Mollification rates for compactly supported Slobodeckij functions

Given: the Axiom of Countable Choice, $d\ge 1$, $0<\theta <1$, $1\le p<\mathrm{\infty }$, a compactly supported $g\in W^{\theta ,p}({\mathbb{R}}^d)$, a radial mollifier $\eta$ with $\int \eta =1$, and $g_{\delta }=g\ast {\eta }_{\delta }$. Write $\omega (t):={\sup }_{\mathrm{\mid }z\mathrm{\mid }\le t}\mathrm{\parallel }{\tau }_{z}g-g{\mathrm{\parallel }}_{L^p({\mathbb{R}}^d)}$ for $t\ge 0$.

Slobodeckij seminorm as a translation integral. Because the diagonal is null and Tonelli's theorem together with the substitution $z=y-x$ applies, $[g{]}_{\theta ,p}^p={\int }_{{\mathbb{R}}^d}\mathrm{\parallel }{\tau }_{z}g-g{\mathrm{\parallel }}_{L^p({\mathbb{R}}^d)}^p\mathrm{\mid }z{\mathrm{\mid }}^{-d-p\theta }dz$. (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a function on ${\mathbb{R}}^n$)

Volume scaling. If $v_d:=\mathrm{\mid }B(0,1)\mathrm{\mid }$, then $\mathrm{\mid }B(0,t)\mathrm{\mid }=v_d{t}^d$ with $0<v_d<\mathrm{\infty }$ for every $t>0$. (Euclidean balls have positive finite Lebesgue measure, A linear map $T$ of ${\mathbb{R}}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with ${\lambda }_n(T[E])=\mathrm{\mid }\det T\mathrm{\mid }{\lambda }_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not)

The translation modulus is subadditive. $\omega$ is nondecreasing, and $\omega (s+t)\le \omega (s)+\omega (t)$ for all $s,t\ge 0$, because ${\tau }_{z}g-g={\tau }_{z_2}({\tau }_{z_1}g-g)+({\tau }_{z_2}g-g)$ when $z=z_1+z_2$ and translations are isometries of $L^p$. (Translation of a function on ${\mathbb{R}}^n$, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space $L^p(\mu )$ as the quotient by null functions)

Minkowski's integral inequality. For measurable $F$ on a product of sigma-finite measure spaces with ${\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{p}d\nu (y)<\mathrm{\infty }$, $\parallel {\int }_{Y}F(\cdot ,y)d\nu (y){\parallel }_p\le {\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{p}d\nu (y)$. (Minkowski's integral inequality)

Young's convolution inequality. $\mathrm{\parallel }f\ast h{\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^1}\mathrm{\parallel }h{\mathrm{\parallel }}_{L^p}$. (Young's convolution inequality under Countable Choice)

Mollifying a locally integrable function. For $g\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ the convolution $g\ast {\eta }_{\delta }$ is smooth with ${\mathrm{\partial }}^{\alpha }(g\ast {\eta }_{\delta })=g\ast ({\mathrm{\partial }}^{\alpha }{\eta }_{\delta })$; an $L^p$ function is locally integrable, and a compactly supported $L^p$ function lies in $L^1$. (Convolution with a mollifier is smooth, and derivatives pass under the integral sign, Holder's inequality for integrals, including the endpoint cases)

Support of a convolution. For Borel representatives of $f,h\in L^1$, $\mathrm{supp}(f\ast h)\subseteq \overline{\mathrm{supp}f+\mathrm{supp}h}$. (The support of a convolution lies in the closure of the support sumset)

The radial mollifier. ${\eta }_{\delta }={\delta }^{-d}\eta (\cdot \mathrm{/}\delta )$ satisfies $\int {\eta }_{\delta }=1$, $\mathrm{supp}{\eta }_{\delta }\subseteq B(0,\delta )$, and ${\int }_{{\mathbb{R}}^d}{\mathrm{\partial }}_j{\eta }_{\delta }=0$ for every $j$, since $\eta$ is radial and its gradient is odd in each coordinate. (A radial mollifier family in Rn)