Let \(u(z)\) be a real-valued harmonic function on a domain \(D\) such that \(u(z) \leq M\) for all \(z \in D\). If \(u\left(z_{0}\right)=M\) for some \(z_{0} \in D\), then \(u(z)=M\) for all \(z \in D\).

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

Let \(h\) be a bounded complex-valued harmonic function on a domain \(D\). If \(|h(z)| \leq M\) for all \(z \in D\), and \(\left|h\left(z_{0}\right)\right|=M\) for some \(z_{0} \in D\), then \(h(z)\) is constant on \(D\).

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 \(D\).

Let \(h(z)\) be a complex-valued harmonic function on a bounded domain \(D\) such that \(h(z)\) extends continuously to the boundary \(\partial D\) of \(D .\) If \(|h(z)| \leq M\) for all \(z \in \partial D\), then \(|h(z)| \leq M\) for all \(z \in D\).

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.

If \(f\) is nonzero on \(\Omega\), then \(f\) attains a minimum on \({{\partial}}\Omega\). This follows from applying the MMP to \(1/f\).

## Exercises

Show that if \(f\) is holomorphic on \({\mathbb{D}}\setminus\left\{{0}\right\}\) and \({\left\lvert {f(z)} \right\rvert} \leq \log\qty{1\over {\left\lvert {z} \right\rvert}}\), then \(f\equiv 0\).

- Claim: \(f\) has a removable singularity at \(z_0=0\).
- Thus \(f\) extends to some holomorphic \(F\) defined on \({\mathbb{D}}\).
- By continuity, \({\left\lvert {F} \right\rvert}\) satisfies the same inequality as \(f\).
- Now \(\lim_{{\left\lvert {z} \right\rvert} \to 1}{\left\lvert {F(z)} \right\rvert}\leq \lim_{{\left\lvert {z} \right\rvert}\to 1} \log\qty{1\over {\left\lvert {z} \right\rvert}} = 0\), so by the MMP on \({\mathbb{D}}\), \(F\equiv 0\) on \({\mathbb{D}}\setminus\left\{{0}\right\}\).