Point-extension counterterms form a finite delta-jet space

By [F2], choose $T_0\in {\mathcal{E}}_{\delta }(t)$. For any $T\in {\mathcal{E}}_{\delta }(t)$ the difference $S=T-T_0$ vanishes on every test supported away from $0$, so $\mathrm{supp}S\subseteq \{0\}$ by [F6]. By [F3], $S={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{c}_{\alpha }{\mathrm{\partial }}^{\alpha }{\delta }_0$ for some finite $m$.

For $S\mathrm{\ne }0$, let $m$ be the largest order with a nonzero coefficient, and choose $\alpha$ of order $m$ with $c_{\alpha }\mathrm{\ne }0$. Choose $\eta \in \mathcal{D}({\mathbb{R}}^n)$ equal to one near $0$ by [F5], and test with ${\psi }_{\alpha }(x)=\eta (x)x^{\alpha }\mathrm{/}\alpha !$. The product rule gives ${\mathrm{\partial }}^{\beta }{\psi }_{\alpha }(0)=0$ for all $\mathrm{\mid }\beta \mathrm{\mid }\le m$ except $\beta =\alpha$, where it equals $1$. Consequently $⟨S_{\rho },{\psi }_{\alpha }⟩=(-1{)}^m{c}_{\alpha }{\rho }^{-n-m}\mathrm{.}$ For every test $\phi$, [F4] likewise gives $⟨S_{\rho },\phi ⟩={\sum }_{\beta }{c}_{\beta }(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }}{\rho }^{-n-\mathrm{\mid }\beta \mathrm{\mid }}{\mathrm{\partial }}^{\beta }\phi (0)$. The finite sum tends to zero after multiplication by ${\rho }^{\omega }$ for every $\omega >n+m$, while the chosen test prevents convergence to zero for $\omega <n+m$. Hence ${\mathrm{sd}}_0(S)=n+m$.

Since $T$ and $T_0$ both have degree $\delta$, for every $\omega >\delta$ both ${\rho }^{\omega }{T}_{\rho }$ and ${\rho }^{\omega }(T_0{)}_{\rho }$ tend to zero. Thus ${\mathrm{sd}}_0(S)\le \delta$. Step 2.1 implies that every nonzero coefficient in $S$ has order at most $\lfloor \delta -n\rfloor$; if $\delta <n$, no nonzero $S$ is possible.

Conversely, let $S$ be any finite sum of ${\mathrm{\partial }}^{\alpha }{\delta }_0$ with $\mathrm{\mid }\alpha \mathrm{\mid }\le k$ when $\delta \ge n$. Step 2.1 gives ${\mathrm{sd}}_0(S)\le n+k\le \delta$, and $T_0+S$ still restricts to $t$. Hence ${\mathrm{sd}}_0(T_0+S)\le \delta$. Restriction to tests supported in ${\mathbb{R}}^n\setminus \{0\}$ gives ${\mathrm{sd}}_0(T_0+S)\ge {\mathrm{sd}}_0(t)=\delta$, so equality holds. Uniqueness of the coefficients in [F3] identifies the displayed finite-dimensional affine space. For $\delta =n$, $k=0$, giving only $c{\delta }_0$.