Maximum Modulus Principle

The idea of the proof is to use the mean value property to show that the set of points for which \(u(z)=M\) is open. Indeed, suppose \(u\left(z_{1}\right)=M\), and express the mean value equality (4.1) in the form \begin{align*} 0=\int_{0}^{2 \pi}\left[u\left(z_{1}\right)-u\left(z_{1}+r e^{i \theta}\right)\right] \frac{d \theta}{2 \pi}, \quad 0 Since the integrand is nonnegative \((\geq 0)\) and continuous, the integral (5.1) can be zero only if the integrand is zero. Thus \(u\left(z_{1}+r e^{i \theta}\right)=u\left(z_{1}\right)=M\) for \(0 \leq \theta \leq 2 \pi\) and \(0<r<\rho\), and the set \(\{u(z)=M\}\) contains a disk centered at each of its points, hence is open. Now, the set \(\{u(z)<M\}\) is also open, since \(u(z)\) is continuous. Since \(D\) is a domain, one of these two sets is empty and the other coincides with all of \(D\). (See Section II.1.) In other words, either \(u(z)<M\) for all \(z \in D\), or \(u(z)=M\) for all \(z \in D\), and this proves the theorem.

This can be derived easily from the real version of the strict maximum principle. We replace \(h(z)\) by \(\lambda h(z)\) for an appropriate unimodular constant \(\lambda\), and we can assume that \(h\left(z_{0}\right)=M\). Let \(u(z)=\operatorname{Re} h(z)\). Then \(u(z)\) is a harmonic function on \(D\) that attains its maximum at \(z_{0}\). By the strict maximum principle for real-valued harmonic functions, \(u(z)=M\) for all \(z \in D\). Since \(|h(z)| \leq M\) and \(\operatorname{Re} h(z)=M\), we must have \(\operatorname{Im} h(z)=0\) for \(z \in D\). Hence \(h(z)\) is constant on \(D\).

Use the extreme value theorem: a continuous function on a compact space attains a maximum absolute value. But then if \(h\) attains its maximum modulus, by the strict maximum principle above, \(h\) is constant.