# Monday February 3rd

## Mean Value Theorem

Theorem (Mean Value for Holomorphic functions)
: 	$$f(z_0) = \frac{1}{\pi r^2} \iint_{D_r(z_0)} f(z) ~dA$$

Proof (of MVT?)
: 	Let $f: \Omega \to \CC$ be holomorphic where $\Omega$ is open and connected.
	Then by Cauchy's integral formula, we have $f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta$ for any $z_0 \in \Omega$.
	
	We can consider $D_r(z_0)$, in which case we have for all $0 < s < r$,

\begin{align*}
f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + se^{i\theta}) ~d\theta \\
\implies s\cdot f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} s\cdot f(z_0 + se^{i\theta}) ~d\theta \\
\implies \cdot f(z_0) \int_0^r s ~ds &= \frac{1}{2\pi} \int_0^{2\pi} \int_0^r f(z_0 + se^{i\theta})\cdot s ~ds~d\theta \\ 
\implies \frac 1 2 r^2 f(z_0) &= \frac{1}{2\pi} \iint_{D_r(z_0)} f(z) ~dA\\
\implies f(z_0) &= \frac{1}{\pi r^2} \iint_{D_r(z_0)} f(z) ~dA \\
\implies f(z_0) &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) ~d\theta
.\end{align*}

Proposition (Maximum in Interior Implies Constant)
: 	Let $f$ be holomorphic on $\Omega$ be open and connected, and suppose that there is a $z_0 \in \Omega$ such that $$\abs{f(z_0)} = \sup_{z\in \Omega} \abs{f(z)},$$ i.e. $z_0$ is a maximal point of $f$.
	Then $f$ is constant on $\Omega$.

	If $\Omega$ is additionally **bounded**, then $f$ is continuous on $\bar \Omega$, then $$\sup_{z \in \bar\Omega} \abs{f(z)} = \max_{z\in\bar\Omega} \abs{f(z)}.$$

Proof
: 	Since $\abs{f}$ is continuous and $\bar \Omega$ is compact, $\abs f$ attains a maximum at some point in $\bar \Omega$.
	We want to show that if $\abs{f(z_0)} = \sup_{z\in \Omega} \abs{f(z)}$, then $f$ is constant.

	Assume that there exists a $z_0 \in \Omega$ such that $f(z) = f(z_0)$.
	Let $O = \theset{ \xi\in \Omega \suchthat f(\xi) = f(z_0)  }$.

	Claim
	:	1. $O$ is not empty, since $z_0 \in O$.
	1. $O$ is closed, since if $\xi_n \to \xi$ then $f(\xi_n) = f(z_0)$ implies $f(\xi) = f(z_0)$ since $f$ is continuous.
	2. (**Claim**) $O$ is open.

	Suppose $\xi_0 \in O$, then there exists a disc $D_\rho(\xi_0) \subset \Omega$ such that $$f(\xi_0) = \frac{1}{\pi\rho^2} \int_{D_\rho(\xi_0)} f(z) dA.$$
	Then (claim) $\abs{f(\xi_0)} \geq \abs{f(z)}$ for all $z\in D_\rho(\xi_0)$, which forces $f(z)  = f(\xi_0)$ for all $z\in D_\rho(\xi_0)$.

Proof (of the claim):
: Suppose that $\sup_{a\in \Omega} \abs{f(z)} = \abs{f(\xi_0)}$ and write $f(\xi_0) = Be^{i\alpha}$ for $B>0$ and $\alpha \in \RR$.
	Then define $g(z) = f(z) e^{-i\alpha}$; then $g(\xi_0) = B$ is real, and thus

	\begin{align*}
	0 = g(\xi_0) - B = \frac{1}{\pi\rho^2} \iint_{D_\rho(\xi_0)} \Re( g(z) - B ) ~dA
	.\end{align*}

	Note that $\Re(g(z) - B) \leq 0$ implies that $\Re(g(z) - B) \equiv 0$ on $D_\rho(z_0)$, so we can write $g(z) = B + iI(z)$ for some real-valued function $I$.

	But then $\abs{g(z)}^2 = B^2 + I(z)^2 = B^2$ by the previous statement, and so $I(z) = 0$, forcing $g(z) = B$ and thus $f(z) = Be^{i\alpha}$.
	This shows that $O$ is open, and thus $O = \Omega$.

