Maximum Modulus Principle

Theorem
Let $D$ be an open region.

Let $f : D \to \C$ be a non-constant holomorphic function.

Then $\cmod f$ does not have any maximum points in the interior of $D$.

That is, for each $z \in D$ and $\delta > 0$, there exists some $\omega \in \map {B_\delta} z \cap D$, such that


 * $\cmod {\map f \omega} > \cmod {\map f z}$.

Proof
Pick some $r > 0$ such that $\map {B_r} z \subset D$.

By the Mean Value Theorem for Holomorphic Functions:
 * $\ds \map f z = \dfrac 1 {2 \pi} \int_0^{2 \pi} \map f {z + r e^{i \theta} } \rd \theta$

Then:

So it must be that there exists $\omega \in \map {C_r} z$ such that:
 * $\cmod {\map f z} \le \cmod {\map f \omega}$

where $\map {C_r} z$ is the circle of radius $r$ centered at $z$.

Note that equality is only obtained when $\cmod f$ is constant on $\map {C_r} z$.

However, since this holds for all sufficiently small $r > 0$, $\cmod f$ would be constant in $\map {B_r} z$.

Then $f$ must be constant in $D$, contradicting our assumption.

It follows that there exists $\omega \in \map {C_r} z$ such that:
 * $\cmod {\map f z} < \cmod {\map f \omega}$