Lemma: The display for R*R

$R\ast R$ is well defined and equals $\mathcal{B}=\frac{1}{8\pi }\Theta (t)\Theta (t^2-|\mathbf{x}{|}^2)=\frac{1}{8\pi }\Theta (t-|\mathbf{x}|)$ almost everywhere. Hence (eq:RR) holds, $\mathcal{B}\in L^{\infty }({\mathbb{R}}^4)$, $\mathrm{supp}\mathcal{B}=J^{+}$, $\square \mathcal{B}=R$, and $\mathcal{B}$ is constant on the interior of $J^{+}$. On $\{t>0\}$, ${\partial }_{\alpha }\mathcal{B}=\frac{1}{4\pi }{x}_{\alpha }\delta (x^2)$, where $x_{\alpha }={\eta }_{\alpha \nu }{x}^{\nu }$ and $x^2=t^2-|\mathbf{x}{|}^2$.

Proof. $R\ast R$ exists since both factors are supported in $J^{+}$, and $\mathrm{supp}(R\ast R)\subseteq J^{+}+J^{+}=J^{+}$, $\square (R\ast R)=(\square R)\ast R=R$. For $\mathcal{B}$: ${\mathcal{F}}_{\mathbf{x}}\mathcal{B}(t,\mathbf{p})=\frac{\Theta (t)}{8\pi }{\int }_{|\mathbf{x}|<t}{e}^{-i\mathbf{p}\cdot \mathbf{x}}{\mathrm{d}}^3\mathbf{x}=\frac{\Theta (t)}{8\pi }\cdot \frac{4\pi }{\rho }{{\int }_0}^{t}r\sin (\rho r)\mathrm{d}r=\Theta (t)\frac{\sin (\rho t)-\rho t\cos (\rho t)}{2{\rho }^3}=:\Theta (t)u_2(t)$, with $u_2(0)=0$, $u_2^{\prime }(t)=t\sin (\rho t)/(2\rho )$, $u_2^{\prime }(0)=0$, and $u_2^{\prime \prime }+{\rho }^2{u}_2=\sin (\rho t)/\rho$. Hence $({{\partial }_t}^2+{\rho }^2){\mathcal{F}}_{\mathbf{x}}\mathcal{B}={\mathcal{F}}_{\mathbf{x}}R$ (no boundary terms; at $\rho =0$, $u_2=t^3/6$), i.e.\ $\square \mathcal{B}=R$. Then $v=\mathcal{B}-R\ast R$ satisfies $\square v=0$, $\mathrm{supp}v\subseteq J^{+}$, so $v=0$ by Lemma lem:R(iii). The last formula is the chain rule for $\Theta (x^2)$ on $\{t>0\}$, where $x\mapsto x^2$ has nonvanishing differential away from $x=0$.