Proposition 3.1.5620D5E4DD6EC

Proof. Fix $0<r<R$. For $\omega \in S^{d-1}$ and small $h\neq 0$, the fundamental theorem of calculus gives

Since $\nabla u$ is uniformly continuous on the compact set $\overline{B(x,R)}$, these difference quotients converge to $\nabla u(x+r\omega )\cdot \omega $ uniformly in $\omega $ as $h\to 0$. Integrating over the finite measure space $(S^{d-1},S)$ and using the definition of surface measure on $\partial B(x,r)$, where $\nu (x+r\omega )=\omega $, we obtain

Applying (3.1.3) on $B(x,r)$ to $F=\nabla u$, which is $C^1$ on $\Omega $ with divergence $\Delta u$, gives (3.1.6).

Now let $u$ be harmonic. Then $M_ u(x,\cdot )$ has zero derivative, and hence is constant, on $(0,R)$. Since $|M_ u(x,r)-u(x)|\leq \max _{|y-x|=r}|u(y)-u(x)|$, which tends to zero as $r\downarrow 0$ because $u$ is continuous at $x$, the constant is $u(x)$. If $\overline{B(x,r)}\subset \Omega $, then the compact set $\overline{B(x,r)}$ has positive distance from the closed set $\mathbb R^ d\setminus \Omega $, so there is $R>r$ with $\overline{B(x,R)}\subset \Omega $, and therefore $u(x)=M_ u(x,r)$; this is the sphere mean-value property. Since $M_ u(x,t)=u(x)$ for $0<t\leq r$, (3.1.2) gives $\int _{B(x,r)}u\, \mathrm dy=s_{d-1}r^ du(x)/d=|B(x,r)|\, u(x)$, which is the ball mean-value property. $\square$