Scaling degree and local extension space

Let $X={\mathcal{D}}_K$ with the metric of [F2]. It is nonempty because it contains $0$, and it is complete by [F3]. AC supplies DC by [F4]–[F6], so the Baire theorem [F7] applies. The proof of [F7] also chooses centers for its recursively constructed balls; that countable family of nonempty balls is covered by AC${}_{\omega }$ from [F6] and [F8].

For fixed $\psi \in X$, set $L_{\rho }(\psi )={\rho }^{\omega }⟨t_{\rho },\psi ⟩$. The exponents $\eta$ for which ${\rho }^{\eta }{t}_{\rho }(\psi )\to 0$ for every fixed test form an upward-closed set: if the limit holds at $\eta$, then for $\omega >\eta$ multiply by ${\rho }^{\omega -\eta }\le 1$. Since its infimum is $\delta <\omega$, some such $\eta$ is below $\omega$, and consequently $L_{\rho }(\psi )\to 0$ as $\rho \downarrow 0$.

On any compact scale interval $[ϵ,1]$ with $ϵ>0$, the tests $\psi (\cdot \mathrm{/}\rho )$ have support in one compact subset of $U$ and vary continuously in every derivative seminorm by the chain rule. Continuity of the distribution $t$ therefore makes $\rho \mapsto L_{\rho }(\psi )$ continuous there. Together with step 1.2 this proves ${\sup }_{0<\rho \le 1}\mathrm{\mid }{L}_{\rho }(\psi )\mathrm{\mid }<\mathrm{\infty }$ for every $\psi \in X$.

For each positive integer $N$, let $E_N=\{\psi \in X:{\sup }_{0<\rho \le 1}\mathrm{\mid }{L}_{\rho }(\psi )\mathrm{\mid }\le N\}$. Each $E_N$ is closed because every $L_{\rho }$ is a continuous linear functional on $X$. Step 2.1 gives $X={\bigcup }_{N\ge 1}{E}_N$. By [F7], some $E_N$ has nonempty interior.

The topology of $X$ is generated by the increasing seminorms in [F2], so for some ${\psi }_0\in E_N$, $m\ge 0$, and $\epsilon >0$, the set ${\psi }_0+\{\zeta :p_m(\zeta )<\epsilon \}$ lies in $E_N$. For every such $\zeta$, subtracting the two bounds for ${\psi }_0+\zeta$ and ${\psi }_0$ gives $\mathrm{\mid }{L}_{\rho }(\zeta )\mathrm{\mid }\le 2N$ for all $\rho$. If $p_m(\psi )>0$, apply this to $\zeta =\epsilon \psi \mathrm{/}(2p_m(\psi ))$; if $p_m(\psi )=0$, then $p_0(\psi )=0$ and hence $\psi =0$. Thus $\mathrm{\mid }{L}_{\rho }(\psi )\mathrm{\mid }\le (4N\mathrm{/}\epsilon )p_m(\psi )$ for every $\psi$ and every $0<\rho \le 1$. Replacing $L_{\rho }$ by its definition proves the claimed uniform order estimate.

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, ${\mathrm{sd}}_0(T)=\delta$.

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 $c{\delta }_0$.

The function $u$ is smooth, hence locally integrable, on $U={\mathbb{R}}^4\setminus \{0\}$. By [F3]–[F6] it defines a distribution $t\in {\mathcal{D}}^{\mathrm{\prime }}(U)$; direct dilation gives $t_{\rho }={\rho }^{-4}t$, so ${\mathrm{sd}}_0(t)=4$.

Let ${\sigma }_3$ be the polar surface measure on $S^3$. By [F7]–[F9] and Tonelli, ${\pi }^2={\int }_{{\mathbb{R}}^4}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2}{d}^{4}x={\sigma }_3(S^3){\int }_0^{\mathrm{\infty }}{e}^{-r^2}{r}^{3}dr=\frac{{\sigma }_3(S^3)}{2}\mathrm{.}$ The last integral is $1\mathrm{/}2$ by $s=r^2$ and integration by parts. Therefore ${\sigma }_3(S^3)=2{\pi }^2$.

Near $0$, $\chi (\mu x)=1$ and $\mathrm{\mid }\phi (x)-\phi (0)\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }{\sup }_B\mathrm{\mid }\mathrm{\nabla }\phi \mathrm{\mid }$ on a compact ball $B$ containing $0$ and the fixed test support. Thus the integrand is bounded by a fixed seminorm times $\mathrm{\mid }x{\mathrm{\mid }}^{-3}$ there, which is integrable in four dimensions. Away from $0$ its support is contained in the compact union of $\mathrm{supp}\phi$ and ${\mu }^{-1}\mathrm{supp}\chi$, and $\mathrm{\mid }x{\mathrm{\mid }}^{-4}$ is bounded on that support; the remaining pairing is bounded by a fixed zeroth-order seminorm, including the evaluation $\mathrm{\mid }\phi (0)\mathrm{\mid }$. These bounds hold uniformly on each fixed-support test space, so $T_{\mu }$ is continuous there and hence a distribution by [F13]. If $\phi$ is supported away from $0$, then $\phi (0)=0$, so $T_{\mu }(\phi )=t(\phi )$.

