# Proving $u(x) = \frac{1}{\omega_n r^{n-1}}\int_{\partial B(x,r)} u\, d\sigma = \frac{n}{\omega_n r^n}\int_{B (x,r)} u\, dV$ for harmonic $u$

I'm having a bit of a problem proving the equality:

$$u(x) = \frac{1}{\omega_n r^{n-1}}\int_{\partial B(x,r)} u\, d\sigma = \frac{n}{\omega_n r^n}\int_{B (x,r)} u\, dV$$

Which is the mean value theorem for Harmonic functions, where $\omega_n$ is the area of $S^n$ and $B(x,r)$ is the ball in $\mathbb{R}^n$ centered at $x$ with radius $r$.

I think that I should use one of Green's formulas, but I'm not sure how.

• Green is only for function define on $\mathbb R^2$
– idm
Jan 10 '15 at 16:31
• Write $$\varphi(r)=\frac{1}{\omega_n r^{n-1}}\int_{\partial B(x,r)} u(y)dy=\frac{1}{\omega_n}\int_{\partial B(0,1)}u(x+ry)dy.$$ Now calculate $\varphi'$ using the divergence theorem. For the second equality just use polar coordinates. Jan 10 '15 at 16:44

## 1 Answer

Set $$\phi(r) := \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y) \, dS(y) = \frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz) \, dS(z).$$ Then $$\phi'(r)=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} \nabla u(x+rz) \cdot z \, dS(z)$$ and consequently, using Green's formulas we obatin

\begin{align}\phi'(r) &= \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \nabla u(y) \cdot \frac{u-x}r dS(y) \\ &= \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \\ &= \frac rn \frac{1}{|B(x,r)|}\int_{B(x,r)} \Delta u(y) \, dy = 0.\end{align}

Hence $\phi$ is constant, that is $\phi(t)= \phi(r)$ for all $r>0$ and all $t>0$ and so by using Lebesgue differentiation theorem we get , $$\phi(r)=\lim_{t \to 0} \phi(t) = \lim_{t \to 0}\frac{1}{|\partial B(x,t)|} \int_{\partial B(x,t)} u(y) \, dS(y) = u(x).$$