Lemma: Joint-kernel control

The kernel $b_f$ satisfies

More precisely, its possible singular branches arise from:

1. the bulk diagonal $x=z$;

2. direct null propagation from $z$ to $x$;

3. $z^0=0$ together with null propagation, with covectors

   $$(\xi ,-\xi +\tau \mathrm{d}t);$$

4. pure boundary conormals

   $$(0,\tau \mathrm{d}t).$$

This is an upper bound, not an assertion that every listed covector occurs.

Proof. The retarded wave kernel has diagonal conormals and, away from the diagonal, the null propagation relation $(x,\xi ;z,-\xi )$. Multiplication by $\theta (z^0)$ is permitted: a wavefront covector of that factor has zero $x$-component, whereas a wavefront covector of the retarded kernel has nonzero $x$-component. Hence opposite covectors cannot sum to zero.

The product theorem yields the direct branches, the boundary sums $(\xi ,-\xi +\tau \mathrm{d}t)$, and the pure boundary conormals. At $z^0=0$, the bulk diagonal does not meet the support of $f$, because $f$ is supported strictly in $M_{+}$.

Integrating the first-order operator $C$ by parts in $z$ gives the kernel of $L$. This can produce a $\delta (z^0)$, but not a ${\delta }^{\prime }(z^0)$. Finally, moving ${\partial }_{\mu }$ from $\varphi$ onto the $x$-kernel and multiplying by $f(x)$ do not enlarge the wavefront set. This proves the assertion.