Normal Families of Functions in C

Let $(M,d)$ be a complete metric space, and $U\subseteq \mathbb{C}$ a region, and the topology of $\mathcal{C}(U,M)$ given in Space of Continuous Functions From a Region in C

Def: A family $\mathcal{F}\subseteq \mathcal{C}(U,M)$ is called normal if every sequence $\{f_n\}\subseteq \mathcal{F}$ there's an $f\in \mathcal{C}(U,M)$, and there's $\{f_{n_k}\}$ is a subsequence of $\{f_n\}$, such that $f_{n_k}\stackrel{\rho }{⟶}f$

Prop: We get that $\mathcal{F}\subseteq \mathcal{C}(U,M)$ is normal iff $\mathcal{F}$ is relatively compact iff $\bar{\mathcal{F}}$ is compact

Th: Let $\mathcal{F}\subseteq \mathcal{C}(U,M)$. Then $\mathcal{F}$ is normal iff $\mathrm{\forall }\delta >0\mathrm{\forall }K\subseteq U$ compact, there's $A\subseteq \mathcal{F}$ such that for any $f\in \mathcal{F}$, there's $g\in A$ such that

Using the prop $\star \star$ of Space of Continuous Functions From a Region in C, we can see that a ball of $(\mathcal{C}(U,M),\rho )$ is basically the same as looking for a compact set $K$ and a $\delta >0$, such that it behaves really closely to being in the ball. So this proposition is basically a reformulation of $\mathcal{F}$ being totally bounded.

Arzelà–Ascoli Theorem

$\mathcal{F}\subseteq \mathcal{C}(U,M)$, where $\mathcal{C}(U,M)$ is complete, $\mathcal{F}$ is normal iff it satisfies:

• for every $z\in U$ the set $A_z=\{f(z)\mid z\in \mathcal{F}\}$ is relatively compact
• $\mathcal{F}$ is equicontinuous on $U$.

Which is probably related to the Arzelà–Ascoli Theorem from Analysis.

Def: Let $U\subseteq \mathbb{C}$ be a region, and $\mathcal{H}(U):=\{f:U\to \mathbb{C}\mid f\text{ is holomorphic}\}$, and $\mathcal{H}(U)\subseteq \mathcal{C}(U,\mathbb{C})$.

Prop: Using the Weierstrass Convergence Theorem, we can see a couple of things:
Let $\{f_n\}\subseteq \mathcal{H}(U)$ be a sequence, such that $f_n\rightrightarrows f$ on every $K\subseteq U$ compact.

• $f\in \mathcal{H}(U)$
• $\mathrm{\forall }k\in \mathbb{N}$, we have that $f_n^{(k)}\stackrel{\rho }{⟶}{f}^{(k)}$

Thus $(\mathcal{H}(U),\rho )$ is complete
The operator $D:\mathcal{H}(U)\to \mathcal{H}(U)$ given by $D(f)=f^{\prime }$, is continuous.

Montel Theorem

Let $\mathcal{F}\subseteq \mathcal{H}(U)$. Then $\mathcal{F}$ is normal iff $\mathcal{F}$ is locally bounded on $U$.

We can reformulate Hurwirtz Theorem

Hurwitz's Theorem

Let $\{f_n\}\subseteq \mathcal{H}(U)$, such that $f_n\stackrel{\rho }{⟶}f$, if there's an $a\in U$ and there's $r>0$ with $\overline{B_r(a)}\subseteq U$ such that for all $\mathrm{\forall }z\in \partial B_r(a)$, $f(z)\ne 0$, then there's $N\in \mathbb{N}$ such that if $n\ge N$ then $f_n$ and $f$ have the same number of zeros on $B_r(a)$, counting multiplicity

Prop: Let $\{f_n\}\subseteq \mathcal{H}(U)$, such that $f_n\stackrel{\rho }{⟶}f$. If $Z(f_n)=\varnothing$ for all $n\in \mathbb{N}$ , then $Z(f)=\varnothing$ or $f\equiv 0$.

Prop: Let $\{f_n\}\subseteq \mathcal{H}(U)$, such that $f_n\stackrel{\rho }{⟶}f$. If every $f_n$ injective, then $f$ is injective or $f$ is constant