Theorem: Criterion for a descending extension

Let $Z\subset X$ be relatively closed and $T^{\circ }\in {\mathcal{D}}^{\prime }(X^{\circ })$. The following are equivalent.

1. There is an extension $T\in {\mathcal{D}}^{\prime }(X)$ of $T^{\circ }$ with $\mathrm{Ann}(T)=\mathrm{Ann}(T^{\circ })$ (equivalently, by Theorem thm:ann, ${d_0}^T=0$).

2. For every compact $K\subset X$ there are $k_K\ge 0$ and $C_K<\infty$ such that

   $$|⟨T^{\circ },\varphi ⟩|\le C_K{p}_{k_K}(\varphi +g)$$

   for all $\varphi \in \mathcal{D}(X^{\circ })$ and $g\in \mathrm{Ann}(T^{\circ })$ with $\mathrm{supp}\varphi ,\mathrm{supp}g\subset K$.

Proof. (i)$\Rightarrow$(ii). Let $T$ be as in (i) and $K$ compact; let $k,C$ be an order and constant for $T$ on tests supported in $K$. For $\varphi ,g$ as in (ii), $⟨T,g⟩=⟨gT,\chi ⟩=0$ with $\chi \in \mathcal{D}(X)$ equal to one on $\mathrm{supp}g$, so $|⟨T^{\circ },\varphi ⟩|=|⟨T,\varphi +g⟩-⟨T,g⟩|=|⟨T,\varphi +g⟩|\le C{p}_k(\varphi +g)$.

(ii)$\Rightarrow$(i). Fix a compact $K$ and let ${\mathcal{D}}_K=\{\psi \in \mathcal{D}(X):\mathrm{supp}\psi \subset K\}$, $S_K=\{\varphi +g\}\subset {\mathcal{D}}_K$ with $\varphi ,g$ as in (ii). Define $\ell (\varphi +g)=⟨T^{\circ },\varphi ⟩$. This is well defined: if $\varphi +g={\varphi }^{\prime }+g^{\prime }$ then $h=\varphi -{\varphi }^{\prime }=g^{\prime }-g\in \mathcal{D}(X^{\circ })\cap \mathrm{Ann}(T^{\circ })$, and $⟨T^{\circ },h⟩=⟨h{T}^{\circ },\chi ⟩=0$ for $\chi \in \mathcal{D}(X^{\circ })$ equal to one on $\mathrm{supp}h$. By (eq:relest), $|\ell |\le C_K{p}_{k_K}$ on $S_K$. The Hahn–Banach theorem for functionals dominated by a seminorm extends $\ell$ to a linear $B_K$ on ${\mathcal{D}}_K$ with $|B_K|\le C_K{p}_{k_K}$.

Choose a locally finite cover of $X$ by relatively compact open sets $V_i$, compacts $K_i\supset \overline{V_i}$ in $X$, and a smooth partition of unity $\{{\chi }_i\}$ with $\mathrm{supp}{\chi }_i\subset V_i$. Define $⟨T,\varphi ⟩={\sum }_i{B}_{K_i}({\chi }_i\varphi )$ for $\varphi \in \mathcal{D}(X)$; the sum is finite and $T$ is a distribution. If $\varphi \in \mathcal{D}(X^{\circ })$ then ${\chi }_i\varphi \in \mathcal{D}(X^{\circ })$, so $B_{K_i}({\chi }_i\varphi )=\ell ({\chi }_i\varphi )=⟨T^{\circ },{\chi }_i\varphi ⟩$, and summing gives $T{|}_{X^{\circ }}=T^{\circ }$. If $g\in \mathrm{Ann}(T^{\circ })$ and $\varphi \in \mathcal{D}(X)$, then ${\chi }_{i}g\varphi \in \mathrm{Ann}(T^{\circ })$ with support in $K_i$, so $B_{K_i}({\chi }_{i}g\varphi )=\ell (0+{\chi }_{i}g\varphi )=0$; hence $⟨gT,\varphi ⟩=0$ and $g\in \mathrm{Ann}(T)$. Thus $\mathrm{Ann}(T^{\circ })\subseteq \mathrm{Ann}(T)$, and the reverse inclusion holds for every extension.