I know how to prove that if $f:A \rightarrow \mathbb{C}$ is a harmonic function (here $A$ is an open connected region in $\mathbb{C}$) such that $|f|$ admits a local maximum on a point $z_0\in A$, then $f$ is constant on $D(z_0,r)$, where $D(z_0,r)$ is a disk centered on $z_0$, $r>0$ being such that the restriction of $|f|$ to $D(z_0,r)$ admits a global maximum on $z_0$.
To do so, I’ve first proved the mean value property for harmonic functions (which you can find for instance here, on page 4), then I followed the proof I knew of the analogous result (the maximum principle) for holomorphic functions, that only used this property. However, I’d like now to extend this to the more general result: $f$ must also be constant on the whole region $A$.
On the proof for holomorphic functions that I have previously studied, this was done by using the principle of analytic continuation. However, this is only valid for holomorphic functions (right?), so I’m not sure what I could apply here. Any help would be welcome.
Best Answer
Here is a standard proof for this result (see, for example, PDEs by L. Evans): Since $\vert f\vert $ attains its maximum at $z_0 \in A$, $f$ must either attain its maximum or minimum at $z_0$. Without loss of generality, we may assume that $f$ attains its maximum there.
Let $U := \{ z\in A \text{ s.t. } f(z) = f(z_0)\} = f^{-1}( \{f(z_0) \} )$. Since $f$ is continuous (harmonic functions are analytic), $U$ is closed in the induced topology on $A$. Moreover, if $z\in U$ and $r>0$ is sufficiently small so that $D(z,r) \subset A$, then $$f(z_0)=f(z) = \frac 1{ \pi r^2} \int_{D(z,r)} f(\tilde z) \, d\tilde z \leqslant f(z_0),$$ so $f(\tilde z)=f(z_0)$ for all $\tilde z \in D(z,r)$, that is, $D(z,r) \subset U$. Hence, $U$ is also open. Thus, since $A$ is connected, $U=A$, which implies that $f$ is a constant on $A$.
As an aside (which is essential for understanding what I wrote above), the paper you cite only seems to give the boundary mean value formula $$u(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0+\rho e^{i\theta})\, d \theta. \tag{$\ast$} $$ There is also a "full" mean value formula $$u(z_0) = \frac{1}{\pi r^2} \int_0^r \int_0^{2\pi} f(z_0+\rho e^{i\theta})\, \rho d \theta \, d\rho = \frac{1}{\pi r^2} \int_{D(z_0,r)} f(z)\, dA \tag{$\ast\ast$} $$ which is what I have used above. One can obtain ($\ast\ast $) from ($\ast$) as follows: \begin{align*} \int_0^r \int_0^{2\pi} f(z_0+\rho e^{i\theta})\, \rho d \theta \, d\rho &=\int_0^r \rho \bigg ( \int_0^{2\pi} f(z_0+\rho e^{i\theta})\, d \theta \bigg ) \, d\rho \\ &= 2\pi f(z_0)\int_0^r \rho \, d\rho\\ &= \pi f(z_0) r^2. \end{align*}