Proposition: The opposite sign is detected

Put ${{\Phi }^{(2)}}_{+}:={{\mathcal{A}}_2}^{\mathrm{sc}}[f,g](\gamma ,0;\varphi )+{\mathcal{C}}_{3,\mathrm{D}}(f,g)[\gamma ,\varphi ]$. Then

Under the causal plateau condition and $A\gamma =0$ on $U$ this is $-2{\mathcal{A}}_1(s_{\xi };\varphi )$, and there are admissible data on which it is nonzero.

Proof. Lemma lem:heldfixed with $k=0$ gives $-{\mathcal{A}}_1(s_{\xi })$ for the first summand; XII-D's ledger (eq:ledgerD) gives ${\Delta }_{\xi }{\mathcal{C}}_{3,\mathrm{D}}=-{\mathcal{A}}_1(s_{\xi })+{\mathcal{A}}_1(K{\rho }_{\xi })-{\mathcal{A}}_1(e_{\xi })$. Add. Under the two hypotheses the last two terms vanish (XII-D's causal plateau proposition and $A\gamma =0$ near $\mathrm{supp}\xi$).

For nonvanishing, take XI-E's evaluated data XIE: $m=0$, $\gamma =t\eta$, $\varphi =t^2+\frac{1}{3}|\mathbf{x}{|}^2$ (so $\square \varphi =0$), ${\xi }_{\lambda }=\lambda a(t){\partial }_t$ with $a\ge 0$, $a\not\equiv 0$, compactly supported above the ramp, in a causal cylinder with $g=1$ and a nonnegative observation after $\mathrm{supp}a$; a compact spatial cutoff and plateau make ${\xi }_{\lambda }$ admissible without changing the evaluation, as XI-E records. For $\gamma =t\eta$, $A\gamma =0$ identically. XI-E computes $I:={\mathcal{A}}_1({\mathcal{L}}_{a{\partial }_t}(t\eta );\varphi )>0$. Since $s_{{\xi }_{\lambda }}=\lambda {\mathcal{L}}_{a{\partial }_t}(t\eta )+\frac{{\lambda }^2}{2}{\mathcal{L}}_{a{\partial }_t}K(a{\partial }_t)$, linearity gives ${\mathcal{A}}_1(s_{{\xi }_{\lambda }};\varphi )=\lambda I+{\lambda }^2{I}^{\prime }$ with $I^{\prime }$ independent of $\lambda$, which is nonzero for all sufficiently small $\lambda \ne 0$.