Maximum modulus principle

Statement

Let $D\subseteq\mathbb C$ be a domain and let $f:D\to\mathbb C$ be holomorphic. If $|f|$ has a local maximum at an interior point of $D$ , then $f$ is constant on $D$ .

Proof

For a closed disc around the maximum point, the Cauchy integral formula expresses $f(a)$ as the average of its boundary values. The triangle inequality gives $|f(a)|$ at most the boundary maximum. Equality forces all boundary values to have the same argument and modulus; applying the same argument on smaller discs makes $f$ constant on a neighborhood of $a$ . The identity theorem then makes it constant on the connected domain.

Boundary form

If $D$ is bounded, $f$ is continuous on $\overline D$ , and holomorphic on $D$ , then

\max_{\overline D}|f|=\max_{\partial D}|f|,

unless $f$ is constant. Compactness of $\overline D$ ensures that the maximum exists; the local principle forces any nonconstant maximum to the boundary.

Minimum modulus

Apply the principle to $1/f$ : if a holomorphic function has no zeros, $|f|$ cannot attain an interior local minimum unless $f$ is constant. A zero is an allowed minimum, so this statement requires the nonvanishing hypothesis.

References

Lars V. Ahlfors, Complex Analysis, 3rd ed., McGraw–Hill, 1979. Relevant: Chapter 4, §3.