Theorem: Supported coefficient-extension equivalence

Suppose $t^{\circ }\in {\mathcal{D}}^{\prime }(U\setminus A)$ is supported on $S\setminus A$ and locally has supported normal order at most $r$. Its unique normal-jet coefficients glue to distributions

The following are equivalent:

1. there exists $t\in {\mathcal{D}}^{\prime }(U)$ supported on $S$, of supported normal order at most $r$, with

   $$t{|}_{U\setminus A}=t^{\circ };$$

2. every ${b_{\gamma }}^{\circ }$ extends to some $b_{\gamma }\in {\mathcal{D}}^{\prime }(X)$.

For any chosen coefficient extensions,

is such a realization. If realizations exist, all of them form an affine space over

Proof. For every $x\in X\setminus Z$, choose a product neighborhood of $(x,0)$ disjoint from $A$. Theorem thm:normal gives its coefficients there. Uniqueness makes these coefficients agree on overlaps, so they glue on $X\setminus Z$.

If (i) holds, extract the coefficients of $t$ by Theorem thm:normal. On every such smaller product neighborhood, uniqueness identifies their restrictions with the coefficients of $t^{\circ }$. This proves (ii).

If (ii) holds, (eq:jetrealization) is a distribution supported on $S$ with the stated normal order. Near every point of $S\setminus A$, its coefficients agree with those of $t^{\circ }$, so the distributions agree by Theorem thm:normal. Away from $S$ both distributions vanish. These local identities establish equality on $U\setminus A$, proving (i).

Finally, two extensions of a coefficient differ by a distribution supported on $Z$. Conversely, any such difference changes no restriction on $U\setminus A$. Coefficient uniqueness proves that (eq:jetambiguity) is the full ambiguity under the stated support and normal-order conditions.