Proposition: Quotient failure at order one

For compact $\xi$ with $K\xi \ne 0$, the ordinary coupled operation does not preserve the weighted free-equation ideal, even through total order $s^2$. Indeed,

and this is nonzero for some compact $g$.

Proof. Formal self-adjointness gives (eq:quotient). If $\mathcal{B}K\xi =0$, then the compactly supported section $K\xi$ solves the normally hyperbolic equation $\mathcal{L}K\xi =0$. Its zero Cauchy data before its support imply $K\xi =0$, contrary to the hypothesis. Nondegeneracy of compact testing supplies $g$. The scalar coefficient is of order $s$, not $s^3$.