Exact four-dimensional Euclidean logarithmic scale anomaly

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.