Subtracting the two definitions gives $T_{\mu }(\phi )-T_{\nu }(\phi )=-\frac{\phi (0)}{16{\pi }^4}{\int }_{{\mathbb{R}}^4\setminus \{0\}}\frac{\chi (\mu x)-\chi (\nu x)}{\mathrm{\mid }x{\mathrm{\mid }}^4}{d}^{4}x\mathrm{.}$ The difference vanishes on a fixed neighborhood of $0$ and outside a fixed compact set, so it is absolutely integrable. Apply [F7] separately to its positive and negative parts; the integral is ${\int }_{S^3}{\int }_0^{\mathrm{\infty }}[\chi (\mu r\omega )-\chi (\nu r\omega )]dr\mathrm{/}rd{\sigma }_3(\omega )$. For $\mu$ in a compact interval $[{\mu }_0,{\mu }_1]\subset (0,\mathrm{\infty })$ choose $0<a<b$ so every first or second derivative of $\chi$ vanishes unless $a\le \mathrm{\mid }y\mathrm{\mid }\le b$. Writing $s=\log \mu$ and $g(s,r,\omega )=\chi (e^{s}r\omega )$, the first and second $s$-derivatives vanish unless $a\mathrm{/}{\mu }_1\le r\le b\mathrm{/}{\mu }_0$. On this annulus, $\mathrm{\mid }{\mathrm{\partial }}_s^{2}g\mathrm{\mid }\le b\mathrm{\parallel }\mathrm{\nabla }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}+b^2\mathrm{\parallel }{D}^2\chi {\mathrm{\parallel }}_{\mathrm{\infty }}=:M$, by the chain rule; these suprema are finite because $\chi$ is smooth and compactly supported. For $s$ in a smaller compact interval and $h$ sufficiently small, the one-variable Taylor estimate gives $\mathrm{\mid }(g(s+h,r,\omega )-g(s,r,\omega ))\mathrm{/}h-{\mathrm{\partial }}_{s}g(s,r,\omega )\mathrm{\mid }\le M\mathrm{\mid }h\mathrm{\mid }\mathrm{/}2$ uniformly in $r,\omega$. The annulus has finite $dr\mathrm{/}r$ measure and $S^3$ has finite measure by step 1.2, so integrating this uniform error proves differentiation under the polar integral. For each $\omega$, ${\int }_0^{\mathrm{\infty }}\frac{\mathrm{\partial }}{\mathrm{\partial }\log \mu }\chi (\mu r\omega )\frac{dr}{r}={\int }_0^{\mathrm{\infty }}\mathrm{\nabla }\chi (s\omega )\cdot \omega ds=\chi (\mathrm{\infty }\omega )-\chi (0)=-1,$ where $s=\mu r$, $\chi =1$ near $0$, and $\chi$ is compactly supported. Since the difference integral vanishes at $\mu =\nu$, its value is $-{\sigma }_3(S^3)\log (\mu \mathrm{/}\nu )$. With step 1.2 this proves the displayed formula for $T_{\mu }-T_{\nu }$.

Changing variables in the defining integral and using $u(\rho x)={\rho }^{-4}u(x)$ gives ${\rho }^4(T_{\mu }{)}_{\rho }=T_{\rho \mu }$. Step 2.2 with $(\rho \mu ,\mu )$ then gives ${\rho }^4(T_{\mu }{)}_{\rho }=T_{\mu }+\frac{1}{8{\pi }^2}\log \rho {\delta }_0\mathrm{.}$ For every $\mathrm{\Omega }>4$, multiplication by ${\rho }^{\mathrm{\Omega }-4}$ makes the right side tend to zero as $\rho \downarrow 0$, so ${\mathrm{sd}}_0(T_{\mu })\le 4$. Restriction to $U$ and step 1.1 give the reverse inequality, hence ${\mathrm{sd}}_0(T_{\mu })=4$.

The exact-degree extension family is nonempty by [F10], and [F11] says every member differs from $T_{\mu }$ by $c{\delta }_0$. Since ${\rho }^4({\delta }_0{)}_{\rho }={\delta }_0$, the scaling identity of step 3.1 yields ${\rho }^4(T_{\mu }+c{\delta }_0{)}_{\rho }=T_{\mu }+c{\delta }_0+(8{\pi }^2{)}^{-1}\log \rho {\delta }_0$, which is not constant in $\rho$. Thus no same-degree extension is exactly homogeneous.