# Monday April 13th

Goal: put together pieces for the Riemann mapping theorem.

Today: normal families.
Recall that normal families on $\Omega$ are holomorphic functions for which every sequence uniformly on every compact subset of $\Omega$.
The family is uniformly bounded iff for every $K$ there exists a $B_K$ bounding $f$ in $K$, and is equicontinuous iff they are continuous with a parameter $\delta$ uniform for the family.

We had Montel's theorem: families that are uniformly bounded on compact subsets are equicontinuous and normal.
This used Arzela-Ascoli and a diagonalization argument.

Theorem (Hurwitz)
: Suppose $g_n$ are holomorphic and nonzero on $\Omega$, then if $g_n \to g$ uniformly on compact subset, then either $g \equiv 0$ or $g$ is nonzero on $\Omega$.

Proof
:   The limit function $g$ is holomorphic on $\Omega$ by the Weierstrass theorem.
    If $g\not\equiv 0$, then the zeros of $g$ are isolated.
    If $\gamma \sim 0$ is a simply curve on which $g\neq 0$, $g'_n \to g'$ and hence $g_n' / g_n \to g'/g$ uniformly on $\gamma$.
    By the argument principle, for $n\gg 0$ the number of zeros of $g_n, g$ enclosed by $\gamma$ are equal.
    Since $g_n \neq 0$, the theorem follows.

Corollary (Hurwitz)
:   If $g_n$ are holomorphic and injective, then either $g$ is injective or constant.

Proof
: Fix $w\in \Omega$ and apply Hurwitz to $g - g(w)$ on $\Omega \setminus \theset w$.

Proof (of part (B) of Riemann Mapping)
:   By Montel's theorem, $\mcf(\Omega)$ is normal since $\abs f < 1$ for all $f\in \mcf$.
    Let $M = \sup{g'(z_0) \suchthat g\in \mcf} \leq M$.
    By definition, we can find a sequence $g_n$ such that $\lim g_n'(z_0) = M$.
    By normality, there exists a subsequence uniformly convergent on compact subsets to some function $f$.

    \

    In particular, $M < \infty$ since otherwise this would yield a simple pole for $f$ (which is holomorphic) and $f'(z_0) = M$.
    By Hurwitz, $f$ is either injective or constant because the subsequence is injective.
    But since $f' = M > 0$, $f$ is not constant.

    \

    Moreover, since $f(z_0) = \lim g_n(z_0) = 0$ and $f'(z_0) = M > 0$, $f$ injective implies $f\in \mcf$.
    This gives the existence of $f\in\mcf$ such that $g' \leq f'$ at $z_0$ for all $g\in \mcf$.

Proof (of part (C) in Riemann Mapping)
:   We want to show that $f(\Omega) = \DD$.
    Toward a contradiction, suppose not, then there exists a $w_0 \in \DD \setminus f(\Omega)$.
    We will construct a function $g_1 \in \mcf$ with $g_1' > f'$ at $z_0$, which is a contradiction.

    The function $-\psi_{w_0}(f(z)) = {f(z) - w_0 \over 1 - \bar w_0 f(z)}: \Omega \to \DD$ is holomorphic, injective (since all such $\psi$ are), and nonvanishing since $w_0 \not\in f(\Omega)$.
    Thus there is a branch of the square root for which $g(z) \definedas \sqrt{-\psi_{w_0}(f(z))}: \Omega \to \DD$ is injective.

    Finally, normalize $g$ so that its derivative at $z_0$ is positive and real by setting
    \[
    g_1(z) = { \abs{g'(z_0)} \over g'(z_0)} \qty{g(z) - g(z_0) \over 1 - \bar{g(z_0)} g(z) }
    .\]

    Then $g_1$ has the same properties as $g$ with $g_1 = 0, g_1' > 0$ at $z_0$.
    The first claim is easy to check, and the second follows from using $\dd{}{z} \qty{z - \alpha \over 1 - \bar \alpha z} = {1 - \abs{alpha}^2 \over \qty{1 - \bar \alpha z}^2 }$ and applying the chain rule.

    We now compare $f'$ and $g'$ at $z_0$.
    Note that $f = 0, f' > 0, g^2 = -w_0$ at $z_0$.
    Take $\dd{}{z} g^2$ to obtain

    \[
    2 g(z) g^{\prime}(z)=\frac{1-\left|w_{0}\right|^{2}}{\left(1-\bar{w}_{0} f(z)\right)^{2}} \cdot f^{\prime}(z)
    .\]

    Evaluate at $z=z_0$ to obtain
    \[
    2 g\left(z_{0}\right) g^{\prime}\left(z_{0}\right)
    =\left(1-\left|w_{0}\right|^{2}\right) f^{\prime}\left(z_{0}\right) \Rightarrow\left|g^{\prime}\left(z_{0}\right)\right|
    =\frac{1-\left|w_{0}\right|^{2}}{2\left|g\left(z_{0}\right)\right|} f^{\prime}\left(z_{0}\right)
    .\]

    Combining this with the previous formula obtained from the chain rule
    \[
    g_{1}^{\prime}\left(z_{0}\right)
    &=\frac{\left|g^{\prime}\left(z_{0}\right)\right|}{1-\left|g\left(z_{0}\right)\right|^{2}}\\
    &=\frac{1-\left|w_{0}\right|^{2}}{2\left|g\left(z_{0}\right)\right|} \cdot \frac{f^{\prime}\left(z_{0}\right)}{1-\left|g\left(z_{0}\right)\right|^{2}} \\
    &=\frac{1-\left|w_{0}\right|^{2}}{2 \sqrt{\left|w_{0}\right|}} \cdot \frac{f^{\prime}\left(z_{0}\right)}{1-\left|w_{0}\right|} \quad \text{ using } g^2(z_0) = - w_0 \\
    &=\frac{1+\left|w_{0}\right|}{2 \sqrt{\left|w_{0}\right|}} f^{\prime}\left(z_{0}\right) \\
    & > f^{\prime}\left(z_{0}\right)
    .\]

    Where we've used the Schwarz inequality,
    \[
    \frac{1+\left|w_{0}\right|}{2 \sqrt{\left|w_{0}\right|}}>1
    .\]

    But then $g_1' > f'$, a contradiction.

Note that this is a long proof in the book -- seven pages!