{\ne }0$.

With $F=K⊛\psi$ as in step 1.3, $F$ is continuous at the origin and $F(0,0)=⟨K,\psi ⟩$. Indeed, for a sequence $(y_n,s_n)\to (0,0)$ with $\mathrm{\mid }{y}_n\mathrm{\mid }\le M$ eventually, every nonzero contribution has $z\in B_{2M}$, and $\psi (y_n+z,s_n+\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }z{\mathrm{\mid }}^{-1}\to \psi (z,\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ for every $z\mathrm{\ne }0$ by continuity of $\psi$; the dominating function $A\mathrm{\mid }z{\mathrm{\mid }}^{-1}{\mathbf{1}}_{B_{2M}}(z)$ is integrable by the polar formula, so dominated convergence gives $F(y_n,s_n)\to \frac{1}{4\pi }\int \psi (z,\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }z{\mathrm{\mid }}^{-1}dz=⟨K,\psi ⟩$; a sequence criterion for continuity at a point gives the claim.

Let $\psi \in \mathcal{D}$ satisfy $\mathrm{supp}\psi \cap C=\mathrm{\varnothing }$. For every $y\mathrm{\ne }0$ the point $(y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)$ lies in $C$, so $\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)=0$; the exceptional set $\{0\}$ has Lebesgue measure zero. Hence the nonnegative measurable function $y\mapsto \mathrm{\mid }\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }\mathrm{\mid }y{\mathrm{\mid }}^{-1}$ vanishes almost everywhere, and its integral is zero by the almost-everywhere criterion; the triangle inequality with the absolute convergence of step 1.1 then gives $⟨K,\psi ⟩=0$.

Let $t_0$ lie below and $T$ above the time support of $h$, so that $h$ vanishes on ${\mathbb{R}}^3\times (-\mathrm{\infty },t_0]$ and on ${\mathbb{R}}^3\times [T,\mathrm{\infty })$ and its restriction to ${\mathbb{R}}^3\times [t_0,T]$ is of class $C^{\mathrm{\infty }}$ in the extension sense. Apply the wave theorem with $a=b=0$ and source $h$: the function $u(x,t)=c^2{\int }_{t_0}^t(t-v)M_{c(t-v)}h(\cdot ,v)(x)dv$ is $C^{\mathrm{\infty }}$ on the slab in the extension sense and $Wu=h$ on ${\mathbb{R}}^3\times (t_0,T)$. Writing the spherical mean as an integral over $S^2$, using Fubini on the compact rectangle $[t_0,t]\times S^2$, and substituting $r=c(t-v)$, with the absolute-derivative one-dimensional change of variables, gives $u(x,t)={\int }_0^{c(t-t_0)}r\frac{1}{4\pi }{\int }_{S^2}h(x+rn,t-r\mathrm{/}c)d\mathrm{\Omega }(n)dr$. Since the chart measure on $S^2$ is $\sigma$, the polar formula applied to the nonnegative Borel function $z\mapsto \mathrm{\mid }h(x+z,t-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }(4\pi \mathrm{\mid }z\mathrm{\mid }{)}^{-1}{\mathbf{1}}_{\mathrm{\mid }z\mathrm{\mid }\le c(t-t_0)}$, followed by linearity for the signed integrand, gives $u(x,t)={\int }_{\mathrm{\mid }z\mathrm{\mid }\le c(t-t_0)}\frac{h(x+z,t-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)}{4\pi \mathrm{\mid }z\mathrm{\mid }}dz$; the omitted range $\mathrm{\mid }z\mathrm{\mid }>c(t-t_0)$ has $t-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c<t_0$, where $h=0$, so the integration extends to all $z$. Finally the translation $y=x+z$, a $C^1$ diffeomorphism of Jacobian determinant one, changes this into $(K\ast h)(x,t)$ as defined in step 1.2. Thus $u=K\ast h$ on the slab, and consequently $K\ast h$ agrees, near each of its points, with a function of class $C^{\mathrm{\infty }}$ in the extension sense.

Since every point $(x_{\ast },t_{\ast })\in Q$ lies in the interior of a slab ${\mathbb{R}}^3\times [t_0,T]$ with $t_0$ below and $T$ above the time support of $h$, step 2.4 shows that $K\ast h$ agrees with a $C^{\mathrm{\infty }}$ function on a neighbourhood of each point; hence $K\ast h$ is of class $C^{\mathrm{\infty }}$ on $Q$. For such a point, $W(K\ast h)(x_{\ast },t_{\ast })=Wu(x_{\ast },t_{\ast })=h(x_{\ast },t_{\ast })$ by the wave theorem applied on that slab, where $W=c^{-2}{\mathrm{\partial }}_t^2-\mathrm{\Delta }$ is evaluated classically; hence $W(K\ast h)=h$ on all of $Q$.

The cone $C=\{(y,t):t=\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c\}$ is closed in $Q$, since $\mathrm{\mid }y\mathrm{\mid }$ is continuous, so $Q\setminus C$ is open, and every test supported there has $\mathrm{supp}\psi \cap C=\mathrm{\varnothing }$ and pairs to zero with $K$ by step 2.3; thus $K$ vanishes on $Q\setminus C$ in the sense of the support definition, and $\mathrm{supp}K\subseteq C$. Conversely let $(y_0,t_0)\in C$ and suppose it were outside $\mathrm{supp}K$; then $Q\setminus \mathrm{supp}K$ would be an open neighbourhood of $(y_0,t_0)$ on which $K$ vanishes, and step 2.1 would produce a test supported there with $⟨K,\psi ⟩\mathrm{\ne }0$, contradicting the vanishing. Hence $C\subseteq \mathrm{supp}K$, and $\mathrm{supp}K=C$.

