Every finite-scaling-degree punctured distribution has an exact-degree extension

Choose $\chi \in \mathcal{D}({\mathbb{R}}^n)$ equal to one near $0$ by [F3], and put $f(x)=\chi (x)-\chi (2x)$. Its support $K$ is compact and avoids $0$, so choose $0<a<b<\mathrm{\infty }$ with $K\subset \{x:a\le \mathrm{\mid }x\mathrm{\mid }\le b\}$. If $\delta <n$, set $k=-1$ and $W\phi =\phi$. If $\delta \ge n$, let $k$ be the integer with $k\le \delta -n<k+1$, put $P_k\phi (x)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le k}{x}^{\alpha }{\mathrm{\partial }}^{\alpha }\phi (0)\mathrm{/}\alpha !$, and define $W\phi =\phi -\chi P_k\phi$.

For $\phi \in \mathcal{D}({\mathbb{R}}^n)$ define $T(\phi ):=⟨t,(1-\chi )W\phi ⟩+{\sum }_{j=0}^{\mathrm{\infty }}⟨t,f(2^j\cdot )W\phi ⟩\mathrm{.}$ Each pairing is with a test supported in $U$: the first factor vanishes near $0$, and $f(2^j\cdot )$ is supported in $2^{-j}K$.

Choose $\omega$ with $\delta <\omega <n+k+1$; this interval is nonempty by the choice of $k$. Put $\lambda =2^{-j}$. By the dilation convention [F1], $⟨t,f(\cdot \mathrm{/}\lambda )W\phi ⟩={\lambda }^n⟨t_{\lambda },fW\phi (\lambda \cdot )⟩\mathrm{.}$ The uniform estimate [F2] bounds this by $C{\lambda }^{n-\omega }{p}_m(fW\phi (\lambda \cdot ))$ for fixed $m,C$. When $k\ge 0$ and $\lambda$ is small enough that $\chi =1$ on $\lambda K$, [F4] gives $p_m(fW\phi (\lambda \cdot ))\le C_{\phi }{\lambda }^{k+1}$. Indeed, after each derivative of order $q\le m$, the Taylor remainder contributes order ${\lambda }^{k+1-q}$ if $q\le k$, and the chain-rule factor contributes ${\lambda }^q$; if $q>k$, the factor ${\lambda }^q$ is already at most ${\lambda }^{k+1}$ for $\lambda \le 1$. The product rule with the fixed smooth $f$ preserves this bound. When $k=-1$, the same seminorm is bounded by $C_{\phi }$. Thus the tail is bounded by a convergent geometric series with exponent $n+k+1-\omega >0$ (or $n-\omega >0$ when $k=-1$). The series in step 2.1 is absolutely convergent.

For tests supported in any fixed compact $C\subset {\mathbb{R}}^n$, the first term of step 2.1 is a continuous functional: its input is supported in a fixed compact subset of $U$, and the cutoff, multiplication, and finite jet maps are continuous by [F5]–[F8]. The geometric bound of step 3.1 controls the small-scale annular tail by one finite derivative seminorm on a compact containing $C$ and $0$; the finitely many larger scales are continuous by [F1] and [F7]. Hence $T$ is continuous on every fixed-support test space and is a distribution by [F6].

If $\phi$ is supported away from $0$, all its jets at $0$ vanish, so $W\phi =\phi$. The partial sums telescope: ${\sum }_{j=0}^{N}f(2^{j}x)=\chi (x)-\chi (2^{N+1}x)\mathrm{.}$ For sufficiently large $N$, $\chi (2^{N+1}x)=0$ on the support of $\phi$. Adding the first term of step 2.1 then gives $T(\phi )=⟨t,\phi ⟩$. Thus $T{\mathrm{\mid }}_U=t$.

