Lemma: The advanced kernel

Let $m\ge 0$, $P_m=\square +m^2$, $\omega =\sqrt{{\rho }^2+m^2}$. The tempered distribution ${E^{\mathrm{A}}}_m$ with

satisfies $P_m{E}_m^{\mathrm{A}}={\delta }_0$ and $\mathrm{supp}{E^{\mathrm{A}}}_m\subseteq J^{-}$, and it is the unique fundamental solution of $P_m$ supported in a half-space $\{t\le c\}$. For $m=0$, ${E^{\mathrm{A}}}_0(t,\mathbf{x})=R(-t,\mathbf{x})=\frac{\Theta (-t)}{2\pi }\delta (t^2-|\mathbf{x}{|}^2)$. For $f\in {C^{\infty }}_c({\mathbb{R}}^4)$, ${E^{\mathrm{A}}}_m\ast f\in C^{\infty }({\mathbb{R}}^4)$ with support in $\mathrm{supp}f+J^{-}$.

Proof. With $u(t)=\sin (\omega t)/\omega$, $u(0)=0$, $u^{\prime }(0)=1$: ${\partial }_t(-\Theta (-t)u)=\delta (t)u(0)-\Theta (-t)u^{\prime }=-\Theta (-t)u^{\prime }$, ${{\partial }_t}^2(-\Theta (-t)u)=\delta (t)u^{\prime }(0)-\Theta (-t)u^{\prime \prime }$, so $({{\partial }_t}^2+{\omega }^2)(-\Theta (-t)u)=\delta (t)-\Theta (-t)(u^{\prime \prime }+{\omega }^{2}u)=\delta (t)$; hence $P_m{E}_m^{\mathrm{A}}={\delta }_0$. Support: fix $t<0$. The function $\mathbf{p}\mapsto \sin (\omega t)/\omega$ is even in $\omega$, hence an entire function of ${\omega }^2=\mathbf{p}\cdot \mathbf{p}+m^2$, hence entire on ${\mathbb{C}}^3$, and $|\sin (\omega t)/\omega |=|t||\sin (\omega t)/(\omega t)|\le |t|e^{|t||\mathrm{Im}\omega |}$ since $\sin z/z={{\int }_0}^1\cos (sz)\mathrm{d}s$. For $\mathbf{p}=\mathbf{a}+i\mathbf{b}$, ${\omega }^2=|\mathbf{a}{|}^2-|\mathbf{b}{|}^2+m^2+2i\mathbf{a}\cdot \mathbf{b}$ and $(\mathrm{Im}\omega {)}^2=\frac{1}{2}(|{\omega }^2|-\mathrm{Re}{\omega }^2)\le |\mathbf{b}{|}^2$, because $|{\omega }^2{|}^2\le (|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+m^2{)}^2$ (the difference of the squares is $4|\mathbf{b}{|}^2{m}^2\ge 0$). By Paley–Wiener–Schwartz Horm (Thm. 7.3.1), the inverse transform in $\mathbf{p}$ is supported in $\{|\mathbf{x}|\le |t|\}$. If $\varphi \in \mathcal{D}({\mathbb{R}}^4)$ vanishes near $J^{-}$, then for each $t$, $\varphi (t,\cdot )$ vanishes near the ball of radius $|t|$, so $⟨{E^{\mathrm{A}}}_m,\varphi ⟩=0$. Uniqueness: as in Lemma lem:R(iii) with $J^{-}$ and $\{t\le c\}$. For $m=0$, ${\mathcal{F}}_{\mathbf{x}}[R(-t,\mathbf{x})]=\Theta (-t)\sin (-\rho t)/\rho$, which is (eq:EA). The last sentence: convolution of a distribution with a test function is smooth, and $\mathrm{supp}({E^{\mathrm{A}}}_m\ast f)\subseteq \mathrm{supp}{E^{\mathrm{A}}}_m+\mathrm{supp}f$.