Lemma: The regulated linear kernel is smooth

For $a\in {C_c}^{\infty }({M_T}^{\circ };T^{\ast }M)$ define ${\mathsf{J}}_{\epsilon }(a)[h]={\int }_{M_T}{a}_{\mu }{{\mathsf{X}}_1}^{\mu }[h]{\mathrm{d}}^{4}x$. Then

The kernel $q_{a,\epsilon }$ is smooth, vanishes on $\{t\le 0\}$, and is supported in $J^{-}(\mathrm{supp}a)\cap \{t\ge 0\}$.

By contrast, the multiplied-equation prescription ${\tilde{\mathsf{X}}}_1={\mathsf{E}}_0({\sigma }_{\epsilon }C[h])$ has kernel $C^{\ast }({\sigma }_{\epsilon }{\mathsf{E}}_{\mathrm{A}}a)={\sigma }_{\epsilon }{C}^{\ast }({\mathsf{E}}_{\mathrm{A}}a)+{\sigma }_{\epsilon }^{\prime }\mathsf{B}({\mathsf{E}}_{\mathrm{A}}a)$, with ${\mathsf{B}}^{\rho \sigma }(v)=-n^{(\rho }{v}^{\sigma )}+\frac{1}{2}{\eta }^{\rho \sigma }{n}_{\mu }{v}^{\mu }$, $n_{\mu }={\partial }_{\mu }t$. The two prescriptions differ by the switch-derivative term ${\sigma }_{\epsilon }^{\prime }\mathsf{B}({\mathsf{E}}_{\mathrm{A}}a)$. Neither is identified with the sharp kernel $\theta C^{\ast }b+\delta (t)\mathsf{B}(b{|}_{\Sigma })$ of VII-B, and no $\epsilon \to 0$ limit is taken here.

Proof. By Lemma lem:Dexp the first coefficient equation is $\square {{\mathsf{X}}_1}^{\mu }=-{\partial }_{\alpha }({\sigma }_{\epsilon }{a_1}^{\alpha \mu }[h])$ with zero data, so ${{\mathsf{X}}_1}^{\mu }=-{\mathsf{E}}_0{\partial }_{\alpha }({\sigma }_{\epsilon }{a_1}^{\alpha \mu }[h])$. Lemma lem:transpose with $b={\mathsf{E}}_{\mathrm{A}}a$ gives

The boundary term at $t=0$ vanishes because ${\sigma }_{\epsilon }(0)=0$; at $t=T$ because $b$ vanishes there; spatially because $b$ is compactly supported in each slice. Inserting (eq:a),

Smoothness and support follow from those of $b$ and ${\sigma }_{\epsilon }$. For the multiplied-equation prescription, $\int a{\mathsf{E}}_0({\sigma }_{\epsilon }Ch)=\int {\sigma }_{\epsilon }bCh=\int C^{\ast }({\sigma }_{\epsilon }b)h$, and the product rule with ${\partial }_{\mu }{\sigma }_{\epsilon }=n_{\mu }{\sigma }_{\epsilon }^{\prime }$ gives the displayed decomposition, exactly as in the VII-B adjoint computation with $\theta$ replaced by ${\sigma }_{\epsilon }$.