Maximum modulus principle

Statement

Let $D \subseteq \mathbb{C}$ be a domain and let $f : D \to \mathbb{C}$ be holomorphic on $D$. Assume that $|f|$ has a local maximum at a point $z_0 \in D$: there is $r > 0$ such that $|f(z)| \le |f(z_0)|$ for each $z$ with $|z - z_0| < r$. Then $f$ is constant on $D$.

Remarks

Proof idea: the mean value formula $f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + \rho e^{i\theta}) \, d\theta$ for small $\rho$ forces $|f| = |f(z_0)|$ near $z_0$. Then $f$ is constant near $z_0$, and on $D$ by the identity theorem. Corollary: if $D$ is bounded and $f$ is continuous on the closure $\overline{D}$ and holomorphic on $D$, then $\max_{\overline{D}} |f|$ is attained at a point of the boundary of $D$. The minimum of $|f|$ can be inside $D$, but only at a zero of $f$. For example, $|z|$ has its minimum on the unit disc at $0$.

Explanations

L1 human

The size $|f|$ of a nonconstant holomorphic function has no highest point inside its domain. For example, on the closed unit disc, $|z^2 + 1|$ is largest on the boundary circle, at $z = \pm 1$, and it equals $2$ there.