Maximum Modulus Principle

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Theorem

A non-constant analytic function $f$ in an open connected set $D$ does not have any interior maximum points.

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

$\left\vert{f \left({\omega}\right)}\right\vert > \left\vert{f \left({z}\right)}\right\vert$.


Proof

Pick some $r > 0$ such that $B_r \left({z}\right) \subset D$.

By the mean value theorem for holomorphic functions:

$\displaystyle f \left({z}\right) = \dfrac 1 {2 \pi} \int_0^{2 \pi} f \left({z + r e^{i \theta} }\right) \, \mathrm d \theta$

Then:

$\displaystyle \left\vert{f \left({z}\right)}\right\vert \le \frac 1 {2 \pi} \int_0^{2 \pi} \left\vert{f \left({z + r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta \le \max_\theta \left\vert{f \left({z + r e^{i \theta} }\right)}\right\vert$

So it must be that:

$\exists\omega \in C_r \left({z}\right): \left\vert{f \left({z}\right)}\right\vert \le \left\vert{f \left({\omega}\right)}\right\vert$

where $C_r \left({z}\right)$ is the circle of radius $r$ centered at $z$.


Note that equality is only obtained when $\left\vert{f}\right\vert$ is constant on $C_r \left({z}\right)$.


However, since this holds for all sufficiently small $r > 0$, $\left\vert{f}\right\vert$ would be constant in $B_r \left({z}\right)$.

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

It follows that:

$\exists\omega \in C_r \left({z}\right): \left\vert{f \left({z}\right)}\right\vert < \left\vert{f \left({\omega}\right)}\right\vert$

$\blacksquare$