% vim: tw=50
% 10/03/2023 10AM

\begin{flashcard}[open-mapping-thm]
\begin{corollary*}[Open mapping theorem]
    \cloze{
    A nonconstant holomorphic function maps open
    sets to open sets.
}
\end{corollary*}

\begin{proof}
    \cloze{
    Want to show that if $f : D \to \CC$ then
    $\forall a \in D$, $\forall r > 0$
    sufficiently small, $f(D(a, r)) \supset
    D(f(a), \eps)$ for some $\eps$. By previous
    theorem\prompt{ (local mapping degree theorem)},
    if $r$ and $\eps$ are sufficiently
    small, then $\forall w \in D(f(a), \eps)$ we
    have that the number of zeroes of $f(z) - w$
    in $D(a, r)$ is $\deg_{z = a} f(z) > 0$.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[rouches-thm]
\begin{theorem*}[Rouch\'e's theorem]
    Let $\gamma$ bound a domain $D$, and $f$, $g$
    \cloze{are \fcemph{holomorphic} on a
    \fcemph{neighbourhood} of $D \cup
    \gamma$. If $|f(z)| > |g(z)|$ for all $z \in
    \gamma$, then $f$ and $f + g$ have the same
    number of zeroes in $D$.}
\end{theorem*}
\end{flashcard}

\begin{hiddenflashcard}[rouches-thm-proof]
\prompt{Proof of Rouch\'e's theorem? \\}
\cloze{
Define $h(z) = \frac{f(z) + g(z)}{f(z)} = 1 +
\frac{g(z)}{f(z)}$. Note $h$ is meromorphic on a
neighbourhood of $D \cup \gamma$ and has no zeroes
or poles on $\gamma$. Also, $h \circ \gamma
\subset D(1, 1)$ so $I(h \circ \gamma; 0) = 0$. So
by argument principle,
\[ \text{\# zeroes of $f + g$ on $D$} -
\text{\# zeroes of $f$ on $D$} = I(h \circ
\gamma; 0) = 0 \]
}
\end{hiddenflashcard}

\begin{proof}
    Define $h(z) = \frac{f(z) + g(z)}{f(z)} = 1 +
    \frac{g(z)}{f(z)}$. Note $h$ is meromorphic on
    a neighbourhood of $D \cup \gamma$. Since
    $|f(z)| > |g(z)|$ $\forall z \in \gamma$< $f +
    g$ and $f$ are nonzero on $\gamma$, so $h$ has
    no zeroes or poles on $\gamma$. By argument
    principle, we have
    \[ \text{\# zeroes of $f + g$ on $D$} -
    \text{\# zeroes of $f$ on $D$} = I(h \circ
    \gamma; 0) \]
    By hypothesis, $h \circ \gamma \subset D(1,
    1)$. So $I(h \circ \gamma; 0)$.
\end{proof}

\begin{example*}
    Consider $p(z) = z^4 + 6z + 3$. If $|z| \ge
    2$, then $\left| z^3 + 6 + \frac{3}{z}
    \right| \ge |z|^3 - 6 - \frac{3}{|z|} > 0$, so
    $p(z) = z \left( z^2 + 6 + \frac{3}{z} \right)
    \neq 0$. We could instead apply Rouch\'e's
    with $\gamma : |z| = 2$, $f(z) = z^4$, $g(z) =
    6z + 3$, so $|z|^4 = 16 > 15 = 6|z| + 3 \ge
    |6z + 3|$. By Rouch\'e's, $p(z)$ has 4 zeroes
    inside $D(0, 2)$. For $|z| = 1$, $|6z| = 6$
    and $|z^4 + 3| \le 4$. So using $\gamma : |z|
    = 1$, $f(z) = 6z$, $g(z) = z^4 + 3$, we see
    $p(z)$ has 1 zero inside $D(0, 1)$. (Note that
    this implies that $p(z)$ has a real root,
    since roots come in conjugate pairs for
    polynomials over $\RR$).
\end{example*}

