Smoothness, parity and zero-radius limits of spherical means

Differentiation, including signed radii. Put $H(x,r):={\omega }_{n-1}^{-1}{\int }_{S^{n-1}}f(x+r\omega )d\sigma (\omega )$ for all real $r$. On a bounded parameter neighbourhood, all points $x+r\omega$ lie in a fixed compact ball. For a coordinate parameter $p$ and a continuous derivative integrand $q(p,\omega )$ with continuous ${\mathrm{\partial }}_{p}q$, [F2] gives $\mid (q(p+h,\omega )-q(p,\omega ))\mathrm{/}h-{\mathrm{\partial }}_{p}q(p,\omega )\mid \le {\sup }_{\mathrm{\mid }s-p\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid },\omega }\mathrm{\mid }{\mathrm{\partial }}_{p}q(s,\omega )-{\mathrm{\partial }}_{p}q(p,\omega )\mathrm{\mid }$. Uniform continuity on the compact ball makes this bound tend to zero uniformly in $\omega$; integration against the probability measure [F1] preserves the bound. Iterating through total order $k$ therefore gives every stated derivative under the integral, with continuity again following from the same uniform estimate. This works for $n=1$ as well, since it requires only a finite measure, not positive-dimensional surface charts.

Part (ii), continuity at $r=0$. Let $K\subseteq {\mathbb{R}}^n$ be compact and choose $R>0$ with $K\subseteq B_R(0)$; the ball $\overline{B_{R+1}(0)}$ is compact, so by [F3] $f$ is bounded there and uniformly continuous on it. Given $\epsilon >0$ choose $\delta \in (0,1)$ such that $\mathrm{\mid }f(u)-f(v)\mathrm{\mid }<\epsilon$ whenever $u,v\in \overline{B_{R+1}(0)}$ and $\mathrm{\mid }u-v\mathrm{\mid }<\delta$. For $x\in K$, $0\le r<\delta$ and $\omega \in S^{n-1}$ one has $x+r\omega \in \overline{B_{R+1}(0)}$ and $\mathrm{\mid }x+r\omega -x\mathrm{\mid }=r<\delta$, so $\mathrm{\mid }{M}_f(x,r)-f(x)\mathrm{\mid }\le {\sup }_{\omega }\mathrm{\mid }f(x+r\omega )-f(x)\mathrm{\mid }<\epsilon$. Hence $M_f(x,r)\to f(x)$ as $r\downarrow 0$ uniformly on compact subsets, and the extension by $M_f(\cdot ,0)=f$ is continuous because $f$ is.

Part (ii), the derivative limit. By part (i) with $\mathrm{\mid }\alpha \mathrm{\mid }=0$, $m=1$, ${\mathrm{\partial }}_r{M}_f(x,r)={\omega }_{n-1}^{-1}{\int }_{S^{n-1}}Df(x+r\omega )\cdot \omega d\sigma (\omega )$ for $r>0$, so adding and subtracting $Df(x)$ inside the integral gives ${\mathrm{\partial }}_r{M}_f(x,r)={\omega }_{n-1}^{-1}{\int }_{S^{n-1}}[Df(x+r\omega )-Df(x)]\cdot \omega d\sigma (\omega )+{\omega }_{n-1}^{-1}Df(x)\cdot {\int }_{S^{n-1}}\omega d\sigma (\omega )$, where the second term vanishes by [F4]; the first is bounded by ${\sup }_{\omega \in S^{n-1}}\mathrm{\mid }Df(x+r\omega )-Df(x)\mathrm{\mid }$, which tends to $0$ as $r\downarrow 0$ uniformly for $x$ in a fixed compact set by uniform continuity of the continuous function $Df$ on a large compact ball, again by [F3]. Hence ${\mathrm{\partial }}_r{M}_f(x,r)\to 0$ uniformly on compact subsets of ${\mathbb{R}}^n$.

Parity and the bound. Reflection preserves $\sigma$ by [F4], hence the substitution $\omega \mapsto -\omega$ gives $H(x,-r)=H(x,r)$. Since $H$ is $C^k$ by step 1.1, its odd-order radial derivatives at zero vanish whenever their orders are at most $k$. For (iv), $\mathrm{\mid }{\mathrm{\partial }}_r{M}_f(x,r)\mathrm{\mid }\le {\omega }_{n-1}^{-1}\int \mathrm{\mid }Df(x+r\omega )\mathrm{\mid }d\sigma (\omega )$. By continuity, each boundary value of $\mathrm{\mid }Df\mathrm{\mid }$ is at most ${\sup }_{B_r(x)}\mathrm{\mid }Df\mathrm{\mid }$, and integration gives the stated bound.

Thus $M_f$ has the differentiated-integral formula, the stated uniform zero-radius limits and gradient bound, and an even $C^k$ signed-radius extension.