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Our last remark before looking at Runge's theorem:

Taylor's theorem is not very helpful in this case.
Consider $\sqrt{z}$ with branch cut negative real
axis. Take the region as sketched:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/0294114020c54adb.png}
\end{center}
Taylor's theorem works up to the first singularity
that you find.

\begin{definition*}
  $E \subseteq \CC$ is path-connected if given $z,
  w \in E$, there exists $\gamma : [0, 1] \to E$
  continuous such that $\gamma(0) = z$, $\gamma(1)
  = w$.
\end{definition*}

\begin{flashcard}[runges-thm]
\begin{theorem*}[Runge's Theorem]
  \cloze{If $\Omega$ is open in $\CC$, $f : \Omega \to
  \CC$ is analytic, $K$ is a compact subset of
  $\Omega$ with $K^c$ path-connected, then we can
  find a sequence $P_n$ of polynomials such that
  $P_n \to f$ uniformly on $K$.}
\end{theorem*}
\end{flashcard}

\vspace{-1em}
(These hypotheses stay with us for the rest of
lecture 7).

\begin{flashcard}[runge-step-1]
\begin{lemma*}
  \prompt{(Step 1 of proof of Runge).}
  Let $K$, $\Omega$, $f$ as stated. We can
  find closed  straight line segments with $l_j
  \subseteq \Omega \setminus K$ such that
  \[ f(z) =  \sum_i \frac{1}{2\pi i} \int_{l_j}
  \frac{f(w)}{w - z} \dd w \]
\end{lemma*}

\fcscrap{In other words, we can approximate the borders
by straight lines like this:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/4b54a2799bbe4be4.png}
\end{center}}

\begin{proof}
  \cloze{Since $K$ is compact, $\Omega^c$ is
  closed and $K \cap \Omega^c = \emptyset$, there
  exists a $\delta > 0$ such that $|z - w| >
  \delta$ for all $z \in K$ and $w \in \Omega^c$
  (proof is that $z \mapsto d(z, \Omega^c)$ is
  continuous $K \to \RR$ and $K$ is compact).
  Choose $N \gg \delta^{-1}$ (for example $N =
  100\delta^{-1} + 100$), and consider the
  collection of squares with vertices
  \[ \frac{r + si}{N}, \frac{(r + 1) + si}{N},
  \frac{(r + 1) + (s + 1)i}{N}, \frac{r + (s + 1)i}{N} \]
  such that
  \[ d \left( \frac{r + si}{N} \right) <
  \frac{\delta}{2} .\]
  If $z \in K$ and $z$ does not lie on the
  boundary of a small square then
  \[ \frac{1}{2\pi i} \int \frac{f(w)}{w - z} \dd w =
  \begin{cases}
    f(z) & \text{if $z$ lies in the small square}
    \\
    0 & \text{otherwise}
  \end{cases} \]
  Summing over the set of squares (which we shall
  call $S$), we get
  \[ \sum_S \frac{1}{2\pi i} \int \frac{f(w)}{w -
  z} \dd w = f(z) \]
  and since interior sides cancel:
  \[ \sum_{l \in L} \frac{1}{2\pi i} \int_l
  \frac{f(w)}{w - z} \dd w = f(z) \]
  where $L$ is the set of edges that are not
  cancelled. Note that the $l$ in the sum lie in
  $\Omega \setminus K$. To extend this formula to
  all $z \in K$ just observe that
  \[ z \mapsto \frac{1}{2\pi i} \int_l
  \frac{f(w)}{w - z} \dd w \]
  is continuous on $\CC \setminus l$.}
\end{proof}
\end{flashcard}

Thus Runge's theorem will follow if we can show
that following: If $\Omega$ open, $K$ compact
subset of $\Omega$, $l \subseteq \Omega
\setminus K$ a closed line segment, and $f :
\Omega \to \CC$ analytic, then we can find a
sequence of polynomials $P_n$ such that $P_n(z)
\to \frac{1}{2\pi i} \int_l \frac{f(w)}{w - z}
\dd w$ uniformly on $K$.

\begin{flashcard}[runge-step-2]
\begin{lemma*}
  \prompt{(Step 2 of proof of Runge).} 
  If $f$ continuous on $K$, $l$ as before
  then given $\eps > $ we can find $N$ and $w_1,
  w_2, \ldots, w_N \in l$ such that
  \[ \left| \int_l \frac{f(w)}{w - z} \dd w - \sum_j
  \frac{A_j}{w_j - z} \right| < \eps \qquad
  \forall z \in K .\]
\end{lemma*}
  
\begin{proof}
  \cloze{$l$ is compact, $K$ is compact so $l
  \times K$ is compact. $G : l \times K \to \CC$,
  $G(z, w) = \frac{f(w)}{w - z}$. Note $G$ is
  continuous, so $G$ is uniformly continuous, so
  if $l$ is given by $\gamma : [0, 1] \to \CC$,
  $\gamma(t) = \alpha + \beta t$ then
  \[ \frac{f(z)}{z - w} - \sum G \left( z,
  \gamma \left( \frac{r}{n} \right) \right)
  \mathbbm{1}_{\{w = \gamma(t), r / n \le t \le r
  + 1 / n\}}(w) \to 0 \]
  uniformly.
  }
\end{proof}
\end{flashcard}

