# Friday April 10th

Continuing the proof from last time.
Existence involved

- The square root function
- Reciprocal functions
- Automorphisms of the disc

Definition
:   Let $\Omega \subset \CC$ be open.
    A family $\mcf(\Omega)$ of holomorphic functions on $\Omega$ if every sequence has a subsequence that converges uniformly on every compact subset of $\Omega$, where the limit need not be in $\mcf(\Omega)$.

Proving that a family of functions is normal is a consequence of

- Uniform boundedness
- Equicontinuity

Definition
:   A family $\mcf$ is said to be uniformly bounded on compact subset iff for each compact $K \subset \Omega$ there exists $B_K > 0$ such that
    \begin{align*}
    \abs{f(z)} \leq B_k \quad\text{for all} \quad z\in K, f\in\mcf
    .\end{align*}

Definition
:   A family $\mcf$ is equicontinuous if for every $\eps > 0$ there exists $\delta > 0$ (not depending on the point) such that for all $z, w\in K$,
    \begin{align*}
    \abs{z-w} < \delta \implies \abs{f(z) - f(w)} < \eps \quad\forall f\in\mcf
    .\end{align*}

Examples:

1. The family $f_n: I \to \CC$ with $\abs{f_n'} \leq M$ for some fixed constant is uniformly bounded and equicontinuous (by the MVT).
2. The family $f_n(x) \definedas x^n$ for $x\in I$ is uniformly bounded but not equicontinuous since $\lim_{n\to\infty} \abs{f_n(1) - f_n(x_0)} = 1$ for any $x_o \in I^\circ$.

Theorem (Montel)
:   Suppose $\mcf(\Omega)$ is uniformly bounded on compact subsets, then

    1. $\mcf$ is equicontinuous on every compact subset,
    2. $\mcf$ is a normal family.

The proof of the theorem consists of two parts:

1.Apply the Cauchy integral formula and use that $\mcf$ is comprised of holomorphic functions.

  > Note the contrast to $\RR$, where $f_n(x) = \sin(nx), \abs{f_n(x)} \leq 1$ is uniformly bounded but not equicontinuous and has no convergent subsequences on *any* compact subinterval of $I^\circ$.

2. Use the fact that uniform bounded + equicontinuous implies existence of convergent subsequences by Arzela-Ascoli (which uses diagonalization).

  > Not complex-analytic, works in $\RR$.

Definition
:   A sequence $\theset{K_\ell}$ of compact subsets is called an *exhaustion* of $\Omega$ iff

    1. $K_\ell \subset (K_{\ell+1})^\circ$ for all $\ell$,
    2. Any compact $K\subset \Omega$ is contained in some $K_\ell$, and $\union_\ell K_\ell = \Omega$


Lemma
: Any open $\Omega \subset \CC$ admits an exhaustion.

Proof
:   If $\Omega$ is bounded, take $K_\ell = \theset{\dist(z, \bd \Omega) > \frac 1 \ell}$.
    Otherwise, take $\tilde K_\ell = K_\ell \intersect B_\ell(0)$.

### Proof of Montel's Theorem

Proving part 1.

Let $K \subset \Omega$ be compact, choose $r> 0$ such that $D_{3r}(z) \subset \Omega$ for all $z\in K$, e.g. $3r < \dist(K, \bd \Omega)$.
Let $z, w\in K$ with $\abs{z-w} < r$ and let $\gamma = \bd D_{2r}(w)$.

Then Cauchy's integral formula yields
\begin{align*}
  f(z)-f(w)=\frac{1}{2 \pi i} \int_{\gamma} f(\zeta)\left[\frac{1}{\zeta-z}-\frac{1}{\zeta-w}\right] d \zeta
.\end{align*}

Then since $\xi \in \gamma$ and $\abs{z-w} < r$, we have

\begin{align*}
  \left|\frac{1}{\zeta-z}-\frac{1}{\zeta-w}\right|=\frac{|z-w|}{|\zeta-z||\zeta-w|} \leq \frac{|z-w|}{r^{2}}
.\end{align*}

Letting $B$ be the uniform bound on $\mcf$ and using $\abs{\gamma} = 4\pi r$, we can apply this estimate to obtain

\begin{align*}
  |f(z)-f(w)| \leq \frac{1}{2 \pi} \frac{2 \pi r}{r^{2}} B|z-w|
.\end{align*}

Then $f$ is uniformly Lipschitz with the constant given above, and the family is equicontinuous.

> Application: show that derivative uniformly bounded implies function uniformly bounded by applying Cauchy's integral formula.

Proving part 2.

Let $\ts{f_n}$ be a sequence in $\mcf$ and $K\subset \Omega$ compact.
Choose a dense sequence $\ts{w_j}$, and use uniform boundedness to obtain a subsequence $\ts{f_{n, 1} }$ such that $\ts{f_{n, 1}(w_1)}$ converges.
Repeat these to get $\ts{f_{n, j}(w_j)}$ all converge, and set $g_n = f_{n, n}$.
The claim is that equicontinuity implies $g_n$ converges uniformly on $K$.

Given $\eps>0$, choose $\delta$ from equicontinuity and note that $K \subset D_\delta(w_1) \union \cdots \union D_\delta(w_J)$ for some $J$ by density of $\ts{w_i}$ and compactness of $K$.
Pick $N \gg 0$ such that
\begin{align*}
\abs{g_m(w_j) - g_n(w_j)} < \eps \forall j=1,2,\cdots, J
.\end{align*}

Then any $z\in K$ is in some $D_\delta(w_j)$, then
\begin{align*}
\begin{aligned}
\left|g_{n}(z)-g_{m}(z)\right| \leq \left|g_{n}(z)-g_{n}\left(w_{j}\right)\right|+\left|g_{n}\left(w_{j}\right)-g_{m}\left(w_{j}\right)\right|+\left|g_{m}\left(w_{j}\right)-g_{m}(z)\right| <& 3 \epsilon
\end{aligned}
,\end{align*}

so $\ts{g_n}$ converges uniformly on $K$.