User:Ovenhouse/sandbox

Theorem
Let $U \subset \C$ be an open set.

Let $f \colon U \to \C$ be analytic on $U$.

For any $r > 0$, and $z_0 \in U$, let $B_r \left( z_0 \right)$ be the open ball around $z_0$ of radius $r$, and $\overline{B_r \left( z_0 \right)}$ its closure.

If $\overline{B_r \left( z_0 \right)} \subset U$, then

$\displaystyle \begin{array}{cl} (a) & f \left( z_0 \right) = \frac{1}{2 \pi} \int_0^{2 \pi} f \left( z_0 + re^{i \theta} \right) \mathrm d \theta \\ (b) & f \left( z_0 \right) = \frac{1}{\pi r^2} \int_{\overline{B_r \left( z_0 \right)}} f \left( x+iy \right) \mathrm d x \mathrm d y \end{array}$

Proof
$(a)$ Using Cauchy Integral Formula, and the parameterization of $\partial B_0$ given by $\gamma \left( t \right) = z_0 + re^{it}$, with $\gamma' \left( t \right) = ire^{it}\mathrm d t$, and $0 \leq t \leq 2 \pi$, we get:

$(b)$ Part $(a)$ holds if we replace $r$ by any $\rho \in (0,r)$.

If we multiply by $2 \pi \rho$ 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
 * $\displaystyle \pi r^2 f \left( z_0 \right) = \int_{B_0} f \left( z \right) \mathrm d x \mathrm d y$

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

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

Let $B_0 = B_r \left( z_0 \right)$.

Let $B_1 = B_s \left( z_1 \right)$.

Thus we have

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