Lemma: Jet-vanishing annihilators produce no defect

Let $T$ extend $T^{\circ }$. Let $K$, $\varrho$, $K^{\prime }$ be as in Lemma lem:taylor, and let $k\ge 0$, $C_T$ be such that $|⟨T,\psi ⟩|\le C_T{p}_k(\psi )$ for all $\psi \in \mathcal{D}(X)$ with $\mathrm{supp}\psi \subset K^{\prime }$, where $p_k(\psi )={\max }_{|\alpha |\le k}\Vert {\partial }^{\alpha }\psi {\Vert }_{\infty }$. If $g\in \mathrm{Ann}(T^{\circ })$ has vanishing $k$-jet at every point of $Z$, then $⟨gT,\varphi ⟩=0$ for every $\varphi \in \mathcal{D}(X)$ with $\mathrm{supp}\varphi \subset K$. Consequently, if $g\in \mathrm{Ann}(T^{\circ })$ is flat on $Z$, then $gT=0$, i.e. $g\in \mathrm{Ann}(T)$.

Proof. Put $\psi =g\varphi$; its $k$-jet vanishes on $Z$ and $\mathrm{supp}\psi \subset K$. Let ${\theta }_{\epsilon }$ be as in Lemma lem:cutoff with $3\epsilon <\varrho$. Then ${\theta }_{\epsilon }\psi \in \mathcal{D}(X^{\circ })$, so

Moreover $(1-{\theta }_{\epsilon })\psi$ is supported in $K\cap \{\mathrm{dist}(\cdot ,Z)\le 3\epsilon \}$, and for $|\alpha |\le k$, by the Leibniz rule, Lemma lem:cutoff, and Lemma lem:taylor applied to $\psi$,

Hence $p_k((1-{\theta }_{\epsilon })\psi )\to 0$, and the order-$k$ estimate for $T$ on $K^{\prime }$ gives $⟨T,\psi ⟩={\lim }_{\epsilon \to 0}⟨T,{\theta }_{\epsilon }\psi ⟩=0$. For flat $g$ every $k$ is admissible and every $\varphi \in \mathcal{D}(X)$ has support in some compact $K$, so $gT=0$.