Thus Runge's theorem follows if I can show that
if $w \notin K$ then there exists a sequence
$P_n$ of polynoials such that $P_n(z) \to
\frac{1}{w - z}$ uniformly on $K$.

\begin{flashcard}[runge-step-3]
\begin{theorem*}
  If $K$ is compact and $K^c$ is path-connected,
  $w \notin K$, then there exists polynomials
  $P_n$ such that $P_n(z) \to \frac{1}{w - z}$
  uniformly for $z \in K$.
\end{theorem*}

\begin{proof}
  \cloze{To prove this call $\Gamma$ the set of $w \notin
  K$ such that the result is true. Observe first
  that if $\ol{B(0, R)} \supseteq K$ then
  \[ \Gamma \supseteq \{w \in \CC : |w| \ge 2R\} .\]
  Proof:
  \[ \frac{1}{w - z} = \frac{1}{w \left( 1 -
  \frac{z}{w} \right)} = \frac{1}{w} \sum_{r =
  1}^\infty \left( \frac{z}{w} \right)^r \]
  and since $\left| \frac{z}{w} \right| \le \half$
  the Weierstrass $M$-test tells us convergence is
  uniform. So
  \[ \frac{1}{w} \sum_{r = 0}^N \left( \frac{z}{w}
  \right)^r \to \frac{1}{w - z} \]
  uniformly.

  Next we observe that if $w \in \Gamma$ and
  $d(w', K) \ge 2\eta$ ($\eta > 0$) then if $w \in
  B(w', \eta)$ then we have $w \in \Gamma$.
  Proof:
  \[ \frac{1}{w - z} = \frac{1}{(w' - z) - (w' -
  w)} = \frac{1}{(w' - z)} \frac{1}{\left( 1 -
  \frac{w' - w}{w' - z} \right)} = \frac{1}{w' - z}
  \sum_n \frac{(w' - w)^n}{w' - z}\]
  convergence is uniform as $\frac{|w' - w|}{|w -
  z|} < \half$. Thus given $\eps > 0$ we can find
  an $N$ such that
  \[ \left| \frac{1}{w - z} - \sum_{n = 0}^N
  \frac{(w' - w)^n}{(w - z)^{n + 1}} \right| <
  \frac{\eps}{2} \]
  for $z \in K$. But we can find polynomials
  $P_m(z) \to \frac{1}{w' - z}$ uniformly on $K$.
  So for large enough $m$,
  \[ \left| \frac{1}{w - z} - \sum_{n = 0}^N (w' -
  w)^n P_m(z)^n \right| < \eps \qquad \forall z
  \in K .\]

  So we have now shown:
  \begin{enumerate}[(i)]
    \item There exists $R$ such that $|w| \ge R$
      implies $w \in \Gamma$.
    \item If $\delta > 0$ and $w' \in \Gamma$,
      $\ol{B(w', 2\delta)} \cap K = \emptyset$,
      then $B(w', \delta) \subseteq \Gamma$.
  \end{enumerate}
  Suppose $w_1 \notin K$. Choose $|w_0| \ge R$. Then
  there exists $\gamma : [0, 1] \to K^c$ such that
  $\gamma$ is continuous and $\gamma(0) = w_1$,
  $\gamma(1) = w_0$. $\gamma([0, 1])$ is compact,
  $\gamma([0, 1]) \cap K = \emptyset$, so there
  exists $\delta > 0$ such that if $t \in [0, 1]$,
  \[ |\gamma(t) - k| \ge 8\delta \qquad \forall k
  \in K .\]
  $\gamma$ is continuous, hence uniformly
  continuous so we can find $N$ such that
  \[ \left| \gamma \left( \frac{r}{N} \right) -
  \gamma \left( \frac{r + 1}{N} \right) \right| <
  \delta .\]
  $\gamma(0) \in \Gamma$, so $\gamma \left(
  \frac{1}{N} \right) \in \Gamma$, so $\gamma
  \left( \frac{2}{N} \right) \in \Gamma$ etc, and
  we deduce $w_0 = \gamma(1) \in \Gamma$.}
\end{proof}
\end{flashcard}