Lemma: Boundary injectivity and noncancellation

The map $v\mapsto \mathcal{B}(v)$ is injective. If $b(0,{\mathbf{x}}_{\ast })\ne 0$, at least one component of $q_a$ has both nonzero normal directions in its wavefront set at $(0,{\mathbf{x}}_{\ast })$.

Proof. With lower components $v_{\mu }$, direct substitution gives

These identities prove injectivity.

Choose a component with nonzero delta coefficient near the point. After localization it has the form

with smooth coefficients and $v({\mathbf{x}}_{\ast })\ne 0$. The Fourier transform of its delta term is independent of the normal frequency and nonzero at some fixed tangential frequency. Integration by parts in $t$ makes the step term $O(|\tau {|}^{-1})$. It cannot cancel the nonzero order-zero term as $\tau \to +\infty$ or $\tau \to -\infty$. These frequency rays lie in arbitrarily narrow cones around the two normals, proving the assertion.