Theorem: Bridge from the boundary-trace certificate alone

Assume (TC), the copied sharp adjoint, the copied Fourier-defined kernel properties, and unrestricted configurations. Then (SB) holds.

Proof. Theorem thm:sphere gives the initial null spheres and $F\cap \Sigma$ as lower supports. Proposition (eq:Wexact) gives ${\mathrm{supp}}_{\mathrm{fun}}{W}_g=G$.

For $t>0$, every observation pair in

satisfies $t-s\le R\le t+s$. Its initial sphere therefore meets the initial ball of $y$. The corresponding actual support point of $B_f$ lies in $J^{-}(y)$, contradicting $B_f\gtrsim W_g$. For $t=0$, the closed slab reduces to $R=s$; the lower support $x\in F\cap \Sigma$ gives the same contradiction.

Thus $F\times G$ misses the closed commutator slab, including its direct cone, reflected boundary, and spatial apex. The copied Fourier support statement and its boundary continuation imply $fg\mathrm{Im}K=0$. No scalar controllability assumption is needed for this narrower conclusion.