## Biholomorphisms of the Open Disc

Proposition (Biholomorphisms of the Open Disc are Contractions (Stein 2.1))
: 	Suppose $f$ is holomorphic on $D_1(0)$ and $\abs{f(z)} \leq 1$ for all $\abs{z} < 1$ with $f(0) = 0$.
	Then $\abs{f(z)} \leq \abs{z}$ for all $\abs{z} < 1$.

	Moreover, there is a point $z_0\in D_1(0)$  such that $\abs{f(z_0)} = \abs{z_0}$ iff $f(z) = c(z)$ for some $c \in S^1$.

Proof
: 	Define

	\begin{align*}
	g(z) =
	\begin{cases}
	\frac{f(z)}{z} & z\neq 0 \\
	f'(0) & z = 0
	\end{cases}
	.\end{align*}

	Then $g$ is holomorphic on $D_1(0)$ and $\abs{g(z)} \leq \frac{1}{\rho}$ for all $\abs{z} < \rho < 1$.
	Now apply the maximum principle: 
	since this is true for all $\rho < 1$, consider the limit $\rho\to 1^-$.
	
	Then $\abs{g(z)} \leq 1$, so $\abs{\frac{f(z)}{z}} \leq 1$ and $\abs{f(z)} \leq \abs{z}$.
	If $\abs{f(z_0)} = \abs{z_0}$ for any point, then $\abs{g(z_0)} = 1$ implies $g(z_0) = c$ and $c\in S^1$. 
	
	Thus $f(z) = cz$ for some $c\in S^1$.

Corollary (Characterization of Biholomorphisms of the Disc)
: 	Recall that $$\Phi_a(z) \definedas \frac{z-a}{1-az}.$$

	If $f: D_1(0) \to D_1(0)$ is a biholomorphism, then $$f(z) = c \Phi_a(z) = e^{i\theta} \Phi_a(z)$$
	So every such function is a rotated form of $\Phi_a$.

Let $\Omega$ be a connected open domain and $f: \Omega \to \CC$ holomorphic with $f\in C^1$.
Then $$\int_\gamma f(z) ~dz = 0$$ for every closed curve $\gamma \subset \Omega$, which implies that $f^{(k)} (z)$ exists for all $k\in \NN$ and $f$ is smooth/holomorphic.

## Morera

Theorem (Morera, Partial Converse to Cauchy's Integral Theorem)
: 	Suppose $g: \Omega \to \CC$ is continuous and $\int_\gamma g(z)~dz = 0$ whenever $\gamma = \bd R$ for some rectangle $R\subset \Omega$ with sides parallel to the axes:

	![Image](figures/2020-02-03-14:15.png)\

	Then $g(z)$ is holomorphic in $\Omega$.

Proof
: 	Fix a point $\alpha = a + ib$ and given $z = x+iy$, construct a rectangle $R$ containing $z$.
	Then by assumption, $\int_{\bd R} g(z) ~dz = 0$.
	Let $\gamma_{az}$ be the path given by traversing the bottom edge of $R$, and $\sigma_{az}$ by the top path.

	![Image](figures/2020-02-03-14:24.png)\

	Let 

	\begin{align*}
	f(z)
	&= \int_{\gamma_{az} } g(z) ~dz \\
	&= \int_a^x g(s+ib) ~ds + i \int_b^y g(x+it) ~dt
	.\end{align*}


	Since $$\int_{\bd R} g(z) ~dz = 0 = \int_{\gamma_{az}} \cdots - \int_{\sigma_{az}} \cdots,$$ we have

	\begin{align*}
	f(z) 
	&= \int_{\sigma_{az}} g(z) ~dz \\
	&= i \int_b^y g(a + it) ~dt + \int_x^a g(s + iy) ~ds
	.\end{align*}

	> Exercise: Apply $\dd{}{y}$ to the first identity and $\dd{}{x}$ to the second.

	This yields $$\dd{f}{x} = g(z) \quad\text{ and }\quad \dd{f}{y} = ig(z) = i \dd{f}{x}$$ by applying the FTC, which are precisely the Cauchy-Riemann equations for $f$.
	So $f$ is holomorphic, and thus $f(z) = g(z)$.