Lemma: Convergent subtracted spectral kernel

For either bounded density $r_i$, the momentum-space integral

exists in ${\mathcal{S}}^{\prime }({\mathbb{R}}^4)$. Its inverse Fourier transform is supported in ${\overline{V}}_{+}$.

Proof. For fixed $u>0$, the first multiplier is the Fourier transform of the retarded Klein–Gordon fundamental solution of mass $\sqrt{u}$. The second is the transform of $-{\delta }_0/u$, so their difference has future causal support.

We verify convergence on Schwartz tests. On $|p{|}_{\mathrm{E}}\le \sqrt{u}/2$, the denominator is bounded away from zero, and subtraction gives a bound $C|p{|}_{\mathrm{E}}^2/u^2$. Its test pairing is therefore $O(u^{-2})$.

On the complementary region, the poles occur at $p^0=\pm \sqrt{u+|\mathbf{p}{|}^2}$. Partial fractions reduce them to one-dimensional principal values and delta terms, with prefactor $1/(2\sqrt{u+|\mathbf{p}{|}^2})$. The principal-value estimate obtained by subtracting the test value at the pole uses only a test and its first derivative. After the rescaling $p=\sqrt{u}k$, Schwartz decay on $|k{|}_{\mathrm{E}}\ge 1/2$ dominates the Jacobian and these finite-order distribution bounds. Choosing a sufficiently high, fixed Schwartz seminorm makes this region $O(u^{-2})$ as well. Thus the whole pairing is bounded by $C{u}^{-2}{p}_N(\psi )$ for some finite Schwartz seminorm $p_N$, uniformly for large $u$.

Since $r_i$ is bounded, integration converges in ${\mathcal{S}}^{\prime }$. Its finite spectral integrals have future causal support, and that support property is preserved by distributional convergence.