Subcritical singularity has a unique same-degree extension

By [A1] and [F2]–[F3], AC${}_{\omega }$ holds, so [F4] applies. On compact subsets of $U$, $g_{\delta }$ is bounded and therefore defines the regular distribution $t_{\delta }$. Put $x_0=(3\mathrm{/}2,0,\dots ,0)$, $K=\bar{B}(x_0,1\mathrm{/}4)$, and $\mathrm{\Omega }=B(x_0,3\mathrm{/}8)\subset \{9\mathrm{/}8<\mathrm{\mid }x\mathrm{\mid }<15\mathrm{/}8\}\subset \{1<\mathrm{\mid }x\mathrm{\mid }<2\}$. By [F10] choose $\phi \in \mathcal{D}(\mathrm{\Omega })$ with $0\le \phi \le 1$ and $\phi =1$ near $K$. For $0<\epsilon <1\mathrm{/}(4\sqrt{n})$ the box $Q={\prod }_{i=1}^n(x_{0,i}-\epsilon ,x_{0,i}+\epsilon )$ lies in $K$, since $\mathrm{\mid }x-x_0\mathrm{\mid }<\sqrt{n}\epsilon <1\mathrm{/}4$ on $Q$. By [F9], ${\lambda }_n(Q)>0$; on $Q$ we have $\phi =1$, $5\mathrm{/}4<\mathrm{\mid }x\mathrm{\mid }<7\mathrm{/}4$, and $g_{\delta }(x)\ge \min \{(5\mathrm{/}4{)}^{-\delta },(7\mathrm{/}4{)}^{-\delta }\}>0$, so $⟨t_{\delta },\phi ⟩>0$. For $\delta <n$, the unit ball integral is ${\int }_{\mathrm{\mid }x\mathrm{\mid }<1}\mathrm{\mid }x{\mathrm{\mid }}^{-\delta }{d}^{n}x=\sigma (S^{n-1}){\int }_0^1{r}^{n-1-\delta }dr=\frac{\sigma (S^{n-1})}{n-\delta }<\mathrm{\infty }\mathrm{.}$ It follows that $g_{\delta }\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$, and [F5]–[F6] give the regular distribution $G_{\delta }$.

The dilation formula [F1] and the change of variables in the regular integral give $(G_{\delta }{)}_{\rho }={\rho }^{-\delta }{G}_{\delta }$. For every $\omega >\delta$, ${\rho }^{\omega }(G_{\delta }{)}_{\rho }={\rho }^{\omega -\delta }{G}_{\delta }\to 0$. For every $\omega <\delta$, the nonnegative annular test $\phi$ from step 1.1 has positive pairing with $G_{\delta }$. Therefore ${\rho }^{\omega }⟨(G_{\delta }{)}_{\rho },\phi ⟩={\rho }^{\omega -\delta }⟨G_{\delta },\phi ⟩$ does not tend to zero. Hence ${\mathrm{sd}}_0(G_{\delta })=\delta$, and its restriction to $U$ is $t_{\delta }$.

Since $\delta <n$, the subcritical uniqueness clause [F8] applies to $t_{\delta }$ and the extension $G_{\delta }$ from step 1.1. Thus every extension of $t_{\delta }$ with scaling degree $\delta$ equals $G_{\delta }$.

At $\delta =n$, [F1] gives $⟨(t_n{)}_{\rho },\phi ⟩={\rho }^{-n}⟨t_n,\phi ⟩$. Thus every $\omega >n$ gives convergence to zero; for $\omega <n$, the nonnegative annular test from step 1.1 has positive pairing with $t_n$, so the rescaled pairing does not tend to zero. Hence ${\mathrm{sd}}_0(t_n)=n$. The polar integral is then $\sigma (S^{n-1}){\int }_0^1{r}^{-1}dr=+\mathrm{\infty }$. Its coefficient is positive: with $c=(3\mathrm{/}4,0,\dots ,0)$, choose a box $Q={\prod }_{i=1}^n(c_i-\epsilon ,c_i+\epsilon )$ for $0<\epsilon <1\mathrm{/}(4\sqrt{n})$. Then $\mathrm{\mid }x-c\mathrm{\mid }<1\mathrm{/}4$ for $x\in Q$, so $Q\subset \{1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<1\}$; [F9] gives ${\lambda }_n(Q)>0$, while [F4] gives ${\lambda }_n(Q)\le \sigma (S^{n-1}){\int }_{1\mathrm{/}2}^1{r}^{n-1}dr$. Hence $\sigma (S^{n-1})>0$. If a locally integrable $h$ induced a regular extension of $t_n$, then $u_h{\mathrm{\mid }}_U=t_n$; the injectivity in [F6] on $U$ gives $h=g_n$ almost everywhere there, so its absolute integral on the punctured unit ball has the same logarithmic divergence. Thus no regular-distribution extension exists. By [F8], the same-degree distributional extension family exists and is $T_0+\mathbb{C}{\delta }_0$.