Maximum Modulus Principle

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Theorem

Let $D$ be an open region of the complex plane $\C$.

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:

$\cmod {\map f z} < \cmod {\map f \omega}$

$\blacksquare$

Also known as

The is also known as the maximum modulus theorem.

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): maximum modulus theorem