Corollary: One dimension: every extension descends

Let $d=1$, $Z=\{z_0\}$, and suppose $z_0\in \overline{\mathrm{supp}{T}^{\circ }}$ (closure in $X$). Then every $g\in \mathrm{Ann}(T^{\circ })$ is flat at $z_0$, and consequently $\mathrm{Ann}(T)=\mathrm{Ann}(T^{\circ })$ and ${d_0}^T=0$ for every extension $T$ of $T^{\circ }$.

Proof. If $g\in \mathrm{Ann}(T^{\circ })$ and $g(p)\ne 0$ for some $p\in \mathrm{supp}{T}^{\circ }$, then $g$ is invertible near $p$ and $T^{\circ }=g^{-1}(g{T}^{\circ })=0$ near $p$, a contradiction; so $g=0$ on $\mathrm{supp}{T}^{\circ }$. By hypothesis there are points $x_n\in \mathrm{supp}{T}^{\circ }$, $x_n\to z_0$, $x_n\ne z_0$; passing to a subsequence on one side of $z_0$, $g$ has infinitely many zeros accumulating at $z_0$, so $g(z_0)=0$ and, by Rolle's theorem applied inductively, $g^{(j)}(z_0)=0$ for all $j$. Lemma lem:flat gives $gT=0$ for every extension $T$.