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 $\map {B_r} {z_0}$ be the open ball around $z_0$ of radius $r$, and $\overline {\map {B_r} {z_0} }$ its closure.

If $\overline {\map {B_r} {z_0} } \subset U$, then


 * $(a) \quad \ds \map f {z_0} = \frac 1 {2 \pi} \int_0^{2 \pi} \map f {z_0 + re^{i \theta} } \rd \theta$


 * $(b) \quad \ds \map f {z_0} = \frac 1 {\pi r^2} \int_{\overline {\map {B_r} {z_0} } } \map f {x + i y} \rd x \rd y$

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 $\rho \in \openint 0 r$.

If we multiply by $2 \pi \rho$ on both sides of the equation, we get:
 * $\ds 2 \pi \rho \map f {z_0} = \int_0^{2 \pi} \map f {z_0 + \rho e^{i \theta} } \rho \rd \theta$

If we integrate with respect to $\rho$ from $0$ to $r$, we get
 * $\ds \int_0^r 2\pi \rho \map f {z_0} \rd \rho = \int_0^r \int_0^{2 \pi} \map f {z_0 + \rho e^{i \theta} } \rho \rd \rho \rd \theta$

Using the change of variables $\rho \rd \rho \rd \theta = \d x \rd y$, we get
 * $\ds \pi r^2 \map f {z_0} = \int_{B_0} \map f z \rd x \rd y$

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 $\map {B_r} {z_0}$ around $z_0$ of radius $r$, is contained in $U$.

If $f$ is not constant on $\map {B_r} {z_0}$, then there is some $z_1 \in \map {B_r} {z_0}$ with $\size {\map f {z_1} } < \size {\map f {z_0} }$.

Let $\epsilon = \size {\map f {z_0} } - \size {\map f {z_1} }$.

Since $\size f$ is continuous, there is an $s > 0$ so that $\map {B_s} {z_1} \subset \map {B_r} {z_0}$ and

whenever $z \in \map {B_s} {z_1}$.

Let $B_0 = \map {B_r} {z_0}$.

Let $B_1 = \map {B_s} {z_1}$.

Thus we have

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