User:Ovenhouse/sandbox

Theorem
Let $U \subset \C$ be a connected open set.

Let $f$ be a holomorphic function on $U$.

Let $z_0 \in U$ be a maximum point for $\left|{f}\right|$ in $U$.

Then $f$ is constant on $U$.

Proof
Let $r>0$ be such that the open ball $B_r \left( z_0 \right)$ around $z_0$ of radius $r$, is contained in $U$.

If $f$ is not constant on $B_r \left( z_0 \right)$, then there is some $z_1 \in B_r \left( z_0 \right)$ with $\left| f \left( z_1 \right) \right| < \left| f \left( z_0 \right) \right|$.

Let $\varepsilon = \left| f \left( z_0 \right) \right| - \left| f \left( z_1 \right) \right|$.

Since $\left| f \right|$ is continuous, there is an $s>0$ so that $B_s \left( z_1 \right) \subset B_r \left( z_0 \right)$ and
 * $\displaystyle \left| f \left( z \right) \right| < \left| f \left( z_1 \right) \right| + \frac{\varepsilon}{2} = \left| f \left( z_0 \right) \right| - \frac{\varepsilon}{2}$

whenever $z \in B_s \left( z_1 \right)$.

Let $B_0 = B_r \left( z_0 \right)$ and $B_1 = B_s \left( z_1 \right)$.

Mean Value Property
Using Cauchy Integral Formula, and the substitution $z = z_0 + re^{i\theta}$, with $\mathrm d z = ire^{i\theta}\mathrm d \theta$, we get:

This is true if we replace $r$ by any $\rho \in (0,r)$.

If we multiply by $\rho$ on both sides of the equation, we get:
 * $\displaystyle 2\pi \rho f \left( z_0 \right) = \int_0^{2 \pi} f \left( z_0 + \rho e^{i\theta} \right) \rho \mathrm d \theta$

If we integrate with respect to $\rho$ from $0$ to $r$, we get
 * $\displaystyle \int_0^r 2\pi \rho f \left( z_0 \right) \mathrm d \rho = \int_0^r \int_0^{2\pi} f \left( z_0 + \rho e^{i\theta} \right) \rho \mathrm d \rho \mathrm d \theta$

Using the change of variables $\rho \mathrm d \rho \mathrm d \theta = \mathrm d x \mathrm d y$, we get
 * $\displaystyle \pi r^2 f \left( z_0 \right) = \int_{B_0} f \left( z \right) \mathrm d x \mathrm d y$

Proof cont...
By Absolute Value of Complex Integral, we have

This is a contradiction, so it must be that $f$ is, in fact, constant.