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\begin{flashcard}[locally-uniform-converging-implies]
\begin{theorem*}
    Let $\{f_n\}$ be a sequence of analytic
    functions on $\mathcal{U}$, converging locally
    uniformly to $f$. Then \cloze{$f$ is holomorphic,
    with $f_n' \to f'$ locally uniformly.}
\end{theorem*}
\end{flashcard}

\begin{hiddenflashcard}[locally-uniformly-converging-implies-proof]
Prove the following: \\
Suppose $f_n \to f$ locally uniformly, $f_n$ holomorphic. Then $f$
holomorphic and $f_n' \to f'$ locally uniformly. \\
\cloze{
    Fix $a \in \mathcal{U}$ and $\ol{D(a, r)}
    \subset \mathcal{U}$. Given $\gamma$ a closed
    curve in $\gamma$ we have
    \[ \int_\gamma f = \lim_{n \to \infty}
    \int_\gamma f_n = 0 \]
    by \fcemph{Cauchy's theorem}. Hence, by
    \fcemph{Morera's theorem}, $f$ is holomorphic
    on $D(a, r)$. By Taylor expansion formula,
    \[ |f'(w) - f_n'(w)| = \frac{1}{2\pi} \left|
    \int_{|z - a| = r} \frac{f(z) - f_n(z)}{(z -
    w)^2} \dd z\right| \]
    Then for $|w - a| \le \rho < r$, we have
    \[ |f'(w) - f_n'(w)| \le r \cdot
    \frac{1}{\rho^2} \cdot \sup_{|z - a| = r}
    |f(z) - f_n(z)| \to 0 \]
    So $f_n' \to f'$ locally uniformly.
}
\end{hiddenflashcard}

\begin{proof}
    Fix $a \in \mathcal{U}$ and $\ol{D(a, r)}
    \subset \mathcal{U}$. For $r \ll 1$, $f_n \to
    f$ uniformly of $\ol{D(a, r)}$. So
    \[ |f(z) - f(w)| = |f(z) - f_n(z) + f_n(z) -
    f_n(w) + f_n(w) - f(w)| \]
    so uniform convergence implies $f$ continuous
    no $\ol{D(a, r)}$. Given $\gamma$ a closed
    curve in $D(a, r)$, we have
    \[ \int_\gamma f = \lim_{n \to \infty}
    \int_\gamma f_n = 0 \]
    by Cauchy's theorem. So Morera's theorem
    implies $f$ is holomorphic on $D(a, r)$. By
    Cauchy's integral formula we have:
    \[ |f'(w) - f_n'(w)| = \frac{1}{2\pi} \left|
    \int_{|z - a| = r} \frac{f(z) - f_n(z)}{(z -
    w)^2} \dd z \right| \]
    for $|w - a| \le \frac{r}{2}$ we have
    \begin{align*}
        |f'(w) - f_n'(w)|
        &\le r \cdot \frac{1}{\left( \frac{r}{2}
        \right)^2} \cdot \sup_{|z - a| = r} |f(z)
        - f_n(z)| \\
        &\to 0
    \end{align*}
    as $n \to \infty$ by uniform convergence. $f_n
    \to f$ uniformly on $\ol{D(a, r)}$ implies
    $|f_n' \to f'|$ uniformly on $D \left( a,
    \frac{r}{2} \right)$.
\end{proof}

\begin{remark*}
    The assumption of locally uniform convergence
    is necessary; a construction with
    non-holomorphic limit can be done via Runge's
    theorem (see Topics in Analysis).
\end{remark*}

\subsubsection*{Application 1: Newton's method and complex dynamics}

Recall Newton's method, an iterative root-finding
algorithm, takes a polynomial $p(z)$ and an
initial $z_0$ for a root of $p(z)$, and compute a
sequence $z_1, z_2, \ldots, z_n \defeq f^n(z_0),
\ldots$ where
\[ f(z) = z - \frac{p(z)}{p'(z)} ;\]
sometimes(??) this sequence limits to a root of
$p$.

