I am in the middle of the proof of the maximum principle for harmonic functions.
Given a harmonic function $u$ on the complex plane and $M_0\in \mathbb{C}$. Take $r>0$ and suppose there is an open arc $\ell$ contained in the circle $\{M_0+re^{it}\colon t\in [0,2\pi)\}$ such that
$$u(M)<u(M_0)\mbox{ for each }M\in \ell.$$
Does it follow from this that
$$u(M_0)\neq \frac{1}{2\pi}\int_0^{2\pi}u(M_0+re^{it})dt?$$
Best Answer
Assume the contrary and consider the function $$ f(s)=\int_0^{s}u(M_0+re^{it})dt, $$ that has the properties $f(0)=0$, $f(2\pi)=2\pi u(M_0)$, and $f'(s)=u(M_0+re^{is})$.