is there a way to prove the Mean-value formulas using complex analysis?

Solution 1:

For $n=2$ one can proceed as follows (this is essentially an elaboration of above comments):

If $u$ is harmonic in $ U \subset \Bbb C$ and $B(z_0, R) \subset U$ then $u = \operatorname{Re} f$ for some holomorphic function $f$ in $B(z_0, R)$. This is true in any simply-connected domain in $\Bbb C$, $f$ can for example be chosen as an anti-derivative of the (holomorphic) function $u_x - i u_y$.

The Cauchy integral formula then states that for $0 < r < R$ and the path $\gamma(t) = z_0 + re^{it}$, $0 \le t \le 2 \pi$, $$ f(z_0) = \frac{1}{2 \pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z_0} \, d\zeta = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + re^{it}) \, dt \, . $$

Taking real parts gives the mean-value formula for $u$: $$ u(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} u(z_0 + re^{it}) \, dt \, . $$

Finally, the “area form” of the mean-value formula follows by integration over the radius: $$ \frac{1}{\pi r^2} \int_{B(z_0, r)} u(x, y)\, dx dy = \frac{1}{\pi r^2} \int_0^r \int_0^{2 \pi} u(z_0 + \rho e^{it}) \rho \, dt \, d\rho = \frac{1}{\pi r^2} \int_0^r 2 \pi u(z_0) \rho \, d\rho = u(z_0) \, . $$