Mean value property of harmonic functions

Without loss of generality centre the disk at the origin ($x_0 = 0$) and define $\phi(r) = \frac{1}{2\pi}\int_0^{2\pi} u(r\cos\theta, r\sin\theta)\, d\theta$ for $0 \le r \le R$, the average of $u$ over the circle of radius $r$. We show $\phi$ is constant by proving $\phi'(r) = 0$.

Differentiating under the integral sign, $\phi'(r) = \frac{1}{2\pi}\int_0^{2\pi} \nabla u(r\cos\theta, r\sin\theta) \cdot (\cos\theta, \sin\theta)\, d\theta = \frac{1}{2\pi}\int_0^{2\pi} \frac{\partial u}{\partial n}\, d\theta$, since $(\cos\theta,\sin\theta)$ is the outward unit normal on the circle of radius $r$. Multiplying and dividing by $r$ turns this into a normalised boundary integral of the normal derivative over the circle $\partial B_r$.

By the divergence theorem, $\int_{\partial B_r} \frac{\partial u}{\partial n}\, ds = \int_{B_r} \nabla^2 u \, dA = 0$ because $u$ is harmonic. Hence $\phi'(r) = 0$ for every $r$ in the domain, so $\phi(r) = \phi(0^+) = u(x_0)$ for all such $r$: the average over any circle centred at $x_0$ equals $u(x_0)$ itself, and averaging this constant value over $0 \le r \le R$ shows the solid-disk average equals $u(x_0)$ as well.