\begin{example*}[Rouch\'e's $\implies$ open
    mapping]
    If $f : D \to \CC$ is holomorphic and
    nonconstant and $a \in D$, we can choose $r >
    0$ such that $D(a, 2r)^\times$ has no zeroes
    of $f(z) - f(a)$. Let $\gamma$ be $|z - a| =
    r$, and let $0 < \eps , \min_{z \in \gamma}
    |f(z) - f(a)|$. Then for $w \in D(f(a),
    \eps)$, $f(z) - w = f(a) - w + f(z) - f(a)$,
    and we have by $|f(a) - w| < \eps < |f(z) -
    f(a)|$ for all $z \in \gamma$. By Rouch\'e's,
    zeroes in $D(a, r)$ of $f(z) - w$ is equal to
    number of zeroes in $D(a, r)$ of $f(z) - f(a)
    > 0$. So $f(D(a, r)) \supset D(f(a), \eps)$.
\end{example*}

\subsubsection*{Uniform limits of holomorphic functions}

\begin{flashcard}[locally-uniformly-defn]
\begin{definition*}[Converging locally uniformly]
    \cloze{
    Let $\mathcal{U} \subset \CC$ be open, and
    $f_n : \mathcal{U} \to \CC$ a sequence of
    functions. Then $f_n \to f$ \emph{converges
    locally uniformly on $\mathcal{U}$} if
    $\forall u \in \mathcal{U}$, $\exists D(a, r)
    \subset \mathcal{U}$ on which $f_n \to f$
    uniformly.
}
\end{definition*}
\end{flashcard}

\begin{example*}
    $f_n(z) = z^n$ on $\mathcal{U} = D(0, 1)$. As
    $n \to \infty$, $f_n$ tends to constant zero
    function pointwise. For $a \in D(0, 1)$,
    $\ol{D\left(a, \frac{1 - |a|}{2}\right)}
    \subset D(0, 1)$, and $f_n \to 0$ uniformly on
    $D \left( a, \frac{1 - |a|}{2} \right)$. So
    $f_n \to 0$ locally uniformly on $D(0, 1)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/63c687bebf3111ed.png}
    \end{center}
    However, for any $\eps > 0$, $|f_n(z)| < \eps
    \iff |z|^n < \eps \iff |z| < \eps^{1/n}$, so
    no uniform bound can hold for all $|z| < 1$.
\end{example*}

\begin{flashcard}[locally-uniformly-convergent-iff]
\begin{proposition*}
    $\{f_n\} : \mathcal{U} \to \CC$ is locally
    uniformly convergent on $\mathcal{U}$ $\iff$
    \cloze{$\{f_n\}$ converges uniformly on any compact
    subset of $\mathcal{U}$.}
\end{proposition*}
\end{flashcard}

\noindent
Recall: $K \subset \CC$ is compact $\iff$ $K$ is
closed and bounded $\iff$ every open cover has a
finite subcover.

\begin{proof}
    If $f_n \to f$ locally uniformly on
    $\mathcal{U}$, and $K \subset \mathcal{U}$ is
    compact, then $\forall a \in K$, there exists
    $r_a > 0$ such that $\{f_n\}$ converges
    uniformly on $D(a, r_a)$. $\bigcup_{a \in K}
    D(a, r_a)$ is an open cover of $K$, so there
    exists $a_1, \ldots, a_l$ such that
    \[ K \subset D(a_1, r_{a_1}) \cup \cdots \cup
    D(a_l, r_{a_l}) .\]
    Taking the max of constants of uniform
    convergence on these discs, $f_n \to f$
    uniformly on $K$.

    \myskip
    If $f_n \to f$ on every compact subset of
    $\mathcal{U}$ then if $a \in \mathcal{U}$,
    find a closed disc $\ol{D(a, r)} \subset
    \mathcal{U}$. Then $f_n \to f$ converges
    uniformly on $D(a, r)$.
\end{proof}