Radial derivative of a spherical average

Compact containment gives a slightly larger ball inside $\mathrm{\Omega }$. Since $u$ is $C^2$ there and $\sigma$ is finite, differentiation under the spherical integral gives $m^{\mathrm{\prime }}(r)={\omega }_{n-1}^{-1}{\int }_{S^{n-1}}\mathrm{\nabla }u(x+r\theta )\cdot \theta d\sigma (\theta )\mathrm{.}$

If $n=1$, [F4] gives $m(t)=(u(x+t)+u(x-t))\mathrm{/}2$ and $\mathrm{\mid }{B}_r(x)\mathrm{\mid }=2r$. Thus $m^{\mathrm{\prime }}(r)=(u^{\mathrm{\prime }}(x+r)-u^{\mathrm{\prime }}(x-r))\mathrm{/}2$. The one-dimensional fundamental theorem of calculus gives ${\int }_{x-r}^{x+r}{u}^{\mathrm{\prime }\mathrm{\prime }}(y)dy=u^{\mathrm{\prime }}(x+r)-u^{\mathrm{\prime }}(x-r)$, exactly the displayed formula.

Suppose $n\ge 2$. The ball is a bounded smooth domain, and $F=\mathrm{\nabla }u$ is $C^1$ on its closure. Its outward normal at $x+r\theta$ is $\theta$. Applying [F1] and using [F2] to identify and scale its surface measure gives ${\int }_{B_r(x)}\mathrm{\Delta }u(y)dy=r^{n-1}{\int }_{S^{n-1}}\mathrm{\nabla }u(x+r\theta )\cdot \theta d\sigma (\theta )={\omega }_{n-1}{r}^{n-1}{m}^{\mathrm{\prime }}(r)\mathrm{.}$ By [F3], ${\omega }_{n-1}{r}^{n-1}=n\mathrm{\mid }{B}_r(x)\mathrm{\mid }\mathrm{/}r$, so division gives the claimed identity.