Fix $\mathrm{\Omega }>\delta$ and choose $\omega$ with $\delta <\omega <\min (\mathrm{\Omega },n+k+1)$. Write ${\phi }_{\rho }(x)={\rho }^{-n}\phi (x\mathrm{/}\rho )$, so $T_{\rho }(\phi )=T({\phi }_{\rho })$. Choose $R$ with $\mathrm{supp}\phi \subset B(0,R)$ and $C>R\mathrm{/}a$, where $a$ is from step 1.1. Choose ${\lambda }_0>0$ so $\chi =1$ on $\lambda K$ for $0<\lambda \le {\lambda }_0$, and take $\rho$ small enough that $C\rho \le {\lambda }_0$ and $\mathrm{supp}{\phi }_{\rho }$ lies where $\chi =1$. For a shell $\lambda K$ with $\lambda >C\rho$, its inner radius is at least $a\lambda >R\rho$, so ${\phi }_{\rho }$ vanishes there. If $k=-1$, the first term in step 2.1 and all such coarse-shell terms are therefore zero. On the remaining shells $\lambda \le C\rho$, the chain rule and boundedness of $\lambda \mathrm{/}\rho$ give $p_m(f{\phi }_{\rho }(\lambda \cdot ))\le C_{\phi }{\rho }^{-n}$; their dyadic sum is $O({\rho }^{-\omega })$ because $n-\omega >0$. If $k\ge 0$, on each fine shell $\lambda \le C\rho$ the cutoff is one. For every derivative order $q\le m$, Taylor's formula for $\phi -P_k\phi$ when $q\le k$, and the chain-rule factor $(\lambda \mathrm{/}\rho {)}^q$ when $q>k$, give $p_m(fW{\phi }_{\rho }(\lambda \cdot ))\le C_{\phi }{\rho }^{-n-k-1}{\lambda }^{k+1}$; the fixed upper bound $\lambda \mathrm{/}\rho \le C$ absorbs the finitely many higher derivative orders. Hence these shells contribute at most $C_{\phi }{\rho }^{-n-k-1}{\sum }_{\begin{array}{c}\lambda =2^{-j},j\ge 0\\ \lambda \le C\rho \end{array}}{\lambda }^{n+k+1-\omega }=O({\rho }^{-\omega }),$ because $n+k+1-\omega >0$ and the dyadic sum is bounded by a constant times ${\rho }^{n+k+1-\omega }$. On the intermediate coarse shells $C\rho <\lambda \le {\lambda }_0$, only the subtracted Taylor polynomial remains. Its degree-$j$ term, $0\le j\le k$, contributes at most $C_{\phi }{\rho }^{-n-j}{\lambda }^{n+j-\omega }$. Since $\omega >\delta \ge n+k\ge n+j$, the exponent $n+j-\omega$ is negative; summing dyadically from the smallest such $\lambda \asymp \rho$ up to ${\lambda }_0$ gives $O({\rho }^{-\omega })$. For the finitely many outer shells ${\lambda }_0<\lambda \le 1$ and the first term of step 2.1, continuity of $t$ on their fixed compact supports bounds the jet contributions by $O({\rho }^{-n-k})=O({\rho }^{-\omega })$, since $\omega >\delta \ge n+k$. Thus in both the fine and coarse regions $\mathrm{\mid }{T}_{\rho }(\phi )\mathrm{\mid }\le C_{\phi }{\rho }^{-\omega }$ for small $\rho$. Therefore $\mathrm{\mid }{T}_{\rho }(\phi )\mathrm{\mid }\le C_{\phi }{\rho }^{-\omega }$ for small $\rho$ in both cases, so ${\rho }^{\mathrm{\Omega }}{T}_{\rho }(\phi )\to 0$. This proves ${\mathrm{sd}}_0(T)\le \delta$.

If an exponent $\mathrm{\Omega }<\delta$ made ${\rho }^{\mathrm{\Omega }}{T}_{\rho }\to 0$, restricting to tests supported in $U$ and using step 5.1 would give ${\rho }^{\mathrm{\Omega }}{t}_{\rho }\to 0$, contrary to ${\mathrm{sd}}_0(t)=\delta$. Hence ${\mathrm{sd}}_0(T)\ge \delta$. Together with step 5.2, ${\mathrm{sd}}_0(T)=\delta$.