= Solution
Write $\omega_n=|B_1|$. If $u$ is a $C^2$ <harmonic function> and $\overline{B_r(x)}\subset\Omega$, its two <mean value properties for harmonic functions> are
$$
\boxed{u(x)=\frac1{n\omega_n r^{n-1}}\int_{\partial B_r(x)}u\,dS
=\frac1{\omega_n r^n}\int_{B_r(x)}u\,dy.}
$$
To prove the spherical property, express the normalized spherical mean on the fixed unit sphere:
$$
M(r)=\frac1{n\omega_n}\int_{S^{n-1}}u(x+r\theta)\,dS_\theta.
$$
Differentiation and the <divergence theorem> give
$$
M'(r)=\frac1{n\omega_n r^{n-1}}\int_{\partial B_r(x)}\partial_\nu u\,dS
=\frac1{n\omega_n r^{n-1}}\int_{B_r(x)}\Delta u\,dy=0.
$$
Since <continuity> gives $M(r)\to u(x)$ as $r\downarrow0$, the spherical mean equals $u(x)$ for every admissible radius. Integrate this identity in radial coordinates:
$$
\int_{B_r(x)}u\,dy=n\omega_n\int_0^r t^{n-1}M(t)\,dt=\omega_n r^n u(x).
$$
This proves the ball mean property as well. For $n=1$, the sphere consists of the two endpoints and the same argument is the <fundamental theorem of calculus>.
Back to article page
