Proposition: Zero-switch identity and a nonzero algebra witness

At $g=0$,

while the bare observable remains

It differs from $\hat{\varphi }(f)$ in general.

Proof. The map is zero because ${\mathsf{L}}_{1,g}=0$, independently of the insertion test. The coordinate formulas do not contain $g$.

For an explicit witness, take $m=0$, choose $f\ge 0$, $f\not\equiv 0$, and let $t_{-}=\min \{t:(t,\mathbf{x})\in K\}>0$. Choose a nonnegative nonzero $\psi \in {C_c}^{\infty }((a,b))$ with $0<a<b<t_{-}$. Choose a smooth compact spatial cutoff equal to one near the spatial projection of ${\mathcal{D}}_f$, and use it to extend $h=\psi (t)\eta$ to a compact smooth background.

Finite propagation makes this cutoff irrelevant on the observation region. There,

Since ${\sigma }_{\epsilon }>0$ for $s>0$, $\Psi >0$ on $K$. Thus $c_1=-{\partial }_t(f\Psi )$, and against the smooth free solution $u=t$,

If $c_1=P_{0}v$, $v\in \mathcal{D}(M)$, this pairing would be zero. More generally $F_{c_1}$ cannot lie in ${\mathfrak{J}}_0$, whose elements vanish on smooth free solutions. Hence $\hat{\varphi }(c_1)\ne 0$.