Let $\psi \in \mathcal{D}(Q)$. By step 3.1, $K\ast h$ is continuous, hence locally integrable, so $⟨K\ast h,\psi ⟩$ is the regular-functional pairing ${\int }_{(x,t)}(K\ast h)\psi$ of the formula in step 1.2. For fixed $(x,t)$ the inner integral defining $K\ast h$ is absolutely convergent by step 1.2, and inserting it gives a triple integral of $h(y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)\psi (x,t)(4\pi \mathrm{\mid }x-y\mathrm{\mid }{)}^{-1}$ over $(y,x,t)$. This triple integral is absolutely convergent: with $\mathrm{supp}h\subseteq {\bar{B}}_{\rho }(0)\times [-\rho ,\rho ]$ and $\mathrm{supp}\psi \subseteq {\bar{B}}_M(0)\times [-M,M]$ it is bounded by $\frac{\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}}{4\pi }{\int }_{[-M,M]}{\int }_{{\bar{B}}_{\rho }}{\int }_{B_M}\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dxdydt<\mathrm{\infty }$, the inner double integral being finite by the polar formula. Fubini therefore lets us integrate first in $(x,t)$ for fixed $y$; the substitution $(x,t)\mapsto (z,s)=(x-y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)$ is a $C^1$ diffeomorphism from $\{(x,t):x\mathrm{\ne }y\}$ onto $({\mathbb{R}}^3\setminus \{0\})\times \mathbb{R}$ with determinant one, the omitted set $\{x=y\}$ being null, so the $L^1$ change-of-variables identity gives ${\int }_{(x,t)}\frac{h(y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)\psi (x,t)}{4\pi \mathrm{\mid }x-y\mathrm{\mid }}dxdt={\int }_{(z,s)}\frac{h(y,s)\psi (y+z,s+\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)}{4\pi \mathrm{\mid }z\mathrm{\mid }}dzds$. Fubini once more and the definition of $K⊛\psi$ yield $⟨K\ast h,\psi ⟩={\int }_{(y,s)}h(y,s)(K⊛\psi )(y,s)dyds=⟨h,K⊛\psi ⟩$.

Let $\eta \in C_c^{\mathrm{\infty }}(Q)$ satisfy $\eta \ge 0$, $\int \eta =1$ and $\mathrm{supp}\eta \subseteq {\bar{B}}_1(0)$: start from [F24] with compact set $\{0\}$ inside $B_1(0)$, giving a cutoff equal to one on a neighbourhood of the origin and supported in $B_1(0)$, which is nonnegative with positive integral over a ball of positive measure, and normalize it. Let ${\eta }_{\epsilon }(u)={\epsilon }^{-4}\eta (u\mathrm{/}\epsilon )$ be the mollifier family, an $L^1$ approximate identity under [A1]. Fix $\psi \in \mathcal{D}$. Since $W(K\ast {\eta }_{\epsilon })={\eta }_{\epsilon }$ pointwise on $Q$ by step 3.1 and $K\ast {\eta }_{\epsilon }$ is $C^{\mathrm{\infty }}$, the classical-compatibility clause of the distributional calculus identifies the distributional $W$ of the regular functional $u_{K\ast {\eta }_{\epsilon }}$ with $u_{W(K\ast {\eta }_{\epsilon })}=u_{{\eta }_{\epsilon }}$; and the pairing identity of step 4.1 gives $⟨K\ast {\eta }_{\epsilon },\psi ⟩=⟨{\eta }_{\epsilon },K⊛\psi ⟩$. With $F=K⊛\psi$, bounded and continuous at the origin with $F(0,0)=⟨K,\psi ⟩$ by steps 1.3 and 2.2, the unit mass and support $\mathrm{supp}{\eta }_{\epsilon }\subseteq {\bar{B}}_{\epsilon }$ give $\mathrm{\mid }⟨{\eta }_{\epsilon },F⟩-F(0,0)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }u\mathrm{\mid }\le \epsilon }\mathrm{\mid }F(u)-F(0,0)\mathrm{\mid }\to 0$ by continuity at the origin; thus $⟨K\ast {\eta }_{\epsilon },\psi ⟩\to F(0,0)=⟨K,\psi ⟩$ without requiring global continuity of $F$. Thus $K\ast {\eta }_{\epsilon }\to K$ in the weak distribution topology; since $W$ is continuous linear for that topology and $⟨{\eta }_{\epsilon },\psi ⟩\to \psi (0,0)=⟨{\delta }_{(0,0)},\psi ⟩$ by the same approximate-identity argument applied to the bounded continuous $\psi$, we obtain $⟨WK,\psi ⟩={\lim }_{\epsilon }⟨W(K\ast {\eta }_{\epsilon }),\psi ⟩={\lim }_{\epsilon }⟨{\eta }_{\epsilon },\psi ⟩=\psi (0,0)$. As $\psi$ was arbitrary, $WK={\delta }_{(0,0)}$.