\begin{example*}
    $p(z) = z^3 - 1$, $f(z) = \frac{2z^3 +
    1}{3z^2}$. In $\RR$:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/09dbfa3ac18a11ed.png}
    \end{center}
    $f^n (z)$ is a sequence of meromorphic
    functions, so if $f^n(z_0)$ approaches a
    limit, for some region $\mathcal{U}$ of
    initial guesses, then $f^n |_{\mathcal{U}}$
    has holomorphic limit.
\end{example*}

\begin{definition*}
    A family $\mathcal{F} = \{f_i\}_{i \in I}$ of
    holomorphic functions on a domain $D$ is
    \emph{normal} if every sequence in
    $\mathcal{F}$ has a locally uniformly
    convergent subsequence. (Note: we allow
    convergence to $\infty$).
\end{definition*}

\noindent
\textbf{Deep theorem} (``Montel's theorem''): If
$\exists a, b, c \in \CC_\infty$ such that
$\forall f \in \mathcal{F}$, $f(D) \cap \{a, b,
c\} = \emptyset$, then $\mathcal{F}$ is a normal
family.

\begin{definition*}
    The \emph{Fatou set} of a rational map $f$ is
    \[ F(f) \defeq \{z \in \CC_\infty : \exists
    \text{ neighbourhood $\mathcal{U}$ of $z$ s.t.
    $\{f^n |_{\mathcal{U}}\}$ forms a normal family} \} \]
\end{definition*}

\subsubsection*{Riemann mapping theorem}

\begin{theorem*}[RMT]
    Lt $\mathcal{U} \subsetneq \CC$ be a nonempty,
    proper, open, simply connected subset of
    $\CC$. Then there exists conformal isomorphism
    $f : \mathcal{U} \to \DD = D(0, 1)$.
\end{theorem*}

\begin{proof}[Sketch of proof]
    Fix $z_0 \in \mathcal{U}$, and consider
    \[ \mathcal{F} \defeq \{f : \mathcal{U} \to
    \DD, \text{$f$ holomorphic, injective and
    $f(z_0) = 0$}\} \]
    Steps:
    \begin{enumerate}[(1)]
        \item $\mathcal{F}$ is non-empty.
        \item Show there exists $g \in
            \mathcal{F}$ such that $|g'(z_0)|$ is
            finite and maximal among elements of
            $\mathcal{F}$.
        \item Prove $g$ is a conformal
            isomorphism.
    \end{enumerate}
    Now we actually prove these claims:
    \begin{enumerate}[(1)]
        \item $\mathcal{U} \neq \CC$ implies
            $\exists a \in \CC \setminus
            \{\mathcal{U}\}$, so by Example Sheet
            2 there exists holomorphic branch of
            the logarithm $\log(z - a)$ on
            $\mathcal{U}$. So there exists
            holomorphic branch $h(z) = \sqrt{z -
            a}$ on $\mathcal{U}$. Show: $h$ is
            injective on $\mathcal{U}$, and
            $h(\mathcal{U}) \cap -h(\mathcal{U}) =
            \emptyset$. By open mapping theorem,
            $h(D)$ contains some $D(h(z_0),
            \eps)$, so $|h(z) + h(z_0)| \ge \eps$
            for all $z \in D$. Can then check
            that:
            \[ f_0(z) = \frac{\eps}{4} \cdot
            \frac{|h'(z_0)|}{|h(z_0)|^2} \cdot
            \frac{h(z_0)}{h'(z_0)} \cdot \frac{h(z) -
            h(z_0)}{h(z) + h(z_0)} \in \mathcal{F} \]
        \item Let $A = \sup_{f \in \mathcal{F}}
            |f'(z_0)|$, and choose $\{f_n\}$ in
            $\mathcal{F}$ such that $f_n'(z_0) \to
            A$. By Montel's, $\mathcal{F}$ is a
            normal family, so there exists
            $f_{n_k}$ converging locally uniformly
            to some $g$, holomorphic. Show $g$ is
            in the family (injectivity requires
            argument).
        \item If $g$ is not surjective then can
            construct an element of $\mathcal{F}$
            violating maximality of $g$: if $c \in
            D(0, 1) \setminus g(\mathcal{U})$,
            then choose (Example Sheet 2) a
            holomorphic branch
            \[ k(z) \defeq \sqrt{\frac{g(z) - c}{1
            - cg(z)}} .\]
            Then
            \[ F(z) = \frac{e^{i\theta} (k(z) -
            k(z_0))}{1 - k(z_0)k(z)}, \qquad
            \frac{k'(z_0)}{|k'(z_0)|} = e^{-i\theta} \]
            is in $\mathcal{F}$, with $|F'(z_0)| >
            |g'(z_0)|$, contradiction. \qedhere
    \end{enumerate}
\end{proof}
