Theorem: Linear domain criterion

For $a\in {C_c}^{\infty }(\{t>0\};T^{\ast }M)$, the following are equivalent:

1. ${\ell }_a\in {\mathcal{P}}_{\mathrm{mc}}$;

2. $q_a=L^{\ast }a$ is smooth;

3. the advanced solution $b=E_{\mathrm{A}}a$ has vanishing full Cauchy data on ${\Sigma }_0$:

   $$b{|}_{{\Sigma }_0}=0,{\partial }_{t}b{|}_{{\Sigma }_0}=0.$$

When these conditions hold, $q_a$ is a smooth compactly supported test tensor.

Proof. For a linear functional, domain membership is exactly the condition $\mathrm{WF}(q_a)\subset {\Xi }_1$. By (eq:q-wf-upper), every possible singular covector of $q_a$ is timelike. Hence domain membership is equivalent to smoothness.

Suppose $q_a$ is smooth. Its delta coefficient must vanish: a distribution supported on the surface cannot cancel a locally bounded step term. Lemma lem:B-injective gives $b{|}_{{\Sigma }_0}=0$. Smoothness of the remaining distribution $\theta C^{\ast }b$ implies

Since the trace of $b$ vanishes identically on the surface, all of its tangential derivatives vanish there. Formula (eq:C-adjoint) then gives

Injectivity again yields ${\partial }_{t}b{|}_{{\Sigma }_0}=0$.

Conversely, choose $ϵ>0$ such that $a=0$ for $t<ϵ$. On that region $b$ solves the homogeneous wave equation. Vanishing Cauchy data imply, by uniqueness, that $b=0$ for $t<ϵ$. Thus $\theta b=b$ is smooth, and $q_a=C^{\ast }b$ is smooth. Its compact support follows from (eq:q-support).